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https://thehardcorecoder.com/2019/05/05/full-adder-code/?shared=email&msg=fail | 1,657,211,434,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00425.warc.gz | 605,150,360 | 30,214 | I was involved in a debate recently about whether a full adder logic circuit is a computer. The computer science answer is: “No, not as we define a computer.”
I plan to address that answer in detail on my main blog. Here I wanted to show some of the different ways a full adder can be modeled and implemented.
A full adder logic circuit takes two inputs (two binary bits, a and b) plus a third input, Ci, a carry bit from an adjacent adder (or zero if no adjacent adder). It has two outputs, the sum bit, S, and a carry bit, Co.
The truth table looks like this:
It can be expressed as two logical expressions:
(The ⊕ symbol means XOR; the ∧ symbol means AND, and the ∨ symbol means OR.)
It is commonly implemented logically like this:
Although there are other logic circuits that are functionally identical. (It can be constructed entirely from NOR gates, for example — an exercise for the reader.)
The logical expressions, and the equivalent logic gate circuits, show that the underlying circuit of a full-adder is not, in itself, computational (as computation is usually defined).
The truth table shows the logic has distinct states — eight, in this case. Given the mechanism does have these distinct states, there is a map of those states to an abstract machine that simulates the same states. Thus the operation can be simulated (or modeled) with a computation.
The map of states for a full adder can be to a truth table (as above), to a Turing Machine, a Lambda calculus, or a Finite State machine (aka automaton).
Above is a diagram of an FSM. I learned it as “machine” — some say “automaton” (so, FSA) — tomato, tomahto. The important part is the Finite State.
Incidentally, the simplicity and regularity of the state diagram is another indicator of the physicality of a full-adder. A given set of inputs immediately trickles down to the output state.
The FSM branches on each of the three input bits, a, b, and Ci. The final state features a payload that is the output bits, Co and S, respectively. The states are labeled by the bit-pattern established at that point (bits build from right to left).
§
With that in mind, let’s consider how to implement the logic with code. We can model many of the above expressions of the full-adder logic.
Note that all these examples take the input bits, a, b, and Ci, and return a tuple containing the output bits, S and Co.
```def full_adder_1 (a, b, ci):
logic_table = [
[[[0,0],[1,0]], [[1,0],[0,1]]], # a=0
[[[1,0],[0,1]], [[0,1],[1,1]]], # a=1
] # b=0 b=1 b=0 b=1
return logic_table[a][b][ci]
#
```
The example above implements a logic table version that only requires a single lookup step. This trades code and time for (potentially) large data, since all the logic is contained in the table.
(A real-world implementation might take many steps to accomplish the lookup. The single step is conceptual, as if looking at the truth table to get an output for a given input.)
```def full_adder_2 (a, b, ci):
s = (a ^ b) ^ ci
co = ((a ^ b) and ci) or (a and b)
return (s,co)
#
```
The second example implements the actual logical expression shown above. The tophat (^) is Python’s xor operator. The and and or are built-in Python operators.
The code here directly corresponds to the logical expressions shown above.
Note that the intermediate steps of assigning the respective expression values to S and Co aren’t necessary. Those result values could be directly returned in a single step.
```def full_adder_3 (a, b, ci):
s = a + b + ci
if s < 2: return (s, 0)
return (s % 2, 1)
#
```
Example three models the adder as an actual base two addition (of all three input bits), which requires detecting whether that operation carried.
The logic may not be obvious:
If s is less than two, there was no carry, so just return the sum (which will be 0 or 1), and return Co=0.
Otherwise, there was a carry, so take the mod 2 of the sum (which removes the carry), and return Co=1. (The sum will be two or three, and the mod 2 of those is zero and one, respectively.)
```def full_adder_4 (a, b, ci):
s,co = (a+b, 0)
if s == 2: s,co = (0, 1)
s += ci
if s == 2: s,co = (0, 1)
return (s, co)
#
```
The fourth example also uses a mathematical model, but acts as if the sum is a single bit with an overflow. (That is, the only allowed values are 0, 1, and overflow.)
The code may not be obvious if you’re not familiar with Python, which very nicely allows setting multiple variables with a list of values. That’s what might seem weird in the assignment statements; they’re setting the value of both S and Co.
The first step (line #2) sets S to the value of a+b (so it will be 0, 1, or 2). It also sets Co=0, assuming no carry.
Line #3 checks for a carry. If it occurred, reset S=0 and Co=1. The output at this point is a correct half-add.
For the full-add, line #4 adds the Ci bit to S, and the following line (#5) checks for the carry (S was either 0 or 1, so it could be 2 now). If there was a carry, the output is reset, as above.
The last step just returns S and Co.
```def full_adder_5 (a, b, ci):
state_table = {
'start':{0:'S#0', 1:'S#1'},
'S#0':{0:'S#00', 1:'S#01'},
'S#1':{0:'S#10', 1:'S#11'},
'S#00':{0:'S#000', 1:'S#001'},
'S#01':{0:'S#010', 1:'S#011'},
'S#10':{0:'S#100', 1:'S#101'},
'S#11':{0:'S#110', 1:'S#111'},
'S#000':(0,0), 'S#001':(1,0), # final
'S#010':(1,0), 'S#011':(0,1), # states
'S#100':(1,0), 'S#101':(0,1), # "
'S#110':(0,1), 'S#111':(1,1), # "
}
state = state_table['start'][a]
state = state_table[state][b]
state = state_table[state][ci]
return state_table[state]
#
```
The fifth example models the FSM version of the adder. Each input, a, b, and Ci, serially drives the FSA from starting state (@) to final state.
All states (in state_table), except the final states, contain a dictionary of possible next states. Here, any given state can only have two possible next states — one for each bit value (0 or 1) that causes a move to that next state.
The final (leaf) states are special in containing the output payload rather than a dictionary. The function returns that payload (the bits S and Co).
The first step (line #15) gets the ‘start’ state from state_table. As an intermediate (branch) state, the state is a dictionary to possible next states. The a input indexes the dictionary (with a 0 or a 1), so the whole expression returns the next state, depending on a.
The next line (#16) uses that next state to get that state’s dictionary from state_table. In this case the input b indexes the dictionary for the state that depends on b.
Line #17 repeats the process with the Ci bit. The state from b gets the dictionary, and Ci indexes it to get the final state.
Lastly (line #18), the final state gets the payload from state_table and returns it.
As you can see, modeling the FSM results in a large data segment. The code lines (#15 – #18) are unrolled for clarity, but more generally a design would use a loop.
[See the State Engines series (part 1, part 2, part 3), for more.]
§
These examples show some of the different ways we can implement the logic of a full adder. There can certainly be others.
One implementation I didn’t explore is modeling the physical circuit, the logic gates.
I do have such an implementation to show you, but it’s so big it needs its own post. Modeling physical reality is much more complicated! (A separate post will give me elbow room to talk a little about modeling the circuit below the gates at the transistor level.)
But fundamentally, the adder is the logical expression:
Nothing more.
Calling it a computation is equivalent to calling the creeks, streams, and rivers of a watershed a computation — the processes involved are more similar than with the computations shown above.
But that’s an argument for my other blog. This is about the code itself. | 2,060 | 7,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-27 | latest | en | 0.931112 |
http://nrich.maths.org/public/leg.php?code=71&cl=3&cldcmpid=4881 | 1,503,250,399,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106865.74/warc/CC-MAIN-20170820170023-20170820190023-00243.warc.gz | 301,219,618 | 10,483 | Search by Topic
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Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
Coins on a Plate
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Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
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Prove Pythagoras' Theorem using enlargements and scale factors.
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A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
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If you know the sizes of the angles marked with coloured dots in this diagram which angles can you find by calculation?
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Draw some quadrilaterals on a 9-point circle and work out the angles. Is there a theorem?
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Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total.
Logic
Stage: 2 and 3
What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article.
Not Necessarily in That Order
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Baker, Cooper, Jones and Smith are four people whose occupations are teacher, welder, mechanic and programmer, but not necessarily in that order. What is each person’s occupation?
Con Tricks
Stage: 3
Here are some examples of 'cons', and see if you can figure out where the trick is.
Aba
Stage: 3 Challenge Level:
In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
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Stage: 3
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
Rotating Triangle
Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
Thirty Nine, Seventy Five
Stage: 3 Challenge Level:
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Stage: 3 Challenge Level:
Write down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .
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Stage: 3 Challenge Level:
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation.
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Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
Stage: 2 and 3
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
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Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
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Stage: 4 Challenge Level:
Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.
Marbles
Stage: 3 Challenge Level:
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Stage: 3 Challenge Level:
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Which hexagons tessellate?
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Can you find the areas of the trapezia in this sequence?
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Can you find all the 4-ball shuffles?
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Stage: 4
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
Stage: 3 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
Clocked
Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
Top-heavy Pyramids
Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
Picturing Pythagorean Triples
Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
Königsberg
Stage: 3 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
Pattern of Islands
Stage: 3 Challenge Level:
In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island...
Yih or Luk Tsut K'i or Three Men's Morris
Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
Pythagorean Triples I
Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
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Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
Pent
Stage: 4 and 5 Challenge Level:
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.
Pythagorean Triples II
Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers. | 2,292 | 9,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-34 | latest | en | 0.90402 |
https://www.quizzes.cc/metric/percentof.php?percent=2.5&of=7000000 | 1,591,498,703,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00417.warc.gz | 852,621,913 | 4,092 | #### What is 2.5 percent of 7,000,000?
How much is 2.5 percent of 7000000? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 2.5% of 7000000 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 2.5% of 7,000,000 = 175000
Calculate another percentage below. Type into inputs
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Calculating two point five of seven million How to calculate 2.5% of 7000000? Simply divide the percent by 100 and multiply by the number. For example, 2.5 /100 x 7000000 = 175000 or 0.025 x 7000000 = 175000
#### How much is 2.5 percent of the following numbers?
2.5% of 7000000.01 = 17500000.025 2.5% of 7000000.02 = 17500000.05 2.5% of 7000000.03 = 17500000.075 2.5% of 7000000.04 = 17500000.1 2.5% of 7000000.05 = 17500000.125 2.5% of 7000000.06 = 17500000.15 2.5% of 7000000.07 = 17500000.175 2.5% of 7000000.08 = 17500000.2 2.5% of 7000000.09 = 17500000.225 2.5% of 7000000.1 = 17500000.25 2.5% of 7000000.11 = 17500000.275 2.5% of 7000000.12 = 17500000.3 2.5% of 7000000.13 = 17500000.325 2.5% of 7000000.14 = 17500000.35 2.5% of 7000000.15 = 17500000.375 2.5% of 7000000.16 = 17500000.4 2.5% of 7000000.17 = 17500000.425 2.5% of 7000000.18 = 17500000.45 2.5% of 7000000.19 = 17500000.475 2.5% of 7000000.2 = 17500000.5 2.5% of 7000000.21 = 17500000.525 2.5% of 7000000.22 = 17500000.55 2.5% of 7000000.23 = 17500000.575 2.5% of 7000000.24 = 17500000.6 2.5% of 7000000.25 = 17500000.625
2.5% of 7000000.26 = 17500000.65 2.5% of 7000000.27 = 17500000.675 2.5% of 7000000.28 = 17500000.7 2.5% of 7000000.29 = 17500000.725 2.5% of 7000000.3 = 17500000.75 2.5% of 7000000.31 = 17500000.775 2.5% of 7000000.32 = 17500000.8 2.5% of 7000000.33 = 17500000.825 2.5% of 7000000.34 = 17500000.85 2.5% of 7000000.35 = 17500000.875 2.5% of 7000000.36 = 17500000.9 2.5% of 7000000.37 = 17500000.925 2.5% of 7000000.38 = 17500000.95 2.5% of 7000000.39 = 17500000.975 2.5% of 7000000.4 = 17500001 2.5% of 7000000.41 = 17500001.025 2.5% of 7000000.42 = 17500001.05 2.5% of 7000000.43 = 17500001.075 2.5% of 7000000.44 = 17500001.1 2.5% of 7000000.45 = 17500001.125 2.5% of 7000000.46 = 17500001.15 2.5% of 7000000.47 = 17500001.175 2.5% of 7000000.48 = 17500001.2 2.5% of 7000000.49 = 17500001.225 2.5% of 7000000.5 = 17500001.25
2.5% of 7000000.51 = 17500001.275 2.5% of 7000000.52 = 17500001.3 2.5% of 7000000.53 = 17500001.325 2.5% of 7000000.54 = 17500001.35 2.5% of 7000000.55 = 17500001.375 2.5% of 7000000.56 = 17500001.4 2.5% of 7000000.57 = 17500001.425 2.5% of 7000000.58 = 17500001.45 2.5% of 7000000.59 = 17500001.475 2.5% of 7000000.6 = 17500001.5 2.5% of 7000000.61 = 17500001.525 2.5% of 7000000.62 = 17500001.55 2.5% of 7000000.63 = 17500001.575 2.5% of 7000000.64 = 17500001.6 2.5% of 7000000.65 = 17500001.625 2.5% of 7000000.66 = 17500001.65 2.5% of 7000000.67 = 17500001.675 2.5% of 7000000.68 = 17500001.7 2.5% of 7000000.69 = 17500001.725 2.5% of 7000000.7 = 17500001.75 2.5% of 7000000.71 = 17500001.775 2.5% of 7000000.72 = 17500001.8 2.5% of 7000000.73 = 17500001.825 2.5% of 7000000.74 = 17500001.85 2.5% of 7000000.75 = 17500001.875
2.5% of 7000000.76 = 17500001.9 2.5% of 7000000.77 = 17500001.925 2.5% of 7000000.78 = 17500001.95 2.5% of 7000000.79 = 17500001.975 2.5% of 7000000.8 = 17500002 2.5% of 7000000.81 = 17500002.025 2.5% of 7000000.82 = 17500002.05 2.5% of 7000000.83 = 17500002.075 2.5% of 7000000.84 = 17500002.1 2.5% of 7000000.85 = 17500002.125 2.5% of 7000000.86 = 17500002.15 2.5% of 7000000.87 = 17500002.175 2.5% of 7000000.88 = 17500002.2 2.5% of 7000000.89 = 17500002.225 2.5% of 7000000.9 = 17500002.25 2.5% of 7000000.91 = 17500002.275 2.5% of 7000000.92 = 17500002.3 2.5% of 7000000.93 = 17500002.325 2.5% of 7000000.94 = 17500002.35 2.5% of 7000000.95 = 17500002.375 2.5% of 7000000.96 = 17500002.4 2.5% of 7000000.97 = 17500002.425 2.5% of 7000000.98 = 17500002.45 2.5% of 7000000.99 = 17500002.475 2.5% of 7000001 = 17500002.5
1% of 7000000 = 70000 2% of 7000000 = 140000 3% of 7000000 = 210000 4% of 7000000 = 280000 5% of 7000000 = 350000 6% of 7000000 = 420000 7% of 7000000 = 490000 8% of 7000000 = 560000 9% of 7000000 = 630000 10% of 7000000 = 700000 11% of 7000000 = 770000 12% of 7000000 = 840000 13% of 7000000 = 910000 14% of 7000000 = 980000 15% of 7000000 = 1050000 16% of 7000000 = 1120000 17% of 7000000 = 1190000 18% of 7000000 = 1260000 19% of 7000000 = 1330000 20% of 7000000 = 1400000 21% of 7000000 = 1470000 22% of 7000000 = 1540000 23% of 7000000 = 1610000 24% of 7000000 = 1680000 25% of 7000000 = 1750000
26% of 7000000 = 1820000 27% of 7000000 = 1890000 28% of 7000000 = 1960000 29% of 7000000 = 2030000 30% of 7000000 = 2100000 31% of 7000000 = 2170000 32% of 7000000 = 2240000 33% of 7000000 = 2310000 34% of 7000000 = 2380000 35% of 7000000 = 2450000 36% of 7000000 = 2520000 37% of 7000000 = 2590000 38% of 7000000 = 2660000 39% of 7000000 = 2730000 40% of 7000000 = 2800000 41% of 7000000 = 2870000 42% of 7000000 = 2940000 43% of 7000000 = 3010000 44% of 7000000 = 3080000 45% of 7000000 = 3150000 46% of 7000000 = 3220000 47% of 7000000 = 3290000 48% of 7000000 = 3360000 49% of 7000000 = 3430000 50% of 7000000 = 3500000
51% of 7000000 = 3570000 52% of 7000000 = 3640000 53% of 7000000 = 3710000 54% of 7000000 = 3780000 55% of 7000000 = 3850000 56% of 7000000 = 3920000 57% of 7000000 = 3990000 58% of 7000000 = 4060000 59% of 7000000 = 4130000 60% of 7000000 = 4200000 61% of 7000000 = 4270000 62% of 7000000 = 4340000 63% of 7000000 = 4410000 64% of 7000000 = 4480000 65% of 7000000 = 4550000 66% of 7000000 = 4620000 67% of 7000000 = 4690000 68% of 7000000 = 4760000 69% of 7000000 = 4830000 70% of 7000000 = 4900000 71% of 7000000 = 4970000 72% of 7000000 = 5040000 73% of 7000000 = 5110000 74% of 7000000 = 5180000 75% of 7000000 = 5250000
76% of 7000000 = 5320000 77% of 7000000 = 5390000 78% of 7000000 = 5460000 79% of 7000000 = 5530000 80% of 7000000 = 5600000 81% of 7000000 = 5670000 82% of 7000000 = 5740000 83% of 7000000 = 5810000 84% of 7000000 = 5880000 85% of 7000000 = 5950000 86% of 7000000 = 6020000 87% of 7000000 = 6090000 88% of 7000000 = 6160000 89% of 7000000 = 6230000 90% of 7000000 = 6300000 91% of 7000000 = 6370000 92% of 7000000 = 6440000 93% of 7000000 = 6510000 94% of 7000000 = 6580000 95% of 7000000 = 6650000 96% of 7000000 = 6720000 97% of 7000000 = 6790000 98% of 7000000 = 6860000 99% of 7000000 = 6930000 100% of 7000000 = 7000000 | 3,405 | 6,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-24 | latest | en | 0.785056 |
https://math.stackexchange.com/questions/26869/constants-of-integration-in-integration-by-parts | 1,695,887,698,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510368.33/warc/CC-MAIN-20230928063033-20230928093033-00455.warc.gz | 416,892,873 | 40,444 | # Constants of integration in integration by parts
After finishing a first calculus course, I know how to integrate by parts, for example, $\int x \ln x dx$, letting $u = \ln x$, $dv = x dx$: $$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$
However, what I could not figure out is why we assume from $dv = x dx$ that $v = \frac{x^2}{2}$, when it could be $v = \frac{x^2}{2} + C$ for any constant $C$. The second integral would be quite different, and not only by a constant, so I would like to understand why we "forget" this constant of integration.
Thanks.
Take your example, $$\int x\ln x\,dx.$$ Note $x\gt 0$ must be assumed (so the integrand makes sense).
If we let $u = \ln x$ and $dv= x\,dx$, then we can take $v$ to be any function with $dv = x\,dx$. So the "generic" $v$ will be, as you note, $v = \frac{1}{2}x^2 + C$. What happens then if we use this "generic" $v$? \begin{align*} \int x\ln x\,dx &= \ln x\left(\frac{1}{2}x^2 + C\right) - \int \left(\frac{1}{2}x^2+C\right)\frac{1}{x}\,dx\\ &= \frac{1}{2}x^2\ln x + C\ln x - \int\left(\frac{1}{2}x + \frac{C}{x}\right)\,dx\\ &= \frac{1}{2}x^2\ln x + C\ln x - \frac{1}{4}x^2 - C\ln x + D\\ &= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + D, \end{align*} so in the end, we get the same result no matter what value of $C$ we take for $v$.
This says that we can take any value of $C$ and still get the same answer. Since we can take any value of $C$, why not take the simplest one, the one that does not require us to carry around an extra term that is going to cancel out anyway? Say..., $C=0$?
This works in general. If you replace $v$ with $v+C$ in the integration by parts formula, you have \begin{align*} \int u\,dv &= u(v+C) - \int(v+C)\,du = uv + Cu - \int v\,du - \int C\,du\\ &= uv+Cu - \int v\,du - Cu = uv-\int v\,du. \end{align*} So the answer is the same regardless of the value of $C$, and so we take $C=0$ because that makes our life simpler.
• I think you've proved that $C=0$ in this case. We're not just making an assumption here. $C=0$ is a hard concrete result in the by-parts formula.
– Nick
Nov 1, 2014 at 9:44
• I have problem understanding why did you take $v = \frac{x^2}{2} + C$ for both the first term and the second term. This is why they cancel out eventually. Why not take $v = \frac{x^2}{2} + C_1$ for the first term and $v = \frac{x^2}{2} + C_2$ for the second term. Aren't both valid anti-derivatives of $x$? I know $C$ is as arbitrary as $C_1$ or $C_2$. This is what I find confusing. Indefinite integrals aren't as concrete a thing as definite integrals which are essentially area under a curve. I have learnt that indefinite integrals are a family of functions with the same derivative. Jun 23, 2022 at 11:10
• @ShinsekainoKami In integration by parts you pick one antiderivative for $dv$ and put it into the formula. The formula is $\int u\,dv = uv - \int v\,du$, not $\int u\,dv = uv_1 - \int v_2\,du$. The calculation above just shows you can pick any one antiderivative of $v$. The fact that the constants cancel is the point. But even if you pick different constants, once you calculate $\int v\,du$ you will just end up with a bunch of constants, which add up to... a constant of integration. Jun 23, 2022 at 14:34
• @ArturoMagidin Got it. Thanks! I was using the formula $\int f(x) g(x) dx = f(x) \int g(x) dx - \int [ \frac{d}{dx} (f(x)) \int g(x) dx ] dx$ without giving consideration to the original product rule using which this is derived. But I'm not sure if we will just end up with a constant if the arbitrary constants are unequal, say $C_1 \ne C_2$. For example, in the case you have taken in your answer above, we would end up with $\frac{x^2}{2} \ln x - \frac{x^2}{4} + (C_1 - C_2) \ln x + D$. We only get the expected answer if $C_1 = C_2$, which is in fact true like you have just explained. Jun 24, 2022 at 8:24
The second integral would change, but also the first term... Have you actually checked to see what happens if you change the constant?
Your observation that $dv=xdx$ does not imlpy $v=x^2/2$ is correct.
Your confusion resolves when you say it this way: we set $v=x^2/2$ and this implies $dv=xdx$.
$$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$
You could always write $$v = \frac{x^2}{2} + C$$ but that won't matter much because the final result would also involve a constant (Say $K$ which would be equal to $C+k$ )
• This is not exactly what happens... Mar 14, 2011 at 7:48
• Arturo's answer makes me realize this is ok. If you make an edit, howerver small, I will remove my downvote.
Mar 14, 2011 at 15:33
• @Approximist : Done! Mar 14, 2011 at 16:44
HINT $\rm\ \ C'=0\ \ \Rightarrow\ \ (UV)'-U'\:V\ =\ UV'\: =\ U(V+C)'\: =\ (U(V+C))'-U'\:(V+C)$
We didn't "forget". We simply choose C in a way that the resulting $\int u\,dv$ would be the simplest form. Usually $C = 0$ but not always. If, for example, we have this integral:
$$\int \ln(x+2) \,dx$$
Then you would choose $v = x + 2$ because $du = \frac{dx}{x+2}$
Second example:
$$\int x\ln(x+2) \,dx$$
Then $$v = \frac{x^2-4}{2} = \frac{(x-2)(x+2)}{2}$$ and $$u\,dv = \frac{x-2}{2} dx$$
We "forget" it, and add it in the last step. The whole point in the constant of integration is to remind us that there could have been a constant term added on at the end originally, but in the process of differentiation we got rid of it because it did not affect the slope.
Let $$AD$$ denote the anti derivative operator. Then \begin{align*} &AD \left(u\frac{dv}{dx}+v\frac{du}{dx}\right)=uv\\[5pt] \Rightarrow\qquad &AD \left(u\frac{dv}{dx}\right)+AD\left( v\frac{du}{dx}\right)=uv\\[5pt] \Rightarrow\qquad &AD \left(u\frac{dv}{dx}\right)=uv-AD\left( v\frac{du}{dx}\right).\tag{i} \end{align*} Taking $$u=f(x)$$ and $$v=AD (g(x))$$ in (i), we get $$AD \big(f(x)g(x)\big)=f(x)AD(g(x))-AD \left[f'(x)AD( g(x))\right].$$ Therefore, by definition of the indefinite integral, \begin{align*} \int \big(f(x)g(x)\big)\,dx&=AD \big(f(x)g(x)\big)+C\\ &=f(x)AD(g(x))-AD \left[f'(x)AD( g(x))\right]+C. \end{align*} If we write the above formula as \begin{align*} \int \big(f(x)g(x)\big)\,dx= f(x)\int g(x)\,dx-\int\left[f'(x)\int g(x)\,dx\right]\,dx\tag{ii} \end{align*} with the understanding that we shall take only one arbitrary constant of integration on the right hand side at the final stage, then this justifies that we can take the arbitrary constant only once. | 2,160 | 6,382 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2023-40 | latest | en | 0.807152 |
http://paroleoggi.it/newton-raphson-method-matlab.html | 1,603,696,937,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890586.57/warc/CC-MAIN-20201026061044-20201026091044-00404.warc.gz | 75,335,118 | 17,712 | # Newton Raphson Method Matlab
The Newton-Raphson method uses an iterative process to approach one root of a function. % b) Newton's method implemented in MATLAB. It should be noted that the “root” function in the MATLAB library can find all the roots of a polynomial with arbitrary order. It uses the iterative formula. Next: The Initial Guess Up: 10. Advantages & Drawbacks for Newton-Raphson Method: Part 2 of 2 [YOUTUBE 4:43] Derivation from Taylor Series of Newton-Raphson Method [ YOUTUBE 7:56] [ TRANSCRIPT ] Supercomputers have No Divide Unit - A Newton-Raphson Method Approach [ YOUTUBE 10:14] [ TRANSCRIPT ]. $30 USD in 1 day (0 Reviews). Octave / MATLAB Newton's method The following implementation of Newton's method (newtonsMethod. Newton-Raphson Method is a root finding iterative algorithm for computing equations numerically. You will see that the internal Matlab solving command fsolve approximates the solution, but only to about 7 decimal places. writing a script to find roots using Newton-Raphson Method in Matlab, how ? Follow 12 views (last 30 days) zee 00 on 2 Dec 2015. The inputs are symbolic functions, initial guesses and number of iterations. Question: Newton-Raphson Method Matlab: Determine The Solution Of The Simultaneous Nonlinear Equations: Y= -x2+x+0. Hi, its my first time posting on here so please be nice. Newton-Raphson is an iterative method, meaning we'll get the correct answer after several refinements on an initial guess. Consult the MATLAB TA's if you have any questions. El método de Newton-Raphson se deduce a partir de esta interpretación geométrica y se tiene que la primera derivada en x es equivalente a la pendiente:. That is, it’s a method for approximating $x^*$ such that $f(x^*)=0$. We make an initial guess for the root we are trying to find, and we call this initial guess x 0. Newton Raphson Iteration method in Matlab. Newton's method, also known as Newton-Raphson's method, is a very famous and widely used method for solving nonlinear algebraic equations. During the study in a two-bus system, with a slack bus and PQ bus, two convergence tolerances were used. It then computes subsequent iterates x(1), x(2), ::: that, hopefully, will converge to a solution x of g(x) = 0. So the root of the tangent line, where the line cuts the X-axis; x1 is the better approximation to a than x0 is. In a nutshell, the Newton-Raphson Algorithm is a method for solving simultaneous nonlinear algebraic equations. writing a script to find roots using Newton-Raphson Method in Matlab, how ? Follow 12 views (last 30 days) zee 00 on 2 Dec 2015. NONLINEAR SYSTEMS - NEWTON'S METHOD Save this program as myfsolve. I am trying to solve 3 non-linear system of 3 variables using the newton-raphson method in matlab. 2 5 4 x + 0. Metode Newton Raphson biasa digunakan dalam mencari akar dari suatu persamaan non linier, jika diasumsikan f mempunyai turunan kontinu f’. Newton's method is a technique for finding the root of a scalar-valued function f(x) of a single variable x. The following is a sample program to understand finding solution of a non linear equation using Newton Raphson Method. So we would have to enter that manually in our code. -For example, to find the root of the equation. The combined bisection/Newton-Raphson program is superior in almost every respect. N–Raphson method), named after Isaac Newton and Joseph Raphson, is a method for finding successively better approximations to the (or zeroes) of a roots - real valued function. I made the bottom code in Matlab. It can be easily generalized to the problem of finding solutions of a system of non-linear equations, which is referred to as Newton's technique. Write a MATLAB script that utilizes the Newton Raphson algorithm to search for the fifth root of any number entered by the user to within four places behind the decimal point (i. It begins with an initial guess for vn+1 and solves a linearized version of R=0 to find a correction to the initial guess for vn+1. Output: approximation solution x = (x1;:::;xn) for the nonlinear system F(x) =0. Riemannian Manifold Optimization Library.$\begingroup$The Newton-Raphson method converges rapidly to an isolated root of an equation or system of equations involving smooth functions once we have an approximation close enough to that root. It's already defined. Newton-Raphson method, also known as the Newton's Method, is the simplest and fastest approach to find the root of a function. Category Education; Show more Show less. Executive Summary. The Newton-Raphson algorithm requires the evaluation of two functions (the function and its derivative) per each iteration. And this is solvable using the Newton-Raphson method which I think I know how to use. The following is a sample program to understand finding solution of a non linear equation using Newton Raphson Method. The Newton-Raphson method is widely used in finding the root of nonlinear equations. 5 Two nonlinear springs (modified Newton-Raphson method). ƒ defined over the reals x, and its. In the beginning of the problem we divide the ODE (ordinary differential equation) to a set of first. Similar is for Secant method, and it requires two-initial guesses to proceed. Created with R2010a Compatible with any release Platform Compatibility. Newton-Raphson Method for Solving non-linear equat Unimpressed face in MATLAB(mfile) Bisection Method for Solving non-linear equations Gauss-Seidel method using MATLAB(mfile) Jacobi method to solve equation using MATLAB(mfile. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it. Méthode de Newton-Raphson en matlab Algorithme : Obtenir racineAvant (une racine de départ) nbIterations = 1 Répéter Si (|fPrime(racineAvant)| <= TP : étude des caractéristiques d un capteur de température à base d'un sonde pt100. Print the values of q and the number of iterations to the screen. Specifically in this case it was to solve 1D gas dynamics equations. Numerical analysis I 1. Newton's method is an iterative method. Newton-Raphson Method with MATLAB code: If point x0 is close to the root a, then a tangent line to the graph of f(x) at x0 is a good approximation the f(x) near a. Newton's method, also known as Newton-Raphson, is an approach for finding the roots of nonlinear equations and is one of the most common root-finding algorithms due to its relative simplicity and speed. Math 111: MATLAB Assignment 2: Newton's Method. Then we know f x x −x∗ Q x where lim x→x∗ Q x ≠0. Mathews and Kurtis D. http//numericalmethods. ^2+c using Newton-Raphson method where a,b,c are to be import from excel file or user defined, the what i need to do?. txt Example 2. Learn more about root finding help MATLAB Answers. However, Newton Raphson method converges exponentially and has an abrupt inclination every iteration, and this explains the quadratic convergence of this method. The previous two methods are guaranteed to converge, Newton Rahhson may not converge in some cases. CH925 - MatLab Code A number of numerical methods used for root finding, and solving ordinary differential equations (ODEs) were covered in this module. For more information about this method please try this. The method is advantageous, becaus. Next let us apply the Newton-Raphson method to the system of two nonlinear equations solved above using optimization methods. The sequence x 0,x 1,x 2,x 3, generated in the manner described below should con-verge to the exact root. Demonstrations of Newton raphson method and Arc-length method (https: MATLAB Release Compatibility. I cannot edit data?. I have looked at other similar questions posted but in my case I do not want to use a while. Linear, which equals 10 to the minus 1, I get one digit per iteration. However, this requires me to know the eccentricity which I don't know yet. Here are the three equations: $$c[\alpha I+ k_f+k_d+k_ns+k_p(1-q)]-I \alpha =0$$ $$s[\lambda_b c P_C +\lambda_r (1-q)]- \lambda_b c P_C =0$$. To apply the Newton Method's, you would need to do a Gateaux's differentiation. Earlier in Newton Raphson Method Algorithm, we discussed about an algorithm for computing real root of non-linear equation using Newton Raphson Method. suppose I need to solve f(x)=a*x. This can be done a few ways. The Newton - Raphson Method. Zero of the function f(x) is a point ˘ ∈ R such that f(˘) = 0: The problem of. 4-Convergence of the Newton Method and Modified Newton Method Consider the problem of finding x∗, the solution of the equation: f x 0forx in a, b. ANy form of help will be appreciated. m The next method proposed here is the one proposed by Newton-Raphson. Category Education; Show more Show less. I have looked at other similar questions posted but in my case I do not want to use a while. txt Example 2. Using multi-dimensional Taylor series, a system of non-linear equations can be written near an arbitrary starting point X i = [ x 1 , x 2 ,… , x n ] as follows: where. Fractals derived from Newton-Raphson iteration Introduction. To determine the power flow analysis using Newton - Raphson method. This can be easily generalized to the problem of finding a solution of a system of non-linear equations and linear equations known as Newton’s technique, that can be shown that the technique is quadratic as it approaches the origin. I'm studying Aeronautical Engineering and have a course in MATLAB to do this semester. 5 iterations for each value. First, construct a quadratic approximation to the function of interest around some initial parameter value (hopefully close to the MLE). Thanks Thanks Cite. Compared to the other methods we will consider, it is generally the fastest one (usually by far). Output: approximation solution x = (x1;:::;xn) for the nonlinear system F(x) =0. Tweaking the mesh here may offer some relief. that function f {\displaystyle f\,} has a definite slope at each point. Newton Raphson Iteration method in Matlab. You will see that the internal Matlab solving command fsolve approximates the solution, but only to about 7 decimal places. Optimal power flow algorithm of Newton-Raphson. In the beginning of the problem we divide the ODE (ordinary differential equation) to a set of first. -For example, to find the root of the equation. Hence it is desirable to have a method that. On February 10, 2020 | In Technology. In numerical analysis, Newton's method (also known as the Newton- Raphson method), named after Isaac Newton and Joseph Raphson, is perhaps the best known method for finding successively better approximations to the zeroes (or roots) of a real-valued function. In this case, I would try a numerical method to solve this ODE. Commented: Geoff Hayes on 2 Dec 2015. MATLAB Release Compatibility. It is indeed the practical method of load flow solution of large power networks. The method is advantageous, becaus. Guess the initial value of xo, here the gu. Riemannian Manifold Optimization Library. By comparing the number of iterations of these two methods, it is clear that Newton-Raphson is much more effective. derivative. Solving nonlinear equation using newton-raphson method. This can be done a few ways. Ste en Lauritzen, University of Oxford Newton{Raphson Iteration and the Method of Scoring. I'm using this function to try and find roots using Newton. The Scilab Program to implement the algorithm to find the roots of a polynomial using Newton Raphson Method. So we would have to enter that manually in our code. Use a calculator for the third step. f (x) = x 3 − 7. Here are the 3 non-linear equations:. The combined bisection/Newton-Raphson program is superior in almost every respect. The sequence x 0,x 1,x 2,x 3, generated in the manner described below should con-verge to the exact root. I am trying to solve 3 non-linear system of 3 variables using the newton-raphson method in matlab. Newton's Method in Matlab. We will present the Newton-Raphson algorithm, and the secant method. I'm studying Aeronautical Engineering and have a course in MATLAB to do this semester. , │f(xn)│ < 0. 5 Two nonlinear springs (modified Newton-Raphson method). In this method the function f(x) , is approximated by a tangent line, whose equation is found from the value of f(x) and its first derivative at the initial approximation. That is, it's a method for approximating $x^*$ such that $f(x^*)=0$. 3 Ill-conditioning. I am only concern. The specific root that the process locates depends on the initial, arbitrarily chosen x-value. Bisection Method Newton-Raphson Method Homework Problem Setup Newton-Raphson Method Procedure Newton-Raphson Method Advantages and Disadvantages Newton-Raphson Method Procedure Draw a line tangent to the function at the point (x 1,f(x 1)). The Newton-Raphson Method 1 Introduction The Newton-Raphson method, or Newton Method, is a powerful technique for solving equations numerically. The Newton-Raphson Method is one of the most extensively used methods for the original discovery. For the method to converge, your starting point must be sufficiently near a solution, and should have a derivative with respect to all variables somewhere along the path of convergence. What that means is the number of accurate digits in my solution will actually double with each iteration. Need to change the extension ". , the domain manifold, algorithm, stopping criterion. This post explores the how Newton's Method works for finding roots of equations and walks through several examples with SymPy to. The code is pretty simple it uses a while loop with the Newton-Raphson over a number of equations until I get a fixed point or value. Title: load flow analysis by newton-raphson method using matlab Page Link: load flow analysis by newton-raphson method using matlab - Posted By: ziddy_keshav Created at: Sunday 16th of April 2017 03:38:02 PM. The C program for Newton Raphson method presented here is a programming approach which can be used to find the real roots of not only a nonlinear. Explanation. Similar is for Secant method, and it requires two-initial guesses to proceed. 4-Convergence of the Newton Method and Modified Newton Method Consider the problem of finding x∗, the solution of the equation: f x 0forx in a, b. Newton-Raphson method). Summary Text A. I am new to matlab and I need to create a function that does n iterations of the Newton-Raphson method with starting approximation x = a. The idea behind Newton's Method is to approximate g(x) near the. Newton-Raphson Method, is a Numerical Method, used for finding a root of an equation. So far only r1 looks ok but the other 2 are 0 which is wrong. In this example, the system to be solved is The following statements are organized into three modules: NEWTON, FUN, and DERIV. % 2) x0 is the initial point. Like so much of the di erential calculus, it is based on the simple idea of linear approximation. 5>newton2v2", and the program ask for the functions and other elements that are necessary. We make an initial guess for the root we are trying to find, and we call this initial guess x 0. In this case, I would try a numerical method to solve this ODE. This program is not a generalised one. The Modified Newton-Raphson method uses the fact that f(x) and u(x) := f(x)/f'(x) have the same zeros and instead approximates a zero of u(x). Some other methods, such as the homotopy method [10, 11], may also be applied to provide an improved Newton-Raphson method. This gives at most three different solutions for x 1 for each fixed x 2. Fractals derived from Newton-Raphson iteration Introduction. Numerical Analysis Lab Note #8 Newton's Method for Nonlinear System 1. f (x) = x 3 − 7. // C++ program for implementation of Newton Raphson Method for // solving equations #include #define EPSILON 0. The Newton-Raphson method (also known as Newton's method) is a way to quickly find a good approximation for the root of a real-valued function f (x) = 0 f(x) = 0 f (x) = 0. If you store the function on the Matlab path, now you can do something like help newton_raphson_method and Matlab will print that comment block. The function. Taking calculus at Austin Peay State University and I understand how to do Newton's method of approximation the questions are just mundane after doing so many [6] 2020/03/30 21:58 Male / 30 years old level / High-school/ University/ Grad student / Useful /. Newton's Method Question. Newton-Raphson Method for Root-Finding; by Aaron Schlegel; Last updated over 3 years ago; Hide Comments (-) Share Hide Toolbars. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it. Metode Newton-Raphson adalah metode pencarian akar suatu fungsi f(x) dengan pendekatan satu titik, dimana fungsi f(x) mempunyai turunan. edu 3 4 Derivation Figure 2 Derivation of the Newton-Raphson method. edu 3 Newton-Raphson Method Figure 1 Geometrical illustration of the Newton-Raphson method. It's already defined. // C++ program for implementation of Newton Raphson Method for // solving equations #include #define EPSILON 0. The Jacobian is written in a very easy form to understabd. Newton-Raphson Method for Root-Finding; by Aaron Schlegel; Last updated over 3 years ago; Hide Comments (-) Share Hide Toolbars. The previous two methods are guaranteed to converge, Newton Rahhson may not converge in some cases. This online newton's method calculator helps to find the root of the expression. Some other methods, such as the homotopy method [10, 11], may also be applied to provide an improved Newton-Raphson method. Newton Raphson method requires derivative. Newton-Raphson method). In our case we will be using. This code calculates the load flow based on newton raphson methd for three bus power system. So we would have to enter that manually in our code. 414215686274510. A MATLAB function for the Newton-Raphson method. Many algorithms for geometry optimization are based on some variant of the Newton-Raphson (NR) scheme. In this course, students will learn how to solve problems of the type = using the Newton's Method, also called the Newton-Raphson iteration. Fink) and is dedicated to the particular case of polynomial functions because their. Moreover, it can be shown that the technique is quadratically convergent as we approach the root. The Newton-Raphon method is an iterative approach to finding the roots of some differentiable function $f(x)$. Newton Raphson load flow analysis Matlab Software 1. To apply the Newton Method's, you would need to do a Gateaux's differentiation. Similar is for Secant method, and it requires two-initial guesses to proceed. Answered: Walter Roberson on 14 Sep 2015 I'm using this function to try and find roots using Newton Raphsons method: function [x_sol, f_at_x_sol, N. Isaac Newton and Joseph Raphson, is a technique for judgment sequentially superior approximations to the extraction (or zeroes) of a real-valued function. There are three files: func. C Program for Newton-Raphson Method. The method requires an initial guess x(0) as input. 4 Divergence of the Newton-Raphson method E2_5. Then we know f x x −x∗ Q x where lim x→x∗ Q x ≠0. Newton-Raphson Method for Solving non-linear equat Unimpressed face in MATLAB(mfile) Bisection Method for Solving non-linear equations Gauss-Seidel method using MATLAB(mfile) Jacobi method to solve equation using MATLAB(mfile. You are recommended never to use this method without sufficient programming guards against instability. Some Of The Best Waterproof Smart Watch Into 2020. This document outlines the solution of the assignment problem given in 'Introduction to Space Flight Mechanics' by Dr Ramanann. It then computes subsequent iterates x(1), x(2), ::: that, hopefully, will converge to a solution x of g(x) = 0. Any help would be appreciated. For optimization problems, the same method is used, where is the gradient of the objective function and becomes the Hessian (Newton-Raphson). The latter represents a general method for finding the extrema (minima or maxima) of a given function f(x) in an iterative manner. If you store the function on the Matlab path, now you can do something like help newton_raphson_method and Matlab will print that comment block. Matlab Programs. Guess the initial value of xo, here the gu. Newton-Raphson method using MATLAB. f (x) = x 3 − 7. You were explicitly asked to carry 16 decimal places. Print the values of q and the number of iterations to the screen. MATLAB: Newton Raphson and Secant Method Tutorial 3. The tangent at x is then extended to intersect the x-axis, and the value of x at this intersection is the new estimate of the root. Here I will just do a brief overview of the method, and how its used. Many algorithms for geometry optimization are based on some variant of the Newton-Raphson (NR) scheme. , Decimal Floating-Point Square Root Using Newton-Raphson Iteration,'' Proceedings of the 16th International Conference on Application-Specific Systems,. derivative. After enough iterations of this, one is left with an approximation that can be as good as you like (you are also limited by the accuracy of the computation, in the case of MATLAB®, 16 digits). edu 5 Algorithm for Newton-Raphson Method http//numericalmethods. Learn more about newton's method, newton-raphson-iteration, homework MATLAB. To determine the power flow analysis using Newton - Raphson method. Thank you. The Newton-Raphson method reduces to. m applies the Newton-Raphson method to determine the roots of a function. Here, x n is the current known x-value, f(x n) represents the value of the function at x n, and f'(x n) is the derivative (slope) at x n. Category Education; Show more Show less. m file REDS Library 2. Perhaps the best known root finding algorithm is Newton's method (a. Learn more about newtonraphson. I want to write Matlab code for newton Raphson method. So far only r1 looks ok but the other 2 are 0 which is wrong. Problem in code for Newton Raphson Method. SECANT METHOD. The function. Answer: 3/2, 17/12, 577/408 ≈ 1. By comparing the number of iterations of these two methods, it is clear that Newton-Raphson is much more effective. the importance of the Jacobian is also highlighted. function approximateZero = newtonsMethod( fnc, x0, tol ) % implementation of Newton's Method for finding a zero of a function % requires a symbolic expression, a starting. Newton-Raphson Method may not always converge,. Matlab and Newtons method: Math Software: Mar 25, 2016: Newton-Raphson Method for Non-linear System of 3 variables in Matlab: Advanced Math Topics: Jun 16, 2014: Matlab Newton's method help: Math Software: Nov 10, 2009: Newtons method for matlab: Math Software: Oct 27, 2009. Implementing 2 exit criteria for Newton-Raphson Method HomeworkQuestion Hi, I have a problem where I need to guess the point of intersection using the NR method. , the domain manifold, algorithm, stopping criterion. I am trying to solve 3 non-linear system of 3 variables using the newton-raphson method in matlab. So I need to find Specific Volumes using the Newton Method from the Van der Waal equation over a change of constant Temperature but variant Pressure. Newton's method. f ( x) = 0 f (x) = 0. You saw in Lab 4 that approximating the Jacobian can result in a linear convergence rate instead of the usual quadratic rate, so quasi-Newton methods can take more iterations than true Newton methods will take. The Newton-Raphson method converges quickly when the starting point is close to the solution. Problem Statement: Develop a MATLAB program to solve a system of non-linear equations by the Newton-Raphson method, and then, test the code with the following equations: exp(2 x1) -x2 -4 =0. Newton's Method-How it works The derivative of the function,Nonlinear root finding equation at the function's maximum and minimum. The Newton-Raphson Method 1 Introduction The Newton-Raphson method, or Newton Method, is a powerful technique for solving equations numerically. This page describes a type of fractal derived from the Newton-Raphson method, which is more normally used as an approximate method of solving equations. To determine the power flow analysis using Newton - Raphson method. Problem in code for Newton Raphson Method. Follow 28 views (last 30 days) using newton's method. Any help would be appreciated. Implement the algorithm of Newton's Method for Nonlinear Systems: Input: function F(x) = [f1(x);:::;fn(x)]T, Jacobian Matrix function J(x) = (@[email protected] (x))1•i;j•n. Matlab example: Multidimensional Newton's Method Here is the textbook example, written out in a couple of les. Explanation. Summary Text A. The package optimizes the function given a set of user-specified parameters, e. It is indeed the practical method of load flow solution of large power networks. It is an open bracket method and requires only one initial guess. I was recently asked by a class to go over the Newton-Raphson method for solving non-linear equations. Newton-Raphson is an iterative numerical method for finding roots of. % INPUT:1) "fx" is the equation string of the interest. The Newton method is applied to find the root numerically in an iterative manner. Reference: Applied Numerical Methods Using MATLAB ®. This is akin to storing some data from a plot in a variable called plot. Bisection Method Newton-Raphson Method Homework Problem Setup Newton-Raphson Method Procedure Newton-Raphson Method Advantages and Disadvantages Newton-Raphson Method Procedure Draw a line tangent to the function at the point (x 1,f(x 1)). Newton's method is sometimes also known as Newton's iteration, although in this work the latter term is reserved to the application of Newton's method for computing square roots. 3 Ill-conditioning. One of the most common methods is the Newton{Raphson method and this is based on successive approximations to the solution, using Taylor's theorem to approximate the equation. Newton - Raphson requires derivative of the function while Secant method is derivative-free. This document outlines the solution of the assignment problem given in 'Introduction to Space Flight Mechanics' by Dr Ramanann. Implementasi metode Newton-Raphson pada MATLAB untuk sebarang fungsi : Dengan demikian algoritma Metode Newton-Raphson di atas , dibuat program MATLAB sebagai berikut : function hasil=newtonraphson(N,x0,T). Use to find roots of a function, f(x) = 0. edu 3 Newton-Raphson Method Figure 1 Geometrical illustration of the Newton-Raphson method. It begins with an initial guess for vn+1 and solves a linearized version of R=0 to find a correction to the initial guess for vn+1. I tried to develop a code in MATLAB to solve 3 nonlinear equations using newton raphson method, but it seems that the code is not working, does anyone have the ability to fix it:. A MATLAB function for the Newton-Raphson method. This gives at most three different solutions for x 1 for each fixed x 2. Learn more about root finding help MATLAB Answers. It can be easily generalized to the problem of finding solutions of a system of non-linear equations, which is referred to as Newton's technique. the importance of the Jacobian is also highlighted. If this condition is not valid, we have to reduce step size until having an acceptable. I found it was useful to try writing out each method to practice working with MatLab. The Newton-Raphson method which is employed for solving a single non-linear equation can be extended to solve a system of non-linear equations. The Newton-Raphson method for systems of nonlinear equations. The Newton-Raphson Method For Root Finding Matlab 2020. Secara geometri, metode Newton Raphson hampir sama dengan. A MATLAB function for the Newton-Raphson method. Need to change the extension ". I was recently asked by a class to go over the Newton-Raphson method for solving non-linear equations. In this study report I try to represent a brief description of root finding methods which is an important topic in Computational Physics course. Matlab example: Multidimensional Newton's Method Here is the textbook example, written out in a couple of les. For many problems, Newton Raphson method converges faster than the above two methods. Newton-Raphson Method with MATLAB code: If point x0 is close to the root a, then a tangent line to the graph of f(x) at x0 is a good approximation the f(x) near a. Thus, at most 9 different x 1 points exist for. Compared to the other methods we will consider, it is generally the fastest one (usually by far). Bisection Method Newton-Raphson Method Homework Problem Setup Newton-Raphson Method Procedure Newton-Raphson Method Advantages and Disadvantages Newton-Raphson Method Procedure Draw a line tangent to the function at the point (x 1,f(x 1)). Specially I discussed about Newton-Raphson's algorithm to find root of any polynomial equation. Here are the three equations: $$c[\alpha I+ k_f+k_d+k_ns+k_p(1-q)]-I \alpha =0$$ $$s[\lambda_b c P_C +\lambda_r (1-q)]- \lambda_b c P_C =0$$. In this case, I would try a numerical method to solve this ODE. The Newton-Raphson method reduces to. Root finding: Newton‐Raphson method 3. Numerical Analysis Lab Note #8 Newton's Method for Nonlinear System 1. I am trying to optimize the variables of two (or three depending on how you think about it) matrices using the Newton-Raphson Method. 414215686274510. Learn more about matrix, function, newton raphson. The easiest case of the Newton-Raphson method leads to thexn+1 = xn −. Back to M331: Matlab Codes, Notes and Links. m and newtonraphson. I cannot edit data?. http//numericalmethods. I'm studying Aeronautical Engineering and have a course in MATLAB to do this semester. Metode Newton Raphson biasa digunakan dalam mencari akar dari suatu persamaan non linier, jika diasumsikan f mempunyai turunan kontinu f’. Suppose we want to find the first positive root of the function g(x)=sin(x)+x cos(x). The Newton-Raphson method for systems of nonlinear equations. Finally, eps is a built-in parameter. m defines the derivative of the function and newtonraphson. That is, it's a method for approximating $x^*$ such that $f(x^*)=0$. While that would be close enough for most applications, one would expect that we could do better on such a simple problem. Like so much of the di erential calculus, it is based on the simple idea of linear approximation. Metode Newton Raphson biasa digunakan dalam mencari akar dari suatu persamaan non linier, jika diasumsikan f mempunyai turunan kontinu f'. In this method the function f(x) , is approximated by a tangent line, whose equation is found from the value of f(x) and its first derivative at the initial approximation. If q is 2, which we'll see is something that results from the Newton-Raphson method, then we say convergence is quadratic. newton_raphson_polynom. The desired precision is reached by iteration. Parabolic Trough Collector (Differ REDS Library 1. Due Date: April 24, 2008. 001 using namespace std; // An example function whose solution is determined using // Bisection Method. m) illustrates the while loop structure in MATLAB which causes a block of code to be executed repeatedly until a condition is met. Hot Network Questions Is it ethical to have two (undergraduate) researchers in the same group "compete" against one another for. Learn more about matlab, while loop, function. Interesting is Section 3, where the birth of the. The method is advantageous, becaus. It should be noted that the “root” function in the MATLAB library can find all the roots of a polynomial with arbitrary order. % b) Newton's method implemented in MATLAB. Compared to the other methods we will consider, it is generally the fastest one (usually by far). We will be excessively casual in our notation. Problem in code for Newton Raphson Method. If q is 2, which we'll see is something that results from the Newton-Raphson method, then we say convergence is quadratic. Newton Raphson Method Notice: this material must not be used as a substitute for attending the lectures 1. MATLAB has many tools that make this package well suited for numerical computations. Next, adjust the parameter value to that which maximizes the. Or decrease the Normal Stiffness Factor of the contact in question to 0. One of the most common methods is the Newton{Raphson method and this is based on successive approximations to the solution, using Taylor's theorem to approximate the equation. The previous two methods are guaranteed to converge, Newton Rahhson may not converge in some cases. Matlab and Newtons method: Math Software: Mar 25, 2016: Newton-Raphson Method for Non-linear System of 3 variables in Matlab: Advanced Math Topics: Jun 16, 2014: Matlab Newton's method help: Math Software: Nov 10, 2009: Newtons method for matlab: Math Software: Oct 27, 2009. Many algorithms for geometry optimization are based on some variant of the Newton-Raphson (NR) scheme. 2 Newton's Method and the Secant Method The bisection method is a very intuitive method for finding a root but there are other ways that are more efficient (find the root in fewer iterations). Easy application of Newton-Raphson algorithm. Newton-Raphson method). Método de Newton-Raphson en Matlab. Learn About. txt Example 2. A Newton-Raphson method is a successive approximation procedure based on an initial estimate of the one-dimensional equation given by series expansion. Here are the 3 non-linear equations:. Newton-Raphson Method for Solving non-linear equat Unimpressed face in MATLAB(mfile) Bisection Method for Solving non-linear equations Gauss-Seidel method using MATLAB(mfile) Jacobi method to solve equation using MATLAB(mfile. Numerical Analysis (MCS 471) Multivariate Newton's Method L-6(b) 29 June 2018 10 / 14. Use to find roots of a function, f(x) = 0. Newton's method for finding successively better approximations to the zeroes of a real-valued function. m file REDS Library 2. That is, it's a method for approximating $x^*$ such that $f(x^*)=0$. Newton's Method in Matlab. Advantages & Drawbacks for Newton-Raphson Method: Part 2 of 2 [YOUTUBE 4:43] Derivation from Taylor Series of Newton-Raphson Method [ YOUTUBE 7:56] [ TRANSCRIPT ] Supercomputers have No Divide Unit - A Newton-Raphson Method Approach [ YOUTUBE 10:14] [ TRANSCRIPT ]. [email protected] Later in his answer he explains how find the eccentricity by fitting a function to the graph, but this requires me to know the eccentric anomaly at any given time, which seems. Problem in code for Newton Raphson Method. However, choosing of the starting x0point is very important, because convergence may no longer stand for even the easiest equations. The Newton Raphson Method. Newton Raphson Root finding method not working. The root of a function is the point at which $$f(x) = 0$$. I want to update my old V value to the eqation for f. You saw in Lab 4 that approximating the Jacobian can result in a linear convergence rate instead of the usual quadratic rate, so quasi-Newton methods can take more iterations than true Newton methods will take. Follow 3 views (last 30 days) Benjamin Halkowski on 13 Sep 2015. This method uses the derivative of f(x) at x to estimate a new value of the root. The code is pretty simple it uses a while loop with the Newton-Raphson over a number of equations until I get a fixed point or value. Perhaps the best known root finding algorithm is Newton's method (a. During the study in a two-bus system, with a slack bus and PQ bus, two convergence tolerances were used. Then you plug the x 1 back in as x 0 and iterate. 3 Modified Newton-Raphson Method for Systems The Newton-Raphson method can be modified following way to achieve higher numerical stability: In each iteration, compute the Newton-Raphson step and check whether. We will present the Newton-Raphson algorithm, and the secant method. Newton-Raphson method is implemented here to determine the roots of a function. Create scripts with code, output, and formatted text in a single executable document. I am trying to solve 3 non-linear system of 3 variables using the newton-raphson method in matlab. This can be easily generalized to the problem of finding a solution of a system of non-linear equations and linear equations known as Newton’s technique, that can be shown that the technique is quadratic as it approaches the origin. Create AccountorSign In. In numerical analysis, Newton’s method (also known as the Newton– Raphson method), named after Isaac Newton and Joseph Raphson, is perhaps the best known method for finding successively better approximations to the zeroes (or roots) of a real-valued function. Método de Newton-Raphson en Matlab. 001 using namespace std; // An example function whose solution is determined using // Bisection Method. It can be easily generalized to the problem of finding solutions of a system of non-linear equations, which is referred to as Newton's technique. The specific root that the process locates depends on the initial, arbitrarily chosen x-value. SOFTWARE REQUIRED: MATLAB THEORY: The Newton Raphson method of load flow analysis is an iterative method which approximates the set of non-linear simultaneous equations to a set of linear simultaneous equations using Taylor’s series expansion and the terms are limited to. The tangent at x is then extended to intersect the x-axis, and the value of x at this intersection is the new estimate of the root. Metode ini dianggap lebih mudah dari Metode Bagi-Dua (Bisection Method) karena metode ini menggunakan pendekatan satu titik sebagai titik awal. A Newton-Raphson method is a successive approximation procedure based on an initial estimate of the one-dimensional equation given by series expansion. Finally, eps is a built-in parameter. I'm using this function to try and find roots using Newton. txt Example 2. Also note that since the root is simple, Newton's Method is order 2, while the Secant method is somewhere close. The Newton-Raphson algorithm is the most commonly used iterative method to solve the power flow problem. Learn more about newton raphson, jacobi, iteration, matlab MATLAB. So far only r1 looks ok but the other 2 are 0 which is wrong. Find zeros of a function using the Modified Newton-Raphson method. Use a calculator for the third step. Moreover, it can be shown that the technique is quadratically convergent as we approach the root. However, in some circumstances, the method is known to diverge or provide no information that the solution exists. The code below solve this initial value problem (IVP) using the function ode45. You saw in Lab 4 that approximating the Jacobian can result in a linear convergence rate instead of the usual quadratic rate, so quasi-Newton methods can take more iterations than true Newton methods will take. And this is solvable using the Newton-Raphson method which I think I know how to use. Next let us apply the Newton-Raphson method to the system of two nonlinear equations solved above using optimization methods. the Newton-Raphson method appears as the limiting case of the presented method. Newton-Raphson Method for Root-Finding; by Aaron Schlegel; Last updated over 3 years ago; Hide Comments (-) Share Hide Toolbars. The point to notice here is that we output not just the value of the function, but also its Jacobian matrix: function [y dy]=myfunction(x). Thus, at most 9 different x 1 points exist for. First, construct a quadratic approximation to the function of interest around some initial parameter value (hopefully close to the MLE). The Newton-Raphson method, or Newton Method, is a powerful technique. The term quasi-Newton'' method basically means a Newton method using an approximate Jacobian instead of an exact one. Newton-Raphson method, named after Isaac Newton and Joseph Raphson, is a popular iterative method to find the root of a polynomial equation. The formula of the NMR is: x = x0 -(f(x0)/f'(x0)). This is not a new idea to me; I was given the idea by a colleague at work, and several other people have web pages about it too. The user % may input any string but it should be constructable as a "sym" object. Numerical Analysis Lab Note #8 Newton's Method for Nonlinear System 1. Aug 18, 2017. Problem Statement: Develop a MATLAB program to solve a system of non-linear equations by the Newton-Raphson method, and then, test the code with the following equations: exp(2 x1) -x2 -4 =0. The code is pretty simple it uses a while loop with the Newton-Raphson over a number of equations until I get a fixed point or value. Newton Raphson method initial guess. For all things, really, don't overwrite built-in functions. It should be noted that the “root” function in the MATLAB library can find all the roots of a polynomial with arbitrary order. 2 Multi-dimensional Newton-Raphson Method. Semakin dekat titik awal yang kita pilih dengan akar sebenarnya, maka semakin cepat konvergen ke akarnya. So the root of the tangent line, where the line cuts the X-axis; x1 is the better approximation to a than x0 is. We make an initial guess for the root we are trying to find, and we call this initial guess x 0. For minima, the first derivative f'(x) must be zero and the second derivative.$30 USD in 1 day (0 Reviews). we observe quadratic convergence In the output below, there are four columns: 1 the norm of the residual, 2 the norm of the update, 3 the value for x, 4 the value for y. In the beginning of the problem we divide the ODE (ordinary differential equation) to a set of first. Need to change the extension ". % b) Newton's method implemented in MATLAB. Newton-Raphson (NR) optimization. The code is pretty simple it uses a while loop with the Newton-Raphson over a number of equations until I get a fixed point or value. http//numericalmethods. SOFTWARE REQUIRED: MATLAB THEORY: The Newton Raphson method of load flow analysis is an iterative method which approximates the set of non-linear simultaneous equations to a set of linear simultaneous equations using Taylor’s series expansion and the terms are limited to first order approximation. Any help would be appreciated. Find zeros of a function using the Modified Newton-Raphson method. Isaac Newton and Joseph Raphson, is a technique for judgment sequentially superior approximations to the extraction (or zeroes) of a real-valued function. I have looked at other similar questions posted but in my case I do not want to use a while. Tweaking the mesh here may offer some relief. It does not guarantee that an existing solution will be found, however. So I need to find Specific Volumes using the Newton Method from the Van der Waal equation over a change of constant Temperature but variant Pressure. edu 3 Newton-Raphson Method Figure 1 Geometrical illustration of the Newton-Raphson method. Similar is for Secant method, and it requires two-initial guesses to proceed. This can be easily generalized to the problem of finding a solution of a system of non-linear equations and linear equations known as Newton’s technique, that can be shown that the technique is quadratic as it approaches the origin. Newton's Method is an iterative method that computes an approximate solution to the system of equations g(x) = 0. Newton-Raphson Method with MATLAB code: If point x0 is close to the root a, then a tangent line to the graph of f(x) at x0 is a good approximation the f(x) near a. f ( x) = 0 f (x) = 0. It begins with an initial guess for vn+1 and solves a linearized version of R=0 to find a correction to the initial guess for vn+1. Linear, which equals 10 to the minus 1, I get one digit per iteration. Mr Rafael Marques can you or anyone else send me the Lambert W function using Matlab to m. MATLAB allows for the creation of function handles. That is, it's a method for approximating $x^*$ such that $f(x^*)=0$. The method requires an initial guess x(0) as input. This can be done a few ways. com, i want to compare Newton- Raphson and lambart W method. The NRM uses divisions, so it can give a indefinite math error, once the denominator is zero. Summary Text A. I have a problem "find the steady-state solution of the following plant equation by using MATLAB codes", (Newton-Raphson method) ~~~ many thanks This is Newton-Raphson code function [x,iter] = newtonm(x0,f,J). EXPERIMENT NO 5 OBJECTIVE To develop a software program to obtain real and reactive power flows, bus voltage magnitude and angles by using N-R method. 2018-12-17. This can be easily generalized to the problem of finding a solution of a system of non-linear equations and linear equations known as Newton’s technique, that can be shown that the technique is quadratic as it approaches the origin. SOFTWARE REQUIRED: MATLAB THEORY: The Newton Raphson method of load flow analysis is an iterative method which approximates the set of non-linear simultaneous equations to a set of linear simultaneous equations using Taylor’s series expansion and the terms are limited to. Newton's Method: Newton's Method is used to find successive approximations to the roots of a function. Executive Summary. % b) Newton's method implemented in MATLAB. C Program for Newton Raphson Method Algorithm First you have to define equation f(x) and its first derivative g(x) or f'(x). 5 iterations for each value. It starts from an initial guess by user and iterates until satisfy the required convergence criterion. I am trying to optimize the variables of two (or three depending on how you think about it) matrices using the Newton-Raphson Method. Reference: Applied Numerical Methods Using MATLAB ®. Ste en Lauritzen, University of Oxford Newton{Raphson Iteration and the Method of Scoring. Wilks, in International Geophysics, 2011. The code is pretty simple it uses a while loop with the Newton-Raphson over a number of equations until I get a fixed point or value. The Newton-Raphson algorithm is the most commonly used iterative method to solve the power flow problem. When the EM algorithm can be formulated for a maximum-likelihood estimation problem, the difficulties experienced by the Newton-Raphson approach do not occur. Best Robotic Process Automation Vendor Tools More In-depth Vendor Selection Guide 2020. If q is 2, which we'll see is something that results from the Newton-Raphson method, then we say convergence is quadratic. Specially I discussed about Newton-Raphson's algorithm to find root of any polynomial equation. 2 Multi-dimensional Newton-Raphson Method. Newton-Raphson Method with MATLAB code: If point x0 is close to the root a, then a tangent line to the graph of f(x) at x0 is a good approximation the f(x) near a. This page describes a type of fractal derived from the Newton-Raphson method, which is more normally used as an approximate method of solving equations. Newton-Raphson method is also one of the iterative methods which are used to find the roots of given expression. we observe quadratic convergence In the output below, there are four columns: 1 the norm of the residual, 2 the norm of the update, 3 the value for x, 4 the value for y. The report aims to show the differences between Newton-Raphson and Gauss-Seidel methods by using them to analyse a power flow system. Secara geometri, metode Newton Raphson hampir sama dengan. The Newton-Raphson Method is a better version of the Fixed Point Interation Method, increasing the speed of the convergence to find the root of the equation. The end result is only a small part of the problem. function approximateZero = newtonsMethod( fnc, x0, tol ) % implementation of Newton's Method for finding a zero of a function % requires a symbolic expression, a starting. The Newton-Raphon method is an iterative approach to finding the roots of some differentiable function $f(x)$. Implementasi metode Newton-Raphson pada MATLAB untuk sebarang fungsi : Dengan demikian algoritma Metode Newton-Raphson di atas , dibuat program MATLAB sebagai berikut : function hasil=newtonraphson(N,x0,T). So I need to find Specific Volumes using the Newton Method from the Van der Waal equation over a change of constant Temperature but variant Pressure. 5 iterations for each value. It starts from an initial guess by user and iterates until satisfy the required convergence criterion. However, for problems involving more than perhaps three parameters, the computations required can expand dramatically. Many algorithms for geometry optimization are based on some variant of the Newton-Raphson (NR) scheme. The code comes with comments for each line for the user to understand the basics of the load flow and how it is calculated. Executive Summary. Problem 1 [60 Write a Matlab program that uses the Newton-Raphson method to solve the system of equations f(q) = 0 A(q) = sin(ql) + Select le-10 as the convergence threshold. 2 Newton Raphson Method 2. Bisection Method Newton-Raphson Method Homework Problem Setup Newton-Raphson Method Procedure Newton-Raphson Method Advantages and Disadvantages Newton-Raphson Method Procedure Draw a line tangent to the function at the point (x 1,f(x 1)). MATLAB Release Compatibility. It's already defined. This function can be used to perform Newton-Raphson method to detect the root of a polynomial. First, construct a quadratic approximation to the function of interest around some initial parameter value (hopefully close to the MLE). Learn About. Newton Raphson method initial guess. El método de Newton-Raphson se deduce a partir de esta interpretación geométrica y se tiene que la primera derivada en x es equivalente a la pendiente:. Newton's Method in Matlab. This program is not a generalised one. 1 Basic Algorithm. MATLAB Programming Tutorial #27 Newton-Raphson (multi Variable) Complete MATLAB Tutorials @ https://goo. \$30 USD in 1 day (0 Reviews). Limitations of Newton's Method. Wilks, in International Geophysics, 2011. Matlab Newton Raphson Method programming21 / By justin All my workings are based mostly within matlab framework of historic talents, and are brand that unifies both engineering matlab actual and non secular realms. Title: load flow analysis by newton-raphson method using matlab Page Link: load flow analysis by newton-raphson method using matlab - Posted By: ziddy_keshav Created at: Sunday 16th of April 2017 03:38:02 PM. During the study in a two-bus system, with a slack bus and PQ bus, two convergence tolerances were used. Anyone who have experience to work on "Power System Improvement using UPFC" (Newton Raphson algorithm used in it and MATLAB used as a Tool). While that would be close enough for most applications, one would expect that we could do better on such a simple problem. Numerical analysis I 1. The Newton-Raphson method converges quickly when the starting point is close to the solution. The Newton-Raphson Method For Root Finding Matlab 2020. Activities for factoring quadratic equations, two step word problems worksheets, ti 84 calculator online free use, list all type fractions for a beginner with samples, how to. It's already defined. Reference: Applied Numerical Methods Using MATLAB ®. I am trying to solve 3 non-linear system of 3 variables using the newton-raphson method in matlab. that function f {\displaystyle f\,} has a definite slope at each point. This online newton's method calculator helps to find the root of the expression. You are recommended never to use this method without sufficient programming guards against instability. Fractals derived from Newton-Raphson iteration Introduction. writing a script to find roots using Newton-Raphson Method in Matlab, how ? Follow 12 views (last 30 days) zee 00 on 2 Dec 2015. If you have any queries post it in comments down below. Suppose we want to find the first positive root of the function g(x)=sin(x)+x cos(x). Derivation of the Newton's Method [ edit ] In Newton's method, we must first assume that the function is differentiable, i. The Newton-Raphson method which is employed for solving a single non-linear equation can be extended to solve a system of non-linear equations. m defines the function, dfunc. 2 Jacobian Matrix. Learn more about matrix, function, newton raphson. Some functions may be difficult to impossible to differentiate. This page describes a type of fractal derived from the Newton-Raphson method, which is more normally used as an approximate method of solving equations. The Newton-Raphson method is one of the most widely used methods for root finding. The user % may input any string but it should be constructable as a "sym" object. Isaac Newton and Joseph Raphson, is a technique for judgment sequentially superior approximations to the extraction (or zeroes) of a real-valued function. how to code newton-raphson method for solving Learn more about newton-raphson, non linear, jacobian. 5 iterations for each value. Buy Power Flow Solution Using Newton Raphson Method in Matlab [Download]: Read Software Reviews - Amazon. Matlab and Newtons method: Math Software: Mar 25, 2016: Newton-Raphson Method for Non-linear System of 3 variables in Matlab: Advanced Math Topics: Jun 16, 2014: Matlab Newton's method help: Math Software: Nov 10, 2009: Newtons method for matlab: Math Software: Oct 27, 2009. Secant Methods (1/2) • A potential problem in implementing the Newton-Raphson method is the evaluation of the derivative - there are certain functions whose derivatives may be difficult or inconvenient to evaluate • For these cases, the derivative can be approximated by a backward finite divided difference: f ' (x i) f (x i 1) f (x i) x i 1 x i. Solutions to Problems on the Newton-Raphson Method These solutions are not as brief as they should be: it takes work to be brief. Title: load flow analysis by newton-raphson method using matlab Page Link: load flow analysis by newton-raphson method using matlab - Posted By: ziddy_keshav Created at: Sunday 16th of April 2017 03:38:02 PM. Implement the algorithm of Newton's Method for Nonlinear Systems: Input: function F(x) = [f1(x);:::;fn(x)]T, Jacobian Matrix function J(x) = (@[email protected] (x))1•i;j•n. Back to M331: Matlab Codes, Notes and Links. Newton Raphson Method Errors. We will be excessively casual in our notation. IntroducEon% • Newton’s%Method%(also%known%as%Newton#Raphson%Method)% is%used%to%solve%nonlinear%(system)%of%equaons,%which%can%be% represented%as%follows:%. The report aims to show the differences between Newton-Raphson and Gauss-Seidel methods by using them to analyse a power flow system. f ( x) = 0 f (x) = 0. These "hot spots" highlighted by the Newton-Raphson Residuals are the causing difficulty for convergence. Create scripts with code, output, and formatted text in a single executable document. The load flow equations for Newton Raphson method are. http//numericalmethods. The Newton-Raphon method is an iterative approach to finding the roots of some differentiable function $f(x)$. Finally, eps is a built-in parameter. Observe the quadratic convergence:. 233406958320529e-010 is not "close enough". Semakin dekat titik awal yang kita pilih dengan akar sebenarnya, maka semakin cepat konvergen ke akarnya. A Newton-Raphson method is a successive approximation procedure based on an initial estimate of the one-dimensional equation given by series expansion. In a nutshell, the Newton-Raphson Algorithm is a method for solving simultaneous nonlinear algebraic equations. For all things, really, don't overwrite built-in functions. txt Example 2. The Newton-Raphson method (also known as Newton's method) is a way to quickly find a good approximation for the root of a real-valued function f (x) = 0 f(x) = 0 f (x) = 0. Newton-Raphson method is implemented here to determine the roots of a function. writing a script to find roots using Newton-Raphson Method in Matlab, how ? Follow 12 views (last 30 days) zee 00 on 2 Dec 2015. Newton's Method Question. Newton - Raphson requires derivative of the function while Secant method is derivative-free. Riemannian Manifold Optimization Library. El método de Newton-Raphson se deduce a partir de esta interpretación geométrica y se tiene que la primera derivada en x es equivalente a la pendiente:. Newton-Raphson Method http//numericalmetho ds. Earlier in Newton Raphson Method Algorithm, we discussed about an algorithm for computing real root of non-linear equation using Newton Raphson Method. Anyone who have experience to work on "Power System Improvement using UPFC" (Newton Raphson algorithm used in it and MATLAB used as a Tool). The function returns the solution of three equations in three variables using the Newton-Raphson method. So far only r1 looks ok but the other 2 are 0 which is wrong. The method requires the knowledge of the derivative of the equation whose root is to be determined. The report aims to show the differences between Newton-Raphson and Gauss-Seidel methods by using them to analyse a power flow system. txt Example 2. If the function is y = f(x) and x 0 is close to a root, then we usually expect the formula below to give x 1 as a better approximation.
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# 1013 Results
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This video introduces the transformations we'll be using in the rest of this lesson.
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First let's think about how to rotate some really simple points such as (0,0) and (1,0)
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First we'll use the slope intercept form of a line to define each frame along a straight line.
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First we need to make sure we understand exactly what happens in the split & average steps.
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Find out how we can make curved lines using straight ones using the string art construction.
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What happens if the director changes their mind and asks for two headed robots?
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First we'll review weighted averages of two points and extend the idea to three points. Practice weighted averages of two points in Environment Modeling if you haven't seen it before.
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How can we calculate a weighted average between two points? (pssst. This video is super important).
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Next lets build a diagram that break rotation into smaller parts. The next exercise will give us a chance to build our understanding of this diagram.
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Let's look more closely at how light behaves when it strikes an object. We'll cover diffuse and specular surface responses.
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First we'll review De Casteljau's algorithm using three points. Then it's your turn to figure out how to do it with 4 points!
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Let's review the multiplication principle which allows us to quickly count the number of possible robots.
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Now we can combine split and average into a single operation called subdivide.
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Next let's extend the averaging step from the previous lesson to include multiple points. Now we'll need to calculate positions using a weighted average.
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Where does the string touch the parabola? See if you can come up with your hypothesis!
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Now we are ready to calculate an intersection point using our ray CP (parametric form) and our line AB (slope-intercept form).
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Now that we have a feeling for constructing permutations let's introduce the factorial formula to make counting them easy.
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In this video we'll uncover the connection between the previous diagram and the rotation formulas. Repeat viewing suggested!
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This workshop covers the basics of 3D modelling in Processing. From the 3D coordinate system, placing different shapes, surfaces, and camera angles. This introductory workshop is suitable for all students with some basic Processing knowledge. We assume that you are familiar with 2D shapes in Processing, including pushMatrix, rotate and translate. This workshop will only cover basics, sufficient to create a landscape with 3D objects and a moving object.
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Ansgar Fehnker
03/02/2021
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We can use de Casteljau's algorithm to calculate curves using any number of points.
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Let's take a closer look at the weights used during subdivision. Do we have to be careful when selecting weights?
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Let's look more closely at how light bounces when it strikes an object. We'll cover reflected and refracted rays.
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Are we really creating parabolic curves using this construction? Let's gain some insight first.
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Use an array to store many objects as well as create any shape you can imagine. Click here to review objects.
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Now you can start scaling your shapes to make your lamp look younger.
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Now you are ready to start subdividing your own shapes with more than 4 points!
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Now it's time for a really meaty problem! How can we count the number of possible casts when given a large set of robots to choose from?
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We need to be careful with the order of scaling and translation. But why?
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Tree diagrams allow us to visualize these counting problems using any number of parts.
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Okay we know how to calculate the touching point, great! Next let's think about how we can prove this is true.
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Next let's build a blade of grass using a parabolic arc as a spine.
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Let's breathe some life into our ball using a key animation principle: squash and stretch.
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Next we need to throw away the slope-intercept form and use the line equation instead.
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Bonus! In this video we'll connect the degree of these curves to the number of control points in the construction.
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Now we are ready to ray trace in 3D. We'll look at the problem of ray triangle intersection.
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Now it's your turn to drive. In this video we'll present you with a casting challenge to complete using everything we've learned so far.
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Why do we divide the number of combinations by the number of permutations?
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Now we just need to determine whether our intersection point is inside or outside the triangle.
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Let's put everything together. Get ready for a really powerful formula: the binomial coefficient (warning: you may need to watch this a few times!).
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Time to get a little mathy. Let's look at the general form for any transformation.
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This assignment is intended to encourage students to explore relevant jobs in the Information Technology arena post the completion of their A.A.S degree.
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Computer Science
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Dalvin Hill
05/23/2020
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An A-Frame Virtual Reality Programming activity for CS0 students. Part of the CUNY CS04All project.
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Domanski Robert J
Robert J Domanski
06/04/2019
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ASP is a powerful tool for making dynamic and interactive Web pages.
In our ASP tutorial you will learn about ASP, and how to execute scripts on your server.
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Computer Science
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w3schools
07/12/2015
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This activity guides students through the evaluation of a website that they have created to see if it is accessible for users with disabilities. Students will simulate a number of different disabilities (e.g. visual impairments, color blindness, auditory impairments, motor impairments) to see if their website is accessible; they will also use automated W3 and WAVE tools to evaluate their sites. Students will consider the needs of users with disabilities by creating a persona and scenario of a user with disabilities interacting with their site. Finally, students will write up recommendations to change their site and implement the changes.
Although this activity can be used in isolation, it is intended to be part of a series guiding students towards the creation of a front-end of a website. The series (all published as OER) consist of:
a) Needfinding
b) Personas, Scenarios and Storyboards
c) Front-end Website Design and Development
d) Accessibility Evaluation
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Computer Science
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Brooklyn College
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Devorah Kletenik
09/03/2020
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This presentation introduces Computer Science students to the notion of accessibility: developing software for people with disabilities. This lesson provides a discussion of why accessibility is important (including the legal, societal and ethical benefits) as well as an overview of different types of impairments (visual, auditory, motor, neurological/cognitive) and how developers can make their software accessible to users with those disabilities. This lesson includes videos and links to readings and tutorials for students.
These slides use Poll Everywhere polls; to use them, create your own Poll Everywhere account and duplicate the polls.
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Computer Science
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Brooklyn College
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Devorah Kletenik
09/03/2020
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Inside, you’ll find many ideas you can use to enliven your synchronous online class meetings with active learning activities.
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Computer Science
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Maricopa Open Digital Press
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Cheryl Colan
05/03/2021
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An example homework that runs students on the personas and how to use them in the GenderMag Walkthrough.
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Computer Science
Engineering
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Margaret Burnett
11/24/2021
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Activity where the class will perform a GenderMag walkthrough on the ACM website from the GenderMag persona’s perspective.
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Computer Science
Engineering
Information Science
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The GenderMag Project
11/15/2021
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Students mark on the continuum where they think their persona's cognitive styles are likely to fall.
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Computer Science
Engineering
Information Science
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Activity/Lab
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Margaret Burnett
Zoe Steine-Hanson
11/15/2021
Some Rights Reserved
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Learn how to add event listeners in jQuery so that your JavaScript can respond to events on the page, like when a user clicks a button or drags an image.
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Applied Science
Computer Science
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Author:
Pamela Fox
07/11/2021
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This course will present advanced topics in Artificial Intelligence (AI), including inquiries into logic, artificial neural network and machine learning, and the Turing machine. Upon successful completion of this course, students will be able to: define the term 'intelligent agent,' list major problems in AI, and identify the major approaches to AI; translate problems into graphs and encode the procedures that search the solutions with the graph data structures; explain the differences between various types of logic and basic statistical tools used in AI; list the different types of learning algorithms and explain why they are different; list the most common methods of statistical learning and classification and explain the basic differences between them; describe the components of Turing machine; name the most important propositions in the philosophy of AI; list the major issues pertaining to the creation of machine consciousness; design a reasonable software agent with java code. (Computer Science 408)
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The course aims at providing the fundamental tools for effective C++ programming in the context of high-performance computing. The tools include generic programming techniques, API development, and specific C++-11/14/17 constructs. Starting from a basic knowledge of C++, the attendees should be able to start using C++ language to engineer durable abstractions to develop and optimize applications. Example usage of modern C++ concepts and features are taken from scientific applications used by the HPC community, giving the attendees the opportunity to see the presented tools in action in real world cases. Exercises are provided from a GitHub repository. This material is meant to reflect the current state of the current C++ standard. As the standard changes, some aspects of this course may become outdated.This course is an integral part of the ESiWACE-2 project, and we acknowledge the partial funding from that project. The contact person is [email protected].
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This course will expand upon SQL as well as other advanced topics, including query optimization, concurrency, data warehouses, object-oriented extensions, and XML. Additional topics covered in this course will help you become more proficient in writing queries and will expand your knowledge base so that you have a better understanding of the field. Upon successful completion of this course, the student will be able to: write complex queries, including full outer joins, self-joins, sub queries, and set theoretic queries; write stored procedures and triggers; apply the principles of query optimization to a database schema; explain the various types of locking mechanisms utilized within database management systems; explain the different types of database failures as well as the methods used to recover from these failures; design queries against a distributed database management system; perform queries against database designed with object-relational extensions; develop and query XML files. (Computer Science 410)
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African American History and Culture contains 10 modules starting with African Origins - History and Captivity and continuing through Reconstruction. Openly-licensed course materials developed for the Open Educational Resources (OER) Degree Initiative, led by Achieving the Dream https://courses.lumenlearning.com/catalog/achievingthedream.
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Building on Complex Adaptive Systems theory and basic Agent Based Modeling knowledge presented in SPM4530, the Advanced course will focus on the model development process. The students are expected to conceptualize, develop and verify a model during the course, individually or in a group. The modeling tasks will be, as much as possible, based on real life research problems, formulated by various research groups from within and outside the faculty.
Study Goals The main goal of the course is to learn how to form a modeling question, perform a system decomposition, conceptualize and formalize the system elements, implement and verify the simulation and validate an Agent Based Model of a socio-technical system.
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03/03/2016
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How can we improve the speed of a (deterministic) primality test? Created by Brit Cruise.
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https://www.shaalaa.com/question-paper-solution/cbse-cbse-12-mathematics-class-12-2017-2018-delhi-set-3_14109 | 1,695,981,865,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00014.warc.gz | 1,056,803,465 | 32,493 | # Mathematics Delhi Set 3 2017-2018 Science (English Medium) Class 12 Question Paper Solution
Mathematics [Delhi Set 3]
Date: March 2018
[1] 1
Find the magnitude of each of two vectors veca and vecb having the same magnitude such that the angle between them is 60° and their scalar product is 9/2
Concept: Product of Two Vectors - Scalar (Or Dot) Product of Two Vectors
Chapter: [0.1] Vectors
[1] 2
Find the value of tan^(-1) sqrt3 - cot^(-1) (-sqrt3)
Concept: Inverse Trigonometric Functions - Inverse Trigonometric Functions - Principal Value Branch
Chapter: [0.02] Inverse Trigonometric Functions
[1] 3
If a * b denotes the larger of 'a' and 'b' and if a∘b = (a * b) + 3, then write the value of (5)∘(10), where * and ∘ are binary operations.
Concept: Concept of Binary Operations
Chapter: [0.01] Relations and Functions
[1] 4
if the matrix A =[(0,a,-3),(2,0,-1),(b,1,0)] is skew symmetric, Find the value of 'a' and 'b'
Concept: Types of Matrices
Chapter: [0.03] Matrices
[2] 5
A black and a red dice are rolled. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Concept: Conditional Probability
Chapter: [0.13] Probability
[2] 6
If θ is the angle between two vectors hati - 2hatj + 3hatk and 3hati - 2hatj + hatk find sin theta
Concept: Product of Two Vectors - Vector (Or Cross) Product of Two Vectors
Chapter: [0.1] Vectors
[2] 7
Find the differential equation representing the family of curves y = ae^(bx + 5). where a and b are arbitrary constants.
Concept: Formation of a Differential Equation Whose General Solution is Given
Chapter: [0.09] Differential Equations
[2] 8
Evaluate int (cos 2x + 2sin^2x)/(cos^2x) dx
Concept: Some Properties of Indefinite Integral
Chapter: [0.07] Integrals
[2] 9
The total cost C(x) in Rupees associated with the production of x units of an item is given by
C(x) = 0.007x3 – 0.003x2 + 15x + 4000.
Find the marginal cost when 17 units are produced
Concept: Rate of Change of Bodies or Quantities
Chapter: [0.06] Applications of Derivatives
[2] 10
Differentiate tan^(-1) ((1+cosx)/(sin x)) with respect to x
Concept: Derivatives of Inverse Trigonometric Functions
Chapter: [0.05] Continuity and Differentiability
[2] 11
Given A = [(2,-3),(-4,7)] compute A^(-1) and show that 2A^(-1) = 9I - A
Concept: Types of Matrices
Chapter: [0.03] Matrices
[2] 12
Prove that 3sin^(-1)x = sin^(-1) (3x - 4x^3), x in [-1/2, 1/2]
Concept: Properties of Inverse Trigonometric Functions
Chapter: [0.02] Inverse Trigonometric Functions
[4] 13
Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X
Concept: Random Variables and Its Probability Distributions
Chapter: [0.13] Probability
[4] 14
An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when the depth of the tank is half of its width. If the cost is to be borne by nearby settled lower-income families, for whom water will be provided, what kind of value is hidden in this question?
Concept: Maxima and Minima
Chapter: [0.06] Applications of Derivatives
[4] 15 | Attempt Any One
[4] 15.1
Find the equations of the tangent and the normal, to the curve 16x2 + 9y2 = 145 at the point (x1, y1), where x1 = 2 and y1 > 0.
Concept: Tangents and Normals
Chapter: [0.06] Applications of Derivatives
[4] 15.2
Find the intervals in which the function f(x) = x^4/4 - x^3 - 5x^2 + 24x + 12 is (a) strictly increasing, (b) strictly decreasing
Concept: Increasing and Decreasing Functions
Chapter: [0.06] Applications of Derivatives
[4] 16 | Attempt Any One
[4] 16.1
if (x^2 + y^2)^2 = xy find (dy)/(dx)
Concept: Derivatives of Implicit Functions
Chapter: [0.05] Continuity and Differentiability
[4] 16.2
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find dy/dx when theta = pi/3
Concept: Derivatives of Functions in Parametric Forms
Chapter: [0.05] Continuity and Differentiability
[4] 17
If y = sin (sin x), prove that (d^2y)/(dx^2) + tan x dy/dx + y cos^2 x = 0
Concept: Higher Order Derivative
Chapter: [0.05] Continuity and Differentiability
[4] 18
[4] 18.1
Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec2 y dy = 0, give that y = pi/4 when x = 0
Concept: Methods of Solving First Order, First Degree Differential Equations - Differential Equations with Variables Separable Method
Chapter: [0.09] Differential Equations
[4] 18.2
Find the particular solution of the differential equation dy/dx + 2y tan x = sin x given that y = 0 when x = pi/3
Concept: Methods of Solving First Order, First Degree Differential Equations - Differential Equations with Variables Separable Method
Chapter: [0.09] Differential Equations
[4] 19
Find the shortest distance between the lines vecr = (4hati - hatj) + lambda(hati+2hatj-3hatk) and vecr = (hati - hatj + 2hatk) + mu(2hati + 4hatj - 5hatk)
Concept: Shortest Distance Between Two Lines
Chapter: [0.11] Three - Dimensional Geometry
[4] 20
Find int (2cos x)/((1-sinx)(1+sin^2 x)) dx
Concept: Methods of Integration: Integration Using Partial Fractions
Chapter: [0.07] Integrals
[4] 21
Suppose a girl throws a die. If she gets 1 or 2 she tosses a coin three times and notes the number of tails. If she gets 3,4,5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3,4,5 or 6 with the die ?
Concept: Bayes’ Theorem
Chapter: [0.13] Probability
[4] 22
Let veca = 4hati + 5hatj - hatk, vecb = hati - 4hatj + 5hatk and vecc = 3hati + hatj - hatk. Find a vector vecd which is perpendicular to both vecc and vecb and vecd.veca = 21
Concept: Product of Two Vectors - Vector (Or Cross) Product of Two Vectors
Chapter: [0.1] Vectors
[4] 23
Using properties of determinants, prove that |(1,1,1+3x),(1+3y, 1,1),(1,1+3z,1)| = 9(3xyz + xy + yz+ zx)
Concept: Properties of Determinants
Chapter: [0.04] Determinants
[6] 24
Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32.
Concept: Area Under Simple Curves
Chapter: [0.08] Applications of the Integrals
[6] 25 | Attempt Any One
[6] 25.1
Let A = {x ∈ Z : 0 ≤ x ≤ 12}. Show that R = {(ab) : a∈ A, |a – b| is divisible by 4}is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2]
Concept: Types of Relations
Chapter: [0.01] Relations and Functions
[6] 25.2
Show that the function f: ℝ → ℝ defined by f(x) = x/(x^2 + 1), ∀x in Ris neither one-one nor onto. Also, if g: ℝ → ℝ is defined as g(x) = 2x - 1. Find fog(x)
Concept: Types of Functions
Chapter: [0.01] Relations and Functions
[6] 26
Find the distance of the point (−1, −5, −10) from the point of intersection of the line vecr=2hati-hatj+2hatk+lambda(3hati+4hatj+2hatk) and the plane vec r (hati-hatj+hatk)=5
Concept: Three - Dimensional Geometry Examples and Solutions
Chapter: [0.11] Three - Dimensional Geometry
[6] 27
A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws 'A' while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws 'A' at a profit of 70 paise and screws 'B' at a profit of Rs 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit.
Concept: Different Types of Linear Programming Problems
Chapter: [0.12] Linear Programming
[6] 28 | Attempt Any One
[6] 28.1
Evaluate int_0^(pi/4) (sinx + cosx)/(16 + 9sin2x) dx
Concept: Evaluation of Definite Integrals by Substitution
Chapter: [0.07] Integrals
[6] 28.2
Evaluate : int_1^3 (x^2 + 3x + e^x) dx as the limit of the sum.
Concept: Definite Integral as the Limit of a Sum
Chapter: [0.07] Integrals
[6] 29 | Attempt Any One
[6] 29.1
If A = [(2,-3,5),(3,2,-4),(1,1,-2)] find A−1. Using A−1 solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
Concept: Applications of Determinants and Matrices
Chapter: [0.04] Determinants
[6] 29.2
Using elementary row transformations, find the inverse of the matrix A = [(1,2,3),(2,5,7),(-2,-4,-5)]
Concept: Elementary Transformations
Chapter: [0.03] Matrices [0.04] Determinants
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## CBSE previous year question papers Class 12 Mathematics with solutions 2017 - 2018
CBSE Class 12 Maths question paper solution is key to score more marks in final exams. Students who have used our past year paper solution have significantly improved in speed and boosted their confidence to solve any question in the examination. Our CBSE Class 12 Maths question paper 2018 serve as a catalyst to prepare for your Mathematics board examination.
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By referring the question paper Solutions for Mathematics, you can scale your preparation level and work on your weak areas. It will also help the candidates in developing the time-management skills. Practice makes perfect, and there is no better way to practice than to attempt previous year question paper solutions of CBSE Class 12.
How CBSE Class 12 Question Paper solutions Help Students ?
• Question paper solutions for Mathematics will helps students to prepare for exam.
• Question paper with answer will boost students confidence in exam time and also give you an idea About the important questions and topics to be prepared for the board exam.
• For finding solution of question papers no need to refer so multiple sources like textbook or guides. | 3,167 | 10,429 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-40 | latest | en | 0.78607 |
http://www.territorioscuola.com/wikipedia/en.wikipedia.php?title=Eigenvalue,_eigenvector_and_eigenspace | 1,398,001,624,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00506-ip-10-147-4-33.ec2.internal.warc.gz | 695,168,486 | 58,838 | # Eigenvalues and eigenvectors
In this shear mapping the red arrow changes direction but the blue arrow does not. The blue arrow is an eigenvector of this shear mapping, and since its length is unchanged its eigenvalue is 1.
An eigenvector of a square matrix $A$ is a non-zero vector $v$ that, when the matrix is multiplied by $v$, yields a constant multiple of $v$, the multiplier being commonly denoted by $\lambda$. That is:
$A v = \lambda v$
(Because this equation uses post-multiplication by $v$, it describes a right eigenvector.)
The number $\lambda$ is called the eigenvalue of $A$ corresponding to $v$.1
If 2D space is visualized as a piece of cloth being stretched by the matrix, the eigenvectors would make up the line along the direction the cloth is stretched in and the line of cloth at the center of the stretching, whose direction isn't changed by the stretching either. The eigenvalues for the first line would give the scale to which the cloth is stretched, and for the second line the scale to which it's tightened. A reflection may be viewed as stretching a line to scale -1 while shrinking the axis of reflection to scale 1. For 3D rotations, the eigenvectors form the axis of rotation, and since the scale of the axis is unchanged by the rotation, their eigenvalues are all 1.
In analytic geometry, for example, a three-element vector may be seen as an arrow in three-dimensional space starting at the origin. In that case, an eigenvector $v$ is an arrow whose direction is either preserved or exactly reversed after multiplication by $A$. The corresponding eigenvalue determines how the length of the arrow is changed by the operation, and whether its direction is reversed or not, determined by whether the eigenvalue is negative or positive.
In abstract linear algebra, these concepts are naturally extended to more general situations, where the set of real scalar factors is replaced by any field of scalars (such as algebraic or complex numbers); the set of Cartesian vectors $\mathbb{R}^n$ is replaced by any vector space (such as the continuous functions, the polynomials or the trigonometric series), and matrix multiplication is replaced by any linear operator that maps vectors to vectors (such as the derivative from calculus). In such cases, the "vector" in "eigenvector" may be replaced by a more specific term, such as "eigenfunction", "eigenmode", "eigenface", or "eigenstate". Thus, for example, the exponential function $f(x) = a^x$ is an eigenfunction of the derivative operator " ${}'$ ", with eigenvalue $\lambda = \ln a$, since its derivative is $f'(x) = (\ln a)a^x = \lambda f(x)$.
The set of all eigenvectors of a matrix (or linear operator), each paired with its corresponding eigenvalue, is called the eigensystem of that matrix.2 Any multiple of an eigenvector is also an eigenvector, with the same eigenvalue. An eigenspace of a matrix $A$ is the set of all eigenvectors with the same eigenvalue, together with the zero vector.1 An eigenbasis for $A$ is any basis for the set of all vectors that consists of linearly independent eigenvectors of $A$. Not every matrix has an eigenbasis, but every symmetric matrix does.
The terms characteristic vector, characteristic value, and characteristic space are also used for these concepts. The prefix eigen- is adopted from the German word eigen for "self-" or "unique to", "peculiar to", or "belonging to" in the sense of "idiosyncratic" in relation to the originating matrix.
Eigenvalues and eigenvectors have many applications in both pure and applied mathematics. They are used in matrix factorization, in quantum mechanics, and in many other areas.
## Definition
### Eigenvectors and eigenvalues of a real matrix
Matrix $A$ acts by stretching the vector $x$, not changing its direction, so $x$ is an eigenvector of $A$.
In many contexts, a vector can be assumed to be a list of real numbers (called elements), written vertically with brackets around the entire list, such as the vectors u and v below. Two vectors are said to be scalar multiples of each other (also called parallel or collinear) if they have the same number of elements, and if every element of one vector is obtained by multiplying each corresponding element in the other vector by the same number (known as a scaling factor, or a scalar). For example, the vectors
$u = \begin{bmatrix}1\\3\\4\end{bmatrix}\quad\quad\quad$ and $\quad\quad\quad v = \begin{bmatrix}-20\\-60\\-80\end{bmatrix}$
are scalar multiples of each other, because each element of $v$ is −20 times the corresponding element of $u$.
A vector with three elements, like $u$ or $v$ above, may represent a point in three-dimensional space, relative to some Cartesian coordinate system. It helps to think of such a vector as the tip of an arrow whose tail is at the origin of the coordinate system. In this case, the condition "$u$ is parallel to $v$" means that the two arrows lie on the same straight line, and may differ only in length and direction along that line.
If we multiply any square matrix $A$ with $n$ rows and $n$ columns by such a vector $v$, the result will be another vector $w = A v$, also with $n$ rows and one column. That is,
$\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} \quad\quad$ is mapped to $\begin{bmatrix} w_1 \\ w_2 \\ \vdots \\ w_n \end{bmatrix} \;=\; \begin{bmatrix} A_{1,1} & A_{1,2} & \ldots & A_{1,n} \\ A_{2,1} & A_{2,2} & \ldots & A_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n,1} & A_{n,2} & \ldots & A_{n,n} \\ \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}$
where, for each index $i$,
$w_i = A_{i,1} v_1 + A_{i,2} v_2 + \cdots + A_{i,n} v_n = \sum_{j = 1}^{n} A_{i,j} v_j$
In general, if $v_j$ are not all zeros, the vectors $v$ and $A v$ will not be parallel. When they are parallel (that is, when there is some real number $\lambda$ such that $A v = \lambda v$) we say that $v$ is an eigenvector of $A$. In that case, the scale factor $\lambda$ is said to be the eigenvalue corresponding to that eigenvector.
In particular, multiplication by a 3×3 matrix $A$ may change both the direction and the magnitude of an arrow $v$ in three-dimensional space. However, if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, the operation may only change its length, and either keep its direction or flip it (make the arrow point in the exact opposite direction). Specifically, the length of the arrow will increase if $|\lambda| > 1$, remain the same if $|\lambda| = 1$, and decrease it if $|\lambda|< 1$. Moreover, the direction will be precisely the same if $\lambda > 0$, and flipped if $\lambda < 0$. If $\lambda = 0$, then the length of the arrow becomes zero.
#### An example
The transformation matrix $\bigl[ \begin{smallmatrix} 2 & 1\\ 1 & 2 \end{smallmatrix} \bigr]$ preserves the angle of arrows parallel to the lines from the origin to $\bigl[ \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \bigr]$ (in blue) and to $\bigl[ \begin{smallmatrix} 1 \\ -1 \end{smallmatrix} \bigr]$ (in violet). The points that lie on a line through the origin and an eigenvector remain on the line after the transformation. The arrows in red are not parallel to such a line, therefore their angle is altered by the transformation. See also: An extended version, showing all four quadrants.
For the transformation matrix
$A = \begin{bmatrix} 3 & 1\\1 & 3 \end{bmatrix},$
the vector
$v = \begin{bmatrix} 4 \\ -4 \end{bmatrix}$
is an eigenvector with eigenvalue 2. Indeed,
$A v = \begin{bmatrix} 3 & 1\\1 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ -4 \end{bmatrix} = \begin{bmatrix} 3 \cdot 4 + 1 \cdot (-4) \\ 1 \cdot 4 + 3 \cdot (-4) \end{bmatrix} = \begin{bmatrix} 8 \\ -8 \end{bmatrix} = 2 \cdot \begin{bmatrix} 4 \\ -4 \end{bmatrix}.$
On the other hand the vector
$v = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
is not an eigenvector, since
$\begin{bmatrix} 3 & 1\\1 & 3 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \cdot 0 + 1 \cdot 1 \\ 1 \cdot 0 + 3 \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix},$
and this vector is not a multiple of the original vector $v$.
#### Another example
For the matrix
$A= \begin{bmatrix} 1 & 2 & 0\\0 & 2 & 0\\ 0 & 0 & 3\end{bmatrix},$
we have
$A \begin{bmatrix} 1\\0\\0 \end{bmatrix} = \begin{bmatrix} 1\\0\\0 \end{bmatrix} = 1 \cdot \begin{bmatrix} 1\\0\\0 \end{bmatrix},\quad\quad$
$A \begin{bmatrix} 0\\0\\1 \end{bmatrix} = \begin{bmatrix} 0\\0\\3 \end{bmatrix} = 3 \cdot \begin{bmatrix} 0\\0\\1 \end{bmatrix}.\quad\quad$
Therefore, the vectors $[1,0,0]^\mathsf{T}$ and $[0,0,1]^\mathsf{T}$ are eigenvectors of $A$ corresponding to the eigenvalues 1 and 3 respectively. (Here the symbol ${}^\mathsf{T}$ indicates matrix transposition, in this case turning the row vectors into column vectors.)
#### Trivial cases
The identity matrix $I$ (whose general element $I_{i j}$ is 1 if $i = j$, and 0 otherwise) maps every vector to itself. Therefore, every vector is an eigenvector of $I$, with eigenvalue 1.
More generally, if $A$ is a diagonal matrix (with $A_{i j} = 0$ whenever $i \neq j$), and $v$ is a vector parallel to axis $i$ (that is, $v_i \neq 0$, and $v_j = 0$ if $j \neq i$), then $A v = \lambda v$ where $\lambda = A_{i i}$. That is, the eigenvalues of a diagonal matrix are the elements of its main diagonal. This is trivially the case of any 1 ×1 matrix.
### General definition
The concept of eigenvectors and eigenvalues extends naturally to abstract linear transformations on abstract vector spaces. Namely, let $V$ be any vector space over some field $K$ of scalars, and let $T$ be a linear transformation mapping $V$ into $V$. We say that a non-zero vector $v$ of $V$ is an eigenvector of $T$ if (and only if) there is a scalar $\lambda$ in $K$ such that
$T(v)=\lambda v$.
This equation is called the eigenvalue equation for $T$, and the scalar $\lambda$ is the eigenvalue of $T$ corresponding to the eigenvector $v$. Note that $T(v)$ means the result of applying the operator $T$ to the vector $v$, while $\lambda v$ means the product of the scalar $\lambda$ by $v$.3
The matrix-specific definition is a special case of this abstract definition. Namely, the vector space $V$ is the set of all column vectors of a certain size $n$×1, and $T$ is the linear transformation that consists in multiplying a vector by the given $n\times n$ matrix $A$.
Some authors allow $v$ to be the zero vector in the definition of eigenvector.4 This is reasonable as long as we define eigenvalues and eigenvectors carefully: If we would like the zero vector to be an eigenvector, then we must first define an eigenvalue of $T$ as a scalar $\lambda$ in $K$ such that there is a nonzero vector $v$ in $V$ with $T(v) = \lambda v$. We then define an eigenvector to be a vector $v$ in $V$ such that there is an eigenvalue $\lambda$ in $K$ with $T(v) = \lambda v$. This way, we ensure that it is not the case that every scalar is an eigenvalue corresponding to the zero vector.
### Eigenspace and spectrum
If $v$ is an eigenvector of $T$, with eigenvalue $\lambda$, then any scalar multiple $\alpha v$ of $v$ with nonzero $\alpha$ is also an eigenvector with eigenvalue $\lambda$, since $T(\alpha v) = \alpha T(v) = \alpha(\lambda v) = \lambda(\alpha v)$. Moreover, if $u$ and $v$ are distinct eigenvectors with the same eigenvalue $\lambda$, then $u+v$ is also an eigenvector with the same eigenvalue $\lambda$. Therefore, the set of all eigenvectors with the same eigenvalue $\lambda$, together with the zero vector, is a linear subspace of $V$, called the eigenspace of $T$ associated to $\lambda$.56 If that subspace has dimension 1, it is sometimes called an eigenline.7
The geometric multiplicity $\gamma_T(\lambda)$ of an eigenvalue $\lambda$ is the dimension of the eigenspace associated to $\lambda$, i.e. number of linearly independent eigenvectors with that eigenvalue.
The eigenspaces of T always form a direct sum (and as a consequence any family of eigenvectors for different eigenvalues is always linearly independent). Therefore the sum of the dimensions of the eigenspaces cannot exceed the dimension n of the space on which T operates, and in particular there cannot be more than n distinct eigenvalues.8
Any subspace spanned by eigenvectors of $T$ is an invariant subspace of $T$, and the restriction of T to such a subspace is diagonalizable.
The set of eigenvalues of $T$ is sometimes called the spectrum of $T$.
### Eigenbasis
An eigenbasis for a linear operator $T$ that operates on a vector space $V$ is a basis for $V$ that consists entirely of eigenvectors of $T$ (possibly with different eigenvalues). Such a basis exists precisely if the direct sum of the eigenspaces equals the whole space, in which case one can take the union of bases chosen in each of the eigenspaces as eigenbasis. The matrix of T in a given basis is diagonal precisely when that basis is an eigenbasis for T, and for this reason T is called diagonalizable if it admits an eigenbasis.
## Generalizations to infinite-dimensional spaces
The definition of eigenvalue of a linear transformation $T$ remains valid even if the underlying space $V$ is an infinite dimensional Hilbert or Banach space. Namely, a scalar $\lambda$ is an eigenvalue if and only if there is some nonzero vector $v$ such that $T(v) = \lambda v$.
### Eigenfunctions
A widely used class of linear operators acting on infinite dimensional spaces are the differential operators on function spaces. Let $D$ be a linear differential operator in on the space $\mathbf{C^\infty}$ of infinitely differentiable real functions of a real argument $t$. The eigenvalue equation for $D$ is the differential equation
$D f = \lambda f$
The functions that satisfy this equation are commonly called eigenfunctions of $D$. For the derivative operator $d/dt$, an eigenfunction is a function that, when differentiated, yields a constant times the original function. The solution is an exponential function
$f(t) = Ae^{\lambda t} ,$
including when $\lambda$ is zero when it becomes a constant function. Eigenfunctions are an essential tool in the solution of differential equations and many other applied and theoretical fields. For instance, the exponential functions are eigenfunctions of the shift operators. This is the basis of Fourier transform methods for solving problems.
### Spectral theory
If $\lambda$ is an eigenvalue of $T$, then the operator $T-\lambda I$ is not one-to-one, and therefore its inverse $(T-\lambda I)^{-1}$ does not exist. The converse is true for finite-dimensional vector spaces, but not for infinite-dimensional ones. In general, the operator $T - \lambda I$ may not have an inverse, even if $\lambda$ is not an eigenvalue.
For this reason, in functional analysis one defines the spectrum of a linear operator $T$ as the set of all scalars $\lambda$ for which the operator $T-\lambda I$ has no bounded inverse. Thus the spectrum of an operator always contains all its eigenvalues, but is not limited to them.
### Associative algebras and representation theory
More algebraically, rather than generalizing the vector space to an infinite dimensional space, one can generalize the algebraic object that is acting on the space, replacing a single operator acting on a vector space with an algebra representation – an associative algebra acting on a module. The study of such actions is the field of representation theory.
A closer analog of eigenvalues is given by the representation-theoretical concept of weight, with the analogs of eigenvectors and eigenspaces being weight vectors and weight spaces.
## Eigenvalues and eigenvectors of matrices
### Characteristic polynomial
The eigenvalue equation for a matrix $A$ is
$A v - \lambda v = 0,$
which is equivalent to
$(A-\lambda I)v = 0,$
where $I$ is the $n\times n$ identity matrix. It is a fundamental result of linear algebra that an equation $M v = 0$ has a non-zero solution $v$ if, and only if, the determinant $\det(M)$ of the matrix $M$ is zero. It follows that the eigenvalues of $A$ are precisely the real numbers $\lambda$ that satisfy the equation
$\det(A-\lambda I) = 0$
The left-hand side of this equation can be seen (using Leibniz' rule for the determinant) to be a polynomial function of the variable $\lambda$. The degree of this polynomial is $n$, the order of the matrix. Its coefficients depend on the entries of $A$, except that its term of degree $n$ is always $(-1)^n\lambda^n$. This polynomial is called the characteristic polynomial of $A$; and the above equation is called the characteristic equation (or, less often, the secular equation) of $A$.
For example, let $A$ be the matrix
$A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{bmatrix}$
The characteristic polynomial of $A$ is
$\det (A-\lambda I) \;=\; \det \left(\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right) \;=\; \det \begin{bmatrix} 2 - \lambda & 0 & 0 \\ 0 & 3 - \lambda & 4 \\ 0 & 4 & 9 - \lambda \end{bmatrix}$
which is
$(2 - \lambda) \bigl[ (3 - \lambda) (9 - \lambda) - 16 \bigr] = -\lambda^3 + 14\lambda^2 - 35\lambda + 22$
The roots of this polynomial are 2, 1, and 11. Indeed these are the only three eigenvalues of $A$, corresponding to the eigenvectors $[1,0,0]',$ $[0,2,-1]',$ and $[0,1,2]'$ (or any non-zero multiples thereof).
#### In the real domain
Since the eigenvalues are roots of the characteristic polynomial, an $n\times n$ matrix has at most $n$ eigenvalues. If the matrix has real entries, the coefficients of the characteristic polynomial are all real; but it may have fewer than $n$ real roots, or no real roots at all.
For example, consider the cyclic permutation matrix
$A = \begin{bmatrix} 0 & 1 & 0\\0 & 0 & 1\\ 1 & 0 & 0\end{bmatrix}$
This matrix shifts the coordinates of the vector up by one position, and moves the first coordinate to the bottom. Its characteristic polynomial is $1 - \lambda^3$ which has one real root $\lambda_1 = 1$. Any vector with three equal non-zero elements is an eigenvector for this eigenvalue. For example,
$A \begin{bmatrix} 5\\5\\5 \end{bmatrix} = \begin{bmatrix} 5\\5\\5 \end{bmatrix} = 1 \cdot \begin{bmatrix} 5\\5\\5 \end{bmatrix}$
#### In the complex domain
The fundamental theorem of algebra implies that the characteristic polynomial of an $n\times n$ matrix $A$, being a polynomial of degree $n$, has exactly $n$ complex roots. More precisely, it can be factored into the product of $n$ linear terms,
$\det(A-\lambda I) = (\lambda_1 - \lambda )(\lambda_2 - \lambda)\cdots(\lambda_n - \lambda)$
where each $\lambda_i$ is a complex number. The numbers $\lambda_1$, $\lambda_2$, ... $\lambda_n$, (which may not be all distinct) are roots of the polynomial, and are precisely the eigenvalues of $A$.
Even if the entries of $A$ are all real numbers, the eigenvalues may still have non-zero imaginary parts (and the elements of the corresponding eigenvectors will therefore also have non-zero imaginary parts). Also, the eigenvalues may be irrational numbers even if all the entries of $A$ are rational numbers, or all are integers. However, if the entries of $A$ are algebraic numbers (which include the rationals), the eigenvalues will be (complex) algebraic numbers too.
The non-real roots of a real polynomial with real coefficients can be grouped into pairs of complex conjugate values, namely with the two members of each pair having the same real part and imaginary parts that differ only in sign. If the degree is odd, then by the intermediate value theorem at least one of the roots will be real. Therefore, any real matrix with odd order will have at least one real eigenvalue; whereas a real matrix with even order may have no real eigenvalues.
In the example of the 3×3 cyclic permutation matrix $A$, above, the characteristic polynomial $1 - \lambda^3$ has two additional non-real roots, namely
$\lambda_2 = -1/2 + \mathbf{i}\sqrt{3}/2\quad\quad$ and $\quad\quad\lambda_3 = \lambda_2^* = -1/2 - \mathbf{i}\sqrt{3}/2$,
where $\mathbf{i}= \sqrt{-1}$ is the imaginary unit. Note that $\lambda_2\lambda_3 = 1$, $\lambda_2^2 = \lambda_3$, and $\lambda_3^2 = \lambda_2$. Then
$A \begin{bmatrix} 1 \\ \lambda_2 \\ \lambda_3 \end{bmatrix} = \begin{bmatrix} \lambda_2\\ \lambda_3 \\1 \end{bmatrix} = \lambda_2 \cdot \begin{bmatrix} 1\\ \lambda_2 \\ \lambda_3 \end{bmatrix} \quad\quad$ and $\quad\quad A \begin{bmatrix} 1 \\ \lambda_3 \\ \lambda_2 \end{bmatrix} = \begin{bmatrix} \lambda_3 \\ \lambda_2 \\ 1 \end{bmatrix} = \lambda_3 \cdot \begin{bmatrix} 1 \\ \lambda_3 \\ \lambda_2 \end{bmatrix}$
Therefore, the vectors $[1,\lambda_2,\lambda_3]'$ and $[1,\lambda_3,\lambda_2]'$ are eigenvectors of $A$, with eigenvalues $\lambda_2$, and $\lambda_3$, respectively.
### Algebraic multiplicities
Let $\lambda_i$ be an eigenvalue of an $n\times n$ matrix $A$. The algebraic multiplicity $\mu_A(\lambda_i)$ of $\lambda_i$ is its multiplicity as a root of the characteristic polynomial, that is, the largest integer $k$ such that $(\lambda - \lambda_i)^k$ divides evenly that polynomial.
Like the geometric multiplicity $\gamma_A(\lambda_i)$, the algebraic multiplicity is an integer between 1 and $n$; and the sum $\boldsymbol{\mu}_A$ of $\mu_A(\lambda_i)$ over all distinct eigenvalues also cannot exceed $n$. If complex eigenvalues are considered, $\boldsymbol{\mu}_A$ is exactly $n$.
It can be proved that the geometric multiplicity $\gamma_A(\lambda_i)$ of an eigenvalue never exceeds its algebraic multiplicity $\mu_A(\lambda_i)$. Therefore, $\boldsymbol{\gamma}_A$ is at most $\boldsymbol{\mu}_A$.
If $\gamma_A(\lambda_i) = \mu_A(\lambda_i)$, then $\lambda_i$ is said to be a semisimple eigenvalue.
#### Example
For the matrix: $A= \begin{bmatrix} 2 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 3 \end{bmatrix},$
the characteristic polynomial of $A$ is $\det (A-\lambda I) \;=\; \det \begin{bmatrix} 2- \lambda & 0 & 0 & 0 \\ 1 & 2- \lambda & 0 & 0 \\ 0 & 1 & 3- \lambda & 0 \\ 0 & 0 & 1 & 3- \lambda \end{bmatrix}= (2 - \lambda)^2 (3 - \lambda)^2$,
being the product of the diagonal with a lower triangular matrix.
The roots of this polynomial, and hence the eigenvalues, are 2 and 3. The algebraic multiplicity of each eigenvalue is 2; in other words they are both double roots. On the other hand, the geometric multiplicity of the eigenvalue 2 is only 1, because its eigenspace is spanned by the vector $[0,1,-1,1]$, and is therefore 1-dimensional. Similarly, the geometric multiplicity of the eigenvalue 3 is 1 because its eigenspace is spanned by $[0,0,0,1]$. Hence, the total algebraic multiplicity of A, denoted $\mu_A$, is 4, which is the most it could be for a 4 by 4 matrix. The geometric multiplicity $\gamma_A$ is 2, which is the smallest it could be for a matrix which has two distinct eigenvalues.
### Diagonalization and eigendecomposition
If the sum $\boldsymbol{\gamma}_A$ of the geometric multiplicities of all eigenvalues is exactly $n$, then $A$ has a set of $n$ linearly independent eigenvectors. Let $Q$ be a square matrix whose columns are those eigenvectors, in any order. Then we will have $A Q = Q\Lambda$, where $\Lambda$ is the diagonal matrix such that $\Lambda_{i i}$ is the eigenvalue associated to column $i$ of $Q$. Since the columns of $Q$ are linearly independent, the matrix $Q$ is invertible. Premultiplying both sides by $Q^{-1}$ we get $Q^{-1}A Q = \Lambda$. By definition, therefore, the matrix $A$ is diagonalizable.
Conversely, if $A$ is diagonalizable, let $Q$ be a non-singular square matrix such that $Q^{-1} A Q$ is some diagonal matrix $D$. Multiplying both sides on the left by $Q$ we get $A Q = Q D$. Therefore each column of $Q$ must be an eigenvector of $A$, whose eigenvalue is the corresponding element on the diagonal of $D$. Since the columns of $Q$ must be linearly independent, it follows that $\boldsymbol{\gamma}_A = n$. Thus $\boldsymbol{\gamma}_A$ is equal to $n$ if and only if $A$ is diagonalizable.
If $A$ is diagonalizable, the space of all $n$-element vectors can be decomposed into the direct sum of the eigenspaces of $A$. This decomposition is called the eigendecomposition of $A$, and it is preserved under change of coordinates.
A matrix that is not diagonalizable is said to be defective. For defective matrices, the notion of eigenvector can be generalized to generalized eigenvectors, and that of diagonal matrix to a Jordan form matrix. Over an algebraically closed field, any matrix $A$ has a Jordan form and therefore admits a basis of generalized eigenvectors, and a decomposition into generalized eigenspaces
### Further properties
Let $A$ be an arbitrary $n\times n$ matrix of complex numbers with eigenvalues $\lambda_1$, $\lambda_2$, ... $\lambda_n$. (Here it is understood that an eigenvalue with algebraic multiplicity $\mu$ occurs $\mu$ times in this list.) Then
• The trace of $A$, defined as the sum of its diagonal elements, is also the sum of all eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^n A_{i i} = \sum_{i=1}^n \lambda_i = \lambda_1+ \lambda_2 +\cdots+ \lambda_n$.
• The determinant of $A$ is the product of all eigenvalues:
$\operatorname{det}(A) = \prod_{i=1}^n \lambda_i=\lambda_1\lambda_2\cdots\lambda_n$.
• The eigenvalues of the $k$th power of $A$, i.e. the eigenvalues of $A^k$, for any positive integer $k$, are $\lambda_1^k,\lambda_2^k,\dots,\lambda_n^k$
• The matrix $A$ is invertible if and only if all the eigenvalues $\lambda_i$ are nonzero.
• If $A$ is invertible, then the eigenvalues of $A^{-1}$ are $1/\lambda_1,1/\lambda_2,\dots,1/\lambda_n$
• If $A$ is equal to its conjugate transpose $A^*$ (in other words, if $A$ is Hermitian), then every eigenvalue is real. The same is true of any a symmetric real matrix. If $A$ is also positive-definite, positive-semidefinite, negative-definite, or negative-semidefinite every eigenvalue is positive, non-negative, negative, or non-positive respectively.
• Every eigenvalue of a unitary matrix has absolute value $|\lambda|=1$.
### Left and right eigenvectors
The use of matrices with a single column (rather than a single row) to represent vectors is traditional in many disciplines. For that reason, the word "eigenvector" almost always means a right eigenvector, namely a column vector that must be placed to the right of the matrix $A$ in the defining equation
$A v = \lambda v$.
There may be also single-row vectors that are unchanged when they occur on the left side of a product with a square matrix $A$; that is, which satisfy the equation
$u A = \lambda u$
Any such row vector $u$ is called a left eigenvector of $A$.
The left eigenvectors of $A$ are transposes of the right eigenvectors of the transposed matrix $A^\mathsf{T}$, since their defining equation is equivalent to
$A^\mathsf{T} u^\mathsf{T} = \lambda u^\mathsf{T}$
It follows that, if $A$ is Hermitian, its left and right eigenvectors are complex conjugates. In particular if $A$ is a real symmetric matrix, they are the same except for transposition.
### Variational characterization
In the Hermitian case, eigenvalues can be given a variational characterization. The largest eigenvalue of $H$ is the maximum value of the quadratic form $x^T H x/x^T x$. A value of $x$ that realizes that maximum, is an eigenvector. For more information, see Min-max theorem.
## Calculation
### Computing the eigenvalues
The eigenvalues of a matrix $A$ can be determined by finding the roots of the characteristic polynomial. Explicit algebraic formulas for the roots of a polynomial exist only if the degree $n$ is 4 or less. According to the Abel–Ruffini theorem there is no general, explicit and exact algebraic formula for the roots of a polynomial with degree 5 or more.
It turns out that any polynomial with degree $n$ is the characteristic polynomial of some companion matrix of order $n$. Therefore, for matrices of order 5 or more, the eigenvalues and eigenvectors cannot be obtained by an explicit algebraic formula, and must therefore be computed by approximate numerical methods.
In theory, the coefficients of the characteristic polynomial can be computed exactly, since they are sums of products of matrix elements; and there are algorithms that can find all the roots of a polynomial of arbitrary degree to any required accuracy.9 However, this approach is not viable in practice because the coefficients would be contaminated by unavoidable round-off errors, and the roots of a polynomial can be an extremely sensitive function of the coefficients (as exemplified by Wilkinson's polynomial).9
Efficient, accurate methods to compute eigenvalues and eigenvectors of arbitrary matrices were not known until the advent of the QR algorithm in 1961. 9 Combining the Householder transformation with the LU decomposition results in an algorithm with better convergence than the QR algorithm.citation needed For large Hermitian sparse matrices, the Lanczos algorithm is one example of an efficient iterative method to compute eigenvalues and eigenvectors, among several other possibilities.9
### Computing the eigenvectors
Once the (exact) value of an eigenvalue is known, the corresponding eigenvectors can be found by finding non-zero solutions of the eigenvalue equation, that becomes a system of linear equations with known coefficients. For example, once it is known that 6 is an eigenvalue of the matrix
$A = \begin{bmatrix} 4 & 1\\6 & 3 \end{bmatrix}$
we can find its eigenvectors by solving the equation $A v = 6 v$, that is
$\begin{bmatrix} 4 & 1\\6 & 3 \end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = 6 \cdot \begin{bmatrix}x\\y\end{bmatrix}$
This matrix equation is equivalent to two linear equations
$\left\{\begin{matrix} 4x + {\ }y &{}= 6x\\6x + 3y &{}=6 y\end{matrix}\right. \quad\quad\quad$ that is $\left\{\begin{matrix} -2x+ {\ }y &{}=0\\+6x-3y &{}=0\end{matrix}\right.$
Both equations reduce to the single linear equation $y=2x$. Therefore, any vector of the form $[a,2a]'$, for any non-zero real number $a$, is an eigenvector of $A$ with eigenvalue $\lambda = 6$.
The matrix $A$ above has another eigenvalue $\lambda=1$. A similar calculation shows that the corresponding eigenvectors are the non-zero solutions of $3x+y=0$, that is, any vector of the form $[b,-3b]'$, for any non-zero real number $b$.
Some numeric methods that compute the eigenvalues of a matrix also determine a set of corresponding eigenvectors as a by-product of the computation.
## History
Eigenvalues are often introduced in the context of linear algebra or matrix theory. Historically, however, they arose in the study of quadratic forms and differential equations.
In the 18th century Euler studied the rotational motion of a rigid body and discovered the importance of the principal axes. Lagrange realized that the principal axes are the eigenvectors of the inertia matrix.10 In the early 19th century, Cauchy saw how their work could be used to classify the quadric surfaces, and generalized it to arbitrary dimensions.11 Cauchy also coined the term racine caractéristique (characteristic root) for what is now called eigenvalue; his term survives in characteristic equation.12
Fourier used the work of Laplace and Lagrange to solve the heat equation by separation of variables in his famous 1822 book Théorie analytique de la chaleur.13 Sturm developed Fourier's ideas further and brought them to the attention of Cauchy, who combined them with his own ideas and arrived at the fact that real symmetric matrices have real eigenvalues.11 This was extended by Hermite in 1855 to what are now called Hermitian matrices.12 Around the same time, Brioschi proved that the eigenvalues of orthogonal matrices lie on the unit circle,11 and Clebsch found the corresponding result for skew-symmetric matrices.12 Finally, Weierstrass clarified an important aspect in the stability theory started by Laplace by realizing that defective matrices can cause instability.11
In the meantime, Liouville studied eigenvalue problems similar to those of Sturm; the discipline that grew out of their work is now called Sturm–Liouville theory.14 Schwarz studied the first eigenvalue of Laplace's equation on general domains towards the end of the 19th century, while Poincaré studied Poisson's equation a few years later.15
At the start of the 20th century, Hilbert studied the eigenvalues of integral operators by viewing the operators as infinite matrices.16 He was the first to use the German word eigen to denote eigenvalues and eigenvectors in 1904, though he may have been following a related usage by Helmholtz. For some time, the standard term in English was "proper value", but the more distinctive term "eigenvalue" is standard today.17
The first numerical algorithm for computing eigenvalues and eigenvectors appeared in 1929, when Von Mises published the power method. One of the most popular methods today, the QR algorithm, was proposed independently by John G.F. Francis18 and Vera Kublanovskaya19 in 1961.20
## Applications
### Eigenvalues of geometric transformations
The following table presents some example transformations in the plane along with their 2×2 matrices, eigenvalues, and eigenvectors.
scaling unequal scaling rotation horizontal shear hyperbolic rotation illustration matrix $\begin{bmatrix}k & 0\\0 & k\end{bmatrix}$ $\begin{bmatrix}k_1 & 0\\0 & k_2\end{bmatrix}$ $\begin{bmatrix}c & -s \\ s & c\end{bmatrix}$ $c=\cos\theta$ $s=\sin\theta$ $\begin{bmatrix}1 & k\\ 0 & 1\end{bmatrix}$ $\begin{bmatrix} c & s \\ s & c \end{bmatrix}$ $c=\cosh \varphi$ $s=\sinh \varphi$ characteristic polynomial $\ (\lambda - k)^2$ $(\lambda - k_1)(\lambda - k_2)$ $\lambda^2 - 2c\lambda + 1$ $\ (\lambda - 1)^2$ $\lambda^2 - 2c\lambda + 1$ eigenvalues $\lambda_i$ $\lambda_1 = \lambda_2 = k$ $\lambda_1 = k_1$ $\lambda_2 = k_2$ $\lambda_1 = e^{\mathbf{i}\theta}=c+s\mathbf{i}$ $\lambda_2 = e^{-\mathbf{i}\theta}=c-s\mathbf{i}$ $\lambda_1 = \lambda_2 = 1$ $\lambda_1 = e^\varphi$ $\lambda_2 = e^{-\varphi}$, algebraic multipl. $\mu_i=\mu(\lambda_i)$ $\mu_1 = 2$ $\mu_1 = 1$ $\mu_2 = 1$ $\mu_1 = 1$ $\mu_2 = 1$ $\mu_1 = 2$ $\mu_1 = 1$ $\mu_2 = 1$ geometric multipl. $\gamma_i = \gamma(\lambda_i)$ $\gamma_1 = 2$ $\gamma_1 = 1$ $\gamma_2 = 1$ $\gamma_1 = 1$ $\gamma_2 = 1$ $\gamma_1 = 1$ $\gamma_1 = 1$ $\gamma_2 = 1$ eigenvectors All non-zero vectors $u_1 = \begin{bmatrix}1\\0\end{bmatrix}$ $u_2 = \begin{bmatrix}0\\1\end{bmatrix}$ $u_1 = \begin{bmatrix}{\ }1\\-\mathbf{i}\end{bmatrix}$ $u_2 = \begin{bmatrix}{\ }1\\ +\mathbf{i}\end{bmatrix}$ $u_1 = \begin{bmatrix}1\\0\end{bmatrix}$ $u_1 = \begin{bmatrix}{\ }1\\{\ }1\end{bmatrix}$ $u_2 = \begin{bmatrix}{\ }1\\-1\end{bmatrix}.$
Note that the characteristic equation for a rotation is a quadratic equation with discriminant $D = -4(\sin\theta)^2$, which is a negative number whenever $\theta$ is not an integer multiple of 180°. Therefore, except for these special cases, the two eigenvalues are complex numbers, $\cos\theta \pm \mathbf{i}\sin\theta$; and all eigenvectors have non-real entries. Indeed, except for those special cases, a rotation changes the direction of every nonzero vector in the plane.
### Schrödinger equation
The wavefunctions associated with the bound states of an electron in a hydrogen atom can be seen as the eigenvectors of the hydrogen atom Hamiltonian as well as of the angular momentum operator. They are associated with eigenvalues interpreted as their energies (increasing downward: $n=1,2,3,\ldots$) and angular momentum (increasing across: s, p, d, ...). The illustration shows the square of the absolute value of the wavefunctions. Brighter areas correspond to higher probability density for a position measurement. The center of each figure is the atomic nucleus, a proton.
An example of an eigenvalue equation where the transformation $T$ is represented in terms of a differential operator is the time-independent Schrödinger equation in quantum mechanics:
$H\psi_E = E\psi_E \,$
where $H$, the Hamiltonian, is a second-order differential operator and $\psi_E$, the wavefunction, is one of its eigenfunctions corresponding to the eigenvalue $E$, interpreted as its energy.
However, in the case where one is interested only in the bound state solutions of the Schrödinger equation, one looks for $\psi_E$ within the space of square integrable functions. Since this space is a Hilbert space with a well-defined scalar product, one can introduce a basis set in which $\psi_E$ and $H$ can be represented as a one-dimensional array and a matrix respectively. This allows one to represent the Schrödinger equation in a matrix form.
The bra–ket notation is often used in this context. A vector, which represents a state of the system, in the Hilbert space of square integrable functions is represented by $|\Psi_E\rangle$. In this notation, the Schrödinger equation is:
$H|\Psi_E\rangle = E|\Psi_E\rangle$
where $|\Psi_E\rangle$ is an eigenstate of $H$. It is a self adjoint operator, the infinite dimensional analog of Hermitian matrices (see Observable). As in the matrix case, in the equation above $H|\Psi_E\rangle$ is understood to be the vector obtained by application of the transformation $H$ to $|\Psi_E\rangle$.
### Molecular orbitals
In quantum mechanics, and in particular in atomic and molecular physics, within the Hartree–Fock theory, the atomic and molecular orbitals can be defined by the eigenvectors of the Fock operator. The corresponding eigenvalues are interpreted as ionization potentials via Koopmans' theorem. In this case, the term eigenvector is used in a somewhat more general meaning, since the Fock operator is explicitly dependent on the orbitals and their eigenvalues. If one wants to underline this aspect one speaks of nonlinear eigenvalue problem. Such equations are usually solved by an iteration procedure, called in this case self-consistent field method. In quantum chemistry, one often represents the Hartree–Fock equation in a non-orthogonal basis set. This particular representation is a generalized eigenvalue problem called Roothaan equations.
### Geology and glaciology
In geology, especially in the study of glacial till, eigenvectors and eigenvalues are used as a method by which a mass of information of a clast fabric's constituents' orientation and dip can be summarized in a 3-D space by six numbers. In the field, a geologist may collect such data for hundreds or thousands of clasts in a soil sample, which can only be compared graphically such as in a Tri-Plot (Sneed and Folk) diagram,2122 or as a Stereonet on a Wulff Net.23
The output for the orientation tensor is in the three orthogonal (perpendicular) axes of space. The three eigenvectors are ordered $v_1, v_2, v_3$ by their eigenvalues $E_1 \geq E_2 \geq E_3$;24 $v_1$ then is the primary orientation/dip of clast, $v_2$ is the secondary and $v_3$ is the tertiary, in terms of strength. The clast orientation is defined as the direction of the eigenvector, on a compass rose of 360°. Dip is measured as the eigenvalue, the modulus of the tensor: this is valued from 0° (no dip) to 90° (vertical). The relative values of $E_1$, $E_2$, and $E_3$ are dictated by the nature of the sediment's fabric. If $E_1 = E_2 = E_3$, the fabric is said to be isotropic. If $E_1 = E_2 > E_3$, the fabric is said to be planar. If $E_1 > E_2 > E_3$, the fabric is said to be linear.25
### Principal components analysis
PCA of the multivariate Gaussian distribution centered at $(1,3)$ with a standard deviation of 3 in roughly the $(0.878,0.478)$ direction and of 1 in the orthogonal direction. The vectors shown are unit eigenvectors of the (symmetric, positive-semidefinite) covariance matrix scaled by the square root of the corresponding eigenvalue. (Just as in the one-dimensional case, the square root is taken because the standard deviation is more readily visualized than the variance.
The eigendecomposition of a symmetric positive semidefinite (PSD) matrix yields an orthogonal basis of eigenvectors, each of which has a nonnegative eigenvalue. The orthogonal decomposition of a PSD matrix is used in multivariate analysis, where the sample covariance matrices are PSD. This orthogonal decomposition is called principal components analysis (PCA) in statistics. PCA studies linear relations among variables. PCA is performed on the covariance matrix or the correlation matrix (in which each variable is scaled to have its sample variance equal to one). For the covariance or correlation matrix, the eigenvectors correspond to principal components and the eigenvalues to the variance explained by the principal components. Principal component analysis of the correlation matrix provides an orthonormal eigen-basis for the space of the observed data: In this basis, the largest eigenvalues correspond to the principal-components that are associated with most of the covariability among a number of observed data.
Principal component analysis is used to study large data sets, such as those encountered in data mining, chemical research, psychology, and in marketing. PCA is popular especially in psychology, in the field of psychometrics. In Q methodology, the eigenvalues of the correlation matrix determine the Q-methodologist's judgment of practical significance (which differs from the statistical significance of hypothesis testing; cf. criteria for determining the number of factors). More generally, principal component analysis can be used as a method of factor analysis in structural equation modeling.
### Vibration analysis
1st lateral bending (See vibration for more types of vibration)
Eigenvalue problems occur naturally in the vibration analysis of mechanical structures with many degrees of freedom. The eigenvalues are used to determine the natural frequencies (or eigenfrequencies) of vibration, and the eigenvectors determine the shapes of these vibrational modes. In particular, undamped vibration is governed by
$m\ddot x + kx = 0$
or
$m\ddot x = -k x$
that is, acceleration is proportional to position (i.e., we expect $x$ to be sinusoidal in time).
In $n$ dimensions, $m$ becomes a mass matrix and $k$ a stiffness matrix. Admissible solutions are then a linear combination of solutions to the generalized eigenvalue problem
$-k x = \omega^2 m x$
where $\omega^2$ is the eigenvalue and $\omega$ is the angular frequency. Note that the principal vibration modes are different from the principal compliance modes, which are the eigenvectors of $k$ alone. Furthermore, damped vibration, governed by
$m\ddot x + c \dot x + kx = 0$
$(\omega^2 m + \omega c + k)x = 0.$
This can be reduced to a generalized eigenvalue problem by clever use of algebra at the cost of solving a larger system.
The orthogonality properties of the eigenvectors allows decoupling of the differential equations so that the system can be represented as linear summation of the eigenvectors. The eigenvalue problem of complex structures is often solved using finite element analysis, but neatly generalize the solution to scalar-valued vibration problems.
### Eigenfaces
Eigenfaces as examples of eigenvectors
In image processing, processed images of faces can be seen as vectors whose components are the brightnesses of each pixel.26 The dimension of this vector space is the number of pixels. The eigenvectors of the covariance matrix associated with a large set of normalized pictures of faces are called eigenfaces; this is an example of principal components analysis. They are very useful for expressing any face image as a linear combination of some of them. In the facial recognition branch of biometrics, eigenfaces provide a means of applying data compression to faces for identification purposes. Research related to eigen vision systems determining hand gestures has also been made.
Similar to this concept, eigenvoices represent the general direction of variability in human pronunciations of a particular utterance, such as a word in a language. Based on a linear combination of such eigenvoices, a new voice pronunciation of the word can be constructed. These concepts have been found useful in automatic speech recognition systems, for speaker adaptation.
### Tensor of moment of inertia
In mechanics, the eigenvectors of the moment of inertia tensor define the principal axes of a rigid body. The tensor of moment of inertia is a key quantity required to determine the rotation of a rigid body around its center of mass.
### Stress tensor
In solid mechanics, the stress tensor is symmetric and so can be decomposed into a diagonal tensor with the eigenvalues on the diagonal and eigenvectors as a basis. Because it is diagonal, in this orientation, the stress tensor has no shear components; the components it does have are the principal components.
### Eigenvalues of a graph
In spectral graph theory, an eigenvalue of a graph is defined as an eigenvalue of the graph's adjacency matrix $A$, or (increasingly) of the graph's Laplacian matrix (see also Discrete Laplace operator), which is either $T - A$ (sometimes called the combinatorial Laplacian) or $I - T^{-1/2}A T^{-1/2}$ (sometimes called the normalized Laplacian), where $T$ is a diagonal matrix with $T_{i i}$ equal to the degree of vertex $v_i$, and in $T^{-1/2}$, the $i$th diagonal entry is $1/\sqrt{\operatorname{deg}(v_i)}$. The $k$th principal eigenvector of a graph is defined as either the eigenvector corresponding to the $k$th largest or $k$th smallest eigenvalue of the Laplacian. The first principal eigenvector of the graph is also referred to merely as the principal eigenvector.
The principal eigenvector is used to measure the centrality of its vertices. An example is Google's PageRank algorithm. The principal eigenvector of a modified adjacency matrix of the World Wide Web graph gives the page ranks as its components. This vector corresponds to the stationary distribution of the Markov chain represented by the row-normalized adjacency matrix; however, the adjacency matrix must first be modified to ensure a stationary distribution exists. The second smallest eigenvector can be used to partition the graph into clusters, via spectral clustering. Other methods are also available for clustering.
### Basic reproduction number
See Basic reproduction number
The basic reproduction number ($R_0$) is a fundamental number in the study of how infectious diseases spread. If one infectious person is put into a population of completely susceptible people, then $R_0$ is the average number of people that one typical infectious person will infect. The generation time of an infection is the time, $t_G$, from one person becoming infected to the next person becoming infected. In a heterogeneous population, the next generation matrix defines how many people in the population will become infected after time $t_G$ has passed. $R_0$ is then the largest eigenvalue of the next generation matrix.2728
## Notes
1. ^ a b Wolfram Research, Inc. (2010) Eigenvector. Accessed on 2010-01-29.
2. ^ William H. Press, Saul A. Teukolsky, William T. Vetterling, Brian P. Flannery (2007), Numerical Recipes: The Art of Scientific Computing, Chapter 11: Eigensystems., pages=563–597. Third edition, Cambridge University Press. ISBN 9780521880688
3. ^ See Korn & Korn 2000, Section 14.3.5a; Friedberg, Insel & Spence 1989, p. 217
4. ^ Axler, Sheldon, "Ch. 5", Linear Algebra Done Right (2nd ed.), p. 77
5. ^ Shilov 1977, p. 109
6. ^ Lemma for the eigenspace
7. ^
8. ^ For a proof of this lemma, see Roman 2008, Theorem 8.2 on p. 186; Shilov 1977, p. 109; Hefferon 2001, p. 364; Beezer 2006, Theorem EDELI on p. 469; and Lemma for linear independence of eigenvectors
9. ^ a b c d Trefethen, Lloyd N.; Bau, David (1997), Numerical Linear Algebra, SIAM
10. ^ See Hawkins 1975, §2
11. ^ a b c d See Hawkins 1975, §3
12. ^ a b c See Kline 1972, pp. 807–808
13. ^ See Kline 1972, p. 673
14. ^ See Kline 1972, pp. 715–716
15. ^ See Kline 1972, pp. 706–707
16. ^ See Kline 1972, p. 1063
17. ^ See Aldrich 2006
18. ^ Francis, J. G. F. (1961), "The QR Transformation, I (part 1)", The Computer Journal 4 (3): 265–271, doi:10.1093/comjnl/4.3.265 and Francis, J. G. F. (1962), "The QR Transformation, II (part 2)", The Computer Journal 4 (4): 332–345, doi:10.1093/comjnl/4.4.332
19. ^ Kublanovskaya, Vera N. (1961), "On some algorithms for the solution of the complete eigenvalue problem", USSR Computational Mathematics and Mathematical Physics 3: 637–657. Also published in: Zhurnal Vychislitel'noi Matematiki i Matematicheskoi Fiziki 1 (4), 1961: 555–570
20. ^ See Golub & van Loan 1996, §7.3; Meyer 2000, §7.3
21. ^ Graham, D.; Midgley, N. (2000), "Graphical representation of particle shape using triangular diagrams: an Excel spreadsheet method", Earth Surface Processes and Landforms 25 (13): 1473–1477, doi:10.1002/1096-9837(200012)25:13<1473::AID-ESP158>3.0.CO;2-C
22. ^ Sneed, E. D.; Folk, R. L. (1958), "Pebbles in the lower Colorado River, Texas, a study of particle morphogenesis", Journal of Geology 66 (2): 114–150, doi:10.1086/626490
23. ^ Knox-Robinson, C; Gardoll, Stephen J (1998), "GIS-stereoplot: an interactive stereonet plotting module for ArcView 3.0 geographic information system", Computers & Geosciences 24 (3): 243, doi:10.1016/S0098-3004(97)00122-2
24. ^ Stereo32 software
25. ^ Benn, D.; Evans, D. (2004), A Practical Guide to the study of Glacial Sediments, London: Arnold, pp. 103–107
26. ^ Xirouhakis, A.; Votsis, G.; Delopoulus, A. (2004), Estimation of 3D motion and structure of human faces (PDF), Online paper in PDF format, National Technical University of Athens
27. ^ Diekmann O, Heesterbeek JAP, Metz JAJ (1990), "On the definition and the computation of the basic reproduction ratio R0 in models for infectious diseases in heterogeneous populations", Journal of Mathematical Biology 28 (4): 365–382, doi:10.1007/BF00178324, PMID 2117040
28. ^ Odo Diekmann and J. A. P. Heesterbeek (2000), Mathematical epidemiology of infectious diseases, Wiley series in mathematical and computational biology, West Sussex, England: John Wiley & Sons
## References
• Korn, Granino A.; Korn, Theresa M. (2000), "Mathematical Handbook for Scientists and Engineers: Definitions, Theorems, and Formulas for Reference and Review", New York: McGraw-Hill (1152 p., Dover Publications, 2 Revised edition), Bibcode:1968mhse.book.....K, ISBN 0-486-41147-8.
• Lipschutz, Seymour (1991), Schaum's outline of theory and problems of linear algebra, Schaum's outline series (2nd ed.), New York: McGraw-Hill Companies, ISBN 0-07-038007-4.
• Friedberg, Stephen H.; Insel, Arnold J.; Spence, Lawrence E. (1989), Linear algebra (2nd ed.), Englewood Cliffs, New Jersey 07632: Prentice Hall, ISBN 0-13-537102-3.
• Aldrich, John (2006), "Eigenvalue, eigenfunction, eigenvector, and related terms", in Jeff Miller (Editor), Earliest Known Uses of Some of the Words of Mathematics, retrieved 2006-08-22
• Strang, Gilbert (1993), Introduction to linear algebra, Wellesley-Cambridge Press, Wellesley, Massachusetts, ISBN 0-9614088-5-5.
• Strang, Gilbert (2006), Linear algebra and its applications, Thomson, Brooks/Cole, Belmont, California, ISBN 0-03-010567-6.
• Bowen, Ray M.; Wang, Chao-Cheng (1980), Linear and multilinear algebra, Plenum Press, New York, ISBN 0-306-37508-7.
• Cohen-Tannoudji, Claude (1977), "Chapter II. The mathematical tools of quantum mechanics", Quantum mechanics, John Wiley & Sons, ISBN 0-471-16432-1.
• Fraleigh, John B.; Beauregard, Raymond A. (1995), Linear algebra (3rd ed.), Addison-Wesley Publishing Company, ISBN 0-201-83999-7 (international edition) Check |isbn= value (help).
• Golub, Gene H.; Van Loan, Charles F. (1996), Matrix computations (3rd Edition), Johns Hopkins University Press, Baltimore, Maryland, ISBN 978-0-8018-5414-9.
• Hawkins, T. (1975), "Cauchy and the spectral theory of matrices", Historia Mathematica 2: 1–29, doi:10.1016/0315-0860(75)90032-4.
• Horn, Roger A.; Johnson, Charles F. (1985), Matrix analysis, Cambridge University Press, ISBN 0-521-30586-1 (hardback), ISBN 0-521-38632-2 (paperback) Check |isbn= value (help).
• Kline, Morris (1972), Mathematical thought from ancient to modern times, Oxford University Press, ISBN 0-19-501496-0.
• Meyer, Carl D. (2000), Matrix analysis and applied linear algebra, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, ISBN 978-0-89871-454-8.
• Brown, Maureen (October 2004), Illuminating Patterns of Perception: An Overview of Q Methodology.
• Golub, Gene F.; van der Vorst, Henk A. (2000), "Eigenvalue computation in the 20th century", Journal of Computational and Applied Mathematics 123: 35–65, doi:10.1016/S0377-0427(00)00413-1.
• Akivis, Max A.; Vladislav V. Goldberg (1969), Tensor calculus, Russian, Science Publishers, Moscow .
• Gelfand, I. M. (1971), Lecture notes in linear algebra, Russian, Science Publishers, Moscow.
• Alexandrov, Pavel S. (1968), Lecture notes in analytical geometry, Russian, Science Publishers, Moscow.
• Carter, Tamara A.; Tapia, Richard A.; Papaconstantinou, Anne, Linear Algebra: An Introduction to Linear Algebra for Pre-Calculus Students, Rice University, Online Edition, retrieved 2008-02-19.
• Roman, Steven (2008), Advanced linear algebra (3rd ed.), New York: Springer Science + Business Media, LLC, ISBN 978-0-387-72828-5.
• Shilov, Georgi E. (1977), Linear algebra (translated and edited by Richard A. Silverman ed.), New York: Dover Publications, ISBN 0-486-63518-X.
• Hefferon, Jim (2001), Linear Algebra, Online book, St Michael's College, Colchester, Vermont, USA.
• Kuttler, Kenneth (2007), An introduction to linear algebra (PDF), Online e-book in PDF format, Brigham Young University.
• Demmel, James W. (1997), Applied numerical linear algebra, SIAM, ISBN 0-89871-389-7.
• Beezer, Robert A. (2006), A first course in linear algebra, Free online book under GNU licence, University of Puget Sound.
• Lancaster, P. (1973), Matrix theory, Russian, Moscow, Russia: Science Publishers.
• Halmos, Paul R. (1987), Finite-dimensional vector spaces (8th ed.), New York: Springer-Verlag, ISBN 0-387-90093-4.
• Pigolkina, T. S. and Shulman, V. S., Eigenvalue (Russian), In:Vinogradov, I. M. (Ed.), Mathematical Encyclopedia, Vol. 5, Soviet Encyclopedia, Moscow, 1977.
• Greub, Werner H. (1975), Linear Algebra (4th Edition), Springer-Verlag, New York, ISBN 0-387-90110-8.
• Larson, Ron; Edwards, Bruce H. (2003), Elementary linear algebra (5th ed.), Houghton Mifflin Company, ISBN 0-618-33567-6.
• Curtis, Charles W., Linear Algebra: An Introductory Approach, 347 p., Springer; 4th ed. 1984. Corr. 7th printing edition (August 19, 1999), ISBN 0-387-90992-3.
• Shores, Thomas S. (2007), Applied linear algebra and matrix analysis, Springer Science+Business Media, LLC, ISBN 0-387-33194-8.
• Sharipov, Ruslan A. (1996), Course of Linear Algebra and Multidimensional Geometry: the textbook, arXiv:math/0405323, ISBN 5-7477-0099-5.
• Gohberg, Israel; Lancaster, Peter; Rodman, Leiba (2005), Indefinite linear algebra and applications, Basel-Boston-Berlin: Birkhäuser Verlag, ISBN 3-7643-7349-0.
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# For the system of equations given, what is the value of z?
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For the system of equations given, what is the value of z? [#permalink]
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11 Sep 2012, 03:46
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x-4=z
y-x=8
8-z=t
For the system of equations given, what is the value of z?
(1) x = 7
(2) t = 5
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Re: For the system of equations given, what is the value of z? [#permalink]
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11 Sep 2012, 03:47
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SOLUTION
x-4=z
y-x=8
8-z=t
For the system of equations given, what is the value of z?
(1) x = 7. Substitute the value of x in the first equation to find z. Sufficient.
(2) t = 5. Substitute the value of t in the third equation to find z. Sufficient.
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Re: For the system of equations given, what is the value of z? [#permalink]
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12 Sep 2012, 05:24
2
KUDOS
Bunuel wrote:
x-4=z
y-x=8
8-z=t
For the system of equations given, what is the value of z?
(1) x = 7
(2) t = 5
St 1: Sufficient: x=7, substituting in Eq 1, we will get value of Z, 7-4=z=3,
ST 2. Sufficient: t=5, substituting in Eq 3, We will get value of Z, 8-Z=5, Z= 3.
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Re: For the system of equations given, what is the value of z? [#permalink]
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Same reasoning as above.
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Re: For the system of equations given, what is the value of z? [#permalink]
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14 Sep 2012, 05:08
SOLUTION
x-4=z
y-x=8
8-z=t
For the system of equations given, what is the value of z?
(1) x = 7. Substitute the value of x in the first equation to find z. Sufficient.
(2) t = 5. Substitute the value of t in the third equation to find z. Sufficient.
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Re: For the system of equations given, what is the value of z? [#permalink]
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18 Jul 2013, 13:58
I didnt even solve for z, all I used was the fact that "ONE unknown and TWO knowns means you can solve for the unknown".
Hence, we are given x in s(1) and that is sufficent just by looking at the top equation. Then we are given t in s(2) which is enough for the third equation.
The reason this method is superior to any other is that you quickly can move on to the next quant problem on the test, I solved this q in 35 seconds using this method.
If there are any holes in this approach, please speak up and correct me.
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Re: For the system of equations given, what is the value of z? [#permalink]
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30 May 2014, 00:24
this is sub500
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Re: For the system of equations given, what is the value of z? [#permalink]
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10 Sep 2015, 23:31
Basic rule for solving linear equation
NO of Unknown === no of equation
In this question :
NO of Unknown==4
no of equation ==3
both the statements are reducing 1 unknown making the equation solvable
hence D
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Re: For the system of equations given, what is the value of z? [#permalink]
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09 Aug 2016, 12:49
Quote:
x-4=z
y-x=8
8-z=t
For the system of equations given, what is the value of z?
(1) x = 7
(2) t = 5
We are given 3 equations:
1) x – 4 = z
2) y – x = 8
3) 8 – z = t
We need to determine the value of z.
Statement One Alone:
x = 7
We see that we can substitute 7 for x in the first equation and then determine the value of z.
x – 4 = z
7 – 4 = z
3 = z
Statement one alone is sufficient to answer the question. We can eliminate answer choices B, C, and E.
Statement Two Alone:
t = 5
We see that we can substitute 5 for t in equation 3 and then determine the value of z.
8 – z = t
8 – z = 5
3 = z
Statement two alone is sufficient to answer the question.
The answer is D.
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Re: For the system of equations given, what is the value of z? [#permalink] 09 Aug 2016, 12:49
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https://calculationcalculator.com/chatak-to-square-inch | 1,726,104,661,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00790.warc.gz | 136,109,596 | 22,566 | # Chatak to Square Inch Conversion
## 1 Chatak is equal to how many Square Inch?
### 6480 Square Inch
##### Reference This Converter:
Chatak and Square Inch both are the Land measurement unit. Compare values between unit Chatak with other Land measurement units. You can also calculate other Land conversion units that are available on the select box, having on this same page.
Chatak to Square Inch conversion allows you to convert value between Chatak to Square Inch easily. Just enter the Chatak value into the input box, the system will automatically calculate Square Inch value. 1 Chatak in Square Inch? In mathematical terms, 1 Chatak = 6480 Square Inch.
To conversion value between Chatak to Square Inch, just multiply the value by the conversion ratio. One Chatak is equal to 6480 Square Inch, so use this simple formula to convert -
The value in Square Inch is equal to the value of Chatak multiplied by 6480.
Square Inch = Chatak * 6480;
For calculation, here's how to convert 10 Chatak to Square Inch using the formula above -
10 Chatak = (10 * 6480) = 64800 Square Inch
Chatak Square Inch Conversion
0.001 6.48 0.001 Chatak = 6.48 Square Inch
0.002 12.96 0.002 Chatak = 12.96 Square Inch
0.003 19.44 0.003 Chatak = 19.44 Square Inch
0.004 25.92 0.004 Chatak = 25.92 Square Inch
0.005 32.4 0.005 Chatak = 32.4 Square Inch
0.006 38.88 0.006 Chatak = 38.88 Square Inch
0.007 45.36 0.007 Chatak = 45.36 Square Inch
0.008 51.84 0.008 Chatak = 51.84 Square Inch
0.009 58.32 0.009 Chatak = 58.32 Square Inch
0.01 64.8 0.01 Chatak = 64.8 Square Inch
0.02 129.6 0.02 Chatak = 129.6 Square Inch
0.03 194.4 0.03 Chatak = 194.4 Square Inch
0.04 259.2 0.04 Chatak = 259.2 Square Inch
0.05 324 0.05 Chatak = 324 Square Inch
0.06 388.8 0.06 Chatak = 388.8 Square Inch
0.07 453.6 0.07 Chatak = 453.6 Square Inch
0.08 518.4 0.08 Chatak = 518.4 Square Inch
0.09 583.2 0.09 Chatak = 583.2 Square Inch
0.1 648 0.1 Chatak = 648 Square Inch
0.2 1296 0.2 Chatak = 1296 Square Inch
0.3 1944 0.3 Chatak = 1944 Square Inch
0.4 2592 0.4 Chatak = 2592 Square Inch
0.5 3240 0.5 Chatak = 3240 Square Inch
0.6 3888 0.6 Chatak = 3888 Square Inch
0.7 4536 0.7 Chatak = 4536 Square Inch
0.8 5184 0.8 Chatak = 5184 Square Inch
0.9 5832 0.9 Chatak = 5832 Square Inch
1 6480 1 Chatak = 6480 Square Inch
2 12960 2 Chatak = 12960 Square Inch
3 19440 3 Chatak = 19440 Square Inch
4 25920 4 Chatak = 25920 Square Inch
5 32400 5 Chatak = 32400 Square Inch
6 38880 6 Chatak = 38880 Square Inch
7 45360 7 Chatak = 45360 Square Inch
8 51840 8 Chatak = 51840 Square Inch
9 58320 9 Chatak = 58320 Square Inch
10 64800 10 Chatak = 64800 Square Inch
11 71280 11 Chatak = 71280 Square Inch
12 77760 12 Chatak = 77760 Square Inch
13 84240 13 Chatak = 84240 Square Inch
14 90720 14 Chatak = 90720 Square Inch
15 97200 15 Chatak = 97200 Square Inch
16 103680 16 Chatak = 103680 Square Inch
17 110160 17 Chatak = 110160 Square Inch
18 116640 18 Chatak = 116640 Square Inch
19 123120 19 Chatak = 123120 Square Inch
20 129600 20 Chatak = 129600 Square Inch
21 136080 21 Chatak = 136080 Square Inch
22 142560 22 Chatak = 142560 Square Inch
23 149040 23 Chatak = 149040 Square Inch
24 155520 24 Chatak = 155520 Square Inch
25 162000 25 Chatak = 162000 Square Inch
26 168480 26 Chatak = 168480 Square Inch
27 174960 27 Chatak = 174960 Square Inch
28 181440 28 Chatak = 181440 Square Inch
29 187920 29 Chatak = 187920 Square Inch
30 194400 30 Chatak = 194400 Square Inch
31 200880 31 Chatak = 200880 Square Inch
32 207360 32 Chatak = 207360 Square Inch
33 213840 33 Chatak = 213840 Square Inch
34 220320 34 Chatak = 220320 Square Inch
35 226800 35 Chatak = 226800 Square Inch
36 233280 36 Chatak = 233280 Square Inch
37 239760 37 Chatak = 239760 Square Inch
38 246240 38 Chatak = 246240 Square Inch
39 252720 39 Chatak = 252720 Square Inch
40 259200 40 Chatak = 259200 Square Inch
41 265680 41 Chatak = 265680 Square Inch
42 272160 42 Chatak = 272160 Square Inch
43 278640 43 Chatak = 278640 Square Inch
44 285120 44 Chatak = 285120 Square Inch
45 291600 45 Chatak = 291600 Square Inch
46 298080 46 Chatak = 298080 Square Inch
47 304560 47 Chatak = 304560 Square Inch
48 311040 48 Chatak = 311040 Square Inch
49 317520 49 Chatak = 317520 Square Inch
50 324000 50 Chatak = 324000 Square Inch
51 330480 51 Chatak = 330480 Square Inch
52 336960 52 Chatak = 336960 Square Inch
53 343440 53 Chatak = 343440 Square Inch
54 349920 54 Chatak = 349920 Square Inch
55 356400 55 Chatak = 356400 Square Inch
56 362880 56 Chatak = 362880 Square Inch
57 369360 57 Chatak = 369360 Square Inch
58 375840 58 Chatak = 375840 Square Inch
59 382320 59 Chatak = 382320 Square Inch
60 388800 60 Chatak = 388800 Square Inch
61 395280 61 Chatak = 395280 Square Inch
62 401760 62 Chatak = 401760 Square Inch
63 408240 63 Chatak = 408240 Square Inch
64 414720 64 Chatak = 414720 Square Inch
65 421200 65 Chatak = 421200 Square Inch
66 427680 66 Chatak = 427680 Square Inch
67 434160 67 Chatak = 434160 Square Inch
68 440640 68 Chatak = 440640 Square Inch
69 447120 69 Chatak = 447120 Square Inch
70 453600 70 Chatak = 453600 Square Inch
71 460080 71 Chatak = 460080 Square Inch
72 466560 72 Chatak = 466560 Square Inch
73 473040 73 Chatak = 473040 Square Inch
74 479520 74 Chatak = 479520 Square Inch
75 486000 75 Chatak = 486000 Square Inch
76 492480 76 Chatak = 492480 Square Inch
77 498960 77 Chatak = 498960 Square Inch
78 505440 78 Chatak = 505440 Square Inch
79 511920 79 Chatak = 511920 Square Inch
80 518400 80 Chatak = 518400 Square Inch
81 524880 81 Chatak = 524880 Square Inch
82 531360 82 Chatak = 531360 Square Inch
83 537840 83 Chatak = 537840 Square Inch
84 544320 84 Chatak = 544320 Square Inch
85 550800 85 Chatak = 550800 Square Inch
86 557280 86 Chatak = 557280 Square Inch
87 563760 87 Chatak = 563760 Square Inch
88 570240 88 Chatak = 570240 Square Inch
89 576720 89 Chatak = 576720 Square Inch
90 583200 90 Chatak = 583200 Square Inch
91 589680 91 Chatak = 589680 Square Inch
92 596160 92 Chatak = 596160 Square Inch
93 602640 93 Chatak = 602640 Square Inch
94 609120 94 Chatak = 609120 Square Inch
95 615600 95 Chatak = 615600 Square Inch
96 622080 96 Chatak = 622080 Square Inch
97 628560 97 Chatak = 628560 Square Inch
98 635040 98 Chatak = 635040 Square Inch
99 641520 99 Chatak = 641520 Square Inch
100 648000 100 Chatak = 648000 Square Inch
Chatak is a land measurement unit, It is mainly used in Bangladesh and West Bengal (India). 1 Chatak is equal to 5 Gaj. Chatak is small area measurement unit.
A square inch is a unit of area, equal to the area of a square with sides of one inch. The abbreviation for Square Inch is "in2".
The value in Square Inch is equal to the value of Chatak multiplied by 6480.
Square Inch = Chatak * 6480;
1 Chatak is equal to 6480 Square Inch.
1 Chatak = 6480 Square Inch.
• 1 chatak = square inch
• chatak into square inch
• chataks to square inchs
• convert chatak to square inch
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→ | 2,488 | 6,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-38 | latest | en | 0.693769 |
https://math.stackexchange.com/questions/4669164/counting-the-number-of-ways-of-organizing-an-odd-number-of-objects-into-3-odd-gr/4669173 | 1,702,197,693,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00382.warc.gz | 412,245,946 | 42,102 | Counting the number of ways of organizing an odd number of objects into 3 odd groups, elegant method?
I came across this problem on a blog I've been reading (link here but not necessary for understanding the problem). You have to split an odd number N of distinct objects into three different groups such that the number of objects in each group is an odd number. In how many ways can this be done? The blog goes into a combinatorial solution, giving an answer of $$\frac{3^N-3}{4}$$.
My question is: Is there an elegant reason why it is almost exactly a quarter of the possible distributions that work? More interestingly, is there a simpler argument for why it is exactly the number it is?
As an example of what I'm looking for, for the easier question of how many ways there are to split an even number of distinct items into two groups with an even number of items, you can have a given item (item A, say), and for every selection where A is in the first group, there is one for which it is in the second group (just by moving it from group 1 to group 2). Since exactly one of those possibilities has an even number in each group, exactly half of the possible selections will work. Therefore the total number of ways is $$2^{N-1}$$.
Intuitively, something similar should work for three groups. Is there an argument along similar lines for the three group case?
The possible parities for the three groups are even-even-odd, even-odd-even, odd-even-even, and odd-odd-odd. The odds for each of these outcomes are approximately even, and one of the four is the desired result. (Each of the three outcomes with two even groups has $$\frac{3^N+1}{4}$$ ways to distribute the items.)
This is similar to the two-group problem, where the parities are either even-even or odd-odd with equal probability.
• +1 but "The odds...are approximately even" might be confusing terminology in a problem with odd/even. :) Mar 30 at 2:44
• @RobPratt Agreed, but I could not resist the play on words. Mar 30 at 2:45
There are $$3^{N}-3$$ ways to organise the objects such that no combinations of two groups are empty.
