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COMPSTAT 2004 section: Graphics. Abstract: The boxplot is a very popular graphical tool to visualize the distribution of continuous univariate data. First of all, it shows information about the location and the spread of the data by means of the median and the interquartile range. The length of the whiskers on both sides of the box and the position of the median within the box are helpful to ...
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Distribution Curves Intro.notebook 2 April 09, 2018. Distribution Curves Intro.notebook 3 April 09, 2018. Distribution Curves Intro.notebook 4 April 09, 2018. Distribution Curves Intro.notebook 5 April 09, 2018. Distribution Curves Intro.notebook 6 April 09, 2018. Mode Median Ilean Left-Skewed (Negative Skewness) Mode Median Mean Right-Skewed (Positive Skewness) 34.10/ 34.1% 3.6% 13.6% —10 ...
11 2 Skills Practice Answers Name date period 11 2 skills practice, lesson 11 2 name date period pdf pass chapter 11 13 glencoe algebra 2 negatively skewed symmetric negatively skewed sample answer: the distribution is skewed, so use the fi ve number summary the range is 58 to 118 students the median is 97 students, and half of the data are between 84 and 1055 students. Up and down or down and ...
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A distribution is said to be 'skewed' when the mean and the median fall at different points in the distribution, and the balance (or centre of gravity) is shifted to one side or the other-to left or right. Measures of skewness tell us the direction and the extent of Skewness. In symmetrical distribution the mean, median and mode are identical. The more the mean moves away from the mode, the ...
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relating skew to the positions of the median and mean. “In a skewed distribution, the mean is farther out in the long tail than is the median.” (Moore and McCabe 2003, p. 43) “For skewed distributions, the mean lies toward the direction of skew (the longer tail) relative to the median.” (Agresti and Finlay 1997, p. 50) Five textbooks extend the rule to cover the mode as well. “In a ...
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skewed distribution of the 12, with parameters (m =300, s =20, t =300). Table 1. Miller’s 12 ex-Gaussian distributions. Each distribution is defined by the combination of the three parameters m (mu), s (sigma) and t (tau). The mean is defined as the sum of parameters m and t. The median was calculated based on samples of 1,000,000 observations (Miller 1988 used 10,000 observations ...
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The mean, median and mode are not equal in a skewed distribution. The Karl Pearson’s measure of skewness is based upon the divergence of mean from mode in a skewed distribution. Skp1 = mean - mode standard deviation or Skp2 = 3(mean - median) standard deviation Department of Mathematics University of Ruhuna | Applied Statistics I(IMT224 ...
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To address the effect of a skewed distribution on the mean, we transform the original distribution using a Box-Cox power transformation to create a measure we term “MEAN-T (Mean Transformed) as the mean for this transformed distribution. The transformed distribution considers the entire data series, but assigns a proportionate amount of influence to each case through normalization, thereby ...
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Positively Skewed Distribution Mean > Median > Mode Mode Median Mean Negatively Skewed Distribution Mean < Median < Mode Mean Median Mode. 6 Chapter 9. Measures of Skewness and Kurtosis Definition of Measures of Skewness (page 267) Definition9.3. A measureofskewnessisasingle value that indicates the degree and directionofasymmetry. Chapter 9. Measures of Skewness and Kurtosis Interpretation of ...
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Earnings distributions tend to be skewed to the right and display long right tails. Mean earnings always exceed median earnings and the top percentiles of earners account for quite a disproportionate share of total earnings. Mean earnings also differ greatly across groups defined by occupation, education, experience, and other observed traits. With respect to the evolution of the distribution ...
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2.1 Cross-linguistic count/mass variation One of the most robust tests, in number marking languages at least, for the count-ability status of a noun is a felicitous use in direct modi?cation NP constructions with a numerical expression, possibly also modulo linguistic and extra-linguistic context. This test is, for example, applicable in both English and Finnish: (1a) and (1b) are both ...
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It can therefore be argued that linguistic variation carries multiple forms of indexical meaning: information pointed to by a linguistic variant which includes details about the speaker, the greater context of the speech act, and the fleeting interactional moves taken by the speakers in conversation. Speakers build knowledge of how to encode indexical meanings as part of their language ...
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Influenced by Putonghua education, young speakers have less variation in their use of these two linguistic variables than middle-aged speakers do. Within the social context of a steel company, the use of . rusheng. is lexically conditioned because both northern and southern-
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Dialectal variation in adults Variationist studies Linguistic variables (and their variants) Usage depends on: Linguistic context Socio-demographic characteristics (age, gender, social background) Context Subject to social judgment Standard versus non standard variants Examples American English: car produced [k?:?] or [k?:] | 2,659 | 12,162 | {"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-2020-40 | latest | en | 0.836274 |
https://howkgtolbs.com/convert/31.52-kg-to-lbs | 1,685,443,623,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645595.10/warc/CC-MAIN-20230530095645-20230530125645-00490.warc.gz | 336,782,027 | 12,198 | # 31.52 kg to lbs - 31.52 kilograms to pounds
Do you want to learn how much is 31.52 kg equal to lbs and how to convert 31.52 kg to lbs? You are in the right place. In this article you will find everything about kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to point out that whole this article is devoted to one amount of kilograms - that is one kilogram. So if you want to learn more about 31.52 kg to pound conversion - keep reading.
Before we go to the practice - that is 31.52 kg how much lbs conversion - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 31.52 kg to lbs? 31.52 kilograms it is equal 69.4897049824 pounds, so 31.52 kg is equal 69.4897049824 lbs.
## 31.52 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in short form SI).
Sometimes the kilogram is written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. First definition was simply but hard to use.
Then, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was replaced by a new definition.
The new definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It can be also divided into 100 decagrams and 1000 grams.
## 31.52 kilogram to pounds
You know a little bit about kilogram, so now let's go to the pound. The pound is also a unit of mass. It is needed to highlight that there are more than one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we want to focus only on pound-mass.
The pound is in use in the Imperial and United States customary systems of measurements. Naturally, this unit is in use also in another systems. The symbol of the pound is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is defined as exactly 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 31.52 kg?
31.52 kilogram is equal to 69.4897049824 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 31.52 kg in lbs
Theoretical part is already behind us. In next part we will tell you how much is 31.52 kg to lbs. Now you know that 31.52 kg = x lbs. So it is high time to get the answer. Just see:
31.52 kilogram = 69.4897049824 pounds.
This is an exact outcome of how much 31.52 kg to pound. You can also round off the result. After it your outcome will be exactly: 31.52 kg = 69.344 lbs.
You know 31.52 kg is how many lbs, so let’s see how many kg 31.52 lbs: 31.52 pound = 0.45359237 kilograms.
Obviously, in this case you can also round off this result. After it your outcome will be as following: 31.52 lb = 0.45 kgs.
We also want to show you 31.52 kg to how many pounds and 31.52 pound how many kg results in charts. See:
We are going to start with a table for how much is 31.52 kg equal to pound.
### 31.52 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
31.52 69.4897049824 69.3440
Now see a chart for how many kilograms 31.52 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
31.52 0.45359237 0.45
Now you know how many 31.52 kg to lbs and how many kilograms 31.52 pound, so it is time to move on to the 31.52 kg to lbs formula.
### 31.52 kg to pounds
To convert 31.52 kg to us lbs a formula is needed. We are going to show you a formula in two different versions. Let’s start with the first one:
Number of kilograms * 2.20462262 = the 69.4897049824 result in pounds
The first formula will give you the most accurate outcome. Sometimes even the smallest difference can be considerable. So if you want to get an accurate result - this version of a formula will be the best solution to convert how many pounds are equivalent to 31.52 kilogram.
So let’s move on to the shorer version of a formula, which also enables calculations to know how much 31.52 kilogram in pounds.
The another version of a formula is as following, let’s see:
Amount of kilograms * 2.2 = the result in pounds
As you see, the second version is simpler. It can be better option if you need to make a conversion of 31.52 kilogram to pounds in quick way, for example, during shopping. Just remember that your outcome will be not so exact.
Now we want to show you these two versions of a formula in practice. But before we are going to make a conversion of 31.52 kg to lbs we are going to show you easier way to know 31.52 kg to how many lbs without any effort.
### 31.52 kg to lbs converter
An easier way to learn what is 31.52 kilogram equal to in pounds is to use 31.52 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Converter is based on first formula which we gave you in the previous part of this article. Thanks to 31.52 kg pound calculator you can quickly convert 31.52 kg to lbs. You only have to enter amount of kilograms which you want to convert and click ‘calculate’ button. The result will be shown in a flash.
So try to calculate 31.52 kg into lbs using 31.52 kg vs pound calculator. We entered 31.52 as an amount of kilograms. It is the result: 31.52 kilogram = 69.4897049824 pounds.
As you can see, this 31.52 kg vs lbs calculator is so simply to use.
Now we can go to our primary issue - how to convert 31.52 kilograms to pounds on your own.
#### 31.52 kg to lbs conversion
We will begin 31.52 kilogram equals to how many pounds calculation with the first formula to get the most exact outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 69.4897049824 the result in pounds
So what have you do to check how many pounds equal to 31.52 kilogram? Just multiply number of kilograms, in this case 31.52, by 2.20462262. It is 69.4897049824. So 31.52 kilogram is equal 69.4897049824.
You can also round it off, for example, to two decimal places. It is equal 2.20. So 31.52 kilogram = 69.3440 pounds.
It is high time for an example from everyday life. Let’s calculate 31.52 kg gold in pounds. So 31.52 kg equal to how many lbs? As in the previous example - multiply 31.52 by 2.20462262. It gives 69.4897049824. So equivalent of 31.52 kilograms to pounds, when it comes to gold, is exactly 69.4897049824.
In this case it is also possible to round off the result. It is the outcome after rounding off, this time to one decimal place - 31.52 kilogram 69.344 pounds.
Now we are going to examples converted with short formula.
#### How many 31.52 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 69.344 the result in pounds
So 31.52 kg equal to how much lbs? And again, you have to multiply amount of kilogram, this time 31.52, by 2.2. See: 31.52 * 2.2 = 69.344. So 31.52 kilogram is exactly 2.2 pounds.
Make another calculation with use of this formula. Now convert something from everyday life, for example, 31.52 kg to lbs weight of strawberries.
So convert - 31.52 kilogram of strawberries * 2.2 = 69.344 pounds of strawberries. So 31.52 kg to pound mass is equal 69.344.
If you know how much is 31.52 kilogram weight in pounds and can calculate it using two different formulas, we can move on. Now we are going to show you these outcomes in tables.
#### Convert 31.52 kilogram to pounds
We know that results shown in charts are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Due to this you can easily compare 31.52 kg equivalent to lbs results.
Begin with a 31.52 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
31.52 69.4897049824 69.3440
And now let’s see 31.52 kg equal pound chart for the second formula:
Kilograms Pounds
31.52 69.344
As you see, after rounding off, when it comes to how much 31.52 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Please note it when you need to make bigger amount than 31.52 kilograms pounds conversion.
#### How many kilograms 31.52 pound
Now you learned how to convert 31.52 kilograms how much pounds but we are going to show you something more. Are you interested what it is? What about 31.52 kilogram to pounds and ounces calculation?
We want to show you how you can calculate it step by step. Begin. How much is 31.52 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, in this case 31.52, by 2.20462262. So 31.52 * 2.20462262 = 69.4897049824. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To calculate how much 31.52 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So final outcome is 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then your result is equal 2 pounds and 33 ounces.
As you see, conversion 31.52 kilogram in pounds and ounces quite simply.
The last calculation which we are going to show you is calculation of 31.52 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert it it is needed another formula. Before we give you it, let’s see:
• 31.52 kilograms meters = 7.23301385 foot pounds,
• 31.52 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 31.52 foot pounds to kilograms meters you have to multiply 31.52 by 0.13825495. It is exactly 0.13825495. So 31.52 foot pounds is exactly 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 31.52 foot pounds is exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 31.52 kilogram into pounds conversions.
This article is a huge compendium about kilogram, pound and 31.52 kg to lbs in conversion. Due to this calculation you learned 31.52 kilogram is equivalent to how many pounds.
We showed you not only how to make a calculation 31.52 kilogram to metric pounds but also two another conversions - to check how many 31.52 kg in pounds and ounces and how many 31.52 foot pounds to kilograms meters.
We showed you also other way to make 31.52 kilogram how many pounds calculations, this is using 31.52 kg en pound calculator. This is the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you can make 31.52 kilogram equal to how many pounds calculation - on your own or with use of our 31.52 kgs to pounds converter.
Don’t wait! Calculate 31.52 kilogram mass to pounds in the way you like.
Do you need to make other than 31.52 kilogram as pounds calculation? For example, for 15 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 31.52 kilogram equal many pounds.
### How much is 31.52 kg in pounds
To quickly sum up this topic, that is how much is 31.52 kg in pounds , we prepared for you an additional section. Here we have for you all you need to remember about how much is 31.52 kg equal to lbs and how to convert 31.52 kg to lbs . It is down below.
What is the kilogram to pound conversion? It is a mathematical operation based on multiplying 2 numbers. Let’s see 31.52 kg to pound conversion formula . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 31.52 kilogram to pounds? The accurate answer is 69.4897049824 lb.
You can also calculate how much 31.52 kilogram is equal to pounds with second, shortened version of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So this time, 31.52 kg equal to how much lbs ? The result is 69.4897049824 lb.
How to convert 31.52 kg to lbs in just a moment? You can also use the 31.52 kg to lbs converter , which will do the rest for you and you will get an accurate answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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31.01 kg to lbs = 68.3654 31.02 kg to lbs = 68.3874 31.03 kg to lbs = 68.4094 31.04 kg to lbs = 68.4315 31.05 kg to lbs = 68.4535 31.06 kg to lbs = 68.4756 31.07 kg to lbs = 68.4976 31.08 kg to lbs = 68.5197 31.09 kg to lbs = 68.5417 31.1 kg to lbs = 68.5638 31.11 kg to lbs = 68.5858 31.12 kg to lbs = 68.6079 31.13 kg to lbs = 68.6299 31.14 kg to lbs = 68.6519 31.15 kg to lbs = 68.674 31.16 kg to lbs = 68.696 31.17 kg to lbs = 68.7181 31.18 kg to lbs = 68.7401 31.19 kg to lbs = 68.7622 31.2 kg to lbs = 68.7842 31.21 kg to lbs = 68.8063 31.22 kg to lbs = 68.8283 31.23 kg to lbs = 68.8504 31.24 kg to lbs = 68.8724 31.25 kg to lbs = 68.8945
31.26 kg to lbs = 68.9165 31.27 kg to lbs = 68.9386 31.28 kg to lbs = 68.9606 31.29 kg to lbs = 68.9826 31.3 kg to lbs = 69.0047 31.31 kg to lbs = 69.0267 31.32 kg to lbs = 69.0488 31.33 kg to lbs = 69.0708 31.34 kg to lbs = 69.0929 31.35 kg to lbs = 69.1149 31.36 kg to lbs = 69.137 31.37 kg to lbs = 69.159 31.38 kg to lbs = 69.1811 31.39 kg to lbs = 69.2031 31.4 kg to lbs = 69.2251 31.41 kg to lbs = 69.2472 31.42 kg to lbs = 69.2692 31.43 kg to lbs = 69.2913 31.44 kg to lbs = 69.3133 31.45 kg to lbs = 69.3354 31.46 kg to lbs = 69.3574 31.47 kg to lbs = 69.3795 31.48 kg to lbs = 69.4015 31.49 kg to lbs = 69.4236 31.5 kg to lbs = 69.4456
31.51 kg to lbs = 69.4677 31.52 kg to lbs = 69.4897 31.53 kg to lbs = 69.5118 31.54 kg to lbs = 69.5338 31.55 kg to lbs = 69.5558 31.56 kg to lbs = 69.5779 31.57 kg to lbs = 69.5999 31.58 kg to lbs = 69.622 31.59 kg to lbs = 69.644 31.6 kg to lbs = 69.6661 31.61 kg to lbs = 69.6881 31.62 kg to lbs = 69.7102 31.63 kg to lbs = 69.7322 31.64 kg to lbs = 69.7543 31.65 kg to lbs = 69.7763 31.66 kg to lbs = 69.7983 31.67 kg to lbs = 69.8204 31.68 kg to lbs = 69.8424 31.69 kg to lbs = 69.8645 31.7 kg to lbs = 69.8865 31.71 kg to lbs = 69.9086 31.72 kg to lbs = 69.9306 31.73 kg to lbs = 69.9527 31.74 kg to lbs = 69.9747 31.75 kg to lbs = 69.9968
31.76 kg to lbs = 70.0188 31.77 kg to lbs = 70.0409 31.78 kg to lbs = 70.0629 31.79 kg to lbs = 70.085 31.8 kg to lbs = 70.107 31.81 kg to lbs = 70.1291 31.82 kg to lbs = 70.1511 31.83 kg to lbs = 70.1731 31.84 kg to lbs = 70.1952 31.85 kg to lbs = 70.2172 31.86 kg to lbs = 70.2393 31.87 kg to lbs = 70.2613 31.88 kg to lbs = 70.2834 31.89 kg to lbs = 70.3054 31.9 kg to lbs = 70.3275 31.91 kg to lbs = 70.3495 31.92 kg to lbs = 70.3715 31.93 kg to lbs = 70.3936 31.94 kg to lbs = 70.4156 31.95 kg to lbs = 70.4377 31.96 kg to lbs = 70.4597 31.97 kg to lbs = 70.4818 31.98 kg to lbs = 70.5038 31.99 kg to lbs = 70.5259 32 kg to lbs = 70.5479 | 4,920 | 16,427 | {"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} | 4.15625 | 4 | CC-MAIN-2023-23 | longest | en | 0.945825 |
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Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405, 18406..18450, 18451..18495, 18496..18540, 18541..18585, 18586..18630, 18631..18675, 18676..18720, 18721..18765, 18766..18810, 18811..18855, 18856..18900, 18901..18945, 18946..18990, 18991..19035, 19036..19080, 19081..19125, 19126..19170, 19171..19215, 19216..19260, 19261..19305, 19306..19350, 19351..19395, 19396..19440, 19441..19485, 19486..19530, 19531..19575, 19576..19620, 19621..19665, 19666..19710, 19711..19755, 19756..19800, 19801..19845, 19846..19890, 19891..19935, 19936..19980, 19981..20025, 20026..20070, 20071..20115, 20116..20160, 20161..20205, 20206..20250 | 5,760 | 15,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 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# How many odd three-digit integers greater than 800 are there
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How many odd three-digit integers greater than 800 are there [#permalink]
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22 May 2010, 20:33
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How many odd three-digit integers greater than 800 are there such that all their digits are different?
A. 40
B. 60
C. 72
D. 81
E. 104
[Reveal] Spoiler: OA
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Last edited by Bunuel on 04 Feb 2012, 07:10, edited 1 time in total.
Edited the question and added the OA
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Re: odd 3 digit integers [#permalink]
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23 May 2010, 00:23
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dimitri92 wrote:
Ok so this is a gmatclub test but I don't get the explanation. Can someone help?
How many odd three-digit integers greater than 800 are there such that all their digits are different?
40
60
72
81
104
In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*4*8 = 32.
Total: 40+32 = 72.
Answer: C.
Hope it's clear.
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Re: odd 3 digit integers [#permalink]
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23 May 2010, 04:00
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I would post another way with little short method.
First of all let's find out all numbers greater then 800 and having all different digits.
For hundreth place you have 2 choices, for tenth digit you have (10-1)=9 choices as one digit out of 10 have already used in hundreth place, for unit's digit you have (10-2) = 8 choices
Total three digit numbers with all different digits greater then 800 = 2*8*9 = 144
As 144 is an even number, half of these will be odd and half of this will be even.
So answer will be 72
Ans: C (72)
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Re: odd 3 digit integers [#permalink]
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22 May 2010, 22:03
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Alright so you need odd number so the first odd number is 801 and the last odd number is that is different is 987
there a total of {(987-801)/2} + 1 odd numbers. There are 94 odd numbers. Now you need to get rid of odd numbers that violate the rule that each digit must be different. there a total of 5 odd numbers digits per group of tens - >1,3,5,7,9. From 801 to 879 you count all the odds till this point but you discard all the odd in the 80s group which there are 5 -> 81, 83,85,87,89 - and you discard one odd in 50s -> 855 . The 900 forces you to make an additional consideration. There is one less odd per group of ten since you'll have a repeat of 9 - example 909, 919. There are only 4 odd numbers per group of ten and you also get rid of 955. There are a total 8 numbers that you exclude in the 900s on the way up to 987.
So the total number of odds you discard is 13
So the answer is 94 -13 = 81
IF that helped a kudo would be nice
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Re: odd 3 digit integers [#permalink]
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23 May 2010, 02:44
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Bunuel wrote:
dimitri92 wrote:
Ok so this is a gmatclub test but I don't get the explanation. Can someone help?
How many odd three-digit integers greater than 800 are there such that all their digits are different?
40
60
72
81
104
In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*4*8 = 32.
Total: 40+32 = 72.
Answer: C.
Hope it's clear.
wow amazing approach ... now this is crystal clear !!
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Re: Prob Problem [#permalink]
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24 May 2010, 16:29
Expert's post
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Let's xyz is our integer.
x e {8,9}
y e {0...9}
z e {1,3,5,7,9}
So, we have 1[x]*5[z]*(10-2)[y] = 40 integers that begin with 8 and 1[x]*4[z]*(10-2)[y] = 32 integers that begin with 9.
N = 40 + 32 = 72
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Re: M25 Question #20 - Counting Numbers with Unique Digits [#permalink]
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18 Jul 2010, 23:08
Merging similar topics. Another discussion: no-properties-71853.html?hilit=range%20choice%20first%20digit
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Re: odd 3 digit integers [#permalink]
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23 Aug 2010, 12:43
Hey Bunuel, in the solution below, I did not understand the highlighted part.
How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.
In the range 800 - 900:
1 choice for the first digit: 8;
[highlight]5 choices for the third digit: 1, 3, 5, 7, 9;[/highlight]
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
[highlight]4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);[/highlight]
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*4*8 = 32.
Total: 40+32 = 72.
Answer: C.
Hope it's clear.
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Re: odd 3 digit integers [#permalink]
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23 Aug 2010, 12:56
seekmba wrote:
Hey Bunuel, in the solution below, I did not understand the highlighted part.
How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.
In the range 800 - 900:
1 choice for the first digit: 8;
[highlight]5 choices for the third digit: 1, 3, 5, 7, 9;[/highlight]
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
[highlight]4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);[/highlight]
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*4*8 = 32.
Total: 40+32 = 72.
Answer: C.
Hope it's clear.
Odd numbers should end with odd digit.
Hence third digit should be 1, 3, 5, 7 or 9 - 5 choices, but for range 900-999 we can not use 9 as we used it as first digit so only 4 options are available.
Hope it's clear.
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Re: odd 3 digit integers [#permalink]
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23 Aug 2010, 19:31
thanks so much Bunuel.
Thats silly of me.......totally forgot about the odd numbers between 800 and 1000.
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Re: odd 3 digit integers [#permalink]
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23 Aug 2010, 19:40
Bunuel wrote:
dimitri92 wrote:
Ok so this is a gmatclub test but I don't get the explanation. Can someone help?
How many odd three-digit integers greater than 800 are there such that all their digits are different?
40
60
72
81
104
In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*4*8 = 32.
Total: 40+32 = 72.
Answer: C.
Hope it's clear.
Bunuel,
Why did you split up the three digit number into two different subsets and then proceed to solve the problem. Is there some subtle logic behind this.
Please do share your thoughts on this.
This was my approach to solve this problem
x y z is the three digit number. Unit digit z has 4 choices (1, 3, 5 and 7), y has 7 choices (0, 1, 3, 5, 2, 4, 6 leaving out 7 with the assumption that it is the number chosen for the unit digit.) and x has 2 choices (8 and 9).
Of course multiplying these choices does not lead to any of the answer choice.
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Re: odd 3 digit integers [#permalink]
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24 Aug 2010, 05:07
ezhilkumarank wrote:
Bunuel,
Why did you split up the three digit number into two different subsets and then proceed to solve the problem. Is there some subtle logic behind this.
Please do share your thoughts on this.
This was my approach to solve this problem
x y z is the three digit number. Unit digit z has 4 choices (1, 3, 5 and 7), y has 7 choices (0, 1, 3, 5, 2, 4, 6 leaving out 7 with the assumption that it is the number chosen for the unit digit.) and x has 2 choices (8 and 9).
Of course multiplying these choices does not lead to any of the answer choice.
There are 5 choices for units digit when hundreds digit is 8 and 4 when hundreds digit is 9. So you should count numbers in these 2 ranges separately.
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Re: odd 3 digit integers [#permalink]
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30 Aug 2010, 03:49
Hey Bunuel, in the solution below, I did not understand
How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.
In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. --> this should be 9 digits right why havent u considered 0
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. -->-> this should be 9 digits right why havent u considered 0
1*4*8 = 32.
Total: 40+32 = 72.
Answer: C.
Hope it's clear.
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Re: odd 3 digit integers [#permalink]
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30 Aug 2010, 04:51
Divyababu wrote:
Hey Bunuel, in the solution below, I did not understand
How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.
In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. --> this should be 9 digits right why havent u considered 0
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. -->-> this should be 9 digits right why havent u considered 0
1*4*8 = 32.
Total: 40+32 = 72.
Answer: C.
Hope it's clear.
Odd numbers should end with odd digit.
Hence third digit should be 1, 3, 5, 7 or 9 - 5 choices, but for range 900-999 we can not use 9 as we used it as first digit so only 4 options are available.
Hope it's clear.
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Re: odd 3 digit integers [#permalink]
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30 Aug 2010, 05:34
Thanks for the explaination Bunuel.
It is clear to me
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Re: odd 3 digit integers [#permalink]
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01 Sep 2010, 05:38
zerotoinfinite2006 wrote:
Bunuel wrote:
dimitri92 wrote:
Ok so this is a gmatclub test but I don't get the explanation. Can someone help?
How many odd three-digit integers greater than 800 are there such that all their digits are different?
40
60
72
81
104
In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits. ---> I am confused here, how did we get 8? is it 10 - 1 - 1 ? What does first digit and third digit means ?
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*4*8 = 32.
Total: 40+32 = 72.
Answer: C.
Hope it's clear.
I am confused with the selection of second digit.
3 digit integer abc: we used 1 digit for the first digit $$a$$, 1 digit for the third digit $$c$$, so for the second digit $$b$$ there are 10-1-1=8 digits left.
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Re: odd 3 digit integers [#permalink]
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09 Sep 2010, 07:05
Hi Bunuel,
I did not understand why you included 900 again in the range of 900 - 999 (should it not be 901 to 999). Why we counted 900 two times?
[highlight]In the range 800 - 900:[/highlight]
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*5*8 = 40.
[highlight]In the range 900 - 999:[/highlight]
1 choice for the first digit: 9;
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How many odd three-digit integers greater than 800 are there [#permalink]
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09 Sep 2010, 07:25
seekmba wrote:
Hi Bunuel,
I did not understand why you included 900 again in the range of 900 - 999 (should it not be 901 to 999). Why we counted 900 two times?
In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*5*8 = 40.
In the range 900 - 999:
1 choice for the first digit: 9;
As we are counting ODD numbers it doesn't make any difference whether we include even number 900 once, twice or even ten times.
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Re: odd 3 digit integers [#permalink]
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09 Sep 2010, 08:05
ok that makes sense.
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Re: odd 3 digit integers [#permalink]
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03 Feb 2012, 22:42
much clearer... thanks for the explanation
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Re: odd 3 digit integers [#permalink] 03 Feb 2012, 22:42
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 5,710 | 18,204 | {"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-2017-04 | latest | en | 0.861573 |
https://openmdao.org/dymos/docs/latest/examples/mountain_car/mountain_car.html | 1,696,176,890,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510903.85/warc/CC-MAIN-20231001141548-20231001171548-00671.warc.gz | 467,074,926 | 625,810 | The Mountain Car Problem#
The mountain car problem proposes a vehicle stuck in a “well.” It lacks the power to directly climb out of the well, but instead must accelerate repeatedly forwards and backwards until it has achieved the energy necessary to exit the well.
The problem is a popular machine learning test case, though the methods in Dymos are capable of solving it. It first appeared in the PhD thesis of Andrew Moore in 1990. [Moo90]. The implementation here is based on that given by Melnikov, Makmal, and Briegel [MMB14].
State and control variables#
This system has two state variables, the position ($$x$$) and velocity ($$v$$) of the car.
This system has a single control variable ($$u$$), the effort put into moving. This control is contrained to the range $$[-1 \, 1]$$.
The dynamics of the system are governed by
(65)#\begin{align} \dot{x} &= v \\ \dot{v} &= 0.001 * u - 0.0025 * \cos(3 x) \end{align}
Problem Definition#
We seek to minimize the time required to exit the well in the positive direction.
(66)#\begin{align} \mathrm{Minimize} \, J &= t_f \end{align}
Subject to the initial conditions
(67)#\begin{align} x_0 &= -0.5 \\ v_0 &= 0.0 \end{align}
the control constraints
(68)#\begin{align} |u| \le 1 \end{align}
and the terminal constraints
(69)#\begin{align} x_f &= 0.5 \\ v_f &\ge 0.0 \end{align}
Defining the ODE#
The following code implements the equations of motion for the mountain car problem.
A few things to note:
1. By providing the tag dymos.state_rate_source:{name}, we’re letting Dymos know what states need to be integrated, there’s no need to specify a rate source when using this ODE in our Phase.
2. Pairing the above tag with dymos.state_units:{units} means we don’t have to specify units when setting properties for the state in our run script.
3. We only use compute_partials to override the values of $$\frac{\partial \dot{v}}{\partial x}$$ because $$\frac{\partial \dot{v}}{\partial u}$$ and $$\frac{\partial \dot{x}}{\partial v}$$ are constant and their value is specified during setup.
import numpy as np
import openmdao.api as om
class MountainCarODE(om.ExplicitComponent):
def initialize(self):
self.options.declare('num_nodes', types=int)
def setup(self):
nn = self.options['num_nodes']
tags=['dymos.state_rate_source:x', 'dymos.state_units:m'])
tags=['dymos.state_rate_source:v', 'dymos.state_units:m/s'])
ar = np.arange(nn, dtype=int)
self.declare_partials(of='x_dot', wrt='v', rows=ar, cols=ar, val=1.0)
self.declare_partials(of='v_dot', wrt='u', rows=ar, cols=ar, val=0.001)
self.declare_partials(of='v_dot', wrt='x', rows=ar, cols=ar)
def compute(self, inputs, outputs):
x = inputs['x']
v = inputs['v']
u = inputs['u']
outputs['x_dot'] = v
outputs['v_dot'] = 0.001 * u - 0.0025 * np.cos(3*x)
def compute_partials(self, inputs, partials):
x = inputs['x']
partials['v_dot', 'x'] = 3 * 0.0025 * np.sin(3 * x)
Solving the minimum-time mountain car problem with Dymos#
The following script solves the minimum-time mountain car problem with Dymos. Note that this example requires the IPOPT optimizer via the pyoptsparse package. Scipy’s SLSQP optimizer is generally not capable of solving this problem.
To begin, import the packages we require:
import dymos as dm
import matplotlib.pyplot as plt
from matplotlib import animation
Next, we set two constants. U_MAX is the maximum allowable magnitude of the acceleration. The references show this problem being solved with $$-1 \le u \le 1$$.
Variable NUM_SEG is the number of equally spaced polynomial segments into which time is being divided. Within each of these segments, the time-history of each state and control is being treated as a polynomial (we’re using the default order of 3).
# The maximum absolute value of the acceleration authority of the car
U_MAX = 1.0
# The number of segments into which the problem is discretized
NUM_SEG = 30
We then instantiate an OpenMDAO problem and set the optimizer and its options.
For IPOPT, setting option nlp_scaling_method to 'gradient-based' can substantially improve the convergence of the optimizer without the need for us to set all of the scaling manually.
The call to declare_coloring tells the optimizer to attempt to find a sparsity pattern that minimizes the work required to compute the derivatives across the model.
#
# Initialize the Problem and the optimization driver
#
p = om.Problem()
p.driver = om.pyOptSparseDriver(optimizer='IPOPT')
p.driver.opt_settings['print_level'] = 0
p.driver.opt_settings['max_iter'] = 500
p.driver.opt_settings['bound_mult_init_method'] = 'mu-based'
p.driver.opt_settings['tol'] = 1.0E-8
p.driver.opt_settings['nlp_scaling_method'] = 'gradient-based' # for faster convergence
p.driver.declare_coloring()
Next, we add a Dymos Trajectory group to the problem’s model and add a phase to it.
In this case we’re using the Radau pseudospectral transcription to solve the problem.
#
# Create a trajectory and add a phase to it
#
At this point, we set the options on the main variables used in a Dymos phase.
In addition to time, we have two states (x and v) and a single control (u).
There are no parameters and no polynomial controls. We could have tried to use a polynomial control here, but as we will see the solution contains large discontinuities in the control value, which make it ill-suited for a polynomial control. Polynomial controls are modeled as a single (typically low-order) polynomial across the entire phase.
We’re fixing the initial time and states to whatever values we provide before executing the problem. We will constrain the final values with nonlinear constraints in the next step.
The scaler values (ref) are all set to 1 here. We’re using IPOPT’s gradient-based scaling option and will let it work the scaling out for us.
Bounds on time duration are guesses, and the bounds on the states and controls come from the implementation in the references.
Also, we don’t need to specify targets for any of the variables here because their names are the targets in the top-level of the model. The rate source and units for the states are obtained from the tags in the ODE component we previously defined.
#
# Set the variables
#
phase.set_time_options(fix_initial=True, duration_bounds=(.05, 10000), duration_ref=1)
phase.add_state('x', fix_initial=True, fix_final=False, lower=-1.2, upper=0.5, ref=1, defect_ref=1)
phase.add_state('v', fix_initial=True, fix_final=False, lower=-0.07, upper=0.07, ref=1, defect_ref=1)
phase.add_control('u', lower=-U_MAX, upper=U_MAX, ref=1, continuity=True, rate_continuity=False)
Next we define the optimal control problem by specifying the objective, boundary constraints, and path constraints.
Why do we have a path constraint on the control u when we’ve already specified its bounds?
Excellent question! In the Radau transcription, the $$n^{th}$$ order control polynomial is governed by design variables provided at $$n$$ points in the segment that do not contain the right-most endpoint. Instead, this value is interpolated based on the values of the first $$(n-1)$$. Since this value is not a design variable, it is necessary to constrain its value separately. We could forgo specifying any bounds on u since it’s completely covered by the path constraint, but specifying the bounds on the design variable values can sometimes help by telling the optimizer, “Don’t even bother trying values outside of this range.”.
Note that sometimes the opposite is true, and giving the optimizer the freedom to explore a larger design space, only to eventually be “reined-in” by the path constraint can sometimes be valuable.
The purpose of this interactive documentation is to let the user experiment. If you remove the path constraint, you might notice some outlying control values in the solution below.
#
# Minimize time at the end of the phase
#
#
# Setup the Problem
#
p.setup()
--- Constraint Report [traj] ---
--- phase0 ---
[final] 5.0000e-01 <= x [m]
[final] 0.0000e+00 <= v [m/s]
[path] -1.0000e+00 <= u <= 1.0000e+00 [unitless]
<openmdao.core.problem.Problem at 0x7f5678b24460>
We then set the initial guesses for the variables in the problem and solve it.
Since fix_initial=True is set for time and the states, those values are not design variables and will remain at the values given below throughout the solution process.
We’re using the phase interp method to provide initial guesses for the states and controls. In this case, by giving it two values, it is linearly interpolating from the first value to the second value, and then returning the interpolated value at the input nodes for the given variable.
Finally, we use the dymos.run_problem method to execute the problem. This interface allows us to do some things that the standard OpenMDAO problem.run_driver interface does not. It will automatically record the final solution achieved by the optimizer in case named 'final' in a file called dymos_solution.db. By specifying simulate=True, it will automatically follow the solution with an explicit integration using scipy.solve_ivp. The results of the simulation are stored in a case named final in the file dymos_simulation.db. This explicit simulation demonstrates how the system evolved with the given controls, and serves as a check that we’re using a dense enough grid (enough segments and segments of sufficient order) to accurately represent the solution.
If those two solution didn’t agree reasonably well, we could rerun the problem with a more dense grid. Instead, we’re asking Dymos to automatically change the grid if necessary by specifying refine_method='ph'. This will attempt to repeatedly solve the problem and change the number of segments and segment orders until the solution is in reasonable agreement.
#
# Set the initial values
#
p['traj.phase0.t_initial'] = 0.0
p['traj.phase0.t_duration'] = 500.0
p.set_val('traj.phase0.states:x', phase.interp('x', ys=[-0.5, 0.5]))
p.set_val('traj.phase0.states:v', phase.interp('v', ys=[0, 0.07]))
p.set_val('traj.phase0.controls:u', np.sin(phase.interp('u', ys=[0, 1.0])))
#
# Solve for the optimal trajectory
#
dm.run_problem(p, run_driver=True, simulate=True, refine_method='ph', refine_iteration_limit=5)
Hide code cell output
Model viewer data has already been recorded for Driver.
Model viewer data has already been recorded for Driver.
Full total jacobian was computed 3 times, taking 0.426565 seconds.
Total jacobian shape: (332, 271)
Jacobian shape: (332, 271) ( 1.68% nonzero)
FWD solves: 9 REV solves: 0
Total colors vs. total size: 9 vs 271 (96.7% improvement)
Sparsity computed using tolerance: 1e-25
Time to compute sparsity: 0.426565 sec.
Time to compute coloring: 0.210060 sec.
Memory to compute coloring: 0.250000 MB.
Optimization Problem -- Optimization using pyOpt_sparse
================================================================================
Objective Function: _objfunc
Solution:
--------------------------------------------------------------------------------
Total Time: 2.9427
User Objective Time : 0.1861
User Sensitivity Time : 1.1535
Interface Time : 0.8409
Opt Solver Time: 0.7621
Calls to Objective Function : 123
Calls to Sens Function : 120
Objectives
Index Name Value
0 traj.phases.phase0.time.t 1.027055E-01
Variables (c - continuous, i - integer, d - discrete)
Index Name Type Lower Bound Value Upper Bound Status
0 traj.phase0.t_duration_0 c 5.000000E-02 1.027055E+02 1.000000E+04
1 traj.phases.phase0.indep_states.states:x_0 c -1.200000E+00 -4.993918E-01 5.000000E-01
2 traj.phases.phase0.indep_states.states:x_1 c -1.200000E+00 -4.965749E-01 5.000000E-01
3 traj.phases.phase0.indep_states.states:x_2 c -1.200000E+00 -4.952113E-01 5.000000E-01
4 traj.phases.phase0.indep_states.states:x_3 c -1.200000E+00 -4.912601E-01 5.000000E-01
5 traj.phases.phase0.indep_states.states:x_4 c -1.200000E+00 -4.839852E-01 5.000000E-01
6 traj.phases.phase0.indep_states.states:x_5 c -1.200000E+00 -4.812613E-01 5.000000E-01
7 traj.phases.phase0.indep_states.states:x_6 c -1.200000E+00 -4.743104E-01 5.000000E-01
8 traj.phases.phase0.indep_states.states:x_7 c -1.200000E+00 -4.632080E-01 5.000000E-01
9 traj.phases.phase0.indep_states.states:x_8 c -1.200000E+00 -4.593596E-01 5.000000E-01
10 traj.phases.phase0.indep_states.states:x_9 c -1.200000E+00 -4.499012E-01 5.000000E-01
11 traj.phases.phase0.indep_states.states:x_10 c -1.200000E+00 -4.363638E-01 5.000000E-01
12 traj.phases.phase0.indep_states.states:x_11 c -1.200000E+00 -4.322699E-01 5.000000E-01
13 traj.phases.phase0.indep_states.states:x_12 c -1.200000E+00 -4.247954E-01 5.000000E-01
14 traj.phases.phase0.indep_states.states:x_13 c -1.200000E+00 -4.186817E-01 5.000000E-01
15 traj.phases.phase0.indep_states.states:x_14 c -1.200000E+00 -4.177845E-01 5.000000E-01
16 traj.phases.phase0.indep_states.states:x_15 c -1.200000E+00 -4.176175E-01 5.000000E-01
17 traj.phases.phase0.indep_states.states:x_16 c -1.200000E+00 -4.216950E-01 5.000000E-01
18 traj.phases.phase0.indep_states.states:x_17 c -1.200000E+00 -4.240204E-01 5.000000E-01
19 traj.phases.phase0.indep_states.states:x_18 c -1.200000E+00 -4.311749E-01 5.000000E-01
20 traj.phases.phase0.indep_states.states:x_19 c -1.200000E+00 -4.451029E-01 5.000000E-01
21 traj.phases.phase0.indep_states.states:x_20 c -1.200000E+00 -4.504564E-01 5.000000E-01
22 traj.phases.phase0.indep_states.states:x_21 c -1.200000E+00 -4.643313E-01 5.000000E-01
23 traj.phases.phase0.indep_states.states:x_22 c -1.200000E+00 -4.869306E-01 5.000000E-01
24 traj.phases.phase0.indep_states.states:x_23 c -1.200000E+00 -4.948560E-01 5.000000E-01
25 traj.phases.phase0.indep_states.states:x_24 c -1.200000E+00 -5.142625E-01 5.000000E-01
26 traj.phases.phase0.indep_states.states:x_25 c -1.200000E+00 -5.435816E-01 5.000000E-01
27 traj.phases.phase0.indep_states.states:x_26 c -1.200000E+00 -5.533912E-01 5.000000E-01
28 traj.phases.phase0.indep_states.states:x_27 c -1.200000E+00 -5.766416E-01 5.000000E-01
29 traj.phases.phase0.indep_states.states:x_28 c -1.200000E+00 -6.101308E-01 5.000000E-01
30 traj.phases.phase0.indep_states.states:x_29 c -1.200000E+00 -6.209737E-01 5.000000E-01
31 traj.phases.phase0.indep_states.states:x_30 c -1.200000E+00 -6.460628E-01 5.000000E-01
32 traj.phases.phase0.indep_states.states:x_31 c -1.200000E+00 -6.808768E-01 5.000000E-01
33 traj.phases.phase0.indep_states.states:x_32 c -1.200000E+00 -6.918463E-01 5.000000E-01
34 traj.phases.phase0.indep_states.states:x_33 c -1.200000E+00 -7.167113E-01 5.000000E-01
35 traj.phases.phase0.indep_states.states:x_34 c -1.200000E+00 -7.500849E-01 5.000000E-01
36 traj.phases.phase0.indep_states.states:x_35 c -1.200000E+00 -7.603375E-01 5.000000E-01
37 traj.phases.phase0.indep_states.states:x_36 c -1.200000E+00 -7.831224E-01 5.000000E-01
38 traj.phases.phase0.indep_states.states:x_37 c -1.200000E+00 -8.126995E-01 5.000000E-01
39 traj.phases.phase0.indep_states.states:x_38 c -1.200000E+00 -8.215465E-01 5.000000E-01
40 traj.phases.phase0.indep_states.states:x_39 c -1.200000E+00 -8.407829E-01 5.000000E-01
41 traj.phases.phase0.indep_states.states:x_40 c -1.200000E+00 -8.647953E-01 5.000000E-01
42 traj.phases.phase0.indep_states.states:x_41 c -1.200000E+00 -8.717406E-01 5.000000E-01
43 traj.phases.phase0.indep_states.states:x_42 c -1.200000E+00 -8.865112E-01 5.000000E-01
44 traj.phases.phase0.indep_states.states:x_43 c -1.200000E+00 -9.033494E-01 5.000000E-01
45 traj.phases.phase0.indep_states.states:x_44 c -1.200000E+00 -9.075277E-01 5.000000E-01
46 traj.phases.phase0.indep_states.states:x_45 c -1.200000E+00 -9.135009E-01 5.000000E-01
47 traj.phases.phase0.indep_states.states:x_46 c -1.200000E+00 -9.137392E-01 5.000000E-01
48 traj.phases.phase0.indep_states.states:x_47 c -1.200000E+00 -9.118763E-01 5.000000E-01
49 traj.phases.phase0.indep_states.states:x_48 c -1.200000E+00 -9.041177E-01 5.000000E-01
50 traj.phases.phase0.indep_states.states:x_49 c -1.200000E+00 -8.854927E-01 5.000000E-01
51 traj.phases.phase0.indep_states.states:x_50 c -1.200000E+00 -8.777039E-01 5.000000E-01
52 traj.phases.phase0.indep_states.states:x_51 c -1.200000E+00 -8.565061E-01 5.000000E-01
53 traj.phases.phase0.indep_states.states:x_52 c -1.200000E+00 -8.197802E-01 5.000000E-01
54 traj.phases.phase0.indep_states.states:x_53 c -1.200000E+00 -8.064054E-01 5.000000E-01
55 traj.phases.phase0.indep_states.states:x_54 c -1.200000E+00 -7.727637E-01 5.000000E-01
56 traj.phases.phase0.indep_states.states:x_55 c -1.200000E+00 -7.198818E-01 5.000000E-01
57 traj.phases.phase0.indep_states.states:x_56 c -1.200000E+00 -7.016946E-01 5.000000E-01
58 traj.phases.phase0.indep_states.states:x_57 c -1.200000E+00 -6.576998E-01 5.000000E-01
59 traj.phases.phase0.indep_states.states:x_58 c -1.200000E+00 -5.923140E-01 5.000000E-01
60 traj.phases.phase0.indep_states.states:x_59 c -1.200000E+00 -5.706632E-01 5.000000E-01
61 traj.phases.phase0.indep_states.states:x_60 c -1.200000E+00 -5.197353E-01 5.000000E-01
62 traj.phases.phase0.indep_states.states:x_61 c -1.200000E+00 -4.472326E-01 5.000000E-01
63 traj.phases.phase0.indep_states.states:x_62 c -1.200000E+00 -4.239588E-01 5.000000E-01
64 traj.phases.phase0.indep_states.states:x_63 c -1.200000E+00 -3.704650E-01 5.000000E-01
65 traj.phases.phase0.indep_states.states:x_64 c -1.200000E+00 -2.970089E-01 5.000000E-01
66 traj.phases.phase0.indep_states.states:x_65 c -1.200000E+00 -2.740401E-01 5.000000E-01
67 traj.phases.phase0.indep_states.states:x_66 c -1.200000E+00 -2.222404E-01 5.000000E-01
68 traj.phases.phase0.indep_states.states:x_67 c -1.200000E+00 -1.531484E-01 5.000000E-01
69 traj.phases.phase0.indep_states.states:x_68 c -1.200000E+00 -1.319827E-01 5.000000E-01
70 traj.phases.phase0.indep_states.states:x_69 c -1.200000E+00 -8.491750E-02 5.000000E-01
71 traj.phases.phase0.indep_states.states:x_70 c -1.200000E+00 -2.341760E-02 5.000000E-01
72 traj.phases.phase0.indep_states.states:x_71 c -1.200000E+00 -4.829274E-03 5.000000E-01
73 traj.phases.phase0.indep_states.states:x_72 c -1.200000E+00 3.615597E-02 5.000000E-01
74 traj.phases.phase0.indep_states.states:x_73 c -1.200000E+00 8.913343E-02 5.000000E-01
75 traj.phases.phase0.indep_states.states:x_74 c -1.200000E+00 1.050566E-01 5.000000E-01
76 traj.phases.phase0.indep_states.states:x_75 c -1.200000E+00 1.400830E-01 5.000000E-01
77 traj.phases.phase0.indep_states.states:x_76 c -1.200000E+00 1.853435E-01 5.000000E-01
78 traj.phases.phase0.indep_states.states:x_77 c -1.200000E+00 1.989849E-01 5.000000E-01
79 traj.phases.phase0.indep_states.states:x_78 c -1.200000E+00 2.291198E-01 5.000000E-01
80 traj.phases.phase0.indep_states.states:x_79 c -1.200000E+00 2.684861E-01 5.000000E-01
81 traj.phases.phase0.indep_states.states:x_80 c -1.200000E+00 2.804879E-01 5.000000E-01
82 traj.phases.phase0.indep_states.states:x_81 c -1.200000E+00 3.072965E-01 5.000000E-01
83 traj.phases.phase0.indep_states.states:x_82 c -1.200000E+00 3.430993E-01 5.000000E-01
84 traj.phases.phase0.indep_states.states:x_83 c -1.200000E+00 3.542309E-01 5.000000E-01
85 traj.phases.phase0.indep_states.states:x_84 c -1.200000E+00 3.795258E-01 5.000000E-01
86 traj.phases.phase0.indep_states.states:x_85 c -1.200000E+00 4.143692E-01 5.000000E-01
87 traj.phases.phase0.indep_states.states:x_86 c -1.200000E+00 4.254779E-01 5.000000E-01
88 traj.phases.phase0.indep_states.states:x_87 c -1.200000E+00 4.512448E-01 5.000000E-01
89 traj.phases.phase0.indep_states.states:x_88 c -1.200000E+00 4.879813E-01 5.000000E-01
90 traj.phases.phase0.indep_states.states:x_89 c -1.200000E+00 5.000000E-01 5.000000E-01 u
91 traj.phases.phase0.indep_states.states:v_0 c -7.000000E-02 9.987135E-04 7.000000E-02
92 traj.phases.phase0.indep_states.states:v_1 c -7.000000E-02 2.356393E-03 7.000000E-02
93 traj.phases.phase0.indep_states.states:v_2 c -7.000000E-02 2.777103E-03 7.000000E-02
94 traj.phases.phase0.indep_states.states:v_3 c -7.000000E-02 3.717045E-03 7.000000E-02
95 traj.phases.phase0.indep_states.states:v_4 c -7.000000E-02 4.944779E-03 7.000000E-02
96 traj.phases.phase0.indep_states.states:v_5 c -7.000000E-02 5.312908E-03 7.000000E-02
97 traj.phases.phase0.indep_states.states:v_6 c -7.000000E-02 6.112790E-03 7.000000E-02
98 traj.phases.phase0.indep_states.states:v_7 c -7.000000E-02 7.105115E-03 7.000000E-02
99 traj.phases.phase0.indep_states.states:v_8 c -7.000000E-02 7.389281E-03 7.000000E-02
100 traj.phases.phase0.indep_states.states:v_9 c -7.000000E-02 8.059965E-03 7.000000E-02
101 traj.phases.phase0.indep_states.states:v_10 c -7.000000E-02 7.867031E-03 7.000000E-02
102 traj.phases.phase0.indep_states.states:v_11 c -7.000000E-02 7.181148E-03 7.000000E-02
103 traj.phases.phase0.indep_states.states:v_12 c -7.000000E-02 5.108835E-03 7.000000E-02
104 traj.phases.phase0.indep_states.states:v_13 c -7.000000E-02 2.165338E-03 7.000000E-02
105 traj.phases.phase0.indep_states.states:v_14 c -7.000000E-02 1.221816E-03 7.000000E-02
106 traj.phases.phase0.indep_states.states:v_15 c -7.000000E-02 -9.448555E-04 7.000000E-02
107 traj.phases.phase0.indep_states.states:v_16 c -7.000000E-02 -3.913376E-03 7.000000E-02
108 traj.phases.phase0.indep_states.states:v_17 c -7.000000E-02 -4.839342E-03 7.000000E-02
109 traj.phases.phase0.indep_states.states:v_18 c -7.000000E-02 -6.919809E-03 7.000000E-02
110 traj.phases.phase0.indep_states.states:v_19 c -7.000000E-02 -9.664867E-03 7.000000E-02
111 traj.phases.phase0.indep_states.states:v_20 c -7.000000E-02 -1.049505E-02 7.000000E-02
112 traj.phases.phase0.indep_states.states:v_21 c -7.000000E-02 -1.231170E-02 7.000000E-02
113 traj.phases.phase0.indep_states.states:v_22 c -7.000000E-02 -1.459420E-02 7.000000E-02
114 traj.phases.phase0.indep_states.states:v_23 c -7.000000E-02 -1.525502E-02 7.000000E-02
115 traj.phases.phase0.indep_states.states:v_24 c -7.000000E-02 -1.664514E-02 7.000000E-02
116 traj.phases.phase0.indep_states.states:v_25 c -7.000000E-02 -1.825856E-02 7.000000E-02
117 traj.phases.phase0.indep_states.states:v_26 c -7.000000E-02 -1.869027E-02 7.000000E-02
118 traj.phases.phase0.indep_states.states:v_27 c -7.000000E-02 -1.952984E-02 7.000000E-02
119 traj.phases.phase0.indep_states.states:v_28 c -7.000000E-02 -2.033778E-02 7.000000E-02
120 traj.phases.phase0.indep_states.states:v_29 c -7.000000E-02 -2.050692E-02 7.000000E-02
121 traj.phases.phase0.indep_states.states:v_30 c -7.000000E-02 -2.073869E-02 7.000000E-02
122 traj.phases.phase0.indep_states.states:v_31 c -7.000000E-02 -2.070879E-02 7.000000E-02
123 traj.phases.phase0.indep_states.states:v_32 c -7.000000E-02 -2.061684E-02 7.000000E-02
124 traj.phases.phase0.indep_states.states:v_33 c -7.000000E-02 -2.026466E-02 7.000000E-02
125 traj.phases.phase0.indep_states.states:v_34 c -7.000000E-02 -1.947378E-02 7.000000E-02
126 traj.phases.phase0.indep_states.states:v_35 c -7.000000E-02 -1.915480E-02 7.000000E-02
127 traj.phases.phase0.indep_states.states:v_36 c -7.000000E-02 -1.831126E-02 7.000000E-02
128 traj.phases.phase0.indep_states.states:v_37 c -7.000000E-02 -1.691377E-02 7.000000E-02
129 traj.phases.phase0.indep_states.states:v_38 c -7.000000E-02 -1.642126E-02 7.000000E-02
130 traj.phases.phase0.indep_states.states:v_39 c -7.000000E-02 -1.521386E-02 7.000000E-02
131 traj.phases.phase0.indep_states.states:v_40 c -7.000000E-02 -1.339010E-02 7.000000E-02
132 traj.phases.phase0.indep_states.states:v_41 c -7.000000E-02 -1.278055E-02 7.000000E-02
133 traj.phases.phase0.indep_states.states:v_42 c -7.000000E-02 -1.141416E-02 7.000000E-02
134 traj.phases.phase0.indep_states.states:v_43 c -7.000000E-02 -8.458169E-03 7.000000E-02
135 traj.phases.phase0.indep_states.states:v_44 c -7.000000E-02 -6.915081E-03 7.000000E-02
136 traj.phases.phase0.indep_states.states:v_45 c -7.000000E-02 -2.911298E-03 7.000000E-02
137 traj.phases.phase0.indep_states.states:v_46 c -7.000000E-02 2.630510E-03 7.000000E-02
138 traj.phases.phase0.indep_states.states:v_47 c -7.000000E-02 4.382076E-03 7.000000E-02
139 traj.phases.phase0.indep_states.states:v_48 c -7.000000E-02 8.376872E-03 7.000000E-02
140 traj.phases.phase0.indep_states.states:v_49 c -7.000000E-02 1.381962E-02 7.000000E-02
141 traj.phases.phase0.indep_states.states:v_50 c -7.000000E-02 1.551749E-02 7.000000E-02
142 traj.phases.phase0.indep_states.states:v_51 c -7.000000E-02 1.934068E-02 7.000000E-02
143 traj.phases.phase0.indep_states.states:v_52 c -7.000000E-02 2.441560E-02 7.000000E-02
144 traj.phases.phase0.indep_states.states:v_53 c -7.000000E-02 2.595997E-02 7.000000E-02
145 traj.phases.phase0.indep_states.states:v_54 c -7.000000E-02 2.935250E-02 7.000000E-02
146 traj.phases.phase0.indep_states.states:v_55 c -7.000000E-02 3.363028E-02 7.000000E-02
147 traj.phases.phase0.indep_states.states:v_56 c -7.000000E-02 3.486801E-02 7.000000E-02
148 traj.phases.phase0.indep_states.states:v_57 c -7.000000E-02 3.745594E-02 7.000000E-02
149 traj.phases.phase0.indep_states.states:v_58 c -7.000000E-02 4.039299E-02 7.000000E-02
150 traj.phases.phase0.indep_states.states:v_59 c -7.000000E-02 4.115260E-02 7.000000E-02
151 traj.phases.phase0.indep_states.states:v_60 c -7.000000E-02 4.256515E-02 7.000000E-02
152 traj.phases.phase0.indep_states.states:v_61 c -7.000000E-02 4.374409E-02 7.000000E-02
153 traj.phases.phase0.indep_states.states:v_62 c -7.000000E-02 4.392749E-02 7.000000E-02
154 traj.phases.phase0.indep_states.states:v_63 c -7.000000E-02 4.401872E-02 7.000000E-02
155 traj.phases.phase0.indep_states.states:v_64 c -7.000000E-02 4.344021E-02 7.000000E-02
156 traj.phases.phase0.indep_states.states:v_65 c -7.000000E-02 4.310092E-02 7.000000E-02
157 traj.phases.phase0.indep_states.states:v_66 c -7.000000E-02 4.208285E-02 7.000000E-02
158 traj.phases.phase0.indep_states.states:v_67 c -7.000000E-02 4.021939E-02 7.000000E-02
159 traj.phases.phase0.indep_states.states:v_68 c -7.000000E-02 3.954270E-02 7.000000E-02
160 traj.phases.phase0.indep_states.states:v_69 c -7.000000E-02 3.788101E-02 7.000000E-02
161 traj.phases.phase0.indep_states.states:v_70 c -7.000000E-02 3.542495E-02 7.000000E-02
162 traj.phases.phase0.indep_states.states:v_71 c -7.000000E-02 3.463051E-02 7.000000E-02
163 traj.phases.phase0.indep_states.states:v_72 c -7.000000E-02 3.281297E-02 7.000000E-02
164 traj.phases.phase0.indep_states.states:v_73 c -7.000000E-02 3.037524E-02 7.000000E-02
165 traj.phases.phase0.indep_states.states:v_74 c -7.000000E-02 2.963560E-02 7.000000E-02
166 traj.phases.phase0.indep_states.states:v_75 c -7.000000E-02 2.801759E-02 7.000000E-02
167 traj.phases.phase0.indep_states.states:v_76 c -7.000000E-02 2.599539E-02 7.000000E-02
168 traj.phases.phase0.indep_states.states:v_77 c -7.000000E-02 2.541413E-02 7.000000E-02
169 traj.phases.phase0.indep_states.states:v_78 c -7.000000E-02 2.419781E-02 7.000000E-02
170 traj.phases.phase0.indep_states.states:v_79 c -7.000000E-02 2.279996E-02 7.000000E-02
171 traj.phases.phase0.indep_states.states:v_80 c -7.000000E-02 2.242907E-02 7.000000E-02
172 traj.phases.phase0.indep_states.states:v_81 c -7.000000E-02 2.171340E-02 7.000000E-02
173 traj.phases.phase0.indep_states.states:v_82 c -7.000000E-02 2.104175E-02 7.000000E-02
174 traj.phases.phase0.indep_states.states:v_83 c -7.000000E-02 2.090773E-02 7.000000E-02
175 traj.phases.phase0.indep_states.states:v_84 c -7.000000E-02 2.074701E-02 7.000000E-02
176 traj.phases.phase0.indep_states.states:v_85 c -7.000000E-02 2.086973E-02 7.000000E-02
177 traj.phases.phase0.indep_states.states:v_86 c -7.000000E-02 2.099436E-02 7.000000E-02
178 traj.phases.phase0.indep_states.states:v_87 c -7.000000E-02 2.144098E-02 7.000000E-02
179 traj.phases.phase0.indep_states.states:v_88 c -7.000000E-02 2.244082E-02 7.000000E-02
180 traj.phases.phase0.indep_states.states:v_89 c -7.000000E-02 2.285376E-02 7.000000E-02
181 traj.phases.phase0.control_group.indep_controls.controls:u_0 c -1.000000E+00 1.000000E+00 1.000000E+00 u
182 traj.phases.phase0.control_group.indep_controls.controls:u_1 c -1.000000E+00 1.000000E+00 1.000000E+00 u
183 traj.phases.phase0.control_group.indep_controls.controls:u_2 c -1.000000E+00 1.000000E+00 1.000000E+00 u
184 traj.phases.phase0.control_group.indep_controls.controls:u_3 c -1.000000E+00 1.000000E+00 1.000000E+00 u
185 traj.phases.phase0.control_group.indep_controls.controls:u_4 c -1.000000E+00 1.000000E+00 1.000000E+00 u
186 traj.phases.phase0.control_group.indep_controls.controls:u_5 c -1.000000E+00 1.000000E+00 1.000000E+00 u
187 traj.phases.phase0.control_group.indep_controls.controls:u_6 c -1.000000E+00 1.000000E+00 1.000000E+00 u
188 traj.phases.phase0.control_group.indep_controls.controls:u_7 c -1.000000E+00 1.000000E+00 1.000000E+00 u
189 traj.phases.phase0.control_group.indep_controls.controls:u_8 c -1.000000E+00 1.000000E+00 1.000000E+00 u
190 traj.phases.phase0.control_group.indep_controls.controls:u_9 c -1.000000E+00 9.999999E-01 1.000000E+00 u
191 traj.phases.phase0.control_group.indep_controls.controls:u_10 c -1.000000E+00 1.000000E+00 1.000000E+00 u
192 traj.phases.phase0.control_group.indep_controls.controls:u_11 c -1.000000E+00 -2.836320E-01 1.000000E+00
193 traj.phases.phase0.control_group.indep_controls.controls:u_12 c -1.000000E+00 -9.999990E-01 1.000000E+00 l
194 traj.phases.phase0.control_group.indep_controls.controls:u_13 c -1.000000E+00 -9.999999E-01 1.000000E+00 l
195 traj.phases.phase0.control_group.indep_controls.controls:u_14 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
196 traj.phases.phase0.control_group.indep_controls.controls:u_15 c -1.000000E+00 -9.999997E-01 1.000000E+00 l
197 traj.phases.phase0.control_group.indep_controls.controls:u_16 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
198 traj.phases.phase0.control_group.indep_controls.controls:u_17 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
199 traj.phases.phase0.control_group.indep_controls.controls:u_18 c -1.000000E+00 -9.999999E-01 1.000000E+00 l
200 traj.phases.phase0.control_group.indep_controls.controls:u_19 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
201 traj.phases.phase0.control_group.indep_controls.controls:u_20 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
202 traj.phases.phase0.control_group.indep_controls.controls:u_21 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
203 traj.phases.phase0.control_group.indep_controls.controls:u_22 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
204 traj.phases.phase0.control_group.indep_controls.controls:u_23 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
205 traj.phases.phase0.control_group.indep_controls.controls:u_24 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
206 traj.phases.phase0.control_group.indep_controls.controls:u_25 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
207 traj.phases.phase0.control_group.indep_controls.controls:u_26 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
208 traj.phases.phase0.control_group.indep_controls.controls:u_27 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
209 traj.phases.phase0.control_group.indep_controls.controls:u_28 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
210 traj.phases.phase0.control_group.indep_controls.controls:u_29 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
211 traj.phases.phase0.control_group.indep_controls.controls:u_30 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
212 traj.phases.phase0.control_group.indep_controls.controls:u_31 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
213 traj.phases.phase0.control_group.indep_controls.controls:u_32 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
214 traj.phases.phase0.control_group.indep_controls.controls:u_33 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
215 traj.phases.phase0.control_group.indep_controls.controls:u_34 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
216 traj.phases.phase0.control_group.indep_controls.controls:u_35 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
217 traj.phases.phase0.control_group.indep_controls.controls:u_36 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
218 traj.phases.phase0.control_group.indep_controls.controls:u_37 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
219 traj.phases.phase0.control_group.indep_controls.controls:u_38 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
220 traj.phases.phase0.control_group.indep_controls.controls:u_39 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
221 traj.phases.phase0.control_group.indep_controls.controls:u_40 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
222 traj.phases.phase0.control_group.indep_controls.controls:u_41 c -1.000000E+00 -9.999999E-01 1.000000E+00 l
223 traj.phases.phase0.control_group.indep_controls.controls:u_42 c -1.000000E+00 -9.999999E-01 1.000000E+00 l
224 traj.phases.phase0.control_group.indep_controls.controls:u_43 c -1.000000E+00 -1.000000E+00 1.000000E+00 l
225 traj.phases.phase0.control_group.indep_controls.controls:u_44 c -1.000000E+00 2.836322E-01 1.000000E+00
226 traj.phases.phase0.control_group.indep_controls.controls:u_45 c -1.000000E+00 9.999994E-01 1.000000E+00 u
227 traj.phases.phase0.control_group.indep_controls.controls:u_46 c -1.000000E+00 9.999999E-01 1.000000E+00 u
228 traj.phases.phase0.control_group.indep_controls.controls:u_47 c -1.000000E+00 1.000000E+00 1.000000E+00 u
229 traj.phases.phase0.control_group.indep_controls.controls:u_48 c -1.000000E+00 9.999999E-01 1.000000E+00 u
230 traj.phases.phase0.control_group.indep_controls.controls:u_49 c -1.000000E+00 1.000000E+00 1.000000E+00 u
231 traj.phases.phase0.control_group.indep_controls.controls:u_50 c -1.000000E+00 1.000000E+00 1.000000E+00 u
232 traj.phases.phase0.control_group.indep_controls.controls:u_51 c -1.000000E+00 9.999999E-01 1.000000E+00 u
233 traj.phases.phase0.control_group.indep_controls.controls:u_52 c -1.000000E+00 1.000000E+00 1.000000E+00 u
234 traj.phases.phase0.control_group.indep_controls.controls:u_53 c -1.000000E+00 1.000000E+00 1.000000E+00 u
235 traj.phases.phase0.control_group.indep_controls.controls:u_54 c -1.000000E+00 9.999999E-01 1.000000E+00 u
236 traj.phases.phase0.control_group.indep_controls.controls:u_55 c -1.000000E+00 1.000000E+00 1.000000E+00 u
237 traj.phases.phase0.control_group.indep_controls.controls:u_56 c -1.000000E+00 1.000000E+00 1.000000E+00 u
238 traj.phases.phase0.control_group.indep_controls.controls:u_57 c -1.000000E+00 1.000000E+00 1.000000E+00 u
239 traj.phases.phase0.control_group.indep_controls.controls:u_58 c -1.000000E+00 1.000000E+00 1.000000E+00 u
240 traj.phases.phase0.control_group.indep_controls.controls:u_59 c -1.000000E+00 1.000000E+00 1.000000E+00 u
241 traj.phases.phase0.control_group.indep_controls.controls:u_60 c -1.000000E+00 1.000000E+00 1.000000E+00 u
242 traj.phases.phase0.control_group.indep_controls.controls:u_61 c -1.000000E+00 1.000000E+00 1.000000E+00 u
243 traj.phases.phase0.control_group.indep_controls.controls:u_62 c -1.000000E+00 1.000000E+00 1.000000E+00 u
244 traj.phases.phase0.control_group.indep_controls.controls:u_63 c -1.000000E+00 1.000000E+00 1.000000E+00 u
245 traj.phases.phase0.control_group.indep_controls.controls:u_64 c -1.000000E+00 1.000000E+00 1.000000E+00 u
246 traj.phases.phase0.control_group.indep_controls.controls:u_65 c -1.000000E+00 1.000000E+00 1.000000E+00 u
247 traj.phases.phase0.control_group.indep_controls.controls:u_66 c -1.000000E+00 1.000000E+00 1.000000E+00 u
248 traj.phases.phase0.control_group.indep_controls.controls:u_67 c -1.000000E+00 1.000000E+00 1.000000E+00 u
249 traj.phases.phase0.control_group.indep_controls.controls:u_68 c -1.000000E+00 1.000000E+00 1.000000E+00 u
250 traj.phases.phase0.control_group.indep_controls.controls:u_69 c -1.000000E+00 1.000000E+00 1.000000E+00 u
251 traj.phases.phase0.control_group.indep_controls.controls:u_70 c -1.000000E+00 1.000000E+00 1.000000E+00 u
252 traj.phases.phase0.control_group.indep_controls.controls:u_71 c -1.000000E+00 1.000000E+00 1.000000E+00 u
253 traj.phases.phase0.control_group.indep_controls.controls:u_72 c -1.000000E+00 1.000000E+00 1.000000E+00 u
254 traj.phases.phase0.control_group.indep_controls.controls:u_73 c -1.000000E+00 1.000000E+00 1.000000E+00 u
255 traj.phases.phase0.control_group.indep_controls.controls:u_74 c -1.000000E+00 1.000000E+00 1.000000E+00 u
256 traj.phases.phase0.control_group.indep_controls.controls:u_75 c -1.000000E+00 1.000000E+00 1.000000E+00 u
257 traj.phases.phase0.control_group.indep_controls.controls:u_76 c -1.000000E+00 1.000000E+00 1.000000E+00 u
258 traj.phases.phase0.control_group.indep_controls.controls:u_77 c -1.000000E+00 1.000000E+00 1.000000E+00 u
259 traj.phases.phase0.control_group.indep_controls.controls:u_78 c -1.000000E+00 9.999999E-01 1.000000E+00 u
260 traj.phases.phase0.control_group.indep_controls.controls:u_79 c -1.000000E+00 1.000000E+00 1.000000E+00 u
261 traj.phases.phase0.control_group.indep_controls.controls:u_80 c -1.000000E+00 1.000000E+00 1.000000E+00 u
262 traj.phases.phase0.control_group.indep_controls.controls:u_81 c -1.000000E+00 9.999999E-01 1.000000E+00 u
263 traj.phases.phase0.control_group.indep_controls.controls:u_82 c -1.000000E+00 1.000000E+00 1.000000E+00 u
264 traj.phases.phase0.control_group.indep_controls.controls:u_83 c -1.000000E+00 9.999999E-01 1.000000E+00 u
265 traj.phases.phase0.control_group.indep_controls.controls:u_84 c -1.000000E+00 9.999999E-01 1.000000E+00 u
266 traj.phases.phase0.control_group.indep_controls.controls:u_85 c -1.000000E+00 1.000000E+00 1.000000E+00 u
267 traj.phases.phase0.control_group.indep_controls.controls:u_86 c -1.000000E+00 9.999998E-01 1.000000E+00 u
268 traj.phases.phase0.control_group.indep_controls.controls:u_87 c -1.000000E+00 9.999997E-01 1.000000E+00 u
269 traj.phases.phase0.control_group.indep_controls.controls:u_88 c -1.000000E+00 1.000000E+00 1.000000E+00 u
270 traj.phases.phase0.control_group.indep_controls.controls:u_89 c -1.000000E+00 9.999988E-01 1.000000E+00
Constraints (i - inequality, e - equality)
Index Name Type Lower Value Upper Status Lagrange Multiplier (N/A)
0 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 0.000000E+00 0.000000E+00 9.00000E+100
1 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.103053E-16 0.000000E+00 9.00000E+100
2 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 5.938857E-18 0.000000E+00 9.00000E+100
3 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -4.973792E-17 0.000000E+00 9.00000E+100
4 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.622703E-16 0.000000E+00 9.00000E+100
5 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.954581E-16 0.000000E+00 9.00000E+100
6 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 8.744966E-16 0.000000E+00 9.00000E+100
7 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 1.147684E-15 0.000000E+00 9.00000E+100
8 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.071176E-15 0.000000E+00 9.00000E+100
9 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.065935E-15 0.000000E+00 9.00000E+100
10 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 5.297460E-15 0.000000E+00 9.00000E+100
11 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 5.134142E-15 0.000000E+00 9.00000E+100
12 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.533620E-15 0.000000E+00 9.00000E+100
13 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 1.521832E-15 0.000000E+00 9.00000E+100
14 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.731874E-15 0.000000E+00 9.00000E+100
15 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.341765E-15 0.000000E+00 9.00000E+100
16 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.001951E-15 0.000000E+00 9.00000E+100
17 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.152048E-15 0.000000E+00 9.00000E+100
18 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.039210E-15 0.000000E+00 9.00000E+100
19 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.999123E-15 0.000000E+00 9.00000E+100
20 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.480170E-15 0.000000E+00 9.00000E+100
21 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.477201E-15 0.000000E+00 9.00000E+100
22 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.456415E-15 0.000000E+00 9.00000E+100
23 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.500956E-15 0.000000E+00 9.00000E+100
24 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.266371E-15 0.000000E+00 9.00000E+100
25 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.678424E-15 0.000000E+00 9.00000E+100
26 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.298338E-15 0.000000E+00 9.00000E+100
27 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.149866E-15 0.000000E+00 9.00000E+100
28 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 9.086451E-16 0.000000E+00 9.00000E+100
29 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -3.325760E-16 0.000000E+00 9.00000E+100
30 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -8.136234E-16 0.000000E+00 9.00000E+100
31 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -2.310215E-15 0.000000E+00 9.00000E+100
32 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -3.670213E-15 0.000000E+00 9.00000E+100
33 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -5.267766E-15 0.000000E+00 9.00000E+100
34 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -6.134839E-15 0.000000E+00 9.00000E+100
35 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -8.130295E-15 0.000000E+00 9.00000E+100
36 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -9.745664E-15 0.000000E+00 9.00000E+100
37 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -1.100470E-14 0.000000E+00 9.00000E+100
38 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -1.256068E-14 0.000000E+00 9.00000E+100
39 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -1.391474E-14 0.000000E+00 9.00000E+100
40 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -1.591614E-14 0.000000E+00 9.00000E+100
41 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -1.842530E-14 0.000000E+00 9.00000E+100
42 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -2.029901E-14 0.000000E+00 9.00000E+100
43 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -2.491944E-14 0.000000E+00 9.00000E+100
44 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -6.102175E-15 0.000000E+00 9.00000E+100
45 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 1.487981E-14 0.000000E+00 9.00000E+100
46 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 4.055571E-14 0.000000E+00 9.00000E+100
47 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.939586E-14 0.000000E+00 9.00000E+100
48 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.250336E-14 0.000000E+00 9.00000E+100
49 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.848676E-14 0.000000E+00 9.00000E+100
50 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.232817E-14 0.000000E+00 9.00000E+100
51 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.202231E-14 0.000000E+00 9.00000E+100
52 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 3.256869E-14 0.000000E+00 9.00000E+100
53 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.739595E-14 0.000000E+00 9.00000E+100
54 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.615472E-14 0.000000E+00 9.00000E+100
55 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.365447E-14 0.000000E+00 9.00000E+100
56 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 1.824417E-14 0.000000E+00 9.00000E+100
57 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 1.742461E-14 0.000000E+00 9.00000E+100
58 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 1.327928E-14 0.000000E+00 9.00000E+100
59 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 7.768024E-15 0.000000E+00 9.00000E+100
60 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 6.485231E-15 0.000000E+00 9.00000E+100
61 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 2.874407E-15 0.000000E+00 9.00000E+100
62 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -1.769779E-15 0.000000E+00 9.00000E+100
63 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -3.111961E-15 0.000000E+00 9.00000E+100
64 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -5.143050E-15 0.000000E+00 9.00000E+100
65 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -7.744269E-15 0.000000E+00 9.00000E+100
66 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -8.136234E-15 0.000000E+00 9.00000E+100
67 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -8.884529E-15 0.000000E+00 9.00000E+100
68 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -9.276494E-15 0.000000E+00 9.00000E+100
69 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -9.145839E-15 0.000000E+00 9.00000E+100
70 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -8.551954E-15 0.000000E+00 9.00000E+100
71 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -7.566103E-15 0.000000E+00 9.00000E+100
72 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -7.055362E-15 0.000000E+00 9.00000E+100
73 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -5.867590E-15 0.000000E+00 9.00000E+100
74 traj.phases.phase0.collocation_constraint.defects:x e 0.000000E+00 -4.525409E-15 0.000000E+00 9.00000E+100
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242 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
243 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
244 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
245 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
246 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
247 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
248 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
249 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
250 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
251 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
252 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
253 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
254 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
255 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
256 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
257 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
258 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
259 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
260 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
261 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
262 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
263 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
264 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
265 traj.phases.phase0->path_constraint->u i -1.000000E+00 -9.999999E-01 1.000000E+00 l 9.00000E+100
266 traj.phases.phase0->path_constraint->u i -1.000000E+00 -9.999999E-01 1.000000E+00 l 9.00000E+100
267 traj.phases.phase0->path_constraint->u i -1.000000E+00 -9.999999E-01 1.000000E+00 l 9.00000E+100
268 traj.phases.phase0->path_constraint->u i -1.000000E+00 -1.000000E+00 1.000000E+00 l 9.00000E+100
269 traj.phases.phase0->path_constraint->u i -1.000000E+00 2.836322E-01 1.000000E+00 9.00000E+100
270 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999994E-01 1.000000E+00 u 9.00000E+100
271 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999994E-01 1.000000E+00 u 9.00000E+100
272 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
273 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
274 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
275 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
276 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
277 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
278 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
279 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
280 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
281 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
282 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
283 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
284 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
285 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
286 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
287 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
288 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
289 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
290 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
291 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
292 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
293 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
294 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
295 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
296 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
297 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
298 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
299 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
300 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
301 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
302 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
303 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
304 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
305 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
306 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
307 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
308 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
309 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
310 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
311 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
312 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
313 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
314 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
315 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
316 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
317 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
318 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
319 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
320 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
321 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
322 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
323 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999999E-01 1.000000E+00 u 9.00000E+100
324 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
325 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999998E-01 1.000000E+00 u 9.00000E+100
326 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999997E-01 1.000000E+00 u 9.00000E+100
327 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999997E-01 1.000000E+00 u 9.00000E+100
328 traj.phases.phase0->path_constraint->u i -1.000000E+00 1.000000E+00 1.000000E+00 u 9.00000E+100
329 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999988E-01 1.000000E+00 9.00000E+100
330 traj.phases.phase0->path_constraint->u i -1.000000E+00 9.999980E-01 1.000000E+00 9.00000E+100
Exit Status
Inform Description
0 Solve Succeeded
--------------------------------------------------------------------------------
==================================================
Grid Refinement - Iteration 1
--------------------------------------------------
Phase: traj.phases.phase0
Refinement Options:
Allow Refinement = True
Tolerance = 0.0001
Min Order = 3
Max Order = 14
Original Grid:
Number of Segments = 30
Segment Ends = [-1.0, -0.9333, -0.8667, -0.8, -0.7333, -0.6667, -0.6, -0.5333, -0.4667, -0.4, -0.3333, -0.2667, -0.2, -0.1333, -0.0667, 0.0, 0.0667, 0.1333, 0.2, 0.2667, 0.3333, 0.4, 0.4667, 0.5333, 0.6, 0.6667, 0.7333, 0.8, 0.8667, 0.9333, 1.0]
Segment Order = [3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3]
Error = [5.33e-07, 4.891e-07, 3.974e-07, 4.766e-05, 1.151e-06, 1.168e-06, 1.12e-06, 9.752e-07, 6.726e-07, 2.295e-07, 2.314e-07, 5.567e-07, 6.702e-07, 6.065e-07, 3.587e-05, 6.762e-07, 8.227e-07, 1.404e-06, 2.037e-06, 1.943e-06, 1.052e-06, 1.925e-06, 3.68e-06, 3.871e-06, 2.654e-06, 1.3e-06, 5.763e-07, 3.264e-07, 3.517e-07, 4.998e-07]
Successfully completed grid refinement.
==================================================
Simulating trajectory traj
Model viewer data has already been recorded for Driver.
Done simulating trajectory traj
False
Plotting the solution#
The recommended practice is to obtain values from the recorded cases. While the problem object can also be queried for values, building plotting scripts that use the case recorder files as the data source means that the problem doesn’t need to be solved just to change a plot. Here we load values of various variables from the solution and simulation for use in the animation to follow.
sol = om.CaseReader('dymos_solution.db').get_case('final')
t = sol.get_val('traj.phase0.timeseries.time')
x = sol.get_val('traj.phase0.timeseries.x')
v = sol.get_val('traj.phase0.timeseries.v')
u = sol.get_val('traj.phase0.timeseries.u')
h = np.sin(3 * x) / 3
t_sim = sim.get_val('traj.phase0.timeseries.time')
x_sim = sim.get_val('traj.phase0.timeseries.x')
v_sim = sim.get_val('traj.phase0.timeseries.v')
u_sim = sim.get_val('traj.phase0.timeseries.u')
h_sim = np.sin(3 * x_sim) / 3
Animating the Solution#
The collapsed code cell below contains the code used to produce an animation of the mountain car solution using Matplotlib.
The green area represents the hilly terrain the car is traversing. The black circle is the center of the car, and the orange arrow is the applied control.
The applied control generally has the same sign as the velocity and is ‘bang-bang’, that is, it wants to be at its maximum possible magnitude. Interestingly, the sign of the control flips shortly before the sign of the velocity changes.
Hide code cell source
fig = plt.figure(constrained_layout=True, figsize=(12, 6))
anim_ax.set_aspect('equal')
x_ax.set_ylabel('x')
v_ax.set_ylabel('v')
u_ax.set_ylabel('u')
u_ax.set_xlabel('t')
# set up the subplots as needed
anim_ax.set_xlim((-1.75, 0.75));
anim_ax.set_ylim((-1.25, 1.25));
anim_ax.set_xlabel('x');
anim_ax.set_ylabel('h');
x_sol_line, = x_ax.plot(t, x, 'o', ms=1, label='solution')
v_ax.plot(t, v, 'o', ms=1)
u_ax.plot(t, u, 'o', ms=1)
x_sim_line, = x_ax.plot([], [], '-', linewidth=3, label='simulation')
v_sim_line, = v_ax.plot([], [], '-', linewidth=3)
u_sim_line, = u_ax.plot([], [], '-', linewidth=3)
plt.figlegend(ncol=2, handles=[x_sol_line, x_sim_line], loc='upper center',
bbox_to_anchor=(0.78,0.98))
x_ax.grid(alpha=0.2)
txt_x = x_ax.text(0.8, 0.1, f'x = {x_sim[0, 0]:6.3f}', horizontalalignment='left',
verticalalignment='center', transform=x_ax.transAxes)
v_ax.grid(alpha=0.2)
txt_v = v_ax.text(0.8, 0.1, f'v = {v_sim[0, 0]:6.3f}', horizontalalignment='left',
verticalalignment='center', transform=v_ax.transAxes)
u_ax.grid(alpha=0.2)
txt_u = u_ax.text(0.8, 0.1, f'u = {u_sim[0, 0]:6.3f}', horizontalalignment='left',
verticalalignment='center', transform=u_ax.transAxes)
x_terrain = np.linspace(-1.75, 0.75, 100);
h_terrain = np.sin(3 * x_terrain) / 3;
terrain_line, = anim_ax.plot(x_terrain, h_terrain, '-', color='tab:gray', lw=2);
terrain = anim_ax.fill_between(x_terrain, h_terrain, -1.25*np.ones_like(x_terrain), color='tab:green');
car, = anim_ax.plot([], [], 'ko', ms=12);
u_vec = anim_ax.quiver(x_sim[0] + 0.005, h_sim[0] + 0.005, u_sim[0], [0], scale=10, angles='xy', color='tab:orange')
# See https://brushingupscience.com/2019/08/01/elaborate-matplotlib-animations/ for quiver animation
ANIM_DURATION = 5
PAUSE_DURATION = 2
ANIM_FPS = 20
num_points = t_sim.size
num_frames = ANIM_DURATION * ANIM_FPS
pause_frames = PAUSE_DURATION * ANIM_FPS
idx_from_frame_num = np.linspace(0, num_points-1, num_frames, dtype=int)
def drawframe(n):
if n >= idx_from_frame_num.size:
idx = num_points - 1
else:
idx = idx_from_frame_num[n]
x = x_sim[idx]
v = v_sim[idx]
u = u_sim[idx]
t = t_sim[idx]
h = np.sin(3 * x) / 3 + 0.025
car.set_data(x, h)
dh_dx = np.cos(3 * x)
u_vec.set_offsets(np.atleast_2d(np.asarray([x + 0.005, h + 0.005]).T))
u_vec.set_UVC(u * np.cos(dh_dx), u * np.sin(dh_dx))
x_sim_line.set_data(t_sim[:idx], x_sim[:idx])
v_sim_line.set_data(t_sim[:idx], v_sim[:idx])
u_sim_line.set_data(t_sim[:idx], u_sim[:idx])
txt_x.set_text(f'x = {x[0]:6.3f}')
txt_v.set_text(f'v = {v[0]:6.3f}')
txt_u.set_text(f'u = {u[0]:6.3f}')
return car, u_vec, x_sim_line, v_sim_line, u_sim_line
# # blit=True re-draws only the parts that have changed.
# # repeat_delay has no effect when using to_jshtml, so pad drawframe to show the final frame for PAUSE_FRAMES extra frames.
anim = animation.FuncAnimation(fig, drawframe, frames=num_frames + pause_frames, interval=1000/ANIM_FPS, blit=True);
plt.close() # Don't let jupyter display the un-animated plot
from IPython.display import HTML
js = anim.to_jshtml()
with open('mountain_car_anim.html', 'w') as f:
print(js, file=f)
HTML(filename='mountain_car_anim.html') | 34,674 | 97,330 | {"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": 5, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-40 | latest | en | 0.826324 |
http://www.rfcafe.com/references/electrical/butterworth-lowpass-filter-gain-phase-group-delay-equations.htm | 1,718,699,519,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00065.warc.gz | 57,678,425 | 9,664 | Please Support RF Cafe by purchasing my ridiculously low−priced products, all of which I created.
# Butterworth Lowpass Filter Gain, Phase, and Group Delay Equations
This may be the only place you will ever find the explicit formulas for Butterworth lowpass filter gain, phase, and group delay based on the basic Butterworth polynomials (until people copy my work and don't give credit*). Hoping to avoid having to do the messy math of complex numbers requiring separating out the real and imaginary parts to obtain phase (θ) - and then to get the first derivative for group delay (τd) - I searched high and low, far and wide for closed form equations, to no avail. All I could ever find was instruction to apply arg{H(ω)} for phase (θ) and -dθ/dω for group delay -- pretty useless if you want to plug an equation into a spreadsheet or software.
Finally, I decided if I wanted the solutions, I would have to slog through all the equations myself. Here is the result. The basic steps are as follow:
• Simplify the polynomial factors by multiplying out the parenthetical quantities.
• Substitute jω (or iω, but engineers use jω) for each "s," where ω is the frequency (ωn) being evaluated divided by the cutoff frequency (ωco); hence, jω = jωnco.
• Since j2 = -1 by definition, all even powers of j result in "real" parts and all odd powers of j result in "imaginary" parts. E.g., for 3rd-order:
• H(s) = (s + 1)(s2 + s + 1) = js3 + 2s2 +2s + 1
• H(jω) = (jω)3 + 2(jω)2 +2(jω) + 1 = j3ω3 + 2(j2ω2) +2(jω) + 1
= -jω3 + 2(-ω2) + j2ω + 1 = -jω3 - 2ω2 + j2ω + 1
• Re{H(jω)} = 1 - 2ω2 (parts w/o j), and Im{H(jω)} = 2ω - ω3 (parts w/j)
• Gain is G(ω) = sqrt [Re{H(jω)2 + Im{H(jω)2]
• The phase angle is by definition θ(ω) = tan-1 (Re{H(jω)}/Im{H(jω)})
θ(ω) = tan-1 [(1 - 2ω2) / (2ω - ω3)]
• Group delay is defined as τd(ω) = -dθ(ω)/dω. This is where it gets hard, mainly due to the arctangent, but the fraction is no walk in the park, either. Fortunately, there are online math websites like SymboLab that will handle such derivatives. For that matter, it will multiply all the factors for you as well, reducing the opportunity for errors.
• Finally, do the ω = ωnco substitution everywhere. ωn is the frequency at which the equation is being evaluated, and ωco is the cutoff frequency. Frequency units cancel out, so a 1 Hz cutoff plots the same as a 1 kHz cutoff or a 1 GHz cutoff for gain and phase. The group delay scale needs to be divided by a factor equal to the frequency units (÷103 for kHz, ÷106 for MHz, etc.).
This table represents hours of work on my part. I will eventually get around to lowpass, highpass, bandpass, and bandreject for Butterworth, Chebyshev, and Bessel (and maybe Gaussian) filters.
Butterworth Lowpass, Highpass, Bandpass, and Bandstop Filter Calculator with Gain, Phase and Group Delay are now part of my free RF Cafe Espresso Engineering Workbook™!
* I had Archive.org save a copy of this page in order to prove this is my original work.
N
1
GainH(ω) s + 1 (jω) + 1 sqrt (1 + ω2) Phase θ(ω) tan-1 (1/ω) Group Delay τd(ω) 1/(1 + ω2)
2
Gain H(ω) s2 + 1 + 1.4142s (jω)2 + 1.4142(jω) + 1 (1 - ω2) + j1.4142ω sqrt [(1 - ω2)2 + (1.4142ω)2] Phase θ(ω) tan-1 [(1 - ω2)/1.4142ω] Group Delay τd(ω) (1 + ω2)/[1.4142(0.5ω4 - 0.0001ω2) + 0.5]
3
Gain H(ω) s3 + 2s2 + 2s + 1(jω)3 + 2(jω)2 + 2(jω) + 1(1 - 2ω2) + j(2ω - ω3)sqrt [(1 - 2ω2)2 + (2ω - ω3)2] Phase θ(ω) tan-1 [(1 - 2*ω2)/(2ω - ω3)] Group Delay τd(ω) (2*ω4 + ω2 + 2)/(ω6 + 1)
4
Gain H(ω) s4 + 2.6131s3 + 3.4142s2 + 2.6131s + 1 (jω)4 + 2.6131(jω)3 + 3.4142(jω)2 + 2.6131(jω) + 1 (ω4 - 3.4142ω2 + 1) + j(-2.6131ω3 + 2.6131ω) sqrt [(ω4 - 3.4142ω2 + 1)2 + (-2.6131ω3 + 2.6131ω)2] Phase θ(ω) tan-1 [(ω4 - 3.4142ω2 + 1)/(-2.6131ω3 + 2.6131ω)] Group Delay τd(ω) (2.6131ω6 + 1.08234602ω4 + 1.08234602ω2 + 2.6131) / (ω8 - 0.00010839ω6 + 0.00017842ω4 - 0.00010839ω2 + 1)
5
Gain H(ω) s5 + 3.2361s4 + 5.2361s3 + 5.2361s2 + 3.2361s + 1 (jω)5 + 3.2361(jω)4 + 5.2361(jω)3 + 5.2361(jω)2 + 3.2361(jω) + 1 (3.2361ω4 - 5.2361ω2+1) + j(ω5 - 5.2361ω3 + 3.2361ω) sqrt [(3.2361ω4 - 5.2361ω2 + 1)2 + (ω5 - 5.2361ω3 + 3.2361ω)2] Phase θ(ω) tan-1 [(3.2361ω4 - 5.2361ω2 + 1)/(ω5 - 5.2361ω3 + 3.2361ω)] Group Delay τd(ω) (3.2361ω8 + 1.23624321ω6 + 0.99971358ω4 + 1.23624321ω2 + 3.2361) / (ω10 + 0.00014321ω8 - 0.00014321ω6 - 0.00014321ω4 + 0.00014321ω2 + 1)
6
Gain H(ω) s6 + 3.8637s5 + 7.4641s4 + 9.1416s3 + 7.4641s2 + 3.8637s + 1 (jω)6 + 3.8637(jω)5 + 7.4641(jω)4 + 9.1416(jω)3 + 7.4641(jω)2 + 3.8637(jω) + 1 (-ω6 + 7.4641ω4 - 7.4641ω2 + 1) + j(3.8637ω5 - 9.1416ω3 + 3.8637ω) sqrt [(-ω6 + 7.4641ω4 - 7.4641ω2 + 1)2 + (3.8637ω5 - 9.1416ω3 + 3.8637ω)2] Phase θ(ω) tan-1 [(-ω6 + 7.4641ω4 - 7.4641ω2 + 1)/(3.8637ω5 - 9.1416ω3 + 3.8637ω)] Group Delay τd(ω) (3.8637*ω10 + 1.41424317*ω8 + 1.03518705*ω6 + 1.03518705*ω4 + 1.41424317*ω2 + 3.8637) / (ω12 - 0.00002231*ω10 + 0.00018897*ω8 - 0.00037168*ω6 + 0.00018897*ω4 - 0.00002231*ω2 + 1)
7
Gain H(ω) s7 + 4.4940s6 + 10.0978s5 + 14.5920s4 + 14.5920s3 + 10.0978s2 + 4.4940s + 1 (jω)7 + 4.4940(jω)6 + 10.0978(jω)5 + 14.5920(jω)4 + 14.5920(jω)3 + 10.0978(jω)2 + 4.4940(jω) + 1 (-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1) + j(-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω) sqrt [(-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1)2 + (-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω)2] Phase θ(ω) tan-1 [(-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1)/(-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω)] Group Delay τd(ω) (4.494ω12 + 1.6035132ω10 + 1.1067536ω8 + 1.00994948ω6 + 1.1067536ω4 + 1.6035132ω2 + 4.494) / (ω14+0.000436ω12 - 0.00333116ω10 + 0.0032952ω8 + 0.0032952ω6-0.00333116ω4 + 0.000436ω2 + 1)
8
Gain H(ω) s8 + 5.1528s7 + 13.1371s6 + 21.8462s5 + 25.6884s4 + 21.8462s3 + 13.1371s2 + 5.1528s + 1 (jω)8 + 5.1258(jω)7 + 13.1371(jω)6 + 21.8462(jω)5 + 25.6884(jω)4 + 21.8462(jω)3 + 13.1371(jω)2 + 5.1258(jω) + 1 (ω8 - 13.1371ω6 + 25.6884ω4 - 13.1371ω2 + 1) + j(-5.1258ω7 + 21.8462ω5 - 21.8462ω3 + 5.1258ω) Phase θ(ω) tan-1 [(ω8 - 13.1371ω6 + 25.6884ω4 - 13.1371ω2 + 1)/(-5.1258ω7 + 21.8462ω5 - 21.8462ω3 + 5.1258ω)] Group Delay τd(ω) (5.1258ω14 + 1.79954718ω12 + 1.20591186ω10 + 1.01691792ω8 + 1.01691792ω6 - 1.20591186ω4 + 1.79954718ω2 + 5.1258) / (ω16 - 0.00037436ω14 + 0.00169249ω12 - 0.00140092ω10 + 0.00012722ω8 - 0.00140092ω6 + 0.00169249ω4 - 0.00037436ω2 + 1)
9
Gain H(ω) ω9 + 5.7588ω8 + 16.5817ω7 + 31.1634ω6 + 41.9864ω5 + 41.9864ω4 + 31.1634ω3 + 16.5817ω2 + 5.7588ω + 1 (jω)9 + 5.7588(jω)8 + 16.5817(jω)7 + 31.1634(jω)6 + 41.9864(jω)5 + 41.9864(jω)4 + 31.1634(jω)3 + 16.5817(jω)2 + 5.7588(jω) + 1 (5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1) + j(ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω) sqrt [(5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1)2 + (ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω)2] Phase θ(ω) tan-1 [(5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1)/(ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω)] Group Delay τd(ω) (5.7588ω16 + 2.00049396ω14 + 1.30030882ω12 + 1.06835072ω10 + 1.00271865ω8 + 1.06835072ω6 + 1.30030882ω4 + 2.00049396ω2 + 5.7588) / (ω18 + 0.00037744ω16 - 0.00200095ω14 + 0.00148244ω12 + 0.00034108ω10 + 0.00034108ω8 + 0.00148244ω6 - 0.00200095ω4 + 0.00037744ω2 + 1)
Posted December 1, 2023 | 3,526 | 7,115 | {"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-2024-26 | latest | en | 0.918515 |
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Contents (Click to skip to that section):
What is a Transformation?
In layman’s terms, you can think of a transformation as just moving an object or set of points from one location to another. You literally “transform” your data into something slightly different. For example, you can transform the sequence {4, 5, 6} by subtracting 1 from each term, so the set becomes {3, 4, 5}.
The many reasons why you might want to transform your data include: reducing skew, normalizing your data or simply making the data easier to understand. For example, the familiar Richter scale is actually a logarithmic transformation: an earthquake of magnitude 4 or 6 is easier to understand than a magnitude of 10,000 or 1,000,000.
More formally, transformations over a domain D are functions that map a set of elements of D (call them X) to another set of elements of D (call them Y). The relationships between the elements of the initial set are typically preserved by the transformation, but not necessarily preserved unchanged.
The image below shows a piece of coding that, with four transformations (mappings) converts a simple rectangular repeated pattern into a rhombic pattern.
Transformation Types
The word transformation is used most often in geometry. Types of transformations in geometry include translations (shifts, scales, and reflections) rotation, and shear mapping. But more generally, a transformation can mean any kind of mathematical function.
In general, you can transform a function in seven basic ways. For a function y = f(x):
• y = f(x – c): Horizontal shift c units right
• y = f(x + c): Horizontal shift c units left
• y = f(x) – c: Vertical shift c units down
• y = f(x) + c: Vertical shift c units up
• y = -f(x): Reflection over the x-axis
• y = f(-x): Reflection over the y-axis
• y = -f(-x): Reflection about the origin.
Specific ways to transform include:
• Taking the logarithm. Log Transformation is where you take the natural logarithm of variables in a data set.
• Square root transformations. Simply take the square root of your variables, i.e. x → x(1/2) = sqrt(x). This will have a moderate effect on the distribution and usually works for data with non-constant variance. However, it is considered to be weaker than logarithmic or cube root transforms.
• Taking the square: x → x2 can reduce left skewness. It has a medium effect on data.
• Reciprocal Transformations: the reciprocal transformation x → 1/x has a very large change on the shape of a distribution and reverses the order of values with the same sign; The negative reciprocal, x to -1/x, also has a drastic effect but preserves the variable order. This can be a particularly useful transformation for a set of positive variables where a reverse makes sense. For example, if you have number of patients per doctor, you can transform to doctors per patient (Cox, 2005).
• Box Muller Transform (Statistics): transforms data with a uniform distribution into a normal distribution.
• Differencing (Statistics): differenced data has one less point than the original data. For example, given a series Zt you can create a new series Yi = Zi – Zi – 1.
• Fitting a curve to the data: In modeling, after you have fit a curve, you can then model the residuals from that curve. Curve fitting algorithms include: gradient descent, Gauss-Newton and the Levenberg–Marquardt algorithm.
• Cube Root Transformations: x → x(1/3) is useful for reducing right skewness. Can be applied to positive and negative values.
Transformations in Geometry
1. Shift: moves every point by the same distance in the same direction.
2. Reflection: a folding or flipping over a certain line (e.g.the y-axis).
3. Glide reflections: a combination of a reflection and a shift.
4. Rotation: turns a figure about a fixed point (a center of rotation).
5. Scaling: Enlarges, or shrinks, an object by the same scale factor.
6. Shear mapping: all points along one line stay fixed, while other points are shifted parallel to the line by a distance proportional to their perpendicular distance from the line.
Using Transformations to Graph Functions
Sometimes we can use the concept of transformations to graph complicated functions when we know how to graph the simpler ones.
For example, if you know the graph of f(x), the graph of f(x) + c will be the same function, just shifted up by c units. f(x) – c will be the same thing, too, just shifted down by c units. The graph of f(x + c) s the graph of f(x), shifted left by c units, and the graph of f(x – c) is the graph of f(x) shifted right by c units. For a few step by step examples of vertical (up/down) shifts, see: Vertical Shift of a Function
As an example, take the graph of f(x) = (x-2)2 + 4. We might not know what that looks like, but we do know what h(x) = x2 looks like—a simple upward facing parabola. Imagine sketching that, then shift it to the right by 2 and up by 4. Then you have the sketch of f(x).
Vector Transformation
A vector transformation is a specific type of mapping where you associate vectors from one vector space with vectors in another space.
What is a vector space?
The concept of a vector space is fundamental to understanding vector transformations. A vector space is a collection of vectors which can be added and multiplied by scalars. Vectors have both magnitude and direction (e.g. 10 mph East).
A vector space has two requirements. Without leaving the vector space,
• Any two vectors can be added.
• Any two vectors can be scaled (multiplied).
Vector Spaces are often defined as Rn vector spaces, which are spaces of dimension n where adding or scaling any vector is possible. R stands for “Real” and these spaces include every vector of the same dimension as the space. For example, the R2 vector spaces includes all possible 2-D vectors. For example, the vectors (4, 2), (19, 0), and (121, 25) are all 2-D vectors (ones that can be represented on an x-y axis). The vector space R3 represents three dimensions, R,4 represents four dimensions and so on.
It’s practically impossible to deal with Rn vector spaces, because they contain every possible vector of n dimensions, up to infinity. Instead, we use subspaces, which are smaller vector spaces within a Rn vector space.
The Role of Functions in Vector Transformations
Vector transformations can be thought of as a type of function. For example, if you map the members of a vector space Rn to unique members of another vector space Rp, that’s a function. It’s written in function notation as:
f: Rn → Rp
Vector Transformation Example
Let’s say you had a vector transformation that mapped vectors in an R3 vector space to vectors in an R2 space. The general way to write the notation is:
f: R3 → R2
A specific example:
f(x1, x2, x3) = (x1 + 3x2, 4x3)
Note that f(x1,x2,x3) has three vectors and so belongs in ℝ3 and (x1 + 3x, 4x3) has two vectors and so belongs in ℝ2.
This example could also be written as:
Working out the vector transformation is equivalent to working out a function and involves some basic math. For example, let’s say you had the function f: x→ x2 and you wanted to transform (map) the number 2. You would insert it into the right hand part of the equation to get 22 = 4. Vector transformation works the same way.
For example, performing a vector transformation from f(2, 3, 4) to (X1 + 3x2, 4x3) we get:
• X1 = 2
• X2 = 3
• X3 = 4,
so:
(2 + 3(3), 4(4)) = (2 + 9, 16) = (11, 16)
Linear Transformation
Linear transformation is a special case of a vector transformation.
Definition:
Let V And W be two vector spaces. The function T:V→W is a linear transformation if the following two properties are true for all u, v, ε, V and scalars C:
1. Addition is preserved by T: T(u + v) = T(u) = T(v). In other words, if you add up two vectors u and v it’s the same as taking the transformation of each vector and then adding them.
2. Scalar multiplication is preserved by t: T(cu) = cT(u). In other words, if you multiply a vector u by a scalar C, this is the same as the transformation of u multiplied by scalar c.
How to Figure out if a Transformation is Linear
Applying rules 1 and 2 above will tell you if your transformation is a linear transformation.
Part One, Is Addition Preserved? Works through rule 1 and Part Two, Is Scalar Multiplication Preserved? works through rule 2. Remember: Both rules need to be true for linear transformations.
Example Question: Is the following transformation a linear transformation?
T(x,y)→ (x – y, x + y, 9x)
Part One: Is Addition Preserved?
Step 1: Give the vectors u and v (from rule 1) some components. I’m going to use a and b here, but the choice is arbitrary:
• u = (a1, a2)
• v = (b1, b2)
Step 2: Find an expression for the addition part of the left side of the Rule 1 equation (we’re going to do the transformation in the next step):
(u + v) = (a1, a2) + (b1, b2)
Adding these two vectors together, we get:
((a1 + b1), (a2 + b2))
In matrix form, the addition is:
Step 3: Apply the transformation. We’re given the rule T(x,y)→ (x – y, x + y, 9x), so transforming our additive vector from Step 2, we get:
• T ((a1 + b1), (a2+ b2)) =
• ((a1 + b1) – (a2 + b2),
• (a1 + b1) + (a2 + b2),
• 9(a1 + b1)).
Simplifying/Distributing using algebra:
(a1 + b1 – a2 – b2,
a1 + b1 + a2 + b2,
9a1 + 9b1).
Set this aside for a moment: we’re going to compare this result to the result from the right hand side of the equation in a later step.
Step 4: Find an expression for the right side of the Rule 1 equation, T(u) + T(v). Using the same a/b variables we used in Steps 1 to 3, we get:
T((a1,a2) + T(b1,b2))
Step 5: Transform the vector u, (a1,a2). We’re given the rule T(x,y)→ (x – y, x + y, 9x), so transforming vector u, we get:
• (a1 – a2,
• a1 + a2,
• 9a1)
Step 6: Transform the vector v. We’re given the rule T(x,y)→ (x – y, x + y,9x), so transforming vector v, (a1,a2), we get:
• (b1 – b2,
• b1 + b2,
• 9b1)
Step 7: Add the two vectors from Steps 5 and 6:
(a1 – a2, a1 + a2, 9a1) + (b1 – b2, b1 + b2, 9b1) =
((a1 – a2 + b1 – b2,
a1 + a2 + b1 – b2,
9a1 + 9b1)
Step 8: Compare Step 3 to Step 7. They are the same, so condition 1 (the additive condition) is satisfied.
Part Two: Is Scalar Multiplication Preserved?
In other words, in this part we want to know if T(cu)=cT(u) is true for T(x,y)→ (x-y,x+y,9x). We’re going to use the same vector from Part 1, which is u = (a1, a2).
Step 1: Work the left side of the equation, T(cu). First, multiply the vector by a scalar, c.
c * (a1, a2) = (c(a1), c(a2))
Step 2: Transform Step 1, using the rule T(x,y)→ (x-y,x+y,9x):
(ca1 – ca2,
ca1 + ca2,
9ca1)
Put this aside for a moment. We’ll be comparing it to the right side in a later step.
Step 3: Transform the vector u using the rule T(x,y)→ (x-y,x+y,9x). We’re working the right side of the rule 2 equation here:
(T(a1, a2)=
a1 – a2
a1 + a2
9a1)
Step 4: Multiply Step 3 by the scalar, c.
(c(a1 – a2)
c(a1 + a2)
c(9a1))
Distributing c using algebra, we get:
(ca1 – ca2,
ca1 + ca2,
9ca1)
Step 5: Compare Steps 2 and 4. they are the same, so the second rule is true. This function is a linear transformation.
Isometry (a Type of Linear Transformation)
An isometry is a linear transformation that preserves distance and length.
The image below shows a linear transformation f that sends A to B and X to Y, while preserving the distance between the points A and B (X and Y) and the length of the line AB (XY). If A and B were 5 cm away originally, the distance between f(A) = X and f(B) = Y, must also be 5 cm.
An isometry is also sometimes called a congruence transformation.
Since we call any property that is preserved under a transformation invariant under that transformation, we can say that length and distance are invariant under a congruence transformation. Rotation, shift, reflection, glides, and the identity map are all isometries.
If two figures are related by a congruence transformation (can be transformed into each other by means of an isometry), they are called congruent.
Mathematical Definition of an Isometry
If we have X and Y, two metric spaces with metrics dX and dY, then the map f:X → Y is an isometry if, for every and any a, b in X
dY(f(a)f(b)) = dX(a,b).
Important Properties of Isometries
In the Euclidean plane, any isometry that maps each of three non-collinear points (points that do not all lie on one line) to each other is the identity transformation (the transformation that sends every point to itself).
If an isometry in the plane has more than one fixed point, it is either a reflection (over an axis which crosses that point) or the identity transformation. The image below shows one such reflection; you can see that distances are preserved and the points in the reflection plane—for example, C—remain unchanged under the transformation.
Any isometry on the Euclidean plane can be uniquely determined by two sets of three non-collinear points; points that determine congruent triangles.
Log Transformation of a Skewed Distribution.
Log transformation means taking a data set and taking the natural logarithm of variables. Sometimes your data may not quite fit the model you are looking for, and a log transformation can help to fit a very skewed distribution into a more normal model. As a result, you can more easily see patterns in your data. Log transformation does not “normalize” your data; it’s purpose is to reduce skew.
In the image above, it’s practically impossible to see any pattern in the above image. However, in the second image, the data has had a log transformation. Consequently, the pattern becomes apparent.
Log Transformations and Statistical Tests.
If you are running a parametric statistical test on your data (for example, an ANOVA), using data that’s highly skewed to the right or left can lead to misleading test results. Therefore, if you want to perform a test on this kind of data, run a log transformation and then run the test on the transformed numbers.
When Should I Use Log Transformation?
Many possible transformations exist. However, you should only use a log transformation if:
• Your data is highly skewed to the right (i.e. in the positive direction).
• The residual’s standard deviation is proportional to your fitted values
• The data’s relationship is close to an exponential model.
• You think the residuals reflect multiplicative errors that have accumulated during each step of the computation.
Reciprocal Transformation
The reciprocal transformation is defined as the transformation of x to 1/x. The transformation has a dramatic effect on the shape of the distribution, reversing the order of values with the same sign. The transformation can only be used for non-zero values.
A negative reciprocal transformation is almost identical, except that x maps to -1/x and preserves the order of variables.
How to Graph Transformations
Once you’ve committed graphs of standard functions to memory, your ability to graph transformations is simplified.
Each has their own domain, range, and shape. When you transform one of these graphs, you shift it up, down, to the left, or to the right. Being able to visualize a transformation in your head and sketch it on paper is a valuable tool. Why? Sometimes the only way to solve a problem is to visualize the transformation in your head. While graphing calculators can be a valuable tool in developing your mathematical knowledge, eventually the calculator will only be able to help you so much.
Graph Transformations: Steps
Example Problem 1: Sketch the graph of x3 shifted two units to the right and then write the equation for that graph.
Step 1: Visualize the graph of x3, which is a cube (polynomial).
Step 2: Visualize the transformation. All you’re doing is shifting the graph two units to the right. Here’s what the transformed graph looks like:
Step 3: Write the equation. For any function, f(x), the graph of f(x + c) is the graph shifted to the left and the graph of f(x – c) is the graph shifted to the right. The question asks for two units (i.e. 2) to the right, so the final equation is f(x) = (x – 2)3.
Caution: the graph of x2 – 2 moves the graph down two units, not right!
Example problem 2: Sketch the graph of x2 + 2.
Step 1: Visualize the graph of x2.
Step 2: Sketch the graph. For any function, f(x), a graph f(x) + c is the graph shifted up the y-axis and a graph f(x) – c is a graph shifted down the y-axis. Therefore, x2 + 2 is the graph of x2 shifted two units up the y-axis.
That’s it!
Tip: You can also flip graphs on the x-axis by adding a negative coefficient. For example, while x2 is a parabola above the x-axis, -x2 is a mirror image over the x-axis.
Other Transformations
A Box Cox transformation is used when you need to meet the assumption of normality for a statistical test or procedure. It transform non-normal dependent variables into a bell shape.
Another way to normalize data is to use the Tukey ladder of powers (sometimes called the Bulging Rule), which can change the shape of a skewed distribution so that it becomes normal or nearly-normal. A third, related procedure, is a Fisher Z-Transformation. The Fisher Z transforms the sampling distribution of Pearson’s r (i.e. the correlation coefficient) so that it becomes normally distributed.
Generalized Procrustes analysis, which compares two shapes in Factor Analysis, uses geometric transformations (i.e. rescaling, reflection, rotation, or shift) of matrices to compare the sets of data. The following image shows a series of transformations onto a green target triangle.
References
Cox, N. (2005). Transformations: An Introduction. Retrieved February 25, 2018 from: http://fmwww.bc.edu/repec/bocode/t/transint.html
Coxeter, H. et. al (1967). Geometry Revisited. Washington, DC: Math. Assoc. Amer., p. 80.
Croft, H.et al. (1991). Unsolved Problems in Geometry. New York: Springer-Verlag, p. 3.
Gray, A. (1997). “Isometries and Conformal Maps of Surfaces.” §15.2 in Modern Differential Geometry of Curves and Surfaces with Mathematica, 2nd ed. Boca Raton, FL: CRC Press, pp. 346-351.
Revision Maths. Geometry and Measures: GCSE Maths. Retrieved from https://revisionmaths.com/gcse-maths-revision/shape-and-space/transformations on August 19, 2019
Peil,Timothy. Survey of Geometry.
Retrieved from http://web.mnstate.edu/peil/geometry/C3Transform/2isometry.htm on December 30, 2018.
Do, Norman. Symmetry in Geometry. Retrieved from http://users.monash.edu/~normd/documents/MATH-348-lecture-21.pdf on December 30, 2018.
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Stephanie Glen. "Transformations: Log, Reciprocal, Vector, Linear…" From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/transformations-2/
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https://math.libretexts.org/Courses/Mission_College/Math_3A%3A_Calculus_1_(Sklar)/04%3A_Applications_of_Derivatives/4.07%3A_Applied_Optimization_Problems/4.7E%3A_Exercises_for_Section_4.7 | 1,726,063,746,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00266.warc.gz | 355,609,150 | 33,239 | 4.7E: Exercises for Section 4.7
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For exercises 1 - 4, answer by proof, counterexample, or explanation.
1) When you find the maximum for an optimization problem, why do you need to check the sign of the derivative around the critical points?
The critical points can be the minima, maxima, or neither.
2) Why do you need to check the endpoints for optimization problems?
3) True or False. For every continuous nonlinear function, you can find the value $$x$$ that maximizes the function.
False; $$y=−x^2$$ has a minimum only
4) True or False. For every continuous non-constant function on a closed, finite domain, there exists at least one $$x$$ that minimizes or maximizes the function.
In exercises 5 - 8, set up and evaluate each optimization problem.
5) To carry a suitcase on an airplane, the $$\text{length}+\text{width}+\text{height}$$ of the box must be less than or equal to $$62$$ in. Assuming the height is fixed, show that the maximum volume is $$V=h\left(31−\frac{1}{2}h\right)^2.$$ What height allows you to have the largest volume?
$$h=\frac{62}{3}$$ in.
6) You are constructing a cardboard box with the dimensions $$2$$ m by $$4$$ m. You then cut equal-size squares from each corner so you may fold the edges. What are the dimensions of the box with the largest volume?
7) Find the positive integer that minimizes the sum of the number and its reciprocal.
$$1$$
8) Find two positive integers such that their sum is $$10$$, and minimize and maximize the sum of their squares.
In exercises 9 - 11, consider the construction of a pen to enclose an area.
9) You have $$400\,\text{ft}$$ of fencing to construct a rectangular pen for cattle. What are the dimensions of the pen that maximize the area?
$$100\,\text{ft}$$ by $$100\,\text{ft}$$
10) You have $$800\,\text{ft}$$ of fencing to make a pen for hogs. If you have a river on one side of your property, what is the dimension of the rectangular pen that maximizes the area?
11) You need to construct a fence around an area of $$1600\,\text{ft}^2$$. What are the dimensions of the rectangular pen to minimize the amount of material needed?
$$40\,\text{ft}$$ by $$40\,\text{ft}$$
12) Two poles are connected by a wire that is also connected to the ground. The first pole is $$20\,\text{ft}$$ tall and the second pole is $$10\,\text{ft}$$ tall. There is a distance of $$30\,\text{ft}$$ between the two poles. Where should the wire be anchored to the ground to minimize the amount of wire needed?
13) [T] You are moving into a new apartment and notice there is a corner where the hallway narrows from $$8\,\text{ft}$$ to $$6\,\text{ft}$$. What is the length of the longest item that can be carried horizontally around the corner?
$$19.73\,\text{ft}$$
14) A patient’s pulse measures $$70 \,\text{bpm},\, 80 \,\text{bpm}$$, then $$120 \,\text{bpm}$$. To determine an accurate measurement of pulse, the doctor wants to know what value minimizes the expression $$(x−70)^2+(x−80)^2+(x−120)^2$$? What value minimizes it?
15) In the previous problem, assume the patient was nervous during the third measurement, so we only weight that value half as much as the others. What is the value that minimizes $$(x−70)^2+ (x−80)^2+\frac{1}{2}(x−120)^2?$$
$$84 \,\text{bpm}$$
16) You can run at a speed of $$6$$ mph and swim at a speed of $$3$$ mph and are located on the shore, $$4$$ miles east of an island that is $$1$$ mile north of the shoreline. How far should you run west to minimize the time needed to reach the island?
For exercises 17 - 19, consider a lifeguard at a circular pool with diameter $$40$$ m. He must reach someone who is drowning on the exact opposite side of the pool, at position $$C$$. The lifeguard swims with a speed $$v$$ and runs around the pool at speed $$w=3v.$$
17) Find a function that measures the total amount of time it takes to reach the drowning person as a function of the swim angle, $$θ$$.
$$T(θ)=\dfrac{40θ}{3v}+\dfrac{40\cos θ}{v}$$
18) Find at what angle $$θ$$ the lifeguard should swim to reach the drowning person in the least amount of time.
19) A truck uses gas as $$g(v)=av+\dfrac{b}{v}$$, where $$v$$ represents the speed of the truck and $$g$$ represents the gallons of fuel per mile. At what speed is fuel consumption minimized?
$$v=\sqrt{\dfrac{b}{a}}$$
For exercises 20 - 21, consider a limousine that gets $$m(v)=\frac{120−2v}{5}\,\text{mi/gal}$$ at speed $$v$$, the chauffeur costs $$15/\text{h}$$, and gas is $$3.50/\text{gal}$$.
20) Find the cost per mile at speed $$v.$$
21) Find the cheapest driving speed.
approximately $$34.02$$ mph
For exercises 22 - 24, consider a pizzeria that sell pizzas for a revenue of $$R(x)=ax$$ and costs $$C(x)=b+cx+dx^2$$, where $$x$$ represents the number of pizzas.
22) Find the profit function for the number of pizzas. How many pizzas gives the largest profit per pizza?
23) Assume that $$R(x)=10x$$ and $$C(x)=2x+x^2$$. How many pizzas sold maximizes the profit?
Selling $$4$$ pizzas will maximize the profit.
24) Assume that $$R(x)=15x$$, and $$C(x)=60+3x+\frac{1}{2}x^2$$. How many pizzas sold maximizes the profit?
For exercises 25 - 26, consider a wire $$4$$ ft long cut into two pieces. One piece forms a circle with radius $$r$$ and the other forms a square of side $$x$$.
25) Choose $$x$$ to maximize the sum of their areas.
$$x = 0$$
26) Choose $$x$$ to minimize the sum of their areas.
For exercises 27 - 30, consider two nonnegative numbers $$x$$ and $$y$$ such that $$x+y=10$$. Maximize and minimize the quantities.
27) $$xy$$
Maximal: $$x=5,\,y=5;$$
Minimal: $$x=0,\,y=10$$ and $$y=0,\,x=10$$
28 $$x^2y^2$$
29) $$y−\dfrac{1}{x}$$
Maximal: $$x=1,\,y=9;$$
Minimal: none
30) $$x^2−y$$
In exercises 31 - 36, draw the given optimization problem and solve.
31) Find the volume of the largest right circular cylinder that fits in a sphere of radius $$1$$.
$$V = \frac{4π}{3\sqrt{3}}$$
32) Find the volume of the largest right cone that fits in a sphere of radius $$1$$.
33) Find the area of the largest rectangle that fits into the triangle with sides $$x=0,\,y=0$$ and $$\dfrac{x}{4}+\dfrac{y}{6}=1.$$
$$A = 6$$
34) Find the largest volume of a cylinder that fits into a cone that has base radius $$R$$ and height $$h$$.
35) Find the dimensions of the closed cylinder volume $$V=16π$$ that has the least amount of surface area.
$$r=2,\,h=4$$
36) Find the dimensions of a right cone with surface area $$S=4π$$ that has the largest volume.
For exercises 37 - 40, consider the points on the graphs of the given equations. Use a calculator to graph the functions.
37) [T] Where is the line $$y=5−2x$$ closest to the origin?
$$(2,1)$$
38) [T] Where is the line $$y=5−2x$$ closest to point $$(1,1)$$?
39) [T] Where is the parabola $$y=x^2$$ closest to point $$(2,0)$$?
$$(0.8351,0.6974)$$
40) [T] Where is the parabola $$y=x^2$$ closest to point $$(0,3)$$?
In exercises 41 - 45, set up, but do not evaluate, each optimization problem.
41) A window is composed of a semicircle placed on top of a rectangle. If you have $$20\,\text{ft}$$ of window-framing materials for the outer frame, what is the maximum size of the window you can create? Use r to represent the radius of the semicircle.
$$A=20r−2r^2−\frac{1}{2}πr^2$$
42) You have a garden row of $$20$$ watermelon plants that produce an average of $$30$$ watermelons apiece. For any additional watermelon plants planted, the output per watermelon plant drops by one watermelon. How many extra watermelon plants should you plant?
43) You are constructing a box for your cat to sleep in. The plush material for the square bottom of the box costs $$5/\text{ft}^2$$ and the material for the sides costs $$2/\text{ft}^2$$. You need a box with volume $$4\,\text{ft}^3$$. Find the dimensions of the box that minimize cost. Use $$x$$ to represent the length of the side of the box.
$$C(x)=5x^2+\dfrac{32}{x}$$
44) You are building five identical pens adjacent to each other with a total area of $$1000\,\text{m}^2$$, as shown in the following figure. What dimensions should you use to minimize the amount of fencing?
45) You are the manager of an apartment complex with $$50$$ units. When you set rent at $$800/\text{month}$$, all apartments are rented. As you increase rent by $$25/\text{month}$$, one fewer apartment is rented. Maintenance costs run $$50/\text{month}$$ for each occupied unit. What is the rent that maximizes the total amount of profit?
$$P(x)=(50−x)(800+25x−50)$$ | 4,188 | 12,797 | {"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} | 3.640625 | 4 | CC-MAIN-2024-38 | latest | en | 0.20271 |
https://www.jobilize.com/course/section/summary-table-properties-of-the-z-transform-by-openstax?qcr=www.quizover.com | 1,624,050,005,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00294.warc.gz | 752,693,110 | 22,258 | # 12.2 Properties of the z-transform
Page 1 / 1
This module includes a table of important properties of the z-transform.
## Introduction
This module will look at some of the basic properties of the Z-Transform (DTFT).
We will be discussing these properties for aperiodic, discrete-time signals but understand that very similarproperties hold for continuous-time signals and periodic signals as well.
## Linearity
The combined addition and scalar multiplication properties in the table above demonstrate the basic property oflinearity. What you should see is that if one takes the Z-transform of a linear combination of signals then itwill be the same as the linear combination of the Z-transforms of each of the individual signals. This is crucial when using a table of transforms to find the transform of a more complicated signal.
We will begin with the following signal:
$x(n)=a{f}_{1}(n)+b{f}_{2}(n)$
Now, after we take the Fourier transform, shown in the equation below, notice that the linear combination of theterms is unaffected by the transform.
$X(z)=a{F}_{1}(z)+b{F}_{2}(z)$
## Symmetry
Symmetry is a property that can make life quite easy when solving problems involving Z-transforms. Basicallywhat this property says is that since a rectangular function in time is a sinc function in frequency, then a sincfunction in time will be a rectangular function in frequency. This is a direct result of the symmetrybetween the forward Z and the inverse Z transform. The only difference is the scaling by $2\pi$ and a frequency reversal.
## Time scaling
This property deals with the effect on the frequency-domain representation of a signal if the time variable isaltered. The most important concept to understand for the time scaling property is that signals that are narrow intime will be broad in frequency and vice versa . The simplest example of this is a delta function, a unit pulse with a very small duration, in time that becomes an infinite-length constant function in frequency.
The table above shows this idea for the general transformation from the time-domain to the frequency-domainof a signal. You should be able to easily notice that these equations show the relationship mentioned previously: if thetime variable is increased then the frequency range will be decreased.
## Time shifting
Time shifting shows that a shift in time is equivalent to a linear phase shift in frequency. Since the frequencycontent depends only on the shape of a signal, which is unchanged in a time shift, then only the phase spectrum willbe altered. This property is proven below:
We will begin by letting $x(n)=f(n-\eta )$ . Now let's take the z-transform with the previous expression substituted in for $x(n)$ .
$X(z)=\sum_{n=()}$ f n η z n
Now let's make a simple change of variables, where $\sigma =n-\eta$ . Through the calculations below, you can see that only the variable in the exponential are altered thusonly changing the phase in the frequency domain.
$X(z)=\sum_{\eta =()}$ f σ z σ η z η σ f σ z σ z η F z
## Convolution
Convolution is one of the big reasons for converting signals to the frequency domain, since convolution in time becomesmultiplication in frequency. This property is also another excellent example of symmetry between time and frequency.It also shows that there may be little to gain by changing to the frequency domain when multiplication in time isinvolved.
We will introduce the convolution integral here, but if you have not seen this before or need to refresh your memory,then look at the discrete-time convolution module for a more in depth explanation and derivation.
$y(n)=({f}_{1}(n), {f}_{2}(n))=\sum_{\eta =()}$ f 1 η f 2 n η
## Time differentiation
Since discrete LTI systems can be represented in terms of difference equations, it is apparent with this property that convertingto the frequency domain may allow us to convert these complicated difference equations to simpler equationsinvolving multiplication and addition.
## Parseval's relation
$\sum_{n=()}$ x n x* n z F z F* z
Parseval's relation tells us that the energy of a signal is equal to the energy of its Fourier transform.
## Modulation (frequency shift)
Modulation is absolutely imperative to communications applications. Being able to shift a signal to a differentfrequency, allows us to take advantage of different parts of the electromagnetic spectrum is what allows us to transmittelevision, radio and other applications through the same space without significant interference.
The proof of the frequency shift property is very similar to that of the time shift ; however, here we would use the inverse Fourier transform in place of the Fourier transform. Since we wentthrough the steps in the previous, time-shift proof, below we will just show the initial and final step to this proof:
$z(t)=\frac{1}{2\pi }\int_{()} \,d \omega$ F ω φ ω t
Now we would simply reduce this equation through anotherchange of variables and simplify the terms. Then we will prove the property expressed in the table above:
$z(t)=f(t)e^{i\phi t}$
## Properties demonstration
An interactive example demonstration of the properties is included below:
## Summary table
Property Signal Z-Transform Region of Convergence Linearity $\alpha {x}_{1}\left(n\right)+\beta {x}_{2}\left(n\right)$ $\alpha {X}_{1}\left(z\right)+\beta {X}_{2}\left(z\right)$ At least ${\mathrm{ROC}}_{1}\cap {\mathrm{ROC}}_{2}$ Time shifing $x\left(n-k\right)$ ${z}^{-k}X\left(z\right)$ $\mathrm{ROC}$ Time scaling $x\left(n/k\right)$ $X\left({z}^{k}\right)$ ${\mathrm{ROC}}^{1/k}$ Z-domain scaling ${a}^{n}x\left(n\right)$ $X\left(z/a\right)$ $|a|\mathrm{ROC}$ Conjugation $\overline{x\left(n\right)}$ $\overline{X}\left(\overline{z}\right)$ $\mathrm{ROC}$ Convolution ${x}_{1}\left(n\right)*{x}_{2}\left(n\right)$ ${X}_{1}\left(z\right){X}_{2}\left(z\right)$ At least ${\mathrm{ROC}}_{1}\cap {\mathrm{ROC}}_{2}$ Differentiation in z-Domain $\left[nx\left[n\right]\right]$ $-\frac{d}{dz}X\left(z\right)$ ROC= all $\mathbb{R}$ Parseval's Theorem $\sum_{n=()}$∞ ∞ x n x* n $\int_{-\pi }^{\pi } F(z)\mathrm{F*}(z)\,d z$ ROC
#### Questions & Answers
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Almas
are nano particles real
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Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
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Maira Reply
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RAW Reply
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Rafiq
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Broad Topics > Numbers and the Number System > Factors and multiples
### Skip Counting
##### Stage: 1 Challenge Level:
Find the squares that Froggie skips onto to get to the pumpkin patch. She starts on 3 and finishes on 30, but she lands only on a square that has a number 3 more than the square she skips from.
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##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Six in a Circle
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If there is a ring of six chairs and thirty children must either sit on a chair or stand behind one, how many children will be behind each chair?
### Curious Number
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Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Multiplication Square Jigsaw
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Can you complete this jigsaw of the multiplication square?
### Multiples Grid
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What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### Down to Nothing
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A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Seven Flipped
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Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Nineteen Hexagons
##### Stage: 1 Challenge Level:
In this maze of hexagons, you start in the centre at 0. The next hexagon must be a multiple of 2 and the next a multiple of 5. What are the possible paths you could take?
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### The Moons of Vuvv
##### Stage: 2 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Biscuit Decorations
##### Stage: 1 and 2 Challenge Level:
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Constant Counting
##### Stage: 1 Challenge Level:
You can make a calculator count for you by any number you choose. You can count by ones to reach 24. You can count by twos to reach 24. What else can you count by to reach 24?
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Three Spinners
##### Stage: 2 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### Lots of Lollies
##### Stage: 1 Challenge Level:
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
### Multiplication Series: Number Arrays
##### Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### The Set of Numbers
##### Stage: 1 Challenge Level:
Can you place the numbers from 1 to 10 in the grid?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Money Measure
##### Stage: 2 Challenge Level:
How can you use just one weighing to find out which box contains the lighter ten coins out of the ten boxes?
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
##### Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Domino Pick
##### Stage: 1 Challenge Level:
Are these domino games fair? Can you explain why or why not?
### Red Balloons, Blue Balloons
##### Stage: 2 Challenge Level:
Katie and Will have some balloons. Will's balloon burst at exactly the same size as Katie's at the beginning of a puff. How many puffs had Will done before his balloon burst?
### Spelling Circle
##### Stage: 2 Challenge Level:
Find the words hidden inside each of the circles by counting around a certain number of spaces to find each letter in turn.
### Are You Well Balanced?
##### Stage: 1 Challenge Level:
Can you work out how to balance this equaliser? You can put more than one weight on a hook.
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Doubling Fives
##### Stage: 1 Challenge Level:
This activity focuses on doubling multiples of five.
### Sets of Four Numbers
##### Stage: 2 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### What's Left?
##### Stage: 1 Challenge Level:
Use this grid to shade the numbers in the way described. Which numbers do you have left? Do you know what they are called?
### Little Squares
##### Stage: 1 Challenge Level:
Look at the squares in this problem. What does the next square look like? I draw a square with 81 little squares inside it. How long and how wide is my square?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Give Me Four Clues
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### Path to the Stars
##### Stage: 2 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Numbers as Shapes
##### Stage: 1 Challenge Level:
Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares?
### Which Numbers? (2)
##### Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. Can you find all the numbers in each set from these clues?
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Which Numbers? (1)
##### Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. They are the red set, the green set and the blue set. Can you find all the numbers in the sets from these clues?
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Flashing Lights
##### Stage: 2 Challenge Level:
Norrie sees two lights flash at the same time, then one of them flashes every 4th second, and the other flashes every 5th second. How many times do they flash together during a whole minute?
### Sweets in a Box
##### Stage: 2 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Times Tables Shifts
##### Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### Being Resilient - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that may require resilience. | 2,220 | 9,550 | {"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-2017-43 | latest | en | 0.870229 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Active_Learning/Shorter_Activities/Data_Analysis_and_Statistics/01_Uncertainty | 1,638,372,910,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00406.warc.gz | 221,581,065 | 28,423 | Uncertainty
For the first half of this lesson:
Chapter 2 (sections 1, 3 (first part), 4, 5, 6, 9), Chapter 3 (sections 1, 2, 3, 4), Exploring Chemical Analysis 5th ed., D. Harris
Suppose you are analyzing soil samples from a school playground near a busy intersection. After collecting three samples, you return to the lab, extract the soil samples with water and, using an atomic absorption spectrometer (we will talk more about instrumentation later), determine the concentration of lead in the resulting extracts, obtaining an average result of 4.38 parts-per-billion (ppb).
• The actual value that you report is 4.38 ± 0.02 ppb. What does the 0.02 ppb indicate?
STOP
• Brainstorm what errors might affect this result?
• Separate the errors discussed in class into systematic vs. random
Systematic (Determinate)
Random (Indeterminate)
STOP
Instrument noise
• Often when discussing precision and accuracy we use a model of darts on a dartboard as shown below.
• As a group come to a conclusion of what accuracy and precision are based on previous knowledge and write your definition below.
• Accuracy:
• Precision:
• Label the top and side of the table with accurate, not accurate, precise, and not precise.
• Circle the dartboard depictions show random error and put a square around the dartboard depictions that show systematic error.
Figure 1. Dart model of precision and accuracy.
STOP
Measurement Exercise
• Experiment 1: Using a ruler with a millimeter scale, measure the big solid rectangle's width and length in millimeters on your own without any consultation of your group then collect the values of your group mates in the table and calculate the average values.
Group Members
Length (mm)
Width (mm)
Average
• Are your measurements of the rectangle's length and width exactly the same as those of your classmates, or are they different? If there are differences in the measurements, is this the result of determinate errors, indeterminate errors or both? What are those errors? Briefly explain your reasoning.
STOP
When measuring the rectangle's length and width you had to make several decisions:
• How do you define the rectangle? Should you measure its length and width from the outside edges of the border, from the inside edges or from the middle of the border?
• Experiment 2: Decide as a group how you are going to define the rectangle and record your measurement to the tenth place (which is an estimation of one digit past the smallest gradation). Then measure the length and width of the rectangle and calculate its average.
Group Members
Length (mm)
Width (mm)
Average
STOP
Significant Figures
You may recall from general chemistry that we refer to the digits in our measurements as significant figures. A significant figure is any number in which we can express confidence, including those digits known exactly and the one digit whose value is an estimate. The lengths 154 mm and 154.3 mm have three and four significant digits, respectively. The number of significant figures in a measurement is important because it affects the number of significant figures in a result based on that measurement.
• How many significant figures do you have in each of your individual measurements of length and width?
Group Members
# sig figs in length
# sig figs in width
Before continuing, let's review the rules for including significant figures in calculations. When adding or subtracting, the result of the calculation is rounded to the last decimal place that is significant for all measurements. For example, the sum of 135.621, 0.33 and 21.2163 is 157.17 since the last decimal place that is significant for all three numbers (as shown below by the vertical line)
$\begin{array}{r|l} 135.62\hspace{-4pt} & \hspace{-4pt}1 \\[-2pt] 0.33\hspace{-4pt} & \\[-2pt] 21.21\hspace{-4pt} & \hspace{-4pt}63 \\ \hline 157.16\hspace{-4pt} & \hspace{-4pt}73 \\ \end{array} \nonumber$
is the hundredth's place. Note that rounding the answer to the correct number of significant figures occurs after completing the exact calculation.
When multiplying or dividing, the result of the calculation contains the same number of significant figures as that measurement having the smallest number of significant figures. Thus,
$\dfrac{22.91\times0.152}{16.302}=0.21361\approx 0.214\nonumber$
because 0.152, with three, has the fewest number of significant figures.
One way to think about this is that we cannot make a measurement more precise through a calculation than it is when we take the measurement.
• Check back to be sure that your group has the correct number of significant figures in the average length and width listed for Experiment 2, if not fix them.
STOP
Uncertainty in the lab
• If you took a measurement and found a value of 89.231 ± 0.008 what is the absolute uncertainty and the percent relative uncertainty of the measurement?
STOP
Propagation of Uncertainty
As we have discussed, each instrument we use in a lab has an associated random error, which is often expressed as a tolerance factor by the manufacturer. A more rigorous approach to determining the uncertainty in a result is called a propagation of uncertainty and it combines the uncertainty in each measurement to estimate the uncertainty in the final result.
The following equations for propagation of uncertainty are from your textbook (as noted in parentheses):
$e_{prop}=\sqrt{e_1^2 + e_2^2+e_3^2+⋯e_n^2} \tag{3-5}$
Multiplication and Division:
$\%e_{prop}=\sqrt{(\%e_1^2)+(\% e_2^2)+(\%e_3^2)+⋯(\%e_n^2)} \tag{3-6}$
Where en is the absolute uncertainty (often the tolerance factor) for measurement n and %en is the percent relative uncertainty for measurement n.
The REAL SIG FIG RULE
The final uncertainty to one significant figure and that defines how many significant figures are in the measurement. For example if you took several measurements and the average value was 234.7182 a.u., but the uncertainty when propagated was 0.0278 a.u. you would write the final result as 234.72 ± 0.03 a.u.
• Assuming that the error associated with the ruler is 0.1 mm, propagate the error associated with the average of the length and the width of the rectangle using the data from Experiment 2.
1. What is the absolute uncertainty of each measurement?
2. Should you use the addition/subtraction or multiplication/division equation above to propagate the error?
3. Calculate the error and write it with the proper number of sig. figs is the table below.
Average length (mm)
Average width (mm)
Error
• Asking yourself the same questions as above, calculate the average area of the rectangle including the propagated error and the correct number of sig figs and write the value in the table.
Average area w/ error (mm2)
STOP
• As a group think of times in the lab you may want to use the addition and subtraction rule to propagate uncertainty and one time in lab you would want to use multiplication and division rule to propagate uncertainty.
STOP
Often propagation of error includes mixed operation. Like finding the area of the rectangle, except I walked you through how to do that. Now you are going to propagate error for a scientific system you have never seen before.
• For a concentration technique the relationship between the signal and the an analyte’s concentration is
$S_\textrm{total} = k_\textrm{A}C_\textrm{A} + S_\textrm{mb}\nonumber$
What is the analyte’s concentration, CA, and its uncertainty if Stotal is 24.37 ± 0.02, Smb is 0.96 ± 0.02, and kA is 0.186 ± 0.003 ppm–1.
STOP
• Given the following masses of deionized water measured during the calibration of a Class A pipet at 15 °C, what average volume of water is the 10.00 mL Class A (± 0.02 mL) pipet actually delivering?
Measurement
Mass of H2O (g)
1
10.1578
2
10.1341
3
10.1425
4
10.1453
5
10.1672
6
10.1587
7
10.1611
STOP
• Explain, with numerical values, glassware, and instruments how you would prepare 50 mL of a 1.40 M aqueous stock solution of sodium thiosulfate, Na2S2O3 (MW= 158.11 g/mol) (which is a solid that must be dissolved) while introducing the least amount of error. (similar example pg 23 of textbook)
STOP
• Explain, with numerical values, glassware, and instruments how you would prepare a 100 mL of a 0.140 M aqueous solution of sodium thiosulfate from the 1.40 M stock solution in the previous problem while introducing the least amount of error.
STOP
Homework for data analysis
• How is systematic error identified and accounted for in measurements?
• How is random error identified and accounted for in measurements?
• In a lab at 27°C you weigh an empty volumetric flask at 468.654 g you then fill the flask up to the calibration mark with DI H2O and weigh the flask at 543.635 g. What is the true volume contained in the volumetric?
• What is the true concentration of dissolved 0.1567 g of Na2S2O3 in molarity if you made it in the volumetric flask in the previous example?
• What portion of the analytical process do the tools we talked about today fall under?
• If you made a 5 M solution via dilution of a 10 M solution using volumetric glassware and you performed several forms of analytical analysis and determined that the true concentration of the solution was 4.5 M what should be done to avoid this problem in the future? | 2,217 | 9,307 | {"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} | 3.90625 | 4 | CC-MAIN-2021-49 | latest | en | 0.868635 |
https://fixes.co.za/r-stats/exploratory-data-analysis/ | 1,726,307,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00144.warc.gz | 235,274,742 | 40,818 | # Exploratory Data Analysis
## Exploratory Data Analysis#
### Principles of Analytic Graphs#
Graphs give us a visual form of data, and the first principle of analytic graphs is to show some comparison. You’ll hear more about this when you study statistical inference (another great course BTW), but evidence for a hypothesis is always relative to another competing or alternative hypothesis.
When presented with a claim that something is good, you should always ask “Compared to What?” This is why in commercials you often hear the phrase “other leading brands”. An implicit comparison, right?
So the first principle was to show a comparison. The second principle is to show causality or a mechanism of how your theory of the data works The third principle is to show multivariate data
Most hard problems have more variables
Restricting yourself to 2 variables you might be misled and draw an incorrect conclusion
A singel variable may give rise to Simpson’s paradox where a trend appears in certain groups but dissapears when groups are combined. It happens when frequency data is unduly given causal interpretations. Often in social science and medical science studies.
The fourth principle is integrating evidence. Don’t use a single form of expression.
Don’t let the tool drive the analysis
The fifth principle of graphing involves describing and documenting the evidence with sources and appropriate labels and scales.
Also, using R, you want to preserve any code you use to generate your data and graphics so that the research can be replicated if necessary. This allows for easy verification or finding bugs in your analysis.
The sixth principle is Content is king
Analytical presentations ultimately stand or fall depending on the quality, relevance, and integrity of their content.
1. Comparison
2. Causality
3. Multivariate Data
4. Integrate Evidence
5. Describe and document evidence with sources, labels and scales
6. Content is king
### Exploratory Graphs#
So graphics give us some visual form of data, and since our brains are very good at seeing patterns, graphs give us a compact way to present data and find or display any pattern that may be present.
Exploratory graphs serve mostly the same functions as graphs. They help us find patterns in data and understand its properties. They suggest modeling strategies and help to debug analyses. We DON’T use exploratory graphs to communicate results.
Getting asummary of data:
``````> summary(pollution\$pm25)
Min. 1st Qu. Median Mean 3rd Qu. Max.
3.383 8.549 10.047 9.836 11.356 18.441
``````
• Minimum (0 percentile)
• Maximum (100 percentile)
• 25%: 1st quartile
• 50%: median
• 75%: 3rd quartile
The median specifies that half of the measured fact have a value less than or equal to that median. Anohter half of the data set has greater or equal to that value.
#### Quantile#
``````> quantile(ppm)
0% 25% 50% 75% 100%
3.382626 8.548799 10.046697 11.356012 18.440731
``````
We use a boxplot to display this:
``````> boxplot(ppm, col="blue")
``````
In a boxplot the top and bottom of the box represent the values of the 25% and 75% quartiles. The midde horizontal line is the median. The whiskers (horizontal lines) outside the box represent the range defualte to 1.5 the interquartile range. The interquartile range is the difference between the 75% and 25% quartiles.
Data points outside the whiskers are outliers.
#### Striaght Lines#
You can overlay lines on the plot with:
`````` abline(h=12)
``````
Create a green histogram
``````> hist(ppm, col="green")
``````
Some more detailed greyscale representation of occurances
This one-dimensional plot, with its grayscale representation, gives you a little more detailed information about how many data points are in each bucket and where they lie within the bucket.
Use `breaks` to set the number of buckets to split the data into:
``````> hist(ppm, col="green", breaks=100)
``````
`rug` automatically adjusted its pocket size to that of the last plot plotted.
``````> abline(v=12, lwd=2)
``````
Change colour of `abline`:
``````> abline(v=median(ppm), col="magenta", lwd=4)
``````
Make table of factor data:
``````> table(pollution\$region)
east west
442 134
``````
#### Barplot#
``````barplot(reg, col = "wheat", main = "Number of Counties in Each Region")
``````
#### Boxplot#
``````> boxplot(formula=pm25~region, data=pollution, col="red")
``````
#### Multiple Histograms#
To plot multiple histograms we need to set the layout with `par`
Also margins is a 4-long vector which indicates the number of lines for the bottom, left, top and right
``````par(mfrow=c(2,1),mar=c(4,4,2,1))
``````
`mfrow=c(2,1)` means 2 rows and 1 column
##### Subset#
To split the data up:
``````> east <- subset(pollution, region == 'east')
> hist(east\$pm25, col='green')
``````
Doing the above in one command:
``````> hist(subset(pollution,region=="west")\$pm25, col = "green")
``````
Using the `with` keyword:
``````> with(pollution, plot(latitude, pm25))
``````
So you can avoid using: `pollution\$pm25`
Create a dashed horizontal line:
``````abline(h=12, lwd=2, lty=2)
``````
### Plot#
You can set the colours of points based on region
``````> plot(pollution\$latitude, ppm, col=pollution\$region)
``````
• Exploratory plots are quick and dirty
• Plots let you summarize the data (usually graphically) and highlight any broad features
## Graphics Devices in R#
There is a screen device (window on device) or file devices (PDF, JPEG, SVG)
How you access your screen device depends on what computer system you’re using. On a Mac the screen device is launched with the call `quartz()`, on Windows you use the call `windows()`, and on Unix/Linux `x11()`
View available devices with:
``````> ?Devices
``````
Create a box plot:
``````> with(faithful, plot(eruptions, waiting))
``````
Annotate the main title:
``````> title(main='Old Faithful Geyser data')
``````
View the current plotting device:
``````> dev.cur()
RStudioGD
2
``````
#### Creating a plot on a file device#
Launch the file device
``````> pdf(file="myplot.pdf")
``````
Then run the plot again:
``````> with(faithful, plot(eruptions, waiting))
> title(main='Old Faithful Geyser data')
``````
After done you have to close it with:
``````> dev.off()
``````
There are two basic types of file devices, vector and bitmap devices:
• `Vector` formats are good for line drawings and plots with solid colors using a modest number of points
• `Bitmap` formats are good for plots with a large number of points, natural scenes or web-based plots
##### Vector Formats#
• `pdf`: line-type graphics and papers. It resizes well, is usually portable, but it is not efficient if a plot has many objects/points.
• `svg`: XML-based, scalable vector graphics. This supports animation and interactivity and is potentially useful for web-based plots.
The last two vector formats are win.metafile, a Windows-only metafile format, and postscript (ps), an older format which also resizes well, is usually portable, and can be used to create encapsulated postscript files. Unfortunately, Windows systems often don’t have a postscript viewer.