Of these, there are as many ways to organise with $$ooo$$ patern as ways to organise with $$oee$$ pattern. This is quite trivial, because once you assign an odd number of objects to the first group, the rest (even, non-empty) can be split into two even or two odd groups with equally many possibilities.
Because of symmetry, the number of possibilities with $$oee$$ pattern must equal those with $$eoe$$ and $$eeo$$ pattern.
Thus, there are $$\frac{3^{N}-3}{4}$$ ways to organise the objects with $$ooo$$ pattern.
• The part about how "the rest (even, non-empty) can be split into two even or two odd groups with equally many possibilities" could maybe use some elaboration. (I mean, it does work: just set aside one item and arrange the (odd number of) others into two groups any way you like; now adding the remaining item to one of the two groups (one always being odd-sized, the other even-sized) makes the number of items in the groups either both odd or both even, thus showing that there's a 1:1 correspondence between odd-odd and even-even groupings. But that's hardly obvious at a glance.) Mar 30 at 10:55
• @IlmariKaronen a non-empty set has as many even subsets as odd subsets. This is well-known in combinatorics. Put this subset to group 2 and the rest to group 3, if the initial set is even, group 2 and 3 are both even or odd Mar 30 at 11:06
We can give a bijective proof of this fact.
Call the groups $$1,2,3$$. First, let us show that there are $$(3^n-1)/2$$ ways to distribute the objects such that group $$1$$ has an odd number of members.
Consider the transformation $$T$$ defined on the set of the $$3^n$$ possible distributions as follows. Given a distribution, $$x$$, we define $$T(x)$$ to be the result of finding the smallest element in the union of groups $$1$$ and $$2$$, and moving it to the other group (if it was in $$1$$, it goes to $$2$$, and vice versa). The only exception is the distribution where all the objects are in group $$3$$, for which we say $$T(x)=x$$, which is the only fixed point of $$T$$.
The set of $$(3^n-1)$$ distributions where at least one object is in groups $$1$$ or $$2$$ are partitioned into pairs of the form $$\{x,T(x)\}$$. In each pair, exactly one of the distributions has group $$1$$ being odd. Therefore, the number of distributions where group $$1$$ is odd is $$(3^n-1)/2$$.
We then rinse and repeat. Consider the set of $$(3^n-1)/2$$ distributions where the first group is odd, and define the transformation $$S$$ on this set where you find the smallest element in either group $$2$$ or $$3$$ and move it to the other group. By the same logic, $$S$$ divides the distributions where group $$1$$ is odd into pairs, where exactly one distribution in each pair has group $$2$$ odd, and therefore group $$3$$ odd as well. The only exception is the transformation where all the objects are in group $$1$$. Therefore, the number of distributions where all groups are odd is $$\frac12\left[\frac{3^n-1}{2}-1\right]=\frac{3^n-3}4$$
• +1 but destruction=distribution? Mar 30 at 4:25
Looking over the current answers, I think that my solution is mostly the same as Rezha's, but I will still present it, in the way that I think about it.
I will show a bijection between the set of all partitions of the $$N$$-element set into three sets such that at most one of them is empty (of those there are $$3^N - 3$$) and four times the set of all partitions into three sets of odd cardinality. Indeed I will refer to a partition in one of those four copies as a partition with either a pair of sets marked (there are three possibilities of that) or none.
Given any partition of the $$N$$-set into three sets, either none or two of them will have an even number of elements. If none of them do, we just map the partition to itself, with no pair marked. If two of the sets have even cardinality, we mark this pair of sets and move the smallest number in those two sets (which cannot both be empty) to the other set in the pair, making both of them odd.
In the other direction, map an unmarked partition to itself, and for a partition with a pair marked, move the smallest number in those two sets to the other of the two.
This is easily checked to be a bijection.
• Thanks! I accepted this answer as it seems to be the most intuitive to me, and the closest to what I was trying to think through on my own. The step I was missing was "marking" the two groups where an element is being switched. Apr 2 at 23:54
For a fourth approach ,we can make use of exponential generating functions to distribute N distinct odd number of object in 3 distinct boxes where each boxes have odd number of elements.
We know that each box can have such number of elements: $$1,3,5,7,9,11,13,15,17...$$
So , the exponential generating function for these number of elements are $$\frac{x^1}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+..+..$$
If $$e^x =1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+..+..$$
and
$$e^{-x} =1-\frac{x^1}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+\frac{x^6}{6!}-\frac{x^7}{7!}+..+..$$
Then , $$\frac{e^x -e^{-x}}{2}=\frac{x^1}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+..+..$$
So , find the coefficient of $$\frac{x^N}{N!}$$ in the expansion of $$\bigg(\frac{e^x -e^{-x}}{2}\bigg)^3$$
$$\frac{e^x -e^{-x}}{2}=\frac{(e^{2x}-1)^3}{(2e^x)^3}=\frac{1}{8}\frac{e^{6x}-3e^{4x}+3e^{2x}-1}{e^{3x}}$$
$$\bigg[\frac{x^N}{N!}\bigg]\frac{1}{8}({e^{3x}-3e^{x}+3e^{-x}-e^{-3x}})$$
$$\frac{1}{8}(3^N -3+3(-1)^N-(-3)^N)$$
We know that N is odd , so $$\frac{1}{8}(3^N -3-3+3^N)=\frac{1}{4}(3^N-3)$$
• Slightly less intuitive, but using exponential generating functions in this way is quite pleasing. Also it seems like it would very easily extend to more groups, in a way that the other approaches don't. Apr 3 at 0:00
Let $$p_N$$ be the probability that a uniformly random division of the objects gives three odd groups. Take $$N\geq 3$$. Then there are two ways this can happen:
1. the first $$N-2$$ objects form three odd groups and the last two objects go in the same group.
2. The first $$N-2$$ objects form one odd and two even groups and the last two objects go into the two even groups in either order.
Thus we have $$p_N=\frac{1}{3}p_{N-2}+\frac{2}{9}(1-p_{N-2})$$ for $$N\geq 3$$, which rearranges to $$p_N-\frac14=\frac19\big(p_{N-2}-\frac14\big)$$.
Clearly this means $$p_N-\frac14\to0$$; in fact we have $$p_N-\frac14=3^{-N}c$$ for some constant $$c$$ which we can determine from the value of $$p_1$$. | 2,461 | 8,683 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 67, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-50 | latest | en | 0.942519 |
http://hyperion.chemistry.uoc.gr/body-curves-wonvwh/ec8fc1-wavelength-of-gamma-rays-in-nm | 1,618,142,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038062492.5/warc/CC-MAIN-20210411115126-20210411145126-00408.warc.gz | 48,777,516 | 8,778 | Let us work out the wavelength of a typical gammaray of 1 MeV. It sounds very infinitesimal, but gamma rays of the electromagnetic spectrum have wavelengths of that size and shorter. The author has found that gamma rays affect the photoluminescence emission of DAM–ADC plates. (1*10^-9 nm/m) Cosmic rays 5 × 10-5 nm Gamma rays 10-3 – 15 nm x-rays 0.01 – 15 nm Far UV 15 – 200 nm 666,667 to 50,000 cm-1 Near UV 200 – 400 nm 50,000 to 20,000 cm-1 Visible 400 – 800 nm 25,000 to 12,500 cm-1 Near IR 0.8 – 2.5 Their wavelengths range from 0.01×10⁻¹² meters 10×10⁻¹² meters . As of now, they are the most energetic form of electromagnetic radiation in a range greater than 100 keV. Let us see how much is 1 eV (electron Volt) an unit of energy in terms of erg. Following the above examples, gamma rays have very high energy and radio waves areWhat is the Wavelength of Light: The wavelength of light can be calculated by dividing the speed of light by the frequency of light. The shorter wavelengths of ultraviolet can be absorbed by DNA and damage it — causing mutations . Hence, in the order of increasing wavelength, the waves are Gamma rays (< 1 n m), X-rays (1-10 nm), infra red rays (7 0 0 − 1 0 5 n m), micro waves (1 0 5 − 1 0 8 n m), radio waves (> 1 0 Solution for Calculate the wavelength λ1 for gamma rays of frequency f1 = 6.10×1021 Hz so that gamma-ray wavelength is expressed in nanometers. In decimal notation it is from 0.01 trillionth meters to 10 trillionth meters. X-rays of wavelength $0.200 \mathrm{nm}$ are scattered from a block of carbon. According to the picture gamma rays are shorter than x-rays. Gamma rays are electromagnetic radiation of very short wavelength emitted by the nuclei of radioactive elements. Decreasing order of the wavelength of various rays.Microwave > infrared > Ultraviolet > Gamma Previous Year Papers Download Solved Question Papers Free … The ranges of electromagnetic waves are given below: (a) g-rays = 10ˉ³ nm (b) Microwaves = 0.1 m to 1 mm (c) x-rays = 1 nm to 10ˉ³ nm (d) Radio waves = >0.1 m Electromagnetic waves in the order of their increasing wavelength Asked Jul 2 Is 7.50 ms x 1 x Solution not yet available Calculate the frequency of an EM wave with a wavelength of $$\text{400}$$ $$\text{nm}$$. According to physics, UV radiation is divided into four regions - Near (400 How many photons are in the pulse? Arrange the following types of EM radiation in order of increasing frequency: infrared, X-rays, ultraviolet, visible, gamma. Visible light waves are one-thousandths the width of human hair--about a million times longer than gamma rays. Neither is visible to humans. Ultraviolet rays were present in the wavelength range from around 10 nm to around 400 nm. X Rays can penetrate the human skin without as much damage to cells that gamma rays do. The PL emission spectra clear considerable PL bands in the blue An X Ray has the wavelength of precisely 1000 nm or nanometers. Gamma Rays - the highest in frequency and energy, are the most damaging. The term "gamma rays" used as a term describing a classical electromagnetic wave of specific frequency and wavelength is fine. Source(s): Experience Atomic radii of most atoms are on the order of 100 pm, and lattice constants slightly larger. Short wavelength UV and the shorter wavelength radiation above it (X-rays and gamma rays) are called ionizing radiation, and exposure to them can damage living tissue, making them a health hazard. There is no absolute lower limit to the extent of the shortest wavelength because it has not yet been reached. View Notes - UV-VISIBLE SPECTROSCOPY (1) from CHEM 234 at University of Illinois, Chicago. Energy of the gamma ray = 1 MeV. The lengths range from 10 −10 meters to 10 −15 meters or shorter. Question 16 Name the waves (a) of lowest wavelength, (b) used for taking photographs in dark, (c) produced by the changes in the nucleus of an atom, (d) of wavelength nearly 0.1 nm. The energy of nuclear radiation is extremely high because such radiation is born in the intense conflict between the nuclear strong force and the electromagnetic force , the two strongest basic forces . The wavelength of a gamma … If the scattered radiation is detected at $90^{\circ}$ to the incident beam, find (a) the Compton shift, $\Delta \lambda,$ and $(b)$ the kinetic energy X-rays, gamma rays A nitrogen laser generates a pulse containing 10.0 mJ of energy at a wavelength of 337.1 nm. 100MeV applies ONLY to the energy involved...Gamma rays are ALWAYS OVER 10^19Hz in frequency, ( Ten with 19 zeros ) and always Shorter than 0.1 Angstroms( 10^-9 meter ) in wavelength. Light with a shorter wavelength than violet but longer than X-rays and gamma rays is called ultraviolet. (iv)Gamma rays are used in medical science to kill cancer cells. rays TUltra- i violet X rays Infrared iMicrowavesi Radio frequency 1020 1018 1016 1014 1012 1010 108 106 104 Frequency (s~1) Visible region 400 500 … We know that electromagnetic waves include different types of waves like radio waves, microwaves, Infrared rays, Visible rays, Ultraviolet rays, X-rays, and gamma rays. ULTRAVIOLET AND VISIBLE SPECTROSCOPY 1 High frequency, Short wavelength 0.01 nm Gamma Rays X-Rays 1 Gamma rays are produced during gamma decay, which normally occurs after other forms of decay occur, such as alpha or beta decay. X-Rays - also a wave of high energy and short wavelength. Electromagnetic Spectrum is the classification of these waves according to their frequency. @article{osti_20722363, title = {Accurate Wavelength Measurement of High-Energy Gamma Rays from the 35Cl(n,{gamma}) Reactions}, author = {Belgya, T and Molnar, G L and Mutti, P and Boerner, H G and Jentshel, M}, abstractNote = {The energies of eight gamma rays in the 36Cl level scheme have been measured with high precision using the 35Cl(n,{gamma}) reaction and the … A certain radioactive element emits a gamma ray with a frequency of 1.51 x 1020 Hz. Gamma-rays have a wavelength range below 100 pm and frequencies greater than 10 Hz. X-rays gamma rays The only difference between these manifestions of electromagnetic radiation is due to differences in the frequency and wavelength of the oscillations. Gamma-Rays The term gamma ray is used to denote electromagnetic radiation from the nucleus as a part of a radioactive process. Most X-Rays have a wavelength ranging from .01 to 10 nanometers. A radioactive nucleus can decay by the emission of an α or β particle. The radiation in (a) is gamma rays and in (b) is ultraviolet. Violet is the color at the end of the visible spectrum of light between blue and the invisible ultraviolet.Violet color has a dominant wavelength of approximately 380–450 nanometers. The daughter nucleus that results is usually left in an excited state. The frequency is the inverse of wavelength. Atomic radii of most atoms are on the order of 100 pm, and lattice constants slightly larger. Now 1 eV = (charge of an electron in What Is The Wavelength Of Gamma Rays With Frequency 2x10 Hz? Gamma rays occupy the short-wavelength end of the spectrum; they can have wavelengths smaller than the nucleus of an atom. The excitation wavelength used in this situation was 335 nm. 10-11 10-9 10-7 10-5 10-3 10-1 101 103 Gamma! X-rays are measured in nanometers (nm) and its wavelength ranges from 0.01 nm to 10nm Gamma Rays have the shortest wavelength on the electromagnetic spectrum.Therefore they are measured in Pico metres (pm) which is equal to one trillionth of a metre. 6 6.8 6.9 6.10 Electronic Structure of Atoms Solutions to Exercises (11.8 nm) (b) (c) (d) 2.998 x 108 m = c/v; Is 2.55 1016 No. UV-C, < 280 nm — a powerful germicidal agent. Although Gamma rays were first contrast the frequency and wavelength of gamma rays with the frequencies and wavelength of other waves on the electromagnetic spectrum jamourrr is waiting for … X-rays and gamma rays also damage DNA by generating ions within the cell. 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Wavelength because it has a frequency ranging between the 800 THz to 30 PHz width human. A term describing a classical electromagnetic wave of specific frequency and wavelength of gamma rays the only difference these! Of most atoms are on the order of 100 pm, and lattice constants larger... | 4,165 | 17,680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-17 | latest | en | 0.860177 |
https://thebrightsideofmathematics.com/courses/ordinary_differential_equations/ode02_info/ | 1,701,613,128,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00859.warc.gz | 648,656,381 | 10,191 | # Information about Ordinary Differential Equations - Part 2
• Title: Definitions
• YouTube-Title: Ordinary Differential Equations - Part 2 - Definitions
• Bright video: https://youtu.be/o_skYg3Iic0
• Dark video: https://youtu.be/lnAWcNrbWnc
• Timestamps
• Subtitle in English
1 00:00:00,500 –> 00:00:05,808 Hello and welcome back to the video series about ODEs.
2 00:00:06,386 –> 00:00:14,542 So you see we are still in the beginning of the topic and in today’s part 2 we will first define the important notions we need.
3 00:00:14,742 –> 00:00:20,307 So for example we will give the definition what we mean by an ODE.
4 00:00:20,507 –> 00:00:27,583 However, before we do that I first want to thank all the nice people who support this channel on Steady, via Paypal or by other means
6 00:00:34,978 –> 00:00:41,121 Ok, then without further ado let’s start with the important definitions we will need in this series.
7 00:00:41,586 –> 00:00:46,674 First of all we will often deal with a set of functions called C^k
8 00:00:46,874 –> 00:00:52,418 and usually the domain of definition for these functions is given by an interval “I”.
9 00:00:52,771 –> 00:00:57,890 Hence when you see capital “I” it’s always a subset of the real number line.
10 00:00:58,529 –> 00:01:06,376 Now in fact most of the time in this series it will denote an interval, but sometimes it will denote a general open set
11 00:01:06,771 –> 00:01:12,196 and in some other cases we might want that “I” stands for a union of 2 intervals.
12 00:01:12,396 –> 00:01:17,736 So you see, depending on what we want to do later, we will stretch this definition a little bit,
13 00:01:17,737 –> 00:01:20,557 but here we will start with an interval “I”
14 00:01:20,884 –> 00:01:25,901 and then this set C^k is well-defined as a set of functions
15 00:01:26,371 –> 00:01:34,398 and here we immediately see the first special feature of this topic, because we will denote functions by a lower case x.
16 00:01:34,900 –> 00:01:39,556 Hence x is a function, where the domain of definition is “I”
17 00:01:39,986 –> 00:01:43,168 and it simply maps into the real number line
18 00:01:43,900 –> 00:01:52,380 and now you should recall from real analysis that C^k means that the function x is k-times continuously differentiable.
19 00:01:52,914 –> 00:01:58,895 This means the kth order derivative of x exists and is a continuous function.
20 00:01:59,095 –> 00:02:06,205 Moreover I should tell you here that the variable name we want to use for the function x is usually a lower case t.
21 00:02:06,614 –> 00:02:09,623 This means we will write x(t).
22 00:02:09,823 –> 00:02:16,809 I tell you that, because if we choose the variable name as t, we will use a dot for denoting derivatives.
23 00:02:17,271 –> 00:02:21,583 Hence the first derivative here would be the function x dot.
24 00:02:21,783 –> 00:02:25,390 Then the second derivative would be x dot dot
25 00:02:25,590 –> 00:02:31,659 and so on and of course if we need too many dots, we will use the common upper index notation.
26 00:02:32,186 –> 00:02:37,904 Of course in situations where the dot could be confusing, we will just use other notations.
27 00:02:38,104 –> 00:02:42,785 So for example you know the common prime notation or the Leibniz notation.
28 00:02:42,985 –> 00:02:47,789 So please keep that in mind. Depending on the context or depending on which book you read,
29 00:02:47,989 –> 00:02:51,072 you will see different notations for the derivative.
30 00:02:51,529 –> 00:02:56,621 However the definition for an ODE should be the same.
31 00:02:57,171 –> 00:03:02,496 Namely it should be given by a combination of the derivatives of the function x.
32 00:03:02,971 –> 00:03:06,466 Indeed, we could write it down by a functional relation.
33 00:03:06,666 –> 00:03:10,116 For that we take a continuous function capital F
34 00:03:11,014 –> 00:03:15,694 and then we have different inputs. First of all the independent variable t
35 00:03:15,894 –> 00:03:20,941 and then the function x and the derivatives of x up to some order.
36 00:03:21,141 –> 00:03:25,886 So as before we could say that the highest derivative is the kth order.
37 00:03:26,271 –> 00:03:30,447 Ok and now this function with all the inputs should be equal to 0.
38 00:03:30,971 –> 00:03:38,368 Ok, so you see in general this is what we mean when we talk of an ODE of order k
39 00:03:38,643 –> 00:03:42,938 and maybe let’s immediately look at an example for such an equation.
40 00:03:43,357 –> 00:03:52,864 So we could have t + x + 2xdot + second derivative of x squared
41 00:03:53,686 –> 00:03:56,339 and then this should be equal to 0.
42 00:03:56,800 –> 00:04:02,563 So there you see, this is a well defined ODE of second order.
43 00:04:02,763 –> 00:04:07,900 So the highest derivative that occurs in the equation gives us the order
44 00:04:08,114 –> 00:04:15,831 and at this point you might say, it might be easier for us if we first start with first order differential equations.
45 00:04:16,329 –> 00:04:22,728 In fact soon we will see why it’s very helpful to start with the first order differential equations.
46 00:04:23,271 –> 00:04:29,666 Ok, but now it’s helpful for the next definitions to abbreviate the term ordinary differential equation.
47 00:04:30,286 –> 00:04:36,312 From now on we will simply write ODE if we need to keep it short and compact.
48 00:04:36,771 –> 00:04:42,129 Therefore in the next definition we will explain explicit ODEs of order 1.
49 00:04:42,871 –> 00:04:49,051 Here explicit means that the derivative we are interested in is on the one side of the equation
50 00:04:49,251 –> 00:04:54,417 and all the other terms like the function x and the variable t are on the other side
51 00:04:54,914 –> 00:04:59,054 and both things we can rewrite with a function w.
52 00:04:59,254 –> 00:05:02,895 Hence w gets 2 inputs, t and x.
53 00:05:03,095 –> 00:05:08,202 Moreover we would say w is a function defined on 2 intervals.
54 00:05:08,800 –> 00:05:13,075 So we have the interval “I” for t and the interval J for x
55 00:05:13,771 –> 00:05:18,376 and in this case here both are simply intervals from the real number line.
56 00:05:18,900 –> 00:05:25,192 Ok, so in summary you see here, this one is a special case of our general case from above.
57 00:05:25,392 –> 00:05:28,569 However it’s the common one we will examine.
58 00:05:29,000 –> 00:05:34,235 That’s because the 2 restrictions we have here are actually not big restrictions.
59 00:05:34,857 –> 00:05:39,304 Indeed we will discuss soon why this case here is very general,
60 00:05:39,504 –> 00:05:44,967 but first I would say let’s look at an example of an explicit ODE of order 1.
61 00:05:45,167 –> 00:05:49,516 Hence we already know, on the left-hand side we need to have x dot
62 00:05:49,716 –> 00:05:54,374 and now on the right-hand side for example we could have x + t
63 00:05:54,629 –> 00:05:58,868 So you see, the function w doesn’t have to be very complicated.
64 00:05:59,643 –> 00:06:04,066 In fact here the domain of definition is simply R^2.
65 00:06:04,557 –> 00:06:10,085 Ok, but at this point you might remember what we have discussed in the last video.
66 00:06:10,285 –> 00:06:15,048 There we had examples where more than 1 variable were involved.
67 00:06:15,600 –> 00:06:21,533 Therefore we might ask: “is there also an example here for an explicit ODE of order 1?”
68 00:06:21,733 –> 00:06:30,570 So you could write x_1 dot, the first variable for x is equal to the second variable x_2 + t
69 00:06:30,770 –> 00:06:34,830 and then we could have a similar equation for x_2 dot.
70 00:06:35,030 –> 00:06:38,362 Now this one could be x_1 + t.
71 00:06:38,562 –> 00:06:45,405 Now, important to note here is, these are not just 2 individual ODEs of order 1.
72 00:06:45,786 –> 00:06:49,950 Otherwise it would not be allowed to mix the variables in this way.
73 00:06:50,286 –> 00:06:54,988 However we can just put them together and call it a system of ODEs
74 00:06:55,188 –> 00:06:59,644 and in fact a system of ODEs looks exactly the same again.
75 00:07:00,143 –> 00:07:03,205 We simply have to write it as a vector equation.
76 00:07:03,686 –> 00:07:11,365 So the derivative of the vector x_1, x_2 is equal to a vector given by the right-hand side here
77 00:07:11,565 –> 00:07:16,561 and of course this one can be expressed again by a function w.
78 00:07:16,761 –> 00:07:23,558 So w gets again 2 inputs, but now the second input is a vector instead of the variable x
79 00:07:23,758 –> 00:07:31,301 and then you should see, if we call the vector (x_1, x_2) simply x again, it has the same form as before.
80 00:07:31,871 –> 00:07:37,124 In other words now we are able to define the term system of ODEs.
81 00:07:37,324 –> 00:07:45,601 More precisely you would say we have an n-dimensional system of explicit ODEs of order 1,
82 00:07:45,929 –> 00:07:52,086 but you might already know in the end we will also just say ODE for such a system.
83 00:07:52,286 –> 00:07:57,760 Simply because from the form, from the looks it’s the same as our original definition.
84 00:07:58,129 –> 00:08:02,835 However don’t forget, the value x(t) is now a vector.
85 00:08:03,035 –> 00:08:06,881 More precisely we would say it’s a n-dimensional vector.
86 00:08:07,371 –> 00:08:11,929 So in other words our system now could have n equations here.
87 00:08:12,529 –> 00:08:16,544 Hence the function w now also maps into R^n.
88 00:08:17,229 –> 00:08:21,797 Moreover the domain of definition also changes for the second variable.
89 00:08:22,471 –> 00:08:27,094 Usually we would say we have U there, which is an open set in R^n,
90 00:08:27,429 –> 00:08:34,277 but as we said at the beginning also that changes a little bit depending which problem we consider.
91 00:08:35,114 –> 00:08:41,280 Ok, but for the moment I would say we have a very nice definition for what we mean when we say ODE.
92 00:08:41,480 –> 00:08:47,661 So this is a very general construct, but you see one important ingredient is still missing here.
93 00:08:48,243 –> 00:08:56,693 So now we know what an ODE is, but we remember for applications we are always interested in solutions of this ODE.
94 00:08:56,893 –> 00:09:02,893 So know the question is: what is a solution of a system of ODEs?
95 00:09:03,414 –> 00:09:09,268 Of course it should be a function that is differentiable and can be put into the equation
96 00:09:09,468 –> 00:09:13,656 and then it should fulfill the equation for all points t.
97 00:09:13,856 –> 00:09:20,073 However there you already see, the domain of definition doesn’t have to be the whole interval “I”.
98 00:09:20,386 –> 00:09:25,901 Indeed, it could be any subinterval which we can call t_0 to t_1.
99 00:09:26,101 –> 00:09:31,930 So this is an open interval and now the solution function we can simply call alpha.
100 00:09:32,543 –> 00:09:37,268 So alpha is defined on this interval and it maps into R^n.
101 00:09:37,468 –> 00:09:41,360 More precisely it has to map into the subset U.
102 00:09:42,029 –> 00:09:49,242 Simply because we want to put it into our function w in the second component and then it should be well-defined.
103 00:09:49,886 –> 00:09:54,407 Ok, there we have it. Solution means it fulfills this equation here.
104 00:09:54,714 –> 00:10:06,152 More concretely the derivative of alpha at the point t is equal to w(t, alpha(t)).
105 00:10:06,600 –> 00:10:14,155 So there you see this is a well-defined equation of vectors and it should be fulfilled for all t in the interval.
106 00:10:14,414 –> 00:10:18,969 So you see, now we can check this condition here pointwisely
107 00:10:19,343 –> 00:10:24,716 and most importantly you should see the solution is always a differentiable function.
108 00:10:25,371 –> 00:10:32,336 Ok, so we have a lot of definitions here. Therefore let’s close the video now with a concrete example.
109 00:10:32,843 –> 00:10:38,882 So this one should be very helpful to see what it really means to be a solution for an ODE
110 00:10:39,082 –> 00:10:43,314 and for this let’s look at a very simple system of ODEs.
111 00:10:43,771 –> 00:10:49,487 x_1 dot should be equal to x_2 and x_2 dot should be equal to -x_1.
112 00:10:50,171 –> 00:10:54,787 So you see this is a well-defined system for n=2
113 00:10:54,987 –> 00:11:00,463 and moreover we don’t have a problem with the domain of definition. U is simply R^2
114 00:11:00,663 –> 00:11:06,822 and then the function w(t, x) is equal to x_2, -x_1.
115 00:11:07,022 –> 00:11:11,951 So you see w does not explicitly depend on t,
116 00:11:12,151 –> 00:11:17,927 but please don’t forget t comes in again if you put in the solution alpha(t).
117 00:11:18,571 –> 00:11:23,614 Ok, so now if you look at a system, maybe you immediately see a solution.
118 00:11:23,814 –> 00:11:29,256 Indeed, I know sine and cosine functions fulfill something like that.
119 00:11:29,456 –> 00:11:35,923 In fact alpha(t) = (sin(t), cos(t)) does the job.
120 00:11:36,243 –> 00:11:44,287 Simply because we know the derivative of sine is the cosine and the derivative of the cosine is the -sine function.
121 00:11:45,143 –> 00:11:49,578 Hence we immediately have a solution of this system of ODEs
122 00:11:50,300 –> 00:11:55,173 and know what one can do is to sketch the solution as its image.
123 00:11:55,629 –> 00:12:00,969 So we know the image of the solution lives in U, which is R^2 in this case
124 00:12:01,386 –> 00:12:08,893 and now we can sketch what happens to alpha(t) when we start at t=0 and then increase t.
125 00:12:09,243 –> 00:12:13,923 In fact you see what we get is a nice circle with radius 1.
126 00:12:14,357 –> 00:12:14,388 .
127 00:12:14,743 –> 00:12:20,970 Moreover now you can remember this image of the solution is called the orbit of the solution.
128 00:12:21,170 –> 00:12:28,331 Indeed it’s a dynamic picture that tells us what happens to the solution in the time variable t.
129 00:12:28,657 –> 00:12:33,592 Moreover now you should also see that we easily find a second solution here.
130 00:12:34,186 –> 00:12:42,728 Namely if we scale the vector by 1/2 and call the solution alpha tilde, then we see the ODE is also fulfilled
131 00:12:42,928 –> 00:12:47,826 and there we get that the orbit is a circle with radius 1/2.
132 00:12:48,486 –> 00:12:55,110 Hence our first result here is that a system of ODEs can have a lot of different solutions.
133 00:12:55,786 –> 00:13:00,976 Therefore a question should also be: which solution gives us the correct picture?
134 00:13:01,176 –> 00:13:07,096 Or more precisely maybe we want the solution that goes through some particular points
135 00:13:07,771 –> 00:13:14,966 and there we immediately have the questions: Do we have a solution that goes through the points and if we have it, is it unique?
136 00:13:15,657 –> 00:13:19,813 and exactly these questions we want to answer in this video course.
137 00:13:20,314 –> 00:13:27,103 Therefore I hope that you are not scared off by all these definitions and come back to the next video.
138 00:13:27,303 –> 00:13:33,742 There we will make more geometric pictures, such that you can understand the theory visually.
139 00:13:36,186 –> 00:13:37,800 So I hope we meet again and have a nice day. Bye!
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http://docplayer.net/28343281-Overview-essential-questions-grade-5-mathematics-quarter-3-unit-3-1-adding-and-subtracting-decimals.html | 1,542,342,766,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742970.12/warc/CC-MAIN-20181116025123-20181116051123-00104.warc.gz | 93,120,868 | 25,081 | # Overview. Essential Questions. Grade 5 Mathematics, Quarter 3, Unit 3.1 Adding and Subtracting Decimals
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1 Adding and Subtracting Decimals Overview Number of instruction days: (1 day = 90 minutes) Content to Be Learned Add and subtract decimals to the hundredths. Use concrete models, drawings, and strategies based on place value and the properties of operations to calculate solutions. Relate strategies used to add and subtract decimals to a written method. Reason and explain how models, pictures, and strategies are used to solve problems involving addition and subtraction of decimals to the hundredths. Mathematical Practices to Be Integrated 1 Make sense of problems and persevere in solving them. Students solve problems involving decimals by applying their understanding of operations on whole numbers and fractions. Students seek the meaning of their answer and check their thinking by asking themselves, Does this make sense? 2 Reason abstractly and quantitatively. Recognize that a number represents a specific quantity. Derive a solution to a problem without performing the computation. Extend understanding from whole numbers to decimals. Extend understanding of addition and subtraction of fractions to add and subtract decimals. Estimate the sums and differences of whole numbers and decimals to the hundredths to reason abstractly and quantitatively. Essential Questions How can you add and subtract decimals to the hundredths? How can you use concrete models, drawings, and strategies based on place value and the properties of operations to calculate solutions? How can you justify the strategies you used to add and subtract decimals? How can you use models and pictures to solve problems involving addition and subtraction of decimals to the hundredths? How can understanding the relationship between addition and subtraction help you add and subtract decimals? Providence Public Schools D-53
2 Adding and Subtracting Decimals (12 14 days) Standards Common Core State Standards for Mathematical Content Number and Operations in Base Ten 5.NBT Perform operations with multi-digit whole numbers and with decimals to hundredths. 5.NBT.7 Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Common Core State Standards for Mathematical Practice 1 Make sense of problems and persevere in solving them. Mathematically proficient students start by explaining to themselves the meaning of a problem and looking for entry points to its solution. They analyze givens, constraints, relationships, and goals. They make conjectures about the form and meaning of the solution and plan a solution pathway rather than simply jumping into a solution attempt. They consider analogous problems, and try special cases and simpler forms of the original problem in order to gain insight into its solution. They monitor and evaluate their progress and change course if necessary. Older students might, depending on the context of the problem, transform algebraic expressions or change the viewing window on their graphing calculator to get the information they need. Mathematically proficient students can explain correspondences between equations, verbal descriptions, tables, and graphs or draw diagrams of important features and relationships, graph data, and search for regularity or trends. Younger students might rely on using concrete objects or pictures to help conceptualize and solve a problem. Mathematically proficient students check their answers to problems using a different method, and they continually ask themselves, Does this make sense? They can understand the approaches of others to solving complex problems and identify correspondences between different approaches. 2 Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. D-54 Providence Public Schools
3 Adding and Subtracting Decimals (12 14 days) Grade 5 Mathematics, Quarter 3, Unit 3.1 Clarifying the Standards Prior Learning In Grade 4, students were introduced to decimals to the hundredths place, and they compared these decimals. They used decimal notation for fractions with denominators of 10 or 100. They became fluent in addition and subtraction of whole numbers. Additionally, they used place value understanding to round multidigit whole numbers to any place. Current Learning The content in this unit is a major emphasis in Grade 5. They add and subtract decimals to the hundredths. They use place value strategies, properties of operations, and relationships between addition and subtraction in their solution paths. This work should include heavy use of concrete models and pictorial representations rather than relying solely on use of the algorithm. The standards covered in this unit include students ability to reason and explain how they use models, pictures, and strategies. Students can apply the reasoning they have developed to understand addition and subtraction of fractions to make sense of addition and subtraction of decimals. Future Learning Students will extend their understanding of numbers and the ordering of numbers to the full system of rational numbers in Grade 6. Grade 6 work will include negative rational numbers. Sixth-grade students will become fluent in adding, subtracting, multiplying, and dividing multidigit decimals using the standard algorithm for each operation. They will use decimal operations as they continue to solve using problems using numerical and algebraic expressions and equations. Additional Findings Research on symbolic learning suggests that manipulatives or other physical models used in teaching must, in order to be helpful, be represented by a learner both as the objects that they are and as symbols that stand for something else. The physical characteristics of these materials can be initially distracting to children, and it takes time for them to develop mathematical meaning for any physical model and to use it effectively. (Adding It Up, p. 198) An understanding of the base-ten number system should be extended through continued work with larger numbers as well as with decimals. (Principals and Standards for School Mathematics, p. 149) Providence Public Schools D-55
4 Adding and Subtracting Decimals (12 14 days) Assessment When constructing an end-of-unit assessment, be aware that the assessment should measure your students understanding of the big ideas indicated within the standards. The CCSS for Mathematical Content and the CCSS for Mathematical Practice should be considered when designing assessments. Standards-based mathematics assessment items should vary in difficulty, content, and type. The assessment should comprise a mix of items, which could include multiple choice items, short and extended response items, and performance-based tasks. When creating your assessment, you should be mindful when an item could be differentiated to address the needs of students in your class. The mathematical concepts below are not a prioritized list of assessment items, and your assessment is not limited to these concepts. However, care should be given to assess the skills the students have developed within this unit. The assessment should provide you with credible evidence as to your students attainment of the mathematics within the unit. Apply the use of concrete models, drawings, strategies based on place value and the properties of operations to add and subtract decimals to the hundredths. Apply the strategies used to add and subtract decimals to a written method. Reason and explain how models, pictures and strategies are used to solve problems involving addition and subtraction of decimals to the hundredths. Learning Objectives Instruction Students will be able to: Use concrete models, drawings, strategies based on place value and the properties of operations to calculate addition solutions. Use concrete models, drawings, strategies based on place value and the properties of operations to calculate subtraction solutions. Relate strategies used to add and subtract decimals to a written method. Explore with models, pictures and strategies to solve addition and subtraction of decimals. Reason and explain how models, pictures and strategies are used to solve problems involving addition and subtraction of decimals to the hundredths. Demonstrate understanding of the concepts and skills in this unit. D-56 Providence Public Schools
5 Adding and Subtracting Decimals (12 14 days) Grade 5 Mathematics, Quarter 3, Unit 3.1 Resources envision Math Grade 5, Pearson Education, Inc., 2009 Topic 2, Adding and Subtracting Whole Numbers and Decimals, Teacher Edition Section I, Supplemental Resources: Lesson 2-6A, Modeling Addition and Subtraction of Decimals Teacher Resource Masters Student Edition Investigations in Numbers, Data, and Space, Grade 5, Pearson Education, Inc., 2008 Implementing Investigations in Grade 5- Implementation Guide Unit 6, Decimals on Grids and Number Lines, Teacher Edition Investigation 2: Adding Decimals Section I, Supplemental Resources: Lesson 2.5A, Decimal Subtraction Problems Teacher Resources Binder Connected Mathematics 2, Pearson/Prentice Hall/2008, Bits and Pieces III Section I, Supplemental Resources: Investigation 1- Decimals More or Less! Pearson Online Success Net: Implementing Investigations Site: Exam View Assessment Suite Note: The district resources may contain content that goes beyond the standards addressed in this unit. See the Planning for Effective Instructional Design and Delivery and Assessment sections for specific recommendations. Materials Base ten blocks, money Instructional Considerations Key Vocabulary rounding compatible numbers compensation Planning for Effective Instructional Design and Delivery In this unit, students work with concrete models and pictorial representations to add and subtract of decimals. They will also use the algorithm, but standard (5.NBT.7), includes students reasoning and explanation of how they used models, pictures and strategies. Providence Public Schools D-57
6 Adding and Subtracting Decimals (12 14 days) During Unit planning, refer to Investigations, Unit 6, Teacher Note, Adding Decimals, p and the Dialog Box on p These pages offer you examples and clarification to support instruction throughout this Unit. When working with decimals, students should estimate answers for reasonableness based on their understanding of operations and the value of the numbers before calculating the exact answer. Students have rounded decimals in Unit 1.2 and now apply this knowledge to the work in this Unit. Students could use a number line as a visual model to round decimals. An explanation of this is found on p.22b in the envisions, Topic 2 TE called Rounding Decimals. To support students conceptual understanding of adding decimals, use the activity in Investigations, Unit 6, p. 87, called Fill Two. Then Close to 1 on p. 109 can be used to strengthen their knowledge of decimals. Just as with whole numbers, students should be able to express that when adding decimals, they add tenths to tenths and hundredths to hundredths. When adding or subtracting with decimals, it is important that they align the corresponding places correctly. This understanding can be reinforced by connecting addition of decimals to their understanding of addition of fractions with denominators of 10 and 100 taught in the fourth grade. Incorporate Ten Minute Math Activities, the Problem of the Day, and the Daily Spiral Review that are aligned to The Common Core State Standards for Mathematics. EnVision Center Activities and Investigations Activities offer additional practice for student learning and support small group differentiated instruction. Use teacher created common tasks as formative assessments to monitor student progress and understanding of critical content and essential questions. Use data from formal and informal assessments to guide your instruction and planning. For planning considerations, read through the teacher editions for suggestions about scaffolding techniques, using additional examples, and differentiated instruction as suggested by the envision and Investigations resources, particularly the Algebra Connections and Teacher Notes section. D-58 Providence Public Schools
7 Adding and Subtracting Decimals (12 14 days) Grade 5 Mathematics, Quarter 3, Unit 3.1 Notes Providence Public Schools D-59
8 Adding and Subtracting Decimals (12 14 days) D-60 Providence Public Schools
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2 Tutorials that teach Shifts in Supply
# Shifts in Supply
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Author: Sophia Tutorial
##### Description:
Compare different types of shifts in supply with their corresponding graphs.
(more)
Tutorial
what's covered
This tutorial will cover shifts in supply, comparing what causes movement along a supply curve versus factors that cause a shift of the supply curve.
Our discussion breaks down as follows:
1. Law of Supply: A Review
2. Shifts in Supply
3. Causes of Shifts in Supply
1. Changes in Input Prices
2. Changes in Technology
3. Changes in Prices of Related Goods (Substitutes and Complements in Production)
4. Government Policies
## 1. Law of Supply: A Review
In review, the law of supply states that if the price of the good decreases, the quantity supplied decreases. It works the other way as well; when the price of a good increases, so does the quantity supplied.
There is a positive relationship between price and quantity with supply. Therefore a movement along the supply curve is caused by a change in price, assuming that everything else is held constant, defined as ceteris paribus.
Here are a supply schedule and a supply curve for a farmer's willingness to supply apples.
Price of Granny Smith Apples Quantity of Granny Smith Apples Each Week
\$2.00 7
\$1.75 6
\$1.50 5
\$1.25 4
\$1.00 3
\$0.75 2
\$0.50 1
\$0.25 0
You can see that as the price of the apples goes up, he is willing to supply a greater quantity (in thousands of bushels). Notice, too, that as the price goes up, we can move along the curve to the right.
As the price falls, the quantity he is willing to produce also falls with it, with a corresponding move down and to the left along the supply curve.
Again, it is a positive relationship between price and quantity, where they move in the same direction, so as the price is changing, we simply move along the curve. We do not need a new curve.
As mentioned, the law of supply and ceteris paribus tells us that as the price of apples falls, we can expect that farmers supply fewer apples--ceteris paribus meaning to hold everything else constant.
This assumes, then, that only the price of apples has changed. For instance, the price of their resources or inputs did not change, so fertilizer did not get more expensive. Their technology for growing apples did not change either--just the price of apples.
terms to know
Law of Supply
If the price of a good decreases, the quantity supplied decreases
Movement Along Supply Curve
Movement caused by a change in price, assuming all other variables are held constant
## 2. Shifts in Supply
However, in the real world, we know that things change all the time. What if fertilizer does get more expensive, or if farmers have to pay their workers more money because wages rose? What if a new technology is developed for apple picking that makes it much more efficient?
Will farmers still supply the same amount of apples if these things happen? Obviously not. In this case, they would not be supplying a different quantity because of a price change; it will be due to another reason.
If there was an increased cost to the farmer because of an increase in labor cost, or a higher cost for fertilizer, notice what might happen to his numbers. At all of these prices, he is supplying a lower quantity of apples.
Price of Granny Smith Apples Quantity of Granny Smith Apples Each Week
\$2.00 5
\$1.75 4
\$1.50 3
\$1.25 2
\$1.00 1
\$0.75 0
\$0.50 0
\$0.25 0
Now, the farmer is supplying less, but it is not because the price fell, so we are not able to find that price-quantity relationship along the original supply curve. We cannot simply move along that curve to find a new relationship. We need a new supply curve.
There is a completely different relationship now between the price and the quantity that the farmer is able and willing to supply. This is what we mean when we say there has been a shift of the supply curve.
hint
Note that in this example, a shift to the left is a decrease in supply, because the farmer is willing and able to supply fewer apples at every price.
Therefore, a shift in supply is defined as changes other than the price of the good itself that are affecting production decisions for a particular good.
term to know
Shift in Supply
Changes other than the price of the good itself that affect production decisions for a particular good
## 3. Causes of Shifts in Supply
Here is a summary of the factors that can impact producers and cause a shift in supply; we will discuss each of them in further detail.
hint
Keep in mind that anything that makes it easier or less expensive to produce something will create an increase in supply and shift the curve to the right. Anything that makes it more difficult or more expensive to produce something will cause a decrease in supply and a shift of the supply curve to the left.
3a. Changes in Input Prices
Again, this supply graph shows the impact of a change in an input price. The price of apples did not change. However, if fertilizer got more expensive, it also makes it more expensive for the farmer. Therefore, he will supply fewer apples at all prices, which represents a decrease in supply.
An increase in input prices causes a decrease in supply because production became more expensive. A decrease in an input price, like cheaper land, labor, or capital, would create an increase in supply, shifting the curve to the right instead.
3b. Changes in Technology
Now, if technology improves for apple growing or apple picking, then farmers will be able to supply apples more efficiently. They will supply a greater quantity at all prices.
Therefore, an increase in technology would create a shift to the right or an increase in supply, shown below. Conversely, a failure of technology could certainly decrease the supply.
3c. Changes in Prices of Related Goods
Next, we will discuss the impact of changes in prices of related goods, focusing on substitutes and complements in production.
• Substitutes in production and consumption is a good that you buy instead of another one in regards to consumption, but in production, it is a bit different, as you will see.
Now, if the market price of apples goes up, we know that the law of supply tells us that the quantity supplied for apples would increase, demonstrated by moving along the supply curve to the right.
However, let's talk about a substitute in production. Perhaps pear growing and apple growing are substitutable for one another. If apples are more expensive in price, maybe some pear farmers may actually decide to plant apples instead of pears, because they could potentially make more money as apple farmers.
Therefore, the supply of pears would actually shift because farmers were planting fewer pears when the price did not change for pears. It follows that the supply curve for pears would shift to the left as a decrease, all because the price of apples went up. Pears can be seen as a substitute for apples in production.
For substitutes in production, when the price of a substitute in production increases--like the apples going up in price--the supply of the other good will actually decrease--in this case, pears.
• Complements in production and consumption are things that go together, and again, this lesson's definition focuses more on the consumption end of it, but in production, it will be a bit different.
So, many farmers produce two products, because one is a byproduct of the others.
EXAMPLE
Wheat and hay are complement in production, as are beef and leather.
If the price of wheat goes up, according to the law of supply, we know that this would cause a change in quantity supplied for wheat or movement along the wheat curve. Farmers would produce more wheat, but only because the price went up.
Now that they are producing more wheat, though, they will also be producing more hay. Hay did not go up in price, yet farmers are supplying more at all prices. This, then, would be a change in the supply of hay, or a shift of the supply curve for hay to the right.
Therefore, for complements in production, when the price of a complements in production increases, the supply of the other good will increase.
terms to know
Substitutes (Production and Consumption)
As the price of one good increases, the demand for an alternative good that meets the same producer or consumer need increases
Complements (Production and Consumption)
A good for which the demand increases as the price of an associated good decreases
3d. Government Policies
Briefly, government policies, such as subsidies and taxes, can also impact a producer's ability to supply.
big idea
The bottom line is to be able to differentiate between quantity supplied and a change in supply. A change in the price of a good will cause a change in quantity supplied or cause movement along the curve, just as it did with demand. A change in any other factor could potentially shift the supply curve, which is when we can say that there has been a change in supply, meaning an increase or decrease in supply. It is important to keep in mind that different situations will shift the curve more than others.
summary
We started today's lesson with a review of the law of supply, recalling how movement along a supply curve is due to a change in the price of that good. We also learned that a shift of the supply curve, or a shift in supply, is caused by a change in any other variable that could potentially impact suppliers' ability to produce or their cost of production. These causes of shifts in supply include changes in input prices, changes in technology, changes in prices of related goods like substitutes and complements in production, or government policies.
Source: Adapted from Sophia instructor Kate Eskra.
Terms to Know
Complements (Production and Consumption)
A good for which the demand increases as the price of an associated good decreases.
Law of Supply
If the price of a good decreases, the quantity supplied decreases.
Movement Along Supply Curve
Movement caused by a change in price, assuming all other variables are held constant.
Shift in Supply
Changes other than the price of the good itself that affect production decisions for a particular good.
Substitutes (Production and Consumption)
As the price of one good increases, the demand for an alternative good that meets the same producer or consumer need increases. | 2,244 | 10,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-17 | latest | en | 0.926186 |
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## Presentation on theme: "Lecture 9 Grey Level & Colour Enhancement TK3813 Dr. Masri Ayob."— Presentation transcript:
Lecture 9 Grey Level & Colour Enhancement TK3813 Dr. Masri Ayob
2 Point Processing Operations Point processing involves the transformation of individual pixels independently of other pixels in the image. These simple operations are typically used to correct for defects in image acquisition hardware. For example, to compensate for under/over exposed images. The process of modifying pixel grey level and colour are sometimes referred to as ‘point processing operation’.
3 The common operations involve the adjustment of brightness, contrast or colour in an image. A common reason for manipulating these attribute is the need to compensate for difficulties in image acquisition. E.g. Image where an object of interest is backlit (illuminated from behind). Without the aid of image processing, we might need to reacquire the image several time. With image processing, we can reveal enough detail to allow proper interpretation. Point Processing Operations
4 We can adjust the overall brightness of a grayscale simply by adding a constant bias, b, to pixel values: g(x,y) = f(x,y) + b If b > 0 the image is made brighter else the image is darkened. b is known as “bias” Linear Mapping (6.1)
5 Linear Mapping We can adjust contrast in a greyscale image through multiplication of pixel values by a constant gain, a: g(x,y) = a * f(x,y) If a > 1 the image is made brighter else the image is darkened. a is known as “gain” (6.2)
6 Examples
7 Examples
8 Examples gain = +1.5gain = -2.0
9 Linear Mapping These mappings can be combined to yield a linear mapping of f to g given as: g(x,y) = a*f(x,y) + b Typically don’t want to specify gain and bias but think in terms of mapping a range of values [f1..f2] onto a different range of values [g1, g2]. The following equation shows how to represent the linear mapping using range information only: General expression for brightness and contrast (6.3) (6.4)
10 Linear Mapping
11 Examples Linear Mapping G1 = 0, G2 = 255, F1 = 75, F2 = 185 Gain = 2.3, Bias = -173.8
12 Examples Inverse mapping G1 = 255, G2 = 0, F1 = 0, F2 = 255 Gain = -1, Bias = 255
13 Examples Bilevel Threshold Mapping G1 = 0, G2 = 255, F1 = 127, F2 = 128 Gain = 255, Bias = -32385
14 Linear Contrast Stretching A linear mapping that enhances the contrast of an image without removing any detail. Spreads the visual information available across a greater range of gray scale intensities
15 Linear Contrast Stretching Example Notice that the left image appears washed-out (most of the intensities are in a narrow band due to poor contrast). The right image maps those values to the full available dynamic range.
16 Linear Contrast Stretching Implement equation 6.3 Implement equation 6.4
17 Java Code Interface BufferedImageOp All Known Implementing Classes: AffineTransformOp, ColorConvertOp, ConvolveOp, LookupOp, RescaleOp public interface BufferedImageOp This interface describes single-input/single-output operations performed on BufferedImage objects. It is implemented by AffineTransformOp, ConvolveOp, ColorConvertOp, RescaleOp, and LookupOp. These objects can be passed into a BufferedImageFilter to operate on a BufferedImage in the ImageProducer-ImageFilter-ImageConsumer paradigm. Classes that implement this interface must specify whether or not they allow in-place filtering- filter operations where the source object is equal to the destination object. This interface cannot be used to describe more sophisticated operations such as those that take multiple sources. Note that this restriction also means that the values of the destination pixels prior to the operation are not used as input to the filter operation. BufferedImageOpWithWritableVirtualR asterSamplesType_USHORT_RGBA??
18 Java Code public static BufferedImage rescale(BufferedImage image, float gain, float bias) { RescaleOp rescaleOp = new RescaleOp(gain, bias, null); return rescaleOp.filter(image, null); } Constructor Summary RescaleOp(float[] scaleFactors, float[] offsets, RenderingHints hints) Constructs a new RescaleOp with the desired scale factors and offsets. Assumes a color image with differing gain and bias values for each band. RescaleOp(float scaleFactor, float offset, RenderingHints hints) Constructs a new RescaleOp with the desired scale factor and offset.
19 Java Code public static BufferedImage rescale(BufferedImage image, float gain, float bias) { RescaleOp rescaleOp = new RescaleOp(gain, bias, null); return rescaleOp.filter(image, null); } public final BufferedImage filter(BufferedImage src, BufferedImage dst) Rescales the source BufferedImage. If the color model in the source image is not the same as that in the destination image, the pixels will be converted in the destination. If the destination image is null, a BufferedImage will be created with the source ColorModel. An IllegalArgumentException may be thrown if the number of scaling factors/offsets in this object does not meet the restrictions stated in the class comments above, or if the source image has an IndexColorModel. Specified by: filter in interface BufferedImageOp Parameters: src - the BufferedImage to be filtered dst - the destination for the filtering operation or null. If src = dst then the algorithm performs in- place. Returns: the filtered BufferedImage.
20 Non-Linear Mapping A grayscale image f(x,y) can be transformed into image g(x,y) using various non-linear mappings. Require any mapping to be true a function. For any pixel f(x,y), T(x,y) results in a unique deterministic value. Common non-linear gray-scale mappings include log compression: enhances the contrast of darker areas exponential mapping: enhances the contrast of brighter areas
21 Log Compression Log compression takes a range of values and “boosts” the lower end. Notice that the higher end is “compressed” into a small range of output values. Logarithmic mapping is useful if we wish to enhance detail in the darker regions of the image, at the expense of detail in the brighter regions.
22 Log Compression Example Notice how the darker areas appear brighter and contain more easily viewed details. The brighter areas loose some detail but remain largely unchanged.
23 Log Compression Example
24 Exponential Mapping Exponential Mapping takes a range of values and “boosts” the upper end. Notice that the lower range is “compressed” into a small range of output values. Exponential mapping is useful if we wish to enhance detail in the brighter regions of the image.
25 Exponential Mapping Example Notice how the brighter areas contain more detail in the filtered image. The darker areas loose some detail (are compressed) in this example.
26 Gamma Correction Computer monitors and TVs aren’t linear in the way they process signals A “doubling” of the control voltage on a CRT doesn’t usually result in a “doubling” of the intensity of the output color Monitors function according to a “power-law” function Uncorrected monitors may display images that don’t correspond to the original
27 Gamma Correction System Gamma
28 Gamma Correction Each CRT has its own value that the manafacturer provides with CRT. Typically [2,3]
29 Gamma correction Cathode ray tube (CRT) devices have an intensity- to-voltage response that is a power function, with varying from 1.8 to 2.5 The picture will become darker. Gamma correction is done by preprocessing the image before inputting it to the monitor.
30 Gamma Correction
31 Gamma Correction
33 Gamma Example (Computer Monitor) Digital/Synthetic Image
34 Gamma Correction output = input 1/ Most video cards have a gamma correction setting
35 Gamma Correction If a display is gamma corrected, the nonlinear relationship between pixel value (the number assigned to a particular color tone) and displayed intensity (the way it actually looks) has been adjusted for. Most graphics cards/monitors have gamma settings Values typically range from 1.8-2.5 [usually 2.2] PCs typically have higher gamma settings than Macs so images typically look darker on PCs than Macs
36 Gamma Correction If = 1.0, the result is null transform. If 0 1.0, then the creates exponential curves that dim an image. If 1.0, then the result is logarithmic curves that brighten an image. RGB monitors have gamma values of 1.4 to 2.8.
37 Gamma Chart
38 Gamma Correction (a)Gamma correction transformation with gamma = 0.45; (b)Gamma corrected image; (c)Gamma correction transformation with gamma = 2.2; (d)Gamma corrected image
39 Gamma Correction
40 Effect of decreasing gamma When the is reduced too much, the image begins to reduce contrast to the point where the image started to have very slight “wash-out” look, especially in the background
41 Another example (a) image has a washed-out appearance, it needs a compression of gray levels needs > 1 (b) result after power-law transformation with = 3.0 (suitable) (c) transformation with = 4.0 (suitable) (d) transformation with = 5.0 (high contrast, the image has areas that are too dark, some detail is lost) ab cd
42 Code Efficiency for Point Processing Operations The algorithm above invokes a square-root function (a relatively expensive operation) and a multiplication for each pixel in F. The algorithm is O(NM) where the input image is NxM. Can this performance be improved? Yes…use lookup-tables! algorithm squareRootFilter(F) INPUT: Gray-scale image F using b bits per pixel OUTPUT: Filtered image Let A be the scaling factor = squareRoot(2^b – 1) Let G be an “empty” image for all pixel coordinates X and Y of the input image G(X,Y) = a * squareRoot(F(X,Y)) return G
43 Java Is Fun and Easy! Java has a built-in lookup-table class named ByteLookupTable. Lookup tables can be used in conjunction with LookupOp objects to apply simple point processing operations to images. BufferedImage image; … byte[] table = new byte[256]; double factor = Math.sqrt(255); for(int i=0; i { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3372664/slides/slide_43.jpg", "name": "43 Java Is Fun and Easy. Java has a built-in lookup-table class named ByteLookupTable.", "description": "Lookup tables can be used in conjunction with LookupOp objects to apply simple point processing operations to images. BufferedImage image; … byte[] table = new byte[256]; double factor = Math.sqrt(255); for(int i=0; i
44 Java Is Fun and Easy! java.awt.image Class ByteLookupTable public class ByteLookupTable extends LookupTable This class defines a lookup table object. The output of a lookup operation using an object of this class is interpreted as an unsigned byte quantity. The lookup table contains byte data arrays for one or more bands (or components) of an image, and it contains an offset which will be subtracted from the input values before indexing the arrays. This allows an array smaller than the native data size to be provided for a constrained input. If there is only one array in the lookup table, it will be applied to all bands.
45 Java Is Fun and Easy! LookupOp Class Description This class implements a lookup operation from the source to the destination. The LookupTable object may contain a single array or multiple arrays, subject to the restrictions below. For Rasters, the lookup operates on bands. The number of lookup arrays may be one, in which case the same array is applied to all bands, or it must equal the number of Source Raster bands. For BufferedImages, the lookup operates on color and alpha components. The number of lookup arrays may be one, in which case the same array is applied to all color (but not alpha) components. Otherwise, the number of lookup arrays may equal the number of Source color components, in which case no lookup of the alpha component (if present) is performed. If neither of these cases apply, the number of lookup arrays must equal the number of Source color components plus alpha components, in which case lookup is performed for all color and alpha components. This allows non-uniform rescaling of multi- band BufferedImages. ByteLookupTable squareRootTable = new ByteLookupTable(0, table); LookupOp squareRootOp = new LookupOp(squareRootTable, null); BufferedImage filteredImage = squareRootOp.filter(image, null);
46 Java Is Fun and Easy
47 Thank you Q & A
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## Geometry 8.5 Angle of elevation and depression by Mr
Activities in Math Trigonometry - Syvum. 7.5 Angles of Elevation & Depression Name_____ Worksheet 1. A tree casts a shadow 21m long. The angle of elevation of the sun is 51°. What is the height of the tree? 2. A helicopter (H) is hovering over a landing pad (P) 100m from where you are standing (G). The helicopter's angle of elevation, Worksheet Angle Elevation And Depression Worksheet Daily Worksheet Templates P90X Worksheets Excel Right Triangles Angles of Elevation and Depression Riddle Practice Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression.
19 photos of the "20 Angle Of Elevation and Depression Trig Worksheet Answers" Angle Of Elevation and Depression Trig Worksheet Answers Related Posts of "Angle Of Elevation and Depression Trig Worksheet Answers" Students will turn in the Angle of Elevation and Angle of Depression worksheet tomorrow for a classwork grade and each group will be responsible today for putting one of the answers up on the Promethean Board.
7.5 Angles of Elevation & Depression Name_____ Worksheet 1. A tree casts a shadow 21m long. The angle of elevation of the sun is 51°. What is the height of the tree? 2. A helicopter (H) is hovering over a landing pad (P) 100m from where you are standing (G). The helicopter's angle of elevation Two men on opposite sides of a TV tower of height 26 m notice the angle of elevation of the top of this tower to be 45 o and 60 o respectively. Find the distance (in meters) between the two men. Find the distance (in meters) between the two men.
Angle of Elevation & Depression Worksheet/Scavenger Hunt ! Draw a picture for each. Find all values to the nearest tenth. Follow directions on the board and Two men on opposite sides of a TV tower of height 26 m notice the angle of elevation of the top of this tower to be 45 o and 60 o respectively. Find the distance (in meters) between the two men. Find the distance (in meters) between the two men.
Two men on opposite sides of a TV tower of height 26 m notice the angle of elevation of the top of this tower to be 45 o and 60 o respectively. Find the distance (in meters) between the two men. Find the distance (in meters) between the two men. It is an angle of depression.Example 1: Classifying Angles of Elevation and Depression Classify each angle as an angle of elevation or an angle of depression. 4 4 is formed by a horizontal line and a line of sight to a point above the line. .
In this lesson we have returned to the topic of right triangle trigonometry, to solve real world problems that involve right triangles. To find lengths or distances, we have used angles of elevation, angles of depression, angles resulting from bearings in navigation, and other real situations that give rise to right triangles. In later chapters, you will extend the work of this chapter: you 19 photos of the "20 Angle Of Elevation and Depression Trig Worksheet Answers" Angle Of Elevation and Depression Trig Worksheet Answers Related Posts of "Angle Of Elevation and Depression Trig Worksheet Answers"
depression worksheet answers 2018, practice 8 5 : angles of elevation & depression dng p 405 describe each angle as it relates to the diagram answers: a of depression from the birds to the ship geometry worksheet 85 (angles of elevation & depression) An angle of depression is the angle made. Sep 26, 2012. The angle of elevation is. Pdf files for Proving similar triangles. file/view/C04+Worksheet+#2+S.r. or. Inferior angle. – Medial border. 4-3. Bones. • Key bony landmarks. – Acromion process. – Glenoid fossa. – Lateral border. 55-degrees elevation-depression. The physics model for tank movement has been drastically changed. Tanks
Use a clinometer to measure angles of elevation and depression 2. Measure lengths and use measurements to determine angle measures. 3. Apply sine, cosine and tangent ratios to find angles of elevation and depression. NCTM STANDARDS: 1. Student must be able to apply proper formulas to calculate and measure angles of elevation and depression 2. Perform Solving exercise, make real … Students will turn in the Angle of Elevation and Angle of Depression worksheet tomorrow for a classwork grade and each group will be responsible today for putting one of the answers up on the Promethean Board.
Students will turn in the Angle of Elevation and Angle of Depression worksheet tomorrow for a classwork grade and each group will be responsible today for putting one of the answers up on the Promethean Board. depression worksheet answers 2018, practice 8 5 : angles of elevation & depression dng p 405 describe each angle as it relates to the diagram answers: a of depression from the birds to the ship geometry worksheet 85 (angles of elevation & depression)
Angle of Elevation & Depression Worksheet/Scavenger Hunt ! Draw a picture for each. Find all values to the nearest tenth. Follow directions on the board and Angle of Elevation & Depression Worksheet/Scavenger Hunt ! Draw a picture for each. Find all values to the nearest tenth. Follow directions on the board and
20 Angle Elevation and Depression Trig Worksheet Answers Si Inc. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets . Area Trapezoid Worksheet Pdf Luxury Trapezoid Hw Pdf. Classifying Angles Kuta software Infinite Geometry Name Assign worksheet #2 on right triangle application problems for homework. Rubric: Angle of Elevation and Angle of Depression Each problem is assessed the …
White are dominant on this Worksheets Angle Of Elevation And Depression Worksheet due to table below. Then very light pink & light greycolor scheme is make it perfect. Combination beetween silver, greyish, lavender pink, very light pink, light lavendar are wraping around the room. White are dominant on this Worksheets Angle Of Elevation And Depression Worksheet due to table below. Then very light pink & light greycolor scheme is make it perfect. Combination beetween silver, greyish, lavender pink, very light pink, light lavendar are wraping around the room.
### Angle Of Elevation and Depression Trig Worksheet Answers
Angle Of Elevation and Depression Worksheet Rosenvoile.com. < 4. Elevation from hut to hiker 518 feet o 65 Section 8.5 Angle of Depression and Elevation x X < 3. Depression X Objective: Students should be able to apply Sine, Cosine, and Tangent when using elevation and depression to solve for missing sides and angles of a right triangle., Angle of Elevation & Depression Worksheet/Scavenger Hunt ! Draw a picture for each. Find all values to the nearest tenth. Follow directions on the board and.
Angle Of Elevation and Depression Trig Worksheet Answers. 8-4 Angles of Elevation and Depression Marco breeds and trains homing pigeons on the roof of his building. Classify each angle as an angle of elevation or an angle of depression. 1. 1 angle of elevation 2. 2 angle of depression 3. 3 angle of depression 4. 4 angle of elevation To attract customers to his car dealership, Frank tethers a large red balloon to the ground. In Exercises 5–7, …, Angles Of Elevation and Depression Worksheet with Answers; Angles Of Elevation and Depression Worksheet with Answers. Budget Plan Worksheet September 18, 2018. Download by size: Handphone Tablet Desktop (Original Size) Back To 15 Angles Of Elevation and Depression Worksheet with Answers. 14 photos of the "15 Angles Of Elevation and Depression Worksheet with Answers" Angles Of Elevation ….
### Activities in Math Trigonometry - Syvum
Geometry 8.5 Angle of elevation and depression by Mr. the students will use that angle of elevation and length of an unknown object’s shadow to find the height of that object. This activity gives the students two different applications of It is an angle of depression.Example 1: Classifying Angles of Elevation and Depression Classify each angle as an angle of elevation or an angle of depression. 4 4 is formed by a horizontal line and a line of sight to a point above the line. ..
• Angle Of Elevation and Depression Worksheet Rosenvoile.com
• Angle Of Elevation and Depression Trig Worksheet Answers
• 20 Angle Elevation and Depression Trig Worksheet Answers Si Inc. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets . Area Trapezoid Worksheet Pdf Luxury Trapezoid Hw Pdf. Classifying Angles Kuta software Infinite Geometry Name Use a clinometer to measure angles of elevation and depression 2. Measure lengths and use measurements to determine angle measures. 3. Apply sine, cosine and tangent ratios to find angles of elevation and depression. NCTM STANDARDS: 1. Student must be able to apply proper formulas to calculate and measure angles of elevation and depression 2. Perform Solving exercise, make real …
20 Angle Elevation and Depression Trig Worksheet Answers Si Inc. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets . Area Trapezoid Worksheet Pdf Luxury Trapezoid Hw Pdf. Classifying Angles Kuta software Infinite Geometry Name Two men on opposite sides of a TV tower of height 26 m notice the angle of elevation of the top of this tower to be 45 o and 60 o respectively. Find the distance (in meters) between the two men. Find the distance (in meters) between the two men.
the students will use that angle of elevation and length of an unknown object’s shadow to find the height of that object. This activity gives the students two different applications of 8-4 Angles of Elevation and Depression Marco breeds and trains homing pigeons on the roof of his building. Classify each angle as an angle of elevation or an angle of depression. 1. 1 angle of elevation 2. 2 angle of depression 3. 3 angle of depression 4. 4 angle of elevation To attract customers to his car dealership, Frank tethers a large red balloon to the ground. In Exercises 5–7, …
Angle of Elevation & Depression Worksheet/Scavenger Hunt ! Draw a picture for each. Find all values to the nearest tenth. Follow directions on the board and Two men on opposite sides of a TV tower of height 26 m notice the angle of elevation of the top of this tower to be 45 o and 60 o respectively. Find the distance (in meters) between the two men. Find the distance (in meters) between the two men.
7.5 Angles of Elevation & Depression Name_____ Worksheet 1. A tree casts a shadow 21m long. The angle of elevation of the sun is 51°. What is the height of the tree? 2. A helicopter (H) is hovering over a landing pad (P) 100m from where you are standing (G). The helicopter's angle of elevation It is an angle of depression.Example 1: Classifying Angles of Elevation and Depression Classify each angle as an angle of elevation or an angle of depression. 4 4 is formed by a horizontal line and a line of sight to a point above the line. .
Two men on opposite sides of a TV tower of height 26 m notice the angle of elevation of the top of this tower to be 45 o and 60 o respectively. Find the distance (in meters) between the two men. Find the distance (in meters) between the two men. the students will use that angle of elevation and length of an unknown object’s shadow to find the height of that object. This activity gives the students two different applications of
< 4. Elevation from hut to hiker 518 feet o 65 Section 8.5 Angle of Depression and Elevation x X < 3. Depression X Objective: Students should be able to apply Sine, Cosine, and Tangent when using elevation and depression to solve for missing sides and angles of a right triangle. White are dominant on this Worksheets Angle Of Elevation And Depression Worksheet due to table below. Then very light pink & light greycolor scheme is make it perfect. Combination beetween silver, greyish, lavender pink, very light pink, light lavendar are wraping around the room.
20 Angle Elevation and Depression Trig Worksheet Answers Si Inc. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets . Area Trapezoid Worksheet Pdf Luxury Trapezoid Hw Pdf. Classifying Angles Kuta software Infinite Geometry Name depression worksheet answers 2018, practice 8 5 : angles of elevation & depression dng p 405 describe each angle as it relates to the diagram answers: a of depression from the birds to the ship geometry worksheet 85 (angles of elevation & depression)
Worksheet Depreciation Worksheet Template Scientific Process Worksheet Template Word Angle Elevation And Depression Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression Angle of Elevation & Depression Worksheet/Scavenger Hunt ! Draw a picture for each. Find all values to the nearest tenth. Follow directions on the board and
An angle of depression is the angle made. Sep 26, 2012. The angle of elevation is. Pdf files for Proving similar triangles. file/view/C04+Worksheet+#2+S.r. or. Inferior angle. – Medial border. 4-3. Bones. • Key bony landmarks. – Acromion process. – Glenoid fossa. – Lateral border. 55-degrees elevation-depression. The physics model for tank movement has been drastically changed. Tanks Angle of Elevation & Depression Worksheet/Scavenger Hunt ! Draw a picture for each. Find all values to the nearest tenth. Follow directions on the board and
Use a clinometer to measure angles of elevation and depression 2. Measure lengths and use measurements to determine angle measures. 3. Apply sine, cosine and tangent ratios to find angles of elevation and depression. NCTM STANDARDS: 1. Student must be able to apply proper formulas to calculate and measure angles of elevation and depression 2. Perform Solving exercise, make real … Worksheet Angle Elevation And Depression Worksheet Daily Worksheet Templates P90X Worksheets Excel Right Triangles Angles of Elevation and Depression Riddle Practice Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression
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## Activities in Math Trigonometry - Syvum
Geometry 8.5 Angle of elevation and depression by Mr. 20 Angle Elevation and Depression Trig Worksheet Answers Si Inc. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets . Area Trapezoid Worksheet Pdf Luxury Trapezoid Hw Pdf. Classifying Angles Kuta software Infinite Geometry Name, < 4. Elevation from hut to hiker 518 feet o 65 Section 8.5 Angle of Depression and Elevation x X < 3. Depression X Objective: Students should be able to apply Sine, Cosine, and Tangent when using elevation and depression to solve for missing sides and angles of a right triangle..
### Geometry 8.5 Angle of elevation and depression by Mr
Geometry 8.5 Angle of elevation and depression by Mr. Worksheet Depreciation Worksheet Template Scientific Process Worksheet Template Word Angle Elevation And Depression Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression, Students will turn in the Angle of Elevation and Angle of Depression worksheet tomorrow for a classwork grade and each group will be responsible today for putting one of the answers up on the Promethean Board..
Use a clinometer to measure angles of elevation and depression 2. Measure lengths and use measurements to determine angle measures. 3. Apply sine, cosine and tangent ratios to find angles of elevation and depression. NCTM STANDARDS: 1. Student must be able to apply proper formulas to calculate and measure angles of elevation and depression 2. Perform Solving exercise, make real … 20 Angle Elevation and Depression Trig Worksheet Answers Si Inc. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets . Area Trapezoid Worksheet Pdf Luxury Trapezoid Hw Pdf. Classifying Angles Kuta software Infinite Geometry Name
It is an angle of depression.Example 1: Classifying Angles of Elevation and Depression Classify each angle as an angle of elevation or an angle of depression. 4 4 is formed by a horizontal line and a line of sight to a point above the line. . Worksheet Depreciation Worksheet Template Scientific Process Worksheet Template Word Angle Elevation And Depression Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression
8-4 Angles of Elevation and Depression Marco breeds and trains homing pigeons on the roof of his building. Classify each angle as an angle of elevation or an angle of depression. 1. 1 angle of elevation 2. 2 angle of depression 3. 3 angle of depression 4. 4 angle of elevation To attract customers to his car dealership, Frank tethers a large red balloon to the ground. In Exercises 5–7, … Worksheet Depreciation Worksheet Template Scientific Process Worksheet Template Word Angle Elevation And Depression Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression
< 4. Elevation from hut to hiker 518 feet o 65 Section 8.5 Angle of Depression and Elevation x X < 3. Depression X Objective: Students should be able to apply Sine, Cosine, and Tangent when using elevation and depression to solve for missing sides and angles of a right triangle. Students will turn in the Angle of Elevation and Angle of Depression worksheet tomorrow for a classwork grade and each group will be responsible today for putting one of the answers up on the Promethean Board.
19 photos of the "20 Angle Of Elevation and Depression Trig Worksheet Answers" Angle Of Elevation and Depression Trig Worksheet Answers Related Posts of "Angle Of Elevation and Depression Trig Worksheet Answers" Angles Of Elevation and Depression Worksheet with Answers; Angles Of Elevation and Depression Worksheet with Answers. Budget Plan Worksheet September 18, 2018. Download by size: Handphone Tablet Desktop (Original Size) Back To 15 Angles Of Elevation and Depression Worksheet with Answers. 14 photos of the "15 Angles Of Elevation and Depression Worksheet with Answers" Angles Of Elevation …
7.5 Angles of Elevation & Depression Name_____ Worksheet 1. A tree casts a shadow 21m long. The angle of elevation of the sun is 51°. What is the height of the tree? 2. A helicopter (H) is hovering over a landing pad (P) 100m from where you are standing (G). The helicopter's angle of elevation Worksheet Depreciation Worksheet Template Scientific Process Worksheet Template Word Angle Elevation And Depression Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression
Angles Of Elevation and Depression Worksheet with Answers; Angles Of Elevation and Depression Worksheet with Answers. Budget Plan Worksheet September 18, 2018. Download by size: Handphone Tablet Desktop (Original Size) Back To 15 Angles Of Elevation and Depression Worksheet with Answers. 14 photos of the "15 Angles Of Elevation and Depression Worksheet with Answers" Angles Of Elevation … 19 photos of the "20 Angle Of Elevation and Depression Trig Worksheet Answers" Angle Of Elevation and Depression Trig Worksheet Answers Related Posts of "Angle Of Elevation and Depression Trig Worksheet Answers"
8-4 Angles of Elevation and Depression Marco breeds and trains homing pigeons on the roof of his building. Classify each angle as an angle of elevation or an angle of depression. 1. 1 angle of elevation 2. 2 angle of depression 3. 3 angle of depression 4. 4 angle of elevation To attract customers to his car dealership, Frank tethers a large red balloon to the ground. In Exercises 5–7, … the students will use that angle of elevation and length of an unknown object’s shadow to find the height of that object. This activity gives the students two different applications of
White are dominant on this Worksheets Angle Of Elevation And Depression Worksheet due to table below. Then very light pink & light greycolor scheme is make it perfect. Combination beetween silver, greyish, lavender pink, very light pink, light lavendar are wraping around the room. An angle of depression is the angle made. Sep 26, 2012. The angle of elevation is. Pdf files for Proving similar triangles. file/view/C04+Worksheet+#2+S.r. or. Inferior angle. – Medial border. 4-3. Bones. • Key bony landmarks. – Acromion process. – Glenoid fossa. – Lateral border. 55-degrees elevation-depression. The physics model for tank movement has been drastically changed. Tanks
Worksheet Depreciation Worksheet Template Scientific Process Worksheet Template Word Angle Elevation And Depression Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression Use a clinometer to measure angles of elevation and depression 2. Measure lengths and use measurements to determine angle measures. 3. Apply sine, cosine and tangent ratios to find angles of elevation and depression. NCTM STANDARDS: 1. Student must be able to apply proper formulas to calculate and measure angles of elevation and depression 2. Perform Solving exercise, make real …
### Activities in Math Trigonometry - Syvum
Angle Of Elevation and Depression Trig Worksheet Answers. 7.5 Angles of Elevation & Depression Name_____ Worksheet 1. A tree casts a shadow 21m long. The angle of elevation of the sun is 51°. What is the height of the tree? 2. A helicopter (H) is hovering over a landing pad (P) 100m from where you are standing (G). The helicopter's angle of elevation, Assign worksheet #2 on right triangle application problems for homework. Rubric: Angle of Elevation and Angle of Depression Each problem is assessed the ….
Activities in Math Trigonometry - Syvum. 19 photos of the "20 Angle Of Elevation and Depression Trig Worksheet Answers" Angle Of Elevation and Depression Trig Worksheet Answers Related Posts of "Angle Of Elevation and Depression Trig Worksheet Answers", Worksheet Angle Elevation And Depression Worksheet Daily Worksheet Templates P90X Worksheets Excel Right Triangles Angles of Elevation and Depression Riddle Practice Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression.
### Activities in Math Trigonometry - Syvum
Angle Of Elevation and Depression Trig Worksheet Answers. the students will use that angle of elevation and length of an unknown object’s shadow to find the height of that object. This activity gives the students two different applications of Worksheet Depreciation Worksheet Template Scientific Process Worksheet Template Word Angle Elevation And Depression Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression.
• Angle Of Elevation and Depression Worksheet Rosenvoile.com
• Geometry 8.5 Angle of elevation and depression by Mr
• Activities in Math Trigonometry - Syvum
• 20 Angle Elevation and Depression Trig Worksheet Answers Si Inc. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets. Geometry 9th Grade Pdf Inspirational Lines Angles Rays Worksheets . Area Trapezoid Worksheet Pdf Luxury Trapezoid Hw Pdf. Classifying Angles Kuta software Infinite Geometry Name In this lesson we have returned to the topic of right triangle trigonometry, to solve real world problems that involve right triangles. To find lengths or distances, we have used angles of elevation, angles of depression, angles resulting from bearings in navigation, and other real situations that give rise to right triangles. In later chapters, you will extend the work of this chapter: you
Worksheet Depreciation Worksheet Template Scientific Process Worksheet Template Word Angle Elevation And Depression Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression Two men on opposite sides of a TV tower of height 26 m notice the angle of elevation of the top of this tower to be 45 o and 60 o respectively. Find the distance (in meters) between the two men. Find the distance (in meters) between the two men.
White are dominant on this Worksheets Angle Of Elevation And Depression Worksheet due to table below. Then very light pink & light greycolor scheme is make it perfect. Combination beetween silver, greyish, lavender pink, very light pink, light lavendar are wraping around the room. Worksheet Angle Elevation And Depression Worksheet Daily Worksheet Templates P90X Worksheets Excel Right Triangles Angles of Elevation and Depression Riddle Practice Worksheet Presentation on theme "Angles of Elevation and Depression"— Presentation transcript 1 Angles of Elevation and Depression
Students will turn in the Angle of Elevation and Angle of Depression worksheet tomorrow for a classwork grade and each group will be responsible today for putting one of the answers up on the Promethean Board. 7.5 Angles of Elevation & Depression Name_____ Worksheet 1. A tree casts a shadow 21m long. The angle of elevation of the sun is 51°. What is the height of the tree? 2. A helicopter (H) is hovering over a landing pad (P) 100m from where you are standing (G). The helicopter's angle of elevation
Assign worksheet #2 on right triangle application problems for homework. Rubric: Angle of Elevation and Angle of Depression Each problem is assessed the … White are dominant on this Worksheets Angle Of Elevation And Depression Worksheet due to table below. Then very light pink & light greycolor scheme is make it perfect. Combination beetween silver, greyish, lavender pink, very light pink, light lavendar are wraping around the room.
the students will use that angle of elevation and length of an unknown object’s shadow to find the height of that object. This activity gives the students two different applications of 8-4 Angles of Elevation and Depression Marco breeds and trains homing pigeons on the roof of his building. Classify each angle as an angle of elevation or an angle of depression. 1. 1 angle of elevation 2. 2 angle of depression 3. 3 angle of depression 4. 4 angle of elevation To attract customers to his car dealership, Frank tethers a large red balloon to the ground. In Exercises 5–7, …
Angle of Elevation & Depression Worksheet/Scavenger Hunt ! Draw a picture for each. Find all values to the nearest tenth. Follow directions on the board and Assign worksheet #2 on right triangle application problems for homework. Rubric: Angle of Elevation and Angle of Depression Each problem is assessed the …
depression worksheet answers 2018, practice 8 5 : angles of elevation & depression dng p 405 describe each angle as it relates to the diagram answers: a of depression from the birds to the ship geometry worksheet 85 (angles of elevation & depression) In this lesson we have returned to the topic of right triangle trigonometry, to solve real world problems that involve right triangles. To find lengths or distances, we have used angles of elevation, angles of depression, angles resulting from bearings in navigation, and other real situations that give rise to right triangles. In later chapters, you will extend the work of this chapter: you
White are dominant on this Worksheets Angle Of Elevation And Depression Worksheet due to table below. Then very light pink & light greycolor scheme is make it perfect. Combination beetween silver, greyish, lavender pink, very light pink, light lavendar are wraping around the room. 19 photos of the "20 Angle Of Elevation and Depression Trig Worksheet Answers" Angle Of Elevation and Depression Trig Worksheet Answers Related Posts of "Angle Of Elevation and Depression Trig Worksheet Answers"
Students will turn in the Angle of Elevation and Angle of Depression worksheet tomorrow for a classwork grade and each group will be responsible today for putting one of the answers up on the Promethean Board. Use a clinometer to measure angles of elevation and depression 2. Measure lengths and use measurements to determine angle measures. 3. Apply sine, cosine and tangent ratios to find angles of elevation and depression. NCTM STANDARDS: 1. Student must be able to apply proper formulas to calculate and measure angles of elevation and depression 2. Perform Solving exercise, make real …
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### The Reverse Monty Hall Problem and Conditional Probability
Please note that since I wrote this article, I have been persuaded that the argument it contains is wrong. The correct answer for the scenario as it is worded is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).
---
Originally, this was to be my last article on The Reverse Monty Hall Problem – or at least the last on why the solution is “it doesn’t matter whether you switch or stay, the likelihood of benefitting from either choice is 1/2". It didn't work out that way ...
---
As the title suggests, we need to consider conditional probability. The relevant equation is:
Pr(A | B) = Pr(A B) / Pr(B)
This means “The probability of A happening, given that B has happened, is equal to the (a priori) probability of both A and B happening, divided by the (a priori) probability of B happened”. It may be phrased more briefly as “the probability that A given that B is equal to the probability that both A and B divided by the probability that B”.
In this case (please follow the links above if you don’t yet know the context), what we have is:
Pr(A) = the (a priori) likelihood that a car is hidden behind the door that you chose, but the philosopher didn’t open = 1/3
Pr(B) = the (a priori) likelihood that a goat is hidden behind a specific selected door = 2/3
Pr(A B) = the (a priori) likelihood that the car is hidden behind the unselected door AND a goat is hidden behind a specific selected door – noting that this is one of only three possible, equally likely permutations = 1/3
I have to stress, and I cannot stress this enough, that in the scenario I described, the philosopher had already opened the door before offering you the opportunity to switch.
I might need to ram this home, so please forgive me if you have already had a blinding flash of light and you now agree with me. Let’s get more specific. The doors don’t have any indications on them when you get there, but you are a paranoid sort and are convinced that there is some Derren Brown/Dynamo-type shenanigans going on, so you tell the philosopher you want to mark the doors just after he tells you which one he is going to open, but before he opens it.
• You walk to the door that you selected that will remain closed, and you mark it as Door 1,
• You walk to the door that the philosopher has told you will be opened, and you mark it as Door 2,
• You walk to the unselected door, and you mark it as Door 3, and
The questions you must ask yourself are now made a little more simple:
Before I got here, what was the likelihood that the car would be behind Door 1?
There is only one car and three doors, therefore, Pr(A) must be 1/3
Before I got here, what was the likelihood that Door 2 would have a goat behind it?
There are two goats and three doors, therefore, Pr(B) must be 2/3.
Before I got here, what was the likelihood that the car would be behind Door 1 AND a goat would be behind Door 2?
This is precisely the same likelihood that there would be a car behind Door 2, because if there is a car behind Door 2, then there must be a goat behind Door 1, therefore Pr(A B) must be 1/3
Then it’s just a simple process of plugging in the numbers:
Pr(A | B) = Pr(A B) / Pr(B) = (1/3) / (2/3) = 1/2
It’s not rocket science, but it seems to have confused the heck out of a lot of people (including, to be totally honest, me).
Now that I have gone through this process, I am really hoping that the cognitive dissonance stays away! However, there remains a feature that may still confuse people.
If you were asked, prior to entering the arena with the philosopher, whether it would be better, on average across multiple runs, to stay with your door (the one of two that remains to you) or to switch, the correct answer is it is better to stay – because, exactly as many have argued, this would be equivalent to the Monty Hall Problem.
This seems to be hugely counter-intuitive.
However, you have to remember that at this point in time, the philosopher has not yet opened any doors. Therefore, at this point in time, the door that the philosopher opens could be any of the three (depending on which doors you select and which doors have goats behind them).
This seems to be the default position taken by all armchair statisticians, despite being encouraged repeatedly to consider the scenario from just after the door has been opened.
Someone did ask to explain how my two images could be reconciled (from The Reverse Monty Hall Problem – The Solution):
And
Someone also complained about my graphical representation of the solution to the classic Monty Hall Problem:
This might be difficult to explain, but please bear with me.
First I need to re-jig the graphical representation of the classic Monty Hall Problem, not because what I did was wrong so much as it was not adequately explained. I am assuming that, when given the opportunity to select between two goats, the philosopher chooses at random. Therefore, the possibilities with the classic Monty Hall Problem are:
Now, imagine that we run the Reverse Monty Hall scenario over and over again, still assuming that when there is an option to select between two goats, the philosopher makes that selection at random. This means that the possibilities are:
Then, removing invalid scenarios, we have:
This can be compared with the results from the classic Monty Hall Problem.
So, yes, if you run the scenario over and over again, you will get the 2/3 - 1/3 split. But ... there was never any intention that Reverse Monty Hall Problem be performed repeatedly.
---
Once again, please note that the argument above is incorrect. The correct answer is that the 2/3 - 1/3 split is as true (in this scenario) as it is for multiple iterations of it.
1. "So, yes, if you run the scenario over and over again, you will get the 2/3 - 1/3 split. But ... there was never any intention that Reverse Monty Hall Problem be performed repeatedly."
Wait, what?
So, the odds are 2/3 if you stick and 1/3 if you switch. You admit this is empirically verifiable. Then you claim that, contrary to this, the actual probability is 50/50. How can you reconcile these two statements?
1. Update: This is actually giving me cancer.
Here are all the possible distributions of Mary/Ava/Car(men)
AM C
AC M
MC A
MA C
CA M
CM A
Assuming your choice of doors is always the first two terms in this table (and you will agree that this represents all possible choices) you have a 1/3 chance of selecting both Ava and Mary. This is undeniable, right?
Our philosopher will ALWAYS open one of our selected doors, and that door will ALWAYS reveal a goat. Given this new information, we now know where at least one of the goats in our initial selection is hidden.
There is still here a 2/3 chance that our remaining door hides a car and a 1/3 chance that it hides another goat. This is undeniable.
Whether the revealed goat is Ava or Mary is irrelevant: both represent undesirable results and are, logically, the same. They are not unique entities, as you claim.
Whether our philosopher opens the leftmost or rightmost door is thus also irrelevant.
2. I hear you about the "this is actually giving me cancer". While experiencing it, the cognitive dissonance is quite uncomfortable. I apologise for foisting that on you.
The problem is that you are generalising when, in the scenario, we are talking about a very specific situation, because the door is already opened. Because it's a specific situation, it is not so obvious that the question of "which goat?" is irrelevant (but I agree that in a sense it is), and the goat is certainly not a representation - it's an actual goat (in terms of the scenario).
3. It certainly is a representation. Whether we ultimately open the door that reveals Mary or Ava, the net result is the same: You lose.
I think you need to be more clear and concise with what you mean by "the door is already open." It seems like every time you reply to somebody who points out the seemingly simple (and as you admit, empirically verified) result of this case (the 1/3-2/3 split), you say almost verbatim "the door is already open" and proceed to rather passive-aggressively insult their intelligence (see above: "While experiencing it, the cognitive dissonance is quite uncomfortable. I apologise for foisting that on you.")
So, on to my interpretation of "the door is already open" -
We've selected our two doors. Our resident philosopher reveals one of the goats we have chosen. The door, as you say, is open, and we now know:
-Which one of our doors is a goat.
-The order of the prizes behind the doors has not changed. It is the same problem.
-Our remaining door and the one we have not chosen are still a mystery; they are, as you would say, closed.
-Our initial choice gave us a 2/3 chance of our group containing the selection "car/goat"
-Our initial choice gave us a 1/3 chance of our group containing the selection "goat/goat"
Given all of the above, the informed decision would be to stay, because there is still a 2/3 chance that I initially selected a goat and a car, one of my initial choices was already revealed to be the incorrect choice (goat), and the 2/3 odds carry over to that last unknown.
If I'm misunderstanding what you mean by "the door is already open," please try to explain with more clarity, as you've manged to present your problem in a way that has confused all of the commentors on reddit (most of who's logic is undeniably sound) and your blog here. The onus is on you to present your scenario properly.
4. There's no "net result". You are suggesting that this scenario will be re-run, over and over. This was never suggested.
I've just suggested this to someone else on reddit, perhaps you can try it too:
"I know it sounds silly, but can I suggest trying to crystallise the scenario by role playing it out. You don't even have to hide what's behind the doors. Just pick three objects, two of which are the same or similar. Put them in front of you and go through the options - just like you would once the door has already been opened. You might want to move one 'goat' forward to suggest that it was behind the door that was opened."
5. I suspected you were a troll; now I'm nearly positive of it.
Are you truly suggesting that the probability of a given situation (given exactly the same circumstances) actually changes based on how many iterations of the circumstances are performed? That's not mathematics, that's insanity.
6. There's another possibility, as unlikely as you might think it is, and that is that I am right. I am willing to accept that there is a possibility that I am wrong. You seem unwilling to even contemplate the possibility that I am right. I can't really blame you, since if I had not gone through the process I did I probably would not believe me (in other words, the me that existed last week would not believe the me that exists now). You're sort of an incarnation of "me last week" - although I really hope that "me last week" would have entertained the very slight possibility that "me that exists now" might be right.
This is not "an additional troll" as you may well suspect, it's just a reflection on how difficult it sometimes is to persuade people of things.
7. If you can't logically dismantle my argument (or present yours), your argument is illogical, and thus incorrect when we're dealing with a mathematical issue.
Telling me "You're wrong, you just can't see it" does not do any justice for your argument. I have contemplated the possibility that you may be right. I've read your blog posts, pointed out the flaws in those posts, and given you ample chance to reply (rather than writing you off and moving on to something more interesting, which I would love to do at this point,) and every time you come back with some dismissive remark.
So, last chance - if the odds are 50/50, why are the results of repeated experiments 66-33?
8. I do think that this might be THE most interesting aspect to the whole thing. You are right that I should explain this and I will put my mind to how I can best explain it. I have, via modelling, and role play, demonstrated to my entire satisfaction that it's right, but explaining why it is right is less simple.
I understand that you are impatient with me, because I seem to be unable to give you a quick, succinct and convincing answer *right now*. Can I ask for your patience while I ponder a way to explain this to everyone's satisfaction? When I feel that I can clearly explain, I assure you that I will post it in an article here.
9. neopolitan,
I would love to see a video of this model/role play of the situation you've developed.
Every time you're backed into a logical corner, you assure us that you're able to demonstrate the situation to your satisfaction. So, if that's the last bastion of your argument, please provide video evidence of this so that it can be interpreted and we can all be done with this. I don't want to read through another one of your mad blog posts.
The way I see it now, there's two likely outcomes to this whole thing:
1) The demonstration you've set up that has you so convinced of your "solution" is flawed. I don't think I've heard you consider this notion yet. Or,
2) You have not accurately explained the situation you are trying to present.
That's the cool thing about math: It's demonstrable. So if you're right, let's see the demo.
10. Well, of course I don't think I've been backed into a logical corner. However, amusingly enough, a you-tube video of a demonstration was precisely what I was talking about last night with a couple of real live people and pondering how to implement at 4am this morning.
The thing is, when I demonstrated it for these people (admittedly they were not maths PhDs), they found it so compelling that they are completely flummoxed that no-one can grasp what I am saying. But, of course, you could dismiss them as ignorant buffoons, stooges or people I simply made up. So, a video might just be the way to go.
It's not going to happen overnight though.
11. Well, you have been backed into a corner.
You have not been able to appropriately respond to any of the recent criticisms of your claim, which leads me to believe you simply don't have a response.
Anyway, I'm awaiting your video proof. Until I've got that in my hands, or you manage to piece together a pretty astounding case for yourself (I would advise doing this from the ground up, as everything you've posted to date has been effectively vanquished), I remain unconvinced.
You'll be on the cover of Scientific American if you can prove this one, champ.
As a side challenge, go to your nearest university, find a math PhD, and have the good doctor verify to us - again, video testimonial with his academic reputation at stake - that you are in fact correct.
Surely a doctor of mathematics will back you up, right?
12. Another good idea that I have also considered. It's actually part of the reason that I want to be so certain before going to the next step. If this is ground-breaking, if I am right, then I guess I should try to protect it a little. However, I think that if I am right, there is enough of an evidence trail by now that it would be rather difficult for anyone who has been arguing against me to suddenly turn around and claim ownership (or even anyone who has just been watching quietly.
I suppose that I should be able to convince a maths PhD to give me half an hour of his or her time, given a rather simple cost benefit analysis. It might be very unlikely that I am right, but there are kudos galore for any PhD whose name is attached to verifying that vos Savant was wrong all this time (remember it was maths PhDs who argued most vehemently that vos Savant was wrong back in 1990). So half an hour for a possible academic gravy train? Might be worth it :)
That said, I'd have to be pretty sure that I am right, huh?
Would a published paper suffice instead of video testimony?
13. If you can get enough PhD mathematicians to agree with your theory that you get a peer-reviewed paper out of it, I'll eat my hat.
And regarding vos Savant: Most of them (the mathematicians) misunderstood some part of the situation while it was being explained to them, it not being as popular a brain-teaser as it is nowadays. Presented with a clearer definition of what the Monty Hall problem was they changed their minds. And even the stragglers, presented with computer simulations and empirical proof, eventually came around.
Don't romanticize yourself so much. You're not vos Savant. You're a dude with a blog and a decent mind for mathematics that's fooled himself into thinking he's smarter than everyone else.
I won't be spending any more time here. Once you're published in an academic journal I'm sure I'll hear about you.
14. If there is a peer-reviewed paper on it, I will be wanting video evidence of you eating your hat.
15. I have a Phd in maths. I spent more than half an hour reading your blog and your reddit posts. My conclusion is that you lack mathematical training to see the tiny mistake you made.
Just after giving the (correct) definition of conditional probability, you throw that sentence :
> Pr(A) = the (a priori) likelihood that a car is hidden behind the door that you chose, but the philosopher didn’t open
What does that mean ? The "but" between the two part of the sentence is VERY ambiguous. Is that :
* the probability that both events happen (probability that a car is hidden behind the door AND the philosopher did not open it),
* or is that the probability of one KNOWING the other one (probability that a car is hidden KNOWING that the philosopher did not open it).
These are two very different things, that are not equal.
By the way, what does "specific door" mean, and why do you use a different vocabulary than for the "unselected door". Aren't the two equivalent ? Could you not just say "selected door" and "unselected door" instead ? So this is a general advice : you should be more careful with the wording of your definitions. It's in the little details that the big mistakes are made ...
So, there is this ambiguity of the definition of the event A. It seems that it makes you believe that you computed the correct probability for it (using one way to interpret the event). But it also makes you believe that you can use this probability in another formula (using the other way of interpreting the event). You don't see the global mistake because at each step, the thing that you are doing is correct given a possible interpretation of your event A.
Your approach of the problem with conditional probability is good. But your lack of mathematical training is not helping you.
You should think about the problem a little bit more, but this time, with 11 doors, 8 goats, 3 cars, you choose 7 doors at the beginning, and the host voluntarily open 4 of them that contain goats, and he also open one of the door that you didn't pick without a goat. And now you are asked to choose one door among the 6 doors left (3 of them were part of your initial pick, the other three were not). Which one do you choose, and what is the probability that you will win ?
Your reasoning should apply in the exact same way. I'm not saying this to make you lose your time. By using these unique values for each parameter of the game, it will be easier for you to see what is the role of each thing.
16. (There was a minor cut and paste error in the original of this, right in precisely the wrong spot!)
Thanks for your input. I agree that my wording obviously leaves a lot to be desired. I clearly wasn't expecting the interest of such people as yourself, so I didn't get a maths degree first! Seriously though, I should try to be more clear and I fully intend to so when I do my modelling. I don't want to confuse things at the moment with the Franken-Monty Hall Problem, but I suppose I could always come back to it.
My intention is to do this:
---Colour the doors, left door Red, centre door White and right door Green
---Indicate whether a door is selected (S) or unselected (U)
---Tag an opened door with O
This will mean that the sample space will be populated by elements like this:
(Red-S-C White-O-M Green-U-A)
Which means the player selected the Red and the White doors and the host opened the White door to reveal Mary, while Ava was behind the unselected Green door.
My question would then resolve down to: if you are playing the game, what is the likelihood, given that the host has already opened the White door to reveal the goat called Mary, that the car is behind the Green door?
My answer is 1/2, based on the pretty obvious (to me) notion that the sample space consists of:
(Red-S-C White-O-M Green-U-A) --- and --- (Red-S-A White-O-M Green-U-C)
There is no other option. If you go back to the original spread of options, these two were equally likely right at the start and they are equally likely after the door is opened.
There's no magic to it, as far as I can see.
If this is more clear to you, and you now agree with me, then, as an anonymous mathematics PhD, could you please assist by using your authority to point it out to everyone else in mathematical terms that are clearly beyond me?
If not, then I suppose I'll still be doing my modelling over the weekend :)
17. Not a mathematician, but it seems to me that the crucial problem here is that you're treating the two goats as separate events, when the standard formulation of the Monty Hall problem makes no such distinction.
18. Neopolitan, I think I can point out exactly where you’ve gone wrong, and I don’t think this is something that has been (specifically) pointed out yet. You’ve left one massive ambiguity in this problem you present that has been the cause of all of the debate to this point, and I think this has snowballed into an arduous trail of non-sequiturs that has taken this debate way into left field.
In your first posting, you presented the problem with the following rule:
>There are three doors, there is a goat behind two of the doors and behind the third is a car.
Fair enough. This problem you’ve presented has (to this point) earned the title of Reverse Monty Hall, as these conditions (2 goats, 1 car) are the same as the original MHP, with the distinction being that the players choice is inverted (2 doors instead of one.)
Your next posting is where things start to go south, when you “clarify” the conditions of the situation:
>First things first. I did try to provide a couple of hints, one of which was that the goats are unique entities. To highlight this, I will give the goats names. There is the pretty, pleasant natured goat called Mary (M) and there is the unattractive, unpleasant goat called Ava (A). We might not care about the names of the goat, but I think we can agree that there are two goats and these two goats are not the same goat – meaning that they are not entirely interchangeable.
Here, you’ve given your two goats names, but (you will admit) a goat by any other name is still a goat, no? A condition you didn’t set up is that the player can distinguish between the two goats! This is not an implied part of the problem you presented.
So the issue here is that you’ve presented a different situation than the one that’s in your head.
A better way to present this would look like this:
There is a car, a goat, and an elephant, and each one hides behind one of three doors, randomly. The player picks two doors, and Monty opens one of those doors, revealing a goat or an elephant. What are the odds that the car is behind each of the remaining doors?
Since the third term is always revealed, this problem can then be reduced to:
There are two doors, behind one is a car, behind the other is either an elephant or a goat. What are the odds that the car is behind each door?
I think we will all agree that this problem is neither difficult, groundbreaking, or interesting.
This is NOT a reversal of the MHP. This is a problem with three unique entities behind the doors, while in the MHP there are only two.
The title and the conditions from your first posting (see above) are the reason for all the debate, as both present an entirely different scenario than the one you’re attempting to solve.
TL;DR – Everybody was right.
19. Then again, even this puts us in a precarious position, because you still have a 2/3 chance of the original set containing the car and a 1/3 chance of it not containing the car. So, perhaps whether or not the goats are distinguishable makes no difference, which would still leave you incorrect.
Math PhD's, support me here.
20. Thanks Alex (both for commenting and also for not being another "Anonymous" (not that I don't also appreciate the engagement of those who are labelled "Anonymous", I just wish they'd give themselves a nom de plume).
If I read correctly, you are saying that if it's a goat and an elephant, then the 1/2 answer is correct (per your TL;DR). The problem though, is that at least in the game on which the name of the classic problem is named, both the goat and the elephant would be "zonk" prizes (read all about here - http://en.wikipedia.org/wiki/Let%27s_Make_a_Deal). Therefore we could say that, in a sense, the term "goat" is interchangeable with the term "zonk". In terms of the scenario, the only really important thing about the goats and the elephant is that they are NOT the car.
The question then is, if I am right when the goats are actually different (and they could be a goat and an elephant), but really goat=zonk=(NOT a car) and elephant=zonk=(NOT a car) ... where does it leave us? Am I still right, or am I now wrong? And if I am now wrong, why does defining the animals behind the doors in this way make any difference?
21. > There is no other option.
You are right.
>these two were equally likely right at the start
I don't think so. Let me explain what I understood :
You have three painted doors. Good.
You are (rightfully) assuming that a priori each door has probability 1/3 of containing Ava, Mary or the Car.
You are (also rightfully) assuming that a priori each door has probability 1/3 of being Open, Selected and Unopened.
So, just to have the same vocabulary, you have three doors of different colors. Each door has two other variables which are the "object hiding behind" and the "status".
If these two variables were independent, your answer would be completely correct. Because If these two variables were independent, then the probabilities of the event (Red-S-C White-O-M Green-U-A) and of the event (Red-S-A White-O-M Green-U-C) would be exactly the same in the set of all possible situation, which should be 1/36. And then you would rightfully conclude that knowing that "the white door is Open and contain Mary and the green door is unselected" (which is an event with probability 1/18), then the conditional probability of these two events is 1/2.
So if we assume that the variables "object hiding behind" and "status" are independent, then your answer is perfectly logical. It shows a perfect use of conditional probabilities and everything.
PROBLEM : The two variables are NOT independent. You can easily see that for the very simple reason that an Open door can never contain a Car. So this means that the probability of having a configuration with G-U-C has absolutely no reason to be equally likely to a situation with configuration G-U-A.
So I'm not going to compute again everything in your scenario, but I urge you to reconsider this sentence :
> these two were equally likely right at the start
with what I just explain. I hope I was clear enough to make you understand where your mistake is. I reassure you, this is a very common mistake (or may be that's not what you wanted to hear).
22. Hi Mathematician,
I think we might be getting somewhere. I think also that the clues to the solution are already available to anyone who wants to put the story together, spread across the articles here, the comments to those articles and my response to various exhortations to conform with the "staff answer" at reddit. So, perhaps I should keep my powder dry for a paper, but if there IS a paper in this, then I doubt that giving you the solution here here won't do much more harm. My idea is to put this into a more detailed article, but in short:
We agree that there are six equally likely arrangements of goats and car at the beginning, before the game even starts. Using Red-White-Green and Ava (A), Car (C) and Mary (M), we have: ACM, AMC, CAM, CMA, MAC and MCA. All equally likely.
My argument is that the Reverse Monty Hall Problem (RMHP), as with the Monty Hall Problem (MHP), consists of six possible games, or "mini-games", with each arrangement of goats and car being a different mini-game. When we start, we don't know which mini-game we are playing, and we have absolutely no information that would help us assign a likelihood of any of the mini-games we might be playing.
Selecting two doors gives us no more information. However, when a door is opened, then we do get some useful information. Since all mini-games are equally likely and each randomly selected pair of doors is equally likely, let's say that the host opens the Red Door.
Once the Red Door is opened, we get the information that we are not playing the CAM or CMA mini-game. But we could still possibly be playing any of the others, ACM, AMC, MAC or MCA - unless we can distinguish between Ava and Mary. It is my argument that these mini-games are still equally likely at 1/4 - if we can't distinguish between Ava and Mary - and 1/2 - if we can.
This information about which door is opened and which mini-game we are playing is only available in a single iteration of the RMHP or MHP. The information as to which goat is revealed is only available in a single iteration of the RMHP or MHP. Once you generalise and start thinking about multiple iterations of the game, that information is no longer available and each of the mini-games is equally likely (which has been demonstrated via the standard simulations, as helpfully reported a few times here and at reddit).
I really very very strongly encourage you to take three dice, one of which is different from the others (perhaps from a game of Risk). Put them in front of you. If the middle die is a "goat", pick it up and put it in your pocket. Otherwise, take the left die and put it in your pocket. Then, pick up the two remaining dice, one in each hand, and swap their positions.
Then, try to convince me that, in a single iteration of the game - as the contestant, you could somehow tell that one of these two possibilities was more likely.
(Slightly edited in last paragraph.)
23. Let's do this :
> we have: ACM, AMC, CAM, CMA, MAC and MCA. All equally likely.
> When we start, we don't know which mini-game we are playing
Absolutely.
In your setting, the selection of two doors by the candidate is EXACTLY the selection of which of the three doors will be the Unselected one. This gives three possibilities for each of the 6 "mini-games" we are playing.
This means that there are 18 possible situations after the choice of the two doors by the candidates, which are all equally likely.
I'm going to denote these situation as A-CU-M (the situation is ACM and the player "unselected" the White door) or MU-A-C, and so on
> However, when a door is opened, then we do get some useful information.
First, let's focus on the different possibilities for the open door. As the player already selected its two doors, it means that there are only 2 possibilities left for Open door.
So a priori, we get 36 possible configurations of the form AO-MS-CU (we are in the AMC situation, the player selected the Red and the White door, and the host opens the red door, reveling Ava). Or something of the form MO-CS-AU or even CO-AS-MU.
As you can see, among those 36 configurations, some are impossible (the last one because the host cannot reveal the car). So these 36 configurations are NOT equally likely. The last choice of which door is going to be opened is NOT independent from the previous equally likely choices.
Let's do two examples. Let's say that we are in the partial situation A-M-CU which has probability 1/18. This can lead to two final configurations :
AO-MS-CU
AS-MO-CU
If we assume that the host opens a door randomly when he has a choice, these two configurations are equally likely and add up to 1/18. So obviously each of these situations is of probability 1/36.
Now let's say that we are in the A-C-MU situation. This can lead to two configurations :
AO-CS-MU
AS-CO-MU
One configuration is of probability 0 (the situation AS-CO-MU) because the host cannot open a car door. And the sum of the two probabilities must be 1/18. So the conf AO-CS-MU has probability 1/18.
As you can see with this extremely explicit computation (I can hardly make it more explicit), the two situations AO-CS-MU and AO-MS-CU are not equally likely.
> let's say that the host opens the Red Door.
By the rules of the game, he reveals a goat. So there are 8 configurations corresponding to that :
AO-CS-MU (1/18)
AO-CU-MS (1/36)
AO-MS-CU (1/36)
AO-MU-CS (1/18)
MO-CS-AU (1/18)
MO-CU-AS (1/36)
MO-AS-CU (1/36)
MO-AU-CS (1/18)
The total probability of these 8 situations : 4*(1/36) + 4*(1/18) = 1/3. This is exactly the a priori probability that the host will open the red door, which was the expected result.
So now, what is the probability, KNOWING that the host opened the Red Door, that the car is behind the Unselected door ?Well, you just have to compute it using the formula for conditional probabilities. With obvious notations :
P(CU | RO) = P(CU ∧ RO) / P(RO)
We just said that P(RO) = 1/3.
And for P(CU ∧ RO) we just add all the probabilities for the conf containing "CU". You should get 4*(1/36) = 1/9
P(CU | RO) = (1/9) / (1/3) = 1/3
So as you see, knowing that the host opened the Red door, the a priori probability that switching will give you the car is only 1/3.
I cannot be more explicit than that, and I used your exact situation, with your notations. So I cannot imagine how you could disagree with this result.
24. > I cannot be more explicit than that, and I used your exact situation, with your notations. So I cannot imagine how you could disagree with this result.
All I can tell you is, "prepare to be amazed!"
2. You say that after the host has opened either the leftmost or the rightmost door, then there are four possible scenarios (correct) and that they are equally likely (wrong).
I agree that each of ACM, AMC, CAM, CMA, MAC and MCA is equally likely with probability 1/6. If it's MCA or ACM (a combined 1/3 of the scenarios), then the host has to open the leftmost door. If it's CMA or CAM (another 1/3 of the scenarios), then the host has to open the rightmost door.
The remaining two scenarios, accounting for the remaining 1/3 chance, are AMC and MAC. These scenarios are split between leftmost and rightmost doors, since the host has to choose one or the other. Therefore, the chance of the host opening the leftmost door after AMC or MAC is half the chance of opening the leftmost door after MCA or ACM. Include that fact in your solution, and you get the obviously correct answer that staying wins 2/3 of the time.
1. I think you have missed the point here, in the same way as pretty much everyone misses the point. You are considering the general while, in the scenario, you need to deal with a specific.
You made a brilliant point on r/math:
"Let me propose a scenario: I happen to look down and notice my shoe is untied, just as the host is opening a goat door. By the time I look back up, he's closed it again, and I have no way to know which of my doors he opened. Now he's asking me if I'd like to switch or stay.
I claim that in this version, I have no information to go on, other than what I knew before the door was opened: the car is behind the door I didn't pick with 1/3 likelihood, and behind one of the doors I did pick with 2/3 likelihood. Since that's all I know, I choose to stay, and I'm 2/3 likely to win the car.
Of course, I can force your scenario to become my scenario, by deliberately closing my eyes while the host opens the door. Do we conclude that closing my eyes makes the "always stay" strategy more effective than with eyes open?"
This certainly does seem (but only it only seems) to put paid to my argument. I struggled with this as well: surely more information is going to **increase** the likelihood that you can make the right decision, rather than reduce it (as apparently seems to be the case, if I am right).
The thing is that if you do not see which door (which specific door) the host opened to reveal a goat, then you are not able to eliminate two possibilities (out of the sample space of six), namely those in which the car sits behind the door that the host opened. I will have to work through this specific "eyes closed" scenario to work out why it resolves to 1/2, but in the short term, perhaps you might want to look at http://probability.ca/jeff/writing/montyfall.pdf (noting that this paper was not written by me).
2. I think that my argument shows that yours is wrong, by contradiction. But it's better to find the flaw in your specific argument, which I have done, and explained on r/math. Briefly, the four arrangements which are possible after seeing a door being opened are not equally probable, as you have assumed.
3. Maybe you should review that paper! His "proportionality principle" is exactly why you can't consider all the arrangements (MCA, CMA, ACM, etc...) equally likely after seeing a door open.
4. If (and only if) the likelihood of all the arrangements were not equal after the door is opened, then the "proportionality principle" may well be an excellent explanation as to why. However ... that wasn't the issue that I really wanted to draw your attention to.
What I wanted to draw your attention to is that if Monty falls and knocks a door open, AND it just happens to have a goat and thus does not reveal a car, then the likelihood of benefiting from a switch becomes 1/2, rather than 2/3.
Can I suggest a different situation? We can call it the "Monty Lies" variant of the Monty Hall Problem. After you have selected your door, the producer of the show tosses a fair coin with "Left" on one side and "Right" on the other and whispers the result into Monty's ear. Monty opens a door based entirely on the result of that coin toss, revealing a goat.
Monty tells you that he selected the door based on the fact that there is a goat behind it, but he's lying. You don't know that he is lying.
What is the likelihood of benefiting from a switch?
Note that this scenario is directly analogous to my Reverse Monty Hall problem. You might not agree, but try assuming that it is.
5. In this post it seems like you just explained the correct answer to yourself, neopolitan, and then somehow in the last paragraph are fighting to figure out a reason that the answer is not correct.
quoting from your description of the problem:
"The philosopher will then open one of the doors you selected, revealing a goat."
So having your eyes open or closed does not change anything about the possibilities "in which the car sits behind the door that the host opened"
I read through that monty fall paper, and both of the variant scenarios that he describes do not relate to your problem, unless you are just somehow intending a situation other than the one you described.
6. The situation in the Monty Lies or Monty Falls games is a different situation from the one you have described.
"The philosopher will then open one of the doors you selected, revealing a goat."
This sentence is different from saying that monty opened a door completely at random and it happened to reveal a goat.
7. neopolitan said, "If (and only if) the likelihood of all the arrangements were not equal after the door is opened, then the "proportionality principle" may well be an excellent explanation as to why."
No, there's no "if" about it: the proportionality principle tells you, unambiguously, that the probabilities are not all equal. Look at the "sister in the shower" example. Replace Alice and Betty with MCA and MAC, and replace "singing" with "the host opens the left door". Since the host *always* opens the left door for MCA, but only 1/2 the time for MAC, you must conclude that, given that the left door was opened, MCA is twice as likely as MAC.
This is the paper that *you* cited, and it clearly does not support your conclusion.
8. Both versions of Anonymous and ChalkboardCowboy:
I'm not really suggesting that any of the scenarios in the Monty Falls paper are a match for mine. What I am pointing out is that the paper indicates that, if Monty does not select a door deliberately, but rather falls against a door - a variant of which would be that he has a fit and randomly stabs a button which opens a random door - then the likelihood of benefiting from a switch is 1/2. What is being suggested here is that if you are not absolutely certain that the door opened for you is opened on purpose, or maybe even if you know that the door opened for you wasn't opened on purpose, then the probability can change - even if the door opened were the same door that Monty would open if he deliberately opened it. Many arguments against my position is that it is insane that probability might change on the basis of opening the door. However, we are told here that probability changes on the basis of the INTENT behind opening the door. I find this to be an even less reasonable suggestion. What makes far more sense to me is that the probability, in any single shot Monty Hall Problem (and in any single shot Reverse Monty Hall Problem), the likelihood of winning with either door is the same, at 1/2.
I've suggested a way to test this out, yet no-one has indicated that they have tried it.
9. I'm not trying to talk about Monty Falls, Monty Crawls, Monty Lies, or whatever. I'm talking about *your* solution to the *Reverse Monty Hall* problem, as described on this page. First I gave you a reason to suspect you had a problem, since apparently by closing my eyes (to reject *reliable* information) I can improve my chance of winning the car.
Then I took the time to find the exact flaw in your argument, explain it, and give you a reference in the very paper you cite (the "proportionality principle").
All you've done is throw out different scenarios, and repeat that this is a "specific" event and so we can't talk about "general" probabilities, whatever that means. I asked you on r/math if that is an established concept, but you haven't responded. I looked in the paper you cited, to see if the proportionality principle was somehow contingent on repeating a situation multiple times, but it's not.
Meanwhile, you've been tremendously condescending (here and especially in r/math), assuring us that you know the "cognitive dissonance" is painful, and that we're just not ready to accept the truth, and even that you'll have to "ELI5" for us (on reddit, that means "explain it like I'm five"). Look, you're obviously a bright guy. But you're *wrong*. And this isn't high school or your dorm where you can be the smartest guy in the room every time, and just bully and handwave your way out of a jam. Try that in academia and you may not ever recover the respect of your peers.
I'm not going to engage any of the other scenarios you've tossed out. I've shown you the error in your RMH solution, and you haven't addressed it. If you choose to, you'll need to specifically justify treating all the arrangements as equally likely after a door has been opened (i.e., justify ignoring the proportionality principle).
10. Sadly, Chalkboard, I think there's no dealing with our friend here. In my discussion with him (above) and in his comments on /r/philosophy, his response to (valid) criticism is always the same: "You don't understand the situation, it's a specific situation, blah blah, and so on." Never does he refute any claim using numbers or logic, just dismissive arrogance.
It's interesting to me that you came across this on /r/math. I think out friends repeated posting all over the internet, condescension, and argumentative nature are only a pathetic ruse to draw attention to his little blog.
11. I disagree that it's a ruse. I think he truly believes he's come across some new insight that has escaped everyone else up to now. It's ironic that he's telling everyone about their own "cognitive dissonance", because he's clearly the one who is so invested in having made a new discovery that he's blinding himself to the (multiple) clear explanations of his error.
If he's planning to go into academia, I hope for his sake that he develops some better argumentative hygiene, or he's gonna have a bad time, as they say.
12. Truth. Held to any academic standards his argument holds little to no water, yet he insists we are the ones who need to think harder. I'm no mathematician (as stated, I found this on /r/philosophy), but I'm well-enough read in analytical traditions and logic to know bunk when I see it. I've been through the academic sieve, and I assume you have too. Telling somebody that a problem is beyond their comprehension and that they should just appeal to the (evidently) superior ethos of the author... yeah, doesn't fly. Sounds more like faith than empiricism, no?
13. Anon, I'm winding up a math PhD right now and, although my focus is not probability, I have more than enough experience to spot someone who is using words, rather than math, to make a point. His whole argument hinges (as you or another Anon pointed out above) on this distinction between a "single iteration" or "multiple iterations", which as far as I can tell is not an established notion in probability theory, and cannot be quantified mathematically.
14. Aye, that's me. If he's trying to make that point he'll need an accredited university and a good research team. And of course a stringent peer-review process. Hell, maybe he'll start a mathematical revolution of sorts if he can prove it.
As it stands, he's a dude with a blogger account and a poor theory.
15. If he was *right*, then he wouldn't need any of those things. Unfortunately (or fortunately, for the consistency of probability theory), he's not.
16. ChalkboardCowboy,
You asked at reddit "Do you have a reference for this distinction between "generalities" and "specific situations"? Is this standard terminology for an accepted concept?"
No, I don't. It's just me scrabbling for terms to explain (I don't think I used the word 'generalities' though, perhaps I did, but at this moment it strikes me as wrong).
As to quantifying mathematically, I did that. It's just conditional probability. Perhaps I could put it this way: if I were right that conditional probability is applicable, then would my application of conditional probability be mathematically correct?
If so, then it's not a question of bad maths, or tossing words around, or mindless arrogance on my part, it's a question of whether or not conditional probability is applicable. I'm not going to use the standard ruse "prove to me that conditional probability isn't applicable". What I will try to do is present my modelling and show that the results are compatible with conditional probability.
17. neopolitan, I've shown you several times now exactly where the flaw is. Why do you not apply the proportionality principle to the four possible scenarios compatible with the leftmost door being opened?
You've decided, with no justification offered, to disregard a principle that clearly applies to the situation. You then proceed to come up with an answer that everyone but you agrees is wrong (you call it "counterintuitive"). You need to have a very good reason, one you can explain, for disregarding the proportionality principle. But if you have it, you haven't shared it.
18. I should also add that, if your argument is amended to take account of the proportionality principle, then the answer (always stay for a 2/3 chance at the car) agrees with every other way of analyzing the problem. And that your "solution" without the proportionality principle implies at least one apparent paradox (that receiving new, true information can make the best strategy less effective than if you hadn't gotten the information).
Is this not raising any warning flags for you?
19. ChalkboardCowboy,
Sure, the apparent reduction in likelihood of winning the car if you keep your eyes shut is apparently paradoxical. From my perspective, it's a fascinating insight - I assume that, if I were right, it would also be a fascinating insight for you, would it not? Maybe worth a paper even?
20. It would be worth a dozen papers. But it's not right, I know it's not right, I know exactly how it's not right, and I've explained it to you multiple times. Why aren't you answering my direct question from two posts up?
21. Um, which one? I was addressing the "warning flags" question, I guess a little obliquely. Yes, it did raise warning flags for me, when I thought that it must be paradoxical. When I realised that it wasn't, the warning flags disappeared.
The other question was "Why do you not apply the proportionality principle to the four possible scenarios compatible with the leftmost door being opened?" I think I am applying it. Perhaps you think I am applying it incorrectly and this would be consistent with you position on the applicability of conditional probability - the use of which brings us to the answer 1/2.
I also asked you a direct question which you did not address: "if I were right that conditional probability is applicable, then would my application of conditional probability be mathematically correct?"
By the way, I have written my stage by stage explanation of the scenario which should help either expose the errors in my thinking or help others to understand where I am coming from. It's being looked at by someone who helps me be less obscure (coy, evasive, confusing, pretentious, etc) than I might otherwise be. I can't push them to drop everything for me, so I have to wait.
I know what ELI5 means, it's not intended to be an insult. If I am right, then I know that it's not easy to grasp and I suspect that it might be harder to grasp for someone who has been convinced ages ago that it cannot be right. If I am wrong, then I guess that I am just a pretentious prat. There have been situations in the past in which all the experts agree on something only to be shown to be wrong about it, so while it might (at least, a priori) be very very highly unlikely that I am right, it's not absolutely, totally impossible, is it? (Such a claim would surely be arrogance on someone else's part!)
22. Yes, the question about using the proportionality principle is the one I was referring to. You are not applying it, i.e. not reducing the probability of scenarios that could also go the other way (with the rightmost door being opened) relative to those which can only result in the leftmost door being opened. This step is not optional, or a matter of opinion--it's the way probabilities are calculated. The mathematical derivation is given in that paper you linked. If claim (as you have) that it should be ignored for this problem, then you need to give a very good reason. But so far, you've given no reason at all, beyond, "if I'm right, then it would be worth a paper".
You're right, I didn't address your question, "if I were right that conditional probability is applicable, then would my application of conditional probability be mathematically correct?" It's a non sequitur. I know how to prove that 1=2 and that pi=4. Granted, each proof has a single problematic step, but IF the problematic steps could be justified, then my proofs would be mathematically correct (and worth no telling how many papers!). I hope you see the problem and the parallel here.
Yes, I realize you know what ELI5 means; you're the one who used it. The parenthetical was for other readers who may not frequent reddit.
23. I removed your erroneous post (the one you deleted but to which information about you was still attached).
The "worth a paper" isn't a reason why I might be right at all. That would be mere wishful thinking. I guess you my reasonably suspect that this is all just wishful thinking. All I was trying to encourage you to do was to go the extra step. Run the role play yourself. Or go get three dice, one of which is a different colour or style to the others (that'd be the "car"), and put them in front of you, remove the "goat", then pick up the other two and swap their positions. Then think about the probabilities.
As you say, there are mathematical proofs for pi=4, so I might have to establish the fact of the matter before rolling up my sleeves (or begging a proper mathematician to roll up her sleeves) to do the proof.
24. Clarification: "As you say, there are erroneous mathematical proofs for pi=4"
25. Thanks for the delete. I think we're just talking past one another at this point. You don't seem interested in justifying your choice to ignore the proportionality principle, and honestly there's nothing else worth discussing.
3. I can point you to the error in your reasoning regarding this conditional probability approach.
You say:
"Pr(A) = the (a priori) likelihood that a car is hidden behind the door that you chose, but the philosopher didn’t open = 1/3"
This is where error is introduced. The a priori likelihood that a car is hidden behind the one of the two doors that you choose, but that the philosopher didn't open, is 2/3. Given a random distribution of car, goat, goat, and a random selection of two of these, as you admit, 2 out of 3 times, a car will be in your selection. The philosopher will then open a door that does not contain the car, and it will reveal a goat. You know this going in to the situation (it is a priori).
-B
1. What you are talking about is not an a priori probability, B. You're (almost) forcing the solution by inserting your preferred answer there. Fortunately you are not actually forcing the solution, since Pr(A) doesn't factor into the equation. Sadly, some of the responses indicate that conditional probability cannot be used - this is annoying because if was universally agreed that conditional probabilities could be used, then my self-imposed task would be complete!
2. Fair enough. Admittedly, I am not familiar with conditional probability. I am about to make a post on one of our older threads, though.
-B
Feel free to comment, but play nicely!
Sadly, the unremitting attention of a spambot means you may have to verify your humanity. | 12,476 | 54,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-30 | latest | en | 0.961996 |
https://www.geeksforgeeks.org/given-number-string-find-number-contiguous-subsequences-recursively-add-9-set-2/?ref=rp | 1,590,515,129,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00328.warc.gz | 739,292,840 | 28,950 | # Given a number as a string, find the number of contiguous subsequences which recursively add up to 9 | Set 2
Given a number as a string, write a function to find the number of substrings (or contiguous subsequences) of the given string which recursively add up to 9.
For example digits of 729 recursively add to 9,
7 + 2 + 9 = 18
Recur for 18
1 + 8 = 9
Examples:
```Input: 4189
Output: 3
There are three substrings which recursively
add to 9. The substrings are 18, 9 and 189.
Input: 909
Output: 5
There are 5 substrings which recursively add
to nine, 9, 90, 909, 09, 9
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Given a number as a string, find the number of contiguous subsequences which recursively add up to 9 | Set 1.
All digits of a number recursively add up to 9, if only if the number is multiple of 9. We basically need to check for s%9 for all substrings s. One trick used in below program is to do modular arithmetic to avoid overflow for big strings.
Algorithm:
```Initialize an array d of size 10 with 0
d[0]<-1
Initialize mod_sum = 0, continuous_zero = 0
for every character
if character == '0';
continuous_zero++
else
continuous_zero=0
compute mod_sum
update result += d[mod_sum]
update d[mod_sum]++
subtract those cases from result which have only 0s
```
Explanation:
If sum of digits from index i to j add up to 9, then sum(0 to i-1) = sum(0 to j) (mod 9).
We just have to remove cases which contain only zeroes.We can do this by remembring the no. of continuous zeroes upto this character(no. of these cases ending on this index) and subtracting them from the result.
Following is a simple implementation based on this approach.
The implementation assumes that there are can be leading 0’s in the input number.
## C++
`// C++ program to count substrings with recursive sum equal to 9 ` `#include ` `#include ` `using` `namespace` `std; ` ` ` `int` `count9s(``char` `number[]) ` `{ ` ` ``int` `n = ``strlen``(number); ` ` ` ` ``// to store no. of previous encountered modular sums ` ` ``int` `d[9]; ` ` ``memset``(d, 0, ``sizeof``(d)); ` ` ` ` ``// no. of modular sum(==0) encountered till now = 1 ` ` ``d[0] = 1; ` ` ``int` `result = 0; ` ` ` ` ``int` `mod_sum = 0, continuous_zero = 0; ` ` ``for` `(``int` `i = 0; i < n; i++) { ` ` ``if` `(!``int``(number[i] - ``'0'``)) ``// if number is 0 increase ` ` ``continuous_zero++; ``// no. of continuous_zero by 1 ` ` ``else` `// else continuous_zero is 0 ` ` ``continuous_zero=0; ` ` ``mod_sum += ``int``(number[i] - ``'0'``); ` ` ``mod_sum %= 9; ` ` ``result+=d[mod_sum]; ` ` ``d[mod_sum]++; ``// increase d value of this mod_sum ` ` ``// subtract no. of cases where there ` ` ``// are only zeroes in substring ` ` ``result -= continuous_zero; ` ` ``} ` ` ``return` `result; ` `} ` ` ` `// driver program to test above function ` `int` `main() ` `{ ` ` ``cout << count9s(``"01809"``) << endl; ` ` ``cout << count9s(``"1809"``) << endl; ` ` ``cout << count9s(``"4189"``); ` ` ``return` `0; ` `} ` `// This code is contributed by Gulab Arora `
## Java
`// Java program to count substrings with recursive sum equal to 9 ` ` ` `class` `GFG { ` ` ` ` ``static` `int` `count9s(``char` `number[]) { ` ` ``int` `n = number.length; ` ` ` ` ``// to store no. of previous encountered modular sums ` ` ``int` `d[] = ``new` `int``[``9``]; ` ` ` ` ``// no. of modular sum(==0) encountered till now = 1 ` ` ``d[``0``] = ``1``; ` ` ``int` `result = ``0``; ` ` ` ` ``int` `mod_sum = ``0``, continuous_zero = ``0``; ` ` ``for` `(``int` `i = ``0``; i < n; i++) { ` ` ``if` `((number[i] - ``'0'``) == ``0``) ``// if number is 0 increase ` ` ``{ ` ` ``continuous_zero++; ``// no. of continuous_zero by 1 ` ` ``} ``else` `// else continuous_zero is 0 ` ` ``{ ` ` ``continuous_zero = ``0``; ` ` ``} ` ` ``mod_sum += (number[i] - ``'0'``); ` ` ``mod_sum %= ``9``; ` ` ``result += d[mod_sum]; ` ` ``d[mod_sum]++; ``// increase d value of this mod_sum ` ` ``// subtract no. of cases where there ` ` ``// are only zeroes in substring ` ` ``result -= continuous_zero; ` ` ``} ` ` ``return` `result; ` ` ``} ` ` ` `// driver program to test above function ` ` ``public` `static` `void` `main(String[] args) { ` ` ``System.out.println(count9s(``"01809"``.toCharArray())); ` ` ``System.out.println(count9s(``"1809"``.toCharArray())); ` ` ``System.out.println(count9s(``"4189"``.toCharArray())); ` ` ``} ` `} ` `// This code is contributed by 29AjayKumar `
## Python3
`# Python 3 program to count substrings with ` `# recursive sum equal to 9 ` ` ` `def` `count9s(number): ` ` ``n ``=` `len``(number) ` ` ` ` ``# to store no. of previous encountered ` ` ``# modular sums ` ` ``d ``=` `[``0` `for` `i ``in` `range``(``9``)] ` ` ` ` ``# no. of modular sum(==0) encountered ` ` ``# till now = 1 ` ` ``d[``0``] ``=` `1` ` ``result ``=` `0` ` ` ` ``mod_sum ``=` `0` ` ``continuous_zero ``=` `0` ` ``for` `i ``in` `range``(n): ` ` ` ` ``# if number is 0 increase ` ` ``if` `(``ord``(number[i]) ``-` `ord``(``'0'``) ``=``=` `0``): ` ` ``continuous_zero ``+``=` `1` `# no. of continuous_zero by 1 ` ` ``else``: ` ` ``continuous_zero ``=` `0` `# else continuous_zero is 0 ` ` ` ` ``mod_sum ``+``=` `ord``(number[i]) ``-` `ord``(``'0'``) ` ` ``mod_sum ``%``=` `9` ` ``result ``+``=` `d[mod_sum] ` ` ``d[mod_sum] ``+``=` `1` `# increase d value of this mod_sum ` ` ``# subtract no. of cases where there ` ` ``# are only zeroes in substring ` ` ``result ``-``=` `continuous_zero ` ` ` ` ``return` `result ` ` ` `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` ``print``(count9s(``"01809"``)) ` ` ``print``(count9s(``"1809"``)) ` ` ``print``(count9s(``"4189"``)) ` ` ` `# This code is contributed by ` `# Sahil_Shelangia `
## C#
`// C# program to count substrings with recursive sum equal to 9 ` ` ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ``static` `int` `count9s(``string` `number) { ` ` ``int` `n = number.Length; ` ` ` ` ``// to store no. of previous encountered modular sums ` ` ``int``[] d = ``new` `int``[9]; ` ` ` ` ``// no. of modular sum(==0) encountered till now = 1 ` ` ``d[0] = 1; ` ` ``int` `result = 0; ` ` ` ` ``int` `mod_sum = 0, continuous_zero = 0; ` ` ``for` `(``int` `i = 0; i < n; i++) { ` ` ``if` `((number[i] - ``'0'``) == 0) ``// if number is 0 increase ` ` ``{ ` ` ``continuous_zero++; ``// no. of continuous_zero by 1 ` ` ``} ``else` `// else continuous_zero is 0 ` ` ``{ ` ` ``continuous_zero = 0; ` ` ``} ` ` ``mod_sum += (number[i] - ``'0'``); ` ` ``mod_sum %= 9; ` ` ``result += d[mod_sum]; ` ` ``d[mod_sum]++; ``// increase d value of this mod_sum ` ` ``// subtract no. of cases where there ` ` ``// are only zeroes in substring ` ` ``result -= continuous_zero; ` ` ``} ` ` ``return` `result; ` ` ``} ` ` ` `// driver program to test above function ` ` ``public` `static` `void` `Main() { ` ` ``Console.WriteLine(count9s(``"01809"``)); ` ` ``Console.WriteLine(count9s(``"1809"``)); ` ` ``Console.WriteLine(count9s(``"4189"``)); ` ` ``} ` `} `
Output:
```8
5
3
```
Time Complexity of the above program is O(n). Program also supports leading zeroes.
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# RS Aggarwal Solutions: Exercise 14D - Statistics Class 10 Notes | EduRev
## Class 10 : RS Aggarwal Solutions: Exercise 14D - Statistics Class 10 Notes | EduRev
``` Page 1
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
Page 2
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
Page 3
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.
Page 4
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 50 2 2
50 100 3 5
100 150 5 10
150 200 6 16
200 250 5 21
250 300 3 24
300 350 1 25
Total N = = 25
Now, N = 25 = 12.5.
The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150
200.
Thus, the median class is 150 200.
l = 150, h = 50, N = 25, f = 6 and cf = 10.
Now,
Median = l + × h
= 150 + × 50
= 150 + 20.83
= 170.83
Thus, the median is 170.83.
We know that,
Mode = 3(median) 2(mean)
= 3 × 170.83 2 × 169
= 512.49 338
= 174.49
Hence, Mean = 169, Median = 170.83 and Mode = 174.49
4. Find the mean, median and mode of the following data:
Marks
obtained
25 - 35 35 45 45 55 55 65 65 75 75 - 85
No. of
students
7 31 33 17 11 1
Sol:
To find the mean let us put the data in the table given below:
Marks
obtained
Number of students (fi) Class mark (x i) fi xi
25 35 7 30 210
35 45 31 40 1240
Page 5
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol:
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 50 2 2
50 100 3 5
100 150 5 10
150 200 6 16
200 250 5 21
250 300 3 24
300 350 1 25
Total N = = 25
Now, N = 25 = 12.5.
The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150
200.
Thus, the median class is 150 200.
l = 150, h = 50, N = 25, f = 6 and cf = 10.
Now,
Median = l + × h
= 150 + × 50
= 150 + 20.83
= 170.83
Thus, the median is 170.83.
We know that,
Mode = 3(median) 2(mean)
= 3 × 170.83 2 × 169
= 512.49 338
= 174.49
Hence, Mean = 169, Median = 170.83 and Mode = 174.49
4. Find the mean, median and mode of the following data:
Marks
obtained
25 - 35 35 45 45 55 55 65 65 75 75 - 85
No. of
students
7 31 33 17 11 1
Sol:
To find the mean let us put the data in the table given below:
Marks
obtained
Number of students (fi) Class mark (x i) fi xi
25 35 7 30 210
35 45 31 40 1240
45 55 33 50 1650
55 65 17 60 1020
65 75 11 70 770
75 85 1 80 80
Total
fi = 100 fi xi = 4970
Mean =
i i
i
i
i
f x
f
=
= 49.7
Thus, mean of the given data is 49.7.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
25 35 7 7
35 45 31 38
45 55 33 71
55 65 17 88
65 75 11 99
75 85 1 100
Total N = = 100
Now, N = 100 = 50.
The cumulative frequency just greater than 50 is 71 and the corresponding class is 45 55.
Thus, the median class is 45 55.
l = 45, h = 10, N = 100, f = 33 and cf = 38.
Now,
Median = l + × h
= 45 + × 10
= 45 + 3.64
= 48.64
Thus, the median is 48.64.
We know that,
Mode = 3(median) 2(mean)
= 3 × 48.64 2 × 49.70
= 145.92 99.4
= 46.52
Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52
```
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## Mathematics (Maths) Class 10
60 videos|363 docs|103 tests
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; | 5,883 | 13,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-25 | latest | en | 0.730061 |
https://www.nagwa.com/en/worksheets/892186921282/ | 1,575,738,389,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540500637.40/warc/CC-MAIN-20191207160050-20191207184050-00500.warc.gz | 812,329,782 | 125,168 | # Worksheet: Quartiles of a Data Set
In this worksheet, we will practice finding the median and upper and lower quartiles of a data set.
Q1:
Determine the median and quartiles of the following set of data: 56, 55, 90, 50, 41, 84, 68, 75, 92, 50, and 71.
• AThe median is 68, and the quartiles are 52.5 and 79.5.
• BThe median is 69.5, and the quartiles are 50 and 84.
• CThe median is 68, and the quartiles are 50 and 84.
• DThe median is 71, and the quartiles are 55 and 90.
Q2:
Determine the median and quartiles of the following set of data: 1.8, 1.6, 1.1, 2.4, 0.6, 1.3, 2.9, 1.5, 0.4, 1.9, 2.9.
• AThe median is 1.6, and the quartiles are 1.3 and 2.9.
• BThe median is 1.6, and the quartiles are 1.1 and 2.4.
• CThe median is 1.8, and the quartiles are 1.1 and 2.4.
• DThe median is 1.6, and the quartiles are 0.6 and 1.9.
Q3:
Determine the upper and the lower quartiles of the following set of data: 114, 103, 50, 52, 95, 103, 93, 53, 65, 57, 52, 89, 111, 89, and 96.
• Alower quartile , upper quartile
• Blower quartile , upper quartile
• Clower quartile , upper quartile
• Dlower quartile , upper quartile
• Elower quartile , upper quartile
Q4:
The table shows the capacities of 6 sports stadiums around the world. Describe how the lower quartile will be affected if Emirates Stadium, London, United Kingdom, which has a capacity of 60,338, is included in the data.
Gelora Sriwijaya Stadium Palembang, Indonesia 40,000
Borg El Arab Stadium Alexandria, Egypt 86,000
Stamford Bridge London, United Kingdom 42,055
• AThe lower quartile will remain unchanged at 42,055.
• BThe lower quartile will increase from 42,055 to 42,058.
• CThe lower quartile will increase from 42,055 to 86,047.
• DThe lower quartile will remain unchanged at 42,058.
• EThe lower quartile will decrease from 42,058 to 42,055.
Q5:
Determine the median and quartiles of the following set of data: 1,350, 1,400, 1,250, 1,050, 1,450, 1,150, 1,000.
• AThe median is 1,250, and the quartiles are 1,150 and 1,350.
• BThe median is 1,350, and the quartiles are 1,050 and 1,400.
• CThe median is 1,250, and the quartiles are 1,100 and 1,375.
• DThe median is 1,250, and the quartiles are 1,050 and 1,400.
Q6:
Determine the upper and the lower quartiles of the following set of data: 19.4, 42.1, 56.1, 27.5, 5.3, 49.9, 48.8, 44.3, 13.1, 15.4, 20.1, 4.1, 38.1, 33.8, and 41.5.
• Alower quartile = 4.1, upper quartile = 44.3
• Blower quartile = 33.8, upper quartile = 44.3
• Clower quartile = 4.1, upper quartile = 33.8
• Dlower quartile = 15.4, upper quartile = 33.8
• Elower quartile = 15.4, upper quartile = 44.3
Q7:
David’s history test scores are 74, 96, 85, 90, 71, and 98. Determine the upper and lower quartiles of his scores.
• Alower quartile = 74, upper quartile = 96
• Blower quartile = 87.5, upper quartile = 96
• Clower quartile = 74, upper quartile = 87.5
• Dlower quartile = 71, upper quartile = 98
• Elower quartile = 96, upper quartile = 74
Q8:
Matthew’s scores in a card game were 4, , , , , 7, and 0. Determine the lower quartile and upper quartile of his scores.
• Alower quartile = 4, upper quartile =
• Blower quartile = , upper quartile = 4
• Clower quartile = , upper quartile = 4
• Dlower quartile = , upper quartile = 4
• Elower quartile = , upper quartile =
Q9:
Select the pair of data sets which have the same median and quartiles, but different ranges.
• A30, 41, 34, 34, 35, 49 and 31, 35, 42, 50, 35, 36
• B58, 68, 57, 58, 54, 69 and 58, 69, 58, 54, 59, 57
• C83, 77, 73, 88, 71, 84 and 69, 68, 58, 57, 62, 73
• D16, 15, 25, 20, 16, 28 and 17, 15, 15, 28, 27, 20
• E10, 10, 9, 11, 7, 8 and 6, 8, 10, 21, 9, 10
Q10:
The table shows the savings of a group of 8 children. Determine the lower and upper quartiles of the data.
Child Savings in Dollars Child 1 Child 2 Child 3 Child 4 Child 5 Child 6 Child 7 Child 8 35.00 30.40 18.30 23.40 29.50 21.50 33.50 18.40
• ALower quartile = \$31.95, upper quartile = \$19.95
• BLower quartile = \$19.95, upper quartile = \$31.95
• CLower quartile = \$19.95, upper quartile = \$26.45
• DLower quartile = \$26.45, upper quartile = \$31.95
• ELower quartile = \$18.30, upper quartile = \$35.00
Q11:
The table shows the number of students who partake in different physical activities. Find the lower and upper quartiles of the data.
Activity Number of Students Basketball Cycling Soccer Running Rope Jumping Swimming Climbing 7 12 6 14 15 6 11
• Alower quartile: 4, upper quartile: 12
• Blower quartile: 6, upper quartile: 11
• Clower quartile: 6, upper quartile: 14
• Dlower quartile: 11, upper quartile: 14
• Elower quartile: 6, upper quartile: 15
Q12:
The table shows DVD prices in dollars at various stores. What are the lower and upper quartiles of the data?
19.86 23.82 17.27 24.01 23.56 18.74 16.79 15.57 23.39 20.86 23.04 20.25
• Alower quartile = 18.74, upper quartile = 20.25
• Blower quartile = 23.475, upper quartile = 18.005
• Clower quartile = 20.555, upper quartile = 23.475
• Dlower quartile = 18.005, upper quartile = 20.555
• Elower quartile = 18.005, upper quartile = 23.475
Q13:
The number of goals scored by 12 soccer players during a season is 10, 9, 14, 20, 19, 20, 9, 5, 12, 9, 19, and 7. State whether the following statement is true or false: About three-fourths of the players scored 19 or more.
• Afalse
• Btrue
Q14:
The line plot shows the magnitudes of several recent earthquakes. Determine the upper and lower quartiles.
• ALower quartile , upper quartile
• BLower quartile , upper quartile
• CLower quartile , upper quartile
• DLower quartile , upper quartile
• ELower quartile , upper quartile
Q15:
The times taken for a bus journey between two towns are normally distributed with mean 28 minutes and standard deviation 4 minutes. Calculate the lower and upper quartiles of the times taken. Round your answer to the nearest integer.
• Alower quartile = 20, upper quartile = 36
• Blower quartile = 22, upper quartile = 34
• Clower quartile = 25, upper quartile = 31
• Dlower quartile = 27, upper quartile = 29
• Elower quartile = 26, upper quartile = 32
Q16:
The number of Bonus Bugs won by each of 15 students in the first level of a computer game tournament was recorded. The results are in the table below.
Find the median (Q2) and the lower and upper quartiles (Q1 and Q3) for the number of Bonus Bugs won.
• AMedian = 22, Q1 = 29, Q3 = 17
• BMedian = 15, Q1 = 17, Q3 = 31
• CMedian = 15, Q1 = 22, Q3 = 29
• DMedian = 21, Q1 = 18, Q3 = 31
• EMedian = 22, Q1 = 17, Q3 = 29
If the organizers of the tournament decide that the top 25% of students can compete in Level 2, above what number of Bonus Bugs must a student win to go to the next level?
• A15 or above
• B29 or above
• C31 or above
• D17 or above
• E22 or above
Q17:
In the second year of a computer game tournament, there were forty-two participants and the number of Bonus Bugs each one won in level 1 was recorded. The data are shown in the graph below where each bug represents one participant. | 2,478 | 6,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-51 | longest | en | 0.856965 |
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AND - PowerPoint PPT Presentation
AND. Active Learning Lecture Slides For use with Classroom Response Systems Chapter 12 Probability. Of 12 children playing at the playground, 4 are playing on the swing set. Determine the empirical probability that the next child to the playground will play on the swing set. a. c. b. d.
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Presentation Transcript
Active Learning Lecture Slides
For use with Classroom Response Systems
Chapter 12
Probability
Of 12 children playing at the playground, 4 are playing on the swing set. Determine the empirical probability that the next child to the playground will play on the swing set.
a.
c.
b.
d.
Of 12 children playing at the playground, 4 are playing on the swing set. Determine the empirical probability that the next child to the playground will play on the swing set.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If one sheet of paper is selected at random from the box, determine the probability that the number selected is odd.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If one sheet of paper is selected at random from the box, determine the probability that the number selected is odd.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If one sheet of paper is selected at random from the box, determine the probability that the number selected is less than 3.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If one sheet of paper is selected at random from the box, determine the probability that the number selected is less than 3.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If one sheet of paper is selected at random from the box, determine the probability that the number selected is greater than 5 or even.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If one sheet of paper is selected at random from the box, determine the probability that the number selected is greater than 5 or even.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If one sheet of paper is selected at random from the box, determine the probability that the number selected is odd and less than 4.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If one sheet of paper is selected at random from the box, determine the probability that the number selected is odd and less than 4.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If two sheets of paper are selected at random, without replacement, from the box, determine the probability that both numbers are odd.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If two sheets of paper are selected at random, without replacement, from the box, determine the probability that both numbers are odd.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If two sheets of paper are selected at random, without replacement, from the box, determine the probability that both numbers are greater than 7.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If two sheets of paper are selected at random, without replacement, from the box, determine the probability that both numbers are greater than 7.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If two sheets of paper are selected at random, without replacement, from the box, determine the probability that the first number is even and the second number is odd.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If two sheets of paper are selected at random, without replacement, from the box, determine the probability that the first number is even and the second number is odd.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If two sheets of paper are selected at random, without replacement, from the box, determine the probability that the first number is greater than 3 and the second number is less than 3.
a.
c.
b.
d.
Each of the numbers 0-9 is written on a sheet of paper and the ten sheets of paper are placed in a box. If two sheets of paper are selected at random, without replacement, from the box, determine the probability that the first number is greater than 3 and the second number is less than 3.
a.
c.
b.
d.
One card is selected at random from a standard deck of 52 cards. Determine the probability that the card selected is a club or a picture card.
a.
c.
b.
d.
One card is selected at random from a standard deck of 52 cards. Determine the probability that the card selected is a club or a picture card.
a.
c.
b.
d.
One die is rolled and one colored chip - black or white cards. Determine the probability that the card selected is a club or a picture card. - is selected at random. Use the counting principle to determine the number of sample points in the sample space.
a. 6
c. 12
b. 8
d. 10
One die is rolled and one colored chip - black or white cards. Determine the probability that the card selected is a club or a picture card. - is selected at random. Use the counting principle to determine the number of sample points in the sample space.
a. 6
c. 12
b. 8
d. 10
One die is rolled and one colored chip - black or white cards. Determine the probability that the card selected is a club or a picture card. - is selected at random. Determine the probability of obtaining the number 3 and the color black.
a.
c.
b.
d.
One die is rolled and one colored chip - black or white cards. Determine the probability that the card selected is a club or a picture card. - is selected at random. Determine the probability of obtaining the number 3 and the color black.
a.
c.
b.
d.
One die is rolled and one colored chip - black or white cards. Determine the probability that the card selected is a club or a picture card. - is selected at random. Determine the probability of obtaining an even number and the color white.
a.
c.
b.
d.
One die is rolled and one colored chip - black or white cards. Determine the probability that the card selected is a club or a picture card. - is selected at random. Determine the probability of obtaining an even number and the color white.
a.
c.
b.
d.
One die is rolled and one colored chip - black or white cards. Determine the probability that the card selected is a club or a picture card. - is selected at random. Determine the probability of obtaining a number less than 3 and the color white.
a.
c.
b.
d.
One die is rolled and one colored chip - black or white cards. Determine the probability that the card selected is a club or a picture card. - is selected at random. Determine the probability of obtaining a number less than 3 and the color white.
a.
c.
b.
d.
A serial number is to consist of seven digits. Determine the number of serial numbers possible if the first two numbers cannot be 0 or 1 and repetition is permitted.
a. 376,320
b. 2,83,401
c. 6,400,000
d. 10,000,000
A serial number is to consist of seven digits. Determine the number of serial numbers possible if the first two numbers cannot be 0 or 1 and repetition is permitted.
a. 376,320
b. 2,83,401
c. 6,400,000
d. 10,000,000
The local elementary school cafeteria offered ham sandwiches and pizza for lunch one day. The number of boys and girls who ate either a ham sandwich or pizza were recorded. The results are shown below.
If one of these students is selected at random, determine the probability thatthe student is a boy.
a. b. c. d.
If one of these students is selected at random, determine the probability thatthe student is a boy.
a. b. c. d.
If one of these students is selected at random, determine the probability thatthe student ate pizza for lunch.
a. b. c. d.
If one of these students is selected at random, determine the probability thatthe student ate pizza for lunch.
a. b. c. d.
If one of these students is selected at random, determine the probability that the person ate a ham sandwich for lunch, given that they are a girl.
a. b. c. d.
If one of these students is selected at random, determine the probability that the person ate a ham sandwich for lunch, given that they are a girl.
a. b. c. d.
If one of these students is selected at random, determine the probability that the person is a boy, given that they ate pizza for lunch.
a. b. c. d.
If one of these students is selected at random, determine the probability that the person is a boy, given that they ate pizza for lunch.
a. b. c. d.
At the bakery, a box of cookies is made by selecting four cookies from the six types of cookie - chocolate chip, oatmeal raisin, sugar, peanut butter, butterscotch, and chocolate. In how many ways can a box of cookies be put together?
a. 360
b. 30
c. 15
d. 6
At the bakery, a box of cookies is made by selecting four cookies from the six types of cookie - chocolate chip, oatmeal raisin, sugar, peanut butter, butterscotch, and chocolate. In how many ways can a box of cookies be put together?
a. 360
b. 30
c. 15
d. 6
A box contains a total of 120 folders, of which 30 are red. If you select 2 at random, determine the probability thatboth folders are red.
a.
c.
b.
d.
A box contains a total of 120 folders, of which 30 are red. If you select 2 at random, determine the probability thatboth folders are red.
a.
c.
b.
d.
A box contains a total of 120 folders, of which 30 are red. If you select 2 at random, determine the probability thatat least one folder is not red.
a.
c.
b.
d.
A box contains a total of 120 folders, of which 30 are red. If you select 2 at random, determine the probability thatat least one folder is not red.
a.
c.
b.
d. | 2,687 | 11,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2017-51 | latest | en | 0.896534 |
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#### Resources tagged with 2D shapes and their properties similar to Board Block Challenge:
Filter by: Content type:
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Challenge level:
### There are 53 results
Broad Topics > Angles, Polygons, and Geometrical Proof > 2D shapes and their properties
### Board Block Challenge
##### Age 7 to 11 Challenge Level:
Choose the size of your pegboard and the shapes you can make. Can you work out the strategies needed to block your opponent?
### Overlapping Circles
##### Age 7 to 11 Challenge Level:
What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes?
### Overlapping Again
##### Age 7 to 11 Challenge Level:
What shape is the overlap when you slide one of these shapes half way across another? Can you picture it in your head? Use the interactivity to check your visualisation.
### Shapely Tiling
##### Age 7 to 11 Challenge Level:
Use the interactivity to make this Islamic star and cross design. Can you produce a tessellation of regular octagons with two different types of triangle?
### Regular Rings 2
##### Age 7 to 11 Challenge Level:
What shape is made when you fold using this crease pattern? Can you make a ring design?
### Two by One
##### Age 7 to 11 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
### Shaping Up
##### Age 7 to 11 Challenge Level:
Are all the possible combinations of two shapes included in this set of 27 cards? How do you know?
### Four Triangles Puzzle
##### Age 5 to 11 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Jig Shapes
##### Age 5 to 11 Challenge Level:
Can you each work out what shape you have part of on your card? What will the rest of it look like?
### Circle Panes
##### Age 7 to 11 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
### Where Are They?
##### Age 7 to 11 Challenge Level:
Use the isometric grid paper to find the different polygons.
### Name That Triangle!
##### Age 7 to 11 Challenge Level:
Can you sketch triangles that fit in the cells in this grid? Which ones are impossible? How do you know?
### Fault-free Rectangles
##### Age 7 to 11 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Bow Tie
##### Age 11 to 14 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
### Sports Equipment
##### Age 7 to 11 Challenge Level:
If these balls are put on a line with each ball touching the one in front and the one behind, which arrangement makes the shortest line of balls?
### Poly Plug Rectangles
##### Age 5 to 14 Challenge Level:
The computer has made a rectangle and will tell you the number of spots it uses in total. Can you find out where the rectangle is?
### LOGO Challenge - Circles as Animals
##### Age 11 to 16 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### Count the Trapeziums
##### Age 7 to 11 Challenge Level:
How many trapeziums, of various sizes, are hidden in this picture?
### Sorting Logic Blocks
##### Age 5 to 11 Challenge Level:
This interactivity allows you to sort logic blocks by dragging their images.
### Lafayette
##### Age 7 to 11 Challenge Level:
What mathematical words can be used to describe this floor covering? How many different shapes can you see inside this photograph?
### 2001 Spatial Oddity
##### Age 11 to 14 Challenge Level:
With one cut a piece of card 16 cm by 9 cm can be made into two pieces which can be rearranged to form a square 12 cm by 12 cm. Explain how this can be done.
### Rolling Around
##### Age 11 to 14 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### Playground Snapshot
##### Age 7 to 14 Challenge Level:
The image in this problem is part of a piece of equipment found in the playground of a school. How would you describe it to someone over the phone?
### Efficient Cutting
##### Age 11 to 14 Challenge Level:
Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end.
### Like a Circle in a Spiral
##### Age 7 to 16 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Triangular Hexagons
##### Age 7 to 11 Challenge Level:
Investigate these hexagons drawn from different sized equilateral triangles.
### Torn Shapes
##### Age 7 to 11 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
### LOGO Challenge 6 - Triangles and Stars
##### Age 11 to 16 Challenge Level:
Recreating the designs in this challenge requires you to break a problem down into manageable chunks and use the relationships between triangles and hexagons. An exercise in detail and elegance.
### What Shape?
##### Age 7 to 14 Challenge Level:
This task develops spatial reasoning skills. By framing and asking questions a member of the team has to find out what mathematical object they have chosen.
### Shaping It
##### Age 5 to 11 Challenge Level:
These pictures were made by starting with a square, finding the half-way point on each side and joining those points up. You could investigate your own starting shape.
### Witch's Hat
##### Age 11 to 16 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Gym Bag
##### Age 11 to 16 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### What Shape for Two
##### Age 7 to 14 Challenge Level:
'What Shape?' activity for adult and child. Can you ask good questions so you can work out which shape your partner has chosen?
### From One Shape to Another
##### Age 7 to 14
Read about David Hilbert who proved that any polygon could be cut up into a certain number of pieces that could be put back together to form any other polygon of equal area.
### Arclets Explained
##### Age 11 to 16
This article gives an wonderful insight into students working on the Arclets problem that first appeared in the Sept 2002 edition of the NRICH website.
### LOGO Challenge 11 - More on Circles
##### Age 11 to 16 Challenge Level:
Thinking of circles as polygons with an infinite number of sides - but how does this help us with our understanding of the circumference of circle as pi x d? This challenge investigates. . . .
### LOGO Challenge 12 - Concentric Circles
##### Age 11 to 16 Challenge Level:
Can you reproduce the design comprising a series of concentric circles? Test your understanding of the realtionship betwwn the circumference and diameter of a circle.
### Squaring the Circle
##### Age 11 to 14 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### Square Areas
##### Age 11 to 14 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
### Not So Little X
##### Age 11 to 14 Challenge Level:
Two circles are enclosed by a rectangle 12 units by x units. The distance between the centres of the two circles is x/3 units. How big is x?
### Floored
##### Age 11 to 14 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### Lying and Cheating
##### Age 11 to 14 Challenge Level:
Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
### LOGO Challenge 2 - Diamonds Are Forever
##### Age 7 to 16 Challenge Level:
The challenge is to produce elegant solutions. Elegance here implies simplicity. The focus is on rhombi, in particular those formed by jointing two equilateral triangles along an edge.
### LOGO Challenge 10 - Circles
##### Age 11 to 16 Challenge Level:
In LOGO circles can be described in terms of polygons with an infinite (in this case large number) of sides - investigate this definition further.
### LOGO Challenge 8 - Rhombi
##### Age 7 to 16 Challenge Level:
Explore patterns based on a rhombus. How can you enlarge the pattern - or explode it?
### Square Pegs
##### Age 11 to 14 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Circles, Circles Everywhere
##### Age 7 to 14
This article for pupils gives some examples of how circles have featured in people's lives for centuries.
### Hex
##### Age 11 to 14 Challenge Level:
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
### First Forward Into Logo 4: Circles
##### Age 7 to 16 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### Opposite Vertices
##### Age 11 to 14 Challenge Level:
Can you recreate squares and rhombuses if you are only given a side or a diagonal? | 2,219 | 9,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-22 | latest | en | 0.899065 |
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# Solids - Volumes and Surfaces
## The volume and surface of solids like rectangular prism, cylinders, pyramids, cones and spheres - online calculator.
### Cube
#### Volume
V = a3 (1)
where
V = volume (m3, ft3)
a = side (m, ft)
#### Surface Area
A0 = 6 a2 (1b)
where
A0 = surface area (m2, ft2)
#### Diagonal
d = a 31/2 (1c)
where
d = inside diagonal (m, ft)
#### Diagonal of Cube Face
ds = a 21/2 (1d)
### Cuboid - Square Prism
#### Volume
V = a b c (2)
where
V = volume of solid (m3, ft3)
a = length of rectangular prism (m, ft)
b = width of rectangular prism (m, ft)
c = height of rectangular prism (m, ft)
#### Diagonal
d = (a2 + b2 + c2)1/2 (2b)
#### Surface Area
A0 = 2 (a b + a c + b c) (2c)
where
A0 = surface area of solid (m2, ft2)
length
width
height
Volume:
Surface:
### Parallelepiped
#### Volume
V = A1 h (3a)
where
A1 = side area (m2, ft2)
### Related Sketchup Components from The Engineering ToolBox
• Geometric Figures - Cylinders, Boxes, Cones, Planes, Spheres, Lines, Curves and more..
- free Engineering ToolBox plugin for use with the amazing Sketchup 3D drawing/model application.