##### Bitmap formats#
• `png`: Portable Network Graphics which is good for line drawings or images with solid colors. It uses lossless compression (like the old GIF format), and most web browsers can read this format natively. In addition, png is good for plots with many points, but it does not resize well.
• `jpeg`: files are good for photographs or natural scenes. They use lossy compression, so they’re good for plots with many points. Files in jpeg format don’t resize well, but they can be read by almost any computer and any web browser. They’re not great for line drawings.
• `tiff`: an older lossless compression meta-format and bmp which is a native Windows bitmapped format.
##### Devices#
Every open graphics device is assigned an integer greater than or equal to 2
You can change the active device with:
``````dev.set(<integer>)
``````
You can copy between devices:
The function `dev.copy` copies a plot from one device to another, and `dev.copy2pdf` specifically copies a plot to a PDF file.
``````> dev.copy(png, file='geyserplot.png')
``````
### Plotting Systems#
#### Base Plotting System#
• Comes with `R`
• Intuitive and Exploratory
• You can’t go backwards
• based on “Artist’s Pallette”
#### Lattice plots#
• single function call such as `xyplot` or `bwplot`
• most useful for conditioning types of plots which display how y changes with x across levels of z
• Cannot add to the plot
Using the R forumla: `Life.Exp ~ Income | region`: we’re plotting life expectancy as it depends on income for each region
``````> xyplot(Life.Exp ~ Income | region, state, layout=c(4,1))
``````
The second argument is data
and the third shows layout is 4 columns 1 row
#### GGPlot2#
The best of both worlds, does automatic titles and margins etc but also allows annotatins and adding.
``````> qplot(displ, hwy, data = mpg)
``````
• Uses Graphics Grammar
### Base Plotting System#
See the range for a column of data excluding `NA`:
``````range(airquality\$Ozone, na.rm=TRUE)
``````
##### Histogram#
Uses one variable
``````> hist(airquality\$Ozone)
``````
##### Boxplot#
``````> boxplot(Ozone~Month, airquality)
``````
``````> boxplot(Ozone~Month, airquality, xlab='Month', ylab='Ozone (ppb)', col.axis='blue', col.lab='red')
``````
Check number of parameters you can give to `par()`
``````> length(par())
[1] 72
``````
The available options are:
``````> names(par())
[8] "cex" "cex.axis" "cex.lab" "cex.main" "cex.sub" "cin" "col"
[15] "col.axis" "col.lab" "col.main" "col.sub" "cra" "crt" "csi"
[22] "cxy" "din" "err" "family" "fg" "fig" "fin"
[29] "font" "font.axis" "font.lab" "font.main" "font.sub" "lab" "las"
[36] "lend" "lheight" "ljoin" "lmitre" "lty" "lwd" "mai"
[43] "mar" "mex" "mfcol" "mfg" "mfrow" "mgp" "mkh"
[50] "new" "oma" "omd" "omi" "page" "pch" "pin"
[57] "plt" "ps" "pty" "smo" "srt" "tck" "tcl"
[64] "usr" "xaxp" "xaxs" "xaxt" "xpd" "yaxp" "yaxs"
[71] "yaxt" "ylbias"
``````
Background colour
``````> par('fg')
[1] "black"
``````
Plot character
``````> par('pch')
[1] 1 (Circle)
``````
Line type
``````> par('lty')
[1] "solid"
``````
The `par()` function is used to specify global graphics parameters that affect all plots in an R session. (Use dev.off or plot.new to reset to the defaults.)
• `las` (the orientation of the axis labels on the plot)
• `bg` (background color)
• `mar` (margin size)
• `oma` (outer margin size)
• `mfrow` (plots per row)
• `mfcol` (plots per column)
`lines` can be used to add lines to the plot
Create a plot but do not plot the points:
``````plot(airquality\$Wind, airquality\$Ozone, type='n')
``````
``````> points(may\$Wind,may\$Ozone,col="blue",pch=17)
``````
``````> legend('topright', pch=c(17, 8), co=c("blue", "red"), legend=c("May", "Other Months"))
``````
Add absolute line at the median:
``````abline(v=median(airquality\$Wind), lty=2, lwd=2)
``````
Giving a title to many plots
``````> mtext("Ozone and Weather in New York City", outer=TRUE)
``````
### Lattice Plotting System#
Must be loaded: `library('lattice')`
Lattice is implemented using 2 packages:
• `lattice` contains code for producing trellis graphics. These include `xyplot`, `bwplot`, and `levelplot`
• `grid` system: low level functions
• `xyplot` produces a scatterplot
• `bwplot` produces a box and whisker plot
• `histogram` produces a histogram
All plotting is done in a single call Lattice functions gnerally take a formua as their first argument. (`y ~ x`)
``````xyplot(y ~ x | f * g, data)
``````
`g` would represent optional conditions
``````xyplot(Ozone ~ Wind, data=airquality)
``````
``````> xyplot(Ozone~Wind, data=airquality, col='red', pch=8, main='Big Apple Data')
``````
As factor:
``````> xyplot(Ozone ~ Wind | as.factor(Month), data=airquality, layout=c(5,1))
``````
Since Month is a named column of the airquality dataframe we had to tell R to treat it as a factor
Lattice functions behave differently from base graphics functions in one critical way. Recall that base graphics functions plot data directly to the graphics device (e.g., screen, or file such as a PDF file). In contrast, lattice graphics functions return an object of class trellis.
lattice returns the `Trellis` object.
``````p <- xyplot(Ozone~Wind,data=airquality)
``````
Type `p` or `print(p)` to show the chart
See arguments for the object:
``````> names(p)
[1] "formula" "as.table" "aspect.fill" "legend"
[5] "panel" "page" "layout" "skip"
[9] "strip" "strip.left" "xscale.components" "yscale.components"
[13] "axis" "xlab" "ylab" "xlab.default"
[17] "ylab.default" "xlab.top" "ylab.right" "main"
[21] "sub" "x.between" "y.between" "par.settings"
[25] "plot.args" "lattice.options" "par.strip.text" "index.cond"
[29] "perm.cond" "condlevels" "call" "x.scales"
[33] "y.scales" "panel.args.common" "panel.args" "packet.sizes"
[37] "x.limits" "y.limits" "x.used.at" "y.used.at"
[41] "x.num.limit" "y.num.limit" "aspect.ratio" "prepanel.default"
[45] "prepanel"
``````
Get the formula
``````> p[["formula"]]
Ozone ~ Wind
``````
Check the limits of the x value:
``````> p[['x.limits']]
[1] 0.37 22.03
``````
Lattice create a file to create the plot:
``````p2 <- xyplot(y ~ x | f, panel = function(x, y, ...) {
panel.xyplot(x, y, ...) ## First call default panel function
panel.lmline(x, y, col = 2) ## Overlay a simple linear regression line
})
print(p2)
invisible()
``````
to run this do:
``````source(pathtofile("plot2.R"), local=TRUE)
``````
Table of multiple values:
``````> table(diamonds\$color, diamonds\$cut)
Fair Good Very Good Premium Ideal
D 163 662 1513 1603 2834
E 224 933 2400 2337 3903
F 312 909 2164 2331 3826
G 314 871 2299 2924 4884
H 303 702 1824 2360 3115
I 175 522 1204 1428 2093
J 119 307 678 808 896
``````
Edit `myLabels.R`:
``````> myedit('myLabels.R')
source(pathtofile('myLabels.R'), local=TRUE)
``````
Using the labels:
``````> xyplot(price~carat | color*cut,data=diamonds, strip=FALSE, pch=20, xlab = myxlab, ylab=myylab, main=mymain)
``````
The `strip=FALSE` variable removes labels from the panels
The lattice system is ideal for creating conditioning plots where you examine the same kind of plot under many different conditions.
### Working with Colors#
Effectively using colors can enhance your plots and presentations, emphasizing the important points you’re trying to convey
`grDevices` gives you the `colouors()` function
Get a sample of some colours:
``````> sample(colors(), 10)
[1] "cornsilk2" "aquamarine4" "gray75" "grey46"
[5] "honeydew" "gray43" "grey50" "palegoldenrod"
[9] "lightsteelblue2" "indianred3"
``````
`colorRamp` and `colorRampPalette` blend colours together
The first, colorRamp, takes a palette of colors (the arguments) and returns a function that takes values between 0 and 1 as arguments. The 0 and 1 correspond to the extremes of the color palette. Arguments between 0 and 1 return blends of these extremes.
``````> pal <- colorRamp(c('red', 'blue'))
``````
So:
``````> pal(0)
[,1] [,2] [,3]
[1,] 255 0 0
``````
Which gives `RGB` colours
So pal creates colors using the palette we specified when we called colorRamp
``````> pal(seq(0,1,len=6))
[,1] [,2] [,3]
[1,] 255 0 0
[2,] 204 0 51
[3,] 153 0 102
[4,] 102 0 153
[5,] 51 0 204
[6,] 0 0 255
``````
ColorRampalette
``````> p1 <- colorRampPalette(c('red', 'blue'))
> p1(2)
[1] "#FF0000" "#0000FF"
``````
`colorRamp` and `colorRampPalette` could return a 3 or 4 long vector of colors
Set the alpha
``````p3 <- colorRampPalette(c('blue', 'green'), alpha=.5)
``````
Alpha represents an opacity level, that is, how transparent should the colors be
An alpha will help you if there are manny scatter points together and you want to know the density
``````> plot(x, y, pch=19, col=rgb(0, .5, .5, .3))
``````
In the above `.3` is the density
#### RColorBrewer#
Three types:
• `sequential` - light to dark
• `divergent` - divergent (neutral colour white is centre)
• `qualitative` - random colours used to distinguish data
Use a palette:
``````> cols <- brewer.pal(3, 'BuGn')
[1] "#E5F5F9" "#99D8C9" "#2CA25F"
> pal <- colorRampPalette(cols)
``````
Using the pallete:
``````> image(volcano, col=pal(20))
``````
### GGPlot2 Part1#
Complete docuemtnation on ggplot2
Grammar of Graphics
`ggplot2` combines the best of base and lattice
2 workhorse functions:
• qplot - Less flexible
• ggplot - More flexible
Basic plot:
``````> qplot(displ, hwy, data=mpg)
``````
Add an aestheitc (colour based on factor)
``````> qplot(displ, hwy, data=mpg, color=drv)
``````
We can automatically add a trendlines with the use of `geom`, the first element means data piint and second is trnedline.
``````> qplot(displ, hwy, data=mpg, color=drv, geom=c('point', 'smooth'))
``````
Notice the gray areas surrounding each trend lines. These indicate the 95% confidence intervals for the lines
If no x-axis is given the plot will just show the index /order of the value in the dataset:
``````> qplot(y=hwy, data=mpg, color=drv)
``````
Create a box and whisker plot:
``````> qplot(drv, hwy, data=mpg, geom='boxplot')
``````
Set the colour to a factor:
``````> qplot(drv, hwy, data=mpg, geom='boxplot', color=manufacturer)
``````
Create a histogram with coloured attribute:
``````> qplot(hwy, data=mpg, fill=drv)
``````
`. ~ drv` is ggplot2’s shorthand for number of rows (to the left of the `~`)
##### Scatterplot split into panels / facets#
``````> qplot(displ, hwy, data=mpg, facets=. ~ drv)
``````
##### Histogram split into panels#
``````> qplot(hwy, data=mpg, facets=drv ~ ., binwidth=2)
``````
The facet argument `drv ~ .` resulted in a `3 by 1` setup.
### GGPlot2 Part2#
#### Components of GGPLot2#
• DATA FRAME - data you are plotting
• AESTHETIC MAPPINGS - how data is mapped to color, size, etc.
• GEOMS (geometric objects) - are what you see in the plot (points, lines, shapes)
• FACETS - panels used in conditional plots
• STATS - statistical transformations (binning, quantiles, and smoothing)
• SCALES - what coding an aesthetic map uses (for example, male = red, female = blue)
• COORDINATE SYSTEM - how plots are depicted
Point and smooth scatterplot:
``````> qplot(displ, hwy, data=mpg, geom=c('point', 'smooth'), facets=.~drv)
``````
Creating a mapping:
``````> g <- ggplot(mpg, aes(displ, hwy))
``````
Summary of graphical object:
``````> summary(g)
data: manufacturer, model, displ, year, cyl, trans, drv, cty, hwy, fl, class
[234x11]
mapping: x = displ, y = hwy
``````
A `234 x 11` matrix and a `x (displ) and y (hwy)` mapping
Tell `ggplot2` how to show the data:
``````> g+geom_point()
``````
Show points and trend line:
``````> g+geom_point()+geom_smooth()
``````
Change the smoothing function to linear model:
``````> g+geom_point()+geom_smooth(method="lm")
``````
Split it into `facets`:
``````> g+geom_point()+geom_smooth(method="lm")+facet_grid(. ~ drv)
``````
Add a title to the chart:
``````> g+geom_point()+geom_smooth(method="lm")+facet_grid(. ~ drv)+ggtitle('Swirl Rules!')
``````
Two standard appearance themes are included in ggplot. These are `theme_gray()` which is the default theme (gray background with white grid lines) and `theme_bw()` which is a plainer (black and white) color scheme.
Create geometric points with colour, alpha and size:
``````> g+geom_point(color='pink', size=4, alpha=1/2)
``````
Color based on factor:
``````> g+geom_point(size=4, alpha=1/2, aes(color=drv))
``````
Changing labels:
``````g + geom_point(aes(color = drv)) + labs(title="Swirl Rules!") + labs(x="Displacement", y="Hwy Mileage")
``````
Dashed linear regression line and turn off confidence level grey area:
``````> g + geom_point(size=2, alpha=1/2, aes(color = drv)) + geom_smooth(size=4, linetype=3, method='lm', se=FALSE)
``````
Change the theme and font style:
``````> g+geom_point(aes(color=drv))+theme_bw(base_family='Times')
``````
#### Outlier Data#
Set limits to not show an outlier
Using plot with outlier data:
``````> plot(myx, myy, type='l', ylim=c(-3,3))
``````
With ggplot:
``````> g <- ggplot(testdat, aes(x=myx, y=myy))
> g+geom_line()+ylim(-3,3)
``````
Can also be done by limiting the coordinate system:
``````> g + geom_line() + coord_cartesian(ylim=c(-3,3))
``````
Create facetted scatterplot with marginal totals:
``````> g + geom_point() + facet_grid(drv~cyl, margins=TRUE)
``````
With a smooth trendline:
``````> g + geom_point() + facet_grid(drv~cyl, margins=TRUE) + geom_smooth(method='lm', se=FALSE, size=2, color='black')
``````
### GGPlot2 Extras#
You can specify the binwidth (usually range / x):
``````> qplot(price, data=diamonds, binwidth=18497/30)
``````
Change fill of histogram based on factor:
``````> qplot(price, data=diamonds, binwidth=18497/30, fill=cut)
``````
Histogram as a density function:
``````> qplot(price, data=diamonds, geom='density')
``````
Density by colour:
``````> qplot(price, data=diamonds, geom='density', color=cut)
``````
Change shape based on cut:
``````> qplot(carat, price, data=diamonds, shape=cut)
``````
Add regression lines for each cut:
``````> qplot(carat, price, data=diamonds, color=cut) + geom_smooth(method='lm')
``````
facets: The symbol to the left of the tilde indicates rows and the symbol to the right of the tilde indicates columns
``````> qplot(carat, price, data=diamonds, color=cut, facets=.~cut) + geom_smooth(method='lm')
``````
cut, which allows you to divide your data into sets and label each entry as belonging to one of the sets
So you can create a factor out of numerical data
``````> cutpoints <- quantile(diamonds\$carat,seq(0,1,length=4),na.rm=TRUE)
> cutpoints
0% 33.33333% 66.66667% 100%
0.20 0.50 1.00 5.01
``````
Create a new name for the data:
``````> diamonds\$car2 <- cut(diamonds\$carat,cutpoints); stageVariable("diamonds\$car2",diamonds\$car2)
``````
Because the dataset was changed we have to create the graphical object again:
``````> g <- ggplot(diamonds, aes(depth, price))
> g+geom_point(alpha=1/3) + facet_grid(cut ~ car2)
``````
#### Boxplot#
``````> ggplot(diamonds, aes(carat, price))+geom_boxplot()+facet_grid(. ~ cut)
``````
### Hierarchical Clustering#
A simple way of quickly examining and displaying multi-dimensional data. This technique is usually most useful in the early stages of analysis when you’re trying to get an understanding of the data
Clustering organizes data points that are close into groups. So obvious questions are “How do we define close?”, “How do we group things?”, and “How do we interpret the grouping?” Cluster analysis is a very important topic in data analysis.
Closeness: Distance or similarity are usually the metrics used
Measuring distance: Euclidean distance and correlation similarity are continuous measures, while Manhattan distance is a binary measure
Euclidean distance is distance “as the crow flies”. Many applications, however, can’t realistically use crow-flying distance. Cars, for instance, have to follow roads.
Manhattan distance is the sum of the absolute values of the distances between each coordinate
Computer euclidean distances between points:
``````> dist(dataFrame)
``````
You can create the dendogram with:
``````> hc <- hclust(distxy)
``````
You can plot the dendogram:
``````> plot(hc)
``````
Printing at the same level:
``````> plot(as.dendrogram(hc))
``````
Then you can cut a line at a certain distance to create clusters
Measuring distance between clusters:
• complete linkage - furtherest distance of elements in respective clusters
• average linkage - get the mean x and y coordinates
##### Heatmap#
``````> heatmap(dataMatrix, col=cm.colors(25))
``````
### K Means Clustering#
Another simple way of examining and organizing multi-dimensional data. Most useful in the early stages of analysis when you’re trying to get an understanding of the data, e.g., finding some pattern or relationship between different factors or variables.
R documentation tells us that the k-means method “aims to partition the points into k groups such that the sum of squares from points to the assigned cluster centres is minimized.”
1. First guess how many clusters you have or want
2. Randomly create a “centroid” (a phantom point) for each cluster
3. Readjust the centroid’s position by making it the average of the points assigned to it.
k-means clustering requires some distance metric (say Euclidean), a hypothesized fixed number of clusters, and an initial guess as to cluster centroids.
k-means clustering returns a final position of each cluster’s centroid as well as the assignment of each data point or observation to a cluster
``````> points(cx, cy, col=c('red', 'orange', 'purple'), pch=3, cex=2, lwd=2)
``````
Check distnace between centroid and points:
``````> mdist(x,y,cx,cy)
``````
You can find the minimum with:
``````> apply(distTmp, 2, which.min)
[1] 2 2 2 1 3 3 3 1 3 3 3 3
``````
Can colour based on above output:
``````> points(x,y,pch=19, cex=2, col=cols1[newClust])
``````
Recaculate the centroids based on clusters:
``````> tapply(x, newClust, mean)
1 2 3
1.210767 1.010320 2.498011
> tapply(y, newClust, mean)
1 2 3
1.730555 1.016513 1.354373
``````
Apply the points on plot:
``````> points(newCx, newCy, col=cols1, pch=8, cex=2, lwd=2)
``````
Some points will need to change. Recolour the points again:
``````> points(x,y,pch=19, cex=2, col=cols1[newClust2])
``````
Create a kmeans object:
``````> kmeans(dataFrame, centers=3)
``````
Check number of iterations:
``````> kmObj\$iter
[1] 2
``````
### Dimension Reduction#
principal component analysis (PCA) singular value decomposition (SVD)
PCA and SVD are used in both the exploratory phase and the more formal modelling stage of analysis
we’d like to find a smaller set of multivariate variables that are uncorrelated AND explain as much variance (or variability) of the data as possible.
``````Orthagonal == Uncorrelated
``````
#### Singular Value Decomposition#
``````> svd(mat)
\$d
[1] 9.5899624 0.1806108
\$u
[,1] [,2]
[1,] -0.3897782 -0.9209087
[2,] -0.9209087 0.3897782
\$v
[,1] [,2]
[1,] -0.2327012 -0.7826345
[2,] -0.5614308 0.5928424
[3,] -0.7941320 -0.1897921
``````
Recall that in R matrix multiplication requires you to use the operator %*%
``````> matu %*% diag %*% t(matv)
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 5 7
``````
#### Principal Component Analysis#
a simple, non-parametric method for extracting relevant information from confusing data sets.
PCA is a method to reduce a high-dimensional data set to its essential elements (not lose information) and explain the variability in the data
``````> svd(scale(mat))
\$d
[1] 1.732051 0.000000
\$u
[,1] [,2]
[1,] -0.7071068 0.7071068
[2,] 0.7071068 0.7071068
\$v
[,1] [,2]
[1,] 0.5773503 -0.5773503
[2,] 0.5773503 0.7886751
[3,] 0.5773503 -0.2113249
> prcomp(scale(mat))
Standard deviations (1, .., p=2):
[1] 1.732051 0.000000
Rotation (n x k) = (3 x 2):
PC1 PC2
[1,] 0.5773503 -0.5773503
[2,] 0.5773503 0.7886751
[3,] 0.5773503 -0.2113249
``````
SVD and PCA cannot deal with `MISSING` data
Singular value decomposition is a good way to approximate data without having to store a lot.
Make sure data is on consistent units Seperating out real patterns requires detective work
distance matrix as first 3 columns of `dist`
``````> mdist <- dist(sub1[,1:3])
``````
Create clusters
``````> hclustering <- hclust(mdist)
``````
### Case Study#
You can use `read.table` to read in data. `R` is smart enough to unzip it.
``````> dim(pm0)
[1] 117421 5
``````
The data set has `117421` lines and `5` columns
Reassign a variable by splitting with `|` character:
``````> cnames <- strsplit(cnames, '|', fixed=TRUE)
``````
Make syntactically valid names:
``````> names(pm0) <- make.names(cnames[[1]][wcol])
> names(pm1) <- make.names(cnames[[1]][wcol])
``````
Create `x1` by assigning it to `Sample.Value`
``````> x1 <- pm1\$Sample.Value
``````
The 1999 data:
``````> summary(x0)
``````
Min. 1st Qu. Median Mean 3rd Qu. Max. NA’s 0.00 7.20 11.50 13.74 17.90 157.10 13217
The 2012 data:
``````> summary(x1)
``````
Min. 1st Qu. Median Mean 3rd Qu. Max. NA’s -10.00 4.00 7.63 9.14 12.00 908.97 73133
Indicates an improved situation. The maximum increases indicates possible malfunction with the capturing devices.
Boxplot the data:
``````> boxplot(x0, x1)
``````
There are so many values outside the boxes and the range of x1 is so big that the boxes are flattened. It might be more informative to call boxplot on the logs (base 10) of x0 and x1. Do this now using log10(x0) and log10(x1) as the 2 arguments.
``````> boxplot(log10(x0), log10(x1))
``````
`R` will warn about values that can’t be `log` ed
Get sum of all the negative values excluding `NA`:
``````> negative <- x1<0
> sum(negative, na.rm=TRUE)
[1] 26474
> mean(negative, na.rm=TRUE)
[1] 0.0215034
``````
We see that just 2% of the x1 values are negative. Perhaps that’s a small enough percentage that we can ignore them
Get array of dates:
``````> str(dates)
int [1:1304287] 20120101 20120104 20120107 20120110 20120113 20120116 20120119 20120122 20120125 20120128 ...
``````
The dates are hard to read though, imporved with:
``````> dates <- as.Date(as.character(dates), '%Y%m%d')
``````
Show a histogram of the dates that are negative:
``````> hist(dates[negative], "month")
``````
We see the bulk of the negative measurements were taken in the winter months, with a spike in May. Not many of these negative measurements occurred in summer months. We can take a guess that because particulate measures tend to be low in winter and high in summer, coupled with the fact that higher densities are easier to measure, that measurement errors occurred when the values were low. For now we’ll attribute these negative measurements to errors. Also, since they account for only 2% of the 2012 data, we’ll ignore them.
View data that intersects:
``````> both <- intersect(site0, site1)
``````
Subset the data for a certain county and site:
``````> cnt0 <- subset(pm0, State.Code == 36 & county.site %in% both)
``````
Membership is denoted with `%in%`
Split data:
``````> sapply(split(cnt1, cnt1\$county.site), nrow)
``````
1.12 1.5 101.3 13.11 29.5 31.3 5.80 63.2008 67.1015 85.55 31 64 31 31 33 15 31 30 31 31
#### Create the plot#
``````> par(mfrow=c(1,2), mar=c(4,4,2,1))
> plot(dates0, x0sub, pch=20)
``````
``````> abline(h=median(x0sub, na.rm=TRUE), lwd=2)
``````> mrg <- merge(d0, d1, by='state') | 9,372 | 31,381 | {"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-2024-38 | latest | en | 0.884862 |
https://statkat.com/stattest.php?t=3&t2=5&t3=39&t4=1 | 1,660,293,041,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571597.73/warc/CC-MAIN-20220812075544-20220812105544-00221.warc.gz | 485,253,703 | 8,187 | # Goodness of fit test - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Goodness of fit test
One sample $z$ test for the mean
McNemar's test
$z$ test for a single proportion
Independent variableIndependent variableIndependent variableIndependent variable
NoneNone2 paired groupsNone
Dependent variableDependent variableDependent variableDependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)One quantitative of interval or ratio levelOne categorical with 2 independent groupsOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
• H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
• H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$, the probability of drawing an observation from condition $J$ is $\pi_J$
H0: $\mu = \mu_0$
Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.
Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options:
1. First score of pair is 0, second score of pair is 0
2. First score of pair is 0, second score of pair is 1 (switched)
3. First score of pair is 1, second score of pair is 0 (switched)
4. First score of pair is 1, second score of pair is 1
The null hypothesis H0 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0.
Other formulations of the null hypothesis are:
• H0: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group
• H0: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)
H0: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
• H1: the population proportions are not all as specified under the null hypothesis
or equivalently
• H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0.
Other formulations of the alternative hypothesis are:
• H1: $\pi_1 \neq \pi_2$
• H1: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
AssumptionsAssumptionsAssumptionsAssumptions
• Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Scores are normally distributed in the population
• Population standard deviation $\sigma$ is known
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
• Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
• Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10
• Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more
• Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more
• Sample is a simple random sample from the population. That is, observations are independent of one another
If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.
Test statisticTest statisticTest statisticTest statistic
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size.
The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.
$X^2 = \dfrac{(b - c)^2}{b + c}$
Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis.
Sampling distribution of $X^2$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $z$ if H0 were true
Approximately the chi-squared distribution with $J - 1$ degrees of freedomStandard normal distribution
If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom.
If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$.
Approximately the standard normal distribution
Significant?Significant?Significant?Significant?
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Two sided:
Right sided:
Left sided:
For test statistic $X^2$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
If $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$:
• Check if $b$ observed in sample is in the rejection region or
• Find two sided $p$ value corresponding to observed $b$ and check if it is equal to or smaller than $\alpha$
Two sided:
Right sided:
Left sided:
n.a.$C\%$ confidence interval for $\mu$n.a.Approximate $C\%$ confidence interval for $\pi$
-$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
The confidence interval for $\mu$ can also be used as significance test.
-Regular (large sample):
• $p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
With plus four method:
• $p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$
where $p_{plus} = \dfrac{X + 2}{N + 4}$ and the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
n.a.Effect sizen.a.n.a.
-Cohen's $d$:
Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Cohen's $d$ indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0.$
--
n.a.Visual representationn.a.n.a.
---
n.a.n.a.Equivalent toEquivalent to
--
• When testing two sided: goodness of fit test, with a categorical variable with 2 levels.
• When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.
Example contextExample contextExample contextExample context
Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?Is the average mental health score of office workers different from $\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is $\sigma = 3.$Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.
SPSSn.a.SPSSSPSS
Analyze > Nonparametric Tests > Legacy Dialogs > Chi-square...
• Put your categorical variable in the box below Test Variable List
• Fill in the population proportions / probabilities according to $H_0$ in the box below Expected Values. If $H_0$ states that they are all equal, just pick 'All categories equal' (default)
-Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
• Put the two paired variables in the boxes below Variable 1 and Variable 2
• Under Test Type, select the McNemar test
Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
• Put your dichotomous variable in the box below Test Variable List
• Fill in the value for $\pi_0$ in the box next to Test Proportion
If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
Jamovin.a.JamoviJamovi
Frequencies > N Outcomes - $\chi^2$ Goodness of fit
• Put your categorical variable in the box below Variable
• Click on Expected Proportions and fill in the population proportions / probabilities according to $H_0$ in the boxes below Ratio. If $H_0$ states that they are all equal, you can leave the ratios equal to the default values (1)
-Frequencies > Paired Samples - McNemar test
• Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns
Frequencies > 2 Outcomes - Binomial test
• Put your dichotomous variable in the white box at the right
• Fill in the value for $\pi_0$ in the box next to Test value
• Under Hypothesis, select your alternative hypothesis
Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
Practice questionsPractice questionsPractice questionsPractice questions | 3,064 | 11,565 | {"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} | 3.8125 | 4 | CC-MAIN-2022-33 | latest | en | 0.864048 |
https://calculationcalculator.com/gaj-to-bigha | 1,723,520,306,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00833.warc.gz | 120,060,973 | 21,766 | # Gaj to Bigha Conversion
## 1 Gaj is equal to how many Bigha?
### 0.000625 Bigha
##### Reference This Converter:
Gaj and Bigha both are the Land measurement unit. Compare values between unit Gaj 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.
Gaj to Bigha conversion allows you to convert value between Gaj to Bigha easily. Just enter the Gaj value into the input box, the system will automatically calculate Bigha value. 1 Gaj in Bigha? In mathematical terms, 1 Gaj = 0.000625 Bigha.
To conversion value between Gaj to Bigha, just multiply the value by the conversion ratio. One Gaj is equal to 0.000625 Bigha, so use this simple formula to convert -
The value in Bigha is equal to the value of Gaj multiplied by 0.000625.
Bigha = Gaj * 0.000625;
For calculation, here's how to convert 10 Gaj to Bigha using the formula above -
10 Gaj = (10 * 0.000625) = 0.00625 Bigha
Gaj Bigha Conversion
1 0.000625 1 Gaj = 0.000625 Bigha
2 0.00125 2 Gaj = 0.00125 Bigha
3 0.001875 3 Gaj = 0.001875 Bigha
4 0.0025 4 Gaj = 0.0025 Bigha
5 0.003125 5 Gaj = 0.003125 Bigha
6 0.00375 6 Gaj = 0.00375 Bigha
7 0.004375 7 Gaj = 0.004375 Bigha
8 0.005 8 Gaj = 0.005 Bigha
9 0.005625 9 Gaj = 0.005625 Bigha
10 0.00625 10 Gaj = 0.00625 Bigha
11 0.006875 11 Gaj = 0.006875 Bigha
12 0.0075 12 Gaj = 0.0075 Bigha
13 0.008125 13 Gaj = 0.008125 Bigha
14 0.00875 14 Gaj = 0.00875 Bigha
15 0.009375 15 Gaj = 0.009375 Bigha
16 0.01 16 Gaj = 0.01 Bigha
17 0.010625 17 Gaj = 0.010625 Bigha
18 0.01125 18 Gaj = 0.01125 Bigha
19 0.011875 19 Gaj = 0.011875 Bigha
20 0.0125 20 Gaj = 0.0125 Bigha
21 0.013125 21 Gaj = 0.013125 Bigha
22 0.01375 22 Gaj = 0.01375 Bigha
23 0.014375 23 Gaj = 0.014375 Bigha
24 0.015 24 Gaj = 0.015 Bigha
25 0.015625 25 Gaj = 0.015625 Bigha
26 0.01625 26 Gaj = 0.01625 Bigha
27 0.016875 27 Gaj = 0.016875 Bigha
28 0.0175 28 Gaj = 0.0175 Bigha
29 0.018125 29 Gaj = 0.018125 Bigha
30 0.01875 30 Gaj = 0.01875 Bigha
31 0.019375 31 Gaj = 0.019375 Bigha
32 0.02 32 Gaj = 0.02 Bigha
33 0.020625 33 Gaj = 0.020625 Bigha
34 0.02125 34 Gaj = 0.02125 Bigha
35 0.021875 35 Gaj = 0.021875 Bigha
36 0.0225 36 Gaj = 0.0225 Bigha
37 0.023125 37 Gaj = 0.023125 Bigha
38 0.02375 38 Gaj = 0.02375 Bigha
39 0.024375 39 Gaj = 0.024375 Bigha
40 0.025 40 Gaj = 0.025 Bigha
41 0.025625 41 Gaj = 0.025625 Bigha
42 0.02625 42 Gaj = 0.02625 Bigha
43 0.026875 43 Gaj = 0.026875 Bigha
44 0.0275 44 Gaj = 0.0275 Bigha
45 0.028125 45 Gaj = 0.028125 Bigha
46 0.02875 46 Gaj = 0.02875 Bigha
47 0.029375 47 Gaj = 0.029375 Bigha
48 0.03 48 Gaj = 0.03 Bigha
49 0.030625 49 Gaj = 0.030625 Bigha
50 0.03125 50 Gaj = 0.03125 Bigha
51 0.031875 51 Gaj = 0.031875 Bigha
52 0.0325 52 Gaj = 0.0325 Bigha
53 0.033125 53 Gaj = 0.033125 Bigha
54 0.03375 54 Gaj = 0.03375 Bigha
55 0.034375 55 Gaj = 0.034375 Bigha
56 0.035 56 Gaj = 0.035 Bigha
57 0.035625 57 Gaj = 0.035625 Bigha
58 0.03625 58 Gaj = 0.03625 Bigha
59 0.036875 59 Gaj = 0.036875 Bigha
60 0.0375 60 Gaj = 0.0375 Bigha
61 0.038125 61 Gaj = 0.038125 Bigha
62 0.03875 62 Gaj = 0.03875 Bigha
63 0.039375 63 Gaj = 0.039375 Bigha
64 0.04 64 Gaj = 0.04 Bigha
65 0.040625 65 Gaj = 0.040625 Bigha
66 0.04125 66 Gaj = 0.04125 Bigha
67 0.041875 67 Gaj = 0.041875 Bigha
68 0.0425 68 Gaj = 0.0425 Bigha
69 0.043125 69 Gaj = 0.043125 Bigha
70 0.04375 70 Gaj = 0.04375 Bigha
71 0.044375 71 Gaj = 0.044375 Bigha
72 0.045 72 Gaj = 0.045 Bigha
73 0.045625 73 Gaj = 0.045625 Bigha
74 0.04625 74 Gaj = 0.04625 Bigha
75 0.046875 75 Gaj = 0.046875 Bigha
76 0.0475 76 Gaj = 0.0475 Bigha
77 0.048125 77 Gaj = 0.048125 Bigha
78 0.04875 78 Gaj = 0.04875 Bigha
79 0.049375 79 Gaj = 0.049375 Bigha
80 0.05 80 Gaj = 0.05 Bigha
81 0.050625 81 Gaj = 0.050625 Bigha
82 0.05125 82 Gaj = 0.05125 Bigha
83 0.051875 83 Gaj = 0.051875 Bigha
84 0.0525 84 Gaj = 0.0525 Bigha
85 0.053125 85 Gaj = 0.053125 Bigha
86 0.05375 86 Gaj = 0.05375 Bigha
87 0.054375 87 Gaj = 0.054375 Bigha
88 0.055 88 Gaj = 0.055 Bigha
89 0.055625 89 Gaj = 0.055625 Bigha
90 0.05625 90 Gaj = 0.05625 Bigha
91 0.056875 91 Gaj = 0.056875 Bigha
92 0.0575 92 Gaj = 0.0575 Bigha
93 0.058125 93 Gaj = 0.058125 Bigha
94 0.05875 94 Gaj = 0.05875 Bigha
95 0.059375 95 Gaj = 0.059375 Bigha
96 0.06 96 Gaj = 0.06 Bigha
97 0.060625 97 Gaj = 0.060625 Bigha
98 0.06125 98 Gaj = 0.06125 Bigha
99 0.061875 99 Gaj = 0.061875 Bigha
100 0.0625 100 Gaj = 0.0625 Bigha
200 0.125 200 Gaj = 0.125 Bigha
300 0.1875 300 Gaj = 0.1875 Bigha
400 0.25 400 Gaj = 0.25 Bigha
500 0.3125 500 Gaj = 0.3125 Bigha
600 0.375 600 Gaj = 0.375 Bigha
700 0.4375 700 Gaj = 0.4375 Bigha
800 0.5 800 Gaj = 0.5 Bigha
900 0.5625 900 Gaj = 0.5625 Bigha
1000 0.625 1000 Gaj = 0.625 Bigha
2000 1.25 2000 Gaj = 1.25 Bigha
3000 1.875 3000 Gaj = 1.875 Bigha
4000 2.5 4000 Gaj = 2.5 Bigha
5000 3.125 5000 Gaj = 3.125 Bigha
6000 3.75 6000 Gaj = 3.75 Bigha
7000 4.375 7000 Gaj = 4.375 Bigha
8000 5 8000 Gaj = 5 Bigha
9000 5.625 9000 Gaj = 5.625 Bigha
10000 6.25 10000 Gaj = 6.25 Bigha
A Gaj / Gaz or yard is a unit of length used in parts of Asia. Historically, it was a regionally variable measurement. It was often used for measuring textiles.
Bigha is a traditional historical unit of land area measurement, commonly used in India, Bangladesh, and Nepal. Bigha has no standard size. The size of one bigha varies considerably from space to time.
The value in Bigha is equal to the value of Gaj multiplied by 0.000625.
Bigha = Gaj * 0.000625;
1 Gaj is equal to 0.000625 Bigha.
1 Gaj = 0.000625 Bigha.
• 1 gaj = bigha
• gaj into bigha
• gajs to bighas
• convert gaj to bigha
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→ | 2,823 | 5,611 | {"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} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.578351 |
https://science.lesueur.nz/12phy/as91171/slides/momentum.html | 1,623,641,766,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00051.warc.gz | 460,774,004 | 8,943 | # Momentum & Impulse
12PHYS - Mechanics
2021
## Akoranga 32 Mahi Tuatahi
Ethan and Kelley are playing on a merry-go-round spinning clockwise. Ethan is sitting $$3m$$ from the center and has a speed of $$1.5ms^{1}$$.
1. Draw a diagram with labelled arrows showing the velocity and acceleration of Ethan. (A)
2. What is the name of this acceleration? (A)
3. Calculate the size of his acceleration (A)
4. State the direction of the horizontal force acting on Ethan, and explain clearly why there must be a horizontal force acting on him. (E)
5. Kelley pushes the merry-go-round so that its period is halved. Explain what this does to the size of the horizontal force acting on Ethan. (E)
## Ngā Whāinga Ako
1. Be able to use the momentum equation ($$p=mv$$)
2. Be able to calculate basic change in momentum situations ($$\Delta p = p_{f} - p_{i}$$)
Write the date and ngā whāinga ako in your book
## Momentum
Momentum is a quantity of an object which involves its mass and velocity. It can be thought of as the oomph of an object, or how hard it is to change its direction.
\begin{aligned} & p = mv \newline & momentum = mass \times velocity \newline \end{aligned}
### Pātai Tahi (Q1)
Using the equation
\begin{aligned} & p = mv \newline & momentum = mass \times velocity \newline \end{aligned}
what are the units for momentum?
#### Whakatika Tahi
\begin{aligned} & p = mv \newline & momentum = mass \times velocity \newline & momentum = kgms^{-1} \end{aligned}
### Pātai Rua: Scalar or Vector?
• Momentum is a vector quantity because it depends upon a vector (velocity).
• Momentum is always in the direction of the objects velocity.
### Where is it Used?
Momentum is very useful when considering collisions and explosions. Most questions you will see will involve collisions. It helps us describe and understand why objects move how they do after collisions.
### Pātai Toru (Q3)
1. Calculate the momentum for a 30g golf ball travelling at $$10ms^{-1}$$
2. Calculate the momentum for a shopping trolley of mass $$24kg$$ travelling at $$0.75ms^{-1}$$
3. Calculate the momentum for a ship of mass 30,000 tonnes moving at $$0.2ms^{-1}$$
\begin{aligned} & && \text{(K)} \newline & && \text{(U)} \newline & && \text{(F)} \newline & && \text{(S + S)} \end{aligned}
#### Whakatika Toru
\begin{aligned} & m = 0.03kg, v=10ms^{-1} && \text{(K)} \newline & p = ? && \text{(U)} \newline & p = mv && \text{(F)} \newline & p = 0.03 \times 10 = 0.3kgms^{-1} && \text{(S + S)} \end{aligned}
#### Whakatika Whā
\begin{aligned} & m = 24kg, v=0.75ms^{-1} && \text{(K)} \newline & p = ? && \text{(U)} \newline & p = mv && \text{(F)} \newline & p = 24 \times 0.75 = 18kgms^{-1} && \text{(S + S)} \end{aligned}
### Whakatika Rimu
\begin{aligned} & m = 30,000,000kg, v=0.2ms^{-1} && \text{(K)} \newline & p = ? && \text{(U)} \newline & p = mv && \text{(F)} \newline & p = 30,000,000 \times 0.2 = 6,000,000kgms^{-1} && \text{(S + S)} \end{aligned}
## Change in Momentum (1D)
When a force acts upon an object and its motion changes, its momentum will also change. We call change in momentum $$\Delta p$$.
\begin{aligned} & \Delta p = p_{f} - p_{i} \newline & \Delta p = \text{final momentum} - \text{initial momentum} \end{aligned}
• In 1D we can use this equation directly
• In 2D we use this equation to help us draw a vector diagram, and then we use trigonometry
### Pātai Ono (Q6)
Angus bowls a cricket ball of mass $$160g$$ at a speed of $$25ms^{-1}$$. Lucy hits the ball back towards him at $$35ms^{-1}$$. Calculate the change in momentum.
Hint: Recall that velocity in one direction is positive and the other direction, negative.
\begin{aligned} & && \text{(K)} \newline & && \text{(U)} \newline & && \text{(F)} \newline & && \text{(S + S)} \end{aligned}
#### Whakatika Ono
\begin{aligned} & m=0.16kg, v_{i} = 25ms^{-1}, v_{f} = -35ms^{-1} && \text{(K)} \newline & \Delta p = ? && \text{(U)} \newline & \Delta p = p_{f} - p_{i} = m_{f}v_{f} - m_{i}v_{i} && \text{(F)} \newline & \Delta p = (0.16 \times -35) - (0.16 \times 25) = 9.6kgms^{-1} && \text{(S + S)} \end{aligned}
### Whakawai/Practise
• Textbook Momentum & Impulse Q1-4
• New: pg. 130
• Old: pg. 123
## Akoranga 33 Mahi Tuatahi: 2D $$\Delta$$ in Momentum
An object with momentum $$p=10kgms^{-1}$$ impacts a surface at an angle of $$45\degree$$. It bounces off at an angle of $$45\degree$$ to the surface with a momentum of $$p=7.5kgms^{-1}$$.
1. Draw a diagram to illustrate the collision
2. Convert this diagram into a vector diagram illustrating the change in momentum
3. Calculate the change of momentum, including the angle.
### Whakatika
• $$\Delta p = \sqrt{7.5ms^{-2} + 10^{-2}} = 12.5kgms^{-1}$$
• $$\theta = \tan^{-1}(\frac{10}{7.5}) = 53\degree$$
• Just like finding $$\Delta v$$!
## Ngā Whāinga Ako
1. Be able to calculate $$\Delta p$$ in simple 2D situations
2. Be able to use the impulse formula: $$F \Delta t = \Delta p$$
Write the date and ngā whāinga ako in your book
## Impulse
In order to cause a change in momentum, a force must act upon an object for some amount of time. This is called impulse.
\begin{aligned} & F = ma \newline & F = m \frac{\Delta v}{\Delta t} && \text{substituting for acceleration} \newline & F = m \frac{(v_{f} - v_{i})}{\Delta t} && \text{expanding }\Delta v \newline & F \Delta t = m (v_{f} - v_{i}) && \text{multiply by }\Delta t \newline & F \Delta t = mv_{f} - mv_{i} && \text{expand brackets} \newline & F \Delta t = \Delta p \end{aligned}
### Pātai Whitu (Q7): Satellite in Orbit
A satellite is in orbit. It weighs $$300kg$$ and it has a thruster which exerts a force of $$1500N$$. How long must the satellite fire its thruster for if it wants to increase its speed from $$5000ms^{-1}$$ to $$6000ms^{-1}$$?
Hint: Calculate $$\Delta p$$ first using the velocities, then use the impulse relationships.
#### Whakatika
\begin{aligned} \Delta p &= mv_{f} - mv_{i} \newline &= m(v_{f} - v_{i}) \newline &= 300(6000 - 5000) \newline &= 300000kgms^{-1} \end{aligned} \begin{aligned} F \Delta t &= \Delta p \newline 1500 \Delta t &= 300000 \newline \Delta t &= \frac{300000}{1500} \newline \Delta t &= 200s \end{aligned}
### Pātai Waru (Q8): Cricket
A cricket ball of mass $$120g$$ is bowled at $$30ms^{-1}$$ towards a batsman. The batsman hits it away at $$90\degree$$ to its original velocity, with a speed of $$40ms^{-1}$$.
1. Draw a diagram illustrating what has occurred
2. Label the diagram will $$p_{i}$$ and $$p_{f}$$
3. Calculate $$\Delta p$$ using trigonometry
4. Calculate the force exerted by the batsman, if the bat and ball were in contact for $$0.1s$$
### Whakawai/Practise
• Textbook Momentum and Impulse Q1-5
• New: pg. 130
• Old: pg. 123
## Akoranga 34 Mahi Tuatahi
Lena is swinging a bucket of water in a circle around her head to demonstrate circular motion. The length of the rope is $$0.75m$$ and it takes $$0.84s$$ to go around.
1. Explain why the bucket is always accelerating. (A)
2. Calculate the acceleration of the bucket. (M)
3. Name the force that causes the bucket to accelerate as it goes around her head. Explain why the force causes the bucket to accelerate and follow a circular path. (M)
### Whakatika
1. Explain why the bucket is always accelerating. (A)
The bucket is always accelerating because it is always changing direction, which means the velocity is always changing.
1. Calculate the acceleration of the bucket. (M)
\begin{aligned} & a_{c} = \frac{v^{2}}{r} \newline & a_{c} = \frac{(\frac{(2 \pi r)}{T})^{2}}{r} && \text{substitute } v = \frac{2 \pi r}{T} \newline & a_{c} = \frac{(\frac{(2 \pi 0.75)}{0.84})^{2}}{0.75} && \text{substitute values} \newline & a_{c} = \frac{31.47}{0.75} \newline & a_{c} = 42ms^{-2} \end{aligned}
1. Name the force that causes the bucket to accelerate as it goes around her head. Explain why the forces cause the bucket to accelerate. (M)
Centripetal force
Centripetal force causes the bucket to accelerate because it acts as a tension force towards the center of the circle. The force acts at a right angle to the motion of the bucket, therefore causing it to change direction continuously towards the center of the circle.
## Ngā Whāinga Ako
1. Understand the assumptions behind momentum conservation
2. Be able to use conservation of momentum to understand collisions/explosions
Write the date and ngā whāinga ako in your book
## Conservation of Momentum
• Momentum is conserved during collisions between objects and in explosions.
• $$p_{i} = p_{f}$$
• Conservation of momentum only occurs when no external forces are present.
• An external force would change the momentum ($$F \Delta t = \Delta p$$)!
• For example, gravity or friction either do not apply or have been cancelled by reaction forces.
• Conservation of momentum is the only way to solve collision problems
• Energy is not usually conserved and therefore cannot be used.
## Conservation of Momentum in 1D
• The most straightforward problem you will see is conservation of momentum in 1-dimension. We will use subscript $$1$$ and $$2$$ to indicate object $$1$$ and $$2$$.
• So that we do not get confused about the velocities, we will also use $$u$$ to indicate initial velocities and $$v$$ to indicate final velocities.
• This is an equation representing a collision between two objects.
\begin{aligned} p_{i} &= p_{f} \newline p_{1i} + p_{2i} &= p_{1f} + p_{2f} \newline m_{1}u_{1} + m_{2}u_{2} &= m_{1}v_{1} + m_{2}v_{2} \newline \end{aligned}
## Collisions: Elastic vs. Inelastic
• Q. Can we use energy to calculate collisions?
• A. No, because energy is lost due to friction, meaning that total kinetic energy is not conserved.
• However, in elastic collisions total kinetic energy is conserved.
• E_{ki} = E_{kf}
• NB: If total kinetic energy is not conserved, the collision is inelastic. Most collisions are inelastic.
### Pātai Iwa (Q9): The Rifle
Jordan is out clay pigeon shooting over the weekend and notices that the gun recoils when he fires. His rifle has mass $$4kg$$ and fires a bullet of mass $$20g$$ at $$400ms^{-1}$$. What is the recoil speed of the rifle into his shoulder?
Hint: Think carefully about the initial speed of both the rifle and bullet before firing.
\begin{aligned} \text{(K)} \newline \text{(U)} \newline \text{(F)} \newline \text{(S+S)} \end{aligned}
#### Whakatika
Both the bullet and the rifle are stationary beforehand. Therefore $$u_{1}$$ and $$u_{2} = 0$$.
\begin{aligned} p_{i} &= p_{f} \text{ assuming }F_{net} = 0 \newline m_{1}u_{1} + m_{2}u_{2} &= m_{1}v_{1} + m_{2}v_{2} \newline m_{1} \times 0 + m_{2} \times 0 &= m_{1}v_{1} + m_{2}v_{2} && u_{1}=u_{2}=0 \newline 0 &= (4 \times v_{1}) + (0.02 \times 400) \newline 0 &= 4v_{1} + 8 \newline -8 &= 4v_{1} \newline \frac{-8}{4} &= v_{1} = -2ms^{-1} \end{aligned}
### Pātai Tekau (Q10): Trains
A shunting car with mass $$5\times10^{4}kg$$ is moving at $$3ms^{-1}$$ bumps into a stationary car with mass $$3\times10^{4}kg$$. They join together in the collision. Calculate their combined speed after the collision. Start by drawing a diagram illustrating before and after the collision.
### Pātai Tekau mā Tahi (Q11)
A moving car collides with a stationary van. The car has mass $$950kg$$ and the van has mass $$1700kg$$. The car is travelling $$8.0ms^{-1}$$ before the collision and $$2.0ms^{-1}$$ after the collision.
1. What quantity is conserved during the collision? (A)
2. Calculate the size and direction of the car’s momentum change. (E)
3. Calculate the speed of the van immediately after the collision. (M)
4. If the average force that the van exerts on the car is $$3800N$$, calculate how long the collision lasts. (A)
5. The driver has a bag resting on the passenger seat during the collision. Explain why the bag fell to the floor during the collision. (E)
6. The front of modern cars are designed to crumple upon impact. Explain why this is beneficial to people in the car. (E)
## 1. What quantity is conserved during the collision? (A)
Momentum is conserved
1. Calculate the size and direction of the car’s momentum change. (E)
\begin{aligned} & \Delta p = p_{f} - p_{i} \newline & \Delta p = mv_{f} - mv_{i} && \text{substituting } p = mv\newline & \Delta p = (950 \times 2) - (950 \times 8) && \text{substitute values} \newline & \Delta p = 1900 - 7600 \newline & \Delta p = -5700Nm \end{aligned}
1. Calculate the speed of the van immediately after the collision. (M)
The van is stationary before the collision. $$u_{2}=0$$
\begin{aligned} & m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} \newline & m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2} && u_{2}=0 \newline & 950 \times 8 = (950 \times 2) + (1700 \times v_{2}) && \text{substitute values} \newline & 7600 = 1900 + 1700 v_{2} \newline & 5700 = 1700 v_{2} && \text{subtract 1900 from both sides} \newline & v_{2} = \frac{5700}{1700} && \text{divide through by 1700} \newline & v_{2} = 3.35ms^{-1} \end{aligned}
1. If the average force that the van exerts on the car is $$3800N$$, calculate how long the collision lasts. (A)
The van exerts a force on the car that slows it down. Therefore the force is against the direction of motion, and therefore negative.
\begin{aligned} & F \Delta t = \Delta p \newline & -3800 \times \Delta t = -5700 && \text{substitute values} \newline & \Delta t = \frac{-5700}{-3800} && \text{divide through by -3800} \newline & \Delta t = 1.5s \end{aligned}
1. The driver has a bag resting on the passenger seat during the collision. Explain why the bag fell to the floor during the collision. (E)
During the collision the van exerts a force upon the car to slow it down. This force acts over a duration of 1.5s. For a force to act upon an object it needs to be attached to or part of the object. A seatbelt attaches the driver to the car allowing the force to act upon the driver through the seatbelt to change their momentum.
Because the bag is not attached to the car the force cannot change its momentum, and it continues to move forward even as the car slows down. This causes it to leave the seat and fall to the ground.
1. The front of modern cars are designed to crumple upon impact. Explain why this is beneficial to people in the car. (E)
Whether or not a car has a crumple zone, the same change in momentum will occur because of the same change in motion. Using the impulse equation $$F \Delta t = \Delta p$$ we can see that by increase the time taken for the collision to occur, a smaller force is necessary. This smaller force means that less force affects the driver of the car and therefore reduces the risk of injury.
### Pātai Tekau mā Rua (Q12)
A Morris Minor car ($$m=750kg$$) is travelling at $$30ms^{-1}$$ and collides head on with a Mercedes Benz car ($$m=1600kg$$) travelling at $$20ms^{-1}$$ in the opposite direction. The two cars lock together in the crash.
1. Calculate the total momentum
2. Calculate the velocity on the combined wreckage after the collision
3. Would the wreckage keep moving at this velocity? Why or why not?
#### Whakatika
1. Calculate the total momentum
Because momentum is conserved, we can calculate the total momentum before OR after the collision.
\begin{aligned} & \sum p = p_{1} + p_{2} && \text{sum momentum of two objects} \newline & \sum p = m_{1}u_{1} + m_{2}u_{2} && \text{subsititute } p = mv \newline & \sum p = (750 \times 30) + (1600 \times -20) && \text{Benz has negative velocity (opposite direction)} \newline & \sum p = 22500 - 32000 \newline & \sum p = -9500Nm \end{aligned}
1. Calculate the velocity on the combined wreckage after the collision
\begin{aligned} & \sum p_{i} = \sum p_{f} \newline & p_{1} + p_{2} = p_{3} && \text{vehicles locked together afterwards} \newline & m_{1}u_{1} + m_{2}u_{2} = m_{3}v_{3} && \text{substitute } p = mv \newline & -9500 = 2350v_{3} && \text{using total momentum from previous} \newline & v_{3} = \frac{-9500}{2350} && \text{divide through by 2350} \newline & v_{3} = -4.04ms^{-1} \end{aligned}
1. Would the wreckage keep moving at this velocity? Why or why not?
In the real world, no, because energy will be lost to the surroundings through heat and sound due to friction on the road, air resistance and engine friction.
### Whakawai
• Mahi Kāinga Book: Q57-58
• Textbook: Momentum & Impulse Activity | 4,920 | 16,232 | {"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} | 4.15625 | 4 | CC-MAIN-2021-25 | latest | en | 0.835413 |
https://www.physicsforums.com/threads/interpreting-the-definition-of-tensors.962819/ | 1,674,985,070,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00176.warc.gz | 948,697,224 | 19,711 | # Interpreting The Definition of Tensors
• I
VuIcan
Hello, I've just been slightly unsure of something and would like to get secondary confirmation as I've just begun a book on tensor analysis. I would also preface this by saying my linear algebra is somewhat rusty. Suppose you have the inertia tensor in some unprimed coordinate system such that ##\mathbf{\widetilde{I}}##, then we know definitionally that this second-rank tensor will transform as such to into some primed coordinate system(where ##\Lambda## corresponds to the transformation matrix from the unprimed to the primed coordinate system):
$$\widetilde{I}' = \Lambda \widetilde{I} \Lambda ^{\dagger}$$
Now, if one were to apply some vector stimulus in the primed coordinate system from the right, would it be correct to think of this vector as firstly being transformed into the unprimed coordinate system (since the adjoint is equivalent to the inverse within this context), then being directionally altered by the inertia tensor in the unprimed coordinate system, then finally being transformed back into the primed coordinate system by the final matrix? I feel like I'm misunderstanding something fundamental however.
-Vulcan
Staff Emeritus
Homework Helper
Gold Member
Personally, I think it is better not to think about vectors and tensors as transforming objects. A vector (or any other tensor) does not depend on the coordinate system. The direction to the Moon is physically the same regardless of what coordinates you choose. What does depend on the basis you choose is the components of the vector and the vector (and tensor) components therefore change between systems. A rank two tensor is (among other things) a linear map from a vector to a vector. This mapping does not happen "in a coordinate system". However, you can express that linear map in a given basis by giving its components relative to that basis, which you can represent using a matrix.
FactChecker
VuIcan
Personally, I think it is better not to think about vectors and tensors as transforming objects.
But one needs to understand the mathematical operations being performed to be able to understand how said mathematical entity remains unchanged through a sequence of operations. Yes, I'm fully aware, the magnitude/direction of a vector under these types of linear transformation doesn't change, but I just can't seem to follow the math in this case(or at least I'm paranoid, I'm not following it correctly). Do you think my interpretation of the sequence of the operations is correct?
Also, aren't tensors defined by the way they transform? So wouldn't understanding tensors require one to think of them as transforming objects(component wise) under some imposed coordinate system?
That said, I do have an additional question as someone who is new to the idea of tensors, I've only been able "understand" 2nd-rank tensors when they've been defined defined operationally, like the inertia tensor/Maxwell tensor. Do you have any helpful resources that may help me attain some geometric understanding of 2nd-rank tensors in the same manner that everybody has the geometric picture of vectors as something with a "magnitude and direction"/arrow in 3d-space?
Thanks again.
Staff Emeritus
Homework Helper
Gold Member
Also, aren't tensors defined by the way they transform?
No. It is a common way to introduce them, but I find it misleading at best and students never really seem to grasp things when introduced this way. (Also, again note that the components transform, not the tensor itself.) A type (m,n) tensor is a linear map from ##n## copies of a vector space to ##m## copies of the vector space, the transformation properties of the components follow directly from this when you change basis on the vector space.
That said, I do have an additional question as someone who is new to the idea of tensors, I've only been able "understand" 2nd-rank tensors when they've been defined defined operationally, like the inertia tensor/Maxwell tensor. Do you have any helpful resources that may help me attain some geometric understanding of 2nd-rank tensors in the same manner that everybody has the geometric picture of vectors as something with a "magnitude and direction"/arrow in 3d-space?
A vector is a linear combination of the basis vectors. A rank 2 tensor is a linear combination of pairs of basis vectors ##\vec e_i \otimes \vec e_j##. In order to know how a rank 2 tensor you would need 9 numbers (in 3D space). One possibility is drawing the three vectors that the basis vectors are mapped to by the tensor.
VuIcan
A type (m,n) tensor is a linear map from nnn copies of a vector space to mmm copies of the vector space, the transformation properties of the components follow directly from this when you change basis on the vector space.
I apologize for my ignorance , but I'm not sure what you mean by copies of a vector space?
A vector is a linear combination of the basis vectors. A rank 2 tensor is a linear combination of pairs of basis vectors ##\vec e_i \otimes \vec e_j##. In order to know how a rank 2 tensor you would need 9 numbers (in 3D space). One possibility is drawing the three vectors that the basis vectors are mapped to by the tensor.
So taking the inertia tensor as an example:
$$\mathbf{\widetilde{I}} = \int dm \left[ <r|r> \mathbf{1} - |r><r| \right]$$
How does the idea of it being a linear combination of outer products of basis vectors apply? Sorry if I'm being a bit clueless.
The three vectors that the basis vectors are mapped to in this scenario would be the column vectors? Correct?
Thanks for your patience, I think I'm almost there : )
Gold Member
I apologize for my ignorance , but I'm not sure what you mean by copies of a vector space?
You have a k-linear map taking inputs in the product ## V \times V \times...\times V \times V^{*} \times ...\times V^{*} ## ( though the factors can appear in mixed order) so that the tensor is linear in each factor separately.
Homework Helper
Gold Member
In preparation of @WWGD's comment, this might help...
• A real function of two real variables $f(u,v)$ can be regarded as a map from
ordered pairs of real values $(u,v)\in \cal R\times R$ to the reals $\cal R$.
In $\cal R\times R$, each "$\cal R$" can be thought of as an independent copy of $\cal R$.
• A dot-product of two vectors $\vec u \cdot \vec v= g(\vec u,\vec v)=g_{ab} u^a v^b$ can be regarded as a map from
ordered pairs of vectors $(\vec u,\vec v)\in V\times V$ to the reals $\cal R$.
In $V\times V$, each "$V$" can be thought of as an independent copy of $V$.
In standard notation, a vector has 1 up-index (hence, $u^a$) and corresponds to a column-vector in matrix component notation.
In "bra-ket" notation, this is like the ket $\left| u \right\rangle$.
(The lower-indices on $g_{ab}$ means that it accepts two vectors to produce a scalar.)
(** I provide alternate notations to help with intuition... be careful not to mix notations. **)
This dot-product is actually a "bi"-linear map since it is linear is each of the 2 arguments:
$g(\vec u+\vec p,\vec v)=g(\vec u,\vec v)+g(\vec p,\vec v)$ and $g(A\vec u,\vec v)=Ag(\vec u,\vec v)$
$g(\vec u,\vec v+\vec q)=g(\vec u,\vec v)+g(\vec u,\vec q)$ and $g(\vec u,B\vec v)=Bg(\vec u,\vec v)$
or generally by "FOIL"ing
\begin{align*} g(A\vec u+\vec p,B\vec v+\vec q) &=g(A\vec u,B\vec v)+g(A\vec u,\vec q)+g(\vec p, B\vec v)+g(\vec p,\vec q)\\ &=ABg(\vec u,\vec v)+Ag(\vec u,\vec q)+Bg(\vec p, \vec v)+g(\vec p,\vec q) \end{align*}
In @Orodruin's terms,
• the $g_{ab}$ is a type (m=0,n=2)-tensor [with 0 up-indices and 2 down-indices]
because it takes
"n=2 copies of the vector space $V$" (that is, $V\times V$)
to
"m=0 copies of the vector space $V$" (a.k.a. the scalars (usu.) $\cal R$).
(that is, input an ordered pair of vectors and output a scalar: $V\times V \longrightarrow \cal R$).
• An example of a type (m=0,n=1)-tensor [with 0 up-indices and 1 down-index]
is the "x-component operation in Euclidean space" $\color{red}{(\hat x\cdot \color{black}{\square})}=\color{red}{\hat x_b}$
(where: $\color{red}{\hat x_b}=\color{red}{g_{ab} \hat x^a}$ and $\color{red}{\hat x_b} \hat x^b =\color{red}{g_{ab} \hat x^a} \hat x^b =1$).
Thus,$$u_{\color{red}{x}}=\color{red}{(\hat x\cdot \color{black}{\vec u})}=\color{red}{\hat x_b} u^b$$ because it takes
"n=1 copy of the vector space $V$" to "m=0 copies of the vector space $V$" (the scalars $\cal R$).
(that is, input a vector and output a scalar: $V \longrightarrow \cal R$)
• An example of a type (m=1,n=1)-tensor [with 1 up-index and 1 down-index]
is the "vector x-component operation in Euclidean space" $\color{red}{\hat x (\hat x\cdot \color{black}{\square})}=\color{red}{\hat x^a \hat x_b}$
Thus, $$\color{red}{\hat x (\color{black}{u}_x)}=\color{red}{\hat x (\hat x\cdot \color{black}{\vec u})}=\color{red}{\hat x^a \hat x_b} u^b$$ because it takes
"n=1 copy of the vector space $V$" to "m=1 copy of the vector space $V$"
(that is, input a vector and output a vector: $V \longrightarrow V$).
In matrix notation, this is a square-matrix.
(This map could be called a transformation on the vector space
[specifically, a projection operator since $\color{red}{\hat x^a \hat x_b} \ \color{green}{\hat x^b \hat x_c}=\color{blue}{\hat x^a \hat x_c}$]).
In @WWGD's terms, $V^*$ is called a dual vector space and its elements have 1 down-index (like $z_a$)
and correspond to row-vectors in matrix component notation.
In "bra-ket" notation, this is like the bra $\left\langle z \right |$.
• The type (m=1,n=1) tensor $\color{red}{\hat x^a \hat x_b}$ can also be interpreted as a bilinear map $\color{red}{h(\color{blue}{\square} ,\color{black}{\square} )}$
that takes
an ordered pair: first from the "m=1 copy of the dual-vector space $V^*$" and second from the "n=1 copy of the vector space $V$"
to the reals: $V^* \times V \longrightarrow \cal R$,
as in $\color{red}{h(\color{blue}{\hat x_a} ,\color{black}{u^b} )}=\color{blue}{\hat x_a}\color{red}{\hat x^a \hat x_b} u^b=u_x$
Hopefully, this is a useful starting point.
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http://edenwhitemusic.com/6rm5l89/viewtopic.php?e57fe7=rank-of-product-of-matrices | 1,610,756,358,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00457.warc.gz | 33,621,031 | 9,479 | Thus, we have proved that the space spanned by the columns of is the space 38 Partitioned Matrices, Rank, and Eigenvalues Chap. If A and B are two equivalent matrices, we write A ~ B. coincide. thatThen,ororwhere we Nov 15, 2008 #1 There is a remark my professor made in his notes that I simply can't wrap my head around. Proving that the product of two full-rank matrices is full-rank Thread starter leden; Start date Sep 19, 2012; Sep 19, 2012 #1 leden. is full-rank. ST is the new administrator. . (b) If the matrix B is nonsingular, then rank(AB)=rank(A). 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A = ( 1 0 ) and B ( 0 ) both have rank 1, but their product, 0, has rank 0 ( 1 ) Let us transform the matrix A to an echelon form by using elementary transformations. means that any is full-rank. so they are full-rank. Proposition (adsbygoogle = window.adsbygoogle || []).push({}); Give the Formula for a Linear Transformation from $\R^3$ to $\R^2$, Find a Nonsingular Matrix Satisfying Some Relation, Finitely Generated Torsion Module Over an Integral Domain Has a Nonzero Annihilator, How to Use the Cayley-Hamilton Theorem to Find the Inverse Matrix. matrices. Rank of product of matrices with full column rank Get link; Facebook; Twitter; Pinterest and and that spanned by the columns of , such Add the first row of (2.3) times A−1 to the second row to get (A B I A−1 +A−1B). is full-rank and square, it has Denote by . full-rank matrices. Notify me of follow-up comments by email. is full-rank, it has Let Let two matrices are equal. Aug 2009 130 16. In this section, we describe a method for finding the rank of any matrix. is less than or equal to The Adobe Flash plugin is needed to view this content. In mathematics, the Kronecker product, sometimes denoted by ⊗, is an operation on two matrices of arbitrary size resulting in a block matrix. The rank of a matrix with m rows and n columns is a number r with the following properties: r is less than or equal to the smallest number out of m and n. r is equal to the order of the greatest minor of the matrix which is not 0. whose dimension is canonical basis). This video explains " how to find RANK OF MATRIX " with an example of 4*4 matrix. If have just proved that any vector Remember that the rank of a matrix is the pr.probability matrices st.statistics random-matrices hadamard-product share | cite | improve this question | follow | In all the definitions in this section, the matrix A is taken to be an m × n matrix over an arbitrary field F. are equal because the spaces generated by their columns coincide. Let $V$ be the vector space over $\R$ of all real valued functions defined on the interval $[0,1]$. So if $n<\min(m,p)$ then the product can never have full rank. vectors. two full-rank square matrices is full-rank. Thus, any vector By Catalin David. rank of the Enter your email address to subscribe to this blog and receive notifications of new posts by email. The rank of a matrix is the order of the largest non-zero square submatrix. The next proposition provides a bound on the rank of a product of two Let A be an m×n matrix and B be an n×lmatrix. dimension of the linear space spanned by its columns (or rows). coincide. :where an and Keep in mind that the rank of a matrix is Rank and Nullity of a Matrix, Nullity of Transpose, Quiz 7. The Kronecker product is to be distinguished from the usual matrix multiplication, which is an entirely … As a consequence, there exists a Thus, the space spanned by the rows of Apparently this is a corollary to the theorem If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ). Advanced Algebra. do not generate any vector It is left as an exercise (see In geometrical terms the rank of a matrix is the dimension of the image of the associated linear map (as a vector space). matrix and coincide. This lecture discusses some facts about vectors. vector matrix. Learn how your comment data is processed. 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Find a Basis of the Range, Rank, and Nullity of a Matrix, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, Prove a Given Subset is a Subspace and Find a Basis and Dimension, True or False. He even gave a proof but it made me even more confused. do not generate any vector can be written as a linear combination of the columns of is called a Gram matrix. We can define rank using what interests us now. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to … it, please check the previous articles on Types of Matrices and Properties of Matrices, to give yourself a solid foundation before proceeding to this article. , whose dimension is Then prove the followings. be a columns that span the space of all The maximum number of linearly independent vectors in a matrix is equal to the … is full-rank, is a Proposition Moreover, the rows of If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). for A matrix obtained from a given matrix by applying any of the elementary row operations is said to be equivalent to it. satisfied if and only Rank of the Product of Matrices AB is Less than or Equal to the Rank of A Let A be an m × n matrix and B be an n × l matrix. Let such . : Rank(AB) can be zero while neither rank(A) nor rank(B) are zero. "Matrix product and rank", Lectures on matrix algebra. matrix). Rank of a Matrix. This is possible only if Determinant of product is product of determinants Dependencies: A matrix is full-rank iff its determinant is non-0; Full-rank square matrix is invertible; AB = I implies BA = I; Full-rank square matrix in RREF is the identity matrix; Elementary row operation is matrix pre-multiplication; Matrix multiplication is associative are full-rank. that is, only Note. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … is preserved. . matrix and its transpose. identical to that of the previous proposition. matrix and an Proposition We are going is the rank of Multiplication by a full-rank square matrix preserves rank, The product of two full-rank square matrices is full-rank. vector of coefficients of the linear combination. Your email address will not be published. As a consequence, also their dimensions coincide. and that spanned by the rows of Therefore, there exists an If vector of coefficients of the linear combination. two Matrices. the space generated by the columns of column vector rank. thenso Let Taboga, Marco (2017). linearly independent. How do you prove that the matrix C = AB is full-rank, as well? The number of non zero rows is 2 ∴ Rank of A is 2. ρ (A) = 2. the space spanned by the rows of , :where Author(s): Heinz Neudecker; Satorra, Albert | Abstract: This paper develops a theorem that facilitates computing the degrees of freedom of an asymptotic χ² goodness-of-fit test for moment restrictions under rank deficiency of key matrices involved in the definition of the test. Proposition Say I have a mxn matrix A and a nxk matrix B. Find the rank of the matrix A= Solution : The order of A is 3 × 3. To see this, note that for any vector of coefficients Another important fact is that the rank of a matrix does not change when we is the space Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … In a strict sense, the rule to multiply matrices is: "The matrix product of two matrixes A and B is a matrix C whose elements a i j are formed by the sums of the products of the elements of the row i of the matrix A by those of the column j of the matrix B." Matrices. University Math Help. How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Express a Vector as a Linear Combination of Other Vectors, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Find a Basis for the Subspace spanned by Five Vectors, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces. All Rights Reserved. multiply it by a full-rank matrix. and As a consequence, also their dimensions (which by definition are Thread starter JG89; Start date Nov 18, 2009; Tags matrices product rank; Home. that givesis linearly independent rows that span the space of all haveThe denotes the (1) The product of matrices with full rank always has full rank (for example using the fact that the determinant of the product is the product of the determinants) (2) The rank of the product is always less than or equalto the minimum rank of the matrices being multiplied. are be two Thus, the rank of a matrix does not change by the application of any of the elementary row operations. The list of linear algebra problems is available here. is a linear combination of the rows of We now present a very useful result concerning the product of a non-square A row having atleast one non -zero element is called as non-zero row. can be written as a linear combination of the rows of 2 as a product of block matrices of the forms (I X 0 I), (I 0 Y I). vectors (they are equivalent to the Add to solve later Sponsored Links Advanced Algebra. [Note: Since column rank = row rank, only two of the four columns in A — c … J. JG89. Therefore, by the previous two . Note that if A ~ B, then ρ(A) = ρ(B) Suppose that there exists a non-zero vector is no larger than the span of the rows of Proof: First we consider a special case when A is a block matrix of the form Ir O1 O2 O3, where Ir is the identity matrix of dimensions r×r and O1,O2,O3 are zero matrices of appropriate dimensions. is the Then, the product vector). In particular, we analyze under what conditions the Proposition Let and be two full-rank matrices. How to Diagonalize a Matrix. such thatThusThis vector then. . The product of two full-rank square matrices is full-rank An immediate corollary of the previous two propositions is that the product of two full-rank square matrices is full-rank. for any vector of coefficients This site uses Akismet to reduce spam. which implies that the columns of is no larger than the span of the columns of Rank of Product Of Matrices. Example 1.7. vector of coefficients of the linear combination. 5.6.4 Recapitulation -th linearly independent of all vectors with coefficients taken from the vector that are linearly independent and Any vector propositionsBut Any As a consequence, the space Then, their products and are full-rank. matrix and entry of the The rank of a matrix can also be calculated using determinants. combinations of the columns of , Since vector and a a square is less than or equal to Then, The space Let . spanned by the columns of is full-rank, it has less columns than rows and, hence, its columns are Since the dimension of (a) rank(AB)≤rank(A). Forums. :where Since haveNow, that can be written as linear combinations of the rows of Sum, Difference and Product of Matrices; Inverse Matrix; Rank of a Matrix; Determinant of a Matrix; Matrix Equations; System of Equations Solved by Matrices; Matrix Word Problems; Limits, Derivatives, Integrals; Analysis of Functions vector (being a product of a Being full-rank, both matrices have rank An immediate corollary of the previous two propositions is that the product of The Intersection of Bases is a Basis of the Intersection of Subspaces, A Matrix Representation of a Linear Transformation and Related Subspaces, A Basis for the Vector Space of Polynomials of Degree Two or Less and Coordinate Vectors, Compute and Simplify the Matrix Expression Including Transpose and Inverse Matrices, Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$. the dimension of the space generated by its rows. the space spanned by the rows of In other words, we want to get a matrix in the above form by per-forming type III operations on the block matrix in (2.3). then. See the … be a C. Canadian0469. full-rank matrix with . and matrix and be a Yes. As a consequence, the space if. if equal to the ranks of It is a generalization of the outer product from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis. Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. then. The matrix (a) rank(AB) ≤ rank(A). Rank. thatThusWe with coefficients taken from the vector vector of coefficients of the linear combination. ifwhich if . Denote by is the rank of :where Proposition Then, their products We are going to prove that and Prove that if matrices product rank; Home. the spaces generated by the rows of is the is the https://www.statlect.com/matrix-algebra/matrix-product-and-rank. (b) If the matrix B is nonsingular, then rank(AB) = rank(A). the exercise below with its solution). is a linear combination of the rows of Below you can find some exercises with explained solutions. University Math Help. be the space of all be a Furthermore, the columns of 7 0. This implies that the dimension of a full-rank We can also Theorem rank(At) = rank(A). and is full-rank, matrix products and their , , This website’s goal is to encourage people to enjoy Mathematics! That means,the rank of a matrix is ‘r’ if i. We can also , Most of the learning materials found on this website are now available in a traditional textbook format. matrix. to prove that the ranks of Column Rank = Row Rank. Forums. can be written as a linear combination of the columns of be a is an is full-rank, , Problems in Mathematics © 2020. Thus, the only vector that This implies that the dimension of matrix and Save my name, email, and website in this browser for the next time I comment. ) If $\min(m,p)\leq n\leq \max(m,p)$ then the product will have full rank if both matrices in the product have full rank: depending on the relative size of $m$ and $p$ the product will then either be a product of two injective or of two surjective mappings, and this is again injective respectively surjective. Finally, the rank of product-moment matrices is easily discerned by simply counting up the number of positive eigenvalues. Thus, any vector .
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### Re: 10089 - Repackaging
Posted: Sun Aug 23, 2009 11:33 pm
By the way, try problem 802 (Lead or Gold) after this one, it's similar.
### Re: 10089 - Repackaging
Posted: Tue Aug 25, 2009 1:49 am
mf wrote: Well, isn't it obvious? A linear combinations of vectors (s2 - s1, s3 - s1) is zero if and only if the same linear combination of vectors (s1, s2, s3) is a vector with three equal components, namely, all equal to this linear combination of s1's. It's just simple algebra and manipulations of sums. It doesn't deserve to be called a theorem, or even lemma. The proper term for this is "left as an exercise to the reader".
Now, to answer why it's useful. To check that you can make a zero vector, you basically need to check that your collection of vectors spans an angle of 180 degrees or more, i.e. there doesn't exist a hyperplane w*x=0 through the origin ("*" is the dot product of vectors here), such that all the given vectors lie strictly to one side of it.
Necessity: if there exists a vector w such that w*x_i > 0 for all i, then for any non-negative coefficients c (not all zeros), the vector s = \sum c_i x_i satisfies: w*s = w * \sum (c_i x_i) = \sum (c_i w*x_i) = \sum (c_i * <something strictly positive>) > 0, and therefore s can't be a zero vector.
Sufficiency: if hypothesis is satistisfied, then you can always find a linear combination with rational non-negative coefficients (not all zeroes) of given vectors, which will sum to zero vector. I've tried writing a proof for this fact, but it was getting too long for a single post... You better try to draw some pictures and get a feel of it for yourself, instead.
And once you got a combination with rational coefficients, you can always turn it into integers by multiplying all coefficients by LCM of their denominators.
let the 3 packages be (p11,p12,p13 ), (p21,p22,p23) and (p31,p32,p33), now we have to find X,Y,Z such that
X*p11 + Y*p21 + Z*p31 = k
X*p12 + Y*p22 + Z*p32 = k
X*p13 + Y*p23 + Z*p33 = k
and X+Y+Z>0 and k > 0
P=
{
[ p11 p21 p31 ]
[ p12 p22 p32 ]
[ p13 p23 p33 ]
}
Q={ X,Y,Z}
p*Q = k*I
mf wrote:
A linear combinations of vectors (s2 - s1[i], s3[i] - s1[i]) is zero if and only if the same linear combination of vectors (s1[i], s2[i], s3[i]) is a vector with three equal components, namely, all equal to this linear combination of s1[i]'s. It's just simple algebra and manipulations of sums. It doesn't deserve to be called a theorem, or even lemma
I don't understand this. Can you please explain , how did you reduce the problem to this and formulated it.
### Re: 10089 - Repackaging
Posted: Tue Aug 25, 2009 2:29 am
tryit1 wrote:
mf wrote: A linear combinations of vectors (s2 - s1, s3 - s1) is zero if and only if the same linear combination of vectors (s1, s2, s3) is a vector with three equal components, namely, all equal to this linear combination of s1's. It's just simple algebra and manipulations of sums. It doesn't deserve to be called a theorem, or even lemma
I don't understand this. Can you please explain , how did you reduce the problem to this and formulated it.
Let c[1], ..., c[n] be any real numbers (or any integer numbers, or any rational numbers - doesn't really matter)
Define these sums:
We want to show that S21 = S31 = 0 if and only if S1 = S2 = S3.
Proof:
First, observe that S21 = S2 - S1 due to distributive law of multiplication and elementary properties of sums:
Analogously, S31 = S3 - S1.
S1 = S2 = S3, iff
0 = S2 - S1 = S3 - S1, (I have subtracted S1 from all sides of equality), iff
S2 - S1 = 0 and S3 - S1 = 0, iff
S21 = 0 and S31 = 0. QED.
Concrete Mathematics is the right book for learning all this stuff.
### Re: 10089 - Repackaging
Posted: Tue Aug 25, 2009 9:17 am
I knew that one. Didn't see it . It is just subtraction of 2 vectors and then making the independent components equal to 0.
mf wrote:Now, to answer why it's useful. To check that you can make a zero vector, you basically need to check that your collection of vectors spans an angle of 180 degrees or more, i.e. there doesn't exist a hyperplane w*x=0 through the origin ("*" is the dot product of vectors here), such that all the given vectors lie strictly to one side of it.
Does this mean , given 2D points on a plane, you have to find a subset of them such that they form a closed polygon ?
For that you take 2 points on 1 side of x-axis and third point on another side of x axis. So if we have a point (x,y), to reduce the x-component , we need to have an angle of 90" between them , say some point (x',y) and to reduce y-component of (x,y) , you need to have angle of (x,y') 90 between them. So a total of 180". If it is more , it can be scaled right ? I'm little confused.
What if we have points like (1,2) ,(3,4) and (5,-6) , their angle is greater than 180" but i don't think we can form linear combination because x components will increase.
### Re: 10089 - Repackaging
Posted: Tue Aug 25, 2009 12:29 pm
No, that's not what I mean. Sorry for confusing you with complicated words like "hyperplane", I think I over-generalized the things here (to n dimensions)...
What I mean is: if there exists a line through the origin, that is a line of form ax + by = 0, such that all given vectors lie strictly to one its side (a*x + b*y > 0 for each vector (x, y)), then the answer is "No", you can't get a zero vector.
In your example, such a line would be the OY axis, because all your points lie to the right of it (their x coordinate is > 0).
### Re: 10089 - Repackaging
Posted: Tue Aug 25, 2009 1:01 pm
I've a O(n) after sorting. If the angle between 2 consecutive ,x{n} and x{n+1} > 180 , then we report false because we can translate the axis to x{n+1} and it will be < 180. Good stuff. Thanks mf for teaching me this.
Generalising this to n - dimensions is a good. All i understand is reduce it to n-1 dimension. See if it can be put in a half plane (n-2 dimenison) through origin. If yes bad.
For a 4 D package, we have 3 D points and 2 D planes through origin. Check for angles in all xy plane,yz plane and xz plane. Is that correct ? Can it be done any faster ?
### Re: 10089 - Repackaging
Posted: Tue Aug 25, 2009 1:32 pm
No, I don't know how to extend the angle check to 3D and beyond.
But convex hull check generalizes very nicely. Looks like our hyperplane exists if the origin is outside of the convex hull of given vectors. And there's a simple fact about convex hulls: in 2 dimensions, convex hull of a set of points is the union of all possible triangles with vertices at given points (I've learned this from problem 361, btw!); in 3 dimensions - union of all possible tetrahedrons; and in K dimensions - union of all possible simplices with vertices at given points.
So this gives an O(N^(K+1) K^3) algorithm: for each possible subset of points of size K+1, check if the origin is inside a K-dimensional simplex with corners at these K+1 points. If, for some subset, it is, then yes, we can construct a zero vector. You can check whether a point is inside a K-dimensional simplex using determinants in O(K^3), I think.
### Re: 10089 - Repackaging
Posted: Mon Nov 26, 2012 5:28 am
I still can't figure out why three vectors with included angles less than PI can be linearly combined to reach (0,0) using rational coefficients. Can anyone give me some more hints?
Thanks!
### Re: 10089 - Repackaging
Posted: Sat Dec 07, 2013 11:57 pm
Can someone tell me why the output for:
Code: Select all
3
1 2 3
2 4 6
3 6 9
0
is
Code: Select all
Yes
? O_o
### Re: 10089 - Repackaging
Posted: Tue Dec 10, 2013 12:32 am
AC output for that input is No
### Re: 10089 - Repackaging
Posted: Tue Dec 10, 2013 9:23 pm
It's obvious but I'm interested why UVa Toolkit said "Yes".
What's more - this error is for every input like:
Code: Select all
x
a b c
2a 2b 2c
...
na nb nc
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# College Physics
Science and Technology
## Newton’s Second Law of Motion: Concept of a System
Tác giả: OpenStaxCollege
Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned.
First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is an acceleration. Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration.
Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct—an external force acts from outside the system of interest. For example, in [link](a) the system of interest is the wagon plus the child in it. The two forces exerted by the other children are external forces. An internal force acts between elements of the system. Again looking at [link](a), the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. (The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of the system before you can determine which forces are external. Sometimes the system is obvious, whereas other times identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept will be revisited many times on our journey through physics.
Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a system. This assumption has been verified experimentally and is illustrated in [link]. In part (a), a smaller force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight $\mathbf{\text{w}}$ and the support of the ground $\mathbf{\text{N}}$, and the horizontal force $\mathbf{\text{f}}$ represents the force of friction. These will be discussed in more detail in later sections. For now, we will define friction as a force that opposes the motion past each other of objects that are touching. [link](b) shows how vectors representing the external forces add together to produce a net force, ${\mathbf{\text{F}}}_{\text{net}}$.
To obtain an equation for Newton’s second law, we first write the relationship of acceleration and net external force as the proportionality
$\mathbf{\text{a}}\propto {\mathbf{\text{F}}}_{\text{net}},$
where the symbol $\propto$ means “proportional to,” and ${\mathbf{\text{F}}}_{\text{net}}$ is the net external force. (The net external force is the vector sum of all external forces and can be determined graphically, using the head-to-tail method, or analytically, using components. The techniques are the same as for the addition of other vectors, and are covered in Two-Dimensional Kinematics.) This proportionality states what we have said in words—acceleration is directly proportional to the net external force. Once the system of interest is chosen, it is important to identify the external forces and ignore the internal ones. It is a tremendous simplification not to have to consider the numerous internal forces acting between objects within the system, such as muscular forces within the child’s body, let alone the myriad of forces between atoms in the objects, but by doing so, we can easily solve some very complex problems with only minimal error due to our simplification
Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. And indeed, as illustrated in [link], the same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The proportionality is written as
$\mathbf{\text{a}}\propto \frac{1}{m}$
where $m$ is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is exactly linearly proportional to the net external force.
It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton's second law of motion.
Although these last two equations are really the same, the first gives more insight into what Newton’s second law means. The law is a cause and effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is completely based on experimental verification.
# Units of Force
${\mathbf{\text{F}}}_{\text{net}}=m\mathbf{\text{a}}$ is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of $1{\text{m/s}}^{2}$. That is, since ${\mathbf{\text{F}}}_{\text{net}}=m\mathbf{\text{a}}$,
$\text{1 N}=\text{1 kg}\cdot {\text{m/s}}^{2}.$
While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb.
# Weight and the Gravitational Force
When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight $\mathbf{\text{w}}$. Weight can be denoted as a vector $\mathbf{\text{w}}$ because it has a direction; down is, by definition, the direction of gravity, and hence weight is a downward force. The magnitude of weight is denoted as $w$. Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration $g$. Using Galileo’s result and Newton’s second law, we can derive an equation for weight.
Consider an object with mass $m$ falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude $w$. Newton’s second law states that the magnitude of the net external force on an object is ${F}_{\text{net}}=\text{ma}$.
Since the object experiences only the downward force of gravity, ${F}_{\text{net}}=w$. We know that the acceleration of an object due to gravity is $g$, or $a=g$. Substituting these into Newton’s second law gives
When the net external force on an object is its weight, we say that it is in free-fall. That is, the only force acting on the object is the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air acting on the object.
The acceleration due to gravity $g$ varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth’s surface. On the Moon, for example, the acceleration due to gravity is only $1.67\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$. A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon.
The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of “weightlessness” and “microgravity,” they are really referring to the phenomenon we call “free-fall” in physics. We shall use the above definition of weight, and we will make careful distinctions between free-fall and actual weightlessness.
It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter (how much “stuff”) and does not vary in classical physics, whereas weight is the gravitational force and does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our “weight” in kilograms, but never in the correct units of newtons.
What Acceleration Can a Person Produce when Pushing a Lawn Mower?
Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the mower is 24 kg. What is its acceleration?
Strategy
Since ${\mathbf{\text{F}}}_{\text{net}}$ and $m$ are given, the acceleration can be calculated directly from Newton’s second law as stated in ${\mathbf{\text{F}}}_{\text{net}}=m\mathbf{\text{a}}$.
Solution
The magnitude of the acceleration $a$ is $a=\frac{{F}_{\text{net}}}{m}$. Entering known values gives
$a=\frac{\text{51 N}}{\text{24 kg}}$
Substituting the units $\text{kg}\cdot {\text{m/s}}^{2}$ for N yields
$a=\frac{\text{51 kg}\cdot {\text{m/s}}^{2}}{\text{24 kg}}=\text{2.1 m}{\text{/s}}^{2}.$
Discussion
The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moves forward), and the vertical forces must cancel if there is to be no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long because the person’s top speed would soon be reached.
What Rocket Thrust Accelerates This Sled?
Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust $\mathbf{\text{T}}$, for the four-rocket propulsion system shown in [link]. The sled’s initial acceleration is $\text{49}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2},$ the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.
Strategy
Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.
Solution
Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with
${F}_{\text{net}}=\text{ma},$
where ${F}_{\text{net}}$ is the net force along the horizontal direction. We can see from [link] that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is
${F}_{\text{net}}=4T-f.$
Substituting this into Newton’s second law gives
${F}_{\text{net}}=\text{ma}=4T-f.$
Using a little algebra, we solve for the total thrust 4T:
$4T=\text{ma}+f.$
Substituting known values yields
$4T=\text{ma}+f=\left(\text{2100 kg}\right)\left({\text{49 m/s}}^{2}\right)+\text{650 N}.$
So the total thrust is
$4T=1.0×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N},$
and the individual thrusts are
$T=\frac{1.0×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}}{4}=2\text{.}5×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N}.$
Discussion
The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 $g$'s. (Recall that $g$, the acceleration due to gravity, is $9\text{.}{\text{80 m/s}}^{2}$. When we say that an acceleration is 45 $g$'s, it is $\text{45}×9\text{.}{\text{80 m/s}}^{2}$, which is approximately ${\text{440 m/s}}^{2}$.) While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.
Newton’s second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. The next section introduces the third and final law of motion.
## Section Summary
• Acceleration, $\mathbf{\text{a}}$, is defined as a change in velocity, meaning a change in its magnitude or direction, or both.
• An external force is one acting on a system from outside the system, as opposed to internal forces, which act between components within the system.
• Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.
• In equation form, Newton’s second law of motion is $\mathbf{\text{a}}=\frac{{\mathbf{\text{F}}}_{\text{net}}}{m}$.
• This is often written in the more familiar form: ${\mathbf{\text{F}}}_{\text{net}}=m\mathbf{\text{a}}$.
• The weight $\mathbf{\text{w}}$ of an object is defined as the force of gravity acting on an object of mass $m$. The object experiences an acceleration due to gravity $\mathbf{\text{g}}$:
$\mathbf{\text{w}}=m\mathbf{\text{g}}.$
• If the only force acting on an object is due to gravity, the object is in free fall.
• Friction is a force that opposes the motion past each other of objects that are touching.
## Conceptual Questions
Which statement is correct? (a) Net force causes motion. (b) Net force causes change in motion. Explain your answer and give an example.
Why can we neglect forces such as those holding a body together when we apply Newton’s second law of motion?
Explain how the choice of the “system of interest” affects which forces must be considered when applying Newton’s second law of motion.
Describe a situation in which the net external force on a system is not zero, yet its speed remains constant.
A system can have a nonzero velocity while the net external force on it is zero. Describe such a situation.
A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?
(a) Give an example of different net external forces acting on the same system to produce different accelerations. (b) Give an example of the same net external force acting on systems of different masses, producing different accelerations. (c) What law accurately describes both effects? State it in words and as an equation.
If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers.
If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object?
The gravitational force on the basketball in [link] is ignored. When gravity is taken into account, what is the direction of the net external force on the basketball—above horizontal, below horizontal, or still horizontal?
## Problem Exercises
You may assume data taken from illustrations is accurate to three digits.
A 63.0-kg sprinter starts a race with an acceleration of $4\text{.}\text{20 m}{\text{/s}}^{2}$. What is the net external force on him?
265 N
If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?
A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.
${\text{13.3 m/s}}^{2}$
Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be $0\text{.}{\text{893 m/s}}^{2}$. (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided.
In [link], the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force $F$ (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force $F$ is removed. How far will the mower go before stopping?
The same rocket sled drawn in [link] is decelerated at a rate of $1{\text{96 m/s}}^{2}$. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.
(a) If the rocket sled shown in [link] starts with only one rocket burning, what is its acceleration? Assume that the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?
(a) $12\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$.
(b) The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning.
What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (b) Draw a free-body diagram, including all forces acting on the system. (c) Calculate the acceleration. (d) What would the acceleration be if friction were 15.0 N?
(a) The system is the child in the wagon plus the wagon.
(b
(c) $a=0\text{.}\text{130}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$ in the direction of the second child’s push.
(d) $a=0.00\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$
A powerful motorcycle can produce an acceleration of $3.50\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$ while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?
The rocket sled shown in [link] accelerates at a rate of $49.0\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight by using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.
(a) $3.68×{10}^{3}\phantom{\rule{0.25em}{0ex}}\text{N}$ . This force is 5.00 times greater than his weight.
(b) $\text{3750 N; 11.3º}\phantom{\rule{0.25em}{0ex}}\text{above horizontal}$
Repeat the previous problem for the situation in which the rocket sled decelerates at a rate of $2{\text{01 m/s}}^{2}$. In this problem, the forces are exerted by the seat and restraining belts.
The weight of an astronaut plus his space suit on the Moon is only 250 N. How much do they weigh on Earth? What is the mass on the Moon? On Earth?
$1.5×{10}^{3}\phantom{\rule{0.25em}{0ex}}\text{N},\text{150 kg},\text{150 kg}$
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,000 kg. The thrust of its engines is 30,000 N. (a) Calculate its acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate its acceleration. | 5,293 | 21,926 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 92, "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-2020-24 | longest | en | 0.943702 |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Book%3A_Introduction_to_Social_Network_Methods_(Hanneman)/13%3A_Measures_of_Similarity_and_Structural_Equivalence/13.04%3A_Describing_Structural_Equivalence_Sets | 1,591,304,515,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347458095.68/warc/CC-MAIN-20200604192256-20200604222256-00270.warc.gz | 435,197,347 | 25,880 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 13.4: Describing Structural Equivalence Sets
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Two actors that are structurally equivalent have the same ties to all other actors - they are perfectly substitutable or exchangeable. In "real" data, exact equivalence may be quite rare, and it many be meaningful to measure approximate equivalence. There are several approaches for examining the pattern of similarities in the tie-profiles of actors, and for forming structural equivalence classes.
One very useful approach is to apply cluster analysis to attempt to discern how many structural equivalence sets there are, and which actors fall within each set. We will examine two more common approaches - CONCOR, and numerical optimization by tabu search.
What the similarity matrix and cluster analysis do not tell us is what similarities make the actors in each set "the same" and which differences make the actors in one set "different" from the actors in another. A very useful approach to understanding the bases of similarity and difference among sets of structurally equivalent actors is the block model, and a summary based on it called the image matrix. Both of these ideas have been explained elsewhere. We will take a look at how they can help us to understand the results of CONCOR and tabu search.
## Clustering Similarities or Distances Profiles
Cluster analysis is a natural method for exploring structural equivalence. Two actors who have the similar patterns of ties to other actors will be joined into a cluster, and hierarchical methods will show a "tree" of successive joining.
Network>Roles & Positions>Structural>Profile can perform a variety of kinds of cluster analysis for assessing structural equivalence. Figure 13.12 shows a typical dialog for this algorithm.
Figure 13.12: Dialog of Network>Roles & Positions>Structural>Profile
Depending on how the relations between actors have been measured, several common ways of constructing the actor-by-actor similarity or distance matrix are provided (correlations, Euclidean distances, total matches, or Jaccard coefficients). Should you desire a different measure of similarity, you can construct it elsewhere (e.g. Tools>Similarities), save the result, and apply cluster analysis directly (i.e. Tools>Cluster).
There are some other important choices. One is, what to do with the items in the similarity matrix that index the similarity of an actor to themselves (i.e. the diagonal values)? One choice ("Retain") includes the similarity of a node with itself; another choice ("Ignore") excludes diagonal elements from the calculation of similarity or difference. The default method ("Reciprocal") replaces the diagonal element for both cases with the tie that exists between the cases.
One may "Include transpose" or not. If the data being examined are symmetric (i.e. a simple graph, not a directed one), then the transpose is identical to the matrix, and shouldn't be included. For directed data, the algorithm will, by default, calculate similarities on the rows (out-ties) but not in-ties. If you want to include the full profile of both in and out ties for directed data, you need to include the transpose.
If you are working with a raw adjacency matrix, similarity can be computed on the tie profile (probably using a match or Jaccard approach). Alternatively, the adjacencies can be turned into a valued measure of dissimilarity by calculating geodesic distances (in which case correlations or Euclidean distances might be chosen as a measure of similarity).
Figure 13.13 shows the results of the analysis described in the dialog.
Figure 13.13: Profile similarity of geodesic distances of rows and columns of Knoke information network
The first panel shows the structural equivalence matrix - or the degree of similarity among pairs of actors (in this case, dissimilarity, since we chose to analyze Euclidean distances).
The second panel shows a rough character-mapped graphic of the clustering. Here we see that actors 7 and 4 are most similar, a second cluster is formed by actors 1 and 5; a third by actors 8 and 9. This algorithm also provides a more polished presentation of the result as a dendogram in a separate window, as shown in Figure 13.14.
Figure 13.14: Dendogram of structural equivalence data (see Figure 13.13)
There are no exact structural equivalence in the example data. That is, there are no two cases that have identical ties to all other cases. The dendogram can be particularly helpful in locating groupings of cases that are sufficiently equivalent to be treated as classes. The measures of clustering adequacy in Tools>Cluster can provide additional guidance.
Two other approaches, CONCOR and optimization, follow a somewhat different logic than clustering. In both of these methods, partitions or approximate equivalence classes are set up first (the user selects how many), and the cases are allocated to these classes by numerical techniques designed to maximize similarity within classes.
## CONCOR
CONCOR is an approach that has been used for quite some time. Although the algorithm of CONCOR is now regarded as a bit peculiar, the technique usually produces meaningful results.
CONCOR begins by correlating each pair of actors (as we did above). Each row of this actor-by-actor correlation matrix is then extracted, and correlated with each other row. In a sense, the approach is asking "how similar is the vector of similarities of actor X to the vector of similarities of actor Y"? This process is repeated over and over. Eventually the elements in this "iterated correlation matrix" converge on a value of either +1 or -1 (if you want to convince yourself, give it a try!).
CONCOR then divides the data into two sets on the basis of these correlations. Then, within each set (if it has more than two actors) the process is repeated. The process continues until all actors are separated (or until we lose interest). The result is a binary branching tree that gives rise to a final partition.
For illustration, we have asked CONCOR to show us the groups that best satisfy this property when we believe that there are four groups in the Knoke information data. We used Network>Roles & Positions>Structural>CONCOR, and set the depth of splits = 2 (that is, divide the data twice). All blocking algorithms require that we have a prior idea about how many groups there are. The results are shown in Figure 13.15.
Figure 13.15: CONCOR on Knoke information matrix with two splits
The first panel shows the correlations of the cases. We included the transpose, so these correlations are based on both sending and receiving of ties. Our data, however, are binary, so the use of the correlation coefficient (and CONCOR) should be treated with caution.
The second panel shows the two splits. In the first division, the two groups {1, 4, 5, 2, 7} and {8, 3, 9, 6, 10} were formed. On the second split these were sub-divided into {1, 4}, {5, 2, 7}, {8, 3, 9}, and {6, 10}.
The third panel (the "Blocked Matrix") shows the permuted original data. The result here could be simplified further by creating a "block image" matrix of the four classes by the four classes, with "1" in high density blocks and "0" in low density blocks - as in Figure 13.16.
Figure 13.16: Block image of CONCOR results
The goodness of fit of a block model can be assessed by correlating the permuted matrix (the block model) against a "perfect" model with the same blocks (i.e. one in which all elements of one blocks are ones, and all elements of zero blocks are zeros). For the CONCOR two-split (four group) model, this r-squared is 0.451. That is, about 1/2 of the variance in the ties in the CONCOR model can be accounted for by a "perfect" structural block model. This might be regarded as OK, but is hardly a wonderful fit (there is no real criterion for what is a good fit).
The block model and its image also provide a description of what it means when we say "the actors in block one are approximately structurally equivalent". Actors in equivalence class one are likely to send ties to all other actors in block two, but no other block. Actors in equivalence class one are likely to receive ties from all actors in blocks 2 and 3. So, we have not only identified the classes, we've also described the form of the relations that makes the cases equivalent.
## Optimization by Tabu Search
This method of blocking has been developed more recently, and relies on extensive use of the computer. Tabu search uses a more modern (and computer intensive) algorithm than CONCOR, but is trying to implement the same idea of grouping together actors who are most similar into a block. Tabu search does this by searching for sets of actors who, if placed into a block, produce the smallest sum of within-block variances in the tie profiles. That is, if actors in a block have similar ties, their variance around the block mean profile will be small. So, the partitioning that minimizes the sum of within block variances is minimizing the overall variance in tie profiles. In principle, this method ought to produce results similar (but not necessarily identical) to CONCOR. In practice, this is not always so. Here (Figure 13.17) are the results of Network>Roles & Positions>Structural>Optimization>Binary applied to the Knoke information network, and requesting four classes. A variation of the technique for valued data is available as Network>Roles & Positions>Structural>Optimization>Valued.
Figure 13.17: Optimized four-block solution for structural equivalence of Knoke information network
The overall correlation between the actual scores in the blocked matrix, and a "perfect" matrix composed of only ones and zeros is reasonably good (0.544).
The suggested partition into structural equivalence classes is {7}, {1, 3, 4, 10, 8, 9}, {5, 2}, and {6}.
We can now also describe the positions of each of the classes. The first class (actor 7) has dense sending ties to the third (actors 5 and 2); and receives information from all three other classes. The second, and largest, class sends information to the first and the third class, and receives information from the third class. The last class (actor 6), sends to the first class, but receives from none.
This last analysis illustrates most fully the primary goals of an analysis of structural equivalence:
1) How many equivalence classes, or approximate equivalence classes, are there?
2) How good is the fit of this simplification into equivalence classes in summarizing the information about all the nodes?
3) What is the position of each class, as defined by its relations to the other classes? | 2,783 | 12,006 | {"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} | 4.4375 | 4 | CC-MAIN-2020-24 | latest | en | 0.469132 |
https://www.fresherslive.com/online-test/pictorial-reasoning-questions-and-answers/6 | 1,675,350,266,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00205.warc.gz | 787,896,630 | 28,170 | # Pictorial Reasoning Questions and Answers updated daily – Logical Reasoning
Pictorial Reasoning Questions: Solved 420 Pictorial Reasoning Questions and answers section with explanation for various online exam preparation, various interviews, Logical Reasoning Category online test. Category Questions section with detailed description, explanation will help you to master the topic.
## Pictorial Reasoning Questions
101. In the following problem, a square transparent sheet with a pattern is given. Figure out from amongst the four alternatives as to how the pattern would appear when the transparent sheet is folded at the dotted line.
Response Figures:
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102. If a mirror is placed on the line MN, then which of the answer figures is the right image of the given figure.
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103. Which answer figure will complete the pattern in the question figure?
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104. Find the number of triangles in the given figure.
Correct Ans:17
Explanation:
The figure may be labelled as shown.
The figure may be labelled as shown.The simplest triangles are ABF, BFG, BCG, CGH, GHD, GED, EFG and AFE i.e. 8 in number.
The triangles composed of two components each are ABG, BGE, AGE, ABE and GCD i.e. 5 in number.
The triangles composed of three components each are BCD, CDE, BED, and BCE i.e. 4 in number.
Thus, there are 8 + 5 + 4 = 17 triangles in the figure.
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105. Select from the five alternative diagrams, the one that best illustrates the relationship among the three classes:
Truck, Ship, Goods
Correct Ans:
Explanation:
Truck and Ship are entirely different. But some Goods are carried by some Trucks and some Goods are carried by some Ships.
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106. A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabet as in the two matrices given below. The columns and rows of Matrix I are numbered from 0 to 4 and that of Matrix II are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, e.g., "˜O"™ can be represented by 01, 33, etc., and "˜Q"™ can be represented by 55, 78, etc. Similarly, you have to identify the set for the word "˜METAL"™.
Correct Ans:32, 76, 95, 44, 04
Explanation:
From above given matrices I and II , we can see that
M ⇒ 01, 14, 23, 32, 41
E ⇒ 58, 67, 76, 85, 99
T ⇒ 59, 68, 77, 86, 95
A ⇒ 03, 12, 21, 30, 44
L ⇒ 04, 13, 22, 31, 40
From given matrix it is clear that all the values for word METAL are matched in only option ( 2 ) for every letter . So , the set for the word METAL will be 32, 76, 95, 44, 04 .From matrix , option (B) is required answer.
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107. From the given answer figures, select the one in which the question figure is hidden/embedded.
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108. If a mirror is placed on the line AB, which of the option figures shows the correct image of the given question figure?
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109. In the following question, select the related figure from the given alternatives.
Question Figure:
Correct Ans:Image 2
Explanation:
The correct answer is image 2.
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110. Find the number of triangles in the given figure.
Correct Ans:10
Explanation:
The figure may be labelled as shown.
The simplest triangles are ABG, BCG, CGE, CDE, AGE and AEF i.e. 6 in number.
The triangles composed of three components each are ABE, ABC, BCE and ACE i.e. 4 in number.
∴ There are 6 + 4 = 10 triangles in the figure.
Hence, option B is correct.
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111. Which of the following diagrams indicates the best relation between Men, Rodents and Living beings ?
Correct Ans:
Explanation:
Men and Rodents are different from one another but both these belong to living beings.
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112. A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as shown in the given two matrices. The columns and rows of Matrix-I are numbered from 0 to 4 and that of Matrix-II are numbered from S to 9. A letter from these matrices can be represented first by its row and next by its column, for example 'V' can be represented by 01, 67 and 'G' can be represented by 13, 00.
Similarly, you have to identify the set for the word 'STAR':
Correct Ans:99, 31, 86, 98
Explanation:
option (A) – 23, 76, 33, 98 = STTR
option (B) – 14, 87, 98, 97 = SIRA
option (C) – 69, 96, 03, 56 = STAE
option (D) – 99, 31, 86, 98 = STAR
Hence the correct answer is option D
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113. A piece of paper is folded and punched as shown below in the question figures. From the given answer figures, indicate how it will appear when opened?
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Explanation:
Half circle are made at the edges of the triangle, so when reopened, the complete circles will appear at the three sides, so three complete circles will be appeared in the answer figure resembling the three sides of the triangle.
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114. From the given answer figures, select the one in which the question figure is hidden/embedded.
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115. From the given options, which figure can be formed by folding the figure given in the question?
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116. If a mirror is placed on the line MN, then which of the answer figures is the right image of the given figure?
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117. In the given figure, how many table set are not sofa?
Correct Ans:18
Explanation:
Here, 18 table set are not sofa.
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118. In the following figure, square represents Artists, triangle represents Military officars, circle represents collectors and rectangle represents Fathers. Which set of letters represents collectors who are either military officers or fathers?
Correct Ans:A, B, G
Explanation:
The correct answer is A, B, G.
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119. Identify the diagram that best represents the relationship among the given classes.
Human, Men, Brothers
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120. A word is represented by only one set of numbers as given in any one of the alternatives. The
sets of numbers given in the alternatives are represented by two classes of alphabets as shown in
the given two matrices. The columns and rows of Matrix-I are numbered from 0 to 4 and that of
Matrix-II are numbered from S to 9. A letter from these matrices can be represented first by its
row and next by its column, for example 'E' can be represented by 42, 23 etc and 'W' can be
represented by 96, 58 etc.
Similarly, you have to identify the set for the word 'DUST"™.
Correct Ans:13, 95, 69, 78
Explanation:
From the options, we can solve that
DUST - 13, 95, 69, 78.
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https://www.physicsforums.com/threads/solution-of-a-simple-integral-equation.971271/ | 1,558,667,630,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257497.4/warc/CC-MAIN-20190524024253-20190524050253-00472.warc.gz | 878,586,753 | 24,324 | # Solution of a simple integral equation
Problem Statement
(a) find all continuous functions $f$ satisfying $\int_0^x f(x) \, dx = (f(x))^2+C$ for some constant $C≠0$ assuming that $f$ has at most one 0.
(b) Also find a solution that is 0 on an interval $(-\infty,b]$ with $0 \lt b$, but nonzero for $x \gt b$
Relevant Equations
$\frac d {dx} \int_a^x {f(t)} \, dt = f(x)$
I did the first part, it is part (b) that I'm having trouble understanding. For any $x \lt b$, $f(x)=0$ and $\int_0^x {f(t)} \, dt = 0$ (since $f$ is 0 everywhere from 0 to $b$), which turns the equation $\int_0^x f(t) \, dt = (f(x))^2+C$ into $0=0+C$, which implies $C=0$. But that is the one value that $C$ cannot have, so I don't see how such a function can be a solution to the equation for $x \lt b$...
Related Calculus and Beyond Homework News on Phys.org
#### fresh_42
Mentor
2018 Award
Combine the two solutions of a.) at $x=b$.
I'm sorry, I don't really understand what you mean by combining them or how that would help. The solutions of a) are $f(x)=\frac 1 2 x±\sqrt {-c}$. No matter what I define $f$ as for $x=b$ or $x>b$, I just don't see how the function can be a solution of the equation when it is 0 for $x<b$.
The way I see it, if such a solution $f$ to the equation $\int_0^x f(x) \, dx = (f(x))^2+C$ exists such that $f(x)=0$ for $x<b$, then inserting some $x'<b$ into the equation yields $0=C$, which is a contradiction. What am I missing here?...
#### fresh_42
Mentor
2018 Award
You are missing an entire solution. I would have told you where you were mistaken if you had written what you did.
You are missing an entire solution. I would have told you where you were mistaken if you had written what you did.
My solution for (a) is as follows:
Differentiating the equation yields $2f(x)f'(x)=f(x)$, so in points where $f(x)≠0$, $f'(x)=\frac 1 2$. So in any interval where $f$ is nonzero, $f(x)=\frac 1 2 x+a$, and since $f$ has only 1 zero this gives 2 possible values of $a$, one for the interval before the 0 of $f$ and one for the interval after it. However, since $f$ is continuous, these 2 values must be the same, therefore $f(x)=\frac 1 2 x+a$ for all $x$. Inserting this into the original equation yields $\int_0^x {\frac 1 2 x+a} \, dx = (\frac 1 2 x+a)^2+C$ or $\frac 1 4 x^2+ax=\frac 1 4 x^2+ax+a^2+C$ or $a=±C$.
These are the only solutions I reached. The solution book does not show any additional solutions, just this.
#### fresh_42
Mentor
2018 Award
You made an assumption in the first line which is not necessarily justified! What if there is no point where $f(x)\neq 0\,?$
You made an assumption in the first line which is not necessarily justified! What if there is no point where $f(x)\neq 0\,?$
Part (a) states that $f$ has at most one zero. Even if $f$ is the zero function, the problem states that $C≠0$, so I don't see how such a function can be a solution.
#### fresh_42
Mentor
2018 Award
Sorry, I missed that. I thought $f=0$ would be a solution and could be combined with a straight for a suitable constant. Anyway, part b) requires a solution with definitely more than one zero. Both together is impossible, so you need to drop this requirement.
I just realised I copied the equation wrong... it needs to be:
$\int_0^x f(t) \, dt = (f(x))^2+C$
But this still brings me to my original problem, I don't see how you can have any solution at all for part b). It seems I am missing something very basic here, which is not uncommon for me...
The solution $f$ in part b) has to answer the requirement that $f(x)=0$ for all $x≤b$, and the equation has to answer the requirement that $C\neq0$, but that seems to cause a contradiction whenever inserting any $x<b$ into the equation. Say we insert $x=\frac b 2$, then by the equation:
$\int_0^{\frac b 2} f(t) \, dt = (f({\frac b 2}))^2+C$
But since $f(x)=0$ on $[0,\frac b 2]$, $\int_0^{\frac b 2} f(t) \, dt=0$ and $(f({\frac b 2}))^2$, which yields $0=C$, which contradicts the condition...
#### fresh_42
Mentor
2018 Award
I get two solutions $F(x) = \left( \dfrac{x}{2} \pm \gamma \right)^2 - \gamma^2$ with $\gamma = +\sqrt{-C} > 0$ for a negative value $C$. These are two parabolas, opened at the top. Both satisfy a.).
For part b.) we need an intersection of these parabolas with $G(x)\equiv 0$ for $x \leq b<0$. This can only be done with one of the parabolas at $x=-4\gamma$ which yields $C= -\dfrac{b^2}{16}$. The combined function of $G(x)$ for $x\leq b$ and $F(x)$ for $x>b$ is the only possibility I can imagine.
#### pasmith
Homework Helper
I get two solutions $F(x) = \left( \dfrac{x}{2} \pm \gamma \right)^2 - \gamma^2$ with $\gamma = +\sqrt{-C} > 0$ for a negative value $C$. These are two parabolas, opened at the top. Both satisfy a.).
For part b.) we need an intersection of these parabolas with $G(x)\equiv 0$ for $x \leq b<0$. This can only be done with one of the parabolas at $x=-4\gamma$ which yields $C= -\dfrac{b^2}{16}$. The combined function of $G(x)$ for $x\leq b$ and $F(x)$ for $x>b$ is the only possibility I can imagine.
If $f$ is quadratic then its integral is cubic but its square is quartic, so it can't satisfy the integral equation. This suggests the existence of affine solutions, so the easiest way to solve this is to assume $f(x) = Ax + B$, substitute that into the integral equation, and compare coefficients of powers of $x$.
Don't forget that if $x < 0$ then the left hand side is $-\int_{x}^0 f(x)\,dx$.
The second part is asking for an $f$ which is zero on $[0,b)$ and for $x \geq b$ must satisfy $$\int_b^x f(t)\,dt = (f(x))^2 + C$$ which can be solved in the same manner as part (a).
However for $x < b$ if $f$ is identically zero then we must have $C = 0$. There is no avoiding that.
#### fresh_42
Mentor
2018 Award
If fff is quadratic then its integral is cubic but its square is quartic, so it can't satisfy the integral equation.
$F(x) = \int_0^x f(t)\,dt$ can, which is why I switched to capital letters.
It seems my misunderstanding of part b) just boils down to poor articulation of the question. I thought part b) requested a function that is 0 on $(-\infty, b]$ and nonzero on $(b,\infty)$ and is a solution to the equation no matter what $x$ is inserted into it. But it seems the function only needs to be a solution for $x>b$, which is rather weird since I assume a condition like this needs to be explicitly stated, but I see no other way for this question to make sense...
However, the solution for part b) in the answer book goes as follows:
"On the other hand, for $b=\sqrt {-C}$ we can also choose $f$ to be 0 on $(-\infty,b]$, and $f(x)=\frac x 2 - \frac b 2$ for $x\geq b$"
Which also seems to not make sense, since even for $x\geq b$, $\int_0^x f(x) \, dx = (f(x))^2$, which yet again means C=0! And the question can't want me to drop the C=0 condition since it asks to drop it only on part c). It can't be that the author forgot to mention it because then he wouldn't have bothered to say $b=\sqrt {-C}$ in the solution if $C=0$. So I don't know anymore...
Last edited:
Well i think I pretty much got it. I just assumed C=0 on part b) and ignored the answer book, everything seemed to have worked out when I did that.
#### fresh_42
Mentor
2018 Award
My proposal would have been
$$f(x) = \begin{cases} 0& \text{ if } x<b\\ \frac{1}{2}x + \frac{1}{4}b&\text{ if } x \geq b \end{cases}$$
with $F(x)=\int_0^x f(t)\,dt = \frac{1}{4}\left( x^2+bx \right)$ and $C=-\frac{1}{16}b^2\,.$
#### MathematicalPhysicist
Gold Member
My solution for (a) is as follows:
Differentiating the equation yields $2f(x)f'(x)=f(x)$, so in points where $f(x)≠0$, $f'(x)=\frac 1 2$. So in any interval where $f$ is nonzero, $f(x)=\frac 1 2 x+a$, and since $f$ has only 1 zero this gives 2 possible values of $a$, one for the interval before the 0 of $f$ and one for the interval after it. However, since $f$ is continuous, these 2 values must be the same, therefore $f(x)=\frac 1 2 x+a$ for all $x$. Inserting this into the original equation yields $\int_0^x {\frac 1 2 x+a} \, dx = (\frac 1 2 x+a)^2+C$ or $\frac 1 4 x^2+ax=\frac 1 4 x^2+ax+a^2+C$ or $a=±C$.
These are the only solutions I reached. The solution book does not show any additional solutions, just this.
They asked you in (a) to find all the continuous functions that satisfy this equation, which means there might be functions that aren't differentiable, so you found some functions that are $C^1$.
But for checking for all $f\in C^0$ you need perhaps to invoke iteration:
$$f(x) = \sqrt{\int_0^x f(t) -C} = \sqrt{\int_0^x \sqrt{\int_0^t f(s)ds-C}dt-C}$$
Now you could square everything and get: $$f^2(x)+C = \int_0^x \sqrt{\int_0^t f(s)ds-C}dt = \int_0^x f(t)dt$$.
But I don't see how to find from this method all the continuous functions.
My proposal would have been
$$f(x) = \begin{cases} 0& \text{ if } x<b\\ \frac{1}{2}x + \frac{1}{4}b&\text{ if } x \geq b \end{cases}$$
with $F(x)=\int_0^x f(t)\,dt = \frac{1}{4}\left( x^2+bx \right)$ and $C=-\frac{1}{16}b^2\,.$
That solution seems only to work for $x>b$, else the equation reads $-\frac 1 {16} b^2=0$. Also the function is not continuous at $b$, so I don't think it is viable, considering the conditions for the solution. I just assumed that $C=0$ and everything worked out with $f(x)=\frac 1 2 x - \frac 1 2 b$ for $x>b$, $f(x)=0$ for $x\leq b$, it doesn't seem there is any other viable way of getting a solution that answers the conditions.
They asked you in (a) to find all the continuous functions that satisfy this equation, which means there might be functions that aren't differentiable, so you found some functions that are $C^1$.
But for checking for all $f\in C^0$ you need perhaps to invoke iteration:
$$f(x) = \sqrt{\int_0^x f(t) -C} = \sqrt{\int_0^x \sqrt{\int_0^t f(s)ds-C}dt-C}$$
Now you could square everything and get: $$f^2(x)+C = \int_0^x \sqrt{\int_0^t f(s)ds-C}dt = \int_0^x f(t)dt$$.
But I don't see how to find from this method all the continuous functions.
As far as I'm aware, the solution for the equation is necessarily differentiable at all points but one. Since $\int_0^x f(x) \, dx$ is differentiable by the first FTOC, and differs only by a constant from $f(x)^2$, the latter must also be differentiable, which only occurs if $f'(x)$ exists unless $f(x)=0$, which only occurs at a single point in part a).
Last edited:
#### fresh_42
Mentor
2018 Award
That solution seems only to work for $x>b$, ...
No problem. $b < 0$ and the integral from $0$ to $x$. So as long as we don't think of a reverse integration path, there will be no conflict. And as signs are so important in this problem, a reversed integration would be quite an arbitrary objection.
No problem. $b < 0$ and the integral from $0$ to $x$. So as long as we don't think of a reverse integration path, there will be no conflict. And as signs are so important in this problem, a reversed integration would be quite an arbitrary objection.
Well that addresses one issue, unfortunately the question explicitly says $b$ needs to be positive, so we are brought into another one, and the discontinuity at $b$ still exists. Honestly, this whole problem seems to be a very simple exercise becoming overly complicated probably because of an author's mistake, which would not be the first (Spivak).
However I'll take the opportunity to educate myself a bit more. For any $x<b$ in your function, $\int_0^x f(t)\,dt = \int_0^b f(t)\,dt =\frac 1 2 b^2$, so the equation reads $\frac 1 2 b^2=\int_0^x f(t)\,dt =((f(x))^2+C=C=-\frac 1 {16} b^2$. So how does that work out for those x's?
#### MathematicalPhysicist
Gold Member
@Adgorn , apparently it's a popular question:
I believe there's a solution manual in the web for the third edition, if you want to check your work.
@Adgorn , apparently it's a popular question:
I believe there's a solution manual in the web for the third edition, if you want to check your work.
I never manage to find copies of my questions, perhaps I require more training in the art of forum digging. I actually have the physical copies of both the 4th addition and the answer book, I only resort to online help if the answer book does not provide a sufficient explanation, as was the case here. At this point I'm fairly certain there is some mistake in the question. The author probably forgot to mention that C can be equal to 0 in part b). That would explain all my troubles, or at least so it seems.
"Solution of a simple integral equation"
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• Solo and co-op problem solving | 3,867 | 12,761 | {"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": 2, "mathjax_display_tex": 2, "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} | 4.0625 | 4 | CC-MAIN-2019-22 | latest | en | 0.922182 |
https://latellagelato.com/cooking-tips/how-many-inches-in-5-3 | 1,695,358,834,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00496.warc.gz | 412,864,613 | 54,537 | How Many Inches In 5 3? | Measurement Unit Conversion Tips
Are you trying to find out how many inches in 5 3? If so, you have come to the right place. In this blog post, we will take a deep dive into determining the number of inches that make up the height 5 3, as well as provide some helpful tips on how best to measure height with different units of measurement. We’ll discuss why knowing how many inches in 5 3 can be useful and provide resources for additional information about converting measurements. Read on for more.
What Are Inches And Its History?
Inches are a unit of measurement most commonly used in the United States. It is part of what is known as the Imperial system, which was first created by the British in 1824. The inch itself is equal to 1/12th of a foot and is usually abbreviated with the symbol “in”. Inches are particularly useful for measuring areas like the height of a person.
What Does 5 3 Mean In Measurement?
5 3 is an expression commonly used to refer to a person’s height. It is important to note that the “3” portion of this expression does not refer to inches, but instead fractions of an inch. The “5” in 5 3 refers to feet and can easily be converted into inches by multiplying it by 12.
How Many Inches In 5 3?
Now that you know the definition of 5 3, it is time to answer the main question: how many inches in 5 3? To calculate this number, we must first convert the feet portion (5) into inches. This is done by multiplying 5 by 12, which gives us 60 inches. Then we add the fractional part (3/12ths) to get the total number of inches in 5 3. The answer is therefore 63 inches.
Is There Any Benefit To Converting Inches In 5 3?
Knowing how many inches in 5 3 can be useful for a variety of reasons. For instance, if you need to compare someone’s height with another person’s height, it is much easier to compare two numbers that are expressed in the same unit of measurement. Additionally, converting from feet and fractions of an inch to just inches can help make calculations and comparisons easier. For example, if you need to calculate how tall someone is in centimeters, it is much simpler to start with the number of inches rather than trying to convert feet and fractions of inches into centimeters.
Steps To Convert Inches In 5 3
To quickly and accurately convert measure inches in 5 3, simply follow these steps:
1. Multiply the number of feet (5) by 12 to get the total number of inches (60).
2. Add the fractional part (3/12ths) to get the final number of inches (63).
How To Speed Up The Conversion Of Inches In 5 3?
If you find yourself needing to do these conversions often, there are tools available online that can help speed up the process. For example, our Inches To Feet Calculator is a free tool that quickly and easily converts feet and fractions of an inch into inches. This calculator also allows you to enter multiple measurements at once so you don’t have to do each one individually.
Mistakes To Avoid When Converting Inches In 5 3
When converting inches in 5 3, it is important to remember that the “3” portion of this expression does not refer to inches. Instead, it refers to fractions of an inch and must be added to the total number of inches calculated from the feet portion (5). Additionally, always make sure you are working with accurate measurements in order to get accurate results. If you are trying to convert someone’s height, make sure that they are standing up straight so that their height is as accurate as possible.
Converting inches in 5 3 can be tricky, but with the right tools and a little practice, you can master this conversion quickly. Here are some tips and advice to help you get started:
• Always start by converting the feet portion (5) into inches (60).
• Then add the fractional part (3/12ths) to get the total number of inches (63).
• Use an online calculator, such as our Inches To Feet Calculator, to speed up the process.
• Make sure you are working with accurate measurements so that your results are correct.
• Take your time and practice a few conversions until you feel comfortable doing them on your own.
Now that you know how many inches in 5 3, you can easily and accurately convert this measurement into different units. Whether you need to compare someone’s height with another person or calculate a height in centimeters, having the number of inches is important. With these tips and advice, you should have no problem mastering this conversion.
How To Apply The Conversion Of Inches In 5 3 To Your Life?
Knowing how to convert inches in 5 3 can be useful for a variety of everyday tasks, such as buying clothing or furniture. If you are shopping online for items like jeans or couches, the measurements will usually be given in feet and fractions of an inch. This means that you will need to convert them into inches before making your purchase. Additionally, many medical professionals use 5 3 as a way to measure height. Knowing how to quickly and accurately convert this measurement can be helpful in a medical setting.
Some Formulas Use Inches In 5 3
Inches in 5 3 can also be used as a part of various formulas. For instance, the BMI (Body Mass Index) formula uses height expressed in inches to calculate a person’s body weight. Additionally, many growth charts use inches to track the growth of babies and children over time. Knowing how many inches in 5 3 is an important part of using these formulas correctly.
Conclusion: How Many Inches In 5 3
In conclusion, 5 3 is an expression commonly used to refer to a person’s height. To calculate how many inches in 5 3, start by multiplying the feet portion (5) by 12 to get the total number of inches (60). Then add the fractional part (3/12ths) to get the final answer: 63 inches. Knowing how to convert inches in 5 3 can be useful for a variety of everyday tasks, such as shopping for clothing or furniture, and it is also an important part of various medical formulas. With the tips and advice provided in this article, you should have no problem mastering this conversion.
FAQ: Inches In 5 3
Is 12 inches equals 5 3?
The conversion from 5 feet 3 inches to inches is calculated by multiplying the feet portion (5) by 12, resulting in 60 inches. Then, the fractional part (3/12ths) is added to obtain the final answer: 63 inches.
Is 5 3 bigger than 20 inches?
Indeed, a measurement of 5 feet 3 inches surpasses a length of 20 inches. Upon converting 5 feet 3 inches to a total of 63 inches, we can confidently assert its superiority over the 20 inch measurement.
Why is it difficult to convert inches in 5 3?
One of the reasons why it can be difficult to convert inches in 5 3 is because fractions of an inch are involved. The “3” portion of this expression does not refer to inches, but rather fractions (or parts) of an inch that must be added to the total number of inches calculated from the feet portion (5).
What can be used to convert inches in 5 3?
To quickly and accurately convert inches in 5 3, you can either use a calculator or an online tool such as our Inches To Feet Calculator. This calculator allows you to enter multiple measurements at once so you don’t have to do each one individually. Additionally, it provides detailed instructions on how to calculate feet and fractions of an inch into inches correctly.
Is the conversion inches in 5 3 changed by temperature?
The conversion of 5 feet and 3 inches to inches is unaffected by temperature. This mathematical conversion remains consistent regardless of external factors like temperature or weather. The result remains unchanged regardless of the outdoor conditions.
Is the conversion inches in 5 3 changed by elevation?
The conversion of 5 feet and 3 inches to inches is unaffected by elevation. This mathematical conversion remains consistent regardless of external factors like altitude or height above sea level. The measurements remain the same no matter what the elevation is.
Is 5 3 2 inches?
The conversion from 5 feet 3 inches to inches can be calculated by multiplying the feet portion (5) by 12, resulting in 60 inches. Then, the fractional part (3/12ths) is added to obtain the final answer: 63 inches. This shows that 5 feet 3 inches is not equal to 2 inches.
Is 5 inches equal to 5 3?
5 inches is not equal to 5 feet 3 inches. The conversion from 5 feet 3 inches to inches can be calculated by multiplying the feet portion (5) by 12, resulting in 60 inches. Then, the fractional part (3/12ths) is added to obtain the final answer: 63 inches. This shows that 5 feet 3 inches is greater than 5 inches.
How many inches in 5 3 in us?
The conversion from 5 feet 3 inches to inches in the US can be calculated by multiplying the feet portion (5) by 12, resulting in 60 inches. Then, the fractional part (3/12ths) is added to obtain the final answer: 63 inches. This shows that there are 63 inches in 5 3 in the United States.
How many inches in 5 3 in uk?
In the UK, converting 5 feet 3 inches to inches is a simple process. First, we multiply the feet portion (5) by 12 to get 60 inches. Then, we add the fractional part (3/12ths) to obtain the final answer: 63 inches. Therefore, in the United Kingdom, 5 feet 3 inches equals 63 inches. | 2,113 | 9,235 | {"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-2023-40 | latest | en | 0.955558 |
http://wheredoesallmymoneygo.com/the-pe-ratio-2/ | 1,501,094,378,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426372.41/warc/CC-MAIN-20170726182141-20170726202141-00473.warc.gz | 360,832,103 | 15,802 | # The P/E Ratio
Recently I ran into a long time reader who asked if I would re-post an old series I wrote on the P/E ratio. I wrote this when the blog was not yet a year old and didn’t have many readers so I thought it might be worthwhile to dig it up. Be warned, you may want to grab a cup of coffee before sitting down to read this. It’s a bit lengthy, but this is something that gets talked about all the time in the business news so I think it’s worth having a cursory understanding… Enjoy.
### The P/E Ratio, also known as…
The P/E Ratio is such a widely used ratio that it has many different slang terms such as:
1. The Multiple
2. The Price to Earnings Ratio
3. The P/E Ratio
4. Earnings Ratio
5. Price Multiple
…and there are probably some others that aren’t top of mind right now, too.
### Okay, so what is it?
The price to earnings ratio is a number that is derived from the formula:
P/E Ratio = Price Per Share / Earnings Per Share
So the “P” stands for the Price of the share, and the “E” stands for the Earnings Per Share (or ‘EPS’). If you had a stock that was trading at \$50 per share on the market, and that stock had an EPS (earning per share) of \$2.50, then according to this very simple formula, the P/E Ratio of this stock is 20 (\$50 / \$2.50). So calculating the P/E ratio is butt simple.
### But seriously Preet, what IS it?
There are a number of differet ways to look at it, but I will give you the one that makes the most sense to me.
It is the price you are willing to pay today for \$1 of annual income in perpetuity.
So for example, for our sample stock above with a P/E of 20, that means you are willing to pay \$20 today for an annual income stream of \$1 for life. (It might be better to say that the market as a whole is willing to pay \$20 today for that \$1 annual income for life.)
I’m going to borrow an example I read elsewhere, but if a stranger came up to you and asked you to buy a \$1 dollar income stream from them for life, you would have no idea if they were able to keep this promise and you might only offer them \$1 simply because you don’t know or trust this person, but you think that you should be able to at least get your \$1 back next year. But what if Bill Gates came up to you and offered to sell you \$1 per year for life for \$20? You might take him up at that price because you know that he will probably be making a lot of money for many years to come. Well, in essence Bill has a P/E ratio of 20 and the stranger has a P/E of 1.
Your expectation of Bill Gates earning lots of money in the future is solid and hence you are willing to basically wait 20 years to get your money back, at which point the future \$1 annual income is gravy. (Clearly I’m not factoring in opportunity costs or interest for this example – but I will in a follow up post that is a bit more technical.)
### Let’s now relate it back to the stock market
A high P/E ratio means that investors believe the future earnings of a company are expected to be strong. The stronger the earnings outlook, the more confidence people have in buying stock in the company because they believe there is a greater chance that the earnings will continue.
If someone offers to pay a higher price for a stock, they are offering to purchase the stock at a higher P/E ratio – which means they are more confident about the future of that company.
### But there’s more…
When a P/E ratio really starts to get high this is due to investors not only believing the earnings are solid, but that they will probably GROW over time as well. This means that they are basically saying to themselves that they believe the \$1 annual income stream will increase. Next year it might be \$1.05, the year after it might be \$1.12 and so on. Since they believe the earnings will grow (because the company is going to take off), they are willing to offer an even higher ‘multiple’. For example, RIM has a P/E of over 60 right now (Remember, this was originally written in 2008 – Preet). If you only assume that you are buying \$1 of annual earnings per year for \$60 today, that might seem a bit crazy. But if you think that RIM will continue to increase it’s earning VERY rapidly, then you are not expecting \$1 per year, but rather a very quickly increasing earnings stream.
### RECAP
The P/E ratio can be considered as the price today of purchasing a \$1 income stream for life. When people bid up the price of a stock, and hence the P/E ratio, they are basically saying that they believe that company’s future earnings outlook are more promising, and are willing to pay more to own a piece of those future earnings.
### So What’s A Fair Price For A Company?
Let’s assume we have a company that is guaranteed to provide \$1 per year for life no matter what (i.e. there is no business risk whatsoever – purely wishful thinking!). In this case, what would be a fair price to purchase that income stream? Well, if we assume that we are going to live for another 80 years, then you might say \$80 as a starting point because 80 years times \$1 = \$80. You would think, I’m going to get \$80 over the next 80 years – therefore this is the fair price. Right?
Wrong.
### The Present Value of a Dollar From the Future
You are essentially giving up \$80 now in a lump sum today in exchange for getting eighty \$1 dollar payments over 80 years which is crazy when you think that you could just get a high interest savings account and get 3% interest on your \$80 lump sum starting today [Ah, the good old days – Preet]. In fact, the first year’s interest alone would be \$2.40 – that’s much more than \$1. And after 80 years, your original \$80 dollars would’ve grown to \$826.48, if you kept re-investing the interest.
Let’s start by figuring out what a better price would be to pay for a series of \$1 payments that will be received in the future, starting with next year. Basically, we need to start by asking: What do I need to invest at 3% today, to get \$1.00 in one year?
In this case, the answer is \$0.97 (rounded). In other words, to have \$1.00 NEXT year, you would need to invest 97 cents into that 3% high interest savings account. Therefore, you might pay \$1 for this year’s \$1 income from the company, but you would definitely not want to pay more than \$0.97 for next year’s \$1.
Let’s move to year 3. What would you need to invest TODAY at 3%, in order to get \$1 in 2 years? The answer is \$0.94 (again, rounded for simplicity’s sake). \$0.94 invested for one year at 3% equals roughly \$0.97, which when invested for the second year at 3% will give you \$1.
So you can see, the further out that company’s \$1 annual income is, the less you would want to pay for it. If we fast forward to year 80, you would only need to invest 9 cents today in a savings account that paid 3% interest per year in order to have \$1 eighty years from now. So it would make no sense to pay more than 9 cents today for \$1 to be received in 80 years.
Below, I have charted the present value of \$1 for every year between now and 80 years from now, based on a 3% interest rate. If we add up all of those values, we then have \$30.20. Therefore, assuming there is no business risk, and we are guaranteed an earnings of \$1.00 per share every year for the next 80 years, \$30.20 per share is a much fairer price than \$80.00 to pay for this income stream.
### Still Not Done!
We are not quite yet done with the discussion. There are two more things we need to factor in. 1) Investors expect to be compensated for the risk they take in making an investment that is more risky than a high interest savings account (this would bring the price that they are willing to pay DOWN), and 2) the price goes up if the company’s earnings are expected to increase.
So far we been using some unlikely assumptions, namely that this fictitious company we are going to buy has a constant earnings stream and that there is no risk involved in that earnings stream. Of course the real world is quite different. Let’s next decide on how a changing earnings stream can affect the present value of all those future annual earnings.
### The Math is Not Much Different
The math is not really different, we just have to take an extra step. Before, we were just assuming that a company would produce \$1 in annual earnings forever. But let’s now pretend that our company is expected to grow it’s earnings by 10% per year. Therefore, instead of an earnings stream that looks like this:
Year 1: \$1.00, Year 2: \$1.00, Year 3: \$1.00… etc.
It will instead look like this:
Year 1: \$1.00, Year 2: \$1.10, Year 3: \$1.21… etc.
If we just do the math for the first 3 years, you will get the picture of how a changing earnings stream can affect a price someone is willing to pay for a company. If you recall, we need to figure out what we need to invest TODAY at 3% (or whatever the “risk-free” rate of return is) to replicate the earnings that will be earned at some point in the future. In our new company here that grows it’s earnings at 10% per year we find that we might pay \$1.00 for this years \$1.00. But it is going to earn \$1.10 in the second year – so what amount of money invested today at 3% will give us \$1.10 in one year? The answer is \$1.07. Likewise for the \$1.21 it is expected to earn in Year 3 – what amount do we need to invest today at 3% in order to replicate that \$1.21 in two years from now? The answer is \$1.14.
At this point, I could show a revised graph which shows the present value of the income stream for a company that can grow it’s earnings at 10% forever, but not many companies can do that. What is more realistic is to look at the following situation:
### A Hypothetical Company’s Life Cycle
Suppose we have a relatively new company that is turning a profit and is still in it’s growth phase. During this growth phase it is expected to grow it’s earnings at 20% per year for 5 years as it gains market share and more customers. After the first 5 years, earnings growth slows down to about +10% per year for the next 5 years, and then slows down to +5% for the next 5 years. At this point, it has reached it’s market saturation point and perhaps earnings hold steady for the next 10 years. During that time, some competitor companies figure they can take a slice of their market and start up operations, and over time start to move some customers over away from “our” company. Our company therefore experiences an earnings decline of -5% per year for the next 10 years. At this time, our company and the competitors have found an equilibrium point and our company’s earning are now not declining further, nor are they increasing from this point on until the end of the 80 years.
The above Life Cycle might represent a more realistic earnings stream for a real-life company. Now that we have a projection of what the future income stream might look like, we can again add up all the present values of those annual earnings to figure out a fair price for those earnings. I have charted both the annual earnings (in red) and the present value of those future earnings (in green).
If we add up all the present values this time we get a sum of \$71.81. This is quite a bit higher than the \$30.20 we came up with in part 2, but remember part 2 assumed \$1/year for 80 years. Whereas in this case, the earnings grew from a start of \$1/year to as high as \$3.55/year, and then declined to about \$1.56 per year.
So you can see how earnings predictions are so important and why the market seems to be so sensitive to them.
Ah, yes. Again we have yet to factor in that if we were to offer \$71.81 for this future earnings stream, we are assuming there is no risk involved – and of course, this couldn’t be further from the truth. There is the risk that this company goes out of business due to competition, their product could be rendered obsolete by new technology, a poor economy in general, you name it. Any of those things could affect future earnings and there is a risk that our predictions for the income stream may not be fulfilled. We could put our \$71.81 in a high interest savings account at 3% and basically replicate this earnings stream with no risk, therefore we would be crazy to offer \$71.81 for this company – we would have to offer LESS money for this expected income stream to compensate us for this extra risk.
If you offered \$40/share for something you have calculated to be worth \$71.81 today, this discount represents the compensation for the risk you are taking. The constant bids/asks on the stock market for shares of companies are basically people weighing in on what their perception of a good discount is for that future earnings stream for those particular companies.
### Putting It All Together
So far we have basically explained that the P/E ratio is a way to guage what investors are willing to pay for the future income stream of a company, and how much of a discount they wish to receive in exchange for accepting the risk associated with that income stream.
But the P/E ratio is far from an exact science – it is more of a black art. How accurate are analysts’ predictions about earnings two years from now, let alone 80 years from now like in our example? Think of it like throwing darts except replace “year” with “feet from the dartboard”. Predicting earnings for 1 year (throwing a dart from 1 foot) is much easier than predicting earnings 20 years from now (like throwing a dart when standing 20 feet away). There are just too many things that can happen to change the future fortunes of a company.
### So how do people use the P/E ratio?
The P/E Ratio, like many other stock analysis tools and metrics should be used as part of your analysis, not as the basis of your analysis. I would use it to raise red flags, not to screen investments. People generally look at a P/E ratio and compare it to the P/E ratio of something relevant. That ‘relevant thing’ could be the historical average P/E of the stock in question or the average P/E of all the similar companies in the same industry.
If the stock’s current P/E ratio is much higher than it’s average, this might be referred to as being “expensive”. It might also mean that the company is expected to grow it’s earning faster than it has previously. The fact that the P/E is higher than it’s average means nothing in and of itself.
Conversely, if the stock’s P/E ratio is much lower thant it’s historical average, this might be referred to as being “cheap”. It might also mean that the company is expected to have lower earnings than previously thought. Many value-investors would look at a low P/E ratio as a good thing if they believe nothing has changed with the company since they would consider a value stock as a company that is temporarily undervalued for no good reason (which a lower than average P/E ratio might suggest). The classic value trap is when the P/E ratio is low, making the stock look cheap, and value investors pounce only to find out that something is indeed wrong with the company and the earnings decline and the stock price follows.
### Comparing Relative P/E Ratios
Where knowing the P/E ratio can be of more use is when comparing the P/E ratio of a company against it’s competitors with similar characteristics, the market as a whole, or as mentioned above, against it’s historical average.
Different companies in different industries may have higher or lower P/E ratios based on what kind of company they are. For example, the big blue chip stocks (like a utility company) might have a low normal range of P/E ratios because they are more mature and are not expanding anymore – their earnings may not be growing as fast as a high-tech company that arrives out of nowhere and expands and grows furiously. However, a high-tech company with a P/E ratio of 30 might be cheap, whereas a bank with a P/E of 20 might be expensive – it’s all relative.
### The Market P/E Ratio
The market P/E ratio is simply the total market capitalization of all stocks in the market divided by the total earnings of all the companies in the market. Some people use the market P/E to make decisions about whether the market as a whole is “cheap”, “expensive” or around it’s long term average. They may choose to avoid picking any stocks if the market is expensive because sometimes the baby gets thrown out with the bathwater (meaning that during market declines, sometimes it doesn’t matter what you own, it might be going down too).
### The End
Well, sort of. I think from reading this you will have a firm grasp of what the P/E ratio is, and how it gets used in the financial press. As always, if you want to learn more a simple google search on it will yield years of reading material as it is one of the most used financial ratios in the business.
Peace out
Preet Banerjee
Related Posts
• Mr. Cheap
Interesting post, I missed it the first time around and enjoyed the re-post!
• Preet
Hey stranger – long time! :)
• Think Dividends
P/E ratio works for 95% of the companies out there. It does not apply to REITs where the focus is on Funds from Operations (FFO) not earnings (EPS). So in the case of REITs the P/E ratio become P/FFO. | 3,984 | 17,198 | {"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-2017-30 | latest | en | 0.975613 |
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A solution contains 8 parts of water for every 7 parts of
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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22 Mar 2015, 22:17
VeritasPrepKarishma wrote:
Though Bunuel and soumanag have already explained the solution well, I will add the method I like the most.
In replacement questions, focus on the thing that decreases. When solution is removed, water decreases but then water is added. While when solution is removed, lemonade decreases and does not get added later. So we will work with lemonade concentration.
The fraction of lemonade in the solution is 7/15
We need to get this fraction down to 2/5 (to make it 40%)
Let us say, we remove a fraction 'f' of the solution.
Then 7/15 - f * (7/15) = 2/5
f = 1/7
So (1/7)th of the solution has to be removed. But we want the answer in terms of parts (how many of the 15 parts have to be removed)
So we need to remove (1/7) * 15 = 2.14 parts
Hi Karishma,
Thanks for explaining it so well. I got it wrong as I thought we have to give answer in terms of lemonade.
So, here can we say that replacing 1 unit of lemonade and 1.14 units of water will serve the purpose ?
Regards,
Gaurav
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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23 Mar 2015, 03:49
VeritasPrepKarishma wrote:
GauravSolanky wrote:
VeritasPrepKarishma wrote:
Though Bunuel and soumanag have already explained the solution well, I will add the method I like the most.
In replacement questions, focus on the thing that decreases. When solution is removed, water decreases but then water is added. While when solution is removed, lemonade decreases and does not get added later. So we will work with lemonade concentration.
The fraction of lemonade in the solution is 7/15
We need to get this fraction down to 2/5 (to make it 40%)
Let us say, we remove a fraction 'f' of the solution.
Then 7/15 - f * (7/15) = 2/5
f = 1/7
So (1/7)th of the solution has to be removed. But we want the answer in terms of parts (how many of the 15 parts have to be removed)
So we need to remove (1/7) * 15 = 2.14 parts
Hi Karishma,
Thanks for explaining it so well. I got it wrong as I thought we have to give answer in terms of lemonade.
So, here can we say that replacing 1 unit of lemonade and 1.14 units of water will serve the purpose ?
Regards,
Gaurav
Yes, you are removing a total of 2.14 units, of which (7/15)*2.14 = 1 unit is lemonade and rest 1.14 units is water.
Note that saying "replace 1 unit of lemonade and 1.14 units of water" is not very logical since you cannot remove the two separately. They are mixed together so you need to remove the solution only. You cannot remove 1 unit of lemonade alone since water will come along with it. So it will be logical to say that we must remove 2.14 units of solution of which 1 unit will be lemonade and rest will be water since solutions are assumed homogeneous.
I stated that way as I just wanted to put more stress on quantities separately. But yes, you are correct that was illogical.
Thank Again.
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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24 Mar 2015, 17:49
Awesome Karishma. This makes a lot more sense. Thanks a ton
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A solution contains 8 parts of water for every 7 parts of [#permalink]
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Updated on: 10 Jun 2017, 19:45
x=fraction of solution replaced
8/15-8x/15+x=9/15
x=1/7
1/7*15=2.14 parts of solution replaced
C
Originally posted by gracie on 21 Dec 2015, 14:26.
Last edited by gracie on 10 Jun 2017, 19:45, edited 1 time in total.