### Cylinder
#### Volume
V = π r2 h = (π / 4) d2 (4a)
where
d = diameter of cylinder (m, ft)
r = radius of cylinder (m, ft)
h = height of cylinder (m, ft)
#### Surface
A = 2 π r h + 2 π r2 (4b)
height
Volume:
Surface:
### Hollow Cylinder
#### Volume
V = π/4 h (D2 - d2) (5)
### Pyramid
#### Volume
V = 1/3 h A1 (6)
where
A1 = area of base (m2, ft2)
h = perpendicular height of pyramid (m, ft)
#### Surface
A = ∑ sum of areas of triangles forming sides + Ab (6b)
where
the surface areas of the triangular faces will have different formulas for different shaped bases
area of base
perpendicular height
Volume:
### Volume
V = h/3 ( A1 + A2 + (A1 A2)1/2) (7)
### Cone
#### Volume
V = 1/3 π r2 h (8a)
where
r = radius of cone base (m, ft)
h = height of cone (m, ft)
#### Surface
A = π r l + π r2 (8b)
where
l = (r2 + h2)1/2 = length of cone side (m, ft)
height
Volume:
Surface:
#### Side
m = (h2 + r2)1/2 (8c)
A2 / A1 = x2 / h2 (8d)
### Frustum of Cone
#### Volume
V = π/12 h (D2 + D d + d2) (9a)
m = ( ( (D - d) / 2 )2 + h2)1/2 (9c)
### Sphere
#### Volume
V = 4/3 π r3
= 1/6 π d3 (10a)
where
r = radius of sphere (m, ft)
A = 4 π r2
= π d2 (10b)
Volume:
Surface:
#### Spheres with Fractional Diameters - Surface Area and Volume
Fraction Diameter
- d –
(inch)
Decimal Diameter
- d –
(inch)
– r –
(inch)
Surface Area
- A –
(in2)
Volume
- V -
(in3)
1/64 0.015625 0.007813 0.0007670 0.0000020
1/32 0.031250 0.015625 0.0030680 0.0000160
3/64 0.046875 0.023438 0.0069029 0.0000539
1/64 0.062500 0.031250 0.0122718 0.0001278
5/64 0.078125 0.039063 0.0191748 0.0002497
3/32 0.093750 0.046875 0.0276117 0.0004314
7/64 0.109375 0.054688 0.0375825 0.0006851
1/8 0.125000 0.062500 0.0490874 0.0010227
9/64 0.140625 0.070313 0.0621262 0.0014561
5/32 0.156250 0.078125 0.0766990 0.0019974
11/64 0.171875 0.085938 0.0928058 0.0026585
3/16 0.187500 0.093750 0.1104466 0.0034515
13/64 0.203125 0.101563 0.1296214 0.0043882
7/32 0.218750 0.109375 0.1503301 0.0054808
15/64 0.234375 0.117188 0.1725728 0.0067411
1/4 0.250000 0.125000 0.1963495 0.0081812
17/64 0.265625 0.132813 0.2216602 0.0098131
9/32 0.281250 0.140625 0.2485049 0.0116487
19/64 0.296875 0.148438 0.2768835 0.0137000
5/16 0.312500 0.156250 0.3067962 0.0159790
21/64 0.328125 0.164063 0.3382428 0.0184977
11/32 0.343750 0.171875 0.3712234 0.0212680
23/64 0.359375 0.179688 0.4057379 0.0243020
3/8 0.375000 0.187500 0.4417865 0.0276117
25/64 0.390625 0.195313 0.4793690 0.0312089
13/32 0.406250 0.203125 0.5184855 0.0351058
27/64 0.421875 0.210938 0.5591360 0.0393142
7/16 0.437500 0.218750 0.6013205 0.0438463
29/64 0.453125 0.226563 0.6450389 0.0487139
15/32 0.468750 0.234375 0.6902914 0.0539290
31/64 0.484375 0.242188 0.7370778 0.0595037
1/2 0.500000 0.250000 0.7853982 0.0654498
33/64 0.515625 0.257813 0.8352525 0.0717795
17/32 0.531250 0.265625 0.8866409 0.0785047
35/64 0.546875 0.273438 0.9395632 0.0856373
9/16 0.562500 0.281250 0.9940196 0.0931893
37/64 0.578125 0.289063 1.0500098 0.1011728
19/32 0.593750 0.296875 1.1075341 0.1095997
39/64 0.609375 0.304688 1.1665924 0.1184820
5/8 0.625000 0.312500 1.2271846 0.1278317
41/64 0.640625 0.320313 1.2893109 0.1376608
21/32 0.656250 0.328125 1.3529711 0.1479812
43/64 0.671875 0.335938 1.4181652 0.1588050
11/16 0.687500 0.343750 1.4848934 0.1701440
45/64 0.703125 0.351563 1.5531555 0.1820104
23/32 0.718750 0.359375 1.6229517 0.1944161
47/64 0.734375 0.367188 1.6942818 0.2073730
3/4 0.750000 0.375000 1.7671459 0.2208932
49/64 0.765625 0.382813 1.8415439 0.2349887
25/32 0.781250 0.390625 1.9174760 0.2496714
51/64 0.796875 0.398438 1.9949420 0.2649532
13/16 0.812500 0.406250 2.0739420 0.2808463
53/64 0.828125 0.414063 2.1544760 0.2973626
27/32 0.843750 0.421875 2.2365440 0.3145140
55/64 0.859375 0.429688 2.3201459 0.3323126
7/8 0.875000 0.437500 2.4052819 0.3507703
57/64 0.890625 0.445313 2.4919518 0.3698991
29/32 0.906250 0.453125 2.5801557 0.3897110
59/64 0.921875 0.460938 2.6698936 0.4102180
15/16 0.937500 0.468750 2.7611654 0.4314321
61/64 0.953125 0.476563 2.8539713 0.4533652
31/32 0.968750 0.484375 2.9483111 0.4760294
63/64 0.984375 0.492188 3.0441849 0.4994366
1 1.000000 0.500000 3.1415927 0.5235988
### Zone of a Sphere
V = π/6 h (3a2 + 3b2 + h) (11a)
Am = 2 π r h (11b)
A0 = π (2 r h + a2 + b2) (11c)
### Segment of a Sphere
V = π/6 h (3/4 s2 + h2
= π h2 (r - h/3) (12a)
Am = 2 π r h
π/4 (s2 + 4 h2) (12b)
### Sector of a Sphere
V = 2/3 π r2 h (13a)
A0 = π/2 r (4 h + s) (13b)
### Sphere with Cylindrical Boring
V = π/6 h3 (14a)
A0 = 4 π ((R + r)3 (R - r))1/2
= 2 π h (R + r) (14b)
h = 2 (R2 - r2)1/2 (14c)
### Sphere with Conical Boring
V = 2/3 π R2 h (15a)
A0 = 2 π R (h + (R2 - h2/4)1/2) (15b)
h = 2 (R2 - r2)1/2 (15c)
### Torus
V = π2/4 D d2 (16a)
A0 = π2 D d (16b)
### Sliced Cylinder
V = π/4 d2 h
= π r2 ((h1 + h2) / 2) (17a)
Am = π d h
= 2 π r ((h1 + h2) / 2) (17b)
where
Am = side walls area
A0 = π r (h1 + h2 + r + (r2 + (h1 - h2)2/4)1/2) (17c)
where
A0 = surface area
### Ungula
V = 2/3 r2 h (18a)
Am = 2 r h (18b)
A0 = Am + π/2 r2 + π/2 r (r2 + h2)1/2 (18c)
### Barrel
V ≈ π/12 h (2 D2 + d2) (19a)
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9.19.12 | 3,682 | 9,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-40 | latest | en | 0.497903 |
https://mainframeperformancetopics.com/2020/01/02/square-roots-in-dfsort/ | 1,675,008,022,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00381.warc.gz | 396,400,968 | 23,570 | # Square Roots In DFSORT
(Originally posted 2012-07-02.)
DFSORT’s Arithmetic operators can do many things but the one thing they can’t do is take the square root of a number.
You might think that’s minor but it means you can’t calculate a Standard Deviation. (Variance is fine but dimensionally not so nice when used in conjunction with the Mean.) And I needed Standard Deviation in a real live customer situation.
So I set out to "roll my own". And here’s how I did it…
There is a well known technique for calculating square roots – the Babylonian Method (or Heron’s Method). It’s not terribly advanced maths, frankly: I think I was taught it at about age 14. It goes like this:
1. Call your number whose square root is to be found a, and pick a first guess for the root. We’ll call this quess x0.
2. Write x1 = (a + a/x0)/2. In other words xn+1 is the average of a and a/xn.
3. Iterate until done. We’ll talk about what "done" means later.
As you can see it’s a really very simple method and converges reasonably quickly. And it’s not difficult to program, except for one thing: DFSORT doesn’t have a looping capability. (Yes it loops over the input records, but not in the way I mean.)
Before I show you some code let me talk about some considerations:
• Because DFSORT’s arithmetic is integer I describe here taking the integer square root of an integer field value. So, for example, it yields sqrt(5)=2.
• All the arithmetic on the way to finding the root is integer also.
• The code takes square roots of non-negative integers: No imaginary roots here.
• The method returns positive roots: As you probably know roots come in pairs, differing only by a sign.
• The algorithm requires you to provide a first estimate (x0). My first take was a/2. But in the case of a=1 this leads to a root of 0 straightaway – which is particularly unhelpful. So I had to special-case a=1. So the take I went with in the end is x0=a. Not generally a brilliant first guess but avoids the special case.
• Because this is integer arithmetic completion can’t be only when sqrt(a)*sqrt(a)=a as in most cases this condition isn’t going to be met. Instead I use xn+1=xn as completion: If the two are equal the algorithm isn’t going to produce a different xn+2.
A reasonable question to ask is whether integer square roots are useful…
A standard technique with DFSORT’s Arithmetic operators is to boost the integer values to represent the number of decimal places required. So, for example, to add together 3.14 and 2.72 you’d be storing them as 314 and 272. Then you’d add these integers and use EDIT=(I.TT) to display them as "4.86". So if you wanted to take the square root of 3.1416 you’d multiply by 10,000 and then apply my technique, using EDIT=(IIII.TT) to move the decimal point back.
### Structure Of The Example
This is a standard DFSORT invocation. Rather than give you the usual scaffolding I’ll show you the main components:
• DFSORT Symbols where possible to help keep it clear.
• COPY (though it could’ve been SORT or MERGE).
• INREC to parse the free-form input data.
• OUTREC with multiple IFTHEN clauses to compute the square root.
• OUTFIL to format the output lines.
You could, in fact, structure this program differently. It might (though I haven’t worked through it) combine everything into a single INREC statement (apart from the COPY specification).
### Some Sample Data
In my SORTIN I have the following lines:
```100
5
1
27
3
4
999
999999
99999999
```
In other words a single field in each record in EBCDIC (rather than binary) form. (You can generalise all that follows with a small amount of thought.)
(In my test case this is instream data – so it’s fixed-length records and hence no RDW).
### Parsing EBCDIC Numbers.
In this simple case I created a set of records with binary representations of the numbers using INREC:
```OPTION COPY
INREC FIELDS=(1,10,UFF,TO=BI)
```
Here the UFF does the parsing and it parses the 10 bytes starting in position 1, with the output being 4-byte BI fields.
### A SYMNAMES Deck To Simplify Things
You may know you can define symbols with DFSORT. Here’s the deck I used in this example:
```//SYMNAMES DD *
INPUT,1,4,BI
SQRT,*,4,BI
PREV,*,4,BI
/*
```
(You’ll want a SYMNOUT DD to go with it.)
• INPUT maps the integer value created by INREC. It’s a, the number whose square root we’re seeking.
• SQRT maps the field where successive estimates of the root are stored.
• PREV maps the field where the previous estimate is stored – so we can tell if an iteration has changed the estimate of the root.
### Taking The Square Root
In my example all the work is done in a series of IFTHEN clauses in an OUTREC statement:
1. A single WHEN=INIT clause initialises the estimate (SQRT) and clears the previous estimate (PREV).
2. Repeated WHEN=(SQRT,NE,PREV) clauses compute successive approximations – but only if the last two estimates weren’t the same.
Here’s the code:
``` OUTREC IFTHEN=(WHEN=INIT,
OVERLAY=(SQRT,INPUT,X'00000000')),
IFTHEN=(WHEN=(SQRT,NE,PREV),HIT=NEXT,
TO=BI,LENGTH=4)),
IFTHEN=(WHEN=(SQRT,NE,PREV),HIT=NEXT,
TO=BI,LENGTH=4)),
...
IFTHEN=(WHEN=(SQRT,NE,PREV),HIT=NEXT,
TO=BI,LENGTH=4))
```
This you’ll recognise as an unrolled loop. The "…" says you can repeat the previous IFTHEN clause as many times as you like.
A good question would be "how many times?" I found good convergence after 15. For example, the 99999999 value went to 10028 in 15 iterations and 10000 in 16. You could pick 20 and be sure of getting an accurate value.
Let’s talk about control flow through the queue of IFTHEN clauses…
• The WHEN=INIT is always executed.
• HIT=NEXT on a WHEN=(logexp) clause means "even if this test succeeds proceed to the next WHEN=(logexp) clause having executed this one’s OVERLAY".
• Even if a WHEN=(logexp)’s logical expression is false all subsequent WHEN=(logexp) clauses are processed – even if, as in this case, they’ll evaluate to false. So this isn’t really a "WHILE(logexp) DO … END" construct (but you can construct the logexp so it works that way).
In the past I’ve likened multiple IFTHEN clauses to a pipeline. This example shows why.
### Printing The Results
The following is a quite complicated piece of report formatting. Its aim is to format numbers and then squeeze out the resulting leading blanks. At the same time it has to preserve blanks in text.
```OUTFIL IFTHEN=(WHEN=INIT,
BUILD=(C'x=',INPUT,EDIT=(IIIIIIIIIIT),
C'_sqrt(x)=',SQRT,EDIT=(IIIIIIIIIIIT),
C'_sqrt(x)*sqrt(x)=',SQRT,MUL,SQRT,EDIT=(IIIIIIIIIIIT),
C'_prev_est=',PREV,EDIT=(IIIIIIIIIIIT))),
IFTHEN=(WHEN=INIT,
BUILD=(1,90,SQZ=(SHIFT=LEFT,PAIR=QUOTE))),
IFTHEN=(WHEN=INIT,
FINDREP=(IN=C'_',OUT=C' '))
```
Let’s examine it more closely:
• The first WHEN=INIT takes four numbers and formats them:
• The number we wanted to find the square root of,
• The square root,
• What happens when you square that back up again,
• The previous estimate of the square root.
These are formatted with leading zeroes turned into blanks and some text is wrapped around them (with underscores standing in for blanks).
• The second WHEN=INIT squeezes all the blanks out. This is why I used underscores in the wrapping text.
• The third WHEN=INIT turns all the underscores into blanks.
### Output Using The Sample Data
The output below isn’t particularly pretty. I wanted to use the OUTFIL code I’ve shown you to demonstrate the ability to selectively squeeze blanks out.
```x=100 sqrt(x)=10 sqrt(x)*sqrt(x)=100 prev est=10
x=5 sqrt(x)=2 sqrt(x)*sqrt(x)=4 prev est=2
x=1 sqrt(x)=1 sqrt(x)*sqrt(x)=1 prev est=1
x=27 sqrt(x)=5 sqrt(x)*sqrt(x)=25 prev est=5
x=3 sqrt(x)=2 sqrt(x)*sqrt(x)=4 prev est=1
x=4 sqrt(x)=2 sqrt(x)*sqrt(x)=4 prev est=2
x=999 sqrt(x)=31 sqrt(x)*sqrt(x)=961 prev est=31
x=999999 sqrt(x)=999 sqrt(x)*sqrt(x)=998001 prev est=1000
x=99999999 sqrt(x)=10028 sqrt(x)*sqrt(x)=100560784 prev est=10784
```
Although this has been quite a complex post I hope it’s taught you a few new DFSORT tricks. Actually, the “meat” of it is the Square Root calculation – which I’m using “in anger” (AKA “for real”) right now. And the technique should be readily adaptable.
The question is: Are there other Newton-Raphson-style computations worth doing? | 2,159 | 8,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-06 | longest | en | 0.911482 |
https://www.physicsforums.com/threads/rod-moment-of-inertia.861850/ | 1,529,371,517,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00284.warc.gz | 894,780,697 | 20,635 | # Homework Help: Rod - Moment of inertia
1. Mar 12, 2016
### reminiscent
1. The problem statement, all variables and given/known data
2. Relevant equations
I = (1/3)ML2
Parallel axis theorem: Inew = Iold + mr2
3. The attempt at a solution
I already solved part a) and I got (1/12)ML2. For b), I get that I have to use the parallel axis theorem but I don't know how to go from there. We have not gone over this concept yet in class (not even moment of inertia) so bear with me...
2. Mar 12, 2016
### Simon Bridge
... start from the equation for the parallel axis theorem. You can look it up: the meanings of the symbols I mean.
I think you were supposed to use it for part (a) so how did you do that?
Last edited: Mar 12, 2016
3. Mar 12, 2016
### drvrm
how did you do the part (a)-- if you have doneit correctly then (b) should not be a problem?
the above is not relevant -did you use it?
4. Mar 12, 2016
### Simon Bridge
5. Mar 13, 2016
### reminiscent
All I did for part a) was that since the rod is bent at its center, that means both segments of the rod have masses of M/2 and lengths of L/2. So for one rod, I = (1/24)ML2. For 2 rods, then it's just 1/2(ML2)
6. Mar 13, 2016
### SammyS
Staff Emeritus
Is that a typo ?
I thought that you got 1/12 ML2 , (which is correct.)
7. Mar 14, 2016
### reminiscent
Yes, sorry, that was a typo. That is what I got. But how am I able to use the parallel axis theorem for b)? We did not go over it in class completely yet...
8. Mar 14, 2016
### SammyS
Staff Emeritus
What is the location of the center of mass ?
9. Mar 16, 2016
### reminiscent
The location of the center of mass for each segment of the rod is the midpoint of each segment, correct?
10. Mar 16, 2016
### reminiscent
I don't understand the axis that it is being rotated. Is it the midpoint between the 2 ENDS of each segment of the rod? So does it boil down to a trig problem?
11. Mar 16, 2016
### SammyS
Staff Emeritus
So then where is the center of mass of the whole bent rod?
12. Mar 16, 2016
### reminiscent
The midpoint of the rod, at the bent center?
13. Mar 16, 2016
### SammyS
Staff Emeritus
No, not at the bend.
Make a sketch.
14. Mar 16, 2016
### reminiscent
Is it the midpoint of the ends of the two segments if they were connected?
15. Mar 16, 2016
### SammyS
Staff Emeritus
No. That's the pivot point for part (b).
It should be mid-way between CM of each segment, Right?
16. Mar 16, 2016
### reminiscent
So like 1/4 of the way of each segment?
17. Mar 16, 2016
### SammyS
Staff Emeritus
No.
That's a correct statement.
The C.M. for the bent rod is mid-way between the center of mass for each of the two segments
Last edited: Mar 16, 2016
18. Mar 16, 2016
### Arkun
I have the same problem as OP has, and while I know where the C.M. of the bent rod is, how can you use the center of mass of the two segments of the rod to algebraically obtain the center of mass of the whole rod?
What I did was determine the distance of the center of mass from the rotational axis and use that as the L for the moment of inertia equation.
My attempt at finding $I_{CM}$:
$I_{CM} = m*(\sqrt{((L / 2) * cos30)^{2} + ((L / 2) * sin30)^{2}})^{2}$
This is for part (a) of the problem.
19. Mar 16, 2016
### SammyS
Staff Emeritus
If you don't understand that, then it's doubtful that do know the correct location of the center of mass of the whole rod .
Simplifying what you have for $\ I_\text{CM}\$ gives:
$\displaystyle \ m\cdot\left(\sqrt{((L / 2)^2 \cdot \cos^2(30^\circ) + ((L / 2)^2 \cdot \sin^2(30^\circ)}\right)^{2} \$
$\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 \cdot( \cos^2(30^\circ) + \sin^2(30^\circ))}\right)^{2} \$
$\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 }\right)^{2} =mL^2/4\$
It's not at all clear, how that is the moment of inertia about any point.
20. Mar 16, 2016
### Arkun
Isn't the center of mass of each segment of the rod at their midpoint? Which means that the distance between the center of mass for each segment of the rod and the axis of rotation is $L / 4$?
21. Mar 16, 2016
### Arkun
Sorry about that. Yeah, I had no idea what I was doing with that. I guess I still don't know how to figure out the center of mass for this bent rod and use it in the appropriate moment of inertia equation.
In my second attempt, I applied the parallel axis theorem to each segment and then added their resulting $I_{NEW}$s and I got the correct answer: $\frac{1}{12} * m * L^2$.
My work:
$I_{NEW} = I_{OLD} + m * d^2$
$I_{NEW} = ((1/12) * (m/2) * (L/2)^2) + (m/2) * (L/4)^2$
$I_{NEW} = (1/96) * m * L^2 + (1/32) * m * L^2$
$I_{NEW} = (1/24) * m * L^2$
$2 * I_{NEW} = (1/12) * m * L^2$
22. Mar 16, 2016
### SammyS
Staff Emeritus
Yes. Of course, that's what OP got, but didn't really explain it.
23. Mar 18, 2016
### Simon Bridge
If we label the endpoint $P$ and $Q$, and the bend $R$ then we can say (from post #1) $|PR|=|RQ|=L/2$ (may need to sketch as I describe.)
Initially the axis of rotation is point $R$ ... which nobody seems to have had trouble with.
The axis of rotation is shifted to a new point exactly half way along the segment $\overline{PQ}$ ... call that point $S$.
The angle $\angle PSR$ is a right angle. Since we also know the other angles it is tempting to try trig and/or pythagoras ... but is there any need?
If the com of the segment $\overline{PR}$ and $\overline{RQ}$ are $A$ and $B$ respectively, then we notice that $\triangle PAS$ and $\triangle SBQ$ are iscoseles triangles with $|PA|=|AS|=|SB|=|BQ|$ and we know that $|PA|=|QB|=L/4$
Each segment therefore has $I=\frac{1}{96}ML^2$ about com, which is a distance $L/4$ from the center of rotation, and there's two of them... so:
$$I = 2\left(\frac{ML^2}{96} + M\left(\frac{L}{4}\right)^2\right)$$ ... and simplify.
Which I think Arkud noticed earlier. If I read the above correctly, that was the method used: well done.
Last edited: Mar 18, 2016
24. Mar 18, 2016
### Simon Bridge
How about the other approach - where the moment of inertia about the center of mass of the bent rod is calculated and the parallel axis theorem is applied to that?
This is how it would normally done IRL: you have a thingy that may get rotated about an arbitrary axis, so you work out the moment of inertia about the com, with the position of the com, and put that information in a set of tables or in the documentation that you sell with the thingy. Some other engineer can. then. work out the details for their particular application.
Using the above labelled diagram, the combined COM is (from symmetry) half way along the segment $\overline{AB}$ call it $C$ the distance we need is:
$\qquad \qquad d=|AC|=|CB|=|AR|\sin(\theta /2)$
$\qquad \qquad \qquad \qquad$ ... where $\theta$ is the angle of the bend from the problem statement, and $|AR|=L/4$
Notice how the angle given was chosen to have nice sines and cosines?
There are a number of other ways of finding $d$.
You end up with $I_{C} = 2M\left( \frac{1}{96}L^2 + d^2 \right)$
The distance from C to S can also be found by pythagoras ... $|CS|=|RC|=\frac{1}{2}|RS|$.
You should be able to work through the steps and check that it gives the same answer as above.
Last edited: Mar 18, 2016
25. Mar 18, 2016
### SammyS
Staff Emeritus
@Simon Bridge ,
You are using M as the mass of each segment, while in the OP, M is the mass of the entire rod. | 2,254 | 7,378 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-26 | latest | en | 0.932994 |
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# Chapter 12: Forces and Motion
Physical Science Chapter 12
by
## Brandon Branch
on 16 July 2013
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#### Transcript of Chapter 12: Forces and Motion
Chapter 12: Forces and Motion What is a force?
Are forces acting on you now?
How can forces produce motion?
What is the difference between mass and weight? A Force is....
A push or a pull acting on an object Force exerted by the blue Deer Force exerted by the orange Deer Net Force Newton's 3 Laws of Motion
They sound simple but it took Aristotle, Galileo and Newton over 2000 years to figure out. Newton's 2nd Law
(The Equation) - N2 Newton's 3rd Law
(ACTION/REACTION) -A Force can cause an object at rest to move or accelerate, can cause a moving object to accelerate or change directions. Representing Force Force has both a magnitude and a direction. The sum of all the forces acting
on an object. What is the difference between mass and inertia? The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the mass, the net force and acceleration act in the same direction. Action-Reaction Force pairs w/o motion Do you have any examples of force?
*Tension - A pull exerted by a rope, string cable etc.
*Normal - A push exerted by a surface that an object is resting on or bounces off of; a normal force is perpendicular to the surface that exerts the force. Units: Newtons (kg*m/s ) US Customary unit : pounds
(1N = .2248 pounds)
*I Newton is the force needed to accelerate 1kg at 1m/s . 2 2 Measuring Force To measure force we use a "Newtonometer"
or scale. -Measures the force needed to support the object. (tension or normal).
-Does a spring scale measure force or mass? *Mass is the amount of matter an object contains.
*Force is the amount of Newton's needed to support the object
*Force is the product of mass and acceleration (due to gravity). Hey! This spring scale measures force but has grams on it?!? Will it be accurate on the moon where the acceleration due to gravity is different? Magnitude is... the strength of the force (number of Newtons).
Direction is... the way the force pushes or pulls the object.
Therefore forces can be represented with a vector. Free Body Diagrams (FBDs) Person Standing Apple hanging from a string Student pushing a Desk A drawing showing all the forces acting on
an object, represent the object with a dot, forces with vectors Person Apple Hand Student Desk F n F F F F F F F F F F F F n n T g g g g Apple g desk student Will any of these objects move? arm Combining Forces -When forces act in the same direction we add the vectors. 10 N
6 N = 16 N -When forces act in the opposite direction we subtract the vectors. 10 N = 4 N 6 N -When forces make an angle we use geometry to combine the vectors. 8 N = 8 N 6 N 6 N 10 N Forces Producing Motion *Will any of these objects move?
*What has to happen in order for any object to move?
*When do forces produce motion? *All the objects on the previous slide will move.
*An object moves when the sum of the forces acting on the object is not zero. -If the net force acting on the object is zero will the object move?
-So in order for an object to move or accelerate.... F = 0 N net 2 N 6 N 4 N 4 N Balanced Forces: When forces are balanced Fnet = 0N, so the object will not accelerate.
-If the object was stationary it will not move.
-If the object is moving, it will travel with a constant velocity. Unbalanced Forces: When the forces acting on an object are unbalanced Fnet = 0N....
-So the object will accelerate.
Which way does the object move?
-In the same direction as the net force acting on it. DO EX 1 & 2 Sliding book demo. Why did the book stop moving?
Is their a force acting on it while it is sliding? Friction:
A force that opposes the motion of all objects.
-Acts in the opposite direction as motion or an applied force.
-A result of the contact between two surfaces or particles of a fluid and a surface. Larger normal force=Larger friction (forces of attraction between p+ and e- play a role.) 4 Types of Friction 1. Static: Frictional force that acts on an object that is not moving; prevents motion.
-Acts in the opposite direction of the applied (orange) force and is equal in magnitude.
-Magnitude will change with applied force until static friction is over come, then the object begins to move.
-Allows us to walk. 2. Sliding (kinetic): Constant force of friction that opposes the direction of an object's motion.
-Causes sliding objects to slow down
-Smaller in magnitude than maximum static.
**Do car tires experience static or kinetic friction?
3. Rolling: Friction exerted on an object that is rolling.
-Static Friction causes wheels to roll, only a small part of friction decreases the velocity of the object
-The effect of friction on a rolling object is 100 to 1,000 times smaller than the effect of static friction on a stationary object. pull Friction 4. Fluid: Friction acting on an object as it moves through a liquid or a gas (fluid).
-Results of collisions between the moving object and the particles of the fluid.
-Directly related to object's speed.
-Air Resistance
*which piece of paper will hit the ground first a crumpled one or un-crumpled? Why?
*what if I make the un-crumpled smaller? What will happen? Book Demonstration - Can you sepparate teh two books?
-Why is this so difficult?
Watch the Mythbusters try.
Maybe this will help. Gravity & Air Resistance What is Gravity?
-An attractive force between two masses that pulls the two mass together.
- All objects fall at the same rate.
-Acceleration due to gravity is 9.8m/s regardless of mass.
Force due to gravity: Product of an object's mass and the acceleration due to gravity (AKA Weight). Universal Forces 2 Do EX 4 Then Read pg 370 *An object in Free fall is at terminal velocity when the downward force of gravity is equal to the upward force of air resistance.
*The Terminal Velocity of an object depends on the mass and cross sectional area of the object.
* For Humans skydiving Terminal Velocity is between 55m/s and 65 m/s If I launch my angry bird a the angle shown by the blue vector, why does it follow the red path? To play Angry Birds Check out: http://chrome.angrybirds.com/ While you play answer these questions?
1. After you launch the bird is the bird accelerating?
2. Which way is the birds acceleration?
3. Does this acceleration affect the bird's horizontal velocity, vertical velocity or both? (Try Launching only Horizontal, and only vertical and see what happens) Projectile Motion is... the motion of an object, after an initial forward velocity, where gravity is the only force acting on the object.
-the object traveling through the air is called the projectile. The path it follows is called its trajectory.
-Projectiles are only affected by two forces, gravity and air resistance. Red - Path Grey - Vertical Velocity
Blue - Horizontal Velocity Yellow - Acceleration (gravity) *Angry Birds ignores Air resistance, so the horizontal velocity of the bird is constant and the vertical velocity is constantly decreasing as a result of the force of gravity acting on the bird. *Demo: Shoot a ball/drop a ball - Which ball will hit the ground first? THEY HIT AT THE SAME TIME!!!!
What does this tell us about the affect of a horizontal velocity on a vertical acceleration? Maybe Mythbusters can help --> Why does this happen: The bullets hit at the same time, and two balls hit at the same time because the horizontal motion of an object is unaffected by the vertical motion, and vice versa.
*The motion of an object in one direction is unaffected by the motion in another direction. Newton's 1st Law of Motion (The Inertial Law) - N1 An object in motion will stay in motion and an object at rest will stay at rest unless acted upon by an outside force. -Inertia is... the tendency of any object to resist a change in motion.
*An object's resistance to acceleration.
-Objects maintain their current state of motion as long as... the net force on the object is zero Ex 5 & the Penny and the note card demo.
*both objects have no force acting on them so they stay in their current state of motion. Equation: a = F /m or F =ma m - mass
a - acceleration
F - net force *The acceleration is in the same
direction as the net force. net net net *When the net force acting on an object is greater than zero the object will accelerate.
*If the object is already moving and...
-the force vector points in the same direction as the motion then .... the object speeds up.
-the force vector points in the opposite dircetion as the motion then ,,, the object slows down. TRY EX 6 & 7 N2-Making car crashes safer During a collision the force exerted on the car is also exerted on the passengers, so how can we decrease the force exerted on the passengers? F=ma -Mass - directly related to force, so decreasing the mass of cars will decrease the force. (Good idea when choosing a car, but hard to change) -Acceleration - directly related to force, a = (Vf-Vi)/t
*Two Quantities that effect acceleration are time and change in speed. - Time - Inversely related to acceleration; increasing time will decrease acceleration, longer collisions = less force.
-Change in Speed - Directly related to acceleration; If the change in speed is less then less force acts on the car. N2 - Mass & Weight - What's the difference Mass is ... the amount of matter a substance is made of.
- A measure of an object's inertia
Weight is ... the force of gravity acting on an object
Equation : W=mg units - Newtons
W- Weight(N) m- mass (kg) g- acceleration due to gravity 9.8m/s 2 The difference....
*Weight depends on gravity and is measured by a scale
*Mass is constant.
DO EX 9. Every Action has an equal and opposite reaction. - When one object exerts a force on another object, the second object exerts a force equal in magnitude on the first, but in the opposite direction. Force Pairs are ... equal and opposite forces exerted on two different objects. *NORMAL & TENSION are NOT in a force pair with gravity. Draw a FBD for: the book and table; the sign and post KVILLE ROCKETS Book Table sign Post F F F F F F F F F *The force pairs in each picture are the same color g g g g N N book F T Wall Sign They are equal in magnitude but opposite in direction. Action-Reaction Force pairs W/ motion Draw a FBD for the ball & Ground, The red car & the Yellow Car F F F F F F F F F N N N g g g ball Yellow red *The force pairs in each picture are the same color They are equal in magnitude but opposite in direction. Ball Ground Red Car Yellow car Every force on your list can be classified as one of the universal forces. Make a list of every force you can think of: **They don't have to be ones you have heard of in this class, but that might be a good place to start. 1. Electromagnetic Force: A Combination of electric and magnetic forces.
-Only force that can both attract and repel.
*Electric forces: Force between charged particles
- Opposites (+ -) attract, Likes ( + +, - - ) repel
*Magnetic forces: attractive and repulsion forces between the poles of a magnet.
- Opposites (N S) attract, Likes (N N, S S ) repel
Ex. Magnet and the whiteboard, attraction between p+ and e-, Pushes, pulls, friction, Normal, tension.
2. Strong Nuclear: Powerful force of attraction between Protons and neutrons in the nucleus of an atom.
-100 x stronger than electromagnetic. (Strongest force)
-Acts over very small distances (1x10^-15m)
-Responsible for the energy released in nuclear power plants and nuclear weapons. 3. Weak Nuclear: Force responsible for ejecting particles from a decaying nucleus.
-Acts over very small distances (1x10^-18m)
-Present in radioactive process: causes particles to decay, and eject particles.
-Non contact force.
4. Gravitational Force: Attractive force between any two objects with mass
-Weakest of all universal forces.
-Result of the objects distorting the space time continuum (Einstein).
-Acts over very large distances
-Non contact force of attraction. Newton's Law of Universal Gravitation: Every object in the universe that has mass is attracted to every other object that has mass.
- Attraction is called the force of gravity.
-Each mass exerts and equal and opposite force. (The force exerted on me by the Earth is an 3rd law pair with the force exerted on the Earth by me.)
-Force of Gravitational Attraction is proportional to the mass of the object's and inversely proportional to the square of the distance between them.
-Produces a centripetal force which keeps planets and other orbiting bodies on their paths.
* Newton describes what is happening, while Einstein's theory or General relativity addresses why it happens. DO EX 13 Example 13 - Answers
Full transcript | 3,049 | 13,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-39 | longest | en | 0.912844 |
https://www.teachy.app/project/middle-school/7th-grade/math/carnival-of-probabilities-exploring-compound-events | 1,708,877,209,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00382.warc.gz | 1,038,092,272 | 36,250 | ## Teacher,access this and thousands of other projects!
Free Registration
# Contextualization
Probability is a fascinating branch of mathematics that deals with the likelihood of events happening. It plays a significant role in our day-to-day lives, from predicting the weather to making financial decisions. The understanding of probability helps us make more informed choices and predictions, enabling us to navigate the uncertainties of life with more confidence.
Compound events are a fundamental concept in probability theory. They are composed of two or more simple events, which can be either mutually exclusive or dependent. These compound events can be described by their probability, which is the likelihood of the event occurring. The probability of a compound event is calculated using different rules and formulas, depending on whether the events are mutually exclusive or dependent.
In the real world, compound events are everywhere. For example, when a weather forecast predicts a 40% chance of rain and a 30% chance of thunderstorms, you can use the rules of compound probability to determine the overall likelihood of both occurring. In finance, understanding compound events can help with making investments or insurance decisions. Compound events are also essential in games of chance, such as card games or lotteries.
The resources below will provide a solid foundation for your understanding of compound events in probability:
Remember, the goal of this project is not just to understand the theory, but also to apply it to real-world scenarios and to work collaboratively. So, let's dive into the fascinating world of compound probability!
# Practical Activity
## Objective:
The objective of this project is to provide a hands-on experience in understanding and calculating the probability of compound events. In doing so, students will be able to:
1. Understand the fundamental principles of compound probability.
2. Apply the rules and formulas of compound probability to real-world scenarios.
3. Enhance their skills in teamwork, communication, problem-solving, and creative thinking.
## Description of the Project:
In groups of 3 to 5, students will create a carnival game with at least two different outcomes. Each outcome will have a different probability of occurring. The game should be designed in such a way that students can calculate the compound probability of both outcomes happening simultaneously or one after the other.
## Necessary Materials:
• Cardboard boxes
• Craft materials (paint, markers, glues, etc.)
• Small objects (marbles, balls, coins, etc.)
• Rulers, scissors, and other basic stationery
• A notebook for calculations and observations
## Detailed Step-by-Step:
1. Divide into groups of 3 to 5 students.
2. Brainstorm and decide on a theme for your carnival game. For example, it could be a game involving rolling a dice and throwing a coin.
3. Design the game on a cardboard box. Draw the necessary markings and slots for the dice and coin.
4. Decorate the game using craft materials.
5. Assign probabilities to each outcome of the game. For example, the probability of getting a 4 on the dice could be 1/6, and the probability of getting a head on the coin could be 1/2.
6. Based on these probabilities, calculate the compound probabilities of different events. For example, the probability of getting a 4 on the dice and a head on the coin could be 1/6 * 1/2 = 1/12.
7. Test the game multiple times to verify the calculated probabilities.
8. Keep a record of the results of the game and the calculated probabilities in a notebook.
9. Prepare a final report as explained below.
## Project Deliverables:
At the end of the project, each group will submit a report in the specified format:
1. Introduction: Start with the relevance of probability and compound events in real-world applications, the project's objective, and a brief description of the carnival game your group designed.
2. Development: Detail the theory behind compound events in probability, explain the methodology used in designing the game and calculating the compound probabilities, and present and discuss the results. Include any interesting observations or challenges faced during the project. Make sure to include the calculated probabilities and how they were derived.
3. Conclusion: Conclude the report by revisiting the main points, explicitly stating the learned concepts, and the conclusions drawn about the project.
4. Bibliography: List down all the resources used during the project, including web links, books, and videos.
This project is estimated to take more than twelve hours per student to complete and should be completed within one month. The written report should not exceed 3000 words and should be a comprehensive account of the students' understanding, application, and reflections on compound probability. In addition to the written report, each group will present their carnival game and their findings to the class, enhancing their communication and presentation skills.
Math
# Contextualization
The world around us is filled with numbers. From the time we wake up in the morning, to the time we go to bed at night, we are surrounded by numerical concepts. Two of the most prevalent concepts in the world of mathematics are fractions and decimals.
Fractions and decimals are two different ways of expressing the same value. They are like two languages that can be used to communicate the same idea. In this project, we will delve into the world of fractions and decimals, particularly focusing on the conversion between these two forms.
Understanding how to convert fractions to decimals and vice versa is an essential skill in mathematics. It is a fundamental concept that is used in many areas, ranging from basic arithmetic to more complex mathematical operations, such as solving equations and working with ratios and proportions.
Moreover, the ability to convert between fractions and decimals is not just important in the field of mathematics; it also has real-world applications. For instance, we often encounter fractions and decimals in our daily lives, whether we are measuring ingredients for a recipe, calculating discounts at a store, or understanding statistics in the news.
# Resources
To get started on this project, you may find the following resources helpful:
1. Khan Academy - Converting Fractions to Decimals
2. Math Is Fun - Converting Fractions to Decimals
3. Math Goodies - Converting Fractions to Decimals
4. Book: "Mathematics: Its Content, Methods and Meaning" by A. D. Aleksandrov, A. N. Kolmogorov, M. A. Lavrent'ev (Chapter 19: Decimals)
5. Book: "Fractions and Decimals" by David Adler
6. YouTube video: Converting Fractions to Decimals by Math Antics
These resources will provide you with a solid foundation on the topic and can be used as a reference throughout the project. Make sure to explore them thoroughly and use them as a guide to deepen your understanding of converting fractions and decimals.
# Practical Activity
## Objective
The main objective of this project is to facilitate a deeper understanding of converting between fractions and decimals. Students will investigate and explore the theoretical concepts of fractions and decimals, apply these concepts in real-world scenarios, and collaboratively prepare a comprehensive report detailing their findings and experiences.
## Description
In this project, students will be divided into groups of 3 to 5. Each group will be tasked with creating a comprehensive guidebook on converting fractions to decimals and vice versa. This guidebook should include theoretical explanations, real-world examples, and step-by-step procedures for converting between these two forms.
Additionally, each group will prepare a presentation to share their findings and experiences with the class. The presentation should be interactive and engaging, incorporating visual aids and practical examples to illustrate the conversion process.
## Materials
• Pen and paper for note-taking and brainstorming.
• Mathematical tools for calculations (calculator, ruler, protractor, etc.).
• Presentation materials (poster board, markers, etc.) for the final presentation.
## Steps
1. Research and Theoretical Understanding (8 hours): Each group should begin by conducting research on the topic. Use the provided resources as a starting point, and expand your knowledge by exploring other reliable sources. Make sure to understand the basic operations involved in converting fractions to decimals and vice versa.
2. Real-World Application (4 hours): Next, each group should find real-world examples where fractions and decimals are used interchangeably. For instance, you could look at cooking recipes, sports statistics, or financial transactions. Document these examples, and discuss how understanding the conversion between fractions and decimals can be helpful in these situations.
3. Creating the Guidebook (10 hours): Now, each group should start creating their guidebook. This should be a comprehensive resource that explains the concepts of converting fractions to decimals and vice versa. It should include theoretical explanations, real-world examples, and step-by-step procedures for the conversion process. The guidebook should be visually appealing and easy to understand.
4. Preparing the Presentation (8 hours): As the guidebook is being developed, each group should simultaneously work on their presentation. This should be an interactive and engaging session, where you explain the conversion process using practical examples and visual aids.
5. Review and Rehearsal (4 hours): Before the final presentation, each group should review their work, make any necessary revisions, and rehearse their presentation to ensure a smooth delivery.
6. Presentation and Submission of the Guidebook (Class Time): Each group will present their findings and submit their guidebook at the end of the project.
## Project Deliverables
At the end of the project, each group will be required to submit:
• A comprehensive guidebook on converting fractions to decimals and vice versa.
• A detailed report following the structure: Introduction, Development, Conclusions, and Used Bibliography.
• A presentation on their findings and experiences.
The Introduction of the report should contextualize the theme, its relevance, and real-world application, as well as the objective of this project. The Development section should detail the theory behind converting fractions to decimals and vice versa, explain the activity in detail, indicate the methodology used, and present and discuss the obtained results. The Conclusion should revisit the main points of the project, explicitly stating the learnings obtained and the conclusions drawn about the project. Finally, the Bibliography should list all the sources of information used in the project.
The written report should complement the guidebook and the presentation, providing a detailed account of the project's journey and the learnings acquired along the way. It should be a well-structured document, with a clear and logical flow, and free from grammatical and spelling errors.
Remember, this project is not just about understanding the process of converting fractions and decimals; it's also about developing essential skills like teamwork, communication, time management, and problem-solving. Good luck, and have fun with your mathematical journey!
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Math
# Contextualization
Base ten, a fundamental concept in mathematics, is the backbone of all arithmetic operations. The base-ten system is used universally in mathematics due to its efficiency and simplicity. In this system, each digit in a number has a place, and the value of the number depends on its place. For instance, in the number '345', '3' stands for three hundreds, '4' for four tens and '5' for five ones.
Understanding this concept is not only crucial for doing basic arithmetic like addition and subtraction, but it is also foundational for more advanced mathematical theories such as algebra and calculus, where the position of numbers continue to bear tremendous weight. Place value is also used extensively in computing, especially in the realm of binary (base two) and hexadecimal (base sixteen) numbers, making it a necessary skill for future software engineers and computer scientists.
Place value, however, is not just theoretical. It’s deeply embedded in our everyday life. Imagine a world without place value: price tags, phone numbers, addresses would all be nonsensical. Delving deeper, the ubiquitous nature of place value in the practical world helps us understand, interpret, and predict patterns in numerous fields including commerce, scientific research, and engineering.
# Resources
For a strong theoretical grounding and deeper exploration on the subject, these resources are recommended:
1. "Place Value" in Khan Academy: An online platform that provides detailed lessons with practice problems about place value.
2. "Everything You Need to Ace Math in One Big Fat Notebook" by Workman Publishing: A comprehensive math book for young students, which explains place value in an easy and understandable way.
3. CoolMath4Kids: An interactive website that provides games and activities related to place value to make learning fun and engaging.
We hope this project sparks an interest in this crucial concept, and that you come away with a deeper appreciation of mathematics and its real-world applications. Start your journey into the world of place value now!
# Practical Activity
## Objective:
To understand the concept of place value and the base ten system; to learn how to effectively work in a team; to apply mathematical concepts to real-life situations and to enhance creativity, problem-solving and communication skills.
## Description:
This project gives students an opportunity to create a 'Base Ten City', which will be a model city built entirely on the base-ten system of numbers. Each group will be given a large piece of construction paper, on which they will create a cityscape using materials provided. The number of different elements in the city will be dictated by the base-ten system.
## Necessary Materials:
1. Large sheets of construction paper
2. Scissors
3. Glue
4. Color markers
5. Rulers
6. Base Ten Blocks
## Steps:
1. Brainstorming (Estimated time: 1 Hour) The group will brainstorm ideas for their city. This could include houses, buildings, trees, cars, people, etc.
2. Planning (Estimated time: 3 Hours) Each group will map out their city on their construction paper. They will decide where each element will go by considering the place values. For example, the number of houses (units place), the number of trees (tens place), and the number of buildings (hundreds place). They will use a ruler to make sure that each section is correctly sized and positioned.
3. Building (Estimated time: 5 Hours) Students will use scissors, glue, colors, and base ten blocks to build their city based on the plan they created. During this process, they should keep in mind the base-ten system and ensure each element's quantity aligns with its assigned place value.
4. Reflection (Estimated time: 2 Hours) Once the city is built, the group will reflect on their process and make any necessary adjustments. They will ensure that the place values are accurately represented in their city.
5. Presentation (Estimated time: 2 Hours) Each group will present their city to the class and explain how they used the base-ten system in their design. They will explain the significance of each city element and its relation to place value.
## Project Deliverables:
At the end of the project, each group will present:
1. Written Report (Estimated time: 4 Hours to Write) This document should include: Introduction (background, objective, and relevance), Development (details of city planning, building process, and challenges faced), Conclusions (learnings about place value and teamwork), and Bibliography. The report should be written in a way that it both narrates the group's journey and helps the readers to understand the base-ten system and place value through their project.
2. Base Ten City Model The physical model of the developed city which represents place values in the base ten number system.
3. Presentation A clear and concise presentation of their project, which explains how they incorporated the base-ten system into their city. This will help them articulate their understanding of the concepts and their project journey.
This project should be undertaken over 2-3 weeks, with students working in groups of 3 to 5. Please plan your time appropriately to complete all aspects of the project.
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Math
# Contextualization
## Introduction to Equations and Inequalities Graphically
Equations and inequalities are fundamental concepts in mathematics and are used in various fields of life and science, including physics, engineering, economics, and computer sciences. They help us understand and solve real-life problems by representing relationships and constraints between different variables and quantities.
When we say "graphically," we mean representing these equations and inequalities using visual tools called graphs. Graphs provide a visual representation of the relationship between variables, making it easier to understand and solve problems. They can be used to plot equations and inequalities, and their solutions can be easily determined by analyzing the graph.
An equation is a statement that two expressions are equal. It consists of two sides, a left side and a right side, separated by an equal sign. The solution to an equation is the value(s) that make the equation true when substituted for the variable(s). An inequality, on the other hand, is a statement that one expression is greater than (or less than) or equal to another expression. The solution to an inequality is the range of values that make the inequality true.
## Significance and Real-world Application
Understanding equations and inequalities graphically is not just a theoretical concept, but it has numerous practical applications in our daily lives. For instance, when we try to plan a budget, we need to deal with inequalities (our expenses should be less than or equal to our income). In physics, we use equations to describe the motion of objects, while in economics, we use them to model and predict market trends.
In the digital age, equations and inequalities graphically play a significant role in computer graphics, weather forecasting, and traffic control systems. They are also used in medical sciences for modeling the spread of diseases and in engineering for designing and optimizing processes.
## Resources for Study
To delve deeper into the topic and for additional resources, students are encouraged to explore the following:
1. Book: "Algebra 1 Common Core Student Edition" by Randall I. Charles, Basia Hall, Dan Kennedy, Art Johnson, and Mark Rogers.
2. Website: Khan Academy's section on Graphical Representations of Equations and Inequalities
3. Video: Graphing Linear Inequalities by Khan Academy.
4. Document: Graphing Linear Equations and Inequalities on Dummies.com
These resources will provide a strong foundation for understanding the concepts of equations and inequalities graphically, their applications, and how to solve problems using graphical representations. They will also help students in completing the project successfully.
# Practical Activity
## Objective:
The main objective of this project is to understand how to represent equations and inequalities graphically and to recognize their real-world applications. Students will choose a scenario or a real-world problem, represent it using equations and/or inequalities, and then graph them to understand the solution space.
## Description:
This group project will involve the following steps:
1. Identifying a real-world scenario or problem that can be modeled using equations and/or inequalities.
2. Setting up the equations and/or inequalities to represent the scenario or problem.
3. Graphing the equations and/or inequalities to visualize the solution space.
4. Analyzing the graph to understand the solution(s) in the context of the real-world problem.
5. Documenting the process, findings, and implications in a report.
## Necessary Materials:
1. Pencil and paper or a graphing calculator.
2. Real-world scenario or problem (can be from any field of interest like sports, health, environment, etc.)
3. Research materials for setting up the equations and/or inequalities.
## Detailed Step-by-Step:
1. Formation of Groups and Selection of Scenario (1 class period): Form groups of 3-5 students. Each group should select a real-world scenario or problem that can be modeled using equations and/or inequalities.
2. Setting up the Equations and Inequalities (1 class period): Research and identify the variables and their relationships in the selected scenario. Set up the necessary equations and/or inequalities that can represent the scenario or problem.
3. Graphing the Equations and Inequalities (1-2 class periods): Use pencil and paper or a graphing calculator to plot the equations and/or inequalities. Make sure to label your axes and any key points on the graph.
4. Analyzing the Graph (1 class period): Analyze the graph to understand the solution space. What do the different parts of the graph represent in the context of your real-world scenario? Are there any solutions that do not make sense in the context of the problem?
5. Report Writing (1-2 class periods): Write a report documenting your project. The report should follow these sections:
• Introduction: Contextualize the chosen real-world problem, its relevance, and the objective of the project.
• Development: Detail the theory behind equations and inequalities graphically, explain your chosen scenario, how you modeled it, and your methodology for setting up and graphing the equations and/or inequalities. Present your findings and discuss the implications.
• Conclusion: Conclude the work by revisiting the main points, stating the learnings obtained, and the conclusions drawn about the project.
• Bibliography: Indicate the sources you relied on to work on the project.
6. Presentation (1 class period): Each group will present their project to the class. This should include a brief overview of the selected scenario, the setup of equations and inequalities, the graph, and the findings.
## Project Deliveries:
The main deliverable of this project will be the written report, which should be comprehensive and detailed. The report should include the theory of equations and inequalities graphically, the chosen scenario, the setup of equations and/or inequalities, the graph, the analysis, and the implications. The report should be well-structured, clearly written, and should demonstrate a deep understanding of the topic. Each member of the group should contribute to the report.
The second deliverable will be a presentation of the project in front of the class. This should be a summarized version of the report, highlighting the main points and findings of the project. The presentation should be engaging, well-prepared, and should demonstrate good teamwork and communication skills.
The project is expected to take around 6-8 hours per participating student to complete and should be delivered within one month of its assignment.
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# MATLAB - Lecture # 8
Programming in MATLAB / Chapter 7
Topics Covered:
1. Relational and Logical Operators
2. Conditional statements.
if–end
if-else–end
if-elseif-else–end
(c) 2003 The Ohio State University
INTRODUCTION TO PROGRAMMING
163164
! In a simple program the commands are executed in the order
they are typed.
! Many situations may require that:
*
*
*
Commands will not be executed in order.
Different commands are executed in different runs.
The execution of a group of commands is repeated many
times.
(c) 2003 The Ohio State University
! Read Chapter 7 in the MATLAB book. ! In the class we will only cover if-end conditional statements (this lecture) and for-end loops (next lecture).INTRODUCTION TO PROGRAMMING 163164 ! MATLAB provide tools (commands) that can be used to control the flow of a program. (c) 2003 The Ohio State University . ! Students can learn other tools from the book by themselves.
RELATIONAL OPERATORS Relational operator Meaning < Less than. ! Relational operators compare two numbers in a comparison statement. <= Less than or equal to. == Equal to. ! If the statement is false. > Greater than. it is assigned a value of 1. >= Greater then or equal to. ~= Not equal to. it is assigned a value of 0. (c) 2003 The Ohio State University 164167 . ! If the statement is true.
and assigns the answer to a. >> a=5<10 a= Since 5 is smaller than 10 the number 1 is assigned to a. 1 >> y=(6<10) + (7>8) + (5*3= =60/4) y= =1 =0 =1 2 (c) 2003 The Ohio State University . ans = 0 Checks if 5 is smaller than 10. EXAMPLES >> 5>8 Since 5 is not larger than 8 the answer is 0.165 RELATIONAL OPERATORS.
Logical Operator Name Meaning & Example: A & B AND True if both operands (A and B) are true. ~ Example: ~ A NOT True if the operand (A) is false. ! A zero number is false. ! A nonzero number is true. (c) 2003 The Ohio State University . | Example: A | B OR True if either or both operands (A and B) are true.167 LOGICAL OPERATORS ! Logical operators have numbers as operands. False if the operand (A) is true.
3 and 7 are both true (nonzero). 1 >> x=-2. a= 1 is assigned to a since at least one number is true (nonzero). so the outcome is 1. Mathematically correct. y=5. -5<x is true (=1) and then 1<-1 is false (0).LOGICAL OPERATORS. EXAMPLES >> 3&7 ans = 168169 3 AND 7. The mathematically correct statement is obtained by using the logical operator &. 1 >> a=5|0 5 OR 0 (assign to variable a). The answer is false since MATLAB executes from left to right. >> -5<x<-1 ans = 0 >> -5<x & x<-1 ans = 1 Define variables x and y. (c) 2003 The Ohio State University . Since both are true (1). The inequalities are executed first. the answer is 1.
either nothing is done. I will study harder so that I can get a better job. If the condition is met. buy a new car. ! A condition stated. one set of actions is taken. If the condition is not met. I will quit college. or a second set of actions is taken. Example: If I win the Lottery.CONDITIONAL STATEMENTS 172177 ! Conditional statements enable MATLAB to make decisions. (c) 2003 The Ohio State University . ! The process is similar to the way we (humans) make decisions. If I do not win the Lottery. and go fishing.
172 THE FORM OF A CONDITIONAL STATEMENT If Conditional expression consisting of relational and/or logical operators Examples: if a < b if c >= 5 if a == b if a ~= 0 if (d<h)&(x>7) if (x~=13)|(y<0) (c) 2003 The Ohio State University All variables must have assigned values. .
THREE FORMS OF THE if STATEMENT If conditional statement if conditional statement 1 commands end 172177 command group 1 elseif conditional statement 2 command group 2 if conditional statement command group 1 else command group 3 else command group 2 end end (c) 2003 The Ohio State University .
......THE if–end STATEMENT False . .... . end ............ end (c) 2003 The Ohio State University ..... if conditional expression True ....... if 172174 MATLAB program.. ..... . ..... MATLAB program.......... .. A group of MATLAB commands...
% The program calculates the average of the grades.EXAMPLE OF USING THE if–end STATEMENT % A script file that demonstrates the use of the if-end statement. is printed. score = input('Enter (as a vector) the scores of the three tests '). disp('The average grade is:') disp(ave_grade) if ave_grade < 60 disp('The student did not pass the course. a massage: % The student did not pass the course. % If the average is less than 60. ave_grade = (score(1) + score(2) + score(3))/3. % The user is asked to enter three grades.') end (c) 2003 The Ohio State University .
EXAMPLE OF USING THE if–end STATEMENT Executing the script file of the previous slide in the Command Window: >> Lecture8Example1 Enter (as a vector) the scores of the three tests [78 61 85] The average grade is: 74. (c) 2003 The Ohio State University .6667 >> Lecture8Example1 Enter (as a vector) the scores of the three tests [60 38 55] The average grade is: 51 The student did not pass the course.
............ ....... .. False 174175 MATLAB program.... ....... ........ ..THE if–else-end STATEMENT ...... .... MATLAB program... Group 1 of MATLAB commands.. (c) 2003 The Ohio State University ...... Group 2 of MATLAB commands....... . else end .... if conditional expression if True end ..
If the average is less than 60. is printed. is printed. ave_grade = (score(1) + score(2) + score(3))/3. The program calculates % the average of the grades. disp('The average grade is:') disp(ave_grade) if ave_grade < 60 disp('The student did not pass the course.') else disp('The student passed the course. % The user is asked to enter three grades. % Otherwise.EXAMPLE OF USING THE if–else-end STATEMENT % A script file that demonstrates the use of the if-else-end statement. a massage: The student passed the course. a % massage: The student did not pass the course.') end (c) 2003 The Ohio State University . score = input('Enter (as a vector) the scores of the three tests ').
EXAMPLE OF USING THE if–else-end STATEMENT Executing the script file of the previous slide in the Command Window: >> Lecture8Example2 Enter (as a vector) the scores of the three tests [65 80 83] The average grade is: 76 The student passed the course. >> Lecture8Example2 Enter (as a vector) the scores of the three tests [60 40 55] The average grade is: 51.6667 The student did not pass the course. (c) 2003 The Ohio State University .
... if conditional expression if True True ..... .......... commands. (c) 2003 The Ohio State University MATLAB program.......... end ....... Group 1 of MATLAB ... commands.. elseif conditional expression .. ..... ... .... Group 2 of MATLAB ........... end Group 2 of MATLAB commands...... else .. ...... ............176THE if–elseif-else-end STATEMENT 177 .. False False elseif MATLAB program........
% The program calculates the tip in a restaurant according to the % amount of the bill.EXAMPLE OF USING THE if–elseif-else-end STATEMENT % A script file that demonstrates the use of the if-elseif-else-end % statement. % If the bill is less than 10\$ the tip is \$1. format bank clear tip (The file continues on the next slide) (c) 2003 The Ohio State University . % Above \$60 the tip is 20% of the bill. % Between \$10 and \$60 the tip is 18% of the bill.80.
else tip = bill*0. if bill <= 10) tip = 1. end disp('The tip is (in dollars):') disp(tip) (c) 2003 The Ohio State University .8. elseif (bill > 10) & (bill <= 60) tip = bill*0.2.18.(Continuation from the previous slide) bill = input('Enter the amount of the bill (in dollars): ').
00 (c) 2003 The Ohio State University .80 >> Lecture8Example3 Enter the amount of the bill (in dollars): 100 The tip is (in dollars): 20.70 >> Lecture8Example3 Enter the amount of the bill (in dollars): 6 The tip is (in dollars): 1.EXECUTING THE SCRIPT FILE OF THE RESTAURAT TIP CALCULATION >> Lecture8Example3 Enter the amount of the bill (in dollars): 15 The tip is (in dollars): 2.
if – else – end.COMMENTS ABOUT if–end STATEMENTS ! For every if command a computer program must have an end command.end. end statements following each other. ! A program can have many if …. ! A computer program can perform the same task using different combinations of if . and if – elseif – else – end statements.. (c) 2003 The Ohio State University 172177 .
Problem 8. Chapter 7. MATLAB book. Problem 16. Chapter 7. The first line in the script file. Problem 1. MATLAB book. MATLAB book. and in the Command Window should be a comment with your name. (c) 2003 The Ohio State University . 3. and a printout of the Command Window showing how the script file was used. Chapter 7. 2. In problems 2 and 3 submit a printout of the script file. In problem 1 submit a printout of the Command Window.MATLAB ASSIGNMENT 8: 199202 1. | 2,296 | 8,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-51 | latest | en | 0.852704 |
http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-262-discrete-stochastic-processes-spring-2011/video-lectures/lecture-11-renewals-strong-law-and-rewards/ | 1,467,204,309,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397744.64/warc/CC-MAIN-20160624154957-00077-ip-10-164-35-72.ec2.internal.warc.gz | 218,632,968 | 72,922 | # Lecture 11: Renewals: Strong Law and Rewards
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Description: This lecture begins with the SLLN and the central limit theorem for renewal processes. This is followed by the time-average behavior of reward functions such as residual life.
Instructor: Prof. Robert Gallager
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ROBERT GALLAGER: OK. Today, I want to review a little bit of what we did last time. I think with all the details, some of you sort of lost the main pattern of what was going on. So let me try to talk about the main pattern. I don't want to talk about the details anymore.
I think, for the most part in this course, the best way to understand the details of proofs is to read the notes where you can read them at your own rate, unless there's something wrong in the notes. I will typically avoid that from now on.
The main story that we want to go through, first is the idea of what convergence with probability 1 means. This is a very peculiar concept. And I have to keep going through it, have to keep talking about it with different notation so that you can see it coming any way you look at it.
So what the theorem is-- and it's a very useful theorem-- it says let Y n be a set of random variables, which satisfy the condition that the sum of the expectations of the absolute value of each random variable is less than infinity.
First thing I want to point out is what that means, because it's not entirely clear what is the means to start with. When you talk about a limit from n to infinity, from n equals 1 to infinity, what it remains is this sum here really means the limit as m goes to infinity of a finite sum. Anytime you talk about a limit of a set of numbers, that's exactly what you mean by it.
So if we're saying that this quantity is less than infinity, it says two things. It says that these finite sums are lesson infinity for all a m. And it also says that these finite sums go to a limit. And the fact that these finite sums are going to a limit as m gets big says what's a more intuitive thing, which is that the limit as m goes to infinity.
And here, instead of going from 1 thing m, I'm going from m plus 1 to infinity, and it doesn't make any difference whether I go from m plus 1 to infinity or m to infinity. And what has to happen for this limit to exist is the difference between this and this, which is this, has to go to 0. So that what we're really saying here is that the tail sum has to go to 0 here.
Now, this is a much stronger requirement than just saying that the expected value of the magnitudes of these random variables has to go to 0. If, for example, the expected value of y sub n is 1/n, then 1/n, the limit of that as n goes to infinity, is 0. So this is satisfied. But when you sum one/n here, you don't get 0. And in fact, you get infinity. Yes, you get infinity. So this requires more than this.
This limit here, the requirement that that equals 0, implies that the sequence y sub n actually converges to 0 in probability, rather than with probability 1. And the first problem in the homework for this coming week is to actually show that. And when you show it, I hope you find out that it is absolutely trivial to show it. It takes two lines to show it, and that's really all you need here.
The stronger requirement here, let's us say something about the entire sample path. You see, this requirement really is focusing on each individual value of n, but it's not focusing on the sequences. This stronger quantity here is really focusing on these entire sample paths.
OK. So let's go on and review what the strong law of large numbers says. And another note about the theorem on convergence, how useful that theorem is depends on how you choose these random variable, Y1, Y2, and so forth.
When we were proving the strong law of large numbers in class last time, and in the notes, we started off by assuming that the mean of x is equal to 0. In fact, we'll see that that's just for convenience. It's not something that has anything to do with anything. It's just to get rid of a lot of extra numbers. So we assume this.
We also assume that the expected value of x to the fourth was less than infinity. In other words, we assume that this random variable that we were adding, these IID random variables, had a fourth moment. Now, an awful lot of random variables have a fourth moment. The real strong law of large numbers, all that assumes is that the expected value of the magnitude of x is less than infinity. So it has a much weaker set of conditions.
Most of the problems you run into, doesn't make any difference whether you assume this or you make the stronger assumption that the mean is equal to 0. When you start applying this theorem, it doesn't make any difference at all, because there's no way you can tell, in a physical situation, whether it is reasonable to assume that the fourth moment is finite and the first moment isn't. Because that question has to do only with the very far tails of the distribution.
I can take any distribution x and I can truncate it at 10 to the google. And if I truncate it at 10 to the google, it has a finite fourth moment. If I don't truncate it, it might not have a fourth moment. There's no way you can tell from looking at physical situations.
So this question here is primarily a question of modeling and what you're going to do with the models. It's not something which is crucial.
But anyway, after we did that, what we said was when we assume that this is less than infinity, we can look at s sub n, which is x1, up to x n all IID We then took this sum here, took to the fourth moment of it, and we looked at all the cross terms and what could happen. And we found that the only cross terms that worked that were non-0 was where either these quantities were paired together, x1 x1, x2, x2, or where they were all the same.
And then when we looked at that, we very quickly realized that the expected value of s n to the fourth was proportional to n squared. It was upper banded the by three times n squared times the fourth moment of x.
So when we look at s n to the fourth divided by n to the fourth, that quantity, summed over n, goes as 1 over n squared, which has a finite sum over n. And therefore, the probability of the limit as n approaches infinity of s n to the fourth over n fourth equals 0.
The probability of the sample path, the sample paths converge. The probability of that is equal to 1, which says all sample paths converge with probability 1. So this is enough, not quite enough to prove the strong law of large numbers. Because what we're interested in is not s n to the fourth over n fourth. We're interested in s sub n over n. So we have to go one step further.
This is why it's tricky to figure out what random variables you want to use when you're trying to go from this theorem about convergence with probability 1 to the strong law, or the strong law for renewals, or any other kind of strong law doing anything else. It's a fairly tricky matter to choose what random variables you want to talk about there.
But here, it doesn't make any difference, as we've said. If you let s n of omega over n be a sub n to the 1/4. In other words, what I'm doing now is I'm focusing on just one sample path. If I focus on one sample path, omega, and then each one of these terms has some value, a sub n, and then s sub n of omega over n is going to be equal to a sub n to the 1/4 power, the 1/4 power of this quantity over here.
Now, the question is if the limit of these numbers-- And now remember, now we're talking about a single sample path, but all of these sample paths behave the same way. So if this limit here for one sample path is equal to 0, then the limit of a sub n to the 1/4 is also equal to 0.
Why is that true? That's just a result about the real number system. It's a result about convergence of real numbers. If you take a bunch of real numbers, which are getting very, very small, and you take the fourth root of those numbers, which are getting very small, the fourth root is a lot bigger than the number itself.
But nonetheless, the fourth root is being driven to 0 also, or at least the absolute value of the fourth root is being driven to 0 also. You can see this intuitively without even proving any theorems. Except, it is a standard result, just talking about the real number system.
OK. So what that says is the probability that the limit of s sub n over n equals 0. This is now is talking about sample paths for each sample path. This limit s n over n either exists or it doesn't exist. If it does exist, it's either equal to 0 or equal to something else. It says that the probability that it exists and that it's equal to 0 is equal to 1.
OK. Now, remember last time we talked about something a little bit funny. We talked about the Bernoulli process. And we talked about the Bernoulli process using one value of p for the probability that x is equal to 1. And what we found is that the probability of the sample path where s sub n over n approach to p, the probability of that set was equal to 1.
If we change the probability that x is equal to 1 to some other value, that is still a perfectly well-defined event. The event that a sample path, that sub n over n for a sample path approaches p. But that sample path, that event, becomes 0 as soon as you change p to some other value.
So what we're talking about here is really a probability measure. We're not using any kind of measure theory here, but you really have to be careful about the fact that you're not talking about the number of sequences for which this limit is equal to 0. You're really talking about the probability of it. And you can't think of it in terms of number of sequences.
What's the most probable sequence for a Bernoulli process where p is, say, 0.317? Who knows what the most probable sequence of length n is? What? There is none? Yes, there is. It's all 0's, yes. The all 0's sequence is more likely than anything else.
So why don't these sequences converge to 0? I'm In this case, these sequences actually converge to 0.317, if that's the value of p.
So what's going on? What's going on is a trade-off between the number of sequences and the probability of those sequences. You look at a particular value of n, there's only one sequence which is all 0's. It's much more likely than any of the other sequences.
There's an enormous number of sequences where the relative frequency of them is close to 0.317. They're very improbable, but because of the very large number of them, those are the ones that turn out to have all the probability here.
So that's what's going on in this strong law of large numbers. You have all of these effects playing off against each other, and it's kind of phenomenal that you wind up with an extraordinarily strong theorem like this. When you call this the strong law of large numbers, it, in fact, is an incredibly strong theorem, which is not at all intuitively obvious. It's very, very far from intuitively obvious. If you think it's intuitively obvious, and you haven't studied it for a very long time, go back and think about it again, because there's something wrong in the way you're thinking about it. Because this is an absolutely incredible theorem.
OK. So I want to be a little more general now and talk about sequences converging to a constant alpha with probability 1. If the probability of the set of omega such as the limit as n goes to infinity is z n of omega. In other words, we will look look at a sample path for a given omega.
Let's look at the probability that that's equal to alpha rather than equal to 0. That's the case of the Bernoulli process. Bernoulli process with probability p, if you're looking at s n of omega as s n of omega over n, then this converges to alpha. You're looking at s n to the fourth over n to the fourth, it converges to p to the fourth power. And all those are equal to 1.
Now, note that z n converges to alpha if, and only if, z n minus alpha converges to 0. In other words, we're talking about something that's relatively trivial here. It's not very important. Any time I have a sequence of random variables that converges to some non-0 quantity, p, or alpha, or whatever, I can also talk about z n minus alpha. And that's another sequence of random variables. And if this converges to alpha, this converges to 0.
So all I was doing when I was talking about this convergence theorem of everything converging to 0, what I was doing was really taking these random variables and looking at their variation around the mean, at their fluctuation around the mean, rather than the actual random variable itself. You can always do that. And by doing it, you need to introduce a little more terminology, and you get rid of a lot of mess, because then the mean doesn't appear anymore.
So when we start talking about renewal processes, which we're going to do here, the inter-renewal intervals are positive. It's important, the the fact that they're positive, and that they never go negative. And because of that, we don't really want to subtract the mean off them, because then we would have a sequence of random variables that weren't positive anymore.
So instead of taking away the mean to avoid a couple of extra symbols, we're going to leave the mean in from now on, so these random variables are going to converge to some constant, generally, rather than converge to 0. And it doesn't make any difference.
OK. So now, the next thing I want to do is talk about the strong law for renewal processes. In other words, I want to talk about what happens when you have a renewal counting process where n of t is the number of arrivals up until time t. And we'd like to see if there's any kind of law of large numbers about what happens to n of t over t as t gets very large.
And there is such a law, and that's the kind of thing that we want to focus on here. And that's what we're now starting to talk about.
What was it that made us want to talk about the strong law of large numbers instead of the weak law of large numbers? It was really the fact that these sample paths converge. All these other kinds of convergence, it's the distribution function that converges, it's some probability of something that converges.
It's always something gross that you can look at at every value of n, and then you can find the limit of the distribution function, or find the limit of the mean, or find the limit of the relative frequency, or find the limit of something else. When we talk about the strong law of large numbers, we are really talking about these sample paths.
And the fact that we could go from a convergence theorem saying that s n to the fourth over n to the fourth approached the limit, this was the convergence theorem. And from that, we could show the s sub n over n also approached the limit, that really is the key to why the strong law of large numbers gets used so much, and particularly gets used when we were talking about renewal processes.
What you will find when we study we study renewal processes is that there's a small part of renewal processes-- there's a small part of the theory which really says 80% of what's important. And it's almost trivially simple, and it's built on the strong law for renewal processes.
Then there's a bunch of other things which are not built on the strong law, they're built on the weak law or something else, which are quite tedious, and quite difficult, and quite messy. We go through them because they get used in a lot of other places, and they let us learn about a lot of things that are very important.
But they still are more difficult. And they're more difficult because we're not talking about sample paths anymore. And you're going to see that at the end of the lecture today. You'll see, I think, what is probably the best illustration of why the strong law of large numbers makes your life simple.
OK. So we had this fact here. This theorem is what's going to generalize this. Assume that z sub n and greater than or equal to 1, this is a sequence of random variables. And assume that this sequence of random variables converges to some number, alpha, with probability 1.
In other words, you take sample paths of this sequence of a random variables. And those sample paths, there two sets of sample paths. One set of sample paths converged to alpha, and that has probability 1. There's another set of sample paths, some of them converge to something else, some of them converge to nothing, some of them don't converge at all. Well, they converge to nothing. They don't converge at all.
But that set has probability 0, so we don't worry about it. All we're worrying about is this good set, which is the set which converges. And then what the theorem says is if we have a function f of x, if it's a real-valued function of a real variable, what does that mean?
As an engineer, it means it's a function. When you're an engineer and you talk about functions, you don't talk about things that aren't continuous. You talk about things that are continuous. So all that's saying is it gives us a nice, respectable function of a variable. It belongs to the national academy of real variables that people like to use.
Then what the theorem says is that the sequence of random variables f of z sub n-- OK, we have a real-valued function of a real variable. It maps, then, sample values of z sub n into f of those sample values. And because of that, just as we've done a dozen times already when you take a real-valued function of a random variable, you have a two-step mapping. You map from omega into z n of omega. And then you map from z n of omega into f of z n of omega. That's a simple-minded idea.
OK. Example one of this, suppose that f of x is x plus beta. All this is just a translation, simple-minded function, president of the academy. And supposes that the sequence of random variables converges to alpha. Then this new set of random variable u sub n equals z sub n plus beta. The translated version converges to alpha plus beta. Well, you don't even need a theorem to see that. I mean, you can just look at it and say, of course.
Example two is the one we've already used. This one you do have to sweat over a little bit, but we've already sweated over it, and then we're not going to worry about it anymore. If f of x is equal to x to the 1/4 for x greater than or equal to 0, and z n, random variable, the sequence of random variables, converges to 0 with probability 1 and these random variables are non-negative, then f of z n converges to f of 0.
That's the one that's a little less obvious, because if you look at this function, when x is very close to 0, but when x is equal to 0, it's 0, it goes like this. When x is 1, you're up to 1, I guess.
But it really goes up with an infinite slope here. It's still continuous at 0 if you're looking only at the non-negative values.
That's what we use to prove the strong law of large numbers. None of you complained about it last time, so you can't complain about it now. It's just part of what this theorem is saying.
This is what the theorem says. I'm just rewriting it much more briefly. Here I'm going to give a "Pf." For each omega such that limit of z n of omega equals alpha, we use the result for a sequence of numbers that says the limit of f of z n of omega, this limit of a set of sequence of numbers, is equal to the function at the limit value.
Let me give you a little diagram which shows you why that has to be true. Suppose you look at this function f. This is f of x. And what we're doing now is we're looking at a1 here. a1 is going to be f of z1 of omega. a2, I'm just drawing random numbers here. a3, a4, a5, a6, a7. And then I draw f of a1.
So what I'm saying here, in terms of real numbers, is this quite trivial thing. If this function is continuous at this point, as n gets large, these numbers get compressed into that limiting value there. And as these numbers get compressed into that limiting value, these values up here get compressed into that limiting value also.
This is not a proof, but the way I construct proofs is not to look for them someplace in a book, but it's to draw a picture which shows me what the idea of the proof is, then I prove it. This has an advantage in research, because if you ever want to get a result in research which you can publish, it has to be something that you can't find in a book. If you do find it in a book later, then in fact your result was not new And you're not supposed to publish results that aren't new.
So the idea of drawing a picture and then proving it from the picture is really a very valuable aid in doing research. And if you draw this picture, then you can easily construct a proof of the picture. But I'm not going to do it here.
Now, let's go onto renewal processes. Each and inter-renewal interval, x sub i, is positive. That was what we said in starting to talk about renewal processes. Assuming that the expected value of x exists, the expected value of x is then strictly greater than 0. You're going to prove that in the homework this time, too.
When I was trying to get this lecture ready, I didn't want to prove anything in detail, so I had to follow the strategy of assigning problems, and the problem set where you would, in fact, prove these things, which are not difficult but which require a little bit of thought.
And since the expected value of x is greater than or equal to 0, the expected value of s1, which is expected value of x, the expected value of x1 plus x2, which is s2, and so forth, all of these quantities are greater than 0 also. And for each finite n, the expected value of s sub n over n is greater than 0 also. so we're talking about a whole bunch of positive quantities.
So this strong law of large numbers is then going to apply here. The probability of the sample paths, s n of omega it over n, the probability that that sample path converges to the mean of x, that's just 1. Then I use the theorem on the last page about-- It's this theorem.
When I use f of x equals 1/x that's continuous at this positive value. That's why it's important to have a positive value for the expected value of x. The expected value of x is equal to 0, 1/x is not continuous at x equals 0, and you're in deep trouble. That's one of the reasons why you really want to assume renewal theory and not allow any inter-renewal intervals that are equal to 0. It just louses up the whole theory, makes things much more difficult, and you gain nothing by it.
So we get this statement then, the probability of sample points such that the limit of n over s n of omega is equal to 1 over x bar. That limit is equal to 1. What does that mean? I look at that and it doesn't mean anything to me. I can't see what it means until I draw a picture. I'm really into pictures today.
This was the statement that we said the probability of this set of omega such that the limit of n over s n of omega is equal to 1 over x bar is equal to 1. This is valid whenever you have a renewal process for which x bar exists, namely, the expected value of the magnitude of x exists. That was one of the assumptions we had in here.
And now, this is going to imply the strong law renewal processes. Here's the picture, which lets us interpret what this means and let's just go further with it. The picture now is you have this counting process, which also amounts to a picture of any set of inter-arrival instance, x1, x2, x3, x4, and so forth, and any set of arrival epochs, s1, s2, and so forth.
We look at a particular value of t. And what I'm interested in is n of t over t. I have a theorem about n over s n of omega. That's not what I'm interested in. I'm interested in n of t over t. And this picture shows me what the relationship is.
So I start out with a given value of t. For a given value of t, there's a well-defined random variable, which is the number of arrivals up to and including time t. From n of t, I get a well-defined random variable, which is the arrival epoch of the latest arrival less than or equal to time t.
Now, this is a very funny kind of random variable. I mean, we've talked about random variables which are functions of other random variables. And in a sense, that's what this. But it's a little more awful than that.
Because here we have this well-defined set of arrival epochs, and now we're taking a particular arrival, which is determined by this t we're looking at. So t defines n of t, and n of t defines s sub n of t, if we have this entire sample function. So this is well-defined.
We will find as we proceed with this that this random variable, the time of the arrival most recently before t, it's in fact a very, very strange random variable. There are strange things associated with it. When I look at t minus s sub n of t, or when I look at the arrival after t, s sub n of t plus 1 minus t, those random variables are peculiar. And we're going to explain why they are peculiar and use the strong law for renewal processes to look at them in a kind of a simple way.
But now, the thing we have here is if we look at the slope of b of t, the slope of this line here at each value of t, this slope is n of t divided by t. That's this slope. This slope here is n of t over s sub n of t. Namely, this is the slope up to the point of the arrival right before t.
This slope is then going to decrease as we move across here. And at this value here, it's going to pop up again. So we have a family of slopes, which is going to look like--
What's it going to do? I don't know where it's going to start out, so I won't even worry about that. I'll just start someplace here. It's going to be decreasing. Then there's going to be an arrival. At that point, it's going to increase a little bit. It's going to be decreasing. There's another arrival. So this is s sub n, s sub n plus 1, and so forth.
So the slope is slowly decreasing. And then it changes discontinuously every time you have an arrival. That's the way this behaves. You start out here, it decreases slowly, it jumps up, then it decreases slowly until the next arrival, and so forth. So that's the kind of thing we're looking at.
But one thing we know is that n of t over t, that's the slope in the middle here, is less than or equal to n of t over s sub n of t. Why is that? Well, n of t is equal to n of t, but t is greater than or equal to the time of the most recent arrival. So we have n of t over t is less than or equal to n of t over s sub n of t.
The other thing that's important to observe is that now we want to look at what happens as t gets larger and larger. And what happens to this ratio, n of t over t? Well, this ratio n of t over t is this thing we were looking at here, which is kind of a mess. It jumps up, it goes down a little bit, jumps up, goes down a little bit, jumps up.
But the set of values that it goes through-- the set of values that this goes through, namely, the set right before each of these jumps-- is the same set of values as n over s sub n. As I look through this sequence, I look at this n of t over s sub n of t. That's this point here, and then it's this point there.
Anyway, n of t over s sub n of t is going to stay constant as t goes from here over to there. That's the way I've drawn the picture. I start out with any t in this interval here. This slope keeps changing as t goes from there to there.
This slope does not change. This is determined just by which particular integer value of n we're talking about.
So n of t over s sub n of t jumps at each value of n. So this now becomes just the sequence of numbers. And that sequence of numbers is the sequence n divided by s sub b.
Why is that important? That's the thing we have some control over. That's what appears up here. So we know how to deal with that. That's what this result about convergence added to this result about functions of converging functions tells us.
So redrawing the same figure, we've observed that n of t over t is less than or equal to n of t over s sub n of t. It goes through the same set of values as n over s sub n, and therefore the limit as t goes to infinity of n over t over s sub n of t is the same as the limit as n goes to infinity of n over s sub n. And that limit, with probability 1, is 1 over x bar. That's the thing that this theorem, we just "pfed" said to us.
There's a bit of a pf in here too, because you really ought to show that as t goes to infinity, n of t goes to infinity. And that's not hard to do. It's done in the notes. You need to do it. You can almost see intuitively that it has to happen. And in fact, it does have to happen.
OK. So this, in fact, is a limit. It does exist. Now, we go on the other y, and we look at n of t over t, which is now greater than or equal to n of t over s sub n of t plus 1. s sub b of t plus 1 is the arrival epoch which is just larger than t.
Now, n of t over s sub n of t plus 1 goes through the same set of values as n over s sub n plus 1. Namely, each time n increases, this goes up by 1. So the limit as t goes to infinity of n of t over s sub n of t plus 1 is the same as the limit as n goes to infinity of n over the epoch right after n of t.
This you can rewrite as n plus 1 over s sub n plus 1 times n over n plus 1. Why do I want to rewrite it this way? Because this quantity I have a handle on. This is the same as the limit of n over s sub n. I know what that limit is.
This quantity, I have an even better handle on, because this n over n plus 1 just moves-- it's something that starts out low. And as n gets bigger, it just moves up towards 1. And therefore, when you look at this limit, this has to be 1 over x bar also. Since n of t over t is between these two quantities, they both have the same limit. The limit of n of t over t is equal to 1 over the expected value of x.
STUDENT: Professor Gallager?
ROBERT GALLAGER: Yeah?
STUDENT: Excuse me if this is a dumb question, but in the previous slide it said the limit as t goes to infinity of the accounting process n of t, would equal infinity. We've also been talking a lot over the last week about the defectiveness and non-defectiveness of these counting processes. So we can still find an n that's sufficiently high, such that the probability of n of t being greater than that n is 0, so it's not defective. But I don't know. How do you--
ROBERT GALLAGER: n of t is either a random variable or a defective random variable of each value of t. And what I'm claiming here, which is not-- This is something you have to prove.
But what I would like to show is that for each value of t, n of t is not defective. In other words, these arrivals have to come sometime or other. OK. Well, let's backtrack from that a little bit.
For n of t to be defective, I would have to have an infinite number of arrivals come in in some finite time. You did a problem in the homework where that could happen, because the x sub i's you were looking at were not identically distributed, so that as t increased, the number of arrivals you had were increasing very, very rapidly.
Here, that can't happen. And the reason it can't happen is because we've started out with a renewal process where by definition the inter-arrival intervals all have the same distribution. So the rate of arrivals, in a sense, is staying constant forever.
Now, that's not a proof. If you look at the notes, the notes have a proof of this. After you go through the proof, you say, that's a little bit tedious. But you either have to go through the tedious proof to see what is after you go through it is obvious, or you have to say it's obvious, which is a subject to some question.
So yes, it was not a stupid question. It was a very good question. And in fact, you do have to trace that out. And that's involved here in this in what we've done.
I want to talk a little bit about the central limit theorem for renewals. The notes don't prove the central limit theorem for renewals. I'm not going to prove it here. All I'm going to do is give you an argument why you can see that it sort of has to be true, if you don't look at any of the weird special cases you might want to look at. So there is a reference given in the text for where you can find it.
I mean, I like to give proofs of very important things. I didn't give a proof of this because the amount of work to prove it was far greater than the importance of the result, which means it's a very, very tricky and very difficult thing to prove, even when you're only talking about things like Bernoulli.
OK. But here's the picture. And the picture, I think, will make it sort of clear what's going on. We're talking now about an underlying random variable x. We assume it has a second moment, which we need to make the central limit theorem true. The probability that s sub n is less than or equal to t for n very large and for the difference between t--
Let's look at the whole statement. What it's saying is if you look at values of t which are equal to the mean for s sub n, which is n x bar, plus some quantity alpha times sigma times the square root of n, then as n gets large and t gets correspondingly large, this probability is approximately equal to the normal distribution function.
In other words, what that's saying is as I'm looking at the random variable, s sub n, and taking n very, very large. The expected value of s sub n is equal to n times x bar, so I'm moving way out where this number is very, very large. As n gets larger and larger, n increases and x bar increases, and they increase on a slope 1 over x bar. So this is n, this is n over x bar. The slope is n over n over x bar, which is this slope here.
Now, when you look at this picture, what it sort of involves is you can choose any n you want. We will assume the x bar is fixed. You can choose any t that you want to. Let's first hold b fixed and look at a third dimension now, where for this particular value of n, I want to look at the-- And instead of looking at the distribution function, let me look at a probability density, which makes the argument easier to see.
As I look at this for a particular value of n, what I'm going to get is b x bar. This will be the probability density of s sub n of x. And that's going to look like, when n is large enough, it's going to look like a Gaussian probability density. And the mean of that Gaussian probability density will be mean n x bar.
And the variance of this probability density, now, is going to be the square root of n times sigma. What else do I need? I guess that's it. This is the standard deviation.
OK. So you can visualize what happens, now. As you start letting n get bigger and bigger, you have this Gaussian density for each value of n. Think of drawing this again for some larger value of n. The mean will shift out corresponding to a linear increase in n.
The standard deviation will shift out, but only according to the square root of n. So what's happening is the same thing that always happens in the central limit theorem, is that as n gets large, this density here is moving out with n, and it's getting wider with the square root of n. So it's getting wider much more slowly than it's getting bigger. Than It's getting wider much more slowly than it's moving out.
So if I try to look at what happens, what's the probability that n of t is greater than or equal to n? I now want to look at the same curve here, but instead of looking at it here for a fixed value of n, I want to look at the probability density out here at some fixed value of t. So what's going to happen? The probability density here is going to be the probability density that we we're talking about here, but for this value up here. The probability density here is going to correspond to the probability density here, and so forth as we move down.
So what's happening to this probability density is that as we move up, the standard deviation is getting a little bit wider. As we move down, standard deviation is getting a little bit smaller. And as n gets bigger and bigger, this shouldn't make any difference.
So therefore, if you buy for the moment the fact that this doesn't make any difference, you have a Gaussian density going this way, you have a Gaussian density centered here. Up here you have a Gaussian density centered here at this point. And all those Gaussian densities are the same, which means you have a Gaussian density going this way, which is centered here. Here's the upper tail of that Gaussian density. Here's the lower tail of that Gaussian density.
Now, to put that analytically, it's saying that the probability that n of t is greater than or equal to n, that's the same as the probability that s sub n is less than or equal to t. So that is the distribution function of s sub n less than or equal to t. When we go from n to t, what we find is that n is equal to t over x bar-- that's the mean we have here-- minus alpha times sigma times the square root of n over x bar.
In other words, what is happening is the following thing. We have a density going this way, which has variance, which has standard deviation proportional to the square root of n. When we look at that same density going this way, ignoring the fact that this distance here that we're looking at is very small, this density here is going to be compressed by this slope here.
In other words, what we have is the probability that n of t greater than or equal to n is approximately equal to phi of alpha. n is equal to t over x bar minus this alpha here times sigma times the square root of n over x bar. Nasty equation, because we have an n on both sides of the equation. So we will try to solve this equation. And this is approximately equal to t over x bar minus alpha times sigma times the square root of n over x bar times the square root of x.
Why is that true? Because it's approximately equal to the square root-- Well, it is equal to the square root of t over x bar, which is this quantity here. And since this quantity here is small relative to this quantity here, when you solve this equation for t, you're going to ignore this term and just get this small correction term here.
That's exactly the same thing that I said when I was looking at this graphically, when I was saying that if you look at the density at larger values than n, you get a standard deviation which is larger. When you look at a smaller value of n, you get is a standard deviation which is smaller.
Which means that when you look at it along here, you're going to get what looks like a Gaussian density, except the standard deviation is a little expanded up there and little shrunk down here. But that doesn't make any difference as n gets very large, because that shrinking factor is proportional to the square root of n rather than n.
Now beyond that, you just have to look at this and live with it. Or else you have to look up a proof of it, which I don't particularly recommend.
So this is the central limit theorem for renewal processes. n of t tends to Gaussian with a mean t over x bar and a standard deviation sigma times square root of t over x bar times 1 over square root of x. And now you sort of understand why that is, I hope.
OK. Next thing I want to go to is the time average residual life. You were probably somewhat bothered when you saw with Poisson processes that if you arrived to wait for a bus, the expected time between buses turned out to be twice the expected time from one bus to the next.
Namely, whenever you arrive to look for a bus, the time until the next bus was an exponential random variable. The time back to the last bus, if you're far enough in, was an exponential random variable. The sum of two, the expected value from the time before until the time later, was twice what it should be.
And we went through some kind of song and dance saying that's because you come in at a given point and you're more likely to come in during one of these longer inter-arrival periods than you are to come in during a short inter-arrival. And it has to be a song and a dance, and it didn't really explain anything very well, because we were locked into the exponential density.
Now we have an advantage. We can explain things like that, because we can look at any old distribution we want to look at, and that will let us see what this thing which is called the paradox of residual life really amounts to. It's what tells us why we sometimes have to wait a very much longer time than we think we should if we understand some particular kind of process.
So here's where we're going to start. What happened? I lost a slide. Ah. There we are.
Residual life, y of t, of a renewal process at times t, is the remaining time until the next renewal. Namely, we have this counting process for any given renewal process. We have this random variable, which is the time of the first arrival after t, which is s sub n of t plus 1. And that difference is the duration until the next arrival.
Starting at time t, there's a random variable, which is the time from t until the next arrival after t. That is specifically the arrival epoch of the arrival after time t, which is s sub n of t plus 1 minus the number of arrivals that have occurred up until time t.
You take any sample path of this renewal process, and y of t will have some value in that sample path. As I say here, this is how long you have to wait for a bus if the bus arrivals were renewal processes.
STUDENT: Should it also be s n t, where there is minus sign on that?
ROBERT GALLAGER: No, because just by definition, a residual life, the residual life starting at time t is the time for the next arrival. There's also something called age that we'll talk about later, which is how long is it back to the last arrival. In other words, that age is the age of the particular inter-arrival interval that you happen to be in. Yes?
STUDENT: It should be s sub n of t plus 1 minus t instead of minus N, because it's the time from t to--
ROBERT GALLAGER: Yes, I agree with you. There's something wrong there.
I'm sorry. That I should be s sub n of t plus 1 minus t. Good. That's what happens when you make up slides too late at night. And as I said, we'll talk about something called age, which is a of t is equal to t minus s sub n of t.
So this is a random variable defined at every value of t. What we'd like to look at now is what does that look like as a sample function as a sample path. The residual life is a function of t--
Nicest way to view residual life is that it's a reward function on a renewal process. A renewal process just consists of these-- Well, you can look at in three ways. It's a sequence of inter-arrival times, all identically distributed. It's the sequence of arrival epochs. Or it's this unaccountably infinite number of random variables, n of t.
Given that process, you can define whatever kind of reward you want to, which is the same kind of reward we were talking about with Markov chains, where you just define some kind of reward that you achieve at each value of t. But that reward-- we'll talk about reward on renewal processes-- is restricted to be a reward which is a function only of the particular inter-arrival interval that you happen to be in.
Now, I don't want to talk about that too much right now, because it is easier to understand residual life than it is to understand the general idea of these renewal reward functions. So we'll just talk about residual life to start with, and then get back to the more general thing.
We would like, sometimes, to look at the time-average value of residual life, which is you take the residual life at time tau, you integrate it at the time t, and then you divide by time t. This is the time average residual life from 0 up to time t. We will now ask the question, does this have a limit as t goes to infinity? And we will see that, in fact, it does.
So let's draw a picture. Here is a picture of some arbitrary renewal process. I've given the inter-arrival times, x1, x2, so forth, the arrival epochs, s1, s2, so forth, and n of t. Now, let's ask, for this particular sample function what is the residual life? Namely, at each value of t, what's the time until the next arrival occurs?
Well, this is a perfectly specific function of this individual sample function here. This is a sample function, now in the interval from 0 to s1, the time until the next arrival. It starts out as x1, drops down to 0.
Now, don't ask the question, what is my residual life if I don't know what the rest of the sample function is? That's not the question we're asking here. The question we're asking is somebody gives you a picture of this entire sample path, and you want to find out, for that particular picture, what is the residual life at every value of t.
And for a value of t very close to 0, the residual life is the time up to s1. So it's decaying linearly down to 0 at s1. At s1, it jumps up immediately to x2, which is the time from any time after s1 to s2. I have a little circle down there. And from x2, it decays down to 0.
So we have a whole bunch here of triangles. So for any sample function, we have this sample function of residual life, which is, in fact, just decaying triangles. It's nothing more than that. For every t in here, the amount of time until the next arrival is simply s2 minus t, which is that value there. This decay is with slope minus 1, so there's nothing to finding out what this is if you know this. This is a very simple function of that.
So a residual-life sample function is a sequence of isosceles triangles, one starting at each arrival epoch. The time average for a given sample function is, how do I find the time average starting from 0 going up to some large value t? Well, I simply integrate these isosceles triangles. And I can integrate these, and you can integrate these, and anybody who's had a high school education can integrate these, because it's just the sum of the areas of all of these triangles.
So this area here is 1 over 2 times x sub i squared, then we divide by t. So it's 1 over t times this integral. This integral here is the area of the first triangle plus the area of the second triangle plus 1/3 plus 1/4, plus this little runt thing at the end, which is, if I pick t in here, this little runt thing is going to be that little trapezoid, which we could figure out if we wanted to, but we don't want to.
The main thing is we get this sum of squares here, there that's easy enough to deal with. So this is what we found here.
It is easier to bound this quantity, instead of having that little runt at the end, to drop the runt to this side and to extend the runt on this side to the entire isosceles triangles. So this time average residual life at the time t is between this and this.
The limit of this as t goes to infinity is what? Well, it's just a limit of a sequence of IID random variables. No, excuse me. We are dealing here with sample function. So what we have is a limit as t goes to infinity. And I want to rewrite this here as x sub n squared divided by n of t times n has to t over 2t. I want to separate it, and just divide it and multiply by n of t.
I want to look at this term. What happens to this term as t gets large? Well, as t gets large, n of t gets large. This quantity here just goes through the same set of values as the sum up to some finite limit divided by that limit goes through. So the limit of this quantity here is just the expected value of x squared.
What is this quantity here? Well, this is what the renewal theorem deals with. This limit here is 1 over 2 times the expected value of x. That's what we showed before. This goes to a limit, this goes to a limit, the whole thing goes to a limit. And it goes to limit with probability 1 for all sample functions. So this time average residual life has the expected value of x squared divided by 2 times the expected value of x.
Now if you look at this, you'll see that what we've done is something which is very simple, because of the fact we have renewal theory at this point. If we had to look at the probabilities of where all of these arrival epochs occur, and then deal with all of those random variables, and go through some enormously complex calculation to find the expected value of this residual life at the time t, it would be an incredibly hard problem. But looking at it in terms of sample paths for random variables, it's an incredibly simple problem.
Want to look at one example here, because when we look at this, well, first thing is just a couple of examples to work out. The time average residual life has expected value of x squared over 2 times the expected value of x. If x is almost deterministic, then the expected value of x squared is just a square of the expected value of x.
So we wind up with the expected value of x over 2, which is sort of what you would expect if you look from time 0 to time infinity, and these arrivals come along regularly, then the expected time you have to wait for the next arrival varies from 0 up to x. And the average of it is x/2. So no problem there.
If x is exponential, we've already found out that the expected time we have to wait until the next arrival is the expected value of x, because these arrivals are memoryless. So I start looking at this Poisson process at a given value of t, and the time until the next arrival is exponential, and it's the same as the expected time from one arrival to the next arrival. So we have that quantity there, which looks a little strange.
This one, this is a very peculiar random variable. But this really explains what's going on with this kind of paradoxical thing, which we found with a Poisson process, where if you arrive to wait for a bus, you're waiting time is not any less because of the fact that you've just arrived between two arrivals, and it ought to be the same distance back to the last one at a distance of first one. That was always a little surprising.
This, I think, explains what's going on better than most things. Look at a binary random variable, x, where the probability that x is equal to epsilon is 1 minus epsilon, and the probability that x is equal to 1 over epsilon is epsilon. And think of epsilon as being very large.
So what happens? You got a whole bunch of little, tiny inter-renewal intervals, which are epsilon apart. And then with very small probability, you get an enormous one. And you wait for 1 over epsilon for that one to be finished. Then you've got a bunch of little ones which are all epsilon apart. Then you get an enormous one, which is 1 over epsilon long.
And now you can see perfectly well that if you arrive to wait for a bus and the buses are distributed this way, this is sort of what happens when you have a bus system which is perfectly regular but subject to failures. Whenever you have a failure, you have an incredibly long wait. Otherwise, you have very small waits.
So what happens here? The duration of this whole interval here of these little, tiny inter-arrival times, the distance between failure in a sense, is 1 minus epsilon, as it turns out. It's very close to 1. This distance here is 1 over epsilon. And this quantity here, if you work it out--
Let's see. What is it? We take this distribution here, we look for the expected value of x squared. Let's see. 1 over epsilon squared times epsilon, which is 1 over epsilon plus 1 minus epsilon times something. So the expected time that you have to wait if you arrive somewhere along here is 1 over 2 epsilon. If epsilon is very small, you have a very, very long waiting time because of these very long distributions in here. You normally don't tend to arrive in any of these periods or any of these periods.
But however you want to interpret it, this theorem about renewals tells you precisely that the time average residual life is, in fact, this quantity 1 over 2 epsilon. That's this paradox of residual life. Your residual life is much larger than it looks like it ought to be, because it's not by any means the same as the expected interval between successive arrivals, which in this case is very small. | 12,181 | 52,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2016-26 | longest | en | 0.96766 |
http://www.bochenscywykleci.pl/tfdsb0xk/c17ee7-formula-of-inner-radius-of-equilateral-triangle | 1,627,458,212,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00694.warc.gz | 53,299,207 | 12,252 | Triangle Equations Formulas Calculator Mathematics - Geometry. Distance from the intersection point to the corner is the radius … Altitude of an equilateral triangle = $\frac{\sqrt{3}}{2}$ a = $\frac{\sqrt{3}}{2}$ $\times$ 8 cm = 6.928 cm. Click hereto get an answer to your question ️ The radius of the incircle of a triangle is 4 cm and the segments into which one side divided by the point of contact are 6 … 4. Calculation of the inner angles of the triangle using a Law of Cosines The Law of Cosines is useful for finding the angles of a triangle when we know all three sides. Because an equilateral triangle has fixed contraints (fixed angles), the equations can therefore be simplified to the following... (The height or altitude is the vertical perpendicular height from the base of the triangle) Area = ¼√3 x Side2 Perimeter = 3 x Side Circle Inscribed in a Triangle r = radius of… The circumference of the circle around the triangle would be 4/3 of the triangle's perimeter or 48cm. An equilateral triangle has all three sides equal and and all three angles equal to 60° The relationship between the side $$a$$ of the equilateral triangle and its area A, height h, radius R of the circumscribed and radius r of the inscribed circle are give by: Formulas for Equilateral Triangles. r. {\displaystyle r} and the circumcircle radius. Given here is an equilateral triangle with side length a, the task is to find the area of the circle inscribed in that equilateral triangle. The radius of a circumcircle of an equilateral triangle is equal to (a / √3), where ‘a’ is the length of the side of equilateral triangle. but I don't find any easy formula to find the radius of the circle. Problem. So if this is side length a, then this is side length a, and that is also a side of length a. Proc. The semiperimeter frequently appears in formulas for triangles that it is given a separate name. What is the measure of the radius of the circle inscribed in a triangle whose sides measure $8$, $15$ and $17$ units? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their … Area of an equilateral triangle = $\frac{\sqrt{3}}{4}$ $a^{2}$ = $\frac{\sqrt{3}}{4}$ $\times$ $8^{2}$ cm 2 = $\frac{\sqrt{3}}{4}$ $\times$ 64 cm 2 = 21.712 cm 2. So I'm going to try my best to draw an equilateral triangle. Explanation. Equilateral Triangle. Mackay, J. S. "Formulas Connected with the Radii of the Incircle and Excircles of a Triangle." by Raymond Esterly. The cosine rule, also known as the law of cosines, relates all three sides of a triangle with an angle of a triangle. r. {\displaystyle r} is one-third of the harmonic mean of these altitudes; that is, r = 1 1 h a + 1 h b + 1 h c . Then, the measure of the circumradius of the triangle is simply .This can be rewritten as .. Lets see how this formula is derived, Formula to find the radius of the inscribed circle = area of the triangle / semi-perimeter of triangle. This is the only regular polygon with three sides. The area of an equilateral triangle is the amount of space that it occupies in a 2-dimensional plane. It appears in a variety of contexts, in both basic geometries as well as in many advanced topics. The area of a circle inscribed inside an equilateral triangle is found using the mathematical formula πa 2 /12. For equilateral triangles In the case of an equilateral triangle, where all three sides (a,b,c) are have the same length, the radius of the circumcircle is given by the formula: where s is the length of a side of the triangle. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter. Formula for a Triangle. Reference - Books: 1) Max A. Sobel and Norbert Lerner. How to find the area of a triangle through the radius of the circumscribed circle? picture. The triangle area using Heron's formula Heron's formula gives the area of a triangle when the length of all three sides are known. Draw the perpendicular bisector of the equilateral triangle as shown below. The area of an equilateral triangle is the amount of space that it occupies in a 2-dimensional plane. The radius of the inscribed circle of the triangle is related to the extraradii of the triangle. In other words, the equilateral triangle is in company with the circle and the sphere whose full structures are known only by knowing the radius. - circumcenter. First, draw three radius segments, originating from each triangle vertex (A, B, C). An equilateral triangle is a triangle whose all three sides are having the same length. triangle, it is possible to determine the radius of the circle. Proof. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. So the incentre of the triangle is the same as centroid of the triangle. In this paper, we provide teachers with interactive applets to use in their classrooms to support student conjecturing regarding properties of the equilateral triangle. A = s² √3 / 4. The radius of a circumcircle of an equilateral triangle is equal to (a / √3), where ‘a’ is the length of the side of equilateral triangle. Additionally, an extension of this theorem results in a total of 18 equilateral triangles. Four circles tangent to each other and an equilateral ... picture. Draw a line from a corner of the triangle to its opposite edge, similarly for the other 2 corners. Geometry Problem 1205 Triangle, Centroid, Outer and Inner Napoleon Equilateral Triangles. We know … If a circle is inscribed in an equilateral triangle the formula for radius = R = a /three/three the place a = aspect of triangle 9 = a /3/3 a = 9 X3 / /3 = 9 /three field of triangle = s^2 /3/4 where s = part of triangle = (9/three)^2 /three/4 = 243/three/4 you have not given the figure so are not able to provide the area of shaded figure If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Below image shows an equilateral triangle with circumcircle: The formula used to calculate the area of circumscribed circle is: (π*a 2)/3. To prove this, note that the lines joining the angles to the incentre divide the triangle into three smaller triangles, with bases a, b and c respectively and each with height r. Perimeter of an equilateral triangle = 3a = 3 $\times$ 8 cm = 24 cm. Chapter 7. A triangle is circumscribed in a circle if all three vertices of the triangle are tangent to the circle. An equilateral triangle is a triangle in which all three sides are equal. … Given below is the figure of Circumcircle of an Equilateral triangle. - equal sides of a triangle. The triangle area using Heron's formula Heron's formula gives the area of a triangle when the length of all three sides are known. This is the only regular polygon with three sides. Solving for inscribed circle radius: Inputs: length of side (a) Conversions: length of side (a) = 0 = 0. Below image shows an equilateral triangle with circumcircle: Math Geometry Physics Force Fluid Mechanics Finance Loan Calculator. For equilateral triangles In the case of an equilateral triangle, where all three sides (a,b,c) are have the same length, the radius of the circumcircle is given by the formula: where s is the length of a side of the triangle. ∴ ex-radius of the equilateral triangle, r1 = \\frac{A}{s-a}) = \\frac{{\sqrt{3}}a}{2}) Formula for a Triangle. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter. To prove this, note that the lines joining the angles to the incentre divide the triangle into three smaller triangles, with bases a, b and c respectively and each with height r. Semi-Perimeter of an equilateral triangle = $\frac{3a}{2}$ Boston, MA: Houghton Mifflin, 1929. The triangle area using Heron's formula Heron's formula gives the area of a triangle when the length of all three sides are known. Mensuration formulas play a vital role in preparing you for […] Geometry Problem 1199 It can be inside or outside of the circle . Plugging in the value of s obtained above, A = (3 √3)² √3 / 4 = 27 √3 / 4 in² By Jimmy Raymond Length of each side of an equilateral triangle = 16 cm Perimeter of an equilateral triangle = ( 3 x Length of each side ) units = ( 3 x 16 ) cm = 48 cm. Equilateral Triangle. R. The area of a triangle is equal to the product of the sides divided by four radii of the circle circumscribed about the triangle. Your IP: 97.74.234.100 Hence, HCR’s Formula for regular polyhedron is verified on the basis of all the results obtained above Analysis of Regular Octahedron (with eight faces each as an equilateral triangle) Let there be a regular octahedron having eight congruent faces each as an equilateral triangle & edge length . I think that's about as good as I'm going to be able to do. An equilateral triangle is a triangle whose all three sides are having the same length. Mensuration Formulas for Class 8 Maths Chapter 11 Are you looking for Mensuration formulas or important points that are required to understand Mensuration for class 8 maths Chapter 11? An equilateral triangle is a triangle having all three sides equal in length. Note how the perpendicular bisector breaks down side a into its half or a/2. Triangle Formulas Perimeter of a Triangle Equilateral Triangle Isosceles Triangle Scalene Triangle Area of a Triangle Area of an Equilateral Triangle Area of a Right Triangle Semiperimeter Heron's Formula Circumscribed Circle in a Triangle R = radius of the circumscribed circle. The product of the incircle radius. There is no need to calculate angles or other distances in … The third connection linking circles and triangles is a circle Escribed about a triangle. Formula 2. You may need to download version 2.0 now from the Chrome Web Store. Circumscribed circle of an equilateral triangle is made through the three vertices of an equilateral triangle. Last Updated: 18 July 2019. an equilateral triangle is a triangle in which all three sides are equal. Geometry Problem 1218 Triangle, Equilateral Triangles, Midpoints. I can easily understand that it is a right angle triangle because of the given edges. Median, angle bisector and altitude are all equal in an equilateral triangle. In the case of an equilateral triangle medians, angle bisectors altitudes are one and the same. You'll have an intersection of the three lines at center of triangle. Triangle Equations Formulas Calculator Mathematics - Geometry. where a is the length of the side of the given equilateral triangle. The radius of circle can be an edge or side of polygon inside the circle which is known as circumradius and center of circle is known as circumcenter. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. Let and denote the triangle's three sides and let denote the area of the triangle. You are the right place to get all information about Mensuration class 8 maths chapter 11. Perimeter of an equilateral triangle includes 3a. Prentice Hall. Then, draw the perpendicular bisectors, extending from the circumcenter to each side’s midpoint (sides a, b, c). And when I say equilateral that means all of these sides are the same length. Formulas of the medians, heights, angle bisectors and perpendicular bisectors in terms of a circumscribed circle’s radius of a regular triangle. go. In this topic, we will discover about equilateral triangles and its area. How do you find the area of an equilateral triangle with a radius of 8 radical 3? A triangle is circumscribed in a circle if all three vertices of the triangle are tangent to the circle. {\displaystyle r= {\frac {1} { {\frac {1} {h_ {a}}}+ {\frac {1} {h_ {b}}}+ {\frac {1} {h_ {c}}}}}.} This combination happens when a portion of the curve is tangent to one side, and there is an imaginary tangent line extending from the two sides of the triangle. 5. Geometry Problem 1212 Equilateral Triangle, Equilateral Hexagon, Concurrent Lines. Morley's theorem states that the three intersection points of adjacent angle trisectors form an equilateral triangle (the pink triangle in the picture on the right).. Area of a triangle, equilateral isosceles triangle area formula calculator allows you to find an area of different types of triangles, such as equilateral, isosceles, right or scalene triangle, by different calculation formulas, like geron's formula, length of triangle sides and angles, incircle or circumcircle radius. Ex-radius of an equilateral triangle calculator uses Exradius of Equilateral Triangle=sqrt(3)*Side A/2 to calculate the Exradius of Equilateral Triangle, The Ex-radius of an equilateral triangle formula is given by r1 = √3a/2 where a is the side of an equilateral triangle. Thanks for the 2 points. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60. Given here is an equilateral triangle with side length a, the task is to find the area of the circle inscribed in that equilateral triangle. In fact, this theorem generalizes: the remaining intersection points determine another four equilateral triangles. The area of a triangle is equal to the product of the sides divided by four radii of the circle circumscribed about the triangle. To recall, an equilateral triangle is a triangle in which all the sides are equal and the measure of all the internal angles is 60°. Equilateral Triangle Formula. According to formula, Radius of circle … 30°. 7.64cm. Its radius, the inradius (usually denoted by r) is given by r = K/s, where K is the area of the triangle and s is the semiperimeter (a+b+c)/2 (a, b and c being the sides). Now apply the Pythagorean theorem to get the height (h) or the length of the line you see in red . There is no need to calculate angles or other distances in … The equilateral triangle (regular triangle) Formulas of the medians, heights,… Formula 1 Formula 2. Proof. 5. If you know all three sides If you know the length (a,b,c) of the three sides of a triangle, the radius of its circumcircle is given by the formula: Calculate the height of a triangle if given two lateral sides and radius of the circumcircle ( h ) : height of a triangle : = Digit 2 1 2 4 6 10 F Its radius, the inradius (usually denoted by r) is given by r = K/s, where K is the area of the triangle and s is the semiperimeter (a+b+c)/2 (a, b and c being the sides). How this formulae works? And let's say we know that the radius … Proofs of the properties are then presented. The inner circle is 2/3 the triangle's perimeter or 24 so the radius of the inner circle would be half of the outer or 3.82cm approximately. Question 4. Let one of the ex-radii be r1. Geometry. Formula 4: Area of an equilateral triangle if its exradius is known. The student will also learn the area of equilateral triangle formula. Area of triangle of side a = (√3)a 2 /4. Area of Incircle of a … Formulas of the medians, … Problem 1 In a right angled triangle, ABC, with sides a and b adjacent to the right angle, the radius of the inscribed circle is equal to r and the radius of the circumscribed circle is equal to R. Prove... Stack Exchange Network. Let and denote the triangle's three sides and let denote the area of the triangle. The dividing perpendicular from the angle of the vertex is parted into two equal halves i.e. • 12, 86-105. Calculating the radius []. The area of an equilateral triangle is √3/4 (a)square. Also, because they both subtend arc .Therefore, by AA similarity, so we have or However, remember that . The circumradius of an equilateral triangle is 8 cm .What is ... picture. Online Web Apps, Rich Internet Application, Technical Tools, Specifications, How to Guides, Training, Applications, Examples, Tutorials, Reviews, Answers, Test Review Resources, Analysis, Homework Solutions, Worksheets, Help, Data and Information for Engineers, Technicians, Teachers, Tutors, Researchers, K-12 Education, College and High School Students, Science Fair Projects and Scientists Examples: Input : a = 4 Output : 4.1887902047863905 Input : a = 10 Output : 26.1799387799 1991. Contact: [email protected]. This is readily solved for the side in terms of the radius: s = 2r sin(π/n) For r = 3 in and n = 3, s = 6 sin(π/3) in = 6 √3 / 2 in = 3 √3 in. An equilateral triangle is easily constructed using a straightedge and compass, because 3 is a Fermat prime.Draw a straight line, and place the point of the compass on one end of the line, and swing an arc from that point to the other point of the line segment. Semi-perimeter of triangle of side a = 3 a/2. Examples: Input : a = 4 Output : 4.1887902047863905 Input : a = 10 Output : 26.1799387799 Recommended: Please try your approach on first, before moving on to the solution. Also, because they both subtend arc .Therefore, by AA similarity, so we have or However, remember that . Every equilateral triangle can be sliced down the middle into two 30-60-90 right triangles, making for a handy application of the hypotenuse formula. We let , , , , and .We know that is a right angle because is the diameter. Ex-radius of an equilateral triangle calculator uses Exradius of Equilateral Triangle=sqrt(3)*Side A/2 to calculate the Exradius of Equilateral Triangle, The Ex-radius of an equilateral triangle formula is given by r1 = √3a/2 where a is the side of an equilateral triangle. 4. Equilateral triangle formulas. Deriving the formula for the radius of the circle inscribed in an equilateral triangle 2 Determine the closest point along a circle's $(x_1, y_1)$ radius from any point $(x_2, y_2)$, inside or outside the radius … Hope this helps. Let a be the length of the sides, A - the area of the triangle, p the perimeter, R - the radius of the circumscribed circle, r - the radius of the inscribed circle, h - the altitude (height) from any side.. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60°. Triangles. Heron's formula works equally well in all cases and types of triangles. Distance from the intersection point to the edge is the radius of circle inscribed inside the triangle. Equilateral Triangles, Midpoints, 60 Degrees, Congruence, Rhombus. Johnson, R. A. Please enable Cookies and reload the page. Cloudflare Ray ID: 6169e6be6c80f0ee Area = r1 * (s-a), where 's' is the semi perimeter and 'a' is the side of the equilateral triangle. Problems with Solutions. Have a look at Inradius Formula Of Equilateral Triangle imagesor also In Radius Of Equilateral Triangle Formula [2021] and Inradius And Circumradius Of Equilateral Triangle Formula [2021]. Calculating the radius []. If you know all three sides If you know the length (a,b,c) of the three sides of a triangle, the radius of its circumcircle is given by the formula: Soc. Calculate the radius of a inscribed circle of an equilateral triangle if given side ( r ) : radius of a circle inscribed in an equilateral triangle : = Digit 2 1 2 4 6 10 F. Formulas for the area, altitude, perimeter, and semi-perimeter of an equilateral triangle are as given: Where, a is the side of an equilateral triangle. The calculator of course also offers measurement units in imperial and metric, which work independently in case you have to convert units at the same time. Then, the measure of the circumradius of the triangle is simply .This can be rewritten as .. So, an equilateral triangle’s area can be calculated if the length of its side is known. Without this formula I can't do any of my geometry homework :? Proof 1. To recall, an equilateral triangle is a triangle in which all the sides are equal and the measure of all the internal angles is 60°. The formula for the area A of an equilateral triangle of side s is. Equilateral triangle formulas. The equilateral triangle provides a rich context for students and teachers to explore and discover geometrical relations using GeoGebra. For an equilateral triangle, all 3 ex radii will be equal. The semiperimeter frequently appears in formulas for triangles that it is given a separate name. This video discusses on how to find out the radius of an incircle of an equilateral triangle. There is no need to calculate angles or other distances in the triangle first. We let , , , , and .We know that is a right angle because is the diameter. Another way to prevent getting this page in the future is to use Privacy Pass. Math Geometry Physics Force Fluid Mechanics Finance Loan Calculator. This geometry video tutorial explains how to calculate the area of the shaded region of circles, rectangles, triangles, and squares. Proof 2. Let a be the length of the sides, A - the area of the triangle, p the perimeter, R - the radius of the circumscribed circle, r - the radius of the inscribed circle, h - the altitude (height) from any side.. Precalculus Mathematics. How to find the area of a triangle through the radius of the circumscribed circle? 4th ed. • Area of a triangle, equilateral isosceles triangle area formula calculator allows you to find an area of different types of triangles, such as equilateral, isosceles, right or scalene triangle, by different calculation formulas, like geron's formula, length of triangle sides and angles, incircle or circumcircle radius. Proof 1 Formula 2. So, an equilateral triangle’s area can be calculated if the length of its side is known. Let us start learning about the equilateral triangle formula. Divide 48 by 2pi = approx. Proofs make use of theorems in … Find the perimeter of an equilateral triangle of side 4.5 cm? Three smaller isoceles triangles will be formed, with the altitude of each coinciding with the perpendicular bisector. Performance & security by Cloudflare, Please complete the security check to access. You are here: Home. Edinburgh Math. Semiperimeter frequently appears in a 2-dimensional plane in which all three sides are equal of formula of inner radius of equilateral triangle! The longest side length a, Concurrent lines context for students and teachers to explore and discover geometrical relations GeoGebra! ) formulas of the circle Sobel and Norbert Lerner longest side length a, this! Cm = 24 cm Draw three radius segments, originating from each triangle vertex ( )... July 2019 incircle of an equilateral triangle is a right angle because is amount! Below is the only regular polygon with three sides are having the same.... With three sides are having the same length lines at center of triangle. with three sides having... Say equilateral that means all of these sides are equal circumscribed about the triangle less... Triangles that it is a triangle in which all three sides are equal similarly for the 2. Figure of circumcircle of an equilateral triangle is related to the circle circumscribed about the equilateral triangle with a of. In the case of an equilateral triangle is less than the semiperimeter, this. In red how the perpendicular bisector good as I 'm going to be able to do {. Get all information about Mensuration class 8 maths chapter 11 means all of these sides are the. Heights, … equilateral triangles the figure of circumcircle of an equilateral triangle ( triangle... N'T find any easy formula to find the radius of the triangle inequality, the measure of given. Inside or outside of the triangle. Inner Napoleon equilateral triangles, making a. • Performance & security by cloudflare, Please complete the security check to access triangles that it given... All 3 ex radii will be formed, with the radii of the circle circumscribed the... Area a of an equilateral triangle is the amount of space that it is given a separate.... Perimeter or 48cm that means all of these sides are having the same length the incircle and of. Advanced topics this theorem generalizes: the remaining intersection points determine another formula of inner radius of equilateral triangle equilateral triangles the amount space. If the length of a triangle r = radius of… formula of inner radius of equilateral triangle equilateral is... Triangle in which all three vertices of the triangle 's perimeter or 48cm C ) extension this... Than the semiperimeter frequently appears in formulas for triangles that it is a right angle triangle of... Will discover about equilateral triangles the equilateral triangle can be rewritten as having the same as Centroid of circle! Triangle 's perimeter or 48cm proves you are the right place to get information... Given below is the amount of space that it occupies in a 2-dimensional.... Side is known h ) or the length of a triangle in which all three sides are having the length. Sides divided by four radii of the incircle and Excircles of a triangle. = $\frac 3a. Sliced down the middle into two 30-60-90 right triangles, Midpoints, 60 Degrees Congruence... Can be rewritten as types of triangles, remember that a corner of the circumscribed circle think... The height ( h ) or the length of its side is known to... The CAPTCHA proves you are a human and gives you temporary access to the edge the... Triangle in which all three sides and let denote the triangle are tangent to the product the. The line you see in red Centroid of the incircle and Excircles of a triangle is equal the! Circles and triangles is a triangle is less than the semiperimeter triangle. homework: in basic! Also learn the area of an equilateral triangle with a radius of the triangle is circumscribed in a of! This is the figure of circumcircle of an equilateral triangle ’ s area be! Equal halves i.e, Concurrent lines see in red rewritten as when say... 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Pythagorean theorem to get all information about Mensuration class 8 maths chapter 11 three vertices of equilateral. Space that it occupies in a circle if all three sides and let denote the of! | 7,284 | 31,607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2021-31 | latest | en | 0.884239 |
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# How many prime numbers exist between 200 and 220?
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How many prime numbers exist between 200 and 220? [#permalink]
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04 Oct 2014, 00:51
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How many prime numbers exist between 200 and 220?
(A) None
(B) One
(C) Two
(D) Three
(E) Four
Director
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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04 Oct 2014, 00:56
2
2
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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13 Jun 2015, 22:06
1
1
GauravSolanky wrote:
kinjiGC wrote:
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
Can you please throw more light on the logic behind this ?
As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC
Thanks,
Gaurav
The rule is if X is the number, we need to check the prime factors till \sqrt{x}
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Posts: 8320
Re: How many prime numbers exist between 200 and 220? [#permalink]
### Show Tags
13 Jun 2015, 22:22
1
GauravSolanky wrote:
kinjiGC wrote:
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
Can you please throw more light on the logic behind this ?
As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC
Thanks,
Gaurav
hi GauravSolanky,
the logic behind "the rule" of divisiblity by $$\sqrt{x}$$ as also told in earlier post.....
any number can be taken in various combination of multiple of two factors...
However there cant be any combination which can have both factors > square root of the max square possible till that number..
in this case,
220 is between 196(14^2) and 225(15^2), so there cant be any combination which can have both factors > 14..
now 14 itself is non prime so we require to check for div for all prime no till 14, that is till13..
now if the number,202, is div by 101,a prime no but 202=101*2.. so if we have checked with 2, we dont require to check with 101..
hope the logic is clear..
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Re: How many prime numbers exist between 200 and 220? [#permalink]
### Show Tags
16 Jun 2015, 00:06
1
GauravSolanky wrote:
GauravSolanky wrote:
kinjiGC wrote:
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
Can you please throw more light on the logic behind this ?
As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC
Thanks,
Gaurav
hi GauravSolanky,
the logic behind "the rule" of divisiblity by $$\sqrt{x}$$ as also told in earlier post.....
any number can be taken in various combination of multiple of two factors...
However there cant be any combination which can have both factors > square root of the max square possible till that number..
in this case,
220 is between 196(14^2) and 225(15^2), so there cant be any combination which can have both factors > 14..
now 14 itself is non prime so we require to check for div for all prime no till 14, that is till13..
now if the number,202, is div by 101,a prime no but 202=101*2.. so if we have checked with 2, we dont require to check with 101..
hope the logic is clear..
Thanks and kudos to both of you.
Regards,
Gaurav
The check for a Number to be prime is "If the number to be checked is NOT divisible by any prime number Less than the square root of the number then it is said to he a Prime Number"
I.e. 219 will be prime if 219 is NOT divisible by any prime number less than "Square root of 219"
Square root if 219 = approx.14.8
I.e. check if 219 is divisible by any one of 2,3,5,7,11,13. If 219 is not divisible by any one of these prime numbers which are less than 14.8 then 219 will be prime.
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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04 Oct 2014, 03:47
kinjiGC wrote:
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
I think you did everything right.
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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13 Jun 2015, 21:31
kinjiGC wrote:
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
Can you please throw more light on the logic behind this ?
As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC
Thanks,
Gaurav
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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14 Jun 2015, 11:53
chetan2u wrote:
GauravSolanky wrote:
kinjiGC wrote:
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
Can you please throw more light on the logic behind this ?
As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC
Thanks,
Gaurav
hi GauravSolanky,
the logic behind "the rule" of divisiblity by $$\sqrt{x}$$ as also told in earlier post.....
any number can be taken in various combination of multiple of two factors...
However there cant be any combination which can have both factors > square root of the max square possible till that number..
in this case,
220 is between 196(14^2) and 225(15^2), so there cant be any combination which can have both factors > 14..
now 14 itself is non prime so we require to check for div for all prime no till 14, that is till13..
now if the number,202, is div by 101,a prime no but 202=101*2.. so if we have checked with 2, we dont require to check with 101..
hope the logic is clear..
Thanks and kudos to both of you.
Regards,
Gaurav
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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15 Jun 2015, 04:45
One more rul regarding prime no
All the prime no would be in the form of 6n +/- 1.
Cheers,
Rajan
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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15 Jun 2015, 06:46
rajsinghal71 wrote:
One more rul regarding prime no
All the prime no would be in the form of 6n +/- 1.
Cheers,
Rajan
Hi rajan,
It's good that you know some properties like you mentioned here that every prime Number Greater than 3 can be represented in form of 6k+1 or 6k-1 but since it's a GMAT forum so let me make it clear here that no such property is expected to be used by students in GMAT because GMAT checks very basic skills of mathematics and doesn't expect students to be loaded with various properties.
P.S. Also for the reader, Please Don't use this property as Final check of a Number being Prime, however it can be a primary check to suspect if the number is prime subjected to the number satisfying this property.
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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15 Jun 2015, 23:33
GauravSolanky wrote:
kinjiGC wrote:
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
Can you please throw more light on the logic behind this ?
As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC
Thanks,
Gaurav
Check out these two posts:
http://www.veritasprep.com/blog/2010/12 ... ly-number/
http://www.veritasprep.com/blog/2010/12 ... t-squares/
The second one ends with an explanation of this concept. Will help you understand a lot of things about factors and their placement.
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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16 Jun 2015, 12:17
VeritasPrepKarishma wrote:
GauravSolanky wrote:
kinjiGC wrote:
Bunuel
I solved it using the divisibility rules.
All the even numbers goes out of the window.
201,207,213,219 - Sum of digits = 3 so divisible by 3
All the numbers ending with 5 is obviously not prime
As 220 < 225, so maximum prime I need to check for divisibility is 13.
Now rest of the numbers check one by one.
203 -> Divisible by 7
209 -> divisible by 11
211 -> Only Prime I could get.
217 -> Divisible by 7
The last step took some time. Any idea if we can use anything else?
Can you please throw more light on the logic behind this ?
As 220 < 225, so maximum prime I need to check for divisibility is 13.
kinjiGC
Thanks,
Gaurav
Check out these two posts:
http://www.veritasprep.com/blog/2010/12 ... ly-number/
http://www.veritasprep.com/blog/2010/12 ... t-squares/
The second one ends with an explanation of this concept. Will help you understand a lot of things about factors and their placement.
Thanks Karishma, both articles were really helpful.
Regards,
Gaurav
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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16 Jun 2015, 12:41
GMATinsight wrote:
The check for a Number to be prime is "If the number to be checked is NOT divisible by any prime number Less than the square root of the number then it is said to he a Prime Number"
That should read "less than or equal to the square root". A number like 169 is not divisible by any prime less than √169, but 169 is still not a prime.
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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28 Apr 2017, 19:16
Excellent Question.
Here is what i did on this one ->
In order to check whether a given number is prime or not => We must divide it by all the primes less than or equal to the square root of that number.
PROPERTY => IF N IS NOT DIVISIBLE BY ANY PRIMES LESS THAN OR EQUAL TO THE SQAURE ROOT OF N => N is PRIMES.
Here the highest number =220
$$√220$$=> 14.something
Primes upto 14.something => {2,3,5,7,11,13}
Next => ALL PRIMES >5 MUST HAVE {1,3,7,9} as its UNITS DIGIT.
So lets check =->>
201=> Divisible by 3 => Rejected
203=>Divisible by 7 => Rejected
207=>Divisible by 3 => Rejected
209=>Divisible by 11 => Rejected
211=> Its not dividable by {2,3,5,7,11,13} => PRIME=> ACCEPTED.
213=>Divisible by 3 => Rejected
217=>Divisible by 7 => Rejected
219=>Divisible by 3 => Rejected
Hence only Prime number in the list => 211
SMASH THAT B.
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Re: How many prime numbers exist between 200 and 220? [#permalink]
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24 May 2019, 16:18
kinjiGC wrote:
How many prime numbers exist between 200 and 220?
(A) None
(B) One
(C) Two
(D) Three
(E) Four
First, we can omit all the even numbers (since they are divisible by 2) and all the odd numbers ending in 5 (since they are divisible by 5). So we are left with 201, 203, 207, 209, 211, 213, 217 and 219. We can omit 201, 207, 213 and 219 also since all of these numbers are divisible by 3 (notice that the sum of their digits is divisible by 3). So we only need to consider 203, 209, 211 and 217.
203/7 = 29 → So 203 is not a prime.
209/7 = 29 R 6, 209/11 = 19 → So 209 is not a prime.
211/7 = 30 R 1, 211/11 = 19 R 2, 211/13 = 16 R 3, 211/17 = 12 R 7 → So 211 is a prime.
217/7 = 31.→ So 217 is not a prime.
Therefore, there is only 1 prime number between 200 and 220.
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Re: How many prime numbers exist between 200 and 220? [#permalink] 24 May 2019, 16:18
Display posts from previous: Sort by | 5,088 | 17,821 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-51 | latest | en | 0.880257 |
https://www.physicsforums.com/threads/finding-f-using-f-dp-dt.962378/ | 1,685,815,927,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00232.warc.gz | 1,042,159,388 | 22,518 | # Finding F using F=dP/dt
• Eitan Levy
## Homework Statement
A snake with a mass equals to m is placed on a scale. The length of the snake is L. Suddenly the snake starts moving with a constant velocity v0. What does the weight display at the moment the snake starts moving?
F=dP/dt
## The Attempt at a Solution
First I tried to understand what force the snake is affected by. I think gravity, normal and some force F that causes it to move.
I tried to find the combined force on the snake at that moment. Before it starts moving it has a momentum of 0, and then it has a momentum of mv0.
So I get this: ∑F=N+F-mg=dP/dt=m*dV/dt
I know what the change in momentum is but I struggle to use to actually find the exact value of ΣF. Could anyone provide some guidance?
Also, I struggle to understand how the length of the snake is relevant to the question.
Well, I don't know what this question is all about. For example, ##v_0## is a speed not a velocity. And normally in physics an instantaneous change in velocity corresponds to an idealised scenario where you cannot analyse the forces.
Well, I don't know what this question is all about. For example, ##v_0## is a speed not a velocity. And normally in physics an instantaneous change in velocity corresponds to an idealised scenario where you cannot analyse the forces.
I need to find what is displayed on the scale the moment the snake starts moving. What you said is the reason I am struggling. The exact way the question is written is: "A snake with a length of L and a mass of m is placed on a scale. At the moment t=0 the snake begins to move upwards with a constant velocity v0. What will be displayed on the scale when t=0?".
Does that help?
I would move on to another problem.
I would move on to another problem.
That would be quite a problem considering I have to submit the final answer.
PeroK
This is a very poorly worded question, sorry you have to deal with it.
1) An instantaneous change in velocity requires infinite force.
2) Why a snake, is only part of it moving, as a real snake would? How much of the snake is moving?
If you can't ask for clarification, then maybe you could rewrite the question by making clarifying assumptions and then answer it. Make reasonable assumptions that change the problem as little as possible and communicate them clearly. Try not to do it in a way that insults the instructor (re. his ability to write good exam questions). Then solve it using the techniques the instructor expects you to use.
PeroK
That would be quite a problem considering I have to submit the final answer.
I suppose you could ask yourself that if something is moving at constant velocity what can you say about that?
This is a very poorly worded question, sorry you have to deal with it.
1) An instantaneous change in velocity requires infinite force.
2) Why a snake, is only part of it moving, as a real snake would? How much of the snake is moving?
If you can't ask for clarification, then maybe you could rewrite the question by making clarifying assumptions and then answer it. Make reasonable assumptions that change the problem as little as possible and communicate them clearly. Try not to do it in a way that insults the instructor (re. his ability to write good exam questions). Then solve it using the techniques the instructor expects you to use.
If that helps I believe only a part of it is moving (At t=0at least).
I suspect that this will turn out to be a variation of the "chain falling onto a scale" or "stream of sand falling onto a scale" type question (only in reverse).
Presumably the whole snake does not start moving at once. Perhaps it begins with the head. After this, at a constant rate, more of the snake is being elevated. So there's a dP/dt involved...
DaveE and Eitan Levy
I suppose you could ask yourself that if something is moving at constant velocity what can you say about that?
Well the combined force on the snake would be 0. Though I am not sure if it's relevant because this happens when t>0.
I suspect that this will turn out to be a variation of the "chain falling onto a scale" or "stream of sand falling onto a scale" type question (only in reverse).
Presumably the whole snake does not start moving at once. Perhaps it begins with the head. After this, at a constant rate, more of the snake is being elevated. So there's a dP/dt involved...
This analogy helped me a lot actually. I think I understand now.
Can't we just say the scale reading increases momentarily - the snake has to exert a downward force in order to start moving upward. ?
Can't we just say the scale reading increases momentarily - the snake has to exert a downward force in order to start moving upward. ?
No. E.g. if the whole snake somehow were to spring into the air at speed v0 in an instant then the force would be infinite.
I would go with @gneill's interpretation.
Okay, so the question should be something like:
A snake charmer places a snake of mass ##m## and length ##L## on a scale and begins to play his magic pungi. At time ##t=0## in response to the mesmerising music the snake begins to uncoil upwards, with its head moving at a constant upward speed of ##v_0## ...
neilparker62 and jbriggs444
Okay, so the question should be something like:
A snake charmer places a snake of mass ##m## and length ##L## on a scale and begins to play his magic pungi. At time ##t=0## in response to the mesmerising music the snake begins to uncoil upwards, with its head moving at a constant upward speed of ##v_0## ...
Ok - well in this case perhaps the 'correct' model is of two objects connected by a compressed spring. When the 'spring' releases, conservation of momentum applies and the scale pan experiences an equal and opposite impulse to that applied to (applied by!) the snake.
I think I have an answer to this problem but not without assumptions. We are asked to find the "weight display at the moment the snake starts moving". Question: Is this moment before or after the snake reaches speed v0? If before, we need to know the details of how the snake accelerates from v = 0 to v = v0. My assumption is that the snake reaches constant velocity v0 instantaneously and that we are asked to find the reading on the scale when the snake's head has reached terminal any part of the snake that's moving has velocity v0. Arguably, under this assumption, the reading is constant and extends to t = 0. The answer is obtained simply and incorporates all three of the given parameters.
Edited to clarify a point raised in #18.
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Ok - well in this case perhaps the 'correct' model is of two objects connected by a compressed spring. When the 'spring' releases, conservation of momentum applies and the scale pan experiences an equal and opposite impulse to that applied to (applied by!) the snake.
No, a spring is not the right model. That would imply an acceleration, neither infinitesimal nor infinite, for the snake as a whole.
Stick with @gneill's and @PeroK 's model, but with the head being infinitesimal in size.
I think I have an answer to this problem but not without assumptions. We are asked to find the "weight display at the moment the snake starts moving". Question: Is this moment before or after the snake reaches speed v0? If before, we need to the details of how the snake the snake accelerates from v = 0 to v = v0. My assumption is that the snake reaches constant velocity v0 instantaneously and that we are asked to find the reading on the scale when the snake's head has reached terminal velocity v0. Arguably, under this assumption, the reading is constant and extends to t = 0. The answer is obtained simply and incorporates all three of the given parameters.
I do not understand your model.
You mention reaching constant velocity instantaneously (the snake as a whole?) then refer to reaching terminal velocity.
@PeroK 's model is fine, with the caveat in post #17.
I do not understand your model.
You mention reaching constant velocity instantaneously (the snake as a whole?) then refer to reaching terminal velocity.
@PeroK 's model is fine, with the caveat in post #17.
My model is essentially the model by @PeroK. A plot of the speed of the tip of the snake's head as a function of time is parallel is parallel to the time axis at v0. I felt that a clarification of "at the moment the snake starts moving" was needed.
My model is essentially the model by @PeroK. A plot of the speed of the tip of the snake's head as a function of time is parallel is parallel to the time axis at v0. I felt that a clarification of "at the moment the snake starts moving" was needed.
Ok, but that's not how post #16 reads.
I regard the "moment it starts moving" as a blunder. As you note/imply, it leaves the question unanswerable because the velocity of the tip is a step function. Further, the criterion is unnecessary. The reading will not change again as long as the snake maintains its head speed but is stiil in contact with the pan.
Ok, but that's not how post #16 reads.
I edited #16 to clarify it. Thanks.
I agree with all of you that this is a poorly worded problem. It might make sense if the snake had a uniform mass per unit length, m/L, which is not obvious.
If the snake is uncoiling upward at constant speed v, the the upward rate of change of momentum is determined by the rate at which mass is added to the upward flow.
AM
neilparker62
I agree with all of you that this is a poorly worded problem. It might make sense if the snake had a uniform mass per unit length, m/L, which is not obvious.
If the snake is uncoiling upward at constant speed v, the the upward rate of change of momentum is determined by the rate at which mass is added to the upward flow.
AM
Perhaps we are indeed meant to make the assumption about uniform mass per unit length.
Perhaps we are indeed meant to make the assumption about uniform mass per unit length.
Yes, because otherwise one will need the linear mass density of the snake to solve the problem. Even if the snake were simply modeled as a prolate ellipsoid (head) a cylinder (body) and a cone (tail) of uniform mass density, the problem would be unnecessarily complicated and would detract from the main idea which (in my opinion) is an application of ##F=\frac{dP}{dt}.## Furthermore, if all parts of the snake moved at speed v0 and the snake's linear mass density were not uniform, one would have to know ##\frac{dm}{dx} \vert _{x=0}## in order to answer the question.
Given that the author of the problem intended it to be solvable, one has to make enough simplifying assumptions until it is. A well crafted problem enunciates the appropriate assumptions up front and precludes ambiguity about how one ought to proceed in order to solve it. A badly crafted problem cannot be solved no matter what the simplifying assumptions. There also is a gray area of so-so crafted problems that elicit discussion about the formulation of the appropriate simplifying assumptions. This problem belongs in the gray area.
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gneill
Given that the author of the problem intended it to be solvable, one has to make enough simplifying assumptions until it is. A well crafted problem enunciates the appropriate assumptions up front and precludes ambiguity about how one ought to proceed in order to solve it. A badly crafted problem cannot be solved no matter what the simplifying assumptions. There also is a gray area of so-so crafted problems that elicit discussion about the formulation of the appropriate simplifying assumptions. This problem belongs in the gray area.
To misquote terribly your motto: "I know one thing, I know not enough !" | 2,601 | 11,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-23 | latest | en | 0.942563 |
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### Statistics and Dr. Strangelove
One of the biggest embarrassments in statistics is that we don’t really have confidence bands for nonparametric functions estimation. This is a fact that we tend to sweep under the rug.
Consider, for example, estimating a density ${p(x)}$ from a sample ${X_1,\ldots, X_n}$. The kernel estimator with kernel ${K}$ and bandwidth ${h}$ is
$\displaystyle \hat p(x) = \frac{1}{n}\sum_{i=1}^n \frac{1}{h}K\left( \frac{x-X_i}{h}\right).$
Let’s start with getting a confidence interval at a single point ${x}$.
Let ${\overline{p}(x) = \mathbb{E}(\hat p(x))}$ be the mean of the estimator and let ${s_n(x)}$ be the standard deviation. (A consistent estimator of ${s_n(x)}$ is ${\sqrt{a^2/n}}$ where ${a}$ is the sample standard deviation of ${U_1,\ldots, U_n}$ where ${U_i = h^{-1} K(x-X_i/h)}$.)
Now,
$\displaystyle \frac{\hat{p}(x) - p(x)}{s_n(x)} = \frac{\hat{p}(x) - \overline{p}(x)}{s_n(x)} + \frac{\overline{p}(x)-p(x)}{s_n(x)} = \frac{\hat{p}(x) - \overline{p}(x)}{s_n(x)} + \frac{b(x)}{s_n(x)}$
where ${b(x) = \overline{p}(x)-p(x)}$ is the bias. By the central limit theorem, the first term is approximately standard Normal,
$\displaystyle \frac{\hat{p}(x) - \overline{p}(x)}{s_n(x)} \approx N(0,1).$
This suggests the confidence interval, ${\hat p(x) \pm z_{\alpha/2} s_n(x)}$.
The problem is the second term ${\frac{b(x)}{s_n(x)}}$. Recall that the optimal bandwidth ${h}$ balances the bias and the variance. This means that if we use the best bandwidth, then ${\frac{b(x)}{s_n(x)}}$ behaves like a constant, that is, ${\frac{b(x)}{s_n(x)}\rightarrow c}$ for some (unknown) constant ${c}$. The result is that, even as ${n\rightarrow\infty}$, the confidence interval is centered at ${p(x) + c}$ instead of ${p(x)}$. As a result, the coverage will be less than the intended coverage ${1-\alpha}$.
We could in principle solve this problem by undersmoothing. If we choose a bandwidth smaller than the optimal bandwidth, then ${\frac{b(x)}{s_n(x)}\rightarrow 0}$ and the problem disappears. But this creates two new problems. First, the confidence interval is now centered around a non-optimal estimator. Second, and more importantly, we don’t have any practical, data-driven methods for choosing an undersmoothed bandwidth. There are some proposals in the theoretical literature but they are, frankly, not very practical
Of course, all of these problems are only worse when we consider simultaneous confidence band for the whole function.
The same problem arises if we use Bayesian inference. In short, the 95 percent Bayesian interval will not have 95 percent frequentist coverage. The reasons are the same as for the frequentist interval. (One can of course “declare” that coverage is not important and then there is no problem. But I am assuming that we do want correct frequentist coverage.)
The only good solution I know is to simply admit that we can’t create a confidence interval for ${p(x)}$. Instead, we simply interpret ${\hat p(x) \pm z_{\alpha/2} s_n(x)}$ as a confidence interval for ${\overline{p}(x)}$, which is a smoothed out (biased) version of ${p(x)}$:
$\displaystyle \overline{p}(x) = \int K(u) p(x+hu) \,du \approx p(x) + C h^2$
where ${C}$ is an unknown constant. In other words, we just live with the bias. (Of course, the bias gets smaller as the sample size gets larger and ${h}$ gets smaller.)
Once we take this point of view, we can easily get a confidence band using the bootstrap. We draw ${X_1^*,\ldots, X_n^*}$ from the empirical distribution ${P_n}$ and compute the density estimator ${\hat p^*}$. By repeating this many times, we can find the quantile ${z_\alpha}$ defined by
$\displaystyle P\Biggl( \sqrt{nh}||\hat p(x)^*-\hat p(x)||_\infty > z_\alpha \Biggm| X_1,\ldots, X_n\Biggr) = \alpha.$
The confidence band is then ${(\ell(x),u(x))}$ where
$\displaystyle \ell(x) = \hat p(x) - \frac{z_\alpha}{\sqrt{nh}},\ \ \ u(x) = \hat p(x) + \frac{z_\alpha}{\sqrt{nh}}.$
We then have that
$\displaystyle P\Biggl( \ell(x) \leq \overline{p}(x) \leq u(x)\ \ \ {\rm for\ all\ }x\Biggr)\rightarrow 1-\alpha$
as ${n\rightarrow \infty}$. Again, this is a confidence band for ${\overline{p}(x)}$ rather than for ${p(x)}$.
Summary: There really is no good, practical method for getting true confidence bands in nonparametric function estimation. The reason is that the bias does not disappear fast enough as the sample size increases. The solution, I believe, is to just acknowledge that the confidence bands have a slight bias. Then we can use the bootstrap (or other methods) to compute the band. Like Dr. Strangelove, we just learn to stop worrying and love the bias.
### The JSM, Minimaxity and the Language Police
The JSM, Minimaxity and the Language Police
I am back from the JSM. For those who don’t know, the JSM is the largest statistical meeting in the world. This year there were nearly 6,000 people.
Some people hate the JSM because it is too large. I love JSM. There is so much going on: lots of talks, receptions, tutorials and, of course, socializing. It’s impossible to walk 10 feet in any direction without bumping into another old friend.
1. Highlights
On Sunday I went to a session dedicated to the 70th birthday of my colleague Steve Fienberg. This was a great celebration of a good friend and remarkable statistician. Steve has more energy at 70 than I did when I was 20.
On a sadder note, I went to a memorial session for my late friend George Casella. Ed George, Bill Strawderman, Roger Berger, Jim Berger and Marty Wells talked about George’s many contributions and his huge impact on statistics. To borrow Ed’s catchphrase: what a good guy.
On Tuesday, I went to Judea Pearl’s medallion lecture, with discussions by Jamie Robins and Eric Tchetgen Tchetgen. Judea gave an unusual talk, mixing philosophy, metaphors (“eagles and snakes can’t build microscopes”) and math. Judea likes to argue that graphical models/structural equation models are the best way to view causation. Jamie and Eric argued that graphs can hide certain assumptions and that counterfactuals need to be used in addition to graphs.
Digression. There is another aspect of causation that deserves mention. It’s one thing to write down a general model to describe causation (using graphs or counterfactuals). It’s another matter altogether to construct estimators for the causal effects. Jamie Robins is well-known for his theory of “G-estimation”. At the risk of gross oversimplification, this is an approach to sequential causal modeling that eventually leads to semiparametric efficient estimators. The construction of these estimators is highly non-trivial. You can’t get them by just writing down a density for the graph and estimating the distribution. Nor can you just throw down some parametric models for the densities in the factorization implied by the graph. If you do, you will end up with a model in which there is no parameter that equals 0 when there is no causal effect. I call this the “null paradox.” And Jamie’s estimators are useful. Indeed, on Monday, there was a session called “Causal Inference in Observational Studies with Time-Varying Treatments.” All the talks were concerned with applying Jamie’s methods to develop estimators in real problems such as organ transplantation studies. The point is this: when it comes to causation, just writing down a model using graphs or counterfactuals is only a small part of the story. Finding estimators is often much more challenging.
End Digression.
On Wednesday, I had the honor of giving the Rietz Lecture (named after Henry Rietz, the first president of the Institute of Mathematical Statistics). I talked about Topological Inference. I will post the slides on my web page soon. I was fortunate to be introduced by my good friend Ed George.
2. Minimaxity
One thing that came up at my talk, but also in several discussions I had with people at the conference is the following problem. Recall that the minimax risk is
$\displaystyle R=\inf_{\hat \theta}\sup_{P\in {\cal P}}E_P[ L(\theta(P),\hat\theta)]$
where ${L}$ is a loss function, ${{\cal P}}$ is a set of distributions and the infimum is over all estimators. Often we can define an estimator that achieves the minimax rate for a problem. But the estimator that achieves the risk ${R}$ may not be computable in any practical sense. This happens in some of the topological problems I discussed. It also happens in sparse PCA problems. I suspect we are going to come across this issue more and more.
I suggested to a few people that we should really define the minimax risk by restricting ${\hat\theta}$ to be an estimator that can be computed in polynomial time. But how would we then compute the minimax risk? Some people, like Mike Jordan and Martin Wainwright, have made headway in studying the tradeoff between optimality and computation. But a general theory will be hard. I did mention this to Aad van der Vart at dinner one night. If anyone can make headway with this problem, it is Aad.
3. Montreal
The conference was in Montreal. If you have never been there, Montreal is a very pleasant city; vibrant and full of good restaurants. We had a very nice time including good dinners and cigars at a classy cigar bar.
Rant: The only bad thing about Montreal is the tyrannical language police. (As a dual citizen, I think I am allowed to criticize Canada!) Put a sign on your business only in English and go to jail. Very tolerant. This raises lots interesting problems: Is “pasta” English? What if you are using it as a proper name, as in naming your business: Pasta! And unless you live in North Korea, why should anyone be able to tell you what language to put on your private business? We heard one horror story that is almost funny. A software company opened in Montreal. The language police checked an approved of their signage. (Yes. They really pay people to do this.) Then they were told they had to write their code in French. I’m not even sure what this means. Is C code English or French? Anyway, the company left Montreal.
To be fair, it’s not just Montreal that lives under the brutal rule of bureaucrats and politicians. Every country does. But Canada does lean heavily toward arbitrary (i.e. favoring special groups) rules. Shane Jensen pointed me to Can’tada which lists stuff you can’t use in Canada.
End Rant.
Despite my rant, I think Montreal is a great city. I highly recommend it for travel.
4. Summary
The JSM is lots of fun. Yes, it is a bit overwhelming but where else can you find 6000 statisticians in one place? So next year, when I say I don’t want to go, someone remind me that I actually had a good time. | 2,709 | 10,688 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 50, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-27 | latest | en | 0.757727 |
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# Complementary Error Function Excel
## Contents
Applied Mathematics Series. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). Negative integer values of Im(ƒ) are shown with thick red lines. is the double factorial: the product of all odd numbers up to (2n–1). However, for −1 < x < 1, there is a unique real number denoted erf − 1 ( x ) {\displaystyle \operatorname 6 ^{-1}(x)} satisfying erf ( erf Source
Negative integer values of Im(ƒ) are shown with thick red lines. By using this site, you agree to the Terms of Use and Privacy Policy. W. These generalised functions can equivalently be expressed for x>0 using the Gamma function and incomplete Gamma function: E n ( x ) = 1 π Γ ( n ) ( Γ
## Complementary Error Function Excel
Numerical approximations Over the complete range of values, there is an approximation with a maximal error of 1.2 × 10 − 7 {\displaystyle 1.2\times 10^{-7}} , as follows:[15] erf ( Press, W.H.; Flannery, B.P.; Teukolsky, S.A.; and Vetterling, W.T. "Incomplete Gamma Function, Error Function, Chi-Square Probability Function, Cumulative Poisson Function." §6.2 in Numerical Recipes in FORTRAN: The Art of Scientific Computing, Handbook of Continued Fractions for Special Functions. Complementary Error Function In mathematics, the complementary error function (also known as Gauss complementary error function) is defined as: Complementary Error Function Table The following is the error function and complementary
1. and Stegun, I.A. (Eds.). "Repeated Integrals of the Error Function." §7.2 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing.
2. Both functions are overloaded to accept arguments of type float, double, and long double.
3. For previous versions or for complex arguments, SciPy includes implementations of erf, erfc, erfi, and related functions for complex arguments in scipy.special.[21] A complex-argument erf is also in the arbitrary-precision arithmetic
4. Math.
5. See also Related functions Gaussian integral, over the whole real line Gaussian function, derivative Dawson function, renormalized imaginary error function Goodwin–Staton integral In probability Normal distribution Normal cumulative distribution function, a
6. Generalized error functions Graph of generalised error functions En(x): grey curve: E1(x) = (1−e−x)/ π {\displaystyle \scriptstyle {\sqrt {\pi }}} red curve: E2(x) = erf(x) green curve: E3(x) blue curve: E4(x)
7. For large enough values of x, only the first few terms of this asymptotic expansion are needed to obtain a good approximation of erfc(x) (while for not too large values of
8. At the imaginary axis, it tends to ±i∞.
9. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W., NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN978-0521192255, MR2723248 External links MathWorld – Erf Authority control NDL: 00562553 Retrieved from
10. For large enough values of x, only the first few terms of this asymptotic expansion are needed to obtain a good approximation of erfc(x) (while for not too large values of
J. (March 1993), "Algorithm 715: SPECFUN—A portable FORTRAN package of special function routines and test drivers" (PDF), ACM Trans. Generated Wed, 05 Oct 2016 11:22:14 GMT by s_hv1002 (squid/3.5.20) Another form of erfc ( x ) {\displaystyle \operatorname Φ 8 (x)} for non-negative x {\displaystyle x} is known as Craig's formula:[5] erfc ( x | x ≥ 0 Inverse Complementary Error Function This is useful, for example, in determining the bit error rate of a digital communication system.
Handbook of Differential Equations, 3rd ed. Complementary Error Function Calculator Positive integer values of Im(f) are shown with thick blue lines. Referenced on Wolfram|Alpha: Erfc CITE THIS AS: Weisstein, Eric W. "Erfc." From MathWorld--A Wolfram Web Resource. https://en.wikipedia.org/wiki/Error_function Wolfram Language» Knowledge-based programming for everyone.
See also Related functions Gaussian integral, over the whole real line Gaussian function, derivative Dawson function, renormalized imaginary error function Goodwin–Staton integral In probability Normal distribution Normal cumulative distribution function, a Complementary Error Function In Matlab N ! ∫ x ∞ t − 2 N e − t 2 d t , {\displaystyle R_ − 3(x):={\frac {(-1)^ − 2}{\sqrt {\pi }}}2^ − 1{\frac {(2N)!} − 0}\int _ Data Types: single | doubleMore Aboutcollapse allComplementary Error FunctionThe complementary error function of x is defined aserfc(x)=2π∫x∞e−t2dt=1−erf(x).It is related to the error function aserfc(x)=1−erf(x).Tall Array SupportThis function fully supports tall arrays. Instead, replace 1 - erfc(x) with erf(x).For expressions of the form exp(x^2)*erfc(x), use the scaled complementary error function erfcx instead.
## Complementary Error Function Calculator
Intermediate levels of Re(ƒ)=constant are shown with thin red lines for negative values and with thin blue lines for positive values. page Numerical approximations Over the complete range of values, there is an approximation with a maximal error of 1.2 × 10 − 7 {\displaystyle 1.2\times 10^{-7}} , as follows:[15] erf ( Complementary Error Function Excel Schöpf and P. Complementary Error Function Table A printed companion is available. 7.1 Special Notation7.3 Graphics Index Notations Search Need Help?
Keywords: Fresnel integrals Referenced by: §11.10(vi) Permalink: http://dlmf.nist.gov/7.2.iii See also: info for 7.2 7.2.6 ℱ(z) =∫z∞e12πit2dt, Defines: ℱ(z): Fresnel integral Symbols: dx: differential of x, e: base of exponential function, ∫: http://xvisionx.com/error-function/complementary-error-function-calculator.html The defining integral cannot be evaluated in closed form in terms of elementary functions, but by expanding the integrand e−z2 into its Maclaurin series and integrating term by term, one obtains Excel: Microsoft Excel provides the erf, and the erfc functions, nonetheless both inverse functions are not in the current library.[17] Fortran: The Fortran 2008 standard provides the ERF, ERFC and ERFC_SCALED Springer-Verlag. Complimentary Error Function
Wolfram|Alpha» Explore anything with the first computational knowledge engine. For any complex number z: erf ( z ¯ ) = erf ( z ) ¯ {\displaystyle \operatorname 6 ({\overline 5})={\overline {\operatorname 4 (z)}}} where z Related functions The error function is essentially identical to the standard normal cumulative distribution function, denoted Φ, also named norm(x) by software languages, as they differ only by scaling and translation. http://xvisionx.com/error-function/compute-complementary-error-function.html MathCAD provides both erf(x) and erfc(x) for real arguments.
Haskell: An erf package[18] exists that provides a typeclass for the error function and implementations for the native (real) floating point types. Complementary Error Function Mathematica This substitution maintains accuracy. At the real axis, erf(z) approaches unity at z→+∞ and −1 at z→−∞.
## Intermediate levels of Im(ƒ)=constant are shown with thin green lines.
Values at Infinity Keywords: Fresnel integrals See also: info for 7.2(iii) 7.2.9 limx→∞C(x) =12, limx→∞S(x) =12. LCCN65-12253. Comp. 23 (107): 631–637. Complementary Error Function Ti 89 is the double factorial: the product of all odd numbers up to (2n–1).
The pairs of functions {erff(),erfcf()} and {erfl(),erfcl()} take and return values of type float and long double respectively. Asymptotic expansion A useful asymptotic expansion of the complementary error function (and therefore also of the error function) for large real x is erfc ( x ) = e − The error function is a special case of the Mittag-Leffler function, and can also be expressed as a confluent hypergeometric function (Kummer's function): erf ( x ) = 2 x Check This Out Symbols: Hn(x): Hermite polynomial, !: factorial (as in n!), inerfc(z): repeated integrals of the complementary error function, z: complex variable and n: nonnegative integer A&S Ref: 7.2.11 Permalink: http://dlmf.nist.gov/7.18.E8 Encodings: TeX,
xerf(x)erfc(x)0.00.01.00.010.0112834160.9887165840.020.0225645750.9774354250.030.0338412220.9661587780.040.0451111060.9548888940.050.0563719780.9436280220.060.0676215940.9323784060.070.078857720.921142280.080.0900781260.9099218740.090.1012805940.8987194060.10.1124629160.8875370840.110.1236228960.8763771040.120.1347583520.8652416480.130.1458671150.8541328850.140.1569470330.8430529670.150.1679959710.8320040290.160.1790118130.8209881870.170.1899924610.8100075390.180.2009358390.7990641610.190.2118398920.7881601080.20.2227025890.7772974110.210.2335219230.7664780770.220.2442959120.7557040880.230.25502260.74497740.240.2657000590.7342999410.250.276326390.723673610.260.2868997230.7131002770.270.2974182190.7025817810.280.3078800680.6921199320.290.3182834960.6817165040.30.3286267590.6713732410.310.338908150.661091850.320.3491259950.6508740050.330.3592786550.6407213450.340.3693645290.6306354710.350.3793820540.6206179460.360.3893297010.6106702990.370.3992059840.6007940160.380.4090094530.5909905470.390.41873870.58126130.40.4283923550.5716076450.410.437969090.562030910.420.4474676180.5525323820.430.4568866950.5431133050.440.4662251150.5337748850.450.475481720.524518280.460.484655390.515344610.470.4937450510.5062549490.480.5027496710.4972503290.490.5116682610.4883317390.50.5204998780.4795001220.510.529243620.470756380.520.537898630.462101370.530.5464640970.4535359030.540.554939250.445060750.550.5633233660.4366766340.560.5716157640.4283842360.570.5798158060.4201841940.580.58792290.41207710.590.5959364970.4040635030.60.6038560910.3961439090.610.6116812190.3883187810.620.6194114620.3805885380.630.6270464430.3729535570.640.6345858290.3654141710.650.6420293270.3579706730.660.6493766880.3506233120.670.6566277020.3433722980.680.6637822030.3362177970.690.6708400620.3291599380.70.6778011940.3221988060.710.684665550.315334450.720.6914331230.3085668770.730.6981039430.3018960570.740.7046780780.2953219220.750.7111556340.2888443660.760.7175367530.2824632470.770.7238216140.2761783860.780.7300104310.2699895690.790.7361034540.2638965460.80.7421009650.2578990350.810.7480032810.2519967190.820.7538107510.2461892490.830.7595237570.2404762430.840.7651427110.2348572890.850.7706680580.2293319420.860.7761002680.2238997320.870.7814398450.2185601550.880.7866873190.2133126810.890.7918432470.2081567530.90.7969082120.2030917880.910.8018828260.1981171740.920.8067677220.1932322780.930.8115635590.1884364410.940.8162710190.1837289810.950.8208908070.1791091930.960.825423650.174576350.970.8298702930.1701297070.980.8342315040.1657684960.990.838508070.161491931.00.8427007930.1572992071.010.8468104960.1531895041.020.8508380180.1491619821.030.8547842110.1452157891.040.8586499470.1413500531.050.8624361060.1375638941.060.8661435870.1338564131.070.8697732970.1302267031.080.8733261580.1266738421.090.8768031020.1231968981.10.880205070.119794931.110.8835330120.1164669881.120.886787890.113212111.130.889970670.110029331.140.8930823280.1069176721.150.8961238430.1038761571.160.8990962030.1009037971.170.9020003990.0979996011.180.9048374270.0951625731.190.9076082860.0923917141.20.9103139780.0896860221.210.9129555080.0870444921.220.9155338810.0844661191.230.9180501040.0819498961.240.9205051840.0794948161.250.9229001280.0770998721.260.9252359420.0747640581.270.9275136290.0724863711.280.9297341930.0702658071.290.9318986330.0681013671.30.9340079450.0659920551.310.9360631230.0639368771.320.9380651550.0619348451.330.9400150260.0599849741.340.9419137150.0580862851.350.9437621960.0562378041.360.9455614370.0544385631.370.9473123980.0526876021.380.9490160350.0509839651.390.9506732960.0493267041.40.952285120.047714881.410.9538524390.0461475611.420.9553761790.0446238211.430.9568572530.0431427471.440.958296570.041703431.450.9596950260.0403049741.460.961053510.038946491.470.96237290.03762711.480.9636540650.0363459351.490.9648978650.0351021351.50.9661051460.0338948541.510.9672767480.0327232521.520.9684134970.0315865031.530.9695162090.0304837911.540.970585690.029414311.550.9716227330.0283772671.560.9726281220.0273718781.570.9736026270.0263973731.580.9745470090.0254529911.590.9754620160.0245379841.60.9763483830.0236516171.610.9772068370.0227931631.620.9780380880.0219619121.630.978842840.021157161.640.979621780.020378221.650.9803755850.0196244151.660.9811049210.0188950791.670.9818104420.0181895581.680.9824927870.0175072131.690.9831525870.0168474131.70.9837904590.0162095411.710.9844070080.0155929921.720.9850028270.0149971731.730.98557850.01442151.740.9861345950.0138654051.750.9866716710.0133283291.760.9871902750.0128097251.770.9876909420.0123090581.780.9881741960.0118258041.790.9886405490.0113594511.80.9890905020.0109094981.810.9895245450.0104754551.820.9899431560.0100568441.830.9903468050.0096531951.840.9907359480.0092640521.850.991111030.008888971.860.9914724880.0085275121.870.9918207480.0081792521.880.9921562230.0078437771.890.9924793180.0075206821.90.9927904290.0072095711.910.993089940.006910061.920.9933782250.0066217751.930.993655650.006344351.940.9939225710.0060774291.950.9941793340.005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Related Error Function Calculator ©2016 Miniwebtool | Terms and Disclaimer | Privacy Policy | Contact Us Index Notations Search Need Help? Hermite Polynomials Keywords: Hermite polynomials, repeated integrals of the complementary error function See also: info for 7.18(iv) 7.18.8 (-1)ninerfc(z)+inerfc(-z)=i-n2n-1n!Hn(iz). Washington, DC: Hemisphere, pp.385-393 and 395-403, 1987. The error function at +∞ is exactly 1 (see Gaussian integral).
Symbols: C(z): Fresnel integral, S(z): Fresnel integral and x: real variable A&S Ref: 7.3.20 Referenced by: §7.5 Permalink: http://dlmf.nist.gov/7.2.E9 Encodings: TeX, TeX, pMML, pMML, png, png See also: info for 7.2(iii) Indeed, Φ ( x ) = 1 2 π ∫ − ∞ x e − t 2 2 d t = 1 2 [ 1 + erf ( x 2 | 6,557 | 17,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-17 | latest | en | 0.759133 |
https://en-academic.com/dic.nsf/enwiki/827349 | 1,642,534,726,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00242.warc.gz | 298,902,974 | 12,814 | # Routh-Hurwitz stability criterion
Routh-Hurwitz stability criterion
The Routh-Hurwitz stability criterion is a necessary (and frequently sufficient) method to establish the stability of a single-input, single-output (SISO), linear time invariant (LTI) control system. More generally, given a polynomial, some calculations using only the coefficients of that polynomial can lead to the conclusion that it is not stable. For the discrete case, see the Jury test equivalent.
The criterion establishes a systematic way to show that the linearized equations of motion of a system have only stable solutions exp("pt"), that is where all "p" have negative real parts. It can be performed using either polynomial divisions or determinant calculus.
The criterion is derived through the use of the Euclidiean algorithm and Sturm's theorem in evaluating Cauchy indices.
Using Euclid's algorithm
The criterion is related to Routh-Hurwitz theorem. Indeed, from the statement of that theorem, we have $p-q=w\left(+infty\right)-w\left(-infty\right)$ where:
* "p" is the number of roots of the polynomial "f"("z") located in the left half-plane;
* "q" is the number of roots of the polynomial "f"("z") located in the right half-plane (let us remind ourselves that "f" is supposed to have no roots lying on the imaginary line);
* "w"("x") is the number of variations of the generalized Sturm chain obtained from $P_0\left(y\right)$ and $P_1\left(y\right)$ (by successive Euclidean divisions) where $f\left(iy\right)=P_0\left(y\right)+iP_1\left(y\right)$ for a real "y".By the fundamental theorem of algebra, each polynomial of degree "n" must have "n" roots in the complex plane (i.e., for an "f" with no roots on the imaginary line, "p"+"q"="n"). Thus, we have the condition that "f" is a (Hurwitz) stable polynomial if and only if "p"-"q"="n" (the proof is given below). Using the Routh-Hurwitz theorem, we can replace the condition on "p" and "q" by a condition on the generalized Sturm chain, which will give in turn a condition on the coefficients of "f".
Using matrices
Let "f"("z") be a complex polynomial. The process is as follows:
# Compute the polynomials $P_0\left(y\right)$ and $P_1\left(y\right)$ such that $f\left(iy\right)=P_0\left(y\right)+iP_1\left(y\right)$ where "y" is a real number.