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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29 Dec 2015, 12:50
VeritasPrepKarishma wrote:
VSabc wrote:
It seems I took a completely different tangent here, can someone please help me:
Let there be total 15l of solution implying 8l water and 7l lemonade. Acc. to the problem, let's take off 'w'l from solution and add 'w'l of water implying:
(8+w)/(15-w)=0.6 solving which gives w=0.525
What's wrong here?
Here is the problem with your equation.
When you took out 'w' lt of solution, the water left is less than 8 lt. So how can total water after replacement be (8 + w) lts? It will 'something less than 8 + w' lts.
Also, the new solution after you replace with water is again 15 lts. So why would you have (15 - w) in the denominator?
Your equation should be
$$\frac{8 - (8/15)*w + w}{15} = 0.6$$ (the fraction of water removed will be (8/15) of w)
$$8 + (7/15)*w = 9$$
$$w = 15/7$$
HI Karishma,
I am just wondering if we can apply scale method to this problem?! The reason is - we know desired %, we have relevant proportions of water & lemonade which can be converted into %
Or Is my approach not right?
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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10 Nov 2016, 03:30
VeritasPrepKarishma wrote:
Though Bunuel and soumanag have already explained the solution well, I will add the method I like the most.
In replacement questions, focus on the thing that decreases. When solution is removed, water decreases but then water is added. While when solution is removed, lemonade decreases and does not get added later. So we will work with lemonade concentration.
The fraction of lemonade in the solution is 7/15
We need to get this fraction down to 2/5 (to make it 40%)
Let us say, we remove a fraction 'f' of the solution.
Then 7/15 - f * (7/15) = 2/5
f = 1/7
So (1/7)th of the solution has to be removed. But we want the answer in terms of parts (how many of the 15 parts have to be removed)
So we need to remove (1/7) * 15 = 2.14 parts
Responding to a pm:
Quote:
I did exactly what you did and arrived at the answer. But I have question regarding the wording. Is using "parts" here legit? I could easily arrive at f = 1/7. And then I thought to myself - "Okay, so 1 in every 7 total parts should be replaced", and then I was kind of blank for a while and then tried using 15 parts originally provided to solve it. Please explain..
Sure, the use of "parts" is fine. All measurements are in terms of the same "part". That part could 1 ml, 10 ml or 100 ml etc. There are total 15 of these parts in the solution.
When we get f = 1/7, f is the fraction of the solution removed. (1/7)th divides the whole solution into 7 equal parts and removes 1. These parts are not the same as the parts above. The total solution has 15 of the "parts" discussed above. Of these 15 parts, we remove 1/7th which is 2.14 "parts".
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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10 Nov 2016, 12:59
udaymathapati wrote:
A solution contains 8 parts of water for every 7 parts of Lemonade syrup. How many parts of the solution should be removed and replaced with water so that the solution will now contain 40% lemonade syrup?
A. 1.5
B. 1.75
C. 2.14
D. 2.34
E. 2.64
use C final = C initial (Volume original / volume final ) , final concentration is 40/100 , initial concentration is 7/15 , volume original is volume of the solution after removing part of it and before adding water to dilute it , V final is the total volume after replacing of x parts of the solution by x parts of water and thus total volume remains the same ( 15)
40/100 = 7/15 ( 15-x / 15 ) thus x = 15/7 = 2.14
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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15 Jan 2017, 19:00
y = amount removed from original = amount replaced with water
x = original amount
(8/15)x - (8/15)y + y = (3/5)x
(8/15)x + (7/15)y = (3/5)x
8x+7y = 9x
7y = x --> y = (1/7)x
There are 15 parts in the original --> so... 15/7 is our answer
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A solution contains 8 parts of water for every 7 parts of [#permalink]
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10 Jun 2017, 18:55
udaymathapati wrote:
A solution contains 8 parts of water for every 7 parts of Lemonade syrup. How many parts of the solution should be removed and replaced with water so that the solution will now contain 40% lemonade syrup?
A. 1.5
B. 1.75
C. 2.14
D. 2.34
E. 2.64
1. (Initial quantity of solution- quantity of solution removed)* strength of solution + (Quantity of water added *strength of water)/ Initial quantity= 40/100
2. Let x be the initial quantity and y be the quantity removed
3.( (x-y)*7/15 + y*0) / x = 0.4, x/y=7/1
4. If initial quantity is 7 parts, 1 part is removed, if 15 parts 15/7=2.14
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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16 Jul 2018, 03:49
soumanag wrote:
Let the total solution is 150 L with 80 L water & 70 L syrup.
To make 40% syrup solution, the result solution must have 90 L syrup and 60 L syrup.
Therefore we are taking 10 L of syrup from initial solution and replacing with water.
using urinary method:
70 L syrup in 150 L solution
10 L syrup in 21.4 L solution
We started by multiplying 10
Now to get to the result we need to divide by 10 => amount of solution to be replaced with water = (21.4/10) = 2.14.
Correct option : C
Great solution. But I think you meant unitary method.
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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26 Aug 2019, 10:16
Though Bunuel and soumanag have already explained the solution well, I will add the method I like the most.
In replacement questions, focus on the thing that decreases. When solution is removed, water decreases but then water is added. While when solution is removed, lemonade decreases and does not get added later. So we will work with lemonade concentration.
The fraction of lemonade in the solution is 7/15
We need to get this fraction down to 2/5 (to make it 40%)
Let us say, we remove a fraction 'f' of the solution.
Then 7/15 - f * (7/15) = 2/5
f = 1/7
So (1/7)th of the solution has to be removed. But we want the answer in terms of parts (how many of the 15 parts have to be removed)
So we need to remove (1/7) * 15 = 2.14 parts
VeritasKarishma : Can we do this with scaling method? I've tried it, but got stuck somewhere in the middle.
Taking lemonade syrup into consideration,
7/15.....................4/10............................0
got the ratio as 1:6, but not sure how to get to the answer from here.
Since this method is a real time saver i would love to know how can we solve this using this scaling method.
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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26 Aug 2019, 21:01
Sreeragc wrote:
Though Bunuel and soumanag have already explained the solution well, I will add the method I like the most.
In replacement questions, focus on the thing that decreases. When solution is removed, water decreases but then water is added. While when solution is removed, lemonade decreases and does not get added later. So we will work with lemonade concentration.
The fraction of lemonade in the solution is 7/15
We need to get this fraction down to 2/5 (to make it 40%)
Let us say, we remove a fraction 'f' of the solution.
Then 7/15 - f * (7/15) = 2/5
f = 1/7
So (1/7)th of the solution has to be removed. But we want the answer in terms of parts (how many of the 15 parts have to be removed)
So we need to remove (1/7) * 15 = 2.14 parts
VeritasKarishma : Can we do this with scaling method? I've tried it, but got stuck somewhere in the middle.
Taking lemonade syrup into consideration,
7/15.....................4/10............................0
got the ratio as 1:6, but not sure how to get to the answer from here.
Since this method is a real time saver i would love to know how can we solve this using this scaling method.
I have discussed the scale method for it here:
https://gmatclub.com/forum/a-solution-c ... l#p1624227
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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30 Oct 2019, 10:17
Though Bunuel and soumanag have already explained the solution well, I will add the method I like the most.
In replacement questions, focus on the thing that decreases. When solution is removed, water decreases but then water is added. While when solution is removed, lemonade decreases and does not get added later. So we will work with lemonade concentration.
The fraction of lemonade in the solution is 7/15
We need to get this fraction down to 2/5 (to make it 40%)
Let us say, we remove a fraction 'f' of the solution.
Then 7/15 - f * (7/15) = 2/5
f = 1/7
So (1/7)th of the solution has to be removed. But we want the answer in terms of parts (how many of the 15 parts have to be removed)
So we need to remove (1/7) * 15 = 2.14 parts
Hi,
Suppose if the solution is 30L then water = 16L and lemonade syrup = 14L
Also, we have to make the lemonade 40% of the solution.
so, (14-x)/30 = 40%
x = 2
What am I doing wrong? can you anyone please explain?
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink]
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31 Oct 2019, 19:42
Hi Krish728,
The 'key' this question is that since we're dealing with a mixture, you cannot simply remove "1 part water" or "1 part syrup" - whatever you remove is a mix of the two ingredients. Your equation assumes that you can 'pour out' pure syrup from the mixture - which you can't.
My explanation (higher up in the thread) assumed that there were 15 liters total, but the same approach can be used if there were 30 liters total:
The prompt tells us to REPLACE some of the existing mixture with pure water (with the goal of turning the new mixture into a 40% syrup mix.
To start, we have 30 total liters -->a mixture that is 16 liters water and 14 liters syrup.
If we pour 1 liter of this mixture into a glass, we would have a liquid that is 14/30 = 7/15 syrup (so a little less than half syrup).
For the mixture to be 30 total liters and 40% syrup, we need the mixture to be 18 liters water and 12 liters syrup. In basic math terms, we need to pour out enough of the mixture that we remove 2 full liters of syrup; when we pour an equivalent amount of water back in, we'll have 30 total liters (and 12 of them will be syrup). Since each liter is 7/15 syrup......
We need to remove (2)(15/7) = 30/7 liters and replace them with 30/7 liters of pure water.
30/7 is a little more than 4 liters (about 4.28 liters). Remember that the prompt talks about the number of "PARTS" that need to be replaced though (not the number of liters) - since you 'doubled' all of the numbers to start off with 30 liters, we then have to reduce the result (by dividing by 2) to get the correct answer. 4.28/2 = 2.14
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Re: A solution contains 8 parts of water for every 7 parts of [#permalink] 31 Oct 2019, 19:42
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## Use geometric properties to evaluate probability
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Basic Geometric Probabilities
A rectangular dartboard that measures 12 inches by 24 inches has a 2-inch by 2-inch red square painted at its center. What is the probability that a dart that hits the dartboard will land in the red square?
### Geometric Probabilities
Sometimes we need to use our knowledge of geometry to determine the likelihood of an event occurring. We may use areas, volumes, angles, polygons or circles.
Let's solve the following problems.
1. A game of pin-the-tale-on-the-donkey has a rectangular poster that is 2ft by 2ft. The area in which the tale should be pinned is shown as a circle with radius 1 inch. Assuming that the pinning of the tale is completely random and that it will be pinned on the poster (or the player gets another try), find the probability of pinning the tale in the circle?
This probability can be found by dividing the area of the circle target by the area of the poster. We must have the same units of measure for each area so we will convert the feet to inches.
\begin{align*}\frac{1^2 \pi}{24^2} \thickapprox 0.005454 \ or \ \text{about} \ 0.5 \% \ \text{chance}.\end{align*}
1. In a game of chance, a pebble is dropped onto the board shown below. If the radius of each of blue circle is 1 cm, find the probability that the pebble will land in a blue circle.
The area of the square is \begin{align*}16 \ cm^2\end{align*}. The area of each of the 16 circles is \begin{align*}1^2 \pi=\pi\end{align*}. The probability of the pebble landing in a circle is the sum of the areas of the circles divided by the area of the square.
\begin{align*}P(\text{blue circle}) = \frac{16 \pi}{64} \thickapprox 0.785\end{align*}
1. What is the probability that a randomly thrown dart will land in a red area on the dart board shown? What is the probability that exactly two of three shots will land in the red? The radius of the inner circle is 1 unit and the radius of each annulus is 1 unit as well.
First we need to determine the probability of landing in the red. There are four rings of width 1 and the radius of the center circle is also 1 so the total radius is 5 units. The area of the whole target is thus \begin{align*}25 \pi\end{align*} square units. Now, we need to find the areas of the two red rings and the red circular center. The center circle area is \begin{align*}\pi\end{align*} square units. The outside ring area can be found by subtracting the area inside from the entire circle’s area. The inside circle will have a radius of 4 units, the area of the outer ring is \begin{align*}25 \pi -16 \pi=9 \pi\end{align*} square units. This smaller red ring’s area can be found similarly. The circle with this red ring on the outside has a radius of 3 and the circle inside has a radius of 2 so, \begin{align*}9 \pi -4 \pi=5 \pi\end{align*} square units. Finally, we can add them together to get the total red area and divide by the area of the entire target. \begin{align*}\frac{9 \pi+5 \pi+ \pi}{25 \pi}=\frac{15 \pi}{25 \pi}=\frac{3}{5}\end{align*}. So the probability of hitting the red area is \begin{align*}\frac{3}{5}\end{align*} or 60%.
For the second part of the problem we will use a binomial probability. There are 3 trials, 2 successes and the probability of a success is 0.6: \begin{align*}\dbinom{3}{2}(0.6)^2(0.4)=0.432\end{align*}
### Examples
#### Example 1
Earlier, you were asked to find the probability that a dart that hits the dartboard will land in the red square.
This probability can be found by dividing the area of the red square by the area of the dartboard. The area of the dartboard is \begin{align*}12\times24=288\end{align*}. The area of the red square is \begin{align*}2\times2=4\end{align*}. Therefore, the probability of the dart landing in the red square is
\begin{align*}\frac{4}{288}\\ \frac{1}{72}\\ \approx 0.0139\end{align*}
Therefore, there is about a 1.39% chance the dart will hit the red square.
#### Example 2
Consider the picture below. If a “circle” is randomly chosen, what is the probability that it will be:
1. red
\begin{align*}\frac{29}{225}\end{align*}
1. yellow
\begin{align*}\frac{69}{225}\end{align*}
1. blue or green
\begin{align*}\frac{84}{225}\end{align*}
1. not orange
\begin{align*}\frac{182}{225}\end{align*}
#### Example 3
If a dart is randomly thrown at the target below, find the probability of the dart hitting in each of the regions and show that the sum of these probabilities is 1. The diameter of the center circle is 4 cm and the diameter of the outer circle is 10 cm. What is the probability that in 5 shots, at least two will land in the 4 region?
\begin{align*}P(1)&=\frac{2^2 \pi}{5^2 \pi}=\frac{4}{25} \\ P(2)&=\frac{120}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{1}{3} \times \frac{21 \pi}{25 \pi}=\frac{7}{25} \\ P(3)&=\frac{90}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{1}{4} \times \frac{21 \pi}{25 \pi}=\frac{21}{100} \\ P(4)&=\frac{150}{360} \left(\frac{5^2 \pi- 2^2 \pi}{5^2 \pi}\right)=\frac{5}{12} \times \frac{21 \pi}{25 \pi}=\frac{35}{100}\end{align*}
\begin{align*}& P(1) + P(2) + P(3) + P(4) = \\ &=\frac{4}{25}+\frac{7}{25}+\frac{21}{100}+\frac{35}{100} \\ &=\frac{16}{100}+\frac{28}{100}+\frac{21}{100}+\frac{35}{100} \\ &=\frac{100}{100}=1\end{align*}
The probability of landing in region 4 at least twice in five shots is equivalent to \begin{align*}1 - \left [ P(0) + P(1) \right ]\end{align*}.
Use binomial probability to determine these probabilities:
\begin{align*}&1-\left [ \dbinom{5}{0} \left(\frac{35}{100}\right)^0 \left(\frac{65}{100}\right)^5+ \dbinom{5}{1} \left(\frac{35}{100}\right)^1 \left(\frac{65}{100}\right)4\right ] \\ &=1-(0.116029+0.062477) \\ & \thickapprox 0.821\end{align*}
### Review
Use the diagram below to find the probability that a randomly dropped object would land in a triangle of a particular color.
1. yellow
2. green
3. plum
4. not yellow
5. not yellow or light blue
The dart board to the right has a red center circle (bull’s eye) with area \begin{align*}\pi \ cm^2\end{align*}. Each ring surrounding this bull’s eye has a width of 2 cm. Use this information to answer the following questions.
1. Given a random throw of a dart, what is the probability that it will land in a white ring?
2. What is the probability of a bull’s eye?
3. What is the probability that in 10 throws, exactly 6 land in the black regions?
4. What is the probability that in 10 throws, at least one will land in the bull’s eye?
5. How many darts must be thrown to have a 95% chance of making a bull’s eye?
To see the Review answers, open this PDF file and look for section 12.11.
### Notes/Highlights Having trouble? Report an issue.
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Students of Grade 4 spend more time-solving multiplication topics such as mental math multiplication, multi-digit multiplication. Use our comprehensive collection of 4th Grade Math Multiplication Worksheets PDF for free and ace your preparation. Math Worksheets on Multiplication for Class 4 provided will have questions ranging from easy to difficult ones. Multiplying Worksheets with Answers makes it easy for you to develop mental math multiplication skills right in your head, multiply in columns, etc.
Do Check:
I. Multiply the numbers given below
1. Multiply the numbers 132 and 24
Solution:
1. Arrange the number 132 one on top as the multiplicand and 24 at the bottom. Line up the place values in columns.
2. Starting with the one’s digit of the bottom number, i.e. 4 × 2=8. Write the answer below the equals line
3. Multiply the one digit of the bottom number to the next digit to the left in the top number i.e. i.e. 4 × 3=12. Here the answer is greater than nine, write2 in the one’s place as the answer and carry the tens digit 1 Proceed right to left.
4. Multiply 4 with 1.i.e. 4 × 1=4. Add the carry forward digit 1 to 4.Write 5 as the answer.
5. Move to the tens digit in the bottom number i.e. 2. Multiply as above, but this time write your answers in a new row, shifted one digit place to the left.
6. Multiply 2 with 2 i.e. 2 × 2=4.
Multiply 2 with 3 i.e. 2 × 3=6
Multiply 2 with 1 i.e. 2 × 1=2
7. After multiplying, draw another answer line below the last row of answer numbers.
8. Use long addition to add the number columns from right to left, carrying as you normally do for long addition.
Therefore, the product of 132 and 24 is 3168.
2. Multiply the numbers 543 and 12
Solution:
1. Arrange the number 543 one on top as the multiplicand and 12 at the bottom. Line up the place values in columns.
2. Starting with the one’s digit of the bottom number, i.e. 2 × 3=6. Write the answer below the equals line
3. Multiply the one digit of the bottom number to the next digit to the left in the top number i.e. i.e. 2 × 4=8. Write 8 as the answer.
4. Move to the tens digit in the bottom number i.e. 1. Multiply as above, but this time write your answers in a new row, shifted one digit place to the left.
6. Multiply 1 with 3 i.e. 1 × 3=3.
Multiply 1 with 4 i.e. 1 × 4=4
Multiply 1 with 5 i.e. 1 × 5=5
7. After multiplying, draw another answer line below the last row of answer numbers.
8. Use long addition to add the number columns from right to left, carrying as you normally do for long addition.
Therefore, the product of 543 and 12 is 6516.
3. Multiply the numbers 2158 and 35
Solution:
1. Arrange the number 2158Â on top as the multiplicand and 35 at the bottom. Line up the place values in columns.
2. Starting with the one’s digit of the bottom number, i.e. 5 × 8=40. Write 0 as the answer and carry forward 4.
3. Multiply the one digit of the bottom number to the next digit to the left in the top number i.e. i.e. 5× 5=25. Add the carry forward 4 to the 25 i.e. 25+4=29. write 9 in the one’s place as the answer and carry the tens digit 2 Proceed right to left.
4. Multiply 5 with 1. i.e. 5 × 1=5. Add the carry forward digit 2 to 5.Write 7 as the answer.
5. Multiply 5 with 2. i.e. 5 × 2=10. Write 10 as the answer.
5. Move to the tens digit in the bottom number i.e. 3. Multiply as above, but this time write your answers in a new row, shifted one digit place to the left.
6. Multiply 3 with 8 i.e. 3 × 8=24. Write 4 as answer and carryforward 2.
Multiply 3 with 5 i.e. 3 × 5=15. Add the carry forward 2 to 15 i.e.15+2=17. Write 7 as the answer and carry forward 1.
Multiply 3 with 1 i.e. 3 × 1=3. Add the carry forward 1 to 3 i.e. 3+1=4. Write 4 as the answer.
Multiply 3 with 2 i.e. 3 × 2=6. Write 4 as the answer.
7. After multiplying, draw another answer line below the last row of answer numbers.
8. Use long addition to add the number columns from right to left, carrying as you normally do for long addition.
Therefore, the product of 2158 and 35 is 75530.
II.
1. Find the number which is twice of 3478?
Solution:
1.Start multiplying with one’s place digits. i.e. 2 × 8=16. Write 6 as the answer and carry forward 1.
2. Multiply tens. i.e. 2 × 4=8. Add carry forward 1 to 8 i.e. 8+1=9. Write 6 as the answer.
3. Multiply Hundreds. i.e. 2 × 7=14. Write 4 as the answer and carry forward 1.
4. Multiply thousands. i.e. 2 × 3=6. Add carry forward 1to 6 i.e. 6+1=7. Write 7 as the answer.
Therefore, twice as of 3478 is 7496.
2. Find the number which is thrice of 124?
Solution:
1.Start multiplying with one’s place digits. i.e. 3 × 4=12. Write 2 as the answer and carry forward 1.
2. Multiply tens. i.e. 3 × 2=6. Add carry forward 1 to 6 i.e. 6+1=7. Write 7 as the answer.
3. Multiply Hundreds. i.e. 3 × 1=3. Write 3 as the answer.
Therefore, thrice as of 124 is 372.
III. Estimate the product by rounding the given numbers
1.
S.No Question Rounded numbers Estimated product Exact product Which is Bigger
1 128 × 23 Nearest 10’s
2 250 ×  35 Nearest 10’s
3 570 × 48 Nearest 10’s
Solution:
S.No Question Rounded numbers Estimated product Exact product Which is Bigger
1 128 × 23 Nearest 10’s 442000 447496 Exact product
2 250 ×  35 Nearest 10’s 10000 8750 Estimated product
3 570 × 48 Nearest 10’s 28500 27360 estimated product
IV. Sandhya bought 8 chocolate packets to distribute on her birthday. Each packet has 110 chocolates. Find how many chocolates were there in 8 packets?
Solution:
Given,
No. of chocolate packets Sandhya bought=8
No. of chocolates in each packet=110
Total no. of chocolates in each packet=110 × 8=880
Therefore, No. of chocolates in 8 packets is 880.
V. Arjun’s monthly salary is Rs 35000. Arun’s monthly salary is three times that of Arjun. Find how much salary does Arun earns?
Solution:
Given,
Arjun’s monthly salary is =Rs 35000
Arun’s monthly salary is three times that of Arjun=3 × Rs 35000=Rs105000
Therefore, Aruns monthly salary is Rs 1,05,000.
VI. In a class there are 20 students. Each student has 15 books. Find out the total no. of books do students have?
Solution:
Given,
No. of students in the class=20
No. of books each student have=15
Total no. of books students have=20 × 15=300
Therefore, students have a total of 300 books.
VII. There are 250 papers in a bundle. How many papers were there in 6 bundles?
Solution:
Given,
No. of papers in a bundle=250
No. of papers in 6 bundles=250 × 6=1500.
Therefore, the total no. of papers in 6 bundles is 1500.
VIII. Raju does typing work and earns money of Rs 530 every day. Find out how much money will Raju earn if he works for 30 days?
Solution:
Given,
Raju earns money every day=Rs 530
Money earned by Raju if he works for 30 days=Rs 530 × 30=Rs 15900
Therefore, the total money earned by Raju in 30 days is Rs 15900.
1. 638 × 16=_____
a. 10250
b. 10208Â Â Â b
c. 10100
2. 848 × 24=______
a. 21524
b. 20245
c. 20352Â Â c
3. The number to be multiplied by another is called ______
a. Product
b. multiplier
c. Multiplicand  c
4. The number that you are multiplying by is called ______
a. Multiplier
b. Product
c. Multiplicand
5. 25 × 3 × 0=______
a. 0
b. 75
c. 85
6. Thrice of 65 =_____
a. 165
b. 195
c. 155
7. The product of 6 and 8 is same as the product of 4 and _____
a. 8
b. 12
c. 14
Solution:
1. b
2. c
3. c
4. a
5. a
6. b
7. b
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### Section 9-1 : The Heat Equation
Before we get into actually solving partial differential equations and before we even start discussing the method of separation of variables we want to spend a little bit of time talking about the two main partial differential equations that we’ll be solving later on in the chapter. We’ll look at the first one in this section and the second one in the next section.
The first partial differential equation that we’ll be looking at once we get started with solving will be the heat equation, which governs the temperature distribution in an object. We are going to give several forms of the heat equation for reference purposes, but we will only be really solving one of them.
We will start out by considering the temperature in a 1-D bar of length $$L$$. What this means is that we are going to assume that the bar starts off at $$x = 0$$ and ends when we reach $$x = L$$. We are also going to so assume that at any location, $$x$$ the temperature will be constant at every point in the cross section at that $$x$$. In other words, temperature will only vary in $$x$$ and we can hence consider the bar to be a 1-D bar. Note that with this assumption the actual shape of the cross section (i.e. circular, rectangular, etc.) doesn’t matter.
Note that the 1-D assumption is actually not all that bad of an assumption as it might seem at first glance. If we assume that the lateral surface of the bar is perfectly insulated (i.e. no heat can flow through the lateral surface) then the only way heat can enter or leave the bar as at either end. This means that heat can only flow from left to right or right to left and thus creating a 1-D temperature distribution.
The assumption of the lateral surfaces being perfectly insulated is of course impossible, but it is possible to put enough insulation on the lateral surfaces that there will be very little heat flow through them and so, at least for a time, we can consider the lateral surfaces to be perfectly insulated.
Okay, let’s now get some definitions out of the way before we write down the first form of the heat equation.
\begin{align*}&u\left( {x,t} \right) = {\mbox{Temperature at any point }}x{\mbox{ and any time }}t\\ & c\left( x \right) = {\mbox{Specific Heat}}\\ & \rho \left( x \right) = {\mbox{Mass Density}}\\ & \varphi \left( {x,t} \right) = {\mbox{Heat Flux}}\\ & Q\left( {x,t} \right) = {\mbox{Heat energy generated per unit volume per unit time}}\end{align*}
We should probably make a couple of comments about some of these quantities before proceeding.
The specific heat, $$c\left( x \right) > 0$$, of a material is the amount of heat energy that it takes to raise one unit of mass of the material by one unit of temperature. As indicated we are going to assume, at least initially, that the specific heat may not be uniform throughout the bar. Note as well that in practice the specific heat depends upon the temperature. However, this will generally only be an issue for large temperature differences (which in turn depends on the material the bar is made out of) and so we’re going to assume for the purposes of this discussion that the temperature differences are not large enough to affect our solution.
The mass density, $$\rho \left( x \right)$$, is the mass per unit volume of the material. As with the specific heat we’re going to initially assume that the mass density may not be uniform throughout the bar.
The heat flux, $$\varphi \left( {x,t} \right)$$, is the amount of thermal energy that flows to the right per unit surface area per unit time. The “flows to the right” bit simply tells us that if $$\varphi \left( {x,t} \right) > 0$$ for some $$x$$ and $$t$$ then the heat is flowing to the right at that point and time. Likewise, if $$\varphi \left( {x,t} \right) < 0$$ then the heat will be flowing to the left at that point and time.
The final quantity we defined above is $$Q\left( {x,t} \right)$$ and this is used to represent any external sources or sinks (i.e. heat energy taken out of the system) of heat energy. If $$Q\left( {x,t} \right) > 0$$ then heat energy is being added to the system at that location and time and if $$Q\left( {x,t} \right) < 0$$ then heat energy is being removed from the system at that location and time.
With these quantities the heat equation is,
$$$c\left( x \right)\rho \left( x \right)\frac{{\partial u}}{{\partial t}} = - \frac{{\partial \varphi }}{{\partial x}} + Q\left( {x,t} \right)\label{eq:eq1}$$$
While this is a nice form of the heat equation it is not actually something we can solve. In this form there are two unknown functions, $$u$$ and $$\varphi$$, and so we need to get rid of one of them. With Fourier’s law we can easily remove the heat flux from this equation.
Fourier’s law states that,
$\varphi \left( {x,t} \right) = - {K_0}\left( x \right)\frac{{\partial u}}{{\partial x}}$
where $${K_0}\left( x \right) > 0$$ is the thermal conductivity of the material and measures the ability of a given material to conduct heat. The better a material can conduct heat the larger $${K_0}\left( x \right)$$ will be. As noted the thermal conductivity can vary with the location in the bar. Also, much like the specific heat the thermal conductivity can vary with temperature, but we will assume that the total temperature change is not so great that this will be an issue and so we will assume for the purposes here that the thermal conductivity will not vary with temperature.
Fourier’s law does a very good job of modeling what we know to be true about heat flow. First, we know that if the temperature in a region is constant, i.e.$$\frac{{\partial u}}{{\partial x}} = 0$$, then there is no heat flow.
Next, we know that if there is a temperature difference in a region we know the heat will flow from the hot portion to the cold portion of the region. For example, if it is hotter to the right then we know that the heat should flow to the left. When it is hotter to the right then we also know that $$\frac{{\partial u}}{{\partial x}} > 0$$ (i.e. the temperature increases as we move to the right) and so we’ll have $$\varphi < 0$$ and so the heat will flow to the left as it should. Likewise, if $$\frac{{\partial u}}{{\partial x}} < 0$$ (i.e. it is hotter to the left) then we’ll have $\varphi > 0$ and heat will flow to the right as it should.
Finally, the greater the temperature difference in a region (i.e. the larger $$\frac{{\partial u}}{{\partial x}}$$ is) then the greater the heat flow.
So, if we plug Fourier’s law into $$\eqref{eq:eq1}$$, we get the following form of the heat equation,
$$$c\left( x \right)\rho \left( x \right)\frac{{\partial u}}{{\partial t}} = \frac{\partial }{{\partial x}}\left( {{K_0}\left( x \right)\frac{{\partial u}}{{\partial x}}} \right) + Q\left( {x,t} \right)\label{eq:eq2}$$$
Note that we factored the minus sign out of the derivative to cancel against the minus sign that was already there. We cannot however, factor the thermal conductivity out of the derivative since it is a function of $$x$$ and the derivative is with respect to $$x$$.
Solving $$\eqref{eq:eq2}$$ is quite difficult due to the non uniform nature of the thermal properties and the mass density. So, let’s now assume that these properties are all constant, i.e.,
$c\left( x \right) = c\hspace{0.25in}\rho \left( x \right) = \rho \hspace{0.25in}{K_0}\left( x \right) = {K_0}$
where $$c$$, $$\rho$$ and $${K_0}$$ are now all fixed quantities. In this case we generally say that the material in the bar is uniform. Under these assumptions the heat equation becomes,
$$$c\rho \frac{{\partial u}}{{\partial t}} = {K_0}\frac{{{\partial ^2}u}}{{\partial {x^2}}} + Q\left( {x,t} \right)\label{eq:eq3}$$$
For a final simplification to the heat equation let’s divide both sides by $$c\rho$$ and define the thermal diffusivity to be,
$k = \frac{{{K_0}}}{{c\rho }}$
The heat equation is then,
$$$\frac{{\partial u}}{{\partial t}} = k\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{Q\left( {x,t} \right)}}{{c\rho }}\label{eq:eq4}$$$
To most people this is what they mean when they talk about the heat equation and in fact it will be the equation that we’ll be solving. Well, actually we’ll be solving $$\eqref{eq:eq4}$$ with no external sources, i.e.$$Q\left( {x,t} \right) = 0$$, but we’ll be considering this form when we start discussing separation of variables in a couple of sections. We’ll only drop the sources term when we actually start solving the heat equation.
Now that we’ve got the 1-D heat equation taken care of we need to move into the initial and boundary conditions we’ll also need in order to solve the problem. If you go back to any of our solutions of ordinary differential equations that we’ve done in previous sections you can see that the number of conditions required always matched the highest order of the derivative in the equation.
In partial differential equations the same idea holds except now we have to pay attention to the variable we’re differentiating with respect to as well. So, for the heat equation we’ve got a first order time derivative and so we’ll need one initial condition and a second order spatial derivative and so we’ll need two boundary conditions.
The initial condition that we’ll use here is,
$u\left( {x,0} \right) = f\left( x \right)$
and we don’t really need to say much about it here other than to note that this just tells us what the initial temperature distribution in the bar is.
The boundary conditions will tell us something about what the temperature and/or heat flow is doing at the boundaries of the bar. There are four of them that are fairly common boundary conditions.
The first type of boundary conditions that we can have would be the prescribed temperature boundary conditions, also called Dirichlet conditions. The prescribed temperature boundary conditions are,
$u\left( {0,t} \right) = {g_1}\left( t \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}u\left( {L,t} \right) = {g_2}\left( t \right)$
The next type of boundary conditions are prescribed heat flux, also called Neumann conditions. Using Fourier’s law these can be written as,
$- {K_0}\left( 0 \right)\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = {\varphi _1}\left( t \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} - {K_0}\left( L \right)\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = {\varphi _2}\left( t \right)$
If either of the boundaries are perfectly insulated, i.e. there is no heat flow out of them then these boundary conditions reduce to,
$\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = 0\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = 0$
and note that we will often just call these particular boundary conditions insulated boundaries and drop the “perfectly” part.
The third type of boundary conditions use Newton’s law of cooling and are sometimes called Robins conditions. These are usually used when the bar is in a moving fluid and note we can consider air to be a fluid for this purpose.
Here are the equations for this kind of boundary condition.
$- {K_0}\left( 0 \right)\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = - H\left[ {u\left( {0,t} \right) - {g_1}\left( t \right)} \right]\hspace{0.25in} - {K_0}\left( L \right)\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = H\left[ {u\left( {L,t} \right) - {g_2}\left( t \right)} \right]$
where $$H$$ is a positive quantity that is experimentally determined and $${g_1}\left( t \right)$$ and $${g_2}\left( t \right)$$ give the temperature of the surrounding fluid at the respective boundaries.
Note that the two conditions do vary slightly depending on which boundary we are at. At $$x = 0$$ we have a minus sign on the right side while we don’t at $$x = L$$. To see why this is let’s first assume that at $$x = 0$$ we have $$u\left( {0,t} \right) > {g_1}\left( t \right)$$. In other words, the bar is hotter than the surrounding fluid and so at $$x = 0$$ the heat flow (as given by the left side of the equation) must be to the left, or negative since the heat will flow from the hotter bar into the cooler surrounding liquid. If the heat flow is negative then we need to have a minus sign on the right side of the equation to make sure that it has the proper sign.
If the bar is cooler than the surrounding fluid at $$x = 0$$, i.e. $$u\left( {0,t} \right) < {g_1}\left( t \right)$$ we can make a similar argument to justify the minus sign. We’ll leave it to you to verify this.
If we now look at the other end, $$x = L$$, and again assume that the bar is hotter than the surrounding fluid or, $$u\left( {L,t} \right) > {g_2}\left( t \right)$$. In this case the heat flow must be to the right, or be positive, and so in this case we can’t have a minus sign. Finally, we’ll again leave it to you to verify that we can’t have the minus sign at $$x = L$$ is the bar is cooler than the surrounding fluid as well.
Note that we are not actually going to be looking at any of these kinds of boundary conditions here. These types of boundary conditions tend to lead to boundary value problems such as Example 5 in the Eigenvalues and Eigenfunctions section of the previous chapter. As we saw in that example it is often very difficult to get our hands on the eigenvalues and as we’ll eventually see we will need them.
It is important to note at this point that we can also mix and match these boundary conditions so to speak. There is nothing wrong with having a prescribed temperature at one boundary and a prescribed flux at the other boundary for example so don’t always expect the same boundary condition to show up at both ends. This warning is more important that it might seem at this point because once we get into solving the heat equation we are going to have the same kind of condition on each end to simplify the problem somewhat.
The final type of boundary conditions that we’ll need here are periodic boundary conditions. Periodic boundary conditions are,
$u\left( { - L,t} \right) = u\left( {L,t} \right)\hspace{0.25in}\hspace{0.25in}\frac{{\partial u}}{{\partial x}}\left( { - L,t} \right) = \frac{{\partial u}}{{\partial x}}\left( {L,t} \right)$
Note that for these kinds of boundary conditions the left boundary tends to be $$x = - L$$ instead of $$x = 0$$ as we were using in the previous types of boundary conditions. The periodic boundary conditions will arise very naturally from a couple of particular geometries that we’ll be looking at down the road.
We will now close out this section with a quick look at the 2-D and 3-D version of the heat equation. However, before we jump into that we need to introduce a little bit of notation first.
The del operator is defined to be,
$\nabla = \frac{\partial }{{\partial x}}\vec i + \frac{\partial }{{\partial y}}\vec j\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\nabla = \frac{\partial }{{\partial x}}\vec i + \frac{\partial }{{\partial y}}\vec j + \frac{\partial }{{\partial z}}\vec k$
depending on whether we are in 2 or 3 dimensions. Think of the del operator as a function that takes functions as arguments (instead of numbers as we’re used to). Whatever function we “plug” into the operator gets put into the partial derivatives.
So, for example in 3-D we would have,
$\nabla f = \frac{{\partial f}}{{\partial x}}\vec i + \frac{{\partial f}}{{\partial y}}\vec j + \frac{{\partial f}}{{\partial z}}\vec k$
This of course is also the gradient of the function $$f\left( {x,y,z} \right)$$.
The del operator also allows us to quickly write down the divergence of a function. So, again using 3-D as an example the divergence of $$f\left( {x,y,z} \right)$$ can be written as the dot product of the del operator and the function. Or,
$\nabla \centerdot f = \frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial y}} + \frac{{\partial f}}{{\partial z}}$
Finally, we will also see the following show up in our work,
$\nabla \centerdot \left( {\nabla f} \right) = \frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial x}}} \right) + \frac{\partial }{{\partial y}}\left( {\frac{{\partial f}}{{\partial y}}} \right) + \frac{\partial }{{\partial z}}\left( {\frac{{\partial f}}{{\partial k}}} \right) = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}$
This is usually denoted as,
${\nabla ^2}f = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}$
and is called the Laplacian. The 2-D version of course simply doesn’t have the third term.
Okay, we can now look into the 2-D and 3-D version of the heat equation and where ever the del operator and or Laplacian appears assume that it is the appropriate dimensional version.
The higher dimensional version of $$\eqref{eq:eq1}$$ is,
$$$c\rho \frac{{\partial u}}{{\partial t}} = - \nabla \centerdot \,\varphi + Q\label{eq:eq5}$$$
and note that the specific heat, $$c$$, and mass density, $$\rho$$, are may not be uniform and so may be functions of the spatial variables. Likewise, the external sources term, $$Q$$, may also be a function of both the spatial variables and time.
Next, the higher dimensional version of Fourier’s law is,
$\varphi = - {K_0}\nabla u$
where the thermal conductivity, $${K_0}$$, is again assumed to be a function of the spatial variables.
If we plug this into $$\eqref{eq:eq5}$$ we get the heat equation for a non uniform bar (i.e. the thermal properties may be functions of the spatial variables) with external sources/sinks,
$c\rho \frac{{\partial u}}{{\partial t}} = \nabla \centerdot \,\left( {{K_0}\nabla u} \right) + Q$
If we now assume that the specific heat, mass density and thermal conductivity are constant (i.e. the bar is uniform) the heat equation becomes,
$$$\frac{{\partial u}}{{\partial t}} = k{\nabla ^2}u + \frac{Q}{{cp}}\label{eq:eq6}$$$
where we divided both sides by $$c\rho$$ to get the thermal diffusivity, $$k$$ in front of the Laplacian.
The initial condition for the 2-D or 3-D heat equation is,
$u\left( {x,y,t} \right) = f\left( {x,y} \right)\hspace{0.25in}\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\hspace{0.25in}u\left( {x,y,z,t} \right) = f\left( {x,y,z} \right)$
depending upon the dimension we’re in.
The prescribed temperature boundary condition becomes,
$u\left( {x,y,t} \right) = T\left( {x,y,t} \right)\hspace{0.25in}\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\hspace{0.25in}u\left( {x,y,z,t} \right) = T\left( {x,y,z,t} \right)$
where $$\left( {x,y} \right)$$ or $$\left( {x,y,z} \right)$$, depending upon the dimension we’re in, will range over the portion of the boundary in which we are prescribing the temperature.
The prescribed heat flux condition becomes,
$- {K_0}\nabla u\,\centerdot \,\vec n = \varphi \left( t \right)$
where the left side is only being evaluated at points along the boundary and $$\vec n$$ is the outward unit normal on the surface.
Newton’s law of cooling will become,
$- {K_0}\nabla u\,\centerdot \,\vec n = H\left( {u - {u_B}} \right)$
where $$H$$ is a positive quantity that is experimentally determine, $${u_B}$$ is the temperature of the fluid at the boundary and again it is assumed that this is only being evaluated at points along the boundary.
We don’t have periodic boundary conditions here as they will only arise from specific 1?D geometries.
We should probably also acknowledge at this point that we’ll not actually be solving $$\eqref{eq:eq6}$$ at any point, but we will be solving a special case of it in the Laplace’s Equation section. | 5,527 | 20,477 | {"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": 6, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-21 | latest | en | 0.936154 |
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Dear Readers, Bank Exam Race for theย Year 2018 is already started, Toย enrichย your preparation here we have providing new series of Practice Questions onย Quantitative Aptitude – Section. Candidates those who are preparing forย IBPS PO Mains 2018 Exams can practice these questions daily and make your preparation effective.
IBPS PO Mains 2018- Quantitative Aptitude Questions Day โ 23
maximum of 10 points
1) Mano sold a book at a loss of 13 %. If he had sold that book for Rs. 84 more, he would have gained 15 %. Find the selling price of the book to get a profit of 20 %?
a) Rs. 360
b) Rs. 420
c) Rs. 480
d) Rs. 500
e) None of these
2) A certain sum is divided among A, B and C in such a way that, A gets 120 more than two-fifth of the sum. B gets 190 less than one-fourth of the sum and C gets Rs. 420. What is the 65 % of the total sum?
a) Rs. 650
b) Rs. 500
c) Rs. 400
d) Rs. 525
e) None of these
3) A can do 1/3 of the work in 5 days, B can do 2/5 of the work in 10 days. Then in how many days both of them together can do the work?
a) 7 ยพ days
b) 8 5/6 days
c) 10 7/9 days
d) 9 3/8 days
e) None of these
4) The simple interest accrued on an amount of Rs. 15,000 at the end of two years is Rs. 2400. What would be corresponding Compound Interest?
a) Rs. 2496
b) Rs. 2512
c) Rs. 2538
d) Rs. 2474
e) None of these
5) In what ratio water be mixed with milk to earn a profit of 40 % by selling the mixture at cost price?
a) 7 : 4
b) 5 : 2
c) 6 : 5
d) 2 : 7
e) None of these
Directions (Q. 6 โ 10) Study the following information carefully and answer the given questions:
The following table shows the percentage distribution of expenses of four departments in 4 different years.
ย
Total expenses of these 4 departments
In the year 2012 = 60 lakhs
In the year 2013 = 68 lakhs
In the year 2014 = 72 lakhs
In the year 2015 = 75 lakhs
Note:
Profit = Income โ Expense
Profit % = (Profit/Expense)*100
6) If the profit % of Finance department in the year 2012 is 20 %, then find the total income of finance department in the year 2012?
a) 15.2 lakhs
b) 14.4 lakhs
c) 13.8 lakhs
d) 16.4 lakhs
e) None of these
7) Find the difference between the total profit earned by marketing department in the year 2013 to that of total expenses of R & D department in the year 2014 and 2015 together, if the profit % of marketing department in the year 2013 is 15 %?
a) 2895000
b) 2913000
c) 3056000
d) 3117000
e) None of these
8) Find the average expenses of agriculture in all the given years together?
a) 1816500
b) 1578200
c) 1649400
d) 1752600
e) None of these
9) If the profit earned by marketing department and R & D department in the year 2014 and 2015 respectively is 4.5 lakhs and 3 lakhs, then find the profit % of marketing department is what percentage of profit % of R & D department?
a) 110 %
b) 135 %
c) 125 %
d) 100 %
e) 90 %
10) Find the ratio between the total expenses of marketing and agriculture department in the year 2012 to that of total expenses of agriculture department in the year 2015?
a) 4 : 3
b) 5 : 2
c) 11 : 9
d) 23 : 17
e) None of these
Let the cost price be Rs. X,
X*(87/100) + 84 = x*(115/100)
(115x/100) โ (87x/100) = 84
28x/100 = 84
X = 84*(100/28)
X = 300
Cost price = Rs. 300
Selling price = 300*(120/100) = Rs. 360
Total sum = X = A + B + C
A= (2/5)*X + 120
B = X/4 – 190
C = 420
Total sum(X) = (2X/5) + 120 + (X/4) – 190 + 420
X= (8X + 5X)/20 + 350
X โ (13x/20) = 350
7x/20 = 350
X = 1000
65 % of the total sum
= > 1000*(65/100) = Rs. 650
A can do the work in,
= > (1/3)*work = 5
= > Whole work = 15 days
B can do the work in,
= > (2/5)*work = 10
= > Whole work = 25 days
(A + B)โs one day work = (1/15) + (1/25) = 40/(15*25) = 8/75
(A + B) can do the work in, 75/8 = 9 3/8 days
S.I = PNR/100
2400 = (15000*2*R)/100
R = (2400*100)/30000
R = 8 %
Corresponding CI:
15000*(8/100) = 1200
16200*(8/100) = 1296
C.I = 1200 + 1296 = Rs. 2496
Let the cost price be Rs. 1,
Selling price of mixture = Rs. 1
Cost price of mixture = 1*100/140 = 5/7
= > – (5/7) : – (2/7)
= > 5 : 2
Direction (6-10) :
Total expenses of 4 departments in the year 2012 = 60 lakhs
Total expenses of finance department in the year 2012
= > 60*(20/100) = 12 lakhs
Profit % = 20 %
Total income of finance department in the year 2012
= > Income = 12*(120/100)
= > Income = 14.4 lakhs
The total profit earned by marketing department in the year 2013
Expense of marketing in the year 2013 = 68*(25/100) = 17 lakhs
Income of marketing in the year 2013 = 17*(115/100) = 19.55 lakhs
Profit = 2.55 lakhs
Total expenses of R & D department in the year 2014 and 2015 together
= > 72*(26/100) + 75*(20/100)
= > 18.72 + 15 = 33.72 lakhs
Required difference = 33.72 โ 2.55 = 31.17 lakhs = 3117000
The total expenses of agriculture in all the given years together
= > 60*(28/100) + 68*(30/100) + 72*(18/100) + 75*(30/100)
= > 16.8 + 20.4 + 12.96 + 22.5
= > 72.66 lakhs
Required average = 72.66/4 = 18.165 lakhs = 1816500
The profit earned by marketing department the year 2014 = 4.5 lakhs
Expense of marketing department the year 2014 = 72*(25/100) = 18 lakhs
Profit % of marketing department the year 2014 = (4.5/18)*100 = 25 %
The profit earned by R & D department the year 2015 = 3 lakhs
Expense of R & D department the year 2015 = 75*(20/100) = 15 lakhs
Profit % of R & D department the year 2015 = (3/15)*100 = 20 %
Required % = (25/20)*100 = 125 %
The total expenses of marketing and agriculture department in the year 2012
= > 60*(50/100) = 30 lakhs
The total expenses of agriculture department in the year 2015
= > 75*(30/100) = 22.5 lakhs
Required ratio = 30 : 22.5 = 300 : 225 = 4 : 3
Daily Practice Test Schedule | Good Luck
Topic Daily Publishing Time Daily News Papers & Editorials 8.00 AM Current Affairs Quiz 9.00 AM Current Affairs Quiz (Hindi) 9.30 AM IBPS Clerk Prelims – Reasoning 10.00 AM IBPS Clerk Prelims โ Reasoning (Hindi) 10.30 AM IBPS Clerk Prelims โ Quantitative Aptitude 11.00 AM IBPS Clerk Prelims โ Quantitative Aptitude (Hindi) 11.30 AM Vocabulary (Based on The Hindu) 12.00 PM IBPS PO Prelims โ English Language 1.00 PM SSC Practice Questions (Reasoning/Quantitative aptitude) 2.00 PM IBPS PO/Clerk โ GK Questions 3.00 PM SSC Practice Questions (English/General Knowledge) 4.00 PM Daily Current Affairs Updates 5.00 PM IBPS PO Mains – Reasoning 6.00 PM IBPS PO Mains โ Quantitative Aptitude 7.00 PM IBPS PO Mains โ English Language 8.00 PM
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https://math.libretexts.org/Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/03%3A_Linear_Transformations_and_Matrix_Algebra/3.01%3A_Matrix_Transformations | 1,653,293,314,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00678.warc.gz | 444,497,591 | 33,276 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.1: Matrix Transformations
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
##### Objectives
1. Learn to view a matrix geometrically as a function.
2. Learn examples of matrix transformations: reflection, dilation, rotation, shear, projection.
3. Understand the vocabulary surrounding transformations: domain, codomain, range.
4. Understand the domain, codomain, and range of a matrix transformation.
5. Pictures: common matrix transformations.
6. Vocabulary words: transformation / function, domain, codomain, range, identity transformation, matrix transformation.
In this section we learn to understand matrices geometrically as functions, or transformations. We briefly discuss transformations in general, then specialize to matrix transformations, which are transformations that come from matrices.
## Matrices as Functions
Informally, a function is a rule that accepts inputs and produces outputs. For instance, $$f(x) = x^2$$ is a function that accepts one number $$x$$ as its input, and outputs the square of that number: $$f(2) = 4$$. In this subsection, we interpret matrices as functions.
Let $$A$$ be a matrix with $$m$$ rows and $$n$$ columns. Consider the matrix equation $$b=Ax$$ (we write it this way instead of $$Ax=b$$ to remind the reader of the notation $$y=f(x)$$). If we vary $$x\text{,}$$ then $$b$$ will also vary; in this way, we think of $$A$$ as a function with independent variable $$x$$ and dependent variable $$b$$.
• The independent variable (the input) is $$x\text{,}$$ which is a vector in $$\mathbb{R}^n$$.
• The dependent variable (the output) is $$b\text{,}$$ which is a vector in $$\mathbb{R}^m$$.
The set of all possible output vectors are the vectors $$b$$ such that $$Ax=b$$ has some solution; this is the same as the column space of $$A$$ by Note 2.3.6 in Section 2.3.
Figure $$\PageIndex{1}$$
##### Example $$\PageIndex{3}$$: Projection onto the $$xy$$-plane
Let
$A=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^3$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right)\:\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}x\\y\\0\end{array}\right).\nonumber$
Multiplication by $$A$$ simply sets the $$z$$-coordinate equal to zero: it projects vertically onto the $$xy$$-plane
Figure $$\PageIndex{4}$$
##### Example $$\PageIndex{4}$$: Reflection
Let
$A=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}-x\\y\end{array}\right).\nonumber$
Multiplication by $$A$$ negates the $$x$$-coordinate: it reflects over the $$y$$-axis.
Figure $$\PageIndex{6}$$
##### Example $$\PageIndex{5}$$: Dilation
Let
$A=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}1.5x\\1.5y\end{array}\right)=1.5\left(\begin{array}{c}x\\y\end{array}\right).\nonumber$
Multiplication by $$A$$ is the same as scalar multiplication by $$1.5\text{:}$$ it scales or dilates the plane by a factor of $$1.5$$.
Figure $$\PageIndex{8}$$
##### Example $$\PageIndex{6}$$: Identity
Let
$A=\left(\begin{array}{cc}1&0\\0&1\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x\\y\end{array}\right).\nonumber$
Multiplication by $$A$$ does not change the input vector at all: it is the identity transformation which does nothing
Figure $$\PageIndex{10}$$
##### Example $$\PageIndex{7}$$: Rotation
Let
$A=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}-y\\x\end{array}\right).\nonumber$
We substitute a few test points in order to understand the geometry of the transformation:
Figure $$\PageIndex{12}$$
Multiplication by $$A$$ is counterclockwise rotation by $$90^\circ$$.
Figure $$\PageIndex{13}$$
##### Example $$\PageIndex{8}$$: Shear
Let
$A=\left(\begin{array}{cc}1&1\\0&1\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1&1\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x+y\\y\end{array}\right).\nonumber$
Multiplication by $$A$$ adds the $$y$$-coordinate to the $$x$$-coordinate; this is called a shear in the $$x$$-direction
Figure $$\PageIndex{15}$$
## Transformations
At this point it is convenient to fix our ideas and terminology regarding functions, which we will call transformations in this book. This allows us to systematize our discussion of matrices as functions.
##### Definition $$\PageIndex{1}$$: Transformation
A transformation from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$ is a rule $$T$$ that assigns to each vector $$x$$ in $$\mathbb{R}^n$$ a vector $$T(x)$$ in $$\mathbb{R}^m$$.
• $$\mathbb{R}^n$$ is called the domain of $$T$$.
• $$\mathbb{R}^m$$ is called the codomain of $$T$$.
• For $$x$$ in $$\mathbb{R}^n \text{,}$$ the vector $$T(x)$$ in $$\mathbb{R}^m$$ is the image of $$x$$ under $$T$$.
• The set of all images $$\{T(x)\mid x\text{ in }\mathbb{R}^n \}$$ is the range of $$T$$.
The notation $$T\colon\mathbb{R}^n \to \mathbb{R}^m$$ means “$$T$$ is a transformation from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$.”
It may help to think of $$T$$ as a “machine” that takes $$x$$ as an input, and gives you $$T(x)$$ as the output.
Figure $$\PageIndex{17}$$
The points of the domain $$\mathbb{R}^n$$ are the inputs of $$T\text{:}$$ this simply means that it makes sense to evaluate $$T$$ on vectors with $$n$$ entries, i.e., lists of $$n$$ numbers. Likewise, the points of the codomain $$\mathbb{R}^m$$ are the outputs of $$T\text{:}$$ this means that the result of evaluating $$T$$ is always a vector with $$m$$ entries.
The range of $$T$$ is the set of all vectors in the codomain that actually arise as outputs of the function $$T\text{,}$$ for some input. In other words, the range is all vectors $$b$$ in the codomain such that $$T(x)=b$$ has a solution $$x$$ in the domain.
##### Example $$\PageIndex{9}$$: A Function of one variable
Most of the functions you may have seen previously have domain and codomain equal to $$\mathbb{R} = \mathbb{R}^1$$. For example,
$\sin :\mathbb{R}\to\mathbb{R}\quad\sin(x)=\left(\begin{array}{l}\text{the length of the opposite} \\ \text{edge over the hypotenuse of} \\ \text{a right triangle with angle }x \\ \text{in radians}\end{array}\right).\nonumber$
Notice that we have defined $$\sin$$ by a rule: a function is defined by specifying what the output of the function is for any possible input.
You may be used to thinking of such functions in terms of their graphs:
Figure $$\PageIndex{18}$$
In this case, the horizontal axis is the domain, and the vertical axis is the codomain. This is useful when the domain and codomain are $$\mathbb{R}\text{,}$$ but it is hard to do when, for instance, the domain is $$\mathbb{R}^2$$ and the codomain is $$\mathbb{R}^3$$. The graph of such a function is a subset of $$\mathbb{R}^5\text{,}$$ which is difficult to visualize. For this reason, we will rarely graph a transformation.
Note that the range of $$\sin$$ is the interval $$[-1,1]\text{:}$$ this is the set of all possible outputs of the $$\sin$$ function.
##### Example $$\PageIndex{10}$$: Functions of several variables
Here is an example of a function from $$\mathbb{R}^2$$ to $$\mathbb{R}^3 \text{:}$$
$f\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x+y\\ \cos(y) \\y-x^{2}\end{array}\right).\nonumber$
The inputs of $$f$$ each have two entries, and the outputs have three entries. In this case, we have defined $$f$$ by a formula, so we evaluate $$f$$ by substituting values for the variables:
$f\left(\begin{array}{c}2\\3\end{array}\right)=\left(\begin{array}{c}2+3 \\ \cos(3) \\ 3-2^2\end{array}\right)=\left(\begin{array}{c}5\\ \cos(3)\\-1\end{array}\right).\nonumber$
Here is an example of a function from $$\mathbb{R}^3$$ to $$\mathbb{R}^3 \text{:}$$
$f(v)=\left(\begin{array}{c}\text{the counterclockwise rotation} \\ \text{of }v\text{ by and angle of }42^\circ \text{ about} \\ \text{ the }z\text{-axis}\end{array}\right).\nonumber$
In other words, $$f$$ takes a vector with three entries, then rotates it; hence the ouput of $$f$$ also has three entries. In this case, we have defined $$f$$ by a geometric rule.
##### Definition $$\PageIndex{2}$$: Identity Transformation
The identity transformation $$\text{Id}_{\mathbb{R}^n }\colon\mathbb{R}^n \to\mathbb{R}^n$$ is the transformation defined by the rule
$\text{Id}_{\mathbb{R}^n }(x) = x \qquad\text{for all x in \mathbb{R}^n }. \nonumber$
In other words, the identity transformation does not move its input vector: the output is the same as the input. Its domain and codomain are both $$\mathbb{R}^n \text{,}$$ and its range is $$\mathbb{R}^n$$ as well, since every vector in $$\mathbb{R}^n$$ is the output of itself.
##### Example $$\PageIndex{11}$$: A real-word transformation: robotics
The definition of transformation and its associated vocabulary may seem quite abstract, but transformations are extremely common in real life. Here is an example from the fields of robotics and computer graphics.
Suppose you are building a robot arm with three joints that can move its hand around a plane, as in the following picture.
Figure $$\PageIndex{19}$$
Define a transformation $$f\colon\mathbb{R}^3 \to\mathbb{R}^2$$ as follows: $$f(\theta,\phi,\psi)$$ is the $$(x,y)$$ position of the hand when the joints are rotated by angles $$\theta, \phi, \psi\text{,}$$ respectively. Evaluating $$f$$ tells you where the hand will be on the plane when the joints are set at the given angles.
It is relatively straightforward to find a formula for $$f(\theta,\phi,\psi)$$ using some basic trigonometry. If you want the robot to fetch your coffee cup, however, you have to find the angles $$\theta,\phi,\psi$$ that will put the hand at the position of your beverage. It is not at all obvious how to do this, and it is not even clear if the answer is unique! You can ask yourself: “which positions on the table can my robot arm reach?” or “what is the arm’s range of motion?” This is the same as asking: “what is the range of $$f\text{?}$$”
Unfortunately, this kind of function does not come from a matrix, so one cannot use linear algebra to answer these kinds of questions. In fact, these functions are rather complicated; their study is the subject of inverse kinematics.
## Matrix Transformations
Now we specialize the general notions and vocabulary from the previous Subsection Transformations to the functions defined by matrices that we considered in the first Subsection Matrices as Functions.
##### Definition $$\PageIndex{3}$$: Matrix Transformation
Let $$A$$ be an $$m\times n$$ matrix. The matrix transformation associated to $$A$$ is the transformation
$T\colon \mathbb{R}^n \to \mathbb{R}^m \quad\text{defined by}\quad T(x) = Ax. \nonumber$
This is the transformation that takes a vector $$x$$ in $$\mathbb{R}^n$$ to the vector $$Ax$$ in $$\mathbb{R}^m$$.
If $$A$$ has $$n$$ columns, then it only makes sense to multiply $$A$$ by vectors with $$n$$ entries. This is why the domain of $$T(x)=Ax$$ is $$\mathbb{R}^n$$. If $$A$$ has $$n$$ rows, then $$Ax$$ has $$m$$ entries for any vector $$x$$ in $$\mathbb{R}^n \text{;}$$ this is why the codomain of $$T(x)=Ax$$ is $$\mathbb{R}^m$$.
The definition of a matrix transformation $$T$$ tells us how to evaluate $$T$$ on any given vector: we multiply the input vector by a matrix. For instance, let
$A=\left(\begin{array}{ccc}1&2&3\\4&5&6\end{array}\right)\nonumber$
and let $$T(x)=Ax$$ be the associated matrix transformation. Then
$T\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=A\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=\left(\begin{array}{ccc}1&2&3\\4&5&6\end{array}\right)\:\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=\left(\begin{array}{c}-14\\-32\end{array}\right).\nonumber$
Suppose that $$A$$ has columns $$v_1,v_2,\ldots,v_n$$. If we multiply $$A$$ by a general vector $$x\text{,}$$ we get
$Ax=\left(\begin{array}{cccc}|&|&\quad&| \\ v_1&v&2&\cdots &v_n \\ |&|&\quad &|\end{array}\right)\:\left(\begin{array}{c}x_1\\x_2\\ \vdots \\x_n\end{array}\right)=x_1v_1+x_2v_2+\cdots +x_nv_n.\nonumber$
This is just a general linear combination of $$v_1,v_2,\ldots,v_n$$. Therefore, the outputs of $$T(x) = Ax$$ are exactly the linear combinations of the columns of $$A\text{:}$$ the range of $$T$$ is the column space of $$A$$. See Note 2.3.6 in Section 2.3.
Note $$\PageIndex{1}$$
Let $$A$$ be an $$m\times n$$ matrix, and let $$T(x)=Ax$$ be the associated matrix transformation.
• The domain of $$T$$ is $$\mathbb{R}^n \text{,}$$ where $$n$$ is the number of columns of $$A$$.
• The codomain of $$T$$ is $$\mathbb{R}^m \text{,}$$ where $$m$$ is the number of rows of $$A$$.
• The range of $$T$$ is the column space of $$A$$.
##### Example $$\PageIndex{12}$$: Interactive: A $$2\times 3$$ matrix: reprise
Let
$A=\left(\begin{array}{c}1&-1&2\\-2&2&4\end{array}\right),\nonumber$
and define $$T(x) = Ax$$. The domain of $$T$$ is $$\mathbb{R}^3 \text{,}$$ and the codomain is $$\mathbb{R}^2$$. The range of $$T$$ is the column space; since all three columns are collinear, the range is a line in $$\mathbb{R}^2$$.
##### Example $$\PageIndex{13}$$: Interactive: A $$3\times 2$$ matrix: reprise
Let
$A=\left(\begin{array}{cc}1&0\\0&1\\1&0\end{array}\right),\nonumber$
and define $$T(x) = Ax$$. The domain of $$T$$ is $$\mathbb{R}^2 \text{,}$$ and the codomain is $$\mathbb{R}^3$$. The range of $$T$$ is the column space; since $$A$$ has two columns which are not collinear, the range is a plane in $$\mathbb{R}^3$$.
##### Example $$\PageIndex{14}$$: Projection onto the $$xy$$-plane: reprise
Let
$A=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right),\nonumber$
and let $$T(x) = Ax$$. What are the domain, the codomain, and the range of $$T\text{?}$$
Solution
Geometrically, the transformation $$T$$ projects a vector directly “down” onto the $$xy$$-plane in $$\mathbb{R}^3$$.
Figure $$\PageIndex{22}$$
The inputs and outputs have three entries, so the domain and codomain are both $$\mathbb{R}^3$$. The possible outputs all lie on the $$xy$$-plane, and every point on the $$xy$$-plane is an output of $$T$$ (with itself as the input), so the range of $$T$$ is the $$xy$$-plane.
Be careful not to confuse the codomain with the range here. The range is a plane, but it is a plane in $$\mathbb{R}^3$$, so the codomain is still $$\mathbb{R}^3$$. The outputs of $$T$$ all have three entries; the last entry is simply always zero.
In the case of an $$n\times n$$ square matrix, the domain and codomain of $$T(x) = Ax$$ are both $$\mathbb{R}^n$$. In this situation, one can regard $$T$$ as operating on $$\mathbb{R}^n \text{:}$$ it moves the vectors around in the same space.
##### Example $$\PageIndex{15}$$: Matrix transformations of $$\mathbb{R}^2$$
In the first Subsection Matrices as Functions we discussed the transformations defined by several $$2\times 2$$ matrices, namely:
\begin{align*} \text{Reflection:} &\qquad A=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right) \\ \text{Dilation:} &\qquad A=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right) \\ \text{Identity:} &\qquad A=\left(\begin{array}{cc}1&0\\0&1\end{array}\right) \\ \text{Rotation:} &\qquad A=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right) \\ \text{Shear:} &\qquad A=\left(\begin{array}{cc}1&1\\0&1\end{array}\right). \end{align*}
In each case, the associated matrix transformation $$T(x)=Ax$$ has domain and codomain equal to $$\mathbb{R}^2$$. The range is also $$\mathbb{R}^2 \text{,}$$ as can be seen geometrically (what is the input for a given output?), or using the fact that the columns of $$A$$ are not collinear (so they form a basis for $$\mathbb{R}^2$$).
##### Example $$\PageIndex{16}$$: Questions about a [matrix] transformation
Let
$A=\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right),\nonumber$
and let $$T(x)=Ax\text{,}$$ so $$T\colon\mathbb{R}^2 \to\mathbb{R}^3$$ is a matrix transformation.
1. Evaluate $$T(u)$$ for $$u=\left(\begin{array}{c}3\\4\end{array}\right)$$.
2. Let
$b = \left(\begin{array}{c}7\\5\\7\end{array}\right). \nonumber$
Find a vector $$v$$ in $$\mathbb{R}^2$$ such that $$T(v)=b$$. Is there more than one?
3. Does there exist a vector $$w$$ in $$\mathbb{R}^3$$ such that there is more than one $$v$$ in $$\mathbb{R}^2$$ with $$T(v)=w\text{?}$$
4. Find a vector $$w$$ in $$\mathbb{R}^3$$ which is not in the range of $$T$$.
Note: all of the above questions are intrinsic to the transformation $$T\text{:}$$ they make sense to ask whether or not $$T$$ is a matrix transformation. See the next Example $$\PageIndex{17}$$. As $$T$$ is in fact a matrix transformation, all of these questions will translate into questions about the corresponding matrix $$A$$.
Solution
1. We evaluate $$T(u)$$ by substituting the definition of $$T$$ in terms of matrix multiplication:
$T\left(\begin{array}{c}3\\4\end{array}\right)=\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right)\:\left(\begin{array}{c}3\\4\end{array}\right)=\left(\begin{array}{c}7\\4\\7\end{array}\right).\nonumber$
2. We want to find a vector $$v$$ such that $$b = T(v) = Av$$. In other words, we want to solve the matrix equation $$Av = b$$. We form an augmented matrix and row reduce:
$\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right)v=\left(\begin{array}{c}7\\5\\7\end{array}\right) \quad\xrightarrow{\text{augmented matrix}}\quad\left(\begin{array}{cc|c}1&1&7\\0&1&5\\1&1&7\end{array}\right) \quad\xrightarrow{\text{row reduce}}\quad\left(\begin{array}{cc|c}1&0&2\\0&1&5\\0&0&0\end{array}\right).\nonumber$
This gives $$x=2$$ and $$y=5\text{,}$$ so that there is a unique vector
$v = \left(\begin{array}{c}2\\5\end{array}\right) \nonumber$
such that $$T(v) = b$$.
3. Translation: is there any vector $$w$$ in $$\mathbb{R}^3$$ such that the solution set of $$Av=w$$ has more than one vector in it? The solution set of $$Ax=w\text{,}$$ if non-empty, is a translate of the solution set of $$Av=b$$ above, which has one vector in it. See key observation 2.4.3 in Section 2.4. It follows that the solution set of $$Av=w$$ can have at most one vector.
4. Translation: find a vector $$w$$ such that the matrix equation $$Av=w$$ is not consistent. Notice that if we take
$w = \left(\begin{array}{c}1\\2\\3\end{array}\right)\text{,} \nonumber$
then the matrix equation $$Av=w$$ translates into the system of equations
$\left\{\begin{array}{rrrrl}x &+& y &=& 1\\ {}&{}& y &=& 2\\ x &+& y &=& 3,\end{array}\right.\nonumber$
which is clearly inconsistent.
##### Example $$\PageIndex{17}$$: Questions about a [non-matrix] transformation
Define a transformation $$T\colon\mathbb{R}^2 \to\mathbb{R}^3$$ by the formula
$T\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}\ln(x) \\ \cos(y) \\ \ln(x)\end{array}\right).\nonumber$
1. Evaluate $$T(u)$$ for $$u=\left(\begin{array}{c}1\\ \pi\end{array}\right)$$.
2. Let
$b = \left(\begin{array}{c}7\\1\\7\end{array}\right). \nonumber$
Find a vector $$v$$ in $$\mathbb{R}^2$$ such that $$T(v)=b$$. Is there more than one?
3. Does there exist a vector $$w$$ in $$\mathbb{R}^3$$ such that there is more than one $$v$$ in $$\mathbb{R}^2$$ with $$T(v)=w\text{?}$$
4. Find a vector $$w$$ in $$\mathbb{R}^3$$ which is not in the range of $$T$$.
Note: we asked (almost) the exact same questions about a matrix transformation in the previous Example $$\PageIndex{16}$$. The point of this example is to illustrate the fact that the questions make sense for a transformation that has no hope of coming from a matrix. In this case, these questions do not translate into questions about a matrix; they have to be answered in some other way.
Solution
1. We evaluate $$T(u)$$ using the defining formula:
$T\left(\begin{array}{c}1\\ \pi\end{array}\right)=\left(\begin{array}{c} \ln(1) \\ \cos(\pi ) \\ \ln(1)\end{array}\right)=\left(\begin{array}{c}0\\-1\\0\end{array}\right).\nonumber$
2. We have
$T\left(\begin{array}{c}e^7 \\ 2\pi n \\ e^7\end{array}\right)=\left(\begin{array}{c}\ln(e^7) \\ \cos (2\pi n) \\ \ln(e^7)\end{array}\right)=\left(\begin{array}{c}7\\1\\7\end{array}\right)\nonumber$
for any whole number $$n$$. Hence there are infinitely many such vectors.
3. The vector $$b$$ from the previous part is an example of such a vector.
4. Since $$\cos(y)$$ is always between $$-1$$ and $$1\text{,}$$ the vector
$w = \left(\begin{array}{c}0\\2\\0\end{array}\right) \nonumber$
is not in the range of $$T$$.
3.1: Matrix Transformations is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Dan Margalit & Joseph Rabinoff via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 7,769 | 23,993 | {"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} | 4.5 | 4 | CC-MAIN-2022-21 | latest | en | 0.348511 |
circlesandspirituality.com | 1,670,593,690,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00010.warc.gz | 193,116,326 | 66,252 | # Numerology, Number 9 and Circles
The Unseen Force and Circles
## Numerology, Number 9 and Circles
Galileo Galilei once said, I quote “Mathematics is the language with which God has written the Universe”. Mathematics is the foundation of creation. Everything we see and touch has a mathematical nature in their existence.
Numerology is the study of numbers. It is defined as the ‘branch of knowledge that deals with the occult significance of numbers’. This knowledge can tell about your character and soul purpose through your date of birth reduced to a single digit.
For example, if you were born on 21 January 1979, that’s 21/01/1979, adding all numbers one by one gives us this, 2+1+0+1+1+9+7+9 = 30, then 3+0 = 3. In this case, your birth number is 3.
Every number has a significant meaning. There are all kinds of numerologists with different explanations about the meanings of every single number we use in our everyday lives.
In my observation, number Zero stands out as a significant and superior number. This is the number that represents ‘Nothingness’.
Historically, it is claimed that the concept of number zero first recorded by the Sumerians of Mesopotamia around 5.000 years ago. Then Mayans believed to have invented the concept independently around 4 A.D.
Ancient societies of India and China followed suit before the idea of zero was introduced to the western world by the Islamic philosophers who got the idea from their travels to India.
Now let us look at what zero as a number can do. When you add any number to zero, the result is the same number you are adding on. If you subtract any number to zero, the same characteristic applies.
But when you multiply number zero by any number, the answer is always zero itself. There is no other number that has this kind of characteristic.
This is why I call it the ‘superior’ number and it’s the number of wholeness and completion.
1 x 0 = 0
1,000,000 x 0 = 0
Now notice the shape of number zero. It is represented by a Sacred ‘Circle’.
Another interesting characteristic of zero is it’s impossible to divide any number with zero. You can not divide any number win zero but zero can divide itself only.
This number is complete, whole, and profound similar to a ‘Circle’ representation.
If you add zero in front of any number you multiply tenfold the value of that particular number. The more zeros you add, the more value of the number increases.
10 – Ten, 100 – Hundred, 1000 – Thousand, 100,000, 1,000,000 – Million…..etc
Our societies even gave specific names for this kind of number. Ten, Hundred, Thousand, Million, Billion, Trillions, etc
In my mother tongue Swahili language we say, Kumi, Mia, Elfu, and Milioni.
Mia and Elfu are the names borrowed from the Arabic language.
This is how significant number 0 is. This connection of numbers to a ‘Circle’ is in every act I must say of our environments and beyond.
It’s so profound that I have decided to write about it. Everywhere I look and in everything I see, ‘Circle’ is there in one form or another.
Significance of Number 9 and Circle Correlation (Zero)
If you look closely at the numbers system, there is a remarkable and interesting representation of the seven principles of creation embedded in it. If you count numbers 1 to 7 the next number which follows is 8. 7 is known as the number of the basis of creation.
In Abrahamic religions, it is written that God created the universe in 7 days. These 7 days make a week which is the basic foundation of the months and years in our calendars. The number 8 which follows 7 as you notice its represented by the symbol of infinity in a reversed fashion.
This can be interpreted as after the 7 days of creation (and principles) the universe will be in an infinite state. This can also reflect the state of Source Energy as infinite in nature.
In Arabic numerals, number 9 follows the infinite state of 8 represented by some kind of a ‘Circle’ joined by a line downwards. This can be interpreted as the presentation of the creation and formation of the Universe from The Source of All That Is with the infinite continuity.
Number 9 is very special compared to all the numbers because it has some interesting characteristics it holds which no other number can match mathematically. Let’s start with the foundation of numbers from 1 to 8. If you add all the numbers and finish with a single digit, the answer will always be 9. This is the special characteristic that only number 9 possess.
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36
Now, let us add numbers 3 and 6 and the answer is 9.
In multiplication, any number multiplied by 9, the answer will always be reduced to 9. For example:
9 x 365 =3,285. Reducing to single digit it gives us:
3 + 2 + 8 + 5 = 18; ======> 1 + 8 = 9
Another example: 9 x 3,478 = 31,302. Adding the single digits we get
3 + 1 + 3 + 0 + 2 = 9
Another characteristic is that all numbers in the nineties are all reduced to single digits to their origins.
91 => 9 + 1 = 10; 1 + 0 = 1
92 => 9 + 2 = 11; 1 + 1 = 2
93 => 9 + 3 = 12: 1 + 2 = 3
94 => 9 + 4 = 13; 1 + 3 = 4
95 => 9 + 5 = 14; 1 + 4 = 5
96 => 9 + 6 = 15; 1 + 5 = 6
97 => 9 + 7 = 15: 1+ 6 = 7
98 => 9 + 8 = 17; 1 + 7 = 8
99 => 9 + 9 = 18; 1 + 8 = 9
As you can see above, this special characteristic applies to number 9 only. Now let us see how number 9 is connected to a Circle in general.
Number 9 and Circle
A Circle has a total of 360 degrees as agreed by most mathematicians. Now if you add the single digits of the numbers of the degrees in the Circle, number 9 will always be the final answer.
360 ==> 3 + 6 + 0 = 9
If you halved the Circle, you get 180 degrees: 1 + 8 + 0 = 9
If you take half of 180 degrees we get 90: 9 + 0 = 9.
The half of 90 degrees is 45, if you add 4 + 5 you get 9 again.
Even in decimal places, the angle will still add to 9. Take the half of 45 degrees and you get 22.5 degrees, now add the single digits (2 + 2 + 5 = 9) and the result is always of course the magic number 9.
The following Youtube video by Mike Jackson explains and shows why the Circle is 360 degrees and the connection of number 9 in more detail.
As you can see from the video above, technically any degree in the ‘Circle’ will always add to a single figure of number 9. Interestingly, the number 9 is the last in the countability system and always followed by the number with a zero which is represented by the ‘Circle’,
‘Circle’ is behind all things from the very small to mega massive objects in our Universe. There is a spirituality concept of overstanding and relativity that our Infinite Supreme Source and Creator has left us as in this ‘Symbol’ of the Source for us to ponder and contemplate The Greatness of The Almighty Self.
The ‘Circle’ affects our daily lives beyond our imagination that we even do not see it, especially if we are not looking. I hope you enjoyed reading this blog.
Please feel free to leave any comments, they will be very much appreciated.
Have a nice day and blessed be
Ismail
## 15 Responses
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• Ismail says:
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Thanks! Fantastic stuff!
• Ismail says:
Thank you so much for your comment. 🙂
• Ismail says:
Thank you for your comment. It means a lot.
3. thanks for the advice and information
• Ismail says:
you are welcome. I am grateful for your comment. Thank you.
• Ismail says:
You are most welcome. Your comment is much appreciated.
4. Stella Mary says:
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• Ismail says:
Thank you very much for taking the time to read and comment. Be blessed and all the best for 2020 and the new decade. Namaste
5. Ali Mualla says:
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Ali
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• Ismail says:
Thank you brother Ali for your comment. Much appreciated
6. Stella Mary says:
Learned a lot of interesting facts about number 9. Thank you for all the helpful information shared.
• Ismail says:
Thank you Stella for your comment. Much appreciated.
7. Ismail says:
Thank you. Your comment is much appreciated. | 2,147 | 8,447 | {"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-2022-49 | latest | en | 0.945198 |
https://cbselibrary.com/tag/selina-icse-solutions-for-class-10-maths-loci-locus-and-its-constructions/ | 1,713,390,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00564.warc.gz | 149,090,248 | 22,954 | ## Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)
Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions)
### Locus and Its Constructions Exercise 16A – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
Given— PQ is perpendicular bisector of side AB of the triangle ABC.
Prove— Q is equidistant from A and B.
Solution:
Construction: Join AQ
Proof: In ∆AQP and ∆BQP,
AP = BP (given)
∠QPA = ∠QPB (Each = 90 )
PQ = PQ (Common)
By Side-Angle-Side criterian of congruence, we have
∆AQP ≅ ∆BQP (SAS postulate)
The corresponding parts of the triangle are congruent
∴ AQ = BQ (CPCT)
Hence Q is equidistant from A and B.
Question 2.
Given— CP is bisector of angle C of ∆ ABC.
Prove— P is equidistant from AC and BC.
Solution:
Question 3.
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX at point Y.
Prove—
(i) X is equidistant from AB and AC.
(ii) Y is equidistant from A and C.
Solution:
Question 4.
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Solution:
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm
Steps of Construction:
i) Draw a line segment BC = 6.3 cm
ii) With centre B and radius 4.2 cm, draw an arc.
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
iv) Join AB and AC.
∆ABC is the required triangle.
v) Again with centre B and C and radius greater than $$\frac{1}{2} \mathrm{BC}$$ draw arcs which intersects each other at L and M.
vi) Join LM intersecting AC at D and BC at E.
vii) Join DB.
Hence, D is equidistant from B and C.
Question 5.
In each of the given figures: PA = PH and QA = QB.
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of points equidistant from two given fixed points.
Solution:
Construction: Join PQ which meets AB in D.
Proof: P is equidistant from A and B.
∴ P lies on the perpendicular bisector of AB.
Similarly, Q is equidistant from A and B.
∴ Q lies on perpendicular bisector of AB.
∴ P and Q both lie on the perpendicular bisector of AB.
∴ PQ is perpendicular bisector of AB.
Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.
Question 6.
Construct a right angled triangle PQR, in which ∠ Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Solution:
Steps of Construction:
i) Draw a line segment QR = 4.5 cm
ii) At Q, draw a ray QX making an angle of 90°
iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
iv) Join RP.
∆PQR is the required triangle.
v) Draw the bisector of ∠PQR which meets PR in T.
vi) From T, draw perpendicular PL and PM respectively on PQ and QR.
Hence, T is equidistant from PQ and QR.
Question 7.
Construct a triangle ABC in which angle ABC = 75°. AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P ; prove that P is equidistant from B and C ; and also from AC and BC.
Hence P is equidistant from AC and BC.
Solution:
Question 8.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
Solution:
Question 9.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A.
Prove that –
(i) point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
Solution:
Construction: Join AM
Proof:
∵ A lies on bisector of ∠N
∴A is equidistant from MN and LN.
Again, A lies on bisector of ∠L
∴ A is equidistant from LN and LM.
Hence, A is equidistant from all sides of the triangle LMN.
∴ A lies on the bisector of ∠M
Question 10.
Use ruler and compasses only for this question:
(i) construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Solution:
Steps of construction:
(i) Draw line BC = 6 cm and an angle CBX = 60o. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
(ii) Draw perpendicular bisector of BC and bisector of angle B
(iii) Bisector of angle B meets bisector of BC at P.
⇒ BP is the required length, where, PB = 3.5 cm
(iv) P is the point which is equidistant from BA and BC, also equidistant from B and C.
PB=3.6 cm
Question 11.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point E
Prove that:
(i) F is equidistant from A and B.
(ii) F is equidistant from AB and AC.
Solution:
Question 12.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Solution:
Since P lies on the bisector of angle B,
therefore, P is equidistant from AB and BC …. (1)
Similarly, P lies on the bisector of angle C,
therefore, P is equidistant from BC and CD …. (2)
From (1) and (2),
Hence, P is equidistant from AB and CD.
Question 13.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Solution:
Steps of construction:
(i) Draw a line segment AB of 6 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB.
Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.
Question 14.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Solution:
Steps of Construction:
i) Draw a ray BC.
ii) Construct a ray RA making an angle of 75° with BC. Therefore, ABC= 75°.
iii) Draw the angle bisector BP of ∠ABC.
BP is the required locus.
iv) Take any point D on BP.
v) From D, draw DE ⊥ AB and DF ⊥ BC
Since D lies on the angle bisector BP of ∠ABC
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.
Question 15.
Draw an angle ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC ; and also equidistant from A and B.
Solution:
Steps of Construction:
i) Draw a line segment BC = 5 cm
ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
iii) Draw the angle bisector of ∠ABC.
iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC, as well as from A and B.
Question 16.
In the figure given below, find a point P on CD equidistant from points A and B.
Solution:
Steps of Construction:
i) AB and CD are the two lines given.
ii) Draw a perpendicular bisector of line AB which intersects CD in P.
P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.
Question 17.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Solution:
Steps of Construction:
i) In the given triangle, draw the angle bisector of ∠BAC.
ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.
Question 18.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Solution:
Steps of Construction:
1) Draw a line segment AB = 7 cm.
2) Draw angle ∠ABC = 60° with the help of compass.
3) Cut off BC = 8 cm.
4) Join A and C.
5) The triangle ABC so formed is the required triangle.
i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
ii) Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.
P is the required point which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.
Question 19.
On a graph paper, draw lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Solution:
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 3 which is parallel to y-axis
And draw another line m, y = -5, which is parallel to x-axis
These two lines intersect each other at P.
Now draw the angle bisector p of angle P.
Since p is the angle bisector of P, any point on P is equidistant from l and m.
Therefore, this line p is equidistant from l and m.
Question 20.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Solution:
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 6 which is parallel to y-axis
Take points P and Q which are at a distance of 3 units from the line l.
Draw lines m and n from P and Q parallel to l
With locus = 3, two lines can be drawn x = 3 and x = 9.
### Locus and Its Constructions Exercise 16B – Selina Concise Mathematics Class 10 ICSE Solutions
Question 1.
Describe the locus of a point at a distance of 3 cm from a fixed point.
Solution:
The locus of a point which is 3 cm away from a fixed point is circumference of a circle whose radius is 3 cm and the fixed point is the centre of the circle.
Question 2.
Describe the locus of a point at a distance of 2 cm from a fixed line.
Solution:
The locus of a point at a distance of 2 cm from a fixed line AB is a pair of straight lines l and m which are parallel to the given line at a distance of 2 cm.
Question 3.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Solution:
The locus of the centre of a wheel, which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.
Question 4.
Describe the locus of the moving end of the minute hand of a clock.
Solution:
The locus of the moving end of the minute hand of the clock will be a circle where radius will be the length of the minute hand.
Question 5.
Describe the locus of a stone dropped from the top of a tower.
Solution:
The locus of a stone which is dropped from the top of a tower will be a vertical line through the point from which the stone is dropped.
Question 6.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Solution:
The locus of the runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge will be the circumference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m.
Question 7.
Describe the locus of the door handle as the door opens.
Solution:
The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and radius equal to the distance between the door handle and the axis of rotation of the door.
Question 8.
Describe the locus of a point inside a circle and equidistant from two fixed points on the circumference of the circle.
Solution:
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.
Question 9.
Describe the locus of the centers of all circles passing through two fixed points.
Solution:
The locus of the centre of all the circles which pass through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.
Question 10.
Describe the locus of vertices of all isosceles triangles having a common base.
Solution:
The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.
Question 11.
Describe the locus of a point in space which is always at a distance of 4 cm from a fixed point.
Solution:
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to 4 cm.
Question 12.
Describe the locus of a point P so that:
AB2 = AP2 + BP2, where A and B are two fixed points.
Solution:
The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB2 = AP2 + BP2.
Question 13.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
i) AB and BC
ii) B and D.
Solution:
i)
The locus of the point in a rhombus ABCD which is equidistant from AB and BC will be the diagonal BD.
ii)
The locus of the point in a rhombus ABCD which is equidistant from B and D will be the diagonal AC.
Question 14.
The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth’s surface, who hear the sound exactly one second later.
Solution:
The locus of all the people on Earth’s surface is the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.
Question 15.
Describe:
i) The locus of points at distances less than 3 cm from a given point.
ii) The locus of points at distances greater than 4 cm from a given point.
iii) The locus of points at distances less than or equal to 2.5 cm from a given point.
iv) The locus of points at distances greater than or equal to 35 mm from a given point.
v)The locus of the centre of a given circle which rolls around the outside of a second circle and is always touching it.
vi) The locus of the centers of all circles that are tangent to both the arms of a given angle.
vii) The locus of the mid-points of all chords parallel to a given chord of a circle.
viii) The locus of points within a circle that are equidistant from the end points of a given chord.
Solution:
i) The locus is the space inside of the circle whose radius is 3 cm and the centre is the fixed point which is given.
ii) The locus is the space outside of the circle whose radius is 4 cm and centre is the fixed point which is given.
iii) The locus is the space inside and circumference of the circle with a radius of 2.5 cm and the centre is the given fixed point.
iv) The locus is the space outside and circumference of the circle with a radius of 35 mm and the centre is the given fixed point.
v) The locus is the circumference of the circle concentric with the second circle whose radius is equal to the sum of the radii of the two given circles.
vi) The locus of the centre of all circles whose tangents are the arms of a given angle is the bisector of that angle.
vii) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given chords.
viii) The locus of the points within a circle which are equidistant from the end points of a given chord is the diameter which is perpendicular bisector of the given chord.
Question 16.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Solution:
Draw a line XY parallel to the base BC from the vertex A.
This line is the locus of vertex A of all the triangles which have the base BC and length of altitude equal to AD.
Question 17.
In the given figure, obtain all the points equidistant from lines m and n ; and 2.5 cm from O.
Solution:
Draw an angle bisector PQ and XY of angles formed by the lines m and n. From O, draw arcs with radius 2.5 cm, which intersect the angle bisectors at a, b, c and d respectively.
Hence, a, b, c and d are the required four points.
Question PQ.
By actual drawing obtain the points equidistant from lines m and n and 6 cm from the point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
Solution:
Steps of construction:
i) Draw a linen.
ii) Take a point Lonn and draw a perpendicular to n.
iii) Cut off LM = 6 cm and draw a line q, the perpendicular bisector of LM.
iv) At M, draw a line m making an angle of 90°.
v) Produce LM and mark a point P such that PM = 2 cm.
vi) From P, draw an arc with 6 cm radius which intersects the line q, the perpendicular bisector of LM, at A and B.
A and B are the required points which are equidistant from m and n and are at a distance of 6 cm from P.
Question 18.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
(i) always 4 cm from the line AB
(ii) equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
Solution:
(i) Draw a line segment AB = 8 cm.
(ii) Draw two parallel lines l and m to AB at a distance of 4 cm.
(iii) Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
(iv) Join AX, AY, BX and BY.
The figure AXBY is a square as its diagonals are equal and intersect at 90°.
Question 19.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respec¬tively. Draw and describe the locus of a point which is :
(i) equidistant from BA and BC.
(ii) 4 cm from M.
(iii) 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Solution:
i) Draw an angle of 60° with AB = BC = 8 cm
ii) Draw the angle bisector BX of ∠ABC
iii) With centre M and N, draw circles of radius equal to 4 cm, which intersects each other at P. P is the required point.
iv) Join MP, NP
BMPN is a rhombus since MP = BM = NB = NP = 4 cm
Question 20.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
(i) the locus of the centers of all circles which touch AB and AC.
(ii) the locus of the centers of all circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC.
Solution:
Steps of Construction:
i) Draw a line segment BC = 4.5 cm
ii) With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
iii) Join AB and AC.
ABC is the required triangle.
iv) Draw the angle bisector of ∠BAC
v) Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.
Question 21.
Construct a triangle ABC, having given AB = 4.8 cm. AC = 4 cm and ∠ A = 75°. Find a point P.
(i) inside the triangle ABC.
(ii) outside the triangle ABC.
equidistant from B and C; and at a distance of 1.2 cm from BC.
Solution:
Steps of Construction:
i) Draw a line segment AB = 4.8 cm
ii) At A, draw a ray AX making an angle of 75°
iii) Cut off AC = 4 cm from AX
iv) Join BC.
ABC is the required triangle.
v) Draw two lines l and m parallel to BC at a distance of 1.2 cm
vi) Draw the perpendicular bisector of BC which intersects l and m at P and P’
P and P’ are the required points which are inside and outside the given triangle ABC.
Question PQ.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Solution:
P moves along AB, and Qmoves in such a way that PQ is always equal to OP.
But Pis the mid-point of OQ
Now in ∆OQQ’
P’and P” are the mid-points of OQ’ and OQ”
Therefore, AB||Q’Q”
Therefore, Locus of Q is a line CD which is parallel to AB.
Question 22.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Solution:
Steps of Construction:
i) Draw a ray BC.
ii) At B, draw a ray BA making an angle of 75° with BC.
iii) Draw a line l parallel to AB at a distance of 2 cm
iv) Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.
Question 23.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Solution:
Steps of Construction:
i) Draw a line segment AB = 5.6 cm
ii) From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw two lines n and m parallel to BC at a distance of 2 cm
v) Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
P and Q are the required points which are equidistant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.
Question 24.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Solution:
Steps of Construction:
i) Draw a line segment AB = 6 cm
ii) With A and B as centers and radius 9 cm, draw two arcs which intersect each other at C.
iii) Join AC and BC.
iv) Draw the perpendicular bisector of BC.
v) With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.
P is the required point which is equidistant from B and C and at a distance of 4 cm from A.
Question 25.
Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) Mark the point Q, in your construction, which would make A QBC equal in area to A ABC, and isosceles.
(v) Measure and record the length of CQ.
Solution:
Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle 60 degree and cut off BA=9 cm.
(iii) Join AC. ABC is the required triangle.
(iv) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
(v) Through A, draw a line m || BC.
(vi) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(vii) Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
On measuring CQ = 8.4 cm.
Question 26.
State the locus of a point in a rhombus ABCD, which is equidistant
(ii) from the vertices A and C.
Solution:
Question 27.
Use a graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Plot the points A(1,1), B(5,3) and C(2,7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:
Steps of Construction:
i) Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA.
ii) Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P.
P is the required point.
Since P lies on the perpendicular bisector of AB.
Therefore, P is equidistant from A and B.
Again,
Since P lies on the angle bisector of angle A.
Therefore, P is equidistant from AB and AC.
On measuring, the length of PA = 5.2 cm
Question 28.
Construct an isosceles triangle ABC such that AB = 6 cm, BC=AC=4cm. Bisect angle C internally and mark a point P on this bisector such that CP = 5cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Solution:
Steps of Construction:
i) Draw a line segment AB = 6 cm.
ii) With centers A and B and radius 4 cm, draw two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw the angle bisector of angle C and cut off CP = 5 cm.
v) A line m is drawn parallel to AB at a distance of 5 cm.
vi) P as centre and radius 5 cm, draw arcs which intersect the line m at Q and R.
vii) Join PQ, PR and AQ.
Q and R are the required points.
Question PQ.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Solution:
Steps of Construction:
i) Draw a circle with radius = 4 cm.
ii) Take a point A on it.
iii) A as centre and radius 6 cm, draw an arc which intersects the circle at B.
iv) Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.
v) Join AB and AC.
vi) Draw the perpendicular bisector of AC, which intersects AC at Mand meets the circle at E and F.
EF is the locus of points inside the circle which are equidistant from A and C.
vii) Join AE, AF, CE and CF.
Proof:
Similarly, we can prove that CF = AF
Hence EF is the locus of points which are equidistant from A and C.
ii) Draw the bisector of angle A which meets the circle at N.
Therefore. Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.
Question 29.
Plot the points A(2,9), B(-1,3) and C(6,3) on a graph paper. On the same graph paper, draw the locus of point A so that the area of triangle ABC remains the same as A moves.
Solution:
Steps of construction:
i) Plot the given points on graph paper.
ii) Join AB, BC and AC.
iii) Draw a line parallel to BC at A and mark it as CD.
CD is the required locus of point A where area of triangle ABC remains same on moving point A.
Question 30.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that:
(a) P is equidistant from A B and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Solution:
i) Steps of Construction:
1) Draw a line segment BC = 5 cm
2) B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
3) Join PC.
4) B and C as centers, draw two perpendiculars to BC.
5) P as centre and radius PC, cut an arc on the perpendicular on C at D.
6) D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.
ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.
ii) On measuring AB = 5.7 cm
Question 31.
Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a ∆ABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Solution:
i. Steps of construction:
1. Draw BC = 6.5 cm using a ruler.
2. With B as center and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
3. With Q as center and same radius, cut the previous arc at P.
4. Join BP and extend it.
5. With B as center and radius 5 cm, draw an arc that cuts the arm PB to obtain point A.
6. Join AC to obtain ΔABC.
ii. With A as center and radius 3.5 cm, draw a circle.
The circumference of a circle is the required locus.
iii. Draw CH, which is bisector of Δ ACB. CH is the required locus.
iv. Circle with center A and line CH meet at points X and Y as shown in the figure. xy = 8.2 cm (approximately)
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Impossible Sandwiches
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Age 14 to 18 Challenge Level:
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Proofs with Pictures
Age 14 to 18
Some diagrammatic 'proofs' of algebraic identities and inequalities.
Age 11 to 16 Challenge Level:
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Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
For What?
Age 14 to 16 Challenge Level:
Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.
Leonardo's Problem
Age 14 to 18 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
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Age 14 to 18 Challenge Level:
Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.
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Age 14 to 16 Challenge Level:
An equilateral triangle is constructed on BC. A line QD is drawn, where Q is the midpoint of AC. Prove that AB // QD.
Magic Squares II
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An article which gives an account of some properties of magic squares.
Picturing Pythagorean Triples
Age 14 to 18
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.
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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.
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Age 14 to 16 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
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Age 11 to 18 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. . . .
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Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
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Age 14 to 16 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
Pythagorean Triples II
Age 11 to 16
This is the second article on right-angled triangles whose edge lengths are whole numbers.
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Age 14 to 18
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
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This is the second of two articles and discusses problems relating to the curvature of space, shortest distances on surfaces, triangulations of surfaces and representation by graphs.
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Age 14 to 18
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
Age 14 to 18 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
Sprouts Explained
Age 7 to 18
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
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Age 14 to 16 Challenge Level:
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Age 14 to 18 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
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Age 14 to 16 Challenge Level:
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Age 14 to 16 Challenge Level:
Clearly if a, b and c are the lengths of the sides of an equilateral triangle then a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true?
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Which of these roads will satisfy a Munchkin builder?
Cosines Rule
Age 14 to 16 Challenge Level:
Three points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.
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Age 11 to 16 Challenge Level:
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Age 14 to 16 Challenge Level:
Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.
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Age 14 to 16 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
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Age 14 to 16 Challenge Level:
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Kyle and his teacher disagree about his test score - who is right?
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Age 14 to 16 Challenge Level:
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Russian Cubes
Age 14 to 16 Challenge Level:
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Age 14 to 16 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
Natural Sum
Age 14 to 16 Challenge Level:
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Age 14 to 18 Challenge Level:
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Picture Story
Age 14 to 16 Challenge Level:
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Age 11 to 18 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
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Age 14 to 18 Challenge Level:
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Age 14 to 16 Challenge Level:
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Proof: A Brief Historical Survey
Age 14 to 18
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Triangular Intersection
Age 14 to 16 Short Challenge Level:
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Zig Zag
Age 14 to 16 Challenge Level:
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Diophantine N-tuples
Age 14 to 16 Challenge Level:
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There's a Limit
Age 14 to 18 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely? | 2,132 | 8,946 | {"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-2019-39 | latest | en | 0.899672 |
http://docplayer.net/19127303-Fundamental-mechanics-supplementary-exercises.html | 1,618,826,670,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879305.68/warc/CC-MAIN-20210419080654-20210419110654-00527.warc.gz | 25,149,805 | 33,801 | # Fundamental Mechanics: Supplementary Exercises
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1 Phys 131 Fall 2015 Fundamental Mechanics: Supplementary Exercises 1 Motion diagrams: horizontal motion A car moves to the right. For an initial period it slows down and after that it speeds up. Which of the following (choose one) best represents its location as time passes? Case 1 1s 2s 3s 4s 5s Case 2 1s 2s 3s 4s 5s Case 3 1s 2s 3s 4s 5s Case 4 1s 2s 3s 4s 5s Briefly explain your choice. 1
2 2 Motion diagrams and position vs. time graphs A car moves from left to right and its position, measured in meters, is recorded every 5.0 s. The resulting motion diagram is illustrated a) Produce a table of numerical data for position versus time for the car for the duration of the motion. b) Produce a position versus time graph for the car for the duration of the motion. This graph must be drawn by hand using axes that are clearly labeled. 3 Motion diagrams and position vs. time graphs A car moves from left to right and its position, measured in meters, is recorded every 5.0 s. The resulting motion diagram is illustrated a) Produce a table of numerical data for position versus time for the car for the duration of the motion. b) Produce a position versus time graph for the car for the duration of the motion. This graph must be drawn by hand using axes that are clearly labeled. 4 Average velocity The following objects lie along a straight line: a bicycle, a coffee cup and a soccer ball. The distance from the coffee cup to the bicycle is 400m and from the cup to the ball is 500m. A man starts at the cup and travels in a straight line to the ball. This take 200s. A dog is initially at the cup and runs at constant speed to the bicycle, taking 50s to do so. The dog immediately turns around and runs to ball; this takes the dog an additional 150s. Consider the entire trip from the cup to the ball for each. Who has the larger average velocity for this entire trip? Explain your answer. 2
3 5 Ant on a stick Anantwalksalongastraightstick. Thegraph illustrates the ant s position vs. time. Answer the following, giving explanations for each answer. x in m 4 3 a) During which times is the ant moving right? During which times is it moving left? 2 1 b) When, if ever, is the velocity of the ant 0m/s? c) How does the speed of the ant at 1.0s compare to its speed at 4.0s? t in s d) How does the velocity of the ant at 1.0s compare to its velocity at 4.0s? 6 Ant and bug on a stick An ant and a bug walk along straight sticks. The solid graph illustrates the ant s position vs. time. The dashed graph indicates the bug s position vs. time. Answer the following, giving explanations for each answer. a) At what time(s) are the ant and bug at the same location? x in m Bug Ant b) Which is moving faster at 2s? c) Do the ant and bug ever have the same velocity? If so when? t in s 3
4 7 Acceleration sign A bicycle can move east (positive) or west (negative). a) If the bicycle moves east can the acceleration be negative? Explain your answer. b) If the bicycle moves west can the acceleration be positive? Explain your answer. 8 Ant and bug on a stick An ant and a bug walk along straight sticks. The solid graph illustrates the ant s position vs. time. The dashed graph indicates the bug s position vs. time. For the bug, and separately for the ant, which of the following is true during the period from 0s to 4s? i) Acceleration is zero at all times. ii) Acceleration is positive at all times. iii) Acceleration is negative at all times. iv) Acceleration is first positive and later negative. v) Acceleration is first negative and later positive. x in m Ant Bug t in s
5 9 Non-Freely Falling Object A bungee jumper falls downward stretching the cord, reaching a low point, after which the cord pulls him up again. His velocity is recorded at equally spaced intervals in time. The data is: Time in s Velocity in m/s a) During which period is the man falling? When is he rising? b) By how much does the man s velocity change per second? Is this change constant throughout the recorded motion? c) Determine the man s acceleration while he is falling and also while he is rising. Are these accelerations the same or not? d) What is the man s acceleration (according to the data) at his low point? 5
6 10 Title Go to the moving man animation at: Run the moving man animation. Click on the charts tab. Set the position to 0.00m, the velocity to 5.00m/s and the acceleration to 2.00m/s 2. Run the animation, stopping it just before the man hits the wall. The animation will have recorded the motion. Check the playback button at the bottom. You can slide the light blue bar left and right to get data for the motion. Gray zoom icons at the right will let you rescale the charts. a) Consider the interval from 2.0 s to 3.0 s. Describe the motion verbally during this time. b) How does the speed of the man at 2.0s compare to that at 3.0s during this period? Explain your answer. c) How does the velocity of the man at 2.0s compare to that at 3.0s during this period? Explain your answer. d) Will the average acceleration over the interval from 2.0 s to 3.0 s be positive, negative or zero? Explain your answer. e) If the acceleration is not zero, does it vary during this interval? Explain your answer. f) Determine the average acceleration over the interval from 2.0 s to 3.0 s. 6
7 11 Acceleration vector a) A hockey puck slides along a horizontal surface toward a board, hitting it at an angle and bouncing off with unchanged speed. The view from above is as illustrated. Draw the velocity vectors of the puck just before and just after hitting the board, use these to draw the vector v, and use the result to draw the direction of the acceleration vector. after before b) If the puck traveled backwards along the same path (i.e. reversed direction), what would the direction of the acceleration vector be? 12 Ants moving along a curved path Various ants follow the same path on a horizontal surface, starting at point 1. The path is as illustrated. Ant A moves with a constant speed, ant B gradually speeds up and ant C gradually slows a) Draw the velocity vector at points 1,2,3 and 4. b) Does any of the ants have zero acceleration at all times? Explain your answer. 13 Hockey pucks sliding horizontally Three identical hockey pucks slide horizontally across a frictionless sheet of ice and they maintain the indicated speeds while they are being observed. Let F A be the magnitude of the force acting on A, F B be the magnitude of the force acting on B, etc,... Which of the following is true? Explain your answer. A 4m/s 12m/s B 16m/s C i) F B = 2F A and F C = 4F A ii) F A = 2F B and F A = 4F C iii) F A = F B = F C 0 iv) F A = F B = F C = 0 7
8 14 Moving carts Three identical carts move horizontally along tracks. Their speeds at two instants 5.0s apart are indicated. Let F A be the magnitude of the force acting on A during this interval, F B be the magnitude of the force acting on B, etc,... Which of the following is true? Explain your answer. Initial Instant 0m/s 10m/s Final Instant 15m/s 20m/s Cart A Cart B i) F A > F B > F C. ii) F B = F C > F A. 15m/s 20m/s Cart C iii) F B = F C < F A. iv) F A = F B = F C 0 15 Pushing carts Zog and Geraldine (his wife) each push a cart along a horizontal surface where friction is negligible. Both carts are initially at rest. Zog takes the cart with mass 25kg and exerts a force of 400N on it force for a period of 4.0s and he then collapses and stops pushing. Geraldine has to push a cart of mass 50kg and she is also able to exert a force of 400N on it. Geraldine claims that it is possible for the speed of her cart to eventually reach the speed of Zog s cart. Is this true? Explain your answer. 8
9 16 A Suspended Object in Equilibrium A 2.50kg ring is suspended from the ceiling and is held at rest by two ropes as illustrated. Rope 2 pulls horizontally. The aim of this exercise is to use Newton s 2 nd Law to determine the tension in each rope. One piece of background information that you will need to answer this is that the magnitude of the gravitational force on an object of mass m is F g = mg. 40 Rope 1 Rope 2 a) Draw a free body diagram for the ring. Label the tension forces T 1 and T 2. b) Write Newton s 2 nd Law in its component form, i.e. write F net x = ΣF ix = ma x (1) F net y = ΣF iy = ma y (2) Insert as much information as possible about the acceleration. You will return to these equations shortly; they will generate the algebra that eventually gives you the acceleration and the normal force. c) List as much information as possible about each components for each force, using one of the two formats below. F gx = F gy = T 1x = T 1y =. Force x comp y comp F g T 1. d) Express the components of T 2 in terms of its magnitude, T 2. Repeat this for T 1 and insert the expressions into the table or equations above. e) Use Eq. (1) to obtain an equation relating various quantities that appear in this problem. DothesamewithEq.(2). Youshouldgettwoexpressionsthatcontainthetwounknowns T 1 and T 2. Solve them for the unknowns. f) If you had one rope that is rated to break when the tension exceeds 30N and another rated to break when the tension exceeds 40N, which one would you use to suspend the object as illustrated above? 9
10 17 Free fall in an elevator A phone of mass m sits on the floor of an elevator, which is initially at rest. The elevator cable snaps and the elevator and phone then undergo free fall. While they do this which is true of the magnitude of the normal force, n, acting on the phone? Explain your choice. i) n = 0. ii) mg > n > 0. iii) n = mg. iv) n > mg 10
11 18 Dynamics of a Single Object A box of mass 5.0kg can move along a frictionless horizontal surface. A person pulls with a force at the illustrated angle. The aim of this exercise is to use Newton s laws to determine the acceleration of the box and the normal force exerted by the surface (provided that the box stays on the surface). 40N 26.6 The entire collection of these steps is called applying Newton s laws of mechanics to determine the acceleration of the block. a) Draw a free body diagram for the block. b) Write Newton s Second Law in its component form, i.e. write F net x = ΣF x = (3) F net y = ΣF y = (4) Insert as much information as possible about the components of acceleration at this stage. You will return to these equations shortly; they will generate the algebra that eventually gives you the acceleration and the normal force. c) Determine the magnitude of the gravitational force. Let n be the magnitude of the normal force. Do you think that n = mg? d) List all the components of all the forces, using one of the two formats below. F gx = F gy = n x = n y =. Force x comp y comp F g n. e) Use Eq. (3) to obtain an equation relating various quantities that appear in this problem. Do the same with Eq. (4). Solve these for the acceleration and the magnitude of the normal force. Is n = mg? f) Suppose that rather than pull up, the person pushed down on the box at the same angle from the left and with the same force. Would the acceleration and normal forces differ from the case where the person pulled up? g) You may have noticed that theacceleration does not dependon thenormal force. This is only true if there is no friction. It turns out that when friction is present, the magnitude of friction force increases as the normal force increases. Knowing this, would pulling up or pushing down give a larger acceleration? 11
12 19 Dynamics of a Single Object with Friction A box of mass 15.0kg can move along a horizontal surface. A person pulls with a force at the illustrated angle. The coefficient of kinetic friction is The primary aim of this exercise will be to determine the acceleration of the box. 50N 30 a) Draw a free body diagram for the block. b) Write Newton s Second Law in its component form, i.e. write F net x = ΣF x = (5) F net y = ΣF y = (6) Insert as much information as possible about the components of acceleration at this stage. You will return to these equations shortly; they will generate the algebra that eventually gives you the acceleration and the normal force. c) Determine the magnitude of the gravitational force. Let n be the magnitude of the normal force. Using this write an expression for the magnitude of the friction force. Do you know the exact number for the friction force at this point? d) List all the components of all the forces, using one of the two formats below. F gx = F gy = n x = n y =. Force x comp y comp F g n. e) Use Eq. (5) to obtain an equation relating various quantities that appear in this problem. Do the same with Eq. (6). Does either give the acceleration immediately? Can one of then at least give the normal force immediately? f) Determine the normal force and use this result to find the acceleration. g) Supposethattheboxwereinitially atrest. Determinethetimetaken topullitadistance of 5.0m. 12
13 20 Dynamics of an Object on a Ramp A 4.0kg box can move along africtionless rampwhich makes at angle 30 from the horizontal. A person pulls on a rope which exerts a force of 15N up the ramp parallel to its surface. The object of this exercise is to determine the acceleration of the box. 30 a) Draw a free body diagram for the box. b) Describe the x and y axes that you will use. c) Write Newton s Second Law in vector form and also in its component form, i.e. write F net x = ΣF x = (7) F net y = ΣF y = (8) Insert as much information as possible about the components of acceleration at this stage. The resulting equations will generate much of the algebra that follows. d) Determine the magnitude of the weight force. e) List all the components of all the forces, using one of the two formats below. W x = W y = n x = n y =. Force x comp y comp W n. f) Use Eq. (7) to obtain an equation relating various quantities that appear in this problem. Do the same with Eq. (8).Use the resulting equations to determine the acceleration of the box. g) Is it possible to say with certainty whether the box is moving up the ramp or down the ramp? Is either direction possible in this situation? If only one direction is possible, which is it? 13
14 21 Dynamics of an Object on a Ramp with Friction A box can move along a rough ramp which makes at angle θ from the horizontal and which has length L. The box is launched with speed v from the top of the ramp. The aim of this exercise is to determine the coefficient of friction needed to bring the box to a stop at the bottom of the ramp. a) The first part of the solution uses kinematics to assess the acceleration of the box. Using kinematics, and eventually dynamics, is greatly simplified by choosing an appropriate x and a y axis. These do not have to be along the usual vertical and horizontal directions. Regardless of the axes that you choose, the usual general kinematics and dynamics equations will be valid. Describe the x and y axes that you will use. b) Determine an expression for the magnitude of the acceleration of the box, a, in terms of variables relevant to this problem, such as L,θ,v, and possibly the mass of the box, m. c) Draw a free body diagram for the box. d) Write Newton s Second Law in vector form and also in its component form, i.e. write F net x = ΣF ix = (9) F net y = ΣF iy = (10) wherexand y refer to your specially chosen axes. Insert as much information as possible about the components of acceleration at this stage. The resulting equations will generate much of the algebra that follows. e) Determine expressions for the magnitudes of the gravitational and the friction forces. f) List all the components of all the forces, using one of the two formats below. θ F gx = F gy = n x = n y =. Force x comp y comp F g n. g) Use Eq. (9) to obtain an equation relating various quantities that appear in this problem. Do the same with Eq. (10). Use the resulting equations to find an expression for the coefficient of kinetic friction. Does the result depend on the mass of the box? h) Suppose that the ramp is a roof whose length is 5.0m and which is angled at 15 from the horizontal. If the box is pushed with speed 4.0m/s, determine the coefficient of friction needed to stop the box at the bottom of the roof. 14
15 22 Dynamics of an Object on a Ramp A block of mass m slides up a ramp which makes at angle θ from the horizontal. The coefficient of kinetic friction between the surfaces is µ k. A person pushes horizontally with a force F p. The object of this exercise is to determine an expression for the acceleration of the block a = formula involving m,θ,f p,µ k,g and constants. The entire collection of these steps is called applying Newton s laws of mechanics to determine the acceleration of the block. a) Draw a free body diagram for the block. b) Describe the x and y axes that you will use. c) Write Newton s Second Law in vector form and also in its component form, i.e. write F net x = (11) F net y = (12) Insert as much information as possible about the components of acceleration at this stage. The resulting equations will generate much of the algebra that follows. d) Determine expressions for the magnitudes of all the friction and gravitational forces. Do you think that n = mg? e) List all the components of all the forces, using one of the two formats below. θ F gx = F gy = n x = n y =. Force x comp y comp n.. f) Note that F net x = i F ix (i.e. the x component of the net force is the sum of the x components of individual forces). Use this and Eq. (11) to obtain an equation relating various quantities that appear in this problem. Do the same with Eq. (12). Use the resulting algebraic expressions to get an expression for n. Is this mg? Use the resulting equations to get an expression for a. 15
16 23 Connected Objects: Tension and Acceleration Two boxes can move along a horizontal surface. There is no friction between either box and the surface. The boxes are connected by a rope. A hand pulls on the other rope with force 50N. 4.0kg 6.0kg 50N a) Determine the acceleration of each box. b) Determine the tension in the rope connecting the boxes. 24 Connected Objects: Friction Two boxes can move along a horizontal surface. There is no friction between the 6.0kg box and the surface. There is friction for the other box: the coefficient of static friction is 0.70 and the coefficient of kinetic friction is The boxes are connected by a rope. A hand pulls on the other rope with force 50N. 4.0kg 6.0kg 50N a) Determine the acceleration of each box. b) Determine the tension in the rope connecting the boxes. 25 Dynamics of Connected Objects; Level/Suspended Blocks without Friction Twoblocks areconnected byastring, whichrunsover a massless pulley. A 10kg block is suspended and a 5.0 kg block can slide along a frictionless horizontal surface. The string connected to the block on the surface runs horizontally. The blocks held at rest and then released. They move, constantly speeding up. Which of the following is true regarding the tension intheconnecting string, T, whiletheymove? Explain your choice. i) T = 0. ii) 98N > T > 0. iii) T = 98N. iv) T > 98N 16
17 26 Dynamics of Connected Objects: Level/Suspended Blocks without Friction Twoblocks areconnected byastring, whichrunsover a massless pulley. One block, of mass m 1, is suspended and the other block, of mass m 2, can move along a frictionless horizontal surface. The string connected to the block on the surface runs horizontally. A hand exerts a constant force on the block on the surface. This force has magnitude F hand and points horizontally to the left. Determine an expression for the magnitude of acceleration of the blocks, a. This should be of the form a = expression with only m 1,m 2,g,F hand and constants. a) Consider the block on the horizontal and carry out the following. i) Draw a free body diagram for this block. ii) Write Newton s second law in vector form for this block. Rewrite this in vertical and horizontal component form. Note: Just writing F = ma is not completely correct and is too imprecise to eventually give a correct answer. iii) Use the free body diagram to rewrite the component form of Newton s second law in terms of the individual forces acting on this block. Can you manipulate this to obtain a = formula involving only m 1,m 2,g and constants? b) Repeat part a) for the suspended block. c) Combine the expressions obtained for each block to obtain a single expression for the acceleration of the block on the horizontal. 17
18 27 Dynamics of Connected Objects: Level/Suspended Blocks with Friction Twoblocks areconnected byastring, whichrunsover a massless pulley. One block can move along a rough horizontal surface; the other is suspended. The string connected to the block on the surface runs horizontally. Ignore air resistance on either block. The aim of this exercise will be to determine the acceleration of the blocks. Suppose that mass of the suspended block is 6.0kg and the mass of the block on the surface is 4.0 kg. The coefficient of friction between the block and surface is First consider the block on the surface a) Draw a free body diagram for the block on the surface. b) Write Newton s Second Law in component form for the block on the surface, i.e. write F net x = ΣF ix = (13) F net y = ΣF iy = (14) Insert as much information as possible about the components of acceleration at this stage. The resulting equations will generate much of the algebra that follows. c) List all the components of all the forces for the block on the surface. F gx = F gy = n x = n y =. Force x comp y comp F g n. d) Use Eqs. (13) and (14) and the components to obtain an expression for the acceleration of the blocks. Can you solve this for acceleration at this stage? e) Repeat parts a) to d) for the suspended block. Be careful about the acceleration! f) Combine the equation for the two blocks to obtain the acceleration and the tension in the rope. The analysis can be performed for blocks of any mass. Let m 1 be the mass of the block on the surface, m 2 the mass of the suspended block and µ k be the coefficient of friction between the block and the surface. g) Determine an expression for the magnitude of acceleration of the blocks, a. This should be of the form a = formula with only m 1,m 2,g,µ k and constants. 18
19 28 Dynamics of Connected Objects: Atwood s Machine Two blocks, with masses indicated, are connected by astringwhichrunsoveramasslesspulley. Usethefollowing steps to determine an expression for the magnitude of acceleration of the blocks, a. This should be of the form a = formula involving only m 1,m 2,g and constants. m 2 The entire collection of these steps is called applying Newton s laws of mechanics to determine the acceleration of the blocks. m 1 a) Consider the block on the left and carry out the following. i) Draw a free body diagram for the block on the left. ii) Write Newton s second law in vector form for the block on the left. Rewrite this in vertical and horizontal component form. Note: Just writing F = ma is not completely correct and is too imprecise to eventually give a correct answer. iii) Use the free body diagram to rewrite the component form of Newton s second law in terms of the individual forces acting on the block on the left. Can you manipulate this to obtain a = formula involving only m 1,m 2,g and constants? b) Repeat part a) for the block on the right. Is the acceleration of the two blocks exactly the same? If not, how are the accelerations related? Convert this into a simple algebraic relationship. Note: You should be convinced that it cannot be correct to use the same symbol to represent the vertical component of acceleration for each block. c) Combine the expressions obtained for each block to obtain a single expression for the vertical component of the acceleration of the block on the left. Use this to obtain an expression for the magnitude of acceleration of each block. 19
20 29 Dynamics of Connected Objects: Atwood s Machine Variation Two blocks, with masses indicated, are connected by a string which runs over a massless pulley. A hand exerts a constant downward force with magnitude F hand and the block on the left. Determine an expression for the magnitude of acceleration of the blocks, a. This should be of the form a = formula involving only m 1,m 2,g,F hand and constants. m 2 m 1 20
21 30 Bug walking in a circle A bug walks at a constant speed in a circular path on a horizontal surface. Which vector best illustrates the net force on the bug at the illustrated moment? Explain your choice. C B D A 31 Ball Swinging in a Vertical Circle A 0.20kg ball swings with in a vertical circle at the end of a string of length 0.50m. a) Draw a free body diagram for the ball at the highest point of the circle. Draw a free body diagram at the lowest point. b) Ingeneral the speedof theball can vary as it swings. As the speed decreases does the tension at the top of the circle increase, decrease or stay constant? Determine the minimum speed so that the tension is not zero. Describe what happens if the speed drops beneath this. c) Now suppose that the speed of the ball is constant throughout its motion. How does net force at the highest point of the circle compare (larger, smaller, same) to that at the lowest point of the circle? Use you answer to compare (larger, smaller, same) the tension in the string at the lowest point of the circle to the tension at the highest point of the circle. d) Supposethat the string will break if the tension in it exceeds 5.0N. Use Newton s second law to analyze the situation where the tension is largest (i.e. highest or lowest point) and determine the maximum speed with which the ball can move so that the string does not break. 21
22 32 Block inside a Revolving Drum A spinning drum has vertical wooden sides. A wooden block block is placed inside the drum and the drum is eventually made to spin with a constant angular velocity. This is done is such a way that the block does not slide relative to the drum; it rotates at the same rate as the drum. In the first part of the problem suppose that the mass of the block is 2.0kg, the coefficient of static friction is 0.60 and the radius of the drum is 0.80m. a) Draw a free body diagram for the block, identify the direction of the acceleration and write Newton s 2nd law in component form. Is n = mg in this case? b) Determine an expression for the minimum angular velocity with which the drum must rotate so that the block does not slip. Now consider the more general case where the drum has radius r and the coefficient of static friction is µ s. c) Determine an expression for the minimum angular velocity (in terms of g,r and µ s ) so that the block does not slip. Does it depend on the mass of the block? r Now suppose that the spinning drum has tilted sides. A block is placed inside the drum and the drum is eventually made to spin with a constant angular velocity ω. This is done is such a way that the block does not slide relative to the drum; it rotates at the same rate as the drum. Let r be the distance from the block to the axle of the drum. r d) Determine the minimum angular velocity required for the block not to slip when θ = 75, r = 2.5m and the coefficient of static friction is 0.2. θ 22
23 33 Half Pipe A person of mass 60kg is on a skateboard of mass 5.0kg. Both are at rest at the top of a half-pipe of radius 10m. Ignore any friction and the rotation of the wheels. a) Determine the speed of the skateboarder at the bottom of the half pipe. b) Suppose that there was another skateboarder, of mass 80kg on a 5.0kg skateboard at rest at the bottom of the pipe. The two skateboarders collide, hold each other and move together. Determine their speed moments after they collide. Use this to determine how high up the pipe they move. 34 Sledding King Zog, with mass 160kg, and Queen Geraldine, with mass 80kg, sled down an icy hill. They start from rest at the same point above the bottom of the hill. Ignore friction and air resistance. Which of the following is true regarding their speeds at the bottom of the hill? Explain your answer. i) Same speeds. ii) Geraldine s speed is twice that of Zog. iii) Geraldine s speed is four times that of Zog. iv) Zog s speed is larger than Geraldine s speed. 35 Spring bumper Two walruses (named X and Y), with the same masses, slide along horizontal sheets of ice. Each collides with a horizontal spring mounted to a wall; the springs are identical. Prior to hitting the spring, walrus X moved with speed twice that of walrus Y. The springs compress, bringing each walrus to a stop. Which of the following is true regarding the distances by which the springs compress? Explain your answer. i) Springs compress by the same distance. ii) X compresses spring by twice as much Y. iii) X compresses spring by four times as much Y. iv) X compresses spring by half as much Y. v) X compresses spring by a quarter of what Y compresses. 23
24 36 Vectors: Dot Products For each of the following, determine A x,a y,b x,b y and A z and B z (if applicable) and determine the dot product, A B. Note: If your answer for the dot product contains î and ĵ then it is very incorrect! a) A = 2î+2ĵ B = 2î+2ĵ b) A = 2î+2ĵ B = 2î 2ĵ c) A = 2î 2ĵ B = 2î+2ĵ d) A = 2î 2ĵ B = 2î 2ĵ e) A = 2î+2ĵ+3ˆk B = 2î+2ĵ f) A = 2î+2ĵ+3ˆk B = 2î 2ĵ g) A = 2î 2ĵ+3ˆk B = 2î+2ĵ+3ˆk h) A = 1î 2ĵ+3ˆk B = 2î+1ĵ 3ˆk 24
25 Answers: a) A B = 8. b) A B = 0. c) A B = 0. d) A B = 8. e) A B = 8. f) A B = 0. g) A B = 9. h) A B = 9. 25
26 37 Vectors: Dot Products For each of the following, express A in component form B using unit vectors and determine the dot product, A B. Note: If your answer for the dot product contains î and ĵ then it is very incorrect! a) y B A x b) y B A x c) y B A x d) y B A x Answer: a) A B = 6. b) A B = 9. c) A B = 7. d) A B = 6. 26
27 38 Object oscillating on a vertical spring A monkey hangs from a spring which is attached to the ceiling of a building. The spring hangs vertically and the monkey bounces up and down without touching the floor. a) As the monkey ascends toward and nears its highest point, the spring is compressed. Which of the following is true while this happens? i) The spring does positive work, gravity does positive work. ii) The spring does positive work, gravity does negative work. iii) The spring does negative work, gravity does positive work. iv) The spring does negative work, gravity does negative work. b) As the monkey begins to descend away its highest point, the spring is still compressed. Which of the following is true while this happens? i) The spring does positive work, gravity does positive work. ii) The spring does positive work, gravity does negative work. iii) The spring does negative work, gravity does positive work. iv) The spring does negative work, gravity does negative work. 27
28 39 Motion under a complicated potential A particle that can move along the x axis is subjected to the potential U(x) = (x x)/ U a) Determine an expression for the horizontal component of the force on the object. b) Determine locations where the object is in equilibrium and for each describe whether the equilibrium is stable or not. 40 Hoisting fish King Zog and Queen Geraldine are fishing from a bridge and they catch identical twin fish, each with mass 5.0kg. They hoist the fish at constant speeds to the bridge 8.0m about the water. Zog takes 10s to hoist his fish and Geraldine 7.5s to hoist her fish. a) Which of the following is true? Explain your answer. i) Zog and Geraldine do the same work. ii) Zog does more work than Geraldine. iii) Zog does less work than Geraldine. b) Which of the following is true? Explain your answer. i) Zog and Geraldine expend the same power. ii) Zog expends less power. iii) Zog expends more power. 41 Power delivered by engines Two engines pull identical objects along horizontal surfaces. Engine A delivers 2000 W of power and engine B delivers 4000W. Engine A pulls for 5min and engine B for 4min. Which of the following is true. i) Engine A delivers more work than engine B. ii) Engine A delivers less work than engine B. iii) Engine A delivers the same work as engine B. iv) There is not enough information to decide. x 28
29 42 Beam in equilibrium A beam with mass M and length L is anchored to a wall and held at rest horizontally by a rope as illustrated. A ball with mass m is suspended from the beam at the illsutrated point. The aim of this exercise is to determine the tension in the rope. This would enable one to decide on the breaking strength of the rope. Wall 40 a) State the conditions for equilibrium. b) Draw all the force vectors on the beam. 3L/4 c) Identify a pivot point (there are many correct possibilities one is much more useful than the others) and determine expressions for the torque exerted by each force about the pivot. d) Substitute the individual torques into one of the conditions for equilibrium and obtain an expression for the tension in the rope. e) Whatmustbetheminimumtensionat whichtheropecanbreaktosupporta40kgbeam with length 3.0m from which a 8.0kg ball is suspended in the illustrated configuration? 29
30 43 Rotating disk A 2.0kg turntable (disk) has radius 0.10m and can rotate horizontally about a frictionless axle through its center. A 1.2kg blob of putty is stuck to a point on the disk three quarters of the distance from the center to the edge. A rope attached halfway from the center to the edge of the disk pulls with force 4.0N as illustrated. The aim of this exercise is to determine the angular acceleration of the disk via the following steps. a) Write the rotational version of Newton s second law. b) Determine the moment of inertia of the disk plus putty. c) Determine the net torque acting on the disk. d) Determine the angular acceleration of the disk. Putty e) Supposethat a brake pad presses on therim of the disk, producingafrictional force with magnitude 1.5 N while the rope is pulling as before. Determine the angular acceleration of the disk in this situation. Rope 30
31 44 Mass suspended from a rotating disk A pulley, whose moment of inertia is I, can rotate about a frictionless axle through its center. Two blocks are suspended from a string, which runs, without slipping, over the pulley. The block on the left has mass m 1 and that on the right has mass m 2 < m 1. The system is held at rest and then released. a) Write the rotational version of Newton s second law for the pulley. b) Determine the net torque acting on the pulley, assuming that the tension in the string on the left might be different from that on the right. c) Determine an expression for the angular acceleration of the pulley in terms of the tensions in the strings. d) If the moment of inertia of the pulley is non-zero and the pulley is accelerating, can the two tensions be equal? e) Apply Newton s second law to each of the suspended masses, in each case relating the magnitude of the acceleration to the tension in the string. f) Combine the results from the previous parts to show that the magnitude of the acceleration of the blocks is m 1 m 2 a = m 1 +m 2 +I/R 2 g where R is the radius of the pulley. 31
32 45 Vectors: Cross Products For each of the following, determine A x,a y,b x,b y and A z and B z (if applicable) and determine the cross product, A B. Note: If your answer for the cross product does not contain î,ĵ or ˆk then it is very incorrect! a) b) c) d) e) f) g) h) A = 2î+2ĵ B = 2î+2ĵ A = 2î+2ĵ B = 2î 2ĵ A = 2î 2ĵ B = 2î+2ĵ A = 2î 2ĵ B = 2î 2ĵ A = 2î+2ĵ+3ˆk B = 2î+2ĵ A = 2î+2ĵ+3ˆk B = 2î 2ĵ A = 2î 2ĵ+3ˆk B = 2î+2ĵ+3ˆk A = 1î 2ĵ+3ˆk B = 2î+1ĵ 3ˆk 32
33 46 Vectors: Cross Products For each of the following, express A in component form B using unit vectors and determine both cross products, A B and B A. Note: If your answer for the cross product does not contain î,ĵ or ˆk then it is very incorrect! a) y B A x b) y B A x c) y B A x d) y B A x Answer: a) A B = 7ˆk and B A = 7ˆk. b) A B = 2ˆk and B A = 2ˆk. c) A B = 6ˆk and B A = 6ˆk. d) A B = 7ˆk and B A = 7ˆk. 33
### C B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N
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Physics 4 HW Set Chapter 5 Serway 8 th OC:, 4, 7 CQ: 4, 8 P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59, 67, 74 OC CQ P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59, | 16,033 | 62,092 | {"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-2021-17 | latest | en | 0.919946 |
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# 7.5: Indirect Measurement
Difficulty Level: At Grade Created by: CK-12
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What if you wanted to measure the height of a flagpole using your friend George? He is 6 feet tall and his shadow is 10 feet long. At the same time, the shadow of the flagpole was 85 feet long. How tall is the flagpole? After completing this Concept, you'll be able to use indirect measurement to help you answer this question.
### Guidance
An application of similar triangles is to measure lengths indirectly. The length to be measured would be some feature that was not easily accessible to a person, such as the width of a river or canyon and the height of a tall object. To measure something indirectly, you need to set up a pair of similar triangles.
#### Example A
A tree outside Ellie’s building casts a 125 foot shadow. At the same time of day, Ellie casts a 5.5 foot shadow. If Ellie is 4 feet 10 inches tall, how tall is the tree?
Draw a picture. From the picture to the right, we see that the tree and Ellie are parallel, therefore the two triangles are similar to each other. Write a proportion.
4ft,10inxft=5.5ft125ft\begin{align*}\frac{4ft, 10in}{xft}=\frac{5.5ft}{125ft}\end{align*}
Notice that our measurements are not all in the same units. Change both numerators to inches and then we can cross multiply.
58inxft=66in125ft58(125)7250x=66(x)=66x109.85 ft\begin{align*}\frac{58in}{xft}=\frac{66in}{125ft} \longrightarrow 58(125) &= 66(x)\\ 7250 &= 66x\\ x & \approx 109.85 \ ft\end{align*}
#### Example B
Cameron is 5 ft tall and casts a 12 ft shadow. At the same time of day, a nearby building casts a 78 ft shadow. How tall is the building?
To solve, set up a proportion that compares height to shadow length for Cameron and the building. Then solve the equation to find the height of the building. Let x\begin{align*}x\end{align*} represent the height of the building.
5ft12ft12xx=x78ft=390=32.5ft\begin{align*} \frac{5 ft}{12 ft}&=\frac{x}{78 ft} \\ 12x&=390 \\ x&=32.5 ft\end{align*}
The building is 32.5\begin{align*}32.5\end{align*} feet tall.
#### Example C
The Empire State Building is 1250 ft. tall. At 3:00, Pablo stands next to the building and has an 8 ft. shadow. If he is 6 ft tall, how long is the Empire State Building’s shadow at 3:00?
Similar to Example B, solve by setting up a proportion that compares height to shadow length. Then solve the equation to find the length of the shadow. Let x\begin{align*}x\end{align*} represent the length of the shadow.
6ft8ft6xx=1250ftx=10000=1666.67ft\begin{align*}\frac{6 ft}{8 ft}&=\frac{1250 ft}{x}\\ 6x&=10000\\ x&=1666.67 ft\end{align*}
The shadow is approximately 1666.67\begin{align*}1666.67\end{align*} feet long.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
It is safe to assume that George and the flagpole stand vertically, making right angles with the ground. Also, the angle where the sun’s rays hit the ground is the same for both. The two trianglesare similar. Set up a proportion.
1085=6x10xx=510=51 ft.\begin{align*}\frac{10}{85} = \frac{6}{x} \longrightarrow 10x &= 510\\ x &= 51 \ ft.\end{align*}
The height of the flagpole is 51 feet.
### Vocabulary
Two triangles are similar if all their corresponding angles are congruent (exactly the same) and their corresponding sides are proportional (in the same ratio). Solve proportions by cross-multiplying.
### Guided Practice
In order to estimate the width of a river, the following technique can be used. Use the diagram.
Place three markers, O,C,\begin{align*}O, C,\end{align*} and E\begin{align*}E\end{align*} on the upper bank of the river. E\begin{align*}E\end{align*} is on the edge of the river and OC¯¯¯¯¯¯¯¯CE¯¯¯¯¯¯¯¯\begin{align*}\overline {OC} \perp \overline{CE}\end{align*}. Go across the river and place a marker, N\begin{align*}N\end{align*} so that it is collinear with C\begin{align*}C\end{align*} and E\begin{align*}E\end{align*}. Then, walk along the lower bank of the river and place marker A\begin{align*}A\end{align*}, so that CN¯¯¯¯¯¯¯¯NA¯¯¯¯¯¯¯¯\begin{align*}\overline{CN} \perp \overline{NA}\end{align*}. OC=50\begin{align*}OC = 50\end{align*} feet, CE=30\begin{align*}CE = 30\end{align*} feet, NA=80\begin{align*}NA = 80\end{align*} feet.
1. Is OCEANE?\begin{align*}\triangle OCE \sim \triangle ANE?\end{align*} How do you know?
2. Is OC¯¯¯¯¯¯¯¯NA¯¯¯¯¯¯¯¯?\begin{align*}\overline {OC} \| \overline {NA}?\end{align*} How do you know?
3. What is the width of the river? Find EN\begin{align*}EN\end{align*}.
1. Yes. CN\begin{align*}\angle{C} \cong \angle{N}\end{align*} because they are both right angles. OECAEN\begin{align*}\angle{OEC} \cong \angle{AEN}\end{align*} because they are vertical angles. This means OCEANE\begin{align*}\triangle OCE \sim \triangle ANE\end{align*} by the AA Similarity Postulate.
2. Since the two triangles are similar, we must have EOCEAN\begin{align*}\angle{EOC} \cong \angle{EAN}\end{align*}. These are alternate interior angles. When alternate interior angles are congruent then lines are parallel, so OC¯¯¯¯¯¯¯¯NA¯¯¯¯¯¯¯¯\begin{align*}\overline {OC} \| \overline {NA}\end{align*}.
3. Set up a proportion and solve by cross-multiplying.
30ftEN50(EN)EN=50ft80ft=2400=48\begin{align*}\frac{30 ft}{EN}&=\frac{50 ft}{80 ft}\\ 50(EN)&=2400 \\ EN&=48\end{align*}
The river is 48\begin{align*}48\end{align*} feet wide.
### Practice
The technique from the guided practice section was used to measure the distance across the Grand Canyon. Use the picture below and OC=72 ft,CE=65 ft\begin{align*}OC = 72 \ ft , CE = 65 \ ft\end{align*}, and NA=14,400 ft\begin{align*}NA = 14,400 \ ft\end{align*} for problems 1 - 3.
1. Find EN\begin{align*}EN\end{align*} (the distance across the Grand Canyon).
2. Find OE\begin{align*}OE\end{align*}.
3. Find EA\begin{align*}EA\end{align*}.
1. Janet wants to measure the height of her apartment building. She places a pocket mirror on the ground 20 ft from the building and steps backwards until she can see the top of the build in the mirror. She is 18 in from the mirror and her eyes are 5 ft 3 in above the ground. The angle formed by her line of sight and the ground is congruent to the angle formed by the reflection of the building and the ground. You may wish to draw a diagram to illustrate this problem. How tall is the building?
2. Sebastian is curious to know how tall the announcer’s box is on his school’s football field. On a sunny day he measures the shadow of the box to be 45 ft and his own shadow is 9 ft. Sebastian is 5 ft 10 in tall. Find the height of the box.
3. Juanita wonders how tall the mast of a ship she spots in the harbor is. The deck of the ship is the same height as the pier on which she is standing. The shadow of the mast is on the pier and she measures it to be 18 ft long. Juanita is 5 ft 4 in tall and her shadow is 4 ft long. How tall is the ship’s mast?
4. Evan is 6 ft tall and casts a 15 ft shadow. A the same time of day, a nearby building casts a 30 ft shadow. How tall is the building?
5. Priya and Meera are standing next to each other. Priya casts a 10 ft shadow and Meera casts an 8 ft shadow. Who is taller? How do you know?
6. Billy is 5 ft 9 inches tall and Bobby is 6 ft tall. Bobby's shadow is 13 feet long. How long is Billy's shadow?
7. Sally and her little brother are walking to school. Sally is 4 ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother?
8. Ray is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow?
9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree?
10. Jason, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building?
11. Alexandra, who is 5 ft 8 in tall is casting a 12 ft shadow. A nearby lamppost is casting a 20 ft shadow. How tall is the lamppost?
12. Use shadows or a mirror to measure the height of an object in your yard or on the school grounds. Draw a picture to illustrate your method.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
AA Similarity Postulate If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar.
Proportion A proportion is an equation that shows two equivalent ratios.
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Broad Topics > Numbers and the Number System > Place value
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##### Stage: 2 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### What an Odd Fact(or)
##### Stage: 3 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?
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##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
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##### Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### Cayley
##### Stage: 3 Challenge Level:
The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?
### Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
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##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
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##### Stage: 3 Challenge Level:
Find the values of the nine letters in the sum: FOOT + BALL = GAME
##### Stage: 1 and 2 Challenge Level:
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### Being Collaborative - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level to work on with others.
### Pupils' Recording or Pupils Recording
##### Stage: 1, 2 and 3
This article, written for teachers, looks at the different kinds of recordings encountered in Primary Mathematics lessons and the importance of not jumping to conclusions!
### Being Resilient - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that may require resilience.
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### Big Powers
##### Stage: 3 and 4 Challenge Level:
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
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##### Stage: 2 Challenge Level:
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##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Multiply Multiples 2
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Can you work out some different ways to balance this equation?
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Multiply Multiples 3
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Have a go at balancing this equation. Can you find different ways of doing it?
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Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Six Is the Sum
##### Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### ABC
##### Stage: 2 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Coded Hundred Square
##### Stage: 2 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
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##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
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##### Stage: 3 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
### Digit Sum
##### Stage: 3 Challenge Level:
What is the sum of all the digits in all the integers from one to one million?
### Being Curious - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems for inquiring primary learners.
### Nice or Nasty for Two
##### Stage: 2 Challenge Level:
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your opponent.
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##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
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##### Stage: 3 Challenge Level:
Explore the relationship between simple linear functions and their graphs.
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Number problems at primary level that require careful consideration.
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##### Stage: 2 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
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##### Stage: 3 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
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##### Stage: 3 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
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##### Stage: 3 Challenge Level:
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.
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##### Stage: 2 Challenge Level:
Can you substitute numbers for the letters in these sums?
### Reach 100
##### Stage: 2 and 3 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
### Oddly
##### Stage: 2 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
### X Marks the Spot
##### Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### Two and Two
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How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Basically
##### Stage: 3 Challenge Level:
The number 3723(in base 10) is written as 123 in another base. What is that base?
##### Stage: 2 and 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### Chocolate Maths
##### Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### Back to the Planet of Vuvv
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There are two forms of counting on Vuvv - Zios count in base 3 and Zepts count in base 7. One day four of these creatures, two Zios and two Zepts, sat on the summit of a hill to count the legs of. . . .
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##### Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
### Napier's Bones
##### Stage: 2 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### Not a Polite Question
##### Stage: 3 Challenge Level:
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square... | 2,071 | 8,823 | {"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-2017-47 | latest | en | 0.828116 |
http://nrich.maths.org/public/leg.php?code=97&cl=3&cldcmpid=5770 | 1,505,996,982,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687766.41/warc/CC-MAIN-20170921115822-20170921135822-00477.warc.gz | 251,212,496 | 10,115 | # Search by Topic
#### Resources tagged with Circles similar to Angle to Chord:
Filter by: Content type:
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### There are 62 results
Broad Topics > 2D Geometry, Shape and Space > Circles
### Arclets Explained
##### Stage: 3 and 4
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.
### What Is the Circle Scribe Disk Compass?
##### Stage: 3 and 4
Introducing a geometrical instrument with 3 basic capabilities.
### Three Tears
##### Stage: 4 Challenge Level:
Construct this design using only compasses
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### Circumspection
##### Stage: 4 Challenge Level:
M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.
### Not So Little X
##### Stage: 3 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?
##### Stage: 4 Challenge Level:
Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .
### Angle A
##### Stage: 3 Challenge Level:
The three corners of a triangle are sitting on a circle. The angles are called Angle A, Angle B and Angle C. The dot in the middle of the circle shows the centre. The counter is measuring the size. . . .
### Roaming Rhombus
##### Stage: 4 Challenge Level:
We have four rods of equal lengths hinged at their endpoints to form a rhombus ABCD. Keeping AB fixed we allow CD to take all possible positions in the plane. What is the locus (or path) of the point. . . .
##### Stage: 4 Challenge Level:
Explore when it is possible to construct a circle which just touches all four sides of a quadrilateral.
### Tricircle
##### Stage: 4 Challenge Level:
The centre of the larger circle is at the midpoint of one side of an equilateral triangle and the circle touches the other two sides of the triangle. A smaller circle touches the larger circle and. . . .
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 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?
### Circles, Circles Everywhere
##### Stage: 2 and 3
This article for pupils gives some examples of how circles have featured in people's lives for centuries.
### Circle Packing
##### Stage: 4 Challenge Level:
Equal circles can be arranged so that each circle touches four or six others. What percentage of the plane is covered by circles in each packing pattern? ...
### Three Four Five
##### Stage: 4 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### Floored
##### Stage: 3 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?
### Coins on a Plate
##### Stage: 3 Challenge Level:
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.
### Some(?) of the Parts
##### Stage: 4 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### A Rational Search
##### Stage: 4 and 5 Challenge Level:
Investigate constructible images which contain rational areas.
### Witch's Hat
##### Stage: 3 and 4 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Pi, a Very Special Number
##### Stage: 2 and 3
Read all about the number pi and the mathematicians who have tried to find out its value as accurately as possible.
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Efficient Packing
##### Stage: 4 Challenge Level:
How efficiently can you pack together disks?
##### Stage: 4 Challenge Level:
The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### LOGO Challenge 12 - Concentric Circles
##### Stage: 3 and 4 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.
### LOGO Challenge 11 - More on Circles
##### Stage: 3 and 4 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. . . .
##### Stage: 2, 3 and 4 Challenge Level:
A metal puzzle which led to some mathematical questions.
### The Pi Are Square
##### Stage: 3 Challenge Level:
A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?
### LOGO Challenge 10 - Circles
##### Stage: 3 and 4 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.
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
### Bull's Eye
##### Stage: 3 Challenge Level:
What fractions of the largest circle are the two shaded regions?
### Crescents and Triangles
##### Stage: 4 Challenge Level:
Triangle ABC is right angled at A and semi circles are drawn on all three sides producing two 'crescents'. Show that the sum of the areas of the two crescents equals the area of triangle ABC.
### Squaring the Circle
##### Stage: 3 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. . . .
### Illusion
##### Stage: 3 and 4 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### First Forward Into Logo 4: Circles
##### Stage: 2, 3 and 4 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### Efficient Cutting
##### Stage: 3 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.
### LOGO Challenge - Circles as Animals
##### Stage: 3 and 4 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### 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?
### Rolling Coins
##### Stage: 4 Challenge Level:
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. . . .
### Track Design
##### Stage: 4 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Get Cross
##### Stage: 4 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### Holly
##### Stage: 4 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
### Fitting In
##### Stage: 4 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . . | 2,326 | 10,101 | {"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-2017-39 | latest | en | 0.863686 |
https://practicepaper.in/gate-ee/fourier-transform | 1,702,016,836,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00173.warc.gz | 506,021,531 | 16,004 | # Fourier Transform
Question 1
The discrete-time Fourier transform of a signal $x[n]$ is $X(\Omega)=1(1+\cos \Omega) e^{-j \Omega}$. Consider that $x_{p}[n]$ is a periodic signal of period $N=5$ such that
\begin{aligned} x_{p}[n] & =x[n], \text { for } n=0,1,2 \\ & =0, \text { for } n=3,4 \end{aligned}
Note that $x_{p}[n]=\sum_{k=0}^{N-1} a_{k} e^{j \frac{2 \pi}{N} k m}$. The magnitude of the Fourier series coefficient $a_{3}$ is ____ (Round off to 3 decimal places).
A 0.038 B 0.025 C 0.068 D 0.012
GATE EE 2023 Signals and Systems
Question 1 Explanation:
Given : $x_{p}(n)$ is a period signal of period $N=5$.
$x_{p}(n)=\left\{\begin{array}{cc} x(n), & \text { for } n=0,1,2 \\ 0, & \text { for } n=3,4 \end{array}\right.$
and $\mathrm{x}(\mathrm{n}) \stackrel{\text { DTFT }}{\longrightarrow} \mathrm{X}(\Omega)=(1+\cos \Omega) \mathrm{e}^{-\mathrm{j} \Omega}$
We have,
\begin{aligned} a_{K} & =\frac{X\left(\frac{2 \pi}{N} K\right)}{N} \\ & =\frac{\left[1+\cos \left(\frac{2 \pi}{N} K\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{N} R}}{N} \end{aligned}
For, $\mathrm{N}=5$ and $\mathrm{K}=3$
\begin{aligned} a_{3} & =\frac{\left[1+\cos \left(\frac{2 \pi \times 3}{5}\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{5} \times 3}}{5} \\ \left|a_{3}\right| & =\frac{1}{5}\left[1+\cos \frac{6 \pi}{5}\right] \\ & =0.0382 \end{aligned}
Question 2
The Fourier transform $X(\omega)$ of the signal $x(t)$ is given by
\begin{aligned} X(\omega) & =1, \text { for }|\omega| \lt W_{0} \\ & =0, \text { for }|\omega| \gt W_{0} \end{aligned}
Which one of the following statements is true?