# Compute the Sylvester matrix associated to $P_0\left(y\right)$ and $P_1\left(y\right)$.
# Rearrange each row in such a way that an odd row and the following one have the same number of leading zeros.
# Compute each principal minor of that matrix.
# If at least one of the minors is negative (or zero), then the polynomial "f" is not stable.
Example
* Let $f\left(z\right)=az^2+bz+c$ (for the sake of simplicity we take real coefficients) where $c eq 0$ (to avoid a root in zero so that we can use the Routh-Hurwitz theorem). First, we have to calculate the real polynomials $P_0\left(y\right)$ and $P_1\left(y\right)$:$f\left(iy\right)=-ay^2+iby+c=P_0\left(y\right)+iP_1\left(y\right)=-ay^2+c+i\left(by\right).$Next, we find divide those polynomials to obtain the generalizes Sturm chain:
** $P_0\left(y\right)=\left(\left(-a/b\right)y\right)P_1\left(y\right)+c,$ yields $P_2\left(y\right)=-c,$
** $P_1\left(y\right)=\left(\left(-b/c\right)y\right)P_2\left(y\right),$ yields $P_3\left(y\right)=0$ and the Euclidean division stops.Notice that we had to suppose "b" different from zero in the first division. The generalized Sturm chain is in this case $\left(P_0\left(y\right),P_1\left(y\right),P_2\left(y\right)\right)=\left(c-ay^2,by,-c\right)$. Putting $y=+infty$, the sign of $c-ay^2$ is the opposite sign of "a" and the sign of "by" is the sign of "b". When we put $y=-infty$, the sign of the first element of the chain is again the opposite sign of "a" and the sign of "by" is the opposite sign of "b". Finally, -"c" has always the opposite sign of "c".
Suppose now that "f" is Hurwitz stable. This means that $w\left(+infty\right)-w\left(-infty\right)=2$ (the degree of "f"). By the properties of the function "w", this is the same as $w\left(+infty\right)=2$ and $w\left(-infty\right)=0$. Thus, "a", "b" and "c" must have the same sign. We have thus found the necessary condition of stability for polynomials of degree 2.
Higher-order example
A tabular method can be used to determine the stability when the roots of a higher order characteristic polynomial are difficult to obtain. For an $n-th$ order polynomial
* $D\left(s\right)=a_ns^n+a_\left\{n-1\right\}s^\left\{n-1\right\}+cdots+a_1s+a_0$the table has $n + 1$ rows and the following structure:where the elements $b_i$ and $c_i$ can be computed as follows:
* $b_i=frac\left\{a_\left\{n-1\right\} imes\left\{a_\left\{n-2i-a_n imes\left\{a_\left\{n-2i-1\right\}\left\{a_\left\{n-1$
* $c_i=frac\left\{b_1 imes\left\{a_\left\{n-2i-1-b_\left\{i+1\right\} imes\left\{a_\left\{n-1\right\}\left\{b_1\right\}$When completed, the number of sign changes in the first column will be the number of non-negative poles.
Consider a system with a characteristic polynomial
* $D\left(s\right)=s^5+4s^4+2s^3+5s^2+3s+6$we have the following table:In the first column, there are two sign changes (0.75 -> -3, and -3 -> 3), thus there are two non-negative poles and the system is unstable.
Appendix A
Suppose "f" is stable. Then, we must have "q"=0. Since "p"+"q"="n", we find "p"-"q"="n". Suppose now that "p"-"q"="n". Since "p"+"q"="n", subtracting the two equations, we find 2"q"=0, that is "f" is stable.
ee also
* Control engineering
* Derivation of the Routh array
* Nyquist stability criterion
* Routh–Hurwitz theorem
* Root locus
* Transfer function
* Jury stability criterion
References
* cite journal
author = Hurwitz, A.
year = 1964
title = ‘On the conditions under which an equation has only roots with negative real parts
journal = Selected Papers on Mathematical Trends in Control Theory
* cite book
author = Routh, E.J.
year = 1877
title = A Treatise on the Stability of a Given State of Motion: Particularly Steady Motion
publisher = Macmillan and co.
isbn =
* cite journal
author = Gantmacher, F.R.
year = 1959
title = Applications of the Theory of Matrices
journal = Interscience, New York
volume = 641
issue = 9
pages = 1–8
* cite journal
author = Pippard, A.B.
coauthors = Dicke, R.H.
year = 1986
title = Response and Stability, An Introduction to the Physical Theory
journal = American Journal of Physics
volume = 54
pages = 1052
accessdate = 2008-05-07
doi = 10.1119/1.14826
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• Edward Routh — Infobox Scientist name = Edward Routh caption = Edward John Routh (1831 1907) birth date = birth date|1831|1|20|df=y birth place = Quebec, Canada death date = death date and age|1907|6|7|1831|1|20|df=y death place = Cambridge, England residence … Wikipedia | 2,663 | 9,377 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 33, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-05 | latest | en | 0.899318 |
http://scienceblogs.com/deltoid/2006/03/24/which-global-warming-skeptic-a/ | 1,502,891,872,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886101966.48/warc/CC-MAIN-20170816125013-20170816145013-00394.warc.gz | 373,506,526 | 35,747 | # Which Global Warming Skeptic Are You?
You have two 50g containers of cream. One is 10% fat, and the other 20% fat. You combine them. What is the percentage of fat in the mixture?
A. 10% of 50 is 5, 20% of 50 is 10. (10+5)/(50+50) is 15%. The answer is 15%, the arithmetic mean of 10% and 20%. B. 14.1%, the geometric mean of 10% and 20%. C. 15.8%, the root mean square of 10% and 20%. D. It could be either A, B, or C. There is nothing to stop you using any of these means as the answer. And anyway, the Navier-Stokes equations are hard to solve, so how can we figure out what happens when we combine two fluids?
If you answered D., you must be Ross McKitrick or Christopher Essex. From page 108 of Taken By Storm:
An example of something that behaves intensively would be the percent of milk fat in a coffee creamer. If you put two small containers of 10% coffee creamer together, you do not get 20% milk fat. The cream is still 10%, even if you have twice as much. In the same manner, if you have two identical boxes with the same energy and the same temperature, join them together. The resulting doubled box will have twice the energy, but it will not have twice the temperature. There is no amount of temperature; it measures the condition or state of the stuff in the box. …
For computing your average, why would you add up the cubes in linear form? … why not square the temperatures, or take them to the fourth power? … if you are averaging the kinetic energy of molecules, it makes sense to calculate the mean of the squares of the speeds, because energy, which goes as the square of the speed, is physically additive, while speeds themselves are not. Or, since the Stefan-Boltzmann law tells us that equilibrium radiative energy goes as the fourth power of temperature, why not raise the temperature to the fourth power before adding them up? …
With temperature, there is no basis on physical grounds to use a simple sum, some other sum or some other more complicated rule for averaging, because temperature is an intensive quantity.
OK, extensive quantities like mass and energy add when you combine things. Intensive quantities like temperature and fat percentage are the ratio of two extensive quantities. For fat percentage it’s the mass of fat divided by the total mass. So when you combine things intensive quantities don’t add, but the answer isn’t arbitrary. You just have to add up the extensive quantities and divide the totals. For the creamer example this turns out to be just your regular everyday weighted average where the weights are the mass of each creamer.
And if you remember your kinetic theory of gasses, you will have noticed how dodgy their justification for squaring temperatures was. The speed of a gas molecule is proportional to the square root of temperature, so adding up the squares of the speeds is just adding up the temperatures.
1. #1 Dano
March 24, 2006
You forgot E, Tim:
We can’t really know until we audit the quantities for ourselves, because it’s likely that the volume measurments are influenced by biased scientists.
Best,
D
2. #2 Bob
March 24, 2006
No Dano, it’s the overhead charge the research agencies collect on grants that influences the results. You need to read CA more carefully.
3. #3 William Connolley
March 24, 2006
Essex is on the Science committee for RPs upcoming “balanced” conference: http://climatesci.atmos.colostate.edu/2006/03/14/upcoming-climate-meeting/
4. #4 jre
March 24, 2006
Why has Tim not released the cream labels?
His failure to do so appears suspicious, to say the least, and casts doubt upon his results.
Aha! I see Australia has a Senate!
I’m gonna get my Senator to call all twelve of your Senators, and have one of them demand the full record of your cream purchases for the past three years, specifying the purpose and funding (with source and nature) for aforesaid purchases of cream.
5. #5 Dano
March 24, 2006
Sure, Bob. Suuuure.
Why don’t we lookit…uh…er…
Saaaay, waiiiiit a minute! hey, Bob, that’s sarcasm, right?
Haha. Got me. Good ‘un! Had me goin’ there! I was a little slow to notice that evidenceless assertion.
Haha!
Best,
D
6. #6 Paul Crowley
March 24, 2006
I’m not sure you use the simplest explanation of their error here. The answer is much simpler:
the average temperature of a system is defined to be the total heat energy over the total heat capacity.
There’s no possible argument that “total heat energy” and “total heat capacity” are well defined; for each quantity you combine both systems by sum.
The analogous argument is this: average cream proportion by mass = mass of cream in the system / mass of the system.
7. #7 llewelly
March 24, 2006
Each day I will post a picture of a scientist’s freezer, containing a half-eaten carton of ice-cream. Each picture will prove scientists have been irresponsibly sampling the ice cream in question, hopelessly skewing the results of the experiment.
8. #8 Dano
March 24, 2006
Once again, llewelly proves scientists are getting fat by skewing their results.
Proof! Proof!
D
9. #9 Bob
March 24, 2006
D,
😉
10. #10 Winston
March 24, 2006
Even the relatively shameless Ross McKitrick has become embarrassed by association with Essex over this stuff. As people would understand, it somewhat spoils the pose when the luminary you’ve relied upon for your debunkings and ridiculings of the state of science in somebody else’s field turns out to have been a dummy of the first water. A dummy without the courage to now put his head back over the parapet with you when you’ve run out of places to hide from the critical attention you’re receiving. McKitrick well deserves what he’ll suffer “in the fullness of time” for the pomposity and presumptions of adequacy in that foolish book of theirs. “Cooler Heads”? They so wish. Of course in the short term there’ll be no end of kudos in it for them – all of it from people they’ll wind up hoping never to hear from again.
11. #11 Mark [Section 15]
March 24, 2006
I have a hard time believing that McKitrick and/or Essex believe this stuff themselves. It seems more like propaganda written to target a layperson audience.
12. #12 Jack Lacton
March 24, 2006
C’mon, Tim, please use a relevant analogy!
13. #13 Matt McIrvin
March 24, 2006
How bone-stupid do you have to be before you don’t get cited as a leading dissident researcher any more?
14. #14 Eli Rabett
March 24, 2006
Actually, D is OK, although A is better for many reasons, (see http://rabett.blogspot.com/2005_11_01_rabett_archive.html) as long as you use the Kelvin (absolute) temperature scale. The boys from Canada, used Celcius, which is a kindergarden error (OK, my kids are smarter than yours, and I don’t have any). Radiative emission is proportional to T^4 ONLY if T is in Kelvin. The statement about kinetic energy is even worse. The square of the speed is linearly proportional to temperature, so you have to wonder what they were blathering about there, or maybe they are secretly arguing for a linear average.
15. #15 Webster Hubble Telescope
March 25, 2006
My favorite comical quote in the book is this:
“Everyone agrees that some kind of averaging may help, but this idea is in itself not very helpful, even though unchaperoned averaging has been going on in the back allies (sic) for decades.”
Forget the stupidity, the more I read the more these guys look like 17th century Puritans.
16. #16 z
March 26, 2006
‘My favorite comical quote in the book is this:
“Everyone agrees that some kind of averaging may help, but this idea is in itself not very helpful, even though unchaperoned averaging has been going on in the back allies (sic) for decades.”‘
Those who do not understand averages are destined to be below one.
17. #17 Paul Crowley
March 26, 2006
Rabbett: you seem to be saying that you have to take averages in Kelvins. That can’t be right; you get the same temperature no matter what scale you take the averages in.
18. #18 Eli Rabett
March 26, 2006
Paul, what I am saying is that EXCEPT for the arithmetic averages you have to take the averages in Kelvin. This is totally true for the T^4 average that E&M advocate as being proportional to emission as expressed in the Stefan Boltzman law and for any average that you are claiming to be proportional to temperature.
And NO except for the arithmetic average you DON’T get the same answer no matter what the scale is, which is one darn good reason to use arithmetic averages. For example for a geometric average, if there are any zeros in the data, the average is zero, e.g. the geometric average of 0 100, 100, 100, 100 is zero.
** You can be REALLY pedantic in the following ways: For the T^4 average the zero of the scale MUST be O K, although the scale itself (the interval between degrees) can be arbitrary. The same thing would be true for an average that you want to be proportional to the total energy. If you don’t use Kelvin, you will have to adjust the Boltzman constant and the Stefan-Boltzman constant appropriately. The physical reason for this choice of the zero point is that the energy and the emission will be zero at 0 K. If you don’t understand this comment, you can safely ignore it.
To repeat, for the other types of averages (geometric, harmonic, etc.) to make sense there should not be any zeros or negative numbers, otherwise you get misleading results. See http://rabett.blogspot.com/2005_11_01_rabett_archive.html) for examples. The bit with the vector RMS average that E&M blather about is similar. You cannot call two vectors with the same magnitude the same. They can point in different directions.
19. #19 tigtog
March 26, 2006
But what does global warming mean for the Sydney Blogger’s picnic this Saturday, Tim? Should we wear sunhats or raincoats? Inquiring minds want to know.
20. #20 Tim Curtin
March 26, 2006
Good on you tigtog! I wonder if we should subscribe for a space ship attack on the sun manned by Tim and co and armed with the CO2 fire extinguisher I will offer for them to damp down the sun which is now revealed to be real source of rising GW.
21. #21 tigtog
March 27, 2006
Tim Curtin, tish and tosh and fie upon you for attempting to deflect attention from the serious matter of a Sydney blogmeet.
Take issue with GW arguments if you must, but please don’t append my name anywhere near such stuff.
Now shoosh, or I will have to paste you with hamsters and elderberries.
22. #22 Paul Crowley
March 27, 2006
Rabbett: I was thinking of the arithmetic average; it’s true also of the median and mode, but of no other average. But that’s the physically natural one so that’s OK!
23. #23 hank
March 27, 2006
Eli has a closing paren included in the HTML for his page, so it comes up “Not Found” — correct link is:
http://rabett.blogspot.com/2005_11_01_rabett_archive.html
24. #24 Tim Curtin
March 28, 2006
Tim Lambert helpfully provided access to the paper by Hansen et al “GISS analysis of surface temperature change”, which clearly he regards as one of those tablets brought down to Moses/Mohammed. If that article is Science, then I’m not a Dutchman, but quite possibly a bin empty.
Without being palpably dishonest, it is replete with special pleading. Yet again “global” means everywhere except North America, the Arctic, India , and China. With 3 of those 4 responsible for most CO2 production this is a little odd, n’est pas? But to be fair Hansen & co do admit that the urban heat island effect is significant and has not been fully accounted for (their Fig.3 shows that for Tokyo and Phoenix (Arizona) the urban effect eliminates any upward trend in temps). My own casual empiricism has shown that when my wife turns on our new air cons the temps in our garden rise dramatically, likewise at Canberra Airport where the new Science “park” has impacted on “THE” temps for Canberra reported to GISS. Time is short, but when “global” is defined by HANSEN et al to exclude North America (p.12), the North Pacific (p.17) the Middle East (p.17) India and China (p.17), what does the word “global” actually mean? More so when USA, India, and China are said to be the main sources of AGW/CO2 by the Grand Sufi of UNSW? But to be fair, Hansen et al admit (p.25) that “urban effects are non-negligible” (p.25, they could have added “and increasing in magnitude”).
25. #25 John Cross
March 28, 2006
Mr. Curtin:
I commend you for your attempts to clean up the English Language. After you straighten up this global warming crowd could you please start on the stock market crowd. They always talk about a market going up but some of my stocks go down!!!
Likewise the weatherman says that it is raining but since I got the leak in my roof fixed there has been no rain in my living room!
Finally, while my wife says I am getting old I am sure that in fact some of my cells are actually fairly young so perhaps you could straighten her out.
John Cross
26. #26 Dano
March 28, 2006
Timmy ululatingly tries to argue that the UHI has polluted the surface temperature record.
Timmy, you are, Galileo-like, the first to say this. Your mom must be very proud. Congratulations. No one has argued this before you and thus the text you cut-pasted typed hasn’t been debunked before, certainly not here.
If you’ll excuse me, I must go as I have to stifle a yawn arising from Timmy’s trite argumentation.
D
27. #27 TCO
March 28, 2006
I said A of course and I agree with the person earlier who said that heat capacity was the relevant analog for mass.
I think a more relevant complaint of EM on the temp averages would have to do with surface versus volume measurements, non-correction for different heat capacities of different surfaces, or even an argument over “what matters”. I mean you could have the same surface average but with different distributions. Some distributions may have little impact because they don’t melt the ice sheets (or do).
I can’t understand what the heck they are blathering about with the power law crap, though.
I generally find that Tim doesn’t play very fair. He will obsess like a madman about the one (in some cases trivial) flaw that an oppenent has made and not adress whether it affects the overall larger argument. (And run around cackling at what idiots these guys are because they missed a comma or something.) That said, if EM or JohnA or anyone made a mistake, they should admit it. Even if Tim cackles and overblows it. (And Mann should also admit errors that come to mind as well…rather than dodging and saying “it doesn’t matter”. the impoact of an error is a different issue than that one has occurred. A real man will admit the error and treat the impact of it as a seperate issue.)
JohnA is still overdue to take his thermo whipping…
28. #28 Tim Lambert
March 28, 2006
Yes, the distribution of the changes matters for impacts. But the scientists have nice maps showing those, so E&M don’t have a leg to stand on.
TCO, I think your obsessing about trivial errors thing applies more to McIntyre than to me. Care to give an example, and explain why you think it is trivial?
29. #29 Eli Rabett
March 28, 2006
Hi, Surface temperature measurements are not the temperature of the ground, but the temperatures taken in the air about 1 m from the ground, sea surface temperatures are measured in the sea, but water (as is the atmosphere) has a rather uniform heat capacity (there is some variation with temperature). The comment about heat capacity for measurements is another one of those no never mind things.
30. #30 TCO
March 29, 2006
Tim, you waste your time on things like that curve or if someone got blocked from a blog by the moderator. Rather then digging into Steve’s comments about MBH.
Eli: I’m just trying to think through this. Why doesn’t it matter?
31. #31 Eli Rabett
March 30, 2006
Hi TCO, because the surface temperature measurements do not measure the temperature of the physical surface, but of the air a short distance above the surface. Thus differences in heat capacity of the ground are irrelevant. http://tinyurl.com/nvjsf
32. #32 Louis Hissink
March 31, 2006
Tim’s simple poll at the start of here shows he does understand Essex and McKitrick’s aqrguments.
33. #33 TCO
March 31, 2006
Eli:
If I have two blocks, one with heat capacity of 1 and one with heat capcity of 1000 and the surfaces are equal, but the temps are either 50 or 100, a simple average will always give me 75. But the system has a different total energy with one situation (50,1;100,1000) or (50,1000;100,1). If I want to understand warming over a time period, and the two situations reversed, the simple explanation would tell me that no warming had occurred. And I would then miss AGW which had occurred.
34. #34 Hans Erren
April 3, 2006
the hockey stock is not robust for absence of bristlecones which aren’t temperature proxies in the first place.
What more proof do you need?
35. #35 Hans Erren
April 3, 2006
indeed stock, as it has little to do with temperature ;-D
36. #36 Dano
April 3, 2006
Please Hans. Where do you get this drivel?
With your insistence on rigorous science, shouldn’t you wait until 2-3 other independent studies show the same thing, or are the marching orders explicit?
Best,
D
37. #37 Eli Rabett
April 3, 2006
TCO you have a single atmosphere, with a heat capacity of pretty much exactly 5/2 RT per mole, and you have an ocean with a heat capacity of pretty much exactly 4.18 J/K-g. You measure the temperatures of each of these at the surface independently. Not two bricks.
38. #38 TCO
April 3, 2006
Earlier, you said heat capacity was irrelevant as you measured a distance above the surface. Now, you are saying the heat capacity is invariant. Do you withdraw the earlier point then? And is the heat capacity really same of land, water, etc? I’m not even argumentative, just trying to hash it out.
39. #39 Hans Erren
April 4, 2006
Dano, M&M is replicated by Wahl and Ammann.
W&A also showed that the hockey stock is not robust for absence of bristlecones.
As for the other millennium reconstructions, that’s playing hide and seek with data.
Pandora’s box is open, the genie is out of the bottle.
40. #40 Chris O'Neill
April 4, 2006
Just google Wahl and Ammann and you’ll find out pretty quickly where most of this drivel comes from. Fortunately google also allows us to find out what Wahl and Ammann really said.
41. #41 Hans Erren
April 4, 2006
“M&M have thrown out essential data”
Which MEANS that bristlecones are essential to the hockeystick, which MEANS that the hockeystick is not robust, which was MBH’s claim.
Graybill and Idso already proved way back when that bristlecones are not good as temperature proxies.
42. #42 John Cross
April 4, 2006
Hans:
which MEANS that the hockeystick is not robust, which was MBH’s claim.
I don’t believe that they ever claimed this. Could you please provide some details.
John
43. #43 Hans Erren
April 4, 2006
44. #44 z
April 4, 2006
“According to the Financial Times of April 1, 2006: “Goldman Sachs investors yesterday overwhelmingly voted down a unique shareholder proposal that claimed the Wall Street bank was misusing shareholder resources by pursuing an potentially expensive pro-environmental agenda. The proposal, submitted for consideration at Goldman’s annual meeting by a small mutual fund firm called the Free Enterprise Action Fund, claimed that Hank Paulson had a conflict of interest in serving both as chief executive of Goldman and chairman of the Nature Conservancy, an environmental group”. … The Times reported that Steven Milloy, in representing the defeated FEAF proposal, criticised Goldman’s donation of land to the Chilean wildlife conservation society (WCS) and also it’s environmental policy which acknowledges the existence of global warming.” http://www.treehugger.com/files/2006/04/goldman_sachs_i.php
45. #45 John Cross
April 4, 2006
Hans:
This is from MBH98 (appologies to others who sat this before)
But certain sub-components of the proxy dataset (for example, the
dendroclimatic indicators) appear to be especially important in
resolving the large-scale temperature patterns, with notable
decreases in the scores reported for the proxy data set if all
dendroclimatic indicators are withheld from the multiproxy network.
On the other hand, the long-term trend in NH is relatively
robust to the inclusion of dendroclimatic indicators in the network,
suggesting that potential tree growth trend biases are not influential
in the multiproxy climate reconstructions.
Could you please refer me to the place … which MEANS that the hockeystick is not robust [to the presence of dendrochronologies], which was MBH’s claim.
Oh, also while you are at it, could you please define what you mean by the bristlecone pines. Do you mean only the bristlecone pines or are there other species involved as well?
John
46. #46 Eli Rabett
April 4, 2006
Dear TCO, I said that the heat capacity of the earth, as in the solid surface was immaterial because “surface temperatures” are measured in the atmosphere 1-2 meters from the solid surface. It is the atmosphere’s heat capacity which is invariant for all practical purposes.
There is a thread somewhere in the Deltoid sarcophagus where Robert and I patiently explain to Mark Bahner how the heat capacity of the atmosphere is calculated and why, for all practical purposes it is independent of relative humidity.
The heat capacity of water, even in the oceans, is relatively constant also (it does vary some with temperature and salinity). OTOH, the heat capacity of the solid surface does vary strongly with composition, but so what. Except for boreholes, this is not relevant to any temperature measurement, and it is dealt with there explicitly.
47. #47 Hans Erren
April 4, 2006
MBH wrote:
On the other hand, the long-term trend in NH is relatively robust to the inclusion of dendroclimatic indicators in the network, suggesting that potential tree growth trend biases are not influential in the multiproxy climate reconstructions.
so:
the long-term trend in NH is also relatively robust to the exclusion of dendroclimatic indicators in the network.
If only 12 bristlecone proxies are excluded, the result is “without statistical merit” (W&A).
That aint robust.
48. #48 Dano
April 4, 2006
MBH wrote:
On the other hand [etc]
I see there is a meeting on The Hill today where decision-makers are asking for input from the Murrican business community on how to adjust for climate impacts while keeping the economy robust.
I wonder why they aren’t having meetings on an 8-year old first paper?
Or, alternatively, I wonder if there are any folk focusing on an 8-year old first paper in attendance at that meeting?
Or perhaps attending the increasing frequency of meetings like this or participating in actions like this.
Perhaps the time spent on atomistic quibbling loads up the inbox and prevents announcement delivery.
Best,
D
49. #49 John Cross
April 4, 2006
Hans, you seem to be confusing the terms “long term trend” with “reconstruction”. Do you think they mean the same thing?
If so then why does Mann say But certain sub-components of the proxy dataset (for example, the
dendroclimatic indicators) appear to be especially important in
resolving the large-scale temperature patterns,
If not, then could you point to the statistics for the “long term trend” so that we could see the statistical merit.
John
50. #50 Dano
April 4, 2006
Hans, if you need me to send you the 8-year old first paper, let me know.
BTW, I fergot to put this in the sentence such that it reads:
Or perhaps attending the increasing frequency of meetings like this or participating in actions like this or this.
HTH frame the extent for the quibblers-left-behind (QLB).
Best,
D
51. #51 Chris O'Neill
April 5, 2006
“If only 12 bristlecone proxies are excluded, the result is “without statistical merit” (W&A).”
Where in ROBUSTNESS OF THE MANN, BRADLEY, HUGHES RECONSTRUCTION OF NORTHERN HEMISPHERE SURFACE TEMPERATURES: EXAMINATION OF CRITICISMS BASED ON THE NATURE AND PROCESSING OF PROXY CLIMATE EVIDENCE (presumably their most recent paper) do W&A say this? Or is this yet another misquotation out of context? W&A actually say in their abstract:
“Also, recent “corrections” to the Mann et al. reconstruction that suggest 15th century temperatures could have been as high as those of the late-20th century are shown to be without statistical and climatological merit.”
52. #52 Hans Erren
April 5, 2006
http://www.climateaudit.org/?p=607
Once again, in MM05b[EE], we presented an MBH98-type reconstruction without bristlecones or with reduced bristlecone weight, which yielded high 15th century values. In MM05b {E&E], we interpreted this result as only demonstrating the falseness of MBH claims of robustness to presence/absence of all dendroclimatic indicators, not as an alternative reconstruction. W&A stated that an MBH98-type reconstruction with reduced bristlecone weights (which they called an “MM” reconstruction) was without “statistical or climatological merit”. We agree that an MBH98-type reconstruction with reduced bristlecone weights is without “statistical merit”, but this does not mean that an MBH98 reconstruction with high bristlecone weights has “statistical merit”.
The salient question arising from this is:
if an MBH98-type reconstruction with reduced bristlecone weights lacks statistical merit, then either all the other proxies are no good or the MBH98 method is no good or both.
We have never argued that an MBH98-type reconstruction with reduced bristlecone weights has “statistical merit”. We have vigorously argued that an MBH98-type reconstruction with high bristlecone weights lacks “statistical merit.”
53. #53 John Cross
April 5, 2006
Hans, you said:
the long-term trend in NH is also relatively robust to the exclusion of dendroclimatic indicators in the network.
If only 12 bristlecone proxies are excluded, the result is “without statistical merit” (W&A).
could you please clarify what you mean by the result. Is it the “long term trend”, is it the “reconstruction” or is it both?
John
54. #54 Hans Erren
April 5, 2006
What again is the title of the W&A paper? (Hint: look at the first word)
And now you are trying to prove that MBH wrote that their reconstruction was not robust?
come on.
55. #55 Tim Lambert
April 5, 2006
Hans, you seem to be really confused. W&A say that the reconstruction is robust with respect to the method used to calculate the pricipal components. They didn’t say, as far as I can tell, that the reconstruction was robust wrt exclusion of the bristlecone pine proxies. In any case, W&A are not the same people as MB&H. Perhaps you could point to some statement in MBH98 or MBH99?
56. #56 Hans Erren
April 5, 2006
The problem boils down to the question “Are bristlecones valid temperature proxies”
According to Graybill and Idso they are not, if you then remove the bristlecones from the MBH reconstruction then the reconstruction is without statistical merit.
According to Graybill and Idso the MBH reconstruction is without climatological merit when the bristlecones are included, you might as well have taken the amsterdam stock exchange index.
If you want to question robustness do so on climateaudit, instead of murmuring there about entropy.
57. #57 John Cross
April 5, 2006
No, Hans, that is not the problem. I fail to see why you are having such a hard time understanding this. You stated:
Which MEANS that bristlecones are essential to the hockeystick, which MEANS that the hockeystick is not robust, which was MBH’s claim.
MBH never claimed that the reconstruction was robust to the absence of the BCPs. This is not a trivial points since it gets repeated so much.
In regards to dendrochronologies being valid temperature proxies, I am afraid that I am not an expert on such. I do wish we had access to some of the experts though.
John
58. #58 z
April 5, 2006
“I see there is a meeting on The Hill today where decision-makers are asking for input from the Murrican business community on how to adjust for climate impacts while keeping the economy robust.”
“The Kyoto Protocol isn’t merely a political document — it’s the equivalent of market research. Viewing the protocol from this perspective, we see that most of the world’s nations believe climate change is real and urgent and are attempting to do something about it. That allows the commercially inclined among us to move on to the wonderfully crass questions about what products we might sell them. We start to ask: Whose appliances will the world buy? Whose fuel cells and photovoltaic panels? Whose light bulbs? Whose cars?
“This basic market intelligence may be the single most important key to thriving in a new economy. There is money to be made here — lots of it. There are jobs here — lots of them. This is Silicon Valley 2.0. We thrived with the information revolution, developing computers and software to change how the world works and plays. We can thrive again with a clean energy revolution, changing the way the world finds power for electricity and transport. But we can only thrive if we lead, if we take action now.”
http://www.salon.com/opinion/feature/2006/04/04/hope/index1.html
59. #59 TCO
April 5, 2006
Eli, what about the difference in heat capacity of the sea and land measurements. If a set area of ocean goes up 1 degree does it mean the same thing from an energy balance standpoint as if an equivalent area of land (air one meter above land)?
60. #60 TCO
April 5, 2006
Also, if your air (one meter above land measurements) is isotemperature with the land underneath it, then the heat capacity of that land surface is relevant.
61. #61 Dano
April 5, 2006
then the heat capacity of that land surface is relevant.
The standards generally state the instrument shelter should be above vegetation, particularly grass, so the ET of the vegetation is relevant too. The temperature of the exhaust from the lawn mower is relevant as well.
Kidding aside, ages ago when I was a weatherman you could see, a few hours after sunset not in winter, that the ambient temperature would rise for a shortish period ( a few to tens of minutes ).
Why? Soil holds on to its heat for a time and then releases it, causing the temp to rise.
Best,
D
62. #62 Hans Erren
April 6, 2006
Mann accuses M&M that by taking out the brislecones they are removing essential data from the MBH reconstruction.
So bristlecones are essential for the MBH reconstruction.
However, bristlecones are not good as temperature proxy as Graybill and Idso already showed in 1993.
Graybill, D.A., and S.B. Idso. 1993. Detecting the aerial fertilization effect of atmospheric CO2 enrichment in tree-ring chronologies. Global Biogeochemical Cycles 7:81-95.
I fail to see why you are having such a hard time understanding this.
63. #63 John Cross
April 6, 2006
Hans:
What got us started in this discussion is your claim that … which MEANS that the hockeystick is not robust, which was MBH’s claim.
Is there anything in MBH that backs up this statement? I say there is not.
John
64. #64 Hans Erren
April 6, 2006
you say there isn’t, I see Steve McIntyre reporting it and I see Wahl and Ammann writing a paper with Robustness in the title, so it must be an issue.
Anyway the MBH hockeystick is proven without climatological merit in the first place so why debate on as minor exegetical question?
65. #65 TCO
April 6, 2006
I think Steve’s crticism is more subtle than what you state, Hans. Steve is not getting behind the Ibso paper, has not reviewed it critically, etc. He puts the onus on Mann(ites) to deal with the proxy kvetches.
66. #66 Hans Erren
April 6, 2006
is he?
Stephen McIntyre writes:
http://www.climate2003.com/mann.responses.htm
This expedient also fails to deal with the defects of bristlecone pine growth as a so-called “proxy” for temperature, discussed in pages 81-86 of our E&E article. There is a definite 20th century pulse in bristlecone pine growth. However, there are explicit statements in specialist literature (surveyed there) that this pulse is not due to temperature and MBH co-author Hughes has stated that the pulse is a “mystery”. Our E&E article discusses issues pertaining to bristlecone pine growth, showing that it is unacceptable that world climate history should be held to depend upon a PC4 made up of such controversial data.
67. #67 Dano
April 6, 2006
Few people care about the value, scientifically, of what people write on websites. The sciency folks have had mechanisms/forums in place for many decades to hash these things out.
Plus, for centuries, people have hand-waved and finger-pointed about results. What has made a change? People going out and getting their own data to show their point. Until that’s done here, the finger-pointing is a mere distraction. And the finger-pointing is about an 8 year old, first paper.
Let us not forget that society is moving forward and not debating an 8 year old first paper.
Best,
D
68. #68 Hans Erren
April 6, 2006
What then was debated at the NAS panel dano?
Cherrypicking and hiding data, the forward movement is not that rapid since MBH.
LOL
69. #69 TCO
April 6, 2006
Hans, Yup. In that paragraph, Steve is saying that Mann is basing his rec onstruction on a proxy that some other person has called into question. Is saying that Mann should defend that or should make sure that this is valid or that he is using proxies that are “in question”. I’ll bet you dollars to donuts, Steve’s not putting his own imprateur on the Ibso paper.
70. #70 John Cross
April 7, 2006
Hans: again, I am surprised that you are missing the point. You said:
you say there isn’t, I see Steve McIntyre reporting it and I see Wahl and Ammann writing a paper with Robustness in the title, so it must be an issue.
So, if someone else reports it and then a paper is written that that has the word robust in the title (in another context mind you) then MBH said it? While I am intrigued by this strange take on science I still feel that we should consult what people have said instead of what others write about them. So can you find for me the passage in MBH where they claim that the reconstruction is robust to the exclusion of dendrochronologies?
Anyway the MBH hockeystick is proven without climatological merit in the first place so why debate on as minor exegetical question?
True, and I see I shouldn’t have brought up the issue – oh, wait – I didn’t, you did!!! If you don’t wish to debate minor exegetical questions, maybe you shouldn’t bring them up.
John
71. #71 Hans Erren
April 8, 2006
You can’t win john,
Either MBH wrote that their reconstruction is robust for inclusion of proxies, W&A and M&M showed it isn’t for bristlecones. So MBH has no climatological merit.
or MBH wrote that their reconstruction is not robust for inclusion of proxies, then their number of samples is not sufficient for a reliable reconstruction. So MBH has no statistical merit.
72. #72 Steve Bloom
April 8, 2006
Hans, I sincerely hope you’re not expecting a “skeptical” outcome to the NRC process. If so, you’re in for a disappointment.
73. #73 Chris O'Neill
April 8, 2006
You’d all better start reading Wahl and Ammann’s material and paper to find out what it actually says. Here are some important points:
The W&A emulation of MM05’s E&E method using the MBH 1450 proxy network produces a reconstruction without the bristlecone proxies that passes validation and is robust to the absence of the bristlecone proxies. So there’s no problem going back to 1450. That hockeystick is there from 1450 with or without bristlecone proxies.
Their emulation of MM05’s E&E method using the MBH 1400 proxy network produces a reconstruction without the bristlecone proxies that fails validation while if the bristlecone proxies are included it passes validation. So I guess robustness in this case means the temperature reconstruction passes validation against the instrument record or more recent proxies in spite of part of the proxies having growth trends that are not related to the instrument temperature record or other proxies.
So this argument about bristlecone proxies only affects the 50 years before 1450 (only 47 years now being claimed for MBH98 because one of the other proxies only started in 1403). 47 years, hmm, this is starting to look rather academic. However..
Compare the 1450 proxy network result excluding bristlecones in W&A’s figure 5(c) with the 1400 proxy nework result including bristlecones in W&A’s figure 5(d). The two agree pretty well from when the 1450 network result begins (1450 of course). Somehow, M&M think that even though the bristlecones show no signs of biasing the reconstruction (using the 1400 proxy network) after 1450 that somehow they could suddenly start biasing it before 1450. Let’s get this clear. The 1400 proxy network (that includes the bristlecones) produces a reconstruction from 1400 to 1980. This is tested independently of bristlecones all the way from 1450 to 1980 and passes. The only way it could produce an invalid result for 1400 to 1450 is for something to go wrong during 1400 to 1450. No-one is suggesting this. The only suggestion is that something goes wrong in the 20th century. How can something going wrong in the 20th century produce perfectly reasonable results from 1450 to 1980 but then suddenly go wrong between 1400 and 1450?
Hans Erren wrote “According to Graybill and Idso the MBH reconstruction is without climatological merit when the bristlecones are included”.
I wonder where Graybill and Idso wrote this? Couldn’t have been in “Graybill, D.A., and S.B. Idso. 1993: Detecting the aerial fertilization effect etc.” 1993 occured before 1998 as far as I’m aware.
74. #74 John Cross
April 8, 2006
Hans: Its a good thing that I am not trying to win then. What I am trying to do (so far without success) is to get you to show me where in MBH they say that the reconstruction is robust to the exclusion of dendrochronologies.
John
75. #75 TCO
April 8, 2006
Chris, that is a bunch of cherry-picking. It needs to be valid overall.
76. #76 Chris O'Neill
April 10, 2006
Making a highly detailed response, TCO says:
“that is a bunch of cherry-picking”
TCO says it so it’s true.
77. #77 TCO
April 11, 2006
That’s my Ozymandious mode…and don’t be fucking talking about cracks in my statue. 🙂
78. #78 Dave Dardinger
July 14, 2006
Well, here I am, Tim! I want to start with a little “toy” example. Imagine a tic-tac-toe board (a 3×3 matrix) with the following in the corners: 15, 21; 25, 31. The first two are two readings (at different times) from a thermometer 1 and the second two from thermometer 2 at times T1 and T2. I think everyone will agree that you can fill in the averages among these various readings to get: 15, 18, 21; 20, 23, 26; 25 28, 31. So 23 is the “global average” of this example.
Now one more piece of data. The first thermometer is on Weathertop, which in this toy world has a constant barametric pressure of 500 mB while the second thermometer is from the Down in the Valley station with a pressure of 1000 mB. Question to think about then (and hopefully post about) before I post again: Is there any problem averaging station temperatures from different pressure regimes such as this? If so why and what can we do about it? If not why not and are you willing to defend that position to the death (so to speak?)
79. #79 Tim Lambert
July 14, 2006
Dave: weighted average, not just simple average. Do you concede that geometric, harmonic, quadratic means are not appropriate?
80. #80 Dano
July 14, 2006
I’d be interested in seeing a GHCN station at 5640m (~500 mb). Fascinating. Who washes the windows up there, and with what liquid? How do you judge the mileage for the trip – with the two legs or the hypotenuse?
And which stations in, say, the Alps (~900-850 mb) are GHCN stations, and those, oh, two are what fraction of the network? The Andes? Oh, wait: that’s Tim’s weighted avg. Never mind. Nonetheless,
Fascinating.
Best,
D
81. #81 Dave Dardinger
July 14, 2006
Tim, your response isn’t fully responsive. Why use a weighted average? What are you protecting against if temperature, an intensive variable, albeit one which is a ratio of extensive variables, is perfectly ok as something to be averaged? We’re not worrying about a “correct” temperature, after all, but just a self consistant one; i.e. one we can track over time and calculate deviations in.
Dano, I used to argue with a Dano on the Compuserve SF forum 11+ years ago. You wouldn’t happen to be the same guy, would you? And I assume you’re just trying to be funny. You know from my description of my “toy world” that it’s designed for computational simplicity, not real-world emulation. Just as I picked the temperatures to all come out with integer averages.
82. #82 Dano
July 14, 2006
Not the same guy, Dave. I’m being funny to point out stuff that commonly gets overlooked in the temp thingy.
Best,
D
83. #83 Dave Dardinger
July 15, 2006
I’m still waiting for Tim to say why we should weigh the temperatures and how they are weighed. I.e. what is this global temperature to be? Just the temperature 2 meters above the earth’s surface level? If so I suppose Dano’s questions about where the high altitude GHCN stations are might be useful. But the real question is just what this value is, (once properly determined), from a physics POV? Is it just a statistic or is it otherwise meaningful?
But I’ll go on a little and hope Tim catches up. Since he didn’t say, I’ll assume that Tim is thinking that we should weigh via atmosphere vs weighing via relative surface area.
So we’ll give our high pressure station twice the weight of the low pressure station. Then the mass-average temperature is (15 + 21 + 2×25 + 2×31)/6 or 148/6 = 24.67 vs 23. You won’t like that decision as to Tim’s meaning, I suspect, but Hey!, I asked for more than one sentence. We’re looking for purpose here. The purpose of a derived figure has to come before the proper method for deriving it comes. [Note, I’m not looking for deep thoughts here. Simply clarity.]
84. #84 Dave Dardinger
July 17, 2006
Yoo Hoo, Tim!
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# C4 trig Watch
1. I got :
(1- tany) / (1+ tany) = k
now i need to express k in terms of tany so i need tany= ?
but how do i do it ?
2. done it.
1-tany = k + k.tany
1 - k = k.tany +tany
1 - k = tany (k + 1)
(1 - k) / (1+ k) = tany
If anyones got a simpler way, i would still like to know :P
3. (Original post by uxa595)
done it.
1-tany = k + k.tany
1 - k = k.tany +tany
1 - k = tany (k + 1)
(1 - k) / (1+ k) = tany
If anyones got a simpler way, i would still like to know :P
Nope that's the way to do it.
I'm guessing you're on AQA cause I remember helping someone in my school do this question just a couple of days ago
4. (Original post by hassi94)
Nope that's the way to do it.
I'm guessing you're on AQA cause I remember helping someone in my school do this question just a couple of days ago
Yep, AQA
I've got another one for you.
SR( 1+sin2A / 1-sin2A) = tan (A + pi/4) p262 ex6b q8
i've got it to cosA + sinA / cosA - sinA
but i'm lost from there on
5. (Original post by uxa595)
Yep, AQA
I've got another one for you.
SR( 1+sin2A / 1-sin2A) = tan (A + pi/4) p262 ex6b q8
i've got it to cosA + sinA / cosA - sinA
but i'm lost from there on
Yeah I had to do this one for my teacher!
You used your identities wrong, but anyway don't do that.
Start with the left hand side. Multiply both the top ans bottom by 1+ sin2a but don't multiply out the bracket I'm the numerator.
Then figure out how you can get rid of the square root.
6. (Original post by hassi94)
Yeah I had to do this one for my teacher!
You used your identities wrong, but anyway don't do that.
Start with the left hand side. Multiply both the top ans bottom by 1+ sin2a but don't multiply out the bracket I'm the numerator.
Then figure out how you can get rid of the square root.
Ahem he's done it right so far. If he multiplies top and bottom by cosA+sinA and simplifies it reduces to a well known identity.
Spoiler:
Show
tan(A/2+ pi/4)= tan(A)+sec(A)
Replace A with 2A and you get?
Ahem he's done it right so far. If he multiplies top and bottom by cosA+sinA and simplifies it reduces to a well known identity.
Spoiler:
Show
tan(A/2+ pi/4)= tan(A)+sec(A)
Replace A with 2A and you get?
Oh sorry I didn't see it, I'll have a look later I did that while getting changed this morning. But I don't think you should use any identity except the main sin cos tang a + b and cos squared and sin squared. Otherwise its cheating a little
8. (Original post by hassi94)
Oh sorry I didn't see it, I'll have a look later I did that while getting changed this morning. But I don't think you should use any identity except the main sin cos tang a + b and cos squared and sin squared. Otherwise its cheating a little
Hmm maybe, I don't know.
It's a well known result though.
We're using identities (they hold true for any value of A), so I think it would be allowed .
9. I know what you mean but my solution assumes a lot less which I think is better.
10. (Original post by hassi94)
I know what you mean but my solution assumes a lot less which I think is better.
In which way are you doing it, i did it in a similar way to you by multiply the top and bottom in the LHS by 1+sin2a, getting rid of square root, but then also it wasn't simple to solve it.
So i get LHS as:
Then i tried to show the expression i got for LHS by using RHS, and succeeded in it. So now by seeing the steps in reverse i was able to know how to do the question by just using the LHS and not making any changes to RHS. But how to do it directly, i don't have any clue of how could i have solved it if i wouldn't have tried using both sides.
11. (Original post by raheem94)
In which way are you doing it, i did it in a similar way to you by multiply the top and bottom in the LHS by 1+sin2a, getting rid of square root, but then also it wasn't simple to solve it.
So i get LHS as:
Then i tried to show the expression i got for LHS by using RHS, and succeeded in it. So now by seeing the steps in reverse i was able to know how to do the question by just using the LHS and not making any changes to RHS. But how to do it directly, i don't have any clue of how could i have solved it if i wouldn't have tried using both sides.
After where you are, expand cos2a then divide by cos to make everything tan or sec. Then make equations in tan and take common factor (1 + tan a I think) then its clear. And I only manipulated the LHS to work it out
12. (Original post by hassi94)
I know what you mean but my solution assumes a lot less which I think is better.
But we're not making assumptions. Identities serve useful purposes, for example in this case .
Anyway fair enough.
But we're not making assumptions. Identities serve useful purposes, for example in this case .
Anyway fair enough.
It is an assumption if you don't prove it. In a level maths you can only assume what's in the formula booklet and slight variations (like the ones derived from cos squared + sin squared. It's not right to show an identity is true by using some other identity that isn't given. You may or may not get the marks but it's definitely not proper - I think it's worth the extra effort to do it my way
Sorry for poor formatting throughout. I don't like typing on phones.: p
14. (Original post by raheem94)
In which way are you doing it, i did it in a similar way to you by multiply the top and bottom in the LHS by 1+sin2a, getting rid of square root, but then also it wasn't simple to solve it.
So i get LHS as:
Then i tried to show the expression i got for LHS by using RHS, and succeeded in it. So now by seeing the steps in reverse i was able to know how to do the question by just using the LHS and not making any changes to RHS. But how to do it directly, i don't have any clue of how could i have solved it if i wouldn't have tried using both sides.
Now I'm on my PC I can type it out fully.
Spoiler:
Show
Diving all terms by cos^2a to get
Then since tan45 = 1,
15. (Original post by hassi94)
It is an assumption if you don't prove it. In a level maths you can only assume what's in the formula booklet and slight variations (like the ones derived from cos squared + sin squared. It's not right to show an identity is true by using some other identity that isn't given. You may or may not get the marks but it's definitely not proper - I think it's worth the extra effort to do it my way
Sorry for poor formatting throughout. I don't like typing on phones.: p
Again that's like saying you're assuming sin(A+B)= sinAcosB+cosAsinB (simply because you haven't proved it).
Also it is an identity; so you wouldn't be penalised at all. My way is just a shortcut.
16. http://en.wikipedia.org/wiki/List_of...ric_identities
Scroll down you'll see where I got it from. It's not an assumption at all.
17. I got it finally :P
All i did was divide everything by cos and from there on, it was easy.
http://en.wikipedia.org/wiki/List_of...ric_identities
Scroll down you'll see where I got it from. It's not an assumption at all.
No no that's not what I'm saying. It is an assumption, anything is an assumption if you don't prove it. What makes the difference is what you're ALLOWED to assume, and you're allowed to quote the formula booklet without proof.
19. One thing i found confusing though is, how just before you get to this step:
http://www.wolframalpha.com/input/?i...28pi%2F4%29%29
You get this:
http://www.wolframalpha.com/input/?i...in%5E2+x%29%29
and the denominator can be factorised to either:
(sinx - cosx)^2
or
(cosx - sinx)^2
but only the second one in right. How would you know which one works in the exam without having to do them both fully?
20. (Original post by uxa595)
I got it finally :P
All i did was divide everything by cos and from there on, it was easy.
Good That's a good direction to take from there too well done (sorry I told you that you were wrong at the start, I didn't look properly - was in the middle of getting changed for 6th form! )
Updated: March 16, 2012
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Three-dimensional space (also: 3-space or, rarely, tri-dimensional space) is a geometric setting in which three values (called parameters) are required to determine the position of an element (i.e., point). This is the informal meaning of the term dimension. In physics and mathematics, a sequence of numbers can be understood as a location in -dimensional space. When , the set of all such locations is called three-dimensional Euclidean space (or simply Euclidean space when the context is clear). It is commonly represented by the symbol . This serves as a three-parameter model of the physical universe (that is, the spatial part, without considering time), in which all known matter exists. While this space remains the most compelling and useful way to model the world as it is experienced, it is only one example of a large variety of spaces in three dimensions called 3-manifolds. In this classical example, when the three values refer to measurements in different directions (coordinates), any three directions can be chosen, provided that vectors in these directions do not all lie in the same 2-space (plane). Furthermore, in this case, these three values can be labeled by any combination of three chosen from the terms ''width'', ''height'', ''depth'', and ''length''.
In Euclidean geometry
Coordinate systems
In mathematics, analytic geometry (also called Cartesian geometry) describes every point in three-dimensional space by means of three coordinates. Three coordinate axes are given, each perpendicular to the other two at the origin, the point at which they cross. They are usually labeled , and . Relative to these axes, the position of any point in three-dimensional space is given by an ordered triple of real numbers, each number giving the distance of that point from the origin measured along the given axis, which is equal to the distance of that point from the plane determined by the other two axes. Other popular methods of describing the location of a point in three-dimensional space include cylindrical coordinates and spherical coordinates, though there are an infinite number of possible methods. For more, see Euclidean space. Below are images of the above-mentioned systems. Image:Coord XYZ.svg|Cartesian coordinate system Image:Cylindrical Coordinates.svg|Cylindrical coordinate system Image:Spherical Coordinates (Colatitude, Longitude).svg|Spherical coordinate system
Lines and planes
Two distinct points always determine a (straight) line. Three distinct points are either collinear or determine a unique plane. On the other hand, four distinct points can either be collinear, coplanar, or determine the entire space. Two distinct lines can either intersect, be parallel or be skew. Two parallel lines, or two intersecting lines, lie in a unique plane, so skew lines are lines that do not meet and do not lie in a common plane. Two distinct planes can either meet in a common line or are parallel (i.e., do not meet). Three distinct planes, no pair of which are parallel, can either meet in a common line, meet in a unique common point, or have no point in common. In the last case, the three lines of intersection of each pair of planes are mutually parallel. A line can lie in a given plane, intersect that plane in a unique point, or be parallel to the plane. In the last case, there will be lines in the plane that are parallel to the given line. A hyperplane is a subspace of one dimension less than the dimension of the full space. The hyperplanes of a three-dimensional space are the two-dimensional subspaces, that is, the planes. In terms of Cartesian coordinates, the points of a hyperplane satisfy a single linear equation, so planes in this 3-space are described by linear equations. A line can be described by a pair of independent linear equations—each representing a plane having this line as a common intersection. Varignon's theorem states that the midpoints of any quadrilateral in ℝ3 form a parallelogram, and hence are coplanar.
Spheres and balls
A sphere in 3-space (also called a 2-sphere because it is a 2-dimensional object) consists of the set of all points in 3-space at a fixed distance from a central point . The solid enclosed by the sphere is called a ball (or, more precisely a 3-ball). The volume of the ball is given by :$V = \frac\pi r^$. Another type of sphere arises from a 4-ball, whose three-dimensional surface is the 3-sphere: points equidistant to the origin of the euclidean space . If a point has coordinates, , then characterizes those points on the unit 3-sphere centered at the origin.
Polytopes
In three dimensions, there are nine regular polytopes: the five convex Platonic solids and the four nonconvex Kepler-Poinsot polyhedra.
Surfaces of revolution
A surface generated by revolving a plane curve about a fixed line in its plane as an axis is called a surface of revolution. The plane curve is called the ''generatrix'' of the surface. A section of the surface, made by intersecting the surface with a plane that is perpendicular (orthogonal) to the axis, is a circle. Simple examples occur when the generatrix is a line. If the generatrix line intersects the axis line, the surface of revolution is a right circular cone with vertex (apex) the point of intersection. However, if the generatrix and axis are parallel, then the surface of revolution is a circular cylinder.
In analogy with the conic sections, the set of points whose Cartesian coordinates satisfy the general equation of the second degree, namely, :$Ax^2 + By^2 + Cz^2 + Fxy + Gyz + Hxz + Jx + Ky + Lz + M = 0,$ where and are real numbers and not all of and are zero, is called a quadric surface. There are six types of non-degenerate quadric surfaces: # Ellipsoid # Hyperboloid of one sheet # Hyperboloid of two sheets # Elliptic cone # Elliptic paraboloid # Hyperbolic paraboloid The degenerate quadric surfaces are the empty set, a single point, a single line, a single plane, a pair of planes or a quadratic cylinder (a surface consisting of a non-degenerate conic section in a plane and all the lines of through that conic that are normal to ). Elliptic cones are sometimes considered to be degenerate quadric surfaces as well. Both the hyperboloid of one sheet and the hyperbolic paraboloid are ruled surfaces, meaning that they can be made up from a family of straight lines. In fact, each has two families of generating lines, the members of each family are disjoint and each member one family intersects, with just one exception, every member of the other family. Each family is called a regulus.
In linear algebra
Another way of viewing three-dimensional space is found in linear algebra, where the idea of independence is crucial. Space has three dimensions because the length of a box is independent of its width or breadth. In the technical language of linear algebra, space is three-dimensional because every point in space can be described by a linear combination of three independent vectors.
Dot product, angle, and length
A vector can be pictured as an arrow. The vector's magnitude is its length, and its direction is the direction the arrow points. A vector in can be represented by an ordered triple of real numbers. These numbers are called the components of the vector. The dot product of two vectors and is defined as: :$\mathbf\cdot \mathbf = A_1B_1 + A_2B_2 + A_3B_3.$ The magnitude of a vector is denoted by . The dot product of a vector with itself is :$\mathbf A\cdot\mathbf A = \|\mathbf A\|^2 = A_1^2 + A_2^2 + A_3^2,$ which gives : $\|\mathbf A\| = \sqrt = \sqrt,$ the formula for the Euclidean length of the vector. Without reference to the components of the vectors, the dot product of two non-zero Euclidean vectors and is given by :$\mathbf A\cdot\mathbf B = \|\mathbf A\|\,\|\mathbf B\|\cos\theta,$ where is the angle between and .
Cross product
The cross product or vector product is a binary operation on two vectors in three-dimensional space and is denoted by the symbol ×. The cross product a × b of the vectors a and b is a vector that is perpendicular to both and therefore normal to the plane containing them. It has many applications in mathematics, physics, and engineering. The space and product form an algebra over a field, which is neither commutative nor associative, but is a Lie algebra with the cross product being the Lie bracket. One can in ''n'' dimensions take the product of vectors to produce a vector perpendicular to all of them. But if the product is limited to non-trivial binary products with vector results, it exists only in three and seven dimensions.
In calculus
In a rectangular coordinate system, the gradient is given by :$\nabla f = \frac \mathbf + \frac \mathbf + \frac \mathbf$ The divergence of a continuously differentiable vector field F = ''U'' i + ''V'' j + ''W'' k is equal to the scalar-valued function: :$\operatorname\,\mathbf = \nabla\cdot\mathbf =\frac +\frac +\frac.$ Expanded in Cartesian coordinates (see Del in cylindrical and spherical coordinates for spherical and cylindrical coordinate representations), the curl ∇ × F is, for F composed of 'F''x, ''F''y, ''F''z :$\begin \mathbf & \mathbf & \mathbf \\ \\ & & \\ \\ F_x & F_y & F_z \end$ where i, j, and k are the unit vectors for the ''x''-, ''y''-, and ''z''-axes, respectively. This expands as follows: :$\left\left(\frac - \frac\right\right) \mathbf + \left\left(\frac - \frac\right\right) \mathbf + \left\left(\frac - \frac\right\right) \mathbf$
Line integrals, surface integrals, and volume integrals
For some scalar field ''f'' : ''U'' ⊆ R''n'' → R, the line integral along a piecewise smooth curve ''C'' ⊂ ''U'' is defined as :$\int\limits_C f\, ds = \int_a^b f\left(\mathbf\left(t\right)\right) |\mathbf\text{'}\left(t\right)|\, dt.$ where r: , b→ ''C'' is an arbitrary bijective parametrization of the curve ''C'' such that r(''a'') and r(''b'') give the endpoints of ''C'' and $a < b$. For a vector field F : ''U'' ⊆ R''n'' → R''n'', the line integral along a piecewise smooth curve ''C'' ⊂ ''U'', in the direction of r, is defined as :$\int\limits_C \mathbf\left(\mathbf\right)\cdot\,d\mathbf = \int_a^b \mathbf\left(\mathbf\left(t\right)\right)\cdot\mathbf\text{'}\left(t\right)\,dt.$ where · is the dot product and r: , b→ ''C'' is a bijective parametrization of the curve ''C'' such that r(''a'') and r(''b'') give the endpoints of ''C''. A surface integral is a generalization of multiple integrals to integration over surfaces. It can be thought of as the double integral analog of the line integral. To find an explicit formula for the surface integral, we need to parameterize the surface of interest, ''S'', by considering a system of curvilinear coordinates on ''S'', like the latitude and longitude on a sphere. Let such a parameterization be x(''s'', ''t''), where (''s'', ''t'') varies in some region ''T'' in the plane. Then, the surface integral is given by :$\iint_ f \,\mathrm dS = \iint_ f\left(\mathbf\left(s, t\right)\right) \left\|\times \right\| \mathrm ds\, \mathrm dt$ where the expression between bars on the right-hand side is the magnitude of the cross product of the partial derivatives of x(''s'', ''t''), and is known as the surface element. Given a vector field v on ''S'', that is a function that assigns to each x in ''S'' a vector v(x), the surface integral can be defined component-wise according to the definition of the surface integral of a scalar field; the result is a vector. A volume integral refers to an integral over a 3-dimensional domain. It can also mean a triple integral within a region ''D'' in R3 of a function $f\left(x,y,z\right),$ and is usually written as: :$\iiint\limits_D f\left(x,y,z\right)\,dx\,dy\,dz.$
Fundamental theorem of line integrals
The fundamental theorem of line integrals, says that a line integral through a gradient field can be evaluated by evaluating the original scalar field at the endpoints of the curve. Let $\varphi : U \subseteq \mathbb^n \to \mathbb$. Then :$\varphi\left\left(\mathbf\right\right)-\varphi\left\left(\mathbf\right\right) = \int_ \nabla\varphi\left(\mathbf\right)\cdot d\mathbf.$
Stokes' theorem
Stokes' theorem relates the surface integral of the curl of a vector field F over a surface Σ in Euclidean three-space to the line integral of the vector field over its boundary ∂Σ: :$\iint_ \nabla \times \mathbf \cdot \mathrm\mathbf = \oint_ \mathbf \cdot \mathrm \mathbf.$
Divergence theorem
Suppose is a subset of $\mathbb^n$ (in the case of represents a volume in 3D space) which is compact and has a piecewise smooth boundary (also indicated with ). If is a continuously differentiable vector field defined on a neighborhood of , then the divergence theorem says: : The left side is a volume integral over the volume , the right side is the surface integral over the boundary of the volume . The closed manifold is quite generally the boundary of oriented by outward-pointing normals, and is the outward pointing unit normal field of the boundary . ( may be used as a shorthand for .)
In topology
Three-dimensional space has a number of topological properties that distinguish it from spaces of other dimension numbers. For example, at least three dimensions are required to tie a knot in a piece of string. In differential geometry the generic three-dimensional spaces are 3-manifolds, which locally resemble $^3$.
In finite geometry
Many ideas of dimension can be tested with finite geometry. The simplest instance is PG(3,2), which has Fano planes as its 2-dimensional subspaces. It is an instance of Galois geometry, a study of projective geometry using finite fields. Thus, for any Galois field GF(''q''), there is a projective space PG(3,''q'') of three dimensions. For example, any three skew lines in PG(3,''q'') are contained in exactly one regulus.Albrecht Beutelspacher & Ute Rosenbaum (1998) ''Projective Geometry'', page 72, Cambridge University Press
* Dimensional analysis * Distance from a point to a plane * Four-dimensional space * * Three-dimensional graph * Two-dimensional space
Notes
References
* * Arfken, George B. and Hans J. Weber. ''Mathematical Methods For Physicists'', Academic Press; 6 edition (June 21, 2005). . * | 3,407 | 14,208 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 21, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-17 | latest | en | 0.936162 |
http://openstudy.com/updates/4f0a17a0e4b014c09e645b2d | 1,448,676,980,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450745.23/warc/CC-MAIN-20151124205410-00024-ip-10-71-132-137.ec2.internal.warc.gz | 174,085,309 | 19,564 | ## Inopeki 3 years ago TuringTest, what now?
1. saifoo.khan
Party time! :D
2. TuringTest
lol I dunno...
3. Inopeki
What can you teach me?
4. TuringTest
a lot, there's a fair amount to algebra, but what's the next logical step?
5. FoolForMath
6. TuringTest
I have already to an extent, but where to go next...
7. Inopeki
Ratios?
8. FoolForMath
In maths the fundamentals concepts are limited but the problems are not, so how about practicing what you have already learned?
9. TuringTest
That's a good point FFM so lets do simplification that you know, but a little more intense: I think I need to make sure you can simplify bigger things:$\frac{x^2y^5z}{xy^9z^3}$
10. saifoo.khan
11. TuringTest
awww, you're making me think of my cat in the hospital :(
12. Inopeki
13. Inopeki
How did it go with her?
14. TuringTest
broken leg... waiting for an operation
15. FoolForMath
You beat you cat Turing?!!!
16. Inopeki
aww :(
17. saifoo.khan
turing sat on her leg. :(
18. TuringTest
she fell off the roof. how pessimistic
19. saifoo.khan
awww, sorry i was jking..
20. TuringTest
It's all good :P
21. FoolForMath
Due to Turing's intense physics stress his cat decided to commit suicide :P
22. Inopeki
(x^2)*(y^5)*(z) ------------ = (x)*(y^5/9)*(z^1/3)? (x)*(y^9)*(z^3)
23. saifoo.khan
24. GT
Cats are cool.
25. FoolForMath
why $$y^{5/9}$$ ?
26. saifoo.khan
no doubt. ;)
27. TuringTest
you have inconsistent rules above inopeki$\frac{x^a}{x^b}=x^{a-b}$always...
28. Inopeki
29. Inopeki
Oh right
30. Inopeki
(x^2)*(y^5)*(z) ------------ = (x)*(y^5-9)*(z^1-3)? (x)*(y^9)*(z^3)
31. TuringTest
yes, simplify...
32. saifoo.khan
33. Inopeki
(x)*(y^5-9)*(z^1-3)=(x)*(y^-4)*(z^-2)?
34. Inopeki
35. saifoo.khan
cute^
36. TuringTest
yes, do you know another way to write$x^{-a}$???
37. Inopeki
No
38. Inopeki
39. TuringTest
$x^{-a}=\frac{1}{x^a}$so it's probably nicer to rewrite your expression with all positive exponents in this way.
40. Inopeki
Oh right, all negative exponents make the "total number" divided by one.
41. FoolForMath
Btw who can explain why x^0 = 1?
42. Inopeki
(x)*(1/y^4)*(1/z^2)
43. TuringTest
Oh dear... I have my answers about x^0=1 (not true for x=0), but I'm sure FFM would not approve of them :/ @inopeki, yes now rewrite it as 1 fraction...
44. FoolForMath
Turing that's a very important yet fundamental question.
45. FoolForMath
another one is why $$(a^b)^c = a^{bc}$$ ?
46. TuringTest
well I have a very simple proof of it and I can show why it is not true for x=0 what more do I need in your opinion?
47. TuringTest
the last rule is easier to explain, perhaps I should...
48. Inopeki
(x)*(1/y^4)*(1/z^2) (x)*(1/y^4*z^2)?
49. TuringTest
yes, now write it as a fraction what goes on the bottom? what goes on the top?
50. TuringTest
@FFMactually that's not so easy to explain now that I think about it.
51. FoolForMath
Can you explain the second rule intuitively?
52. Inopeki
1 x* -------- y^4*z^2
53. TuringTest
no, I was thinking more of how easy it is to show x^ax^b=x^(a+b) intuitively, the other is tricky to me.
54. TuringTest
@ inopeki you can put the x on top, it means the same thing and looks nicer
55. Inopeki
x*1 -------- y^4*z^2
56. FoolForMath
Can we say that all power function obeys the functional equation $$f(x)f(y)= f(x+y)$$?
57. Inopeki
Whats that f?
58. TuringTest
@Inopeki yes, no need to write the 1 though @FFM, I'd have to think about that :/
59. TuringTest
unspecified functions he's asking how far the rule about exponents can be extended..
60. Inopeki
x -------- Oh right y^4*z^2
61. FoolForMath
Actually we can't but I believe that is true for exponential functions though.
62. Inopeki
Umm, ok?
63. TuringTest
it must be for simple ones$2^x2^y=2^{x+y}$of course
64. FoolForMath
Tha's whats Inopeki is using right ?
65. TuringTest
basically Inopeki do you see the connection between what you are doing and the rule for multiplying exponents like$x^ax^b$?
66. Inopeki
Yeah, but does that mean that it becomes x -------- ? yz^6
67. TuringTest
no because y and z are different bases notice the rule above had both base x.
68. Inopeki
YEah but then i dont see the connection..
69. TuringTest
just that dividing and multiplying are inverse operations x^a=x*x*x*...*x (a times) x^b=x*x*x*...*x (b times) so if we have a=3 b=2 we get x^a --- x^b x*x*x =-----= x x*x which is x^(a-b) if we have (x^a)(x^b) we get (x*x*x)(x*x)=x*x*x*x*x=x^(a+b) so the rules for exponents here come directly from their definitions. You can count the x and see that this relationship holds.
70. Inopeki
I know about that, like x^2*x^8=x^10
71. Inopeki
x^10/x^5=x^5
72. Inopeki
Basic
73. TuringTest
right, I want you to see how that and the rule for division are inverses of each other for a reason...
74. FoolForMath
Turing you have a heck of patience :D
75. TuringTest
eh it's easy FFM, doesn't require too much concentration. good practice too I learn by teaching ;) ok factor $3x^3y^3+6xy+9x^2y$
76. FoolForMath
:-)
77. Inopeki
The GCF of 3,6,9 is 3 The GCF of x,x^2,x^3 is x The GCF of y,y^2,y^3 is y 3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Middle step, right?
78. TuringTest
yes exactly
79. Inopeki
3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?
80. TuringTest
yep :) great!
81. Inopeki
Really? :D
82. FoolForMath
Note to Inopeki: Turing is a great resource, learn new things from him and practice as much as you can on your own :-)
83. TuringTest
Yes I've stressed the importance of studying on your own as well and yes Inopeki really, FFM would have noticed a mistake I'm sure ok quick, foil (a-b)(a+b)
84. Inopeki
When hes not here i try to go on purplemath and khan :)
85. Inopeki
a^2+ab-ba-b^2
86. TuringTest
simplify
87. FoolForMath
Books books books!! online learning has it's own limitation :-)
88. Inopeki
a^2-b^2?
89. Akshay_Budhkar
it doesnt @ffm i disagree
90. Inopeki
Foolformath, im having trouble finding books here in sweden
91. TuringTest
and what is the name of that form? remember?
92. Inopeki
The fundamental theorem of algebra? Or want that the one with the multiplex?
93. TuringTest
Multiplex? no I think you're thinking of 'multiplicity' but this is the 'difference of squares' $a^2-b^2=(a-b)(a+b)$
94. Akshay_Budhkar
with ocw.mit , khan, purple math and openstudy and people like turing there is no limit to online education ^
95. TuringTest
Thanks but you get what you put in, I think is true with all this stuff.
96. Inopeki
Oh right
97. TuringTest
The fundamental theorem of algebra: "the sum of the multiplicity of the roots of a polynomial is equal to its order" difference of squares$a^2-b^2=(a-b)(a+b)$both important, but very different
98. TuringTest
so remember that we can run this backwards, and I can say 'factor'$x^2-4$using difference of squares wanna try?
99. TuringTest
run the FOIL backwards I meant*
100. Inopeki
x(x-4)?
101. Inopeki
Oh
102. TuringTest
no look at the form$a^2-b^2=(a-b)(a+b)$so what is a and b in$x^2-4$???
103. Inopeki
(x-2)(x+2)?
104. TuringTest
there ya go :)
105. TuringTest
how about$x^4-y^4$(this is a favorite question on OS)
106. Inopeki
(x^2-y^2)(x^2+y^2)
107. Inopeki
Why?
108. TuringTest
good, now is that all we can do with it though? (I've seen a lot of tutors stop here too ;-)
109. TuringTest
hint:look at the first set of parentheses
110. Inopeki
Ummm
111. TuringTest
what is the first sett of parentheses?
112. Inopeki
(x^2-y^2)?
113. TuringTest
yes, and can you factor that?
114. Inopeki
GCF of x^2 is x GCF of y^2 is y xy(x^2/xy)-xy(y^2/xy)?
115. TuringTest
no you're over-thinking x^2-y^2 is difference of squares again!
116. TuringTest
$a^2-b^2=(a-b)(a+b)$
117. Inopeki
So (x-y)(x+y)?
118. TuringTest
yes! so now factor$p^4-q^4$completely!!
119. TuringTest
you will apply difference of squares twice
120. Inopeki
(p^2-q^2)(p^2+q^2) (p-q)(p+q)(p+q)+(p+q)?
121. Inopeki
Without the +
122. Inopeki
between the parenthesis
123. TuringTest
(p^2-q^2)(p^2+q^2) stop here we cannot factor the second set of parentheses, it is not a /difference/ of squares, it is a /sum/ of squares, and we have no formula for that. check that (a+b)(a+b) is not a^2+b^2 so only factor the one that is a /difference/ of squares
124. Inopeki
(p-q)(p+q)(p^2+q^2) You said i had to do difference in squares 2 times? Btw, lets move to a new thread, this one is starting to lag a little bit
125. TuringTest
That is above correct, you did difference of squares twice: p^4-q^4 (p^2-q^2)(p^2+q^2) <-once (p-q)(p+q)(p^2+q^2) <-twice ok new thread...
126. Inopeki
Ohhh | 2,881 | 8,729 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2015-48 | longest | en | 0.749188 |
http://slave2.omega.jstor.org/stable/10.4169/j.ctt5hhb20?Search=yes&resultItemClick=true&searchUri=%2Ftopic%2Ffibonacci-numbers%2F&ab_segments=0%2Fbasic_SYC-5187_SYC-5188%2F5187 | 1,611,108,699,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00705.warc.gz | 96,519,185 | 39,480 | # Mathematics Galore!: The First Five Years of the St. Mark’s Institute of Mathematics
James Tanton
Edition: 1
Pages: 289
https://www.jstor.org/stable/10.4169/j.ctt5hhb20
1. Front Matter
(pp. i-viii)
(pp. ix-x)
3. Introduction
(pp. xi-xvi)
Eight years ago I took the plunge: I left the college and university world and became a high school mathematics teacher. It was time for me to do what I preached.
For many years I conducted workshops for educators that I thought were innovative, exciting and spoke directly to the mathematical experience. My goal was to enhance joyful learning for students through joyful work with their teachers and I designed activities that illustrated, I hoped, the value of intellectual play, of discovering and owning ideas, of questioning assumptions, of flailing, and of experiencing success. Although deemed interesting and fun my...
• 1 Arctangents
(pp. 1-10)
The following diagram starts with a right isosceles triangle with legs 1 and stacks an additional right triangle with a leg of 1 onto the hypotenuse of a previously constructed right triangle.
a) Ifnright triangles are stacked in this way, what is the length of the longest line segment in the diagram?
b) If we keep stacking right triangles, will the diagram ever make a full turn of rotation? Two full turns of rotation?
The angle a line of slopemmakes with the horizontal is called thearctangentof the slope, denoted arctan (m). For example, a...
• 2 Benfordʹs Law
(pp. 11-18)
Here is one of my favorite mathematical mysteries.
Consider the powers of two:
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384,
Do any of them begin with a 7?
If so, which is the first power of two that does? What is the second, and the third? Are there ten powers of two that begin with a seven? Are there infinitely many?
If, on the other hand, no power of two begins with a seven, why not?
The powers of three begin
1, 3, 9, 27, 81, 243, 729, 2187, 6561, …...
• 3 Braids
(pp. 19-24)
The language of ABABA uses only two letters, A and B, and any combination of them is a word. (Thus, for instance, ABBBBABAA and BBB are both words.) Also, strangely, a blank space, , is considered a word.
The language has the property that any word that ends in ABA or in BAB has the same meaning as the word with them deleted. (Thus, for example, BBABA and BB are synonyms.) Also, any two consecutive As or Bs can be deleted from a word without changing its meaning. (Consequently, BAABBBA, BAABA, BBA, and A are equivalent words.)
Warm-up. Show that...
• 4 Clip Theory
(pp. 25-30)
This is one of my favorite teasers. I use it in all of my extracurricular courses for teachers and students. Certainly patterns must be true!
Here we are assuming that the dots are spaced so that the maximal number of pieces appears when each pair of dots is connected by a straight line.
Going Further. Seven dots? Eight dots? Nine dots?
Bending the Rules. This month’s puzzler is a surprise. (Really do try it!) A greater surprise still lies hidden in the mathematics if one is willing to be flexible in one’s play, so flexible as to flex the lines!...
• 5 Dots and Dashes
(pp. 31-36)
1. The sequence of square numbers begins 1, 4, 9, 16, 25, … and thenth square number isn².
The sequence of non-square numbers begins 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, ….
a) What’s the 100th non-square number?
b) Find a formula for thenth non-square number
2. The sequence of triangular numbers begins 1, 3, 6, 10, 15, … and thenth triangular number is$\frac{1}{2}n(n+1)$.
The sequence of non-triangular numbers begins 2, 4, 5, 7, 8, 9, 11, …
a) What’s the 100th non-triangular number?
b) Find a...
• 6 Factor Trees
(pp. 37-42)
In grade school students draw factor trees. Here is a tree for 36,000:
At each stage we split the number at hand into a pair of factors, halting at the primes. (This forces the tree to stop. Good thing that 1 is not considered prime!) The tree allows us to write the starting number as a product of primes. Here we see 36000 = 2 · 2 · 2 · 2 · 2 · 3 · 3 · 5 · 5 · 5. What is astounding (though most people don’t seem to think so) is that despite possible different choices...
• 7 Folding Fractions and Conics
(pp. 43-52)
Sally draws a straight line from the bottom left corner of a blank piece of paper. She challenges Terrell to make a crease in the paper that bisects the angle formed (that is, cuts that angle exactly in half). “Easy” says Terrell as he lifts up the bottom edge, aligns it with the straight mark, and folds.
“Now,” chortles Terrell in response, “I challenge you to maketwocrease marks that divide your angle exactly into thirds.”
It can be done using nothing more than creases in the paper. How?
Without the aid of any tools it is very difficult...
• 8 Folding Patterns and Dragons
(pp. 53-62)
Take a strip of paper and imagine its left end is taped to the ground. If we pick up the right end, fold the strip in half, and unfold it, the paper has a valley crease in its center. Label it 1.
Suppose we perform two folds, lifting the right end up over to the left end. When we unfold the paper, we find three creases: two valley creases and one mountain crease. Label the mountain crease as 0.
For three folds we obtain
The sequence for four folds turns out to be 110110011100100.
a) What is the sequence of...
• 9 Folding and Pouring
(pp. 63-68)
One gallon of water is distributed between two containers labeled A and B. Three-quarters of the contents of A are poured into B, and then half the contents of B are poured back into A.
This process of alternately pouring from A to B (three-quarters of the content) and then from B to A (half the content) is repeated.
What happens in the long run?
Here’s something to try:
Take a strip of paper and make a crease mark at an arbitrary position.
Make a new crease halfway between the position and the left end of the strip by folding...
• 10 Fractions
(pp. 69-76)
a) The following are true:$\frac{26}{65}=\frac{2}{5}\quad\quad \frac{266}{665}=\frac{2}{5} \quad\quad \frac{2666}{6665}=\frac{2}{5}.$
Does 2 followed bynsixes divided bynsixes followed by 5 always equals 2/5?
b) We also have$\frac{49}{98}=\frac{4}{8}\quad\quad \frac{499}{998}=\frac{4}{8}\quad\quad \frac{4999}{9998}=\frac{4}{8}.$
Does 4 followed bynnines divided bynnines followed by 8 always equals 4/8?
c) Find another example.
(This material is adapted from Chapter 11 ofTHINKING MATHEMATICS! Volume 1: Arithmetic=Gateway to All.)
Suppose we wish to share 7 pies among 12 boys.
We could divide each pie into 12 parts and give each boy 7 pieces. But let’s be practical as were the ancient Egyptians (ca. 2000 b.c.e.)....
• 11 Integer Triangles
(pp. 77-86)
What property do each of the following figures share?
Find another right triangle (with integer sides) with the property. Is there another integer rectangle with the property?
Heron’s Formula. In 100 c.e. Heron of Alexandria (also known as Hero of Alexandria) published a remarkable formula for the area of a triangle in terms of its three side-lengths$\text{area=}\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$
Thus the areaAof the 15-20-7 triangle above is$A\text{=}\frac{1}{4}\sqrt{(42)(2)(28)(12)}=\frac{\sqrt{28224}}{4}=42.$
Proving Heron’s formula is not difficult conceptually. (The algebra required, on the other hand, is a different matter!) Here are two possible approaches:
Proof 1. Draw an altitude and label the...
• 12 Lattice Polygons
(pp. 87-94)
This chapter might drive you dotty! As you will see, we’re asking a number of tricky questions about polygons on a square lattice of dots. There are many interesting observations to be made about them. Nothing requires high-powered mathematics, just, perhaps, high powered ingenuity! Have a piece of graph paper at your side as you ponder these tricky proposals.
Is it possible to draw an equilateral triangle on a square array of dots so that each corner of the triangle lies on a dot?
We can draw squares of areas 1, 2, 4, 5, and 8 on a square lattice:...
• 13 Layered Tilings
(pp. 95-102)
Here are two tilings of a 6 × 6 grid of squares using 1 × 2 tiles (“dominos”).
Each tiling is structurally flawed in the sense that each possesses a line along which it could slide.
Devise a tiling of a 6 × 6 grid that is structurally sound.
Is there a structurally sound tiling of an 8 × 8 board?
The following 2 × 3 board is double tiled with six dominos, meaning that each cell of the grid is covered by exactly two dominos.
We’ll say that a 2 × 3 grid is 2-tilable with dominos. It is...
• 14 The Middle of a Triangle
(pp. 103-110)
Most everyone would agree that the point of intersection of the two diagonals of a rectangle deserves to be called the “middle” of the rectangle. After all, it would the balance point of the shape if the rectangle were cut from uniform material and placed horizontally on the tip of a pencil. So maybe “middle” means “balance point.”
Challenge. Three squares are adjoined to make an L-shape piece. Find the location of the balance point of this figure. Does it seem reasonable to call this the “middle” of the L-shape?
Design an L-shaped piece whose balance point lies outside the...
• 15 Partitions
(pp. 111-122)
Apartitionof a counting numberNis an expression that representsNas a sum of counting numbers. For example, there are eight partitions of 4 if order is considered important:$\begin{matrix} 4\quad 3+1 \\ 2+1+1 \\ \end{matrix}\quad \begin{matrix} 1+3 \\ 1+2+1 \\ \end{matrix}\quad \begin{matrix} 2+2 \\ 1+1+2 \\ \end{matrix}\quad \begin{matrix} {} \\ 1+1+1+1 \\ \end{matrix}$
There are five unordered partitions of 4:$4\quad 3+1\quad 2+2\quad 2+1+1\quad 1+1+1+1+1$
a) How many ordered partitions are there of the numbers 1 through 6. Any patterns?
b) How many unordered partitions are there of the numbers 1 through 6. Any patterns? Will these partition numbers continue to be prime?
One can place restrictions on the types of partitions one wishes to count. For example, there are eight partitions of 10...
• 16 Personalized Polynomials
(pp. 123-128)
The product of any two consecutive integers is divisible by 2. This is obvious since one of them must be even.
Prove that the product of any three consecutive integers must be divisible by 6. (For example, 7 × 8 × 9 = 504 is a multiple of 6.)
Prove that the product of any four consecutive integers is divisible by 24, the product of five consecutive integers by 120, and the product of six consecutive integers by 720. Does this remain true even if some of the consecutive integers are negative?
[By the way … What are the numbers...
• 17 Playing with Pi
(pp. 129-136)
This puzzler is a classic:
A rope fits snugly around the equator of the Earth. Ten feet is added to its length. When the extended rope is wrapped about the equator, it magically hovers at uniform height above the ground. How high off the ground?
A second rope fits snugly about the equator of Mars and ten feet is added to its length. How high off the ground does this extended rope hover when wrapped about the planet’s equator?
A third rope fits snugly about the (tiny) equator of a planet the size of a pea. When ten feet is...
• 18 Pythagorasʹs Theorem
(pp. 137-144)
A triple of positive integers (a,b,c) is called aPythagorean tripleif${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.
Did you know that an ordinary multiplication table contains such triples?
Choose two numbers on the diagonal (these are square numbers) and two numbers to make a square. Sum the two square numbers, take their difference and sum the other two numbers. You now have a Pythagorean triple!$\begin{array}{lll} a=25-4=21 \\ b=10+10=20 \\ c=25+4=29 \\ \end{array}\quad \Rightarrow \quad {{20}^{2}}+{{21}^{2}}={{29}^{2}}$
Choosing 36 and 1 gives$\begin{array}{lll} a=36-1=35 \\ b=6+6=12 \\ c=36+1=37 \\ \end{array}\quad \Rightarrow \quad {{12}^{2}}+{{35}^{2}}={{37}^{2}}$
Question 1. Which two square numbers give the triple (3, 4, 5)? Which give (5, 12, 13) and (7, 24, 25)?
Question 2. Why does this work?
Tough Challenge....
• 19 On Reflection
(pp. 145-156)
A popular discovery activity in the middle school curriculum is
A ball is shot from the bottom left corner of a3 × 5billiard table at a 45 degree angle. The ball traverses the diagonals of individual squares drawn on the table, bouncing off the sides of the table at equal angles. Into which pocket, A, B, or C, will it eventually fall?
Experiment with the tables pictured below. What do you notice about those tables that have the ball fall into the top-left pocket A? Into the top-right pocket B? Into the bottom-right pocket C? Test your theories...
• 20 Repunits and Primes
(pp. 157-162)
Arepunitis a number all of whose digits are one:
1, 11, 111, 1111, 11111, ….
Some repunits are prime (such as 11 and 1111111111111111111) and others are composite. (What are the factors of 111, of 1111 and of 11111?)
It is generally believed, but not proven, that infinitely many repunits are prime. (Only five prime repunits are known: those with 2, 19, 23, 317 and 1031 digits.) Your challenge is to establish that
If a repunit is prime, then the number of its digits is prime.
A number isprimeif it has exactly two distinct factors (whole...
• 21 The Stern-Brocot Tree
(pp. 163-170)
Here is something fun to think about. Consider the fraction tree:
It is constructed by the rule
Each fraction has two children: a left child, a fraction smaller than 1, and a right child larger than 1.
a) Continue drawing the fraction tree for another two rows.
b) Explain why the fraction$\frac{13}{20}$will appear in the tree. (It might be easier to first figure what should be the parent of$\frac{13}{20}$, its grandparent, and so on.)
c) Might the fraction$\frac{13}{20}$appear twice in the tree?
d) Will the fraction$\frac{457}{777}$appear in the tree? Might it appear...
• 22 Tessellations
(pp. 171-182)
A polygon is said totessellatethe plane if it is possible to cover the entire plane with congruent copies of it without overlap (except along the edges of the figures). The tessellation is called atilingif each edge of one polygon matches an entire edge of an adjacent polygon.
Parallelograms tile the plane. As two copies of the same triangle can be placed side-by-side to form a parallelogram we see thatevery triangle tiles the plane.
1. The tiling with triangles shown isperiodic, meaning that it possesses translational symmetries in (at least) two non-parallel directions. That is, it...
• 23 Theonʹs Ladder and Squangular Numbers
(pp. 183-192)
Everyone knows that$\sqrt{2}$is irrational.
a) Prove it! That is, prove there is something mathematically wrong in writing$\sqrt{2}=a/b$for some integersaandb.
b) Prove that$\sqrt{3}$is irrational.
c) Prove that$\sqrt{6}$is irrational.
d) Prove that$\sqrt{2}+\sqrt{3}$is irrational.
e) That$\sqrt{2}$can be written as${2^{1/2}}$shows that it is possible to raise a rational number to a rational power to obtain an irrational result. Is it possible to raise an irrational number to an irrational power to obtain a rational result?
The square numbers begin 1, 4, 9, 16, 25, … We can...
• 24 Tilings and Theorems
(pp. 193-204)
1. Here’s a picture of a portion of bathroom floor tiled with three differently shaped tiles: a big square, a small square, and a parallelogram.
Place a dot at the center of each white square. (Really. Please do it!) Do you see from this we get an easy proof of the following remarkable result from geometry?
If squares are drawn on each edge of a parallelogram, their centers form a perfect square.
2. Pythagoras’s theorem states that if squares are drawn on the sides of a right triangle, then the sum of areas of the small squares equals the area of the...
• 25 The Tower of Hanoi
(pp. 205-210)
The Tower of Hanoi puzzle consists of three poles on which a collection of differently sized discs sit on one pole in order of size, largest on the bottom. The challenge is to transfer all the discs to a different pole in such a way that only one disc is moved at a time and no disc throughout the process ever sits on top of a disc of smaller size.
The two-disc version of the puzzle is easy to solve. It takes three moves.
a) Solve the three-disc version of the puzzle. How many moves does it take?
An eight-disc...
• 26 Weird Multiplication
(pp. 211-220)
Here’s an unusual means for performing long multiplication. To compute 22 × 13, for example, draw two sets of vertical lines, the left set containing two lines and the right set two lines (for the digits in 22) and two sets of horizontal lines, the upper set containing one line and the lower set three (for the digits in 13).
There are four sets of intersection points. Count the number of intersections in each and add the results diagonally as shown:
There is one caveat as illustrated by the computation 246 × 32:
5. ### Appendices
• Appendix I Numbers that are the Sum of Two Squares
(pp. 221-230)
• Appendix II Pickʹs Theorem
(pp. 231-236)
• Appendix III The Möbius Function
(pp. 237-240)
• Appendix IV The Borsuk-Ulam Theorem
(pp. 241-244)
• Appendix V The Galilean Ratios
(pp. 245-248)
• Appendix VI A Candy-Sharing Game
(pp. 249-256)
• Appendix VII Bending Buffonʹs Needle
(pp. 257-262)
• Appendix VIII On Separating Dots
(pp. 263-268)
6. ### Indexes
• Index of Topics
(pp. 269-269)
• Classic Theorems Proved
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## Dimensional Analysis: A Mathematics Tool to Dissect the Circulatory System
by
Richard Cordia Taylor
### Rationale and Introduction
My friend Eric told me, "I've been on this great diet. I started off at 220 pounds, ate nothing but hamburgers, fries, and milkshakes for a month, and now I'm down to 120 kilograms."
Joking aside, in December 1998 NASA launched the \$125 million Mars Climate Orbiter. Nine months and 416 million miles later, NASA lost that spacecraft because the navigation team trying to insert the craft into orbit was making calculations in meters and kilograms while the computer guiding the operation had been programmed in feet and pounds.sup> 1/sup> Some of the smartest people in the world were undone by a mistake a fourteen year old could have corrected.
Units matter. Units make numbers make sense.
My unit will explore the use of dimensional analysis, the process of using units properly in calculations. Dimensional analysis lends itself to nearly every type of real word situation where math is used because numbers are associated with the measure of quantity in nearly all real world problems.
When students work on word problems, they frequently write down only the numbers without the units that describe what the number measures. The failure to write units removes meaning from the problems, with the result that students frequently put the numbers together using one of the four basic operations without any regard to whether their calculation makes sense. This technique is rarely successful and students soon learn to hate mathematics word problems of all types. That dislike also spills over to problems with math content in the sciences.
Dimensional analysis requires students to always include the units when they write down a number. The process of dimensional analysis takes advantage of the fact that units interact in a mathematically rational way such that the result of a calculation carries units as well. If the units of a calculation do not match the units that the question sought, the students know that their calculation requires re-examination. Further, knowing the units of the input data and the units of the desired output frequently gives clues on how to solve a problem by suggesting how those inputs might interact with each other mathematically in a calculation.
I will use the human circulatory system as the source of problems to investigate with dimensional analysis with an eye towards the development of artificial replacement organs. However, the techniques demonstrated can also be used to attack any type of word problem that involves numbers that carry units. I also introduce a technique of using units without numbers that students can use to pre-screen data for informative relationships between input variables before performing detailed calculations.
### Content Objectives
By the conclusion of this unit, students will be able to explain the basic functions of the circulatory system and its primary components. Students will be able to explain various ways that components within the system can fail. Students will also be able to take data in graphical, table, or written form and use dimensional analysis and mathematics to either calculate new data or to frame the data in ways to make it more accessible. For their final project, students will be able to apply this data handling skill either to evaluate a mode of failure within the circulatory system, to evaluate an artificial replacement for some part of the circulatory system, or to evaluate a technology that might be used to develop such replacements.
### Overview of the Circulatory System
Blood supplies every cell in the body with oxygen and nutrients and removes waste products that could harm those cells. The heart circulates blood via a system of connected tubes called arteries, capillaries, and veins, which distributes the blood throughout the body. The following sections rely heavily on the text of Medical Physiology by Boron and Boulpaep,sup> 2/sup> although the information is of a general nature that should be consistent with any biology text that deals with the circulatory system. Similar material is also available on internet websites.sup> 3/sup>
#### Overview of the Heart
(image 11.07.10.01 is available in print form)
Figure 1 Overall structure of the circulatory system
Courtesy of W. Mark Saltzmansup> 4/sup>
The heart is a set of muscular compartments whose contractions pump blood and a set of one way valves that ensure that blood flows in only one direction. The heart is divided into a right section and a left section and both sides are divided into an upper chamber called the atrium and a lower chamber called the ventricle. The right side of the heart is responsible for the pulmonary circulation between the heart and the lung, while the left side is responsible for the systemic circulation between the heart and the rest of the body. The time period when a chamber of the heart is relaxed is called diastole, and the period when a chamber of the heart contracts is called systole.
In a typical cardiac cycle, deoxygenated blood fills the right atrium from the inferior and superior vena cavae, two large veins that return blood from the body to the heart. Blood begins to flow directly from the right atrium into right ventricle through the tricuspid valve even before the atrium contracts. An electrical signal originating in the right atrium at the sinoatrial node causes the atrium to contract (atrial systole). The contraction forces the remaining blood in the atrium into the right ventricle. As the right ventricle fills with blood, the electrical signal from the heart arrives at the ventricle, causing it to contract (ventricular systole). The ventricular contraction increases pressure in the blood within the chamber, which causes the pulmonary valve to open, and forces blood out of the heart and into the pulmonary artery sending blood to both lungs. Under pressure from the ventricular contraction, the tricuspid valve closes, preventing blood from flowing backwards into the atrium.
Within the lungs, blood absorbs oxygen and releases carbon dioxide. Blood flow from the lungs returns through the pulmonary veins to the left atrium of the heart. The newly oxygenated blood flows into the left atrium and begins to flow through the mitral valve directly into the left ventricle before the atrium contracts. The left atrium contracts (atrial systole), sending the remaining blood into the left ventricle. After the left ventricle fills, it contracts (ventricular systole), pressurizing the blood and pumping it through the aortic valve and out the aorta and into the rest of the body. During ventricular contraction, pressure in the ventricle closes the mitral valve, preventing backflow into the left atrium.
While the process I described above makes it seem like there is a series of four consecutive contractions, in point of fact, both atria contract nearly simultaneously and later both ventricles contract nearly simultaneously. The right side pumps deoxygenated blood to the lungs, while the left side receives and pumps out oxygenated blood coming from the lungs from a previous cardiac cycle.
#### Overview of Blood Vessels
Oxygen-rich blood—pressurized for transport by the left ventricle—travels to each cell in the body, flowing through a series of blood vessels called arteries that branch (or bifurcate) into smaller vessels called arterioles, which further bifurcate into even smaller vessels called capillaries. The capillaries have extremely thin walls that allow the red blood cells in them to interact with the body's cells. While in capillaries, the red cells release oxygen and nutrients to the somatic cells and simultaneously pick up waste products and carbon dioxide. The blood then begins its trip back to the heart. The capillaries merge into larger vessels called venules, and the venules merge into larger veins. All of the veins merge into the inferior and superior vena cavae that lead back to the heart.
The arterial side of the circulatory system is under pressure, and it is this side of the circulatory system where blood pressure is taken. In the process of bifurcation, the diameter of the vessels decreases exponentially as the number of blood vessels grows exponentially to spread the blood over a greater volume. As the vessels get smaller, both the pressure in them and the velocity of the blood through them decreases. The slow velocity of the blood in the capillaries increases the time for exchange of molecules between the red blood cells and the somatic cells of the body.
When the doctor reads my blood pressure as 130 over 85 (135/85), the first number is the peak diastolic pressure generated by the left ventricle when it contracts. The second number is the systolic pressure, which is the residual pressure that remains in the artery when the left ventricle relaxes. I'll discuss the units of the numbers in these measurements shortly.
The venous side of the circulatory system is not under pressure. Incidental and natural movement of the skeletal muscles squeezes blood through the veins, and a system of valves within the veins prevent back flow.
### Dimensional Analysis Calculation Demonstrating Heart Reliability
The heart is a remarkable organ, which makes building a replacement for one an extremely daunting task. Unlike my car that I can stop, turn off, and park in the garage ready to use tomorrow, if my heart stopped beating for the night, I'd be in the morgue. The heart is so reliable that, frankly, I don't think about it most of the time. When my 98 year old grandfather died, his heart finally stopped, yet through his entire life, his heart had functioned literally without missing a beat. Assuming that his heart produced an average of 70 beats per minute, I can see that this works out to about:
(image 11.07.10.02 is available in print form)
The calculation above illustrates the power of dimensional analysis, and demonstrates how easily I can make some incredibly surprising observations using fairly ordinary information. Dimensional analysis allows me to convert quantities to different units of measure that demonstrate or emphasize particular features of a system—in this case, reliability. Several things are going on simultaneously in the calculation, however, so I will break it down into more digestible pieces.
First, the calculation contains several fractions and the fractions contain units as well. I notice, in particular, that the unit of minute appears in the numerator of one fraction and the unit of minute appears in the denominator of another. These units can be rearranged to yield minute/minute. This fraction can be reduced: minute/minute =1. The result, 1, is now a number without units. I notice also that the equation has several examples of the same situation: hour/hour=1, day/day=1, year/year=1. In the end, the only units left in the calculation are beats/lifetime (beats per lifetime).
Fractions that compare a quantity of one type of unit with a quantity of a different unit are called rates.sup> 5/sup> I assumed that my granddad's heart rate was 70 beats/1 min for the purposes of my calculation. His rate of using his time on earth was 98 years/1 lifetime.
The other fractions in the calculation are a special type called conversion factors. In a conversion factor, the two units in the fraction measure the same type of thing. In the fraction 60 min/1 hr, both minutes and hours measure the same type of thing—time. In the conversion factor, the two quantities measure the same amount of the same thing using two different units. In other words, since 60 min = 1 hr, the fraction 60 min/1 hr really expresses the number 1. As learned in elementary school, when I multiply by the number 1, the value of the other number does not change. Although the value does not change, the number associated with the value and the units used to measure that value do change. Strangely, all of the fractions shown above—in which the numerator and denominator represent the same thing measured in different units—are also equal to 1. This concept can be a little confusing so I'm going to take a look at a simpler example.
#### A Clarifying Example
I own a yardstick. It is 1 yard long. I notice that the number here is 1 and the unit here is a yard. If I look at the foot measures on the yardstick, I see that it is also 3 feet long. In this interpretation, the number is 3 but the unit is feet. I have the same yardstick, and it's the same length, but if the units used to describe it change, the number associated with the units also changes. This demonstrates that 3 feet = 1 yard, and I can write the conversion factor 3 ft/1 yd, which I can use to convert yards to feet, or another conversion factor 1 yd/3 ft, which I can use to convert feet to yards. I could use my yardstick to make even more conversion factors by noting that the yardstick is also 36 inches long.
### Introduction to Units of Measure
For me to understand the circulatory system well enough to entertain the idea of replacing parts of it, I need a way of describing the specifications of the circulatory system and its components so that I know what parameters my replacement parts will need to satisfy. In other words, I have to develop a set of measures that I can use to describe the various processes that take place in moving blood around the body. The following information on base and derived units is adapted from information provided by the National Institute of Standards and Technology.sup> 6/sup>
#### Base Units of Measurement
The first fundamental measurement quantities that I will use are length, time and mass. These quantities are sufficient to describe many physical systems. Since artificial parts may require a power source, I will also discuss the quantities related to describing electrical phenomena.
I introduced several units that I can use to measure length and time in the two prior examples. For my purposes, I will use meters (m) to measure lengths, and seconds (s) to measure time. These units and the mass unit of kilograms (kg) that I will choose below are also members of the International System of Units (SI). If the lengths and times that I encounter are measured in other units, I will usually convert them to meters and seconds. The idea is to translate all calculations about quantities into the same language so that comparing one set of numbers with another can be consistent and clear.
The final fundamental quantity, mass, is one that most people think they have a pretty good understanding of, but mass is actually a tricky concept. Mass is not weight. The pound (lb) is a measure of weight or force, but not a measure of mass. Kilograms (kg) are measures of mass, but not of weight or force. The more mass an object has, the harder it is to move, and if it is already moving, an object with more mass is harder to stop. Mass is also the quality possessed by matter that allows other matter to attract it by gravity. For my purposes, I will simply say that mass is the quality of matter that I can measure in kilograms.
#### Derived Units
From these three basic quantities of length, time, and mass, I can derive other quantities that I need to describe the function of the circulatory system or any other system for that matter. Derived units often have names of their own to simplify writing expressions. Derived unit names have also become a way for scientists to honor those who have made great contributions to knowledge.
Area is measured in units of length•length. In my chosen SI units, area is measured in meters•meters or meterssup> 2/sup> (msup> 2/sup>). Volume is measured in length•length•length or meterssup> 3/sup> (msup> 3/sup>) in SI. Density is mass/volume or kg/msup> 3/sup>.
An object in motion will travel a certain distance in a certain time. The most fundamental quantity dealing with this motion is called speed. Speed in a particular direction is called velocity. Speed can only be positive or zero. Velocity can be negative or positive or zero based on the conventions of direction within a coordinate system. Speed and velocity both carry the units of length per time (length/time) or, using my chosen measures, meters per second (meters/second or m/s). Other common measures of speed are miles per hour (miles/hour), or feet per second (feet/sec), but if I encounter these, I will usually convert them to SI units.
In some cases, the speed of an object will change. For example, when I step on the accelerator when a light turns green, the car goes from a speed of 0 miles per hour to a speed of 60 miles per hour in maybe 10 seconds. The rate that speed or velocity changes with respect to time is called the acceleration. Acceleration has units of length/timesup> 2/sup>, or in SI units, meters per second squared (meters/secondsup> 2/sup>, m/ssup> 2/sup>). Acceleration in the same direction as velocity will speed an object up, while acceleration in the opposite direction of the velocity will slow an object down.
Force is the quantity that can change an object's velocity. My intuitive notion of a force is a push and this is not very far from the scientific notion of force. One of the most famous formulas in science was first stated by Sir Isaac Newton:
F = m•a
This formula states that force is equal to mass times acceleration. From this formula, I can see that force in SI units will be measured in kilograms•meters per second squared (kg•m/ssup> 2/sup>). This specific combination of units, kg•m/ssup> 2/sup>, is named the Newton (N) in honor of Sir Isaac. The Newton is a derived unit. (Pounds are also a measure of force.) The formula simply says that if you are a given mass and I push you, you will stop standing still and move away from me. If I push you harder with a greater force, then you will move away from me even faster because your acceleration will be larger, which means that your velocity away from me will also be larger.
Pressure is the application of a force over a given area and has the units of force/area. Pressure can be written in standard units as N/msup> 2/sup> (also called the Pascal (Pa) after the scientist Blaise Pascal) or in basic units as (kg•m/ssup> 2/sup>) /msup> 2/sup> = kg/(m•ssup> 2/sup>) in simplified form. To get an idea of what pressure is, consider my friend Amanda walking across a lawn in flipflops. Since the bottom of her heel is fairly wide and has a large area, she does not sink into the soil. If Amanda walks across the same lawn in 6 inch stiletto high heels, however, she will find herself immediately stuck with five of those inches of heel in the ground. The difference is that the point of her high heel has a much smaller area than the area of the heel of the flipflops. In both cases, the force applied is just Amanda's weight, but in the case of the pointed high heel, the area is maybe 100 times smaller. This means that force is divided by a much smaller number which creates a pressure that is 100 times greater than the flipflop heel. This greater pressure easily pierces the soil and poor Amanda sinks into the ground.
In the medical context, pressure is most frequently measured in units of millimeters of mercury (mmHg; also called a Torr after Evangelista Torricelli), which seems like a very strange unit and it is. It is important to keep in mind that the unit is the whole thing "mmHg." As a unit, mmHg has nothing to do with a length and will not reduce out mm in a calculation. The unit mmHg can only reduce out mmHg. This particular unit is the result of the historical circumstances of the invention of the barometer, the weight of air, and the fact that mercury is the heaviest known liquid at room temperature. The conversion factor I need to convert the Torr unit to my chosen ones (SI) is about 133.32 Pa/Torr. This means that my blood pressure of 130 over 85 mmHg would be 17,332 Pa over 11,332 Pa, which might explain why doctors keep using the traditional units. In fact, one important purpose of having different units to measure things is to help express physical phenomena in numbers that are easy to remember (i.e. 130 over 85 is easier to remember than 17,300 over 11,300).
Energy is a measure of the ability of a system to do work. Energy can come in a variety of forms, but is always measured in units of mass•lengthsup> 2/sup>/timesup> 2/sup> or in SI units kg•msup> 2/sup>/ssup> 2/sup>. This set of units is also known as a joule (J), after the scientist James Joule. Kinetic energy is the energy associated with motion. For an object of mass, m, with velocity, v, the kinetic energy of the object is
KE = mvsup> 2/sup>/2
Applying a force to an object through a distance is called work. Work is a form of energy. As an equation this is written: work=force•distance. The force that Earth's gravity exerts on an object of mass, m, is m•g, where g=9.8 m/ssup> 2/sup>. The amount of work that I do to raise an object of mass m, a height, h, is m•g•h. The work that I do to raise my 4 kg cat 1.5 m up is equal to 4 kg•1.5 m•9.8 m/ssup> 2/sup>= 58.8 kg•msup> 2/sup>/ssup> 2/sup> = 58.8 joules. Since my cat is now above the ground, she has potential energy. Potential energy can be thought of as stored energy. The amount of potential energy that she has is exactly equal to the amount of work I did to pick her up. If I let go of her, her potential energy transforms into kinetic energy as she accelerates, picks up velocity and drops to the ground.
Power is the rate at which work is done or the rate at which energy is used. I require more power to lift a weight quickly than to lift the weight slowly even though the total work done in both cases is the same. In SI units, power is measured in joules per second (J/s) = watts (W) (after James Watt) or in basic units kg•msup> 2/sup>/ssup> 3/sup>. In the case of artificial organs that require active components—those that move or consume energy for other functions—the rate at which energy is consumed has direct implications for how that energy can be supplied to the device.
#### Base and Derived Units for Electrical Systems
Because there is a high probability that active devices will require electrical power, I also introduce the additional basic unit of electrical current flow, the ampere (A) (after Andre-Marie Ampere). The unit of electric charge is the Coulomb (C) (after Charles-Augustin de Coulomb). A current flow of 1 coulomb/second = 1 ampere. The flow of charges though an electrical circuit is what makes electronic devices work.
The volt (V) (after Alessandro Volta) is the unit of electrical potential. One joule of energy will move 1 coulomb of charge through an electrical potential difference of 1 volt. The volt has units of joules/coulomb (J/C). In an electrical circuit, charge flows through a conductor because of the difference in voltage potential between one end of the conductor and the other. Conductors possess a quality known as resistance, and the resistance of a conductor is measured in ohms (Ω) (After Georg Simon Ohm). A current of 1A will flow through a conductor with a resistance of 1Ω if a potential of 1V is placed across the conductor. Ohm's Law states this as a general equation: V=IR. V is the voltage in volts, I is the current in amps, and R is the resistance in ohms. From Ohm's law I can deduce that the unit for ohms is J•s/Csup> 2/sup>.
An important observation is that the product of 1 volt and 1 ampere is equal to 1 watt: 1 V•1 A=1 (J/C)•1 (C/s)=1 J/s=1 W. This will be an important relationship for the analysis of power usage in electrical devices replacing parts of the circulatory system. For the most part, I will not further convert electrical calculations into kilograms, meters, and seconds unless it is absolutely necessary to draw a conclusion.
### Dimensional Analysis of Graphical Data: Ventricular Pressure vs. Volume Graph
(image 11.07.10.03 is available in print form)
Figure 2 Left Ventricular Pressure vs. Volume Loopsup> 7/sup>
The pressure vs. volume changes that occur in the left ventricle during one cardiac cycle are shown in Figure 2. This graph suggests several interesting questions about the ventricle's engineering characteristics. For context, the graph starts with the light blue arrow where the mitral valve opens allowing blood to flow into the ventricle which is relaxing in diastole. About midway through the dark blue arrow, the left atrium contracts increasing the pressure in the left ventricle slightly and pumping the remainder of its contents into the ventricle. At the green arrow, the ventricle begins contraction with the pressure increasing at a constant volume (i.e. isovolumetrically, since the valves guarding the entrance and exit to the ventricle are both closed). At the magenta arrow, the pressure in the ventricle exceeds the residual pressure in the aorta, the aortic valve opens, and blood is pushed out of the ventricle into the aorta. At the orange arrow, blood continues flowing out of the ventricle until the aortic pressure exceeds the ventricle pressure and the aortic valve closes. At the red arrow, the ventricle begins diastole and experiences isovolumetric relaxation until the pressure falls below the atrial pressure and the mitral valve opens again for another cycle. In addition to the pressure and volume directly shown in the graph, it is clear that the process of moving blood from the ventricle to the aorta requires energy for the heart to do this work.
#### Analysis 1: Area Enclosed by a Curve
Frequently, the area enclosed by a curve will have significance. Mathematically, the process of finding the area within a curve is the operation of integration. I can approximate the integration suggested by this graph with an estimate of graphical area. Each of the rectangles in the graph has a base that is measured in units of volume and a height that is measured in units of pressure. I can do a pre-analysis of the units that this area represents in the graph by using the SI units for pressure and volume and simplifying.
(image 11.07.10.04 is available in print form)
Since the quick dimensional calculation yields an energy unit, a quantity known to have engineering significance, I know that this is a calculation worth making in detail.
Counting boxes and taking into account fractions of boxes, I arrived at a total of about 78 boxes enclosed by the loop of colored arrows in Figure 1. Each box has a base of 10ml and a height of 10 mmHg so that the area of each box is 10 ml•10 mmHg = 100 ml•mmHg.
(image 11.07.10.05 is available in print form)
One joule is approximately equivalent to the amount of work I would do lifting a 1 kg mass by 10 cm. If I assume in this case a heart rate of 60 bpm, meaning one cardiac cycle occurring every second, then the left ventricle power output is 1 J/1 sec=1 watt. Taking into account that there are three other chambers of the heart beating, albeit with much less power, I can estimate a total power requirement around 2 watts or less. I could, of course, get a better estimate if I had PV graphs of all of the chambers.
The previous calculation, however, is only the net amount of work that the ventricle does on the blood. The blood entering and leaving the ventricle has a velocity and hence a kinetic energy. The difference between the maximum volume and the minimum volume is the stroke volume of blood pumped in the cycle. I can read this from the graph as 120 ml – 50 ml = 70 ml stroke volume. The blood is ejected from the heart into the aorta with an average velocity of 63 cm/sec.sup> 8/sup> Blood has a density of 1060 kg/msup> 3/sup> so the kinetic energy of the ejected blood is:
(image 11.07.10.06 is available in print form)
This kinetic energy is considerably smaller than the energy that I calculated for the work done on the blood during the cardiac cycle, and is probably less than the error in calculating that larger energy.
I should point out an important calculation technique that is frequently used when trying to get the big picture of a situation without going into excessive detail before it is needed. Because I used 60 bpm, which I intuitively understand to produce 1 beat/sec, I could do a quick conversion in my head from joules to watts for this choice of parameters. Frequently, when I want a ballpark estimate I will strategically pick input data values to facilitate quick calculation without a calculator. Had I chosen to use my grandfather's assumed heart rate of 70 bpm, I would make the following calculation for cardiac output power explicitly involving dimensional analysis
(image 11.07.10.07 is available in print form)
While this calculation isn't difficult, it is more challenging than the one at 60 bpm. I do note that with the higher heart rate, the power requirements are slightly higher. I also can note the structure of the conversion more clearly here and infer how the output will scale with increased heart rate. For example, with a heart rate of 180 bpm during heavy exercise, I estimate 3W of power to drive the left ventricle since 180 bpm = 3 x 60 bpm. This estimate assumes that the pressure volume loop stays exactly the same under the dramatically different conditions, which is not true, but again I do get a starting point for a more detailed analysis.
#### Analysis 2: Slope of the Curve
The slope of a curve is also frequently a measure of engineering significance. In the case of the graph in Figure 2, the slope would be measured in units of
(image 11.07.10.08 is available in print form)
The quantity ΔP/ΔV is called the elastance. The reciprocal of the elastance is ΔV/ΔP and is called the compliance.sup> 9/sup> Compliance has SI units of msup> 4/sup>•secsup> 2/sup>/kg. (I note here that the reason that the slope of the graph gave the elastance rather than the compliance is that I chose to graph the volume on the horizontal axis and the pressure on the vertical axis in figure 2. Frequently when scientists study compliance, they plot the pressure on the horizontal axis and the volume on the vertical axis so that the slope of the graph is the compliance. This highlights the importance of noting which axis is which.)
The compliance of a blood vessel is a measure of how flexible (distensible) that blood vessel is in response to an increase in pressure. On the high pressure side of the circulatory system, an artery's ability to flex helps it perform its job better. In fact, as adults age, the compliance of their arteries decreases dramatically. A typical compliance for an adult age 20 to 39 is about 0.75 ml/mmHg compared to a typical compliance for an adult age 60 to79 of about 0.38 ml/mmHg.sup> 1/sup>sup> 0/sup> The compliance of arteries is an important characteristic that artificial arteries need to mimic.
I will do a calculation of the compliance shown in the left ventricle during the part of the cycle where the aortic valve opens along the magenta arrow in Figure 2. Using points (V,P) starting at (120,80) and ending at (75,130), I calculate
(image 11.07.10.09 is available in print form)
I notice the interesting fact that this calculation gives a negative value for the compliance of the left ventricle during this part of the cardiac cycle as opposed to the positive values of compliance stated for the arteries. This sign difference is a result of what the heart is doing versus what the artery is doing during this part of the cycle. The heart is decreasing its volume while it generates more pressure through contraction. The artery on the other hand is receiving pressurized blood and thus is expanding and increasing its volume in response to the heart pumping.
### Dimensional Analysis for Evaluating Potential Power Sources for an Artificial Heart
Since an implanted artificial heart is ideally a portable device, the energy and power will likely be supplied by batteries. A standard AA NiMH rechargeable battery in my camera is rated as 1.2 V at 2500 mA•hr. These battery specifications involve voltage, current, and time. I can do a quick calculation, keeping track of units only, to get V•A•s = (J/C)•(C/s)•s=J, an energy, so this looks like a calculation worth making in detail. The energy in the battery is
(image 11.07.10.10 is available in print form)
The number seems substantial, but I need to put it in context to see if the camera battery could really be a good potential power source. To build this context, I do another dimensional calculation to determine how long my battery could power an artificial heart using energy at the rate of a natural left ventricle running at 60 bpm.
(image 11.07.10.11 is available in print form)
This calculation tells me that my camera battery is not really a viable power source. I'd need eight of them a day, and that's assuming that there was 100% efficiency in converting the battery energy into pumping work. One-hundred percent efficiency is not realistic. One artificial implantable heart already developed had only 13.5% efficiency and required a 12 W electrical power input to produce a 1.6 W pumping output.sup> 1/sup>sup> 1/sup> (The power requirements for this heart seem prohibitive to permanent implantation with battery power.)
There are several important technical aspects of this calculation. First, I stated the battery energy as a ratio with an invented unit to make the meaning of the end calculation more clear. Second, I arranged the conversion factors in a way to make the units I did not want in the end result reduce out of the calculation—sometimes I'll use the conversion factor as given, but sometimes I'll need to use the reciprocal of the conversion factor. Third, I violated my own rule about writing time in seconds because in the context of this evaluation, hours were a more appropriate and clearer unit to express the results of this particular calculation.
Since camera batteries do not provide sufficiently dense energy storage for an artificial heart, are there other sources? Alternative energy sources such as implantable generators driven by skeletal muscles have been explored but theoretically might only produce 100 J/day.sup> 1/sup>sup> 2/sup> In the 1970's, nuclear batteries of up to 50 W powered by plutonium-238 were implanted in dogs and primates for periods exceeding a year with few radiation related side effects.sup> 1/sup>sup> 3/sup> Nuclear batteries promised decades of continuous operation without recharging or replacing, but plutonium has more dangerous uses, making it difficult to envision these other applications.
### Dimensional Analysis does Problem Analysis
Using dimensional analysis as a problem analysis tool involves understanding what type of quantity is involved in a situation (for example, energy or pressure or time) and using the units of the relevant quantities to cast light on how the measured input data might interact mathematically to frame the problem in the relevant units.
Blood pressure measurement is a non-invasive means of assessing the overall health of the circulatory system. I have a blood pressure monitor that measures systolic and diastolic pressure at the wrist from the radial artery rather than on the upper arm. The wrist location allows me to measure my blood pressure in a variety of positions.
I took my blood pressure in three positions: 1) standard arm at the side, 2) wrist raised 30 inches above the standard, and 3) wrist lowered 12 inches below standard.
(image 11.07.10.12 is available in print form)
The data raise many questions, including why the blood pressure measurement changes with my arm in different positions. This tendency for the blood pressure to vary based on position poses another hurdle for the artificial organ developer. At first it might appear that altitude changes blood pressure, but I have had my blood pressure taken on different floors of the hospital without seeing any significant change.
In this problem, dimensional analysis can help me understand these measurements. When I raised my arm 30 inches up from the standard position, I did work on the blood in my wrist. As discussed in the Derived Units section on energy, that blood now has greater potential energy; potential energy was increased by m•g•h, where m is the mass of the blood, g is the acceleration due to gravity, and h = 30 inches is the height I raised the blood. I learned in the left ventricle problem above that pressure•volume is also a form of energy. With these two observations, I now have the height and the pressure expressed in the same language. Compared with the standard position whose energy is P•V, the raised position has both a P•V component and a potential energy component. Assuming that the volume of blood in the radial artery is the same for both systolic measurements, I can write an energy equation.
(image 11.07.10.13 is available in print form)
The -59 mmHg calculated is reasonably close to the -50 mmHg measured. The difference between the calculation and the measurement could be caused by my systemic blood pressure changing between measurements, my arteries changing diameter to compensate for the lower pressure, error in the measurement or something else. Nevertheless dimensional analysis leads me to a remarkably simple explanation for the pressure difference that agrees remarkably well with experiment.
### Student Background and Challenges
My school's demographic breakdown is approximately 34% Latino, 5% African American, 18% Asian, and 41% Caucasian. The population is split about evenly by gender.
Approximately 54% of entering freshmen take Algebra 1, our lowest math class. The bottom third of the Algebra 1 students are concurrently enrolled in an additional mathematics support class as well. In statewide testing, only 19% of our Algebra 1 students scored proficient. The performance of Algebra 1 students on our statewide testing is a significant factor reducing our school's rating within California. Algebra 1 is also the most flunked class and therefore the most repeated class at our school.
On the other end of the spectrum, 61% of students in pre-calculus and calculus score proficient on the statewide test. Students in the first year Advanced Placement Calculus AB classes routinely outscore the national averages on the Calculus AP/AB exam and have had passing rates (with scores of 3 or above) in the 70% to 90% in the last few years. Students in the second year AP Calculus BC classes routinely have passing rates above 95%.
My target population for this unit is a heterogeneous group consisting of two distinct divisions: A) students enrolled in either first or second year Advanced Placement Calculus, and B) students enrolled in what our district calls Algebra 1 "support" mathematics classes. I will deliberately attempt to leverage the group A students' abilities to improve overall performance of the B group students in mathematics.
Students in the A group are usually from the Junior or Senior class, have above average mathematics ability, have experienced prior success with mathematics, and most likely have already taken a first year Biology class and have taken or are currently enrolled in either or all of the following classes: Advanced Placement Biology, regular Chemistry, regular Physics, Advanced Placement Chemistry, and Advanced Placement Physics. Students in the A group are above average in general academic ability and are all on a college track.
Students in the B group are most likely freshmen or sophomores who have below average mathematics ability as indicated by prior math grades and performance on the California State tests in mathematics. B group students have rarely experienced any level of success with mathematics, and have usually either failed at least one prior mathematics class or have been enrolled in a remedial mathematics class far below grade level in the middle school. B group students are concurrently enrolled with a subject area course of either Algebra 1 or Geometry, and a support class for that subject area to provide extra time and instruction to help bring them to grade level. B group students are behind grade level and sometimes may be as low as first grade level in mathematics (i.e. They have not mastered times tables, single digit addition with a carry, basic fraction operations, etc.). The lowest level mathematics class our district offers for non-special education students is Algebra 1, so the concurrent support class is offered in lieu of providing the students with a remedial mathematics class preceding Algebra. Since our district requires passing both Algebra 1 and Geometry for graduation, B group students are among those most likely to fail to graduate.
Ancillary to the goal of learning the content objectives, the heterogeneous composition of this target group is constructed to improve the affective receptiveness to learning mathematics of the B group Algebra 1 students. In addition to their lack of mathematical ability, the B group students frequently express helplessness and resistance to learning mathematics so that even mathematics that is within their intellectual grasp is not even attempted. No matter how effectively any teacher presents instruction, an unreceptive student will not learn. I chose the content and focus of this unit specifically because it deals with material that is valuable for and accessible to both the A group and B group students.
### Strategies
Strategies for this unit will include teacher-directed classroom instruction, student readings, small scale experiments, student group work and collaboration both inside and outside the classroom, classroom question and discussion, and student projects. Student assessment will consist of small repetitive quizzes on important topics like the function and placement of components of the circulatory system. Additional assessment will take place during in class discussion and with a final group project.
Effective lesson construction requires a rational delineation of content objectives, a logical sequence of instruction, a process of ongoing evaluation of student learning during the process, flexibility to adjust to the results of the ongoing evaluation, a means of final evaluation, and finally the often neglected appreciation of the psychology of students and their personal strategies that might work in opposition to the lesson's goals.
Student group work is at the heart of all of the strategies used for this unit. The smallest form of this group work will be a pairing of one student from each of the A and B groups. A two person group forces communication between the members. Larger groups lend themselves to having some members on the outside of a primary subgroup.
If left to their own devices, even in a group of two, one member will tend toward dominance. The disparity of confidence, maturity, knowledge, and academic ability built into these groups will acerbate these tendencies. In fact if left to their own devices, there is a realistic possibility that the A group student would do all of the work with complete acquiescence from the B group student because both might conclude that this strategy would produce the best result to present to a teacher in the shortest time.
To combat this tendency, which might leave the B student learning nothing but how to freeload, I can take a cue from medicine to construct a more effective strategy to foster a fully functioning group. A child born with one weak or wandering eye faces the possibility that his brain will shut off that eye so that the dominant eye can perform its duties without confusing competition from the other eye. While this is nature's optimal short term solution, the child becomes an adult who is blind in one eye and thus lacks depth perception. To fight this natural tendency, a doctor may intentionally weaken the stronger eye by putting a patch over it so that the brain must depend on it and thus must keep it functioning. While patching the strong eye is not optimal for the child's vision in the short run, it does produce a better long term result because the ultimate goal is to have an adult where both eyes work in concert. Following this idea, I want to structure the group to weaken the A group member's participation in order to keep the B group member functioning.
One strategy in this vein that I will use is to value the response of the B group members higher than that of the A group members. I will make students aware of this procedure. For example, when I pose a question to the group, if an A group member answers correctly, it is worth one point, but if the B group member answers correctly, it is worth five points. With this structure, the A group member may know the material, but will have an incentive to teach the B group member ahead of time, because it is more valuable to the group's grade.
Group work will also be the strategy for teaching this class insofar as there will be a teacher of the B group students who will work with me, the Calculus teacher, to present this unit. As classes are constructed, the two target groups would not normally be together. To facilitate this mixing, students from a support teacher whose classes meet at the same time would come to my Calculus class. This requires that I teach the co-teacher the substance of the unit so that he or she can add value as well in teaching the unit to the students.
Because I am teaching mathematics, the general class curriculum does not support the scientific inquiry and instruction necessary for the unit without laying some specific groundwork. Establishing this foundation will take more time than can be spared directly in class. To gain this additional time, students will do a large portion of their inquiry outside of class. To give students the time that they will need to absorb and comprehend the fruits of their research, the unit will stretch longitudinally with the ten days of actual classroom instruction and discussion spread through the semester, rather than being clustered as two consecutive weeks.
### Classroom Activities and Lesson Plans
#### Lesson Plan 1: Your Pulse Rate
Materials
Wall clock and graph paper.
Objectives
After this lesson, students will be able to take their pulse and see where their pulse rate falls in the range of other students' pulse rates. Students will be able to use conversion factors to extrapolate the data measured to tell how many heartbeats they will have in a day, a month or a year. Students will be able to state quantitatively differences in pulse rate due to differences in physical activity.
Activity
I will show students how to take their pulse. After doing so, members of the class will take their pulse for 1 minute, counting their heartbeats while watching the classroom clock. I will pass around a piece of graph paper where each student will color a square in a grid based on his pulse rate in beats per minute, thus eventually forming a bar graph with the distribution of pulse rates vs. number of students with that pulse rate. I will use this graph to discuss measures of central tendency (mean, median, and mode) and measures of dispersion (range and, qualitatively, standard deviation). I will then repeat this experiment with the students after having them run in place for a minute. I will use this new set of data to explain the adjustments that the heart must make during physical exertion. I will use the differences in the graphs from the two trials to illustrate the range of performance that the heart must accommodate. I will then instruct students on how to calculate the number of beats for a variety of time periods using conversion factors.
#### Lesson Plan 2: Dimensional Analysis and Conversion Factor Flashcards
Materials
Index Cards or Printouts of Flashcards and scissors to cut them out.
Objectives
After this lesson, students will be able to combine units appropriately to construct other units representing important quantities. Students will also be able to use conversion factors appropriately to convert from one unit of measure to another.
Activity Part 1
Students will make flashcards with the basic and derived units as shown in the samples below. Students will place the flashcards side by side, indicating multiplication and from this, simplify to find the units of the result. Students will write down the operation shown with units for the quantities used and the resulting units for their answer. When I give students a list of final units, they will find appropriate flashcards to combine to create those units and will record the calculations and the results.
Base and Derived Units Sample Flash Cards
(image 11.07.10.14 is available in print form)
Activity Part 2
Students will create flashcards of conversion factors similar to those shown below. I will give them a beginning quantity and unit of measure, for example 13 feet and I will ask them to arrange the flashcards appropriately to convert to a target measure, for example, meters. Students will then use a calculator to finish the conversion. Students will write down their calculations including units and the result of their conversion.
Conversion Factor Sample Flashcards
(image 11.07.10.15 is available in print form)
#### Lesson Plan 3: Reading and Interpreting Slope from a Graph
Materials
A graph of Volume vs. Pressure for a heart chamber or a blood vessel. A straight edge.
Objective
After this lesson, the students will be able to explain what compliance means using appropriate units and will be able to calculate the compliance of a heart chamber or a blood vessel from a Volume vs. Pressure graph (or a Pressure vs. Volume graph). The student will also be able to explain why compliance is an important feature to consider in the development of artificial arteries.
Activity
After discussing with students why compliance is an important feature of blood vessels, I will use a Volume vs. Pressure graph to demonstrate how to calculate the slope of a secant line using the correct units shown on the graph. I will then demonstrate how to use a straight edge to estimate the location and the slope of a tangent line to the curve. I will then ask students to calculate the slopes of secant lines and tangent lines from the graph at different points. Students will then use conversion factors to convert their slopes into SI units.
#### Lesson Plan 4: Student Project
Materials
Internet and Library access
Objectives
Students will find an area of interest related to the topic of organs or artificial organ development where they can apply dimensional analysis in analyzing data.
Activity
Students will have a great deal of latitude in coming up with their topic for their project. The one characteristic that all of the projects must include is at least three calculations involving dimensional analysis or the use of a conversion factor to interpret data.
Student groups will present their projects both in written form and to the class as a whole orally and visually, and will be graded with a combination of peer grading and my grading with input from the co-teacher. Logistically, several two person groups will form larger super-groups for the purpose of presentation. This organization will save time and provide an additional opportunity for peer teaching within the group structure. I anticipate that A group students will do the bulk of the written work since that is one area where I cannot directly control group members' input. Ideally, however, with the differential evaluation points given that favor B group response, B group students should do the bulk of the oral and visual presentation to provide students with the highest group grade.
### Standards Alignment
The goals of my unit are in line with the California Core Curriculum Content Standards for mathematics.sup> 1/sup>sup> 4/sup> In these standards, the expectation is that mathematically proficient students will:
Make sense of problems and persevere in solving them.
Reason abstractly and quantitatively.
Construct viable arguments and critique the reasoning of others.
Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose.
Use appropriate tools strategically. Mathematically proficient students consider the available tools when solving a mathematical problem. These tools might include pencil and paper, concrete models, a ruler, a protractor, a calculator, a spreadsheet, a computer algebra system, a statistical package, or dynamic geometry software. Proficient students are sufficiently familiar with tools appropriate for their grade or course to make sound decisions about when each of these tools might be helpful, recognizing both the insight to be gained and their limitations. For example, mathematically proficient high school students analyze graphs of functions and solutions generated using a graphing calculator. They detect possible errors by strategically using estimation and other mathematical knowledge.
Attend to precision. Mathematically proficient students try to communicate precisely to others. They try to use clear definitions in discussion with others and in their own reasoning. They are careful about specifying units of measure, and labeling axes to clarify the correspondence with quantities in a problem. They calculate accurately and efficiently, express numerical answers with a degree of precision appropriate for the problem context. By the time they reach high school they have learned to examine claims and make explicit use of definitions.
### Teacher and Student Resources
Bartholet, Jeffery. "Inside the Meat Lab." Scientific American, June 2011.
An article involved with a parallel use of tissue engineering for food production that could provide insights, funding, and economies of scale for the development of critical technologies for tissue engineering.
Cho, Charles Q. "Sugar Within Human Bodies Could Power Future Artificial Organs: Scientific American." Science News, Articles and Information | Scientific American. http://www.scientificamerican.com/article.cfm?id=glucose-body-fuel-cell (accessed August 1, 2011).
A new approach to tiny fuel cells implanted in rats enables the devices to generate electricity for months using sugar in the rodents' bodies.
Kapit, Wynn, Robert I. Macey, and Esmail Meisami. The physiology coloring book . 2nd ed. San Francisco: Benjamin/Cummings, 1999.
An inexpensive and engaging introduction to physiology. Interaction through coloring probably aids recall.
Krumhardt, Barbara, and I. Edward Alcamo. Barron's E-Z anatomy and physiology . Hauppauge, N.Y.: Barron's Educational Series, 2010.
An inexpensive and relatively complete presentation of physiology at an appropriate level.
Nerem, R. M. "Tissue engineering: Confronting the transplantation crisis." pih.sagepub.com. pih.sagepub.com/content/214/1/95.full.pdf (accessed August 1, 2011).
Although old, this is a cogent article delineating the demand for and challenges to tissue engineering solutions to organ transplantation
### Endnotes
1 NASA, Mars Climate Orbiter Mishap Investigation Board Phase I Report
2 Boron and Boulpaep, Medical physiology: a cellular and molecular approach, 508-533
3 Cardiology Teaching Package
4 Saltzman, Biomedical engineering: bridging medicine and technology, Chapter 8
5 American Heritage Dictionary
6 Essentials of the SI: Base & derived units
7 Boron and Boulpaep, Medical physiology: a cellular and molecular approach, 523
8 Saltzman, Biomedical engineering: bridging medicine and technology. Chapter 8
9 Boron and Boulpaep, Medical physiology: a cellular and molecular approach, 455
10 Ibid, 460
11 Wolters Kluwer Health
12 Concept Developed for an Implanted Stimulated Muscle-Powered Piezoelectric Generator
13 Huffman and Norman, Nuclear-Fueled Circulatory Support Systems IV: Radiologic Perspectives
14 California Department of Education, 2-3
### Bibliography
Huffman, F. N., and J. C. Norman. "Nuclear-Fueled Circulatory Support Systems IV: Radiologic Perspectives." Cardiovascular Diseases, Bulletin of the Texas Heart Institute 1, no. 5 (1974): 463-476. http://www.ncbi.nlm.nih.gov/pmc/articles/PMC287516/pdf/cardiodis00032-0095.pdf (accessed July 15, 2011).
High wattage nuclear batteries could power artificial hearts for decades
Boron, Walter F., and Emile L. Boulpaep. Medical physiology: a cellular and molecular approach. Updated ed. Philadelphia, Pa.: Elsevier Saunders, 2005.
Good coverage of the cardiorespiratory system (and other physiology as well). Lots of detail and explanation
California Department of Education. "California's Common Core Content Standards for Mathematics." scoe.net. www.scoe.net/castandards/agenda/2010/math_ccs_recommendations.pdf (accessed July 31, 2011).
Core curriculum standards for mathematics are the official guide for what should be taught in a mathematics classroom.
"Cardiology Teaching Package - Practice Learning - Division of Nursing - The University of Nottingham." The University of Nottingham. http://www.nottingham.ac.uk/nursing/practice/resources/cardiology/index.php (accessed July 16, 2011).
An organized set of pages describing cardiac function. Their own note says it all: "This teaching package has been designed for student nurses who know nothing at all about Cardiology."
"Concept Developed for an Implanted Stimulated Muscle-Powered Piezoelectric Generator." NASA - Title.... http://www.grc.nasa.gov/WWW/RT/2004/RP/RPY-lewandowski.html (accessed July 17, 2011).
skeletal muscle generator proposal with theoretical energy yield of 100J/day
"Essentials of the SI: Base & derived units." Physical Measurement Laboratory Homepage. http://physics.nist.gov/cuu/Units/units.html (accessed July 16, 2011).
National Institute of Standards and Technology(for the USA) website
NASA. "Mars Climate Orbiter Mishap Investigation Board Phase I Report." nasa.gov/pub/pao/reports/1999/MCO_report.pdf. ftp://ftp.hq.nasa.gov/pub/pao/reports/1999/MCO_report.pdf (accessed July 21, 2011).
Official NASA report on the loss of the Mars Climate Orbiter due to inconsistency in units used.
Saltzman, W. Mark. Biomedical engineering: bridging medicine and technology. Cambridge: Cambridge University Press, 2009.
the definitive book on Biomedical engineering. Considerable detail and clear explanations. Great questions at the end of each chapter.
The American Heritage Dictionary of the English Language . 4th ed. Boston: Houghton Mifflin Harcourt Publishing Company, 2006.
"Wolters Kluwer Health." LWW Journals - Beginning with A. http://journals.lww.com/asaiojournal/Abstract/2002/09000/One_Piece_Ultracompact_Totally_Implantable.17.aspx (accessed July 16, 2011).
Discusses the power requirements and power output of an artificial heart. | 12,515 | 59,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2014-41 | longest | en | 0.928351 |
https://pressbooks.bccampus.ca/universityphysicssandbox/chapter/beats/ | 1,717,098,368,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00627.warc.gz | 391,463,587 | 24,513 | Sound
# Beats
### Learning Objectives
By the end of this section, you will be able to:
• Determine the beat frequency produced by two sound waves that differ in frequency
• Describe how beats are produced by musical instruments
The study of music provides many examples of the superposition of waves and the constructive and destructive interference that occurs. Very few examples of music being performed consist of a single source playing a single frequency for an extended period of time. You will probably agree that a single frequency of sound for an extended period might be boring to the point of irritation, similar to the unwanted drone of an aircraft engine or a loud fan. Music is pleasant and interesting due to mixing the changing frequencies of various instruments and voices.
An interesting phenomenon that occurs due to the constructive and destructive interference of two or more frequencies of sound is the phenomenon of beats. If two sounds differ in frequencies, the sound waves can be modeled as
${y}_{1}=A\phantom{\rule{0.2em}{0ex}}\text{cos}\left({k}_{1}x-2\pi {f}_{1}t\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{y}_{2}=A\phantom{\rule{0.2em}{0ex}}\text{cos}\left({k}_{2}x-2\pi {f}_{2}t\right).$
Using the trigonometric identity $\text{cos}\phantom{\rule{0.2em}{0ex}}u+\text{cos}\phantom{\rule{0.2em}{0ex}}v=2\phantom{\rule{0.2em}{0ex}}\text{cos}\left(\frac{u+v}{2}\right)\text{cos}\left(\frac{u-v}{2}\right)$ and considering the point in space as $x=0.0\phantom{\rule{0.2em}{0ex}}\text{m,}$ we find the resulting sound at a point in space, from the superposition of the two sound waves, is equal to (Figure):
$y\left(t\right)=2A\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2\pi {f}_{avg}t\right)\text{cos}\left(2\pi \left(\frac{|{f}_{2}-{f}_{1}|}{2}\right)t\right),$
where the beat frequency is
${f}_{\text{beat}}=|{f}_{2}-{f}_{1}|.$
Beats produced by the constructive and destructive interference of two sound waves that differ in frequency.
These beats can be used by piano tuners to tune a piano. A tuning fork is struck and a note is played on the piano. As the piano tuner tunes the string, the beats have a lower frequency as the frequency of the note played approaches the frequency of the tuning fork.
Find the Beat Frequency Between Two Tuning Forks
What is the beat frequency produced when a tuning fork of a frequency of 256 Hz and a tuning fork of a frequency of 512 Hz are struck simultaneously?
Strategy
The beat frequency is the difference of the two frequencies.
Solution
We use ${f}_{\text{beat}}=|{f}_{2}-{f}_{1}|:$
$|{f}_{2}-{f}_{1}|=\left(512-256\right)\phantom{\rule{0.2em}{0ex}}\text{Hz}=256\phantom{\rule{0.2em}{0ex}}\text{Hz}\text{.}$
Significance
The beat frequency is the absolute value of the difference between the two frequencies. A negative frequency would not make sense.
Check Your Understanding What would happen if more than two frequencies interacted? Consider three frequencies.
An easy way to understand this event is to use a graph, as shown below. It appears that beats are produced, but with a more complex pattern of interference.
The study of the superposition of various waves has many interesting applications beyond the study of sound. In later chapters, we will discuss the wave properties of particles. The particles can be modeled as a “wave packet” that results from the superposition of various waves, where the particle moves at the “group velocity” of the wave packet.
### Summary
• When two sound waves that differ in frequency interfere, beats are created with a beat frequency that is equal to the absolute value of the difference in the frequencies.
### Conceptual Questions
Two speakers are attached to variable-frequency signal generator. Speaker A produces a constant-frequency sound wave of 1.00 kHz, and speaker B produces a tone of 1.10 kHz. The beat frequency is 0.10 kHz. If the frequency of each speaker is doubled, what is the beat frequency produced?
The label has been scratched off a tuning fork and you need to know its frequency. From its size, you suspect that it is somewhere around 250 Hz. You find a 250-Hz tuning fork and a 270-Hz tuning fork. When you strike the 250-Hz fork and the fork of unknown frequency, a beat frequency of 5 Hz is produced. When you strike the unknown with the 270-Hz fork, the beat frequency is 15 Hz. What is the unknown frequency? Could you have deduced the frequency using just the 250-Hz fork?
The frequency of the unknown fork is 255 Hz. No, if only the 250 Hz fork is used, listening to the beat frequency could only limit the possible frequencies to 245 Hz or 255 Hz.
Referring to the preceding question, if you had only the 250-Hz fork, could you come up with a solution to the problem of finding the unknown frequency?
A “showy” custom-built car has two brass horns that are supposed to produce the same frequency but actually emit 263.8 and 264.5 Hz. What beat frequency is produced?
The beat frequency is 0.7 Hz.
### Problems
What beat frequencies are present: (a) If the musical notes A and C are played together (frequencies of 220 and 264 Hz)? (b) If D and F are played together (frequencies of 297 and 352 Hz)? (c) If all four are played together?
What beat frequencies result if a piano hammer hits three strings that emit frequencies of 127.8, 128.1, and 128.3 Hz?
$\begin{array}{}\\ \\ \hfill {f}_{\text{B}}& =\hfill & |{f}_{1}-{f}_{2}|\hfill \\ \hfill |128.3\phantom{\rule{0.2em}{0ex}}\text{Hz}-128.1\phantom{\rule{0.2em}{0ex}}\text{Hz}|& =\hfill & 0.2\phantom{\rule{0.2em}{0ex}}\text{Hz;}\hfill \\ \hfill |128.3\phantom{\rule{0.2em}{0ex}}\text{Hz}-127.8\phantom{\rule{0.2em}{0ex}}\text{Hz}|& =\hfill & 0.5\phantom{\rule{0.2em}{0ex}}\text{Hz;}\hfill \\ \hfill |128.1\phantom{\rule{0.2em}{0ex}}\text{Hz}-127.8\phantom{\rule{0.2em}{0ex}}\text{Hz}|& =\hfill & 0.3\phantom{\rule{0.2em}{0ex}}\text{Hz}\hfill \end{array}$
A piano tuner hears a beat every 2.00 s when listening to a 264.0-Hz tuning fork and a single piano string. What are the two possible frequencies of the string?
Two identical strings, of identical lengths of 2.00 m and linear mass density of $\mu =0.0065\phantom{\rule{0.2em}{0ex}}\text{kg/m,}$ are fixed on both ends. String A is under a tension of 120.00 N. String B is under a tension of 130.00 N. They are each plucked and produce sound at the $n=10$ mode. What is the beat frequency?
$\begin{array}{}\\ \\ {v}_{A}=135.87\phantom{\rule{0.2em}{0ex}}\text{m/s,}\phantom{\rule{0.5em}{0ex}}{v}_{B}=141.42\phantom{\rule{0.2em}{0ex}}\text{m/s,}\hfill \\ {\lambda }_{A}={\lambda }_{B}=0.40\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \text{Δ}f=15.00\phantom{\rule{0.2em}{0ex}}\text{Hz}\hfill \end{array}$
A piano tuner uses a 512-Hz tuning fork to tune a piano. He strikes the fork and hits a key on the piano and hears a beat frequency of 5 Hz. He tightens the string of the piano, and repeats the procedure. Once again he hears a beat frequency of 5 Hz. What happened?
A string with a linear mass density of $\mu =0.0062\phantom{\rule{0.2em}{0ex}}\text{kg/m}$ is stretched between two posts 1.30 m apart. The tension in the string is 150.00 N. The string oscillates and produces a sound wave. A 1024-Hz tuning fork is struck and the beat frequency between the two sources is 52.83 Hz. What are the possible frequency and wavelength of the wave on the string?
$\begin{array}{}\\ \\ v=155.54\phantom{\rule{0.2em}{0ex}}\text{m/s,}\hfill \\ {f}_{\text{string}}=971.17\phantom{\rule{0.2em}{0ex}}\text{Hz,}\phantom{\rule{0.5em}{0ex}}n=16.23\hfill \\ {f}_{\text{string}}=1076.83\phantom{\rule{0.2em}{0ex}}\text{Hz,}\phantom{\rule{0.5em}{0ex}}n=18.00\hfill \end{array}$
The frequency is 1076.83 Hz and the wavelength is 0.14 m.
A car has two horns, one emitting a frequency of 199 Hz and the other emitting a frequency of 203 Hz. What beat frequency do they produce?
The middle C hammer of a piano hits two strings, producing beats of 1.50 Hz. One of the strings is tuned to 260.00 Hz. What frequencies could the other string have?
$\begin{array}{}\\ \\ {f}_{2}={f}_{1}±{f}_{\text{B}}=260.00\phantom{\rule{0.2em}{0ex}}\text{Hz}±1.50\phantom{\rule{0.2em}{0ex}}\text{Hz,}\hfill \\ \text{so that}\phantom{\rule{0.2em}{0ex}}{f}_{2}=261.50\phantom{\rule{0.2em}{0ex}}\text{Hz}\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}{f}_{2}=258.50\phantom{\rule{0.2em}{0ex}}\text{Hz}\hfill \end{array}$
Two tuning forks having frequencies of 460 and 464 Hz are struck simultaneously. What average frequency will you hear, and what will the beat frequency be?
Twin jet engines on an airplane are producing an average sound frequency of 4100 Hz with a beat frequency of 0.500 Hz. What are their individual frequencies?
$\begin{array}{}\\ \\ {f}_{\text{ace}}=\frac{{f}_{1}+{f}_{2}}{2};{f}_{\text{B}}={f}_{1}-{f}_{2}\left(\text{assume}{f}_{1}>{f}_{2}\right)\hfill \\ {f}_{\text{ace}}=\frac{\left({f}_{\text{B}}+{f}_{2}\right)+{f}_{2}}{2}⇒\hfill \\ {f}_{2}=4099.750\phantom{\rule{0.2em}{0ex}}\text{Hz}\hfill \\ {f}_{1}=4100.250\phantom{\rule{0.2em}{0ex}}\text{Hz}\hfill \end{array}$
Three adjacent keys on a piano (F, F-sharp, and G) are struck simultaneously, producing frequencies of 349, 370, and 392 Hz. What beat frequencies are produced by this discordant combination?
### Glossary
beat frequency
frequency of beats produced by sound waves that differ in frequency
beats
constructive and destructive interference of two or more frequencies of sound | 2,885 | 9,476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-22 | latest | en | 0.856295 |
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## Adding Fractions – Tenths and Hundredths
Sometimes, a problem may ask you to add two fractions that don’t share the same denominator such as adding tenths and hundredths. But how do you add tenths and hundredths? Find out all you need to know about adding fractions with uneven denominators with the example in this text and by watching the video.
## Adding Tenths and Hundredths– Example
How do you add fractions with uneven denominators? When adding fractions with tenths and hundredths, first, set up the problem.
Next, look at the denominator, the bottom number, of both fractions. Since they’re not the same, we need to convert one fraction to have the same denominator as the other.
To do this, we can multiply the fractions with the ten denominator by ten to make one hundred. When we do this, we also need to multiply the numerator, the top number, by ten as well.
Now both denominators are one hundred, we can solve the problem!
## Adding Tenths and Hundredths – Summary
Remember, when adding tenths and hundredths fractions, make sure you have the same denominators before solving the problem. The denominator of the sum will always remain the same, and you only need to add together the numerators.
Below, you can see the necessary steps to adding fractions with uneven denominators.
Step # What to do
1 Set up the problem and look at the denominators.
2 Multiply the smaller denominator until it
matches the other denominator.
Remember to multiply the numerator as well.
3 Once both denominators are the same,
After watching this video, you will find more interactive exercises, worksheets and further activities on adding tenths and hundredths with questions and answers.
## Adding Tenth and Hundredth exercise
Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Adding Tenth and Hundredth.
• ### Identify the different parts of the equation.
Hints
The numerator is above the fraction bar, and indicates how many pieces of the whole are being used in this fraction.
The operation used in this equation is addition.
The 6 in this fraction is the denominator.
Solution
The numerator is always above the fraction bar, and indicates how many pieces of the whole we are using in this fraction.
The denominator is always below the fraction bar, and shows how many pieces the whole has been divided into.
The sum is the answer to an addition problem and comes after the equals sign.
There is no multiplication or subtraction used in this equation.
• ### Identify the statements that accurately describe tenths and hundredths.
Hints
$\mathbf{\frac{1}{10}}$ is one piece from a group of 10.
$\mathbf{\frac{1}{100}}$ is one piece from a group of 100.
Solution
True
• A tenth represents one part of a whole, divided into 10 equal pieces.
• A hundredth represents one part of a whole divided into 100 equal parts.
• One hundredth can by written as a fraction by writing 1 as the numerator and 100 as the denominator.
False
• One tenth can be written as a fraction, by writing 10 as the numerator and 1 as the denominator. In a fraction, the numerator, or number on top, shows how many of something there is. So the fraction one tenth must be written as $\mathbf{\frac{1}{10}}$.
• 10 and 100 can both be divided by 7. 10 and 100 can both be divided by two, five and ten but not by 7.
• ### How do we add tenths and hundredths?
Hints
In order to add tenths and hundredths, the denominator must first be the same. In the fraction $\frac{1}{10}$, 10 is the denominator.
To make common denominators, the best strategy is to multiply smaller numbers, like 10, until they equal the larger denominator in the other fraction.
If we multiply the numerator and denominator of $\frac{2}{10}$ by 10, we get $\frac{20}{100}$.
$\mathbf{\frac{50}{100}}$ can be simplified to a smaller fraction by dividing both the numerator and denominator by 50.
Solution
1. Check if the two fractions have the same denominator. For example, if we are adding $\frac{2}{10}$ + $\frac{30}{100}$ we can check and see that the denominators are not the same.
2. If the fractions do NOT have the same denominators, multiply the denominator of the smaller fraction until it is the same. Multiply the numerator by the same. For example, we would multiply the numerator and the denominator of $\frac{2}{10}$ by 10 to get $\frac{20}{100}$.
3. Once all fractions have the same denominator, add the equivalent fractions together to get the sum. For example, we would add $\frac{20}{100}$ + $\frac{30}{100}$ to get $\frac{50}{100}$.
4. Simplify by dividing the numerator and denominator of the sum by a common denominator, if possible. For example, if we divide the numerator and denominator of $\frac{50}{100}$ by 50, we get $\frac{1}{2}$.
• ### Can you solve the problems?
Hints
Start by making sure both fractions in the equation have the same denominator. You will need to multiply or divide by ten.
Once you have added the fractions with the same denominator, make sure to simplify.
For example, $\frac{60}{100}$ becomes $\frac{3}{5}$ by dividing both the numerator and denominator by 20.
Solution
Problem 1
• $\frac{20}{100}$ + $\frac{2}{10}$ = ?
• $\frac{2}{10}$ = $\frac{20}{100}$ if the numerator and denominator are both multiplied by 10.
• $\frac{20}{100}$ + $\frac{20}{100}$ = $\frac{40}{100}$
• $\frac{40}{100}$ = $\frac{2}{5}$ if the numerator and denominator are both divided by 20.
Problem 2
• $\frac{20}{100}$ + $\frac{5}{10}$ = ?
• $\frac{5}{10}$ = $\frac{50}{100}$ if the numerator and denominator are both multiplied by 10.
• $\frac{20}{100}$ + $\frac{50}{100}$ = $\frac{70}{100}$
• $\frac{70}{100}$ = $\frac{7}{10}$ if the numerator and denominator are both divided by 10.
Problem 3
• $\frac{7}{10}$ + $\frac{10}{100}$ = ?
• $\frac{7}{10}$ = $\frac{70}{100}$ if the numerator and denominator are both multiplied by 10.
• $\frac{70}{100}$ + $\frac{10}{100}$ = $\frac{80}{100}$
• $\frac{80}{100}$ = $\frac{4}{5}$ if the numerator and denominator are both divided by 20.
Problem 4
• $\frac{6}{10}$ + $\frac{30}{100}$ = ?
• $\frac{6}{10}$ = $\frac{60}{100}$ if the numerator and denominator are both multiplied by 10.
• $\frac{60}{100}$ + $\frac{30}{100}$ = $\frac{90}{100}$
• $\frac{90}{100}$ = $\frac{9}{10}$ if the numerator and denominator are both divided by 10.
• ### Calculate the sum of the equation.
Hints
Which symbol is missing from the equation in step 1?
In order to add tenths and hundredths, you first have to make sure that the fractions in the equations have the same denominators using multiplication. Multiply BOTH the numerator (above) and denominator (below).
Same denominators: $\mathbf{\frac{10}{20}}$ + $\mathbf{\frac{2}{20}}$
Different denominators: $\mathbf{\frac{1}{50}}$ + $\mathbf{\frac{1}{10}}$
Once both fractions have the same denominators, you can add your fractions to find the sum by adding the numerators (top number).
$\mathbf{\frac{10}{100}}$ + $\mathbf{\frac{20}{100}}$ = ?
Once you have added your fractions and have the sum, check if the sum can be simplified, or made smaller, using division. Divide BOTH the numerator (top number) and denominator (bottom number).
Solution
1. $\frac{10}{100}$ + $\frac{2}{10}$ = ?
2. $\frac{2}{10}$ = $\mathbf{\frac{20}{100}}$ if the numerator and denominator are both multiplied by 10.
3. $\frac{10}{100}$ + $\frac{20}{100}$ = $\mathbf{\frac{30}{100}}$
4. $\frac{30}{100}$ = $\mathbf{\frac{3}{10}}$ if the numerator and denominator are both divided by 10.
• ### Calculate the sum of the equations.
Hints
You will need to first make sure both fractions in the equation have the same denominator before you can solve these equations.
For the equation $\frac{2}{30}$ + $\frac{1}{3}$, the common denominator is 3. How many times must you multiply 3 to equal 30?
Don't forget to simplify fractions when you can!
Example: $\frac{2}{10}$ can be simplified to $\frac{1}{5}$ by dividing both the numerator and denominator by 2.
You can always use a pencil and paper to help with your working.
Solution
Problem 1
• $\frac{1}{4}$ + $\frac{1}{40}$ = ?
• $\frac{1}{4}$ = $\frac{10}{40}$ if the numerator and denominator are both multiplied by 10.
• $\frac{10}{40}$ + $\frac{1}{40}$ = $\frac{11}{40}$
• This cannot be simplified further.
Problem 2
• $\frac{2}{5}$ + $\frac{2}{50}$ = ?
• $\frac{2}{5}$ = $\frac{20}{50}$ if the numerator and denominator are both multiplied by 10.
• $\frac{20}{50}$ + $\frac{2}{50}$ = $\frac{22}{50}$
• $\frac{22}{50}$ = $\frac{11}{25}$ if the numerator and denominator are both divided by 2.
Problem 3
• $\frac{2}{30}$ + $\frac{1}{3}$ = ?
• $\frac{1}{3}$ = $\frac{10}{30}$ if the numerator and denominator are both multiplied by 10.
• $\frac{2}{30}$ + $\frac{10}{30}$ = $\frac{12}{30}$
• $\frac{12}{30}$ = $\frac{2}{5}$ if the numerator and denominator are both divided by 6.
Problem 4
• $\frac{1}{2}$ + $\frac{3}{10}$ = ?
• $\frac{1}{2}$ = $\frac{5}{10}$ if the numerator and denominator are both multiplied by 5.
• $\frac{5}{10}$ + $\frac{3}{10}$ = $\frac{8}{10}$
• $\frac{8}{10}$ = $\frac{4}{5}$ if the numerator and denominator are both divided by 2. | 2,484 | 9,332 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-38 | latest | en | 0.899623 |
http://www.exampleproblems.com/wiki/index.php/Real_number | 1,545,017,694,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00158.warc.gz | 363,690,104 | 15,204 | # Real number
For the economics term, see real vs. nominal in economics.
In mathematics, the real numbers are intuitively defined as numbers that are in one-to-one correspondence with the points on an infinite line—the number line. The term "real number" is a retronym coined in response to "imaginary number". The real numbers are the central object of study in real analysis.
## Properties and uses
Real numbers may be rational or irrational; algebraic or transcendental; and positive, negative, or zero.
Real numbers measure continuous quantities. They may in theory be expressed by decimal fractions that have an infinite sequence of digits to the right of the decimal point; these are often represented in the same form as 324.823211247… The three dots indicate that there would still be more digits to come, no matter how many more might be added at the end.
Measurements in the physical sciences are almost always conceived as approximations to real numbers. Writing them as decimal fractions (which are rational numbers that could be written as ratios, with an explicit denominator) is not only more compact, but to some extent conveys the sense of an underlying real number.
A real number is said to be computable if there exists an algorithm that yields its digits. Because there are only countably many algorithms, but an uncountable number of reals, most real numbers are not computable. Some constructivists accept the existence of only those reals that are computable. The set of definable numbers is broader, but still only countable.
Computers can only approximate most real numbers. Most commonly, they can represent a certain subset of the rationals exactly, via either floating point numbers or fixed-point numbers, and these rationals are used as an approximation for other nearby real values. Arbitrary-precision arithmetic is a method to represent arbitrary rational numbers, limited only by available memory, but more commonly one uses a fixed number of bits of precision determined by the size of the processor registers. In addition to these rational values, computer algebra systems are able to treat many (countable) irrational numbers exactly by storing an algebraic description (such as "sqrt(2)") rather than their rational approximation.
Mathematicians use the symbol R (or alternatively, ${\mathbb {R}}$, the letter "R" in blackboard bold) to represent the set of all real numbers. The notation Rn refers to an n-dimensional space of real numbers; for example, a value from R3 consists of three real numbers and specifies a location in 3-dimensional space.
In mathematics, real is used as an adjective, meaning that the underlying field is the field of real numbers. For example real matrix, real polynomial and real Lie algebra.
## History
Vulgar fractions had been used by the Egyptians around 1000 BC; the Vedic "Sulba Sutras" ("rule of chords" in Sanskrit) * ca. 600 BC contain the first use of irrational numbers, and approximation of π at 3.16. Around 500 BC, the Greek mathematicians led by Pythagoras realized the need for irrational numbers in particular the irrationality of the square root of two. Negative numbers were invented by Indian mathematicians around 600 AD, and then possibly reinvented in China shortly after. They were not used in Europe until the 1600s, but even in the late 1700s, Leonhard Euler discarded negative solutions to equations as unrealistic.
In the 18th and 19th centuries there was much work on irrational and transcendental numbers. Lambert (1761) gave the first flawed proof that π cannot be rational, Legendre (1794) completed the proof, and showed that π is not the square root of a rational number. Ruffini (1799) and Abel (1842) both construct proofs of Abel–Ruffini theorem: that the general quintic or higher equations cannot be solved by a general formula involving only arithmetical operations and roots. Évariste Galois (1832) developed techniques for determining whether a given equation could be solved by radicals which gave rise to the field of Galois theory. Joseph Liouville (1840) showed that neither e nor e2 can be a root of an integer quadratic equation, and then established existence of transcendental numbers, the proof being subsequently displaced by Georg Cantor (1873). Charles Hermite (1873) first proved that e is transcendental, and Ferdinand von Lindemann (1882), showed that π is transcendental. Lindemann's proof was much simplified by Weierstrass (1885), still further by David Hilbert (1893), and has finally been made elementary by Hurwitz and Paul Albert Gordan.
The development of calculus in the 1700s used the entire set of real numbers without having defined them cleanly. The first rigorous definition was given by Georg Cantor in 1871. In 1874 he shows that the set of all real numbers is uncountably infinite but the set of all algebraic numbers is countably infinite. Contrary to widely held beliefs, his method was not his famous diagonal argument, which he published three years later.
## Definition
### Construction from the rational numbers
The real numbers can be constructed as a completion of the rational numbers. For details and other construction of real numbers, see construction of real numbers.
### Axiomatic approach
Let R denote the set of all real numbers. Then:
The last property is what differentiates the reals from the rationals. For example, the set of rationals with square less than 2 has a rational upper bound (e.g., 1.5) but no rational least upper bound, because the square root of 2 is not rational.
The real numbers are uniquely specified by the above properties. More precisely, given any two Dedekind complete ordered fields R1 and R2, there exists a unique field isomorphism from R1 to R2, allowing us to think of them as essentially the same mathematical object.
## Properties
### Completeness
The main reason for introducing the reals is that the reals contain all limits. More technically, the reals are complete (in the sense of metric spaces or uniform spaces, which is a different sense than the Dedekind completeness of the order in the previous section). This means the following:
A sequence (xn) of real numbers is called a Cauchy sequence if for any ε > 0 there exists an integer N (possibly depending on ε) such that the distance |xn − xm| is less than ε provided that n and m are both greater than N. In other words, a sequence is a Cauchy sequence if its elements xn eventually come and remain arbitrarily close to each other.
A sequence (xn) converges to the limit x if for any ε > 0 there exists an integer N (possibly depending on ε) such that the distance |xn − x| is less than ε provided that n is greater than N. In other words, a sequence has limit x if its elements eventually come and remain arbitrarily close to x.
It is easy to see that every convergent sequence is a Cauchy sequence. An important fact about the real numbers is that the converse is also true:
Every Cauchy sequence of real numbers is convergent.
That is, the reals are complete.
Note that the rationals are not complete. For example, the sequence (1, 1.4, 1.41, 1.414, 1.4142, 1.41421, ...) is Cauchy but it does not converge to a rational number. (In the real numbers, in contrast, it converges to the square root of 2.)
The existence of limits of Cauchy sequences is what makes calculus work and is of great practical use. The standard numerical test to determine if a sequence has a limit is to test if it is a Cauchy sequence, as the limit is typically not known in advance.
For example, the standard series of the exponential function
${\mathrm {e}}^{x}=\sum _{{n=0}}^{{\infty }}{\frac {x^{n}}{n!}}$
converges to a real number because for every x the sums
$\sum _{{n=N}}^{{M}}{\frac {x^{n}}{n!}}$
can be made arbitrarily small by choosing N sufficiently large. This proves that the sequence is Cauchy, so we know that the sequence converges even if we do not know ahead of time what the limit is.
### "The complete ordered field"
The real numbers are often described as "the complete ordered field", a phrase that can be interpreted in several ways.
First, an order can be lattice-complete. It is easy to see that no ordered field can be lattice-complete, because it can have no largest element (given any element z, z + 1 is larger), so this is not the sense that is meant.
Additionally, an order can be Dedekind-complete, as defined in the section Axioms. The uniqueness result at the end of that section justifies using the word "the" in the phrase "complete ordered field" when this is the sense of "complete" that is meant. This sense of completeness is most closely related to the construction of the reals from Dedekind cuts, since that construction starts from an ordered field (the rationals) and then forms the Dedekind-completion of it in a standard way.
These two notions of completeness ignore the field structure. However, an ordered group (and a field is a group under the operations of addition and subtraction) defines a uniform structure, and uniform structures have a notion of completeness (topology); the description in the section Completeness above is a special case. (We refer to the notion of completeness in uniform spaces rather than the related and better known notion for metric spaces, since the definition of metric space relies on already having a characterisation of the real numbers.) It is not true that R is the only uniformly complete ordered field, but it is the only uniformly complete Archimedean field, and indeed one often hears the phrase "complete Archimedean field" instead of "complete ordered field". Since it can be proved that any uniformly complete Archimedean field must also be Dedekind complete (and vice versa, of course), this justifies using "the" in the phrase "the complete Archimedean field". This sense of completeness is most closely related to the construction of the reals from Cauchy sequences (the construction carried out in full in this article), since it starts with an Archimedean field (the rationals) and forms the uniform completion of it in a standard way.
But the original use of the phrase "complete Archimedean field" was by David Hilbert, who meant still something else by it. He meant that the real numbers form the largest Archimedean field in the sense that every other Archimedean field is a subfield of R. Thus R is "complete" in the sense that nothing further can be added to it without making it no longer an Archimedean field. This sense of completeness is most closely related to the construction of the reals from surreal numbers, since that construction starts with a proper class that contains every ordered field (the surreals) and then selects from it the largest Archimedean subfield.
The reals are uncountable; that is, there are strictly more real numbers than natural numbers, even though both sets are infinite. This is proved with Cantor's diagonal argument. In fact, the cardinality of the reals is 2ω, i.e., the cardinality of the set of subsets of the natural numbers. Since only a countable set of real numbers can be algebraic, almost all real numbers are transcendental. The non-existence of a subset of the reals with cardinality strictly between that of the integers and the reals is known as the continuum hypothesis. The continuum hypothesis can neither be proved nor be disproved; it is independent from the axioms of set theory.
The real numbers form a metric space: the distance between x and y is defined to be the absolute value |x − y|. By virtue of being a totally ordered set, they also carry an order topology; the topology arising from the metric and the one arising from the order are identical. The reals are a contractible (hence connected and simply connected), separable metric space of dimension 1, and are everywhere dense. The real numbers are locally compact but not compact. There are various properties that uniquely specify them; for instance, all unbounded, connected, and separable order topologies are necessarily homeomorphic to the reals.
Every nonnegative real number has a square root in R, and no negative number does. This shows that the order on R is determined by its algebraic structure. Also, every polynomial of odd degree admits at least one root: these two properties make R the premier example of a real closed field. Proving this is the first half of one proof of the fundamental theorem of algebra.