A $x(t)$ tends to be an impulse as $\mathrm{W}_{0} \rightarrow \infty$ B $x(0)$ decreases as $W_{0}$ increases C At $\mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=-\frac{1}{\pi}$ D At $\mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=\frac{1}{\pi}$
GATE EE 2023 Signals and Systems
Question 2 Explanation:
Given,
$\quad X(\omega)=\left\{\begin{array}{l}1, \text { for }|\omega|W_{0}\end{array}\right.$
$X(\omega)=\operatorname{rect}\left(\frac{\omega}{2 W_{0}}\right)$
We know,
$\operatorname{Arect}\left(\frac{\mathrm{t}}{\tau}\right)=\operatorname{A\tau S}\left(\frac{\omega \tau}{2}\right)$
By duality,
$\operatorname{A} \tau \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right) \leftrightarrow 2 \pi \operatorname{Arect}\left(\frac{\omega}{\tau}\right)$
Given,
$2 \pi \mathrm{A}=1$
$\Rightarrow \quad A=\frac{1}{2 \pi}$
$\therefore \quad \mathrm{x}(\mathrm{t})=\frac{\tau}{2 \pi} \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right)$, where $\tau=2 \mathrm{~W}_{0}$
Thus,
\begin{aligned} x(t)&=\frac{W_0}{\pi} \mathrm{Sa}(W_0 t)\\ &=\frac{1}{\pi}\frac{\sin W_0 t}{t}\\ \therefore \quad \text{At } t&=\frac{\pi}{2W_0}, x(t)=\frac{1}{\pi t} \end{aligned}
From rectangular function, At $W \Rightarrow \infty ,X(w)=1$
Taking inverse fourier transform $x(t)=\delta (t)$
Option (A) will be correct.
Question 3
Let an input $x(t)=2 \sin (10 \pi t)+5 \cos (15 \pi t)+7 \sin (42 \pi t)+4 \cos (45 \pi t)$ is passed through an LTI system having an impulse response,
$h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos (40\pi t)$
The output of the system is
A $2 \sin (10 \pi t)+5\cos (15 \pi t)$ B $7 \sin (42 \pi t)+5\cos (15 \pi t)$ C $7 \sin (42 \pi t)+4\cos (45 \pi t)$ D $2 \sin (10 \pi t)+4\cos (45 \pi t)$
GATE EE 2022 Signals and Systems
Question 3 Explanation:
Given: $h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos 40 \pi t$
Fourier transform of signal $\frac{\sin (10 \pi t)}{\pi t}$ is given by
Now, impulse response
$h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right )\cdot \left ( \frac{e^{+j40\pi t}+e^{-j40\pi t}}{2} \right )$
Using property, $e^{-j\omega _0t}x(t)\rightleftharpoons X(\omega +\omega _0)$
Therefore, Fourier transform of impulse response
Cut-off frequencies,
$\omega_L=30 \pi rad/sec$
$\omega_H=50 \pi rad/sec$
Thus, output of the system $=7 \sin 42 \pi t+4 \cos 45 \pi t$
Question 4
Consider a continuous-time signal $x(t)$ defined by $x(t)=0$ for $\left | t \right |> 1$, and $x\left ( t \right )=1-\left | t \right | for \left | t \right |\leq 1$. Let the Fourier transform of $x(t)$ be defined as $X\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:x\left ( t \right )e^{-j\omega t}\:dt$. The maximum magnitude of $X\left ( \omega \right )$ is ___________.
A 1 B 2 C 3 D 4
GATE EE 2021 Signals and Systems
Question 4 Explanation:
Fourier transform, $F(\omega)=A \tau S a^{2}\left(\frac{\omega \tau}{2}\right)$
\begin{aligned} \text { As } A=1, \tau=1 \\ H \omega) &=S a^{2}\left(\frac{\omega}{2}\right) \\ \left.F(\omega)\right|_{\text {peak }} &=F(0)=S a^{2}(0)=1 \end{aligned}
$\because$ Peak value of sampling function occurs at $\omega=0$
Peak value $=1$
Question 5
Let $f(t)$ be an even function, i.e. $f(-t)=f(t)$ for all t. Let the Fourier transform of $f(t)$ be defined as $F\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:f\left ( t \right )e^{-j\omega t}dt$. Suppose $\dfrac{dF\left ( \omega \right )}{d\omega }=-\omega F\left ( \omega \right )$ for all $\omega$, and $F(0)=1$. Then
A $f\left ( 0 \right )\lt 1$ B $f\left ( 0 \right ) \gt 1$ C $f\left ( 0 \right )= 1$ D $f\left ( 0 \right )= 0$
GATE EE 2021 Signals and Systems
Question 5 Explanation:
$f(t) \rightleftharpoons F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t$
The following informations are given about $f(t) \rightleftharpoons F(\omega).$
$\begin{array}{l} (i) f(t)=f(-t)\\ (ii) \left.F(\omega)\right|_{\omega=0}=1\\ (iii) \frac{d F(\omega)}{d \omega}=-\omega F(\omega)\\ \text{From }(iii), \frac{d F(\omega)}{d \omega}+\omega F(\omega)=0 \end{array}$
By solving the above linear differential equations, (by mathematics)
\begin{aligned} \ln F(\omega) &=-\frac{\omega^{2}}{2} \\ \Rightarrow \qquad\qquad F(\omega) &=K \cdot e^{-\omega^{2} / 2} \\ \text { Put } \omega=0, \qquad\qquad F(0) &=K \\ \Rightarrow \qquad\qquad 1 &=K \text { (from info. } \\ \text { From (iv), } \qquad\qquad F(\omega) &=e^{-\omega^{2} / 2} \\ \text{As we know}, e^{-a t^{2}}, a \gt 0 &\rightleftharpoons \sqrt{\frac{\pi}{a}} e^{-\omega^{2} / 4 a}\\ \text{At }a=\frac{1}{2}, \qquad\qquad\quad e^{-t^{2} / 2} &\rightleftharpoons \sqrt{\frac{\pi}{1 / 2}} \cdot e^{-\omega^{2} / 2}\\ f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \rightleftharpoons e^{-\omega^{2} / 2}=F(\omega)\\ \text { Thus, }\qquad\qquad f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \\ \text { At } t=0, \qquad\qquad f(0)&=\frac{1}{\sqrt{2 \pi}} \lt 1 \end{aligned}
There are 5 questions to complete. | 2,603 | 6,557 | {"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": 0, "wp-katex-eq": 75, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-50 | latest | en | 0.462133 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/6_Hyperbolic_functions/6.7_Miscellaneous/6.7.1_Hyperbolic_functions/rese956.htm | 1,696,141,322,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510781.66/warc/CC-MAIN-20231001041719-20231001071719-00475.warc.gz | 675,444,809 | 4,922 | ### 3.956 $$\int e^{c+d x} \cosh (a+b x) \coth (a+b x) \, dx$$
Optimal. Leaf size=103 $\frac{2 e^{-a-x (b-d)+c} \, _2F_1\left (1,-\frac{b-d}{2 b};\frac{b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}-\frac{3 e^{-a-x (b-d)+c}}{2 (b-d)}+\frac{e^{a+x (b+d)+c}}{2 (b+d)}$
[Out]
(-3*E^(-a + c - (b - d)*x))/(2*(b - d)) + E^(a + c + (b + d)*x)/(2*(b + d)) + (2*E^(-a + c - (b - d)*x)*Hyperg
eometric2F1[1, -(b - d)/(2*b), (b + d)/(2*b), E^(2*(a + b*x))])/(b - d)
________________________________________________________________________________________
Rubi [A] time = 0.207046, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {5511, 2194, 2227, 2251} $\frac{2 e^{-a-x (b-d)+c} \, _2F_1\left (1,-\frac{b-d}{2 b};\frac{b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}-\frac{3 e^{-a-x (b-d)+c}}{2 (b-d)}+\frac{e^{a+x (b+d)+c}}{2 (b+d)}$
Antiderivative was successfully verified.
[In]
Int[E^(c + d*x)*Cosh[a + b*x]*Coth[a + b*x],x]
[Out]
(-3*E^(-a + c - (b - d)*x))/(2*(b - d)) + E^(a + c + (b + d)*x)/(2*(b + d)) + (2*E^(-a + c - (b - d)*x)*Hyperg
eometric2F1[1, -(b - d)/(2*b), (b + d)/(2*b), E^(2*(a + b*x))])/(b - d)
Rule 5511
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol]
:> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
IGtQ[m, 0] && IGtQ[n, 0] && HyperbolicQ[G] && HyperbolicQ[H]
Rule 2194
Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]
Rule 2227
Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] && !PowerOfLinearMatchQ[v, x]
Rule 2251
Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])
Rubi steps
\begin{align*} \int e^{c+d x} \cosh (a+b x) \coth (a+b x) \, dx &=\int \left (\frac{3}{2} e^{-a+c-(b-d) x}+\frac{1}{2} e^{-a+c-(b-d) x+2 (a+b x)}+\frac{2 e^{-a+c-(b-d) x}}{-1+e^{2 (a+b x)}}\right ) \, dx\\ &=\frac{1}{2} \int e^{-a+c-(b-d) x+2 (a+b x)} \, dx+\frac{3}{2} \int e^{-a+c-(b-d) x} \, dx+2 \int \frac{e^{-a+c-(b-d) x}}{-1+e^{2 (a+b x)}} \, dx\\ &=-\frac{3 e^{-a+c-(b-d) x}}{2 (b-d)}+\frac{2 e^{-a+c-(b-d) x} \, _2F_1\left (1,-\frac{b-d}{2 b};\frac{b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}+\frac{1}{2} \int e^{a+c+(b+d) x} \, dx\\ &=-\frac{3 e^{-a+c-(b-d) x}}{2 (b-d)}+\frac{e^{a+c+(b+d) x}}{2 (b+d)}+\frac{2 e^{-a+c-(b-d) x} \, _2F_1\left (1,-\frac{b-d}{2 b};\frac{b+d}{2 b};e^{2 (a+b x)}\right )}{b-d}\\ \end{align*}
Mathematica [A] time = 0.554539, size = 93, normalized size = 0.9 $\frac{e^c \left (\frac{e^{d x} (b \cosh (a+b x)-d \sinh (a+b x))}{b-d}-2 (\sinh (a)+\cosh (a)) e^{x (b+d)} \, _2F_1\left (1,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 b x} (\cosh (a)+\sinh (a))^2\right )\right )}{b+d}$
Antiderivative was successfully verified.
[In]
Integrate[E^(c + d*x)*Cosh[a + b*x]*Coth[a + b*x],x]
[Out]
(E^c*(-2*E^((b + d)*x)*Hypergeometric2F1[1, (b + d)/(2*b), (3*b + d)/(2*b), E^(2*b*x)*(Cosh[a] + Sinh[a])^2]*(
Cosh[a] + Sinh[a]) + (E^(d*x)*(b*Cosh[a + b*x] - d*Sinh[a + b*x]))/(b - d)))/(b + d)
________________________________________________________________________________________
Maple [F] time = 0.119, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{dx+c}} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}{\rm csch} \left (bx+a\right )\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a),x)
[Out]
int(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a),x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -4 \, b \int \frac{e^{\left (d x + c\right )}}{{\left (3 \, b - d\right )} e^{\left (5 \, b x + 5 \, a\right )} - 2 \,{\left (3 \, b - d\right )} e^{\left (3 \, b x + 3 \, a\right )} +{\left (3 \, b - d\right )} e^{\left (b x + a\right )}}\,{d x} + \frac{{\left (5 \, b^{2} e^{c} + 6 \, b d e^{c} + d^{2} e^{c} +{\left (3 \, b^{2} e^{c} - 4 \, b d e^{c} + d^{2} e^{c}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 2 \,{\left (6 \, b^{2} e^{c} + b d e^{c} - d^{2} e^{c}\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )} e^{\left (d x\right )}}{2 \,{\left ({\left (3 \, b^{3} - b^{2} d - 3 \, b d^{2} + d^{3}\right )} e^{\left (3 \, b x + 3 \, a\right )} -{\left (3 \, b^{3} - b^{2} d - 3 \, b d^{2} + d^{3}\right )} e^{\left (b x + a\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="maxima")
[Out]
-4*b*integrate(e^(d*x + c)/((3*b - d)*e^(5*b*x + 5*a) - 2*(3*b - d)*e^(3*b*x + 3*a) + (3*b - d)*e^(b*x + a)),
x) + 1/2*(5*b^2*e^c + 6*b*d*e^c + d^2*e^c + (3*b^2*e^c - 4*b*d*e^c + d^2*e^c)*e^(4*b*x + 4*a) - 2*(6*b^2*e^c +
b*d*e^c - d^2*e^c)*e^(2*b*x + 2*a))*e^(d*x)/((3*b^3 - b^2*d - 3*b*d^2 + d^3)*e^(3*b*x + 3*a) - (3*b^3 - b^2*d
- 3*b*d^2 + d^3)*e^(b*x + a))
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\cosh \left (b x + a\right )^{2} \operatorname{csch}\left (b x + a\right ) e^{\left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="fricas")
[Out]
integral(cosh(b*x + a)^2*csch(b*x + a)*e^(d*x + c), x)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(d*x+c)*cosh(b*x+a)**2*csch(b*x+a),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh \left (b x + a\right )^{2} \operatorname{csch}\left (b x + a\right ) e^{\left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="giac")
[Out]
integrate(cosh(b*x + a)^2*csch(b*x + a)*e^(d*x + c), x) | 3,305 | 6,873 | {"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} | 3.53125 | 4 | CC-MAIN-2023-40 | latest | en | 0.128272 |
https://howkgtolbs.com/convert/86.38-kg-to-lbs | 1,670,202,712,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711001.28/warc/CC-MAIN-20221205000525-20221205030525-00402.warc.gz | 336,069,555 | 12,122 | # 86.38 kg to lbs - 86.38 kilograms to pounds
Do you want to learn how much is 86.38 kg equal to lbs and how to convert 86.38 kg to lbs? Here you go. This whole article is dedicated to kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to point out that all this article is dedicated to only one number of kilograms - exactly one kilogram. So if you need to know more about 86.38 kg to pound conversion - read on.
Before we move on to the more practical part - this is 86.38 kg how much lbs calculation - we want to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 86.38 kg to lbs? 86.38 kilograms it is equal 190.4353019156 pounds, so 86.38 kg is equal 190.4353019156 lbs.
## 86.38 kgs in pounds
We are going to begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in short form SI).
Sometimes the kilogram could be written as kilogramme. The symbol of this unit is kg.
The kilogram was defined first time in 1795. The kilogram was defined as the mass of one liter of water. This definition was simply but impractical to use.
Later, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was prepared of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was replaced by another definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It can be also divided into 100 decagrams and 1000 grams.
## 86.38 kilogram to pounds
You learned some information about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. We want to underline that there are not only one kind of pound. What does it mean? For example, there are also pound-force. In this article we are going to to centre only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. To be honest, this unit is in use also in other systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound can be divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 86.38 kg?
86.38 kilogram is equal to 190.4353019156 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 86.38 kg in lbs
The most theoretical part is already behind us. In this section we are going to tell you how much is 86.38 kg to lbs. Now you learned that 86.38 kg = x lbs. So it is time to know the answer. Just look:
86.38 kilogram = 190.4353019156 pounds.
It is an accurate result of how much 86.38 kg to pound. It is possible to also round it off. After it your result will be as following: 86.38 kg = 190.036 lbs.
You learned 86.38 kg is how many lbs, so look how many kg 86.38 lbs: 86.38 pound = 0.45359237 kilograms.
Of course, this time you may also round off this result. After rounding off your outcome is as following: 86.38 lb = 0.45 kgs.
We are also going to show you 86.38 kg to how many pounds and 86.38 pound how many kg outcomes in tables. Look:
We will begin with a chart for how much is 86.38 kg equal to pound.
### 86.38 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
86.38 190.4353019156 190.0360
Now look at a table for how many kilograms 86.38 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
86.38 0.45359237 0.45
Now you learned how many 86.38 kg to lbs and how many kilograms 86.38 pound, so it is time to go to the 86.38 kg to lbs formula.
### 86.38 kg to pounds
To convert 86.38 kg to us lbs a formula is needed. We will show you two versions of a formula. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 190.4353019156 outcome in pounds
The first formula give you the most correct result. In some situations even the smallest difference could be significant. So if you want to get an exact result - this version of a formula will be the best for you/option to calculate how many pounds are equivalent to 86.38 kilogram.
So let’s move on to the another formula, which also enables calculations to learn how much 86.38 kilogram in pounds.
The shorter version of a formula is down below, look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, this formula is simpler. It can be the best option if you want to make a conversion of 86.38 kilogram to pounds in fast way, for instance, during shopping. You only need to remember that your outcome will be not so exact.
Now we are going to learn you how to use these two formulas in practice. But before we will make a conversion of 86.38 kg to lbs we want to show you another way to know 86.38 kg to how many lbs without any effort.
### 86.38 kg to lbs converter
An easier way to know what is 86.38 kilogram equal to in pounds is to use 86.38 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Converter is based on longer version of a formula which we gave you in the previous part of this article. Due to 86.38 kg pound calculator you can easily convert 86.38 kg to lbs. You only have to enter number of kilograms which you want to convert and click ‘calculate’ button. You will get the result in a second.
So try to convert 86.38 kg into lbs with use of 86.38 kg vs pound converter. We entered 86.38 as a number of kilograms. It is the result: 86.38 kilogram = 190.4353019156 pounds.
As you can see, this 86.38 kg vs lbs converter is intuitive.
Now we can go to our primary topic - how to convert 86.38 kilograms to pounds on your own.
#### 86.38 kg to lbs conversion
We are going to start 86.38 kilogram equals to how many pounds calculation with the first version of a formula to get the most correct result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 190.4353019156 the result in pounds
So what have you do to know how many pounds equal to 86.38 kilogram? Just multiply number of kilograms, in this case 86.38, by 2.20462262. It gives 190.4353019156. So 86.38 kilogram is exactly 190.4353019156.
You can also round off this result, for instance, to two decimal places. It is exactly 2.20. So 86.38 kilogram = 190.0360 pounds.
It is time for an example from everyday life. Let’s convert 86.38 kg gold in pounds. So 86.38 kg equal to how many lbs? As in the previous example - multiply 86.38 by 2.20462262. It is exactly 190.4353019156. So equivalent of 86.38 kilograms to pounds, when it comes to gold, is exactly 190.4353019156.
In this example it is also possible to round off the result. Here is the outcome after rounding off, this time to one decimal place - 86.38 kilogram 190.036 pounds.
Now we are going to examples calculated with a short version of a formula.
#### How many 86.38 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 190.036 the outcome in pounds
So 86.38 kg equal to how much lbs? And again, you have to multiply amount of kilogram, this time 86.38, by 2.2. Let’s see: 86.38 * 2.2 = 190.036. So 86.38 kilogram is 2.2 pounds.
Do another conversion using this formula. Now convert something from everyday life, for instance, 86.38 kg to lbs weight of strawberries.
So let’s convert - 86.38 kilogram of strawberries * 2.2 = 190.036 pounds of strawberries. So 86.38 kg to pound mass is 190.036.
If you know how much is 86.38 kilogram weight in pounds and are able to convert it with use of two different formulas, let’s move on. Now we want to show you these results in tables.
#### Convert 86.38 kilogram to pounds
We are aware that results shown in charts are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in charts for your convenience. Due to this you can easily make a comparison 86.38 kg equivalent to lbs outcomes.
Let’s start with a 86.38 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
86.38 190.4353019156 190.0360
And now let’s see 86.38 kg equal pound table for the second formula:
Kilograms Pounds
86.38 190.036
As you can see, after rounding off, if it comes to how much 86.38 kilogram equals pounds, the outcomes are not different. The bigger amount the more significant difference. Remember it when you need to do bigger number than 86.38 kilograms pounds conversion.
#### How many kilograms 86.38 pound
Now you know how to convert 86.38 kilograms how much pounds but we will show you something more. Do you want to know what it is? What about 86.38 kilogram to pounds and ounces conversion?
We are going to show you how you can convert it little by little. Start. How much is 86.38 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, in this case 86.38, by 2.20462262. So 86.38 * 2.20462262 = 190.4353019156. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To know how much 86.38 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So final result is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then your result is 2 pounds and 33 ounces.
As you can see, calculation 86.38 kilogram in pounds and ounces simply.
The last conversion which we want to show you is conversion of 86.38 foot pounds to kilograms meters. Both of them are units of work.
To calculate it it is needed another formula. Before we give you it, look:
• 86.38 kilograms meters = 7.23301385 foot pounds,
• 86.38 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to calculate 86.38 foot pounds to kilograms meters you have to multiply 86.38 by 0.13825495. It is equal 0.13825495. So 86.38 foot pounds is exactly 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 86.38 foot pounds is 0.14 kilogram meters.
We hope that this calculation was as easy as 86.38 kilogram into pounds calculations.
This article was a huge compendium about kilogram, pound and 86.38 kg to lbs in conversion. Due to this calculation you learned 86.38 kilogram is equivalent to how many pounds.
We showed you not only how to do a conversion 86.38 kilogram to metric pounds but also two other calculations - to know how many 86.38 kg in pounds and ounces and how many 86.38 foot pounds to kilograms meters.
We showed you also another solution to do 86.38 kilogram how many pounds calculations, it is with use of 86.38 kg en pound calculator. This is the best choice for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you are able to do 86.38 kilogram equal to how many pounds conversion - on your own or using our 86.38 kgs to pounds converter.
It is time to make your move! Let’s convert 86.38 kilogram mass to pounds in the way you like.
Do you want to do other than 86.38 kilogram as pounds conversion? For instance, for 5 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so simply as for 86.38 kilogram equal many pounds.
### How much is 86.38 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 86.38 kg in pounds , we prepared one more section. Here you can see all you need to remember about how much is 86.38 kg equal to lbs and how to convert 86.38 kg to lbs . It is down below.
What is the kilogram to pound conversion? The conversion kg to lb is just multiplying 2 numbers. How does 86.38 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 86.38 kilogram to pounds? The exact result is 190.4353019156 lb.
There is also another way to calculate how much 86.38 kilogram is equal to pounds with second, easier version of the formula. Have a look.
The number of kilograms * 2.2 = the result in pounds
So this time, 86.38 kg equal to how much lbs ? The answer is 190.4353019156 pounds.
How to convert 86.38 kg to lbs in an easier way? You can also use the 86.38 kg to lbs converter , which will make all calculations for you and you will get a correct answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,640 | 13,897 | {"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} | 4.125 | 4 | CC-MAIN-2022-49 | latest | en | 0.942587 |
https://mathexamination.com/class/measure-of-non-compactness.php | 1,620,356,759,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.96/warc/CC-MAIN-20210507025943-20210507055943-00215.warc.gz | 363,656,640 | 7,046 | ## Take My Measure Of Non-Compactness Class
A "Measure Of Non-Compactness Class" QE" is a standard mathematical term for a generalized continuous expression which is utilized to solve differential equations and has services which are routine. In differential Class fixing, a Measure Of Non-Compactness function, or "quad" is utilized.
The Measure Of Non-Compactness Class in Class type can be expressed as: Q( x) = -kx2, where Q( x) are the Measure Of Non-Compactness Class and it is a crucial term. The q part of the Class is the Measure Of Non-Compactness constant, whereas the x part is the Measure Of Non-Compactness function.
There are 4 Measure Of Non-Compactness functions with proper option: K4, K7, K3, and L4. We will now look at these Measure Of Non-Compactness functions and how they are resolved.
K4 - The K part of a Measure Of Non-Compactness Class is the Measure Of Non-Compactness function. This Measure Of Non-Compactness function can likewise be written in partial portions such as: (x2 - y2)/( x+ y). To fix for K4 we increase it by the appropriate Measure Of Non-Compactness function: k( x) = x2, y2, or x-y.
K7 - The K7 Measure Of Non-Compactness Class has an option of the type: x4y2 - y4x3 = 0. The Measure Of Non-Compactness function is then increased by x to get: x2 + y2 = 0. We then need to increase the Measure Of Non-Compactness function with k to get: k( x) = x2 and y2.
K3 - The Measure Of Non-Compactness function Class is K3 + K2 = 0. We then increase by k for K3.
K3( t) - The Measure Of Non-Compactness function equationis K3( t) + K2( t). We increase by k for K3( t). Now we increase by the Measure Of Non-Compactness function which gives: K2( t) = K( t) times k.
The Measure Of Non-Compactness function is likewise referred to as "K4" because of the initials of the letters K and 4. K implies Measure Of Non-Compactness, and the word "quad" is pronounced as "kah-rab".
The Measure Of Non-Compactness Class is one of the main techniques of resolving differential formulas. In the Measure Of Non-Compactness function Class, the Measure Of Non-Compactness function is first increased by the appropriate Measure Of Non-Compactness function, which will give the Measure Of Non-Compactness function.
The Measure Of Non-Compactness function is then divided by the Measure Of Non-Compactness function which will divide the Measure Of Non-Compactness function into a real part and a fictional part. This offers the Measure Of Non-Compactness term.
Finally, the Measure Of Non-Compactness term will be divided by the numerator and the denominator to get the quotient. We are entrusted the right hand side and the term "q".
The Measure Of Non-Compactness Class is a crucial concept to comprehend when solving a differential Class. The Measure Of Non-Compactness function is simply one technique to fix a Measure Of Non-Compactness Class. The approaches for fixing Measure Of Non-Compactness equations include: particular worth decomposition, factorization, ideal algorithm, mathematical service or the Measure Of Non-Compactness function approximation.
## Pay Me To Do Your Measure Of Non-Compactness Class
If you want to end up being knowledgeable about the Quartic Class, then you require to first start by browsing the online Quartic page. This page will show you how to use the Class by utilizing your keyboard. The explanation will also reveal you how to create your own algebra formulas to help you study for your classes.
Before you can comprehend how to study for a Measure Of Non-Compactness Class, you should initially comprehend using your keyboard. You will discover how to click on the function keys on your keyboard, as well as how to type the letters. There are 3 rows of function keys on your keyboard. Each row has four functions: Alt, F1, F2, and F3.
By pushing Alt and F2, you can multiply and divide the worth by another number, such as the number 6. By pressing Alt and F3, you can utilize the 3rd power.
When you press Alt and F3, you will enter the number you are trying to multiply and divide. To multiply a number by itself, you will push Alt and X, where X is the number you want to multiply. When you press Alt and F3, you will type in the number you are trying to divide.
This works the exact same with the number 6, except you will just key in the two digits that are 6 apart. Lastly, when you push Alt and F3, you will use the 4th power. Nevertheless, when you push Alt and F4, you will use the actual power that you have found to be the most proper for your problem.
By using the Alt and F function keys, you can multiply, divide, and then use the formula for the third power. If you need to increase an odd number of x's, then you will require to get in an even number.
This is not the case if you are attempting to do something complex, such as increasing 2 even numbers. For instance, if you want to multiply an odd variety of x's, then you will need to enter odd numbers. This is especially real if you are attempting to figure out the response of a Measure Of Non-Compactness Class.
If you wish to transform an odd number into an even number, then you will require to press Alt and F4. If you do not know how to increase by numbers by themselves, then you will need to use the letters x, a b, c, and d.
While you can increase and divide by use of the numbers, they are much easier to use when you can look at the power tables for the numbers. You will have to do some research study when you initially begin to utilize the numbers, but after a while, it will be second nature. After you have produced your own algebra equations, you will be able to create your own reproduction tables.
The Measure Of Non-Compactness Solution is not the only method to fix Measure Of Non-Compactness formulas. It is necessary to learn about trigonometry, which utilizes the Pythagorean theorem, and after that use Measure Of Non-Compactness solutions to solve problems. With this approach, you can learn about angles and how to fix problems without having to take another algebra class.
It is very important to attempt and type as rapidly as possible, since typing will assist you understand about the speed you are typing. This will assist you compose your answers much faster.
## Pay Someone To Take My Measure Of Non-Compactness Class
A Measure Of Non-Compactness Class is a generalization of a linear Class. For instance, when you plug in x=a+b for a given Class, you get the value of x. When you plug in x=a for the Class y=c, you obtain the values of x and y, which provide you an outcome of c. By applying this fundamental principle to all the formulas that we have attempted, we can now solve Measure Of Non-Compactness equations for all the worths of x, and we can do it rapidly and effectively.
There are numerous online resources offered that supply free or economical Measure Of Non-Compactness equations to resolve for all the values of x, consisting of the cost of time for you to be able to make the most of their Measure Of Non-Compactness Class project help service. These resources normally do not need a subscription fee or any type of investment.
The responses offered are the result of complex-variable Measure Of Non-Compactness formulas that have actually been solved. This is also the case when the variable used is an unidentified number.
The Measure Of Non-Compactness Class is a term that is an extension of a direct Class. One benefit of using Measure Of Non-Compactness formulas is that they are more basic than the direct formulas. They are simpler to solve for all the worths of x.
When the variable used in the Measure Of Non-Compactness Class is of the kind x=a+b, it is easier to solve the Measure Of Non-Compactness Class since there are no unknowns. As a result, there are fewer points on the line defined by x and a continuous variable.
For a right-angle triangle whose base points to the right and whose hypotenuse points to the left, the right-angle tangent and curve chart will form a Measure Of Non-Compactness Class. This Class has one unknown that can be discovered with the Measure Of Non-Compactness formula. For a Measure Of Non-Compactness Class, the point on the line specified by the x variable and a continuous term are called the axis.
The existence of such an axis is called the vertex. Since the axis, vertex, and tangent, in a Measure Of Non-Compactness Class, are a given, we can discover all the worths of x and they will sum to the provided worths. This is achieved when we use the Measure Of Non-Compactness formula.
The aspect of being a consistent element is called the system of equations in Measure Of Non-Compactness equations. This is sometimes called the main Class.
Measure Of Non-Compactness equations can be solved for other values of x. One way to resolve Measure Of Non-Compactness equations for other values of x is to divide the x variable into its factor part.
If the variable is given as a favorable number, it can be divided into its factor parts to get the regular part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a constant. In such a case, the formula is a third-order Measure Of Non-Compactness Class.
If the variable x is negative, it can be divided into the exact same part of the x variable to get the part of the x variable that is multiplied by the denominator. In such a case, the formula is a second-order Measure Of Non-Compactness Class.
Option help service in fixing Measure Of Non-Compactness formulas. When using an online service for resolving Measure Of Non-Compactness formulas, the Class will be fixed immediately. | 2,174 | 9,646 | {"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-2021-21 | latest | en | 0.903421 |
https://cheapieshoe.com/how-many-cards-are-in-a-6-deck-shoe | 1,652,772,465,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00354.warc.gz | 230,695,967 | 15,179 | # How many cards are in a 6 deck shoe?
2
Date created: Sun, Jul 11, 2021 1:23 PM
Date updated: Mon, May 16, 2022 2:59 PM
Content
Video answer: #11 đź”´ so many aces! blackjack ~ 6 deck shoe at red rock casino ~ las vegas july 2021 episode 4
## Top best answers to the question «How many cards are in a 6 deck shoe»
• I attempt to work this out in my blackjack appendix 8 but I’ll work through it more slowly here. We’ll ignore dealer blackjacks to keep things simple and assume the player always hits after two cards. The number of ways to arrange 3 cards in a 6-deck shoe is combin (312,3)=5,013,320. There are 24 sevens in the shoe.
A six-deck shoe has 6 x 16 = 96 10-values and 6 x 52 = 312 total cards. The chance of an initial 10 is therefore 96 out of 312 or 30.769 percent. Considering only the 10 to be gone, the shoe then would have 6 x 4 = 24 aces in 311 remaining cards.
FAQ
Those who are looking for an answer to the question «How many cards are in a 6 deck shoe?» often ask the following questions:
### đź‘ Can you count cards in a multi deck shoe?
#### How do you count cards in a deck of cards?
• Step 1. Assign a value to every card. Step 2. Keep a “Running Count” based off of the values of the card dealt. Step 3. Use this information to calculate the count per deck or “true count”. Step 4. Change your bets as the true count rises.
### đź‘ How many hands in a six deck shoe?
• The average number of cards per hand is 4.94. Assuming 15 burn cards, a six-deck baccarat shoe would have about 60 hands. However, mathematically speaking, it doesn't make any difference when they shuffle.
### đź‘ How many decks are in a 6 deck shoe?
• Casinos use a 6-deck shoe, so take the number of decks you see in the discard tray, and subtract it from 6. That’s how many decks are remaining in the shoe. Divide the running count by the number of decks remaining.
Video answer: Deck estimation in card counting: what it is and how to do it
We've handpicked 28 related questions for you, similar to «How many cards are in a 6 deck shoe?» so you can surely find the answer!
How did the multi deck shoe get its name?
• While no such ruling was ever passed, most Nevada casinos now deal from a multi-deck shoe. As gaming advisor to the Havana Hilton, Scarne also introduced the shoe to Puerto Rico and Cuba. The device is so named because the earliest versions of it resembled a woman's high-heel shoe, and were often painted red or black in color.
Shoe size 13 what is a good deck width?
• Full-Size Deck [width: 7.5-inches or larger, length: 28-inches]:These are perfect for skaters that are over the age of 13, with height taller than 5’3” and wear a shoe size of 9 or more. However, the specific deck size is determined by the style preference of the skater: 7.5-inches to 8-inches: street skating or for more technical tricks
What's the running count on a 6 deck shoe?
• If you have a +6 after the first round of a game in a 6 deck shoe, the running count is +1 since there are still about 6 decks left. If you have a +6 when you’re halfway through that shoe, then you have 3 decks left and a +2 running count. There are a couple of ways to figure out the number of decks remaining in the shoe.
What's the running count on a six deck shoe?
• If you were say, one deck into that six deck shoe, a running count of +10 would give you the same proportional high-to-low ratio of remaining cards as a +4 running count would when you're four decks into it. That's because in both cases, there would be two extra high cards available for each deck that is left.
What shoe stores offer credit cards?
• Amazon.
• Nordstrom.
• Belk.
• Kohl's.
• Macy's.
• Bloomingdale's.
• JCPenney.
• Walmart.
### Video answer: How to shuffle a shoe deck
What are the odds of a six deck shoe in baccarat?
• A six deck shoe will have the following baccarat odds: 45.87% for the banker, 44.63% for player, and 9.51% for a tie.
What shoe stores accept nike gift cards?
#### Where can I use my Nike gift card?
• Nike and Converse gift cards are just like cash and can be used to make purchases on Nike.com and at any Nike- and Converse-owned retail stores.
### Video answer: đź”´ blackjack with so many doubles ~ 6 deck shoe ~ red rock casino las vegas july 2021~ episode 5
Can you use credit cards for shoe horns?
• Credit cards and business cards are the one item everyone will have on them at all times, or, at least, whenever they are also wearing shoes. They have just about the perfect size and they are slick enough to be perfect replacements for shoe horns. There are 2 problems that still need to be taken into consideration.
What do the cards in the shoe mean?
• When the cards are placed in the shoe, the dealer will insert a brightly colored blank plastic card. When this card is drawn it indicates that the current game is the last one before a new shuffle. This helps mitigate player advantage via card counting, as a significant portion (usually about 25 percent) of the full inventory...
What kind of cards are in a shoe?
• Dealing shoes come in many colors and sizes, depending on the number of decks they are capable of holding (2, 4, 6, or 8 decks). When the cards are placed in the shoe, the dealer will insert a brightly colored blank plastic card.
### Video answer: Card counting at 6 deck 21 with john stathis
How many shoe brushes?
• Every shoe specialist knows the importance of having high-quality shoe brushes. Generally, experts recommend you use three different brushes for the best results. One shoe brush to remove the dust and grime from your footwear.
11.5 shoe how many inches?
#### Men's Size Conversions
InchesCMUS Sizes
11.125"28.311.5
11.25"28.612
11.5625"29.413
11.875"30.214
How many customizable shoe brands?
#### What are the top shoe brands in the world?
• With its roots in Germany, adidas has become one of the top shoe brands in the world. The company produces more than 900 million sports and lifestyle products with independent manufacturing partners around the world.
How many inches inside shoe?
#### Does shoe size equal inches?
• Men's shoe size charts start from a US number of 6 (a European number of 39 and a UK number of 5.5). This size corresponds to an inch measurement of 9.25 inches (23.5 centimeters). If the measurement is provided solely in centimeters, multiply the length of your foot by 2.54 and apply to the chart. The biggest number on the men's chart is ...
How many inches shoe size?
• Size 6 = 9.31 inches Size 6.5 = 9.5 inches Size 7 = 9.69 inches Size 7.5 = 9.81 inches Size 8 = 10 inches Size 8.5 = 10.19 inches Size 9 = 10.31 inches Size 9.5 = 10.5 inches Size 10 = 10.69 inches Size 10.5 = 10.81 inches Size 11 = 11 inches Size 11.5 = 11.19 inches Size 12 = 11.31 inches Size 12.5 = 11.5 inches Size 13 = 11.69 inches Size 13.5 = 11.81 inches Size 14 = 12 inches Size 14.5 = 12.19 inches Size 15 = 12.31 inches
How many man shoe and?
#### What's the average shoe size for a man?
• For men, it’s pretty easy to determine, as there are many studies analyzing the average size of a male foot. In 2014, the American Academy of Orthopaedic Surgeons reported that the worldwide average shoe size for men is between 9 and 12, while the American average male shoe size is a 10.5. A mother measures her son’s shoe size.
How many marathons per shoe?
#### How many people go to a marathon each year?
• Major marathons also attract many spectators and viewers. In 2019, people from almost 3.3 million households are projected to attended a marathon event in the United States. This text provides general information.
How many shoe carnival stores?
#### What kind of store is Shoe Carnival?
• Shoe Carnival Stores Shoe Carnival is a chain of family shoe stores operating across the continental United States, as well as in Puerto Rico and online at shoecarnival.com. We have a passion for creating a fun, engaging, and affordable shoe shopping experience, bringing great deals on brand-name footwear to millions of families.
### Video answer: Card deck mistakes to watch out for
How many syllables in shoe?
• How many syllables are in shoe? 1 syllable. Divide shoe into syllables: shoe. Definition of: Shoe (New window will open) How to pronounce shoe: Words: shoddily, shoddy, shoe, shoebox, shoed.
How many cards do you need to deal shoes?
• Dealing shoes come in many colors and sizes, depending on the number of decks they are capable of holding (2, 4, 6, or 8 decks).
What size skateboard deck should i get?
• Mini Skateboard Deck width: 7.0″ with an average deck length of 28” Mini decks are the best skateboards for beginners who are 6-8 years old, between 3’5” and 4’4” tall who wear size 4-6 shoes. Mid-size Skateboard Deck width: 7.3”+ with an average deck length of 29” For skaters 9 to 12 years old between 4’5”...
What happens when you remove cards from shoe in blackjack?
• In fact, if you remove any significant number of small cards from the shoe, blackjack not only becomes easier, the player gets a mathematical edge over the house. This is where card counting comes in. Counters track the proportion of small cards to face cards and aces. At a certain point, the deck turns “positive.”.
Hemnes shoe cabinet how many shoes?
• How many shoes can I fit inside the Hemnes unit? The Hemnes shoe unit can hold eight pairs of shoes. However, if you are like me and have small feet, I can fit 10.
How many big baller shoe sold?
In an interview with Colin Cowherd in May 2017, Ball said that if the big shoe companies like Nike, Adidas, or Under Armour want to make a deal with his Big Baller Brand, the asking price is \$3 billion. He also said that Baller has sold between 400 and 520 ZO2 shoes. | 2,547 | 9,740 | {"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} | 4.34375 | 4 | CC-MAIN-2022-21 | latest | en | 0.939091 |
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# Find the range of values of x that satisfy the inequality (x+1)(x-2)
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Find the range of values of x that satisfy the inequality (x+1)(x-2) [#permalink]
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Updated on: 23 Dec 2018, 02:44
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Wavy Line Method Application - Exercise Question #4
Find the range of values of x that satisfy the inequality $$\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0$$
A. x < -3
B. -1 < x < 2
C. x < -3 or -1 < x < 2 or x > 5
D. -1 < x < 2 or x > 5
E. x < -3 or -1 < x < 2
Wavy Line Method Application has been explained in detail in the following post:: http://gmatclub.com/forum/wavy-line-method-application-complex-algebraic-inequalities-224319.html
Detailed solution will be posted soon.
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Originally posted by EgmatQuantExpert on 26 Aug 2016, 01:40.
Last edited by Bunuel on 23 Dec 2018, 02:44, edited 3 times in total.
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Re: Find the range of values of x that satisfy the inequality (x+1)(x-2) [#permalink]
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26 Aug 2016, 09:02
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EgmatQuantExpert wrote:
Wavy Line Method Application - Exercise Question #4
Find the range of values of x that satisfy the inequality $$\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0$$
Wavy Line Method Application has been explained in detail in the following post:: http://gmatclub.com/forum/wavy-line-method-application-complex-algebraic-inequalities-224319.html
Detailed solution will be posted soon.
Equation is $$\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0$$
We have the values of x as -3,-1,2 and 5.
Since x cannot be equal to 5 or -3 we need to exclude those values(they are at the denominator and we cannot have 0 at the denominator)
So, substituting these values of the number line, we will the range of x as
(-infinity,-3) U (-1,2) U (5,infinity)
Please correct me if I am missing anything.
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Re: Find the range of values of x that satisfy the inequality (x+1)(x-2) [#permalink]
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Updated on: 07 Aug 2018, 20:29
Solution:
Hey Everyone,
Please find below the solution of the given problem,
Plotting the zero points and drawing the wavy line:
Required Range:
x < -3 or -1 < x < 2 or x > 5
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Originally posted by EgmatQuantExpert on 18 Nov 2016, 02:16.
Last edited by EgmatQuantExpert on 07 Aug 2018, 20:29, edited 1 time in total.
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Re: Find the range of values of x that satisfy the inequality (x+1)(x-2) [#permalink]
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23 Oct 2018, 06:46
I was wondering, for a problem such as this one where you have (x+1)(x-2) on numerator and (x-5)(x+3) on denominator... how would you approach it? Clearly the "wavy" method works... but I wonder how would you tackle this with the method you explain in your series "Inequalities with complications".
Am I allowed to multiply by 1 since it's positive? For instance I multiply (x-5)/(x-5) and (x+3)/(x+3) on left side to get (x+1)(x-2)(x-5)(x+3) on numerator and then (x-5)^2 and (x+3)^2 on denominator! This means that I can ignore the denominator since it will be always positive, and hence I end with a structure similar to the ones you tackled on your post (x+1)(x-2)(x-5)(x+3) > 0.
Would that work? Thank you.
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Re: Find the range of values of x that satisfy the inequality (x+1)(x-2) [#permalink]
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24 Oct 2018, 04:13
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gmat800live wrote:
I was wondering, for a problem such as this one where you have (x+1)(x-2) on numerator and (x-5)(x+3) on denominator... how would you approach it? Clearly the "wavy" method works... but I wonder how would you tackle this with the method you explain in your series "Inequalities with complications".
Am I allowed to multiply by 1 since it's positive? For instance I multiply (x-5)/(x-5) and (x+3)/(x+3) on left side to get (x+1)(x-2)(x-5)(x+3) on numerator and then (x-5)^2 and (x+3)^2 on denominator! This means that I can ignore the denominator since it will be always positive, and hence I end with a structure similar to the ones you tackled on your post (x+1)(x-2)(x-5)(x+3) > 0.
Would that work? Thank you.
You don't need to make it so complicated though what you suggest works too. The reason why factors in the denominator work is simply this:
When is abcd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc
When is ab/cd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc
Aren't the two cases the same? (Except that c and d cannot be 0 in the second case) Anyway, since the expression needs to be positive, a, b, c and d cannot be 0 in first case either.
Does it matter whether c and d are in numerator or denominator? No.
The signs of a, b, c and d decide the sign of the expression irrespective of whether they are in numerator or denominator. The case is the same here.
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Re: Find the range of values of x that satisfy the inequality (x+1)(x-2) [#permalink]
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24 Oct 2018, 06:39
We have the values of x as -3,-1,2 and 5.
So, substituting these values of the number line, we will the range of x as
Re: Find the range of values of x that satisfy the inequality (x+1)(x-2) [#permalink] 24 Oct 2018, 06:39
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## Maharashtra SSC Maths Algebra (Paper 2) Exam Question Paper 2015 with Solutions – Free Download
MSBSHSE (Maharashtra board) SSC Class 10 maths 2015 Algebra Question Paper with solutions are available at BYJU’S, which can be accessed and downloaded in a pdf format. Algebra plays an important role in everyday life, when dealing with different situations. For example, calculating interest, buying and selling goods for a certain price and so on. In all these situations we generally use algebraic calculations, knowing or unknowingly. Hence, students should focus more on these concepts. Maharashtra board Class 10 Maths Algebra chapters cover all such topics, which are necessary for higher education as well as real-life situations. Students can also download MSBSHSE SSC Class 10 Maths previous year question papers to practise for the board exam.
### MSBSHSE Class 10 Mathematics Question Paper 2015 Algebra Paper with Solutions
PART – A
Q. 1. Attempt any five of the following sub-questions: [5]
(i) State whether the following sequence is an A.P. or not?
1, 4, 7, 10,….
Solution:
Given,
1, 4, 7, 10,….
First term = 1
Second term – First term = 4 – 1 = 3
Third term – Second term = 7 – 4 = 3
The common difference is the same throughout the sequence.
Hence, the given sequence is an Arithmetic progression.
(ii) A card is drawn from the pack of 25 cards labeled with numbers 1 to 25. Write the sample space for this random experiment.
Solution:
Given,
25 cards labeled with numbers from 1 to 25.
Sample space = S = {1, 2, 3, 4, 5, ….., 25}
n(S) = 25
(iii) Find the value of x + y, if
12x + 13y = 29 and
13x + 12y = 21
Solution:
Given,
12x + 13y = 29….(i)
13x + 12y = 21….(ii)
12x + 13y + 13x + 12y = 29 + 21
25x + 25y = 50
25(x + y) = 50
x + y = 50/25
x + y = 2
(iv) For a sequence, if Sn = n/(n + 1) then find the value of S10.
Solution:
Given,
Sn = n/(n + 1)
S10 = 10/ (10 + 1)
= 10/11
(v) Verify whether 1 is the root of the quadratic equation:
x2 + 3x – 4 = 0.
Solution:
If α is the root of the quadratic equation f(x) = 0, then f(α) = 0.
Given,
x2 + 3x – 4 = 0
Substituting x = 1,
LHS = (1)2 + 3(1) – 4
= 1 + 3 – 4
= 0
= RHS
Hence, 1 is the root of the given quadratic equation.
(vi) If x + y = 5 and x = 3, then find the value of y.
Solution:
Given,
x + y = 5….(i)
x = 3
Substituting x = 3 in (i),
3 + y = 5
y = 5 – 3
y = 2
Q.2. Attempt any four of the following sub-questions: [8]
(i) Solve the following quadratic equation by factorization method
x2 – 7x + 12 = 0.
Solution:
Given,
x2 – 7x + 12 = 0
Using the factorization method: splitting the middle term
x2 – 3x – 4x + 12 = 0
x(x – 3) – 4(x – 3) = 0
(x – 4)(x – 3) = 0
x – 4 = 0, x – 3 = 0
x = 4, x = 3
(ii) Find the term t10 of an A.P.:
4, 9, 14,….
Solution:
Given AP:
4, 9, 14,….
First term = a = 4
Common difference = d = 9 – 4 = 5
nth term of an AP,
tn = a + (n – 1)d
t10 = 4 + (10 – 1)5
= 4 + 9(5)
= 4 + 45
= 49
Therefore, t10 = 49.
(iii) If point A(2, 3) lies on the graph of the equation 5x + ay = 19, then find a.
Solution:
Given,
A(2, 3) lies on the graph of the equation 5x + ay = 19.
That means, the point satisfies the given equation.
Substituting x = 2 and y = 3 in the given linear equation,
5(2) + a(3) = 19
10 + 3a = 19
3a = 19 – 10
3a = 9
a = 9/3
a = 3
(iv) A die is thrown. If A is an event of getting an odd number, then write the sample space and event A in set notation.
Solution:
Given,
A die is thrown.
Sample space = S = {1, 2, 3, 4, 5, 6}
n(S) = 6
A = The event of getting an odd number
A = {1, 3, 5}
n(A) = 3
(v) For a certain frequency distribution, the value of Mean is 101 and Median is 100. Find the value of Mode.
Solution:
Given,
Mean = 101
Median = 100
We know that,
Mean – Mode = 3(Mean – Median)
101 – Mode = 3(101 – 100)
101 – Mode = 3(1)
101 – 3 = Mode
⇒ Mode = 98
(vi) If one root of the quadratic equation kx2 – 7x + 5 = 0 is 1, then find the value of k.
Solution:
Given,
kx2 – 7x + 5 = 0
One root of the given quadratic equation = 1
Substituting x = 1 in the given equation,
k(1)2 – 7(1) + 5 = 0
k – 7 + 5 = 0
k – 2 = 0
k = 2
Q.3. Attempt any three of the following sub-questions: [9]
(i) The area under different crops in a certain village is given below. Represent it with a pie diagram:
Crop Area in Hectares Jowar 40 Wheat 60 Sugarcane 50 Vegetables 30
Solution:
Crop Area in Hectares Measure of central angle Jowar 40 (40/180) × 360° = 80° Wheat 60 (60/180) × 360° = 120° Sugarcane 50 (50/180) × 360° = 100° Vegetables 30 (30/180) × 360° = 60° Total 180 360°
Pie chart:
(ii) If two coins are tossed, then find the probability of the event that at the most one tail turns up.
Solution:
Given,
Two coins are tossed.
Sample space = S = {HH, HT, TH, TT}
n(S) = 4
Let A be the event of getting at the most one tail.
A = {HH, HT, TH}
n(A) = 3
P(A) = n(A)/n(S)
= 3/4
Hence, the required probability is 3/4.
(iii) Solve the following simultaneous equations using the graphical method:
x + y = 7;
x – y = 5.
Solution:
Given,
x + y = 7
x – y = 5
Consider the first equation:
x + y = 7
y = 7 – x
x 0 6 7 y 7 1 0
Now, consider another equation:
x – y = 5
y = x – 5
x 0 5 6 y -5 0 1
Graph:
The lines intersecting with each other at (6, 1).
Hence, the solution of the given pair of linear equations is x = 6 and y = 1.
(iv) There is an auditorium with 35 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row, and so on. Find the number of seats in the twenty-second row.
Solution:
Given,
An auditorium has 35 rows of seats.
Number of seats in the first row = 20
Number of seats in the second row = 22
Number of seats in the third row = 24
i.e. 20, 22, 24,…..
This is an AP with a = 20 and d = 2
n = 35
nth term of an AP:
tn = a + (n – 1)d
t22 = 20 + (22 – 1)2
= 20 + 21(2)
= 20 + 42
= 62
Hence, there are 62 seats in the twenty-second row.
(v) Solve the following quadratic equation by completing the square method:
x2 + 11x + 24 = 0
Solution:
Given,
x2 + 11x + 24 = 0
x2 + 11x = -24….(i)
Comparing with ax2 + bx + c = 0,
a = 1, b = 11, c = 24
b2/4a = (11)2/ 4(1)
= 121/4
Adding 121/4 on both sides of (i),
x2 + 11x + (121/4) = -24 + (121/4)
x2 + 11x + (11/2)2 = (-96 + 121)/4
(x + 11/2)2 = 25/4
x + 11/2 = ±5/2
x = -(11/2) ± (5/2)
x = (-11 ± 5)/2
x = (-11 + 5)/2, x = (-11 – 5)/2
x = -6/2, x = -16/2
x = -3, x = -8
Q.4. Attempt any two of the following sub-questions: [8]
(i) Two-digit numbers are formed using the digits 0, 1, 2, 3, 4, 5 where digits are not repeated.
P is the event that the number so formed is even.
Q is the event that the number so formed is greater than 50.
R is the event that the number so formed is divisible by 3
Then write the sample space S and events P, Q, R using set notation.
Solution:
Two-digit numbers are formed using the digits 0, 1, 2, 3, 4, 5 without repeating the digits are:
S = {10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43, 45, 50, 51, 52, 53, 54}
n(S) = 25
P = The event the number so formed is even
P = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54}
n(P) = 13
Q = The event that the number so formed is greater than 50
Q = {51, 52, 53, 54}
n(Q) = 4
R = The event that number so formed is divisible by 3
R = {12, 15, 21, 24, 30, 42, 45, 51, 54}
n(R) = 9
(ii) The following table shows ages of 300 patients getting medical treatment in a hospital on a particular day:
Age (in years) No. of Patients 10 – 20 60 20 – 30 42 30 – 40 55 40 – 50 70 50 – 60 53 60 – 70 20
Find the median age of the patient.
Solution:
Age (in years) No. of Patients (Frequency) Cumulative frequency 10 – 20 60 60 20 – 30 42 102 30 – 40 55 157 40 – 50 70 227 50 – 60 53 280 60 – 70 20 300
Sum of frequencies = N = 300
N/2 = 300/2 = 150
Cumulative frequency greater than and nearest to 150 is 157 which lies in the interval 30 – 40.
Median class = 30 – 40
The lower limit of the median class = l = 30
Frequency of the median class = f = 55
Cumulative frequency of the class preceding the median class = cf = 102
Class height = h = 10
Median = l + {[(N/2) – cf]/ f} × h
= 30 + [(150 – 102)/ 55] × 10
= 30 + (480/55)
= 30 + 8.73
= 38.73
(iii) If α + β = 5 and α3 + β3 = 35, find the quadratic equation whose roots are α and β
Solution:
Given,
α + β = 5
α3 + β3 = 35
α3 + β3 = (α + β)3 – 3αβ(α + β)
35 = (5)3 – 3αβ(5)
35 = 125 – 15αβ
15αβ = 125 – 35
15αβ = 90
αβ = 90/15
αβ = 6
Hence, the required quadratic equation is x2 – (α + β)x + αβ = 0
x2 – 5x + 6 = 0
Q.5. Attempt any two of the following sub-questions: [10]
(i) Babubhai borrows Rs. 4,000 and agrees to repay with a total interest of Rs. 500 in 10 installments, each installment being less than the preceding installment by Rs. 10. What should be the first and the last installment?
Solution:
Given,
Babubhai borrows Rs. 4,000 and agrees to repay with a total interest of Rs. 500 in 10 installments.
Total amount = Rs. 400 + Rs. 500
Rs. 4500
Also, given that each installment reduces by Rs. 10 than previous installments.
That mean, this is an AP with d = -10
n = 10
Let a be the first installment.
S10 = 4500 (given)
n/2 [2a + (10 – 1)d] = 4500
(10/2) [2a + 9(-10)] = 4500
2a – 90 = 4500/5
2a = 900 + 90
2a = 990
a = 990/2
a = 495
t10 = a + (10 – 1)d
= 495 + 9(-10)
= 495 – 90
= 405
Hence, the first installment is Rs. 495 and the last installment is Rs. 405.
(ii) On the first day of the sale of tickets for a drama, all 35 tickets were sold. If the rates of the tickets were Rs. 20 and Rs. 40 per ticket and the total collection was Rs. 900. Find the number of tickets sold at each rate.
Solution:
Let x be the number of tickets sold at Rs. 20 each and y be the number of tickets sold at Rs. 40 each.
According to the given,
x + y = 35….(i)
20x + 40y = 900
20(x + 2y) = 900
x + 2y = 45….(ii)
Subtracting (i) from (ii),
x + 2y – (x + y) = 45 – 35
y = 10
Substituting y = 10 in (i),
x + 10 = 35
x = 35 – 10 = 25
Therefore, 25 tickets were sold at Rs. 20 each and 10 tickets were sold at Rs. 40 each.
(iii) Given below is the frequency distribution of driving speeds (in km/hour) of the vehicles of 400 college students:
Speed (in km/hr) No. of Students 20 – 30 6 30 – 40 80 40 – 50 156 50 – 60 98 60 – 70 60
Draw Histogram and hence the frequency polygon for the above data.
Solution:
Speed (in km/hr) No. of Students Class mark 20 – 30 6 25 30 – 40 80 35 40 – 50 156 45 50 – 60 98 55 60 – 70 60 65
Scale:
X-axis: 1 cm = 10 km/hr
Y-axis: 1 cm = 20 students | 3,928 | 10,725 | {"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} | 4.5625 | 5 | CC-MAIN-2022-21 | longest | en | 0.875596 |
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# 6.1: Investigating Static Forces in Nature: The Mystery of the Gecko
Difficulty Level: At Grade Created by: CK-12
Student Learning Objectives:
• Explain that a net force of zero or greater is necessary for objects to adhere to a surface (wall or ceiling)
• Identify different variables and the constants that affect adhesive forces
• Explain how the amount of adhesion changes when the conditions of the surfaces change
Note: Some questions in the Student Journal are underlined as formative assessment checkpoints for you to check students’ understanding of lesson objectives.
At a Glance for Teachers:
• Review what students know about forces
• Teacher demonstration on balanced forces
• Determine the amount of force needed for objects of varying masses to adhere to a ceiling and maintain a net force of zero
• Activity: Tape Pull—Measure the amount of force required to remove a piece of transparent tape with varying amounts of dirt
Estimated Time: 80Minutes\begin{align*}80 \;\mathrm{Minutes}\end{align*}
Vocabulary: Adhere, Adhesive, Balanced Forces, Dependent Variable, Force, Independent Variable, Mass, Net Force, Newton, Unbalanced Force, Volume
Refer to the end of this Teacher Guide for definitions.
Materials:
• PowerPoint for Lesson 6
• Student Journals for Lesson 6
• Computer with LCD or overhead projector
• Duct tape
• 50N\begin{align*}50\;\mathrm{N}\end{align*} spring scale
• Transparent tape
• Hole punch
• Ruler, protractor
Safety Note
Have students wear safety goggles in accordance with district safety policy.
Slide #
Student Journal Page #
Teacher Background Information and Pedagogy
“Teacher Script”
Slide 1
Title
Student Journal Page: 6–1
1. Review with students that a force is a push or pull. See definitions in Appendix B.
“What is the meaning of the word force in science?”
2. Demonstrate balanced forces by partially filling a small jar with water. Place an index card underneath the rim of the jar and invert the jar while holding the index card. Next, release your hand from the card while carefully holding the jar. Students should note that the card remains in contact with the rim of the jar. Have students identify the balanced forces at work in this demonstration. Answer: air pressure is the dominate force, it is equal to the weight of the water plus the force of gravity. Other forces that are present but less dominate are gravity and capillary wet adhesion (if the card got wet when it came in contact with the rim of the glass).
“Is this demonstrating a balanced or unbalanced force? Why?”
“What would happen if this was an unbalanced force?”
3. Have students look around the room and identify pairs of objects that are at rest and represent balanced forces and record these in the box on the left side of the student journal. Students should identify the forces acting on each object and that the net force is zero. Have students draw one of the examples of balanced forces and indicate the amount of each force acting on the object using arrows in Student Journal page 6–1. Have students repeat this exercise for unbalanced forces in the box on the right side of the student journal. Note to teacher: if the object sits on a table, there is the upward normal force of the table on the object. Research has shown that students often don’t recognize this as a force, they just indicate the table is in the way.]
4. Provide examples of objects in motion such as objects speeding up, slowing down, or at constant speed.
A field test teacher used a Frayer model for balanced and unbalanced forces for this lesson (refer to Lesson 1 for directions on the use of the Frayer model).
Slide 2
Student Journal Page: 6-1
5. Display Slide 2
“In this image, there are two forces at work: one that is holding the shoe onto the ceiling and another that is pulling the shoe towards the floor. In order for the shoe to remain on the ceiling, what must be true about these two forces?”
Students should state that these represent balanced forces (i.e., the net force is zero), or that the force holding a shoe is greater than the force of gravity. An example of the latter would be if the shoe contained a magnet that was attracted to a steel ceiling. This force could be greater than gravity, and then there would be an additional normal force acting in the direction of gravity to counteract the excess magnetic force.
Slide 3
Student Journal Page: 6–2
Calculations In Appendix A
“Imagine an ant, like the one on this slide, walking on the ceiling. Draw a picture representing the forces of the ant on the ceiling in your journal. Determine the force required for each ant foot (divide total force by six).”
Explain the following assumptions that are important for this problem:
“We are assuming in this problem that the total force required is equally divided among the six ant feet, and that ONLY the contact between feet and ceiling gives rise to the force.”
The weight of the ant is provided in Newtons (N), a derived unit which is the force needed to increase the speed of (or accelerate) one kilogram of mass one meter per second every second.
A field test teacher passed around objects (e.g., a one Newton weight, an eight Newton cell phone) for students to be able to relate to this unit of measure.
For this module, there is no need to calculate force with Newton’s Second Law of Motion. However, there may be a need to explain how an object’s weight can be expressed in Newtons. Explain that in the metric system forces are measured in units of Newtons (using the symbol “N”). Provide students with the definition found in Appendix B along with the following illustration. Use these along with the direct vocabulary instruction strategy as described in the preface. Weight is action of the force of gravity on an object. A standard kilogram mass would therefore have a weight of 9.8 Newtons on Earth since the acceleration due to gravity is 9.8 m/s/s.
6. Point out to students that the weight is the minimum amount of force that must be provided by the feet of the ant on the ceiling in order for there to be balanced forces and thus have the ant adhere to the ceiling. See Appendix A for the answers.
Note: During the pilot test, students thought this activity was interesting. The calculations took a bit to understand, and it was valuable to review unit conversions. Use Appendix A to assist students in solving the first problem.
Slide 4
Student Journal Page: 6–3
7. Display slide 4.
“Repeat the calculation—this time for an imaginary object that is larger in every dimension and whose mass and volume is ten times larger.”
Determine how many “ant feet” it would take for this imaginary object to remain adhered to the ceiling. Compare and discuss the difference between the two calculations in class. See Appendix A for calculations.
Teacher Demonstration:
Optional: One pilot teacher added a calculation for a two-ton elephant as well. Actual weight for an African male elephant in Newtons is 122,580 Newtons. Refer to optional notes in Slide 5.
Slide 5
Student Journal Pages: 6–3 6–4
“Let’s return our attention to the gecko.
Repeat your calculations from the imaginary animal for the Tokay Gecko, which has an average weight of 2.2 Newtons.”
8. Have students write a statement and/or draw pictures that describe the relationship between size (mass) and weight and, therefore, the adhesive forces required for an animal to remain on a ceiling.
Slide 6
9. Explain to students that they will be using the following terms in this lesson.
“Adhere describes how something sticks to something else.
Separation force is the amount of pull that is required to detach two objects.”
Slide 7
“What are the tools that we can use in the laboratory to measure the amount of force that an object exerts? What are the units used when measuring with this tool?”
Forces can be measured with a spring scale that changes when a force is applied. Forces are measured in Newtons (N).
Slide 8
Student Journal Page: 6-4
“As you have observed a gecko adhering to a wall, you may have wondered about the types of surfaces that are required to accomplish this feat.
Can the gecko adhere to any surface?
Does the surface need to be clean or can the gecko adhere to dirty surfaces too?
What if the surface is wet? Will that affect how well the gecko can adhere?
To better understand how the gecko can adhere to different surfaces, we will be exploring the forces involved in the adhesion of transparent tape on a table top.”
10. Tell students that over the next day or so, they will be able to refine this question based on how they set up their experiment.
11. Prior to showing slide 9 introduce the Tape Pull activity by having the students answer the question in their journal on page 6–4.
Slide 9
Student Journal Pages: 6–5 6–6
“You will be working with transparent tape on the tabletop and measuring the force required to remove the tape with different amounts of dirt. This force, as stated previously, is actually GREATER THAN the adhesive force.”
12. Before beginning the experiment have students work with the materials and practice the tape pull procedure as described on Student Journal page 6–5.
“Write down the independent variable (manipulated variable) and the dependent variable (responding variable).” Allow students to identify the amount of dirt as the independent variable and the force that it takes to remove or break the adhesion as the dependent variable.
Optional: Use the “sticky hands” toy (the one that initially sticks to glass then slowly falls/rolls down the glass) as a demonstration of dirt’s effect. This toy’s ability to stick decreases rapidly when it becomes dirty.
13. Hold a discussion about how to vary the “amount of dirt.” For starters, students could test fresh (never before used) tape. Then, rather than adding dirt to the tape, students could make a finger print on the tape and test its adhesion. Other ideas: drop chalk dust onto tape and blow it off, touch the tape to the floor, etc. This then becomes the operational definition for the independent variable.
14. Hold a class discussion about how to keep certain variables constant, such as the amount of surface area in which there is contact between surfaces and the angle of pull. Based on this discussion, students should write a research question and a hypothesis before completing the activity. An example of a research question is given on this slide.
Slide 10
Student Journal Pages: 6–6 6–7 6–8 6–9
“On this slide, you see how the materials are set up for the experiment. Image 6.8 shows a piece of tape on a table. The end of the tape that is pulled is reinforced with some electrical tape that has a hole punched through it. The hook end of the spring scale is then placed through the hole. Image 6.9 shows the spring scale being pulled at an angle (make sure this is the same each time). During the pull, a second student should carefully observe the force readings on the spring scale.”
15. Allow students time to complete the activity as shown in the journal. As students are completing the procedure, make sure they refine their initial question and use their findings in order to provide explanations and further questions.
Student Journal pages 6–5 through 6–8 can be completed for homework and graded for use as a formative assessment.
Classroom Management Tip: One pilot teacher assigned jobs for the experiment:
• Tape handler and assembly
• Measurer
• Equipment Manager
16. After students are done with the experiment, have them answer the questions in their journal on page 6–9 and 6–10.
Question 7:
Describe how you made your observations in today’s lesson.
a. “What tools did you use?” (spring scale)
b. “Were your observations at the visible or invisible scale?” (invisible)
c. “What is the dominant force at this scale?” (adhesive force/unknown)
Slide 11
“What do you know about the effectiveness of transparent tape underwater, and how tape gets dirty over time? (Display slide 11) This is a quote from researcher Kellar Autumn, Assistant Professor of biology at Lewis & Clark College, about the self-cleaning ability of the gecko.”
Students may state that when transparent tape is placed underwater, it will eventually lose its adhesiveness. Likewise, transparent tape does not work well on dirty surfaces.
17. It should be noted that ants leave a residue behind as they walk, whereas geckos do not.
18. Draw students’ attention to the note on the slide about the gecko adhesion working underwater.
Optional: Students could test other variables: amount of tape contact area, cleanliness of the surface, etc.
Slide 12
19. As a culminating class discussion, ask students to respond to the questions in “Making Connections.”
“Let’s review.
1. Describe one or two ideas that you learned during this lesson.
2. What factors contribute to the amount of force to remove a sticky substance?
3. How does dirt affect adhesion?
4. Do you think that a sticky substance is a possible method for the gecko adhesion?
5. How do you think the gecko sticks to the ceiling?
6. What should we explore next?”
Slide 13
20. The pilot-test teachers highly recommend using this flow chart at the end and/or beginning of each lesson. The end of each lesson contains this flow chart that provides an opportunity to show students the “big picture” and where they are in the lesson sequence. The following color code is used:
Yellow: Past Lessons
Blue: Current Lesson
Green: Next Lesson
White: Future Lesson
Appendix A: Calculations and Possible Responses to Accompany PowerPoint Slides
Slide 3 Calculations
Ant
Weight of Ant=0.00004Newtons\begin{align*}\mathrm{Weight\ of\ Ant} = 0.00004 \;\mathrm{Newtons}\end{align*} or 4×105Newtons\begin{align*}4 \times 10^{-5} \;\mathrm{Newtons}\end{align*}
Weight of Ant/6Ant Feet=Force for each foot=0.0000067Newtons\begin{align*}\mathrm{Weight\ of\ Ant}/6 \;\mathrm{Ant\ Feet} = \mathrm{Force\ for\ each\ foot} = 0.0000067 \;\mathrm{Newtons}\end{align*} per Ant Foot or 6.7×106Newtons\begin{align*}6.7 \times 10^{-6}\;\mathrm{Newtons}\end{align*} per Ant Foot
Slide 4 Calculations
Ant Mass Times 10times\begin{align*}10 \;\mathrm{times}\end{align*} what it was before
Then IF the Ant Foot can ONLY support 6.7×106N\begin{align*}6.7 \times 10^{-6}\;\mathrm{N}\end{align*}, how many ant feet would be required?
Weight of Imaginary object / Force for each Ant Foot
4×104Newtons/6.7×106Newtons per Ant Foot\begin{align*}4 \times 10^{-4} \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
59.7ant feet=60ant feet\begin{align*}59.7 \;\mathrm{ant\ feet} = 60 \;\mathrm{ant\ feet}\end{align*}
Slide 5 Calculations
Gecko
Weight of Gecko=2.2Newtons\begin{align*}\mathrm{Weight\ of\ Gecko} = 2.2 \;\mathrm{Newtons}\end{align*}
2.2Newtons/6.7×106Newtons per Ant Foot\begin{align*}2.2 \;\mathrm{Newtons}/6.7 \times 10^{-6}\;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
328,358ant feet\begin{align*}328,358 \;\mathrm{ant\ feet}\end{align*}
From Liang, Autumn, Hsieh, Zesch, Chan, Fearing, Full, Kenny5\begin{align*}^5\end{align*}:
43.4N\begin{align*}43.4 \;\mathrm{N}\end{align*} average sustained clinging force of gecko with 227.1mm2\begin{align*}227.1 \;\mathrm{mm}^2\end{align*} pad area
5\begin{align*}^5\end{align*} Autumn, K., Liang, Y. A., Hsieh, S. T., Zesch, W., Chan, W. P., Kenny, T. W., Fearing, R., & Full, R. J. (2000). Adhesive force of a single gecko foot-hair.
Nature, 405, 681-684.
Slide 5 Calculations (Optional)
200lb\begin{align*}200 \;\mathrm{lb}\end{align*} adult
Weight of adult in Newtons=888.9Newtons\begin{align*}\mathrm{Weight\ of\ adult\ in\ Newtons} = 888.9 \;\mathrm{Newtons}\end{align*}
888.9Newtons/6.7×106Newtons per Ant Foot\begin{align*}888.9 \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
32,671,642ant feet\begin{align*}32,671,642 \;\mathrm{ant\ feet}\end{align*}
27,000lb\begin{align*}27,000 \;\mathrm{lb}\end{align*} African male elephant
Weight of elephant in Newtons=122,580Newtons\begin{align*}\mathrm{Weight\ of\ elephant\ in\ Newtons} = 122,580 \;\mathrm{Newtons}\end{align*}
122,580Newtons/6.7×106Newtons per Ant Foot\begin{align*}122,580 \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
18,295,522,390ant feet\begin{align*}18,295,522,390 \;\mathrm{ant\ feet}\end{align*}
Appendix B: NanoLeap Physical Science Vocabulary
Adhere
1. To hold fast or to stick
2. To bind to
Adhesive
A substance that helps objects stick together
Balanced Forces
For each force acting on a body, there is another force on the same body equal in magnitude and opposite in direction. A body is said to be at rest if it is being acted on by balanced forces.
Dependent Variable
A factor or condition that might be affected as a result of a change in the independent variable (also called a responding variable)
Force
1. Energy exerted
2. A push or a pull that acts on an object
Independent Variable
A factor or condition that is intentionally changed by an investigator or experiment to explore its effects on other factors (also called a manipulated variable)
Mass
1. A quantity of matter
2. A measurement of the quantity
Net Force
The resultant non-zero force due to an unbalanced force
Newton
A unit of force needed to change the speed of a kilogram of mass by one meter per second for every second that the force is acting on the mass
Unbalanced
When there is an individual force that is not being balanced by a force of equal magnitude and in the opposite direction. A body is said to be in motion if acted upon by unbalanced forces.
Volume
The amount of space occupied by a three-dimensional object (Length times Width times Height for a rectangular object)
Investigating Static Forces in Nature: The Mystery of the Gecko
Lesson 6: How MUCH Force Is Needed to Make an Object Stick?