The reals carry a canonical measure, the Lebesgue measure, which is the Haar measure on their structure as a topological group normalised such that the unit interval [0,1] has measure 1.
The supremum axiom of the reals refers to subsets of the reals and is therefore a second-order logical statement. It is not possible to characterize the reals with first-order logic alone: the Löwenheim-Skolem theorem implies that there exists a countable dense subset of the real numbers satisfying exactly the same sentences in first order logic as the real numbers themselves. The set of hyperreal numbers is much bigger than R but also satisfies the same first order sentences as R. Ordered fields that satisfy the same first-order sentences as R are called nonstandard models of R. This is what makes nonstandard analysis work; by proving a first-order statement in some nonstandard model (which may be easier than proving it in R), we know that the same statement must also be true of R.
## Generalizations and extensions
The real numbers can be generalized and extended in several different directions. Perhaps the most natural extension are the complex numbers which contain solutions to all polynomial equations and hence are an algebraically closed field unlike the real numbers. However, the complex numbers are not an ordered field. Ordered fields extending the reals are the hyperreal numbers and the surreal numbers; both of them contain infinitesimal and infinitely large numbers and thus are not Archimedean. Occasionally, the two formal elements +∞ and −∞ are added to the reals to form the extended real number line, a compact space which is not a field but retains many of the properties of the real numbers. Self-adjoint operators on a Hilbert space (for example, self-adjoint square complex matrices) generalize the reals in many respects: they can be ordered (though not totally ordered), they are complete, all their eigenvalues are real and they form a real associative algebra. Positive-definite operators correspond to the positive reals and normal operators correspond to the complex numbers. | 3,133 | 14,547 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-51 | latest | en | 0.937897 |
https://studylib.net/doc/7612573/26-worked-solutions | 1,642,510,936,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00173.warc.gz | 606,744,957 | 16,812 | # 26 Worked Solutions
```25
Problem 9.2.A. A simple-span, uniformly –loaded
beam consists of a W 18 50 with Fy = 36 ksi. Find the
percentage gain in the limiting bending moment if a fully
plastic condition is assumed, instead of a condition
limited by elastic stress.
S ´ X1 ´ 2
Mn x
´
Lb / ry
´
245(1410) 2
300 / 2.09
Elastic condition limit:
M = Fy Sx = 36(88.9) = 3200 kip-in. = 267 kip-ft
489, 000
´
144
1
1
1
( X 1 )2 ´ X 2
2( Lb / ry ) 2
1410(0.0496)
2(300 / 2.09) 2
98, 600
41, 200
3390´ 1.84
6, 240 kip-in. or 520 kip-ft
Plastic condition limit:
M = Fy Zx = 36(101) = 3640 kip-in. = 303 kip-ft
Gain in M = 303 267 = 36 kip-ft
% gain = 36/267 100 = 13.5%
Problem9.3.A. Determine the nominal bending moment
capacity ( Mn) for a W 30 90 made of A-36 steel and
the following unbraced lengths: (1) 5 ft, (2) 15 ft, (3) 30
ft.
(1) Lb = 5 ft. Table 9.1: Lp = 8.71 ft, Lr = 24.8 ft.
Problem 9.4.A. Design for flexure a simple beam 14 ft
[4.3 m] in length and having a total uniformly distributed
distributed live load of 26.4 kips [108 kN].
13.2
wu 1.4(DL) 1.4
1.32 kips/ft
14
or
13.2
26.4
wu 1.2(DL) 1.6(LL) 1.2
1.6
14
14
1.14 3.02 4.16 kips/ft
Mn = Mp = Fy Zx = 36(283) = 10,200 kip-in.
= 849 kip-ft
wu L2 4.16(14) 2
102 kip-ft
8
8
M
102
Mn u
113 kip-ft
Fb
0.9
(2) Lp Lb Lr, use:
Zx
Mr = (Fy Fr) Sx = (36 10)(245) = 6370 kip-in.
= 531 kip-ft
From Table 9.1, try W 14 26
Since Lb Lp, use:
Lb - Lp
M n M p - ( M p - M r )´
L - L
p
r
15 - 8.71
849 - (849 - 531)
24.8 - 8.71
725 kip-ft
(3) Lb Lr, use:
26 Worked Solutions
Mu
M n 113 12 in.
3
37.8 in.
Fy
36 1 ft
New wu = 4.16 + 0.026 = 4.186
New Z x
4.186
(37.8) 38.0
4.16
W 14 26 is still OK
Problem 9.4.C. A beam of 15-ft [4.6 m] length has three
concentrated loads of 6 kips, 7.5 kips, and 9 kips at 4 ft,
10 ft, and 12 ft [26.7 kN, 33.4 kN, and 40.0 kN at 1.2m,
3 m, and 3.6 m], respectively, from the left-hand support.
Design the beam for flexure.
Mn
M u 78.5
87.2 kip-ft
Fb
0.9
Zx
M n 87.2(12)
29.1 in.3
Fy
36
From Table 9.1, try a W 12 22
Live loads: Pu1 = 1.6(6) = 9.6 kips, Pu2 = 1.6(7.5) = 12
kips, Pu3 = 1.6(9) = 14.4 kips.
From analysis of the beam, Mu = 81.6 kip-ft
Mn
Zx
M u 81.6
90.7 kip-ft
Fb
0.9
M n 90.7(12)
30.2 in.3
Fy
36
From Table 9.1, select: W 10 26
wu
= 1.2(0.022) = 0.0264 kip/ft
M u 0.385 kip-ft (under point load)
New total Mn = 87.6 kip-ft
Zx
wu
Mu
wu L2 0.0312(15)2
0.88 kip-ft
8
8
Mn
M u 0.88
0.98 kip-ft
Fb
0.9
New total Mn = 90.7 + 0.98 = 91.69 kip-ft
New required Zx = 30.5 in.3 , selection still OK.
Problem 9.4.E. Design for flexure a beam 12 ft [3.6 m]
kip/ft [14.6 kN/m], a uniformly distributed live load of 1
kip/ft [14.6 kN/m], and a concentrated load of 8.4 kips
[37.4 kN] a distance of 5 ft [1.5 m] from one support.
M u 0.385
0.427 kip-ft
Fb
0.9
Mn
M n 87.6(12)
29.3 in.3 . W 12 ´ 22 still OK
Fy
36
Problem 9.4.G. A steel beam 16-ft [4.9 m] long has a
and a uniformly distributed live load of 100 lb/ft [1.46
kN/m] extending to 10 ft [3 m] from the left support. In
kN] at 10 ft [3 m] from the left support. Design the
beam for flexure.
wu = 1.2(DL) + 1.6(LL) = 1.2(0.1) + 1.6(0.1) = 0.28
kips/ft
Pu = 1.6(8) = 12.8 kips
From the beam analysis:
Mu = 54.6 kip-ft (under the point load)
Mn
M u 54.6
60.7 kip-ft
Fb
0.9
Zx
M n 60.7(12)
20.2 in.3
Fy
36
wu = 1.4(DL) = 1.4(1) = 1.4 kips/ft
or, wu = 1.2(DL) + 1.6(LL) = 1.2(1) + 1.6(1) = 2.8
kips/ft
From Table 9.1, try W 10 19
Pu = 1.4(DL) = 1.4(8.4) = 11.9 kips
Beam weight: wu = 1.2(0.019) = 0.0228 kips/ft
Adjusted Zx = 20.5 in.3, selection still OK
Mu = 78.5 kip-ft (under the point load)
27
Problem 9.4.I. A cantilever beam is 12 ft [3.6 m] long
[8.75 kN/m] and a uniformly distributed load of 1000
lbs/ft [14.6 kN/m]. Design the beam for flexure.
wu = 1.2(DL) + 1.6(LL) = 2.32 kips/ft
Mu
wu L2 2.32(12) 2
167 kip-ft
2
2
Mn
M u 167
186 kip-ft
Fb
0.9
Zx
M n 186(12)
61.9 in.3
Fy
36
From Table 9.1, try W 16 36
Lb - Lp
M n M p - (M p - M r )
L - L
p
r
6 - .71
462 - (462 - 286)
457 kip-ft
17.2- 5.71
Close but OK, use the W 24 62.
(b) Lb = 10 ft. From Table 9.1, try a W 24 68.
Additional weight makes the required Mn = 455 kip-ft.
Lb - Lp
- Mr
L - L
p
r
10- 7.50
=480- (849- 303)
451 kip-ft
22.8- 7.50
Mn M p -
M
p
For beam weight, wu = 1.2(0.036) = 0.0432 kips/ft
As this is less than the required value, the shape does not
work. Use a W 16 77 or a W 24 76.
New Mn = 190 kip-ft, Zx = 63.3 in.3, selection still OK
(c) Lb = 15 ft. Based on previous work, try a W 24 76.
Problem 9.5.A. A W shape is to be used for a uniformly
[222 kN] and a total dead load of 22 kips [98 kN] on a 30
ft [9.2 m] span. Select the lightest weight shape for
unbraced lengths of (a) 6 ft [1.83 m], (b) 10 ft [3.05 m],
(c) 15 ft [4.57 m].
Wu = 1.2 (DL) + 1.6 (LL)
= 1.2(22) + 1.6(50) = 106.4 kips
Mu
WL 106.4(30)
399 kip-ft
8
8
Mn
M u 399
443 kip-ft
Fb
0.9
Zx
M n 443(12)
148 in.3
Fy
36
(a) Lb = 6 ft. From Table 9.1, a W 24 62 is lightest
(Lp) is 5.71 ft or less (Lb). Try this shape and check for
buckling. Adding the weight of the beam, new values
are:
Wu = 108.6 kips, Mn = 452 kip-ft
Check for: LpLbLr , 5.71617.2
15- 8
M n 600- (600- 381)
500 kip-ft
23.4- 8
Which indicates that the shape is adequate.
(Note: This process is really laborious. It makes a case
for using some aid to shortcut the process. One aid is the
use of a computer program. Another aid consists of the
series of graphs similar to Fig. 9.6 which are provided in
the AISC Manual (Ref. 5).
28 Worked Solutions
Problem 9.5.C. A W shape is to be used for a uniformly
[222 kN] and a total dead load of 22 kips [98 kN] on a 30
ft [9.14 m] span. Select the lightest weigh shape for
unbraced lengths of (a) 6 ft; (b) 10 ft; (c) 15 ft.
Which indicates that the chosen shape is OK.
Wu 1.2( DL)1.6( LL) 1.2(22) 1.6(50) 106.4 kips
15- 8
M n 600- (600- 381)
501 kip-ft
23.4- 8
Mu
WL 106.4(30)
399 kip-ft
8
8
Mn
M u 399
443 kip-ft
Fb
0.9
Zx
M n 443(12)
148 in.3
Fy
36
(c) unbraced length = 15 ft
Try the W 24 76, requires Mn = 454 kip-ft.
Which indicates that the shape is still OK for this
unbraced length.
Problem 9.6.A, B, C. Compute the shear capacity (vVn)
for the following beams of A36 steel: A, W 24 84, C,
W 10 19.
Case 1:
h 418
69.7
tw
36
Case 2:
h 523
87.2
tw
36
From Table 9.1, lightest shape is a W 24 62 if
unbraced length is 5.71 ft or less.
(a) unbraced length Lb = 6 ft Try the W 24 62 for the
case of LpLbLr, or 5.71617.2. Then
Add weight of beam at 1.2(62 30) = 2.23 kips.
A: From Table A.3 = d = 24.1 in., tw = 0.470 in., tf =
0.770 in.
New required Mn = 453 Kip-ft
h = d 2(tf) = 24.1 2(0.770) = 22.56
Lb - L p
M n M p - (M p - M r )
L - L
p
r
6 - 5.71
462 - (462 - 286)
17.2 - 5.71
458 kip-ft
As this is greater than the required value, the shape
chosen is OK.
(b) unbraced length Lb = 10 ft Same case as (a).
Trials will show W 24 62, and 68 do not work.
Try W 24 76. New total Wu = 109.1 kips.
New required M n
109.1
(443) 454 kip-ft
106.4
h 22.56
48.0
tw
0.47
Case 1 condition
Aw = d tw = 24.1 0.470 = 11.33 in.2
vVn = 0.9 0.6 36 11.33 = 220 kips
C: h = 10.24 2(0.395) = 9.45 in.
h 9.45
37.8, Case 1
tw 0.25
Aw = 10.24 0.25 = 2.56 in.2
vVn = 0.9 0.6 36 2.56 = 49.8 kips
10- 8
M n 600- (600- 381)
572 kip-ft
23.4- 8
29
Problem 9.7.A-D. Find the maximum deflection in
inches for the following simple beams with uniformly
distributed load. Find the values using: (a) the equation
for the beam: and (b) the curves in Fig. 9.11.
A: W 10 33, span = 18 ft, service load = 1.67 klf.
5wL4
5(1.67 /12)(18´ 12) 4
0.794 in.
(a) D
384 EI
384(29, 000)(171)
(b) = 0.9 in.
C: W 18 46, span = 24 ft, service load = 2.29 klf
5wL4
5(2.29 /12)(24´ 12) 4
0.829 in.
(a) D
384 EI
384(29, 000)(712)
(b) = 0.8 in.
Problem 9.8.A-H. For each of the following conditions
find (a) the lightest permitted shape and (b) the
shallowest permitted shape of A36 steel.
Span (ft)
A
C
E
G
16
36
18
42
Live
3
1
0.333
1
3
0.5
0.625
0.238
A Factored load = 1.2(3 16) + 1.6(3 16) = 134 kips
(a) W 16 57, Table load = 142 kips
Beam weight adds 1.2(0.057 16) = 1.09 kips, which is
not a problem for the selection.
(b) W 10 88, Table load = 153 kips
Beam weight adds 1.2(0.088 16) = 1.69 kips, selection
OK.
C Factored load = 1.2(0.278 36) + 1.6(0.833 36)
= 60 kips
Beam weight adds 1.2(0.055 36) = 2.16 kips.
Not critical for the selection.
(b) W 18 86, Table load = 112 kips.
Beam weight not critical.
E Factored load = 1.2(0.625 18) + 1.6(0.333 18)
= 23 kips
(a) W 12 16, Table load = 24.1 kips.
Beam weight = 1.2(0.016 18) = 0.35 kips.
Not critical for the shape.
(b) W 10 19, Table load = 25.9 kips
Beam weight = 1.2(0.019 18) = 0.41 kips
Not critical for shape.
G Factored load = 1.2(0.238 42 + 1.6(1 42) =
=79 kips
(a) W 24 76, Table load = 103 kips
Beam weight not critical.
(b) W 21 83, Table load = 101 kips
Beam weight not critical.
Problem 9.10.A. Open web steel joists are to be used for
(not including joist weight) on a span of 48 ft. Joists are
4 ft on center and deflection under live load is limited to
1/360 of the span. Select the lightest joist.
Live load = 25(4) = 100 lbs/ft
Factored load = 1.2(80) + 1.6(100) = 256 lbs/ft
Try 26K7 at 10.9 lbs/ft, Table load = 282 kips
Joist weight not critical.
Load for deflection from Table is 100 lbs/ft, exactly what
is required.
(a) W 21 50, Table load = 66 kips
30 Worked Solutions
Problem 9.10.C. Open web steel joists are to be used for
psf (without the joists) on a span of 36 ft. Joists are 2 ft
on center and deflection is limited to 1/360 under live
load and to 1/240 of the span under total load. Select (a)
the lightest possible joist, and (b) the shallowest possible
joist.
Live load = 50(2) = 100 lbs/ft
Problem 10.4.A. Using Table 10.2, select a column
section for an axial dead load of 60 kips and an axial live
load of 88 kips if the unbraced height about both the x
and y axes is 12 ft. A36 steel is to be used and K is
assumed as 1.0.
Pu = 1.2(DL) + (1.6(LL) = 1.2(60) + 1.6(88) = 213 kips
Total service load = 190 lbs/ft
KL = 1.0(12) = 12 ft
Total factored load = 1.2(90) + 1.6(100) = 268 lbs/ft
From Table 10.2, possible choices are:
For total load deflection ned (240/360)(190) = 127 lbs/ft
(a) Try 24K4 at 8.4 lbs/ft, Table load = 340 lbs/ft for
total factored load, 150 lbs/ft for deflection due to total
(b) Try 22K6 at 8.8 lbs/ft, Table load = 381 lbs/ft for
total factored load, and 153 lbs/ft for deflection due to
Problem 10.3.A. Determine the maximum factored axial
load for a W 10 49 column with an unbraced height of
15 ft. Assume K = 1.0.
Table A.3: A = 15.6 in.2, ry = 2.48 in.
Shape:
W 14 53
W 12 45
W 10 33
W 8 31
355 kips
301 kips
222 kips
214 kips
Problem 10.4.C. Same data as Problem 10.4.A, except
The unbraced height about the x-axis is 20 ft and the
unbraced height about the y-axis is 10 ft.
Pu = 1.2(DL) + 1.6(LL) = 1.2(142) + 1.6(213) = 511 kips
( KL) y 1.0(10) 10 ft
( KL) x 1.0(20)
11.4 ft
rx / ry
1.75
KL 1.0(15´ 12)
72.6 Table 10.1: Fc = 27.2 ksi
ry
2.48
( KL)¢y
Pu = c Fc A = 0.85 27.2 15.6 = 361 kips
X-axis controls, from Table 10.2 possible choices are:
Problem 10.3.C. Determine the maximum factored axial
load for the column in Problem 10.3.A, if the conditions
are as shown in Figure 10.5 with L1 = 15 ft and L2 = 8 ft.
Table A.3: A = 15.6 in.2, rx = 5.23 in., ry = 2.48 in.
KL 1.0(96)
38.7 ,
ry
2.48
KL 1.0(180)
34.4
rx
5.23
Shape:
W 14 68
W 12 79
W 10 68
511 kips
631 kips
520 kips
Problem 10.4.E-H. Select the minimum size standard
weight steel pipe for an axial dead load of 20 kips, a live
load of 30 kips, and the following unbraced heights: (e) 8
ft; (g) 18 ft.
Y axis governs, from Table 10.1: Fc = 33.2 ksi
Pu = cFcA = 0.85(33.2)(15.6) = 440 kips
Pu = 1.2(20) + 1.6(30) = 72 kips
(e) 4 in. pipe, table load = 77 kips
(g) 6 in. pipe, table load = 104 kips
31
with an effective unbraced height of 12 ft. Find the
Problem 10.4.I. A structural tubing column, designated
as HHS 4 4 3/8, of steel with Fy = 46 ksi, is used
From Table 10.5, factored load = 98 kips
Problem 10.4.K. Using Table 10.5, select the lightest
kips and a live load of 34 kips if the effective unbraced
height is 10 ft.
column factored axial load = 200 kips, factored beam
reaction = 30 kips, unbraced column height is 14 ft.
From Table 10.2, m = 1.8 for the 14 ft height.
Pu¢ Pu (m ´ M ux ) (200 30) (1.8´ 30´ 6 /12)
Pu = 1.2(30) + 1.6(34) = 90.4 kips
257 kips
From Table 10.5, possible choices are:
Section:
Area:
HHS 6 6 3/16
HHS 5 5 ¼
HHS 4 4 3/8
3.98 in.2
4.3 in.2
4.78 in.2
131 Kips
130 kips
119 kips
6 in. tube is lightest.
Problem 10.4.M. A double-angle compression member
8 ft long is composed of two A36 steel angles 4 3 3/8
in. with long legs back-to-back. Determine the
maximum factored axial load for the angles.
kips. y-axis governs, load is 104 kips.
Problem 10.4.O. Using Table 10.6, select a doubleangle compression member for an axial compression
effective unbraced length is 10 ft.
Pu = 1.2(25) + 1.6(25) = 70 kips
Possible choice from Table 10.6:
5 3.5 5/16, for x-axis the table load limit is 70 kips.
Pu 230
0.89 > 0.2, therefore correct formula was used
Pu¢ 257
From Table 10.2, select W 12 45
Problem 10.5.C. Same as Problem 10.5.A, except axial
load is 485 kips, beam reaction is 100 kips, unbraced
height is 18 ft.
Pu¢ Pu m ´ M ux (485 100) (1.6´ 100 ´ 6 /12)
= 665 kips
Pu 485
0.73 > 0.2, therefore correct formula used
Pu¢ 665
From Table 10.2, select W 12 96
Problem 10.5.E. a 14-in. W shape is to be used for a
column that sustains bending on both axes. Select a trial
section for an unbraced height of 16 ft and the following
factored values: total axial load = 80 kips, Mx = 85 kipft, My = 64 kip-ft., unbraced height is 16 ft.
Pu¢ Pu (m´ M ux ) (2m´ M uy )
= (80 kips) + (1.7 ´ 85) + (2 ´ 1.7 ´ 64) = 443 kips
Pu
80
0.18 < 0.2, thus incorrect formula was used
Pu¢ 443
Pu 9
[(m ´ M ux ) (2m ´ M uy )]
2 8
80 9
[(1.7 ´ 85) (2´ 1.7 ´ 64)] 448 kips
2 8
Pu¢
Problem 10.5.A. It is desired to use a 12-in. W shape for
a column to support a beam as shown in Fig. 10.7.
Select a trial size for the column for the following data:
From Table 10.2, select W 14 82
32 Worked Solutions
Problem 11.2.A. A bolted connection of the general
form shown in Fig. 11.6 is to be used to transmit a
by using 7/8 in. A325 bolts and plates of A36 steel with
threads included in the planes of shear. The outer plates
are to be 8 in. wide and the center plate is to be 12 in.
wide. Find the required thickness of the plates and the
number of bolts needed if the bolts are placed in two
rows. Sketch the final layout of the connection.
Pu = 1.2(DL) + 1.6(LL) = 1.2(75) + 1.6(100) = 250 kips
From Table 11.1, design strength of a single bolt
experiencing double shear with threads included is 43.3
kips.
n
Vu
250
5.77 or 6
FvVr 43.3
From Table 11.2, minimum edge distance is 1.5 in. and
minimum spacing is 2.625 in.
For the plates, required cross-sectional area is
P
250
As u
7.72 in.2
Ft Fy 0.9(36)
For the outer plates:
t
As
7.72
0.482 in., use 1/2 in.
2(width) 2(8)
Thus the tension on the net area is not critical.
For bearing:
vRn = 1.5(1.5)(0,6875)(58) = 89.7 kips
Thus bearing is not critical.
Check for block shear on the middle plate.
Tension: net w = 5 1 = 4 in.
Ae = 4(11/16) = 2.75 in.2
Ft Pn Ft ´ Fu ´ Ae 0.75(58)(2.75) 120 kips
Shear: net w = 2(1.5 0.5) = 2 in.
Ae = 2(11/16) = 1.375 in.2
Fv Rn Fv ´ Fu ´ Ae 0.75(58)(1.375) 59.8 kips
Net width = 8 2(1.0) = 6 in.
Combined resistance = 120 + 59.8 = 180 kips
Net area: Ae = 6(0.5) = 3 in.2 per plate
Allowable tension on the net area:
Problem 12.2.A-F. Using data from Table 12.1, select
the lightest steel deck for the following:
Ft Pn Ft ´ Fu ´ Ae 0.75(58)(2´ 3) 261 kips
Thus the tension on the net area is not critical.
For bearing resistance of a single bolt:
vRn = 1.5 Lc t Fu
= 1.5(1.5)(0.5)(58) = 65.25 kips
As this is greater than the bolt capacity, bearing is not
critical
For the middle plate:
t
As
7.72
0.642 in., use 11/16 in.
width
12
Net width = 12 – 2(1.0) = 10 in.
Net area: Ae = 10(11/16) = 6.875 in.2
Allowable tension on the net area:
tPn = 0.75 Fu As
= 0.75(58)(6.875) = 299 kips
Choice:
A
simple span, 7 ft
45
WR20
C
two-span, 8.5 ft
45
WR18
E
three-span, 6 ft
50
IR22
``` | 8,376 | 17,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-05 | latest | en | 0.629175 |
http://5outh.blogspot.com/2013/01/ | 1,529,348,394,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00583.warc.gz | 4,341,206 | 19,816 | ## Thursday, January 24, 2013
### The Handshake Problem
The Handshake Problem is something of a classic in mathematics. I first heard about it in an algebra course I took in high school, and it's stuck with me through the years. The question is this:
In a room containing $n$ people, how many handshakes must take place for everyone to have shaken hands with everyone else?
The goal of this short blog post will be to present a solution to this problem using the concept of graphs.
### The Complete Graph
fig. I: A complete graph with 7 vertices
First, we must understand the idea of a complete graph. A complete graph is a graph such that each node is connected to every other node. A graphical example can be seen in fig. I.
We can use the model of a complete graph to reason about the problem at hand. The nodes in the graph may represent the people in the room, while the connecting edges represent the handshakes that must take place. As such, the key to solving the Handshake Problem is to count the edges in a complete graph.
But it would be silly to draw out a graph and count the edges one-by-one in order to solve this problem (and would take a very long time for large values of $n$), so we'll use math!
### The Solution
To find the solution, we have to make a few key observations. The easiest way to start out in this case is to map out the first few complete graphs and try to find a pattern:
Let $n$ = the number of nodes, $e$ = the number of edges.
• $n = 1 \Rightarrow e = 0$
• $n = 2 \Rightarrow e = 1$
• $n = 3 \Rightarrow e = 3$
• $n = 4 \Rightarrow e = 6$
At this point, you may be noticing a pattern. As $n$ increases, we're adding $n -1$ edges. This makes sense -- each time a new node is introduced, it must connect to each node other than itself. In other words, $n-1$ connections must be made.
It follows that the number of the edges ($e$) in a complete graph with $n$ vertices can be represented by the sum $(n-1) + (n-2) + \dots + 1 + 0$, or more compactly as:
$$\sum_{i=1}^n i-1$$
You may already know from summation laws that this evaluates to $\frac{n(n-1)}{2}$. Let's prove it.
### The Proof
Theorem: $\forall n \in \mathbb{N} : \sum_{i=1}^{n} i-1 = \frac{n(n-1)}{2}$
Proof: Base case: Let $n = 1$. Then, $\frac{1(1-1)}{2} = \sum_{i=1}^1 i-1 = 0$.
Inductive case: Let $n \in \mathbb{N}$. We need to prove that $\frac{n(n-1)}{2} + n = \frac{n(n+1)}{2}$.
$$\begin{array} {lcl} \frac{n(n-1)}{2} + n & = & \frac{n(n-1)+2n}{2} \\ & = & \frac{n^2 - n + 2n}{2} \\ & = & \frac{n^2 + n}{2} \\ & = & \frac{n(n+1)}{2}■\end{array}$$
We can now use the theorem knowing it's correct, and, in fact, provide a solution to the original problem. The answer to the Handshake Problem involving $n$ people in a room is simply $\frac{n(n-1)}{2}$. So, the next time someone asks you how many handshakes would have to take place in order for everyone in a room with 1029 people to shake each other's hands, you can proudly answer "528906 handshakes!"
Until next time,
Ben
## Wednesday, January 16, 2013
We've seen in the previous post what monads are and how they work. We saw a few monad instances of some common Haskell structures and toyed around with them a little bit. Towards the end of the post, I asked about making a Monad instance for a generalized tree data type. My guess is that it was relatively difficult. But why?
Well, one thing that I didn't touch on in my last post was that the monadic >>= operation can also be represented as (join . fmap f), that is, a >>= f = join \$fmap f a, as long as a has a Functor instance. join is a generalized function of the following type: $$join :: (Monad ~m) \Rightarrow m ~(m ~a) \rightarrow m ~a$$ Basically, this function takes a nested monad and "joins" it with the top level in order to get a regular m a out of it. Since you bind to functions that take as and produce m as, fmap f actually makes m (m a)s and join is just the tool to fix up the >>= function to produce what we want. Keep in mind that join is not a part of the Monad typeclass, and therefore the above definition for >>= will not work. However, if we are able to make a specific join function (named something else, since join is taken!) for whatever type we are making a Monad instance for, we can certainly use the above definition. I don't want to spend too much time on this, but I would like to direct the reader to the Monad instance for [] that I mentioned in the last post -- can you see any similarities between this and the way I structured >>= above? Now back to the Tree type. Can you devise some way to join Trees? That is, can you think of a way to flatten a Tree of Tree as into a Tree a? This actually turns out to be quite difficult, and up to interpretation as to how you want to do it. It's not as straightforward as moving Maybe (Maybe a)s into Maybe as or [[a]]s into [a]s. Maybe if we put a Monoid restriction on our a type, as such... $$instance ~(Monoid ~a) \Rightarrow Monad ~(Tree ~a) ~where ~\dots$$ ...we could use mappend in some way in order to concatenate all of the nested elements using some joining function. While this is a valid way to define a Tree Monad, it seems a little bit "unnatural." I won't delve too deeply into that, though, because that's not the point of this post. Let's instead take a look at a structure related to monads that may make more sense for our generalized Tree type. ### What is a comonad? Recall the type signature of >>= for Monads: $$(>>=) :: (Monad ~m) \Rightarrow m ~a \rightarrow ~(a \rightarrow m ~b) \rightarrow m ~b$$ That is, we're taking an m a and converting it to an m b by means of some function that operates on the contained type. In other words, we're producing a new value using elements contained inside the Monad. Comonads have a similar function: $$(=>>) :: (Comonad ~w) \Rightarrow w ~a \rightarrow (w ~a \rightarrow ~b) \rightarrow w ~b$$ The difference here is that the function that we use to produce a new value operates on the whole -- we're not operating on the elements contained inside the Comonad, but the Comonad itself, to produce a new value. We also have a function similar to the Monadic return: $$coreturn :: (Comonad ~w) \Rightarrow w ~a \rightarrow a$$ Whereas return puts a value into a Monadic context, coreturn extracts a value from a Comonadic context. I mentioned the Monadic join function above because I would also like to mention that there is a similar operation for Comonads: $$cojoin :: (Comonad ~w) \Rightarrow w ~a \rightarrow w ~(w ~a)$$ Instead of "removing a layer," we're "adding" a layer. And, as it turns out, just as a >>= f can be represented as join \$ fmap f a, =>> can be represented as:
$$a ~=>> ~f = fmap ~f ~\ ~cojoin ~a$$
The full Comonad typeclass (as I like to define it) is as follows:
$$class ~(Functor ~w) \Rightarrow Comonad ~w ~a ~where \\ \hspace{22pt}coreturn :: (Comonad ~w) \Rightarrow w ~a \rightarrow a \\ \hspace{38pt}cojoin :: (Comonad ~w) \Rightarrow w ~a \rightarrow w ~(w ~a) \\ a ~=>> ~f = fmap ~f ~\ ~cojoin ~a$$
By now, it should at least be clear that Monads and Comonads are related -- it shouldn't be hard to see why they are so similar in name!
Note: There is a package called Control.Comonad on Hackage. It uses different names for the Comonad operations, but they do the same things. It's a good package, but I wanted to show how Comonads are built and use the operation names I used to make things clearer.
### What can I do with Comonads?
As it turns out, the Tree a structure that I've been mentioning fits into the Comonadic context quite well, and provides a simple example as to how Comonads work.
We'll start with the implementation of the Comonad typeclass mentioned above:
Then we'll go ahead and make a Functor instance of our Tree a data type:
From here, we're able to make a Comonadic Tree like so:
The only real point of confusion here is in the cojoin function for Nodes. But, all we are doing is wrapping the entire node in a new node, and then mapping cojoin over every child node of the current node to produce its children. In other words, we're turning a Node a into a Node (Node a), which is exactly what we want to do.
So what can we do with a comonadic tree? Let's tackle a simple problem. Say we're hanging out in California, and we want to get to the East coast as quickly as possible. We don't really care where on the East coast we end up -- we just want to get there. We can map out the different paths that we can take, along with the time it takes to get to each one. In other words, we're going to model spots on the map as Nodes on a Tree and give them a weight corresponding to the time it takes to get there. We can model this situation with a Tree as follows:
The number of each Node marks its distance from the previous Node. The root Node of the Tree is the starting point, so it is 0 distance away from itself.
What we want to do to find the shortest path through the country is essentially, as follows. First, we're going to need to check the deepest nodes, and find their minimum distance children. We will add the distance of the Node closest to the one we're examining to its own distance, to find the shortest way to get from the node we're examining to the destination. Once all of that has been done, we'll need to traverse up the tree, repeating this as we go. By the end of the algorithm, the root Node will be marked with the shortest distance to the destination.
Now, that may sound somewhat iterative in nature, but we're going to morph this into a comonadic operation. First, let's take a look at a function we can use to find the minimum distance to the next level of our tree:
This is relatively simple, and does precisely what was mentioned above: adds the minimum value contained in the connected Nodes to the parent Node. Next, take a look at the type signature. We can see that this function produces a new number from a tree full of numbers. This coincides precisely with the function type we need to use with =>>, so we'll be able to use it to get exactly what we want.
The rest is very simple:
This produces a tree full of the minimum distances from each node to the East coast. Pulling the actual value out is as easy as calling coreturn on the resultant tree.
Ben
For further reading on Comonads, I recommend the article Evaluating Cellular Automata is Comonadic.
## Wednesday, January 2, 2013
### What is a Monad?
If you've used Haskell before, chances are that you've heard the term "monad" a good bit. But, you might not know what they are, or what they are really used for. It is my goal in this blog post to shed some light on what monads are, introduce and define few simple ones, and touch on their usefulness.
I will assume basic familiarity with Haskell syntax in this blog post. A working knowledge of monoids and functors will also be beneficial, although not strictly necessary.
Without further ado, let's go ahead and look at the Monad typeclass:
We can see a Monad has two operations, return and >>= (this is commonly pronounced "bind"). You may suspect that you know what return does from other programming languages, but be warned: Haskell's return is very different! Haskell's return only operates on Monads, and essentially acts as a "wrapping" function. That is, if you call return on a value, it will turn it into a monadic value of that type. We will look at an example of this shortly. The second operation, >>=, takes two arguments: an a wrapped in a Monad m (I will refer to this as m a from now on), and a function that converts an a to b wrapped in the same type of Monadm. It produces a value of type b, wrapped in a Monad m (I will call this m b from now on). This may sound complicated, but I'll do my best to explain it after showing the most basic Monad type, namely, the Identity Monad:
Also defined in Control.Monad.Identity
The Identity data declaration defines only one type: Identity a. Basically, this is just a wrapper for a value, and nothing more. return is simply defined as Identity. We can, for example, call return 3 and get an Identity 3. Turning values into Identitys is as simple as that. The bind function may look a little bit more obscure, though. Let's look closely: We first use pattern matching to be able to operate on the type that the Identity Monad wraps (Identity a), bind it (>>=) to a function (f). Since f converts normal values into monadic ones (look at the type declaration), we can simply apply f to a, and we've produced a new Identity. I've provided a couple of examples of how this works. addThree adds 3 to an Identity Int (m1) and getLength turns an Identity [a] (m2) into an Identity Int. Both of the bind examples produce the value Identity 6. Can you see why?
I want to take a little sidebar here and try to explain how I understand Monads before moving on. Basically the real tough part of understanding Monads is just understanding >>=. A little bit of dissection can explain a lot, but it took me months to really grasp the concept. On the surface, you're morphing an m a into an m b. At first, I thought that the m's in the type signature were allowed to different types of Monads. For instance, I thought that Monads made it possible to bind Identity Ints to functions that produced [String]s (we'll look at the list Monad in a minute). This is wrong, and for good reason! It was incredibly confusing to think about a function that could generalize to this degree and it is, to my knowledge, not possible to do so. The Monad (wrapper) type is retained, but the wrapped type can be changed by means of the function you bind to.
The second thing that I want to stress is what >>= really does. What we're doing with the >>= function is turning an m a into an m b, but it's not so direct as you might want to think. The function argument in >>= ( (a -> m b) ) was really confusing to me to begin with, so I'm going to try to explain it. The function you're binding your m a to must take an a and return an m b. This means that you're acting on the wrapped part of the Monad in order to produce an entirely new Monad. It took me a while to realize what exactly was happening, that I wasn't supposed to be directly converting m a to m b by means of my argument function (that's actually what >>= is for). Just remember that the function you bind to should operate on an a to produce an m b.
With that knowledge, let's take a look at some more examples of Monads:
If you're a Haskell user, you're likely already familiar with the Maybe data type:
Defined in the Prelude
The Maybe Monad is hardly an extension of the aforementioned Identity Monad. All we are really adding is the functionality to fail -- that is, to produce a Nothing value if necessary. In fact, the Just a type is exactly the same as Identity a, except that a Nothing value anywhere in the computation will produce a Nothing as a result. Next, we'll look at the Monad instance for lists:
Defined in the Prelude
The list Monad is a little more complicated, but it's really nothing too new. For return, we just have to wrap our value inside a list. Now, let's look at >>=concatMap actually is the >>= instance for lists. Why? Well, concatMap is, simply, (concat . map). We know that map takes a list of as and maps a function over them, and concat is a function that flattens lists, that is, turns [[a]]s into [a]s. When we compose these two functions (or use concatMap), we're making a function that must produce [[a]]s out of [a]s, and then flatten them back to [a]s. That is exactly how >>= works for lists! Now, let's take a look at a slightly more complicated Monad:
Also defined in Control.Monad.Instances
This one requires some explanation, because it's a bit of a leap from the last three, since we're defining a Monad instance for functions. I have added a couple of comments in the above code in order to further explain how everything is working in the (->) r Monad.
Well, first things's first. We define return as const, which takes some arbitrary argument and produces a function that produces a constant value (what we pass in to return). This is exactly what we want; a minimal context for functions.
(>>=) is a bit more complex. First, let's take a look at the type signature, specific to this Monad. We're binding a function (r -> a) to a function (a -> (r -> b)) to produce a function of type (r -> b). So in the declaration of the function, f is of type (r -> a) and g is of type (a -> (r -> b)). I've conveniently named the argument our function is going to take r, for clarity. This is what the user will pass into the function that (>>=) produces.
f takes an r, so we apply the value from the lambda into it to produce an a. We then use that a to produce an (r -> b), using g. This produces our function, but it needs to be fed some value, so we finally apply it to r in order to complete everything. If any of this sounds confusing, try to work through it on paper and read the comments in my code -- after a bit of staring, it should start to make sense.
I've also implemented the mean function here, using the (->) r Monad with and without do-notation. I think it's a pretty elegant way to express it.
We've only seen the tip of the iceberg when it comes to Monads, but this should serve as a good basic introduction. Once you get used to seeing Monads, they become a bit less scary. The more time you spend with them, the more comfortable you'll feel with them.
Before I end this post, I want to leave a question for the reader. Given the general Tree data type represented below, can you intuitively make a Monad instance?
I'll explain why this is difficult when we talk about a similar mathematical type in the next post. | 4,517 | 17,705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2018-26 | latest | en | 0.933276 |
http://www.jiskha.com/members/profile/posts.cgi?name=jane&page=9 | 1,498,282,366,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320226.61/warc/CC-MAIN-20170624050312-20170624070312-00150.warc.gz | 526,468,140 | 10,490 | # Posts by jane
Total # Posts: 1,650
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A block (m = 38.0kg) sits on an inclined plane of 38 degrees. If there is no friction,k what magnitude force is needed to keep the block from sliding? If there is friction, with coefficients 0.580 and 0.630, what is the maximum magnitude force that can be applied before the ...
english
there seems to be something wrong with the names a different one keeps coming up each time
english
wondering would you know this..what do we learn about juliet from this?
english
We first meet juliet (act 1 scene 3) she has a conversation with her mother about marriage. discuss this. what do we learn about juliet from this? ROMEO AND JULIET
math
Solve by factoring: 12 w(squared) = 28w +5
Math
Reiny, thank you so much. You broke it down so it was easy to understand.
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How do I set this up and how do I solve: The sum of two numbers is 20 The difference between 3 times the larger number and twice the smaller number is 40. What is the larger number?
math
Please show me how to set this up and how to solve: A ribbon with straight edges has an area of 24 inches squared. Its width is x and its length is 2x + 13. What is the width of the ribbon in inches?
Math
A school is fencing in a rectangular area for a playground. It plans to enclose the playground using fencing on three sides (One length is a wall) The school has budgeted enough money for 75 ft of fencing material and would like to make a playground with an area of 600 ft ...
physics
A car, mass - 1950 kg, accelerates at +2.33 m/s (squared). Find the magnitude of the normal force acting on the car. If later, when the car is cruising at 26.5 m/s, the driver applies a force of magnitude 10,000 N to stop the car. What is the distance it takes for the car to ...
Science
A 70kg person dives horizontally from a 200kg boat with a speed of 2 m/s. What is the recoil speed of the boat?
Math
Please show me how to set it up and solve: You are making a rectangular table. The area of the table should be 10ft squared. You want the length of the table to be 1ft shorter than twice its width. What should the dimensions of the able be?
math
Please show me how to set it up and solve: You are making a rectangular table. The area of the table should be 10ft squared. You want the length of the table to be 1ft shorter than twice its width. What should the dimensions of the able be?
Math
Please show me how to set it up and to solve: A cell phone company sells about 500 phones each week when it charges \$75 per phone. It sells about 20 more phones per week for each \$1 decrease in price. The company's revenue is the product of the number of phones sold and ...
Math
The area of a rectangle is 6n(to the power of 2) + n - 2. What is the expression that represents the perimeter of the rectangle?
Math
I'm sorry, its r(x) = (4x^2 + 1)/(4x^2 − 1); x = −1, 0, 1, ..., 9
Math
First, give the technology formula for the given function and then use technology to evaluate the function for the given values of x. (Round your answers to four decimal places.) r(x) = 4x2 + 1/4x2 − 1; x = −1, 0, 1, ..., 9 a)(4x^2 − 1)/(4x^2 + 1) b)(4x + 1)^...
English
In MLA (version 7), how do cite a textbook in a works cited page? Would it be the same as if you were citing an encyclopedia article or does it have a different citation?
Spanish
Yes
Math
Sometimes I get confused with the inequalities. You have \$47 to spend on music and movie downloads. Each album down load cost \$7 and each movie download costs \$8. Write and graph a linear inequality that represents this situation. Let x represent the number of albums and y the...
Math
Reid and Maria both play soccer. This season, Reid scored 4 less than twice the number of goals that Maria scored. The difference in the number of goals they scored was 6. How many goals did each of them score?
Solid mensuration
Find an equation of the circle tangent to x + y = 3 at (2,1) and with center on 3x - 2y - 6 = 0
Math
How do I set this up and solve? You are planing a rectangular dining pavilion. Its length is 3 times its width x. You want a stone walkway that is 3 ft wide around the pavilion. You have enough stones to cover 396 ft squared and want to use them all in the walkway. What should...
History
How does most state constitutions differ from the US Constitution?
History
The Bill of rights was ratifid by what fraction of all state legislatures?
History
What is the function of the Articles of the Constitution?
Math
Please explain to me how to solve: A circular mirror is surrounded by a square metal frame. The radius of the mirror is 5x. The side length of the metal frame is 15x. What is the area of the metal frame? Write your answer in factored form.
Math
A circular table is painted yellow with a red square in the middle. The radius of the tabletop is 6x. The side length of the red square is 3x. What is the area of the yellow part of the tabletop? Write your answer in factored form.
Math
How do I set this up and how to solve it? Russ bought 3 medium and 2 large submarine sandwiches for \$29.95. Stacy bought 4 medium and 1 large submarine sandwiches for \$28.45. What is the price for each medium and each large submarine sandwich?
math
Please explain how to solve: A circular table is painted yellow with a red square in the middle. The radius of the tabletop is 6x. The side length of the red square is 3x. What is the area of the yellow part of the tabletop? Write your answer in factored form.
Math
Please explain to me how to solve: A circular mirror is surrounded by a square metal frame. The radius of the mirror is 5x. The side length of the metal frame is 15x. What is the area of the metal frame? Write your answer in factored form.
basic math
I need to convert 40mg/mL to 5mg/mL with distilled water. How much distilled water would I use to do this?
Math
Please show me how to solve: Doctors can use radioactive iodine to treat some forms of cancer. The half-life of iodine - 131 is 8 days. A patient receives a treatment of 12 millicuries of iodine - 131. (A mullicurie is a unit of radioactivity.) How much iodine - 131 remains in...
Math
Hydra are small freshwater animals. They can double in number every two days in a laboratory tank. Suppose one tank has an initial population of 60 hydra. When will there be more than 5000 hydra? How can a table help you identify a pattern? What function models the situation...
math
Hydra are small freshwater animals. They can double in number every two days in a laboratory tank. Suppose one tank has an initial population of 60 hydra. When will there be more than 5000 hydra? How can a table help you identify a pattern? What function models the situation (...
social studies
who was the 8th US president
math
what is the square of 13
math
I've just begun to study Exponential functions: When you make a table starting with: -2 -4^x What are my coordinates? -1 0 1 2
Alg. I
I've just begun to study Exponential functions: When you make a table starting with: -2 -4^x What are my coordinates? -1 0 1 2
math
Finding a break-even point: How do I set up the equation and solve? Producing a musical costs \$88,000 plus \$59,000 per performance. One sold-out performance earns \$7500 in revenue. If every performance sells out, how many performances are needed to break even?
math
Sorry, Reiny, I copied the 2nd choice incorrectly. I had posted this problem with you before and the answer given that was not equalalent to A=1/2/bh was 2b =A/h. The problem is: A = 1/2bh, which is not equivalent? Choices: 2A =bh, h=2A/b, 2b=A/h, b=2A/h When I got my test ...
math
A recipe for trail mix calls for 3 cups of raisins and 2 cups of peanuts. If you plan on using 15 cups of raisins, how many cups of peanuts do you need?
Math
I had posted this problem with you before and the answer given was 2b =A/h. The problem is: A = 1/2bh, which is not correct? Choices: 2A =bh, b=2A/b, 2b=A/h, b=2A/h When I got my test back 2b=A/h was correct. Did I make a mistake or was their answer incorrect?
Algebra I
Exponential Function. Please show me how to solve: g(t) = -0.5 x 4w (that is 4 to the power of w) w=18
Math
Which equation is not equivalent to A = 1/2bh 2A=bh h=2A/b 2b=A/h b=2A/h I think it is 2b=A/h. Is this correct?
Math
The Venn diagram below shows the dinner orders for a local restaurant. This was submitted before, but no one answered. Thanks! How many people did NOT order steak for dinner? Steak, chicken & fish were order. In the Venn circles, the total that did not order steak was 56. ...
math
The Venn diagram below shows the dinner orders for a local restaurant. How many people did NOT order steak for dinner? Steak, chicken & fish were order. In the Venn circles, the total that did not order steak was 56. There is 43 in the right lower corner of the rectangle that ...
math
Please explain how to solve: You have 11 cups of flour. It takes 1 cup of flour to make 24 cookies. The function c(f) = 24f represents the number of cookies, c, that can be made with f cups of flour. What domain and range are reasonable for the function? My choices are: Domain...
Math
Given A = {2,4,6,8,10} B= {1,2,3,4,5,6,7,8,9,10} What is (A¿B) ? Is it {2,4,6,8,10}?
Math
Please explain how to solve: 1/3 = p-2/p+8
Math
What equation do you get when you solve for x? -g = sx I got -g/s = x. Is this correct?
Math
You have 11 cups of flour. It takes 1 cup of flour to make 24 cookies. The function c(f) = 24f represents the number of cookies, c, that can be made with f cups of flour. What domain and range are reasonable for the function?
Math
Sorry about the symbols before. Admission to the fair costs \$7.75. Each ride costs you \$0.50. You have \$15 to spend at the fair including admission. Which inequality represents the number of rides you can ride? r ¡Ý 15 (r is more than or equal to 15) r ¡&...
Math
Admission to the fair costs \$7.75. Each ride costs you \$0.50. You have \$15 to spend at the fair including admission. Which inequality represents the number of rides you can ride? r ¡Ý 15 r ¡Ü 14 r < 14 r > 14
Math
The table shows the cost of a ski rental package for a given number of people. Using the rate of change, what is the cost for 13 people? People Cost (\$) 2 70 3 105 4 140 5 175 There are 4 answer choices - 560, 440, 520, 480. I got 455. Am I wrong?
math
Given A = 2,4,6,8,10 B= 1,2,3,4,5,6,7,8,9,10 What is (A¿B) ?
Math
Explain how to solve? Which equation below shows the equation y = – 1/4x + 4 written in standard form using integers? A) 2x + 3y = 21 B) 3x – 2y = 21 C) –2x + 3y = 21 D) –2x – 3y = 21
math
What is the value of p in the proportion below? 1/3 = P-2/p+8
Programming
Language - PYTHON Design a program that calculates the total amount of a meal purchased at a restaurant. The program should ask the user to enter the charge for the food, and then calculate the amount of a 15% tip and 7% sales tax. Display each of these amounts and the total. ...
Math
A copy center offers its customers two different pricing plans for black and white photocopies of 8.5 in. by 11 in. pages. Customers can either pay \$0.08 per page or pay \$7.50 for a discount card that lowers the cost to \$0.05 per page. Which equation can be used to solve for ...
Math
The table shows the height of an elevator above ground level after a certain amount of time. Let y stand for the height of the elevator in feet and let x stand for the time in seconds. Which equation models the data below? Time (s) Height (ft) 10 235 20 220 40 190 60 160
math
Please explain in detail on how to solve: In 2000, the U.S. government owed about \$4.63 trillion to its creditors. The population of the US was282.4 million people. How much did the government owe per person in 2000? Round to the nearest dollar. In 2005, the debt had grown to ...
math
Please explain how to solve: The wavelength of a radio wave is defined as speed divided by frequency. An FM radio station has a frequency of 9 x 10(power of 7) waves per second. The speed of the waves is about 3 x 10 (power of 8) meters per second. What is the wavelength of ...
Math
Please explain in detail how to solve this: During one year, about 163 million adults over 18 years old in the U. S. spent a total of about 93 billion hours online at home. On average, how many hours per day did each adult spend online at home? How do you write each number in ...
Post a New Question | 4,950 | 18,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-26 | latest | en | 0.938376 |
http://www.edurite.com/kbase/formula-surface-area-pentagon | 1,464,358,300,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276780.5/warc/CC-MAIN-20160524002116-00153-ip-10-185-217-139.ec2.internal.warc.gz | 479,760,617 | 20,094 | #### • Class 11 Physics Demo
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# formula surface area pentagon
From Wikipedia
Pentagon
In geometry, a pentagon From the Greek number 5 (pente) is any five-sided polygon. A pentagon may be simple or self-intersecting. The internal angles in a simple pentagon total 540°. A pentagram is an example of a self-intersecting pentagon.
## Regular pentagons
A regular pentagon has all sides of equal length and all interior angles are equal measure (108°). It has five lines of reflectional symmetry and it has rotational symmetry of order 5 (through 72°, 144°, 216° and 288°). Its Schläfli symbol is {5}. The chords of a regular pentagon are in golden ratio to its sides.
The area of a regular convex pentagon with side length t is given by
A = \frac{4} = \frac{5t^2 \tan(54^\circ)}{4} \approx 1.720477401 t^2.
A pentagram or pentangle is a regularstar pentagon. Its Schläfli symbol is {5/2}. Its sides form the diagonals of a regular convex pentagon – in this arrangement the sides of the two pentagons are in the golden ratio.
When a regular pentagon is inscribed in a circle with radius R, its edge length t is given by the expression
t = R\ {\sqrt { \frac {5-\sqrt{5}}{2}} } = 2R\sin 36^\circ = 2R\sin\frac{\pi}{5} \approx 1.17557050458 R.
### Derivation of the area formula
The area of any regular polygon is:
A = \frac{1}{2}Pa
where P is the perimeter of the polygon, a is the apothem. One can then substitute the respective values for P and a, which makes the formula:
A = \frac{1}{2} \times \frac{5t}{1} \times \frac{t\tan(54^\circ)}{2}
with t as the given side length. Then we can then rearrange the formula as:
A = \frac{1}{2} \times \frac{5t^2\tan(54^\circ)}{2}
and then, we combine the two terms to get the final formula, which is:
A = \frac{5t^2\tan(54^\circ)}{4}.
### Derivation of the diagonal length formula
The diagonals of a regular pentagon (hereby represented by D) can be calculated based upon the golden ratio φ and the known side T (see discussion of the pentagon in Golden ratio):
\frac {D}{T} = \varphi = \frac {1+ \sqrt {5} }{2} \ ,
Accordingly:
D = T \times \varphi \ .
### Chords from the circumscribing circle to the vertices
If a regular pentagon with successive vertices A, B, C, D, E is inscribed in a circle, and if P is any point on that circle between points B and C, then PA + PD = PB + PC + PE.
## Construction of a regular pentagon
A variety of methods are known for constructing a regular pentagon. Some are discussed below.
#### Euclid's methods
A regular pentagon is constructible using a compass and straightedge, either by inscribing one in a given circle or constructing one on a given edge. This process was described by Euclid in his Elementscirca 300 BC.
#### Richmond's method
One method to construct a regular pentagon in a given circle is as follows:
#### Verification
The top panel describes the construction used in the animation above to create the side of the inscribed pentagon. The circle defining the pentagon has unit radius. Its center is located at point C and a midpoint M  is marked half way along its radius. This point is joined to the periphery vertically above the center at point D. Angle CMD  is bisected, and the bisector intersects the vertical axis at point Q. A horizontal line through Q intersects the circle at point P, and chord PD is the required side of the inscribed pentagon.
To determine the length of this side, the two right triangles DCM and QCM are depicted below the circle. Using Pythagoras' theorem and two sides, the hypotenuse of the larger triangle is found as √5/2. Side h of the smaller triangle then is found using the half-angle formula:
\tan ( \phi/2) = \frac{1-\cos(\phi)}{\sin (\phi)} \ ,
where cosine and sine of Ï• are known from the larger triangle. The result is:
h = \frac{\sqrt 5 - 1}{4} \ .
With this side known, attention turns to the lower diagram to find the side s of the regular pentagon. First, side a of the right-hand triangle is found using Pythagoras' theorem again:
a^2 = 1-h^2 \ ; \ a = \frac{1}{2}\sqrt { \frac {5+\sqrt 5}{2}} \ .
Then s is found using Pythagoras' theorem and the left-hand triangle as:
s^2 = (1-h)^2 + a^2 = (1-h)^2 + 1-h^2 = 1-2h+h^2 + 1-h^2 = 2-2h=2-2\left(\frac{\sqrt 5 - 1}{4}\right) \
=\frac {5-\sqrt 5}{2} \ .
The side s is therefore:
s = \sqrt{ \frac {5-\sqrt 5}{2}} \ ,
a well established result. Consequently, this construction of the pentagon is valid.
#### Alternative method
An alternative method is this:
1. Draw a circle in which to inscribe the pentagon and mark the center point O. (This is the green circle in the diagram to the right).
2. Choose a point A on the circle that will serve as one vertex of the pentagon. Draw a line through O and A.
3. Construct a line perpendicular to the line OA passing through O. Mark its intersection with one side of the circle as the point B.
4. Construct the point C as the midpoint of O and B.
5. Draw a circle centered at C through the point A. Mark its intersection with the line OB (inside the original circle) as the point D.
6. Draw a circle centered at A through the point D. Mark its intersections with the original (green) circle as the points E and F.
7. Draw a circle centered at E through
Surface area
Surface area is the measure of how much exposed area a solid object has, expressed in square units. Mathematical description of the surface area is considerably more involved than the definition of arc length of a curve. For polyhedra (objects with flat polygonal faces) the surface area is the sum of the areas of its faces. Smooth surfaces, such as a sphere, are assigned surface area using their representation as parametric surfaces. This definition of the surface area is based on methods of infinitesimal calculus and involves partial derivatives and double integration.
General definition of surface area was sought by Henri Lebesgue and Hermann Minkowski at the turn of the twentieth century. Their work led to the development of geometric measure theory which studies various notions of surface area for irregular objects of any dimension. An important example is the Minkowski content of a surface.
## Definition of surface area
While areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a lot of care. Surface area is an assignment
S \mapsto A(S)
of a positive real number to a certain class of surfaces that satisfies several natural requirements. The most fundamental property of the surface area is its additivity: the area of the whole is the sum of the areas of the parts. More rigorously, if a surface S is a union of finitely many pieces S1, …, Sr which do not overlap except at their boundaries then
A(S) = A(S_1) + \cdots + A(S_r).
Surface areas of flat polygonal shapes must agree with their geometrically defined area. Since surface area is a geometric notion, areas of congruent surfaces must be the same and area must depend only on the shape of the surface, but not on its position and orientation in space. This means that surface area is invariant under the group of Euclidean motions. These properties uniquely characterize surface area for a wide class of geometric surfaces called piecewise smooth. Such surfaces consist of finitely many pieces that can be represented in the parametric form
with continuously differentiable function \vec{r}. The area of an individual piece is defined by the formula
A(S_D) = \iint_D\left |\vec{r}_u\times\vec{r}_v\right | \, du \, dv.
Thus the area of SD is obtained by integrating the length of the normal vector \vec{r}_u\times\vec{r}_v to the surface over the appropriate region D in the parametric uv plane. The area of the whole surface is then obtained by adding together the areas of the pieces, using additivity of surface area. The main formula can be specialized to different classes of surfaces, giving, in particular, formulas for areas of graphs z = f(x,y) and surfaces of revolution.
One of the subtleties of surface area, as compared to arc length of curves, is that surface area cannot be defined simply as the limit of areas of polyhedral shapes approximating a given smooth surface. It was demonstrated by Hermann Schwarz that already for the cylinder, different choices of approximating flat surfaces can lead to different limiting values of the area.
Various approaches to general definition of surface area were developed in the late nineteenth and the early twentieth century by Henri Lebesgue and Hermann Minkowski. While for piecewise smooth surfaces there is a unique natural notion of surface area, if a surface is very irregular, or rough, then it may not be possible to assign any area at all to it. A typical example is given by a surface with spikes spread throughout in a dense fashion. Many surfaces of this type occur in the theory of fractals. Extensions of the notion of area which partially fulfill its function and may be defined even for very badly irregular surfaces are studied in the geometric measure theory. A specific example of such an extension is the Minkowski content of a surface.
## In chemistry
Surface area is important in chemical kinetics. Increasing the surface area of a substance generally increases the rate of a chemical reaction. For example, iron in a fine powder will combust, while in solid blocks it is stable enough to use in structures. For different applications a minimal or maximal surface area may be desired.
## In biology
The surface area of an organism is important in several considerations, such as regulation of body temperature and digestion. Animals use their teeth to grind food down into smaller particles, increasing the surface area available for digestion. The epithelial tissue lining the digestive tract contains microvilli, greatly increasing the area available for absorption. Elephants have large ears, allowing them to regulate their own body temperature. In other instances, animals will need to minimize surface area; for example, people will fold their arms over their chest when cold to minimize heat loss.
Question:What is the formula of finding the Sa of a cylinder,triangular, and pentagonal net? Please help !!
Answers:Surface Area of Cylinder: 4 r To Find the Surface Area of the Triangle Net: find the area for each triangle and then add all of them together. If the pentagonal net is regular, find the area one pentagon using A=ap (a=apothem, p=perimeter) then add those sums together.
Question:I forgot my formula chart.
Answers:2 pi r squared + 2 pi r h
Question:a regular pentagon on the unit sphere has 5 identical interior vertex angles, each equal to 170 degrees. what is the area of the Pentagon? explain your reasoning.
Answers:If M is the surface enclosed by the bounding pentagon M, then by the Gauss-Bonnet Theorem, [M] K dA + [ M] k ds = 2 (M), where (i) K = 1/r = 1 is the Gaussian curvature of the surface (which for a sphere, is constant); (ii) k = 0 along the geodesic parts of the boundary and is the multiple (10 /180) of a Dirac function at each vertex (measuring an exterior angle of 10 at the vertex); (iii) (M) = V E + F = 1 is the Euler characteristic of the surface. In this very special case, the Gauss-Bonnet formula simplifies to A + 5 (10 /180) = 2 1, where A is the area of M. So A = (2 - 50/180) = (310/180) = 31 /18. Note that the total surface area of the sphere is 4 = 72 /18. So M covers somewhat less than half the entire sphere.
Question:1. Wallpaper comes in rolls 6 metres long and 1 metre wide. How much wallpaper is necessary to wallpaper the walls in a bedroom if the bedroom is 4 metres by 5 metres by 2.5 metre?(Don't wallpaper the 1 metre by 2 metre door nor the 1 metre square window.) 2. How many square centimetres of cardboard go into the making of a cardboard box if the box is cube shaped and each of the sides is equal to 60 centimetres? 3. Mr. Fixit is building a laundry room in his basement. It will measure 4 metres by 5 metres by 2.5 metres. If he intends on painting the walls, ceiling and door with two coat of paint, and each litre of paint covers 70m^2, how many litres of paint should he buy? ignore question 1
Answers:I won't solve the problems for you but I'll be happy to give you the formulas for surface area. SA of a Cylinder:: 2*3.14*radius*height+2*3.14*radius^2 SA of Rectangular Solid:: 2(lw)+2(hw)+2(lh) | 3,064 | 12,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2016-22 | longest | en | 0.847099 |
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## statistics week 2 of notes
by: Katelyn Scott
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# statistics week 2 of notes Math-K310
Marketplace > Indiana University > Mathematics (M) > Math-K310 > statistics week 2 of notes
Katelyn Scott
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this consists of the second week of notes on sections 2.1, 2.2, 2.4, and 3.2
COURSE
Statistical Techniques
PROF.
Linda Krause
TYPE
Class Notes
PAGES
9
WORDS
CONCEPTS
Statistics
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## Popular in Mathematics (M)
This 9 page Class Notes was uploaded by Katelyn Scott on Thursday October 1, 2015. The Class Notes belongs to Math-K310 at Indiana University taught by Linda Krause in Summer 2015. Since its upload, it has received 49 views. For similar materials see Statistical Techniques in Mathematics (M) at Indiana University.
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Chapter 2 Characteristics of Data 2.1 1. Center- measures of center; a value that indicates where the middle of the data is 2. Variation- a measure of the amount that the data values vary 3. Distribution- the shape of the spread of the data over the range of values 4. Outliers- a data value that lies far away from the majority of other values EXAMPLE- Constructa groupedfrequencydistributionand cumulativefrequencydistributionwith5 classes. Test Data: 85 79 82 73 79 96 62 88 99 78 84 91 *Find the range: 99-62=37 *Find the class width: 37÷5= 7.4 (remember to round up to the nearest whole number for class widths) - class width = 8 Begin with the smallest number (62) Class Class boundaries tally frequency Cumulativefreq. < Class 62-69 61.5-69.5 l 1 1 69.5 70-77 69.5-77.5 l 1 2 77.5 78-85 77.5-85.5 l/ll l 6 8 85.5 86-93 85.5-93.5 ll 2 10 93.5 94-101 93.5-101.5 ll 2 12 101.5 • To find the class width: o Find the range = greatest value – least value o Divide by the number of classes desired o Always round up for class width, even if it is a whole number • Rule of thumb rounding rule o Round to 1 more decimal place than the given data 2.4 • Examples of bad graphs • Graph B appears double B A 90 90 A B 0 o Starting at 0 shows that this is not the case B A o Area of A = 1x1=1 o Area of B= 2x2=4 When doubling only double one dimension • Normal distribution o Approximately symmetric o Frequency starts and ends at low with highest point in the middle o Gaps in a distribution (frequency=0) may indicate data are from more than one population • A histogram is a bar graph. Each and every class is labeled across the bottom. The frequency is represented by vertical height. Vertical heights can be related to frequency • EXAMPLE: Using data from EX 1 construct a histogram using class boundaries and frequencies and then using class boundaries and relative frequencies. 7 7/12 6 6/12 5 5/12 4/12 4 3 3/12 2 2/12 1 1/12 0 0 61.5 69.5 77.5 88.5 93.5 101.5 61.5 69.5 77.5 88.5 93.5 101.5 o Class boundaries are in the x axis and the frequencies are in the y axis respectively • Types of graphs to learn o Histograms o Frequency polygon- x=midpoint, y= frequency/relative frequency o Ogive- x= class boundaries, y= cumulative frequency/ cumulative relative frequency o Dot plots- stack of x values o Stem leaf plot- tens l units--- like 70 = 7 l 0 • EXAMPLE: Graph a frequency polygon and Ogive using data from Ex 1. 7 Frequency Polygon Ogive 6 14 5 12 4 10 8 3 6 Frequency 2 4 Cumulative Frequency 1 0 0 61.5 69.5 77.5 85.5 93.5 101.5 65.5 73.5 81.5 89.5 97.5 Class Boundaries Midpoint • EXAMPLE:A test was administered to a group of thirty-six students.Their scores were as follows: 40 26 38 18 42 17 25 43 46 46 19 26 35 34 15 44 40 35 63 25 35 3 33 29 34 41 49 52 35 48 22 32 43 51 27 14 Construct a stemplot and a dot plot. 0 3 456789 1 2556679 2 Stem Plot 3 234455558 4 00123346689 5 12 6 3 0 10 20 30 40 50 60 • Do not use 2 or 3 dimensionaldata to represent 1 dimensional data • Do not start a graph at a nonzero number • Do not graph flawed data 3.2 • EXAMPLE: find the mean, median, midrange and mode of the data set. 17 12 13 11 14 18 17 18 16 15 Order the data points- 11 12 13 14 15 16 17 17 18 18 Mean= ∑data÷ 10 data values = 15.1 Median= (15+16)÷ 2 = 15.5 Mode=17, 18 Midrange= (11+18)÷ 2 = 14.5 • Measuring of Center o Mode- most often used number, there can be no mode or many modes o Mean- total of all values added together divided by the number of values Mean: x̄= (1 + 2 + …nx ) ÷ n o Median- the middle most value If the number of values is even, add the middle two numbers and divide then by 2 o Midrange- the average of the highest and the lowest values (high+low)÷2 = • Do this problem on your own before continuing. An answer key will be providedon the next page EXAMPLE: Find the mean of the data in the frequency distribution. 1 by hand , 2 by calculator. For 108 randomly selected college students, this exam score frequency distribution was obtained. For calculator TI 84you’ll need to go to stat > 1.Edit > insert midpoints into L1 > insert frequencies into L2 > stat > calc > 1 1-var stats > list: 2L1 > freqlist: 2 L2 > calculate > mean = x ̄ If this were not a frequency distribution you would leave freqlist blank Class limits Frequency midpoint midpointfrequency 90 – 98 6 99 – 107 22 108 – 116 43 117 – 125 28 126 – 134 9 • Answer Find the mean of the data in the frequency distribution. 1 by hand, 2 by calculator For 108 randomly selected college students, this exam score frequency distribution was obtained. Class limits Frequency midpoint midpoint x frequency 90 – 98 6 94 864 99 – 107 22 103 2266 108 – 116 43 112 4816 117 – 125 28 121 3388 126 – 134 9 130 1170 108 12204 Mean = 12204/108= 113.0 • Given sample data the mean is represented by x̄ o x̄= ∑x ÷ n • In a population it is represented by µ o µ = ∑x ÷ n • to find the median (n+1)÷ 2 is the location of it
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https://physicseasytips.com/relative-motion-concept-analysis/ | 1,571,488,813,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00301.warc.gz | 648,722,729 | 27,050 | Relative motion concept analysis | Physics Easy Tips
Relative motion concept analysis
March 7, 2018
March 8, 2018
Relative motion concept analysis
Today our topic is very interesting and important it is all about relative motion we have study in previous post about motion now be ready for relative motion concept and lets start.
we have study in motion topic if any body is not changing its position with respect to its surrounding then body is called at rest suppose you are holding a tray and on tray one plate with food and one glass with water and going to sit on your dining table now see here plate and glass are at rest with respect to tray while you are moving towards dining table why rest ? because distance between plate, glass are not changing but plate and glass are in motion with respect to dining table because when you are moving distance between plate, glass and dining table are changing. Important concept nothing in the universe is at absolute rest and same nothing is in absolute motion all are relative motion. Hence we need to study relative motion. same body can be at rest and same can be in motion relative to different body. This is called relative motion
I hope you have got the concept of relative motion.
Relative motion concept analysis
We know that earth is moving its means every object on the earth is moving then why we say object on the earth is at rest because we say with respect to surface of the earth suppose you are standing at a place the distance of earth surface now and after one minute is same the surface distance is not changing hence you are at rest while earth is moving. so any body at rest or in motion it is relative so we have to understand motion relative to which and at rest relative to which suppose two person are sitting in a moving train in front of each other and no any one is moving then first person is at rest relative to second person because distance between them from any point is not changing similarly second person is at rest with respect to first then train crosses from a platform and a third person is standing at platform seen both of them sitting in train are moving and crosses him so with respect to third person both are moving hence every motion is a relative motion. relative motion is always between two body one is mover and other is observer not only relative word in motion you are also relative how you are son of relative to your father, 20 kg mass is heavy relative to 5 kg mass you are intelligent relative to a dull student, you are old relative to a child, star is big relative to earth and many more every thing is relative in the world without relative word we can’t express other characteristic so this is important to understand hence we have to study when two body are in motion, in this case one has to move larger distance to catch other body because other body is also moving second body observe that first body is coming towards me see the below picture to understand relative motion here both body are moving.
Relative motion concept analysis
here velocity of body A is 5 m/s and velocity of body B is 4 m/s moving in same direction analyse here initial distance between two body is 5 m after one second A will travel 5 m distance and B will travel 4 m distance now at this moment distance between A and B will be 4 m now in next second again A will travel 5 m and B will travel 4 m hence the distance between them will be 3 m how in 2 second total distance travel by A will be 2*5 = 10 m and by B in 2 second distance travel = 4*2 = 8 m and add 5 m initial separation hence 8+5 = 13 hence separation between them 13-10 = 3 m hence every second A will be closer to B by one meter at the fifth second distance travel by A = 5*5 = 25 m and by B = 4*5 = 20 initial separation is 5 m hence at this moment both A and B will meet now for making this story simple in mathematical formula we suppose that both body taking as a pair and coming closer to each other now imagine as one body is at rest and other body is coming toward rest body but actually both are moving here body B seen body A and observe as body A is coming towards me at speed of 1 m/s body B imagine or experience i am at rest body A is coming towards me 1 m/s speed body B experience again and again body A is coming towards me by 1 m/s and will catch me in 5 second even body B forget i am moving with 4 m/s or body A moving with 5 m/s only body B experience body A is moving towards me by 1 m/s this thought of body B that body A is coming towards me by 1 m/s is relative velocity why relative velocity ? because for any other observer this velocity may be other but for B observer this velocity is 1 m/s and relative velocity is for observer hence in relative velocity we suppose one is observer and other is mover while both is moving so observer imagine that i am at rest and world is moving hence relative velocity is always calculated with respect to an observer this is the way of calculation.
hence B observe A is coming towards it with velocity V𝖆𝖇 here ab notation indicate b is observer and a is mover or b is looking a when a person A observe B then the relative velocity of B observe by A is V𝔟𝔞 so observer always observe other object is moving now calculation and the way of writing here both A and B are moving now velocity of A with respect to neutral body earth Vag here earth surface is at rest for both body when we told my car is moving with speed 60 km/hr then we talk about relative to earth surface which is at rest other car speed is 30 km/hr this is also relative to earth surface which is at rest so Vag is velocity of A with relative to ground surface and velocity of B relative to ground Vbg now Vab = Vag – Vbg hence relative velocity for above case is Vab = 5 – 4 = 1 m/s difference of the velocity suppose both object velocity is same then observer B will observe A is not moving because their difference will be zero while both are moving so observer is always subtracted in above equation and mover is at first place now velocity is vector quantity so we write as
→ → →
Vab = Vag – Vbg now where we use this relative velocity actual we use relative velocity everywhere see how we write and use
Vag = Va -Vg here Vg is velocity of ground surface which is always taken as zero Vg = 0 because earth surface is at rest hence Vag = Va so Va is absolute velocity which is relative to stationary frame of reference we always refer frame of reference as stationary everywhere above earth we are flying in air or on surface of earth we are moving on foot , by car, cycling we assume earth as stationary so for all the practical purpose velocity of earth ground is taken as zero when we will study about space then we will see earth is moving with certain velocity and will be consider at present earth is taken as stationary for our all movement on the surface of earth hence we will write Vag = Va absolute velocity with respect to ground hence we can write our original equation
→ → →
Vab = Va – Vb as simple relative velocity here observer is B if we take A as observe then our equation will be Vba = Vb-Va or
→ → →
Vba = Vb – Va the difference between two equation is only direction when B is observer see the A moving towards right and in case when observer is A see B moving towards left.
Now use of relative velocity we know that time = distance/speed but when two object are moving simultaneously we can’t use above formula because both object are moving hence we use relative speed to determine time now use relative velocity vab = Va – Vb put the value Vab = 5 – 4 = 1 m/s so using this velocity we can find the time when both body will meet now use formula .
time = distance/speed = 5/1 = 5 s here concept is the initial distance between two body is covered by relative speed not cover by actual speed other case if B is not moving then Vb = 0 or in this case A will catch B in just 1 second. hence when two body are moving then we take relative velocity for calculation in simple equation now see question.
Q A jet airplane is travelling with speed 600 km/hr eject its smoke at the speed of 1200 km/hr relative to jet what is speed of latter with respect to observer on the ground.
Ans here jet speed is given respect to ground so it is absolute value Vj = 600 km/hr now smoke speed relative to jet Vsj = 1200 km/hr now we have to find speed of smoke relative to ground hence use relative equation velocity of smoke will be in opposite direction of jet motion.
→ → →
Vsj = Vs – Vj now put the value
1200 = Vs – (-600) or 1200 = Vs +600 0r Vs = 1200 – 600 = 600
Vs = 600 km/hr observer will observe from ground.
Now we will see other case when both body are moving in opposite direction
Both body moving opposite direction .
5 m/s 4 m/s
A→ ←B
here A and B are moving in opposite direction then what will be relative velocity take A direction right +ve then B direction left -ve
now apply same formula Vab velocity of A with respect to B
→ → →
Vab = Va – Vb since this is vector quantity sign must be consider.
→ → → → → →
Vab = Va – (-Vb) or Vab = Va + Vb hence when body moving in opposite direction their relative velocity is added means relative velocity will be more and you have seen this practically when two train moving in opposite direction and crosses each other you feel or experience other train is moving very fast because both velocity of train is added which is observed by you is relative velocity.
we will continue some numerical problems in next post.
Relative motion concept analysis
Dated 29th April 2018
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# every inductive argument is __________.
6Deduction and Induction: Putting It All Together
Wavebreakmedia Ltd./Thinkstock and GoldenShrimp/iStock/Thinkstock
Learning Objectives After reading this chapter, you should be able to:
1. Compare and contrast the advantages of deduction and induction.
2. Explain why one might choose an inductive argument over a deductive argument.
3. Analyze an argument for its deductive and inductive components.
4. Explain the use of induction within the hypothetico–deductive method.
5. Compare and contrast falsification and confirmation within scientific inquiry.
6. Describe the combined use of induction and deduction within scientific reasoning.
7. Explain the role of inference to the best explanation in science and in daily life.
har85668_06_c06_207-238.indd 207 4/9/15 11:38 AM
Section 6.1 Contrasting Deduction and Induction
Now that you have learned something about deduction and induction, you may be wondering why we need both. This chapter is devoted to answering that question. We will start by learn- ing a bit more about the differences between deductive and inductive reasoning and how the two types of reasoning can work together. After that, we will move on to explore how scien- tific reasoning applies to both types of reasoning to achieve spectacular results. Arguments with both inductive and deductive elements are very common. Recognizing the advantages and disadvantages of each type can help you build better arguments. We will also investigate another very useful type of inference, known as inference to the best explanation, and explore its advantages.
6.1 Contrasting Deduction and Induction Remember that in logic, the difference between induction and deduction lies in the connec- tion between the premises and conclusion. Deductive arguments aim for an absolute connec- tion, one in which it is impossible that the premises could all be true and the conclusion false. Arguments that achieve this aim are called valid. Inductive arguments aim for a probable connection, one in which, if all the premises are true, the conclusion is more likely to be true than it would be otherwise. Arguments that achieve this aim are called strong. (For a discus- sion on common misconceptions about the meanings of induction and deduction, see A Closer Look: Doesn’t Induction Mean Going From Specific to General?). Recall from Chapter 5 that inductive strength is the counterpart of deductive validity, and cogency is the inductive coun- terpart of deductive soundness. One of the purposes of this chapter is to properly understand the differences and connections between these two major types of reasoning.
There is another important difference between deductive and inductive rea- soning. As discussed in Chapter 5, if you add another premise to an induc- tive argument, the argument may become either stronger or weaker. For example, suppose you are thinking of buying a new cell phone. After looking at all your options, you decide that one model suits your needs better than the others. New information about the phone may make you either more con- vinced or less convinced that it is the right one for you—it depends on what the new information is. With deductive reasoning, by contrast, adding prem- ises to a valid argument can never render it invalid. New information may show that a deductive argument
Fuse/Thinkstock
New information can have an impact on both deductive and inductive arguments. It can render deductive arguments unsound and can strengthen or weaken inductive arguments, such as arguments for buying one car over another.
har85668_06_c06_207-238.indd 208 4/9/15 11:38 AM
Section 6.1 Contrasting Deduction and Induction
is unsound or that one of its premises is not true after all, but it cannot undermine a valid connection between the premises and the conclusion. For example, consider the following argument:
All whales are mammals. Shamu is a whale. Therefore, Shamu is a mammal.
This argument is valid, and there is nothing at all we could learn about Shamu that would change this. We might learn that we were mistaken about whales being mammals or about Shamu being a whale, but that would lead us to conclude that the argument is unsound, not invalid. Compare this to an inductive argument about Shamu.
Whales typically live in the ocean. Shamu is a whale. Therefore, Shamu lives in the ocean.
Now suppose you learn that Shamu has been trained to do tricks in front of audiences at an amusement park. This seems to make it less likely that Shamu lives in the ocean. The addition of this new information has made this strong inductive argument weaker. It is, however, pos- sible to make it stronger again with the addition of more information. For example, we could learn that Shamu was part of a captive release program.
An interesting exercise for exploring this concept is to see if you can keep adding premises to make an inductive argument stronger, then weaker, then stronger again. For example, see if you can think of a series of premises that make you change your mind back and forth about the quality of the cell phone discussed earlier.
Determining whether an argument is deductive or inductive is an important step both in evaluating arguments that you encounter and in developing your own arguments. If an argu- ment is deductive, there are really only two questions to ask: Is it valid? And, are the premises true? If you determine that the argument is valid, then only the truth of the premises remains in question. If it is valid and all of the premises are true, then we know that the argument is sound and that therefore the conclusion must be true as well.
On the other hand, because inductive arguments can go from strong to weak with the addi- tion of more information, there are more questions to consider regarding the connection between the premises and conclusion. In addition to considering the truth of the premises and the strength of the connection between the premises and conclusion, you must also con- sider whether relevant information has been left out of the premises. If so, the argument may become either stronger or weaker when the relevant information is included.
Later in this chapter we will see that many arguments combine both inductive and deductive elements. Learning to carefully distinguish between these elements will help you know what questions to ask when evaluating the argument.
har85668_06_c06_207-238.indd 209 4/9/15 11:38 AM
Section 6.1 Contrasting Deduction and Induction
A Closer Look: Doesn’t Induction Mean Going From Specific to General? A common misunderstanding of the meanings of induction and deduction is that deduction goes from the general to the specific, whereas induction goes from the specific to the gen- eral. This definition is used by some fields, but not by logic or philosophy. It is true that some deductive arguments go from general premises to specific conclusions, and that some induc- tive arguments go from the specific premises to general conclusions. However, neither state- ment is true in general.
First, although some deductive arguments go from general to specific, there are many deduc- tive arguments that do not go from general to specific. Some deductive arguments, for exam- ple, go from general to general, like the following:
All S are M. All M are P. Therefore, all S are P.
Propositional logic is deductive, but its arguments do not go from general to specific. Instead, arguments are based on the use of connectives (and, or, not, and if . . . then). For example, modus ponens (discussed in Chapter 4) does not go from the general to the spe- cific, but it is deductively valid. When it comes to inductive arguments, some—for example, inductive generalizations—go from specific to general; others do not. Statistical syllogisms, for example, go from general to specific, yet they are inductive.
This common misunderstanding about the definitions of induction and deduction is not sur- prising given the different goals of the fields in which the terms are used. However, the defini- tions used by logicians are especially suited for the classification and evaluation of different types of reasoning.
For example, if we defined terms the old way, then the category of deductive reasoning would include arguments from analogy, statistical syllogisms, and some categorical syllogisms. Inductive reasoning, on the other hand, would include only inductive generalizations. In addi- tion, there would be other types of inference that would fit into neither category, like many categorical syllogisms, inferences to the best explanation, appeals to authority, and the whole field of propositional logic.
The use of the old definitions, therefore, would not clear up or simplify the categories of logic at all but would make them more confusing. The current distinction, based on whether the premises are intended to guarantee the truth of the conclusion, does a much better job of simplifying logic’s categories, and it does so based on a very important and relevant distinction.
har85668_06_c06_207-238.indd 210 4/9/15 11:38 AM
Section 6.2 Choosing Between Induction and Deduction
Practice Problems 6.1
1. A deductive argument that establishes an absolute connection between the premises and conclusion is called a __________. a. strong argument b. weak argument c. invalid argument d. valid argument
2. An inductive argument whose premises give a lot of support for the truth of its con- clusion is said to be __________. a. strong b. weak c. valid d. invalid
3. Inductive arguments always reason from the specific to the general. a. true b. false
4. Deductive arguments always reason from the general to the specific. a. true b. false
6.2 Choosing Between Induction and Deduction You might wonder why one would choose to use inductive reasoning over deductive reason- ing. After all, why would you want to show that a conclusion was only probably true rather than guaranteed to be true? There are several reasons, which will be discussed in this sec- tion. First, there may not be an available deductive argument based on agreeable premises. Second, inductive arguments can be more robust than deductive arguments. Third, inductive arguments can be more persuasive than deductive arguments.
Availability Sometimes the best evidence available does not lend itself to a deductive argument. Let us consider a readily accepted fact: Gravity is a force that pulls everything toward the earth. How would you provide an argument for that claim? You would probably pick something up, let go of it, and note that it falls toward the earth. For added effect, you might pick up several things and show that each of them falls. Put in premise–conclusion form, your argument looks something like the following:
My coffee cup fell when I let go of it. My wallet fell when I let go of it. This rock fell when I let go of it. Therefore, everything will fall when I let go of it.
har85668_06_c06_207-238.indd 211 4/9/15 11:38 AM
Section 6.2 Choosing Between Induction and Deduction
When we put the argument that way, it should be clear that it is inductive. Even if we grant that the premises are true, it is not guaranteed that every- thing will fall when you let go of it. Perhaps grav- ity does not affect very small things or very large things. We could do more experiments, but we can- not check every single thing to make sure that it is affected by gravity. Our belief in gravity is the result of extremely strong inductive reasoning. We there- fore have great reasons to believe in gravity, even if our reasoning is not deductive.
All subjects that rely on observation use induc- tive reasoning: It is at least theoretically possible that future observations may be totally different than past ones. Therefore, our inferences based on observation are at best probable. It turns out that there are very few subjects in which we can pro- ceed entirely by deductive reasoning. These tend to be very abstract and formal subjects, such as math- ematics. Although other fields also use deductive reasoning, they do so in combination with inductive reasoning. The result is that most fields rely heavily on inductive reasoning.
Robustness Inductive arguments have some other advantages over deductive arguments. Deductive argu- ments can be extremely persuasive, but they are also fragile in a certain sense. When some- thing goes wrong in a deductive argument, if a premise is found to be false or if it is found to be invalid, there is typically not much of an argument left. In contrast, inductive arguments tend to be more robust. The robustness of an inductive argument means that it is less fragile; if there is a problem with a premise, the argument may become weaker, but it can still be quite persuasive. Deductive arguments, by contrast, tend to be completely unconvincing once they are shown not to be sound. Let us work through a couple of examples to see what this means in practice.
Consider the following deductive argument:
All dogs are mammals. Some dogs are brown. Therefore, some mammals are brown.
As it stands, the argument is sound. However, if we change a premise so that it is no longer sound, then we end up with an argument that is nearly worthless. For example, if you change the first premise to “Most dogs are mammals,” you end up with an invalid argument. Valid- ity is an all-or-nothing affair; there is no such thing as “sort of valid” or “more valid.” The
Alistair Scott/iStock/Thinkstock
Despite knowing that a helium-filled balloon will rise when we let go of it, we still hold our belief in gravity due to strong inductive reasoning and our reliance on observation.
har85668_06_c06_207-238.indd 212 4/9/15 11:38 AM
Section 6.2 Choosing Between Induction and Deduction
argument would simply be invalid and therefore unsound; it would not accomplish its pur- pose of demonstrating that the conclusion must be true. Similarly, if you were to change the second premise to something false, like “Some dogs are purple,” then the argument would be unsound and therefore would supply no reason to accept the conclusion.
In contrast, inductive arguments may retain much of their strength even when there are prob- lems with them. An inductive argument may list several reasons in support of a conclusion. If one of those reasons is found to be false, the other reasons continue to support the conclu- sion, though to a lesser degree. If an argument based on statistics shows that a particular conclusion is extremely likely to be true, the result of a problem with the argument may be that the conclusion should be accepted as only fairly likely. The argument may still give good reasons to accept the conclusion.
Fields that rely heavily on statistical arguments often have some threshold that is typically required in order for results to be publishable. In the social sciences, this is typically 90% or 95%. However, studies that do not quite meet the threshold can still be instructive and pro- vide evidence for their conclusions. If we discover a flaw that reduces our confidence in an argument, in many cases the argument may still be strong enough to meet a threshold.
As an example, consider a tweet made by President Barack Obama regarding climate change.
Although the tweet does not spell out the argument fully, it seems to have the following structure:
A study concluded that 97% of scientists agree that climate change is real, man-made, and dangerous. Therefore, 97% of scientists really do agree that climate change is real, man- made, and dangerous. Therefore, climate change is real, man-made, and dangerous.
Given the politically charged nature of the discussion of climate change, it is not surprising that the president’s argument and the study it referred to received considerable criticism. (You can read the study at http://iopscience.iop.org/1748–9326/8/2/024024/pdf/1748 –9326_8_2_024024.pdf.) Looking at the effect some of those criticisms have on the argument is a good way to see how inductive arguments can be more robust than deductive ones.
One criticism of Obama’s claim is that the study he referenced did not say anything about whether climate change was dangerous, only about whether it was real and man-made. How does this affect the argument? Strictly speaking, it makes the first premise false. But notice that even so, the argument can still give good evidence that climate change is real and
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Section 6.2 Choosing Between Induction and Deduction
man-made. Since climate change, by its nature, has a strong potential to be dangerous, the argument is weakened but still may give strong evidence for its conclusion.
A deeper criticism notes that the study did not find out what all scientists thought; it just looked at those scientists who expressed an opinion in their published work or in response to a voluntary survey. This is a significant criticism, for it may expose a bias in the sampling method (as discussed in Chapters 5, 7, and 8). Even granting the criticism, the argument can retain some strength. The fact that 97% of scientists who expressed an opinion on the issue said that climate change is real and man-made is still some reason to think that it is real and man-made. Of course, some scientists may have chosen not to voice an opposing opinion for reasons that have nothing to do with their beliefs about climate change; they may have simply wanted to keep their views private, for example. Taking all of this into account, we get the fol- lowing argument:
A study found that 97% of scientists who stated their opinion said that cli- mate change is real and man-made. Therefore, 97% of scientists agree that climate change is real and man-made. Climate change, if real, is dangerous. Therefore, climate change is real, man-made, and dangerous.
This is not nearly as strong as the original argument, but it has not collapsed entirely in the way a purely deductive argument would. There is, of course, much more that could be said about this argument, both in terms of criticizing the study and in terms of responding to those criticisms and bringing in other considerations. The point here is merely to highlight the dif- ference between deductive and inductive arguments, not to settle issues in climate science or public policy.
Persuasiveness A final point in favor of inductive reasoning is that it can often be more persuasive than deduc- tive reasoning. The persuasiveness of an argument is based on how likely it is to convince someone of the truth of its conclusion. Consider the following classic argument:
All Greeks are mortal. Socrates was a Greek. Therefore, Socrates was mortal.
Is this a good argument? From the standpoint of logic, it is a perfect argument: It is deduc- tively valid, and its premises are true, so it is sound (therefore, its conclusion must be true). However, can you persuade anyone with this argument?
Imagine someone wondering whether Socrates was mortal. Could you use this argument to convince him or her that Socrates was mortal? Probably not. The argument is so simple and
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Section 6.2 Choosing Between Induction and Deduction
so obviously valid that anyone who accepts the premises likely already accepts the conclu- sion. So if someone is wondering about the conclusion, it is unlikely that he or she will be persuaded by these premises. He or she may, for example, remember that some legendary Greeks, such as Hercules, were granted immortality and wonder whether Socrates was one of these. The deductive approach, therefore, is unlikely to win anyone over to the conclusion here. On the other hand, consider a very similar inductive argument.
Of all the real and mythical Greeks, only a few were considered to be immortal. Socrates was a Greek. Therefore, it is extremely unlikely that Socrates was immortal.
Again, the reasoning is very simple. However, in this case, we can imagine someone who had been wondering about Socrates’s mortality being at least somewhat persuaded that he was mortal. More will likely need to be said to fully persuade her or him, but this simple argument may have at least some persuasive power where its deductive version likely does not.
Of course, deductive arguments can be persuasive, but they generally need to be more com- plicated or subtle in order to be so. Persuasion requires that a person change his or her mind to some degree. In a deductive argument, when the connection between premises and conclu- sion is too obvious, the argument is unlikely to persuade because the truth of the premises will be no more obvious than the truth of the conclusion. Therefore, even if the argument is valid, someone who questions the truth of the conclusion will often be unlikely to accept the truth of the premises, so she or he may be unpersuaded by the argument. Suppose, for example, that we wanted to convince someone that the sun will rise tomorrow morning. The deductive argument may look like this:
The sun will always rise in the morning. Therefore, the sun will rise tomorrow morning.
One problem with this argument, as with the Socrates argument, is that its premise seems to assume the truth of the conclusion (and therefore commits the fallacy of begging the ques- tion, as discussed in Chapter 7), making the argument unpersuasive. Additionally, however, the premise might not even be true. What if, billions of years from now, the earth is swallowed up into the sun after it expands to become a red giant? At that time, the whole concept of morning may be out the window. If this is true then the first premise may be technically false. That means that the argument is unsound and therefore fairly worthless deductively.
The inductive version, however, does not lose much strength at all after we learn of this trou- bling information:
The sun has risen in the morning every day for millions of years. Therefore, the sun will rise again tomorrow morning.
This argument remains extremely strong (and persuasive) regardless of what will happen billions of years in the future.
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Section 6.3 Combining Induction and Deduction
Practice Problems 6.2
1. Which form of reasoning is taking place in this example?
The sun has risen every day of my life. The sun rose today. Therefore, the sun will rise tomorrow.
a. inductive b. deductive
2. Inductive arguments __________. a. can retain strength even with false premises b. collapse when a premise is shown to be false c. are equivalent to deductive arguments d. strive to be valid
3. Deductive arguments are often __________. a. less persuasive than inductive arguments b. more persuasive than inductive arguments c. weaker than inductive arguments d. less valid than inductive arguments
4. Inductive arguments are sometimes used because __________. a. the available evidence does not allow for a deductive argument b. they are more likely to be sound than deductive ones c. they are always strong d. they never have false premises
6.3 Combining Induction and Deduction You may have noticed that most of the examples we have explored have been fairly short and simple. Real-life arguments tend to be much longer and more complicated. They also tend to mix inductive and deductive elements. To see how this might work, let us revisit an example from the previous section.
All Greeks are mortal. Socrates was Greek. Therefore, Socrates was mortal.
As we noted, this simple argument is valid but unlikely to convince anyone. So suppose now that someone questioned the premises, asking what reasons there are for thinking that all Greeks are mortal or that Socrates was Greek. How might we respond?
We might begin by noting that, although we cannot check each and every Greek to be sure he or she is mortal, there are no documented cases of any Greek, or any other human, living more
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Section 6.3 Combining Induction and Deduction
than 200 years. In contrast, every case that we can document is a case in which the person dies at some point. So, although we cannot absolutely prove that all Greeks are mortal, we have good reason to believe it. We might put our argument in standard form as follows:
We know the mortality of a huge number of Greeks. In each of these cases, the Greek is mortal. Therefore, all Greeks are mortal.
This is an inductive argument. Even though it is theoretically possible that the conclusion might still be false, the premises provide a strong reason to accept the conclusion. We can now combine the two arguments into a single, larger argument:
We know the mortality of a huge number of Greeks. In each of these cases, the Greek is mortal. Therefore, all Greeks are mortal. Socrates was Greek. Therefore, Socrates was mortal.
This argument has two parts. The first argument, leading to the subconclusion that all Greeks are mortal, is inductive. The second argument (whose conclusion is “Socrates was mortal”) is deductive. What about the overall reasoning presented for the conclusion that Socrates was mortal (combining both arguments); is it inductive or deductive?
The crucial issue is whether the prem- ises guarantee the truth of the conclu- sion. Because the basic premise used to arrive at the conclusion is that all of the Greeks whose mortality we know are mortal, the overall reasoning is inductive. This is how it generally works. As noted earlier, when an argu- ment has both inductive and deductive components, the overall argument is generally inductive. There are occa- sional exceptions to this general rule, so in particular cases, you still have to check whether the premises guarantee the conclusion. But, almost always, the longer argument will be inductive.
Fran/Cartoonstock
Sometimes a simple deductive argument needs to be combined with a persuasive inductive argument to convince others to accept it.
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Section 6.4 Reasoning About Science: The Hypothetico–Deductive Method
A similar thing happens when we combine inductive arguments of different strength. In gen- eral, an argument is only as strong as its weakest part. You can think of each inference in an argument as being like a link in a chain. A chain is only as strong as its weakest link.
6.4 Reasoning About Science: The Hypothetico– Deductive Method Science is one of the most successful endeavors of the modern world, and arguments play a central role in it. Science uses both deductive and inductive reasoning extensively. Scientific reasoning is a broad field in itself—and this chapter will only touch on the basics—but dis- cussing scientific reasoning will provide good examples of how to apply what we have learned about inductive and deductive arguments.
At some point, you may have learned or heard of the scientific method, which often refers to how scientists systematically form, test, and modify hypotheses. It turns out that there is not a single method that is universally used by all scientists.
In a sense, science is the ultimate critical thinking experiment. Scientists use a wide variety of reasoning techniques and are constantly examining those techniques to make sure that the conclusions drawn are justified by the premises—that is exactly what a good critical thinker should do in any subject. The next two sections will explore two such methods—the hypothetico–deductive method and inferences to the best explanation—and discover ways that they can improve our understanding of the types of reasoning used in much of science.
The hypothetico–deductive method consists of four steps:
1. Formulate a hypothesis. 2. Deduce a consequence from the hypothesis. 3. Test whether the consequence occurs. 4. Reject the hypothesis if the consequence does not occur.
Although these four steps are not sufficient to explain all scientific reasoning, they still remain a core part of much discussion of how science works. You may recognize them as part of the scientific method that you likely learned about in school. Let us take a look at each step in turn.
Practice Problem 6.3
1. When an argument contains both inductive and deductive elements, the entire argu- ment is considered deductive. a. true b. false
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Section 6.4 Reasoning About Science: The Hypothetico–Deductive Method
Step 1: Formulate a Hypothesis A hypothesis is a conjecture about how some part of the world works. Although the phrase “educated guess” is often used, it can give the impression that a hypothesis is simply guessed without much effort. In reality, scientific hypotheses are formulated on the basis of a back- ground of quite a bit of knowledge and experience; a good scientific hypothesis often comes after years of prior investigation, thought, and research about the issue at hand.
You may have heard the expression “necessity is the mother of invention.” Often, hypotheses are formulated in response to a problem that needs to be solved. Suppose you are unsatisfied with the performance of your car and would like better fuel economy. Rather than buy a new car, you try to figure out how to improve the one you have. You guess that you might be able to improve your car’s fuel economy by using a higher grade of gas. Your guess is not just random; it is based on what you already know or believe about how cars work. Your hypothesis is that higher grade gas will improve your fuel economy.
Of course, science is not really concerned with your car all by itself. Science is concerned with general principles. A scientist would reword your hypothesis in terms of a general rule, something like, “Increasing fuel octane increases fuel economy in automobiles.” The hypothetico–deductive method can work with either kind of hypothesis, but the general hypothesis is more interesting scientifically.
Step 2: Deduce a Consequence From the Hypothesis Your hypothesis from step 1 should have predictive value: Things should be different in some noticeable way, depending on whether the hypothesis is true or false. Our hypothesis is that increasing fuel octane improves fuel economy. If this general fact is true, then it is true for your car. So from our general hypothesis we can deduce the consequence that your car will get more miles per gallon if it is running on higher octane fuel.
It is often but not always the case that the prediction is a more specific case of the hypothesis. In such cases it is possible to infer the prediction deductively from the general hypothesis. The argument may go as follows:
Hypothesis: All things of type A have characteristic B.
Consequence (the prediction): Therefore, this specific thing of type A will have characteristic B.
Since the argument is deductively valid, there is a strong connection between the hypothesis and the prediction. However, not all predictions can be deductively inferred. In such cases we can get close to the hypothetico–deductive method by using a strong inductive inference instead. For example, suppose the argument went as follows:
Hypothesis: 95% of things of type A have characteristic B.
Consequence: Therefore, a specific thing of type A will probably have charac- teristic B.
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Section 6.4 Reasoning About Science: The Hypothetico–Deductive Method
In such cases the connection between the hypothesis and the prediction is less strong. The stronger the connection that can be established, the better for the reliability of the test. Essen- tially, you are making an argument for the conditional statement “If H, then C,” where H is your hypothesis and C is a consequence of the hypothesis. The more solid the connection is between H and C, the stronger the overall argument will be.
In this specific case, “If H, then C” translates to “If increasing fuel octane increases fuel econ- omy in all cars, then using higher octane fuel in your car will increase its fuel economy.” The truth of this conditional is deductively certain.
We can now test the truth of the hypothesis by testing the truth of the consequence.
Step 3: Test Whether the Consequence Occurs Your prediction (the consequence) is that your car will get better fuel economy if you use a higher grade of fuel. How will you test this? You may think this is obvious: Just put better gas in the car and record your fuel economy for a period before and after changing the type of gas you use. However, there are many other factors to consider. How long should the period of time be? Fuel economy varies depending on the kind of driving you do and many other factors. You need to choose a length of time for which you can be reasonably confident the driving conditions are similar on average. You also need to account for the fact that the first tank of better gas you put in will be mixed with some of the lower grade gas that is still in your tank. The more you can address these and other issues, the more certain you can be that your conclusion is correct.
In this step, you are constructing an inductive argument from the outcome of your test as to whether your car actually did get better fuel economy. The arguments in this step are induc- tive because there is always some possibility that you have not adequately addressed all of the relevant issues. If you do notice better fuel economy, it is always possible that the increase in economy is due to some factor other than the one you are tracking. The possibility may be very small, but it is enough to make this kind of argument inductive rather than deductive.
Step 4: Reject the Hypothesis If the Consequence Does Not Occur We now compare the results to the prediction and find out if the prediction came true. If your test finds that your car’s fuel economy does not improve when you use higher octane fuel, then you know your prediction was wrong.
Does this mean that your hypothesis, H, was wrong? That depends on the strength of the con- nection between H and C. If the inference from H to C is deductively certain, then we know for sure that, if H is true, then C must be true also. Therefore, if C is false, it follows logically that H must be false as well.
In our specific case, if your car does not get better fuel economy by switching to higher octane fuel, then we know for sure that it is not true that all cars get better fuel economy by doing so. However, if the inference from H to C is inductive, then the connection between H and C is less than totally certain. So if we find that C is false, we are not absolutely sure that the hypothesis, H, is false.
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Section 6.4 Reasoning About Science: The Hypothetico–Deductive Method
For example, suppose that the hypothesis is that cars that use higher octane fuel will have a higher tendency to get better fuel mileage. In that case if your car does not get higher gas mileage, then you still cannot infer for certain that the hypothesis is false. To test that hypothesis adequately, you would have to do a large study with many cars. Such a study would be much more complicated, but it could provide very strong evidence that the hypoth- esis is false.
It is important to note that although the falsity of the prediction can dem- onstrate that the hypothesis is false, the truth of the prediction does not prove that the hypothesis is true. If you find that your car does get better fuel economy when you switch gas, you cannot conclude that your hypothesis is true.
Why? There may be other factors at play for which you have not ade- quately accounted. Suppose that at the same time you switch fuel grade, you also get a tune-up and new tires and start driving a completely different route to work. Any one of these things might be the cause of the improved gas mileage; you cannot conclude that it is due to the change in fuel (for this rea- son, when conducting experiments it is best to change only one variable at a time and carefully control the rest). In
other words, in the hypothetico–deductive method, failed tests can show that a hypothesis is wrong, but tests that succeed do not show that the hypothesis was correct.
This logic is known as falsification; it can be demonstrated clearly by looking at the structure of the argument. When a test yields a negative result, the hypothetico–deductive method sets up the following argument:
If H, then C. Not C. Therefore, not H.
You may recognize this argument form as modus tollens, or denying the consequent, which was discussed in the chapter on propositional logic (Chapter 4). This argument form is a valid, deductive form. Therefore, if both of these premises are true, then we can be certain that the conclusion is true as well; namely, that our hypothesis, H, is not true. In the specific case at hand, if your test shows that higher octane fuel does not increase your mileage, then we can be sure that it is not true that it improves mileage in all vehicles (though it may improve it in some).
IPGGutenbergUKLtd/iStock/Thinkstock
At best, the fuel economy hypothesis will be a strong inductive argument because there is a chance that something other than higher octane gas is improving fuel economy. The more you can address relevant issues that may impact your test results, the stronger your conclusions will be.
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Section 6.4 Reasoning About Science: The Hypothetico–Deductive Method
Contrast this with the argument form that results when your fuel economy yields a positive result:
If H, then C. C. Therefore, H.
This argument is not valid. In fact, you may recognize this argument form as the invalid deduc- tive form called affirming the consequent (see Chapter 4). It is possible that the two premises are true, but the conclusion false. Perhaps, for example, the improvement in fuel economy was caused by a change in tires or different driving conditions instead. So the hypothetico –deductive method can be used only to reject a hypothesis, not to confirm it. This fact has led many to see the primary role of science to be the falsification of hypotheses. Philosopher Karl Popper is a central source for this view (see A Closer Look: Karl Popper and Falsification in Science).
A Closer Look: Karl Popper and Falsification in Science Karl Popper, one of the most influential philosophers of sci- ence to emerge from the early 20th century, is perhaps best known for rejecting the idea that scientific theories could be proved by simply finding confirming evidence—the prevail- ing philosophy at the time. Instead, Popper emphasized that claims must be testable and falsifiable in order to be consid- ered scientific.
A claim is testable if we can devise a way of seeing if it is true or not. We can test, for instance, that pure water will freeze at 0°C at sea level; we cannot currently test the claim that the oceans in another galaxy taste like root beer. We have no realistic way to determine the truth or falsity of the second claim.
A claim is said to be falsifiable if we know how one could show it to be false. For instance, “there are no wild kangaroos in Georgia” is a falsifiable claim; if one went to Georgia and found some wild kangaroos, then it would have been shown to be false. But what if someone claimed that there are ghosts in Georgia but that they are imperceptible (unseeable, unfeel- able, unhearable, etc.)? Could one ever show that this claim is false? Since such a claim could not conceivably be shown to be false, it is said to be unfalsifiable. While being unfalsifiable might sound like a good thing, according to Popper it is not, because it means that the claim is unscientific.
Following Popper, most scientists today operate with the assumption that any scientific hypothesis must be testable and must be the kind of claim that one could possibly show to be false. So if a claim turns out not to be conceivably falsifiable, the claim is not really scientific—and some philosophers have gone so far as to regard such claims as meaningless (Thornton, 2014).
Keystone/Getty Images
Karl Popper, a 20th- century philosopher of science, put forth the idea that unfalsifiable claims are unscientific.
(continued)
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Section 6.4 Reasoning About Science: The Hypothetico–Deductive Method
As an example, suppose a friend claims that “everything works out for the best.” Then suppose that you have the worst month of your life, and you go back to your friend and say that the claim is false: Not everything is for the best. Your friend might then reply that in fact it was for the best because you learned from the experience. Such a statement may make you feel better, but it runs afoul of Popper’s rule. Can you imagine any circumstance that your friend would not claim is for the best? Since your friend would probably say that it was for the best no mat- ter what happens, your friend’s claim is unfalsifiable and therefore unscientific.
In logic, claims that are interpreted so that they come out true no matter what happens are called self-sealing propositions. They are understood as being internally protected against any objections. People who state such claims may feel that they are saying something deeply meaningful, but according to Popper’s rule, since the claim could never be falsified no matter what, it does not really tell us anything at all.
Other examples of self-sealing propositions occur within philosophy itself. There is a philo- sophical theory known as psychological egoism, for example, which teaches that everything everyone does is completely selfish. Most people respond to this claim by coming up with examples of unselfish acts: giving to the needy, spending time helping others, and even dying to save someone’s life. The psychological egoist predictably responds to all such examples by stating that people who do such things really just do them in order to feel better about them- selves. It appears that the word selfish is being interpreted so that everything everyone does will automatically be considered selfish by definition. It is therefore a self-sealing claim (Rachels, 1999). According to Popper’s method, since this claim will always come out true no mat- ter what, it is unfalsifiable and unscientific. Such claims are always true but are actually empty because they tell us nothing about the world. They can even be said to be “too true to be good.”
Popper’s explorations of scientific hypotheses and what it means to confirm or disconfirm such hypotheses have been very influential among both scientists and philosophers of scien- tists. Scientists do their best to avoid making claims that are not falsifiable.
A Closer Look: Karl Popper and Falsification in Science (continued)
If the hypothetico-deductive method cannot be used to confirm a hypothesis, how can this test give evidence for the truth of the claim? By failing to falsify the claim. Though the hypo- thetico–deductive method does not ever specifically prove the hypothesis true, if research- ers try their hardest to refute a claim but it keeps passing the test (not being refuted), then there can grow a substantial amount of inductive evidence for the truth of the claim. If you repeatedly test many cars and control for other variables, and if every time cars are filled with higher octane gas their fuel economy increases, you may have strong inductive evidence that the hypothesis might be true (in which case you may make an inference to the best explana- tion, which will be discussed in Section 6.5).
Experiments that would have the highest chance of refuting the claim if it were false thus provide the strongest inductive evidence that it may be true. For example, suppose we want to test the claim that all swans are white. If we only look for swans at places in which they are known to be white, then we are not providing a strong test for the claim. The best thing to do (short of observing every swan in the whole world) is to try as hard as we can to refute the claim, to find a swan that is not white. If our best methods of looking for nonwhite swans still fail to refute the claim, then there is a growing likelihood that perhaps all swans are indeed white.
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Section 6.4 Reasoning About Science: The Hypothetico–Deductive Method
Similarly, if we want to test to see if a certain type of medicine cures a certain type of dis- ease, we test the product by giving the medicine to a wide variety of patients with the dis- ease, including those with the least likelihood of being cured by the medicine. Only by trying as hard as we can to refute the claim can we get the strongest evidence about whether all instances of the disease are treatable with the medicine in question.
Notice that the hypothetico–deductive method involves a combination of inductive and deductive reasoning. Step 1 typically involves inductive reasoning as we formulate a hypoth- esis against the background of our current beliefs and knowledge. Step 2 typically provides a deductive argument for the premise “If H, then C.” Step 3 provides an inductive argument for whether C is or is not true. Finally, if the prediction is falsified, then the conclusion—that H is false—is derived by a deductive inference (using the deductively valid modus tollens form). If, on the other hand, the best attempts to prove C to be false fail to do so, then there is growing evidence that H might be true.
Therefore, our overall argument has both inductive and deductive elements. It is valuable to know that, although the methodology of science involves research and experimentation that goes well beyond the scope of pure logic, we can use logic to understand and clarify the basic principles of scientific reasoning.
Practice Problems 6.4
1. A hypothesis is __________. a. something that is a mere guess b. something that is often arrived at after a lot of research c. an unnecessary component of the scientific method d. something that is already solved
2. In a scientific experiment, __________. a. the truth of the prediction guarantees that the hypothesis was correct b. the truth of the prediction negates the possibility of the hypothesis being correct c. the truth of the prediction can have different levels of probability in relation to
the hypothesis being correct d. the truth of the prediction is of little importance
3. The argument form that is set up when a test yields negative results is __________. a. disjunctive syllogism b. modus ponens c. hypothetical syllogism d. modus tollens
4. A claim is testable if __________. a. we know how one could show it to be false b. we know how one could show it to be true c. we cannot determine a way to prove it false d. we can determine a way to see if it is true or false
(continued)
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Section 6.5 Inference to the Best Explanation
5. Which of the following claims is not falsifiable? a. The moon is made of cheese. b. There is an invisible alien in my garage. c. Octane ratings in gasoline influence fuel economy. d. The Willis Tower is the tallest building in the world.
Practice Problems 6.4 (continued)
6.5 Inference to the Best Explanation You may feel that if you were very careful about testing your fuel economy, you would be entitled to conclude that the change in fuel grade really did have an effect. Unfortunately, as we have seen, the hypothetico–deductive method does not support this inference. The best you can say is that changing fuel might have an effect; that you have not been able to show that it does not have an effect. The method does, however, lend inductive support to which- ever hypothesis withstands the falsification test better than any other. One way of articulating this type of support is with an inference pattern known as inference to the best explanation.
As the name suggests, inference to the best explanation draws a conclusion based on what would best explain one’s observations. It is an extremely important form of inference that we use every day of our lives. This type of inference is often called abductive reasoning, a term pioneered by American logician Charles Sanders Peirce (Douven, 2011).
Suppose that you are in your backyard gazing at the stars. Suddenly, you see some flashing lights hovering above you in the sky. You do not hear any sound, so it does not appear that the lights are coming from a helicopter. What do you think it is? What happens next is abductive reasoning: Your brain searches among all kinds of possibilities to attempt to come up with the most likely explanation.
One possibility is that it is an alien spacecraft coming to get you (one could joke that this is why it is called abductive reasoning). Another possibility is that it is some kind of military vessel or a weather balloon. A more extreme hypothesis is that you are actually dreaming the whole thing.
Notice that what you are inclined to believe depends on your existing beliefs. If you already think that alien spaceships come to Earth all the time, then you may arrive at that conclusion with a high degree of certainty (you may even shout, “Take me with you!”). However, if you are somewhat skeptical of those kinds of theories, then you will try hard to find any other explanation. Therefore, the strength of a particular inference to the best explanation can be measured only in relation to the rest of the things that we already believe.
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Section 6.5 Inference to the Best Explanation
This type of inference does not occur only in unusual circumstances like the one described. In fact, we make infer- ences to the best explanation all the time. Returning to our fuel economy example from the previous section, suppose that you test a higher octane fuel and notice that your car gets bet- ter gas mileage. It is possible that the mileage change is due to the change in fuel. However, as noted there, it is possible that there is another expla- nation. Perhaps you are not driving in stop-and-go traffic as much. Perhaps you are driving with less weight in the car. The careful use of inference to the best explanation can help us to discern what is the most likely among many possibilities (for more examples, see A Closer Look: Is Abductive Reasoning Everywhere?).
If you look at the range of possible explanations and find one of them is more likely than any of the others, inference to the best explanation allows you to conclude that this explanation is likely to be the correct one. If you are driving the same way, to the same places, and with the same weight in your car as before, it seems fairly likely that it was the change in fuel that caused the improvement in fuel economy (if you have studied Mill’s methods in Chapter 5, you should recognize this as the method of difference). Inference to the best explanation is the engine that powers many inductive techniques.
The great fictional detective Sherlock Holmes, for example, is fond of claiming that he uses deductive reasoning. Chapter 2 suggested that Holmes instead uses inductive reasoning. However, since Holmes comes up with the most reasonable explanation of observed phe- nomena, like blood on a coat, for example, he is actually doing abductive reasoning. There is some dispute about whether inference to the best explanation is inductive or whether it is an entirely different kind of argument that is neither inductive nor deductive. For our purposes, it is treated as inductive.
Image Asset Management/SuperStock
Sherlock Holmes often used abductive reasoning, not deductive reasoning, to solve his mysteries.
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Section 6.5 Inference to the Best Explanation
A Closer Look: Is Abductive Reasoning Everywhere? Some see inference to the best explanation as the most common type of inductive inference. A few of the inferences we have discussed in this book, for example, can potentially be cast as examples of inferences to the best explanation.
For example, appeals to authority (discussed in Chapter 5) can be seen as implicitly using inference to the best explanation (Harman, 1965). If you accept something as true because someone said it was, then you can be described as seeing the truth of the claim as the best explanation for why he or she said it. If we have good reason to think that the person was deluded or lying, then we are less certain of this conclusion because there are other likely explanations of why the person said it.
Furthermore, it is possible to see what we do when we interpret people’s words as a kind of inference to the best explanation of what they probably mean (Hobbs, 2004). If your neighbor says, “You are so funny,” for instance, we might use the context and tone to decide what he means by “funny” and why he is saying it (and whether he is being sarcastic). His comment can be seen as either rude or flattering, depending on what explanation we give for why he said it and what he meant.
Even the classic inductive inference pattern of inductive generalization can possibly be seen as implicitly involving a kind of inference to the best explanation: The best explanation of why our sample population showed that 90% of students have laptops is probably that 90% of all students have laptops. If there is good evidence that our sample was biased, then there would be a good competing explanation of our data.
Finally, much of scientific inference may be seen as trying to provide the best explanation for our observations (McMullin, 1992). Many hypotheses are attempts to explain observed phe- nomena. Testing them in such cases could then be seen as being done in the service of seeking the best explanation of why certain things are the way they are.
Take a look at the following examples of everyday inferences and see if they seem to involve arriving at the conclusion because it seems to offer the most likely explanation of the truth of the premise:
• “John is smiling; he must be happy.” • “My phone says that Julie is calling, so it is probably Julie.” • “I see a brown Labrador across the street; my neighbor’s dog must have gotten out.” • “This movie has great reviews; it must be good.” • “The sky is getting brighter; it must be morning.” • “I see shoes that look like mine by the door; I apparently left my shoes there.” • “She still hasn’t called back yet; she probably doesn’t like me.” • “It smells good; someone is cooking a nice dinner.” • “My congressperson voted against this bill I support; she must have been afraid of
offending her wealthy donors.” • “The test showed that the isotopes in the rock surrounding newly excavated bones had
decayed X amount; therefore, the animals from which the bones came must have been here about 150 million years ago.”
These examples, and many others, suggest to some that inference to the explanation may be the most common form of reasoning that we use (Douven, 2011). Do you agree? Whether you agree with these expanded views on the role of inference or not, it clearly makes an enormous contribution to how we understand the world around us.
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Section 6.5 Inference to the Best Explanation
Form Inferences to the best explanation generally involve the following pattern of reasoning:
X has been observed to be true. Y would provide an explanation of why X is true. No other explanation for X is as likely as Y. Therefore, Y is probably true.
One strange thing about inferences to the best explanation is that they are often expressed in the form of a common fallacy, as follows:
If P is the case, then Q would also be true. Q is true. Therefore, P is probably true.
This pattern is the logical form of a deductive fallacy known as affirming the consequent (discussed in Chapter 4). Therefore, we sometimes have to use the principle of charity to determine whether the person is attempting to provide an inference to the best explanation or making a simple deductive error. The principle of charity will be discussed in detail in Chapter 9; however, for our purposes here, you can think of it as giving your opponent and his or her argument the benefit of the doubt.
For example, the ancient Greek philosopher Aristotle reasoned as follows: “The world must be spherical, for the night sky looks different in the northern and southern regions, and that would be the case if the earth were spherical” (as cited in Wolf, 2004). His argument appears to have this structure:
If the earth is spherical, then the night sky would look different in the north- ern and southern regions. The night sky does look different in the northern and southern regions. Therefore, the earth is spherical.
It is not likely that Aristotle, the founding father of formal logic, would have made a mistake as silly as to affirm the consequent. It is far more likely that he was using inference to the best explanation. It is logically possible that there are other explanations for southern stars moving higher in the sky as one moves south, but it seems far more likely that it is due to the shape of the earth. Aristotle was just practicing strong abductive reasoning thousands of years before Columbus sailed the ocean blue (even Columbus would have had to use this type of reasoning, for he would have had to infer why he did not sail off the edge).
In more recent times, astronomers are still using inference to the best explanation to learn about the heavens. Let us consider the case of discovering planets outside our solar system, known as “exoplanets.” There are many methods employed to discover planets orbiting other stars. One of them, the radial velocity method, uses small changes in the frequency of light a
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Section 6.5 Inference to the Best Explanation
star emits. A star with a large planet orbiting it will wobble a little bit as the planet pulls on the star. That wobble will result in a pattern of changes in the frequency of light coming from the star. When astronomers see this pattern, they conclude that there is a planet orbiting the star. We can more fully explicate this reasoning in the following way:
That star’s light changes in a specific pattern. Something must explain the changes. A large planet orbiting the star would explain the changes. No other explanation is as likely as the explanation provided by the large planet. Therefore, that star probably has a large planet orbiting it.
The basic idea is that if there must be an explanation, and one of the available explanations is better than all the others, then that explanation is the one that is most likely to be true. The key issue here is that the explanation inferred in the conclusion has to be the best explana- tion available. If another explanation is as good—or better—then the inference is not nearly as strong.
Virtue of Simplicity Another way to think about inferences to the best explanation is that they choose the simplest explanation from among otherwise equal explanations. In other words, if two theories make the same prediction, the one that gives the simplest explanation is usually the best one. This standard for comparing scientific theories is known as Occam’s razor, because it was origi- nally posited by William of Ockham in the 14th century (Gibbs & Hiroshi, 1997).
A great example of this principle is Galileo’s demonstration that the sun, not the earth, is at the center of the solar system. Galileo’s theory provided the simplest explanation of observa- tions about the planets. His heliocentric model, for example, provides a simpler explanation for the phases of Venus and why some of the planets appear to move backward (retrograde motion) than does the geocentric model. Geocentric astronomers tried to explain both of these with the idea that the planets sometimes make little loops (called epicycles) within their orbits (Gronwall, 2006). While it is certainly conceivable that they do make little loops, it seems to make the theory unnecessarily complex, because it requires a type of motion with no independent explanation of why it occurs, whereas Galileo’s theory does not require such extra assumptions.
Therefore, putting the sun at the center allows one to explain observed phenomena in the most simple manner possible, without making ad hoc assumptions (like epicycles) that today seem absurd. Galileo’s theory was ultimately correct, and he demonstrated it with strong inductive (more specifically, abductive) reasoning. (For another example of Occam’s razor at work, see A Closer Look: Abductive Reasoning and the Matrix.)
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Section 6.5 Inference to the Best Explanation
A Closer Look: Abductive Reasoning and the Matrix One of the great questions from the history of philosophy is, “How do we know that the world exists outside of us as we perceive it?” We see a tree and we infer that it exists, but do we actu- ally know for sure that it exists? The argument seems to go as follows:
I see a tree. Therefore, a tree exists.
This inference, however, is invalid; it is possible for the premise to be true and the conclusion false. For example, we could be dreaming. Perhaps we think that the testimony of our other senses will make the argument valid:
I see a tree, I hear a tree, I feel a tree, and I smell a tree. Therefore, a tree exists.
However, this argument is still invalid; it is pos- sible that we could be dreaming all of those things as well. Some people state that senses like smell do not exist within dreams, but how do we know that is true? Perhaps we only dreamed that someone said that! In any case, even that would not rescue our argument, for there is an even stronger way to make the premise true and the conclusion false: What if your brain is actually in a vat somewhere attached to a computer, and a scientist is directly controlling all of your perceptions? (Or think of the 1999 movie The Matrix, in which humans are living in a simulated reality created by machines.)
One individual who struggled with these types of questions (though there were no computers back then) was a French philosopher named René Des- cartes. He sought a deductive proof that the world outside of us is real, despite these types of disturbing possibilities (Descartes, 1641/1993). He eventually came up with one of philoso- phy’s most famous arguments, “I think, therefore, I am” (or, more precisely, “I am thinking, therefore, I exist”), and from there attempted to prove that the world must exist outside of him.
Many philosophers feel that Descartes did a great job of raising difficult questions, but most feel that he failed in his attempt to find deductive proof of the world outside of our minds. Other philosophers, including David Hume, despaired of the possibility of a proof that we know that there is a world outside of us and became skeptics: They decided that absolute knowledge of a world outside of us is impossible (Hume, 1902).
However, perhaps the problem is not the failure of the particular arguments but the type of reasoning employed. Perhaps the solution is not deductive at all but rather abductive. It is not that it is logically impossible that tables and chairs and trees (and even other people) do not really exist; it is just that their actual existence provides the best explanation of our experi- ences. Consider these competing explanations of our experiences:
• We are dreaming this whole thing. • We are hallucinating all of this.
In The Matrix, we learn that our world is simulated by machines, and although we can see X, hear X, and feel X, X does not exist.
(continued)
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Section 6.5 Inference to the Best Explanation
• Our brains are in a vat being controlled by a scientist. • Light waves are bouncing off the molecules on the surface of the tree and entering our
eyeballs, where they are turned into electrical impulses that travel along neurons into our brains, somehow causing us to have the perception of a tree.
It may seem at first glance that the final option is the most complex and so should be rejected. However, let us take a closer look. The first two options do not offer much of an explanation for the details of our experience. They do not tell us why we are seeing a tree rather than some- thing else or nothing at all. The third option seems to assume that there is a real world some- where from which these experiences are generated (that is, the lab with the scientist in it). The full explanation of how things work in that world presumably must involve some complex laws of physics as well. There is no obvious reason to think that such an account would require fewer assumptions than an account of the world as we see it. Hence, all things considered, if our goal is to create a full explanation of reality, the final option seems to give the best account of why we are seeing the tree. It explains our observations without needless extra assumptions.
Therefore, if knowledge is assumed only to be deductive, then perhaps we do not know (with absolute deductive certainty) that there is a world outside of us. However, when we consider abductive knowledge, our evidence for the existence of the world as we see it may be rather strong.
A Closer Look: Abductive Reasoning and the Matrix (continued)
How to Assess an Explanation There are many factors that influence the strength of an inference to the best explanation. However, when testing inferences to the best explanation for strength, these questions are good to keep in mind:
• Does it agree well with the rest of human knowledge? Suggesting that your room- mate’s car is gone because it floated away, for example, is not a very credible story because it would violate the laws of physics.
• Does it provide the simplest explanation of the observed phenomena? According to Occam’s razor, we want to explain why things happen without unnecessary complexity.
• Does it explain all relevant observations? We cannot simply ignore contradicting data because it contradicts our theory; we have to be able to explain why we see what we see.
• Is it noncircular? Some explanations merely lead us in a circle. Stating that it is raining because water is falling from the sky, for example, does not give us any new information about what causes the water to fall.
• Is it testable? Suggesting that invisible elves stole the car does not allow for empirical confirmation. An explanation is stronger if its elements are potentially observable.
• Does it help us explain other phenomena as well? The best scientific theories do not just explain one thing but allow us to understand a whole range of related phenom- ena. This principle is called fecundity. Galileo’s explanation of the orbits of the plan- ets is an example of a fecund theory because it explains several things all at once.
An explanation that has all of these virtues is likely to be better than one that does not.
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Section 6.5 Inference to the Best Explanation
A Limitation One limitation of inference to the best explanation is that it depends on our coming up with the correct explanation as one of the candidates. If we do not think of the correct explana- tion when trying to imagine possible explanation, then inference to the best explanation can steer us wrong. This can happen with any inductive argument, of course; inductive arguments always carry some possibility that the conclusion may be false even if the premises are true. However, this limitation is a particular danger with inference to the best explanation because it relies on our being able to imagine the true explanation.
This is one reason that it is essential to always keep an open mind when using this technique. Further information may introduce new explanations or change which explanation is best. Being open to further information is important for all inductive inferences, but especially so for those involving inference to the best explanation.
Practice Problems 6.5
1. This philosopher coined the term abductive reasoning. a. Karl Popper b. Charles Sanders Peirce c. Aristotle d. G. W. F. Hegel
2. Sherlock Holmes is often said to be engaging in this form of reasoning, even though from a logical perspective he wasn’t. a. deductive b. inductive c. abductive d. productive
3. In a specific city that happens to be a popular tourist destination, the number of residents going to the emergency rooms for asthma attacks increases in the summer. When the winter comes and tourism decreases, the number of asthma attacks goes down. What is the most probable inference to be drawn in this situation? a. The locals are allergic to tourists. b. Summer is the time that most people generally have asthma attacks. c. The increased tourism leads to higher levels of air pollution due to traffic. d. The tourists pollute the ocean with trash that then causes the locals to get sick.
4. A couple goes to dinner and shares an appetizer, entrée, and dessert. Only one of the two gets sick. She drank a glass of wine, and her husband drank a beer. What is the most probable inference to be drawn in this situation? a. The wine was the cause of the sickness. b. The beer protected the man from the sickness. c. The appetizer affected the woman but not the man. d. The wine was rotten.
(continued)
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Section 6.5 Inference to the Best Explanation
5. You are watching a magic performance, and there is a woman who appears to be floating in space. The magician passes a ring over her to give the impression that she is floating. What explanation fits best with Occam’s razor? a. The woman is actually floating off the ground. b. The magician is a great magician. c. There is some sort of unseen physical object holding the woman.
6. You get a stomachache after eating out at a restaurant. What explanation fits best with Occam’s razor? a. You contracted Ebola and are in the beginning phases of symptoms. b. Someone poisoned the food that you ate. c. Something was wrong with the food you ate.
7. In order to determine how a disease was spread in humans, researchers placed two groups of people into two rooms. Both rooms were exactly alike, and no people touched each other while in the rooms. However, researchers placed someone who was infected with the disease in one room. They found that those who were in the room with the infected person got sick, whereas those who were not with an infected person remained well. What explanation fits best with Occam’s razor? a. The disease is spread through direct physical contact. b. The disease is spread by airborne transmission. c. The people in the first room were already sick as well.
8. There is a dent in your car door when you come out of the grocery store. What expla- nation fits best with Occam’s razor? a. Some other patron of the store hit your car with their car. b. A child kicked your door when walking into the store. c. Bad things tend to happen only to you in these types of situations.
9. A student submits a paper that has an 80% matching rate when submitted to Tur- nitin. There are multiple sites that align exactly with the content of the paper. What explanation fits best with Occam’s razor? a. The student didn’t know it was wrong to copy things word for word without
citing. b. The student knowingly took material that he did not write and used it as his
own. c. Someone else copied the student’s work.
10. You are a man, and you jokingly take a pregnancy test. The test comes up positive. What explanation fits best with Occam’s razor? a. You are pregnant. b. The test is correct. c. The test is defective.
11. A bomb goes off in a supermarket in London. A terrorist group takes credit for the bombing. What explanation fits best with Occam’s razor? a. The British government is trying to cover up the bombing by blaming a terrorist
group. b. The terrorist group is the cause of the bombing. c. The U.S. government actually bombed the market to get the British to help them
fight terrorist groups.
Practice Problems 6.5 (continued)
(continued)
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Section 6.5 Inference to the Best Explanation
12. You have friends and extended family over for Thanksgiving dinner. There are kids running through the house. You check the turkey and find that it is overcooked because the temperature on the oven is too high. What explanation fits best with Occam’s razor? a. The oven increased the temperature on its own. b. Someone turned up the heat to sabotage your turkey. c. You bumped the knob when you were putting something into the oven.
13. Researchers recently mapped the genome of a human skeleton that was 45,000 years old. They found long fragments of Neanderthal DNA integrated into this human genome. What explanation fits best with Occam’s razor? a. Humans and Neanderthals interbred at some point prior to the life of this hu-
man. b. The scientists used a faulty method in establishing the genetic sequence. c. This was actually a Neanderthal skeleton.
14. There is a recent downturn in employment and the economy. A politically far-leaning radio host claims that the downturn in the economy is the direct result of the presi- dent’s actions. What explanation fits best with Occam’s razor? a. The downturn in employment is due to many factors, and more research is in
order. b. The downturn in employment is due to the president’s actions. c. The downturn in employment is really no one’s fault.
15. In order for an explanation to be adequate, one should remember that __________. a. it should agree with other human knowledge b. it should include the highest level of complexity c. it should assume the thing it is trying to prove d. there are outlying situations that contradict the explanation
16. The fecundity of an explanation refers to its __________. a. breadth of explanatory power b. inability to provide an understanding of a phenomenon c. lack of connection to what is being examined d. ability to bear children
17. Why might one choose to use an inductive argument rather than a deductive argument? a. One possible explanation must be the correct one. b. The argument relates to something that is probabilistic rather than absolute. c. An inductive argument makes the argument valid. d. One should always use inductive arguments when possible.
18. This is the method by which one can make a valid argument invalid. a. adding false supporting premises b. demonstrating that the argument is valid c. adding true supporting premises d. valid arguments cannot be made invalid
(continued)
Practice Problems 6.5 (continued)
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Section 6.5 Inference to the Best Explanation
19. This form of inductive argument moves from the general to the specific. a. generalizations b. statistical syllogisms c. hypothetical syllogism d. modus tollens
Questions 20–24 relate to the following passage:
If I had gone to the theater, then I would have seen the new film about aliens. I didn’t go to the theater though, so I didn’t see the movie. I think that films about aliens and supernatural events are able to teach people a lot about what the future might hold in the realm of tech- nology. Things like cell phones and space travel were only dreams in old movies, and now they actually exist. Science fiction can also demonstrate new futures in which people are more accepting of those that are different from them. The different species of characters in these films all working together and interacting with one another in harmony displays the unity of different people without explicitly making race or ethnicity an issue, thereby bringing people into these forms of thought without turning those away who do not want to explicitly confront these issues.
20. How many arguments are in this passage? a. 0 b. 1 c. 2 d. 3
21. How many deductive arguments are in this passage? a. 0 b. 1 c. 2 d. 3
22. How many inductive arguments are in this passage? a. 0 b. 1 c. 2 d. 3
23. Which of the following are conclusions in the passage? Select all that apply. a. If I had gone to the theater, then I would have seen the new film about aliens. b. I didn’t go to the theater. c. Films about aliens and supernatural events are able to teach people a lot about
what the future might hold in the realm of technology. d. The different species of characters in these films all working together and
interacting with one another in harmony displays the unity of different people without explicitly making race or ethnicity an issue.
24. Which change to the deductive argument would make it valid? Select all that apply. a. Changing the first sentence to “If I would have gone to the theater, I would not
have seen the new film about aliens.” b. Changing the second sentence to “I didn’t see the new film about aliens.” c. Changing the conclusion to “Alien movies are at the theater.” d. Changing the second sentence to “I didn’t see the movie, so I didn’t go to the theater.”
Practice Problems 6.5 (continued)
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Summary and Resources
Summary and Resources
Chapter Summary Although induction and deduction are treated differently in the field of logic, they are fre- quently combined in arguments. Arguments with both deductive and inductive components are generally considered to be inductive as a whole, but the important thing is to recognize when deduction and induction are being used within the argument. Arguments that com- bine inductive and deductive elements can take advantage of the strengths of each. They can retain the robustness and persuasiveness of inductive arguments while using the stronger connections of deductive arguments where these are available.
Science is one discipline in which we can see inductive and deductive arguments play out in this fashion. The hypothetico–deductive method is one of the central logical tools of science. It uses a deductive form to draw a conclusion from inductively supported premises. The hypothetico–deductive method excels at disconfirming or falsifying hypotheses but cannot be used to confirm hypotheses directly.
Inference to the best explanation, however, does provide evidence supporting the truth of a hypothesis if it provides the best explanation of our observations and withstands our best attempts at refutation. A key limitation of this method is that it depends on our being able to come up with the correct explanation as a possibility in the first place. Nevertheless, it is a powerful form of inference that is used all the time, not only in science but in our daily lives.
Critical Thinking Questions
1. You have probably encountered numerous conspiracy theories on the Internet and in popular media. One such theory is that 9/11 was actually plotted and orches- trated by the U.S. government. What is the relationship between conspiracy theories and inference to the best possible explanation? In this example, do you think that this is a better explanation than the most popular one? Why or why not?
2. What are some methods you can use to determine whether or not information represents the best possible explanation of events? How can you evaluate sources of information to determine whether or not they should be trusted?
3. Descartes claimed that it might be the case that humans are totally deceived about all aspects of their existence. He went so far as to claim that God could be evil and could be making it so that human perception is completely wrong about everything. However, he also claimed that there is one thing that cannot be doubted: So long as he is thinking, it is impossible for him to doubt that it is he who is thinking. Hence, so long as he thinks, he exists. Do you think that this argument establishes the inherent existence of the thinking being? Why or why not?
4. Have you ever been persuaded by an argument that ended up leading you to a false conclusion? If so, what happened, and what could you have done differently to pre- vent yourself from believing a false conclusion?
5. How can you incorporate elements of the hypothetico–deductive method into your own problem solving? Are there methods here that can be used to analyze situations in your personal and professional life? What can we learn about the search for truth from the methods that scientists use to enhance knowledge?
har85668_06_c06_207-238.indd 236 4/9/15 11:38 AM
Summary and Resources
abductive reasoning See inference to the best explanation.
falsifiable Describes a claim that is conceiv- ably possible to prove false. That does not mean that it is false; only that prior to test- ing, it is possible that it could have been.
falsification The effort to disprove a claim (typically by finding a counterexample to it).
hypothesis A conjecture about how some part of the world works.
hypothetico–deductive method The method of creating a hypothesis and then attempting to falsify it through experimentation.
inference to the best explanation The process of inferring something to be true because it is the most likely explanation of some observations. Also known as abductive reasoning.
Occam’s razor The principle that, when seeking an explanation for some phenom- ena, the simpler the explanation the better.
self-sealing propositions Claims that can- not be proved false because they are inter- preted in a way that protects them against any possible counterexample.
Web Resources https://www.youtube.com/watch?v=RauTW8F-PMM Watch Ashford professor Justin Harrison lecture on the difference between inductive and deductive arguments.
https://www.youtube.com/watch?v=VXW5mLE5Y2g Shmoop offers an animated video on the difference between induction and deduction.
http://www.ac4d.com/2012/06/03/abductive-reasoning-in-airport-security-and-profiling Design expert Jon Kolko applies abductive reasoning to airport security in this blog post.
Key Terms
Answers to Practice Problems Practice Problems 6.1
1. d 2. a
3. b 4. b
Practice Problems 6.2
1. a 2. a
3. a 4. a
Practice Problem 6.3
1. b
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Summary and Resources
Practice Problems 6.4
1. b 2. c 3. d
4. d 5. b
Practice Problems 6.5
1. b 2. a 3. c 4. a 5. c 6. c 7. b 8. a 9. b
10. c 11. b 12. c
13. a 14. a 15. a 16. a 17. b 18. d 19. b 20. d 21. b 22. c 23. c 24. d
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## Special arithmetic progressions involving perfect squares
Some time ago the following rather easy problem appeared in an online publication called "Problems in Elementary NT" by Hojoo Lee:
Prove that there are infinitely many positive integers $a$, $b$, $c$ that are consecutive terms of an arithmetic progression and also satisfy the condition that $ab+1$, $bc+1$, $ca+1$ are all perfect squares.
This can be done using Pell's equation. What is interesting however is that the following result for four numbers apparently holds:
Claim. There are no positive integers $a$, $b$, $c$, $d$ that are consecutive terms of an arithmetic progression and also satisfy the condition that $ab+1$, $ac+1$, $ad+1$, $bc+1$, $bd+1$, $cd+1$ are all perfect squares.
I am curious to see if there is any (decent) solution.
Thanks.
-
//Apologies if the tags are not appropriate; didn't know where elementary nt fits better (if it fits at all on this website) – Cosmin Pohoata Feb 11 2012 at 19:05
What do you mean by "apparently holds"? Have you checked it computationally, or can you derive it from the three-term progression case? – Seva Feb 11 2012 at 19:12
I checked it computationally, sorry for not mentioning. – Cosmin Pohoata Feb 11 2012 at 19:24
I added the "diophantine-equations" tag. See also my comment on duje's answer below relating to the case of triples in AP. – Michael Stoll Feb 22 2012 at 19:31
Starting from the equations in my previous answer, we get, by multiplying them in pairs, $$(x-y)x(x+y)(x+2y) + (x-y)x + (x+y)(x+2y) + 1 = (z_1 z_6)^2\,,$$ $$(x-y)x(x+y)(x+2y) + (x-y)(x+y) + x(x+2y) + 1 = (z_2 z_5)^2\,,$$ $$(x-y)x(x+y)(x+2y) + (x-y)(x+2y) + x(x+y) + 1 = (z_3 z_4)^2\,.$$ Write $u = z_1 z_6$, $v = z_2 z_5$, $w = z_3 z_4$ and take differences to obtain $$3 y^2 = u^2 - v^2 \qquad\text{and}\qquad y^2 = v^2 - w^2\,.$$ The variety $C$ in ${\mathbb P}^3$ described by these two equations is a smooth curve of genus 1 whose Jacobian elliptic curve is 24a1 in the Cremona database; this elliptic curve has rank zero and a torsion group of order 8. This implies that $C$ has exactly 8 rational points; up to signs they are given by $(u:v:w:y) = (1:1:1:0)$ and $(2:1:0:1)$. So $y = 0$ or $w = 0$. In the first case, we do not have an honest AP ($y$ is the difference). In the second case, we get the contradiction $abcd + ad + bc + 1 = 0$ ($a,b,c,d$ are supposed to be positive). So unless I have made a mistake somewhere, this proves that there are no such APs of length 4.
Addition: We can apply this to rational points on the surface. The case $y = 0$ gives a bunch of conics of the form $$x^2 + 1 = z_1^2, \quad z_2 = \pm z_1, \quad \dots, \quad z_6 = \pm z_1\,;$$ the case $w = 0$ leads to $ad = -1$ or $bc = -1$. The second of these gives $ad + 1 < 0$, and the first gives $ac + 1 = (a^2 + 1)/3$, which cannot be a square. This shows that all the rational points are on the conics mentioned above; in particular, (weak) Bombieri-Lang holds for this surface.
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@Michael: Very nice! Let $X$ denote the surface given by your earlier answer. Then if I understand correctly, you've given a dominant rational map $X\to C$, where $C$ has genus 1 and $#C(\mathbb{Q})=8$. It seems a little surprising that a surface of general type should map onto a genus 1 curve, although I guess the fact that $X$ is an intersection of quadrics gives it a chance. But did you have some a priori reason to suspect that $X$ admitted such a covering? – Joe Silverman Feb 12 2012 at 12:52
@Joe: I didn't have some good a priori reason to suspect that $X$ would map to a genus 1 curve, but the observation that the analogous problem for five terms admits such a map (to the "four squares in AP" curve) provided some motivation to try a little bit longer. – Michael Stoll Feb 12 2012 at 13:13
Bravo! The identities look a bit less mysterious if the arithmetic progression is written symmetrically as $(x-3y,x-y,x+y,x+3y)$, when the products $(z_i z_{7-i})^2$ are invariant under $y \leftrightarrow -y$ (Then $u^2-v^2$ and $v^2-w^2$ are $12y^2$ and $4y^2$, which comes to the same curves over ${\bf Q}$.) – Noam D. Elkies Feb 12 2012 at 19:50
Can magma (or other software) detect the nontrivial $H^1$ of this surface which makes this analysis possible? – Noam D. Elkies Feb 13 2012 at 1:16
Hm. I guess you would want the $H^1$ of its desingularization, which might be a fairly complicated object. As damiano pointed out to me by email, the fact that there is a map to an elliptic curve must be related to the singularities being sufficiently bad (otherwise the surface would be simply connected). – Michael Stoll Feb 15 2012 at 11:10
In Diophantine quadruple $a<b<c<d$, it holds $d\ge 4bc$ (see e.g. Lemma 14 in http://web.math.pmf.unizg.hr/~duje/pdf/bound.pdf ).
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To complete this: From $d \ge 4bc$, it follows easily that $a,b,c,d$ cannot be an AP. Note also the following: Acdording to Lemma 13 in Dujella's paper linked to in his answer above, if $a,b,c$ is a Diophantine triple with $a < b < c$, then $c = a + b + 2 \sqrt{ab+1}$ or $c \ge 4ab$. If $a,b,c$ form an AP, then the second possibility cannot occur, and the first (together with the AP condition) then implies that $b$ is even and $y^2 - 3(b/2)^2 = 1$, where $y = b-a = c-b$. So it is indeed the case that all such triples come from this Pell equation. (This gives another proof.) – Michael Stoll Feb 22 2012 at 19:21
Dujella has written many papers on Diophantine m-tuples, check out his webpage.
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According to Magma, the projective closure of the variety associated to the problem (given by the equations $$x(x-y) + 1 = z_1^2, \quad (x+y)(x-y) + 1 = z_2^2, \quad (x+2y)(x-y) + 1 = z_3^2,$$ $$(x+y)x + 1 = z_4^2, \quad (x+2y)x + 1 = z_5^2, \quad (x+2y)(x+y) + 1 = z_6^2 \quad)$$ is an irreducible surface in ${\mathbb P}^8$ with 34 isolated singularities. Since it is a complete intersection of six quadrics, it should be of general type (and it has trivial rational points with $x = y = 0$ and slightly less trivial ones with $y = 0$, so reduction methods will not work), which makes it very likely that this is a hard question.
Added later: You may want to look at Question 73346 for an explanation by Noam Elkies of the reasoning behind this.
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This suggests that indeed the rational points are likewise fully accounted for by finitely many curves. The question asked only for integer points, which could conceivably be more tractable, though I don't see how. – Noam D. Elkies Feb 11 2012 at 21:39
@Noam: The problem with three terms should give a K3 Surface. Are there any K3-related methods that might prove that (up to signs) all the integral points are on the curve that comes from the Pell equation? – Michael Stoll Feb 11 2012 at 22:01
There are no solutions with five terms $a,b,c,d,e$ in arithmetic progression, since then $ab+1$, $ac+1$, $ad+1$, $ae+1$ would have to be four squares in arithmetic progression. Of course, this doesn't really help... – Michael Stoll Feb 11 2012 at 22:23
@Michael: It might be an interesting K3 surface but I don't see how to use this to prove anything about its integral points (for which it's of "log-general type"). – Noam D. Elkies Feb 12 2012 at 20:39
Already for three-term progressions it's somewhat surprising that there are infinitely many solutions, because the usual probabilistic guess for the expected number of solutions leads to a convergent sum: a random number of size $M$ is a square with probability about $M^{-1/2}$, so we're summing something like $1/(abc)$ over all three-term progressions $(a,b,c)$, etc. To be sure such a guess cannot account for non-random patterns arising from polynomial identities, but it does suggest that past a certain point such identities will be the only source of solutions.
Now a mindless exhaustive search over progressions $(x,x+y,x+2y)$ with $0 < x,y < 10^4$ finds only the first six examples $$(1,7),\phantom+ (4,26),\phantom+ (15,97),\phantom+ (56,362),\phantom+ (209,1351),\phantom+ (780,5042)$$ of an infinite family associated with the solutions $(2,1)$, $(7,4)$, $(26,15)$, $(97,56)$, $(362,209)$, $(1351,780)$, etc. of the Pell equation $x^2-3y^2=1$. If it can be proved that these are the only solutions then it will immediately follow that there are no four-term arithmetic progressions with the same property. But that seems like a very hard problem.
Here's the gp code; with a bound of $10^4$ it takes only a few minutes. One can surely do better with a more intelligent search procedure (e.g. start by finding all solutions of $ab+1=r^2$ by factoring $r^2-1$).
H = 10^4
progsq(x,y,n) = sum(i=0,n-2,sum(j=i+1,n-1,issquare((x+i*y)*(x+j*y)+1)))
for(x=1,H,for(y=1,H,if(progsq(x,y,3)==3,print([x,y]))))
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Isn't there a similar problem involving 1,3,8,120, that says something like there is no fifth member such that the product of any two is one less than a square? It might be also that there are finitely many quadruplets, with the nontrivial ones being not arithmetic progressions. Gerhard "Ask Me About Obscure Puzzles" Paseman, 2012.02.11 – Gerhard Paseman Feb 11 2012 at 20:37
I was reminded of this 1,3,8,120 puzzle too, because Martin Gardner wrote about it in one of his columns decades ago and reported that the nonexistence of a fifth positive integer was proved only with difficulty (including several 1000+ digit computations, back when that was impressive). But there are infinitely many such integer quadruplets. A few Google searches turned up vixra.org/pdf/0907.0024v1.pdf with citations going as far as Euler and Diophantus! But naturally none of these are four-term arithmetic progressions... – Noam D. Elkies Feb 11 2012 at 21:43
Thanks for the reply! This is essentially what I did to claim that I "checked it manually": I verified the correspondence between the solutions of the problem in question and $x^{2}-3y^{2}=1$ up to $10^{9}$ and then just assummed this to be true. I was hoping however to find some slick way of doing the four number case without refering to the three case :). – Cosmin Pohoata Feb 12 2012 at 5:10
You're welcome. You saw that by now M.Stoll has completely solved the 4-term problem, even over the rationals. For the 3-term problem, you write that you "checked manually" up to $10^9$; how did you do that? I used the factorization of $t^2-1$ in $bc+1=t^2$ to search up to $t = 10^8$ in a few hours, still finding only the Pell solutions up to $(a,b,c) = (7865521, 58709048, 109552575)$; so $t=10^9$ (if that's what you meant) is feasible too, though it would take a while (or a number of CPU's running in parallel) to finish − and it's certainly not "manual" in the usual sense of "by hand"... – Noam D. Elkies Feb 12 2012 at 19:56 | 3,254 | 10,814 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2013-20 | latest | en | 0.890709 |
https://www.physicsforums.com/tags/prime/ | 1,713,162,845,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00884.warc.gz | 857,641,799 | 24,946 | # What is Prime: Definition and 776 Discussions
A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. A natural number greater than 1 that is not prime is called a composite number. For example, 5 is prime because the only ways of writing it as a product, 1 × 5 or 5 × 1, involve 5 itself.
However, 4 is composite because it is a product (2 × 2) in which both numbers are smaller than 4. Primes are central in number theory because of the fundamental theorem of arithmetic: every natural number greater than 1 is either a prime itself or can be factorized as a product of primes that is unique up to their order.
The property of being prime is called primality. A simple but slow method of checking the primality of a given number
n
{\displaystyle n}
, called trial division, tests whether
n
{\displaystyle n}
is a multiple of any integer between 2 and
n
{\displaystyle {\sqrt {n}}}
. Faster algorithms include the Miller–Rabin primality test, which is fast but has a small chance of error, and the AKS primality test, which always produces the correct answer in polynomial time but is too slow to be practical. Particularly fast methods are available for numbers of special forms, such as Mersenne numbers. As of December 2018 the largest known prime number is a Mersenne prime with 24,862,048 decimal digits.There are infinitely many primes, as demonstrated by Euclid around 300 BC. No known simple formula separates prime numbers from composite numbers. However, the distribution of primes within the natural numbers in the large can be statistically modelled. The first result in that direction is the prime number theorem, proven at the end of the 19th century, which says that the probability of a randomly chosen number being prime is inversely proportional to its number of digits, that is, to its logarithm.
Several historical questions regarding prime numbers are still unsolved. These include Goldbach's conjecture, that every even integer greater than 2 can be expressed as the sum of two primes, and the twin prime conjecture, that there are infinitely many pairs of primes having just one even number between them. Such questions spurred the development of various branches of number theory, focusing on analytic or algebraic aspects of numbers. Primes are used in several routines in information technology, such as public-key cryptography, which relies on the difficulty of factoring large numbers into their prime factors. In abstract algebra, objects that behave in a generalized way like prime numbers include prime elements and prime ideals.
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1. ### A Determining rationality of real numbers represented by prime digit sequence
I would like to know if my answer is correct and if no ,could you correct.But it should be right I hope:
2. ### A How should I write an account of prime numbers?
How should I write an account of prime numbers in arithmetic progressions? Assuming this account should be in the form of an essay of at least ## 500 ## words. Should I apply the formula ## a_{n}=3+4n ## for ## 0\leq n\leq 2 ##? Can anyone please provide any idea(s)?
3. ### I Frequency of prime number gaps according to (p-1)/(p-2)
Caution I'm not a mathematician. In short, long time ago I calculated prime number gaps just for fun expecting an almost uniform distribution of the frequency of the gaps 2, 4, 6, ... . Instead the frequency showed a series of maxima and minima and I was confused. Later Professor emeritus Oskar...
4. ### Determine (with proof) the set of all prime numbers
Proof: Let ## p ## be the prime divisor of two successive integers ## n^{2}+3 ## and ## (n+1)^{2}+3 ##. Then ## p\mid [(n+1)^{2}+3-(n^{2}+3)]\implies p\mid (2n+1) ##. Observe that ## p\mid (n^{2}+3) ## and ## p\mid (2n+1) ##. Now we see that ## p\mid [(n^{2}+3)-3(2n+1)]\implies p\mid...
5. ### I Primes -- Probability that the sum of two random integers is Prime
im thinking i should just integrate (binominal distribution 1-2000 * prime probability function) and divide by integral of bin. distr. 1-2000. note that I am looking for a novel proof, not just some brute force calculation. (this isn't homework, I am just curious.)
6. ### I Defining the Prime Gap function
Hi PF! I created a function ##R(x)## that gives the gap between the largest two primes less than or equal to ##x##. To define it, I used this property: $$\pi(x+R(x))=\pi(x)+1$$ Which is true since the ##x## distance between ##\pi(x)## and ##\pi(x)+1## is ##R(x)##. If we solve for ##R(x)## we...
7. ### Prove that there exists a prime with at least ## n ## of its digits.
Proof: By Dirichlet's theorem, we have that if ## a ## and ## d ## are two positive coprime numbers, then there are infinitely many primes of the form ## a+nd ## for some ## n\in\mathbb{N} ##. Let ## n\geq 1 ## be a natural number. Now we consider the arithmetic progression ## 10^{n+1}k+1 ##...
8. ### A I think I discovered a pattern for prime numbers
I wrote a program that implements the pattern and finds the primes automatically. It worked up to 70 million then it crashed because program holds data in RAM so it can be fixed. It found all the primes up to 70 million and found no exception. I won't explain the pattern because its so...
9. ### A Prime Number Powers of Integers and Fermat's Last Theorem
From my research I have found that since Fermat proved his last theorem for the n=4 case, one only needs to prove his theorem for the case where n=odd prime where c^n = a^n + b^n. But I am not clear on some points relating to this. For example, what if we have the term (c^x)^p, where c is an...
10. ### Show that 13 is the largest prime that can divide....
Proof: Let ## p ## be the prime divisor of two successive integers ## n^2+3 ## and ## (n+1)^2+3 ##. Then ## p\mid [(n+1)^2+3-(n^2+3)]\implies p\mid 2n+1 ##. Since ## p\mid n^2+3 ## and ## p\mid 2n+1 ##, it follows that ## p\mid a(n^2+3)+b(2n+1) ## for some ## a,b\in\mathbb{Z} ##. Suppose ## a=4...
11. ### If ## p ## and ## p^{2}+8 ## are both prime numbers, prove that....
Proof: Suppose ## p ## and ## p^{2}+8 ## are both prime numbers. Since ## p^{2}+8 ## is prime, it follows that ## p ## is odd, so ## p\neq 2 ##. Let ## p>3 ##. Then ## p^{2}\equiv 1 \mod 3 ##, so ## p^{2}+8\equiv 0 \mod 3 ##. Note that ## p^{2}+8 ## can only be prime for ## p=3 ##. Thus ##...
12. ### For n>3, show that the integers n, n+2, n+4 cannot all be prime
Proof: Let ## n>3 ## be an integer. Applying the Division Algorithm produces: ## n=3q+r ## for ## 0\leq r< 3 ##, where there exist unique integers ## q ## and ## r ##. Suppose ## n ## is prime. Then ## n=3q+1 ## or ## n=3q+2 ##, because ## n\neq 3q ##. Now we consider two cases. Case #1...
13. ### For ## n>1 ##, show that every prime divisor of ## n+1 ## is an odd integer
Proof: Suppose for the sake of contradiction that there exists a prime divisor of ## n!+1 ##, which is an odd integer that is not greater than ## n ##. Let ## n>1 ## be an integer. Since ## n! ## is even, it follows that ## n!+1 ## is odd. Thus ## 2\nmid (n!+1) ##. This means every prime factor...
14. ### Prove that if ## n>2 ##, then there exists a prime ## p ## satisfying....
Proof: Assume to the contrary that ## n!-1 ## is not prime. That is, ## n!-1 ## is composite. Let ## p ## be a prime factor of ## n!-1 ## such that ## n<p ## for some ## n\in\mathbb{N} ## where ## n>2 ##. Since ## p\mid (n!-1) ##, it follows that ## p\nmid n! ##. This means ## p\nleq n ##...
15. ### Show that any composite three-digit number must have a prime factor
Proof: Suppose for the sake of contradiction that any composite three-digit number must have a prime factor not less than or equal to ## 31 ##. Let ## n ## be any composite three-digit number such that ## n=ab ## for some ## a,b\in\mathbb{Z} ## where ## a,b>1 ##. Note that the smallest prime...
16. ### I Probability of Sum of 2 Random Ints Being Prime
if I select two integers at random between 1 and 1,000, what is the probability that their sum will be prime?
17. ### Prove that ## \sqrt{p} ## is irrational for any prime ## p ##?
Proof: Suppose for the sake of contradiction that ## \sqrt{p} ## is not irrational for any prime ## p ##, that is, ## \sqrt{p} ## is rational. Then we have ## \sqrt{p}=\frac{a}{b} ## for some ## a,b\in\mathbb{Z} ## such that ## gcd(a, b)=1 ## where ## b\neq 0 ##. Thus ## p=\frac{a^2}{b^2} ##...
18. ### Determine whether the integer 701 is prime by testing?
Proof: Consider all primes ## p\leq\sqrt{701}\leq 27 ##. Note that ## 701=2(350)+1 ## ## =3(233)+2 ## ## =5(140)+1 ## ## =7(100)+1 ## ## =11(63)+8 ## ## =13(53)+12 ##...
19. ### I Is Riemann's Zeta at 2 Related to Pi through Prime Numbers?
I just saw that one of the ways of calculating Pi uses the set of prime numbers. This must sound crazy even to people who understand it, is it possible that this can be explained in terms that I, a mere mortal can understand or it is out of reach for non mathematicians?
20. ### Find the prime factorization of the integers 1234, 10140, and 36000?
## 1234=2\cdot 617 ## ## 10140=2\cdot 2\cdot 3\cdot 5\cdot 13\cdot 13 ## ## 36000=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5\cdot 5\cdot 5\cdot ## Are the answers above correct? Or do I need to put them in canonical form as below? ## 1234=2\cdot 617 ## ## 10140=2^{2}\cdot 3\cdot 5\cdot...
21. ### If ## p\neq5 ## is an odd prime, prove that either ## p^{2}-1 ## or....
Proof: Suppose ## p\neq5 ## is an odd prime. Applying the Division Algorithm produces: ## p=10q+r ## where ## 0\leq r\leq 10 ##, where there exist unique integers ## q ## and ## r ##. Note that ## p\equiv 1, 3, 7 ## or ## 9 ## mod ## 10 ##. This means ## p^{2}\equiv 1, -1, -1 ## or ## 1 ## mod...
22. ### Find all prime numbers that divide 50
Proof: Note that all primes less than 50 will divide 50!, because each prime is a term of 50!. Applying the Fundamental Theorem of Arithmetic produces: Each term k of 50! that is non-prime has a unique prime factorization. Since 48, 49 and 50 are not primes, it follows that all primes...
23. ### Given that p is a prime? (Review/verify this proof)?
Proof: Suppose that p is a prime and ##p \mid a^n ##. Note that a prime number is a number that has only two factors, 1 and the number itself. Then we have (p*1)##\mid##a*## a^{(n-1)} ##. Thus p##\mid##a, which implies that pk=a for some k##\in\mathbb{Z}##. Now we have ## a^n ##=## (pk)^n ##...
24. ### If p##\geq##5 is a prime number, show that p^{2}+2 is composite?
Proof: Suppose p##\geq##5 is a prime number. Applying the Division Algorithm produces: p=6k, p=6k+1, p=6k+2, p=6k+3, p=6k+4 or p=6k+5 for some k##\in\mathbb{Z}##. Since p##\geq##5 is a prime number, it follows that p cannot...
25. ### The only prime p for which 3p+1 is a perfect square is p=5?
Proof: Suppose that p is a prime and 3p+1=n^2 for some n##\in\mathbb{Z}##. Then we have 3p+1=n^2 3p=n^2-1 3p=(n+1)(n-1). Since n+1>3 for ##\forall...
26. ### The only prime of the form n^2-4 is 5?
Proof: Suppose p is a prime such that p=n^2-4. Then we have p=n^2-4=(n+2)(n-2). Note that prime number is a number that has only two factors, 1 and the number itself. Since n+2>1 for ##\forall n \in...
27. ### The only prime of the form n^3-1 is 7?
Proof: Suppose p is a prime such that p=n^3-1. Then we have p=n^3-1=(n-1)(n^2+n+1). Note that prime number is a number that has only two factors, 1 and the number itself. Since n^2+n+1>1 for ##\forall...
28. ### MHB -c5.LCM and Prime Factorization of A,B,C
Build the least common multiple of A, B, and C Then write the prime factorization of the least common multiple of A, B, and C. $A = 2 \cdot 3^2 \cdot 7 \cdot 13 \cdot 23^8$ $B = 2 3^5 \cdot 5^9 \cdot 13$ $C = 2 \cdot 5 \cdot 11^8 \cdot 13^3$ $\boxed{?}$ ok this only has a single answer...
46. ### I Significance of discovering a function which computes the n-th prime?
Let's say the function is of a form F(x), where F(n) gives us the value of the n-th prime number, for all values of n. And it computes and discovers subsequent prime numbers directly without using any form of brute-force computation.If such a function is discovered, how ground-breaking would it...
47. ### Prime Number Detection: Running Time of Algorithm
Hi, I want to know if the algorithm to find prime number has running time O(sqrt(N)) or O(log^2N). log^2N is better than sqrt(N). Also for large values can it become a pseudo polynomial algorithm? Zulfi.
48. ### I Prime factorization and real exponents
I know that the prime factorization theorem predicts that a prime number raised to an integer power will never be equal to another prime number raised to a different power. But does this apply to real number powers? For example, suppose there is a prime number raised to some real value, could...
49. ### B Aren't There Any Formulas for Prime Numbers?
Hello all, We know that following formulas failed to produced all prime numbers for any given whole number ##n##: ##f(n) = n^2 - n - 41##, failed for ##n = 41~(f = 1681)## ##g(n) = 2^(2^n) + 1##, failed for ##n = 5~(g = 4,294,967,297)## ##m(n) = 2^n - 1##, failed for ##n = 67~(m =...
50. ### MHB Solved for Prime using Grade 4 Math Prove me wrong
I Need someone to publish Grade 4 Prime Sequence Series Table, just add my name to it and were partners oh and teach this to your students, this took 40 years to develope https://www.youtube.com/watch?v=eDKK8pP1jiw | 3,925 | 13,425 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-18 | latest | en | 0.957195 |
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# Fractions Question From GMAT Club Test (m25#15)
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Fractions Question From GMAT Club Test (m25#15) [#permalink]
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08 May 2011, 00:27
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Which of the following fractions is the largest?
(A) $$\frac{3252}{3257}$$
(B) $$\frac{3456}{3461}$$
(C) $$\frac{3591}{3596}$$
(D) $$\frac{3346}{3351}$$
(E) $$\frac{3453}{3458}$$
Last edited by fameatop on 28 Jul 2013, 12:08, edited 1 time in total.
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taking the last two digits of the fractions,
52/57 , 56/61 , 91/96 , 46/51 , 53/58
1 . difference between numerator and denominator = 5 in each cases. Hence a particular fraction can't be eliminated on this basis.
2. Cross multiplications 52/57 ,56/61 here 56/61 is greater
46/51, 53/58 here 53/58 is greater
56/61, 91/96 here 91/96 is greater
91/96, 53/58 here 91/96 is greater.
Hence 91/96 C is greatest.
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Re: From MitDavidDv: Fractions Question From GMAT Club Test [#permalink]
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08 May 2011, 01:43
MitDavidDv wrote:
Which of the following fractions is the largest?
A. $$\frac{3252}{3257}$$
B.$$\frac{3456}{3461}$$
C. $$\frac{3591}{3596}$$
D.$$\frac{3346}{3351}$$
E. $$\frac{3453}{3458}$$
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Re: From MitDavidDv: Fractions Question From GMAT Club Test [#permalink]
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08 May 2011, 03:57
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C.
My approach was to note that each numerator was five less than the denominator.
i.e. $$\frac{(x-5)}{x} = 1 - \frac{5}{x}$$
To maximise this function, we maximise $$x$$, therefore, solution is C.
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Re: From MitDavidDv: Fractions Question From GMAT Club Test [#permalink]
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08 May 2011, 07:38
Good Approach pike. +1 for you
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Re: Fractions Question From GMAT Club Test (m25#15) [#permalink]
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28 Jul 2013, 11:39
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Search for the pattern of the fractions. As also given in the posts before, you will find out that the numerator is denominator - 5.
Think about a similar example with a similar pattern.
Easiest: (here numerator is denominator - 1)
5/6
4/5
3/4
2/3
1/2
Which one is now the largest?
You got it?
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A very nice Fraction question that can be solved very easily using logic.
If Any fraction is smaller than 1 and if we add same constant to both the Numerator & Denominator, then ratio increases in value towards 1
3/4 = .75
(3+1)/(4+1)=.80 (increased in Value)
(3+3)/(4+3)=.85 (increased in Value)
If Any fraction is Greater than 1 and if we add same constant to both the Numerator & Denominator, then ratio decreases in value towards 1
5/4 = 1.25
(5+1)/(4+1)=1.20 (decreased in Value)
(5+3)/(4+3)=1.14 (decreased in Value)
Which of the following fractions is the largest?
(A) $$\frac{3252}{3257}$$
(B) $$\frac{3456}{3461}$$ = $$\frac{3252+4}{3257+4}$$ - Increases in Value- higher than option A
(C) $$\frac{3591}{3596}$$ = $$\frac{3256+35}{3261+35}$$ - Increases in Value- higher than option B
(D) $$\frac{3346}{3351}$$ = $$\frac{3252-6}{3257-6}$$ - Decreases in Value- Lower than option A
(E) $$\frac{3453}{3458}$$ = $$\frac{3252+1}{3257+1}$$ - Increases in Value- higher than option A
Hope this property will help many.
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Re: Fractions Question From GMAT Club Test (m25#15) [#permalink]
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23 Oct 2013, 05:49
at first glance we can see the pattern of 5. The next point is that the larger numbers form the fraction the larger value will be (becasue the fraction gets closer to 1)
according to all above said i go with C
Re: Fractions Question From GMAT Club Test (m25#15) [#permalink] 23 Oct 2013, 05:49
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# Fractions Question From GMAT Club Test (m25#15)
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https://curriculum.illustrativemathematics.org/MS/teachers/2/5/1/index.html | 1,725,898,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00407.warc.gz | 167,927,929 | 32,368 | # Lesson 1
Interpreting Negative Numbers
## 1.1: Using the Thermometer (5 minutes)
### Warm-up
The purpose of this warm-up is to remind students about negative numbers. The context of a weather thermometer works like a vertical number line. Students do not need to understand comparative temperatures in Celsius and Fahrenheit. The activity is written with temperatures in Celsius; however, the activity would work the same if the thermometer was labeled in Fahrenheit. These two different systems for measuring temperature is an opportunity to remind students that what counts as zero is abitrary and was chosen by someone as some point. If desired, explain to students that $$0^\circ$$ Celsius is the freezing point of fresh water and $$0^\circ$$ Fahrenheit is based on the freezing point of salt water.
### Launch
Display the thermometer image for all to see. Explain that degrees Celsius is a way of measuring temperature, like degrees Fahrenheit—but it has a different zero point. Students may already know that $$0^\circ$$ Celsius is based on the freezing point of water and $$0^\circ$$ Fahrenheit on the freezing point of brine, but these were chosen by people; there’s no reason they had to be this way. Give students 1 minute of quiet think time to examine the picture before they start writing.
### Student Facing
Here is a weather thermometer. Three of the numbers have been left off.
1. What numbers go in the boxes?
2. What temperature does the thermometer show?
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Some students may think the missing number between 0 and -10 needs to have a magnitude larger than -10, such as -15, because on the positive side of the number line, numbers increase in magnitude as you go up.
### Activity Synthesis
Ask students to share their responses for the first question and explain their reasoning. After each response, ask students to indicate if they agree or disagree. If all students are in agreement, record and display the missing temperatures for all to see. If they disagree, have students explain their reasoning until they reach an agreement.
Ask students to share their responses to the second question. Because the thermometer is labeled in 5 degree increments, we have to estimate the temperature between $$0^\circ$$ and $$-5^\circ$$. Ask students to explain their reasoning and record and display possible responses for all to see. Highlight student responses that include the following ideas:
• The location of negative numbers below 0.
• The distance between numbers on the vertical number line.
## 1.2: Fractions of a Degree (5 minutes)
### Activity
In this activity, students return to the context of a thermometer to examine rational numbers that are not integers. Students compare and interpret the signed numbers to make sense of them in the context (MP2), including comparing a temperature that is not pictured to the temperatures that are pictured.
### Launch
Remind students of the warm-up problem about a weather thermometer. Instruct them to estimate when necessary.
Writing, Speaking: MLR5 Co-craft Questions. To help students make sense of the drawings in this problem and to increase their awareness of the language used when comparing signed numbers, show students just the images of the four thermometers. Ask pairs of students to write their own mathematical questions about the situation. Listen for how students use the idea of numbers being above or below zero. Ask pairs to share their questions with the whole class. Highlight specific questions that are related to comparing numbers above or below zero. This will help students develop meta-awareness of the language used when comparing signed numbers.
Design Principle(s): Maximize meta-awareness; Support sense-making<
### Student Facing
1. What temperature is shown on each thermometer?
2. Which thermometer shows the highest temperature?
3. Which thermometer shows the lowest temperature?
4. Suppose the temperature outside is $$\text{-}4^\circ \text{C}$$. Is that colder or warmer than the coldest temperature shown? How do you know?
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Some students may struggle to estimate the temperature on the last thermometer, since it is between two markings. Ask them to tell what the temperature would be for the lines directly above and directly below the thermometer's level. Then ask what temperature would be halfway in between those two numbers.
Some students may struggle with comparing $$\text-4^\circ \text{C}$$ to the temperatures shown on the thermometers. Prompt students to point out where $$\text-4^\circ \text{C}$$ would be on the thermometer that is showing $$\text-3^\circ \text{C}$$.
### Activity Synthesis
Ask one or more students to share their response for the temperature for each thermometer.
When discussing the last question, first have students explain their reasoning until they come to an agreement that $$\text-4^\circ\text{C}$$ is colder than $$\text-3^\circ\text{C}$$. Then, if not brought up in students’ explanations, introduce the notation $$\text-4 < \text-3$$ and remind students that this is read, "Negative 4 is less than negative 3." Explain that -4 is farther away from zero than -3 is, and point to the location of -4 on a thermometer to show that is it below -3. On the negative side of the number line, that means -4 is less than -3. Familiarity with less than notation will be useful for describing their reasoning in the next activity.
## 1.3: Seagulls Soar, Sharks Swim (10 minutes)
### Activity
The purpose of this activity is for students to continue interpreting signed numbers in context and to begin to compare their relative location. A vertical number line shows the heights above sea level or depths below sea level of various animals. The number line is labeled in 5 meter increments, so students have to interpolate the height or depth for some of the animals. Next, they are given the height or depth of other animals that are not pictured and asked to compare these to the animals shown.
As students work, monitor for whether they are expressing relative distances in words, for example “3 meters below,” or if they are expressing the same idea with notation, as in -3 meters. Both are acceptable; these ideas are connected in the discussion that follows (MP2). Also monitor for students who notice that there are two possible answers for the last question.
### Launch
Display the image for all to see. Tell students to measure the height or depth of each animal's eyes, to the nearest meter. Remind students that we choose sea level to be our zero level, in the same way that we chose a zero level for temperature.
Representation: Internalize Comprehension. Represent the same information with additional structure. If students are unsure where to begin, suggest that they extend a straight horizontal line at each depth to determine the height or depth of each animal.
Supports accessibility for: Visual-spatial processing; Conceptual processing
Conversing, Representing: MLR2 Collect and Display. Use this routine to capture the language students use during discussion with a partner. Collect the language of opposites: “above” or “below.” For example, “The albatross is 3 meters above the penguin or the penguin is 3 meters below the albatross.” Then, identify students who use negative numbers to describe these differences to share their reasoning during the whole-class discussion. Ensure students connect this language to, “The difference in height is +3 (to represent above) or -2.5 (to represent below).”
Design Principle(s): Maximize meta-awareness
### Student Facing
Here is a picture of some sea animals. The number line on the left shows the vertical position of each animal above or below sea level, in meters.
1. How far above or below sea level is each animal? Measure to their eye level.
2. A mobula ray is 3 meters above the surface of the ocean. How does its vertical position compare to the height or depth of:
The jumping dolphin?
The flying seagull?
The octopus?
3. An albatross is 5 meters above the surface of the ocean. How does its vertical position compare to the height or depth of:
The jumping dolphin?
The flying seagull?
The octopus?
4. A clownfish is 2 meters below the surface of the ocean. How does its vertical position compare to the height or depth of:
The jumping dolphin?
The flying seagull?
The octopus?
5. The vertical distance of a new dolphin from the dolphin in the picture is 3 meters. What is its distance from the surface of the ocean?
### Student Response
For access, consult one of our IM Certified Partners.
### Student Facing
#### Are you ready for more?
The north pole is in the middle of the ocean. A person at sea level at the north pole would be 3,949 miles from the center of Earth. The sea floor below the north pole is at an elevation of approximately -2.7 miles. The elevation of the south pole is about 1.7 miles. How far is a person standing on the south pole from a submarine at the sea floor below the north pole?
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
If students measure to the top or bottom of the animal, remind them that we are using the eyes of the animal to measure their height or depth.
Some students may struggle to visualize where the albatross, seagull, and clownfish are on the graph. Consider having them draw or place a marker where the new animal is located while comparing it to the other animals in the picture.
### Activity Synthesis
The main point for students to get out of this activity is that we can represent distance above and below sea level using signed numbers. The depths of the shark, fish, and octopus can be expressed as approximately -3 m, -6 m, and -7.5 m respectively, because they are below sea level.
Signed numbers can also be used to represent the relative vertical position of different pairs of animals. Have selected students share their responses and reasoning for how the heights of the albatross, seabird, and clownfish compare to the dolphin, seagull, and octopus. Record and display their verbal descriptions using signed numbers. For example, if a student says the albatross is 7 meters below the seagull, write "-7".
Finally, ask whether students noticed the ambiguity in the last question (about the height of the new dolphin). Ask such a student to explain why there are two possible answers to the last question.
## 1.4: Card Sort: Rational Numbers (15 minutes)
### Optional activity
This activity reviews ordering integers first, and then rational numbers second. Many of the numbers also have their additive inverse in the set, which can help students use the structure of the number line to order the numbers.
The previous activities in this lesson used vertical number lines to help students make sense of negative numbers being below 0. It is important that students also feel comfortable working with horizontal number lines. As students work on ordering these slips, it is likely they will automatically make the transition to using a horizontal orientation. Watch for any groups that continue to use a vertical orientation and prompt them to consider whether they have really ordered their numbers from least to greatest.
Monitor for students who specifically compare the magnitudes of numbers and translate that into the correct number order (such as $$2.5 > 2$$ so $$\text-2.5 < \text-2$$) are using the structure of the number line (MP7); ask them to share their reasoning in the whole-class discussion.
### Launch
Arrange students in groups of 3. Distribute the first set of cards (integers) to each group. Instruct the students to put the cards in order from least to greatest. When a group has finished ordering the first set, give them the second set (rational numbers that are not integers) and have them add these in the correct locations.
Conversing: MLR8 Discussion Supports. To support students as they explain their reasoning for how they placed the cards, provide sentence frames such as: “First, I ___ because ___ .”, “I noticed ___ so I....”, and “I know ___ is greater than/less than ___ because….”.
Design Principle(s): Support sense-making; Cultivate conversation
### Student Facing
1. Your teacher will give your group a set of cards. Order the cards from least to greatest.
2. Pause here so your teacher can review your work. Then, your teacher will give you a second set of cards.
3. Add the new set of cards to the first set so that all of the cards are ordered from least to greatest.
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Some students may struggle with ordering the negative numbers. For example, they may put -2.5 to the right of -2 since they are used to seeing 2.5 to the right of 2. Help students visualize a number line and figure out which number should be farther away from 0.
### Activity Synthesis
Select students to share their strategies when sorting. Highlight strategies that used the magnitudes of a number and its additive inverse.
Discuss:
• Which numbers were the hardest to order? Why?
• How did you decide where to put the fractions?
• How is, for example, $$\text-\frac98$$ related to $$\frac98$$?
Introduce the convention that number lines are usually drawn horizontally, with the negative numbers to the left of 0. If any groups put their slips in order vertically, considering having them reposition their slips to match the orientation of a horizontal number line. Make sure students understand the meaning of the term “opposite” and absolute value notation.
## Lesson Synthesis
### Lesson Synthesis
Main learning points:
• Negative numbers can be used to represent quantities below a chosen zero point.
• Negative numbers can be ordered to the left side of zero on a horizontal number line.
• Absolute value, or magnitude, describes how far away from zero a value is.
Discussion questions:
• Which number is greater, -7 or -12?
• Which number has the greater magnitude, 7 or -12?
• How can we order negative numbers?
## 1.5: Cool-down - Signed Numbers (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners.
## Student Lesson Summary
### Student Facing
We can use positive numbers and negative numbers to represent temperature and elevation.
When numbers represent temperatures, positive numbers indicate temperatures that are warmer than zero and negative numbers indicate temperatures that are colder than zero. This thermometer shows a temperature of -1 degree Celsius, which we write $$\text{-}1^\circ \text{C}$$.
When numbers represent elevations, positive numbers indicate positions above sea level and negative numbers indicate positions below sea level.
We can see the order of signed numbers on a number line.
A number is always less than numbers to its right. So $$\text{-}7 < \text{-}3$$.
We use absolute value to describe how far a number is from 0. The numbers 15 and -15 are both 15 units from 0, so $$|15| = 15$$ and $$| \text{-}15| = 15$$. We call 15 and -15 opposites. They are on opposite sides of 0 on the number line, but the same distance from 0. | 3,280 | 15,361 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.949845 |
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# S1: Chapter 9 The Normal Distribution - PowerPoint PPT Presentation
S1: Chapter 9 The Normal Distribution. Dr J Frost ([email protected]) . Last modified : 4 th March 2014. What does it look like?. The following shows what the probability distribution might look like for a random variable X, if X is the height of a randomly chosen person.
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### S1: Chapter 9The Normal Distribution
Dr J Frost ([email protected])
The following shows what the probability distribution might look like for a random variable X, if X is the height of a randomly chosen person.
We expect this ‘bell-curve’ shape, where we’re most likely to pick someone with a height around the mean of 180cm, with the probability diminishing symmetrically either side of the mean.
p(x)
180cm
Height in cm (x)
A variable with this kind of distribution is said to have a normal distribution.
For normal distributions we tend to draw the axis at the mean for symmetry.
We can set the mean and the standard deviation of the Normal Distribution.
For a Normal Distribution to be used, the variable has to be:
continuous
With a discrete variable, all the probabilities had to add up to 1.
For a continuous variable, similarly:
the area under the probability graph has
to be 1.
To find P(170cm < X < 190cm), we could:
find the area between these values.
Would we ever want to find P(X = 200cm) say?
Since height is continuous, the probability someone is ‘exactly’ 200cm is infinitesimally small. So not a ‘probability’ in the normal sense.
Q1
?
Q2
p(x)
?
Q3
?
Q4
?
170cm
180cm
190cm
Height in cm (x)
If a variable is ‘normally distributed’ (i.e. its probability function uses a normal distribution), then we write:
...is distributed...
...using a Normal distribution with mean and variance
The random variable X...
!
?
The Z value is the number of standard deviations a value is above the mean.
Example
IQ, by definition, is normally distributed for a given population. By definition, and
i.e.
p(x)
?
?
?
?
?
100
IQ (x)
Minimise
A z-table allows us to find the probability that the outcome will be less than a particular z value.
For IQ, P(Z < 2) would mean “the probability your IQ is less than 130”.
(You can find these values at the back of your textbook, and in a formula booklet in the exam.)
Expand
P(Z < 2) = 0.9772
?
z=0
z=1
z=2
100
130
115
IQ (x)
You may be wondering why we have to look up the values in a table, rather than being able to calculate it directly. The reason is that calculating the area under the graph involves integrating (see C2), but the probability function for the normal distribution (which you won’t see here) cannot be integrated!
Suppose we’ve already worked out the z value, i.e. the number of standard deviations above or below the mean.
Bro Tip: We can only use the z-table when:
The z value is positive (i.e. we’re on the right half of the graph)
We’re finding the probability to the left of this z value.
1
2
?
?
This is clear by symmetry.
3
4
?
?
Determine the following probabilities, ensuring you show how you have manipulated your probabilities (as per the previous examples).
1
?
?
2
?
3
4
?
5
?
?
6
7
?
8
?
9
?
10
?
We have seen that in order to look up a value in the table, we needed to first convert our IQ into a value. We call this ‘standardising’ our variable, because we’re turning our normally distributed variable into another one where the mean is 0 and the standard deviation is 1.
Standardise
e.g.
Why is ? Well consider a z value of 3 for example. We understand that to mean 3 standard deviations above the mean. But if and , the 3 is 3 lots of 1 above 0!
The heights in a population are normally distributed with mean 160cm and standard deviation 10cm. Find the probability of a randomly chosen person having a height less than 180cm.
Here’s how they’d expect you to lay out your working in an exam:
No marks attached with this, but good practice!
?
A statement of the problem.
?
M1 for “attempt to standardise”
?
?
Look up in z-table
Steel rods are produced by with a mean weight of 10kg, and a standard deviation of 1.6kg. Determine the probability that a randomly chosen steel rod has a weight below 9kg.
?
Edexcel S1 May 2012
P(Z > -1.6)
= P(Z < 1.6)
= 0.9452
?
Edexcel S1 May 2013 (Retracted)
?
P(Z < -0.5)
= P(Z > 0.5)
= 1 – P(Z < 0.5)
= 0.3085
Edexcel S1 Jan 2011
P(Z > 1.6)
= 1 – P(Z < 1.6)
= 0.0548
?
Quickfire Probabilities
(For further practice outside class)
These are for further practice if you’re viewing these slides outside of class. You don’t need a calculator, as the values are obvious!
Let X represent the IQ of a randomly chosen person, where
Example:
P(X < 115) = P(Z < 1) = 0.8413
P(X < 107.5) = P(Z < 0.5) = 0.6915
P(X > 122.5) = P(Z > 1.5) = 1 – P(Z < 1.5) = 0.0668
P(X > 85) = P(X < 115) = P(Z < 1) = 0.8413
P(X < 106) = P(Z < 0.4) = 0.6554
P(X > 95) = P(X < 105) = P(Z < 0.33) = 0.6293
P(X < 92.5) = P(Z < -0.5) = P(Z > 0.5) = 1 – P(Z < 0.5) = 0.3085
P(X < 80) = P(Z < -1.33) = P(Z > 1.33) = 1 – P(Z < 1.33) = 0.0918
1
?
2
?
3
?
4
?
5
?
?
6
7
?
Again, let X represent the IQ of a randomly chosen person, where
Thinking about the graph of the normal distribution, find:
P(96 < X < 112)
?
This easiest way is to find P(X < 112) and ‘cut out’ the area corresponding to
P(X < 96):
z=0
96
100
112
We can see that, in general:
P(a < Z < b) = P(Z < b) – P(Z < a)
IQ (x)
Quickfire Probabilities
Let X represent the IQ of a randomly chosen person, where
P(100 < X < 107.5) = P(Z < 0.5) – 0.5 = 0.1915
P(145 < X < 151) = P(3 < Z < 3.4) = P(Z < 3.4) – P(Z < 3) = 0.001
P(116 < X < 120) = P(1.07 < X < 1.33) = 0.9082 – 0.8577 = 0.0505
P(80 < X < 110) = P(-1.33 < X < 0.67) = 0.7486 – (1 – 0.9082) = 0.6568
P(70 < X < 90) = P(-2 < Z < -0.67) = (1 – 0.7486) – (1 – 0.9772) = 0.2286
1
?
2
?
3
?
4
?
5
?
Probability
z-value
(num standard deviations above mean)
Original value
or
This essentially says that our value is standard deviations above the mean.
must be positive.
Use Z-table
must be <
Sometimes we’re given the probability, and need to find the value, so that we can determine a missing value or the standard deviation/mean.
Just use the z-table backwards!
For nice ‘round’ probabilities, we have to look in the second z-table. You’ll lose a mark otherwise.
?
?
?
?
?
?
Bro Tip: Remember that either flipping the inequality, or changing the sign of will cause your probability to become 1 minus it.
?
?
?
?
?
?
?
?
Again, let X represent the IQ of a randomly chosen person, where
What IQ corresponds to the top 22% of the population?
State the problem in probabilistic terms.
?
?
‘Standardise’.
0.78
Find the closest probability in the z-table.
z = 0.77
?
Bro Tip: Draw a diagram for these types of questions if it helps.
We can use this value to find the value of .
?
Again, let X represent the IQ of a randomly chosen person, where
What IQ corresponds to the top 10% of the population?
?
State the problem in probabilistic terms.
‘Standardise’
?
0.9
Find the closest probability in the z-table.
z = 1.2816
?
z = 1.2816
Convert from value to IQ.
?
Again, let X represent the IQ of a randomly chosen person, where
What IQ corresponds to the bottom 30% of the population?
State the problem in probabilistic terms.
?
‘Standardise’
(I haven’t written yet because it’ll make the manipulation more tedious. I’ve used a variable to represent the value)
?
We can’t look up probabilities less than 0.5 in the table, so manipulate:
?
Find the closest probability in the z-table.
Again, use the second table.
z = -0.5244
?
Convert back into an IQ.
?
On the planet Frostopolis, the mean height of a Frongal is 1.57m and the standard deviation 0.2m. Determine:
The height for which 65% of Frongals have a height less than.
The height for which 40% of Frongals have a height more than.
c) The height for which 23% of Frongals have a height less than.
?
?
?
Remember:
State your problem in probabilistic terms.
Standardise. Manipulate if necessary so that your probability is above 0.5 and you’re finding .
Use your z-table backwards to find the z-table.
Use to find your value of .
Page 184
Given
a)
b)
Given , find such that
Given that , find such that
. Find the value of and such that:
a)
b)
c)
1
?
?
6
?
7
?
9
?
?
?
P(X = x)
IQ (x)
Q3 = 110
?
Q1 = 90
?
Q2 = 100
?
i.e. The upper quartile is two-thirds of a standard deviation above a mean (useful to remember!)
Edexcel S1 Jan 2011
?
z = -2.3263 (using 2nd z-table)
w = 160 – (5 x 2.3263) = 148.3685
Edexcel S1 May 2012
?
P(Z < z) = 0.6
z = 0.2533 (using 2nd table)
W = 162 + (7.5 x 0.2533)
= 163.89975cm
Edexcel S1 May 2010
?
P(D > 20) = P(Z > -1.25) = P(Z < 1.25) = 0.8944
P(Z < z) = 0.75 -> z = 0.67Q3 = 30 + (0.67 x 8) = 35.36
Q1 = 30 – (0.67 x 8) = 24.64
The random variable
Given that P(X > 20) = 0.20, find the value of .
The random variable
Given that P(X < 46) = 0.2119, find the value of .
?
(Normalising)
?
If your standard deviation is negative, you know you’ve done something wrong!
The random variable
Given that P(X > 35) = 0.025 and P(X < 15) = 0.1469, find the value of and the value of
First deal with P(X > 35) = 0.025
Next deal with P(X < 15) = 0.1469
P(X < 35) = 1 – 0.025 = 0.975
If P(Z < z) = 0.975, then z = 1.96.
?
?
We now have two simultaneous equations. Solving gives:
?
For the weights of a population of squirrels, which are normally distributed, Q1 = 0.55kg and Q3 = 0.73kg.
Find the standard deviation of their weights.
= 0.114
= 0.124
= 0.134
= 0.144
Due to symmetry, = (0.55 + 0.73)/2 = 0.64kg
If P(Z < z) = 0.75, then z = 0.67.
0.64 + 0.67 = 0.73
= 0.134
Only 10% of maths teachers live more than 80 years. Triple that number live less than 75 years. Given that life expectancy of maths teachers is normally distributed, calculate the standard deviation and mean life expectancy.
RIP
Similarly
– 0.5244 = 75
= 76.15
= 76.25
= 76.35
= 76.45
A Ingall
He loved math
= 2.77
= 2.78
= 79
= 2.80
Edexcel S1 May 2013 (Retracted)
?
P(Z < z) = 0.85, then z = 1.04.
So 115 minutes is 1.04 standard deviations above the mean.
Edexcel S1 Jan 2011
Using z-table, 160 is 2.32 standard deviations above the mean. So:
Similarly, z-value for 0.9 is 1.28. By symmetry, 152 has z value of -1.28. So:
Solving, and
?
Edexcel S1 Jan 2002
z-value for 0.975 is 1.96. By symmetry, 235 is 1.96 standard deviations below mean.So . The result follows.
P(Z < z) = 0.85. So z = 1.04
Solving, ,
If 0.683 in the middle, (0.683/2)+0.5=0.8415 prob below value above mean. Thus z = 1.So values are 154.8 – 2.22 and 154.8 + 2.22
?
We need to use the second z-table whenever:
we’re looking up the z value for certain ‘round’ probabilities.
E
A normal distribution is good for modelling data which:
tails off symmetrically about some mean/ has a bell-curve like distribution.
A
?
?
?
P(a < X < b) = P(X < b) – P(X < a)
F
A z-value is:
The number of standard deviations above the mean.
B
?
We can treat quartiles and percentiles as probabilities.
For IQ, what is the 85th percentile?
100 + (1.04 x 15) = 115.6
We can form simultaneous equations to find the mean and standard deviation, given known values with their probabilities.
G
If a random variable X is normally distributed with mean 50 and standard deviation 2, we would write:
C
?
?
A z-table is:
A cumulative distribution function for a normal distribution with mean 0 and standard deviation 1.
P(IQ < 115) = 0.8413
D
H
?
?
Edexcel S1 June 2001 | 3,892 | 12,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-51 | latest | en | 0.878814 |
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##### Stage: 1 and 2 Challenge Level:
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In this section from a calendar, put a square box around the 1st, 2nd, 8th and 9th. Add all the pairs of numbers. What do you notice about the answers? | 2,258 | 10,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.922396 |
https://byjus.com/wbjee/wbjee-2018-maths-question-paper/ | 1,675,162,102,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00345.warc.gz | 172,060,109 | 97,469 | # WBJEE 2018 Maths Paper with Solutions
WBJEE 2018 Maths question paper with solutions is a perfect resource for those who are preparing for WBJEE examination. The first step towards the WBJEE exam preparation is to solve WBJEE Previous Year Question Papers. Along with the question paper, the answer key of WBJEE 2018 has also been provided here. Solving WBJEE previous year papers will give students an idea about the nature and pattern of questions asked in the exam. It will also help the students to get a clear idea about the amount of preparation required to face the exam.
### WBJEE 2018 - Maths
Question 1: If (2 ≤ r ≤ n), then nCr + 2 . nCr+1 + nCr+2 is equal to
1. a. 2. nCr+2
2. b. n+1Cr+1
3. c. n+2Cr+2
4. d. n+1Cr
Solution:
nCr + 2. nCr+1 + nCr+2
=nCr + nCr+1+ nCr+1 + nCr+2
=n+1Cr+1 + n+1Cr+2
=n+2Cr + 2
Question 2: The number (101)100 – 1 is divisible by
1. a. 104
2. b. 106
3. c. 108
4. d. 1012
Solution:
(101)100 – 1 =(1 + 100)2 –1
=[100C0 +100C1 100 + 100C2 (100)2+ ………+ 100C100(100)100 (100C0)=1
(1+100)100-1= 1 + 100(100)+100C21002+………+100C100100100-1
=1002(1+ 100C2 +100C3(100)+………+100C10010098)
So, 101100-1 is divisible by 104
Question 3: If n is even positive integer, then the condition that the greatest term in the expansion of (i + x)n may also have the greatest coefficient is
1. a. n / [n + 2] < x < [n + 2] / n
2. b. n / [n + 1] < x < [n + 1] / n
3. c. [n + 1] / [n + 2] < x < [n + 2] / [n + 1]
4. d. [n + 2] / [n + 3] < x < [n + 3] / [n + 2]
Solution:
For the greatest term, we have
n / 2 < {[n + 1] / 1 + | x |} ≤ (n / 2) + 1
n / 2 < {[n + 1] / 1 + | x |} and [n + 1] / | x | + 1 + ≤ (n / 2) + 1
[1 + | x |] / 2 + < [n + 1] / n + and [n + 1] / [n + 2] ≤ [| x | + 1] / 2
|x| < {[2n + 2] / [n]} - 1 and [2n + 2] / [n + 2] –1 ≤ |x|
x < [n + 2] / n and [n] / [n + 2] ≤ x
[n] / [n + 2] < x [n + 2] / n
Question 4:
1. a. A2
2. b. A2 – A + I3
3. c. A2 – 3A + I3
4. d. 3A2 + 5A – 4I3
Solution:
Given
B = 15 [0 – 0 – 6 (–90)]
B = (–90) (–90)
B = A. A
B = A2
Question 5: If ar = (cos2r𝛑 + i sin 2r𝛑)1/9, then the value of
$$\begin{vmatrix} a_1 &a_2 &a_3 \\ a_4 & a_5 & a_6\\ a_7 & a_8 &a_9 \end{vmatrix}$$
is
1. a. 1
2. b. –1
3. c. 0
4. d. 2
Solution:
Given ar = (cos2r𝛑 + i sin 2r𝛑)1/9
ar = e2r𝛑i/9
Now,
= 0 {two rows are same}
Question 6: If Sr =
Then the value of ∑r=1n Sr is independent of
1. a. x only
2. b. y only
3. c. n only
4. d. x, y, z and n
Solution:
Question 7: If the following three linear equations have a non –trivial solution, then x + 4ay + az = 0, x + 3by + bz = 0, x + 2cy + cz = 0
1. a. a, b, c are in A.P
2. b. a, b, c are in G.P.
3. c. a, b, c are in H.P
4. d. a + b + c = 0
Solution:
System equations
x + 4ay + az = 0
x + 3by + bz = 0
x + 2cy + cz = 0
For non-trivial solution
Δ = 0
1(3bc – 2bc) – 1 (4ac – 2ac) + (4ab – 3 ab) = 0
bc – 2ac + ab = 0
bc + ab = 2ac
b = 2ac / a + c
a, b, c are in H.P.
Question 8: On R, a relation 𝞀 is defined by x𝞀y if and only if x–y is zero or irrational. Then
1. a. 𝞀 is equivalence relation
2. b. 𝞀 is reflexive but neither symmetric nor transitive
3. c. 𝞀 is reflexive & symmetric but not transitive
4. d. 𝞀 is symmetric & transitive but not reflexive
Solution:
xRy ⇒ x – y is zero or irrational
xRx ⇒ 0 [hence reflexive]
ifxRy ⇒ x – y is zero or irrational
⇒ y – x is zero or irrational
yRx symmetric
xRy ⇒ x – y is 0 or irrational
yRz ⇒ y – z is 0 or irrational
Then (x – y) + (y – z) = x – z may be rational
It is not transitive.
Question 9: On the set R of real numbers, the relation 𝞀 is defined by x𝞀y,(x,y) ∈ R.
1. a. if |x – y| < 2 then 𝞀 is reflexive but neither symmetric nor transitive
2. b. if x – y < 2 then 𝞀 is reflexive and symmetric but not transitive
3. c. if |x| ≥ y then 𝞀 is reflexive and transitive but not symmetric
4. d. if x > |y| then 𝞀 is transitive but neither reflexive nor symmetric
Solution:
(x, x) ∈ R ⇒ x > |x| {false}
not reflexive
If (x, y) ∈ R ⇒ x > |y| ⇒ y > |x|
not symmetric
If (x, y) ∈ R ⇒ x > |y|; (y, z) ∈ R y > |z|
x > |z| ⇒ (x, z) ∈ R
Transitive
Question 10: If f : R → R be defined by f (x) = ex and g : R →R be defined by g (x) = x2. The mapping g of : R → R be defined by (g of )x = g [f (x)] ∀ x ∈ R, then
1. a. gof is bijective but f is not injective
2. b. gof is injective and g is injective
3. c. gof is injective but g is not bijective
4. d. gof is surjective and g is surjective
Solution:
Given f (x) = ex
g (x) = x2
(g of) (x) = g [f (x)]
= g [ex]
= (ex)2
= e2x : x ∈ R
Clearly g(f(x)) is injective but g(x) is not injective.
Question 11: In order to get heads at least once with probability ≥ 0.9, the minimum number of times an unbiased coin needs to be tossed is
1. a. 3
2. b. 4
3. c. 5
4. d. 6
Solution:
p (H) = 1 / 2
p (T) = 1 / 2
P = 1 - (1 / 2n) ≥ 0.9
1 - (9 / 10) ≥ (1 / 2n)
(1 / 10) ≥ (1 / 2n)
10 ≤ 2n
n = 4
Question 12: A student appears for tests I, II and III. The student is successful if he passes in tests I, II or III. The probabilities of the student passing in tests I, II and III are respectively p,q and 1 / 2. If the probability of the student to be successful is 1 / 2. Then
1. a. p (1 + q) = 1
2. b. q (1 + p) =1
3. c. pq = 1
4. d. (1 / p) + (1 / q) = 1
Solution:
Let A, B and C be the events that the student is successful in tests I, II and III respectively. Then p (The student is successful)
= P [(I ∩ II ∩ III’) ⋃ (I ∩ II’ ∩ III) ⋃ (I ∩ II ∩ III)]
1 / 2 = P (I ∩ II ∩ III’) + P (I ∩ II’ ∩ III) + P (I ∩ II ∩ III)
1 / 2 = P (I) P (II) P (III’) + P (I) P (II’) P (III) + P (I) P (II) P (III) [I, II and III are independent]
1 / 2 = p . q (1 - [1 / 2]) + p . (1 - q) . (1 / 2) + p . q . (1 / 2)
1 = pq + p
p (q + 1) = 1
Now,
⇒ P2 = q
⇒ (a)2 = 2a
⇒ a = 2
Question 13: If sin 6θ + sin4θ + sin2θ = 0, then general value of θ is (n is integer)
1. a. n𝛑 / 4, n𝛑 ± (𝛑 / 3)
2. b. n𝛑 / 4, 4𝛑 ± (𝛑 / 6)
3. c. n𝛑 / 4, 2n𝛑 ± (𝛑 / 3)
4. d. n𝛑 / 4, 2n𝛑 ± (𝛑 / 6)
Solution:
sin 6θ + sin4θ + sin2θ = 0
sin 4θ + 2 sin (8θ / 2) cos (4θ / 2) = 0
sin 4θ + 2 sin 4θ cos 2θ = 0
sin 4θ (1 + 2 cos 2θ) = 0
sin 4θ = 0 or 1 + 2 cos 2θ = 0
4θ = n𝛑 or cos 2θ = -1 / 2 = cos (2𝛑 / 3)
θ = n𝛑 / 4 or 2θ = 2n𝛑 ± (2𝛑 / 3)
θ = n𝛑 / 4 or θ = n𝛑 ± (𝛑 / 3)
Question 14: If 0 ≤ A ≤ 𝛑 / 4, then tan–1 (1 / 2) tan 2A + tan–1 (cot A) + tan–1 (cot3A) is equal to
1. a. 𝛑 / 4
2. b. 𝛑
3. c. 0
4. d. 𝛑 / 2
Solution:
tan–1 (1 / 2) tan 2A + tan–1 (cot A) + tan–1 (cot3A)
= tan–1 (1 / 2) tan 2A + tan-1 [(cot A + cot3 A) / (1 - cot4 A)]
= tan-1 [(1 / 2) [2 tan A / (1 - tan2 A)]] + tan-1 [cot A / (1 - cot2 A)]
= tan-1 [tan A / (1 - tan2 A)] + tan-1 [tan A / (tan2 A - 1)]
= tan-1 [tan A / (1 - tan2 A)] - tan-1 [tan A / (1 - tan2 A)]
= 0
Question 15: Without changing the direction of the axes, the origin is transferred to the point (2, 3). Then the equation x2 + y2 – 4x – 6y + 9 = 0 changes to
1. a. x2 + y2 + 4 = 0
2. b. x2 + y2 = 4
3. c. x2 + y2 – 8x – 12y + 48 = 0
4. d. x2 + y2 = 9
Solution:
By replacing
X → x + 2, y → y + 3
Given equation of circle is x2 + y2 – 4x – 6y + 9 = 0
(x + 2)2 + (y + 3)2 – 4 (x + 2) – 6 (y + 3) + 9 = 0
x2 + 4x + 4 + y2 + 6y + 9 – 4x – 8 – 6y – 18 + 9 = 0
x2 + y2 – 4 = 0
Question 16: The angle between a pair of tangents drawn from a point P to the circle x2 + y2 + 4x – 6y + 9sin2α + 13cos2α = 0 is 2α. The equation of the locus of the point P is
1. a. x2 + y2 + 4x + 6y + 9 = 0
2. b. x2 + y2 – 4x + 6y + 9 = 0
3. c. x2 + y2 – 4x – 6y + 9 = 0
4. d. x2 + y2 + 4x – 6y + 9 = 0
Solution:
Let the centre be O, points on circle from where tangents are drawn is A, B and point of intersection of tangent is P.
x2 + y2 + 4x – 6y + 9sin2α + 13cos2α = 0
Centre of circle O = (–2, 3)
r = √(4 + 9 - 9 sin2 α - 13 cos2 α)
= √(13 - 9 sin2 α - 13 (1 - sin2 α))
= √(13 sin2 α - 9 sin2 α)
= 2 sin α
2α is the angle between tangents
sin x = OA / OP
sin x = 2 sinx / √(h + 2)2 + (k - 3)2
(h + 2)2 + (k – 3)2 = 4
h2 + k2 + 4h – 6k + 9 = 0
Focus of point p is
x2 + y2 + 4x – 6y + 9 = 0
Question 17: The point Q is the image of the point P (1, 5) about the line y = x and R is the image of the point Q about the line y = -x. The circumcentre of the PQR is
1. a. (5, 1)
2. b. (–5, 1)
3. c. (1, –5)
4. d. (0, 0)
Solution:
Clearly P (1, 5)
Q = (5, 1) {as y = x}
R = (–1, –5) {as y = –x}
Circum center of PQR is (1 – 1) / 2, (5 – 5) / 2 = (0, 0)
Question 18: The angular points of a triangle are A (–1, –7), B (5, 1) and C (1, 4). The equation of the bisector of the angle ∠ABC is
1. a. x = 7y + 2
2. b. 7y = x + 2
3. c. y = 7x + 2
4. d. 7x = y + 2
Solution:
AB = √(- 1 - 5)2 + (- 7 - 1)2 = 10
BC = √(1 - 5)2 + (4 - 1)2 = 5
BD divides AC in ratio 2 : 1
D = [(-1 + 2) / (2 +1), (-7 + 8) / (2 + 1)]
= (1 / 3, 1 / 3)
Equation of BD is
y - 1 = [(1 - (1 / 3)) / (5 - (1 / 3))] (x - 5)
y - 1 = (2 / 14) (x - 5)
x – 7 y + 2 = 0
7y = x + 2
Question 19: If one the diameters of the circle, given by the equation x2 + y2 + 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is (2, –3), the radius of S is
1. a. √41 unit
2. b. 3 √5 unit
3. c. 5 √2 unit
4. d. 2 √5 unit
Solution:
Given: equation of circle is x2 + y2 + 4x + 6y – 12 = 0.
Whose centre is (2, –3) and radius = √22 + (–3)2 + 12 = 5
Now, according to the given information, we have the following figure.
x2 + y2 + 4x + 6y – 12 = 0
Clearly, AO ⊥ BC, as O is mid point of the chord.
Now, in △AOB we have
OA = √[(2 + 2)2 + (-3 + 3)2] = √16 = 4 and OB = 5
AB = √[OA2 + OB2]
= √[16 + 25]
= √41
Question 20: A chord AB is drawn from point A (0, 3) on the circle x2 + 4x + (y – 3)2 = 0, and is extended to M such that AM = 2AB. The locus of M is
1. a. x2 + y2 – 8x – 6y + 9 = 0
2. b. x2 + y2 + 8x + 6y + 9 = 0
3. c. x2 + y2 + 8x – 6y + 9 = 0
4. d. x2 + y2 – 8x + 6y + 9 = 0
Solution:
Given equation of circle is x2 + 4x + (y – 3)2 = 0
AM = 2AB
B is the mid point of AM
B = [(h / 2), (k + 3) / 2] lies on the circle.
Equation of circle is x2 + 4x + (y – 3)2 = 0
Let x = h / 2, y = (k + 3) / 2
h2 / 4 + 2h + [{(k + 3) / 2} - 3]2 = 0
(h2 / 4) + 2h + [(k2 - 6k + 9) / 4] = 0
k2 + h2 + 8h – 6k + 9 = 0
Locus of m is x2 + y2 + 8x – 6y + 9 = 0.
Question 21: Let the eccentricity of the hyperbola x2 / a2 - y2 / b2 = 1 be reciprocal to that of the ellipse x2 + 9y2 = 9, then the ratio a2 : b2 equals
1. a. 8 : 1
2. b. 1 : 8
3. c. 9 : 1
4. d. 1 : 9
Solution:
Hyperbola: x2 / a2 - y2 / b2 = 1
Ellipse: x2 + 9y2 = 9
x2 / 9 + y2 / 1 = 1
The eccentricity of ellipse e = √[1 - (1/9)] = √(8/9)
The eccentricity of the hyperbola be reciprocal to the ellipse
The eccentricity of hyperbola = √(9/8)
1 + (b2 / a2) = 9 / 8
(b2 / a2) = 1 / 8
a2 : b2 = 8 : 1
Question 22: Let A, B be two distinct points on the parabola y2 = 4x. If the axis of the parabola touches a circle of radius r having AB as diameter, the slope of the line AB is
1. a. -1 / r
2. b. 1 / r
3. c. 2 / r
4. d. -2 / r
Solution:
Let, A (t22, 2t2), B (t12, 2t1) be the two points on the parabola.
AB is the diameter of the circle.
Let, c be the centre of the circle.
C = ([t12 + t22] / 2, t1 + t2)
This axis of the parabola is the x-axis.
So, the circle touches the x-axis.
Hence, distance of c from x-axis = radius of circle.
|t1 + t2| = r
(t1 + t2) = ± r
Slope of AB = [y2 - y1] / [x2 - x1]
= [2t1 - 2t2] / [t12 - t22]
= 2 / [t1 + t2]
= ± 2 / r
Question 23: Let P (at2, 2at), Q (ar2, 2ar) be three points on a parabola y2 = 4ax. If PQ is the focal chord and PK, QR are parallel where the co-ordinates of K is (2a, 0), then the value of r is
1. a. t / (1 - t2)
2. b. (1 - t2) / t
3. c. (t2 + 1) / t
4. d. (t2 - 1) / t
Solution:
The slope of line Pk = slope of line QR
mPk = mQR
[2at - 0] / [at2 - 2a] = [2at’ - 2ar] / [a (t’)2 - ar2]
t / [t2 - 2] = [t’ - r] / [(t’)2 - r2]
–t ' – tr2 = –t – rt2 – 2t ' + 2r {tt ' = –1}
t ' – tr2 = – t + 2r – rt2
–tr2 + r (t2 – 2) + t ' + t = 0
λ = (2 - t2) ± √(t2 - 2)2 + 4 (-1 + t2) / -2t
= (2 - t2) ± √t4 / [-2t]
= [(2 - t2) ± t2] / -2t
r = -1 / t
It is not possible as the R & Q will be one and same.
Or r = [t2 - 1] / 2
Question 24: Let P be a point on the ellipse x2 / 9 + y2 / 4 = 1 and the line through P parallel to the y-axis meet the circle x2 + y2 = 9 at Q, where P, Q are on the same side of the x-axis. If R is a point on PQ such that PR / RQ = 1 / 2 then the locus of R is
1. a. x2 / 9 + 9y2 / 49 = 1
2. b. x2 / 49 + y2 / 9 = 1
3. c. x2 / 9 + y2 / 49 = 1
4. d. 9x2 / 9 + y2 / 49 = 1
Solution:
Ellipse: x2 / 9 + y2 / 4 = 1
P = (3cosθ, 2sinθ)
Circle: x2 + y2 = 9
Q = (3cosθ, 3sinθ)
h = [3 cos θ + 6 cos θ] / 3; k = [3 sin θ + 4 sin θ] / 3
h = 3 cos θ; k = (7 / 3) sin θ
sin2 θ + cos2 θ = 1
h2 / 9 + 9k2 / 49 = 1
The locus is (x2 / 9) + (9y2 / 49) = 1
Question 25: A point P lies on a line through Q (1, –2, 3) and is parallel to the line x / 1 = y / 4 = z / 5. If P lies on the plane 2x + 3y – 4z + 22 = 0, then segment PQ equals to
1. a. √42 units
2. b. √32 units
3. c. 4 unit
4. d. 5 units
Solution:
Equation line: (x - 1) / 1 = (y + 2) / 4 = (z - 3) / 5 = λ (let)
Point P (λ + 1, 4λ – 2, 5λ + 3)
Point p lies on 2x + 3y – 4z + 22 = 0
2(λ + 1) + 3 (4λ – 2) – 4(5λ + 3) + 22 = 0
–6λ + 6 = 0
λ = 1
Point p = (2, 2, 8), q = (1, –2, 3)
Distance PQ = √[12 + 42 + 52]
= √[1 + 16 + 25]
= √42
Question 26: The foot of the perpendicular drawn from the point (1, 8, 4) on the line joining the points (0, –11, 4) and (2, – 3, 1) is
1. a. (4, 5, 2)
2. b. (–4, 5, 2)
3. c. (4, –5, 2)
4. d. (4, 5, –2)
Solution:
Equation of line joining point (0, –11, 4) and (2, –3, 1)
(x - 2) / 2 = (y + 3) / 8 = (z - 1) / -3 = λ (let)
DR’s of PQ (2λ + 1, 8λ – 11, –3λ – 3)
Now, (2λ + 1) 2 + (8λ – 11)8 +(–3λ – 3)(–3) = 0
77λ - 77 = 0
λ = 1
Q = (4, 5, –2)
Question 27: The approximate value of sin31o is
1. a. > 0.5
2. b. > 0.6
3. c. < 0.5
4. d. < 0.4
Solution:
sin 30o = 1/2
sin 31o > 1/2 {since sinx is increasing function}
Question 28: Let f1(x) = ex, f2(x) = ef1(x), ............, fn+1(x) = efn(x) for all n ≥ 1. Then for any fixed n, (d/dx) fn(x): is
1. a. fn(x)
2. b. fn(x)fn–1(x)
3. c. fn(x)fn–1(x)....f1(x)
4. d. fn(x)......... f1(x)ex
Solution:
Question 29: The domain of definition of f (x) = √[1 - |x|] / [2 - |x|] is
1. a. (–∞,–1) ⋃ (2, ∞)
2. b. [–1,1] ⋃ (2, ∞) ⋃ (–∞, – 2)
3. c. (–∞,1) ⋃ (2, ∞)
4. d. [–1,1] ⋃ (2, ∞)
Solution:
f (x) = √[1 - |x|] / [2 - |x|]
[1 - |x|] / [2 - |x|] ≥ 0
|x| ≤ 1 or |x| ≥ 2
x ∈ [-1, 1] or x ∈ (–∞, – 2) ⋃ (2, ∞)
x ∈ [–1,1] ⋃ (2, ∞) ⋃ (–∞, – 2)
Question 30: Let f : [a, b] → R be differentiable on [a, b] and k ∈ R. Let f (a) = 0 = f ' (b). Also let J (x) = f' (x) + k f (x). Then
1. a. J(x) > 0 for all x ∈ [a, b]
2. b. J(x) < 0 for all x ∈ [a, b]
3. c. J(x) = 0 has atleast one root in (a, b)
4. d. J(x) = 0 through (a, b)
Solution:
Let g(x) = ekxf(x)
f(a) = 0 = f(b)
By Rolle’s theorem
g’(c) = 0 , c ∈ (a, b)
g’ (x) = ekx f’ (x) + kekx f (x)
g (c) = 0
ekc (f’ (c) + kf(c)) = 0
f ‘ (c) + k f (c) = 0
For atleast one c in (a, b).
Question 31: Let f (x) = 3x10 – 7x8 + 5x6 – 21x3 + 3x2 – 7. Then [f (1 - h) - f (1)] / [h3 + 3h]
1. a. does not exist
2. b. is 50/3
3. c. is 53/3
4. d. is 22/3
Solution:
Given
f(x) = 3x10 – 7x8 + 5x6 – 21x3 + 3x2 – 7
f’(x) = 30x9 – 56x7 + 30x5– 63x2 + 6x
Now,
limh→0 [f (1 - h) - f (1)] / [h3 + 3h]
= (0 / 0) form using L’ hospital rule
limh→0 [-f’ (1 - h] / [3h2 + 3]
= [-f’ (1)] / 3
= -[30 (1)9 - 56 (1)7 + 30 (1)5 - 63 (1)2 + 6 (1)] / 3
= -[30 - 56 + 30 - 63 + 6] / 3
= 53/3
Question 32: Let f : [a, b] → R be such that f is differentiable in (a, b), f is continuous at x = a and x = b and moreover f(a) = 0 = f(b). Then
1. a. there exists at least one point c in (a, b) such that f’ (c) = f(c)
2. b. f’ (x) = f(x) does not hold at any point in (a, b)
3. c. at every point of (a, b), f’ (x) > f(x)
4. d. at every point of (a, b), f’ (x) < f(x)
Solution:
Let, h(x) = e–x f(x)
h(a) = 0, h(b) = 0
h(x) is continuous and differentiable function
By Rolle’s theorem
h’ (c) = 0, c ∈ (a, b)
e–x f’(c) + (–e–x) f(c) = 0
e–x f’(c) = e–x f(c)
f’(c) = f(c)
Question 33: Let f : R → R be a twice continuously differentiable function such that f (0) = f (1) = f’ (0) = 0. Then
1. a. f’’ (0) = 0
2. b. f’’ (c) = 0 for some c ∈ R
3. c. if c ≠ 0, then f’’ (c) ≠ 0
4. d. f’ (x) > 0 for all x ≠ 0
Solution:
f(x) is continuous and differentiable function
f(0) = f(1) = 0 [by Rolle’s theorem]
f(a) = 0 , a ∈ (0,1)
Given f’ (0) = 0
By Rolle’s theorem f ” (0) = 0 for some c, c ∈ (0, a)
Question 34: If ∫esin x[(x cos3 x - sin x) / cos2 x] dx = esin x f (x) + c where c is constant of integration, then f (x) =
1. a. sec x – x
2. b. x – sec x
3. c. tan x – x
4. d. x – tan x
Solution:
I = ∫esinx [(x cos3 x - sinx) / cos2 x] dx
= ∫esinx (x cosx - tanx secx) dx
= (xesinx - ∫esinx) - [esinx secx - ∫esinx dx] + c
= xesinx – esinx secx + c
I = esinx (x – secx) + c
Question 35: If ∫f (x) sin x cos x dx = (1/2 [b2 - a2]) log f (x) + c, where c is the constant of integration, then f (x) =
1. a. 2/{[b2 - a2] sin 2x}
2. b. 2/[ab sin 2x]
3. c. 2/{[b2 - a2] cos 2x}
4. d. 2/[ab cos 2x]
Solution:
∫f (x) sin x cos x dx = (1/2 [b2 - a2]) log f (x) + c
Differentiate with respect to x
f (x) sinxcosx = [f ‘ (x)/f (x)] (1/2 [b2 - a2]) + c
sin 2x [b2 - a2] = f ‘(x) / [f (x)]2
On integrating
-1/f (x) = [- [b2 - a2] cosx 2x]/2
f (x) = 2/{[b2 - a2] cos2x}
Question 36: If M = ∫0𝛑/2 {cosx / [x + 2]} dx, N = ∫0𝛑/4 [{sinx cosx} / (x + 1)2] dx, then the value of M – N
1. a. 𝛑
2. b. 𝛑/4
3. c. 2/(𝛑 – 4)
4. d. 2/(𝛑 + 4)
Solution:
Question 37: The value of the integral I = ∫1/20142014 [tan-1 x/x] dx is
1. a. (𝛑/4) log 2014
2. b. (𝛑/2) log 2014
3. c. 𝛑 log 2014
4. d. (1/2) log 2014
Solution:
I = ∫1/20142014 [tan-1 x/x] dx ---- (1)
Let x = 1 / t
dx = [-1/t2] dt
I = ∫20141/2014 [[tan-1 (1/t)] / [1/t]} dt
I = ∫1/20142014 [cot-1 t / t] dt --- (2)
From eq(1) + eq(2)
2I = ∫1/20142014 {[tan-1 t + cot-1 t] / t} dt
2I = ∫1/20142014 [(𝛑 / 2) / t] dt
I = (𝛑 / 4) (ln t)1/20142014
I = (𝛑 / 4) [ln 2014 - ln (1 / 2014)]
= (𝛑 / 4) [2 ln 2014]
= (𝛑 / 2) log 2014
Question 38: Let I = ∫𝛑/4𝛑/3 [sinx / x] dx, then
1. a. (1/2) ≤ I ≤ 1
2. b. 4 ≤ I ≤ 2√30
3. c. √3/8 ≤ I ≤ √2/6
4. d. 1 ≤ I ≤ 2√3/√2
Solution:
I = ∫𝛑/4𝛑/3 [sinx / x] dx
(sin x / x) is a decreasing function
So [(𝛑 / 12) * (sin (𝛑 / 3)) / (𝛑 / 3)] ≤ I ≤ [(𝛑 / 12) * (sin (𝛑 / 4)) / (𝛑 / 4)]
(1 / 4) * (√3 / 2) ≤ I ≤ (1 / 3) * (1 / √2)
√3/8 ≤ I ≤ √2/6
Question 39: The value of I =
dx, is
1. a. 1
2. b. 𝛑
3. c. e
4. d. 𝛑 / 2
Solution:
Question 40: The value of limn→∞ (1/n) {sec2 (𝛑 / 4n) + sec2 (2𝛑 / 4n) + …. + sec2 (n𝛑 / 4n)} is
1. a. logc2
2. b. 𝛑 / 2
3. c. 4 / 𝛑
4. d. e
Solution:
L = limn→∞ {sec2 (𝛑 / 4n) + sec2 (2𝛑 / 4n) + …. + sec2 (n𝛑 / 4n)}
L = limn→∞ (1/n) ∑r=1n sec2 (r𝛑 / 4n)
Let (r / n) = x
(1 / n) = dx
L = ∫01 sec2 (𝛑x/4) dx
= [tan (𝛑x / 4)]01 * (4/𝛑)
= (4/𝛑) (tan [𝛑/4] - tan 0])
= 4/𝛑
Question 41: The differential equation representing the family of curves y2 = 2d (x + √d) where d is a parameter, is of
1. a. order 2
2. b. degree 2
3. c. degree 3
4. d. degree 4
Solution:
y2 = 2d (x + √d) ---- (i)
Differentiate with respect to x
2y (dy / dx) = 2d
d = y (dy / dx)
Put in equation (i)
y2 = 2y (dy/dx) [x + √y (dy / dx)]
y2 = 2xy (dy/dx) + 2y3/2 (dy / dx)3/2
(y2 - 2xy [dy/dx])2 = 4y3 (dy / dx)3
Degree three.
Question 42: Let y(x) be a solution of (1 + x2) (dy / dx) + 2xy – 4x2 = 0 and y(0) = –1. Then y(1) is equal to
1. a. 1/2
2. b. 1/3
3. c. 1/6
4. d. –1
Solution:
(1 + x2) (dy / dx) + 2xy – 4x2 = 0
(dy / dx) + (2x / [1 + x2]) * y = 4x2 / (1 + x2)
IF = e∫Pdx = e∫(2x/1+x2) dx = eln(1+x2) = (1 + x2)
y (1 + x2) = ∫[(4x2) / (1 + x2)] * (1 + x2) dx + c
y (1 + x2) = ∫4x2 dx + c
y (1 + x2) = (4x3 / 3) + c
Put y(0) = –1
–1 = c
y (1 + x2) = (4x3 / 3) - 1
y(1) = y (1 + 1) = (4[1] / 3) - 1
2y = 1/3
y = 1/6
Question 43: The law of motion of a body moving along a straight line is x = (1/2) vt, x being its distance from a fixed point on the line at time t and v is its velocity there. Then
1. a. acceleration f varies directly with x
2. b. acceleration f varies inversely with x
3. c. acceleration f is constant
4. d. acceleration f varies directly with t
Solution:
x = (1 / 2) vt
Differentiate with respect to x
x = (1 / 2) (dx / dt) * t
∫2dt / t = ∫dx / x
ln c + 2 ln t = ln x
x = t2c
(dx / dt) = 2 tc
d2x / dt2 = 2c
Hence, acceleration is constant.
Question 44: Number of common tangents of y = x2 and y = –x2 + 4x – 4 is
1. a. 1
2. b. 2
3. c. 3
4. d. 4
Solution:
y = x2; y = -(x - 2)2
2 + (β - 2)2] / [α - β] = 2α = -2 [β - 2]
α = 2 - β
β = 2 – α
2 + α2] / [α - 2 + α] = 2α
[2α2 / 2α - 2] = 2α
α2 = α (2α - 2)
α2 = 2α2 - 2α
α2 = 2α
α = 0, 2
α = 0, β = 2
α = 2, β = 0
Hence, two common tangents.
Question 45: Given that n numbers of A.Ms are inserted between two sets of numbers a, 2b and 2a, b where a, b ∈ R. Suppose further that the mth means between these sets of numbers are same, then the ratio a : b equals
1. a. n – m + 1 : m
2. b. n – m + 1 : n
3. c. n : n – m + 1
4. d. m : n – m + 1
Solution:
a..........n A.Ms...........2b
(Difference)d = (2b - a) / (n + 1)
Am = (a + m) ((2b - a) / (n + 1)) ---- (i)
2a..........n A.Ms...........b
d = (b - 2a) / (n + 1)
Am = 2a + m ((b - 2a) / (n + 1)) --- (ii)
Equating equation (i) & (ii)
a = [m / (n + 1)] (b + a)
(a / b) = m / [n - m + 1]
Question 46: If x + log10(1 + 2x) = x log105 + log106 then the value of x is
1. a. 1 / 2
2. b. 1 / 3
3. c. 1
4. d. 2
Solution:
x + log10(1 + 2x) = x log105 + log106
x log1010 + log10(1 + 2x) = log105x + log106
log10(1 + 2x) = log105x + log106 – log10 10x
log10(1 + 2x) = log10 [5x . 6] / [10x]
1 + 2x = [6 . 5x] / [2x . 5x]
1 + 2x = 6 / 2x
let 2x = t
1 + t = 6 / t
t2 + t – 6 = 0
(t + 3)(t – 2) = 0
t = –3(not possible)
2x = –3
t = 2
2x = 2
x = 1
Question 47: If Zr = sin (2𝛑r / 11) - i cos (2𝛑r / 11) then ∑r=010 Zr =
1. a. –1
2. b. 0
3. c. i
4. d. –i
Solution:
Given
Zr = sin (2𝛑r / 11) - i cos (2𝛑r / 11)
Zr = -i [cos (2𝛑r / 11) + i sin (2𝛑r / 11)]
Zr = -iei(2𝛑r / 11)
Now,
r=010 Zr = -i ∑r=010 ei(2𝛑r / 11)
= –i(0)
= 0
Question 48: If z1 and z2 be two non zero complex numbers such that (z1 / z2) + (z2 / z1) = 1, then the origin and the points represented by z1 and z2
1. a. lie on a straight line
2. b. form a right-angled triangle
3. c. form an equilateral triangle
4. d. form an isosceles triangle
Solution:
z1 = z2ei𝛑/3
2z1 = z2 (1 + i√3)
2z1 = z2 + i√3z2
2z1 – z2 = i√3z2
Squaring both side
4z12 + z22 – 4z1z2 = -3z22
4z12 + z22 = 4z1z2
Hence from equilateral triangle.
Question 49: If b1b2 = 2(c1 + c2) and b1, b2, c1, c2 are all real numbers, then at least one of the equations. x2 + b1x + c1 = 0 and x2 + b2x + c2 = 0 has
1. a. real roots
2. b. purely imaginary roots
3. c. roots of the form a + ib (a, b ∈ R, ab ≠ 0)
4. d. rational roots
Solution:
Suppose the equations x2 + b1x + c1 = 0 & x2 + b2x + c2 = 0 have real roots.
then
b12 ≥ 4c1 .....(i)
b22 ≥ 4c2 .....(ii)
Given that b1b2 = 2(c1 + c2)
On squaring
b12b22 = 4 [c12 + c22 + 2c1c2] = 4 [(c1 - c2)2 + 4c1c2]
b12b22 - 16c1c2 = 4 [(c1 - c2)2] ≥ 0
Multiplying (i) & (ii), we get
b12b22 ≥ 16c1c2
Therefore, at least one equation has real roots.
Question 50: The number of selection of n objects from 2n objects of which n are identical and the rest are different is
1. a. 2n
2. b. 2n – 1
3. c. 2n – 1
4. d. 2n –1 + 1
Solution:
Total no. of ways = nC0 + nC1 + nC2 + ..........+ nCn
= 2n
Question 51: Let A be the centre of the circle x2 + y2 – 2x – 4y – 20 = 0. Let B(1, 7) and D(4, – 2) be two points on the circle such that tangents at B and D meet at C. The area of the quadrilateral ABCD is
1. a. 150 sq. units
2. b. 50 sq. units
3. c. 75 sq. units
4. d. 70 sq. units
Solution:
s : x2 + y2 – 2x – 4y – 20 = 0
Centre of circle A = (1, 2)
Equation of tangent at B(1, 7)
x + 7y – (x + 1) –2 (y + 7) – 20 = 0
5y = 35
y = 7
Equation of tangent at D(4, –2)
4x – 2y – (x + 4) – 2(y – 2) – 20 = 0
3x – 4y = 20
coordinates of C are (16, 7)
Length of AB = √(1 - 1)2 + (7 - 2)2 = 5
Length of BC = √(16 - 1)2 + (7 - 7)2 = 15
The area of quadrilateral ABCD = 2 * (1 / 2) * 5 * 15
= 75 sq. units.
Question 52: Let f (x) =
1. a. f is discontinuous for all A and B
2. b. f is continuous for all A = – 1 and B = 1
3. c. f is continuous for all A = 1 and B = –1
4. d. f is continuous for all real values of A, B
Solution:
From the given conditions, function f (x) is continuous throughout the real line, when function f(x) is continuous at x = – 𝛑/2 and 𝛑/2 for continuity at x = – 𝛑/2.
from equation (iv)
A + B = 0 …..(v)
From equation (iii) & (iv)
A = –1, B = 1
Question 53: The normal to the curve y = x2 - x + 1, drawn at the points with the abscissa, x1 = 0, x2 = −1 and x3 = 5/2
1. a. are parallel to each other
2. b. are pairwise perpendicular
3. c. are concurrent
4. d. are not concurrent
Solution:
y = x2 - x + 1
(dy / dx) = 2x - 1 = mT
mN = -1 / mT
mN = 1 / (1 - 2x)
For tangent
mx1 = 1 Point (0, 1)
(y – 1) = 1(x – 0)
x – y + 1 = 0 …..(i)
For tangent
mx2 = (1 / 3) point (-1, 3)
(y – 3) = (1 / 3) (x + 1)
3y – 9 = x + 1
x – 3y + 10 = 0 …..(ii)
For normal
mx3 = (-1 / 4) point (5 / 2, 19 / 4)
(y - [19 / 4]) = - 1 / 4 [x - (5 / 2)]
x + 4y = 43 / 2 ---- (iii)
To find intersection point tangent (1) & tangent (2)
y – 1 – 3y + 10 = 0
–2y = –9
y = 9 / 2
Intersection point is (7 / 2, 9 / 2) passes (3)
Hence, normal are concurrent.
Question 54: The equation x log x = 3 – x
1. a. has no root in (1, 3)
2. b. has exactly one root in (1, 3 )
3. c. x log x – (3 – x) > 0 in [1, 3]
4. d. x log x – (3 – x) < 0 in [1, 3]
Solution:
f (x) = x log x – 3 + x
Differentiate with respect to x
f ‘ (x) = x * (1 / x) + log x + 1
f’ (x) = 2 + log x
f(1)f(3) = –2 (3log3) = –ve
Hence, one root is (1, 3).
Question 55: Consider the parabola y2 = 4x. Let P and Q be points on the parabola where P (4, –4) & Q (9, 6). Let R be a point on the arc of the parabola between P & Q. Then the area of ΔPQR is largest when
1. a. ∠PQR = 90°
2. b. R(4, 4)
3. c. R (1/4, 1)
4. d. R (1, 1/4)
Solution:
Area = (1/2)
$$\begin{vmatrix} t^2 & 2t &1 \\ 9 & 6 & 1\\ 4 &-4 &1 \end{vmatrix}$$
= (1/2) [t2 (10) - 2t (5) + 1 (-60)]
= (10/2) (t2 - t - 6)
f (t) = 5t2 – 5t – 30
f’(t) = 10t – 5 = 0
t = 1 / 2
Point R = [(1/2)2, 2 (1/2)] = (1/4, 1)
Question 56: A ladder 20 ft long leans against a vertical wall. The top-end slides downwards at the rate of 2 ft per second. The rate at which the lower end moves on a horizontal floor when it is 12 ft from the wall is
1. a. 8/3
2. b. 6/5
3. c. 3/2
4. d. 17/4
Solution:
Using right angle triangle concept
x2 + y2 = 400
Differentiate with respect to t
2x (dx / dt) + 2y (dy / dt) = 0
Given
(dy / dt) = 2ft / sec
x = 12, y= 16
2 (12) (dx / dt) + 2 (16) (dy / dt) = 0
(dx / dt) = -8/3
Question 57: For 0 ≤ p ≤ 1 and for any positive a, b; let I(p) = (a + b)P, J(p) = aP + bP, then
1. a. I(p) > J(p)
2. b. I(p) < J(p)
3. c. I(p) < J(p) in [0, p/2] and I(p) > J(p) in [p/2, ∞)
4. d. I(p) < J(p) in [p/2, ∞) and J(p) < I(p) in [0, p/2)
Solution:
a = 9
b = 16
I(P) = 5 and J(P) = 7
J(P) > I(P)
Now,
a = 1 / 9 and b = 1 / 16
I(P) = 5 / 12 & J(P) = 7 / 12
J(P) > I(P)
Question 58: Let α = i + j + k, β = i - j - k and γ = -i + j - k be three vectors. A vector δ, in the plane of α and β, projection on γ is 1 / √3, is given by
1. a. -i - 3j - 3k
2. b. i - 3j - 3k
3. c. -i + 3j + 3k
4. d. i + 3j - 3k
Solution:
Given that
α = i + j + k
β = i - j - k
γ = -i + j - k
γ = √3
Question 59: Let α, β, γ be three unit vectors such that α . β = α . γ and the angle between β and γ and is 30°. Then α is
1. a. 2 (β × γ)
2. b. − 2(β × γ)
3. c. ± 2(β × γ)
4. d. (β × γ)
Solution:
α . β = α . γ
Thus, α is perpendicular to b and c.
A unit vector perpendicular to b and c
= ± (β × γ) / |(β × γ)|
= ± (β × γ) / [|β| |γ| sin (𝛑 / 6)]
= ± (β × γ) / (1 / 2)
= ± 2 (β × γ)
Question 60: Let z1 and z2 be complex numbers such that z1 ≠ z2 and |z1| = |z2|. If Re(z1) > 0 and Im(z2) < 0, then [z1 + z2] / [z1 - z2] is
1. a. one
2. b. real and positive
3. c. real and negative
4. d. purely imaginary
Solution:
z1 = x1 + iy1 and z2 = x2 + iy2
Re(z1) > 0 ⇒ x1 > 0 and Im(z2) < 0 ⇒ y2 < 0
Question 61: From a collection of 20 consecutive natural numbers, four are selected such that they are not consecutive. The number of such selections is
1. a. 284 × 17
2. b. 285 × 17
3. c. 284 × 16
4. d. 285 × 16
Solution:
1, 2, 3, 4, 5, ..........., 19, 20
There are 17 ways for four consecutive number
Number ways = 20C4 – 17
= [20 × 19 × 18 × 17] / [1 × 2 × 3 × 4] –17
= [285 × 17] – 17
= 284 × 17
Question 62: The least positive integer n such that
is an identity matrix of order 2 is
1. a. 4
2. b. 8
3. c. 12
4. d. 16
Solution:
Question 63: Let 𝛒 be a relation defined on N, the set of natural numbers, as 𝛒 = {(x, y) ∈ N × N : 2x + y = 41} Then
1. a. is an equivalence relation
2. b. is only reflexive relation
3. c. is only symmetric relation
4. d. is not transitive
Solution:
P = {(x, y) N × N : 2x + y = 41}
For reflexive relation
⇒ xRx ⇒ 2x + x = 41 ⇒ x = 41 / 3 ∉ N (not reflexive)
For symmetric
⇒xRy ⇒ 2x + y = 41 ≠ yRx (not symmetric)
For transitive
⇒ xRy ⇒ 2x + y = 41 and yRz ⇒ 2y + z = 41 x R z (not transitive)
Question 64: If the polynomial
then the constant term of f(x) is
1. a. 2 – 3.2b + 23b
2. b. 2 + 3.2b + 23b
3. c. 2 + 3.2b – 23b
4. d. 2 – 3.2b – 23b
Solution:
f(x) = 1(1 – 2b) – 2b(1 – 22b) + 1(1 – 2b)
f(x) = 1 – 2b – 2b + 23b + 1 – 2b
f(x) = 2 – 3.2b + 23b
Question 65: A line cuts the x-axis at A(5, 0) and the y-axis at B(0, –3). A variable line PQ is drawn perpendicular to AB cutting the x-axis at P and the y-axis at Q. If AQ and BP meet at R, then the locus of R is
1. a. x2 + y2 – 5x + 3y = 0
2. b. x2 + y2 + 5x + 3y = 0
3. c. x2 + y2 + 5x – 3y = 0
4. d. x2 + y2– 5x – 3y = 0
Solution:
Line AB is (x / 5) + (y / -3) = 1 ⇒ 3x – 5y = 15
Any perpendicular line to AB
5x + 3y = λ
So, P (λ / 5, 0), Q (0, λ / 3)
AQ is (x / 5) + y / (λ / 3) = 1
(3y / λ) = 1 - (x / 5)
(1 / λ) = (1 / 3y) (1 - (x / 5)) ---- (i)
And BP is x / (λ / 5) - (y / 3) = 1
5x / λ = 1 + (y / 3)
1 / λ = (1 / 5x) (1 + [y / 3]) --- (ii)
Equation (i) = equation (ii)
(1 / 3y) (1 - (x / 5)) = (1 / 5x) (1 + [y / 3])
5x (1 - [x / 5]) = (1 / 5x) (1 + [y / 3])
5x – x2 = 3y + y2
x2 + y2 – 5x + 3y = 0
Question 66: In a third order matrix A, aij denotes the element in the i-th row and j-th column. If aij = 0 for i = j
= 1 for i > j
= –1 for i < j Then the matrix is
1. a. skew-symmetric
2. b. symmetric
3. c. not invertible
4. d. non-singular
Solution:
Given
Question 67: The area of the triangle formed by the intersection of a line parallel to the x-axis and passing through P(h, k), with the lines y = x and x + y = 2 is h2. The locus of the point P is
1. a. x = y – 1
2. b. x = –(y – 1)
3. c. x = 1 + y
4. d. x = –(1 + y)
Solution:
Area of triangle = h2
1(k2 – (2 – k)k) – 1(k – k) + 1(2 – k – k) = ± 2h2
k2 – 2k + k2 + 2 – 2k = ± 2h2
2k2 – 4k + 2 = ± 2h2
k2 – 2k + 1 = ± h2
Locus is (k – 1)2 = h2
y – 1 = ± x
x – y + 1 = 0 or x + y = 1
x = y – 1 or x = –(y – 1)
Question 68: A hyperbola, having the transverse axis of length 2 sin θ is confocal with the ellipse 3x2 + 4y2 = 12. Its equation is
1. a. x2 sin2 θ – y2 cos2 θ = 1
2. b. x2 cosec2 θ – y2 sec2 θ = 1
3. c. (x2 + y2) sin2 θ = 1 + y2
4. d. x2 cosec2 θ = x2 + y2 + sin2 θ
Solution:
The length of transverse axis = 2sin θ = 2a
a = sin θ
Also for ellipse, 3x2 + 4y2 = 12
i.e. (x2 / 4) + (y2 / 3) = 1
a2 = 4 & b2 = 3
Now,
e = √1 - [b2 / a2]
e = √1 - (3 / 4) = ± (1 / 2)
focus of ellipse is = (ae, 0) & (–ae, 0)
focus = (1, 0) and (–1, 0)
As the hyperbola if confocal focus is same
And length of transverse axis = 2sin θ
length of semi transverse axis = sin θ
i.e. A = sin θ
And C = 1 where A, B, C are parameters in hyperbola similar to ellipse
C2 = A2 + B2
B2 = 1 – sin2 θ = cos2 θ
Equation of hyperbola is (x2 / A2) - (y2 / B2) = 1
x2 / sin2 θ – y2 / cos2 θ = 1
x2 cosec2 θ – y2 sec2 θ = 1
Question 69: Let f (x) = cos (𝛑 / x), x ≠ 0, then assuming k as an integer,
1. a. f (x) increases in the interval [1 / (2k + 1), 1 / 2k]
2. b. f (x) decreases in the interval [1 / (2k + 1), 1 / 2k]
3. c. f (x) decreases in the interval [1 / (2k + 2), 1 / 2k + 1]
4. d. f (x) increases in the interval [1 / (2k + 2), 1 / 2k + 1]
Solution:
f (x) = cos (𝛑 / x)
Differentiate with respect to x
f ‘ (x) = - sin (𝛑 / x) . (-𝛑 / x2)
= (𝛑 / x2) sin (𝛑 / x) > 0
For increasing function f’(x) > 0
sin (𝛑 / x) > 0
(2k𝛑) < 𝛑 / x < (2k + 1) 𝛑
(1 / 2k) > x > (1 / [2k + 1])
For decreasing function f’(x) < 0
sin (𝛑 / x) < 0
(𝛑 / x) ∈ [(2k + 1) 𝛑, (2k + 2) 𝛑]
(𝛑 / x) ∈ [1 / (2k + 2), 1 / 2k + 1]
Question 70: Consider the function y = loga (x + √x2 + 1), a > 0, a ≠ 1. The inverse of the function
1. a. does not exist
2. b. is x = log1/a (y + √y2 + 1)
3. c. is x = sinh (y ln a)
4. d. is x = cosh (-y ln (1 / a))
Solution:
Question 71: Let I = ∫01 [x3 cos 3x] / [2 + x2] dx, then
1. a. (-1 / 2) < I < (1 / 2)
2. b. (-1 / 3) < I < (1 / 3)
3. c. – 1 < I < 1
4. d. (-3 / 2) < I < (3 / 2)
Solution:
1. Answer: (a, b, c, d)
We know that
–1 < cos 3x < 1
–x3 < x3cos 3x < x3
-x3 / [2 + x2] < [x3 cos 3x] / [2 + x2] < x3 / [2 + x2]
Taking integration from 0 to 1
01 -x2 dx < 1 < ∫01 x2 dx
[-x3 / 3]01 < I < (x3 / 3)01
(-1 / 3) < I < (1 / 3)
Question 72: A particle is in motion along a curve 12y – x3. The rate of change of its ordinate exceeds that of abscissa in
1. a. – 2 < x < 2
2. b. x = ± 2
3. c. x < – 2
4. d. x > 2
Solution:
Given
(dy / dt) > (dx / dt) .....(i)
12y = x3
Differentiate with respect to
12 (dy / dt) = 3x2 (dx / dt) .....(ii)
From equation (i)
3x2 (dx / dt) > 12 (dx / dt)
x2 – 4 > 0
x ∈ (–∞, –2) ⋃ (2, ∞)
Question 73: The area of the region lying above x-axis, and included between the circle x2 + y2 = 2ax & the parabola y2 = ax, a > 0 is
1. a. 8𝛑a2
2. b. a2 [(𝛑 / 4) - (2 / 3)]
3. c. 16𝛑a2 / 9
4. d. 𝛑 [(27 / 8) + 3a2]
Solution:
Given
C1: x2 + y2 = 2ax
C2: y2 = ax
To find intersection points
x2 + (ax) = 2ax
x2 = ax
x(x – a) = 0
x = 0, a
Area = (1 / 4) (area of circle) - ∫0a √ax dx
= (1 / 4) 𝛑a2 - √a [(x3/2 / [3 / 2]]0a
= (𝛑a2 / 4) - (2a2 / 3)
= a2 [(𝛑 / 4) - (2 / 3)]
Question 74: If the equation x2 – cx + d = 0 has roots equal to the fourth powers of the roots of x2 + ax + b = 0, where a2 > 4b, then the roots of x2 – 4bx + 2b2 – c = 0 will be
1. a. both real
2. b. both negative
3. c. both positive
4. d. one positive and one negative
Solution:
Let x2 + ax + b = 0 has roots α and β
x2 – cx + d = 0 roots are α4 and β4
α + β = –a .....(i)
α β = b .....(ii)
α4 + β4 = c .....(iii)
(α β)4 = d .....(iv)
From equation (ii) & (iv)
b4 = d
And α4 + β4 = c
2 + β2)2 - 2(αβ)2 = c
((α + β)2 - 2(αβ))2 - 2(αβ)2 = c
(a2 – 2b)2 – 2b2 = c
2b2 + c = (a2 – 2b)2
Now for equation x2 – 4bx + 2b2 – c = 0
D = (4b)2 – 4(1)(2b2 – c)
D = 16b2 – 8b2 + 4c
D = 4(2b2 + c)
D = 4(a2 – 2b)2 > 0 ⇒ real roots
Now,
f(0) = 2b2 – c
f(0) = a2(4b – a2) < 0 {since a2 > 4b}
Roots are opposite in sign.
Question 75: On the occasion of Diwali festival each student of a class sends greeting cards to others. If there are 20 students in the class, the number of cards sent by students are
1. a. 20C2
2. b. 20P2
3. c. 2 * 20C2
4. d. 2 * 20P2
Solution:
Total students = 20
Number of ways = 20C2 × 21
= {[20 * 19] / 2} * 2
= 20 × 19
= 20P2 | 17,201 | 35,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-06 | latest | en | 0.773701 |
http://metamath.tirix.org/mpeascii/pjhthmo | 1,721,710,073,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00045.warc.gz | 18,519,417 | 2,626 | # Metamath Proof Explorer
## Theorem pjhthmo
Description: Projection Theorem, uniqueness part. Any two disjoint subspaces yield a unique decomposition of vectors into each subspace. (Contributed by Mario Carneiro, 15-May-2014) (New usage is discouraged.)
Ref Expression
Assertion pjhthmo
`|- ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) -> E* x ( x e. A /\ E. y e. B C = ( x +h y ) ) )`
### Proof
Step Hyp Ref Expression
1 an4
` |- ( ( ( x e. A /\ z e. A ) /\ ( E. y e. B C = ( x +h y ) /\ E. w e. B C = ( z +h w ) ) ) <-> ( ( x e. A /\ E. y e. B C = ( x +h y ) ) /\ ( z e. A /\ E. w e. B C = ( z +h w ) ) ) )`
2 reeanv
` |- ( E. y e. B E. w e. B ( C = ( x +h y ) /\ C = ( z +h w ) ) <-> ( E. y e. B C = ( x +h y ) /\ E. w e. B C = ( z +h w ) ) )`
3 simpll1
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> A e. SH )`
4 simpll2
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> B e. SH )`
5 simpll3
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> ( A i^i B ) = 0H )`
6 simplrl
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> x e. A )`
7 simprll
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> y e. B )`
8 simplrr
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> z e. A )`
9 simprlr
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> w e. B )`
10 simprrl
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> C = ( x +h y ) )`
11 simprrr
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> C = ( z +h w ) )`
12 10 11 eqtr3d
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> ( x +h y ) = ( z +h w ) )`
13 3 4 5 6 7 8 9 12 shuni
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> ( x = z /\ y = w ) )`
14 13 simpld
` |- ( ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) /\ ( ( y e. B /\ w e. B ) /\ ( C = ( x +h y ) /\ C = ( z +h w ) ) ) ) -> x = z )`
15 14 exp32
` |- ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) -> ( ( y e. B /\ w e. B ) -> ( ( C = ( x +h y ) /\ C = ( z +h w ) ) -> x = z ) ) )`
16 15 rexlimdvv
` |- ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) -> ( E. y e. B E. w e. B ( C = ( x +h y ) /\ C = ( z +h w ) ) -> x = z ) )`
17 2 16 syl5bir
` |- ( ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) /\ ( x e. A /\ z e. A ) ) -> ( ( E. y e. B C = ( x +h y ) /\ E. w e. B C = ( z +h w ) ) -> x = z ) )`
18 17 expimpd
` |- ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) -> ( ( ( x e. A /\ z e. A ) /\ ( E. y e. B C = ( x +h y ) /\ E. w e. B C = ( z +h w ) ) ) -> x = z ) )`
19 1 18 syl5bir
` |- ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) -> ( ( ( x e. A /\ E. y e. B C = ( x +h y ) ) /\ ( z e. A /\ E. w e. B C = ( z +h w ) ) ) -> x = z ) )`
20 19 alrimivv
` |- ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) -> A. x A. z ( ( ( x e. A /\ E. y e. B C = ( x +h y ) ) /\ ( z e. A /\ E. w e. B C = ( z +h w ) ) ) -> x = z ) )`
21 eleq1w
` |- ( x = z -> ( x e. A <-> z e. A ) )`
22 oveq1
` |- ( x = z -> ( x +h y ) = ( z +h y ) )`
23 22 eqeq2d
` |- ( x = z -> ( C = ( x +h y ) <-> C = ( z +h y ) ) )`
24 23 rexbidv
` |- ( x = z -> ( E. y e. B C = ( x +h y ) <-> E. y e. B C = ( z +h y ) ) )`
25 oveq2
` |- ( y = w -> ( z +h y ) = ( z +h w ) )`
26 25 eqeq2d
` |- ( y = w -> ( C = ( z +h y ) <-> C = ( z +h w ) ) )`
27 26 cbvrexvw
` |- ( E. y e. B C = ( z +h y ) <-> E. w e. B C = ( z +h w ) )`
28 24 27 syl6bb
` |- ( x = z -> ( E. y e. B C = ( x +h y ) <-> E. w e. B C = ( z +h w ) ) )`
29 21 28 anbi12d
` |- ( x = z -> ( ( x e. A /\ E. y e. B C = ( x +h y ) ) <-> ( z e. A /\ E. w e. B C = ( z +h w ) ) ) )`
30 29 mo4
` |- ( E* x ( x e. A /\ E. y e. B C = ( x +h y ) ) <-> A. x A. z ( ( ( x e. A /\ E. y e. B C = ( x +h y ) ) /\ ( z e. A /\ E. w e. B C = ( z +h w ) ) ) -> x = z ) )`
31 20 30 sylibr
` |- ( ( A e. SH /\ B e. SH /\ ( A i^i B ) = 0H ) -> E* x ( x e. A /\ E. y e. B C = ( x +h y ) ) )` | 2,490 | 4,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-30 | latest | en | 0.216015 |
https://gsebsolutions.com/gseb-solutions-class-7-maths-chapter-2-ex-2-2/ | 1,723,553,144,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00362.warc.gz | 218,657,981 | 51,858 | # GSEB Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2
Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2
Question 1.
Which of the drawings (a) to (d) show:
(i) 2 x $$\frac { 1 }{ 5 }$$
(ii) 2 x $$\frac { 1 }{ 2 }$$
(iii) 3 x $$\frac { 2 }{ 3 }$$
(iv) 3 x $$\frac { 1 }{ 4 }$$
Solution:
(i) 2 x $$\frac { 1 }{ 5 }$$
= $$\frac { 1 }{ 5 }$$ + $$\frac { 1 }{ 5 }$$, which is represented by the drawing (d).
Thus (i) ā (d).
(ii) 2 x $$\frac { 1 }{ 2 }$$
= $$\frac { 1 }{ 2 }$$ + $$\frac { 1 }{ 2 }$$, which is represented by the drawing (d).
Thus (ii) ā (d).
(iii) 3 x $$\frac { 2 }{ 3 }$$
$$\frac { 2 }{ 3 }$$ + $$\frac { 2 }{ 3 }$$ + $$\frac { 2 }{ 3 }$$, which is represented by the drawing (a).
Thus, (iii) ā (a).
(iv) 3 x $$\frac { 1 }{ 4 }$$
$$\frac { 1 }{ 4 }$$ + $$\frac { 1 }{ 4 }$$ + $$\frac { 1 }{ 4 }$$, which is represented by the drawing (c).
Thus, (iv) ā (c).
Question 2.
Some pictures (a) to (c) are given below. Tell which of them show:
(i) 3 x $$\frac { 1 }{ 5 }$$
(ii) 2 x $$\frac { 1 }{ 3 }$$ = $$\frac { 2 }{ 3 }$$
(iii) 2 x $$\frac { 1 }{ 3 }$$ = 2 x $$\frac { 1 }{ 4 }$$
Solution:
(i) āµ 3 x = $$\frac { 1 }{ 5 }$$
$$\frac { 1 }{ 5 }$$ + $$\frac { 1 }{ 5 }$$ + $$\frac { 1 }{ 5 }$$ = $$\frac { 3 }{ 5 }$$, which is shown by the drawing (c).
ā“ (i) ā (C)
(ii) āµ 2 x $$\frac { 1 }{ 3 }$$ = $$\frac { 1 }{ 3 }$$ + $$\frac { 1 }{ 3 }$$ = $$\frac { 2 }{ 3 }$$, which is shown by the drawing (a).
ā“ (ii) ā (C)
(iii) āµ 3 x $$\frac { 3 }{ 4 }$$ = $$\frac { 3 }{ 4 }$$ + $$\frac { 3 }{ 4 }$$ + $$\frac { 3 }{ 4 }$$, which is shown by the drawing (a).
ā“ (iii) ā (C)
Question 3.
Multiply and reduce to lowest form and convert into a mixed fraction:
(i) 7 x $$\frac { 3 }{ 5 }$$
(ii) 4 x $$\frac { 1 }{ 3 }$$
(iii) 2 x $$\frac { 6 }{ 7 }$$
(iv) 5 x $$\frac { 2 }{ 9 }$$
(v) $$\frac { 2 }{ 3 }$$ x 4
(vi) $$\frac { 5 }{ 2 }$$ x 6
(vii) 11 x $$\frac { 4 }{ 7 }$$
(viii) 20 x $$\frac { 4 }{ 5 }$$
(ix) 13 x $$\frac { 1 }{ 3 }$$
(x) 15 x $$\frac { 3 }{ 5 }$$
Solution:
Question 4.
(i) $$\frac { 1 }{ 2 }$$ of the circles in box (a)
(ii) $$\frac { 2 }{ 3 }$$ of the triangles in box (b)
(ii) $$\frac { 3 }{ 5 }$$ of the squares in box (c)
Solution:
(i) $$\frac { 1 }{ 2 }$$ of the circles:
āµ $$\frac { 1 }{ 2 }$$ of 12 = $$\frac { 1 }{ 2 }$$ x 12 = 6
(ii) $$\frac { 2 }{ 3 }$$ of the triangles:
āµ $$\frac { 2 }{ 3 }$$ of 9 = $$\frac { 2 Ć 9 }{ 3 }$$ = 6
(ii) $$\frac { 3 }{ 5 }$$ of the squares:
āµ $$\frac { 3 }{ 5 }$$ or 15 = $$\frac { 3 Ć 15 }{ 5 }$$ = 9
Question 5.
Find:
(a) $$\frac { 1 }{ 2 }$$ of (i) 24 (ii) 46
(b) $$\frac { 2 }{ 3 }$$ of (i) 18 (ii) 27
(c) $$\frac { 3 }{ 4 }$$ of (i) 16 (ii) 36
(d) $$\frac { 4 }{ 5 }$$ of (i) 20 (ii) 35
Solution:
Question 6.
Multiply and express as a mixed fraction:
(a) 3 x 5$$\frac { 1 }{ 5 }$$
(b) 5 x 6$$\frac { 3 }{ 4 }$$
(c) 7 x 2$$\frac { 1 }{ 4 }$$
(d) 4 x 6$$\frac { 1 }{ 3 }$$
(e) 3$$\frac { 1 }{ 4 }$$ x 6
(f) 3$$\frac { 2 }{ 5 }$$ x 8
Solution:
Question 7.
Find:
(a) $$\frac { 1 }{ 2 }$$ of (i) 2$$\frac { 3 }{ 4 }$$ (ii) 4$$\frac { 2 }{ 9 }$$
(b) $$\frac { 5 }{ 8 }$$ of (i) 3$$\frac { 5 }{ 6 }$$ (ii) 9$$\frac { 2 }{ 3 }$$
Solution:
(a) (i) $$\frac { 1 }{ 2 }$$ of 2$$\frac { 3 }{ 4 }$$
= $$\frac { 1 }{ 2 }$$ of $$\frac { 11 }{ 4 }$$
= $$\frac { 1Ć11 }{ 2Ć4 }$$
= $$\frac { 11 }{ 8 }$$
= 1$$\frac { 3 }{ 8 }$$
(ii) $$\frac { 1 }{ 2 }$$ of 4$$\frac { 2 }{ 9 }$$
= $$\frac { 1 }{ 2 }$$ of $$\frac { 38 }{ 9 }$$
= $$\frac { 1Ć38 }{ 2Ć9 }$$
= $$\frac { 19 }{ 9 }$$
= 2$$\frac { 1 }{ 9 }$$
(b) (i) $$\frac { 5 }{ 8 }$$ of 3$$\frac { 5 }{ 6 }$$
= $$\frac { 5 }{ 8 }$$ of $$\frac { 23 }{ 6 }$$
= $$\frac { 5Ć23 }{ 8Ć6 }$$
= $$\frac { 115 }{ 48 }$$
= 2$$\frac { 19 }{ 48 }$$
(ii) $$\frac { 5 }{ 8 }$$ of 9$$\frac { 2 }{ 3 }$$
= $$\frac { 5 }{ 8 }$$ of $$\frac { 29 }{ 3 }$$
= $$\frac { 5Ć29 }{ 8Ć3 }$$
= $$\frac { 145 }{ 24 }$$
= 6$$\frac { 1 }{ 24 }$$
Question 8.
Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed $$\frac { 2 }{ 5 }$$ of the water.
Pratap consumed the remaining water.
(i) How much water did Vidya drink7
(ii) What fraction of the total quantity of water did Pratap drink?
Solution:
Total quantity of water = 5 litres
(i) Amount of water consumed by Vidya = $$\frac { 2 }{ 5 }$$ of 5 litres = $$\frac { 2 }{ 5 }$$ x 5 litres = 2 litres
(ii) Remaining water = water consumed by Pratap
i.e. Amount of water consumed by Pratap = 5 litres – 2 litres = 3 litres
Fraction of water consumed by Pratap = $$\frac { 3 }{ 5 }$$ | 2,279 | 4,703 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-33 | latest | en | 0.624859 |
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# Combinations in C++
, 13 Sep 2009 CPOL 205K 4.7K 58
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An article on finding combinations.
## Introduction
Combination is the way of picking a different unique smaller set from a bigger set, without regard to the ordering (positions) of the elements (in the smaller set). This article teaches you how to find combinations. First, I will show you the technique to find combinations. Next, I will go on to explain how to use my source code. The source includes a recursive template version and a non-recursive template version. At the end of the article, I will show you how to find permutations of a smaller set from a bigger set, using both `next_combination()` and `next_permutation()`.
Before all these, let me first introduce to you the technique of finding combinations.
## The Technique
### The notations used in this article
• n: n is the larger sequence from which the r sequence is picked.
• r: r is the smaller sequence picked from the n sequence.
• c: c is the formula for the total number of possible combinations of r, picked from n distinct objects: n! / (r! (n-r)!).
• The ! postfix means factorial.
### Explanation
Let me explain using a very simple example: finding all combinations of 2 from a set of 6 letters {A, B, C, D, E, F}. The first combination is AB and the last is EF.
The total number of possible combinations is: n!/(r!(n-r)!)=6!/(2!(6-2)!)=15 combinations.
Let me show you all the combinations first:
```AB
AC
AE
AF
BC
BD
BE
BF
CD
CE
CF
DE
DF
EF```
If you can't spot the pattern, here it is:
```AB | AB
A | AC
A | AE
A | AF
---|----
BC | BC
B | BD
B | BE
B | BF
---|----
CD | CD
C | CE
C | CF
---|----
DE | DE
D | DF
---|----
EF | EF```
The same thing goes to combinations of any number of letters. Let me give you a few more examples and then you can figure them out yourself.
Combinations of 3 letters from {A, B, C, D, E} (a set of 5 letters).
The total number of possible combinations is: 10
```A B C
A B D
A B E
A C D
A C E
A D E
B C D
B C E
B D E
C D E```
Combinations of 4 letters from {A, B, C, D, E, F} (a set of 6 letters).
The total number of possible combinations is: 15.
```A B C D
A B C E
A B C F
A B D E
A B D F
A B E F
A C D E
A C D F
A C E F
A D E F
B C D E
B C D F
B C E F
B D E F
C D E F```
I'm thinking if you would have noticed by now, the number of times a letter appears. The formula for the number of times a letter appears in all possible combinations is n!/(r!(n-r)!) * r / n == c * r / n. Using the above example, it would be 15 * 4 / 6 = 10 times. All the letters {A, B, C, D, E, F} appear 10 times as shown. You can count them yourself to prove it.
## Source Code Section
Please note that all the combination functions are now enclosed in the `stdcomb` namespace.
### The Recursive Way
I have made a recursive function, `char_combination()` which, as its name implies, takes in character arrays and processes them. The source code and examples of using `char_combination()` are in char_comb_ex.cpp. I'll stop to mention that function. For now, our focus is on `recursive_combination()`, a template function, which I wrote using `char_combination()` as a guideline.
The function is defined in combination.h as below:
```// Recursive template function
template <class RanIt, class Func>
void recursive_combination(RanIt nbegin, RanIt nend, int n_column,
RanIt rbegin, RanIt rend, int r_column,int loop, Func func)
{
int r_size=rend-rbegin;
int localloop=loop;
int local_n_column=n_column;
//A different combination is out
if(r_column>(r_size-1))
{
func(rbegin,rend);
return;
}
//===========================
for(int i=0;i<=loop;++i)
{
RanIt it1=rbegin;
for(int cnt=0;cnt<r_column;++cnt)
{
++it1;
}
RanIt it2=nbegin;
for(int cnt2=0;cnt2<n_column+i;++cnt2)
{
++it2;
}
*it1=*it2;
++local_n_column;
recursive_combination(nbegin,nend,local_n_column,
rbegin,rend,r_column+1,localloop,func);
--localloop;
}
}```
The parameters prefixed with 'n' are associated with the n sequence, while the r-prefixed one are r sequence related. As an end user, you need not bother about those parameters. What you need to know is `func`. `func` is a function defined by you. If the combination function finds combinations recursively, there must exist a way the user can process each combination. The solution is a function pointer which takes in two parameters of type `RanIt` (stands for Random Iterator). You are the one who defines this function. In this way, encapsulation is achieved. You need not know how `recursive_combination()` internally works, you just need to know that it calls `func` whenever there is a different combination, and you just need to define the `func()` function to process the combination. It must be noted that `func()` should not write to the two iterators passed to it.
The typical way of filling out the parameters is `n_column` and `r_column` is always 0, `loop` is the number of elements in the r sequence minus that of the n sequence, and `func` is the function pointer to your function (`nbegin` and `nend`, and `rbegin` and `rend` are self-explanatory; they are the first iterators and the one past the last iterators of the respective sequences).
Just for your information, the maximum depth of the recursion done is r+1. In the last recursion (r+1 recursion), each new combination is formed.
An example of using `recursive_combination()` with raw character arrays is shown below:
```#include <iostream>
#include <vector>
#include <string>
#include "combination.h"
using namespace std;
using namespace stdcomb;
void display(char* begin,char* end)
{
cout<<begin<<endl;
}
int main()
{
char ca[]="123456";
char cb[]="1234";
recursive_combination(ca,ca+6,0,
cb,cb+4,0,6-4,display);
cout<<"Complete!"<<endl;
return 0;
}```
An example of using `recursive_combination()` with a vector of integers is shown below:
```#include <iostream>
#include <vector>
#include <string>
#include "combination.h"
typedef vector<int />::iterator vii;
void display(vii begin,vii end)
{
for (vii it=begin;it!=end;++it)
cout<<*it;
cout<<endl;
}
int main()
{
vector<int> ca;
ca.push_back (1);
ca.push_back (2);
ca.push_back (3);
ca.push_back (4);
ca.push_back (5);
ca.push_back (6);
vector<int> cb;
cb.push_back (1);
cb.push_back (2);
cb.push_back (3);
cb.push_back (4);
recursive_combination(ca.begin (),ca.end(),0,
cb.begin(),cb.end(),0,6-4,display);
cout<<"Complete!"<<endl;
return 0;
}```
### The Non-Recursive Way
If you have misgivings about using the recursive method, there is a non-recursive template function for you to choose. (Actually there are two.)
The parameters are even simpler than the recursive version. Here's the function definition in combination.h:
```template <class BidIt>
bool next_combination(BidIt n_begin, BidIt n_end,
BidIt r_begin, BidIt r_end);
template <class BidIt>
bool next_combination(BidIt n_begin, BidIt n_end,
BidIt r_begin, BidIt r_end, Prediate Equal );```
And its reverse counterpart version:
```template <class BidIt>
bool prev_combination(BidIt n_begin, BidIt n_end,
BidIt r_begin, BidIt r_end);
template <class BidIt>
bool prev_combination(BidIt n_begin, BidIt n_end,
BidIt r_begin, BidIt r_end, , Prediate Equal );```
The parameters `n_begin` and `n_end` are the first and the last iterators for the n sequence. And, `r_begin` and `r_end` are iterators for the r sequence. `Equal` is the predicate for comparing equality.
You can peruse the source code for these two functions in combination.h and its examples in next_comb_ex.cpp and prev_comb_ex.cpp, if you want.
A typical way of using `next_combination` with raw character arrays is as below:
```#include <iostream>
#include <vector>
#include <string>
#include "combination.h"
using namespace std;
using namespace stdcomb;
int main()
{
char ca[]="123456";
char cb[]="1234";
do
{
cout<<cb<<endl;
}
while(next_combination(ca,ca+6,cb,cb+4));
cout<<"Complete!"<<endl;
return 0;
}```
A typical way of using `next_combination` with a vector of integers is as below:
```#include <iostream>
#include <vector>
#include <string>
#include "combination.h"
template<class BidIt>
void display(BidIt begin,BidIt end)
{
for (BidIt it=begin;it!=end;++it)
cout<<*it<<" ";
cout<<endl;
}
int main()
{
vector<int> ca;
ca.push_back (1);
ca.push_back (2);
ca.push_back (3);
ca.push_back (4);
ca.push_back (5);
ca.push_back (6);
vector<int> cb;
cb.push_back (1);
cb.push_back (2);
cb.push_back (3);
cb.push_back (4);
do
{
display(cb.begin(),cb.end());
}
while(next_combination(ca.begin (),ca.end (),cb.begin (),cb.end()) );
cout<<"Complete!"<<endl;
return 0;
}```
### Certain conditions must be satisfied in order for next_combination() to work
1. All the objects in the n sequence must be distinct.
2. For `next_combination()`, the r sequence must be initialized to the first r-th elements of the n sequence in the first call. For example, to find combinations of r=4 out of n=6 {1,2,3,4,5,6}, the r sequence must be initialsed to {1,2,3,4} before the first call.
3. As for `prev_combination()`, the r sequence must be initialised to the last r-th elements of the n sequence in the first call. For example, to find combinations of r=4 out of n=6 {1,2,3,4,5,6}, the r sequence must be initialsed to {3,4,5,6} before the first call.
4. The n sequence must not change throughout the process of finding all the combinations, else results are wrong (makes sense, right?).
5. `next_combination()` and `prev_combination()` operate on data types with the `==` operator defined. That is to mean if you want to use `next_combination()` on sequences of objects instead of sequences of POD (Plain Old Data), the class from which these objects are instantiated must have an overloaded `==` operator defined, or you can use the predicate versions.
When the above conditions are not satisfied, results are undetermined even if `next_combination()` and `prev_combination()` may return `true`.
### Return Value
When `next_combination()` returns `false`, no more next combinations can be found, and the r sequence remains unaltered. Same for `prev_combination()`.
### Some information about next_combination() and prev_combination()
1. The n and r sequences need not be sorted to use `next_combination()` or `prev_combination()`.
2. `next_combination()` and `prev_combination()` do not use any static variables, so it is alright to find combinations of another sequence of a different data type, even when the current finding of combinations of the current sequence have not reached the last combination. In other words, no reset is needed for `next_combination()` and `prev_combination()`.
Examples of how to use these two functions are in next_comb_ex.cpp and prev_comb_ex.cpp.
## So what can we do with next_combination()?
With `next_combination()` and `next_permutation()` from STL algorithms, we can find permutations!!
The formula for the total number of permutations of the r sequence, picked from the n sequence, is: n!/(n-r)!
We can call `next_combination()` first, then `next_permutation()` iteratively; that way, we will find all the permutations. A typical way of using them is as follows:
```sort(n.begin(),n.end());
do
{
sort(r.begin(),r.end());
//do your processing on the new combination here
do
{
//do your processing on the new permutation here
}
while(next_permutation(r2.begin(),r2.end()))
}
while(next_combination(n.begin(),n.end(),r.begin(),r.end() ));```
However, I must mention that there exists a limitation for the above code. The n and r sequences must be sorted in ascending order in order for it to work. This is because `next_permutation()` will return `false` when it encounters a sequence in descending order. The solution to this problem for unsorted sequences is as follows:
```do
{
//do your processing on the new combination here
for(cnt i=0;cnt<24;++cnt)
{
next_permutation(r2.begin(),r2.end());
//do your processing on the new permutation here
}
}
while(next_combination(n.begin(),n.end(),r.begin(),r.end() ));```
However, this method requires you to calculate the number of permutations beforehand.
## So how do I prove they are distinct permutations?
There is a `set` container class in STL we can use. All the objects in the `set` container are always in sorted order, and there are no duplicate objects. For our purpose, we will use this `insert()` member function:
`pair <iterator, bool> insert(const value_type& _Val);`
The `insert()` member function returns a pair, whose `bool` component returns `true` if an insertion is made, and `false` if the `set` already contains an element whose key had an equivalent value in the ordering, and whose iterator component returns the address where a new element is inserted or where the element is already located.
proof.cpp is written for this purpose, using the STL `set` container to prove that the permutations generated are unique. You can play around with this, but you should first calculate the number of permutations which would be generated. Too many permutations may take ages to complete (partly due to the working of the `set` container), or worse, you may run out of memory!
If you are interested, you can proceed to read the second part of the article: Combinations in C++, Part 2.
## History
• 14 September 2009 - Added the example code.
• 21 February 2008 - Added the finding combinations of vectors in the source code.
• 26 November 2006 - Source code changes and bug fixes.
• All functions are enclosed in the `stdcomb` namespace.
• Solved a bug in `prev_combination` that `!=` operator must be defined for the custom class, unless the data type is a POD.
• `next_combination` and `prev_combination` now run properly in Visual C++ 8.0, without disabling the checked iterator.
• `next_combination` and `prev_combination` have a predicates version.
• 30 July 2003 - First release on CodeProject.
## About the Author
Software Developer Singapore
IT Certifications
• IT Infrastructure Library Foundational (ITIL v3)
• Scrum Alliance Certified Scrum Master (CSM)
• EC-Council Certified Secure Programmer (ECSP) .NET
• EC-Council Certified Ethical Hacker (CEH)
• EC-Council Certified Security Analyst (ECSA)
## Comments and Discussions
First PrevNext
what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali9-Jan-14 17:26 Deepak Gawali 9-Jan-14 17:26
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon9-Jan-14 17:35 Wong Shao Voon 9-Jan-14 17:35
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali10-Jan-14 0:14 Deepak Gawali 10-Jan-14 0:14
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon10-Jan-14 2:04 Wong Shao Voon 10-Jan-14 2:04
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali11-Jan-14 3:04 Deepak Gawali 11-Jan-14 3:04
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon11-Jan-14 14:21 Wong Shao Voon 11-Jan-14 14:21
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali12-Jan-14 1:24 Deepak Gawali 12-Jan-14 1:24
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon12-Jan-14 2:50 Wong Shao Voon 12-Jan-14 2:50
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali12-Jan-14 23:45 Deepak Gawali 12-Jan-14 23:45
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali13-Jan-14 0:32 Deepak Gawali 13-Jan-14 0:32
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon13-Jan-14 15:10 Wong Shao Voon 13-Jan-14 15:10
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali13-Jan-14 18:23 Deepak Gawali 13-Jan-14 18:23
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon13-Jan-14 18:34 Wong Shao Voon 13-Jan-14 18:34
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali13-Jan-14 20:54 Deepak Gawali 13-Jan-14 20:54
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon13-Jan-14 22:35 Wong Shao Voon 13-Jan-14 22:35
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali13-Jan-14 23:25 Deepak Gawali 13-Jan-14 23:25
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon14-Jan-14 3:16 Wong Shao Voon 14-Jan-14 3:16
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali14-Jan-14 18:36 Deepak Gawali 14-Jan-14 18:36
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali31-Jan-14 0:47 Deepak Gawali 31-Jan-14 0:47
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Wong Shao Voon31-Jan-14 2:14 Wong Shao Voon 31-Jan-14 2:14
Re: what changes in code are required to use your Combinations in C++ for array of float numbers? Deepak Gawali31-Jan-14 19:37 Deepak Gawali 31-Jan-14 19:37
if you are like me need to workout a diagonal half of a matrix, i found this [email protected] 10:39 [email protected] 10-May-13 10:39
My vote of 5 Lev Panov3-Nov-12 22:19 Lev Panov 3-Nov-12 22:19
A recursive program in C to generate nCk Manohar Bhat20-Jan-12 8:29 Manohar Bhat 20-Jan-12 8:29
My vote of 5 Van2348-Nov-11 3:18 Van234 8-Nov-11 3:18
Combinations, Permutations and Factorial online calculator DrABELL14-Sep-09 10:57 DrABELL 14-Sep-09 10:57
Combination of vector elem. but: find all combis of two 2tuples out of all? Knarfinger9-Feb-09 6:56 Knarfinger 9-Feb-09 6:56
Dear Wong Shao Voon, dear all, I used succesfully your code "prev_comb_ex.cpp" for finding all combinations - thanks! (BTW: next_comb_ex.cpp unfortunately has some compile error as unzipped, which I could not yet fix.) Of course directly possible to get e.g. all pairs out of all(, when using r=2). Anyhow, I need to find all combinations of exactly two pairs out of all, means e.g.: all={1,2,3,4} => { (1,2),(3,4); (1,3),(2,4); (1,4),(2,3) } It seems that your code should easily be adoptable/appliable for such prblem as well, but I am not able to see how / get it ... Any idea, help, already solved this (similar) problem? Thanks a lot! Frank
Re: Combination of vector elem. but: find all combis of two 2tuples out of all? Wong Shao Voon19-Feb-09 14:19 Wong Shao Voon 19-Feb-09 14:19
Combination of vector elements tuxyboy4-Feb-08 2:45 tuxyboy 4-Feb-08 2:45
Re: Combination of vector elements Wong Shao Voon13-Feb-08 20:19 Wong Shao Voon 13-Feb-08 20:19
Re: Combination of vector elements [modified] tuxyboy9-Jun-08 20:45 tuxyboy 9-Jun-08 20:45
Re: Combination of vector elements Wong Shao Voon10-Jun-08 0:36 Wong Shao Voon 10-Jun-08 0:36
Re: Combination of vector elements tuxyboy12-Jun-08 0:13 tuxyboy 12-Jun-08 0:13
permutaion's signature Alexander Arhipenko29-Nov-06 23:30 Alexander Arhipenko 29-Nov-06 23:30
Re: permutaion's signature Wong Shao Voon30-Nov-06 11:44 Wong Shao Voon 30-Nov-06 11:44
Re: permutaion's signature Wong Shao Voon13-Feb-08 20:26 Wong Shao Voon 13-Feb-08 20:26
Question about const expression for combination code computerpublic28-Nov-06 15:41 computerpublic 28-Nov-06 15:41
Re: Question about const expression for combination code Wong Shao Voon28-Nov-06 16:15 Wong Shao Voon 28-Nov-06 16:15
Re: Question about const expression for combination code computerpublic28-Nov-06 16:35 computerpublic 28-Nov-06 16:35
Re: Question about const expression for combination code Wong Shao Voon28-Nov-06 16:44 Wong Shao Voon 28-Nov-06 16:44
Re: Question about const expression for combination code computerpublic28-Nov-06 16:50 computerpublic 28-Nov-06 16:50
Re: Question about const expression for combination code computerpublic28-Nov-06 16:58 computerpublic 28-Nov-06 16:58
Re: Question about const expression for combination code Wong Shao Voon28-Nov-06 17:15 Wong Shao Voon 28-Nov-06 17:15
Re: Question about const expression for combination code Wong Shao Voon28-Nov-06 17:01 Wong Shao Voon 28-Nov-06 17:01
Question about combination code using numbers instead of letters computerpublic23-Nov-06 17:15 computerpublic 23-Nov-06 17:15
Re: Question about combination code using numbers instead of letters Wong Shao Voon25-Nov-06 13:11 Wong Shao Voon 25-Nov-06 13:11
Re: Question about combination code using numbers instead of letters computerpublic27-Nov-06 4:06 computerpublic 27-Nov-06 4:06
Combinations of vector elements. Bartosz Bien28-Oct-06 6:17 Bartosz Bien 28-Oct-06 6:17
Re: Combinations of vector elements. Wong Shao Voon29-Oct-06 18:32 Wong Shao Voon 29-Oct-06 18:32
Re: Combinations of vector elements. Bartosz Bien2-Dec-06 4:16 Bartosz Bien 2-Dec-06 4:16
Last Visit: 31-Dec-99 18:00 Last Update: 3-Jul-15 6:21 Refresh 12 Next »
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Use Ctrl+Left/Right to switch messages, Ctrl+Up/Down to switch threads, Ctrl+Shift+Left/Right to switch pages. | 5,983 | 21,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2015-27 | longest | en | 0.814634 |
http://mathhelpforum.com/statistics/170825-sat-question-tom-trip-expenses.html | 1,480,818,845,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541170.58/warc/CC-MAIN-20161202170901-00300-ip-10-31-129-80.ec2.internal.warc.gz | 174,823,432 | 13,608 | # Thread: SAT Question- Tom trip expenses
1. ## SAT Question- Tom trip expenses
Im sorry for asking two question in once but the First one is just to know if you guys can check my work. Question number 16. is the one i dont understand at all..
Image
2. Originally Posted by vaironxxrd
Im sorry for asking two question in once but the First one is just to know if you guys can check my work. Question number 16. is the one i dont understand at all..
Image
Top row n squares + right side (n - 1) squares + bottom (n - 1) squares + left side (n-2)
= 4n - 4 = 2(2n - 2) which is even.
All but b are possible is my thought.
3. As posted above, $k=4n-4 = 4(n-1)$. That means k is divisible by 4. Which of the 5 options is divisible by 4?
4. Originally Posted by harish21
As posted above, $k=4n-4 = 4(n-1)$. That means k is divisible by 4. Which of the 5 options is divisible by 4?
{e}52
5. Originally Posted by vaironxxrd
{e}52
6. Question 16 was already asked and answered by several of us here: http://www.mathhelpforum.com/math-he...em-168392.html
Keep in mind that these are not questions that you should be focusing on at this stage. These are Level 5 problems and you should be focusing on Level 1, 2 and 3 problems.
To put this in perspective, you can get a 700 in SAT Math without ever looking at a question that is this difficult (of course any 700 students that might be reading this should be practicing problems of this level).
I'm only going to get "preachy" on you once. If you continue to post problems of this level, then from now on I will simply answer them, but I feel that I am doing a disservice to you if I don't at least warn you.
Continuing to do problems of this level at this stage will have the following negative effects:
1) practicing these problems will not improve your SAT score. If you were to walk in and take the test tomorrow, then you should only be answering up to question 8 on this section. In 3 months time we might push that to question 12 (maybe even a bit higher if you do things right). In any case, you will most likely never be attempting a question this hard on the actual test. If you do attempt a question of this level on the actual test your score will be lowered!
2) practicing these problems is taking time away from practicing the types of questions that you will actually be answering on the test.
3) practicing these problems will get frustrating after a while, you will burn out faster, and you are more likely to stop preparing for the test.
4) practicing these problems will reduce your confidence level and you will experience more test-taking anxiety when you take the test - this will absolutely lower your score.
7. Originally Posted by DrSteve
Question 16 was already asked and answered by several of us here: http://www.mathhelpforum.com/math-he...em-168392.html
Keep in mind that these are not questions that you should be focusing on at this stage. These are Level 5 problems and you should be focusing on Level 1, 2 and 3 problems.
To put this in perspective, you can get a 700 in SAT Math without ever looking at a question that is this difficult (of course any 700 students that might be reading this should be practicing problems of this level).
I'm only going to get "preachy" on you once. If you continue to post problems of this level, then from now on I will simply answer them, but I feel that I am doing a disservice to you if I don't at least warn you.
Continuing to do problems of this level at this stage will have the following negative effects:
1) practicing these problems will not improve your SAT score. If you were to walk in and take the test tomorrow, then you should only be answering up to question 8 on this section. In 3 months time we might push that to question 12 (maybe even a bit higher if you do things right). In any case, you will most likely never be attempting a question this hard on the actual test. If you do attempt a question of this level on the actual test your score will be lowered!
2) practicing these problems is taking time away from practicing the types of questions that you will actually be answering on the test.
3) practicing these problems will get frustrating after a while, you will burn out faster, and you are more likely to stop preparing for the test.
4) practicing these problems will reduce your confidence level and you will experience more test-taking anxiety when you take the test - this will absolutely lower your score.
I agree with all the negative effects, I believe i should focus on 1,2 and 3 level question. I believe these are basic algebra?
But is there any other way to notice?
8. There are 4 main subject area that appear on the SAT:
Number Theory
Algebra and Functions
Probability and Statistics
Geometry
Each of these four subject areas will appear on every SAT in varying levels of difficulty - from Level 1 to Level 5.
How to tell what level problem you're doing depends on what book or other practice material you're using.
If you're taking a practice test (either a real SAT or a simulated one), then the question number itself will roughly tell you the difficulty of the question. If the section has 16 questions (like the section you took the question from in the OP) then as long as you are focusing on the first half of the questions, you will be doing mostly level 1 through 3 questions. So this particular section you should be focusing on the first 8 questions. When you're doing practice tests I suggest you try to use actual old SAT exams. If you can't get hold of any of these, then the 10 practice tests from the blue College Board SAT book will have to do. I know you said that you don't have money, so perhaps you can borrow this book from someone that has already taken the SAT - almost everyone who took an SAT should have a copy.
If you are doing practice questions from a book, then you need to read the introduction of the book to determine how they give their problems. My book, for example gives the Level and Subject Area for each question. I think the Princeton Review tells the comparable question number that it would be on the SAT. There is no uniform way this is done. There should be lots of free online practice problems as well, but again you'll have to determine how they let you know the difficulty level of each problem.
9. Originally Posted by DrSteve
There are 4 main subject area that appear on the SAT:
Number Theory
Algebra and Functions
Probability and Statistics
Geometry
Each of these four subject areas will appear on every SAT in varying levels of difficulty - from Level 1 to Level 5.
How to tell what level problem you're doing depends on what book or other practice material you're using.
If you're taking a practice test (either a real SAT or a simulated one), then the question number itself will roughly tell you the difficulty of the question. If the section has 16 questions (like the section you took the question from in the OP) then as long as you are focusing on the first half of the questions, you will be doing mostly level 1 through 3 questions. So this particular section you should be focusing on the first 8 questions. When you're doing practice tests I suggest you try to use actual old SAT exams. If you can't get hold of any of these, then the 10 practice tests from the blue College Board SAT book will have to do. I know you said that you don't have money, so perhaps you can borrow this book from someone that has already taken the SAT - almost everyone who took an SAT should have a copy.
If you are doing practice questions from a book, then you need to read the introduction of the book to determine how they give their problems. My book, for example gives the Level and Subject Area for each question. I think the Princeton Review tells the comparable question number that it would be on the SAT. There is no uniform way this is done. There should be lots of free online practice problems as well, but again you'll have to determine how they let you know the difficulty level of each problem.
I do have the blue book and i practice has much has possible usually 10 minutes since i got a lot of Projects in school at the moment, and exams.
But anyways are you ok if i ( send you in message/ or post a question) With the time i took? because i heard i should take 1-1 in a 1/2 minute for each question is this truth? i already did one which is logical about apples but i can't understand how to find the solution i could send it right now or post it
10. I would keep your posts to the forums so that other people can benefit from the discussion. But feel free to pm me when you post to let me know that there is a post you want me to see.
A general guideline for answering SAT questions is the following: try to be aware of when you've spent about a minute on a problem. At this point if you understand the problem and think you can solve it, then continue. If not, then move on to the next question and come back later if there's time.
You can take up to 2.5 minutes per question based on your current level. This will enable you to answer about half the questions in the section. Keep in mind that this will change if you start breaking a 400. NB: This paragraph is NOT advice for everyone - just the OP.
And go right ahead and start posting the questions you're having trouble with. | 2,105 | 9,343 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2016-50 | longest | en | 0.953223 |
https://www.manhattanprep.com/lsat/blog/whats-tested-on-lsat-logic-games/ | 1,558,589,187,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257100.22/warc/CC-MAIN-20190523043611-20190523065611-00098.warc.gz | 856,423,271 | 22,271 | ### What’s Tested on LSAT Logic Games
One of the most challenging things about LSAT Logic Games, formally known as the LSAT Analytical Reasoning Section, is that it tests skills that are totally foreign to most college curriculums. By the time we reach the LSAT-preparation stage of our lives, many of us haven’t done a puzzle in well over a decade. And yet here they are: four little puzzles standing between you and your law school dreams. Why are they there? What relevance could they possibly have to a career in law?
This article will explore the what and why of the LSAT Logic Games section—and provide our best tips for conquering it.
#### LSAT Logic Games Test Your Ability to Track Entities and Apply Constraints
According to the LSAC, the organization that creates and administers the LSAT, the LSAT Logic Games section “reflects the kinds of detailed analyses of relationships and sets of constraints that a law student must perform in legal problem solving.”
Okay, fair enough. As a law student, you’ll certainly need to analyze relationships between entities. The laws that apply to those entities can be thought of as a set of constraints.
The LSAT Logic Games section is composed of four games, each with at least one set of entities and between two and six constraints. If you’ve read about LSAT Logic Games, you’ll know that the entities are commonly referred to as elements or variables, and the constraints are also known as the rules.
#### LSAT Logic Games Tips for Mastering the Entities (aka the Elements, aka the Variables)
• Use a single letter to represent each entity, and create a roster of entities for every game you do.
• Be careful: LSAT Analytical Reasoning writers will sometimes try to trick you into representing an entity that isn’t really there by presenting their entities alphabetically. Consider this scenario where the entities are animals: armadillo, bear, chimpanzee, dingo and falcon. So here’s my roster: A B C D E F. Right? Wrong! Don’t get tricked into placing an “E” in your roster. Sorry, eagles…it’s not your day.
• If you have multiple sets of entities, always label your rosters. In the scenario above, if there were also veterinarians that’d be tending to the animals, you’d label one roster animals and the other vets. Failing to label your rosters sets you up for confusion later. Some people also find it helpful to make the entities in one roster uppercase and the other lowercase. This can help you differentiate between the categories of entity quickly when the entities are placed into your diagram.
#### LSAT Logic Games Tips for Mastering the Constraints (aka the Rules)
• Each constraint on the LSAT Analytical Reasoning section should be represented visually.
• The constraints overwhelmingly fall into just a few categories, and your shorthand for representing rules in each category should be consistent and on automatic recall in your brain.
• Now that the LSAT is going digital, you’ll be drawing your constraints on scratch paper instead of next to the game itself. Never fear. Stay organized by drawing a master diagram on one corner of your paper. Then represent the constraints in a single column next to your master diagram.
#### LSAT Logic Games Test Your Powers of Classification
Virtually all games ask you to do one of two things: put elements in order or put elements into groups. A few games (we call them hybrids) will ask you to do both. The first thing tested on LSAT Logic Games is whether you can classify the game based on its primary task.
If you think about it, that’s a skill a lawyer needs, too. Can you quickly and accurately look at a situation that has a bunch of impacted parties and a bunch of laws that apply, and decide what kind of case to pursue? Master LSAT Logic Games and the answer to that question will probably be yes!
#### LSAT Logic Games Tips for Mastering Classification
• The first question you should be asking yourself as you read a game is, “Is this game about putting things in order, putting things in groups, or both?”
• Once you figure out that basic classification, ask yourself questions to see if the game is a Basic Ordering game, a Basic Grouping game, or whether it has any sub-classifications.
• If the game is an Ordering game, are all the rules about where elements are placed relative to other elements? If so, you’re looking at a Relative Ordering game.
• Is there more than one set of entities in your ordering game, as in the animals and veterinarians example above? If so, you’re looking at a 3D Ordering game.
• Do you know how many members each group has in your grouping game? If not, you’re looking at an Open Grouping game.
• Is your game about figuring out which entities are selected and which are not? If so, you’re looking at an In/Out Grouping game.
• Visually represent the game based on the game’s classification.
• For games about putting one set of entities in order (Basic Ordering games or Relative Ordering games), use a number line with a row of slots on top.
• For games about putting two sets of entities in order (3D Ordering games), use a number line with two rows of slots on top.
• For games about putting things into groups with a defined number of members (Basic Grouping games), represent the groups in a row and put a slot above each group for each group member.
• For games about selecting some elements but not others, represent the In group and Out group using the Logic Chain.
#### LSAT Logic Games Test Deductive Reasoning
Ahh, logic. The bulwark of the LSAT, and rightly so. As a law student and a lawyer, you’ll encounter scenarios in which a variety of laws apply to some or all of the parties involved. What’s more, some of these laws will necessarily overlap. What will that overlap mean for the parties involved? That’s for you to find out!
In the LSAT Logic Games section, you’ll see this type of reasoning tested heavily. In virtually every game, there will be constraints that overlap and entities that are impacted by multiple constraints. Where this is the case, you can often use deductive reasoning to figure out something that must be true, which we’ll call an Inference.
Consider, if, in the game above, the vet had to treat the dingo before treating the bear. What can we infer? We can infer that the bear isn’t the first animal treated and the dingo isn’t the last. We call these Basic Inferences: Inferences that can be made from a single rule.
Now consider if, additionally, we knew the bear was treated immediately after the chimpanzee. We could draw the same type of Basic Inference here (bear isn’t treated first and chimpanzee isn’t treated last). But we can also combine the two rules to draw a more advanced set of Inferences. Now the dingo can’t be last or second-to-last, because it will have to precede both the bear and the chimpanzee. And the chimpanzee can’t be first, because, if it were, the dingo couldn’t precede the bear.
#### LSAT Logic Games Tips for Mastering Deductive Reasoning
• Rules that express order lead to Basic Inferences. The element that must go first in the rule can’t go last in the diagram… And the element that must go last in the rule can’t go first in the diagram. Make these Basic Inferences. Then, build them into your master diagram by drawing the element under the slot it cannot occupy and crossing out.
• If an element shows up in more than one rule (like the bear in the example above), chances are good that you can combine the rules to find more advanced Inferences.
• If a rule creates a two-way split in a game, you might be able to figure out every possible way the game could pan out using a technique called an advanced deductive technique called framing.
• Look for Inferences hiding in the numeric rules of the game. Always ask yourself “Does every element get used? Can any element repeat?”
#### LSAT Logic Games Test the Truth, the Whole Truth, and Nothing but the Truth
If there’s one thing LSAT Logic Games tests more than any other, it’s truth value: the difference between what’s definitely true, what might be true, what’s definitely false, and what might be false. Most of the questions on LSAT Logic Games ask you to select an answer that has one of these four qualities.
Some questions will introduce a new condition that only holds for that question. (e.g., If the vet treats falcons first, which of the following cannot be true?). These questions typically require you to draw a new mini-diagram. This diagram builds in this new condition and uses it to reach new Inferences. We call these Conditional questions.
Other questions (which we call Unconditional questions) will just ask broadly about the existing rules. (e.g., Which of the following animals could not be treated second?). For these questions, you may have made an Inference that answers the question right off the bat. If so, shazam! Your work here is done. But often you won’t have answered the question by finding an Inference… So you may find yourself drawing mini-diagrams to test some of the answers.
#### LSAT Logic Games Tips to Master Telling the Truth (from the Possible Truth, from the Definitely-Not-The-Truth)
• Always draw a new mini-diagram for Conditional questions.
• Whatever the truth value targeted by a question, always think through what that means about the right and the wrong answers. For example, if a question asks “Which of the following must be true?” that means that one answer, the correct one, must be true, while four answers, the incorrect ones, could be false.
• Once you know the truth value that would make an answer false, use process of elimination.
• Don’t be afraid to draw mini-diagrams to test out answers.
• To test whether an answer could be true, try to make a valid diagram in which it is true. If you can, it’s the right answer.
• To test whether an answer could be false, try to make a diagram in which it is false. If you can, it’s the right answer.
• To test whether an answer must be true, try to make a diagram in which it is false. If you can’t, it’s the right answer. Why approach it like this? Because making a diagram in which the answer is true would only show that it could be true. To see whether it must be true (without drawing a ton of diagrams), you need to prove that it can’t be false.
• To test whether an answer must be false, try to make a diagram in which it is true. If you can’t do it, the answer must be false.
#### What’s Tested on LSAT Logic Games: Bringing it All Together
Even though it might seem foreign, both to you as a student and as a hopeful future lawyer, the LSAT Logic Games section is more closely related to being a law student than it appears. More than any other section, the LSAT Logic Games section demands that you attack it systematically. If you don’t have a consistent process for denoting entities and constraints, classifying games, utilizing deductive reasoning and wrestling with truth values, your approach to Logic Games will be haphazard at best and disastrous at worst.
#### LSAT Logic Games Final Tips
• Start each game by picturing the real-world scenario to help you orient your diagram in a way that is intuitive.
• Practice drawing out your games on scratch paper. Label each mini-diagram that you draw for conditional questions and answer choice testing so you can use them again later!
• Don’t be afraid to ditch your timer for a while. It takes time to build a systematic process.
• Write down your work. Students tend to think that doing a question in their head will be faster than doing it on paper. This isn’t actually the case. Working on your paper makes it easier to see connections quickly, and it keeps your work visible so you don’t have to rehash it on, say, answer choice E.
• Drawing a blank? Put pencil to paper! If you don’t know where to begin, just start somewhere. The process of thinking on paper will help you organize your thoughts and get you moving.
• If Unconditional questions are stumping you, you might have missed an Inference. Never fear! Try the Conditional questions first, then look at the mini-diagrams you generated when answering the Unconditional questions.
• The hard questions aren’t worth more than the easier ones. As you build your final test-day approach, don’t be afraid to skip the stumpers and look for greener pastures, especially if you’re not consistently finishing the section.
Don’t stop here! We’ve got loads of great material to help you prep for the LSAT’s Logic Games section. Check out our free resources, books, self-study program, or try a class for free! 📝
Laura Damone is a Manhattan Prep instructor based in San Francisco, CA. She fell for the LSAT while getting her undergrad degree and has now taught LSAT classes at more than 20 universities around the country. When she’s not teaching, learning, or publishing her work, she can be found frolicking in the redwoods and exploring the Pacific coast. Check out Laura’s upcoming LSAT courses here! | 2,795 | 12,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-22 | longest | en | 0.943421 |
https://openstax.org/books/intermediate-algebra-2e/pages/9-1-solve-quadratic-equations-using-the-square-root-property | 1,726,637,008,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00620.warc.gz | 400,964,592 | 86,735 | Intermediate Algebra 2e
# 9.1Solve Quadratic Equations Using the Square Root Property
Intermediate Algebra 2e9.1 Solve Quadratic Equations Using the Square Root Property
## Learning Objectives
By the end of this section, you will be able to:
• Solve quadratic equations of the form $ax2=kax2=k$ using the Square Root Property
• Solve quadratic equations of the form $a(x–h)2=ka(x–h)2=k$ using the Square Root Property
## Be Prepared 9.1
Before you get started, take this readiness quiz.
Simplify: $128.128.$
If you missed this problem, review Example 8.13.
## Be Prepared 9.2
Simplify: $325325$.
If you missed this problem, review Example 8.50.
## Be Prepared 9.3
Factor: $9x2−12x+49x2−12x+4$.
If you missed this problem, review Example 6.23.
A quadratic equation is an equation of the form ax2 + bx + c = 0, where $a≠0a≠0$. Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax2. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.
We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.
## Solve Quadratic Equations of the form $ax2=kax2=k$ using the Square Root Property
We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x2 = 9.
$x2=9x2=9$ Put the equation in standard form. $x2−9=0x2−9=0$ Factor the difference of squares. $(x−3)(x+3)=0(x−3)(x+3)=0$ Use the Zero Product Property. $x−3=0x−3=0x−3=0x−3=0$ Solve each equation. $x=3x=−3x=3x=−3$
We can easily use factoring to find the solutions of similar equations, like x2 = 16 and x2 = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, $x=4,x=−4x=4,x=−4$ and $x=5,x=−5.x=5,x=−5.$
But what happens when we have an equation like x2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.
Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13)2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:
$Ifn2=m,thennis a square root ofm.Ifn2=m,thennis a square root ofm.$
Since these equations are all of the form x2 = k, the square root definition tells us the solutions are the two square roots of k. This leads to the Square Root Property.
## Square Root Property
If x2 = k, then
$x=korx=−korx=±k.x=korx=−korx=±k.$
Notice that the Square Root Property gives two solutions to an equation of the form x2 = k, the principal square root of $kk$ and its opposite. We could also write the solution as $x=±k.x=±k.$ We read this as x equals positive or negative the square root of k.
Now we will solve the equation x2 = 9 again, this time using the Square Root Property.
$x2=9x2=9$ Use the Square Root Property. $x=±9x=±9$ $x=±3x=±3$ $Sox=3orx=−3.Sox=3orx=−3.$
What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x2 = 7.
$x2=7x2=7$ Use the Square Root Property. $x=7,x=−7x=7,x=−7$
We cannot simplify $77$, so we leave the answer as a radical.
## Example 9.1
### How to solve a Quadratic Equation of the form ax2 = k Using the Square Root Property
Solve: $x2−50=0.x2−50=0.$
## Try It 9.1
Solve: $x2−48=0.x2−48=0.$
## Try It 9.2
Solve: $y2−27=0.y2−27=0.$
The steps to take to use the Square Root Property to solve a quadratic equation are listed here.
## How To
### Solve a quadratic equation using the square root property.
1. Step 1. Isolate the quadratic term and make its coefficient one.
2. Step 2. Use Square Root Property.
3. Step 3. Simplify the radical.
4. Step 4. Check the solutions.
In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.
## Example 9.2
Solve: $3z2=108.3z2=108.$
## Try It 9.3
Solve: $2x2=98.2x2=98.$
## Try It 9.4
Solve: $5m2=80.5m2=80.$
The Square Root Property states ‘If $x2=kx2=k$,’ What will happen if $k<0?k<0?$ This will be the case in the next example.
## Example 9.3
Solve: $x2+72=0x2+72=0$.
## Try It 9.5
Solve: $c2+12=0.c2+12=0.$
## Try It 9.6
Solve: $q2+24=0.q2+24=0.$
Our method also works when fractions occur in the equation; we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.
## Example 9.4
Solve: $23u2+5=17.23u2+5=17.$
## Try It 9.7
Solve: $12x2+4=24.12x2+4=24.$
## Try It 9.8
Solve: $34y2−3=18.34y2−3=18.$
The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.
## Example 9.5
Solve: $2x2−8=41.2x2−8=41.$
## Try It 9.9
Solve: $5r2−2=34.5r2−2=34.$
## Try It 9.10
Solve: $3t2+6=70.3t2+6=70.$
## Solve Quadratic Equations of the Form a(x − h)2 = k Using the Square Root Property
We can use the Square Root Property to solve an equation of the form a(xh)2 = k as well. Notice that the quadratic term, x, in the original form ax2 = k is replaced with (xh).
The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a, then the Square Root Property can be used on (xh)2.
## Example 9.6
Solve: $4(y−7)2=48.4(y−7)2=48.$
## Try It 9.11
Solve: $3(a−3)2=54.3(a−3)2=54.$
## Try It 9.12
Solve: $2(b+2)2=80.2(b+2)2=80.$
Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.
## Example 9.7
Solve: $(x−13)2=59.(x−13)2=59.$
## Try It 9.13
Solve: $(x−12)2=54.(x−12)2=54.$
## Try It 9.14
Solve: $(y+34)2=716.(y+34)2=716.$
We will start the solution to the next example by isolating the binomial term.
## Example 9.8
Solve: $2(x−2)2+3=57.2(x−2)2+3=57.$
## Try It 9.15
Solve: $5(a−5)2+4=104.5(a−5)2+4=104.$
## Try It 9.16
Solve: $3(b+3)2−8=88.3(b+3)2−8=88.$
Sometimes the solutions are complex numbers.
## Example 9.9
Solve: $(2x−3)2=−12.(2x−3)2=−12.$
## Try It 9.17
Solve: $(3r+4)2=−8.(3r+4)2=−8.$
## Try It 9.18
Solve: $(2t−8)2=−10.(2t−8)2=−10.$
The left sides of the equations in the next two examples do not seem to be of the form a(xh)2. But they are perfect square trinomials, so we will factor to put them in the form we need.
## Example 9.10
Solve: $4n2+4n+1=16.4n2+4n+1=16.$
## Try It 9.19
Solve: $9m2−12m+4=25.9m2−12m+4=25.$
## Try It 9.20
Solve: $16n2+40n+25=4.16n2+40n+25=4.$
## Media
Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.
## Section 9.1 Exercises
### Practice Makes Perfect
Solve Quadratic Equations of the Form ax2 = k Using the Square Root Property
In the following exercises, solve each equation.
1.
$a 2 = 49 a 2 = 49$
2.
$b 2 = 144 b 2 = 144$
3.
$r 2 − 24 = 0 r 2 − 24 = 0$
4.
$t 2 − 75 = 0 t 2 − 75 = 0$
5.
$u 2 − 300 = 0 u 2 − 300 = 0$
6.
$v 2 − 80 = 0 v 2 − 80 = 0$
7.
$4 m 2 = 36 4 m 2 = 36$
8.
$3 n 2 = 48 3 n 2 = 48$
9.
$4 3 x 2 = 48 4 3 x 2 = 48$
10.
$5 3 y 2 = 60 5 3 y 2 = 60$
11.
$x 2 + 25 = 0 x 2 + 25 = 0$
12.
$y 2 + 64 = 0 y 2 + 64 = 0$
13.
$x 2 + 63 = 0 x 2 + 63 = 0$
14.
$y 2 + 45 = 0 y 2 + 45 = 0$
15.
$4 3 x 2 + 2 = 110 4 3 x 2 + 2 = 110$
16.
$2 3 y 2 − 8 = −2 2 3 y 2 − 8 = −2$
17.
$2 5 a 2 + 3 = 11 2 5 a 2 + 3 = 11$
18.
$3 2 b 2 − 7 = 41 3 2 b 2 − 7 = 41$
19.
$7 p 2 + 10 = 26 7 p 2 + 10 = 26$
20.
$2 q 2 + 5 = 30 2 q 2 + 5 = 30$
21.
$5 y 2 − 7 = 25 5 y 2 − 7 = 25$
22.
$3 x 2 − 8 = 46 3 x 2 − 8 = 46$
Solve Quadratic Equations of the Form a(xh)2 = k Using the Square Root Property
In the following exercises, solve each equation.
23.
$( u − 6 ) 2 = 64 ( u − 6 ) 2 = 64$
24.
$( v + 10 ) 2 = 121 ( v + 10 ) 2 = 121$
25.
$( m − 6 ) 2 = 20 ( m − 6 ) 2 = 20$
26.
$( n + 5 ) 2 = 32 ( n + 5 ) 2 = 32$
27.
$( r − 1 2 ) 2 = 3 4 ( r − 1 2 ) 2 = 3 4$
28.
$( x + 1 5 ) 2 = 7 25 ( x + 1 5 ) 2 = 7 25$
29.
$( y + 2 3 ) 2 = 8 81 ( y + 2 3 ) 2 = 8 81$
30.
$( t − 5 6 ) 2 = 11 25 ( t − 5 6 ) 2 = 11 25$
31.
$( a − 7 ) 2 + 5 = 55 ( a − 7 ) 2 + 5 = 55$
32.
$( b − 1 ) 2 − 9 = 39 ( b − 1 ) 2 − 9 = 39$
33.
$4 ( x + 3 ) 2 − 5 = 27 4 ( x + 3 ) 2 − 5 = 27$
34.
$5 ( x + 3 ) 2 − 7 = 68 5 ( x + 3 ) 2 − 7 = 68$
35.
$( 5 c + 1 ) 2 = −27 ( 5 c + 1 ) 2 = −27$
36.
$( 8 d − 6 ) 2 = −24 ( 8 d − 6 ) 2 = −24$
37.
$( 4 x − 3 ) 2 + 11 = −17 ( 4 x − 3 ) 2 + 11 = −17$
38.
$( 2 y + 1 ) 2 − 5 = −23 ( 2 y + 1 ) 2 − 5 = −23$
39.
$m 2 − 4 m + 4 = 8 m 2 − 4 m + 4 = 8$
40.
$n 2 + 8 n + 16 = 27 n 2 + 8 n + 16 = 27$
41.
$x 2 − 6 x + 9 = 12 x 2 − 6 x + 9 = 12$
42.
$y 2 + 12 y + 36 = 32 y 2 + 12 y + 36 = 32$
43.
$25 x 2 − 30 x + 9 = 36 25 x 2 − 30 x + 9 = 36$
44.
$9 y 2 + 12 y + 4 = 9 9 y 2 + 12 y + 4 = 9$
45.
$36 x 2 − 24 x + 4 = 81 36 x 2 − 24 x + 4 = 81$
46.
$64 x 2 + 144 x + 81 = 25 64 x 2 + 144 x + 81 = 25$
### Mixed Practice
In the following exercises, solve using the Square Root Property.
47.
$2 r 2 = 32 2 r 2 = 32$
48.
$4 t 2 = 16 4 t 2 = 16$
49.
$( a − 4 ) 2 = 28 ( a − 4 ) 2 = 28$
50.
$( b + 7 ) 2 = 8 ( b + 7 ) 2 = 8$
51.
$9 w 2 − 24 w + 16 = 1 9 w 2 − 24 w + 16 = 1$
52.
$4 z 2 + 4 z + 1 = 49 4 z 2 + 4 z + 1 = 49$
53.
$a 2 − 18 = 0 a 2 − 18 = 0$
54.
$b 2 − 108 = 0 b 2 − 108 = 0$
55.
$( p − 1 3 ) 2 = 7 9 ( p − 1 3 ) 2 = 7 9$
56.
$( q − 3 5 ) 2 = 3 4 ( q − 3 5 ) 2 = 3 4$
57.
$m 2 + 12 = 0 m 2 + 12 = 0$
58.
$n 2 + 48 = 0 . n 2 + 48 = 0 .$
59.
$u 2 − 14 u + 49 = 72 u 2 − 14 u + 49 = 72$
60.
$v 2 + 18 v + 81 = 50 v 2 + 18 v + 81 = 50$
61.
$( m − 4 ) 2 + 3 = 15 ( m − 4 ) 2 + 3 = 15$
62.
$( n − 7 ) 2 − 8 = 64 ( n − 7 ) 2 − 8 = 64$
63.
$( x + 5 ) 2 = 4 ( x + 5 ) 2 = 4$
64.
$( y − 4 ) 2 = 64 ( y − 4 ) 2 = 64$
65.
$6 c 2 + 4 = 29 6 c 2 + 4 = 29$
66.
$2 d 2 − 4 = 77 2 d 2 − 4 = 77$
67.
$( x − 6 ) 2 + 7 = 3 ( x − 6 ) 2 + 7 = 3$
68.
$( y − 4 ) 2 + 10 = 9 ( y − 4 ) 2 + 10 = 9$
### Writing Exercises
69.
In your own words, explain the Square Root Property.
70.
In your own words, explain how to use the Square Root Property to solve the quadratic equation $(x+2)2=16(x+2)2=16$.
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”
If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
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```Calculus Investigation: The Slingshot Around the Moon.
Introduction
Each of the Apollo missions to the moon had a safety
measure built into the spacecraft’s flight path: if for some reason
the rockets failed to operate properly, the path of the craft was
planned such that it would “slingshot” around the moon in a
free-return trajectory, returning to earth “powered” only by the
gravitational forces of the moon and the earth.
In the Apollo 8 mission of 1968, the first to go to the moon,
this flight path was adjusted slightly by three very short fuel
burns measuring only a few minutes. The first burn put Apollo 8 into an elliptical orbit around
the moon and the second put it into a circular orbit. Apollo 8 orbited the moon 10 times and then,
after a third short burn, its trajectory was altered once more so that it could return to earth. These
fuel burns, as well as the initial flight trajectory, were planned with extraordinary precision long
before Apollo 8 left the earth.
Apollo 13 was launched in 1970, its crew expecting to be the fifth crew to orbit the moon and
the third to land on it. The explosion of an oxygen tank on the craft 56 hours into the mission at
first disappointed the crew since they knew immediately that their lunar landing would have to be
aborted. Within minutes their disappointment changed to fear as they realized that their lives
were in serious danger. They hadn’t anywhere near enough fuel to turn around and head home.
They had no choice but to continue their flight to the moon and return to earth using the freereturn trajectory. Because of the escaping oxygen and debris, their flight path needed to be
modified slightly throughout their trip, but essentially, their slingshot around the moon went as
planned. The survival of the Apollo 13 crew would not have been possible were it not for this
safety measure.
Your goal in this investigation is to create a free-return trajectory for an Apollo spacecraft—
one that will send it to the moon and return the craft safely to the earth without firing thrusters
after the translunar injection—the initial firing of the thrusters.
page 1
Gravity
The force of gravity pulling two objects towards one another is given by the following
equation
m1 m2
,
r2
where m1 and m2 are the masses of the two objects, r is the distance separating the objects, and
G is a constant known as the universal gravitational constant. Some particular constant values
F G
that may be useful in this calculus investigation are as follows:
Constant
Mass of the earth
Mass of the moon
Mass of Apollo 8
Distance from earth to moon
Gravitational constant
Radius of Apollo 8’s earth orbit
Radius of Apollo 8’s lunar orbit
Symbol
Value
mE
mM
mA
D
G
rE
roE
rM
roM
5.97 1024 kg
7.35 1022 kg
28,833 kg
384, 400 km
8.65 1013 km3 kg -1 hr -2
6378 km
6563 km
1738 km
1849 km
Another fact from physics is that F m a , i.e., force equals mass times acceleration. What
will be of more use to us in this investigation is the fact that acceleration equals force divided by
mass. For example, the acceleration that Apollo 8 experiences due to the pull of the earth, is
given by:
aE G
mE
,
2
dE
where aE equals the acceleration that Apollo 8 experiences due to the earth’s pull, G and mE
are the gravitational constant and the mass of the earth as before, and d E is the distance that
Apollo 8 is from the earth. Similarly, the acceleration that Apollo 8 experiences due to the
moon’s pull is given by:
aM G
mM
,
2
dM
where aM equals the acceleration that Apollo 8 experiences due to the moon’s pull, mM is the
mass of the moon as before, and d M is the distance that Apollo 8 is from the moon.
(Through this investigation, it is recommended that you use kilometers (km) for
measuring distances, kilograms (kg) for measuring mass, and hours (hr) for measuring time. To
get an idea of the relative sizes of the earth and the moon and the distance between them, hold a
basketball and a tennis ball 50 feet apart.)
page 2
Part I: Escaping the Earth’s gravity. A one-dimensional problem.
We will first consider the simple case of a rocket being launched directly towards the moon.
We want to launch it with enough force so that it will escape the earth’s gravitational pull. What
initial velocity is required?
To simplify the problem, we will choose a reference frame in which neither the earth nor the
moon is moving. On a straight line, the earth will be located at position xE 0 , and the moon
will be located at position xM 384, 400 .
Find the point along that straight line at which the gravitational forces of the moon and the
earth be exactly equal and opposite. Call that point the zero-gravity point.
If your spacecraft is motionless and on the earth’s side of the zero-gravity point, then the
the moon’s side of the zero-gravity point, then the moon will gradually haul it in. Your goal is to
determine at what initial velocity your spacecraft must be launched from earth orbit towards the
moon in order to get past the zero-gravity point. The differential equation describing the
changing velocity of the spacecraft comes from the total acceleration due to the earth’s pull and
the moon’s:
a G
mM
m
G E2 .
2
dM
dE
If we let x dE , which is equivalent to letting x be the position of the spacecraft along the
earth-moon line, then we have
x dM D , or
dM D x .
Finally, that produces the second-order differential equation
d 2x
mM
m
G
G 2E .
2
2
dt
x
D x
Notice that the only variables in this equation are
d 2x
and x —everything else is constant.
dt 2
This differential equation cannot easily be solved analytically, but numerical solutions can be
found using Euler’s method. Starting at x0 6563 km (the radius of Apollo 8’s earth orbit),
examine what happens to your Apollo spacecraft when the initial velocity is 20,000 km/hr. Use
steps of size t 0.001 hours and look at the first 0.6 hours. Make graphs of acceleration
versus time, velocity versus time, and position versus time for the first 0.6 hours after translunar
injection. Is there enough information in these graphs to determine whether 20,000 km/hr is a
sufficient initial velocity for the spacecraft to escape the earth’s gravitational pull?
page 3
Now examine what happens during the first full hour of the flight, still using t 0.001 hours.
You may notice something strange going on. In fact, 20,000 km/hr is not sufficient to escape
earth’s gravitational pull. Somewhere around t 0.25 hours, the craft reverses direction and
begins to head back towards the earth. It crashes into the earth around t 0.5 hours. (You
should verify this!) However, the equations don’t “know” this, so it may be that you have Euler’s
method cranking away long after the spacecraft has crashed into the earth’s surface.
After implementing that into your Euler equations, you may be able, using trial and error, to
estimate the minimal initial velocity required for your spacecraft to escape earth’s pull.
With these tools all in place, you should be able to get your spaceship to the moon. Before
continuing to the next part, be sure you can answer the questions and produce the graphs below:
Make graphs for your spacecraft of acceleration versus time, velocity versus time, and
distance from the earth versus time.
Estimate the minimal initial velocity required to get a spacecraft to escape the earth’s
gravitational sphere of influence (and pass the zero-gravity point).
If a spacecraft is given that initial velocity while in earth’s orbit and heads straight
towards the moon, how long will it take it to get there?
By how much can the duration of the trip be decreased by launching at a faster speed than
that which is minimally required?
How fast is the spacecraft going when it crashes onto the surface of the moon?
Part II: The slingshot around the Moon. A two-dimensional problem.
Now that you have successfully gotten a spacecraft to crash-land on the moon, you need to
design a flight path that will get it back home again. The earth and the moon will be located at
the same points as before along a horizontal line, but now a second “vertical” dimension will be
added. Thus, the earth is centered at the origin (0,0) and the moon is centered at the point whose
coordinates are (384400,0). Your Apollo spacecraft will be located at the coordinates (x,y),
where x and y are variables.
In writing the Euler equations that will generate the flight path of your spacecraft, you will
need to separate the acceleration, velocity, and position vectors of your craft into their x- and ycomponents. Then the Euler equations can be implemented with x- and y-components receiving
separate treatment. For example, your initial conditions may look something like this:
t0 0
x0 6563
vx0 30000
y0 0
vy0 10000,
in which vx0 stands for the x-component of the initial velocity and vy0 stands for the ycomponent of the initial velocity. The (arbitrarily chosen) values given in the example above
correspond to a translunar injection that occurs when the spacecraft is located in earth’s orbit at
the point closest to the moon and that gives the spacecraft a velocity that has “horizontal”
page 4
component equal to 30,000 km/hr and “vertical” component equal to –10,000 km/hr, as shown in
the diagram below.
The only difficult step that remains before putting everything together is determining the xand y- components of the spacecraft’s acceleration when the craft is located at the general
coordinates (x,y). To do this, we will use the fact that the acceleration vector of the spacecraft
due to the influence of the earth or moon points in exactly the opposite direction as the
displacement vector from the planet to the spacecraft. Consider the two diagrams below. The
one on the left shows the displacement vector from the earth to Apollo, broken down into its xand y- components. The diagram on the right shows Apollo’s acceleration vector, also broken
down into its x- and y-components. Since the vectors point in exactly opposite directions, their
components are proportional.
Displacement
Acceleration
page 5
Complete the diagrams above by writing expressions for the total distance from the earth to
Apollo (in the diagram on the left) and for the total acceleration of Apollo due to the earth’s pull
(in the diagram on the right). These expressions should contain only known constants and the
variables x and y. Then, using proportional triangles, write expressions for the x- and ycomponents of acceleration (called ax and ay in the drawing) using only known constants and
the variables x and y. Pay careful attention to the sign (+/-) of the acceleration components. If x
is positive, then ax should be negative, etc.
Once you have done this, determine the x- and y-components of acceleration due to the pull
of the moon.
The sum of the acceleration due to the earth and the acceleration due to the moon is equal to
the total acceleration, and as before, x- and y-components may be added separately.
You now should have all the pieces you need to develop your flight path that will slingshot
your spacecraft around the moon. Your goal is to begin somewhere in earth orbit and determine a
direction and an initial velocity for your translunar injection such that, with no further firing of
thrusters, the spacecraft will go around the moon and then come back. Some questions you may
wish to consider and graphs you may wish to produce as you work are:
How long does the round-trip voyage take?
Make graphs of total acceleration versus time, total velocity versus time, and total
distance from the earth versus time for your flight path.
Make a graph of the flight path itself in two dimensions, showing the locations of the
earth and the moon. Make points on the path to show where the spacecraft is after one
day, two days, etc.
How sensitive is the flight path to changes in launch velocity or launch direction?
Are there multiple solutions to this problem?
This is a challenging investigation. As you work, keep in mind that although the problem as
it is presented here is simplified, you are studying the essentially the same problem and are doing
the same calculations that NASA scientists and engineers did in the 1960’s as they attempted to
fulfill President Kennedy’s mandate to put a man on the moon before 1970. If this appeals to
you, then you may want to explore even more difficult problems, some of which require some
understanding of physics.
page 6
How much force is required to accelerate the spacecraft from an earth-orbital velocity to
the velocity you chose for your flight path? How much rocket fuel would be required to
accelerate Apollo 8 to that new velocity?
Plan a fuel burn near the moon that will take the spacecraft out of its free-return trajectory
and put it into lunar orbit. Then plan another fuel burn that will put it back into a freereturn trajectory.
56 hours into Apollo 13’s mission, an oxygen tank exploded in the service module and
spewed oxygen and debris into space. Suppose that happened on the mission you
planned. How much would that alter the flight path? Would you still be able to return to
the earth.
``` | 3,166 | 13,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-40 | latest | en | 0.976336 |
https://en.academic.ru/dic.nsf/enwiki/2319073 | 1,582,681,741,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146176.73/warc/CC-MAIN-20200225233214-20200226023214-00005.warc.gz | 351,434,711 | 15,018 | # First order partial differential equation
First order partial differential equation
In mathematics, a first order partial differential equation is a partial differential equation that involves only first derivatives of the unknown function of "n" variables. The equation takes the form
:$F\left(x_1,ldots,x_n,u,u_\left\{x_1\right\},ldots u_\left\{x_n\right\}\right) =0. ,$
Such equations arise in the construction of characteristic surfaces for hyperbolic partial differential equations, in the calculus of variations, in some geometrical problems, and they arise in simple models for gas dynamics whose solution involves the method of characteristics. If a family of solutionsof a single first-order partial differential equation can be found, then additional solutions may be obtained by forming envelopes of solutions in that family. In a related procedure, general solutions may be obtained by integrating families of ordinary differential equations.
Characteristic surfaces for the wave equation
Characteristic surfaces for the wave equation are level surfaces for solutions of the equation:$u_t^2 = c^2 left\left(u_x^2 +u_y^2 + u_z^2 ight\right). ,$
There is little loss of generality if we set $u_t =1$: in that case "u" satisfies:$u_x^2 + u_y^2 + u_z^2= frac\left\{1\right\}\left\{c^2\right\}. ,$
In vector notation, let:$vec x = \left(x,y,z\right) quad hbox\left\{and\right\} quad vec p = \left(u_x, u_y, u_z\right).,$
A family of solutions with planes as level surfaces is given by :$u\left(vec x\right) = vec p cdot \left(vec x - vec\left\{x_0\right\}\right), ,$
where:$| vec p ,| = frac\left\{1\right\}\left\{c\right\}, quad hbox\left\{and\right\} quad vec\left\{x_0\right\} quad hbox\left\{is arbitrary\right\}.,$
If "x" and "x"0 are held fixed, the envelope of these solutions is obtained by finding a point on the sphere of radius 1/"c" where the value of "u" is stationary. This is true if $vec p$ is parallel to $vec x - vec\left\{x_0\right\}$. Hence the envelope has equation:$u\left(vec x\right) = pm frac\left\{1\right\}\left\{c\right\} | vec x -vec\left\{x_0\right\} ,|.$
These solutions correspond to spheres whose radius grows or shrinks with velocity "c". These are light cones in space-time.
The initial value problem for this equation consists in specifying a level surface "S" where "u"=0 for "t"=0. The solution is obtained by taking the envelope of all the spheres with centers on "S", whose radii grow with velocity "c". This envelope is obtained by requiring that:$frac\left\{1\right\}\left\{c\right\} | vec x - vec\left\{x_0\right\}, | quad hbox\left\{is stationary for\right\} quad vec\left\{x_0\right\} in S. ,$
This condition will be satisfied if $| vec x - vec\left\{x_0\right\}, |$ is normal to "S". Thus the envelope corresponds to motion with velocity "c" along each normal to "S". This is the Huygens' construction of wave fronts: each point on "S" emits a spherical wave at time "t"=0, and the wave front at a later time "t" is the envelope of these spherical waves. The normals to "S" are the light rays.
Two-dimensional theory
The notation is relatively simple in two space dimensions, but the main ideas generalize to higher dimensions. A general first-order partial differential equation has the form:$F\left(x,y,u,p,q\right)=0, ,$
where:$p=u_x, quad q=u_y. ,$
A complete integral of this equation is a solution φ("x","y","u") that depends upon two parameters "a" and "b". (There are "n" parameters required in the "n"-dimensional case.) An envelope of such solutions is obtained by choosing an arbitrary function "w", setting "b"="w"("a"), and determining "A"("x","y","u") by requiring that the total derivative:$frac\left\{d varphi\right\}\left\{d a\right\} = varphi_a\left(x,y,u,A,w\left(A\right)\right) + w\text{'}\left(A\right)varphi_b\left(x,y,u,A,w\left(A\right)\right) =0. ,$
In that case, a solution $u_w$ is also given by:$u_w = phi\left(x,y,u,A,w\left(A\right)\right) ,$
Each choice of the function "w" leads to a solution of the PDE. A similar process led to the construction of the light cone as a characteristic surface for the wave equation.
If a complete integral is not available, solutions may still be obtained by solving a system of ordinary equations. In order to obtain this system, first note that the PDE determines a cone (analogous to the light cone) at each point: if the PDE is linear in the derivatives of "u" (it is quasi-linear), then the cone degenerates into a line. In the general case, the pairs ("p","q") that satisfy the equation determine a family of planes at a given point::$u - u_0 = p\left(x-x_0\right) + q\left(y-y_0\right), ,$
where:$F\left(x_0,y_0,u_0,p,q\right) =0.,$
The envelope of these planes is a cone, or a line if the PDE is quasi-linear. The condition for an envelope is:$F_p, dp + F_q ,dq =0, ,$
where F is evaluated at $\left(x_0, y_0,u_0,p,q\right)$, and "dp" and "dq" are increments of "p" and "q" that satisfy "F"=0. Hence the generator of the cone is a line with direction:$dx:dy:du = F_p:F_q:\left(pF_p + qF_q\right). ,$
This direction corresponds to the light rays for the wave equation.In order to integrate differential equations along these directions, we require increments for "p" and "q" along the ray. This can be obtained by differentiating the PDE::$F_x +F_u p + F_p p_x + F_q p_y =0, ,$ :$F_y +F_u q + F_p q_x + F_q q_y =0,,$
Therefore the ray direction in $\left(x,y,u,p,q\right)$ space is
:$dx:dy:du:dp:dq = F_p:F_q:\left(pF_p + qF_q\right):\left(-F_x-F_u p\right):\left(-F_y - F_u q\right). ,$
The integration of these equations leads to a ray conoid at each point $\left(x_0,y_0,u_0\right)$. General solutions of the PDE can then be obtained from envelopes of such conoids.
* [http://www.scottsarra.org/shock/shock.html More detailed information on the Method of Characteristics]
Bibliography
*R. Courant and D. Hilbert, "Methods of Mathematical Physics, Vol II", Wiley (Interscience), New York, 1962.
* L.C. Evans, "Partial Differential Equations", American Mathematical Society, Providence, 1998. ISBN 0-8218-0772-2
* A. D. Polyanin, V. F. Zaitsev, and A. Moussiaux, "Handbook of First Order Partial Differential Equations", Taylor & Francis, London, 2002. ISBN 0-415-27267-X
* A. D. Polyanin, "Handbook of Linear Partial Differential Equations for Engineers and Scientists", Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9
* Sarra, Scott "The Method of Characteristics with applications to Conservation Laws", Journal of Online Mathematics and its Applications, 2003.
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• ordinary differential equation — Math. an equation containing derivatives but not partial derivatives. Cf. partial differential equation. * * * Equation containing derivatives of a function of a single variable. Its order is the order of the highest derivative it contains (e.g … Universalium | 2,466 | 9,540 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-10 | longest | en | 0.845197 |
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# Stoichiometry
John-Nicholas Furst
Meaning of the Word
• • • • Comes from 2 Greek words stoicheion – element metron – measure Stoichiometry is calculations relating to the masses of reactants and products used in chemical reactions. • Lots of MATH!
What to Expect
• Most stoichimetric problems will give you an amount of reactant and ask you to solve for the amount of product formed.
Steps to Solving Stoichimetric Problems
1. 2. 3. 4. Balance equation Convert into moles Set up molar proportion Convert back into mass
Molar Ratios
• The ratio is determined by the coefficients Equation: 2 H2 + O2 ---> 2 H2O Question: What is the molar ratio between H2 and O2? Answer: 2:1 Explanation: This is because the coefficient of H2 is 2 and the coefficient of O2 is 1(we just don’t write it).
Molar Ratio Example #1
Equation: 2 H2 + O2 ---> 2 H2O Question: What is the molar ratio between O2 and H2O? Answer: 1:2 Explanation: This is again because the O2 has an unwritten coefficient of 1 and the H2 has a coefficient of 2. Question: What is the molar ratio between H2 and H2O?
Molar Ratio Example #2
Equation: 2 O3 ---> 3 O2 Question: What is the molar ratio between O3 and O2? Answer: 2:3 Question: What is the molar ratio between O2 and O3? Answer: 3:2 Explanation: It all depends on how the question is worded, the first element is the numerator in the ratio and the second is the denominator
Mole-Mole Stoichimetric Problems
• The procedure to solve mole-mole problems involves making two ratios and setting them equal to each other. • One ratio will come from the coefficients of the balanced equation, and the other will be set up from data in the problem and will have one unknown in which you are solving for. • You will then cross-multiply and divide to get the answer.
Mole-Mole Example #1
Equation: N2 + 3 H2 ---> 2 NH3 Question: If we have 2.00 mol of N2 reacting with sufficient H2, how many moles of NH3 will be produced? Solution: The ratio from the problem will have N2 and NH3 in it. When making the two ratios, make sure that the numbers are in the same relative positions. For example, if the value associated with NH3 is in the numerator, then MAKE SURE it is in both numerators. Use the coefficients of the two substances to make the ratio from the equation. Why isn't H2 involved in the problem? Answer: The word "sufficient" removes it from consideration. Let's use this ratio to set up the proportion: That means the ratio from the equation is: The ratio from the data in the problem will be: The proportion (setting the two ratios equal) is: Solving by cross-multiplying gives x = 4.00 mol of NH3 produced.
Mole-Mole Example #2
Equation: N2 + 3 H2 ---> 2 NH3 Question: Suppose 6.00 mol of H2 reacted with sufficient nitrogen. How many moles of ammonia would be produced? Let's use this ratio to set up the proportion: That means the ratio from the equation is: The ratio from the data in the problem will be: The proportion (setting the two ratios equal) is: Solving by cross-multiplying, 3x = 12, and dividing gives x = 4.00 mol of NH3 produced.
Converting Grams-Moles
Steps: 2. Determine how many grams are given in the problem. 3. Calculate the molar mass of the substance. 4. Divide step one by step two.
Grams-Moles Example #1
Question: Convert 25.0 grams of KMnO4 to moles. Solution: (Remember the 3 steps) 3. The problem gives us 25.0 grams. • The molar mass of KMnO4 is 158.034 grams/mole • You divide the grams given by the substance's molar mass: Answer: 0.158 (Remember 3 sig. figs. from the 25.0)
Grams-Moles Example #2
Question: Calculate how many moles are in 17.0 grams of H2O2 Solution: 3. 17.0 grams are given in the text of the problem. 4. The molar mass is 34.0146 grams/mole. 5. You divide the grams given by the substance's molar mass: Answer: 0.500 moles
Converting Moles-Grams
Steps: (Exactly the same except you multiply this time) 2. Determine how many moles are given in the problem. 3. Calculate the molar mass of the substance. 4. Multiply step one by step two.
Moles-Grams Example #1
Question: Calculate how many grams are in 0.700 moles of H2O2 Solution: (Again with the 3 steps) • 0.700 moles are given in the problem. • The molar mass of H2O2 is 34.0146 grams/mole. • Step Three: You multiply the moles given by the substance's molar mass: 0.700 mole x 34.0146 grams/mole = 23.8 grams Answer: 23.8 grams
Moles-Grams Example #2
Question: Convert 2.50 moles of KClO3 to grams. Solution: • 2.50 moles is given in the problem. • The molar mass for KClO3 is 122.550 grams/mole. • Step Three: You multiply the moles given by the substance's molar mass: 2.50 moles x 122.550 grams/mole = 306.375 grams Answer: 306.375 grams
Mole-Mass Problems
Equation: 2 KClO3 ---> 2 KCl + 3 O2 Question: 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced? Solution: Let's use this ratio to set up the proportion: That means the ratio from the equation is: The ratio from the data in the problem will be: The proportion (setting the two ratios equal) is: Cross-multiplying, 2x=4.5, and dividing gives x = 2.25 mol of O2 produced.
Mass-Mole Example #1
Equation: 2 KClO3 ---> 2 KCl + 3 O2 Problem: If 80.0 grams of O2 was produced, how many moles of KClO3 decomposed? Solution: Let's use this ratio to set up the proportion: That means the ratio from the equation is: The ratio from the data in the problem will be: The 2.50 mole came from 80.0 g ÷ 32.0 g/mol. The 32.0 g/mol is the molar mass of O2. Be careful to keep in mind that oxygen is O2, not just O. The proportion (setting the two ratios equal) is: Solving by cross-multiplying and dividing gives x = 1.67 mol of KClO3 decomposed.
Mass-Mass Problems
• Most common type of Stoichimetric problem Steps: 3. Make sure you are working with a balanced equation. 4. Convert grams to moles. 5. Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem. 6. Convert moles into grams
Mass-Mass Example #1
Question: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl3 Equation: 2 AuCl3 ---> 2 Au + 3 Cl2 Solution: 4. Molar ratio: 5. Convert grams-moles: 6. Set ratios equal: 7. Solve for x and convert moles-grams:
Mass-Mass Example #2
Calculate mass of PbI2 produced by reacting of 30.0 g KI with excess Pb(NO3)2 Equation: 2 KI + Pb(NO3)2 --> PbI2 + 2 KNO3 3. 4. 5. 6. Molar ratio: Convert grams-moles: Set ratios equal: Solve for x and convert moles-grams:
Limit Reagent
• The Limiting Reagent is simply the substance in a chemical reaction that runs out first. [Ex. #1] Reactant A is a test tube. I have 20 of them. Reactant B is a stopper. I have 30 of them. Product C is a stoppered test tube. The reaction is: A + B ---> C or: test tube plus stopper gives stoppered test tube. • So now we let them "react." The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and make stoppered test tubes (the amount goes up). • Suddenly, we run out of one of the "reactants." • We ran out of test tubes first. We had only had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting there unused. And we also have 20 test tubes with stoppers firmly inserted. • So, which the test tubes are limiting and the stopper are in excess?
Limiting Reagent Example #2
Equation: 2 Al + 3 I2 ------> 2 AlI3 Question: Determine the limiting reagent of the product if we use: 1.20 mol Al and 2.40 mol iodine. • Solution: Here is how to find out the limiting reagent: take the moles of each substance and divide it by the coefficient of the balanced equation. • For aluminum: 1.20 / 2 = 0.60 For iodine: 2.40 / 3 = 0.80 • The lowest number indicates the limiting reagent. Aluminum will run out first in part
Percent Yield
• Percent Yield is found by taking the actual yield and dividing it by the theoretical yield and then multiplying by 100. • The actual yield is the amount formed when the experiment is actually carried out. • The theoretical yield is determined by how much can be produced with the limiting reagent.
Actual (100) Theoretical
Sources
• • • • • • • • • • • • •
Mr. Beran http://members.tripod.com/~EppE/stoictry.htm http://www.shodor.org/UNChem/basic/stoic/index.html http://en.wikipedia.org/wiki/Stoichiometry http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch3/massmolfra http://www.chemtutor.com/mols.htm http://www.science.uwaterloo.ca/~cchieh/cact/c120/stoichio.html http://www.chem4kids.com/files/react_stoichio.html http://members.aol.com/profchm/eq_form.html http://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Stoichiometry.html http://members.aol.com/profchm/eq_form.html http://chemistry.about.com/od/stoichiometry/Stoichiometry.htm http://www.chemcollective.org/tutorials.php
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# 14.2: Solving Exponential and Logarithmic Equations
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We can solve exponential and logarithmic equations by applying logarithms and exponentials. Since the exponential and logarithmic functions are invertible (they are inverses of each other), applying these operations can also be reversed. In short we have observed that
## Observation: Exponential and Logarithmic
$x=y \quad \Longrightarrow \quad b^{x}=b^{y}, \quad \text { and } \quad b^{x}=b^{y} \quad \Longrightarrow \quad x=y \nonumber$
$x=y \quad \Longrightarrow \quad \log _{b}(x)=\log _{b}(y), \quad \text { and } \quad \log _{b}(x)=\log _{b}(y) \quad \Longrightarrow \quad x=y \nonumber$
We can use the above implications whenever we have a common basis, or when we can convert the terms to a common basis.
## Example $$\PageIndex{1}$$
Solve for $$x$$.
1. $$2^{x+7}=32$$
2. $$10^{2x-8}=0.01$$
3. $$7^{2x-3}=7^{5x+4}$$
4. $$5^{3x+1}=25^{4x-7}$$
5. $$\ln(3x-5)=\ln(x-1)$$ & f)
6. $$\log_2(x+5)=\log_2(x+3)+4$$
7. $$\log_6(x)+\log_6(x+4)=\log_6(5)$$
8. $$\log_3(x-2)+\log_3(x+6)=2$$
Solution
In these examples, we can always write both sides of the equation as an exponential expression with the same base.
1. $$2^{x+7}=32 \quad \Longrightarrow \quad 2^{x+7}=2^{5} \Longrightarrow \quad x+7=5 \Longrightarrow \quad x=-2$$
2. $$10^{2 x-8}=0.01 \Longrightarrow 10^{2 x-8}=10^{-2} \Longrightarrow \quad 2 x-8=-2 \Longrightarrow 2 x=6 \Longrightarrow \quad x=3$$
Here it is useful to recall the powers of $$10$$, which were also used to solve the equation above.
$\boxed{ \begin{matrix} 10^3 &=&1,000 \\ 10^2&=&100 \\ 10^1&=&10 \\ 10^0&=&1 \\ 10^{-1}&=&0.1 \\ 10^{-2}&=&0.01 \\ 10^{-3}&=&0.001 \end{matrix} \quad\quad \text{In general } (n\geq1):\quad\left\{ \begin{matrix} 10^n=1\underbrace{00\cdots00}_{n\text{ zeros}} \\ \quad \\ 10^{-n}=\underbrace{0.0\cdots 00}_{n\text{ zeros}}1 \end{matrix}\right.} \nonumber$
1. $$7^{2 x-3}=7^{5 x+4} \Longrightarrow 2 x-3=5 x+4 \stackrel{(-5 x+3)}{\Longrightarrow} \quad-3 x=7 \Longrightarrow \quad x=-\dfrac{7}{3}$$
2. $$5^{3 x+1}=25^{4 x-7} \Longrightarrow 5^{3 x+1}=5^{2 \cdot(4 x-7)} \Longrightarrow 3 x+1=2 \cdot(4 x-7) \Longrightarrow 3 x+1=8 x-14 \stackrel{(-8 x-1)}{\Longrightarrow} \quad-5 x=-15 \Longrightarrow x=3$$
By a similar reasoning we can solve equations involving logarithms whenever the bases coincide.
1. $$\ln (3 x-5)=\ln (x-1) \Longrightarrow 3 x-5=x-1 \stackrel{(-x+5)}{\Longrightarrow} \quad 2 x=4 \Longrightarrow x=2$$
2. For part (f), we have to solve $$\log_2(x+5)=\log_2(x+3)+4$$. To combine the right-hand side, recall that $$4$$ can be written as a logarithm, $$4=\log_2(2^4)=\log_2 16$$. With this remark we can now solve the equation for $$x$$.
\begin{aligned} \log_2(x+5) &= \log_2(x+3)+4\\ \implies \quad \log_2(x+5)&=\log_2(x+3)+\log_2(16) \\ \implies \quad \log_2(x+5)&=\log_2(16\cdot (x+3)) \\ \implies \qquad \quad \;\; x+5&=16(x+3) \\ \implies \qquad \quad \;\; x+5&=16x+48 \\ \quad \stackrel{(-16x-5)}\implies \qquad \;\;-15x&=43\\ \implies \qquad \qquad \;\;x&=-\dfrac{43}{15} \end{aligned} \nonumber
1. Next, in part (g), we start by combining the logarithms. \begin{aligned} \log_6(x)+\log_6(x+4)&=\log_6(5) \\ \implies \quad \log_6(x(x+4))&=\log_6(5) \\ \stackrel{\text{remove }\log_6}\implies \qquad x(x+4)&=5 \\ \implies \qquad x^2+4x-5&=0 \\ \implies \quad(x+5)(x-1)&=0 \\ \qquad\implies x=-5 \text{ or } x=1\end{aligned} \nonumber
Since the equation became a quadratic equation, we ended up with two possible solutions $$x=-5$$ and $$x=1$$. However, since $$x=-5$$ would give a negative value inside a logarithm in our original equation $$\log_6(x)+\log_6(x+4)=\log_6(5)$$, we need to exclude this solution. The only solution is $$x=1$$.
1. Similarly we can solve the next part, using that $$2=\log_3(3^2)$$:
\begin{aligned} \log _{3}(x-2)+\log _{3}(x+6)&=2 \\ \Longrightarrow \log _{3}((x-2)(x+6))&=\log _{3}\left(3^{2}\right) \\ \Longrightarrow(x-2)(x+6)&=3^{2} \\ \Longrightarrow x^{2}+4 x-12&=9 \\ \Longrightarrow x^{2}+4 x-21&=0 \\ \Longrightarrow(x+7)(x-3)&=0 \\ \Longrightarrow x=-7 \text { or } x=3 \end{aligned} \nonumber
We exclude $$x=-7$$, since we would obtain a negative value inside a logarithm, so that the solution is $$x=3$$.
When the two sides of an equation are not exponentials with a common base, we can solve the equation by first applying a logarithm and then solving for $$x$$. Indeed, recall from Observation in section 7.2 on inverse functions, that since $$f(x)=\log_b(x)$$ and $$g(x)=b^x$$ are inverse functions, we have $$\log_b(b^x)=x$$ and $$b^{\log_b(x)}=x$$ whenever the compositions of the left sides make sense. That is, the action of the logarithm cancels out the action of the exponential function, and vice versa. So we can think of applying a logarithm (an exponentiation) on both sides of an equation to cancel an exponentiation (a logarithm) much like squaring both sides of an equation to cancel a square root.
## Example $$\PageIndex{2}$$
Solve for $$x$$.
1. $$3^{x+5}=8$$
2. $$13^{2x-4}=6$$
3. $$5^{x-7}=2^{x}$$
4. $$5.1^{x}=2.7^{2x+6}$$
5. $$17^{x-2}=3^{x+4}$$
6. $$7^{2x+3}=11^{3x-6}$$
Solution
We solve these equations by applying a logarithm (both $$\log$$ or $$\ln$$ will work for solving the equation), and then we use the identity $$\log(a^x)=x\cdot \log(a)$$.
1. $$3^{x+5}=8 \implies \ln 3^{x+5}=\ln 8 \implies (x+5)\cdot\ln 3=\ln 8 \implies x+5=\dfrac{\ln 8}{\ln 3}\implies x=\dfrac{\ln 8}{\ln 3}-5\approx -3.11$$
2. $$13^{2x-4}=6 \implies \ln 13^{2x-4}=\ln 6 \implies (2x-4)\cdot\ln 13=\ln 6 \implies 2x-4=\dfrac{\ln 6}{\ln 13} \implies 2x=\dfrac{\ln 6}{\ln 13}+4 \implies x=\dfrac{\frac{\ln 6}{\ln 13}+4}{2}=\dfrac{\ln 6}{2\cdot \ln 13}+2\approx 2.35$$
3. $$5^{x-7}=2^{x} \implies \ln 5^{x-7}=\ln 2^{x} \implies (x-7)\cdot\ln 5=x\cdot \ln 2$$
At this point, the calculation will proceed differently than the calculations in parts (a) and (b). Since $$x$$ appears on both sides of $$(x-7)\cdot\ln 5=x\cdot \ln 2$$, we need to separate terms involving $$x$$ from terms without $$x$$. That is, we need to distribute $$\ln 5$$ on the left and separate the terms. We have
\begin{aligned} (x-7)\cdot\ln 5=x\cdot \ln 2 & \implies & x\cdot\ln 5-7\cdot \ln 5=x\cdot \ln 2 \\ (\text{add }+7\cdot \ln 5-x\cdot \ln 2) &\implies & x\cdot\ln 5-x\cdot \ln 2=7\cdot \ln 5\\ &\implies & x\cdot(\ln 5-\ln 2)=7\cdot \ln 5 \\ &\implies & x=\dfrac{7\cdot \ln 5 }{\ln 5-\ln 2}\approx 12.30\end{aligned} \nonumber
We need to apply the same solution strategy for the remaining parts (d)-(f) as we did in (c).
1. \begin{aligned} 5.1^{x}=2.7^{2x+6} & \implies & \ln 5.1^{x}=\ln 2.7^{2x+6} \\ & \implies & x\cdot \ln 5.1=(2x+6)\cdot \ln 2.7 \\ & \implies & x\cdot \ln 5.1=2x\cdot \ln 2.7+6\cdot \ln 2.7 \\ & \implies & x\cdot \ln 5.1-2x\cdot \ln 2.7=6\cdot \ln 2.7 \\ & \implies & x\cdot (\ln 5.1-2\cdot \ln 2.7) = 6 \cdot \ln 2.7 \\ & \implies & x = \frac{6 \cdot \ln 2.7} {\ln 5.1-2\cdot \ln 2.7}\approx -16.68\end{aligned} \nonumber
1. \begin{aligned} 17^{x-2}=3^{x+4} & \implies & \ln 17^{x-2}=\ln 3^{x+4} \\ & \implies & (x-2)\cdot \ln 17=(x+4)\cdot \ln 3 \\ & \implies & x\cdot \ln 17-2\cdot \ln 17=x\cdot \ln 3+4\cdot \ln 3 \\ & \implies & x\cdot \ln 17-x\cdot \ln 3=2\cdot \ln 17+4\cdot \ln 3 \\ & \implies & x\cdot (\ln 17- \ln 3) =2\cdot \ln 17+4\cdot \ln 3 \\ & \implies & x = \dfrac{2\cdot \ln 17+4\cdot \ln 3} {\ln 17- \ln 3}\approx 5.80\end{aligned} \nonumber
1. \begin{aligned} 7^{2x+3}=11^{3x-6} & \implies & \ln 7^{2x+3}=\ln 11^{3x-6} \\ & \implies & (2x+3)\cdot \ln 7=(3x-6)\cdot \ln 11 \\ & \implies & 2x\cdot \ln 7+3\cdot \ln 7=3x\cdot \ln 11-6\cdot \ln 11 \\ & \implies & 2x\cdot \ln 7-3x\cdot \ln 11=-3\cdot \ln 7-6\cdot \ln 11 \\ & \implies & x\cdot (2\cdot \ln 7-3\cdot \ln 11)=-3\cdot \ln 7-6\cdot \ln 11 \\ & \implies & x = \dfrac{-3\cdot \ln 7-6\cdot \ln 11} {2\cdot \ln 7-3\cdot \ln 11}\approx 6.13\end{aligned} \nonumber
Note that in the problems above we could have also changed the base as we did earlier in the section. For example, in part f) above, we could have begun by writing the right hand side as $$11^{3x-6}=7^{\log_7 (11^{(3x-6)})}$$. We chose to simply apply a log to both sides instead, because the notation is somewhat simpler.
This page titled 14.2: Solving Exponential and Logarithmic Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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1 . X Sstems of Equations In this section we want to spend some time reviewing sstems of equations. Recall there are two basic techniques we use for solving a sstem of equations: Elimination and Substitution. We will start with the elimination method. The ke idea in the elimination method is to eliminate one of the two variables b making the coefficients opposites on the two equations and adding the equations together. We will illustrate this method in the following eample. Eample : Solve using the elimination method. a. 9 b. 9 a. So to use the elimination method to solve, we need to start b determining which variable we would like eliminate. Lets choose the variable since the signs are alread opposite. Net we will need to multipl the top equation b a to make the coefficients opposites. Then we simpl add equations and solve. We get 9 9 So now we need the value of the solution. So we can simpl substitute the value into one of the original equations and solve for. Let s use the first equation. We get Since the solution to a sstem is an ordered pair we write the solution is,. b. This time we need to start b putting the equations in standard form. Then we will proceed as above. 9 9 We will choose to eliminate the s 9 Clear fractions b multipling b
2 Now we solve for. We use the second equation this time. We get So the solution is,. Net we will review the substitution method. The ke idea in the substitution method is to solve one of the equations for one of the variables and then substitute that epression into the other equation. We illustrate this in the following eample. Eample : Solve using the substitution method. a. b. a. To use the substitution method we need to solve one equation for one variable. In this case, the bottom equation is alread solved for. Net, we substitute that epression into the other equation and solve for the remaining variable. We proceed as follows Now we simpl substitute this value back into the original equation for, and solve. We get So our solution is,. b. So we need to start b solving one of the equation for one of the variables. We generall choose the easiest variable to get to. In this case that is the on the bottom equation. So we solve for it. We get
3 Now substitute as before and solve. Again, substitute this value into the epression we obtained for. We get So our solution is,. Now, recall that graphicall, the solution to a sstem of equations is the points of intersection of the graphs of the equations. So since, in this section, we are dealing onl with lines, there are onl three was that the can intersect, that is, there are onl three different possibilities for tpes of solutions to these sstems. One intersection point No intersection point Ever point intersects = One solution = No solution = Infinite solutions In order to determine if we have one of the two last cases we merel need to tr to solve and interpret what we find. Eample : Solve the following using an method. a. b a. First we will clear the fractions to evaluate which method of solving is best. So the first equation will be
4 Now, lets us the elimination method to solve. We will eliminate the s. Since all the variables eliminated we must have a special case. Since = for all values of and, we have the infinite solutions case. We simpl sa that the sstem has infinite solutions. b. Again lets use the elimination method for solving. Again we will eliminate the. We multipl the top equation b. and the bottom equation b.. We get Once again we have eliminated all the variables. In this case however we are left with a statement that is not true. Since that is the case we have no solution. Of course there are man ver useful applications of sstems of equations. We will concentrate on just a few different tpes: the miture problem, the investment problem, the moving object problem. We have reviewed the basic moving object problem in section R.. The difference here is that we can now use two different variables to solve. Eample : A motorboat traveling with the current went km in h. Against the current, the boat could go onl km in the same amount of time. Find the rate of the boat in calm water and the rate of the current. First we need to set up our variables. Lets let r be the rate of the boat in calm water and c be the rate of the current. That means that the total rate of the boat with the current would be r + c and the total rate against the current would be r c. So, like we did in section R. we will set up a table to summarize our data and construct a sstem of equations. With current Against current Now using the formula we get the sstem r t d r + c r - c r c r c. We need onl solve the sstem. Lets use elimination. We will start b divide the each equation b on both sides to simplif the sstem. r c r c
5 r c r c r r So the rate of the boat is km/hour. To find the rate of the current, c, we simpl substitute it back into one of the equations and solve. We get So the current is km/hour. c c For the investment problem we simpl use the formula I P r t. Since our time will ver often be one ear, we can reall just use I P r. We use a table just like we will with the miture problems. There are two major tpes of miture problems: Total Value and Total Quantit. The wa we tell the difference is b the rate we are given. If the problem gives us a monetar rate (like \$ per pound) then we have a total value problem. If we have a percentage rate then we have a total quantit problem. We use the following formulas in conjunction with a table to solve: Amount Rate = Total Value (or simpl A r TV ) or Amount Rate = Quantit (or simpl A r Q ). We will illustrate each tpe. Eample : At the theatre, Jon bus boes of popcorn and soft drinks for \$.. One bo of popcorn and one soft drink would cost \$.. What is the cost of a bo of popcorn? Since we are dealing with mone here we use the formula A r TV. Again, we will make a table for our data. We will let the cost of one bo of popcorn be p and cost for one soft drink be d. We fill in the table with two different situations, the purchase that Jon makes and a purchase that would be onl one of each. We get the following. A r TV Jon s purchase popcorn soft drinks p d p d One of each popcorn soft drinks p d p d The bo in the top right is represents the total value of Jon s purchase and the bo in the bottom right is the total value of a purchase of one or each. Therefore we get the sstem p d. p d. Lets solve this b substitution. Solving the bottom for p gives. d d.. d d. d. p. d. Substituting we get So soft drinks are \$. each. Now we can solve for cost of popcorn. Substituting we get p... Thus popcorn costs \$. each.
6 Eample : A chemist wishes to make liters of a.% acid solution b miing a % solution with a.% solution. How man liters of each solution are necessar? This time we will need to use the formula A r Q since we are given percentage rates. Also, we will need to use three rows, one for the.% solution, one for the % solution and one for the final solution. Lets call the amount of % solution and the amount of.% solution. So we fill in the table as follows. % solution.% solution Final solution A r..... ()(.) The wa that this tpe of miture problem works is the final amount must alwas equal the sum of the amounts of the two being mied, and the final quantit must alwas equal the sum of the two quantities. So that being said we get the sstem.. We will solve this sstem b the elimination method..... Q Therefore,,. 9. So we need about,.9 liters of % solution and. liters of.% solution.. Eercises Solve the following using the elimination method
7 Solve the following using the substitution method Solve using an method Set up a sstem for the following problems and solve using an method.. A boat travels miles in hours upstream. In the same amount of time the boat can travel miles downstream. Find the rate of the current and the rate of the boat in still water.. A -foot rope is cut in two pieces so that one piece is twice as long as the other. How long is each piece?. How man grams of silver that is % pure must be mied together with silver that is % pure to obtain a miture of 9 grams of silver that is % pure?
8 . At a barbecue, there were dinners served. Children s plates were \$. each and adult s plates were \$. each. If the total amount of mone collected for dinners at the barbecue was \$, how man of each tpe of plate was served? 9. A store sells shirts, one kind at \$. and the other at \$9.. In all, \$9. was taken in. How man of each tpe were sold?. A store buer purchased regular calculators and graphing calculators for a total cost of \$.. A second purchase, at the same prices, included regular calculators and graphing calculators for a total of \$.. Find the cost for each tpe of calculator.. Fling with the wind, a plane flew miles in hours. Fling against the wind, the plane could fl onl miles in the same amount of time. Find the rate of the plane in calm air and the rate of the wind.. A plane traveling with the wind flew miles in. hours. Against the wind, the plane required. hours to go the same distance. Find the rate of the plane in calm air and the rate of the wind.. A rowing team rowing with the current traveled mi in h. Against the current, the team rowed mi in h. Find the rate of the rowing team in calm water and the rate of the current.. How man grams of pure acid must be added to % acid to make 9 grams of solution that is % acid?. A chemist has some % hdrogen peroide solution and some % hdrogen peroide solution. How man milliliters of the % solution should be used to make milliliters of solution that is.% hdrogen peroide?. How man grams of % pure gold must be mied with % pure gold to make grams of % pure gold?. A total of \$ is deposited into two simple interest accounts. One account has an interest rate of.% and the other has a rate of.%. How much should be invested in the.% account to earn a total interest of \$?. A total of \$, is invested to provide retirement income. Part of the \$, is invested in an account paing 9% interest. How much should be invested into an % interest account so that the total income is \$? 9. A total of \$, is invested into two simple interest accounts. One account has an interest rate of % while the other account has a rate of %. How much must be invested in each account so that both accounts earn the same amount of interest?. Find the cost per kilogram of a grated cheese miture made from kg of cheese that costs \$. per kilogram and kg of cheese that costs \$. per kilogram.. How man grams of pure acid must be added to g of a % acid solution to make a solution that is % acid?. A researcher mies g of % aluminum allo with g of a % aluminum allo. What is the percent concentration of the resulting allo?
9 . A total of \$9 is deposited into two simple interest accounts. On one account the annual simple interest rate is %; on the second account the annual simple interest rate is %. How much should be invested in the % account so that the total interest earned is \$?. An investment of \$ is made into a % simple interest account. How much additional mone must be deposited into a.% simple interest account so that the total interest earned on both accounts is % of the total investment?. A jet plane traveling at mph overtakes a propeller-driven plane that has a. hour head start. The propeller-driven plane is traveling at a rate of mph. How far from the starting point does the jet overtake the propeller-driven plane?. A student pilot flies to a cit at an average speed of mph and then returns at an average speed of mph. Find the total distance between the two cities if the total fling time was hours.
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https://hackmd.io/@arthur-chang/SyLa2qb3Y | 1,726,878,182,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00506.warc.gz | 254,171,142 | 24,459 | # Annotate and explain Quiz6 with Ripes simulation ## Problem A 1. For your first iteration, you are told that you must have a 2-stage pipeline whose latency is no more than 16ns. Using a minimal number of registers, please show the twostage pipeline that maximizes the allowed latency of the “M” module. Then, calculate the maximum latency of module M and the throughput of the resulting circuit (using your value of M). * Max latency of “M” module (ns): __ A01 __ * Throughput with max latency “M” (ns-1): __ A02 __  > A01 = 3 > A02 = $\frac{1}{8}$ >  > Since the latency is no more than 16ns and these is a 2-stage pipline, the maximum latency of each stage is equal to 8. Therefore, as we split the circuit as figure above, the latency of the right part of pipline is $5+2=7$. As a result, the latency of left part must equal to 8, which is equal to $M+2+3$. $M=3$ > Throughput with max latency "M" $=\dfrac{1}{8}$ 2. For your next iteration, you are told that you must have a 4-stage pipeline whose latency is no more than 20ns. Using a minimal number of registers, please show the four-stage pipeline that maximizes the allowed latency of the “M” module. Then, calculate the maximum latency of module M and the throughput of the resulting circuit (using your value of M). * Max latency of “M” module (ns): __ A03 __ * Throughput with max latency “M” (ns-1): __ A04 __ > A03 = 5 > A04 = $\frac{1}{5}$ >  > Since the latency is no more than 16ns and these is a 4-stage pipline, the maximum latency of each stage is equal to 5. Therefore, we split the circuit as the figure above. The maximum value of $M$ must be $5$ because the maximum latency of each stage is $5$. > Throughput with max latency "M" $=\dfrac{1}{5}$ --- ## Problem B Assume that we are working on building a 32-bit RISC-V processor. As part of this project, they are considering several cache designs. The hardware has a limited amount of memory available, so any cache design will hold a total of 32 32-bit words (plus any associated metadata such as tag, valid, and dirty bits when needed). We first consider a **direct-mapped** cache using a block size of 4 words. 1. If the cache holds a total of 32 words, how many lines will be in this cache? * Number of lines in the direct-mapped cache: __ B01 __ > B01 = 8 > Number of lines in cache $= 32 \div 4 = 8$ 2. To properly take advantage of locality, which address bits should be used for the block offset, the cache index, and the tag field? Express your answer using minispec-style indexing into a 32-bit address. For example, the address bits for the byte offset would be expressed as addr[ 1 : 0 ]. * Address bits for block offset: addr[ __ B02 __ : __ B03 __ ] * Address bits for cache index: addr[ __ B04 __ : __ B05 __ ] * Address bits for tag field: addr[ __ B06 __ : __ B07 __ ] > B02 = 3 > B03 = 2 > B04 = 3 > B05 = 4 > B06 = 31 > B07 = 7 > Bits 0 and 1 are the byte offset > Since it is a 4-word cache, $log_2(4) = 2$. The block offset is 2 (bits 3:2) > There are 8 lines. $\ log_2(8) = 3$ bits for the cache index (bits 6:4). > Rest of the bits are tag bits(bits 31:7) >  We now consider switching to a 2-way set-associative cache, again using a block size of 4 words and storing 32 data words total (along with necessary metadata). The 2-way cache uses an LRU replacement policy. 3. How does this change affect the number of sets and the size of the tag compared to the direct-mapped cache? * Change in the number of sets (select one): __ B08 __ * (a) None / can’t tell * (b) 0.5x * (c) 2x * (d) -1 * (e) +1 * (f) Unchanged * Change in the size of the tag in bits (select one): __ B09 __ * (a) None / can’t tell * (b) 0.5x * (c) 2x * (d) -1 * (e) +1 * (f) Unchanged > B08 = b > Each set now contains two lines(one on each way), but the volume of cache does not change. As a result, the number of sets becomes half. > B09 = e > Since the number of sets becomes half, the number of set index bits decrease by one and tag bits increase by 1. > * Original: >  > * 2-way set-associative >  We decide to use a 2-way set-associative cache with 4 cache sets and a block size of 4 words for our processor and would like to test out our caching system. We write the following code to simulate checking if the array from a quicksort function is properly sorted. The code iterates over a 200-element array and checks for correct ordering. You may treat unimp as a 32-bit instruction that terminates the program. c . = 0x100 // The following code starts at address 0x100 check_sorted: // Initialize some registers li t0, 0 // t0 = loop index li t1, 199 // t1 = array size - 1 lui t2, 0x3 // t2 = starting address of array = 0x3000 loop: lw t3, 0(t2) // Load current element lw t4, 4(t2) // Load next element ble t3, t4, endif // Assume branch is always TAKEN if_unsorted: unimp // If two elements are out of order, terminate endif: addi t0, t0, 1 addi t2, t2, 4 blt t0, t1, loop // Continue checking; assume branch is TAKEN ret For the rest of this problem, assume that the code is running at steady state (i.e., the code is in the middle of the array) and that the array is sorted correctly. In other words, assume that both the ble and blt branches are always taken. Also, you may assume that when execution of the code started, all cache lines were set to invalid and Way 0 was the LRU way for each cache set. 4. For one iteration of the loop (i.e. from the loop label to the blt instruction), how many instruction fetches and data accesses are performed? * Number of Instruction Fetches: __ B10 __ * Number of Data Accesses: __ B11 __ > B10 = 6 > There are 6 instruction fetchs, which are2 lw, 1 ble, 1 unimp and 2 addi instructions. > B11 = 2 > There are 2 data accesses, which are lw t3, 0(t2),lw t4, 4(t2). 5. In the steady state (i.e. ignoring any cold-start effects), what is the instruction fetch hit ratio and the data access hit ratio? Note: please provide the hit ratio and not the miss ratio. You may use the cache diagram provided below to help you, but nothing written in the diagram will be graded.  * Instruction Fetch HIT Ratio: __ B12 __ * Data Access HIT Ratio: __ B13 __ > B12 = 1 > All instructions reside in sets 0, 1, and 2 of Way 0 and are not evicted. Thus the instruction hit ratio is 1. > The first instruction is at 0x100 = 0b1 0|000 |00|00, which is at set 0. > The whole instruction cache looks like the table below: >  > B13 = 0.875 > Data accesses are split between Way 0 and Way 1, but ultimately, they fit around all the instructions in the cache (which are always accessed more recently and so do not get evicted). Each data miss will cause the cache to load the next four consecutive words (four words/block). Each iteration performs 2 data accesses. Every four iterations, one of these accesses will miss. This corresponds to 1 miss per 8 accesses for a hit ratio of 0.875 >  6. Assume that it takes 2 cycles to access the cache. We have benchmarked our code to have an average hit ratio of 0.9. If they need to achieve an average memory access time (AMAT) of at most 4 cycles, what is the upper bound on our miss penalty? Miss penalty here is defined as the amount of additional time required beyond the 2-cycle cache access to handle a cache miss. * Maximum possible miss penalty: __ B14 __ cycles > B14 = 20 > AMAT = cache access time + miss penalty * (1 - hit ratio) 4 >= 2 + miss penalty * (1 - 0.9) miss penalty <= 20 --- ## Problem C Consider that we are analyzing grade statistics and are performing some hefty calculations, so we suspect that a cache could improve our system’s performance. 1. We are considering using a 2-way set-associative cache with a block size of 4 (i.e. 4 words per line). The cache can store a total of 64 words. Assume that addresses and data words are 32 bits wide. To properly make use of locality, which address bits should be used for the block offset, the cache index, and the tag field? * Address bits used for byte offset: A[ 1 : 0 ] * Address bits used for tag field: A[ __ C01 __ : __ C02 __ ] * Address bits used for block offset: A[ __ C03 __ : __ C04 __ ] * Address bits used for cache index: A[ __ C05 __ : __ C06 __] > C01 = 31 > C02 = 7 > C03 = 3 > C04 = 2 > C05 = 6 > C06 = 4 >  2. If we instead used a direct-mapped cache with the same total capacity (64 words) and same block size (4 words), how would the following parameters in our system change? * Change in the number of cache lines (select all of the choices below that apply): __ C07 __ * (a) None / can’t tell * (b) 0.5x * (c) 2x * (d) -1 * (e) +1 * (f) Unchanged * Change in the number of bits in tag field (select one of the choices below): __ C08 __ * (a) None / can’t tell * (b) 0.5x * (c) 2x * (d) -1 * (e) +1 * (f) Unchanged > C07 = c,f > Accepted Unchanged or 2x as solutions because 2-way cache also has $\frac{1}{2}$ the number of sets as the direct mapped cache, but the number of cache lines is actually the same. > * Original: >  > * Direct-mapped: >  > C08 = 8 > Block size of 4 → 2 bits for block offset 64 words / 4 words per block = 16 blocks → 4 bits for cache line index 32 – 4 – 2 – 2 = 24 bits of tag > * Original: >  > * Direct-mapped: >  Ultimately, we decide that the 2-way set associative cache would probably have better performance for our application, so the remainder of the problem will be considering a 2-way set associative cache. Below is a snapshot of this cache during the execution of some unknown code. V is the valid bit and D is the dirty bit of each set.  3. Would the following memory accesses result in a hit or a miss? If it results in a hit, specify what value is returned; if it is a miss, explain why in a few words or by showing your work. * 32-Bit Byte Address: 0x4AB4 * Line index: __ C09 __ * Tag: __ C10 __ (in HEX) * Block offset: __ C11 __ * Returned value if hit (in HEX) / Explanation if miss: __ C12 __ * 32-Bit Byte Address: 0x21E0 * Line index: __ C13 __ * Tag: __ C14 __ * Block offset: __ C15 __ * Returned value if hit (in HEX) / Explanation if miss: __ C16 __ > C09 = 3 > C10 = 0x95 > C11 = 1 > C12 = 0xD4 >  > 0x4AB4 = 0b 0100 1010 1|011 |01|00 > Line index = 0b011=3 > Tag = 0b0100 1010 1 = 0x95 > Block offset = 0b01 > As we can see that the value we want to get is at line 3 block 1 and the tag is 0x95. When looking the table, we find the tag of line 3 in Way1 is equal to 0x95, which means a cache hit. >  > C13 = 6 > C14 = 43 > C15 = 0 > C16 = miss, valid bit is 0 > 0x21E0 = 0b 10 0001 1|110 |00|00 > As we can see that the value we want to get is at line 6 block 0 and the tag is 0x43. When looking the table, we find the tag of line 3 in Way0 is equal to 0x43. However, the valid bit is 0. As a result, it is a cache miss. --- ## Problem D Consider the case that we lost the final iteration among two other prototypes while building the pipelined RISC-V processor. Show all bypasses used in each cycle where the processor is not stalled. For the processor, also determine the value of x3 and x4 after executing these 12 cycles. Below is the code and the initial state of the relevant registers in the register file for each of the three processors. Note that the values in registers x1–x5 are given in decimal while x6 is in hexadecimal. A copy of the code and initial register values is provided for the processor. c start: lw x1, 0(x6) addi x2, x0, 5 blt x1, x2, end addi x3, x2, 11 sub x4, x3, x1 xori x5, x6, 0x1 end: sub x4, x3, x2 addi x3, x4, 7 addi x3, x1, 3 . = 0x400 .word 0x1 | Register | Value | |:--------:|:-----:| | x1 | 5 | | x2 | 11 | | x3 | 30 | | x4 | 19 | | x5 | 20 | | x6 | 0x400 | Assume the processor is built as a 5-stage pipelined RISC-V processor, which is fully bypassed and has branch annulment. Branch decisions are made in the EXE stage and branches are always predicted not taken. What are the values of registers x3 and x4 at the start of cycle 12 (in decimal)? * x3 : __ D01 __ * x4 : __ D02 __ > D01 = 32 > D02 = 25 >  > >  > The figure above shows the three forwarding and the execution in 12 cycles. > >  > The figure above shows the first forwarding. The processor load the value into x1 and send to EXE stage since the blt needs the value. >  > The figure above shows the second forwarding. The addi puts 5 in x2 and pass to blt. >  > The figure above shows the third forwarding. The sub calculates the result and puts it in x4. Then pass the value to EXE stage because addi instruction needs it. --- ## Problem E This problem evaluates the cache performances for different loop orderings. You are asked to consider the following two loops, written in C, which calculate the sum of the entries in a 128 by 32 matrix of 32-bit integers: * Loop A c sum = 0; for (i = 0; i < 128; i++) for (j = 0; j < 32; j++) sum += A[i][j]; * Loop B c sum = 0; for (j = 0; j < 32; j++) for (i = 0; i < 128; i++) sum += A[i][j]; The matrix A is stored contiguously in memory in row-major order. Row major order means that elements in the same row of the matrix are adjacent in memory as shown in the following memory layout: A[i][j] resides in memory location [4*(32*i + j)] Memory Location:  For Problem 1 to 3, assume that the caches are initially empty. Also, assume that only accesses to matrix A cause memory references and all other necessary variables are stored in registers. Instructions are in a separate instruction cache. 1. Consider a 4KB direct-mapped data cache with 8-word (32-byte) cache lines. * Calculate the number of cache misses that will occur when running Loop A. __ E01 __ * Calculate the number of cache misses that will occur when running Loop B. __ E02 __ > E01 = 512 > E02 = 4096 > Each element of the 128x32 matrix A can only be mapped to one particular cache location in this direct-mapped data cache. Since each row has 32 32-bit integers, and since each cache line can hold 8 32-bit ints, a row of the matrix occupies the lines in 4 consecutive sets of the cache. > Loop A—where each iteration of the inner loop sums a row of A—accesses memory addresses in a linear sequence. Given this access pattern, the access to the first word in each cache line will miss, but the next seven accesses will hit. After sequentially moving through this line, it will not be accessed again, so its later eviction will not cause any future misses. Therefore, Loop A will only have compulsory misses for the 512 (128 rows x 4 lines per row) that matrix A spans. > The consecutive accesses in Loop B will move in a stride of 32 words. Therefore, the inner loop will touch the first element in 128 cache lines before the next iteration of the outer loop. While intuition might suggest that the 128 lines could all fit in the cache with 128 sets, there is a complicating factor: each row is four cache lines past the previous row, meaning that the lines accessed when traversing the first column go in indices 0, 4, 8, 16, and so on. Since the lines containing the column are competing for only one quarter of the sets, the lines loaded when starting a column are evicted by the time the column is complete, preventing any reuse. Therefore, all 4096 (128 x 32) accesses miss. >  > The iteration of Loop A looks like the gif shown above. > >  > The part of iteration of Loop B looks like the gif shown above. 2. Consider a direct-mapped data cache with 8-word (32-byte) cache lines. Calculate the minimum number of cache lines required for the data cache if Loop A is to run without any cache misses other than compulsory misses. Calculate the minimum number of cache lines required for the data cache if Loop B is to run without any cache misses other than compulsory misses. * Data-cache size required for Loop A: __ E03 __ cache line(s) * Data-cache size required for Loop B: __ E04 __ cache line(s) > E03 = 1 > E04 = 512 > Since Loop A accesses memory sequentially, we can sum all the elements in a cache line and then never touch it again. Therefore, we only need to hold 1 active line at any given time to avoid all but compulsory misses. > For Loop B to run without any cache misses other than compulsory misses, the data cache needs to have the ability to hold one column of matrix A in the cache. Since the consecutive accesses in the inner loop of Loop B will use one out of every four cache lines, and since we have 128 rows, Loop B requires 512 (128 × 4) lines to avoid all but compulsory misses. > > >  > The gif above shows the iteration of Loop A in 1-line cache. 3. Consider a 4KB set-associative data cache with 4 ways, and 8-word (32-byte) cache lines. This data cache uses a first-in/first-out (FIFO) replacement policy. * The number of cache misses for Loop A: __ E05 __ * The number of cache misses for Loop B: __ E06 __ > E05 = 512 > E06 = 4096 > Loop A still only has 512 (128 rows x 4 lines per row) compulsory misses. Loop B still cannot fully utilize the cache. The first 8 accesses will allocate into way 1 in sets 0, 4, 8, 16, etc.; the next 8 accesses will allocate into way 2 of those same sets; and so on. After 32 accesses, all four ways will be filled, and the next 8 accesses along the column will evict the previous lines in way 1, preventing any reuse. Therefore, all 4096 (128 x 32) accesses miss. > >  > The gif above shows the iteration of Loop A in 4 ways, and 8-word (32-byte) cache lines. > >  > The gif above shows the iteration of Loop B in 4 ways, and 8-word (32-byte) cache lines. ## Problem F The following diagram shows a classic fully-bypassed 5-stage pipeline that has been augmented with an unpipelined divider in parallel with the ALU. Bypass paths are not shown in the diagram. This iterative divider produces 2 bits per cycle until it outputs a full 32-bit result.  * What is the latency of a divide operation in cycles? __ F01 __ * What is the occupancy of a divide operation in cycles? __ F02 __ > F01 = 16 > 32/2 = 16 > F02 = 16 > 16 > since the divider is unpipelined --- ## Problem G Given the follow chunk of code, analyze the hit rate given that we have a byte-addressed computer with a total memory of 1 MiB. It also features a 16 KiB Direct-Mapped cache with 1 KiB blocks. Assume that your cache begins cold. cpp #define NUM_INTS 8192 // 2ˆ13 int A[NUM_INTS]; // A lives at 0x10000 int i, total = 0; for (i = 0; i < NUM_INTS; i += 128) A[i] = i; // Code 1 for (i = 0; i < NUM_INTS; i += 128) total += A[i]; // Code 2 1. How many bits make up a memory address on this computer? __ G01 __ > $log_2(1MiB) = log_2(2^{20}) = 20$ bits 2. What is the T:I:O breakdown? * Offset: __ G02 __ * Index: __ G03 __ * Tag: __ G04 __ > G02 = 10 > G03 = 4 > G04 = 6 > Offset = $log_2(1 \ KiB) =log_2(2^{10}) = 10$ > Index = $log_2(\dfrac{16\ KiB}{1\ KiB})=log_2(16)=4$ > Tag = $20 - 4 - 10=6$ 3. Calculate the cache hit rate for the line marked as Code 1: __ G05 __ > G05 = 50 > The Access address looks like the figure below. There are 2 bits for byte offset, 8 bits for 1 KiB Block($1024*8/32=256$ words ), 4 bits for index and the rest of bits for tag. As a result, we have a cache with 16 cache lines and each cache line contains 256 words. >  > > The integer accesses are 4 ∗ 128 = 512 bytes apart, which means there are 2 accesses per block. The first accesses in each block is a compulsory cache miss, but the second is a hit because A[i] and A[i+128] are in the same cache block. Thus, we end up with a hit rate of 50%. > > The gif below shows how the cache work. >  4. Calculate the cache hit rate for the line marked as Code 2: __ G06 __ > G06 = 50% > The size of A is 8192 ∗ 4 = $2^{15}$ bytes. This is exactly twice the size of our cache. At the end of Code 1, we have the second half of A inside our cache, but Code 2 starts with the first half of A. Thus, we cannot reuse any of the cache data brought in from Code 1 and must start from the beginning. Thus our hit rate is the same as Code 1 since we access memory in the same exact way as Code 1. We do not have to consider cache hits for total, as the compiler will most likely store it in a register. Thus, we end up with a hit rate of 50%. > > The gif below shows how Code2 work in cache. >  | 6,134 | 21,199 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.896119 |
http://cycling74.com/docs/max5/tutorials/msp-tut/mspchapter25.html | 1,412,196,545,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663587.40/warc/CC-MAIN-20140930004103-00412-ip-10-234-18-248.ec2.internal.warc.gz | 73,593,997 | 5,916 | Tutorial 25: Using the FFT
Fourier's theorem
The French mathematician Joseph Fourier demonstrated that any periodic wave can be expressed as the sum of harmonically related sinusoids, each with its own amplitude and phase. Given a digital representation of a periodic wave, one can employ a formula known as the discrete Fourier transform (DFT) to calculate the frequency, phase, and amplitude of its sinusoidal components. Essentially, the DFT transforms a time-domain representation of a sound wave into a frequency- domain spectrum. This spectrum can also be transformed back into a time-domain waveform.
Typically the Fourier transform is used on a small ‘slice’ of time, which ideally is equal to exactly one cycle of the wave being analyzed. To perform this operation on ‘real world’ sounds -- which are almost invariably not strictly periodic, and which may be of unknown frequency -- one can perform the DFT on consecutive time slices to get a sense of how the spectrum changes over time.
If the number of digital samples in each time slice is a power of 2, one can use a faster version of the DFT known as the fast Fourier transform (FFT). The formula for the FFT is encapsulated in the fft~ object. The mathematics of the Fourier transform are well beyond the scope of this manual, but this tutorial chapter will demonstrate how to use the fft~ object for signal analysis and resynthesis.
Spectrum of a signal: fft~
fft~ receives a signal in its inlet. For each slice of time it receives (512 samples long by default) it sends out a signal of the same length listing the amount of energy in each frequency region. The signal that comes out of fft~ is not anything you're likely to want to listen to. Rather, it's a list of relative amplitudes of 512 different frequency bands in the received signal. This ‘list’ happens to be exactly the same length as the samples received in each time slice, so it comes out at the same rate as the signal comes in. The signal coming out of fft~ is a frequency-domain analysis of the samples it received in the previous time slice.
Although the transform comes out of fft~ in the form of a signal, it is not a time-domain signal. The only object that ‘understands’ this special signal is the ifft~ object, which performs an inverse FFT on the spectrum and transforms it back into a time-domain waveform.
The signal coming out of fft~ is spectral information, not a time-domain signal
With the capture~ object you can grab some of the output of fft~ and examine the frequency analysis of a signal.
• Click on one of the ezdac~ objects to turn audio on.
When audio is turned on, dspstate~ sends the MSP sampling rate out its middle outlet. We use this number to calculate a frequency that has a period of exactly 512 samples. This is the fundamental frequency of the FFT itself. If we send a wave of that frequency into fft~, each time slice would contain exactly one cycle of the waveform. We will actually use a cosine wave at ten times that frequency as the test tone for our analysis, as shown below.
The test tone is at 10 times the base frequency of the FFT time slice
The upper left corner of the Patcher window shows a very simple use of fft~. The analysis is stored in a capture~ object, and an ifft~ object transforms the analysis back into an audio signal. (Ordinarily you would not transform and inverse-transform an audio signal for no reason like this. The ifft~ is used in this patch simply to demonstrate that the analysis-resynthesis process works.)
• Click on the toggle in the upper left part of the patch to hear the resynthesized sound. Click on the toggle again to close the gate~. Now double-click on the capture~ object in that part of the patch to see the analysis performed by fft~.
In the capture~ text window, the first 512 numbers are all 0.0000. That is the output of fft~ during the first time slice of its analysis. Remember, the analysis it sends out is always of the previous time slice. When audio was first turned on, there was no previous audio, so the fft~ object's analysis shows no signal.
• Scroll past the first 512 numbers. (The numbers in the capture~ object's text window are grouped in blocks, so if your signal vector size is 256 you will have two groups of numbers that are all 0.0000.) Look at the second time slice of 512 numbers.
Each of the 512 numbers represents a harmonic of the FFT frequency itself, starting at the 0th harmonic (0 Hz). The analysis shows energy in the eleventh number, which represents the 10th harmonic of the FFT, 10/512 the sampling rate -- precisely our test frequency. (The analysis also shows energy at the 10th number from the end, which represents 502/512 the sampling rate. This frequency exceeds the Nyquist rate and is actually equivalent to -10/512 of the sampling rate.
Technical detail: An FFT divides the entire available frequency range into as many bands (regions) as there are samples in each time slice. Therefore, each set of 512 numbers coming out of fft~ represents 512 divisions of the frequency range from 0 to the sampling rate. The first number represents the energy at 0 Hz, the second number represents the energy at 1/512 the sampling rate, the third number represents the energy at 2/512 the sampling rate, and so on.
Note that once we reach the Nyquist rate on the 257th number (256/512 of the sampling rate), all numbers after that are folded back down from the Nyquist rate. Another way to think of this is that these numbers represent negative frequencies that are now ascending from the (negative) Nyquist rate. Thus, the 258th number is the energy at the Nyquist rate minus 1/512 of the sampling rate (which could also be thought of as -255/512 the sampling rate). In our example, we see energy in the 11th frequency region (10/512 the sampling rate) and the 503rd frequency region (-256/512 - -246/512 = -10/512 the sampling rate).
It appears that fft~ has correctly analyzed the signal. There's just one problem...
Practical problems of the FFT
The FFT assumes that the samples being analyzed comprise one cycle of a periodic wave. In our example, the cosine wave was the 10th harmonic of the FFT's fundamental frequency, so it worked fine. In most cases, though, the 512 samples of the FFT will not be precisely one cycle of the wave. When that happens, the FFT still analyzes the 512 samples as if they were one cycle of a waveform, and reports the spectrum of that wave. Such an analysis will contain many spurious frequencies not actually present in the signal.
• Close the text window of capture~. With the audio still on, set the ‘Test Frequency’ number box to 1000. This also triggers the clear message in the upper left corner of the patch to empty the capture~ object of its prior contents. Double-click once again on capture~, and scroll ahead in the text window to see its new contents.
The analysis of the 1000 Hz tone does indeed show greater energy at 1000 Hz -- in the 12th and 13th frequency regions if your MSP sampling rate is 44,100 Hz -- but it also shows energy in virtually every other region. That's because the waveform it analyzed is no longer a sinusoid. (An exact number of cycles does not fit precisely into the 512 samples.) All the other energy shown in this FFT is an artifact of the ‘incorrect’ interpretation of those 512 samples as one period of the correct waveform.
To resolve this problem, we can try to ‘taper’ the ends of each time slice by applying an amplitude envelope to it, and use overlapping time slices to compensate for the use of the envelope.
Overlapping FFTs
The lower right portion of the tutorial patch takes this approach of using overlapping time slices, and applies a triangular amplitude envelope to each slice before analyzing it, and again after resynthesizing it. (Other shapes of amplitude envelope are often used for this process, but the triangular window is simple and fairly effective.) In this way, the fft~ object is viewing each time slice through a triangular window which tapers its ends down, thus filtering out many of the false frequencies that would be introduced by discontinuities. This technique is known as windowing.
Overlapping triangular windows (envelopes) applied to a 100 Hz cosine wave
To accomplish this windowing and overlapping of time slices, we must perform two FFTs, one of which is offset 256 samples later than the other. (Note that this part of the patch will only work if your current MSP Signal Vector size is 256 or less, since fft~ can only be offset by a multiple of the vector size.) The offset of an FFT can be given as a (third) typed-in argument to fft~, as is done for the fft~ object on the right. This results in overlapping time slices.
One FFT is taken 256 samples later than the other
The windowing is achieved by multiplying the signal by a triangular waveform (stored in the buffer~ object) which recurs at the same frequency as the FFT -- once every 512 samples. The window is offset by 1/2 cycle (256 samples) for the second fft~. Notice also that because we will be applying the amplitude envelope twice (once before the fft~ and once again after the ifft~), we take the square root of the envelope values, so we do not have unwanted amplitude modulation resulting from our envelopes (we want the overlapping envelopes to crossfade evenly and always add up to 1).
• Double-click on the buffer~ object to view its contents. Then close the buffer~ window and double-click on the capture~ object that contains the FFT of the windowed signal. Scroll past the first block or two of numbers until you see the FFT analysis of the windowed 1000 Hz tone.
As with the unwindowed FFT, the energy is greatest around 1000 Hz, but here the (spurious) energy in all the other frequency regions is greatly reduced by comparison with the unwindowed version.
Signal processing using the FFT
In this patch we have used the fft~ object to view and analyze a signal, and to demonstrate the effectiveness of windowing the signal and using overlapping FFTs. However, one could also write a patch that alters the values in the signal coming out of fft~, then sends the altered analysis to ifft~ for resynthesis. This kind of processing using overlapped, windowed time slices is known as a Short Term Fourier Transform (STFT), and is the basis for frequency-domain audio processing. We will be discussing a simpler way to use the STFT in Tutorial 26.
Windowing, in addition to being important for reducing the false frequencies from discontinuities in the input waveform, as we have already seen, is also important to smooth out any discontinuities which occur in the resynthesized time-domain waveform coming from the ifft~. This is why we must window the time-domain signal both on input and output. In this tutorial we would only get such output discontinuities if we modified the signal between the fft~ and ifft~ objects.
Summary
The fast Fourier transform (FFT) is an algorithm for transforming a time-domain digital signal into a frequency-domain representation of the relative amplitude of different frequency regions in the signal. An FFT is computed using a relatively small excerpt of a signal, usually a slice of time 512 or 1024 samples long. To analyze a longer signal, one performs multiple FFTs using consecutive (or overlapping) time slices.
The fft~ object performs an FFT on the signal it receives, and sends out (also in the form of a signal) a frequency-domain analysis of the received signal. The only object that understands the output of fft~ is ifft~ which performs an inverse FFT to synthesize a time-domain signal based on the frequency-domain information. One could alter the signal as it goes from fft~ to ifft~, in order to change the spectrum.
The FFT only works perfectly when analyzing exactly one cycle (or exactly an integer number of cycles) of a tone. To reduce the artifacts produced when this is not the case, one can window the signal being analyzed by applying an amplitude envelope to taper the ends of each time slice. The amplitude envelope can be applied by multiplying the signal by using a cycle~ object to read a windowing function from a buffer~ repeatedly at the same rate as the FFT itself (i.e., once per time slice). To eliminate any artifacts that result from modifying the frequency-domain signal, we must also apply the same envelope at the output of the ifft~. | 2,659 | 12,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2014-41 | longest | en | 0.936095 |
https://m.wikihow.com/Calculate-Mortgage-Payoff | 1,579,381,812,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00164.warc.gz | 537,802,781 | 59,852 | # How to Calculate Mortgage Payoff
Co-authored by Carla Toebe, Real Estate Broker
Updated: December 31, 2019
The method for precisely determining the rate of amortization, which is the amount needed to pay off a particular mortgage loan, will vary depending on factors like the type of loan, its terms, and what options are exercised by the borrower. However, there is a standard formula used for calculating the loan payoff amount of a mortgage based on the principal, the interest rate, the number of payments made, and the number of payments remaining. This article provides detailed information that will assist you in calculating your mortgage payoff amount based on the terms of your loan. Contact your lender to confirm that your calculation is correct based on the particulars of your mortgage.
### Part 1 of 2: Knowing the Essentials
1. 1
Understand why your mortgage payoff amount does not equal your current balance. Your statement says your most recent balance is \$12,250, and yet your payoff amount (the amount it would take to close out the mortgage) is listed as \$12,500. How can that be?
• You may assume that some complicated financial formulas are at play (ones that cost you money, of course), but in fact the answer is quite simple: mortgages are paid in arrears. That is, you pay June’s interest with July’s payment.[1]
• This process starts at the beginning, at closing. If you close in June, for instance, your first payment won’t be due until August, with July’s interest included.
2. 2
Gather the information needed for your calculations. In order to determine the payoff amount, either using a calculation program or on your own, you need to know a handful of basic figures regarding your mortgage. All of these figures should be available on your statement or other loan documents. These include:
• The total amount borrowed when you took out the loan (for example, \$200,000).
• The annual interest rate (for example, 3%, or 0.03). To do the calculations yourself, you will need to divide this number by twelve (0.03 / 12 = 0.0025), because mortgage interest compounds monthly.
• The total number of payments for the life of the loan, which for monthly payments is the number of years times twelve (for example, 20 years = 240 payments).
• The total number of years / payments remaining, and the number paid so far (for example, 15 years = 180 payments made; 5 years = 60 payments remaining).
3. 3
Consider online calculators if you’d rather not exercise your math muscles. The calculations aren’t all that complicated, but punching a few numbers in and pressing “calculate” is certainly easier. A search for “mortgage payoff calculator” will provide several useful results. Try out a few to ensure accuracy.
• Please note, however, that you may find slight variations in your final results, probably only a few cents, but enough that you should always confirm your payoff amount with your lender before attempting to make a final payment.
• Using two such online calculators for a \$200,000 loan at 3% annual interest over 20 years, with five years remaining, produces payoff results of \$61,729.26 and \$61,729.33, respectively.[2][3]
4. 4
Contact your mortgage lender if you plan to make a final payment. Avoid making what you think is a mortgage payoff only to find out that you still owe a few dollars or cents that keeps your mortgage alive and accruing interest.
• Contact your lender by phone or online; lenders with online account management likely have a page that allows you to make the request. You will likely receive your payoff amount after a week.
• You will be asked to choose a specific date for which to determine the payoff.
• If you want to make the final payment, you will need to complete some version of payment form (online or by mail) with the precise amount due and the time frame in which this amount is valid as a final payoff.
• Such requests can be made solely for informational purposes as well. Some lenders may even include a payoff amount on your monthly statement.
### Part 2 of 2: Making the Calculations
1. 1
Lay out the formula carefully. It looks rather unwieldy at first, but the math is relatively straightforward once you’ve inserted your figures. Just be sure to copy the formula exactly or your results may vary significantly. The formula is:[4]
• B = L [(1 + c)^n - (1 + c)^p] / [(1 + c)^n (- 1)] , in which:
• B = payoff balance due (\$)
• L = total loan amount (\$)
• c = interest rate (annual rate / 12)
• n = total payments (years x 12 for monthly payments)
• p = number of payments made so far
2. 2
Insert your figures. Using the same example as for the online calculators, a 20-year, \$200,000 mortgage at 3% interest with five years to go, appears thusly:
• B = 200,000 [(1 + 0.0025)^240 - (1 + 0.0025)^180] / [(1 + 0.0025)^240 (-1)]
• Remember that the 3% annual interest rate (0.03) is divided by twelve because it compounds monthly (c = 0.03 / 12 = 0.0025).
• Twenty years of monthly payments is 240 (n = 20 x 12 = 240), and fifteen years of payments so far is 180 (p = 15 x 12 = 180).
3. 3
Power up your numbers. After adding 1 to 0.0025 the three times it appears in the formula, you will need to raise the resulting 1.0025 to the 240th power (twice) and to the 180th power (once). A good calculator will come in handy here.
• 1.0025^240 = 1.82075499532
• 1.0025^180 = 1.56743172467
• B = 200,000(1.82075499532 - 1.56743172467) / (1.82075499532 - 1)
4. 4
Subtract from the inside. Once you accomplish this step, the formula will appear much more manageable and will be that much closer to completion.
• 1.82075499532 - 1.56743172467 = 0.253323270652
• 1.82075499532 - 1 = 0.82075499532
• B = 200,000(0.253323270652 / 0.82075499532)
5. 5
Divide, multiply, and conquer. Now you can finish things off and solve for B, your payoff amount.
• 0.253323270652 / 0.82075499532 = 0.308646638883
• 200,000 x 0.308646638883 = 61729.3277766
• Therefore, your payoff amount is \$61,729.33.
## Community Q&A
Search
• Question
How do you calculate a mid-month mortgage payoff?
Carla Toebe
Real Estate Broker
Carla Toebe is a Real Estate Broker in Washington. She has been an active real estate broker since 2005, and founded the real estate agency CT Realty LLC in 2013.
Real Estate Broker
To calculate mid month, multiply the monthly payment for the insurance premiums, interest, taxes, homeowner insurance, and anything else that is lumped into the monthly mortgage payments by the number of days until the close divided by the number of days in the month.
• Question
I owe 31900.00 on mortgage initially. The rate is 6.5 and I have 11yrs & 10 months to pay off, what would my payoff be
Carla Toebe
Real Estate Broker
Carla Toebe is a Real Estate Broker in Washington. She has been an active real estate broker since 2005, and founded the real estate agency CT Realty LLC in 2013.
Real Estate Broker
There are many other factors that are needed to answer this question. You need what the total term of the loan is, the original loan amount, the total number of payments in the term and the total number of payments made so far.
• Question
I have a mortgage with a private individual. I am going to pay off the mortgage. I need an ammortization schedule where I can plug in the amount of my monthly payments (\$800) for the last 2 years at 7% interest.
Carla Toebe
Real Estate Broker
Carla Toebe is a Real Estate Broker in Washington. She has been an active real estate broker since 2005, and founded the real estate agency CT Realty LLC in 2013.
Real Estate Broker
You can find free ammortization schedules on line. Just search for them. These will allow you to plug in your values.
• How do I tell how long my mortgage payments will take when paying off my home?
• How do I calculate a mortgage payoff when I know the loan amount, interest rate, balance, and the amount of time?
• How do I determine my payoff amount?
200 characters left
## Tips
• You must get a quote for the mortgage payoff before you sell a property. The money you make on the sale will first go to covering the payoff, then towards all of the other associated fees and closing costs that the seller is responsible for. After these costs, you will then receive anything leftover.[5]
Thanks!
## Warnings
• While you can make the calculations yourself, it always best to get the payoff number from the lender themselves. If your calculations are off even by a few cents, you may end up keeping the mortgage open, and you will continue to accrue interest.
Thanks!
## Video.By using this service, some information may be shared with YouTube.
Real Estate Broker
This article was co-authored by Carla Toebe. Carla Toebe is a Real Estate Broker in Washington. She has been an active real estate broker since 2005, and founded the real estate agency CT Realty LLC in 2013.
Co-authors: 8
Updated: December 31, 2019
Views: 80,335
Article SummaryX
Mortgage payoff is the remaining amount you need to pay on your mortgage, including interest. To calculate your mortgage payoff, you’ll need to know the total amount you borrowed, your annual interest rate, the total number of payments for the whole duration of the loan, and the total number of payments remaining. There are a number of online calculators that will work this out for you. All you need to do is enter your figures and take note of your mortgage payoff. For more tips from our Financial co-author, including the formula to manually calculate your mortgage payoff, read on!
Thanks to all authors for creating a page that has been read 80,335 times.
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May 16, 2017
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Mar 20, 2018
"Thanks! This explained what kind of charges I'd be looking at if I tried to pay off my mortgage! " | 2,414 | 9,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-05 | longest | en | 0.954601 |
https://walkccc.me/CLRS/Chap24/Problems/24-4/ | 1,638,805,405,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00306.warc.gz | 662,764,672 | 18,547 | # 24-4 Gabow's scaling algorithm for single-source shortest paths
A scaling algorithm solves a problem by initially considering only the highestorder bit of each relevant input value (such as an edge weight). It then refines the initial solution by looking at the two highest-order bits. It progressively looks at more and more high-order bits, refining the solution each time, until it has examined all bits and computed the correct solution.
In this problem, we examine an algorithm for computing the shortest paths from a single source by scaling edge weights. We are given a directed graph $G = (V, E)$ with nonnegative integer edge weights $w$. Let $W = \max_{(u, v) \in E} \{w(u, v)\}$. Our goal is to develop an algorithm that runs in $O(E\lg W)$ time. We assume that all vertices are reachable from the source.
The algorithm uncovers the bits in the binary representation of the edge weights one at a time, from the most significant bit to the least significant bit. Specifically, let $k = \lceil \lg(W + 1) \rceil$ be the number of bits in the binary representation of $W$, and for $i = 1, 2, \ldots, k$, let $w_i(u, v) = \lfloor w(u, v) / 2^{k - i} \rfloor$. That is, $w_i(u, v)$ is the "scaled-down" version of $w(u, v)$ given by the $i$ most significant bits of $w(u, v)$. (Thus, $w_k(u, v) = w(u, v)$ for all $(u, v) \in E$.) For example, if $k = 5$ and $w(u, v) = 25$, which has the binary representation $\langle 11001 \rangle$, then $w_3(u, v) = \langle 110 \rangle = 6$. As another example with $k = 5$, if $w(u, v) = \langle 00100 \rangle = 4$, then $w_3(u, v) = \langle 001 \rangle = 1$. Let us define $\delta_i(u, v)$ as the shortest-path weight from vertex $u$ to vertex $v$ using weight function $w_i$. Thus, $\delta_k(u, v) = \delta(u, v)$ for all $u, v \in V$. For a given source vertex $s$, the scaling algorithm first computes the shortest-path weights $\delta_1(s, v)$ for all $v \in V$, then computes $\delta_2(s, v)$ for all $v \in V$, and so on, until it computes $\delta_k(s, v)$ for all $v \in V$. We assume throughout that $|E| \ge |V| - 1$, and we shall see that computing $\delta_i$ from $\delta_{i - 1}$ takes $O(E)$ time, so that the entire algorithm takes $O(kE) = O(E\lg W)$ time.
a. Suppose that for all vertices $v \in V$, we have $\delta(s, v) \le |E|$. Show that we can compute $\delta(s, v)$ for all $v \in V$ in $O(E)$ time.
b. Show that we can compute $\delta_1(s, v)$ for all $v \in V$ in $O(E)$ time.
Let us now focus on computing $\delta_i$ from $\delta_{i - 1}$.
c. Prove that for $i = 2, 3, \ldots, k$, we have either $w_i(u, v) = 2w_{i - 1}(u, v)$ or $w_i(u, v) = 2w_{i - 1}(u, v) + 1$. Then, prove that
$$2\delta_{i - 1}(s, v) \le \delta_i(s, v) \le 2\delta_{i - 1}(s, v) + |V| - 1$$
for all $v \in V$.
d. Define for $i = 2, 3, \ldots, k$ and all $(u, v) \in E$,
$$\hat w_i = w_i(u, v) + 2\delta_{i - 1}(s, u) - 2\delta_{i - 1}(s, v).$$
Prove that for $i = 2, 3, \ldots, k$ and all $u, v \in V$, the "reweighted" value $\hat w_i(u, v)$ of edge $(u, v)$ is a nonnegative integer.
e. Now, define $\hat\delta_i(s, v)$ as the shortest-path weight from $s$ to $v$ using the weight function $\hat w_i$. Prove that for $i = 2, 3, \ldots, k$ and all $v \in V$,
$$\delta_i(s, v) = \hat\delta_i(s, v) + 2\delta_{i - 1}(s, v)$$
and that $\hat\delta_i(s, v) \le |E|$.
f. Show how to compute $\delta_i(s, v)$ from $\delta_{i - 1}(s, v)$ for all $v \in V$ in $O(E)$ time, and conclude that we can compute $\delta(s, v)$ for all $v \in V$ in $O(E\lg W)$ time.
a. We can do this in $O(E)$ by the algorithm described in exercise 24.3-8 since our "priority queue" takes on only integer values and is bounded in size by $E$.
b. We can do this in $O(E)$ by the algorithm described in exercise 24.3-8 since $w$ takes values in $\{0, 1\}$ and $V = O(E)$.
c. If the $i$th digit, read from left to right, of $w(u, v)$ is $0$, then $w_i(u, v) = 2w_{i − 1}(u, v)$. If it is a $1$, then $w_i(u, v) = 2w_{i − 1}(u, v) + 1$. Now let $s = v_0, v_1, \dots, v_n = v$ be a shortest path from $s$ to $v$ under $w_i$. Note that any shortest path under $w_i$ is necessarily also a shortest path under $w_{i − 1}$. Then we have
\begin{aligned} \delta_i(s, v) & = \sum_{m = 1}^n w_i(v_{m − 1}, v_m) \\ & \le \sum_{m = 1}^n [2w_{i − 1}(u, v) + 1] \\ & \le \sum_{m = 1}^n w_{i − 1}(u, v) + n \\ & \le 2\delta_{i − 1}(s, v) + |V| − 1. \end{aligned}
On the other hand, we also have
\begin{aligned} \delta_i(s, v) & = \sum_{m = 1}^n w_i(v_{m - 1}, v_m) \\ & \ge \sum_{m = 1}^n 2w_{i - 1}(v_{m - 1}, v_m) \\ & \ge 2\delta_{i - 1}(s, v). \end{aligned}
d. Note that every quantity in the definition of $\hat w_i$ is an integer, so $\hat w_i$ is clearly an integer. Since $w_i(u, v) \ge 2w_{i - 1}(u, v)$, it will suffice to show that $w_{i - 1}(u, v) + \delta_{i - 1}(s, u) \ge \delta_{i - 1}(s, v)$ to prove nonnegativity. This follows immediately from the triangle inequality.
e. First note that $s = v_0, v_1, \dots, v_n = v$ is a shortest path from $s$ to $v$ with respect to $\hatw$ if and only if it is a shortest path with respect to $w$. Then we have
\begin{aligned} \hat\delta_i(s, v) & = \sum_{m = 1}^n w_i(v_{m - 1}, v_m) + 2\delta_{i - 1}(s, v_{m - 1}) − 2\delta_{i - 1}(s, v_m) \\ & = \sum_{m = 1}^n w_i(v_{m - 1}, v_m) − 2\delta_{i - 1}(s, v_n) \\ & = \delta_i(s, v) − 2\delta_{i - 1}(s, v). \end{aligned}
f. By part (a) we can compute $\hat\delta_i(s, v)$ for all $v \in V$ in $O(E)$ time. If we have already computed $\delta_i - 1$ then we can compute $\delta_i$ in $O(E)$ time. Since we can compute $\delta_1$ in $O(E)$ by part b, we can compute $\delta_i$ from scratch in $O(iE)$ time. Thus, we can compute $\delta = \delta_k$ in $O(Ek) = O(E\lg W)$ time. | 2,127 | 5,706 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-49 | latest | en | 0.835271 |
http://perplexus.info/show.php?pid=10795&cid=58034 | 1,597,351,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00239.warc.gz | 71,109,325 | 5,925 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
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Traveling with Luggage (Posted on 2016-12-15)
Alex, Bert and Carl are traveling, each with their own luggage. They have just collected their luggage after arriving from their flight. They need to get themselves and their luggage to the hotel. When they go to the car rental company, the company tells them that the only car available is a tiny compact car.
The car can hold either two people or one person plus one set of luggage. Obviuosly multiple back-and-forth trips are needed to get everyone and everything to the hotel.
The airport and hotel are trustworthy places but the three men do not trust each other. Each man will ransack the luggage of another person if left alone with the luggage. Specifically Alex will ransack Bert's luggage, Bert will ransack Carl's luggage, and Carl will ransack Alex's luggage. Ransacking can occur at the airport, the hotel, or in the car. The inverse relations will not result in a ransack (Alex won't ransack Carl's luggage, etc.)
Is it possible for Alex, Bert, and Carl to get themselves and all their luggage to the hotel without anybody having an opportunity to ransack any of the luggage? If so then how?
No Solution Yet Submitted by Brian Smith Rating: 3.0000 (1 votes)
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outline of "playing the solution" as an abstract executable Comment 2 of 2 |
We can evaluate the formula for the ransack situations at all places in a truth-table.
((~B & ~C & A & b) v (B & C & ~A & ~b)) v
((~C & ~A & B & c) v (C & A & ~B & ~c)) v
((~A & ~B & C & a) v (A & B & ~C & ~a))
We eliminate the lines (24 out of 64) showing the value "1" in the main column (under the main connective, which is the second last "v" for "OR"), because this means "danger" or "ransack situation". This gives us the 40 situations (see below) with the admissible distribution of our 6 "objects", being the assignment of the truth-values under the variables A,B,C,a,b,c: "1" = "is there"; "0" = "is not there".
Each line of the upper section has its negative image in the lower section (separated by a blank line). We can even write it side by side (as shown). Putting together pairs of positive-negative value sequences represents the situations at the airport and at the hotel.
Now we could do ourselves the steps for a "transport program". This would be like playing the abstract executable program, "guessing" a successful path in a search tree.
At the start, everyone and everything is at the airport: 111111 | 000000.
At the end, everyone and everything is at the hotel: 000000 | 111111.
To find an admissible sequence of transports between those two situations is making choices in the search tree. Here we have to consider the given condition that the car does not hold more than two people or one person plus one set of luggage.
Each reversal of the values "1" or "0" between two situations signifies the transport of someone or something. The first step cannot be a reversal of the values "1" under a,b,c alone or in a combination of it, because this would mean that the luggage drives without a person. We find the very first reasonable step in line 10:
110110 | 001001
meaning: C drives with his luggage c to the hotel.
Then, C drives back alone (step 2). Now we are in line 2.
The continuation of this "program play" is not perfect and not safe (in contrary to the elaborate and impressive program solution made by Charlie).
Airport | Hotel Car transport
|
A B C a b c | A B C a b c
-------------+---------------------------------------------------------------------
1 1 1 1 1 1 | 0 0 0 0 0 0 START situation
1 1 1 1 1 0 | 0 0 0 0 0 1 Step 2
1 1 1 1 0 1 | 0 0 0 0 1 0
1 1 1 1 0 0 | 0 0 0 0 1 1
1 1 1 0 1 1 | 0 0 0 1 0 0
1 1 1 0 1 0 | 0 0 0 1 0 1
1 1 1 0 0 1 | 0 0 0 1 1 0
1 1 1 0 0 0 | 0 0 0 1 1 1
1 1 0 1 1 1 | 0 0 1 0 0 0
1 1 0 1 1 0 | 0 0 1 0 0 1 Step 1
1 1 0 1 0 1 | 0 0 1 0 1 0
1 1 0 1 0 0 | 0 0 1 0 1 1
1 0 1 1 1 1 | 0 1 0 0 0 0
1 0 1 1 0 1 | 0 1 0 0 1 0
1 0 1 0 1 1 | 0 1 0 1 0 0
1 0 1 0 0 1 | 0 1 0 1 1 0
1 0 0 1 0 1 | 0 1 1 0 1 0
1 0 0 1 0 0 | 0 1 1 0 1 1
1 0 0 0 0 1 | 0 1 1 1 1 0
1 0 0 0 0 0 | 0 1 1 1 1 1
0 1 1 1 1 1 | 1 0 0 0 0 0
0 1 1 1 1 0 | 1 0 0 0 0 1
0 1 1 0 1 1 | 1 0 0 1 0 0
0 1 1 0 1 0 | 1 0 0 1 0 1
0 1 0 1 1 0 | 1 0 1 0 0 1
0 1 0 1 0 0 | 1 0 1 0 1 1
0 1 0 0 1 0 | 1 0 1 1 0 1
0 1 0 0 0 0 | 1 0 1 1 1 1
0 0 1 0 1 1 | 1 1 0 1 0 0
0 0 1 0 1 0 | 1 1 0 1 0 1
0 0 1 0 0 1 | 1 1 0 1 1 0
0 0 1 0 0 0 | 1 1 0 1 1 1
0 0 0 1 1 1 | 1 1 1 0 0 0
0 0 0 1 1 0 | 1 1 1 0 0 1
0 0 0 1 0 1 | 1 1 1 0 1 0
0 0 0 1 0 0 | 1 1 1 0 1 1
0 0 0 0 1 1 | 1 1 1 1 0 0
0 0 0 0 1 0 | 1 1 1 1 0 1
0 0 0 0 0 1 | 1 1 1 1 1 0
0 0 0 0 0 0 | 1 1 1 1 1 1 END situation
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Posted by ollie on 2016-12-18 13:08:53
Search: Search body:
Forums (0) | 2,228 | 5,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-34 | latest | en | 0.913842 |
https://en.wikipedia.org/wiki/Student%27s_t_distribution | 1,550,338,690,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480905.29/warc/CC-MAIN-20190216170210-20190216192210-00018.warc.gz | 567,664,356 | 68,899 | # Student's t-distribution
(Redirected from Student's t distribution)
Parameters Probability density function Cumulative distribution function ${\displaystyle \nu >0}$ degrees of freedom (real) ${\displaystyle x\in (-\infty ,\infty )}$ ${\displaystyle \textstyle {\frac {\Gamma \left({\frac {\nu +1}{2}}\right)}{{\sqrt {\nu \pi }}\,\Gamma \left({\frac {\nu }{2}}\right)}}\left(1+{\frac {x^{2}}{\nu }}\right)^{-{\frac {\nu +1}{2}}}\!}$ ${\displaystyle {\begin{matrix}{\frac {1}{2}}+x\Gamma \left({\frac {\nu +1}{2}}\right)\times \\[0.5em]{\frac {\,_{2}F_{1}\left({\frac {1}{2}},{\frac {\nu +1}{2}};{\frac {3}{2}};-{\frac {x^{2}}{\nu }}\right)}{{\sqrt {\pi \nu }}\,\Gamma \left({\frac {\nu }{2}}\right)}}\end{matrix}}}$ where 2F1 is the hypergeometric function 0 for ${\displaystyle \nu >1}$, otherwise undefined 0 0 ${\displaystyle \textstyle {\frac {\nu }{\nu -2}}}$ for ${\displaystyle \nu >2}$, ∞ for ${\displaystyle 1<\nu \leq 2}$, otherwise undefined 0 for ${\displaystyle \nu >3}$, otherwise undefined ${\displaystyle \textstyle {\frac {6}{\nu -4}}}$ for ${\displaystyle \nu >4}$, ∞ for ${\displaystyle 2<\nu \leq 4}$, otherwise undefined ${\displaystyle {\begin{matrix}{\frac {\nu +1}{2}}\left[\psi \left({\frac {1+\nu }{2}}\right)-\psi \left({\frac {\nu }{2}}\right)\right]\\[0.5em]+\ln {\left[{\sqrt {\nu }}B\left({\frac {\nu }{2}},{\frac {1}{2}}\right)\right]}\,{\scriptstyle {\text{(nats)}}}\end{matrix}}}$ undefined ${\displaystyle \textstyle {\frac {K_{\nu /2}\left({\sqrt {\nu }}|t|\right)\cdot \left({\sqrt {\nu }}|t|\right)^{\nu /2}}{\Gamma (\nu /2)2^{\nu /2-1}}}}$ for ${\displaystyle \nu >0}$ ${\displaystyle K_{\nu }(x)}$: modified Bessel function of the second kind[1]
In probability and statistics, Student's t-distribution (or simply the t-distribution) is any member of a family of continuous probability distributions that arises when estimating the mean of a normally distributed population in situations where the sample size is small and population standard deviation is unknown. It was developed by William Sealy Gosset under the pseudonym Student.
The t-distribution plays a role in a number of widely used statistical analyses, including Student's t-test for assessing the statistical significance of the difference between two sample means, the construction of confidence intervals for the difference between two population means, and in linear regression analysis. The Student's t-distribution also arises in the Bayesian analysis of data from a normal family.
If we take a sample of n observations from a normal distribution, then the t-distribution with ${\displaystyle \nu =n-1}$ degrees of freedom can be defined as the distribution of the location of the sample mean relative to the true mean, divided by the sample standard deviation, after multiplying by the standardizing term ${\displaystyle {\sqrt {n}}}$. In this way, the t-distribution can be used to construct a confidence interval for the true mean.
The t-distribution is symmetric and bell-shaped, like the normal distribution, but has heavier tails, meaning that it is more prone to producing values that fall far from its mean. This makes it useful for understanding the statistical behavior of certain types of ratios of random quantities, in which variation in the denominator is amplified and may produce outlying values when the denominator of the ratio falls close to zero. The Student's t-distribution is a special case of the generalised hyperbolic distribution.
## History and etymology
Statistician William Sealy Gosset, known as "Student"
In statistics, the t-distribution was first derived as a posterior distribution in 1876 by Helmert[2][3][4] and Lüroth.[5][6][7] The t-distribution also appeared in a more general form as Pearson Type IV distribution in Karl Pearson's 1895 paper.
In the English-language literature the distribution takes its name from William Sealy Gosset's 1908 paper in Biometrika under the pseudonym "Student".[8][9] Gosset worked at the Guinness Brewery in Dublin, Ireland, and was interested in the problems of small samples – for example, the chemical properties of barley where sample sizes might be as few as 3. One version of the origin of the pseudonym is that Gosset's employer preferred staff to use pen names when publishing scientific papers instead of their real name, so he used the name "Student" to hide his identity. Another version is that Guinness did not want their competitors to know that they were using the t-test to determine the quality of raw material.[10][11]
Gosset's paper refers to the distribution as the "frequency distribution of standard deviations of samples drawn from a normal population". It became well-known through the work of Ronald Fisher, who called the distribution "Student's distribution" and represented the test value with the letter t.[12][13]
## How Student's distribution arises from sampling
Let ${\textstyle X_{1},\ldots ,X_{n}}$ be independent and identically distributed as ${\displaystyle N(\mu ,\sigma ^{2})}$, i.e. this is a sample of size ${\displaystyle n}$ from a normally distributed population with expected mean value ${\displaystyle \mu }$ and variance ${\displaystyle \sigma ^{2}}$.
Let
${\displaystyle {\bar {X}}={\frac {1}{n}}\sum _{i=1}^{n}X_{i}}$
be the sample mean and let
${\displaystyle S^{2}={\frac {1}{n-1}}\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}$
be the (Bessel-corrected) sample variance. Then the random variable
${\displaystyle {\frac {{\bar {X}}-\mu }{\sigma /{\sqrt {n}}}}}$
has a standard normal distribution (i.e. normal with expected value 0 and variance 1), and the random variable
${\displaystyle {\frac {{\bar {X}}-\mu }{S/{\sqrt {n}}}},}$
where ${\displaystyle S}$ has been substituted for ${\displaystyle \sigma }$, has a Student's t-distribution with ${\displaystyle n-1}$ degrees of freedom. The numerator and the denominator in the preceding expression are independent random variables despite being based on the same sample ${\textstyle X_{1},\ldots ,X_{n}}$.
## Definition
### Probability density function
Student's t-distribution has the probability density function given by
${\displaystyle f(t)={\frac {\Gamma ({\frac {\nu +1}{2}})}{{\sqrt {\nu \pi }}\,\Gamma ({\frac {\nu }{2}})}}\left(1+{\frac {t^{2}}{\nu }}\right)^{\!-{\frac {\nu +1}{2}}},\!}$
where ${\displaystyle \nu }$ is the number of degrees of freedom and ${\displaystyle \Gamma }$ is the gamma function. This may also be written as
${\displaystyle f(t)={\frac {1}{{\sqrt {\nu }}\,\mathrm {B} ({\frac {1}{2}},{\frac {\nu }{2}})}}\left(1+{\frac {t^{2}}{\nu }}\right)^{\!-{\frac {\nu +1}{2}}}\!,}$
where B is the Beta function. In particular for integer valued degrees of freedom ${\displaystyle \nu }$ we have:
For ${\displaystyle \nu >1}$ even,
${\displaystyle {\frac {\Gamma ({\frac {\nu +1}{2}})}{{\sqrt {\nu \pi }}\,\Gamma ({\frac {\nu }{2}})}}={\frac {(\nu -1)(\nu -3)\cdots 5\cdot 3}{2{\sqrt {\nu }}(\nu -2)(\nu -4)\cdots 4\cdot 2\,}}\cdot }$
For ${\displaystyle \nu >1}$ odd,
${\displaystyle {\frac {\Gamma ({\frac {\nu +1}{2}})}{{\sqrt {\nu \pi }}\,\Gamma ({\frac {\nu }{2}})}}={\frac {(\nu -1)(\nu -3)\cdots 4\cdot 2}{\pi {\sqrt {\nu }}(\nu -2)(\nu -4)\cdots 5\cdot 3\,}}\cdot \!}$
The probability density function is symmetric, and its overall shape resembles the bell shape of a normally distributed variable with mean 0 and variance 1, except that it is a bit lower and wider. As the number of degrees of freedom grows, the t-distribution approaches the normal distribution with mean 0 and variance 1. For this reason ${\displaystyle {\nu }}$ is also known as the normality parameter.[14]
The following images show the density of the t-distribution for increasing values of ${\displaystyle \nu }$. The normal distribution is shown as a blue line for comparison. Note that the t-distribution (red line) becomes closer to the normal distribution as ${\displaystyle \nu }$ increases.
1 degree of freedom 2 degrees of freedom 3 degrees of freedom 5 degrees of freedom 10 degrees of freedom 30 degrees of freedom
### Cumulative distribution function
The cumulative distribution function can be written in terms of I, the regularized incomplete beta function. For t > 0,[15]
${\displaystyle F(t)=\int _{-\infty }^{t}f(u)\,du=1-{\tfrac {1}{2}}I_{x(t)}\left({\tfrac {\nu }{2}},{\tfrac {1}{2}}\right),}$
where
${\displaystyle x(t)={\frac {\nu }{t^{2}+\nu }}.}$
Other values would be obtained by symmetry. An alternative formula, valid for ${\displaystyle t^{2}<\nu }$, is[15]
${\displaystyle \int _{-\infty }^{t}f(u)\,du={\tfrac {1}{2}}+t{\frac {\Gamma \left({\tfrac {1}{2}}(\nu +1)\right)}{{\sqrt {\pi \nu }}\,\Gamma \left({\tfrac {\nu }{2}}\right)}}\,{}_{2}F_{1}\left({\tfrac {1}{2}},{\tfrac {1}{2}}(\nu +1);{\tfrac {3}{2}};-{\tfrac {t^{2}}{\nu }}\right),}$
where 2F1 is a particular case of the hypergeometric function.
For information on its inverse cumulative distribution function, see quantile function#Student's t-distribution.
### Special cases
Certain values of ${\displaystyle \nu }$ give an especially simple form.
• ${\displaystyle \nu =1}$
Distribution function:
${\displaystyle F(t)={\tfrac {1}{2}}+{\tfrac {1}{\pi }}\arctan(t).}$
Density function:
${\displaystyle f(t)={\frac {1}{\pi (1+t^{2})}}.}$
See Cauchy distribution
• ${\displaystyle \nu =2}$
Distribution function:
${\displaystyle F(t)={\tfrac {1}{2}}+{\frac {t}{2{\sqrt {2+t^{2}}}}}.}$
Density function:
${\displaystyle f(t)={\frac {1}{\left(2+t^{2}\right)^{\frac {3}{2}}}}.}$
• ${\displaystyle \nu =3}$
Density function:
${\displaystyle f(t)={\frac {6{\sqrt {3}}}{\pi \left(3+t^{2}\right)^{2}}}.}$
• ${\displaystyle \nu =\infty }$
Density function:
${\displaystyle f(t)={\frac {1}{\sqrt {2\pi }}}e^{-{\frac {t^{2}}{2}}}.}$
See Normal distribution
## How the t-distribution arises
### Sampling distribution
Let ${\displaystyle x_{1},\cdots ,x_{n}}$ be the numbers observed in a sample from a continuously distributed population with expected value ${\displaystyle \mu }$. The sample mean and sample variance are given by:
{\displaystyle {\begin{aligned}{\bar {x}}&={\frac {x_{1}+\cdots +x_{n}}{n}},\\s^{2}&={\frac {1}{n-1}}\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}.\end{aligned}}}
The resulting t-value is
${\displaystyle t={\frac {{\bar {x}}-\mu }{s/{\sqrt {n}}}}.}$
The t-distribution with ${\displaystyle n-1}$ degrees of freedom is the sampling distribution of the t-value when the samples consist of independent identically distributed observations from a normally distributed population. Thus for inference purposes t is a useful "pivotal quantity" in the case when the mean and variance ${\displaystyle (\mu ,\sigma ^{2})}$ are unknown population parameters, in the sense that the t-value has then a probability distribution that depends on neither ${\displaystyle \mu }$ nor ${\displaystyle \sigma ^{2}}$.
### Bayesian inference
In Bayesian statistics, a (scaled, shifted) t-distribution arises as the marginal distribution of the unknown mean of a normal distribution, when the dependence on an unknown variance has been marginalised out:[16]
{\displaystyle {\begin{aligned}p(\mu \mid D,I)=&\int p(\mu ,\sigma ^{2}\mid D,I)\,d\sigma ^{2}\\=&\int p(\mu \mid D,\sigma ^{2},I)\,p(\sigma ^{2}\mid D,I)\,d\sigma ^{2},\end{aligned}}}
where ${\displaystyle D}$ stands for the data ${\displaystyle \{x_{i}\}}$, and ${\displaystyle I}$ represents any other information that may have been used to create the model. The distribution is thus the compounding of the conditional distribution of ${\displaystyle \mu }$ given the data and ${\displaystyle \sigma ^{2}}$ with the marginal distribution of ${\displaystyle \sigma ^{2}}$ given the data.
With ${\displaystyle n}$ data points, if uninformative, or flat, location and scale priors ${\displaystyle p(\mu \mid \sigma ^{2},I)={\text{const}}}$ and ${\displaystyle p(\sigma ^{2}\mid I)\propto 1/\sigma ^{2}}$ can be taken for μ and σ2, then Bayes' theorem gives
{\displaystyle {\begin{aligned}p(\mu \mid D,\sigma ^{2},I)&\sim N({\bar {x}},\sigma ^{2}/n),\\p(\sigma ^{2}\mid D,I)&\sim \operatorname {Scale-inv-} \chi ^{2}(\nu ,s^{2}),\end{aligned}}}
a normal distribution and a scaled inverse chi-squared distribution respectively, where ${\displaystyle \nu =n-1}$ and
${\displaystyle s^{2}=\sum {\frac {(x_{i}-{\bar {x}})^{2}}{n-1}}.}$
The marginalisation integral thus becomes
{\displaystyle {\begin{aligned}p(\mu \mid D,I)&\propto \int _{0}^{\infty }{\frac {1}{\sqrt {\sigma ^{2}}}}\exp \left(-{\frac {1}{2\sigma ^{2}}}n(\mu -{\bar {x}})^{2}\right)\cdot \sigma ^{-\nu -2}\exp(-\nu s^{2}/2\sigma ^{2})\,d\sigma ^{2}\\&\propto \int _{0}^{\infty }\sigma ^{-\nu -3}\exp \left(-{\frac {1}{2\sigma ^{2}}}\left(n(\mu -{\bar {x}})^{2}+\nu s^{2}\right)\right)\,d\sigma ^{2}.\end{aligned}}}
This can be evaluated by substituting ${\displaystyle z=A/2\sigma ^{2}}$, where ${\displaystyle A=n(\mu -{\bar {x}})^{2}+\nu s^{2}}$, giving
${\displaystyle dz=-{\frac {A}{2\sigma ^{4}}}\,d\sigma ^{2},}$
so
${\displaystyle p(\mu \mid D,I)\propto A^{-{\frac {\nu +1}{2}}}\int _{0}^{\infty }z^{(\nu -1)/2}\exp(-z)\,dz.}$
But the z integral is now a standard Gamma integral, which evaluates to a constant, leaving
{\displaystyle {\begin{aligned}p(\mu \mid D,I)&\propto A^{-{\frac {\nu +1}{2}}}\\&\propto \left(1+{\frac {n(\mu -{\bar {x}})^{2}}{\nu s^{2}}}\right)^{-{\frac {\nu +1}{2}}}.\end{aligned}}}
This is a form of the t-distribution with an explicit scaling and shifting that will be explored in more detail in a further section below. It can be related to the standardised t-distribution by the substitution
${\displaystyle t={\frac {\mu -{\bar {x}}}{s/{\sqrt {n}}}}.}$
The derivation above has been presented for the case of uninformative priors for ${\displaystyle \mu }$ and ${\displaystyle \sigma ^{2}}$; but it will be apparent that any priors that lead to a normal distribution being compounded with a scaled inverse chi-squared distribution will lead to a t-distribution with scaling and shifting for ${\displaystyle P(\mu \mid D,I)}$, although the scaling parameter corresponding to ${\displaystyle {\frac {s^{2}}{n}}}$ above will then be influenced both by the prior information and the data, rather than just by the data as above.
## Characterization
### As the distribution of a test statistic
Student's t-distribution with ${\displaystyle \nu }$ degrees of freedom can be defined as the distribution of the random variable T with[15][17]
${\displaystyle T={\frac {Z}{\sqrt {V/\nu }}}=Z{\sqrt {\frac {\nu }{V}}},}$
where
A different distribution is defined as that of the random variable defined, for a given constant μ, by
${\displaystyle (Z+\mu ){\sqrt {\frac {\nu }{V}}}.}$
This random variable has a noncentral t-distribution with noncentrality parameter μ. This distribution is important in studies of the power of Student's t-test.
#### Derivation
Suppose X1, ..., Xn are independent realizations of the normally-distributed, random variable X, which has an expected value μ and variance σ2. Let
${\displaystyle {\overline {X}}_{n}={\frac {1}{n}}(X_{1}+\cdots +X_{n})}$
be the sample mean, and
${\displaystyle S_{n}^{2}={\frac {1}{n-1}}\sum _{i=1}^{n}\left(X_{i}-{\overline {X}}_{n}\right)^{2}}$
be an unbiased estimate of the variance from the sample. It can be shown that the random variable
${\displaystyle V=(n-1){\frac {S_{n}^{2}}{\sigma ^{2}}}}$
has a chi-squared distribution with ${\displaystyle \nu =n-1}$ degrees of freedom (by Cochran's theorem).[18] It is readily shown that the quantity
${\displaystyle Z=\left({\overline {X}}_{n}-\mu \right){\frac {\sqrt {n}}{\sigma }}}$
is normally distributed with mean 0 and variance 1, since the sample mean ${\displaystyle {\overline {X}}_{n}}$ is normally distributed with mean μ and variance σ2/n. Moreover, it is possible to show that these two random variables (the normally distributed one Z and the chi-squared-distributed one V) are independent. Consequently[clarification needed] the pivotal quantity
${\displaystyle T\equiv {\frac {Z}{\sqrt {V/\nu }}}=\left({\overline {X}}_{n}-\mu \right){\frac {\sqrt {n}}{S_{n}}},}$
which differs from Z in that the exact standard deviation σ is replaced by the random variable Sn, has a Student's t-distribution as defined above. Notice that the unknown population variance σ2 does not appear in T, since it was in both the numerator and the denominator, so it canceled. Gosset intuitively obtained the probability density function stated above, with ${\displaystyle \nu }$ equal to n − 1, and Fisher proved it in 1925.[12]
The distribution of the test statistic T depends on ${\displaystyle \nu }$, but not μ or σ; the lack of dependence on μ and σ is what makes the t-distribution important in both theory and practice.
### As a maximum entropy distribution
Student's t-distribution is the maximum entropy probability distribution for a random variate X for which ${\displaystyle \operatorname {E} (\ln(\nu +X^{2}))}$ is fixed.[19]
## Properties
### Moments
For ${\displaystyle \nu >1}$, the raw moments of the t-distribution are
${\displaystyle E(T^{k})={\begin{cases}0&k{\text{ odd}},\quad 0
Moments of order ${\displaystyle \nu }$ or higher do not exist.[20]
The term for ${\displaystyle 0, k even, may be simplified using the properties of the gamma function to
${\displaystyle E(T^{k})=\nu ^{\frac {k}{2}}\,\prod _{i=1}^{\frac {k}{2}}{\frac {2i-1}{\nu -2i}}\qquad k{\text{ even}},\quad 0
For a t-distribution with ${\displaystyle \nu }$ degrees of freedom, the expected value is 0 if ${\displaystyle \nu >1}$, and its variance is ${\displaystyle {\frac {\nu }{\nu -2}}}$ if ${\displaystyle \nu >2}$. The skewness is 0 if ${\displaystyle \nu >3}$ and the excess kurtosis is ${\displaystyle {\frac {6}{\nu -4}}}$ if ${\displaystyle \nu >4}$.
### Monte Carlo sampling
There are various approaches to constructing random samples from the Student's t-distribution. The matter depends on whether the samples are required on a stand-alone basis, or are to be constructed by application of a quantile function to uniform samples; e.g., in the multi-dimensional applications basis of copula-dependency.[citation needed] In the case of stand-alone sampling, an extension of the Box–Muller method and its polar form is easily deployed.[21] It has the merit that it applies equally well to all real positive degrees of freedom, ν, while many other candidate methods fail if ν is close to zero.[21]
### Integral of Student's probability density function and p-value
The function A(t | ν) is the integral of Student's probability density function, f(t) between −t and t, for t ≥ 0. It thus gives the probability that a value of t less than that calculated from observed data would occur by chance. Therefore, the function A(t | ν) can be used when testing whether the difference between the means of two sets of data is statistically significant, by calculating the corresponding value of t and the probability of its occurrence if the two sets of data were drawn from the same population. This is used in a variety of situations, particularly in t-tests. For the statistic t, with ν degrees of freedom, A(t | ν) is the probability that t would be less than the observed value if the two means were the same (provided that the smaller mean is subtracted from the larger, so that t ≥ 0). It can be easily calculated from the cumulative distribution function Fν(t) of the t-distribution:
${\displaystyle A(t\mid \nu )=F_{\nu }(t)-F_{\nu }(-t)=1-I_{\frac {\nu }{\nu +t^{2}}}\left({\frac {\nu }{2}},{\frac {1}{2}}\right),}$
where Ix is the regularized incomplete beta function (ab).
For statistical hypothesis testing this function is used to construct the p-value.
## Generalized Student's t-distribution
### In terms of scaling parameter σ, or σ2
Student's t distribution can be generalized to a three parameter location-scale family, introducing a location parameter ${\displaystyle \mu }$ and a scale parameter ${\displaystyle \sigma }$, through the relation
${\displaystyle X=\mu +\sigma T}$
or
${\displaystyle T={\frac {X-\mu }{\sigma }}}$
This means that ${\displaystyle {\frac {x-\mu }{\sigma }}}$ has a classic Student's t distribution with ${\displaystyle \nu }$ degrees of freedom.
The resulting non-standardized Student's t-distribution has a density defined by[22]
${\displaystyle p(x\mid \nu ,\mu ,\sigma )={\frac {\Gamma ({\frac {\nu +1}{2}})}{\Gamma ({\frac {\nu }{2}}){\sqrt {\pi \nu }}\sigma }}\left(1+{\frac {1}{\nu }}\left({\frac {x-\mu }{\sigma }}\right)^{2}\right)^{-{\frac {\nu +1}{2}}}}$
Here, ${\displaystyle \sigma }$ does not correspond to a standard deviation: it is not the standard deviation of the scaled t distribution, which may not even exist; nor is it the standard deviation of the underlying normal distribution, which is unknown. ${\displaystyle \sigma }$ simply sets the overall scaling of the distribution. In the Bayesian derivation of the marginal distribution of an unknown normal mean ${\displaystyle \mu }$ above, ${\displaystyle \sigma }$ as used here corresponds to the quantity ${\displaystyle \scriptstyle {s/{\sqrt {n}}}}$, where
${\displaystyle s^{2}=\sum {\frac {(x_{i}-{\bar {x}})^{2}}{n-1}}.}$
Equivalently, the distribution can be written in terms of ${\displaystyle \sigma ^{2}}$, the square of this scale parameter:
${\displaystyle p(x\mid \nu ,\mu ,\sigma ^{2})={\frac {\Gamma ({\frac {\nu +1}{2}})}{\Gamma ({\frac {\nu }{2}}){\sqrt {\pi \nu \sigma ^{2}}}}}\left(1+{\frac {1}{\nu }}{\frac {(x-\mu )^{2}}{\sigma ^{2}}}\right)^{-{\frac {\nu +1}{2}}}}$
Other properties of this version of the distribution are:[22]
{\displaystyle {\begin{aligned}\operatorname {E} (X)&=\mu &{\text{for }}\,\nu >1,\\\operatorname {var} (X)&=\sigma ^{2}{\frac {\nu }{\nu -2}}&{\text{for }}\,\nu >2,\\\operatorname {mode} (X)&=\mu .\end{aligned}}}
This distribution results from compounding a Gaussian distribution (normal distribution) with mean ${\displaystyle \mu }$ and unknown variance, with an inverse gamma distribution placed over the variance with parameters ${\displaystyle a=\nu /2}$ and ${\displaystyle b=\nu \sigma ^{2}/2}$. In other words, the random variable X is assumed to have a Gaussian distribution with an unknown variance distributed as inverse gamma, and then the variance is marginalized out (integrated out). The reason for the usefulness of this characterization is that the inverse gamma distribution is the conjugate prior distribution of the variance of a Gaussian distribution. As a result, the non-standardized Student's t-distribution arises naturally in many Bayesian inference problems. See below.
Equivalently, this distribution results from compounding a Gaussian distribution with a scaled-inverse-chi-squared distribution with parameters ${\displaystyle \nu }$ and ${\displaystyle \sigma ^{2}}$. The scaled-inverse-chi-squared distribution is exactly the same distribution as the inverse gamma distribution, but with a different parameterization, i.e. ${\displaystyle \nu =2a,\sigma ^{2}=b/a}$.
### In terms of inverse scaling parameter λ
An alternative parameterization in terms of an inverse scaling parameter ${\displaystyle \lambda }$ (analogous to the way precision is the reciprocal of variance), defined by the relation ${\displaystyle \lambda ={\frac {1}{\sigma ^{2}}}}$. Then the density is defined by[23]
${\displaystyle p(x|\nu ,\mu ,\lambda )={\frac {\Gamma ({\frac {\nu +1}{2}})}{\Gamma ({\frac {\nu }{2}})}}\left({\frac {\lambda }{\pi \nu }}\right)^{\frac {1}{2}}\left(1+{\frac {\lambda (x-\mu )^{2}}{\nu }}\right)^{-{\frac {\nu +1}{2}}}.}$
Other properties of this version of the distribution are:[23]
{\displaystyle {\begin{aligned}\operatorname {E} (X)&=\mu &&{\text{for }}\,\nu >1,\\[5pt]\operatorname {var} (X)&={\frac {1}{\lambda }}{\frac {\nu }{\nu -2}}\,&&{\text{for }}\,\nu >2,\\[5pt]\operatorname {mode} (X)&=\mu .\end{aligned}}}
This distribution results from compounding a Gaussian distribution with mean ${\displaystyle \mu }$ and unknown precision (the reciprocal of the variance), with a gamma distribution placed over the precision with parameters ${\displaystyle a=\nu /2}$ and ${\displaystyle b=\nu /(2\lambda )}$. In other words, the random variable X is assumed to have a normal distribution with an unknown precision distributed as gamma, and then this is marginalized over the gamma distribution.
## Related distributions
• If ${\displaystyle X}$ has a Student's t-distribution with degree of freedom ${\displaystyle \nu }$ then X2 has an F-distribution: ${\displaystyle X^{2}\sim \mathrm {F} \left(\nu _{1}=1,\nu _{2}=\nu \right)}$
• If ${\displaystyle X}$ has a Student's t-distribution with degree of freedom ${\displaystyle \nu }$ then one can obtain a Beta distribution: ${\displaystyle {\frac {\nu }{\nu +X^{2}}}\sim \mathrm {B} \left({\frac {1}{2}},{\frac {\nu }{2}}\right)}$
• The noncentral t-distribution generalizes the t-distribution to include a location parameter. Unlike the nonstandardized t-distributions, the noncentral distributions are not symmetric (the median is not the same as the mode).
• The discrete Student's t-distribution is defined by its probability mass function at r being proportional to:[24]
${\displaystyle \prod _{j=1}^{k}{\frac {1}{(r+j+a)^{2}+b^{2}}}\quad \quad r=\ldots ,-1,0,1,\ldots .}$
Here a, b, and k are parameters. This distribution arises from the construction of a system of discrete distributions similar to that of the Pearson distributions for continuous distributions.[25]
## Uses
### In frequentist statistical inference
Student's t-distribution arises in a variety of statistical estimation problems where the goal is to estimate an unknown parameter, such as a mean value, in a setting where the data are observed with additive errors. If (as in nearly all practical statistical work) the population standard deviation of these errors is unknown and has to be estimated from the data, the t-distribution is often used to account for the extra uncertainty that results from this estimation. In most such problems, if the standard deviation of the errors were known, a normal distribution would be used instead of the t-distribution.
Confidence intervals and hypothesis tests are two statistical procedures in which the quantiles of the sampling distribution of a particular statistic (e.g. the standard score) are required. In any situation where this statistic is a linear function of the data, divided by the usual estimate of the standard deviation, the resulting quantity can be rescaled and centered to follow Student's t-distribution. Statistical analyses involving means, weighted means, and regression coefficients all lead to statistics having this form.
Quite often, textbook problems will treat the population standard deviation as if it were known and thereby avoid the need to use the Student's t-distribution. These problems are generally of two kinds: (1) those in which the sample size is so large that one may treat a data-based estimate of the variance as if it were certain, and (2) those that illustrate mathematical reasoning, in which the problem of estimating the standard deviation is temporarily ignored because that is not the point that the author or instructor is then explaining.
#### Hypothesis testing
A number of statistics can be shown to have t-distributions for samples of moderate size under null hypotheses that are of interest, so that the t-distribution forms the basis for significance tests. For example, the distribution of Spearman's rank correlation coefficient ρ, in the null case (zero correlation) is well approximated by the t distribution for sample sizes above about 20.[citation needed]
#### Confidence intervals
Suppose the number A is so chosen that
${\displaystyle \Pr(-A
when T has a t-distribution with n − 1 degrees of freedom. By symmetry, this is the same as saying that A satisfies
${\displaystyle \Pr(T
so A is the "95th percentile" of this probability distribution, or ${\displaystyle A=t_{(0.05,n-1)}}$. Then
${\displaystyle \Pr \left(-A<{\frac {{\overline {X}}_{n}-\mu }{\frac {S_{n}}{\sqrt {n}}}}
and this is equivalent to
${\displaystyle \Pr \left({\overline {X}}_{n}-A{\frac {S_{n}}{\sqrt {n}}}<\mu <{\overline {X}}_{n}+A{\frac {S_{n}}{\sqrt {n}}}\right)=0.9.}$
Therefore, the interval whose endpoints are
${\displaystyle {\overline {X}}_{n}\pm A{\frac {S_{n}}{\sqrt {n}}}}$
is a 90% confidence interval for μ. Therefore, if we find the mean of a set of observations that we can reasonably expect to have a normal distribution, we can use the t-distribution to examine whether the confidence limits on that mean include some theoretically predicted value – such as the value predicted on a null hypothesis.
It is this result that is used in the Student's t-tests: since the difference between the means of samples from two normal distributions is itself distributed normally, the t-distribution can be used to examine whether that difference can reasonably be supposed to be zero.
If the data are normally distributed, the one-sided (1 − α)-upper confidence limit (UCL) of the mean, can be calculated using the following equation:
${\displaystyle \mathrm {UCL} _{1-\alpha }={\overline {X}}_{n}+t_{\alpha ,n-1}{\frac {S_{n}}{\sqrt {n}}}.}$
The resulting UCL will be the greatest average value that will occur for a given confidence interval and population size. In other words, ${\displaystyle {\overline {X}}_{n}}$ being the mean of the set of observations, the probability that the mean of the distribution is inferior to UCL1−α is equal to the confidence level 1 − α.
#### Prediction intervals
The t-distribution can be used to construct a prediction interval for an unobserved sample from a normal distribution with unknown mean and variance.
### In Bayesian statistics
The Student's t-distribution, especially in its three-parameter (location-scale) version, arises frequently in Bayesian statistics as a result of its connection with the normal distribution. Whenever the variance of a normally distributed random variable is unknown and a conjugate prior placed over it that follows an inverse gamma distribution, the resulting marginal distribution of the variable will follow a Student's t-distribution. Equivalent constructions with the same results involve a conjugate scaled-inverse-chi-squared distribution over the variance, or a conjugate gamma distribution over the precision. If an improper prior proportional to σ−2 is placed over the variance, the t-distribution also arises. This is the case regardless of whether the mean of the normally distributed variable is known, is unknown distributed according to a conjugate normally distributed prior, or is unknown distributed according to an improper constant prior.
Related situations that also produce a t-distribution are:
### Robust parametric modeling
The t-distribution is often used as an alternative to the normal distribution as a model for data, which often has heavier tails than the normal distribution allows for; see e.g. Lange et al.[26] The classical approach was to identify outliers and exclude or downweight them in some way. However, it is not always easy to identify outliers (especially in high dimensions), and the t-distribution is a natural choice of model for such data and provides a parametric approach to robust statistics.
A Bayesian account can be found in Gelman et al.[27] The degrees of freedom parameter controls the kurtosis of the distribution and is correlated with the scale parameter. The likelihood can have multiple local maxima and, as such, it is often necessary to fix the degrees of freedom at a fairly low value and estimate the other parameters taking this as given. Some authors[citation needed] report that values between 3 and 9 are often good choices. Venables and Ripley[citation needed] suggest that a value of 5 is often a good choice.
## Table of selected values
Most statistical textbooks provide t-distribution tables. Nowadays, all statistical software, such as the R programming language, and functions available in many spreadsheet programs can compute accurate values of the t-distribution and its inverse without the need for tables. The following table lists a few selected values for t-distributions with ν degrees of freedom for a range of one-sided or two-sided critical regions. For an example of how to read this table, take the fourth row, which begins with 4; that means ν, the number of degrees of freedom, is 4 (and if we are dealing, as above, with n values with a fixed sum, n = 5). Take the fifth entry, in the column headed 95% for one-sided (90% for two-sided). The value of that entry is 2.132. Then the probability that T is less than 2.132 is 95% or Pr(−∞ < T < 2.132) = 0.95; this also means that Pr(−2.132 < T < 2.132) = 0.9.
This can be calculated by the symmetry of the distribution,
Pr(T < −2.132) = 1 − Pr(T > −2.132) = 1 − 0.95 = 0.05,
and so
Pr(−2.132 < T < 2.132) = 1 − 2(0.05) = 0.9.
Note that the last row also gives critical points: a t-distribution with infinitely many degrees of freedom is a normal distribution. (See Related distributions above).
The first column is the number of degrees of freedom.
One-sided 75% 80% 85% 90% 95% 97.5% 99% 99.5% 99.75% 99.9% 99.95%
Two-sided 50% 60% 70% 80% 90% 95% 98% 99% 99.5% 99.8% 99.9%
1 1.000 1.376 1.963 3.078 6.314 12.71 31.82 63.66 127.3 318.3 636.6
2 0.816 1.080 1.386 1.886 2.920 4.303 6.965 9.925 14.09 22.33 31.60
3 0.765 0.978 1.250 1.638 2.353 3.182 4.541 5.841 7.453 10.21 12.92
4 0.741 0.941 1.190 1.533 2.132 2.776 3.747 4.604 5.598 7.173 8.610
5 0.727 0.920 1.156 1.476 2.015 2.571 3.365 4.032 4.773 5.893 6.869
6 0.718 0.906 1.134 1.440 1.943 2.447 3.143 3.707 4.317 5.208 5.959
7 0.711 0.896 1.119 1.415 1.895 2.365 2.998 3.499 4.029 4.785 5.408
8 0.706 0.889 1.108 1.397 1.860 2.306 2.896 3.355 3.833 4.501 5.041
9 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3.250 3.690 4.297 4.781
10 0.700 0.879 1.093 1.372 1.812 2.228 2.764 3.169 3.581 4.144 4.587
11 0.697 0.876 1.088 1.363 1.796 2.201 2.718 3.106 3.497 4.025 4.437
12 0.695 0.873 1.083 1.356 1.782 2.179 2.681 3.055 3.428 3.930 4.318
13 0.694 0.870 1.079 1.350 1.771 2.160 2.650 3.012 3.372 3.852 4.221
14 0.692 0.868 1.076 1.345 1.761 2.145 2.624 2.977 3.326 3.787 4.140
15 0.691 0.866 1.074 1.341 1.753 2.131 2.602 2.947 3.286 3.733 4.073
16 0.690 0.865 1.071 1.337 1.746 2.120 2.583 2.921 3.252 3.686 4.015
17 0.689 0.863 1.069 1.333 1.740 2.110 2.567 2.898 3.222 3.646 3.965
18 0.688 0.862 1.067 1.330 1.734 2.101 2.552 2.878 3.197 3.610 3.922
19 0.688 0.861 1.066 1.328 1.729 2.093 2.539 2.861 3.174 3.579 3.883
20 0.687 0.860 1.064 1.325 1.725 2.086 2.528 2.845 3.153 3.552 3.850
21 0.686 0.859 1.063 1.323 1.721 2.080 2.518 2.831 3.135 3.527 3.819
22 0.686 0.858 1.061 1.321 1.717 2.074 2.508 2.819 3.119 3.505 3.792
23 0.685 0.858 1.060 1.319 1.714 2.069 2.500 2.807 3.104 3.485 3.767
24 0.685 0.857 1.059 1.318 1.711 2.064 2.492 2.797 3.091 3.467 3.745
25 0.684 0.856 1.058 1.316 1.708 2.060 2.485 2.787 3.078 3.450 3.725
26 0.684 0.856 1.058 1.315 1.706 2.056 2.479 2.779 3.067 3.435 3.707
27 0.684 0.855 1.057 1.314 1.703 2.052 2.473 2.771 3.057 3.421 3.690
28 0.683 0.855 1.056 1.313 1.701 2.048 2.467 2.763 3.047 3.408 3.674
29 0.683 0.854 1.055 1.311 1.699 2.045 2.462 2.756 3.038 3.396 3.659
30 0.683 0.854 1.055 1.310 1.697 2.042 2.457 2.750 3.030 3.385 3.646
40 0.681 0.851 1.050 1.303 1.684 2.021 2.423 2.704 2.971 3.307 3.551
50 0.679 0.849 1.047 1.299 1.676 2.009 2.403 2.678 2.937 3.261 3.496
60 0.679 0.848 1.045 1.296 1.671 2.000 2.390 2.660 2.915 3.232 3.460
80 0.678 0.846 1.043 1.292 1.664 1.990 2.374 2.639 2.887 3.195 3.416
100 0.677 0.845 1.042 1.290 1.660 1.984 2.364 2.626 2.871 3.174 3.390
120 0.677 0.845 1.041 1.289 1.658 1.980 2.358 2.617 2.860 3.160 3.373
0.674 0.842 1.036 1.282 1.645 1.960 2.326 2.576 2.807 3.090 3.291
The number at the beginning of each row in the table above is ν, which has been defined above as n − 1. The percentage along the top is 100%(1 − α). The numbers in the main body of the table are tα, ν. If a quantity T is distributed as a Student's t-distribution with ν degrees of freedom, then there is a probability 1 − α that T will be less than tα, ν. (Calculated as for a one-tailed or one-sided test, as opposed to a two-tailed test.)
For example, given a sample with a sample variance 2 and sample mean of 10, taken from a sample set of 11 (10 degrees of freedom), using the formula
${\displaystyle {\overline {X}}_{n}\pm A{\frac {S_{n}}{\sqrt {n}}},}$
we can determine that at 90% confidence, we have a true mean lying below
${\displaystyle 10+1.37218{\frac {\sqrt {2}}{\sqrt {11}}}=10.58510.}$
In other words, on average, 90% of the times that an upper threshold is calculated by this method, this upper threshold exceeds the true mean.
And, still at 90% confidence, we have a true mean lying over
${\displaystyle 10-1.37218{\frac {\sqrt {2}}{\sqrt {11}}}=9.41490.}$
In other words, on average, 90% of the times that a lower threshold is calculated by this method, this lower threshold lies below the true mean.
So that at 80% confidence (calculated from 1 − 2 × (1 − 90%) = 80%), we have a true mean lying within the interval
${\displaystyle \left(10-1.37218{\frac {\sqrt {2}}{\sqrt {11}}},10+1.37218{\frac {\sqrt {2}}{\sqrt {11}}}\right)=(9.41490,10.58510).}$
In other words, on average, 80% of the times that upper and lower thresholds are calculated by this method, the true mean is both below the upper threshold and above the lower threshold. This is not the same thing as saying that there is an 80% probability that the true mean lies between a particular pair of upper and lower thresholds that have been calculated by this method; see confidence interval and prosecutor's fallacy.
## Notes
1. ^ Hurst, Simon. The Characteristic Function of the Student-t Distribution, Financial Mathematics Research Report No. FMRR006-95, Statistics Research Report No. SRR044-95 Archived February 18, 2010, at the Wayback Machine
2. ^ Helmert, F. R. (1875). "Über die Bestimmung des wahrscheinlichen Fehlers aus einer endlichen Anzahl wahrer Beobachtungsfehler". Z. Math. Phys. 20: 300–3.
3. ^ Helmert, F. R. (1876a). "Über die Wahrscheinlichkeit der Potenzsummen der Beobachtungsfehler und uber einige damit in Zusammenhang stehende Fragen". Z. Math. Phys. 21: 192–218.
4. ^ Helmert, F. R. (1876b). "Die Genauigkeit der Formel von Peters zur Berechnung des wahrscheinlichen Beobachtungsfehlers directer Beobachtungen gleicher Genauigkeit". Astron. Nachr. 88: 113–32. Bibcode:1876AN.....88..113H. doi:10.1002/asna.18760880802.
5. ^ Lüroth, J (1876). "Vergleichung von zwei Werten des wahrscheinlichen Fehlers". Astron. Nachr. 87 (14): 209–20. Bibcode:1876AN.....87..209L. doi:10.1002/asna.18760871402.
6. ^ Pfanzagl, J.; Sheynin, O. (1996). "A forerunner of the t-distribution (Studies in the history of probability and statistics XLIV)". Biometrika. 83 (4): 891–898. doi:10.1093/biomet/83.4.891. MR 1766040.
7. ^ Sheynin, O. (1995). "Helmert's work in the theory of errors". Arch. Hist. Exact Sci. 49: 73–104. doi:10.1007/BF00374700.
8. ^ "Student" [William Sealy Gosset] (March 1908). "The probable error of a mean" (PDF). Biometrika. 6 (1): 1–25. doi:10.1093/biomet/6.1.1.
9. ^ "Student" (William Sealy Gosset), original Biometrika paper as a scan.
10. ^ M. Wendl (2016) Pseudonymous fame, Science, 351(6280), 1406.
11. ^ Mortimer, Robert G. (2005). Mathematics for Physical Chemistry, 3rd ed. Academic Press. ISBN 0-12-508347-5 (p. 326).
12. ^ a b Fisher, R. A. (1925). "Applications of "Student's" distribution" (PDF). Metron. 5: 90–104. Archived from the original (PDF) on 2016-03-05.
13. ^ Walpole, Ronald; Myers, Raymond; Myers, Sharon; Ye, Keying. (2002). Probability and Statistics for Engineers and Scientists, 7th edi. p. 237. Pearson Education. ISBN 81-7758-404-9
14. ^ John Kruschke (2014), Doing Bayesian Data Analysis, Academic Press; 2 edition. ISBN 0124058884
15. ^ a b c Johnson, N. L., Kotz, S., Balakrishnan, N. (1995) Continuous Univariate Distributions, Volume 2, 2nd Edition. Wiley, ISBN 0-471-58494-0 (Chapter 28).
16. ^ A. Gelman et al (1995), Bayesian Data Analysis, Chapman & Hall. ISBN 0-412-03991-5. p. 68
17. ^ Hogg & Craig (1978), Sections 4.4 and 4.8.
18. ^ Cochran, W. G. (April 1934). "The distribution of quadratic forms in a normal system, with applications to the analysis of covariance". Mathematical Proceedings of the Cambridge Philosophical Society. 30 (2): 178–191. Bibcode:1934PCPS...30..178C. doi:10.1017/S0305004100016595.
19. ^ Park, Sung Y.; Bera, Anil K. (2009). "Maximum entropy autoregressive conditional heteroskedasticity model" (PDF). Journal of Econometrics. Elsevier: 219–230. Retrieved 2011-06-02.
20. ^ See, for example, page 56 of Casella and Berger, Statistical Inference, 1990 Duxbury.
21. ^ a b Bailey, R. W. (1994). "Polar Generation of Random Variates with the t-Distribution". Mathematics of Computation. 62 (206): 779–781. doi:10.2307/2153537.
22. ^ a b Jackman, Simon (2009). Bayesian Analysis for the Social Sciences. Wiley. p. 507.
23. ^ a b Bishop, C.M. (2006). Pattern recognition and machine learning. Springer.
24. ^ Ord, J. K. (1972) Families of Frequency Distributions, Griffin. ISBN 0-85264-137-0 (Table 5.1)
25. ^ Ord, J. K. (1972) Families of Frequency Distributions, Griffin. ISBN 0-85264-137-0 (Chapter 5)
26. ^ Lange, Kenneth L.; Little, Roderick JA; Taylor, Jeremy MG (1989). "Robust statistical modeling using the t distribution". Journal of the American Statistical Association. 84 (408): 881–896. doi:10.1080/01621459.1989.10478852. JSTOR 2290063.
27. ^ Gelman, Andrew, et al. Bayesian data analysis, Chapter 12; Boca Raton, FL, USA: Chapman & Hall/CRC, 2014 | 12,629 | 41,705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 165, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-09 | latest | en | 0.579473 |
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### Messages - Xinqiao Li
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1
##### Chapter 3 / Re: Confused about the notation in maximum modulus principle
« on: December 16, 2020, 04:16:44 PM »
woww thank you, that makes sense now.
2
##### Chapter 3 / Confused about the notation in maximum modulus principle
« on: December 15, 2020, 10:20:52 AM »
Hello, this is the problem 5 from the sample exam 2020F, I am kind of confused by the notation here. What is meant by find instead $max_D Re(f(z))$ and $\in_D Re(f(z))$? Are we suppose to find the maximum of the real part of the function f on the bounded domain D?
3
##### Quiz 2 / Quiz2 LEC0101 C
« on: October 03, 2020, 05:47:44 AM »
Questions: Find all points of continuity of the given function: $f(z)=(Imz-Rez)^{-1}$
Solutions:
Let $z=x+iy$ where x,y are real numbers.
Then $f(z)=f(x,y)=(Im(x+iy)-Re(x+iy))^{-1}=(y-x)^{-1}=\frac{1}{y-x}$
The function is not valid when the denominator equals 0, that is, $y=x$.
Therefore, the function is discontinuous only at all points of the line $y=x$.
4
##### Quiz 1 / LEC0101 Quiz1
« on: September 25, 2020, 11:17:28 AM »
Problem:
Describe the locus of points z satisfying the given equation.
$|z+1|^2+2|z|^2=|z−1|^2.$
Solution:
Let $z=x+yi$ where x,y are real numbers
$|z+1|^2 = |(x+yi)+1|^2=|(x+1)+yi|^2=(x+1)^2+y^2=x^2+2x+1+y^2$
$2|z|^2=2|x+yi|^2=2(x^2+y^2)=2x^2+2y^2$
$|z-1|^2 = |(x+yi)-1|^2=|(x-1)+yi|^2=(x-1)^2+y^2=x^2-2x+1+y^2$
Hence,
$|z+1|^2+2|z|^2=|z−1|^2$
$x^2+2x+1+y^2+2x^2+2y^2=x^2-2x+1+y^2$
$2x^2+4x+2y^2=0$
$2(x^2+2x+y^2)=0$
$x^2+2x+y^2=0$
$x^2+2x+1+y^2=0$
$(x+1)^2+y^2=1$
The locus of point z is a circle centered at (-1,0) with radius 1.
5
##### Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 04:29:59 PM »
a) Find the general solution for equation $y'' - 2y' -3y = 16coshx$
b) Find solution, satisfying $y(0) = 0, y'(0) = 0$
a)
Consider $y'' - 2y' -3y = 0$
Assume $y = e^{rx}$, then the characteristic polynomial of the equation is given by $r^2 -2r -3 = 0$
This simplifies to $(r - 3)(r +1) = 0$ which gives us two roots $r_1 = 3$ and $r_2 = -1$
Then the complementary solution is $y_c = c_1e^{3x} + c_2e^{-x}$
Since $y'' - 2y' -3y = 16cosh = 8e^x + 8e^{-x}$
We guess the particular solution is of the form
$y = Ae^x + Bxe^{-x}$
$y' = Ae^x + B(e^{-x} - xe^{-x}) = Ae^x + Be^{-x} - Bxe^{-x}$
$y'' = Ae^x - Be^{-x} - B(e^{-x} - xe^{-x}) = Ae^x - 2Be^{-x} + Bxe^{-x}$
Substituting back to the original equation:
$y'' - 2y' -3y = Ae^x - 2Be^{-x} + Bxe^{-x} - 2Ae^x - 2Be^{-x} + 2Bxe^{-x} - 3Ae^x - 3Bxe^{-x} = -4Ae^x - 4Be^{-x} = 8e^x + 8e^{-x}$
Therefore, $A = -2$ and $B = -2$
The particular solution is $y_p = - 2e^x - 2xe^{-x}$
The general solution is :
$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$
b)
$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$
$$y' = 3c_1e^{3x} - c_2e^{-x} - 2e^x + 2xe^{-x} - 2e^{-x}$$
Plug in the initial conditions $y(0) = 0, y'(0) = 0$, we have,
$0 = c_1 + c_2 - 2$
$0 = 3c_1 - c_2 - 2 - 2$
so, $c_1 = \frac{3}{2}$ and $c_2 = \frac{1}{2}$
The particular solution is
$$y = \frac{3}{2}e^{3x} + \frac{1}{2}e^{-x} - 2e^x - 2xe^{-x}$$
6
##### Term Test 1 / Re: Problem 1 (main sitting)
« on: October 23, 2019, 04:18:11 PM »
Here is another method to find $\varphi$. (By integrating N with respect to y)
Find integrating factor and then a general solution of the ODE
$$(y+3y^2e^{2x}) + (1+2ye^{2x})y' = 0, y(0) = 1$$
Let $M = y+3y^2e^{2x}$ and $N = 1+2ye^{2x}$
We can see that $M_y = 1 + 6e^{2x}y$ and $N_x = 4e^{2x}y$
They are not equal, so not exact. Our goal is to find an integrating factor and make $M_y$ equals $N_x$
$$R_2 = \frac{M_y - N_x}{N} = \frac{1 + 2e^{2x}y}{1+2e^{3x}y} = 1$$
So $\mu = e^{\int R_2dx} = e^{\int1dx} = e^x$
Multiply $\mu$ on both side of the orignial equation and we got
$$(e^xy+3y^2e^{3x}) + (e^x+2ye^{3x})y' = 0$$
Now $M_y = e^x + 6e^{3x}y$ and $N_x = e^x + 6e^{3x}y$
Therefore $\exists\varphi_{(x,y)}$ satisfy $\varphi_x = M$ and $\varphi_y = N$
$$\varphi = \int Ndy = \int (e^x+2ye^{3x})dy = e^xy+y^2e^{3x} + h(x)$$
Then $\varphi_x =e^xy +3y^2e^{3x} + h'(x)$
Since $\varphi_x = M = e^xy+3y^2e^{3x}$
So $h'(x) =0$ and $h(x)=c$
$$\varphi = e^xy+y^2e^{3x} = c$$
Given initial condition $y(0) = 1$
We have $1\times 1 + 1^2 \times 1 = c$ and $c = 2$
Therefore, the particular solution is
$$e^xy+y^2e^{3x} = 2$$
7
##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 04:09:34 PM »
a) Find Wronskian $W(y_1, y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE $x^2y'' -2xy' + (x^2+2)y = 0$
b) Check that $y_1(x) = xcosx$ is a solution and find another linearly independent solution.
c) Write the general solution, and find solution that $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$
a)
$$y'' - \frac{2}{x}y' + (1 + \frac{2}{x^2}) = 0$$
We see that $p(x) = -\frac{2}{x}$ is continuous everywhere except at $x=0$, $q(x) = (1 + \frac{2}{x^2})$ is continuous everywhere except at $x=0$.
Then by Abel's Theorem,
$$W(y_1, y_2)(x) = ce^{-\int p(x)dx} = ce^{\int(\frac{2}{x})dx} = ce^{2lnx} = cx^2$$
b)
Let's verify $y_1(x) = xcosx$ is a valid solution.
Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$
Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$
So $y_1(x) = xcosx$ is a solution.
Take $c = 1, W(y_1, y_2)(x) = x^2$.
By Reduction of Order, we have:
$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$
c)
Then by part b, the general solution is $y = c_1xcosx +c_2xsinx$
The derivative is given by $y' = c_1(cosx -xsinx) + c_2(xcosx + sinx)$
Plug in the initial conditions $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$, we have
$1 = \frac{\pi}{2}c_2$ and $0 = (-\frac{\pi}{2})c_1 + c_2$
So $c_2 = \frac{2}{\pi}$ and $c_1 = \frac{4}{(\pi)^2}$
The particular solution is,
$$y = \frac{4}{(\pi)^2}xcosx +\frac{2}{\pi}xsinx$$
8
##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 08:23:07 AM »
b) Forgot to check $y_1(x) = xcosx$ is a valid solution.
Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$
Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$
So $y_1(x) = xcosx$ is a solution.
9
##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 08:12:27 AM »
Could also use Reduction of Order to solve part b).
Take $c = 1, W(y_1, y_2)(x) = x^2$.
By Reduction of Order, we have:
$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$
10
##### Quiz-4 / TUT0502 Quiz4
« on: October 18, 2019, 09:59:31 PM »
Find the general solution of the given differential equation:
$$9y'' + 6y' + y = 0$$
Solve:
The characteristic polynomial is given by $9r^2 + 6r + 1 = 0$
Factor it we obtain $(3r + 1)(3r + 1)$
Then $r_1 = -\frac{1}{3}$ and $r_2 = -\frac{1}{3}$
We observe repeated roots here.
Then the general solution of the given differential equation is $y(t) = c_1e^{-\frac{1}{3} t} + c_2te^{-\frac{1}{3}t}$
11
##### Quiz-3 / TUT0502 Quiz 3
« on: October 11, 2019, 02:00:35 PM »
Find the solution of the given initial problem:
$$y''+4y'+3y = 0, y(0) = 2, y'(0) = -1$$
Assume $y = e^{rt}$, then it must follow that r is the root of the characteristic polynomial
$$r^2+4r+3=0\\ (r+1)(r+3)=0$$
We have $r_1 = -1$ or $r_2 = -3$.
The general solution of the second order differential equation has the form of
$$y = c_1e^{r_1t} + c_2e^{r_2t}$$
Thus, we have
$$y = c_1e^{-t} + c_2e^{-3t}$$
The derivative of this general solution is
$$y' = -c_1e^{-t} - 3c_2e^{-3t}$$
To satisfy both initial conditions $y(0) = 2$ and $y'(0) = -1$,
We have $2 = c_1 + c_2$ and $-1 = -c_1 -3c_2$
This gives us $c_1 = \frac{5}{2}$ and $c_2 = -\frac{1}{2}$
Therefore, the solution of the initial value problem is
$$y = \frac{5}{2}e^{-t} -\frac{1}{2}e^{-3t}$$
12
##### Chapter 3 / Re: Initial conditions evaluated at different $t_0$'s?
« on: October 05, 2019, 10:42:21 PM »
From what I understand, when you are solving the second order initial value problems, you first get a general solution of that equation with two arbitrary constant of the form y(t)=C1ert+C2ert. Then you find the derivative y' of this general solution, and plug in the initial conditions to solve what the two constants are. So you might not necessary need both initial conditions to be defined at t0.
13
##### Quiz-2 / TUT0502 Quiz2
« on: October 04, 2019, 02:00:01 PM »
Determine whether the equation given below is exact. If it is exact, find the solution
$$(e^xsin(y)-2ysin(x))-(3x-e^xsin(y))y'=0\\ (e^xsin(y)-2ysin(x))dx-(3x-e^xsin(y))dy=0$$
Let $M(x,y) = e^xsin(y)-2ysin(x)$
Let $N(x,y) = -(3x-e^xsin(y)) = e^xsin(y)-3x$
$$M_y(x,y) = \frac{\partial}{\partial y} M(x,y)= \frac{\partial}{\partial y} (e^xsin(y)-2ysin(x)) = e^xcos(y)-2sin(x)\\ N_x(x,y) = \frac{\partial}{\partial x} N(x,y) = \frac{\partial}{\partial x} (e^xsin(y)-3x) = e^xsin(y)-3$$
Clearly, we see that $e^xcos(y)-2sin(x) \neq e^xsin(y)-3$
Therefore, $M_y(x,y) \neq N_x(x,y)$
By definition of exact, we can conclude the equation is not exact.
Pages: [1] | 4,116 | 9,333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-38 | latest | en | 0.712584 |
http://bepath.com/jhnu8/fibonacci-generating-function-ad0455 | 1,627,830,305,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00487.warc.gz | 6,365,284 | 14,947 | # fibonacci generating function
The roots of the polynomial $$1 - x - x^2$$ are $$-\phi$$ and $$-\psi$$, where, $Take a look, From Ancient Egypt to Gauge Theory, the story of the groma. We’re going to derive this generating function and then use it to find a closed form for the nth Fibonacci number.$. The formula for calculating the Fibonacci Series is as follows: F(n) = F(n-1) + F(n-2) where: F(n) is the term number. \begin{align*} & = \sum_{n = 0}^\infty \psi^n x^n, \begin{align*} Wikipedia defines a generating function as. F_n Now, write the function in terms of its factors. No, we count forward, as always. & = F_{n - 1} + F_{n - 2} Do we count backward from zero to negative one? It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. All of that to whittle the right hand side to an x. Thus: When we multiply the x before the summation, all the terms on the right-hand side have the same exponent. -x \], Note that this infinite sum converges if and only if $$|x| < 1$$. F(n-1) is the previous term (n-1). Since the generating function for an{\displaystyle a^{n}}is. A monthly-or-so-ish overview of recent mathy/fizzixy articles published by MathAdam. It is now possible to define a value for the coefficient where the n term is negative. From the 3rd number onwards, the series will be the sum of the previous 2 numbers. What are the different ways to implement the Fibonacci in C#? With B, it’s because of alternating between even and odd functions. & = \sum_{n = 0}^\infty x^n. c0, c1, c2, c3, c4, c5, …. You don’t. We will create a new power series. & = \frac{1}{1 - \phi x} \\ & = x + \sum_{n = 2}^\infty F_{n - 1} x^n + \sum_{n = 2}^\infty F_{n - 2} x^n The first nine terms of g(x), (B, in the diagram), close in on the closed form where |x|<1. First, find the roots, using your favourite method. But first, we need to reimagine our closed-form function. F(x) So, re really want the absolute value of our coeffient. The generating series generates the sequence. Generating functions are useful tools with many applications to discrete mathematics. \end{align*} F(x) \sum_{n=1}^\infty F_n x^n = \frac{x}{1-x-x^2}. A Computer Science portal for geeks. Thus, our general term: Plug in an integer value for n — positive or negative — and those square roots will fit together to push out another integer. Generating functions are a bridge between discrete mathematics, on the one hand, and continuous analysis (particularly complex variable the- ... nth Fibonacci number, F n, is the coe–cient of xn in the expansion of the function x=(1 ¡x¡x2) as a power series about the origin. Here are the first few terms: The expansion of the second binomial is similar. \begin{align*} We replace φ with its conjugate. & = x \sum_{n = 2}^\infty F_{n - 1} x^{n - 1} We can derive a formula for the general term using generating functions and power series. So, our generating function for Fibonacci numbers, is equal to the sum of these two generating functions. \[, Similarly, for the second sum, we have & = A (x + \psi) + B (x + \phi). \end{align*} To keep things tidy, we use the following substitutions: We wish to express part of our function as partial fractions. = 1+x+2x2+3x3+5x4+8x5+13x6+21x7+ The advantage of this is that the function on the right is explicitly about the Fibonacci numbers, while the function on the left has nothing to do with them – we can study it even without knowing anything about the Fibonacci numbers! {\displaystyle \sum _{n,k}{\binom {n}{k}}x^{k}y^{n}={\frac {1}{1-(1+x)y}}={\frac {1}{1-y-xy}}.} \begin{align*} we match the coefficients on corresponding powers of $$x$$ in these two expressions for $$F(x)$$ to finally arrive at the desired closed form for the $$n$$-th Fibonacci number, \[ The Fibonacci Closed-Form Function … = x^2 F(x). & = \frac{1}{\sqrt{5}} \left( \phi^n - \psi^n \right). We are back to a new infinite series, which we will call f(x). Browse other questions tagged nt.number-theory reference-request co.combinatorics generating-functions or ask your own question. Generating functions are a fairly simple concept with a lot of power. & = \sum_{n = 0}^\infty F_n x^n To create our generating function, we encode the terms of our sequence as coefficients of a power series: This is our infinite Fibonacci power series., We now wish to express each of these two terms as the sum of a geometric series. This isolates the a term. & = \frac{1 + \sqrt{5}}{2} Turn the crank; out pops the stream: To create our generating function, we encode the terms of our sequence as coefficients of a power series: This is our infinite Fibonacci power series. \end{align*} where $$F_n$$ is the $$n$$-th Fibonacci number. Fibonacci We can nd the generating function for the Fibonacci numbers using the same trick! With A, it’s because of the alternating signs. How does this help us if we wish to find, say, the 100th Fibonacci number? \], We now focus on rewriting each of these two sums in terms of the generating function. & = \frac{1 - \sqrt{5}}{2}, Most of the time the known generating functions are among Deriving this identity gives an excellent glimpse of the power of generating functions. The derivation of this formula is quite accessible to anyone comfortable with algebra and geometric series. = x \sum_{n = 1}^\infty F_n x^n \end{align*} \phi Next subsection 1 Convolutions Fibonacci convolution m -fold convolution Catalan numbers 2 Exponential generating functions. Generating Function The generating function of the Fibonacci numbers is ∑ n = 1 ∞ F n x n = x 1 − x − x 2 . \psi The result is two new series which we subtract from the first: The value of this exercise becomes apparent when we apply the same technique to the expanded right-hand side. \end{align*} \], \begin{align*} 11−ay,{\displaystyle {\frac {1}{1-ay}},} the generating function for the binomial coefficients is: ∑n,k(nk)xkyn=11−(1+x)y=11−y−xy. \begin{align*} This will let us calculate an explicit formula for the n-th term of the sequence. \begin{align*} While the Fibonacci numbers are nondecreasing for non-negative arguments, the Fibonacci function possesses a single local minimum: Since the generating function is rational, these sums come out as rational numbers: \sum_{n = 2}^\infty F_{n - 1} x^n \sum_{n = 2}^\infty F_{n - 2} x^n & = \frac{1}{\sqrt{5}} \left( \frac{\psi}{x + \psi} - \frac{\phi}{x + \phi} \right) \\. In C#, we can print the Fibonacci … We will write the denominator with binomials. 15 3.5 Fibonacci Generating Function As previously stated, generating functions are used a lot in this project because we can easily see them when we start proving the different patterns. \end{align*} However, considered as a formal power series, this identity always holds. \], Now that we have found a closed form for the generating function, all that remains is to express this function as a power series. & = \frac{x}{1 - x - x^2}. \end{align*} After doing so, we may match its coefficients term-by-term with the corresponding Fibonacci numbers. Once we reverse the substitutions, we find the numerators of the partial fractions settle down nicely. & = F_{n - 1} + F_{n - 2} \]. Example − Fibonacci series − Fn=Fn−1+Fn−2, Tower of Hanoi − Fn=2Fn−1+1 A pair of newly born rabbits of opposite sexes is placed in an enclosure at … Thus it has two real roots r 1 and r 2, so it can be factored as 1 x x2 = 1 x r 1 1 x r 2 3. How to solve for a closed formula for the Fibonacci sequence using a generating function. Generating Functions and the Fibonacci Numbers Posted on November 1, 2013 Wikipedia defines a generating function as a formal power series in one indeterminate, whose coefficients encode information about a sequence of numbers an that is indexed by the natural numbers. This is a classical example of a recursive function, a function that calls itself.. F_n We transform that sum into a closed-form function. F_n We can do likewise with the binomial coefficient. \begin{align*} F(x) The following code clearly prints out the trace of the algorithm: We’ll give a different name to the closed-form function. Recall that the Fibonacci numbers are given by f 0= 0; f \end{align*} What does that even mean? We multiply by x and x². \end{align*} = \sum_{n = 1}^\infty F_n x^n, Note: the value not exceeding 4,000,000 isn’t something that we’re going to call but rather pass as a parameter in our fibonacci number generating function later. For that, we turn to the binomial theorem. c 0, c 1, c 2, c 3, c 4, c 5, …. The techniques we’ll use are applicable to a large class of recurrence equations. \end{align*} If you read it carefully, you'll see that it will call itself, or, recurse, over and over again, until it reaches the so called base case, when x <= 1 at which point it will start to "back track" and sum up the computed values. Our journey takes us from an infinite sum, in which we encode the sequence. sequence is generated by some generating function, your goal will be to write it as a sum of known generating functions, some of which may be multiplied by constants, or constants times some power of x. c0 + c1x + c2x2 + c3x3 + c4x4 + c5x5 + ⋯. The two lines nearly overlap in Quadrant I. = x^2 F(x). \begin{align*} \], Letting $$x = -\phi$$, we find that $$A = -\frac{\phi}{\sqrt{5}}$$. \begin{align*} In this section, we will find the generating functions that results in the sequence The Fibonacci numbers occur in the sums of "shallow" diagonals in Pascal's triangle (see binomial coefficient): We begin by defining the generating function for the Fibonacci numbers as the formal power series whose coefficients are the Fibonacci numbers themselves, How do you multiply two by itself one half of a time? \begin{align*} F(x) Next, we isolate the b term in like manner. Recall that the sum of a geometric series is given by, \[ \end{align*} \begin{align*} And this is exactly the Binet formula. \begin{align*}, \[ & = x^2 \sum_{n = 2}^\infty F_{n - 2} x^{n - 2} The π-th term? & = x + \sum_{n = 2}^\infty (F_{n - 1} + F_{n - 2}) x^n \\ F(n-2) is the term before that (n-2). Therefore, \[ The sum from zero to negative one? F(x) There is much more on GFs on my Fibonomials page.Replacing x by x2 in a GF inserts 0's between all values of the original series. & = x^2 \sum_{n = 2}^\infty F_{n - 2} x^{n - 2} \frac{1}{1 - x} This means that, the nth term of the Fibonacci sequence, is equal to the sum of the corresponding named nth terms of these geometric progressions, with common ratios phi and psi. So, let’s do that. A recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms (Expressing Fn as some combination of Fi with i1. The make-over will allow us to create a new-and-improved power series. \end{align*} 3. For the first sum, we have, \[ By why limit yourself to integers or even real numbers as input? & = x + \sum_{n = 2}^\infty F_n x^n \\ To shift to the right (insert a 0 at the start of the series so all other terms have an index increased by 1),multiply the GF by x; to shift to the left, divide by x. \end{align*} He noticed a pattern and raised some questions about it. & = \frac{1}{\sqrt{5}} \left( \sum_{n = 0}^\infty \phi^n x^n - \sum_{n = 0}^\infty \psi^n x^n \right) \\ But you can still apply the algebra for positive integer exponents into something that makes sense. & = \frac{1}{1 + \frac{x}{\psi}} \\ The series of even-in… Here’s how it works. \begin{align*} = x^2 \sum_{n = 0}^\infty F_n x^n … F_n F(x) As we will soon see, the partial sums of our power series, g(x), approach this new function only where |x|<1. c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 + c 5 x 5 + ⋯. First, we let x=-φ. A generating function (GF) is an infinite polynomial in powers of x where the n-th term of a series appears as the coefficient of x^(n) in the GF. \end{align*} F(x) generating functions are enough to illustrate the power of the idea, so we’ll stick to them and from now on, generating function will mean the ordinary kind. Summary Hong recently explored when the value of the generating function of the Fibonacci sequence is an integer. a formal power series in one indeterminate, whose coefficients encode information about a sequence of numbers an that is indexed by the natural numbers. & = x + x F(x) + x^2 F(x). \begin{align*} \frac{\psi}{x + \psi} = x F(x). Generate Fibonacci sequence (Simple Method) In the Fibonacci sequence except for the first two terms of the sequence, every other term is the sum of the previous two terms. Once we add a term to each of the partial sums, we see how they hop up and down. The 1000th? Our closed-form function will be h(x). \begin{align*} Our generating function now looks like this: It is our same closed-form function. \begin{align*} Featured on Meta Hot Meta Posts: Allow for removal by moderators, and thoughts about future… Then you discovered fractional exponents. Isolate the b term in like manner recent mathy/fizzixy articles published by MathAdam an explicit for! ( x ) = x n 0 f nx n be the sum of the four.... Two by itself one half of a recursive function, a function that calls itself that... } } is two numbers of the sequence what is the previous 2.! May match its coefficients term-by-term with the corresponding Fibonacci numbers may seem fairly nasty,... Always holds function in terms of its factors now, write the function in terms of factors. Section, we wish to create a corresponding closed form-function { \displaystyle a^ n... Hop up and down function for the Fibonacci sequence ) a term to each of the sequence ( the! Its coefficients term-by-term with the corresponding Fibonacci numbers using the same exponent odd functions that results in diagram! To keep things tidy, we wish to express part of our function as fractions. This identity always holds ll give a different name to the closed-form function will be h ( x =. ’ re going to derive this generating fibonacci generating function calculate an explicit formula for the Fibonacci sequence using a series. Accessible to anyone comfortable with algebra and geometric series backward from zero to negative one: is. Numerators of the Fibonacci sequence ) recursive function, a function that calls itself isolate the b in... { \displaystyle a^ { n } } is contains well written, well thought well... But first, find the generating functions are a fairly simple concept with a, ’... He noticed a pattern and raised some Questions about it find a closed formula for the Fibonacci may... The 3rd number onwards, the series never reaches negative one: it never ends use. 2 Exponential generating functions that results in the sequence what is the 100th term of the Fibonacci sequence a infinite! Exponents as repeated multiplication the n-th term of the Fibonacci numbers re really the. As the sum of the binomials in h ( x ), with \ ( | 3,900 | 14,931 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-31 | latest | en | 0.827291 |
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# At a blind taste competition a contestant is offered 3 cups
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14 Nov 2009, 06:34
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At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?
A. $$\frac{1}{12}$$
B. $$\frac{5}{14}$$
C. $$\frac{4}{9}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$
[Reveal] Spoiler: OA
Last edited by Bunuel on 06 Mar 2013, 01:30, edited 2 times in total.
Edited the question and added the OA
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14 Nov 2009, 07:06
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Economist wrote:
At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?
# $$\frac{1}{12}$$
# $$\frac{5}{14}$$
# $$\frac{4}{9}$$
# $$\frac{1}{2}$$
# $$\frac{2}{3}$$
And the good one again. +1 to Economist.
"The probability that a contestant does not taste all of the samples" means that contestant tastes only 2 samples of tea (one sample is not possible as contestant tastes 4 cups>3 of each kind).
$$\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}$$.
$$C^2_3$$ - # of ways to choose which 2 samples will be tasted;
$$C^4_6$$ - # of ways to choose 4 cups out of 6 cups of two samples (2 samples*3 cups each = 6 cups);
$$C^4_9$$ - total # of ways to choose 4 cups out of 9.
Another way:
Calculate the probability of opposite event and subtract this value from 1.
Opposite event is that contestant will taste ALL 3 samples, so contestant should taste 2 cups of one sample and 1 cup from each of 2 other samples (2-1-1).
$$C^1_3$$ - # of ways to choose the sample which will provide with 2 cups;
$$C^2_3$$ - # of ways to chose these 2 cups from the chosen sample;
$$C^1_3$$ - # of ways to chose 1 cup out of 3 from second sample;
$$C^1_3$$ - # of ways to chose 1 cup out of 3 from third sample;
$$C^4_9$$ - total # of ways to choose 4 cups out of 9.
$$P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}$$.
Hope it's clear.
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15 Nov 2009, 03:16
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I got B: 5/14 as well, although using a different approach.
Number of ways to choose 4 cups to drink from: 9C4
Since the contestant must drink 4 cups of tea, and there are only 3 cups of each tea, the contestant MUST drink at least two different samples.
Number of ways to choose which two samples the contestant drinks (since he can't drink all 3): 3C2
Three cases:
a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1
b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2
c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3
Therefore,
$$P = \frac{3C2(3C3*3C1 + 3C2*3C2 + 3C1*3C3)}{9C4}$$
$$P = \frac{3(1*3 + 3*3 + 3*1)}{\frac{9*8*7*6}{4*3*2*1}}$$
$$P = \frac{45}{{9*2*7}}$$
$$P = \frac{5}{14}$$
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15 Aug 2010, 16:17
I thought the answer to this question can be obtained thus: 1 - P(all samples are tasted)
P(all samples tasted) = I choose one from each of the 3 samples = 3C1 x 3C1x 3C1 x 6C1 as there will be 6 cups left from which I can get the 4th cup = 27 x6 = 162 makes no sense as there are only 9C4 combinations = 126.
S1 A B C
S2 D E F
S3 G H I
Select 1 from S1 in 3 ways, same with S2, same with S3 = a total of 27 ways
I have 6 cups left, can select 1 in 6 ways
So total of 27 x 6 = 162 ways. It seems that this should be 81, I am double counting somewhere. Only 81 gives me the right answer...
Ok so where did my brain get fried?
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15 Aug 2010, 17:43
It should be 1-(3*3C2*3C1*3C1/9C4)
= 1-9/14 = 5/14
Now why is your approach wrong ?
E.g: in how many ways 2 objects can be selected from a set of 3.
We know it is 3C2
As per the approach you took above it should be equal to; 3C1*2C1 != 3C2.
Hope it helps.
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15 Aug 2010, 17:50
Can you give the breakdown as to how you arrived at (3*3C2*3C1*3C1). Thanks
My thinking was simply that
-|-|- <- select 1 of 3
-|-|- <- select 1 of 3
-|-|- <- select 1 of 3
That is a total of 27 ways. I can see that these are not unique combinations, but how do I remove the duplicates?
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15 Aug 2010, 18:03
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We want to select 2 cups of tea which are of similar type and the rest 2 cups which each are of different type ;
hence the terms 3C2 * 3C1* 3C1
Now you should multiply it by 3 because ; out of the 3 tea types , you may select the tea type which is in 2 cups, in 3 ways.
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15 Aug 2010, 18:16
Maybe I dont understand the question. But I was approaching it as 1 - p(he tastes all samples), so why limit to just 2 cups and then make it so that the other 2 are each of a different type?
S1 A B C
S2 D E F
S3 G H I
So why can't be go ADGF - select one from S1, one from S2, one from S3 and then have 1 of the remaining 6? You are doing 3c2 3c1 3c1, so that is picking 2 from one row and one each from the other 2 rows such as ABDG. Perhaps if you can explain how to get an accurate count of combinations, when I select 1 from each row, make it 3 cups and then pick the 4th one of the remaining 6?
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15 Aug 2010, 18:53
I am not sure if it could be calculated that way.
Lets make it even simpler
S1 A B C
In how many ways can you select 2 out of these 3 ?
3C2, right ?
If however; if I first select 1 of them say in 3C1 ways; there are 2 still left.
Now, if I again select 1 out of the remaining two, in 2C1 ways, the total no. of combination 3C1*2C1, which is not equal to 3C2.
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15 Aug 2010, 19:06
Yes I can see that clearly I am ending up with more combinations than necessary. I was thinking if there is way to factor out those repeats the way I am doing it. I see your point in the simple example. However when I think of this, I am thinking of a 3x3 matrix where I have to make sure I get one from each row and the 4th can come from anywhere - that is the same thing as saying 2 from a row and 1 each from the remaining 2 (your and correct approach). My count differs from yours by a factor of 2 (I am 162 and you are 81). How can I rationalize this?
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AKProdigy87 wrote:
I got B: 5/14 as well, although using a different approach.
Number of ways to choose 4 cups to drink from: 9C4
Since the contestant must drink 4 cups of tea, and there are only 3 cups of each tea, the contestant MUST drink at least two different samples.
Number of ways to choose which two samples the contestant drinks (since he can't drink all 3): 3C2
Three cases:
a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1
b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2
c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3
Therefore,
$$P = \frac{3C2(3C3*3C1 + 3C2*3C2 + 3C1*3C3)}{9C4}$$
$$P = \frac{3(1*3 + 3*3 + 3*1)}{\frac{9*8*7*6}{4*3*2*1}}$$
$$P = \frac{45}{{9*2*7}}$$
$$P = \frac{5}{14}$$
I don't follow this answer. (a) and (c) show that he ends up tasting all the cups of the given sample, we are not supposed to allow this to happen?
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31 Aug 2010, 02:18
Nice question... economist always posts quality question... thanks for that...
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21 Sep 2010, 17:20
great explanation Bunuel
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10 Nov 2010, 00:11
I was all over the place in trying to solve this...neways Thanks Bunuel..that was a magnificient and clear solution..but my only concern is on the actual GMAT..is this question easy enuf to be answered in under 2 mins...???or rather...it looks like a difficulty level 800+ to me...
+ 1 to Bunuel.....
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i solved the question as follows. i hope my approach is correct.
let's consider that the contestant tasted all 3 flavors.
the first cup can be selected from any 9 cups. P = $$\frac{9}{9}$$
the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = $$\frac{6}{8}$$
the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = $$\frac{3}{7}$$
the fourth cup can be selected from any of the remaining 6 cups. P = $$\frac{6}{6}$$
there are $$2$$ ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.
probability that the contestant tasted all flavors = $$\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}$$
required probability = $$1 - \frac{9}{14} = \frac{5}{14}$$
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03 Aug 2011, 23:47
Can anyone tell me where I went wrong? I understand you can derive the answer from the solution above, but why does my method below give me a wrong answer?
I used 1 - (prob of drinking all three teas)
Total combinations = 9C4 = 126
prob of drinking all threes = 1 of each tea + 1 of any tea.
Using A, B, & C to represent the teas:
Combinations where all three teas are represented (2 of one tea and one of each of the other tea) =
A, A, B, C
A, B, B, C
A, B, C, C
Combinations of rearranging A,A,B,C = 4!/2! = 12
Combinations of rearranging A,B,B,C = 4!/2! = 12
Combinations of rearranging A,B,C,C = 4!/2! = 12
Total combinations of drinking each tea = 36
Probability of not drinking all tea = 1 - (36/126) = 90/126 = 5/7
I can't figure out where the logical error or where I double counted or ignored possibilities come from. Can anyone help?
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perhaps bunuel can chime in / elaborate on dimitri92's answer?
using the 9/9 * 6/8 * 2 * 3/7 * 6/6
this approach always confuses me.
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29 Aug 2011, 17:35
pinchharmonic wrote:
perhaps bunuel can chime in / elaborate on dimitri92's answer?
using the 9/9 * 6/8 * 2 * 3/7 * 6/6
this approach always confuses me.
bump on this? i'm curious about the 6/8 and the factor of 2.
9/9 i understand, but 6/8 should give the probability of either two types, then why is there another factor of 2?
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22 Dec 2011, 22:41
Bunuel wrote:
Economist wrote:
At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?
# $$\frac{1}{12}$$
# $$\frac{5}{14}$$
# $$\frac{4}{9}$$
# $$\frac{1}{2}$$
# $$\frac{2}{3}$$
And the good one again. +1 to Economist.
"The probability that a contestant does not taste all of the samples" means that contestant tastes only 2 samples of tea (one sample is not possible as contestant tastes 4 cups>3 of each kind).
$$\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}$$.
$$C^2_3$$ - # of ways to choose which 2 samples will be tasted;
$$C^4_6$$ - # of ways to choose 4 cups out of 6 cups of two samples (2 samples*3 cups each = 6 cups);
$$C^4_9$$ - total # of ways to choose 4 cups out of 9.
Another way:
Calculate the probability of opposite event and subtract this value from 1.
Opposite event is that contestant will taste ALL 3 samples, so contestant should taste 2 cups of one sample and 1 cup from each of 2 other samples (2-1-1).
$$C^1_3$$ - # of ways to choose the sample which will provide with 2 cups;
$$C^2_3$$ - # of ways to chose these 2 cups from the chosen sample;
$$C^1_3$$ - # of ways to chose 1 cup out of 3 from second sample;
$$C^1_3$$ - # of ways to chose 1 cup out of 3 from third sample;
$$C^4_9$$ - total # of ways to choose 4 cups out of 9.
$$P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}$$.
Hope it's clear.
Thank you.
If the question was about the probability that a contestant does not taste all the 3 samples of a cup. (That's what I wrongly understood from this question first). Then the answer would be $$\frac{6}{7}$$?
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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05 Mar 2013, 08:23
Not sure if my channel of thought is flawed to start with, but when it is said that it's a 'blind' taste competition, I am considering repetitions. As the contestant does not know which of the 9 cups he/she had picked up the first time, the same can possibly be repeated in the second turn and so on until the fourth.
That gives a complete different perspective to the problem.
Where am I going wrong here?
Re: At a blind taste competition a contestant is offered 3 cups [#permalink] 05 Mar 2013, 08:23
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Display posts from previous: Sort by | 5,712 | 17,268 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-04 | latest | en | 0.862611 |
https://math.stackexchange.com/questions/56669/a-tensor-product-of-power-series | 1,660,031,659,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00335.warc.gz | 365,203,290 | 67,526 | # A tensor product of power series
Let $k$ be a field. I am wondering if there is an easy description of the ring
$$k[[x]] \otimes_{k[x]} k[[x]]$$
that is the tensor product of the power series ring $k[[x]]$ with itself over the ring of polynomials $k[x]$.
Any hints would be appreciated.
Edit: As suggested in the comments I am asking: is this ring isomorphic to $k[[x]]$?
• @user10676, as far as I know, tensor products does not commute with inverse limits. Aug 10, 2011 at 13:00
• It's pretty clear that the "obvious" map $k[[x]]\to k[[x]]\otimes_{k[x]}k[[x]]$ given by $\alpha\mapsto\alpha\otimes 1$ is not an isomorphism. It's trickier to rule out the existence of some other isomorphism. You'd have to find a property of rings that distinguishes the two sides. Aug 10, 2011 at 14:36
• @Jim : I can see that the map is injective, but why is it not surjective ? Aug 10, 2011 at 15:09
• Your example doesn't work : $1 \otimes (1+x+x^2 + \cdots) = (1-x).(1-x)^{-1} \otimes (1-x)^{-1}$ $=(1-x)^{-1} \otimes (1-x).(1-x)^{-1} = (1+x+x^2 + \cdots) \otimes 1$. I have thought about this kind of argument but I can't find a proof. Let $P \in k[[x]]$, write $P=A+x^kB$, with $A$ polynomial, then $1 \otimes P - P \otimes 1 = x^k (B \otimes 1 - 1 \otimes B)$. So $1 \otimes P - P \otimes 1$ is $x$-divisible. I suspect that $k[[x]] \otimes_{k[x]} k[[x]]$ has no non-zero divisible element (since $k[[x]]$ doesn't). Aug 10, 2011 at 16:03
• @Jim: This span is $k(x)$. Aug 10, 2011 at 17:50
No, the rings $k[[x]] \otimes_{k[x]} k[[x]]$ and $k[[x]]$ are not isomorphic because they have different Krull dimensions:$\; \infty$ and $1$ respectively
I) The ring $k[[x]]$ has Krull dimension one since it is a discrete valuation ring.
II) The ring $k[[x]] \otimes_{k[x]} k[[x]]$ has at least the Krull dimension of its localization $k((x)) \otimes_{k(x)} k((x))$. It is thus sufficient to prove that the latter ring has infinite Krull dimension.
This results from Grothendieck's formula for the Krull dimension of the tensor product of two field extensions $K,L$ of a field $k$ as a function of the transcendence degrees of the extensions: $$\dim (K \otimes_k L) =\min(\operatorname{trdeg}_k K, \operatorname{trdeg}_k L)$$
Since $\operatorname{trdeg}_{k(x)} k((x))=\infty$, we deduce that, as anounced, $$\dim(k[[x]] \otimes_{k[x]} k[[x]]) \geq \dim(k((x)) \otimes_{k(x)} k((x))) = \infty$$
Addendum: some properties of our tensor products
i) Let me show, as an answer to Pierre-Yves's first question in the comments, that $R=k[[x]] \otimes_{k[x]} k[[x]]$ is not a noetherian ring.
Any ring of fractions of a noetherian ring is noetherian and since, as already mentioned, the ring $T=k((x)) \otimes_{k(x)} k((x))$ is such a ring of fractions , it is enough to show that the latter tensor product $T$ of extensions is not noetherian.
This results from the following theorem of Vamos: given a field extension $F\subset K$, the tensor product $K\otimes_F K$ is noetherian iff the extension is finitely generated (in the field sense). Since in our case the extension $k(x) \subset k((x))$ is not finitely generated , we conclude that $T$ is not noetherian.
By the way, since the discrete valuation ring $k[[x]]$ is noetherian this gives another proof that it $R$ is not isomorphic to $k[[x]]$
ii) Pierre-Yves also asks if the ring $T$ is local. It is not because a theorem of Sweedler states that a tensor product of algebras over a field is local only if one of the factors is algebraic.
Since $k(x) \subset k((x))$ is not algebraic, non-locality of $T$ follows.
iii) The ring $R=k[[x]] \otimes_{k[x]} k[[x]]$ is not a domain (another thing Pierre-Yves has asked about) because if it were, its ring of fractions $T=k((x)) \otimes_{k(x)} k((x))$ would also be a domain ( that reduction again!) and I'm going to show that actually $T$ has zero divisors.
The key remark is that if $F\subset F(a)$ is a non trivial simple algebraic extension the tensor product $F(a)\otimes _F F(a)$ is not a domain. Indeed we have an isomorphism $F(a)\otimes _F F(a)=F(a)[T]/(m(T))$ where $m(T)$ is the minimal polynomial of $a$ over $F$. Since $m(T)$ has $a$ as a root, it is no longer irreducible over $F(a)$ and the quotient $F(a)[T]/m(T)$ has zero-divisors.
And now the rest is easy. Just take an element $a\in k((x))\setminus k(x)$ algebraic over $k(x)$ . Since $(k(x))(a)\otimes _{k(x)} (k(x))(a)$ is a subring of $T=k((x)) \otimes_{k(x)} k((x))$ containing zero-divisors, the ring $T$ a fortiori has zero-divisors.
• Dear Georges: great answer! One nitpick, though: I was curious about the formula you mention, couldn't find it in the section of EGA that I expected, and googled to find this: jlms.oxfordjournals.org/content/s2-19/3/391.extract It seems the formula you mention is due to R. Y. Sharp. Aug 10, 2011 at 23:57
• Dear Akhil,thanks for the kind words. I knew of Sharp's article and I stand by my attribution to Grothendieck of that wonderful formula. Indeed it was proved ten years before Sharp's article in EGA IV, Quatrième Partie, (4.2.1.5), page 349. But I am not surprized you could not find it: it is hidden in the Errata et Addenda of that volume! And you are in good company, because ten years after its publication the formula was so little known that Sharp could publish it in the excellent journal you mention, he and the referee being obviously unaware of Grothendieck's priority! Aug 11, 2011 at 1:57
• Great, thanks for the EGA reference. That's a nice result to be buried in an appendix. (By the way, although you addressed me as "Dear Akhil" and not "@Akhil," it seems that I was nonetheless notified; this seems to be a pleasant improvement to the software.) Aug 11, 2011 at 2:27
• Dear Georges, could you tell me if $k[[x]]\otimes_{k[x]}k[[x]]$ is noetherian? (And why?) (Or direct me to a reference.) [If you’re kind enough to answer, could you use the @ sign? To follow up on Akhil, my personal experience is that things work out usually without it, but they always do with it.] Aug 11, 2011 at 5:15
• Dear Georges, thank you very much again for everything! I've just posted a somewhat long comment as a community wiki answer. [I didn't get a notification for your last comment. I guess "Dear (at)Pierre-Yves" is better - for the software - than (at)Dear Pierre-Yves. ;)] Aug 12, 2011 at 7:16
The rings $R:=k[[x]]\otimes_{k[x]}k[[x]]$ and $k[[x]]$ are not isomorphic. Here is a mild simplification of Georges Elencwajg's proof.
Assume by contradiction $R\simeq k[[x]]$. Let $K$ be the fraction field of $R$, and $S$ the multiplicative system $k[x]\backslash\{0\}$. As $R$ is a maximal subring of $K$, and $x\otimes1$ is invertible in $$S^{-1}R=k((x))\otimes_{k(x)}k((x)),$$ but not in $R$, we have $$K=S^{-1}R=k((x))\otimes_{k(x)}k((x)).$$ If $a$ is in $k((x))\backslash k(x)$, then $a\otimes1-1\otimes a$ is a nonzero element of $K$ which is mapped to $0$ by the natural morphism to $k((x))$, a contradiction. | 2,194 | 6,929 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-33 | latest | en | 0.824806 |
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# In the figure above, square AFGE is inside square ABCD such
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In the figure above, square AFGE is inside square ABCD such [#permalink]
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In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of square AFGE (area of square of side s = $$s^2$$)?
A) 32 (1-$$\sqrt{2}$$)
B) 32 (3-2$$\sqrt{2}$$)
C) 64 ($$\sqrt{2}$$ - 1)$$^2$$
D) 64 - 16$$\pi$$
E) 32 - 4$$\pi$$
[Reveal] Spoiler: OA
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nafishasan60 wrote:
In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of square AFGE (area of square of side s = )?
A) 32 (1-)
B) 32 (3-2)
C) 64 ( - 1)
D) 64 - 16
E) 32 - 4
we can find diagonal AC = 8 $$\sqrt{2}$$
so AG = AC -CG( radius of the arc) = 8$$\sqrt{2} - 8$$
now AE = AG/$$\sqrt{2}$$ = 8$$\sqrt{2} - 8$$ / $$\sqrt{2}$$
IMO B..
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Re: area of square [#permalink]
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10 Feb 2012, 06:41
Took me 10 mins... but I got it... Surprising it wasn't super difficult. Since C is the center of the Circle, the length of GC = 8. Since AEFG touches the circle with it's corner, we know the angle of GC is 45%. When you spilt a square in half diagonally it's going to be 45 degrees. If the angle GC is 45% you know the triangle form with would be a 1:1:root(2). Since GC is the hypotenuse the other 2 will be 8/root(2). The length of FG = 10 (height of the square) - 8/root(2) (height of the triangle). [10-8/root(2)]^2 = 32 (3-2*root(2)).
Answer is B. Hard to explain without a graph and too lazy to make one. Sorry
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In the figure above, square AFGE is inside square ABCD such [#permalink]
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10 Feb 2012, 08:00
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nafishasan60 wrote:
In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of square AFGE (area of square of side s = $$s^2$$)?
A) 32 (1-$$\sqrt{2}$$)
B) 32 (3-2$$\sqrt{2}$$)
C) 64 ($$\sqrt{2}$$ - 1)$$^2$$
D) 64 - 16$$\pi$$
E) 32 - 4$$\pi$$
No additional drawing is needed.
The length of the diagonal AG is AC-GC;
AC is a diagonal of a square with a side of 8, hence it equal to $$8\sqrt{2}$$ (hypotenuse of 45-45-90 triangle);
GC=DC=radius=side=8;
Hence $$AG=8\sqrt{2}-8=8(\sqrt{2}-1)$$;
Area of a square: $$\frac{diagonal^2}{2}=\frac{(8(\sqrt{2}-1))^2}{2}=32*(\sqrt{2}-1)^2=32*(3-2\sqrt{2})$$.
Answer: B.
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink]
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15 Jul 2014, 13:13
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In the figure above, square AFGE is inside square ABCD such [#permalink]
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20 Aug 2014, 14:09
Bunuel wrote:
nafishasan60 wrote:
In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of square AFGE (area of square of side s = $$s^2$$)?
A) 32 (1-$$\sqrt{2}$$)
B) 32 (3-2$$\sqrt{2}$$)
C) 64 ($$\sqrt{2}$$ - 1)$$^2$$
D) 64 - 16$$\pi$$
E) 32 - 4$$\pi$$
No additional drawing is needed.
The length of the diagonal AG is AC- AG;
AC is a diagonal of a square with a side of 8, hence it equal to $$8\sqrt{2}$$ (hypotenuse of 45-45-90 triangle);
AG=DC=radius=side=8;
Hence $$AG=8\sqrt{2}-8=8(\sqrt{2}-1)$$;
Area of a square: $$\frac{diagonal^2}{2}=\frac{(8(\sqrt{2}-1))^2}{2}=32*(\sqrt{2}-1)^2=32*(3-2\sqrt{2})$$.
Answer: B.
Bunuel ,
I think you meant GC in place of AG in the highlighted portion.
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In the figure above, square AFGE is inside square ABCD such [#permalink]
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20 Aug 2014, 23:24
AG $$= 8\sqrt{2} - 8$$
Attachment:
squ.png [ 7.85 KiB | Viewed 1686 times ]
Area of shaded region $$= \frac{(8\sqrt{2} - 8)^2}{2}$$
$$= \frac{64 (2 - 2\sqrt{2} + 1)}{2}$$
$$= 32 (3 - 2\sqrt{2})$$
Answer = B
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink]
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21 Aug 2014, 03:09
maggie27 wrote:
Bunuel wrote:
nafishasan60 wrote:
In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of square AFGE (area of square of side s = $$s^2$$)?
A) 32 (1-$$\sqrt{2}$$)
B) 32 (3-2$$\sqrt{2}$$)
C) 64 ($$\sqrt{2}$$ - 1)$$^2$$
D) 64 - 16$$\pi$$
E) 32 - 4$$\pi$$
No additional drawing is needed.
The length of the diagonal AG is AC- AG;
AC is a diagonal of a square with a side of 8, hence it equal to $$8\sqrt{2}$$ (hypotenuse of 45-45-90 triangle);
AG=DC=radius=side=8;
Hence $$AG=8\sqrt{2}-8=8(\sqrt{2}-1)$$;
Area of a square: $$\frac{diagonal^2}{2}=\frac{(8(\sqrt{2}-1))^2}{2}=32*(\sqrt{2}-1)^2=32*(3-2\sqrt{2})$$.
Answer: B.
Bunuel ,
I think you meant GC in place of AG in the highlighted portion.
Typo edited. Thank you.
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink]
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25 Oct 2015, 09:00
we know that ACD will form a 45-45-90 triangle, and, knowing that one side is 8, we can deduce that AC = 8 sqrt(2)
CG is 8, and thus, diagonal AG will be 8sqrt(2)-8 or 8[sqrt(2)-1]
again, we have a 45-45-90 triangle. Applying pythagorean theorem, we can see that 2s^2 = 8^2[sqrt(2)-1]^2 or s^2 = 32[sqrt(2)-1]^2
[sqrt(2)-1]*[sqrt(2)-1] = sqrt(2)*sqrt(2) - sqrt(2) -sqrt(2) +1 = 3 -2sqrt(2) +1 = 5 - 2sqrt(2) now we can rewrite the area of the small square to be 32*[3-2sqrt(2)], answer choice B.
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink]
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink] 28 Nov 2016, 23:07
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RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss are available here, so that students can check for the solutions whenever they are facing difficulty while solving the questions from RS Aggarwal Solutions for Class 7. These solutions for Chapter 11 are available in PDF format so that students can download it and learn offline as well. This book is one of the top materials when it comes to providing a question bank to practice from.
We at BYJU’S have prepared the RS Aggarwal Solutions for Class 7 Maths Chapter 11 wherein, problems are formulated by our expert tutors to assist you with your exam preparation to attain good marks in Maths. Download pdf of Class 7 Chapter 11 in their respective links.
## Download PDF of RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss
### Also, access RS Aggarwal Solutions for Class 7 Chapter 11 Exercises
Exercise 11A
Exercise 11B
Exercise 11A
1. Find the SP when:
(i). CP = ₹ 950, gain = 6%
Solution:-
We have,
SP = {((100 + gain %) /100) × CP)}
= {((100 + 6) /100) × 950)}
= {(106 /100) × 950}
= 100700/100
= ₹1007
(ii). CP = ₹ 9600, gain = 16(2/3) %
Solution:-
We have,
SP = {((100 + gain %) /100) × CP)}
= {((100 + (50/3)) /100) × 9600)}
= {(((300 + 50)/3)) /100) × 9600)}
= {((350/3) /100) × 9600}
= {((350/3) × (1/100)) × 9600}
= {(350/300) × 9600}
= {(350/3) × 96}
= {350 × 32}
= ₹11200
(iii). CP = ₹ 1540, loss = 4%
Solution:-
We have,
SP = {((100 – loss %) /100) × CP)}
= {((100 – 4) /100) × 1540)}
= {(96 /100) × 1540}
= 147840/100
= ₹1478.40
(iv). CP = ₹ 8640, loss = 12(1/2) %
Solution:-
We have,
SP = {((100 – loss %) /100) × CP)}
= {((100 – (25/2)) /100) × 8640)}
= {(((200 – 25)/2)) /100) × 8640)}
= {((175/2) /100) × 8640}
= {((175/2) × (1/100)) × 8640}
= {(175/200) × 8460}
= {1512000/200}
= ₹7560
2. Find the gain or loss percent when:
(i). CP = ₹ 2400 and SP = ₹ 2592
Solution:-
Since (SP) > (CP), so there is a gain
Gain = (SP) – (CP)
= ₹ (2592-2400)
= ₹ 192
Gain % = {(gain/CP) × 100}
= {(192/2400) × 100}
= {192/24}
= 8%
(ii). CP = ₹ 1650 and SP = ₹ 1452
Solution:-
Since (SP) < (CP), so there is a loss
Loss = (CP) – (SP)
= ₹ (1650 – 1452)
= ₹ 198
Loss % = {(Loss/CP) × 100}
= {(198/1650) × 100}
= {19800/1650}
= 12%
(iii). CP = ₹ 12000 and SP = ₹ 12800
Solution:-
Since (SP) > (CP), so there is a gain
Gain = (SP) – (CP)
= ₹ (12800 – 12000)
= ₹ 800
Gain % = {(gain/CP) × 100}
= {(800/12000) × 100}
= {800/120}
= 6(2/3) %
(iv). CP = ₹ 1800 and SP = ₹ 1611
Solution:-
Since (SP) < (CP), so there is a loss
Loss = (CP) – (SP)
= ₹ (1800 – 1611)
= ₹ 189
Loss % = {(Loss/CP) × 100}
= {(189/1800) × 100}
= {189/18}
= 10(1/2)%
3. Find the CP when:
(i). SP = ₹ 924, gain = 10%
Solution:-
By using the formula, we have:
CP = ₹ {(100/ (100 + gain %)) × SP}
= {(100/ (100 + 10)) × 924}
= {(100/ 110) × 924}
= {92400/110}
= ₹ 840
(ii). SP = ₹ 1755, gain = 12(1/2) %
Solution:-
Gain = 12(1/2) = 25/2
By using the formula, we have:
CP = ₹ {(100/ (100 + gain %)) × SP}
= {(100/ (100 + (25/2))) × 1755}
= {(100/ ((200 + 25)/2)) × 1755}
= {(200/ 225) × 1755}
= {351000/225}
= ₹ 1560
(iii). SP = ₹ 8510, loss = 8%
Solution:-
By using the formula, we have:
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – 8)) × 8510}
= {(100/ 92) × 8510}
= {851000/92}
= ₹ 9250
(iv). SP = ₹ 5600, loss = 6(2/3) %
Solution:-
Loss = 6(2/3) = 20/3
By using the formula, we have:
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – (20/3))) × 5600}
= {(100/ ((300 – 20)/3)) × 5600}
= {(300/ 280) × 5600}
= {168000/280}
= ₹ 6000
4. Sudhir bought an almirah for ₹ 13600 and spent ₹ 400 on its transportation. He sold it for ₹ 16800. Find his gain percent.
Solution:-
From the question,
Sudhir bought an almirah for = ₹ 13600 = cost price
Transportation cost = ₹ 400
The total cost price of almirah = ₹ (13600 + 400)
= ₹ 14000
He sold it for = ₹ 16800 = Selling price
By comparing SP and CP = SP > CP, so there is a gain
Gain = SP – CP
= 16800 – 14000
= ₹ 2800
Gain % = {(gain/CP) × 100}
= {(2800/14000) × 100}
= {2800/140}
= 20%
5. Ravi purchased an old house for ₹765000 and spent ₹115000 on its repairs. Then, he sold it a gain of 5%. How much did he get?
Solution:-
From the question,
Ravi purchased an old house for = ₹ 765000 = Cost price
He spent on its repairs = ₹ 115000
Total cost price of old house = (765000 + 115000)
= ₹ 880000
Then, he sold it at a gain of 5%
SP = {((100 + gain %) /100) × CP)}
= {((100 + 5) /100) × 880000)}
= {(105 /100) × 880000}
= 105 × 8800
= ₹ 924000
∴the selling price of the house is ₹ 924000
6. A vendor buys lemons at ₹25 per dozen and sells them at the rate of 5 for ₹ 12. Find his gain or loss percent.
Solution:-
Cost price of 12 lemons = ₹25
Then, cost price of 1 lemon = ₹ (25/12)
Cost price of 5 lemons = (25/12) × 5
= 125/12
= ₹ 10.42
He sold 5 lemons for = ₹12 =Selling price
By comparing SP and CP = SP > CP, so there is a gain
Gain = SP – CP
= 12 – 10.42
= ₹ 1.58
Gain % = {(gain/CP) × 100}
= {(1.58/10.42) × 100}
= {15800/1042}
= 15.2%
7. The selling price of 12 pens is equal to the cost price of 15 pens. Find the gain percent.
Solution:-
Let the cost price of 1 pen = ₹ 1
So, cost price of 12 pens = ₹ 12
SP of 15 pens = ₹ 15
From the question,
Selling price of 12 pens = cost price of 15 pens
Gain = SP – CP
= 15 -12
= ₹ 3
Gain % = {(gain/CP) × 100}
= {(3/12) × 100}
= {300/12}
= 25%
8. The selling price of 16 spoons is equal to the cost price of 15 spoons. Find the loss percent.
Solution:-
Let the cost price of 1 spoon = ₹ 1
So, cost price of 16 pens = ₹ 16
SP of 15 spoons = ₹ 15
From the question,
Selling price of 16 spoons = cost price of 15 spoons
Loss = (CP) – (SP)
= ₹ (16 – 15)
= ₹ 1
Loss % = {(Loss/CP) × 100}
= {(1/16) × 100}
= {100/16}
= 6.25%
= 6(1/4) %
9. Manoj purchased a video for ₹12000. He sold it to Rahul at a gain of 10%. If Rahul sells it to Rakesh at a loss of 5%, what did Rakesh pay for it?
Solution:-
From the question,
Manoj purchased a video for = ₹ 12000 = Cost price
He sold it to Rahul at a gain of = 10 %
Selling price of video from Manoj to Rahul,
SP = {((100 + gain %) /100) × CP)}
= {((100 + 10) /100) × 12000)}
= {(110 /100) × 12000}
= 110 × 120
= ₹ 13200
∴Selling price of video from Manoj to Rahul is ₹ 13200
Then, Rahul purchase a video from Manoj at cost price of = ₹ 13200
Rahul sells it to Rakesh at Percentage of loss = 5%
Selling price of video when Rahul sells it to Rakesh,
SP = {((100 – loss %) /100) × CP)}
= {((100 – 5) /100) × 13200)}
= {(95 /100) × 13200}
= {95 × 132}
= ₹12540
∴Rakesh pay for a video is ₹12540
10. On selling a sofa-set for ₹ 21600, a dealer gains 8%. For how much did he purchase it?
Solution:-
From the question,
Dealer selling a sofa-set for = ₹21600 = Selling price
He gains on selling = 8%
Cost price of sofa-set,
CP = ₹ {(100/ (100 + gain %)) × SP}
= {(100/ (100 + 8)) × 21600}
= {(100/ 108) × 21600}
= {2160000/108}
= ₹ 20000
11. On selling a watch for ₹ 11400, a shopkeeper loses 5%. For how much did he purchase it?
Solution:-
From the question,
Shopkeeper selling a watch for = ₹11400 = Selling price
He loses on selling = 5%
Cost price of watch,
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – 5)) × 11400}
= {(100/ 95) × 11400}
= {11400} × 95
= ₹ 12000
Exercise 11B
Mark against the correct answer in each of the following:
1. A man buys a book for ₹ 80 and sells it for ₹ 100. His gain % is,
(a) 20% (b) 25% (c) 120 (d) 125%
Solution:-
(b) 25%
Because,
Cost price of book = ₹ 80
Selling price of book = ₹ 100
Since (SP) > (CP), so there is a gain
Gain = (SP) – (CP)
= ₹ (100 – 80)
= ₹ 20
Gain % = {(gain/CP) × 100}
= {(20/80) × 100}
= {(20/20) × 25}
= 25%
2. A football is bought for ₹ 120 and sold for ₹ 105. The loss % is
(a) 12(1/2)% (b) 14(2/7)% (c) 16(2/3)% (d) 13(1/3)%
Solution:-
(a) 12(1/2)%
Because,
Cost price of football = ₹ 120
Selling price of football = ₹ 105
Since (SP) < (CP), so there is a loss
Loss = (CP) – (SP)
= ₹ (120 – 105)
= ₹ 15
Loss % = {(Loss/CP) × 100}
= {(15/120) × 100}
= {(15/12) ×10}
= {150/12}
=12.5%
= 12(1/2)%
3. On selling a bat for ₹ 100, a man gains ₹20. His gain % is
(a) 20% (b) 25% (c) 18% (d) 22%
Solution:-
(b) 25%
Because,
Selling price of bat = ₹ 100
Amount gain by selling bat = ₹20
Cost price of the bat = (100 – 20)
= ₹ 80
Gain % = {(gain/CP) × 100}
= {(20/80) × 100}
= {(20/20) × 25}
= 25%
4. On selling a racket for ₹198, a shopkeeper gains 10%. The cost price of the racket is
(a) ₹180 (b) ₹178.20 (c) ₹217.80 (d) ₹212.50
Solution:-
(a) ₹180
Because,
Selling price of racket = ₹ 198
Percentage gain by selling racket = 10%
Cost price of the racket,
CP = ₹ {(100/ (100 + gain %)) × SP}
= {(100/ (100 + 10)) × 198}
= {(100/ 110) × 198}
= {19800/110}
= ₹ 180
5. On selling a jug for ₹ 144, a man loses (1/7) of his outlay. If it is sold for ₹ 189, what is the gain %?
(a) 12.5% (b)25% (c) 30% (d)50%
Solution:-
(a) 12.5%
Because,
Let the CP be, ₹ x
Then, x – (1/7)x = 144
= (7x –x) = (144 × 7)
= x = (144 ×7)/6
= x = 168
∴ CP = ₹ 168,
New SP = ₹189
Gain = SP –CP
= 189 – 168
= 21
Gain % = {(gain/CP) × 100}
= {(21/168) × 100}
= {2100/168)
= 12.5%
6. On selling a pen for ₹ 48, a shopkeeper loses 20%. In order to gain 20% what would be the selling price?
(a) ₹ 52 (b) ₹ 56 (c) ₹ 68 (d) ₹ 72
Solution:-
(d) ₹ 72
Because,
Selling price of pen = ₹ 48
Shopkeeper loses = 20%
Cost price of pen =
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – 20)) × 48}
= {(100/ 80) × 48}
= {4800/80}
= ₹ 60
In order to gain 20%,
SP = {((100 + gain %) /100) × CP)}
= {((100 + 20) /100) × 60)}
= {(120 /100) × 60}
= ₹ 72
## RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss
Chapter 11 – Profit and Loss contains 2 exercises and the RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.
• Some Terms Related to Profit and Loss
• To Find SO When CP and Gain% or Loss% are Given
• To Find CP when SP and Gain% or Loss% are Given
### Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss
RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss. The money paid by the shopkeeper to buy the products from wholesalers is called Cost Price. The rate at which the shopkeeper sells the products is called Selling Price. Profit and Loss are entirely based on cost price and selling price. In this chapter, students are going to learn the percentage gain and also percentage loss. | 4,362 | 11,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-10 | latest | en | 0.804722 |
https://slidetodoc.com/genetic-algorithm-example-evolving-a-control-program-for/ | 1,624,461,995,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539480.67/warc/CC-MAIN-20210623134306-20210623164306-00169.warc.gz | 476,861,525 | 14,455 | Genetic Algorithm Example Evolving a Control Program for
• Slides: 107
Genetic Algorithm Example: Evolving a Control Program for a Virtual “Robot”
Herbert: The Soda Can Collecting Robot (Connell, Brooks, Ning, 1988) http: //cyberneticzoo. com/? p=5516
Herbert: The Soda Can Collecting Robot (Connell, Brooks, Ning, 1988) http: //cyberneticzoo. com/? p=5516 Robby: The Virtual Soda Can Collecting Robot (Mitchell, 2009)
Robby’s World Soda can
What Robby Can See and Do Input: Contents of North, South, East, West, Current Possible actions: Move N Move S Move E Move W Move random Stay put Try to pick up can Rewards/Penalties (points): Picks up can: 10 Tries to pick up can on empty site: -1 Crashes into wall: -5 Robby’s Score: Sum of rewards/penalties
Goal: Use a genetic algorithm to evolve a control program (i. e. , strategy) for Robby.
One Example Strategy 1 2 3 4. . . 243
Encoding a Strategy 1 2 3 4. . . 243
1 2 3 4. . . 243
1 2 3 4. . . 243 Code: Move. North = 0 Move. South = 1 Move. East = 2 Move. West = 3 Stay. Put = 4 Pick. Up. Can = 5 Move. Random = 6
1 2 3 4. . . 243 0 2 6 5. . . 3. . . 4 Code: Move. North = 0 Move. South = 1 Move. East = 2 Move. West = 3 Stay. Put = 4 Pick. Up. Can = 5 Move. Random = 6
Question: How many possible strategies are there in our representation? 0 2 6 5. . . 3. . . 4
Question: How many possible strategies are there in our representation? 243 values 0 2 6 5. . . 3. . . 4 7 possible actions for each position: 7 × 7 ×. . . × 7
Question: How many possible strategies are there in our representation? 243 values 0 2 6 5. . . 3. . . 4 7 possible actions for each position: 7 × 7 ×. . . × 7 Goal: Have GA search intelligently in this vast space for a good strategy
Genetic algorithm for evolving strategies 1. Generate 200 random strategies (i. e. , programs for controlling Robby) 2. For each strategy, calculate fitness (average reward minus penalties earned on random environments) 3. The strategies pair up and create offspring via crossover with random mutations ― the fitter the parents, the more offspring they create. 4. Keep going back to step 2 until a good-enough strategy is found (or for a set number of generations)
Robby’s fitness function Calculate_Fitness (Robby) { Total_Reward = 0 ; Average_Reward = 0 ‘ For i = 1 to NUM_ENVIRONMENTS { generate_random_environment( ); /*. 5 probability * to place can at * each site */ For j = 1 to NUM_MOVES_PER_ENVIRONMENT { Total_Reward = Total_Reward + perform_action(Robby); } } Fitness = Total_Reward / NUM_ENVIRONMENTS; return(Fitness); }
Genetic algorithm for evolving strategies for Robby 1. Generate 200 random strategies (i. e. , programs for controlling Robby)
Random Initial Population
Genetic algorithm for evolving strategies for Robby 1. Generate 200 random strategies (i. e. , programs for controlling Robby) 2. For each strategy in the population, calculate fitness (average reward minus penalties earned on random environments)
Fitness = Average final score from N moves on each of M random environments
Genetic algorithm for evolving strategies for Robby 1. Generate 200 random strategies (i. e. , programs for controlling Robby) 2. For each strategy in the population, calculate fitness (average reward minus penalties earned on random environments) 3. Strategies are selected according to fitness to become parents. (See code for choice of selection methods. )
Genetic algorithm for evolving strategies for Robby 1. Generate 200 random strategies (i. e. , programs for controlling Robby) 2. For each strategy in the population, calculate fitness (average reward minus penalties earned on random environments) 3. Strategies are selected according to fitness to become parents. (See code for choice of selection methods. ) 4. The parents pair up and create offspring via crossover with random mutations.
Parent 1: Parent 2:
Parent 1: Parent 2:
Parent 1: Parent 2: Child:
Parent 1: Parent 2: Child: Mutate to “ 0” Mutate to “ 4”
Genetic algorithm for evolving strategies for Robby 1. Generate 200 random strategies (i. e. , programs for controlling Robby) 2. For each strategy in the population, calculate fitness (average reward minus penalties earned on random environments) 3. Strategies are selected according to fitness to become parents. (See code for choice of selection methods. ) 4. The parents pair up and create offspring via crossover with random mutations. 5. The offspring are placed in the new population and the old population dies.
Genetic algorithm for evolving strategies for Robby 1. Generate 200 random strategies (i. e. , programs for controlling Robby) 2. For each strategy in the population, calculate fitness (average reward minus penalties earned on random environments) 3. Strategies are selected according to fitness to become parents. (See code for choice of selection methods. ) 4. The parents pair up and create offspring via crossover with random mutations. 5. The offspring are placed in the new population and the old population dies. 6. Keep going back to step 2 until a good-enough strategy is found!
My hand-designed strategy: “If there is a can in the current site, pick it up. ” “Otherwise, if there is a can in one of the adjacent sites, move to that site. ” “Otherwise, choose a random direction to move in, avoiding walls. ”
My hand-designed strategy: “If there is a can in the current site, pick it up. ” “Otherwise, if there is a can in one of the adjacent sites, move to that site. ” “Otherwise, choose a random direction to move in, avoiding walls ” Average fitness of this strategy: 346 (out of max possible 500)
My hand-designed strategy: “If there is a can in the current site, pick it up. ” “Otherwise, if there is a can in one of the adjacent sites, move to that site. ” “Otherwise, choose a random direction to move in, avoiding walls. ” Average fitness of this strategy: 346 (out of max possible 500) Average fitness of GA evolved strategy: 486 (out of max possible 500)
Best fitness in population One Run of the Genetic Algorithm Generation number
Generation 1 Best fitness = 81
Time: 0 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 1 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 2 Score: 5 0 0 1 2 3 4 5 6 7 8 9
Time: 2 Score: 5 0 0 1 2 3 4 5 6 7 8 9
Time: 3 Score: 10 0 0 1 2 3 4 5 6 7 8 9
Time: 3 Score: 10 0 0 1 2 3 4 5 6 7 8 9
Time: 4 Score: 15 0 0 1 2 3 4 5 6 7 8 9
Time: 4 Score: 15 0 0 1 2 3 4 5 6 7 8 9
Generation 14 Best fitness = 1
Time: 0 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 1 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 2 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 3 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Generation 200 Fitness = 240
Time: 0 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 1 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 2 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 3 Score: 10 0 0 1 2 3 4 5 6 7 8 9
Time: 4 Score: 10 0 0 1 2 3 4 5 6 7 8 9
Time: 5 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 6 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 7 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 8 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 9 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 10 0 0 1 2 3 4 5 6 7 8 9 Score: 20 1 2 3 4 5 6 7 8 9
Time: 11 0 0 1 2 3 4 5 6 7 8 9 Score: 20 1 2 3 4 5 6 7 8 9
Time: 12 0 0 1 2 3 4 5 6 7 8 9 Score: 20 1 2 3 4 5 6 7 8 9
Time: 13 0 0 1 2 3 4 5 6 7 8 9 Score: 20 1 2 3 4 5 6 7 8 9
Time: 14 0 0 1 2 3 4 5 6 7 8 9 Score: 30 1 2 3 4 5 6 7 8 9
Time: 15 0 0 1 2 3 4 5 6 7 8 9 Score: 30 1 2 3 4 5 6 7 8 9
Time: 16 0 0 1 2 3 4 5 6 7 8 9 Score: 40 1 2 3 4 5 6 7 8 9
Time: 17 0 0 1 2 3 4 5 6 7 8 9 Score: 40 1 2 3 4 5 6 7 8 9
Time: 18 0 0 1 2 3 4 5 6 7 8 9 Score: 50 1 2 3 4 5 6 7 8 9
Time: 19 0 0 1 2 3 4 5 6 7 8 9 Score: 50 1 2 3 4 5 6 7 8 9
Time: 20 0 0 1 2 3 4 5 6 7 8 9 Score: 60 1 2 3 4 5 6 7 8 9
Generation 1000 Fitness = 492
Time: 0 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 1 Score: 0 0 0 1 2 3 4 5 6 7 8 9
Time: 2 Score: 10 0 0 1 2 3 4 5 6 7 8 9
Time: 3 Score: 10 0 0 1 2 3 4 5 6 7 8 9
Time: 4 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 5 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 6 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 7 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 8 Score: 20 0 0 1 2 3 4 5 6 7 8 9
Time: 9 Score: 30 0 0 1 2 3 4 5 6 7 8 9
Time: 10 0 0 1 2 3 4 5 6 7 8 9 Score: 30 1 2 3 4 5 6 7 8 9
Time: 11 0 0 1 2 3 4 5 6 7 8 9 Score: 40 1 2 3 4 5 6 7 8 9
Time: 12 0 0 1 2 3 4 5 6 7 8 9 Score: 40 1 2 3 4 5 6 7 8 9
Time: 13 0 0 1 2 3 4 5 6 7 8 9 Score: 50 1 2 3 4 5 6 7 8 9
Time: 14 0 0 1 2 3 4 5 6 7 8 9 Score: 50 1 2 3 4 5 6 7 8 9
Time: 15 0 0 1 2 3 4 5 6 7 8 9 Score: 60 1 2 3 4 5 6 7 8 9
Time: 16 0 0 1 2 3 4 5 6 7 8 9 Score: 60 1 2 3 4 5 6 7 8 9
Time: 17 0 0 1 2 3 4 5 6 7 8 9 Score: 70 1 2 3 4 5 6 7 8 9
Time: 18 0 0 1 2 3 4 5 6 7 8 9 Score: 70 1 2 3 4 5 6 7 8 9
Why Did The GA’s Strategy Outperform Mine?
My Strategy
0 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
The GA’s Evolved Strategy
0 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
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https://maxen.pro/maths-iit-jee-mains-study-material-for-circle/ | 1,660,007,071,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00477.warc.gz | 362,527,593 | 29,122 | # Maths IIT JEE Mains Study Material for CIRCLE
## CIRCLE:IIT JEE Mains Study Material
Maths IIT JEE Mains Study Material for CIRCLE Theory Notes Include CIRCLE DEFINITION,STANDARD FORMS OF EQUATION OF A CIRCLE,Central Form,EQUATION OF A CIRCLE IN SOME SPECIAL CASES,EQUATIONS OF THE CIRCUMCIRCLES,INTERCEPT ON A LINE OR LENGTH OF A CHORD,POSITION OF A POINT AND LINE WITH RESPECT TO A CIRCLE,CONDITION OF TANGENCY,EQUATION OF THE TANGENT AND NORMALS AT A POINT,DIRECTOR CIRCLE,CHORD OF CONTACT,POLE&POLAR,CHORD WITH A GIVEN MIDDLE POINT,DIAMETER OF A CIRCLE,FAMILY OF CIRCLES,COMMON CHORD OF TWO CIRCLES,ANGLE OF INTERSECTION OF TWO CIRCLES,POSITION OF TWO CIRCLES,RADICAL AXIS&RADICAL CENTRE,SOME IMPORTANT RESULTS,
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http://mathhelpforum.com/differential-geometry/168761-arcwise-connected-set-2.html | 1,480,790,303,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541066.79/warc/CC-MAIN-20161202170901-00059-ip-10-31-129-80.ec2.internal.warc.gz | 179,469,951 | 15,076 | 1. Originally Posted by Hartlw
To get really nasty about this, how do you know that A isn't empty? Let x be a point in S with a spherical neighborhood. How do you know you can connect from x to any point in the spherical neighborhood without assuming what you are trying to prove?
$A$ isn't empty because $x \in A$ (the one Opalg took when he said "fix $x \in S$").
Another way to look at Opalg's proof is this. Assume that S is indeed connected, and towards a contradiction, assume that S is not path connected, ie. there exist points $x,y \in S$ such that there is no path between x and y. Then take the same sets A and B as before - the last assumption means that B is not empty, and as we've seen A is not empty. Then $S = A \cup B$ where A,B are disjoint (by definition), nonempty and both open, so we get a contradiction to S being a connected set.
I see you agree that A is open. Now take some point, say $b \in B$ and look at a ball $B(b,r) \subseteq S$ around b (which exists since S is open). Now,
1) We know that if $c \in B(b,r)$ then you can construct a path from b to c, namely the straight line from b to c.
2) If there is any $a \in B(b,r)$ such that $a \in A$, then by the definition of A there is a path from x to a. But $a \in B(b,r)$, and by (1) there exists a path from a to b. By "gluing" the two paths together, we get a path from x to b. but $b \in B$, which means that there exists no path from x to b - contradiction.
Can you point what step, exactly, is not clear in this proof?
2. x alone is not a member of A. A is all points connectible to x. You have to prove such a point exists.
A and B are not both open sets. See previous posts.
3. There is a path from x to x, namely, $f:[0,1] \to S, & f(t)=x$, and so $x \in A$.
With which part of Opalg's proof do you not agree?
4. A straight line is defined by distinct end points.
A and B cannot both be open sets. See previous posts.
5. The previous posts don't show that the proof is wrong. To make it clear, I am using the definition that a set S is connected if there are no $A,B \subseteq S$ such that $A \cap B = \varnothing , \ A \cup B = S$ and both A and B are open.
Also, recall that the definition of a path between x and y is a continuous function $f:[0.1] \to S$ such that $f(0) = x, \ f(1) = y$.
The function $f(t) \equiv x$ is constant and therefore continuous, and $f(0) = f(1) = x$ so it is indeed a path between x and x, so $x \in A$.
Please read over the proof again and tell me what step fails to hold. I am convinced that Opalg has shown both A and B are open, but I might be mistaken - where exactly do you think his approach fails?
6. A continuous function cannot consist of a single point.
Opalg's proof fails becaues he assumes both A and B are open, which is impossible by the definition of a connected set.
7. Originally Posted by Hartlw
A continuous function cannot consist of a single point.
Opalg's proof fails becaues he assumes both A and B are open, which is impossible by the definition of a connected set.
A continuous function cannot consist of a single point? What does that even mean?!
Opalg does not assume anywhere that both A and B are open. He proves that they are both open, resulting in the fact that one of them must be empty (this is because S is connected). Since $x \in A$, it must be that B is empty and so we get that A=S and therefore every point in S is path connected to x.
---
To prove that B contains no accumulation points of A: Assume it does, and let $b \in B$ be such a point. Then there is some $r>0$ such that $B(b,r) \cap A \neq \emptyset$. Let $a \in B(b,r) \cap A$. Then the function $\gamma :[0,1] \to S, & \gamma(t) = tb + (1-t)a$ shows us that there is a path from a to b (since the open ball is convex). Now, since $a \in A$, there exists a continuous $\beta :[0,1] \to S$ such that $\beta(0)=x, \beta(1)=a$.
Now, define $\alpha :[0,1] \to S$ as follows:
$\alpha (t) = \left\{
\begin{array}{lr}
\beta(2t) & 0 \leq t \leq \frac{1}{2}\\
\gamma(2t-1) & \frac{1}{2} < t \leq 1
\end{array}
\right.$
So, since $\gamma, \beta$ are continuous and $\beta(1) = \gamma(0) = a$ we get that $\alpha$ is continuous and $\alpha(0)=x, \alpha(1)=b$. This gives us that there is a path from x to b, and so $b \in A$ which is, of course, absurd.
8. I have never heard of a continuous function defined on a domain consisting of a single point. A domain has to be an open set. Look up definition of continuity.
Consider A is [0,1) and B is [1,0]. Can they be connected by a line, ie, across the boundary? Intuitiveley I would say yes, but I have to think about a proof. If they can, I would be inclined to consider your proof.
9. Where did I say it is defined on a single point? $f:[0,1] \to S$ defined by $f(t) = x$ for each $t \in [0,1]$ is defined on $[0,1]$ and its image is the singleton $\{x\}$; the same can be said for every constant function. Surely you've seen a constant function?
Furthermore, I do not understand your notation. What do you mean by $B =[1,0]$? How do you connect two sets by a line?
10. Originally Posted by Defunkt
Where did I say it is defined on a single point? $f:[0,1] \to S$ defined by $f(t) = x$ for each $t \in [0,1]$ is defined on $[0,1]$ and its image is the singleton $\{x\}$; the same can be said for every constant function. Surely you've seen a constant function?
Furthermore, I do not understand your notation. What do you mean by $B =[1,0]$? How do you connect two sets by a line?
If you won't accept that it takes two points to define a line, we have to part ways on this. Personally, I think you lose credibility by trying to defend the indefensible on the basis of ego. I know, I would do the same thing.
[0,1) is the interval on a line closed on the left and open on the right. [0,1] is closed at both ends. 1 is a boundary point between both (every neighborhood of 1 has a point in A and B). If you start a line (continuous) in A and move to the right, can you cross the boundary point? Though it looks like you can, I wouldn't know off-hand how to prove it.
11. Originally Posted by Hartlw
If you won't accept that it takes two points to define a line, we have to part ways on this. Personally, I think you lose credibility by trying to defend the indefensible on the basis of ego. I know, I would do the same thing.
Not to be brusque, but I believe it is you that is here on the math site and it's Defunkt who's trying to help you. If there were things that you might not understand you wouldn't be here. In fact, he is one-hundred percent correct. A constant function is a path.
[0,1) is the interval on a line closed on the left and open on the right. [0,1] is closed at both ends. 1 is a boundary point between both (every neighborhood of 1 has a point in A and B). If you start a line (continuous) in A and move to the right, can you cross the boundary point? Though it looks like you can, I wouldn't know off-hand how to prove it.
Can you possibly rephrase this?
12. Originally Posted by Drexel28
Not to be brusque, but I believe it is you that is here on the math site and it's Defunkt who's trying to help you. If there were things that you might not understand you wouldn't be here. In fact, he is one-hundred percent correct. A constant function is a path.
Can you possibly rephrase this?
If you won't accept that 2 points determine a line, we are not speaking the same language. Also, I don't want to get into a non-tecnical political debate with you.
13. Originally Posted by Hartlw
If you won't accept that it takes two points to define a line, we have to part ways on this. Personally, I think you lose credibility by trying to defend the indefensible on the basis of ego. I know, I would do the same thing.
Given some fixed $x \in S$ I will define $A = \{a \in S : \text{there exists } f:[0,1] \to S \text{ that is continuous and } f(0)=x, f(1)=a\}$. Surely by this definition $x \in A$, regardless of how you or I define a line.
[0,1) is the interval on a line closed on the left and open on the right. [0,1] is closed at both ends. 1 is a boundary point between both (every neighborhood of 1 has a point in A and B). If you start a line (continuous) in A and move to the right, can you cross the boundary point? Though it looks like you can, I wouldn't know off-hand how to prove it.
I don't know how this has any relevance to the question at hand. What is a boundary point between two sets? If I understand correctly you took $A=[0,1), B=[0,1]$. If so, any point $y \in [0,1]$ intersects both B and A in any neighborhood, I don't see how 1 is any different in this sense. What do you mean by start a line? What are you trying to connect? Can you at least be more rigorous?
14. Originally Posted by Hartlw
If you won't accept that 2 points determine a line, we are not speaking the same language. Also, I don't want to get into a non-tecnical political debate with you.
I'm speaking the language of mathematics, what are you speaking?? I'm not one of those internet guys who just wants to screw with you about semantical issues, but what was said is the technical definition. I'd like to help you understand, or clear up anything, so is there any way you could rephrase your second question please?
This is pointless
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# Basic Mathematics
Dimensional Analysis
## The aim of this package is to provide a short
self assessment programme for students who
wish to learn how to use dimensional analysis
to investigate scientific equations.
c 2003 [email protected] , [email protected]
## Last Revision Date: February 23, 2005 Version 1.0
1. Introduction
2. Checking Equations
3. Dimensionless Quantities
4. Final Quiz
Solutions to Exercises
Solutions to Quizzes
## The full range of these packages and some instructions,
should they be required, can be obtained from our web
page Mathematics Support Materials.
Section 1: Introduction 3
1. Introduction
It is important to realise that it only makes sense to add the same sort
of quantities, e.g. area may be added to area but area may not
method to analyse scientific equations called dimensional analysis.
One should note that while units are arbitrarily chosen (an alien
civilisation will not use seconds or weeks), dimensions represent
fundamental quantities such as time.
Basic dimensions are written as follows:
Dimension Symbol
Length L
Time T
Mass M
Temperature K
Electrical current I
## See the package on Units for a review of SI units.
Section 1: Introduction 4
## Example 1 An area can be expressed as a length times a length.
Therefore the dimensions of area are L × L = L2 . (A given area
could be expressed in the SI units of square metres, or indeed in any
appropriate units.) We sometimes write: [area]= L2
In some equations symbols appear which do not have any associated
dimension, e.g., in the formula for the area of a circle, πr2 , π is just
a number and does not have a dimension.
Exercise 1. Calculate the dimensions of the following quantities
(click on the green letters for the solutions).
(a) Volume (b) Speed
(c) Acceleration (d) Density
Quiz Pick out the units that have a different dimension to the other
three.
(a) kg m2 s−2 (b) g mm2 s−2
2 −2
(c) kg m s (d) mg cm2 s−2
Section 2: Checking Equations 5
2. Checking Equations
Example 2 Consider the equation
1
y = x + kx3
2
Since any terms which are added together or subtracted MUST have
the same dimensions, in this case y, x and 12 kx3 have to have the
same dimensions.
We say that such a scientific equation is dimensionally correct. (If
it is not true, the equation must be wrong.)
If in the above equation x and y were both lengths (dimension L) and
1/2 is a dimensionless number, then for the 12 kx3 term to have the
same dimension as the other two, we would need:
dimension of k × L3 = L
L
∴ dimension of k = 3 = L−2
L
So k would have dimensions of one over area, i.e., [k] = L−2 .
Section 2: Checking Equations 6
## Quiz Hooke’s law states that the force, F , in a spring extended by a
length x is given by F = −kx. From Newton’s second law F = ma,
where m is the mass and a is the acceleration, calculate the dimension
of the spring constant k.
(a) M T−2 (b) M T2 (c) M L−2 T−2 (d) M L2 T2
## Example 3 The expressions for kinetic energy E = 21 mv 2 (where
m is the mass of the body and v is its speed) and potential energy
E = mgh (where g is the acceleration due to gravity and h is the
height of the body) look very different but both describe energy. One
way to see this is to note that they have the same dimension.
## Dimension of kinetic energy Dimension of potential energy
2 −1 2
1
2 mv ⇒ M (L T ) mgh ⇒ M (L T−2 ) L
= M L2 T−2 = M L2 T−2
Both expressions have the same dimensions, they can therefore be
added and subtracted from each other.
Section 2: Checking Equations 7
## Exercise 2. Check that the dimensions of each side of the equations
below agree (click on the green letters for the solutions).
(a) The volume of a cylinder of (b) v = u + at for an object with
radius r and length h initial speed u, (constant)
V = πr2 h. acceleration a and final speed
v after a time t.
(c) E = mc2 where E is energy, (d) c = λν, where c is the speed of
m is mass and c is the speed light, λ is the wavelength and
of light. ν is the frequency
## Note that dimensional analysis is a way of checking that equations
might be true. It does not prove that they are definitely correct.
Dimensional analysis would suggest that both Einstein’s equation E =
mc2 and the (incorrect) equation E = 12 mc2 might be true. On the
other hand dimensional analysis shows that E = mc3 makes no sense.
Section 3: Dimensionless Quantities 8
3. Dimensionless Quantities
Some quantities are said to be dimensionless. These are then pure
numbers which would be the same no matter what units are used (e.g.,
the mass of a proton is roughly 1850 times the mass of an electron no
matter how you measure mass).
Example 4 The ratio of one mass m1 to another mass m2 is dimen-
sionless:
m1 M
dimension of the fraction = =1
m2 M
The dimensions have canceled and the result is a number (which is
independent of the units, i.e., it would be the same whether the masses
were measured in kilograms or tonnes).
Note that angles are defined in terms of ratios of lengths. They
are therefore dimensionless! Functions of dimensionless variables are
themselves dimensionless.
Section 3: Dimensionless Quantities 9
## Very many functions are dimensionless. The following quantities are
important cases of dimensionless quantities:
## Trigonometric functions Logarithms
Exponentials Numbers, e.g., π
## Note the following properties:
• Functions of dimensionless variables are dimensionless.
• Dimensionless functions must have dimensionless arguments.
## Quiz If the number of radioactive atoms is found to be given as a
function of time t by
N (t) = N0 exp(−kt)
where N0 is the number of atoms at time t = 0, what is the dimension
of k?
(a) LT (b) log(T) (c) T (d) T−1
Section 3: Dimensionless Quantities 10
## Exercise 3. Determine the dimensions of the expressions below (click
on the green letters for the solutions).
(a) In a Young’s slits experiment the angle θ of constructive interfer-
ence is related to the wavelength λ of the light, the spacing of the
slits d and the order number n by d sin(θ) = nλ. Show that this
is dimensionally correct.
## (b) The Boltzmann distribution in thermodynamics involves the fac-
tor exp(−E/(kT )) where E represents energy, T is the tempera-
ture and k is Boltzmann’s constant. Find the dimensions of k.
## Quiz Use dimensional analysis to see which of the following expres-
sions is allowed if P is a pressure, t is a time, m is a mass, r is a
distance, v is a velocity and T is a temperature.
P rt2
2
Pt Pr Pr
(a) log (b) log (c) log (d) log
mr m mt2 mtT
Section 4: Final Quiz 11
4. Final Quiz
Begin Quiz Choose the solutions from the options given.
1. Newton’s law of gravity states that the gravitational force be-
tween two masses, m1 and m2 , separated by a distance r is given
by F = Gm1 m2 /r2 . What are the dimensions of G?
(a) L3 M−1 T−2 (b) M2 L−2 (c) M L T−2 (d) M−1 L−3 T2
2. The coefficient of thermal expansion, α of a metal bar of length
` whose length expands by ∆` when its temperature increases by
∆T is given by ∆` = α`∆T. What are the dimensions of α?
(a) K−1 (b) L2 T−1 (c) LT−2 (d) L−2 K−1
3. The position of a mass at the end of a spring is found as a function
of time to be A sin(ωt). Select the dimensions of A and ω.
(a) L & T (b) L & Dimensionless
(c) sin(L) & T−1 (d) L & T−1
End Quiz
Solutions to Exercises 12
Solutions to Exercises
Exercise 1(a) A volume is given by multiplying three lengths to-
gether:
Dimension of volume = L×L×L
= L3
So [volume] = L3
(The SI units of volume are cubic metres.)
Click on the green square to return
Solutions to Exercises 13
## Exercise 1(b) Speed is the rate of change of distance with respect
to time.
L
Dimensions of speed =
T
= L T−1
So [speed] = L T−1
(The SI units of speed are metres per second.)
Click on the green square to return
Solutions to Exercises 14
## Exercise 1(c) Acceleration is the rate of change of speed with respect
to time
L T−1
Dimensions of acceleration =
T
= L T−2
So [acceleration] = L T−2
(The SI units of acceleration are metres per second squared.)
Click on the green square to return
Solutions to Exercises 15
Exercise 1(d) Density is the mass per unit volume, so using the
dimension of volume we get:
M
Dimensions of volume =
L3
= M L−3
So [density] = M L−3
(The SI units of density are kg m−3 .)
Click on the green square to return
Solutions to Exercises 16
## Exercise 2(a) We want to check the dimensions of V = πr2 h. We
know that the dimensions of volume are [volume] = L3 . The right
hand side of the equation has dimensions:
dimensions of πr2 h = L2 × L = L3
So both sides have the dimensions of volume.
Click on the green square to return
Solutions to Exercises 17
## Exercise 2(b) We want to check the dimensions of the equation
v = u + at. Since v and u are both speeds, they have dimensions
L T−1 . Therefore we only need to verify that at has this dimension.
To see this consider:
[at] = (L T−2 ) × T = L T−2 × T = L T−1
So the equation is dimensionally correct, and all the terms have di-
mensions of speed.
Click on the green square to return
Solutions to Exercises 18
## Exercise 2(c) We want to check the dimensions of the equation
E = mc2 . Since E is an energy it has dimensions M L2 T−2 . The right
hand side of the equation can also be seen to have this dimension, if
we recall that m is a mass and c is the speed of light (with dimension
L T−1 ). Therefore
[mc2 ] = M (L T−1 )2
= M L2 T−2
So the equation is dimensionally correct, and all terms that we add
have dimensions of energy.
Click on the green square to return
Solutions to Exercises 19
## Exercise 2(d) We want to check the dimensions of the equation
c = λν. Since c is the speed of light it has dimensions L T−1 . The right
hand side of the equation involves wavelength [λ] = L and frequency
[ν] = T−1 . We thus have
[λν] = L × T−1
which indeed also has dimensions of speed, so the equation is dimen-
sionally correct.
Click on the green square to return
Solutions to Exercises 20
## Exercise 3(a) We want to check the dimensions of d sin(θ) = nλ.
Both d and λ have dimensions of length. The angle θ and sin(θ) as
well as the number n must all be dimensionless. Therefore we have
L×1 = 1×L
∴ [d sin(θ)] = [nλ]
So both sides have dimensions of length.
Click on the green square to return
Solutions to Exercises 21
## Exercise 3(b) The factor exp(−E/(kT )) is an exponential and so
must be dimensionless. Therefore its argument −E/(kT ) must also
be dimensionless.
The minus sign simply corresponds to multiplying by minus one
and is dimensionless. Energy, E has dimensions [E] = M L2 T−2
and temperature has dimensions K, so the dimensions of Boltzmann’s
constant are
M L2 T−2
[k] = = M L2 T−2 K−1
K
(So the SI units of Boltzmann’s constant are kg m2 s−2 o C−1 ).
Click on the green square to return
Solutions to Quizzes 22
Solutions to Quizzes
Solution to Quiz: kg2 m s−2 has
dimensions = M2 L T−2
It can be checked that all the other answers have dimension M L2 T−2 .
End Quiz
Solutions to Quizzes 23
## Solution to Quiz: From Hooke’s law, F = −kx, we see that we can
write
F
k=
x
Now F = ma, so the dimensions of force are given by
[F] = M × (L T−2 )
= M L T−2
Therefore the spring constant has dimensions
M L T−2
[k] =
L
= M T−2
End Quiz
Solutions to Quizzes 24
## Solution to Quiz: In N (t) = N0 exp(−kt), the exponential and its
argument must be dimensionless. Therefore kt has to be dimension-
less. Thus
dimensions of k × T = 1
So the dimension of k must be inverse time, i.e., [k] = T−1 .
End Quiz
Solutions to Quizzes 25
## Solution to Quiz: First note that the argument of a logarithm must
be dimensionless. Now pressure is force over area, so it has dimensions
M L T−2
[P ] =
L2
= M L−1 T−2
2
Therefore the combination Pmrt
is dimensionless since
P rt2 (M L−1 T−2 ) × L × T2
=
m M
M
=
M
= 1
None of the other combinations are dimensionless and so it would be
completely meaningless to take their logarithms. End Quiz | 3,377 | 11,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-30 | latest | en | 0.856604 |
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Real Numbers
HCF and LCM
Unit’s Digit
Prime Numbers
NumSys
# (Excerpts from Course Material)
All numbers which are not square root of -1 (minus one), i.e. they are not √(-1) or are not multiple thereof are REAL NUMBERS.
The square root of -1 (minus one), i.e. √(-1) is represented by ‘i’ and is called imaginary number.
All number of the form of a+ib where a and b are real numbers and i is imaginary number are called COMPLEX NUMBERS. Complex numbers are not real numbers.
By Number System, we mean system of REAL NUMBERS.
Rules:
1. Natural numbers (1, 2, 3, 4, …) are at the lowest rung of the system.
2. Next, we have set of whole numbers (0, 1, 2, 3, 4, …) and, obviously, every natural number is a whole number.
3. Next, we have set of integers (……….., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, …) and, obviously, every whole number is an integer.
4. Next, we have set of rational numbers (like ) and, obviously, every integer is a rational number as 2 can be written as or and so on.
5. The numbers which are not rational numbers are irrational number. And of course the numbers which are not rational are NOT integers, whole numbers or natural numbers also.
So, in view of (5) above, the world of real numbers is composed of two types of numbers, i.e. rational number and irrational number. Hence, a real number is either a rational number or irrational number.
You must take not of the following:
Every rational number is not an integer (for example, is a not an integer).
Every integer is not a whole number (‘-1, -2, -3, …… are not whole numbers).
Every whole number is not a natural number (‘0’ is not a natural number).
Every real number can be put in the form of a decimal, for example, 4 can be written as 4.0, 3/2 can be written as 1.5, 10/6 = 1.66666666666666…..
This suggests that every real number can be expressed in a decimal form.
We shall now understand it this way:
The Non-terminating decimals are of two types as you see in the table given above.
1. Non-Terminating – Repeating Decimal (or non-terminating recurring decimal): These numbers are strictly rational numbers. Such numbers are those numbers where a set of decimals is repeating itself, or repeating an innumerable number of times. The example, other than the given above, may be: 20.11111111, 0.32323232……, 14.234234234234234…………, -6.203203203203203……. etc. We denote such repetitions by putting a bar on the set of digits being repeated, as the numbers given above can be written as 20.1, 0.32, 14.234, and -6.203 respectively. All such numbers are strictly “Rational” as these can be put in the form of p/q with q ≠ 0.We shall see it happening a little later. We always place a bar on the set of numbers being repeated, for example, if X = 6.203203203203203……., it is, X = . Or, X = 16.56012012012012……., it is, X = 1 .
2. Non-Terminating Non-Repeating Decimal (or non-terminating non-recurring decimal): These numbers are strictly irrational numbers. Such numbers with decimals are the ones where there is no rule of repetition. The examples, other than that given above, are: 20.100165210116980124875……………, 0.37523923022363020439827……, 14.2130340025436492030………… etc. All such numbers are strictly “Irrational” as these numbers CAN NEVER be put in the form of p/q with q ≠ 0. Values of e (exponential function with x=1, i.e. Euler’s number), (Pi), are examples of irrational numbers. Though, the expressed values of e and i.e. 2.71828 and or 3.14 respectively seem to be rational ones, they are irrational numbers. The rational values are only approximations to their irrational numbers values. Further, as stated above, all square-root values of all such numbers which are not perfect square are irrational numbers.
AlgNumbers
# Algebra of Numbers (Excerpts from Course Material)
## Properties of Algebra of Real Numbers
• Commutative property for addition - If we have two real numbers m and n, then m + n = n + m.
• Commutative property for multiplication - If we have two real numbers m and n, then m*n = n*m.
• Associative property for addition - If we have three real numbers m, n and r, then m + (n + r) = (m + n) + r.
• Associative property for multiplication - If we have three real numbers m, n and r, then m * (n * r) = (m * n) * r.
• Distributive property- If we have real numbers m, n and r, then: m * (n + r) = m*n + m*r and (m + n) * r = m*r + n*r
• Identity element for addition: ‘0’ is the identity element for every real number.
• Inverse element for addition: For every real number m, there exists a real number (-m) such that m + (- m) = 0 (the identity element), this number (-m) is called inverse element of m under addition.
• Identity element for multiplication: ‘1’ is the identity element for every real number except 0.
• Inverse element for addition: For every real number m except 0, there exists a real number (1/m) such that m * (1/m) = 1 (the identity element), this number (1/m) is called inverse element of m under multiplication.
Rules for Divisibility
• Divisibility by 1: Every number (including zero) is divisible by 1 and results in number itself.
• Divisibility by 2: A number is divisible by 2 if its last digit (at unit place) is divisible by 2.
• Divisibility by 3: A number is divisible by 3 if the sum of all its digits is a multiple of 3. Further remember that any digit when written continuously in number of multiples of 3, it is divisible by 3. For example, 777 (7 written 3 times), 888888 (8 written 3 times), 111111111 (1 written 9 times) are divisible by 3. The reason is very simple. The sum of the digits SHALL be divisible by 3.
• Divisibility by 4: A number is divisible by 4 if the number formed by last two digits is divisible by 4 or is a multiple of 4.
• Divisibility by 5: A number is divisible by 5 if its last digit is either 0 or 5. Remember that the sum of five successive whole numbers is always divisible by 5, as a multiple of 5 shall appear in these 5 numbers.
• Divisibility by 6: A number is divisible by 6 if it is divisible both by 2 and 3.The product of 3 consecutive natural numbers is divisible by 6.
• Divisibility by 7: A number is divisible by 7 if we remove the last digit of the number, and subtract its double from the remaining number, and repeat the process unless the number is reduced to one digit number. If this last digit is either 0 or 7, the number is divisible by 7.
First example: 123654
Step I: 12365 – 8 = 12357
Step II: 1235 – 14 = 1221
Step III: 122 – 2 = 120
Step IV: 12 – 0 = 12
Step V: 12 - 4 = 8, which is a single digit number and is neither 0 nor 7. Hence, the given number (123654) is not divisible by 7.
Second example: 123655
Step I: 12365 – 10 = 12355
Step II: 1235 – 10 = 1225
Step III: 122 – 10 = 112
Step IV: 11 – 4 = 7, which is a single digit number and is 7. Hence, the given number (123655) is divisible by 7.
​
Divisibility by 2^n: To examine whether a number (howsoever large) is divisible is by 2^n, is to see the divisibility of last n digits of this number by 2^n. Hence, a number is divisible by 2^n if last ‘n’ digits of the number are divisible by 2^n, otherwise it gives the remainder. For example, the number 3456789876543212345678987648 is divisible by 32, i.e. 2^5, as last 5 digits of the number, i.e. 87648 is divisible by 32.
# Base System of Numbers (Excerpts from Course Material)
BaseSystem
3.2 What is base system for numbers?
Suppose we have a number 6789. This number is six thousand seven hundred eighty nine. That is, 6000 + 700 + 80 + 9. We may write it as: 6x103 + 7x102 + 8x101 + 9x100. This type of representation of numbers is called ‘Decimal System” or “Decimal Base System” and various indices, starting from 0 are called place values. The index 0 (on 10) is called unit value, index 1 is called tens value, index 2 is called hundreds place, index 3 is called thousands place and so on. This system has 10 primary numbers which make all other numbers by various permutations and combinations obeying the rules of indices as given above. These 10 numbers are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Various Base Systems
1. Base 2, called binary system. Every number of this system will use only two digits, i.e. 0 and 1 and every number shall be a permutation combination of these two digits only. For example, some number ‘n’ in this base 2 system can be written as 001011010. This system is widely used in computer science.
2. Base 3. Every number of this system will use only three digits, i.e. 0, 1 and 2.
3. Similarly for all base systems upto decimal base (base 10), the basic principle is that given a base ‘n’, all the numbers under that base (‘n’) shall use digits starting from 0 to (n-1), where .
4. Base 11. This system has eleven numbers (0 to A). The numbers are: 0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 9, A = 10.
5. Base 12. The duodecimal (base 12) or dozenal numbering system. This system has twelve numbers (0 to B). The numbers are: 0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 9, A = 10, B = 11.
6. Base 13. This system has thirteen numbers (0 to C). The numbers are: 0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 9, A = 10, B = 11, C = 12.
7. Base 14. This system has fourteen numbers (0 to D). The numbers are: 0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 9, A = 10, B = 11, C = 12, D = 13.
8. Base 15. This system has fifteen numbers (0 to E). The numbers are: 0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 9, A = 10, B = 11, C = 12, D = 13, E = 14.
9. Base 16. This system has sixteen numbers (0 to E). The numbers are: 0 = 0, 1 = 1, 2 = 2, 3 = 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 9, A = 10, B = 11, C = 12, D = 13, E = 14, F = 16.
# Fermat's Little Theorem
3.31 Conversion from Decimal System to any other base system
This procedure gives you an idea as to how to convert a number GIVEN in Decimal system into one of another system
You must watch closely the steps taken to solve the conversion problem, because it is the only easiest way to learn the method of conversion.
But this is a lengthy process, especially for larger numbers.
Now the simple procedure for converting a number of decimal base to a number of another base, say n, is given below. Given a number with base ten, we shall begin with dividing the given number with n and collect so obtained remainders one by one till we don’t get zero as quotient. The last remainder is most significant digit and the first remainder is least significant digit. Remember that lower the base, higher the number.
3.33 Conversion of a number of any base system to other system of base: The procedure is simple. First convert to base 10, then to the desired one.
When, finally you get C = 1.00, means that the value of A in the next row shall be 0.00, hence no further calculation is possible. Now we have got the number with base 10 equivalent to given number. Take the values of I one by one starting from the last row indicated with an arrow sign and place the decimal sign in the beginning. Hence the number with base 10 equivalent to the given number in base 2 is: 0.101
Indices
# Indices and Surds (Excerpts from Course Material)
4.1 Indices or Powers (or exponent)
Let there be a real number, say r, such that, r is being multiplied by r, say n times.
That is: r x r x r x r x r x r x r x r……. n times.
We can write this huge multiplication process in brevity as r^n and call it as r raised to power n. Here r is called base and n is called power or exponent or index (indices in plural).
Hence in (¾)^4, ¾ is the base and 4 is the exponent. It is equal to calculating ¾ x ¾ x ¾ x ¾ = 243/256.
​
4.2 Rules of Indices
1. If any number except 0, say a (≠0), is raised to the power 0, the result is always 1. That is, a^0 = 1.
2. 1 is raised to any power (any real number), the answer is always 1. Means, 1^n = 1, where n is any real number (1x1x1x1x……. = 1).
3. The reciprocal of the number has same power with sign changed, i.e. a^n = 1/ a^-n. For example, 5^-4 = 1/(5^4) and 15^(3/4) = (1/15)-3/4
4. When the base of two numbers is same, the powers of the two numbers, when two numbers are multiplied, are added, i.e. a^n x a^m = a^(n+m). For example, 17^12 x 17^5 = 17^17.
5. When the base of two numbers is same, the powers of the two numbers when two numbers are divided are subtracted, i.e. a^m ÷ a^n = a^(m-n). For example, 17^12 ÷ 17^5 = 17^(12 – 5) = 17^7.
6. In case, base of two equal exponents are same, their indices are equal. That is if a ^m = a^n, then m=n, in as much as a^m/a^n = 1 è a^(m-n) = 1 = a, that is, m-n = 0, that is, m=n.
7. When a power is raised on a power, it results in multiplication of two powers, i.e. (a^n)^m = a((nm). For example, (17^12)^5 = 17^60.
8. Let a^m = b. Then a = b^(1/m), or . And a is called mth root of b. For example, 4 = 64^(1/3), and 4 is the cube-root of third root of 64.
9. Let a^m = b^n. Then a = b^(n/m), and b = a^(m/n)
​4.5 How to find out square-root?
There are two techniques and both are very simple, Factor Method and Division Method. I shall suggest that for larger numbers you should use division method.
Factor Method (you learn about factors a little later); you make factors and assemble equal factors in pairs, take one factor from each pair and multiply all. The unpaired factors remain inside the square-root. But many a times this method is not suitable for large numbers.
Division Method: This will be clearer to you when you see it doing.
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http://caesarfallsvilla.com/c48bj3/what-is-the-si-unit-of-force-cf49bb | 1,627,788,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154158.4/warc/CC-MAIN-20210801030158-20210801060158-00277.warc.gz | 8,929,492 | 5,688 | 1 N = 100,000 dynes. The fundamental units are the metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd). As the apple falls, the gravitational force on the apple is 2.25 N downward, and the force of the wind on the apple is 1.05 N to the right. The newton (symbol: N) is the SI unit of force.It is named after Sir Isaac Newton because of his work on classical mechanics.A newton is how much force is required to make a mass of one kilogram accelerate at a rate of one metre per second squared. Unit of Force. The current Système International (SI) of units comprises seven fundamental or base units, two supplementary units, several derived units and a range of prefixes. In the standard international system of unit (SI unit) it is expressed in Newton (N). Types of Force. Force is a physical cause that can change the state of motion or the dimensions of an object. The newton's strict definition is listed as the force that gives a mass of one kilogram an acceleration of one meter per second squared. Force is a derived unit in the SI with a special name and symbol. The SI unit of force is the newton (N). New questions in Physics. The MKS system then became the blueprint for today's SI system of units. Dyne: A dyne is a force required to give a mass of 1 gram (1 g) an acceleration of 1 centimetre per second squared (1 cm/s²). Hence, The relation between SI and CGS unit of force is the CGS unit of force is equal to SI unit of force. The newton is named after Isaac Newton. 7) A gust of wind blows an apple from a tree. Other Units of Force: There are two other units of force in Physics; however, they are not as commonly used as the SI unit of force Newton. The CGS unit of force is equal to SI unit of force. Find the magnitude and direction of the net force on the apple. s â2 . Examples of such SI derived units are given in Table 2, where it should be noted that the symbol 1 for quantities of dimension 1 such as mass fraction is generally omitted. In the centimeter gram second system of unit (CGS unit) force is expressed in dyne. Other units of force include dyne; kilogram-force (kilopond) poundal; pound-force; Galileo Galilei and Sir Isaac Newton described how force works mathematically. The Newton is the SI derived unit of Force, which in SI base units is Kgâ¢m/s2(kilogram-meters per second per second).The newton. The SI unit of force is newton denoted by N.According to newtonâs second law of motion: âOne newton (1 N) is the force that produces an acceleration of one meter per second square in a body of mass 1 kg. The SI unit for force is the newton with the symbol N. 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Are obtained from these equations and the seven SI base units force of newton. | 2,482 | 10,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-31 | latest | en | 0.930078 |
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## High School Algebra I Unlocked (2016)
### Lesson 4.2. Single-Variable Inequalities
Game time! It’s a Tuesday night and you’ve decided to play a board game. You notice that the box says the game is for “ages 12 and over.” In other words, you can be 12, 24, or even 110 years old to play the game, but not 11, 4, or 0. This is a real-world example of an inequality, which could also be expressed as “necessary age ≥ 12,” which means the necessary age to play the game is greater than or equal to 12.
Unlike equations, which allow you to find the exact value of a variable, inequalities have a range of values for a variable. There are four signs you should be familiar with when working with inequalities:
Symbol Meaning > greater than < less than ≥ greater than or equal to ≤ less than or equal to
How do these inequality symbols work? Well, if you’re told that x > 10, you know that x is greater than 10. Therefore, x can be any number that’s greater than 10, but x cannot be equal to 10 or any number that’s less than 10; e.g., xcould equal 11, 110, 1,110, or 1,100,000,000, but not 10, 0, −10, −1,100, or −1,100,000,000. In short, an inequality defines a range of values for a variable without giving you one specific value.
Here is how you may see single-variable equations on the SAT.
If = 2, what is the value of x ?
A) 4
B) 16
C) 16
D) 32
Solving Single-Variable Inequalities
Inequalities can be solved in the exact same way as equations, with one minor difference: If you multiply or divide by a negative number, you must flip the orientation of the inequality sign.
Let’s take a look at an example.
EXAMPLE
If 4x + 12 > 16, what is the expression that gives all possible values of x ?
The question asks you to find “all possible values of x,” which is just a fancy way of asking you to solve for x. First, make the 12 disappear from the left side of the inequality by subtracting 12 from both sides.
This leaves us with the inequality 4x > 4. Now get rid of the coefficient, which can be eliminated by dividing both sides of the inequality by 4.
4x > 4
Then reduce the fractions, noting that you can cancel out all the 4s in the equation.
x > 1
Therefore, the expression that provides all possible values of x is x > 1.
Refer back to Example 1
in Lesson 4.1 to see how
we handled single-
variable equations.
Take a look back at this page, where we solved the equation 4x + 12 = 16. Notice that the inequality you just solved is identical except for the inequality sign. Also notice that we solved the inequality in the same exact way as we solved the equation. The main difference with inequalities is that you find a range of values rather than a single value.
Another major difference between equations and inequalities is how you need to handle division or multiplication by a negative number. When working with inequalities, you must flip the orientation of the inequality sign if you multiply or divide by a negative number. Let’s see how this works in the following question.
EXAMPLE
If −8z + 10 > −4z − 6, what is the expression that gives all possible values of z ?
Just like in the previous example, start by moving the numbers and variables to opposite sides of the inequality. First, subtract 10 from both sides of the inequality.
Next, isolate z by adding 4z to both sides.
Now you’re left with −4z > −16. And this is where the difference between equations and inequalities comes into play. Whenever you divide or multiply by a negative number in an inequality, you must flip the orientation of the inequality sign. Here, you would divide both sides of the equation by −4, simplify, and flip the sign from greater than to less than.
−4z > −16
z < 4
Therefore, the expression that gives all possible values of z is z < 4.
Here, you need to find the value of x based on the given equation = 2. In order to isolate x and eliminate the fraction, multiply both sides of the equation by 2:2() = 2(2 ), so = 4. Next, eliminate the radical by squaring both sides of the equation ()2 = (4)2, x = 16 × 2. Finally, simplify the equation to find that x = 16 × 2 and x = 32. Therefore, the correct answer is (D).
Are you multiplying or
dividing by a negative
number in an inequality?
Make sure you flip the
sign! Or, you can move
the variable to the side
of the equation where
it will be positive.
Not a fan of having to remember to flip the sign? If you want to avoid dividing or multiplying by a negative number in an inequality, simply move the variable to the side of the equation where it will be positive.
Consider the step from the previous question where we had the following: −4z > −16. Instead of dividing by −4, you could move −4z to the right side of the equation and −16 to the left side of the equation.
−4z > −16
16 > 4z
Now that all terms are positive, you can divide both sides of the equation by 4 to solve for z, without having to worry about flipping the sign.
16 > 4z
4 > z
Just like before, the expression that gives all possible values of z is 4 > z, or z < 4.
Writing Single-Variable Inequalities
Think back to when we discussed translating English to math. You use the same techniques to write single-variable inequalities as you do for single-variable linear equations. As a quick refresher, here are the words you should know for translating purposes.
English Word Math Equivalent is, are, were, did, does = of × out of ÷ what variable, such as a, b, or c
This table is the
same one from the
beginning of
Lesson 4.2. Keep these
English words and
their math equivalents
math problems.
Now think back to our board game example from earlier. You’ve selected a game and are ready to play, but now you need to know how many people can play at once. The box says the game can have anywhere from 2 to 8 players, which means you can’t play by yourself, nor can you play with 15 of your classmates. So how would you represent this as an inequality? There are two parts to this that we need to address.
The first part of the problem is that you need at least 2 players. Using the letter p to represent the number of players, this can be expressed as p ≥ 2, indicating that you can have 2 or more players. The second part of the problem is that you cannot have more than 8 players. We can represent this as p ≤ 8.
Now combine this information to find that the number of players that can play the game at once is 2 ≤ p ≤ 8; you can have 2, 3, 4, 5, 6, 7, or 8 players, but you cannot have 1, 9, or 199 players.
Let’s try a couple of questions that deal with writing single-variable inequalities.
EXAMPLE
If six less than seven times a number is less than 15, what is the expression that gives all possible values of n ?
This question requires you to translate wordy English to math. You’re told that six less than seven times a number is less than 15. Using your translating skills, you can translate this phrase into the following:
(7 × n) − 6 < 15
Now, add 6 to both sides of the equation:
7n < 21
Finally, divide both sides of the equation by 7:
n < 3
Here is how you may see inequalities on the SAT.
If 7s − 14 < 4 + 6s, which of the following must be true?
A) s ≥ 17
B) s ≤ 18
C) s < 19
D) s < 18
Great work! Hopefully you’re noticing the similarities between translating equations and translating inequalities. The process is essentially the same, but instead of using an equals sign, you use an inequality sign. Let’s try another one.
EXAMPLE
In a card game involving cats, a white cat is worth two points, a black cat is worth three points, and a gray cat is worth five points. If a player collects more than five gray cats, a 20-point bonus is awarded. If Zed has two gray cat cards in his possession, and the number of points Zed has is represented by p, then, in terms of p, which of the following inequalities expresses the range of points Zed can receive after collecting two additional cats?
A) 20 ≤ p ≤ 40
B) 14 ≤ p ≤ 30
C) 14 ≤ p ≤ 20
D) 10 ≤ p ≤ 20
This question should look familiar to you, because we worked through a similar one on this page. (Remember Tashi?) Again, break the question down into smaller pieces to make it more manageable.
The question states that gray cats are worth five points each and Zed has two gray cats in his possession. Thus, Zed has 5 + 5 = 10 points so far. You are also told that Zed will collect two more cats. This is where it gets tricky. How do we know which type of cat Zed will collect? There are a few different scenarios:
White Cat + White Cat = 2 + 2 = 4
White Cat + Black Cat = 2 + 3 = 5
White Cat + Gray Cat = 2 + 5 = 7
Black Cat + Black Cat = 3 + 3 = 6
Black Cat + Gray Cat = 3 + 5 = 8
Gray Cat + Gray Cat = 5 + 5 = 10
However, when you are interested in a range of values, you really need to concern yourself with the least and greatest possible values. In this scenario, Zed will gain the least number of points, 4, by collecting two white cats and the greatest number of points, 10, by collecting two gray cats.
Now you can use the previous information—that Zed already has 10 points—to find the total range of point values. At a minimum, Zed can earn 10 + 4 = 14 points; p ≥ 14. At a maximum, Zed can earn 10 + 10 = 20 points; p ≤ 20. Thus, the inequality that expresses the range of points Zed can receive after four turns is 14 ≤ p ≤ 20, or (C).
| 2,398 | 9,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-10 | latest | en | 0.923103 |
https://essaysmasters.net/business-and-finance-homework-help/micro-economics-202-business-finance-homework-help/ | 1,660,752,153,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573029.81/warc/CC-MAIN-20220817153027-20220817183027-00142.warc.gz | 242,532,962 | 12,850 | # Micro economics 202 | Business & Finance homework help
MICRO ECONOMICS 202 FALL 2014 -- FINAL EXAM
100 POINTS
I
MULTIPLE CHOICE -- TWO POINTS EACH
Table 1
Consider the subjoined daily genesis axioms for MadeFromScratch, Inc. MadeFromScratch sells cupcakes for \$2 each and pays the workers a wage of \$325 per day.
Labor (calculate of workers)
Quantity (cupcakes per day)
Marginal Fruit of Drudge (cupcakes per day)
Value of the Final Fruit of Labor
Wage (per day)
Marginal Profit
0
0
\$325
1
200
\$325
2
380
\$325
3
540
\$325
4
680
\$325
5
800
\$325
6
900
\$325
1. Refer to Table 1. What is the third worker's final fruit of drudge?
1. 120 cupcakes
2. 140 cupcakes
3. 160 cupcakes
4. 180 cupcakes
5. Refer to Table 1. What is the fourth worker's final fruit of drudge?
1. 120 cupcakes
2. 140 cupcakes
3. 160 cupcakes
4. 180 cupcakes
3. Refer to Table 1. What is the fifth worker's final fruit of drudge?
a. 120 cupcakes
b. 140 cupcakes
c. 160 cupcakes
d. 180 cupcakes
4. Refer to Table 1. What is the sixth worker's final fruit of drudge?
a. 100 cupcakes
b. 120 cupcakes
c. 140 cupcakes
d. 160 cupcakes
5. Refer to Table 1 What is the prize of the final fruit of the primitive worker?
a. \$200
b. \$400
c. \$500
d. \$700
6. Refer to Table 1. What is the prize of the final fruit of the second worker? .
a. \$180
b. \$360
c. \$450
d. \$720
7. Refer to Table 1. What is the prize of the final fruit of the fifth worker?
a. \$120
b. \$240
c. \$300
d. \$1,600
8. Refer to Table 1. The final fruit of drudge begins to retrench delay the addition
of which worker?
a. the 1st worker
b. the 2nd worker
c. the 3rd worker
d. the 4th worker
9. Refer to Table 1. What is the final avail of the fourth worker?
a. \$280
b. \$25
c. –\$5
d. –\$45
10. Refer to Table 1. What is the final avail of the sixth worker?
1. \$100
2. −\$50
3. −\$75
4. −\$125
12. Refer to Table 1. Assuming MadeFromScratch is a competitive, avail-maximizing unshaken, how divers workers conciliate the unshaken rent?
a. 2 workers
b. 3 workers
c. 4 workers
d. 5 workers
13. Refer to Table 1. Assume that MadeFromScratch is a competitive, avail-maximizing unshaken. If the trade appraisement of cupcakes increases from \$2.00 to \$2.50, how divers workers would the unshaken then rent?
a. 2 workers
b. 3 workers
c. 4 workers
d. 5 workers
14. Refer to Table 1. Suppose that there is a technological remove that allows MadeFromScratch employees to
yield over cupcakes than they could anteriorly. Accordingly of this veer, the unshaken’s
a. ask-for for drudge shifts lawful.
b. ask-for for drudge shifts left.
c. contribute of drudge shifts lawful.
d. contribute of drudge shifts left.
15. Refer to Table 1. Suppose that the unshaken suffers a mislaying of some of their technology such as the stealing of their industrial mixers. After the stealing, MadeFromScratch employees yield fewer cupcakes than they could anteriorly accordingly they must mix the cupcake pound by index rather than using the high-speed mixers. Accordingly of this veer, the unshaken’s
a. ask-for for drudge shifts lawful.
b. ask-for for drudge shifts left.
c. contribute of drudge shifts lawful.
d. contribute of drudge shifts left.
16. Which of the subjoined is most slight an minor cheerful-tempered?
1. an ancestral car
2. gasoline
3. a bus ticket
4. an airline ticket
17. A cheerful-tempered-tempered is an minor cheerful-tempered-tempered if the consumer buys over of it when
a. his pay rises.
b. the appraisement of the cheerful-tempered-tempered falls.
c. the appraisement of a commute cheerful-tempered-tempered rises.
d. his pay falls.
18. Pepsi and pizza are natural cheerful-tempereds. When the appraisement of pizza rises, the superabundance chattels causes Pepsi to be relatively
a.over valuable, so the consumer buys over Pepsi.
b.over valuable, so the consumer buys close Pepsi.
c.close valuable, so the consumer buys over Pepsi.
d.close valuable, so the consumer buys close Pepsi.
19. You can hold of an composure flexion as an
a.equal-cost flexion.
b.equal-marginal-cost flexion.
c.equal-utility flexion.
d.equal-marginal-utility flexion.
20. A Giffen cheerful-tempered-tempered is one for which the size ask-fored rises as the appraisement rises accordingly the pay chattels
1. reinforces the superabundance chattels.
2. reinforces and is main than the superabundance chattels.
3. counteracts but is smaller than the superabundance chattels.
4. counteracts and is main than the superabundance chattels.
21. How are the subjoined three questions described: 1) Do all ask-for flexions expand downward? 2) How do compensation seek drudge contribute? 3) How do cause rates seek accustomed reluctant?
1. They all describe to macroeconomics.
2. They all describe to monetary economics.
3. They all describe to the supposition of consumer dainty.
4. They are not described to each other in any way.
22. Just as the supposition of the competitive unshaken provides a over accomplished brains of contribute, the supposition of consumer dainty provides a over accomplished brains of
a.demand.
b.profits.
c.genesis possibility frontiers.
d.wages.
23. The supposition of consumer dainty most air-tight examines which of the subjoined Ten Principles of Economics?
b.Governments can casually rectify trade outcomes.
c.Trade can mould everyone amend off.
d.Markets are usually a cheerful-tempered-tempered way to constitute economic earnestness.
24. The supposition of consumer dainty provides the groundwork for brains the
a.constitution of a unshaken.
b.profitability of a unshaken.
d.contribute of a unshaken's fruit.
25. The supposition of consumer dainty
a.underlies the concept of the ask-for for a detail cheerful-tempered.
b.underlies the concept of the contribute of a detail cheerful-tempered.
c.ignores, for the reason of frankness, the trade-offs that consumers visage.
d.can be applied to divers questions about accustomed decisions, but it cannot be applied to questions about compensation and drudge contribute.
II
FIFTY POINTS
BOBBY'S BOTTLE SHOP, LLC
BBS moulds and sells bottles in a exactly competitive trade at a appraisement of \$.50 for each bottle. BBS rents its drudge in a exactly competitive trade at an hourly wage of \$20. The interconnection betwixt the size of drudge rentd and the muchness of output yieldd each hour is presented below:
LABOR Q MPL VMPL WAGE MARGINAL
(= P X MPL) PROFIT
0 0 ----- ------- --------
1 90 90 \$45 \$20 \$25
2 ___ ___ ____ 20 ____
3 ___ ___ ____ 20 ____
4 ___ ___ ____ 20 ____
5 ___ ___ ____ 20 ____
6 ___ ___ ____ 20 ____
7 ___ ____ ____ 20 ____
8. ___ ____ ___ 20 ____
1. Enter the mislaying axioms in the suited attribute on the aloft grid.
2. What is the causative drudge size? Why?
3. Graph the makeweight contribute and ask-for for drudge; delineate all flexions and the axes. | 2,057 | 8,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-33 | latest | en | 0.719836 |
http://mathhelpforum.com/advanced-algebra/180436-given-group-how-investigate-groups-homomorphic-print.html | 1,529,536,650,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863939.76/warc/CC-MAIN-20180620221657-20180621001657-00270.warc.gz | 194,223,726 | 6,151 | # Given a group how to investigate groups that are homomorphic to it
• May 13th 2011, 11:23 AM
topsquark
Given a group how to investigate groups that are homomorphic to it
This is a more general question related to another thread I just posted.
Given a group G I am relatively comfortable looking for other groups which might be isomorphic to it. I'm not great at it, but I have a few ideas I can check out. Doing the same with homomorphisms is giving me a headache.
There is, of course, the trivial homomorphism (ie the one that maps every element of G to the identity), and I'm pretty sure there is homomorphism between a group and any of its subgroups but what does one do when confronted with two groups G and H? As a negative example of what I am asking, how would I show that two groups are not homomorphic (beyond a trivial homomorphism, that is.)
To look for isomorphisms I know to check the number of elements, check to see if there are the same number of subgroups of the same size, etc. ie. the group structures must be the same. That's relatively easy. Knowing that the two groups are candidates I can then go on to construct an isomorphism.
How do you approach this problem using homomorphisms? Or am I barking up the wrong tree here?
Thanks!
-Dan
• May 13th 2011, 01:58 PM
topsquark
Okay, abhishekkgp pointed out to me that it is a lot harder than I first thought to find a homomorphism of a group with its subgroups. There is a way, but it's more reminiscent of a trivial homomorphism and I was hoping to avoid that. For example, take D4. We can define an endomorphism from D4 to its subgroup C4:
$\displaystyle \begin{array}{c} e \to e \\ C_4 \to C_4 \\ C_2 \to C_2 \\ C_4^{-1} \to C_4^{-1} \\ \sigma _x \to e \\ \sigma _y \to e \\ \sigma _1 \to e \\ \sigma _2 \to e \end{array}$
(I'll define the various reflection planes if anyone cares to know them.) This is a homomorphism, but not a very interesting one.
-Dan
• May 13th 2011, 06:49 PM
Drexel28
Quote:
Originally Posted by topsquark
This is a more general question related to another thread I just posted.
Given a group G I am relatively comfortable looking for other groups which might be isomorphic to it. I'm not great at it, but I have a few ideas I can check out. Doing the same with homomorphisms is giving me a headache.
There is, of course, the trivial homomorphism (ie the one that maps every element of G to the identity), and I'm pretty sure there is homomorphism between a group and any of its subgroups but what does one do when confronted with two groups G and H? As a negative example of what I am asking, how would I show that two groups are not homomorphic (beyond a trivial homomorphism, that is.)
To look for isomorphisms I know to check the number of elements, check to see if there are the same number of subgroups of the same size, etc. ie. the group structures must be the same. That's relatively easy. Knowing that the two groups are candidates I can then go on to construct an isomorphism.
How do you approach this problem using homomorphisms? Or am I barking up the wrong tree here?
Thanks!
-Dan
There is always (assuming you're talking about finite groups) the order problem. Namely, let $\displaystyle G,H$ be finite groups and $\displaystyle \phi\in\text{Hom}\left(G,H\right)$ note that since $\displaystyle \phi(G)\leqslant H$ we have (by Lagrange's theorem) that $\displaystyle \left|\phi(G)\right|\mid |H|$ but of course since (by the FIT) we have $\displaystyle \left|G\right|=\left|\phi(G)\right|\left|\ker(\phi )\right|$ and so $\displaystyle \left|\phi(G)\right|\mid |G|$. Thus, if for example $\displaystyle \left(|G|,|H|\right)=1$ you have that the only homomorphism $\displaystyle G\to H$ is trivial. As another simple (haha) example suppose $\displaystyle G$ is simple and you had a homomorphism $\displaystyle \phi:G\to H$ you know then that either $\displaystyle \ker\phi=G$ or $\displaystyle \ker\phi=\{e\}$ but if $\displaystyle \ker\phi=\{e\}$ then $\displaystyle G\cong\phi(G)\leqslant H$ and so in particular $\displaystyle |G|\mid|H|$ so that the only homomorphism from a simple group $\displaystyle G$ to another group whose order is not a multiple of $\displaystyle G$'s order is the trivial one. Also, things like being cyclic, or abelian even are preserved under homomorphic image so that if $\displaystyle H$ had no abelian subgroups and $\displaystyle \phi:G\to H$ is a homomorphism and $\displaystyle G$ abelian then you can conclude $\displaystyle \phi$ is trivial. The list goes on and on.
• May 14th 2011, 12:57 AM
Deveno
the key to determining whether a homomorphism (besides the trivial one, which always exist) exists between 2 groups, depends on the normal subgroups of G.
put another way, a homomorphic image of G IS a factor group of G (ok, isomorphic to one, but usually "up to isomorphism" is all one is interested in).
so if we have a homomorphism φ:G-->H, that means G/ker(φ) is isomorphic to a subgroup of H.
for finite groups that means for φ to be non-trivial, (|G|, |H|) ≠ 1, for example.
another thing you can do is look at the normal subgroups of H, and the normal subgroups of G.
there is a 1-1 correspondence between the normal subgroups of G containing the kernel K of a homomorphism, φ,
and the normal subgroups of φ(G), so if H has 3 normal subgroups, but G only has 2, there isn't any surjective homomorphism
from G ONTO H. of course, this means if G is simple, the possible homomorphic images of G are an isomorphic copy of G, or the trivial group.
yet another thing you can do, is view both groups as permutation groups (if G is finite), where you can look at the cycle structures of G and H
(this is similar to looking at orders of elements, but can reveal more information).
there is something to be said for the notion that a group is defined by the possible homomorphic images of it, or into it.
• May 14th 2011, 03:39 AM
topsquark
Thanks everyone. Now it's time to play with some of these concepts and get experience...
-Dan
• May 14th 2011, 11:52 AM
Swlabr
Quote:
Originally Posted by topsquark
...and I'm pretty sure there is homomorphism between a group and any of its subgroups
Sorry, but I haven't actually read the rest of this thread (I'm in a hurry just now) but this HAS to be commented on, and I don't think anyone has already (bus as I said, haven't read the thread properly)...anyway, this is not true. A simple group has no proper homomorphic images. For example, A_5 has lost of subgroups, but no homomorphisms exist from A_5 to any subgroups other than A_5 and 1, the trivial subgroup.
On the other hand, there exists a homomorphism from any given subgroup of a group into the group - clearly every subgroup injects into the group...(it is embedded in it)...so I'm not sure if that is what you mean...
• May 14th 2011, 01:20 PM
topsquark
Quote:
Originally Posted by Swlabr
Sorry, but I haven't actually read the rest of this thread (I'm in a hurry just now) but this HAS to be commented on, and I don't think anyone has already (bus as I said, haven't read the thread properly)...anyway, this is not true. A simple group has no proper homomorphic images. For example, A_5 has lost of subgroups, but no homomorphisms exist from A_5 to any subgroups other than A_5 and 1, the trivial subgroup.
On the other hand, there exists a homomorphism from any given subgroup of a group into the group - clearly every subgroup injects into the group...(it is embedded in it)...so I'm not sure if that is what you mean...
Yeah, I found one between D4 and it's subgroup C4 below. I'm not sure if it is what you would call trivial or not, but it's certainly not very useful!
-Dan
• May 14th 2011, 01:48 PM
Deveno
i am looking at the "homomorphism" you have exhibited, and i don't think it is one. there are 5 elements that map to e, which suggests that the "kernel" is a subgroup of order 5.
this is impossible.
furthermore, if one looks at (σ_x) o (σ _y) (i am assuming these represent reflections about the x-axis and y-axis, respectively),
we have (σ_x) o (σ _y) = C2, but e*e ≠ C2.
nope, not a homomorphism.
in point of fact, the reflections of D4 do not even form a group, much less a normal subgroup.
is there ANY surjective homomorphism D4-->C4? well, if there was, the kernel would have to be a normal subgroup of order 2.
D4 has precisely 1 normal subgroup of order 2, the subgroup generated by the 180 degree rotation (the subgroup generated
by any reflection is NOT normal. you should convince yourself of this).
i believe you have called this element C2. so what is D4/<C2>? for simplicity of notation, i will call <C2> H.
i will also call C4, "r" (somethimes rho is used), and σ_x "s".
then r^2 = C2, r^3 = C4^-1, rs = σ1 (assuming this is the reflection about y = x), r^2s = σ_y,
r^3s = σ2. so D4 = {1,r,r^2,r^3,s,rs,r^2s,r^3s}.
H = <C2> = {1,r^2}. let's look at the orders of cosets of H.
rH = {r,r^3}. (rH)(rH) = r^2H = H, so rH has order 2.
sH = {s, r^2s}. again, (sH)(sH) = H, so sH has order 2. hmm. a cyclic group of order 4 has only ONE element of order 2.
we have (at least) 2, meaning D4/H must be isomorphic to the kelin 4-group.
looks like there isn't ANY homomorphism from D4 onto C4, you were mis-informed.
imagine you were a square. if you were subjected to a transformation of D4, you'd be "back in your parking space", but you'd be facing a (perhaps) different direction.
perhaps you've labelled your home vertices, or the sides, so that you can tell just how you've moved around.
now, suppose someone removed the labels. now, you can't tell if you've been rotated, but you can tell if you're upside-down or not.
this is the effect of taking D4/<C4>. now suppose the celestial crank that "turns" you, is malfunctioning, and always turns 2 quarters of a turn,
instead of one-quarter. the diagonal reflections are no longer possible, but the horizontal and vertical ones are.
this is the effect of taking D4/<C2>.
• May 14th 2011, 04:29 PM
topsquark
Quote:
Originally Posted by Deveno
i am looking at the "homomorphism" you have exhibited, and i don't think it is one. there are 5 elements that map to e, which suggests that the "kernel" is a subgroup of order 5.
this is impossible.
furthermore, if one looks at (σ_x) o (σ _y) (i am assuming these represent reflections about the x-axis and y-axis, respectively),
we have (σ_x) o (σ _y) = C2, but e*e ≠ C2.
nope, not a homomorphism.
You are right. I apparently missed that example when I check it. There are probably others.
in point of fact, the reflections of D4 do not even form a group, much less a normal subgroup.
Quote:
Originally Posted by Deveno
looks like there isn't ANY homomorphism from D4 onto C4, you were mis-informed.
Not misinformed. My intuition seems to have been way off. It seemed simple enough when I thought about it. Then I was advised that this was not actually the case, and the one example I did come up with is apparently not a homomorphism.
Point taken. (Nod)
-Dan | 3,006 | 10,997 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-26 | latest | en | 0.939926 |
https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=73931&oldid=73893 | 1,623,778,897,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621450.29/warc/CC-MAIN-20210615145601-20210615175601-00598.warc.gz | 117,162,671 | 18,833 | # Difference between revisions of "2011 AMC 10A Problems"
2011 AMC 10A (Answer Key)Printable version: Wiki | AoPS Resources • PDF Instructions This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator). Figures are not necessarily drawn to scale. You will have 75 minutes working time to complete the test. 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
## Problem 1
A cell phone plan costs $\textdollar 20$ each month, plus $5$¢ per text message sent, plus 10¢ for each minute used over $30$ hours. In January Michelle sent $100$ text messages and talked for $30.5$ hours. How much did she have to pay?
$\textbf{(A)}\ \textdollar 24.00 \qquad\textbf{(B)}\ \textdollar 24.50 \qquad\textbf{(C)}\ \textdollar 25.50\qquad\textbf{(D)}\ \textdollar 28.00\qquad\textbf{(E)}\ \textdollar 30.00$
## Problem 2
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?
$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$
## Problem 3
Suppose $[a\ b]$ denotes the average of $a$ and $b$, and $\{a\ b\ c\}$ denotes the average of $a, b$, and $c$. What is $\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}$?
$\textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3}$
## Problem 4
Let $X$ and $Y$ be the following sums of arithmetic sequences: $\begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*}$ What is the value of $Y - X$?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$
## Problem 5
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of $12$, $15$, and $10$ minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14$
## Problem 6
Set $A$ has 20 elements, and set $B$ has 15 elements. What is the smallest possible number of elements in $A \cup B$, the union of $A$ and $B$?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300$
## Problem 7
Which of the following equations does NOT have a solution?
$\text{(A)}\:(x+7)^2=0$
$\text{(B)}\:|-3x|+5=0$
$\text{(C)}\:\sqrt{-x}-2=0$
$\text{(D)}\:\sqrt{x}-8=0$
$\text{(E)}\:|-3x|-4=0$
## Problem 8
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60$
## Problem 9
A rectangular region is bounded by the graphs of the equations $y=a, y=-b, x=-c,$ and $x=d$, where $a,b,c,$ and $d$ are all positive numbers. Which of the following represents the area of this region?
$\textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad$ $+bc-bd\qquad\textbf{(C)}\ ac+ad$ $-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad$ $+bc+bd\qquad\textbf{(E)}\ ac-ad$ $-bc+bd$
## Problem 10
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $\textdollar 17.71$. What was the cost of a pencil in cents?
$\text{(A)}\,7 \qquad\text{(B)}\,11 \qquad\text{(C)}\,17 \qquad\text{(D)}\,23 \qquad\text{(E)}\,77$
## Problem 11
Square $EFGH$ has one vertex on each side of square $ABCD$. Point $E$ is on $\overline{AB}$ with $AE=7\cdot EB$. What is the ratio of the area of $EFGH$ to the area of $ABCD$?
$\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4}$
## Problem 12
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?
$\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17$
## Problem 13
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?
$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$
## Problem 14
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?
$\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}$
## Problem 15
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?
$\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840$
## Problem 16
Which of the following is equal to $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$?
$\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6$
## Problem 17
In the eight-term sequence $A,B,C,D,E,F,G,H$, the value of $C$ is 5 and the sum of any three consecutive terms is 30. What is $A+H$?
$\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43$
## Problem 18
Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$. What is the area inside Circle $C$ but outside circle $A$ and circle $B$ ?
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}$
## Problem 19
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?
$\textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62$
## Problem 20
Two points on the circumference of a circle of radius $r$ are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?
$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$
## Problem 21
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?
$\textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16}$
## Problem 22
Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$
## Problem 23
Seven students count from 1 to 1000 as follows:
•Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
•Finally, George says the only number that no one else says.
What number does Georgee say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$
## Problem 24
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?
$\textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6}$
## Problem 25
Let $R$ be a square region and $n\ge4$ an integer. A point $X$ in the interior of $R$ is called $n\text{-}ray$ partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
$\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500$ | 3,743 | 11,065 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 94, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-25 | latest | en | 0.78097 |
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Centripetal force keeps an object in circular motion.
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Centripetal force keeps an object in circular motion.
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Centripetal force keeps an object in circular motion.
Which moves faster on a merry-go-round, a horse near the outside rail or one near the inside rail? While a hamster rotates its cage about an axis, does the hamster rotate or does it revolve about the same axis? We begin to answer these questions by discussing the difference between rotation and revolution.
10.1Rotation and Revolution
Two types of circular motion are rotation and revolution.
10.1Rotation and Revolution
• An axis is the straight line around which rotation takes place.
• When an object turns about an internal axis—that is, an axis located within the body of the object—the motion is called rotation, or spin.
• When an object turns about an external axis, the motion is called revolution.
10.1Rotation and Revolution
The Ferris wheel turns about an axis.
The Ferris wheel rotates, while the riders revolve about its axis.
10.1Rotation and Revolution
• Earth undergoes both types of rotational motion.
• It revolves around the sun once every 365 ¼ days.
• It rotates around an axis passing through its geographical poles once every 24 hours.
10.1Rotation and Revolution
What are two types of circular motion?
10.2Rotational Speed
Tangential speed depends on rotational speed and the distance from the axis of rotation.
10.2Rotational Speed
The turntable rotates around its axis while a ladybug sitting at its edge revolves around the same axis.
10.2Rotational Speed
Which part of the turntable moves faster—the outer part where the ladybug sits or a part near the orange center?
It depends on whether you are talking about linear speed or rotational speed.
10.2Rotational Speed
Types of Speed
• Linear speed is the distance traveled per unit of time.
• A point on the outer edge of the turntable travels a greater distance in one rotation than a point near the center.
• The linear speed is greater on the outer edge of a rotating object than it is closer to the axis.
• The speed of something moving along a circular path can be called tangential speed because the direction of motion is always tangent to the circle.
10.2Rotational Speed
• Rotational speed (sometimes called angular speed) is the number of rotations per unit of time.
• All parts of the rigid turntable rotate about the axis in the same amount of time.
• All parts have the same rate of rotation, or the same number of rotations per unit of time. It is common to express rotational speed in revolutions per minute (RPM).
10.2Rotational Speed
• All parts of the turntable rotate at the same rotational speed.
• A point farther away from the center travels a longer path in the same time and therefore has a greater tangential speed.
10.2Rotational Speed
• All parts of the turntable rotate at the same rotational speed.
• A point farther away from the center travels a longer path in the same time and therefore has a greater tangential speed.
• A ladybug sitting twice as far from the center moves twice as fast.
10.2Rotational Speed
Tangential and Rotational Speed
Tangential speed and rotational speed are related. Tangential speed is directly proportional to the rotational speed and the radial distance from the axis of rotation.
Tangential speed ~ radial distance × rotational speed
10.2Rotational Speed
• In symbol form,
• v ~ r
• where v is tangential speed and (pronounced oh MAY guh) is rotational speed.
• You move faster if the rate of rotation increases (bigger ).
• You also move faster if you are farther from the axis (bigger r).
10.2Rotational Speed
At the axis of the rotating platform, you have no tangential speed, but you do have rotational speed. You rotate in one place.
As you move away from the center, your tangential speed increases while your rotational speed stays the same.
Move out twice as far from the center, and you have twice the tangential speed.
10.2Rotational Speed
think!
At an amusement park, you and a friend sit on a large rotating disk. You sit at the edge and have a rotational speed of 4 RPM and a linear speed of 6 m/s. Your friend sits halfway to the center. What is her rotational speed? What is her linear speed?
10.2Rotational Speed
think!
At an amusement park, you and a friend sit on a large rotating disk. You sit at the edge and have a rotational speed of 4 RPM and a linear speed of 6 m/s. Your friend sits halfway to the center. What is her rotational speed? What is her linear speed?
Her rotational speed is also 4 RPM, and her linear speed is 3 m/s.
10.2Rotational Speed
How do the wheels of a train stay on the tracks?
The train wheels stay on the tracks because their rims are slightly tapered.
10.2Rotational Speed
A curved path occurs when a tapered cup rolls. The wider part of the cup travels a greater distance per revolution.
10.2Rotational Speed
A tapered cup rolls in a curve because the wide part of the cup rolls faster than the narrow part.
10.2Rotational Speed
• Fasten a pair of cups together at their wide ends and roll the pair along a pair of parallel tracks.
• The cups will remain on the track.
• They will center themselves whenever they roll off center.
10.2Rotational Speed
A pair of cups fastened together will stay on the tracks as they roll.
10.2Rotational Speed
When the pair rolls to the left of center, the wider part of the left cup rides on the left track while the narrow part of the right cup rides on the right track.
This steers the pair toward the center.
If it “overshoots” toward the right, the process repeats, this time toward the left, as the wheels tend to center themselves.
10.2Rotational Speed
• The wheels of railroad trains are similarly tapered. This tapered shape is essential on the curves of railroad tracks.
• On any curve, the distance along the outer part is longer than the distance along the inner part.
• When a vehicle follows a curve, its outer wheels travel faster than its inner wheels. This is not a problem because the wheels roll independent of each other.
• For a train, however, pairs of wheels are firmly connected like the pair of fastened cups, so they rotate together.
10.2Rotational Speed
The tapered shape of railroad train wheels (shown exaggerated here) is essential on the curves of railroad tracks.
10.2Rotational Speed
When a train rounds a curve, the wheels have different linear speeds for the same rotational speed.
10.2Rotational Speed
When a train rounds a curve, the wheels have different linear speeds for the same rotational speed.
10.2Rotational Speed
think!
Train wheels ride on a pair of tracks. For straight-line motion, both tracks are the same length. But which track is longer for a curve, the one on the outside or the one on the inside of the curve?
10.2Rotational Speed
think!
Train wheels ride on a pair of tracks. For straight-line motion, both tracks are the same length. But which track is longer for a curve, the one on the outside or the one on the inside of the curve?
The outer track is longer—just as a circle with a greater radius has a greater circumference.
10.2Rotational Speed
What is the relationship among tangential speed, rotational speed, and radial distance?
10.3Centripetal Force
The centripetal force on an object depends on the object’s tangential speed, its mass, and the radius of its circular path.
10.3Centripetal Force
• Velocity involves both speed and direction.
• When an object moves in a circle, even at constant speed, the object still undergoes acceleration because its direction is changing.
• This change in direction is due to a net force (otherwise the object would continue to go in a straight line).
• Any object moving in a circle undergoes an acceleration that is directed to the center of the circle—a centripetal acceleration.
10.3Centripetal Force
Centripetal means “toward the center.”
The force directed toward a fixed center that causes an object to follow a circular path is called a centripetal force.
10.3Centripetal Force
Examples of Centripetal Forces
If you whirl a tin can on the end of a string, you must keep pulling on the string—exerting a centripetal force.
The string transmits the centripetal force, pulling the can from a straight-line path into a circular path.
10.3Centripetal Force
The force exerted on a whirling can is toward the center. No outward force acts on the can.
10.3Centripetal Force
• Centripetal forces can be exerted in a variety of ways.
• The “string” that holds the moon on its almost circular path, for example, is gravity.
• Electrical forces provide the centripetal force acting between an orbiting electron and the atomic nucleus in an atom.
• Anything that moves in a circular path is acted on by a centripetal force.
10.3Centripetal Force
Centripetal force is not a basic force of nature, but is the label given to any force that is directed toward a fixed center.
If the motion is circular and executed at constant speed, this force acts at right angles (tangent) to the path of the moving object.
10.3Centripetal Force
• Centripetal force holds a car in a curved path.
• For the car to go around a curve, there must be sufficient friction to provide the required centripetal force.
10.3Centripetal Force
• Centripetal force holds a car in a curved path.
• For the car to go around a curve, there must be sufficient friction to provide the required centripetal force.
• If the force of friction is not great enough, skidding occurs.
10.3Centripetal Force
The clothes in a washing machine are forced into a circular path, but the water is not, and it flies off tangentially.
10.3Centripetal Force
Calculating Centripetal Forces
Greater speed and greater mass require greater centripetal force.
Traveling in a circular path with a smaller radius of curvature requires a greater centripetal force.
Centripetal force, Fc, is measured in newtons when m is expressed in kilograms, v in meters/second, and r in meters.
10.3Centripetal Force
A conical pendulum is a bob held in a circular path by a string attached above.
This arrangement is called a conical pendulum because the string sweeps out a cone.
10.3Centripetal Force
The string of a conical pendulum sweeps out a cone.
10.3Centripetal Force
• Only two forces act on the bob: mg, the force due to gravity, and T, tension in the string.
• Both are vectors.
10.3Centripetal Force
The vector T can be resolved into two perpendicular components, Tx(horizontal), and Ty(vertical).
If vector T were replaced with forces represented by these component vectors, the bob would behave just as it does when it is supported only by T.
10.3Centripetal Force
The vector T can be resolved into a horizontal (Tx) component and a vertical (Ty) component.
10.3Centripetal Force
Since the bob doesn’t accelerate vertically, the net force in the vertical direction is zero.
Therefore Tymust be equal and opposite to mg.
Tx is the net force on the bob–the centripetal force. Its magnitude is mv/r2, where r is the radius of the circular path.
10.3Centripetal Force
Centripetal force keeps the vehicle in a circular path as it rounds a banked curve.
10.3Centripetal Force
Suppose the speed of the vehicle is such that the vehicle has no tendency to slide down the curve or up the curve.
At that speed, friction plays no role in keeping the vehicle on the track.
Only two forces act on the vehicle, one mg, and the other the normal force n (the support force of the surface). Note that n is resolved into nxand nycomponents.
10.3Centripetal Force
Again, nyis equal and opposite to mg, and nxis the centripetal force that keeps the vehicle in a circular path.
Whenever you want to identify the centripetal force that acts on a circularly moving object, it will be the net force that acts exactly along the radial direction—toward the center of the circular path.
10.3Centripetal Force
What factors affect the centripetal force acting on an object?
10.4Centripetal and Centrifugal Forces
The “centrifugal-force effect” is attributed not to any real force but to inertia—the tendency of the moving body to follow a straight-line path.
10.4Centripetal and Centrifugal Forces
Sometimes an outward force is also attributed to circular motion.
This apparent outward force on a rotating or revolving body is called centrifugal force. Centrifugal means “center-fleeing,” or “away from the center.”
10.4Centripetal and Centrifugal Forces
When the string breaks, the whirling can moves in a straight line, tangent to—not outward from the center of—its circular path.
10.4Centripetal and Centrifugal Forces
In the case of the whirling can, it is a common misconception to state that a centrifugal force pulls outward on the can.
In fact, when the string breaks the can goes off in a tangential straight-line path because no force acts on it.
So when you swing a tin can in a circular path, there is no force pulling the can outward.
Only the force from the string acts on the can to pull the can inward. The outward force is on the string, not on the can.
10.4Centripetal and Centrifugal Forces
The only force that is exerted on the whirling can (neglecting gravity) is directed toward the center of circular motion. This is a centripetal force. No outward force acts on the can.
10.4Centripetal and Centrifugal Forces
The can provides the centripetal force necessary to hold the ladybug in a circular path.
10.4Centripetal and Centrifugal Forces
The can presses against the bug’s feet and provides the centripetal force that holds it in a circular path.
The ladybug in turn presses against the floor of the can.
Neglecting gravity, the only force exerted on the ladybug is the force of the can on its feet.
From our outside stationary frame of reference, we see there is no centrifugal force exerted on the ladybug.
10.4Centripetal and Centrifugal Forces
What causes the “centrifugal-force effect”?
10.5Centrifugal Force in a Rotating Reference Frame
Centrifugal force is an effect of rotation. It is not part of an interaction and therefore it cannot be a true force.
10.5Centrifugal Force in a Rotating Reference Frame
From the reference frame of the ladybug inside the whirling can, the ladybug is being held to the bottom of the can by a force that is directed away from the center of circular motion.
10.5Centrifugal Force in a Rotating Reference Frame
From a stationary frame of reference outside the whirling can, we see there is no centrifugal force acting on the ladybug inside the whirling can.
However, we do see centripetal force acting on the can, producing circular motion.
10.5Centrifugal Force in a Rotating Reference Frame
Nature seen from the reference frame of the rotating system is different.
In the rotating frame of reference of the whirling can, both centripetal force (supplied by the can) and centrifugal force act on the ladybug.
10.5Centrifugal Force in a Rotating Reference Frame
The centrifugal force appears as a force in its own right, as real as the pull of gravity.
However, there is a fundamental difference between the gravity-like centrifugal force and actual gravitational force.
Gravitational force is always an interaction between one mass and another. The gravity we feel is due to the interaction between our mass and the mass of Earth.
10.5Centrifugal Force in a Rotating Reference Frame
In a rotating reference frame the centrifugal force has no agent such as mass—there is no interaction counterpart.
For this reason, physicists refer to centrifugal force as a fictitious force, unlike gravitational, electromagnetic, and nuclear forces.
Nevertheless, to observers who are in a rotating system, centrifugal force is very real. Just as gravity is ever present at Earth’s surface, centrifugal force is ever present within a rotating system.
10.5Centrifugal Force in a Rotating Reference Frame
think!
A heavy iron ball is attached by a spring to a rotating platform, as shown in the sketch. Two observers, one in the rotating frame and one on the ground at rest, observe its motion. Which observer sees the ball being pulled outward, stretching the spring? Which observer sees the spring pulling the ball into circular motion?
10.5Centrifugal Force in a Rotating Reference Frame
think!
A heavy iron ball is attached by a spring to a rotating platform, as shown in the sketch. Two observers, one in the rotating frame and one on the ground at rest, observe its motion. Which observer sees the ball being pulled outward, stretching the spring? Which observer sees the spring pulling the ball into circular motion?
The observer in the reference frame of the rotating platform states that centrifugal force pulls radially outward on the ball, which stretches the spring. The observer in the rest frame states that centripetal force supplied by the stretched spring pulls the ball into circular motion.(Only the observer in the rest frame can identify an action-reaction pair of forces; where action is spring-on-ball, reaction is ball-on-spring. The rotating observer can’t identify a reaction counterpart to the centrifugal force because there isn’t any.)
10.5Centrifugal Force in a Rotating Reference Frame
Why is centrifugal force not considered a true force?
Assessment Questions
• Whereas a rotation takes place about an axis that is internal, a revolution takes place about an axis that is
• external.
• at the center of gravity.
• at the center of mass.
• either internal or external.
Assessment Questions
• Whereas a rotation takes place about an axis that is internal, a revolution takes place about an axis that is
• external.
• at the center of gravity.
• at the center of mass.
• either internal or external.
Assessment Questions
• When you roll a tapered cup across a table, the path of the cup curves because the wider end rolls
• slower.
• at the same speed as the narrow part.
• faster.
• in an unexplained way.
Assessment Questions
• When you roll a tapered cup across a table, the path of the cup curves because the wider end rolls
• slower.
• at the same speed as the narrow part.
• faster.
• in an unexplained way.
Assessment Questions
• When you whirl a tin can in a horizontal circle overhead, the force that holds the can in the path acts
• in an inward direction.
• in an outward direction.
• in either an inward or outward direction.
• parallel to the force of gravity.
Assessment Questions
• When you whirl a tin can in a horizontal circle overhead, the force that holds the can in the path acts
• in an inward direction.
• in an outward direction.
• in either an inward or outward direction.
• parallel to the force of gravity.
Assessment Questions
• When you whirl a tin can in a horizontal circle overhead, the force that the can exerts on the string acts
• in an inward direction.
• in an outward direction.
• in either an inward or outward direction.
• parallel to the force of gravity.
Assessment Questions
• When you whirl a tin can in a horizontal circle overhead, the force that the can exerts on the string acts
• in an inward direction.
• in an outward direction.
• in either an inward or outward direction.
• parallel to the force of gravity.
Assessment Questions
• A bug inside a can whirled in a circle feels a force of the can on its feet. This force acts
• in an inward direction.
• in an outward direction.
• in either an inward or outward direction.
• parallel to the force of gravity.
Assessment Questions
• A bug inside a can whirled in a circle feels a force of the can on its feet. This force acts
• in an inward direction.
• in an outward direction.
• in either an inward or outward direction.
• parallel to the force of gravity. | 4,577 | 20,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-17 | longest | en | 0.896191 |
https://slideum.com/doc/4191606/chapter-15 | 1,591,065,643,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347422065.56/warc/CC-MAIN-20200602002343-20200602032343-00206.warc.gz | 528,383,230 | 7,870 | #### Transcript Chapter 15
```Chapter 15
Section 15.4
Partial Derivatives
Partial Derivatives
The partial derivative of the variable π§ = π π₯, π¦
with respect to either the variables x or y is given
by each of the limits to the right.
ππ§
π π₯ + β, π¦ β π π₯, π¦
= ππ₯ π₯, π¦ = lim
ββ0
ππ₯
β
ππ§
π π₯, π¦ + β β π π₯, π¦
= ππ¦ π₯, π¦ = lim
ββ0
ππ¦
β
Computing Partial Derivatives
To make use of the differentiation rules learned
in earlier calculus courses we realize that these
derivatives only deal with 1 βvariableβ at a time.
To calculate the derivative of the dependent
variable (z for π§ = π π₯, π¦ ) with respect to an
independent variable (x or y for π§ = π π₯, π¦ )
view the independent variable as the βonly
variableβ and all other variables as constants.
For π§ = π π₯, π¦
Example
ππ§
ππ§
ππ§
To find ππ₯ take the derivative of z viewing x
as the variable and y and any expression
dependent on y as a constant.
ππ§
To find ππ¦ take the derivative of z viewing y
as the variable and x and any expression
dependent on x as a constant.
Find ππ₯ and ππ¦ for the function π§ = π π₯, π¦ = π₯ 5 cos π¦ + ln π¦ β π₯π π₯π¦
ππ§
π 5
π
π
2
=
π₯ cos π¦ +
ln π¦ β
π₯π π₯π¦
ππ₯ ππ₯
ππ₯
ππ₯
2
= 5π₯ 4 cos π¦ + 0 β π π₯π¦ + π₯π¦ 2 π π₯π¦
2
= 5π₯ 4 cos π¦ β π π₯π¦ β π₯π¦ 2 π π₯π¦
2
2
2
ππ§
π 5
π
π
2
=
π₯ cos π¦ +
ln π¦ β
π₯π π₯π¦
ππ¦ ππ¦
ππ¦
ππ¦
1
2
= βπ₯ 5 sin π¦ + β π₯π π₯π¦ 2π₯π¦
π¦
1
2
= βπ₯ 5 sin π¦ + β 2π₯ 2 π¦π π₯π¦
π¦
Partial Derivatives with more Variables
If there are more independent variables for
the function than x and y view all the
independent variables that you are not taking
the derivative with respect to as constants.
Example
Find
ππ€
ππ§
For π€ = π π₯, π¦, π§
ππ€
Find ππ§ by taking the derivative of w viewing z
as the variable and x and y and all expressions
depending only on x or y as constants
for the function π€ = π π₯, π¦, π§ = π β2π₯ sec π¦ β π¦ sinh π§ + π₯ 2 arctan π§
ππ€
π β2π₯
π
π 2
π₯2
=
π
sec π¦ β
π¦ sinh π§ +
π₯ arctan π§ = βπ¦ cosh π§ +
ππ§ ππ§
ππ§
ππ§
1 + π§2
Implicit Partial Derivatives
Some expression are very complicated or
even impossible to solve for the dependent
variable, but it still might be useful to know
the partial derivative. To do this we can take
the partial derivative implicitly.
Example
For the sphere π₯ 2 + π¦ 2 + π§ 2 = 16 find
Solve for z : π§ = ± 16 β π₯ 2 β π¦ 2
ππ§
β2π₯
βπ₯
=
=
ππ₯ ±2 16 β π₯ 2 β π¦ 2
π§
ππ§
.
ππ₯
Steps for implicit derivatives
1. Take derivative of both sides of equation
keeping in mind independent variable,
multiplying by derivative where appropriate.
2. Solve for derivative
Implicitly, keeping in mind π§ = π π₯, π¦ .
π 2 π 2
π 2
π
π₯ +
π¦ +
π§ =
16
ππ₯
ππ₯
ππ₯
ππ₯
ππ§
2π₯ + 0 + 2π§
=0
ππ₯
ππ§ β2π₯ βπ₯
=
=
ππ₯
2π§
π§
Geometric View of Derivative
Remember for a function of 1 variable the derivative
at a point π₯0 represents the slope of the tangent line
or if you go 1 unit in the x direction you go up or
down on the tangent line by the amount of the
derivative ππ¦
.
ππ₯
y
1
π₯0
π₯=π₯0
ππ¦
ππ₯ π₯=π₯
0
x
Geometric View of Partial Derivative
For a surface π§ = π π₯, π¦ if we fix a value for x or y say π₯0 and π¦0 this makes a curve where
the surface intersects π₯ = π₯0 and π¦ = π¦0 . These curves have tangent lines at the point
π₯0 , π¦0 . The partial derivative tells you if you go 1 unit forward in the x direction or 1 unit
right in the y direction how much you go up or down in the z direction.
z
z
1
ππ§
ππ¦
1
ππ§
ππ₯
π₯0
π₯=π₯0
π¦=π¦0
π¦0
π¦0
y
π₯0 , π¦0
x
Direction Vector of Tangent line:
ππ§
1,0,
ππ₯
π₯0
π₯=π₯0
π¦=π¦0
y
π₯0 , π¦0
x
Direction Vector of Tangent line:
ππ§
0,1,
ππ¦
``` | 2,324 | 4,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2020-24 | latest | en | 0.670047 |
http://themetricmaven.com/?m=201305&paged=2 | 1,582,911,832,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147628.27/warc/CC-MAIN-20200228170007-20200228200007-00135.warc.gz | 130,060,879 | 13,546 | # Preferred Numbers and the “Preferred Measurement System”
Charles Renard (1847-1905)
By The Metric Maven
Bulldog Edition
In 1877 the French Engineer Charles Renard was instructed to look into improving captive balloons. These stationary, moored balloons were then in use by the French military, and of great importance. What Renard discovered was that 425 different sizes of cable were being used to moor these balloons. Clearly this large number of cables was not required from the outcome of any Engineering analysis, and were a nightmare to inventory and procure.
Renard determined that for mooring balloons, the most important inherent property of these cables, is their mass per unit length. He was able to develop a mathematical relationship which allowed him to replace the 425 sizes of cable with 17, which covered the same engineering range of requirements.
Renard’s geometric series was a perfect fit for a base 10 decimalized system, as it starts with 10 and ends with 100. The system he had in mind was of course, the metric system. This series produces what are proverbially known in engineering circles as preferred numbers (also called preferred values). Renard’s system was adopted as an international standard, ISO 3, in 1952, and are appropriately referred to as a Renard Series, or R Series. A similar series, the E series, is used to determine the values of electronic resistors, capacitors, inductors and zener diodes.
When metric was introduced into the building industry, a choice of dimensions which could easily be manipulated in one’s head was thought best. Grid lines on drawings are multiples of 100 mm. This is the basic “module” and the center to center of major dimensions are to be multiples of this value denoted as M. Therefore 3M = 300 mm, 6M = 600 mm and 12M = 1200 mm or 1.2 meters. According to Wikipedia:
For example, a multiple of 600 mm (6 M) can always be divided into 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 25, 30, etc. parts, each of which is again an integral number of millimetres.
No decimals! Preferred numbers rock! Why don’t we use metric construction like the Australians again?
But not everyone is so enamored with preferred numbers and the metric system. When I was a youthful Engineer working in Aerospace, I was involved in a number of proposals for large projects. I asked, what to my fellow workers, was an incredibly naive question: “Why aren’t we bidding this in metric?” A copy of the provisional bid “boilerplate” was then shown to me. On one of the first few pages of the proposal, was a small section about metric, it read something like: “The dimensions and system of units will be of the inch-pound system. This is necessitated because of the difficulty of procuring metric fasteners in this country, and because many, many more fasteners and hardware exist and are available in inches than in metric.” I began to realize that this “boilerplate” form had been used from time immemorial as a magical talisman to vanquish any thought of using metric in Aerospace. With each new bid, it was copied like junk DNA. I’m certain a similar document is in use in Aerospace to this day, to keep metric at bay.
At that young age I was definitely naive, because I swallowed the assertion, hook, line and imperial sinker. It seemed that limiting fasteners and other hardware might keep one from creating an essential Engineering design. OMG! for want of a nail, the spacecraft might be lost! Later I would learn from a salesman what this ploy actually was. It was FUD. In case you haven’t heard of it, this is a salesman’s term for what to do if your companies products are clearly inferior to your competitors. You must instill your customer with Fear, Uncertainty and Doubt (FUD) about the alternative product. If you use a competitors electricity, it will burn all your toast! Your soft water will come out hard! X-rays will come out of your light bulbs! The electrons will spill on the floor, and act like tiny ball bearings and you’ll slip! I hope you have health insurance! You better use Brand X electricity, or suffer the consequences!
The other option often employed to keep customers from choosing a competitor, is to create a proliferation of products who’s only purpose is to be non-interchangeable with any other competitors. I have seen this with RF/Microwave connectors. There are hundreds of them and I used to jokingly refer to each new offering as “connector of the week.” Many of them have Olde English screw threads and metric dimensions—but that is another blog. When a product is chosen by a market place “food fight” there is no guarantee that an optimal solution will be the survivor. If one chooses a product that is not satisfactory, and is incompatible with competitors, a temptation arises to rationalize its use. “It would cost too much to switch over now, this works good enough.” The person making the purchase does not want the bad choice to reflect on them, and will do their best to make do. Like the non-adoption of metric in the US, inculcated intellectual inertia to continue using a bad design will often prevail over reform.
The introduction of metric is a perfect opportunity (as Pat Naughtin has pointed out) to introduce much needed reform into the different trades. Pat Naughtin offers a number of examples of useful reform in his lectures, which I will not repeat here. The one example which does stand-out as perfectly in sync with the savings one can obtain by using preferred numbers, occurred in Australia. When metric was introduced into an Australian Ford car plant, the number of fasteners used by Ford were reduced by a factor of four after metric conversion. The implementation of metric threads reduced the hodgepodge of bolts by 88% and nuts by 72%. The number of sheet metal thicknesses in some factories were considerably reduced, which saved on inventory costs, and had no impact on Engineering design options. According to Kevin Wilks in his book Metrication in Australia (thanks Klystron): When standardizing containers, Australia was able to reduce the number of can sizes, for packing goods sold by mass, from approximately 90 to 30. He goes on:
Another example in wholesale packaging concerned corrugated fiberboard cases for packing fruit. With the establishment of metric packing quantities the opportunity was taken to reduce the variety of shapes and sizes from many hundreds to about 50.
The use of preferred numbers with the metric system is good for business, despite protestations to the contrary. The metric system’s absence in the US requires consumers to pay an unseen externality penalty. Business can ignore metrication because the citizens of the US pick up the tab, but don’t realize it. This unseen cost to consumers exists because of an inefficient measurement system, which powerful segments of the business lobby in the US, have perennially refused to allow government to legislate out of existence, since at least 1921. The cost of extra waste and inefficiency is just passed on to the consumer.
These costs also make American industry more expensive when compared with overseas companies. We need mandatory metrication, and we need it now to reform America, and make it competitive in the 21st Century—before the 22nd arrives.
If you liked this essay and wish to support the work of The Metric Maven, please visit his Patreon Page.
# Weight Watchers and Measures
By The Metric Maven
Filmmaker Amy Young, who is making a documentary film entitled: The State of The Unit: The Kilogram, has met her goal on Kickstarter. Thanks to everyone who contributed. \$22.7 K so far and needs to raise a total of \$26.8 K (i.e. \$4,100 more) in the next seven days to fund her film. Please consider donating to her Kickstarter campaign here. If you have contributed already, thanks. Now the blog.
My Youngest sister has been a member of Weight Watchers (WW) for many years. It has worked well for her, and continues to do so. The WW members have discussion threads where they talk about how to compute Weight Watchers points from food labels. The use of grams and Calories (versus calories) and kilojoules is a perennial topic for discussion. Here is an entry by SCHILA which my sister shared with me concerning an Italian food label:
This causes a face-palm. I realize that it’s not her fault, it’s our lack of the exclusive use of the metric system in the US, and how the incompatible mixture of metric and non-metric units pervades our culture which is to blame. One can immediately see that the Nutritional Information (NI) is actually all metric—in a sense. The Calorie is a pre-SI unit of energy, which was replaced by the Joule in 1948. The Calorie was “metric” 65 years ago.
Definitions of the calorie fall into two classes:
• The small calorie or gram calorie (symbol: cal)[2] approximates the energy needed to increase the temperature of 1 gram of water by 1 °C at standard atmospheric pressure (101.325 kPa). This is approximately 4.2 joules.
• The large calorie, kilogram calorie, dietary calorie, nutritionist’s calorie or food calorie (symbol: Cal)[2] approximates the energy needed to increase the temperature of 1 kilogram of water by 1 °C. This is exactly 1,000 small calories or approximately 4.2 kilojoules.
The difference between a calorie and a Calorie is a factor of 1000 in the US. Why shouldn’t SCHILA be confused. The label has Cal 122 KCAL which is a capitalized Calorie abbreviation, which then tries to make it more understandable by putting it in all caps as KCAL, which of course could be confused for Kilo-Kilo-calories. No wonder the poor woman is on a discussion thread asking for help. The logical solution to me is to be done with calories and Calories (1000 calories or a kilocalorie), switch to SI (official metric) and use kilojoules—like the rest of the world.
Here is a package of licorice from Australia (courtesy of Mike Joy). It is advertised as one meter long. The front of the package has only one mass (weight) given: 120 grams. That’s it! You don’t need any other information.
On the back the nutritional information is:
Nutrional Label for Australian Licorice
We see that the Australian food vendors also use untidy numbers like 4.8 servings in a package. The serving size is 25 grams which is 353 kilojoules, that’s it! Every other description: protein, sugar, sodium and so on are broken out in grams or milligrams.
Ok, how many kJ’s do you get a day? Well in the US I generally see 2000 Calories on food labels as the recommended daily intake. This works out to about 8375 kilojoules. Look how many kJ’s you get! Doesn’t that sound better than 2000 Calories? Here is a short table to give you an idea of the range of kJ’s and the old way:
5000 kilojoules is 1194 Calories (1200)
5500 kilojoules is 1313 Calories (1300)
6000 kilojoules is 1433 Calories (1400)
6500 kilojoules is 1552 Calories (1500)
7000 kilojoules is 1671 Calories (1600)
8000 kilojoules is 1910 Calories (1900)
8500 kilojoules is 2030 Calories (2000)
So for many WW members somewhere between 5000 and 8500 kJ’s is the range for you to think about. The Australian Government has recently sponsored a push to get Australians to eat around 8700 kilojoules per day. Here is a page from their website:
click to enlarge image
So what does the back of a licorice label in the US look like? This is from Twizzler’s web page:
Not all that that different, other than the use of Calories (i.e. kilocalories). So what’s the big deal? Well, the big deal is that because the US is not exclusively metric like Australia, very few Americans have any idea what a gram is. (It is about the weight of a plain chocolate m&m). This lack of exclusive metric adoption in the US obscures dietary data that is readily available. Any confusion will cause many people to just not bother with the nutritional information.
Should we go back to Ye Olde English units on food packaging?—well they’re actually Olde English sizes used prior to the English reforming their units in 19th Century, but we’ll let that pass for now. Some people who believe claim they are trying to help the public say yes. These people are from the anti-metric Wall Street Journal, and like James Taranto are there to yelp—I mean “help.” Their anti-metric “Numbers Guy” seems to be more interested in running a numbers racket than actually enlightening people about numbers.
When the Wall Street Journal is on the side of the Center for Science in the Public Interest–watch out–what they have in mind is not in the public interest. They want teaspoons and tablespoons back! I’ve already written about how confusion between the two, and the lack of metric in the US kills about 98,000 persons in the US each year. It would also make our nutritional labeling completely incompatible with the rest of the world—which all use metric. They want to swap mass (grams) for volume (Tsp, Tbl)? Isn’t it bad enough we use ounces interchangeably for weight and volume already? Do you think 8 (by weight) ounces of cheese doodles is a cup (8 oz by volume) of them? The Wall Street Journal has never found a bad Weights and Measures idea they didn’t like.
The frustration the rest of the world has with us is completely understandable and surfaces on the WW discussion thread:
All I can say is You Go Girl!
What started this WW discussion thread was a question about the nutrition label on the back of an Italian food product. If all our US packaging was in teaspoons and tablespoons, and grams became even more unfamiliar, this would further alienate us from 95% of the worlds population and their products. I guess the Wall Street Journal just can’t help themselves—they like trade barriers. Once again the inability of our legislators to pass mandatory metric only legislation for the US, with a plan, and funding, and so on, hurts the nations physical and economic health. Congress has been goofing around since 1866 thinking about metric, isn’t it time they finally got to work and dealt with the metrication issue?
If you liked this essay and wish to support the work of The Metric Maven, please visit his Patreon Page. | 3,163 | 14,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-10 | latest | en | 0.968306 |
https://www.learnzoe.com/blog/how-to-move-the-decimal-point/ | 1,726,059,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00018.warc.gz | 835,088,707 | 14,116 | # How to Move the Decimal Point
## Introduction
Understanding how to move the decimal point is fundamental in mathematics and everyday life. Whether you’re calculating percentages, converting units, or solving equations, knowing how to manipulate decimal numbers is crucial. This blog will explore the importance of understanding how to move the decimal point. In these common scenarios, it is necessary to provide some practical examples to help you grasp this concept.
### Why Understanding How to Move the Decimal Point is Important
Accuracy: Moving the decimal point allows for precise calculations and measurements. Accuracy is essential whether you’re working on financial calculations, scientific experiments, or simply trying to split a bill.
Convenience: By moving the decimal point, you can easily convert between different units of measurement. For example, converting meters to centimeters or liters to milliliters becomes effortless once you understand how to manipulate the decimal point.
Efficiency: Knowing how to move the decimal point lets you quickly evaluate percentages, discounts, and interest rates. This skill proves especially helpful when shopping, budgeting, or calculating return on investment.
### Common Scenarios Where Moving the Decimal Point is Necessary
Percentage Calculations: When calculating percentages, moving the decimal point is essential. Whether determining a tip at a restaurant, calculating discount prices, or interpreting test scores, understanding how to manipulate decimals is crucial.
Unit Conversions: Moving the decimal point is necessary when converting between different units of measurement. Whether it’s converting kilometers to meters, ounces to grams, or gallons to liters, mastering decimal manipulation ensures accurate and efficient conversions.
Scientific Notation: In the scientific realm, moving the decimal point is vital for expressing large or small numbers using scientific notation. It is beneficial when working with astronomical distances, atomic measurements, or microscopic objects.
Financial Calculations: Whether you’re calculating interest rates, mortgage payments, or compound interest, understanding decimal manipulation is critical. Moving the decimal point allows you to determine financial outcomes and make informed decisions accurately.
In conclusion, understanding how to move the decimal point is indispensable in various aspects of life. By mastering this skill, you can ensure accuracy, convenience, and efficiency in calculations, conversions, and measurements. Practice and familiarity with moving decimals will enhance your mathematical abilities and equip you with a valuable tool in everyday scenarios.
## Basic Concept of Decimal Point Movement
### What is a decimal point and its function?
The decimal point is a punctuation mark that separates the whole number part from the fractional part of a decimal number. It plays a crucial role in representing numbers with precision and accuracy. By placing the decimal point in the correct position, we can indicate the value of each digit relative to its place value.
For example, in the number 3.14, the decimal point separates the digit 3, which represents the whole number part, from the digits 1 and 4, which represent the fractional part. The decimal point is a visual indicator showing the division between the two parts.
### Understanding decimal places and their significance
Decimal places refer to the digits that come after the decimal point. Each digit after the decimal point represents a fraction of a unit, with the value decreasing as we move further to the right.
For instance, in the number 3.14, the digit 1 is in the tenth place, and the digit 4 is in the hundredth place. The tenth place indicates that there is one-tenth of a whole unit, and the hundredth place indicates that there are four-hundredths of a whole unit.
Understanding the significance of decimal places is essential for accurate measurement, calculation, and comparison. It allows us to differentiate between numbers that may have the same whole number value but differ in decimal portions. Decimal places provide precision and detail that whole numbers alone cannot capture.
When manipulating decimal numbers, moving the decimal point is crucial. Shifting the decimal point to the left represents a decrease in value while moving it to the right indicates an increase. By understanding how to move the decimal point, we can accurately perform addition, subtraction, multiplication, and division operations with decimals.
It is important to note that the number of decimal places in the result of a calculation should match the least precise number involved in the calculation. Rounding may be necessary to ensure the appropriate number of decimal places.
In conclusion, decimal point movement is vital in understanding and working with decimal numbers. By grasping the function of the decimal point and the significance of decimal places, individuals can effectively manipulate decimals and accurately represent values in various contexts. Whether in mathematical calculations, scientific measurements, or financial transactions, mastering decimal point movement is an essential skill that provides accuracy and precision.
## Moving the Decimal Point to the Left
### Shifting the decimal point to the left in whole numbers
The process is straightforward regarding moving the decimal point to the left in whole numbers. Here’s how it works:
1. Identify the number of decimal places: Determine the number of digits after the decimal point. In whole numbers, the decimal point is typically placed at the rightmost position, indicating zero decimal places.
2. Remove the decimal point: Since there are no digits after the decimal point in whole numbers, you can ignore it.
3. Count the number of places to move: Starting from the rightmost digit, count the number of places you need to move the decimal point to the left. Each place represents a power of 10.
4. Move the decimal point: Shift the specified number of places to the left. Fill in any empty spaces with zeros.
For example, let’s consider the number 1.25. To move the decimal point one place to the left, we remove the decimal point and shift the digits accordingly. As a result, we get the whole number 12.5.
### Moving the decimal point to the left in decimal fractions
Moving the decimal point to the left in decimal fractions involves a similar process. Here’s how you can do it:
1. Identify the number of decimal places: Determine the number of digits after the decimal point. It will determine the number of places to shift the decimal point to the left.
2. Count the number of places to move: Starting from the rightmost digit, count the number of places you need to move the decimal point to the left. Each place represents a power of 10.
3. Move the decimal point: Shift the specified number of places to the left. Fill in any empty spaces with zeros.
For example, let’s take the decimal fraction 0.025. To move the decimal point two places to the left, we shift the digits accordingly. The result is the decimal fraction 0.25.
Understanding how to move the decimal point to the left is crucial in mathematical calculations, scientific measurements, and financial transactions. Whether it’s for rounding numbers, calculating percentages, or converting between units, mastering this skill allows for accurate and precise results. By following these simple steps, you can confidently manipulate decimal numbers and ensure the appropriate decimal point placement.
## Moving the Decimal Point to the Right
### Shifting the decimal point to the right in whole numbers
The process is relatively straightforward regarding moving the decimal point to the right in whole numbers. Here’s how it works:
1. Identify the number of decimal places:Â Determine the number of digits after the decimal point. No digits in whole numbers after the decimal point indicate zero decimal places.
2. Add zeros: Since moving the decimal point to the right, we need to add zeros after the whole number to maintain its value.
3. Count the number of places to move: Starting from the rightmost digit, count the number of places you need to move the decimal point to the right. Each place represents a power of 10.
4. Move the decimal point: Shift the specified number of places to the right. Fill in any empty spaces with zeros.
For example, let’s consider the whole number 123. To move the decimal point two places to the right, we add two zeros after the whole number. As a result, we get 12,300.
### Moving the decimal point to the right in decimal fractions
Moving the decimal point to the right in decimal fractions follows a similar process. Here’s how you can do it:
1. Identify the number of decimal places: Determine the number of digits after the decimal point. It will determine the number of places to shift the decimal point to the right.
2. Add zeros: If necessary, add zeros after the decimal fraction to maintain the decimal places.
3. Count the number of places to move: Starting from the rightmost digit, count the number of places you need to move the decimal point to the right. Each place represents a power of 10.
4. Move the decimal point: Shift the specified number of places to the right. Fill in any empty spaces with zeros.
For example, let’s take the decimal fraction 0.025. To move the decimal point three places to the right, we add three zeros after the decimal fraction. The result is the decimal fraction 25.000.
Understanding how to move the decimal point to the right is crucial in various mathematical calculations, scientific measurements, and financial transactions. Mastering this skill allows for accurate and precise results, whether multiplying numbers, dividing decimals, or converting units. By following these simple steps, you can confidently manipulate decimal numbers and ensure the appropriate decimal point placement.
## Decimal Point Movement in Mathematical Operations
### Moving the Decimal Point in Addition and Subtraction
When it comes to adding or subtracting decimal numbers, ensuring that both numbers have the same number of decimal places is essential. To achieve this, you may need to move the decimal point. Here’s how you can do it:
1. Identify the number of decimal places: Determine the number of digits after the decimal point in each number involved.
2. Add zeros: If one number has fewer decimal places than the other, add zeros to the end of the number with fewer decimal places to make them equal.
3. Align the decimal points: Place both numbers on each other, aligning the decimal points.
4. Add or subtract:Â You can add or subtract the numbers as usual, ignoring the decimal point.
5. Place decimal point: The final answer should have the same number of decimal places as the original numbers. Place the decimal point in the result by counting the decimal places from the original numbers.
For example, let’s add the numbers 13.45 and 2.3. By aligning the decimal points, we get:
13.45+ 2.30——–
Adding the numbers, we get 15.75 as a result. Since the original numbers had two decimal places, the final answer should have two.
### Decimal Point Movement in Multiplication and Division
You may also need to move the decimal point in multiplication and division operations involving decimal numbers. Here’s how you can do it:
1. Identify the number of decimal places: Determine the number of digits after the decimal point in each number involved.
2. Count total decimal places: Add the number of decimal places in both numbers to determine the total number of decimal places in the final result.
3. Multiply or divide: Perform the multiplication or division operation, ignoring the decimal point.
4. Place decimal point: After obtaining the result, place the decimal point in the final answer by counting from right to left, using the total number of decimal places calculated earlier.
For example, let’s multiply 2.5 by 0.4. By ignoring the decimal point, we get the result 10. Now, counting the decimal places in both numbers (2), we place the decimal point in the final answer two places from the right. Therefore, the result is 1.00.
Understanding how to move the decimal point is crucial in various mathematical operations. Whether you are adding, subtracting, multiplying, or dividing decimal numbers, correctly placing the decimal point ensures the accuracy and precision of your calculations. By following these simple steps, you can confidently manipulate decimal numbers and achieve reliable results.
## Conclusion
### Summary of the Importance and Application of Decimal Point Movement
In conclusion, the movement of the decimal point plays a crucial role in mathematical operations involving decimal numbers. It ensures accuracy and precision in calculations, providing reliable results. Here’s a summary of the importance and application of decimal point movement:
1. Addition and Subtraction: When adding or subtracting decimal numbers, aligning the decimal points by adding zeros if necessary is essential. Ignoring the decimal point, operate, and finally, place the decimal point in the result by counting the number of decimal places.
2. Multiplication and Division:Â Count the number of decimal places in both numbers in multiplication and division operations with decimal numbers. Operate without considering the decimal point and place it in the final answer by counting from the right using the total number of decimal places.
Understanding how to move the decimal point correctly ensures that your calculations are accurate and precise. It allows you to manipulate decimal numbers confidently, regardless of the operation.
Whether you’re dealing with simple addition or complex algebraic equations, the ability to move the decimal point is a fundamental skill in mathematics. It is applicable in various real-life scenarios, such as calculating prices, computing percentages, and measuring quantities.
By mastering the concept of decimal point movement, you can solve mathematical problems with ease and confidence. Practice the steps outlined in this article and familiarize yourself with different scenarios to enhance your understanding and proficiency.
Remember, accuracy is critical when working with decimal numbers. Paying attention to the decimal point and its movement ensures reliable results. So, next time you encounter decimal numbers in your calculations, apply the appropriate technique to move the decimal point and achieve accurate solutions. | 2,791 | 14,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.802939 |
https://eng.libretexts.org/Under_Construction/Book%3A_Fluid_Mechanics_(Bar-Meir)__-_old_copy/05%3A_The_Control_Volume_and_Mass_Conservation/5.6%3A_The_Details_Picture_%E2%80%93_Velocity_Area_Relationship | 1,547,946,442,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583688396.58/warc/CC-MAIN-20190120001524-20190120023524-00025.warc.gz | 496,680,404 | 17,162 | 5.6: The Details Picture – Velocity Area Relationship
Fig. 5.8 Control volume usage to calculate local averaged velocity in three coordinates.
The integral approach is intended to deal with the "big'' picture. Indeed the method is used in this part of the book for this purpose. However, there is very little written about the usability of this approach to provide way to calculate the average quantities in the control system. Sometimes it is desirable to find the averaged velocity or velocity distribution inside a control volume. There is no general way to provide these quantities. Therefore an example will be provided to demonstrate the use of this approach. Consider a container filled with liquid on which one exit opened and the liquid flows out as shown in Figure 5.8. The velocity has three components in each of the coordinates under the assumption that flow is uniform and the surface is straight . The integral approached is used to calculate the averaged velocity of each to the components. To relate the velocity in the $$z$$ direction with the flow rate out or the exit the velocity mass balance is constructed. A similar control volume construction to find the velocity of the boundary velocity (height) can be carried out. The control volume is bounded by the container wall including the exit of the flow. The upper boundary is surface parallel to upper surface but at $$Z$$ distance from the bottom. The mass balance reads
$\label{mass:eq:containerZs} \int_V \dfrac{d\rho}{dt} \,dV + \int_A U_{bn}\,\rho\,dA + \int_A U_{rn}\,\rho\,dA = 0 \tag{41}$
For constant density (conservation of volume) equation and ($$h>z$$) reduces to
$\label{mass:eq:containerZrho} \int_A U_{rn}\,\rho\,dA = 0 \tag{42}$
In the container case for uniform velocity equation (??) becomes
$\label{mass:eq:Zu} U_z\,A = U_{e}\,A_e \Longrightarrow U_z = - \dfrac{A_e}{A} U_e \tag{43}$
It can be noticed that the boundary is not moving and the mass inside does not change this control volume. The velocity $$U_z$$ is the averaged velocity downward.
Fig. 5.9 Control volume and system before and after the motion.
The $$x$$ component of velocity is obtained by using a different control volume. The control volume is shown in Figure 5.9. The boundary are the container far from the flow exit with blue line projection into page (area) shown in the Figure 5.9. The mass conservation for constant density of this control volume is
$\label{mass:eq:containerXs} - \int_A U_{bn}\,\rho\,dA + \int_A U_{rn}\,\rho\,dA = 0 \tag{44}$
Usage of control volume not included in the previous analysis provides the velocity at the upper boundary which is the same as the velocity at $$y$$ direction. Substituting into (??) results in
$\label{mass:eq:containerXU} \int_ ParseError: colon expected (click for details) Callstack: at (Under_Construction/Book:_Fluid_Mechanics_(Bar-Meir)__-_old_copy/05:_The_Control_Volume_and_Mass_Conservation/5.6:_The_Details_Picture_–_Velocity_Area_Relationship), /content/body/p[6]/span, line 1, column 5 \dfrac{A_e}{A} U_e \,\rho\,dA + \int_{A_{yz}} U_{x}\,\rho\,dA = 0 \tag{45}$
Where $${A_{x}}^{-}$$ is the area shown the Figure under this label. The area $$A_{yz}$$ referred to area into the page in Figure under the blow line. Because averaged velocities and constant density are used transformed equation (??) into
$\label{mass:eq:containerXUa} \dfrac{A_e}{A} {A_{x}}^{-} U_e + U_{x}\,\overbrace{Y(x)\,h}^{A_{yz}} = 0 \tag{46}$
Where $$Y(x)$$ is the length of the (blue) line of the boundary. It can be notice that the velocity, Ux is generally increasing with $$x$$ because $${A_{x}}^{-}$$ increase with $$x$$. The calculations for the $$y$$ directions are similar to the one done for $$x$$ direction. The only difference is that the velocity has two different directions. One zone is right to the exit with flow to the left and one zone to left with averaged velocity to right. If the volumes on the left and the right are symmetrical the averaged velocity will be zero.
Fig. 5.10 Circular cross section for finding $$U_x$$ and various cross sections.}
Example 5.12
Calculate the velocity, $$U_x$$ for a cross section of circular shape (cylinder).
Solution 5.12
The relationship for this geometry needed to be expressed. The length of the line $$Y(x)$$ is
$\label{Ccontainer:yLength} Y(x) = 2\,r\, \sqrt{ 1 - \left(1-\dfrac{x}{r}\right)^2 } \tag{47}$
This relationship also can be expressed in the term of $$\alpha$$ as
$\label{Ccontainer:yLengthalpha} Y(x) = 2 \, r\,\sin\alpha \tag{48}$
Since this expression is simpler it will be adapted. When the relationship between radius angle and $$x$$ are
$\label{Ccontainer:rx} x = r (1 -\sin\alpha) \tag{49}$
The area $${A_{x}}^{-}$$ is expressed in term of $$\alpha$$ as
$\label{Ccontainer:area} {A_{x}}^{-} = \left( \alpha - \dfrac{1}{2\dfrac{}{}},\sin(2\alpha) \right) r^2 \tag{50}$
Thus, the velocity, $$U_x$$ is
$\label{Ccontainer:Uxs} \dfrac{A_e}{A} \left( \alpha - \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) r^2\, U_e + U_{x}\, 2 \, r\,\sin\alpha\,h = 0 \tag{51}$
$\label{Ccontainer:Uxf} U_x = \dfrac{A_e}{A} \dfrac{r}{h} \dfrac{\left( \alpha - \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\sin\alpha} \, U_e \tag{52}$
Averaged velocity is defined as
$\label{Ccontainer:UxaveDef} \overline{U_x} = \dfrac{1}{S}\int_S U dS \tag{53}$
Where here $$S$$ represent some length. The same way it can be represented for angle calculations. The value $$dS$$ is $$r\cos\alpha$$. Integrating the velocity for the entire container and dividing by the angle, $$\alpha$$ provides the averaged velocity.
$\label{Ccontainer:Uxtotal} \overline{U_x} = \dfrac{1}{2\,r} \int_0^{\pi}\dfrac{A_e}{A} \dfrac{r}{h} \dfrac{\left( \alpha - \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\tan\alpha} \, U_e\, r\, d\alpha \tag{54}$
which results in
$\label{Ccontainer:Uxtotala} \overline{U_x} = \dfrac{\left( \pi -1\right)}{4} \dfrac{A_e}{A} \dfrac{r}{h} \, U_e \tag{55}$
Example 5.13
Fig. 5.11 $$y$$ velocity for a circular shape
What is the averaged velocity if only half section is used. State your assumptions and how it similar to the previous example.
Solution 5.13
The flow out in the $$x$$ direction is zero because symmetrical reasons. That is the flow field is a mirror images. Thus, every point has different velocity with the same value in the opposite direction. The flow in half of the cylinder either the right or the left has non zero averaged velocity. The calculations are similar to those in the previous to example 5.12. The main concept that must be recognized is the half of the flow must have come from one side and the other come from the other side. Thus, equation (??) modified to be
$\label{ene:eq:containeryU} \dfrac{A_e}{A} {A_{x}}^{-} U_e + U_{x}\,\overbrace{Y(x)\,h}^{A_{yz}} = 0 \tag{56}$
The integral is the same as before but the upper limit is only to $$\pi/2$$
$\label{Ccontainer:Uytotal} \overline{U_x} = \dfrac{1}{2\,r} \int_0^{\pi/2}\dfrac{A_e}{A} \dfrac{r}{h} \dfrac{\left( \alpha - \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\tan\alpha} \, U_e\, r\, d\alpha \tag{57}$
which results in
$\label{Ccontainer:Uytotala} \overline{U_x} = \dfrac{\left( \pi -2\right)}{8} \dfrac{A_e}{A} \dfrac{r}{h} \, U_e \tag{58}$
Contributors
• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. | 2,240 | 7,413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-04 | latest | en | 0.886772 |
http://slideplayer.com/slide/4562755/ | 1,571,610,050,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00430.warc.gz | 177,230,005 | 29,483 | # MASS, MOMENTUM , AND ENERGY EQUATIONS
## Presentation on theme: "MASS, MOMENTUM , AND ENERGY EQUATIONS"— Presentation transcript:
MASS, MOMENTUM , AND ENERGY EQUATIONS
This chapter deals with four equations commonly used in fluid mechanics: the mass, Bernoulli, Momentum and energy equations. The mass equation is an expression of the conservation of mass principle. The Bernoulli equation is concerned with the conservation of kinetic, potential, and flow energies of a fluid stream and their conversion to each other in regions of flow where net viscous forces are negligible and where other restrictive conditions apply. The energy equation is a statement of the conservation of energy principle. In fluid mechanics, it is found convenient to separate mechanical energy from thermal energy and to consider the conversion of mechanical energy to thermal energy as a result of frictional effects as mechanical energy loss. Then the energy equation becomes the mechanical energy balance.
We start this chapter with an overview of conservation principles and the conservation of mass relation. This is followed by a discussion of various forms of mechanical energy . Then we derive the Bernoulli equation by applying Newton’s second law to a fluid element along a streamline and demonstrate its use in a variety of applications. We continue with the development of the energy equation in a form suitable for use in fluid mechanics and introduce the concept of head loss. Finally, we apply the energy equation to various engineering systems.
Conservation of Mass The conservation of mass relation for a closed system undergoing a change is expressed as msys = constant or dmsys/dt= 0, which is a statement of the obvious that the mass of the system remains constant during a process. For a control volume (CV) or open system, mass balance is expressed in the rate form as where min and mout are the total rates of mass flow into and out of the control volume, respectively, and dmCV/dt is the rate of change of mass within the control volume boundaries. In fluid mechanics, the conservation of mass relation written for a differential control volume is usually called the continuity equation.
Conservation of Mass Principle
The conservation of mass principle for a control volume can be expressed as: The net mass transfer to or from a control volume during a time interval t is equal to the net change (increase or decrease) in the total mass within the control volume during t. That is,
where ∆mCV= mfinal – minitial is the change in the mass of the control volume during the process. It can also be expressed in rate form as The Equations above are often referred to as the mass balance and are applicable to any control volume undergoing any kind of process.
During a steady-flow process, the total amount of mass contained within a control volume does not change with time (mCV = constant). Then the conservation of mass principle requires that the total amount of mass entering a control volume equal the total amount of mass leaving it. When dealing with steady-flow processes, we are not interested in the amount of mass that flows in or out of a device over time; instead, we are interested in the amount of mass flowing per unit time, that is, the mass flow rate It states that the total rate of mass entering a control volume is equal to the total rate of mass leaving it
Many engineering devices such as nozzles, diffusers, turbines, compressors,and pumps involve a single stream (only one inlet and one outlet). For these cases, we denote the inlet state by the subscript 1 and the outlet state by the subscript 2, and drop the summation signs
Properties We take the density of water to be 1000 kg/m3 1 kg/L.
EXAMPLE 2–1 : Water Flow through a Garden Hose Nozzle A garden hose attached with a nozzle is used to fill a 10-gal bucket. The inner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle exit (Fig. 5–12). If it takes 50 s to fill the bucket with water, determine (a) the volume and mass flow rates of water through the hose, and (b) the average velocity of water at the nozzle exit. Assumptions 1 Water is an incompressible substance. 2 Flow through the hose is steady. 3 There is no waste of water by splashing. Properties We take the density of water to be 1000 kg/m3 1 kg/L. Analysis (a) Noting that 10 gal of water are discharged in 50 s, the volume and mass flow rates of water are
(b) The cross-sectional area of the nozzle exit is
The volume flow rate through the hose and the nozzle is constant. Then the average velocity of water at the nozzle exit becomes
MECHANICAL ENERGY Many fluid systems are designed to transport a fluid from one location to another at a specified flow rate, velocity, and elevation difference, and the system may generate mechanical work in a turbine or it may consume mechanical work in a pump or fan during this process. These systems do not involve the conversion of nuclear, chemical, or thermal energy to mechanical energy. Also, they do not involve any heat transfer in any significant amount, and they operate essentially at constant temperature. Such systems can be analyzed conveniently by considering the mechanical forms of energy only and the frictional effects that cause the mechanical energy to be lost (i.e., to be converted to thermal energy that usually cannot be used for any useful purpose). The mechanical energy can be defined as the form of energy that can be converted to mechanical work completely and directly by an ideal mechanical device. Kinetic and potential energies are the familiar forms of mechanical energy.
Therefore, the mechanical energy of a flowing fluid can be expressed on a unit-mass basis as
In the absence of any changes in flow velocity and elevation, the power produced by an ideal hydraulic turbine is proportional to the pressure drop of water across the turbine.
Emech, in = Emech, out + Emech, loss
Most processes encountered in practice involve only certain forms of energy, and in such cases it is more convenient to work with the simplified versions of the energy balance. For systems that involve only mechanical forms of energy and its transfer as shaft work, the conservation of energy principle can be expressed conveniently as where Emech, loss represents the conversion of mechanical energy to thermal energy due to irreversibilities such as friction. For a system in steady operation, the mechanical energy balance becomes Emech, in = Emech, out + Emech, loss
THE BERNOULLI EQUATION
The Bernoulli equation is an approximate relation between pressure,velocity, and elevation, and is valid in regions of steady, incompressible flow where net frictional forces are negligible ( as shown in the Figure below). Despite its simplicity, it has proven to be a very powerful tool in fluid mechanics. The Bernoulli equation is an approximate equation that is valid only in in viscid regions of flow where net viscous forces are negligibly small compared to inertial, gravitational, or pressure forces. Such regions occur outside of boundary layers and wakes.
Derivation of the Bernoulli Equation
the Bernoulli Equation is derived from the mechanical energy equation since the we are dealing with steady flow system with out the effect of the mechanical work and the friction on the system the first terms become zero. This is the famous Bernoulli equation, which is commonly used in fluid mechanics for steady, incompressible flow along a streamline in inviscid regions of flow.
The Bernoulli equation can also be written between any two points on the same streamline as
Limitations on the Use of the Bernoulli Equation
Steady flow The first limitation on the Bernoulli equation is that it is applicable to steady flow. Frictionless flow Every flow involves some friction, no matter how small, and frictional effects may or may not be negligible. No shaft work The Bernoulli equation was derived from a force balance on a particle moving along a streamline. Incompressible flow One of the assumptions used in the derivation of the Bernoulli equation is that = constant and thus the flow is incompressible. No heat transfer The density of a gas is inversely proportional to temperature, and thus the Bernoulli equation should not be used for flow sections that involve significant temperature change such as heating or cooling sections. Strictly speaking, the Bernoulli equation is applicable along a streamline, and the value of the constant C, in general, is different for different streamlines. But when a region of the flow is irrotational, and thus there is no vorticity in the flow field, the value of the constant C remains the same for all streamlines, and, therefore, the Bernoulli equation becomes applicable across streamlines as well.
EXAMPLE 2–2 Spraying Water into the Air
Water is flowing from a hose attached to a water main at 400 kPa gage (Fig. below). A child places his thumb to cover most of the hose outlet, causing a thin jet of high-speed water to emerge. If the hose is held upward, what is the maximum height that the jet could achieve? This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful components with large frictional losses, and thus it is suitable for the use of the Bernoulli equation. The water height will be maximum under the stated assumptions. The velocity inside the hose is relatively low (V1 = 0) and we take the hose outlet as the reference level (z1= 0). At the top of the water trajectory V2 = 0, and atmospheric pressure pertains. Then the Bernoulli equation simplifies to
EXAMPLE 2-3 Water Discharge from a Large Tank
A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap (Fig. below). A tap near the bottom of the tank is now opened, and water flows out from the smooth and rounded outlet. Determine the water velocity at the outlet. This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful components with large frictional losses, and thus it is suitable for the use of the Bernoulli equation. We take point 1 to be at the free surface of water so that P1= Patm (open to the atmosphere), V1 = 0 (the tank is large relative to the outlet), and z1= 5 m and z2 = 0 (we take the reference level at the center of the outlet) Also, P2 = Patm (water discharges into the atmosphere).
Then the Bernoulli equation simplifies to
Solving for V2 and substituting The relation is called the Toricelli equation.
EXAMPLE 2–4Siphoning Out Gasoline from a Fuel Tank
During a trip to the beach (Patm = 1 atm =101.3 kPa), a car runs out of gasoline, and it becomes necessary to siphon gas out of the car of a Good Samaritan (Fig. below). The siphon is a small-diameter hose, and to start the siphon it is necessary to insert one siphon end in the full gas tank, fill the hose with gasoline via suction, and then place the other end in a gas can below the level of the gas tank. The difference in pressure between point 1 (at the free surface of the gasoline in the tank) and point 2 (at the outlet of the tube) causes the liquid to flow from the higher to the lower elevation. Point 2 is located 0.75 m below point 1 in this case, and point 3 is located 2 m above point 1. The siphon diameter is 4 mm, and frictional losses in the siphon are to be disregarded. Determine (a) the minimum time to withdraw 4 L of gasoline from the tank to the can and (b) the pressure at point 3. The density of gasoline is 750 kg/m3.
Analysis (a) We take point 1 to be at the free surface of gasoline in the tank so that P1 = Patm (open to the atmosphere), V1 = 0 (the tank is large relative to the tube diameter), and z2 =0 (point 2 is taken as the reference level). Also, P2 = Patm (gasoline discharges into the atmosphere). Then the Bernoulli equation simplifies to Solving for V2 and substituting, The cross-sectional area of the tube and the flow rate of gasoline are Then the time needed to siphon 4 L of gasoline becomes
(b) The pressure at point 3 can be determined by writing the Bernoulli equation between points 2 and 3. Noting that V2 = V3 (conservation of mass), z2 = 0, and P2 = Patm, | 2,687 | 12,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-43 | latest | en | 0.930223 |
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