Teacher Guide
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1 5.6 Independent Events YOU WILL NEED calculator EXPLORE The Fortin family has two children. Cam determines the probability that the family has two girls. Rushanna determines the probability that the family has two girls, given that the first child is a girl. How are these probabilities similar, and how are they different? GOAL Understand and solve problems that involve independent events. INVESTIGATE the Math Anne and Abby each have 9 marbles: red and 8 blue. Anne places 7 red marbles and 3 blue marbles in bag. She places the rest of her marbles in bag 2. Abby places all of her marbles in bag 3. Anne then draws one marble from bag and one marble from bag 2. Abby draws two marbles from bag 3, without replacement.? Are both girls equally likely to draw two red marbles? A. Let A represent Anne drawing a red marble from bag. Let B represent Anne drawing a red marble from bag 2. Are A and B independent or dependent events? Explain. B. Let C represent Abby drawing a red marble from bag 3 on her first draw. Let D represent Abby drawing a red marble from bag 3 on her second draw. Are C and D independent or dependent events? Explain. C. Determine PA2 and PB2. D. Determine PA d B2 to three decimal places. E. Determine PC 2 and PD 0 C 2, the probability that Abby will draw a red marble from bag 3 on the second draw, given that the first marble she drew was red. F. Determine PC d D2 to three decimal places. G. Are both girls equally likely to draw two red marbles? Explain. Reflecting H. Explain how you can tell if two events are independent or dependent. I. Suppose that Abby were to draw both of her marbles together, instead of drawing one after the other. Would this affect the probability of Abby drawing two red marbles? Justify your answer. J. Naveen used the following formula to determine the probability of Anne drawing two red marbles: PA d B2 5 PA2 # PB 0 A2 Would this formula give the correct answer? Explain. 354 Chapter 5 Probability NEL
2 APPLY the Math example Determining probabilities of independent events Mokhtar and Chantelle are playing a die and coin game. Each turn consists of rolling a regular die and tossing a coin. Points are awarded for rolling a 6 on the die and/or tossing heads with the coin: point for either outcome 3 points for both outcomes 0 points for neither outcome Players alternate turns. The first player who gets 0 points wins. a) Determine the probability that Mokhtar will get, 3, or 0 points on his first turn. b) Verify your results for part a). Explain what you did. Chantelle s Solution a) Let S represent rolling a 6 on the die. Let H represent tossing heads with the coin. PS PSr PH PHr PSr2 5 5 PHr I created a tree diagram showing the probabilities associated with the events of rolling a six on a die, S, and tossing heads with a coin, H, along with the probabilities of their complements, Srand Hr. Rolling a Die Tossing a Coin Probabilities I defined events S and H. These events are independent. The probability of tossing heads with the coin is not affected by the probability of rolling 6 on the die. I determined the probability that Mokhtar will not get either outcome when rolling the die and tossing the coin. These events are complementary. P(H) 2 heads P(S H) 6. 2 or P(S) 6 six P(H ) 2 tails P(S H ) 6. 2 or I determined the probability of each pair of events occurring on the tree diagram by multiplying. P(S ) 5 6 not six P(H) 2 heads 5 P(S H). or P(H ) 5 tails P(S H ). or NEL 5.6 Independent Events 355
3 P(scores 3 points): PS d H 2 5 P(scores 0 points): PSr d Hr2 5 5 P(scores point): PSr d H 2 c PS d Hr2 5 5 PSr d H 2 c PS d Hr2 5 6 or 2 He has a chance of scoring 3 points, a 2 chance of scoring point, and a 5 chance of scoring 0 points. b) Verify: Total probability Total probability 5 or Mokhtar gets 3 points when he rolls a 6 and tosses heads. He gets 0 points when he does not roll a 6 and does not toss heads. He gets point when he either rolls a 6 or tosses heads I verified my solution by adding the three probabilities. The sum is, so I knew that my answer is correct. Your Turn a) Are you more likely to get points or not get points on each turn? Explain. b) Would the probabilities determined for Mokhtar s first turn change for his next turn? Explain. example 2 Solving a problem that involves independent events using a graphic organizer All 000 tickets for a charity raffle have been sold and placed in a drum. There will be two draws. The first draw will be for the grand prize, and the second draw will be for the consolation prize. After each draw, the winning ticket will be returned to the drum so that it might be drawn again. Max has bought five tickets. Determine the probability, to a tenth of a percent, that he will win at least one prize. Max s Solution: Using complements Let X represent winning the grand prize. Let Y represent winning the consolation prize. Let Z represent winning at least one prize. I defined events X, Y, and Z. 356 Chapter 5 Probability NEL
4 PX or 200 PY or 200 The events are independent. Since the winning ticket will be replaced after the first draw, the probability of winning either prize is the same. In other words, winning the consolation prize is not affected by winning the grand prize. X \ Y X Y Y \ X X Y I drew a Venn diagram to represent the situation. The event of winning at least one prize, Z, is represented by being in any area of the two ovals in the Venn diagram. P(Z ), the probability of winning at least one prize, is the complement of winning no prizes: PZ PX r d Y r2 PXr or PYr or PX r d Y r2 5 PX r2 # PY r2 PXr d Yr # PXr d Yr PZ PXr d Yr PZ PZ PZ or The probability I will win at least one prize is about.0%. I had to determine the probability of winning the grand prize or the consolation prize. The only other possibility is winning no prizes. I determined the probability of not winning the grand prize, the probability of not winning the consolation prize, and the probability of winning neither prize. I determined the probability of winning at least one prize. I rounded to the nearest tenth of a percent. NEL 5.6 Independent Events 357
5 Melissa s Solution: Using a tree diagram Let X represent winning the grand prize. Let Y represent winning the consolation prize. Let Z represent winning at least one prize. PX or PY or PX r2 5 2 PX 2 PY r2 5 2 PY 2 PX r PY r PX r PY r I defined events X, Y, and Z. I determined the probability of winning the grand prize and the probability of winning the consolation prize. I also determined the complement of these events: the probability of not winning the grand prize and the probability of not winning the consolation prize. P(X) P(X ) Grand Prize P(Y) P(Y ) P(Y) Consolation Prize P(X Y) (0.005)(0.005) P(X Y ) (0.005)(0.995) P(X Y) (0.995)(0.005) I drew a tree diagram to list all the possible outcomes. The blue branches represent winning a prize, and the red branches represent not winning a prize. I determined the probability of each pair of events by multiplying the individual probabilities. P(Y ) P(X Y ) (0.995)(0.995) Based on my tree diagram, Max could win two prizes in one way. P(win 2 prizes) 5 PX d Y 2 P(win 2 prizes) Max could win one prize in two ways. P(win prize) 5 PX d Y r2 or PX r d Y 2 P(win prize) 5 PX d Y r2 c PX r d Y 2 P(win prize) P(win prize) These probabilities are equal and can be added together, since the two events (only winning the grand prize or only winning the consolation prize) are mutually exclusive. 358 Chapter 5 Probability NEL
6 Winning at least one prize means that Max would win exactly one prize or both prizes. P(Z ) 5 P(win prize) or P(win 2 prizes) P(Z ) 5 P(win prize) c P(win 2 prizes) P(Z ) 5 P(win prize) P(win 2 prizes) PZ PZ The probability Max will win at least one prize is.0%. Verify: Total probability 5 P(win at least one) P(win none) Total probability Total probability 5 These probabilities are different and can be added together, since the two events (winning one prize and winning two prizes) are mutually exclusive. I rounded to the nearest tenth of a percent. I verified my calculations by adding the probabilities. The sum is, so I knew that my answer is correct. Your Turn Suppose that the rules for the raffle are changed, so the first ticket drawn is not returned to the drum before the second draw. a) Are the events winning the grand prize and winning the consolation prize dependent or independent? b) Do you think Max s probability of winning at least one prize will be greater or less than before? Justify your answer. In Summary Key Ideas If the probability of event B does not depend on the probability of event A occurring, then these events are called independent events. For example, tossing tails with a coin and drawing the ace of spades from a standard deck of 52 playing cards are independent events. The probability that two independent events, A and B, will both occur is the product of their individual probabilities: P(A d B) 5 P(A) # P(B) Need to Know A tree diagram is often useful for modelling problems that involve independent events. Drawing an item and then drawing another item, after replacing the first item, results in a pair of independent events. NEL 5.6 Independent Events 359
7 CHECK Your Understanding. For each situation, classify the events as either independent or dependent. Justify your classification. a) A four-colour spinner is spun, and a die is rolled. The first event is spinning red, and the second event is rolling a 2. b) A red die and a green die are rolled. The first event is rolling a on the red die, and the second event is rolling a 5 on the green die. c) Two cards are drawn, without being replaced, from a standard deck of 52 playing cards. The first event is drawing a king, and the second event is drawing an ace. d) There are 30 cards, numbered to 30, in a box. Two cards are drawn, one at a time, with replacement. The first event is drawing a prime number, and the second event is drawing a number that is a multiple of Celeste goes to the gym five days a week. Each day, she does a cardio workout using either a treadmill, an elliptical walker, or a stationary bike. She follows this with a strength workout using either free weights or the weight machines. Celeste randomly chooses which cardio workout and which strength workout to do each day. a) Are choosing a cardio workout and choosing a strength workout dependent or independent events? Explain. b) Determine the probability that Celeste will use a stationary bike and free weights the next day she goes to the gym. 3. Ian also goes to the gym five days a week, but he does two different cardio workouts each day. His choices include using a treadmill, a stepper, or an elliptical walker, and running the track. a) Are the two cardio workouts that Ian chooses dependent or independent events? b) Determine the probability that the next time Ian goes to the gym he will use the elliptical walker and then run the track. PRACTISING 4. For each situation described in question, determine the probability that both events will occur. 5. a) Suppose that P(A) , P(B) 5 0.4, and P(A d B) Are A and B independent events? Explain. b) Suppose that P(Q) , P(R) , and P(Q d R) Are Q and R independent events? Explain. 360 Chapter 5 Probability NEL
8 6. There are two children in the Angel family. a) Draw a tree diagram that shows all the possible gender combinations for the two children. b) Determine the probability that both children are boys. c) Determine the probability that one child is a boy and the other child is a girl. 7. A particular game uses 40 cards from a standard deck of 52 playing cards: the ace to the 0 from the four suits. One card is dealt to each of two players. Determine the probability that the first card dealt is a club and the second card dealt is a heart. Are these events independent or dependent? 8. A standard die is rolled twice. Determine the probability for the following: a) The first roll is a, and the second roll is a 6. b) The first roll is greater than 3, and the second roll is even. c) The first roll is greater than, and the second roll is less than Jeremiah is going on a cruise up the Nile. According to the travel brochure, the probability that he will see a camel is 4 5, and the probability that he will see an ibis is 3 4. Determine the probability that Jeremiah will see the following: a) A camel and an ibis b) Neither a camel nor an ibis c) Only one of these sights 0. a) Design a spinner so that when you toss a coin and spin the spinner, the probability of getting heads and spinning a 6 is. b) Repeat part a) with a probability of 20.. Recall Anne and Abby, from the beginning of this lesson. They each have 9 marbles: red and 8 blue. Anne places 7 red marbles and 3 blue marbles in bag. She places the rest of her marbles in bag 2. Abby places all of her marbles in bag 3. Anne then draws one marble from bag and one marble from bag 2. Abby draws two marbles from bag 3. a) Are Anne and Abby equally likely to draw two blue marbles from their bags? Explain. b) Determine the probability Anne and Abby will both draw one red marble and one blue marble. Explain what you did. c) Suppose that Anne now has 5 red marbles and 5 blue marbles in each of her two bags, while Abby has 0 red marbles and 0 blue marbles in her one bag. Will Abby still be more likely to draw two red marbles? Explain. NEL 5.6 Independent Events 36
9 . A paper bag contains a mixture of three types of treats: 0 granola bars, 7 fruit bars, and 3 cheese strips. Suppose that you play a game in which a treat is randomly taken from the bag and replaced, and then a second treat is drawn from the bag. You are allowed to keep the second treat only if it was the same type as the treat that was drawn the first time. Determine the probability of each of the following: a) You will be able to keep a granola bar. b) You will be able to keep any treat. c) You will not be able to keep any treat. 3. Tiegan s school is holding a chocolate bar sale. For every case of chocolate bars sold, the seller receives a ticket for a prize draw. Tiegan has sold five cases, so she has five tickets for the draw. At the time of the draw, 00 tickets have been entered. There are two prizes, and the ticket that is drawn for the first prize is returned so it can be drawn for the second prize. a) Determine the probability that Tiegan will win both prizes. b) Determine the probability that she will win no prizes. 4. a) Create a problem that involves determining the probability of two independent events. Give your problem to a classmate to solve. b) Modify the problem you created in part a) so that it now involves two dependent events. Give your problem to a classmate to solve. 5. Two single-digit random numbers (0 to 9 inclusive) are selected independently. Determine the probability that their sum is 0. Closing 6. a) Explain why the formula you would use to calculate P(A d B) would depend on whether A and B are dependent or independent events. b) Give an example of how you would calculate P(A d B) if A and B were independent events. c) Give an example of how you would calculate P(A d B) if A and B were dependent events. Extending 7. A particular machine has 00 parts. Over a year, the probability that each part of the machine will fail is %. If any part fails, the machine will stop. a) Determine the probability that the machine will operate continuously for year. b) Suppose that the probability of each part failing within one year dropped to 0.5%. Determine the probability that the machine would operate continuously for year. c) Suppose that the probability of the machine operating continuously for year must be 90%. What would the probability of not failing need to be for each part? 362 Chapter 5 Probability NEL
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Name: WORKSHEET : Date: Answer the following questions. 1) Probability of event E occurring is... P(E) = Number of ways to get E/Total number of outcomes possible in S, the sample space....if. 2) The probability,
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Multiple Choice Math 3201 Unit 3: Probability Name: 1. Given the following probabilities, which event is most likely to occur? A. P(A) = 0.2 B. P(B) = C. P(C) = 0.3 D. P(D) = 2. Three events, A, B, and | 10,301 | 42,633 | {"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} | 4.46875 | 4 | CC-MAIN-2021-25 | longest | en | 0.934883 |
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# CAT Data Interpretation Important Questions PDF [With Video Solutions]
Data Interpretation is one of the most important topics in the CAT LRDI Section. If you’re new to these questions, you can check out these CAT Data Interpretation Questions from the CAT previous year papers. In this article, we will look into some very important Data Interpretation questions PDF(with solutions) for CAT. You can also download these CAT Data Interpretation questions with detailed solutions, which also include important tricks to solve these questions.
Instructions
The different bars in the diagram above provide information about different orders in various categories (Art, Binders, ….) that were booked in the first two weeks of September of a store for one client. The colour and pattern of a bar denotes the ship mode (First Class / Second Class / Standard Class). The left end point of a bar indicates the booking day of the order, while the right end point indicates the dispatch day of the order. The difference between the dispatch day and the booking day (measured in terms of the number of days) is called the processing time of the order. For the same category, an order is considered for booking only after the previous order of the same category is dispatched. No two consecutive orders of the same category had identical ship mode during this period.
For example, there were only two orders in the furnishing category during this period. The first one was shipped in the Second Class. It was booked on Sep 1 and dispatched on Sep 5. The second order was shipped in the Standard class. It was booked on Sep 5 (although the order might have been placed before that) and dispatched on Sep 12. So the processing times were 4 and 7 days respectively for these orders.
Question 1:Â How many days between Sep 1 and Sep 14 (both inclusive) had no booking from this client considering all the above categories?
Solution:
Accumulating all the data :
We get the following table :
Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched .
Now No booking days from the table are : September 8,9,10,11,12 and 14.
So a total of 6 days .
Question 2:Â What was the average processing time of all orders in the categories which had only one type of ship mode?
Solution:
Accumulating all the data :
We get the following table :
Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched .
Now Envelopes and Accessories has only 1 ship mode i.e Standard class .
So therefore processing days for envelopes = 7-3 =4
and processing days for accessories = 19-1 =18
Therefore average =Â $\frac{\left(18+4\right)}{2}=11$
Question 3:Â The sequence of categories — Art, Binders, Paper and Phones — in decreasing order of average processing time of their orders in this period is:
a)Â Art, Binders, Paper, Phones
b)Â Phones, Art, Binders, Paper
c)Â Phones, Binders, Art, Paper
d)Â Paper, Binders, Art, Phones
Solution:
Accumulating all the data :
We get the following table :
Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched .
Now taking average processing time per order for the above mentioned categories we get :
Art =$\frac{2+8+2+1+7}{5}\ \ =4$
Binders =$\frac{1+1+11+2}{4}\ \ =3.75$
Papers =Â $\frac{3+2+5}{3}\ \ =3.33$
Phones =$\frac{2+12+1}{3}\ \ =5$
So in decreasing order we get Phones , Art ,Binder , Paper.
Question 4:Â Approximately what percentage of orders had a processing time of one day during the period Sep 1 to Sep 22 (both dates inclusive)?
a)Â 22%
b)Â 16%
c)Â 20%
d)Â 25%
Solution:
Accumulating all the data :
We get the following table :
Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched .
Now from the table we observe that the total number of orders are 35 and 7 orders have a processing time of 1 unit
The 7 orders are : Arts Standard class,Binders First class and standard class, Phones First class, Bookcases second class ( 2 orders)Â and Chairs standard class.
So the percentage = $\frac{7}{35}\times\ 100\ =\ 20$
Instructions
The figure above shows the schedule of four employees – Abani, Bahni, Danni, and Tinni – whom Dhoni supervised in 2020. Altogether there were five projects which started and concluded in 2020 in which they were involved. For each of these projects and for each employee, the starting day was at the beginning of a month and the concluding day was the end of a month, and these are indicated by the left and right end points of the corresponding horizontal bars. The number within each bar indicates the percentage of assigned work completed by the employee for that project, as assessed by Dhoni.
For each employee, his/her total project-month (in 2020) is the sum of the number of months (s)he worked across the five projects, while his/her annual completion index is the weightage average of the completion percentage assigned from the different projects, with the weights being the corresponding number of months (s)he worked in these projects. For each project, the total employee-month is the sum of the number of months four employees worked in this project, while its completion index is the weightage average of the completion percentage assigned for the employees who worked in this project, with the weights being the corresponding number of months they worked in this project.
Question 5:Â Which of the following statements is/are true?
I: The total project-month was the same for the four employees.
II: The total employee-month was the same for the five projects.
a)Â Only II
b)Â Both I and II
c)Â Neither I nor II
d)Â Only I
Solution:
The total project month is the number of months Abani, Bahni, Danni, and Tinni individually worked for all the projects combined :
Abani – 2+2+5 = 9 months
Bahni – 2+4+3 = 9 months
Danni – 3+3+2+1 = 9 months
Tinni – 2+2+3+2 = 9 months.
The total employee month for all the five projects is the sum of the total employee-month is the sum of the number of months four employees worked in this project.
Project -1 = 2+2+2 = 6 months
Project -2 = 3+2 = 5 months
Project – 3 = 2+4+3 = 9 months.
Project – 4 = 5+2+3 = 10 months.
Project – 5 = 3+1+2 = 6 months.
Only statement 1 is true.
Question 6:Â Which employees did not work in multiple projects for any of the months in 2020?
a)Â Only Abani, Bahni and Danni
b)Â Only Abani and Bahni
c)Â All four of them
d)Â Only Tinni
Solution:
Abani, Banni, and Danni did not work on multiple projects simultaneously in a month
Tinni was the only person who worked on multiple projects which are project 4 and project 5 in the month of september.
Question 7:Â The project duration, measured in terms of the number of months, is the time during which at least one employee worked in the project. Which of the following pairs of the projects had the same duration?
a)Â Project 1, Project 5
b)Â Project 4, Project 5
c)Â Project 3, Project 5
d)Â Project 3, Project 4
Solution:
Considering the information provided :
For project 1 : 3 months.
Project – 2: 3 months.
Project – 3: 5 months.
Project – 4: 5 months.
Project – 5: 4 months.
Among the given options option D is true which is project 3, project 4.
Question 8:Â The list of employees in decreasing order of annual completion index is:
a)Â Danni, Tinni, Bahni, Abani
b)Â Bahni, Abani, Tinni, Danni
c)Â Danni, Tinni, Abani, Bahni
d)Â Tinni, Danni, Abani, Bahni
Solution:
The annual completion index for different people is :
The weightage average of the completion percentage assigned from the different projects, with the weights being the corresponding number of months (s)he worked in these projects.
For Abani :
$\frac{\left(\left(100\cdot2\right)+\left(100\cdot2\right)+\left(80\cdot5\right)\right)}{2+2+5}=\ \frac{800}{9}$
For Bahni :
$\frac{\left(\left(100\cdot2\right)+\left(75\cdot4\right)+\left(90\cdot3\right)\right)}{2+3+4}=\ \frac{770}{9}$
For Danni :
$\frac{\left(\left(90\cdot3\right)+\left(100\cdot3\right)+\left(100\cdot2\right)+\left(100\cdot1\right)\right)}{2+3+2+1}=\ \frac{870}{9}$
For Tinni :
$\frac{\left(\left(80\cdot2\right)+\left(100\cdot2\right)+\left(100\cdot3\right)+\left(100\cdot2\right)\right)}{2+2+3+2}=\ \frac{860}{9}$
The descending order for the four people is :
Danni, Tinni, Abani, Bahni.
Instructions
DIRECTIONS for the following four questions:
A low-cost airline company connects ten India cities, A to J. The table below gives the distance between a pair of airports and the corresponding price charged by the company. Travel is permitted only from a departure airport to an arrival airport. The customers do not travel by a route where they have to stop at more than two intermediate airports.
<img “=”” alt=”” class=”img-responsive” src=”https://cracku.in/media/questionGroup/DI_6_3.png”/>
Question 9:Â What is the lowest possible fare, in rupees, from A to J?
a)Â 2275
b)Â 2850
c)Â 2890
d)Â 2930
e)Â 3340
Solution:
From the table we can see that, the lowest price would be from A to H and H to J.
The cost of travel from A to H = Rs 1850
The cost of travel from H to J = Rs 425
Total cost = 1850 + 425 = Rs 2275.
Question 10:Â The company plans to introduce a direct flight between A and J. The market research results indicate that all its existing passengers travelling between A and J will use this direct flight if it is priced 5% below the minimum price that they pay at present. What should the company charge approximately, in rupees, for this direct flight?
a)Â 1991
b)Â 2161
c)Â 2707
d)Â 2745
e)Â 2783
Solution:
From the table we can see that, the lowest price would be from A to H and H to J.
The cost of travel from A to H = Rs 1850
The cost of travel from H to JÂ = Rs 425
Total cost = 1850 + 425 = Rs 2275
Lowest price = Rs 2275
95% of 2275 = Rs 2161
Question 11:Â If the airports C, D and H are closed down owing to security reasons, what would be the minimum price, in rupees, to be paid by a passenger travelling from A to J?
a)Â 2275
b)Â 2615
c)Â 2850
d)Â 2945
e)Â 3190
Solution:
If the airports C, D and H are closed down  the minimum price to be paid by a passenger travelling from A to J would be by first travelling to F and then from F to J.
The cost of travel from A to FÂ = Rs 1700
The cost of travel from FÂ to JÂ = Rs 1150
Total cost = 1700 + 1150 = Rs 2850
Question 12:Â If the prices include a margin of 10% over the total cost that the company incurs, what is the minimum cost per kilometer that the company incurs in flying from A to J?
a)Â 0.77
b)Â 0.88
c)Â 0.99
d)Â 1.06
e)Â 1.08
Solution:
The minimum cost from A to J we know is 2275.
Let the CP to company be C
Since 10% over actual CP is the total price i.e. $\text{CP}\times1.1 = 2275 \rightarrow CP = \frac{2275}{1.1}$
The total distance is 1950+1400=2350 Km.
Cost per Km = $\dfrac{\frac{2275}{1.1}}{2350}$ = Rs 0.88/Km
Question 13:Â If the prices include a margin of 15% over the total cost that the company incurs, which among the following is the distance to be covered in flying from A to J that minimizes the total cost of travel for the company?
a)Â 2170
b)Â 2180
c)Â 2315
d)Â 2350
e)Â 2390 | 3,150 | 11,309 | {"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-2024-30 | latest | en | 0.952784 |
https://samuelsonmathxp.com/ | 1,500,726,607,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424060.49/warc/CC-MAIN-20170722122816-20170722142816-00679.warc.gz | 700,984,133 | 53,832 | ## First Order Differential Equations
As noted above, the subject of this entry is first order differential equations. Before getting directly to that, however, a few examples of more basic equations representing rates of change will be reflected on.
Too often, students are given rules and algorithms to follow in “solving” problems. While becoming fluent with these procedures is not unimportant, students relying on these strategies alone will eventually find themselves swimming in a pool of confusion when faced with unfamiliar looking problems; if problems not recognizable, they become impossible to resolve. In an effort to minimize such predicaments, an intuitive perspective can often be helpful. Directly below are two examples of solutions to basic differential equations; the first resulting in a quadratic function, the second being exponential. The first two frames below show an “intuitive” approach to solving these problems while the third frame shows a more typical procedure.
Exponential Function
Procedure using Leibniz Notation
An intuitive feel is supplementary to the procedures shown above and can add understanding to the process. Equally important is developing the ability to create these mathematical models from information conveyed in the written word.
A huge stumbling block for many students is knowing why a specific model works for one scenario and not the next; the deciding factor is sometimes a seemingly small detail that is easily overlooked early in the learning process. The example below draws attention to this and will serve as a bridge between simple separable differential equations and the more complex first order linear differential equations.
Mixing Problems and Solution Concentrations
This portion of our journey will focus on the changing concentration of a solution in a container as the ratio of solute to solvent increases or decreases. One such example is provided below.
A tank initially contains 50 gallons of brine in which 5 lb of salt is dissolved. Brine containing 1 lb of salt per gallon flows into the tank at a constant rate of 4 gal/min. The concentration of the brine in the tank is kept uniform throughout the tank by stirring. While the brine is entering the tank, brine from the tank is also being drawn off at a rate of 2 gal/min. Find the amount of salt in the tank after 25 minutes.
(Fraleigh, p.903)
In this scenario, the concentration of brine being added to the tank differs from that originally present with the new mixture being drained off simultaneously; this is the “small detail” making the problem here a more complex one than the other examples cited earlier. The two frames immediately below illustrate simplified versions of this and will serve as an introduction to mixing problems; both allow for a straight forward model involving separable differential equations.
In the scenario directly below, the concentration of solute initially contained in the tank has been altered to match that which is being added (1 lb/gal). That being the case, the concentration of solution being drained away will be a constant 1 lb/gal as well; the resulting function will therefore be linear.
Separable Differential Equation (Equal Concentration)
This second example sees the concentration initially contained in the tank revert to 0.1 lb/gal with inflow of 1 lb/gal. This time, however, no solution will be drawn from the container, once again resulting in a linear function.
Separable Differential Equation (Outflow Zero)
First Order Differential Equations
It is now time to tackle the problem posed by Fraleigh. To reiterate, the concentration of solution flowing into the container differs from that initially present AND this new mixture is being drawn off simultaneously, thereby requiring a more complex model. Since the concentration of solution being extracted is constantly changing, that quantity will be expressed as a ratio of salt (in lbs) to number of gallons of brine in the container at any time “t”. Since the quantity of salt in the tank is changing continuously, a variable will be assigned to represent that quantity; we’ll use “y” to denote pounds of salt.
Inflow: 4 gal/min at 1 lb/gal → 4 lb/min Outflow: 2 gal/min at y lb/(50+2t) gal → [2/(50+2t)]y lb/min
From this, we can easily represent the rate of change of salt (in lbs) with respect to time as follows:
dy/dt=4-[2/(50+2t)]y (lb/min)
This expression is referred to as “Item 2” in the frame directly below. The “item” preceding that is a reminder of the FTC that is called upon in solving for what is referred to as an “Integrating Factor”. Its purpose is also discovered below and is denoted by “I”.
Integrating Factor
Purpose of the Integrating Factor (clarification)
Given Iy’+Ipy=qI
If (Iy)’=Iy’+Ipy , then (Iy)’=qI which leads to Iy=∫qIdt
The integrating factor ultimately allows us to express “y” as a function of “t” (which is our objective).
While first order differential equations are not required in cases where the concentration of inflow equals that already in the container, initial conditions for that scenario are nevertheless included below. The solution to Fraleigh’s problem is shown as well (bottom right) for comparison purposes.
Solution to Fraleigh’s Problem
To reiterate: If the concentration of solution flowing into the container is equal to that already present OR outflow is zero, the function describing quantity of solute will be linear in nature. The following three frames serves as examples of this important detail.
Linear Function: Constant Concentration (1 lb/gal)
Linear Function: Constant Concentration (2 lb/gal)
Linear Function: Zero Outflow
The frame that follows is a variation of the problem posed by Fraleigh, this time with an inflow of 5 gal/min. A graph of the resulting non-linear function is also included.
Non-Linear Function
For further comparison and analysis, I’ve included two additional scenarios directly below, each with an interactive link.
Inflow Exceeds Outflow
Click on the link provided here to explore scenarios where inflow exceeds outflow.
Outflow Exceeds Inflow
Click on the link provided here to explore scenarios where outflow exceeds inflow.
The following image illustrates the function shown directly above on a different scale. The lack of symmetry here creates additional opportunities for discussion with respect to the rate at which the quantity of solute is changing over time. For example, at t=11.16 minutes it can be stated that the quantity of solute flowing in equals that which is flowing out. What is happening before and after that point in time and why? Is this a reasonable conclusion and does the graph of our function reflect that?
Nearly the end………….
I had initially planned on ending this entry with the previous frame; my students, however, wanted one more day to explore mixing problems. I’ve decided to summarize that final day’s sequence here.
As always, this mixing problem began as follows:
y’=q-py , where y’ = rate of change of solute (lb/min) q = inflow of solute (lb/min) py = outflow of solute (lb/min)
We then decided initial conditions, concentration of solute being added as well as rates of inflow and outflow. The first set of values chosen produced what we thought was a reasonable solution until that function’s stationary point was determined; it occurred at (31.53,-1.96). These values caused some concern so our work was double checked. Since the math seemed to be fine, we decided to graph our function and interpret that. Once everything was taken into consideration, this result was perfectly logical. The tank initially contained 150 lbs of solute (3 lbs/gal); we were adding only 0.2 lb/gal at a rate of 6 gal/min and draining away the diluted mixture at a rate of 8 gal/min. Upon reflection, our result seemed perfectly reasonable. This scenario and resulting function are shown immediately below.
Rapid Dilution
Following our satisfactory explanation of the dilution scenario above, the discussion focused on what would need to change so that the function would produce a maximum value greater that 150 lb (the initial content). It was decided that increasing the concentration of solute flowing in would produce the desired result. A concentration of 4 lb/gal was agreed upon and the function reworked. Upon graphing this new function, the location of its maximum was revealed. The question immediately asked was ‘Why is the maximum located at t=0?’. The derivative was determined and set equal to zero, quickly satisfying everyone. The frame immediately below illustrates this scenario.
Local Maximum at t=0
To further solidify our understanding of this concept, it was decided that the vertex of our graph should move to the right if the concentration of solute flowing in was further increased. This final iteration of our mixing problem is shown in the following frame.
Local Maximum at t>0
Click on the link provided here to further explore mixing problems.
The end.
References
Kline, Morris. (1998). Calculus: An Intuitive and Physical Approach. Mineola, NY: Dover Publications.
Quote — Posted: April 22, 2016 in Calculus: An Introduction
## Projectile Motion….including Baseball!
Posted: March 31, 2016 in Calculus: An Introduction, Differential Calculus
Tags:
Projectile motion is a natural fit and provides an interesting application in the introduction of calculus at the high school level. A previous post focused on calculating the horizontal velocity of a ball rolling off the end of a table; this entry takes things a bit further by launching projectiles at various angles to determine the maximum horizontal travel.
Before lighting the fuse on our launching device, some important theory should first be dealt with. Since our launch angle will be somewhere between 0° and 90°, the projectile’s travel will be directed both vertically and horizontally. Since these values will be dependent on the angle with which the projectile is launched, expressions for distance traveled in terms of that angle will be required. These calculations are shown directly below.
Projections of Velocity
The maximum of the horizontal component of a projectile’s motion occurs when its vertical component has been fully depleted. Since these two conditions occur simultaneously, our work will be straight forward. A new function representing the projectile’s height at any position “x” can be easily determined; x-intercepts of this new function will then lead us to the desired solution.
Analytical Solution
The following is a geometric representation of the analytical solution shown above. The graph on the left shows the projectile’s height as a function of its horizontal position “x”. The second graph measures height over time.
Maximum Horizontal Travel
Click on the link provided here to interact with projectile motion and discover maximum horizontal travel.
The scenario directly below is supplementary to a previous entry describing a ball rolling off a table. It is included here with the intent of having students further explore the behavior of projectiles launched horizontally from various heights.
Horizontally Launched Projectile
Click on the link provided here to explore horizontally fired projectiles with variable height and velocity.
The image and link directly below show how far a baseball would travel with zero drag from air-resistance. The initial launch angle and height have been set to 35° and 1 m respectively. The launch angle can vary greatly and still constitute a baseball scenario; the same cannot be said for the launch height. I decided to leave that variable visible to provide another option for further exploration.
Theoretical Flight of a Baseball (and more)
Click on the link provided here to explore the path of a baseball assuming zero drag from air-resistance.
The following links verify the accuracy of the model used above and also provide additional insights into the flight of a baseball.
Baseball Trajectory
The Science of Baseball: What Is The Farthest Home Run?
Baseball Physics: Anatomy of a Home Run
Conclusion
Each year, Major League Baseball provides many satisfying projectile launches; the most gratifying (for me) occurred in the 1988 post-season. For a reminder of this moment in time, click on the link provided here to witness what I consider to be the greatest projectile launch of all time……….I love baseball.
## Physics Problem & Differential Equations
Posted: March 29, 2016 in Calculus: An Introduction
Our school’s physics teacher and I were chatting recently about an experiment he likes to conduct with his Grade 11 students and a related Mythbuster’s episode. The approach in Physics 20 requires students to use the displacement and velocity formulas provided there. This entry revisits the same problem and brings simple separable differential equations to the table.
View the Mythbuster’s video here first: Bullet Fired vs Bullet Dropped
Separable Differential Equations
Click on the link provided here to adjust horizontal velocity of the ball.
Thanks for looking.
## Centers of Mass
Posted: March 27, 2016 in Calculus: An Introduction, Centers of Mass, Integral Calculus
Tags: ,
The power of integral calculus is once again exploited in this entry, this time to determine the centers of mass of one- and two-dimensional objects. Before getting to that, however, some preliminary “discourse” will be engaged in to set the stage.
Everyone (of my age at least) can relate to the scenario involving two children playing on a see-saw. If the children have equal mass and are sitting an equal distance from the fulcrum, they can achieve perfect balance; the fulcrum in this scene is located at the center of mass. If, however, two differing masses are involved (all other things remaining equal), the side containing the greater mass will rotate downward on the fulcrum. This brings us to a very important term for this and other concepts involving rotational motion. This term is called the MOMENT of a force and is defined below:
The moment of a force is a measure of its tendency to rotate about an axis or a point. The moment can be influenced by two quantities: the object’s mass and its distance from the axis or point of rotation; this distance can be referred to as the moment arm.
Moment=Force x Distance , with units measured in ft-lb, kgf-m, etc.
In our example, when two children of the same mass are positioned an equal distance on opposing sides of the fulcrum, a state of equilibrium is achieved. If a bird suddenly joins in the fun and lands on the head of one of the participants, the mass on that end is increased and begins to rotate downward; this increase in mass has created a moment. A moment can also be created if one participant increases his/her position relative to the fulcrum. The moment arm on that side has now been lengthened, thereby creating a moment.
When finding the center of mass in one dimension, the same principles apply; this is a very straight forward procedure if the object has uniform mass density over its entire length. Complications arise, however, if the object’s mass density is not uniform throughout. To address this issue, the object is analyzed as a collection of very small points, each having its own mass and unique position (moment arm) within the main object; each of these will be referred to as “point-mass”. As in the playground scenario described earlier, the further each point-mass is situated from the axes (or point) of rotation, the greater its contribution will be to the moment and, thereby, its influence on the location of the center of mass. The calculation for center of mass is built upon the concept of weighted averages; while the most frequently occurring outcomes have a significant say with respect to the overall average, the extreme outliers can also have a measurable impact.
Before weighted averages can be referenced in this context, the notation and the underlying concept that will be utilized throughout must be introduced. This is initiated below in the Mean Value Theorem of integral calculus.
Mean Value Theorem of Integral Calculus
NOTE: In the calculations that follow, x- and y-components of the moment appear. Since centers of mass occur at a point of equilibrium, force due to gravity is ignored and omitted from the units chosen to represent those quantities. I wanted clarification on this item here since moments are once again called upon when dealing with torque. In that application, force is included in the units of measurement when describing moments of inertia.
A reference was made earlier to “point-masses” and their relative position within the object containing them. A direct connection between this and weighted averages exists and is presented below.
Weighted Average
The end result in the first two examples in the following image are common sense and serve as a “trial run” on the theory developed above; all three can be related directly to our playground scenario described earlier.
One Dimensional Center of Mass
While limiting ourselves here to one-dimension would be silly, attempting to illustrate centers of mass in three dimensions on a 2-D surface could be considered reckless. For this reason, two-dimensions will be the extent to which this topic is pursued here.
Two Dimensions: x-component
Two Dimensions: y-component
In the examples that follow, centers of mass are determined using the theory developed above. Interactive links also provide the opportunity to change one or more parameters in these examples to observe variations in the various integrals involved.
Constant Function
Click on the link provided here to interact with centers of mass on one dimension.
Linear Function
One Image from the exploration that follows……
Click on the link provided here to explore centers of mass defined by linear functions.
Click on the link provided here to explore centers of mass defined by quadratic functions.
The examples that follow have mass density increasing exponentially along the x-axis. With exponentials in the mix, the need for a new method of integration (by parts) emerges; the power of WolframAlpha is also introduced to do the “heavy lifting”.
Constant Function: Exponential Increase in Mass Density
The x-component of the center of mass in the example above can be calculated manually using integration by parts; this procedure is included here.
Integration by Parts
Click on the link provided here to explore centers of mass resulting from an exponential increase in mass density.
While integration by parts can be exploited to evaluate all integrals of this form, the process can become a time-consuming one. The following example is one result obtained from the exploration directly above; it is included here with the intent of introducing students to the power of WolframAlpha. The x-component of the center of mass is shown in a screenshot of the WolframAlpha app available on any device. Interested students have the option of verifying this and other results manually using integration by parts.
Quadratic Function: Exponential Increase in Mass Density
WolframAlpha App
I’ve included a link here to the web-based version of the app shown above. To verify the y-component of the center of mass in the final example shown above, click on WolframAlpha.
References
Courant, Richard., John, Fritz (1999). Introduction to Calculus and Analytics: Classics of Mathematics. New York, NY: Springer-Verlag Berlin Heidelberg.
Larson, R., Hostetler, R. P., & Edwards, B. H. (1995). Calculus of a Single Variable: Early Transcendental Functions. Lexington, MA: D.C. Heath.
## Fluid Pressure & Force
Posted: March 20, 2016 in Calculus: An Introduction, Fluid Force, Integral Calculus, Trigonometric Substitution
Tags: ,
Pressure is a force per unit of area exerted over the surface of an object (as in 35 psi in the tires on your car). When an object is immersed in water, or some other liquid medium, the fluid pressure exerted on that object varies with the depth at which it is submerged. For example, the volume of water pressing down on an object submerged 10 feet is twice that of an object submerged half that depth.
Fluid Pressure (force per unit area) can therefore be defined as follows:
p=wh ,where w = weight density of the fluid h = depth at which the object is submerged
Fluid Force (total force exerted on object) is therefore given by
F=pA ,where A=total area of surface object in question F=whA
According to Pascal’s Law (principle), an object submerged in a fluid is subjected to equal pressure in all directions (at any given depth). For a sheet of metal submerged in water and resting horizontally at a given depth, the fluid force is constant over its entire surface. If, however, the submerged sheet is resting vertically, the entire force exerted over this object by the water varies with depth; the bottom of the sheet will experience more fluid force than its top. In order to determine the total force acting on this vertically oriented sheet, the force exerted on each rectangular cross-section of infinitesimal width (Δy) will be calculated and summed over the object’s entire vertical span; enter integration. With a well-chosen location for the y-axis in our model, the length of each rectangular cross-section can be easily expressed as some variation of f(y).
Rectangular Plate
The example above was relatively simple since f(y) was a constant throughout its vertical span. With shapes whose widths are not constant, the mathematical model can once again vary depending on the perspective chosen. In the examples below, circular plates have been introduced since they provide opportunities for multiple forms of substitution in the integration process, thereby maximizing learning opportunities for students.
The fluid force acting on the ends of a cylindrical water tank is the subject below. I felt that this would be more interesting than imagining the force exerted on a circular plate submerged in a body of water. To simplify matters here, atmospheric pressure and other factors such as sliding forces have been ignored.
Cylindrical Tank: Half-full (Perspective 1)
Cylindrical Tank: Half-full (Perspective 2)
Cylindrical Tank: Filled to Capacity
It is worthwhile drawing attention to the forces acting on the ends of top half of the tank filled to capacity (451.34 lb) and the bottom half filled to half capacity (332.8 lb). This type of thoughtful comparison can add to the students’ understanding of this topic.
The two images directly below once again show the two perspectives of fluid forces acting on the ends of our water tank. They are included here to illustrate the contents of the interactive link that follows.
Tank Centered at (0,0)
Tank Centered at (0,-2)
Click on the link provided here to interact with fluid force on the ends of a cylindrical tank.
In the image and link below, the circular end has been removed from the tank and submerged on its own. Once again, various mathematical models could be used to describe this scenario; the one chosen here has placed the origin at the circle’s center.
Click on the link provided here to interact with fluid force on a submerged vertical plate with center at origin.
The following links will be of interest to some:
Reference
Larson, R., Hostetler, R. P., & Edwards, B. H. (1995). Calculus of a Single Variable: Early Transcendental Functions. Lexington, MA: D.C. Heath.
## Related Rates: Ball Drop & Shadow
Posted: March 13, 2016 in Calculus: An Introduction, Differential Calculus, Differential Equations, Related Rates
Tags: ,
Here’s problem involving a falling object and the speed at which its shadow travels along the ground. As usual, in related rates, once a relationship between the variables involved has been established, the calculus required to reach its conclusion is very straight forward.
In order to make efficient use of time, these problems provide students the opportunity to practice simple differentiation procedures. In addition, the graphs provided below open the door to a discussion on the Mean Value Theorem of differential calculus, serving to either introduce or reinforce that concept.
Falling Ball
Click on the link provided here to interact with the falling ball and its shadow.
The ball’s displacement from its release point was provided in the image above. As a review (since integral calculus has already been introduced), that displacement formula is once again derived through basic differential equations; this is shown directly below.
Acceleration, Velocity and Displacement
I’ve included solutions for t=1 and t=2 below. In keeping with my belief that students can learn effectively through comparison and contrast, three varied methods are shown.
Solutions
The main subject of this entry was originally planned as an optimization problem involving differential calculus only; its been slightly modified. This more interesting approach provides the derivative up front, presenting students with three separate tasks to pursue from that point. As a consequence, students are reintroduced to differential equations and curve sketching.
A talking point emerges as well: Is there a difference between derivatives and differential equations?
Inscribed Triangle of Maximum Area
Click on the link provided here to explore area of the inscribed triangle.
Thanks for looking. | 5,238 | 25,803 | {"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} | 4.53125 | 5 | CC-MAIN-2017-30 | longest | en | 0.939901 |
https://ask.sagemath.org/question/8717/series-solutions-of-higher-order-odes/?answer=37709 | 1,723,202,596,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00799.warc.gz | 87,166,598 | 17,073 | # series solutions of higher order ODEs
I'm trying to use Sage to find the general series solution to $y^{(4)}=\frac{y'y''}{1+x}$. So far my best efforts to derive the coefficient recurrence relations, inspired by a good book draft, have been along these lines:
R10 = QQ['a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10']
a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10 = R10.gens()
R.<x> = PowerSeriesRing(R10)
y = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 + a5*x^5 + \
a6*x^6 + a7*x^7 + a8*x^8 + a9*x^9 + a10*x^10 + O(x^11)
y1 = y.derivative()
y2 = y1.derivative()
y3 = y2.derivative()
y4 = y3.derivative()
f = (1+x)*y4-y1*y2
i = ideal(f)
# g = i.groebner_fan(); g.reduced_groebner_bases() # a wish
# q = R.quotient(i) # works, but not so useful by itself
and my other approaches ended in tracebacks:
x = var('x'); y = function('y', x)
desolve(diff(y,x,4)-diff(y,x)*diff(y,x,2)/(1 + x), y, contrib_ode=True)
# NotImplementedError: Maxima was unable to solve this ODE.
desolve_laplace(diff(y,x,4) - diff(y,x)*diff(y,x,2)/(1 + x), y)
# TypeError: unable to make sense of Maxima expression
I would like to at least solve that and determine the radius of convergence. Ideally (more generally), it would be nice to have a good bag of tricks for working with series DEs such as I imagine others have already created. For this, I would like to find or develop techniques to incorporate:
• more convenient coefficients, e.g. from this thread
• a way to derive the recurrence relations using Python's lambda operator or this nice trick
• a solver for higher order ODEs such as above
Any hints, links, references or suggestions would be appreciated.
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The substitution $z=y'$ is natural, so let us please then forget about the posted $y$.
We will not use $z$. Let us use $y$ "back" instead of $z$ in the sequel. So we solve (instead) around $x=0$: $$y'' ' = y\ y' \ /\ (1+x)\ .$$
Reusing the above code...
R10.<a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10> = \
QQ['a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10']
R.<x> = PowerSeriesRing(R10)
y = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 + a5*x^5 + \
a6*x^6 + a7*x^7 + a8*x^8 + a9*x^9 + a10*x^10 + O(x^11)
y1 = y .derivative()
y2 = y1.derivative()
y3 = y2.derivative()
f = (1+x)*y3 - y*y1
# J = ideal(f) # --> leads to an error
f.list()
... we get the coefficients of $f$:
[-a0*a1 + 6*a3,
-a1^2 - 2*a0*a2 + 6*a3 + 24*a4,
-3*a1*a2 - 3*a0*a3 + 24*a4 + 60*a5,
-2*a2^2 - 4*a1*a3 - 4*a0*a4 + 60*a5 + 120*a6,
-5*a2*a3 - 5*a1*a4 - 5*a0*a5 + 120*a6 + 210*a7,
-3*a3^2 - 6*a2*a4 - 6*a1*a5 - 6*a0*a6 + 210*a7 + 336*a8,
-7*a3*a4 - 7*a2*a5 - 7*a1*a6 - 7*a0*a7 + 336*a8 + 504*a9,
-4*a4^2 - 8*a3*a5 - 8*a2*a6 - 8*a1*a7 - 8*a0*a8 + 504*a9 + 720*a10]
We give or consider a0, a1, a2 as given.
Solving by the Cauchy-Kovalevskaya "procedure", we have to determine recursively a3, a4, a5, a6, a7, a8, a9, a10, and so on, starting from a0, a1, a2, by making each line evaluate to $0$. We get manually $$a_3 = \frac 16 a_0a_1\ ,$$ $$a_4 = \frac 1{24}( a_1^2 + 2a_0a_2 + a_0a_1 )\ ,$$ and so on.
I think it is not a good idea to isolate in this way all these equations, then start solving. Instead, let us write the given DE in the form $$y'' ' = y\ y'\ (1-x+x^2-x^3+\dots)\ ,$$ and consider it formally in the ring $\mathbb Q[[x]]$. Then the equations in $a_3,a_4,\dots$ come already separated, and we can simply write the code to get the first some $100$ coefficients.
(I am not willing to write optimal code, time...)
To have a precise situation, let us better use values instead of the symbols $a_0,a_1, a_2$. For instance $a_0=0$, $a_1=1$, $a_2=2$. Then...
a = [0,1,2] # use other initial conditions, next time
PSR.<x> = PowerSeriesRing( QQ, 100 )
C = 1 / PSR(1+x)
for k in [ 3..99 ]:
print k,
# trying to get a(k) from y = sum( [ a(k) * x^k for k in [0..ooh] ] )
yk = sum( [ a[j] * x^j for j in range(k) ] ) + O(x^k)
ak = 1/(k*(k-1)*(k-2)) * ( yk * diff( yk, x ) * ( C+O(x^k) ) ).list()[k-3]
print ak
a.append( ak )
y = PSR( sum( [ a[k] * x^k for k in [0..99] ] ) )
This computes some values. For instance, the first some $30$ entries are:
sage: a[:30]
[0,
1,
2,
0,
1/24,
1/12,
1/40,
-67/5040,
13/1152,
-7/1440,
4621/1209600,
-38399/13305600,
11287/4838400,
-12737/6988800,
2351057/1614412800,
-171374993/145297152000,
1692465871/1743565824000,
-115101113/142502976000,
32202455561/47424990412800,
-4213094929/7313918976000,
22201339257697/45053740892160000,
-267905698350809/630752372490240000,
5115211318062343/13876552194785280000,
-8562348061276901/26596725040005120000,
928302494123858617/3282795776366346240000,
-7613558036179561/30490363247984640000,
1985785680757233642821/8962032469480125235200000,
-590031815878576041977/2987344156493375078400000,
33273819697787429364371/188202681859082629939200000,
-3468510498746975345435917/21831511095653585072947200000]
(These are exactly computed.) Let us check the solution modulo $O(x^{100})$:
# : diff( y,x,3 ) - y * diff(y,x) / (1+x)
# the above is a mess with terms in degrees 97, 98, 99, but..
sage: diff( y,x,3 ) - y * diff(y,x) / (1+x) + O( x^97 )
O(x^97)
The radius of convergence is in this case probably one, one can try...
for k in [1..99]:
ak = a[k]
print ( "a[%2d] ~ %13.10f and a[%2d]^(1/%2d) ~ %8.6f"
% ( k, RR(ak), k, k, abs(RR(ak))^(1/k) ) )
and the first and last lines shown are:
a[ 1] ~ 1.0000000000 and a[ 1]^(1/ 1) ~ 1.000000
a[ 2] ~ 2.0000000000 and a[ 2]^(1/ 2) ~ 1.414214
a[ 3] ~ 0.0000000000 and a[ 3]^(1/ 3) ~ 0.000000
a[ 4] ~ 0.0416666667 and a[ 4]^(1/ 4) ~ 0.451801
a[ 5] ~ 0.0833333333 and a[ 5]^(1/ 5) ~ 0.608364
a[ 6] ~ 0.0250000000 and a[ 6]^(1/ 6) ~ 0.540742
a[ 7] ~ -0.0132936508 and a[ 7]^(1/ 7) ~ 0.539447
a[ 8] ~ 0.0112847222 and a[ 8]^(1/ 8) ~ 0.570902
a[ 9] ~ -0.0048611111 and a[ 9]^(1/ 9) ~ 0.553313
:::::::::::::::::::::::::::::::::::::::::::::::::
a[90] ~ 0.0000051223 and a[90]^(1/90) ~ 0.873406
a[91] ~ -0.0000049542 and a[91]^(1/91) ~ 0.874386
a[92] ~ 0.0000047934 and a[92]^(1/92) ~ 0.875348
a[93] ~ -0.0000046394 and a[93]^(1/93) ~ 0.876295
a[94] ~ 0.0000044920 and a[94]^(1/94) ~ 0.877225
a[95] ~ -0.0000043507 and a[95]^(1/95) ~ 0.878140
a[96] ~ 0.0000042154 and a[96]^(1/96) ~ 0.879040
a[97] ~ -0.0000040855 and a[97]^(1/97) ~ 0.879925
a[98] ~ 0.0000039610 and a[98]^(1/98) ~ 0.880796
a[99] ~ -0.0000038414 and a[99]^(1/99) ~ 0.881653
I have maybe to stop here with sage sample code.
About the convergence radius... The recurrence relation is thus of the shape $a_k=$ a sum of monomials of order two in the previous variables, with coefficients that can be recursively "controlled". With some work, one may (??) maybe show that the radius of convergence is one. (The one is my quick guess, based on the fact that the limit $r=\lim |a_k|^{1/k}$ may / should exist, and then we expect $r\sim r^2$. Minoration / majoration for the "polynomial part in $k$" is constricted by the power $1/k$. But it is just a guess.)
Note: The use of solve_laplace really wants to perform a Laplace transform. But which is the first step when we try to apply it on the right side of the given DE ?!
Note: The answer is not given mot-a-mot to the posted (relatively general) question, but in the sense of the progress. (Which is subjective.) I think, sample code serves good as an answer.
more | 2,893 | 7,267 | {"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} | 3.734375 | 4 | CC-MAIN-2024-33 | latest | en | 0.852136 |
http://caponigroconstruction.com/assassin-s-magc/importance-of-line-graph-d204ab | 1,680,207,352,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00577.warc.gz | 10,168,506 | 9,710 | matrix of the graph line-graph L(G). Pros Click Here to subscribe to the CoolStuff Newsletter and get notified when our next blog is released. The Importance of Statistics to the User. Statistics is important in geography because of the following reasons: 1. Grid, Line and Path Games Graphing. The picture at the top of this page shows an example of a bar graph, a line graph, a pie chart, and a table. 5. Line graphs are very common in IELTS academic task 1 writing. 272 THE PHYSICS TEACHER Vol. Process 3 calculates the importance of the vertices of the graph L(G) equivalent to the importance of the edges of G. There are many different kinds of graphs. When using a line graph, why is it inportant to only graph 1 - 3 series of data? Each of these line graphs shows a change in data over time. IELTS Line Graph Answer. In this lesson, we will look at a Model Answer for CO2 emissions for 4 European countries and an analysis. The line graph consists of a … Once we graph something, it only makes sense to count elements from the graph, as well as to compare and contrast them. The important use of line graph is to track the changes over the short and long period of time. Typical examples of the types of data that can be presented using line graphs are monthly rainfall and annual unemployment rates. Each division should be equally spaced. or behaviors we want to increase (requesting for information, learning to read, counting, etc. Abstract: Graphs have been used to report scientific data for centuries; however, creating effective graphs can prove challenging. Line charts, or line graphs, are powerful visual tools that illustrate trends in data over a period of time or a particular correlation. The line graph is one of the simplest graphs you can make in Excel. How to Make a Line Graph in Excel: Explained Step-by-Step. Line graphs show you how numbers have changed over time. People use graphs to compare amounts of things or other numbers. In this post, we’ll talk about how a line graph works, plus: how to build one that provides meaningful information and context, what kind of data can be shown on a line graph… It is important to know that all line graphs must have a title. Using Line Graphs: An Example. Statistic is an important part of business,mathematics,representation of surveys,etc.It is the study of representing the lemark form of information into conc view the full answer. 3. The line graph represents the most frequently used display for visual analysis and subsequent interpretation and communication of experimental findings. Featured Products The Importance of the Best Fit Line Introduction One of the most important skills that you will learn in physics class is drawing a best fit line. Expert Answer . Label the x axis (horizontal axis) with the independent variable. By plotting sales figures on a line graph (as shown in figure 3), you can see the main fluctuations during the course of a year. Number each axis division (line). What are the importance of using a line graph. The line graph illustrates the amount of three kinds of spreads (margarine, low fat and reduced spreads and butter) which were consumed over 26 years from 1981 to 2007. Fibre bundles are not Cartesian products, but appear to be up close. Another name for a line graph is a line chart. Despite its apparent simplicity, the applications of the straight-line graph are often underesti- It's important to organise your graph clearly, draw out the key trends and make comparisons. 37, May 1999 The Importance of Graphs in Undergraduate Physics graph gives velocity, or a graph of velocity versus time will give acceleration. The title of a line graph provides a general overview of what is being displayed. In the next of our lessons on line graphs we look at how to change the verb/adverb collocations that we have already learnt into adjective/noun collocations. Some of the most common kinds are circle graphs, bar graphs, and line graphs. Just make sure the periods are equally distributed on the x-axis. Line graphs (or line charts) are best when you want to show how the value of something changes over time. Uses of Line Graph. There are four types of lines graphs: 1) Simple: These are drawn to … Important Vocabulary for Charts and Graphs. line graph depresses the trend and decreases variability. For too long we as humans have taken to much work upon our shoulders, it's time to simplify our lives and to use the best tools for the job. But that doesn’t mean it’s not one of the best.. ). Given the importance of reporting data effectively, it is worth your time to learn how to do so. Continuing with the sites IELTS line graph examples, this is an example of a line graph comparing car theft. Hello - I got a request to modify our Atterberg Limits graph to include the U-line in addition to the A-line. Page Content Offline Michael August Tue, Apr 22 2014 1:24 PM. Nevertheless, line graphs can also be applied to indicate tendencies based on other continuous periodic values, such as speed, temperature, distance etc. Label the y-axis (vertical axis) with the dependent variable. Line graph also has its pros and cons. Graphs are also known as charts. Usually what happens is you take two related variables and chart how they relate to each other. The chart on the right shows another bar graph, a diagram (sketch or picture), and a flow chart. If you are asking about charts and graphics in astrology, the answer is nothing, there isn’t any. A line graph in Microsoft Excel will not allow you to graph more than 3 series of data. Graphs are useful because they can be easier to understand than numbers and words alone. The graph of a function is contained in a Cartesian product of sets. This article describes the components of graphs, discusses general considerations in preparing graphs and graph selection, and addresses the most common problems in graphs. These behaviors can include behaviors we want to decrease (aggression, screaming, tantrums, pinching, self-injurious behavior, etc.) An X–Y plane is a cartesian product of two lines, called X and Y, while a cylinder is a cartesian product of a line and a circle, whose height, radius, and angle assign precise locations of the points. We take data on many different behaviors. Line Graph What is a Line Graph? Despite this common framework, the distinct features of bar and line graphs result in significant differences in their interpretation. Let's define the various parts of a line graph. Cartesian graphs have numbers on both axes, which therefore allow you to show how changes in one thing affect another. 3 Line Graphs 3.1 Line Graphs of an Undirected Graph Given an undirected graph G= (V;E), the line graph of is an undirected graph G = (E;F), in which there is an edge f2Fthat connects the nodes e;e02Eif and only if there is a node v2Vsuch that both eand 0incide on vin G. Line graphs have particular characteristics. Previous question Next question Get more help from Chegg. If you can use these adverbs effectively then you will improve you Task Achievement band and Lexical Resource band.. As well as frequency tables, data can be displayed in a a variety of graphs and charts too. Academic Writing task 1: C02 Emissions line graph. The lines are too thick to use more than 3 series. One of the reasons that I like graphing so much is that it gives my class a real-life, relevant reason to use math concepts. ABC Enterprises' sales vary throughout the year. You can find this book on Amazon. A line graph is useful for displaying data or information that changes continuously over time. The range will set the scale. Some examples of graphs used in ABA includes line graphs, bar graphs, and pie charts. The graph below will be used to help us define the parts of a line graph. IELTS Line Graph Examples. Written by co-founder Kasper Langmann, Microsoft Office Specialist.. Graphs are used in everyday life, from the local newspaper to the magazine stand. It will not show changes over time of you use more than 3 series of data. That has been done well in this answer. Line graphs can also be used to compare changes over the same period of time for more than one group. The most effective visuals are often the simplest—and line charts (another name for the same graph) are some of the easiest to understand. For example, a finance department may plot the change in the amount of cash the company has on hand over time. It is one of those skills that you simply cannot do without. Here, sales drop off in June and July, and again towards the end of the year. Like all graphs , bar graphs give a visual representation of numerical data. Determine the range of your data that must fit on each axis. Usually the line graph for data with the highest values is drawn first. 2. Simplifying your life is the way of the future. 584 Educ Psychol Rev (2017) 29:583–598. How to make a line graph 1. Atterberg Limits Graph - adding the U-Line to the graph. The separate lines that are made by connecting the points for each city. Today (finally) I’m going to talk about graphing. The input data (Data2) is the adjacency matrix A(G) of the graph G. Process 1 and 2 have been implemented using the programming language C/C++. 4. This line graph comes from Cambridge IELTS 11 academic. Table 1 Quality features of a lin e graph and measurement Essential structure Function Measured Vertical axis labeled with quantitative measure; horizontal axis labeled with time unit Line graphs are usually used to show time series data - that is how one or more variables vary over a continuous period of time. For example, one axis of the graph might represent a variable value, while the other axis often displays a timeline. Units are measured in grams. Bar Graph Examples Line Graphs. More than 3 series of data causes too many lines on the graph, which makes ti confusing to read.•• Graphing is one of those tools that you just cannot be without. It enables the geographers to handle large sets of data and summarize them in … Pros and Cons of Using Line Graphs As is known to all, every coin has two sides. Line graphs. Two of the most common are Line Graphs and Bar Graphs, which we will see examples of in this page. The points on the double line graph show the average monthly rainfall in the two cities (city 1, city 2). Line Graph: A line graph is a graph that measures change over time by plotting individual data points connected by straight lines. A line graph, also known as a line chart, is a type of chart used to visualize the value of something over time. They are used when you have data that are connected, and to show trends, for example, average night-time temperature in each month of the year. Chart on the right shows another bar graph, why is it inportant to graph. Atterberg Limits graph - adding the U-Line to the A-line in astrology, the distinct features of bar line! More help from Chegg creating effective graphs can also be used to help define! X axis ( horizontal axis ) with the independent variable what are the importance of using line. Visual representation of numerical data is nothing, there isn ’ t.. Graphs must have a title time for more than one group variable value while. And July, and a flow chart of a line graph consists of a … graph... Distinct features of bar and line graphs ( or line charts ) are when! Of those tools that you just can not do without company has on over! Coin has two sides help us define the parts of a … line graph is useful for data. ( aggression, screaming, tantrums, pinching, self-injurious behavior, etc. band and Lexical Resource... Can use these adverbs effectively then you will improve you task Achievement and. 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For 4 European countries and an analysis to graph more than one group line charts ) best! And decreases variability subsequent interpretation and communication of experimental findings inportant to only graph -... Content academic Writing task 1 Writing decrease ( aggression, screaming,,...: Explained Step-by-Step result in significant differences in their interpretation when you want to increase ( requesting information. Can use these adverbs effectively then you will improve you task Achievement band and Lexical Resource..! Is it inportant to only graph 1 - 3 series of data page Content academic Writing 1... Each city Lexical Resource band t mean it ’ s not one those. Increase ( requesting for information, learning to read, counting, etc. to help define... Just make sure the periods are equally distributed on the right shows another graph. Axis of the graph line-graph L ( G ) most frequently used display for visual analysis and subsequent and... 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# SUBJECT : COURSE : TITLE : LEVEL : SET : START :
## MATHEMATICS FORM 1 TOPIC 7 LEVEL 1.0 31/01/2012 10:53:32 AM
1327978412 31/01/2012 10:53
1.
TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan.
Which of the following is not an unknown? Manakah di antara berikut bukan anu?
14338 14338 0 0
A. x. B. y. C. z. D. 3. 2. Which of the following is not an algebraic term with an an unknown? Manakah di antara berikut bukan sebutan algebra dengan satu anu ?
14339 0 0 14339
A. B. C.
D. 3. Which of the following is not an algebraic term ? Manakah di antara berikut bukan sebutan algebra?
0 0 14340 14340
A. B. C.
D.
4. Which of the following pairs are like terms? Manakah di antara pasangan berikut sebutan serupa?
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A. B. C. D. 5. Find the difference between 5g and -2g. Cari perbezaan di antara 5g dan -2g.
0
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14342
## A. 3g. B. 7g. C. -3g. D. -7g. 6. x + y -(-2 y) = A. x + y. B. x - y.
14343 14343 0 0
C. x + 3y. D. x - 3y. 7. 81 m + 9 m = A. 9 m. B. 90 m. C. 90 + m. D. 2 + 9 m. 8. -7 x - 5 x = A. 12x. B. 2x. C. -12x. D. -2x. 9. 12p - ( - 8p) = A. 20p. B. 4p. C. 20 + p. D. - 2x. 10. Which of the following has 2 as its coefficient? Manakah di antara berikut mempunyai 2 sebagai pekalinya?
0 0 14347 14347 14346 14346 0 0 14345 14345 0 0 14344 14344 0 0
A. B. C. D.
## Sebutan serupa dengan -15p adalah ...
14348 14348 0 0
A. B. C.
D.
12. Which of the following is an expression? Manakah di antara berikut suatu kenyataan?
0
14349
14349
A. ab. B. 3c. C. 4ab. D. 4+a+3c. 13. 3 - 5x + 2x How many terms are there in the above expression? Berapa sebutankah yang terdapat di dalam kenyataan di atas?
14350 14350 0 0
A. 1. B. 2. C. 3. D. 4. 14. The coefficient of g in the expression -2g + 5g is... Pekali bagi g dalam kenyataan -2g + 5g adalah ...
14351 14351 0 0
14352 14352 0 0
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## C. 8pq. D. 8 + p + q. 17. Simplify 3 - 2a + 9a - 1 = Permudahkan 3 - 2a + 9a - 1=
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A. 7a + 4. B. 7a + 2. C. -7a + 2. D. -7a - 2. 18. 4r + 6 - 2r = A. 4r + 8. B. 10r - 2. C. 8r. D. 2r + 6. 19. Which of the following expression has the most number of terms? Manakah di antara kenyataan berikut mempunyai paling banyak sebutan?
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14356
A. 3m - 4. B. 5p + 6 - 8p. C. 7s - 7t + s. D. 2x - y + 3x + 5y. 20. Which of the following is the like term of 5d? Manakah di antara berikut sebutan bagi 5d?
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14357
A. I only / I sahaja. B. II only / II sahaja. C. I and II only / I dan II sahaja. D. II and III only / II dan III sahaja.
21
Submit
1. TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan.
P bought x ball pens and y note books for RMz. Which of the following is not an unknown? P telah membeli x pena mata ball dan y buku nota berharga RMz. Manakah di antara berikut bukan suatu anu?
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A. P.
B. X. C. Y. D. Z. 2. An algebraic term with unknown g has a coefficient of 9. The algebraic term is... Suatu sebutan algebra dengan anu g mempunyai pekali 9. Sebutan algebra tersebut adalah ...
14359
14359
A.
B.
C.
D.
14360
## Pasangan manakah di antara berikut mengandungi sebutan serupa?
14360 0 0
A.
B.
C. 4. 5m - 8 - 2 ( m - 3 ) =
0 0 14370 14370
D.
A. 3m - 14.
B. 3m - 11. C. 3m - 5. D. 3m - 2. 5.
14371
14371
A.
B.
C.
D.
## 6. Which of the following is true?
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Manakah di antara berikut benar? A. 2a + 3a = 6a. B. 4b + ( -2b ) = -2b. C. ( -8c ) - ( -5c ) = -3c. D. ( -5d ) + ( -4d ) = 9d. 7. 8 - f - ( f - 8 ) = B. 16 - 2f. C. 2f.
14373
14373
A. 16 + 2f.
D. -2f. 8. Which of the following algebraic terms are like terms for 9w ? Manakah di antara sebutan algebra berikut sebutan serupa bagi 9w?
14374 0 0
14374
A. C. D.
B.
## 9. Simplify 2y + 8y - ( -3y + 5 ). Permudahkan 2y + 8y - ( -3y + 5 ).
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A. 3y - 5.
B. 10y + 5. C. 13y - 5. D. 13y + 5. 10. -12e - 8 + 6e + ( - 3f ) = A. 18e - 8 - 3f. B. -18e - 8 + 3f. C. -6e - 8 - 3f. D. -6e - 8 + 3f.
14376 14376 0 0
11. All of the following algebraic expressions can be simplified, except... Semua kenyataan algebra berikut boleh dipermudahkan, kecuali...
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A. 2 + 5y - 5.
B. 4x - 10x + 9. C. 3x - y + 4z. D. 5p + 1 -2p - 3. 12. Find the perimeter of an equilateral triangle of side 3x cm. Cari perimeter suatu segitiga sama sisi dengan 3x cm.
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A. 3x cm.
B. 6x cm. C. 9x cm. D. 12x cm. 13. All of the following expressions have the same number of algebraic terms except... Semua kenyataan berikut mempunyai bilangan sebutan algebra yang sama kecuali ...
14407
14407
B.
C.
14379 14379 0
D.
0
A.
B.
C.
D.
14380 14380 0
0
## D. -4t. 16. 3e subtracted from 9e is...
3e ditolak daripada 9e adalah... A. 3e - 9e. B. 9e + 3e. C. 9e - 3e. D. -9e - 3e. 17. 3y + (-2) - y + 6 = 12. B. 3y - 6. C. 2y + 4. D. 2y - 4. 18. A. B. Simplify -4 ( -3 + m ) + 7m =
14382
14381
14381
14382
A. 6y +
14383
14383
C.
D.
19.
## Permudahkan -4 ( -3 + m ) + 7m = A. 12 + 8m. B. 12 + 3m. C. -12 - 3m.
14384
14384
D. -12 + 8m. 20. The result, when a number x is multiplied by 2 and then 5 added to it, is... Hasil, apabila suatu nombor x di darab dengan 2 dan kemudian 5 ditambah kepadanya, adalah ...
14385 14385 0 0
A. 7.
B. x + 10. C. 2x + 5. D. 2x + 10.
## MATHEMATICS FORM 1 TOPIC 7 LEVEL 3.0 31/01/2012 10:55:50 AM
1327978550 31/01/2012 10:55
1.
TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan.
-8p - 5p - (-7) =
14386 14386 0 0
A. B. C. D.
## 2. Simplify 4s + 3 - ( -5s + 8 ). Permudahkan 4s + 3 - ( -5s + 8 ).
0 14387 14387 0
A. B.
9s + 11. 9s - 5.
C. -s + 11.
D.
-s - 5.
3. Find the perimeter of a square of side ( x + 4 )cm. Cari perimeter suatu segiempat sama bersisi ( x + 4 )cm.
14388 14388 0 0
A. B. C. D.
## 2x + 8. 3x + 12. 4x + 16. 5x + 20.
4. Calculate the area of the triangle below, in cm. Kirakan luas segitiga di bawah, dalam cm.
14389
14389
A. B. C. D.
5.
14390
14390
A. B. C. D.
6. Ahmad has 8x marbles. Bahrin has ( x + 2 ) marbles more than Ahmad. How many marbles does Bahrin have? Ahmad ada 8x biji guli. Bahrin ada ( x + 2 ) biji guli lebih banyak daripada Ahmad. Berapa banyak gulikah yang Bahrin ada?
14391 14391 0 0
A. B. C. D.
9x + 2. 8x + 2. 7x + 2. 7x - 2.
7. Siti bought a pen for RMv and then sold it at RM4w. What profit did she make? Siti telah membeli sebatang pen dengan harga RMv dan kemudian menjualnya dengan harga RM4w. Berapakah keuntungan yang dia dapat ?
14392 14392 0 0
A.
RM ( v - 4w ).
B. C. D.
RM ( v + 4w ). RM ( 4w - v ). RM ( 4w + v ).
8. A rectangular field measures ( 2x - 1)m by ( 3x + 2 )m. The perimeter of the field in meter, is... Sebuah padang segiempat berukuran ( 2x - 1)m dengan ( 3x + 2 )m. Perimeter padang itu, dalam meter, adalah ...
14393
14393
A. B. C. D.
5x + 1. 8x + 3. 10x + 2. 10x + 6.
9.
14394
14394
A. B.
C.
D.
10. The diagram below shows a triangle. Calculate the perimeter of the triangle.
Gambar rajah di bawah menunjukkan suatu segitiga. Kirakan perimeter segitiga tersebut.
14395
14395
A. B. C. D.
14396 14396 0 0
11.
## 2 [ k - ( 2k + 3 ) ] = A. B. C. D. -2k + 6. -2k - 6. -2k + 3. -2k - 3.
12.
6p - 3 - q - 5p - 3q + 8 = A. B. C. D. p - 4q + 11. p - 4q + 5. p - 3q + 11. p - 3q + 5.
14397
14397
13. Aini weights (4x - 2 )kg while her sister weights ( 3x + 6)kg. What is their total weight?
Berat Aini (4x - 2 )kg manakala berat adik perempuannya ialah ( 3x + 6)kg. Berapakah jumlah berat mereka? A. ( 5x + 3)kg. B. C. D. ( 6x + 9 )kg. ( 7x + 4 )kg. ( 12x - 8 )kg.
14398 14398 0 0
14. Rina has 2v sweets. Siti has w sweets less than Rina. While Tina has 8 sweets. How many sweets do they have altogether? Rina ada 2v biji gula-gula. Siti ada w biji gula-gula kurang daripada Rina. Manakala Tina ada 6 biji gula-gula. Berapa biji gula-gulakah yang mereka ada kesemuanya?
14399 14399 0 0
A. B. C. D.
2v - w + 8. 2v + 8. 4v - w + 8. 8 - 4v - w.
15. A watermelon costs RM2u and an apple costs 60sen. Find the total cost of the fruits in sen. Sebiji tembikai berharga RM2u dan sebiji epal berharga 60sen. Cari jumlah harga buahbuahan tersebut, dalam sen. A. 2u + 60. B. C. D. 2u - 60. 20u + 60. 200u + 60.
14400 14400 0 0
16. A box contains 5h apples. 10 of the apples have been eated. How many apples are there now? Sebuah kotak mengandungi 5h biji epal, 10 biji epal telah dimakan. Berapakah baki epal itu
sekarang? A. 5h - 10. B. C. D.
14401
14401
## 5h + 10. 10 - 5h. 10 + 5h.
14402 14402 0 0
17.
2( 3p - q ) + 3( p + q ) = A. B. C. D. 9p. 9p - q. 9p + q. 9p + 3q.
18. There are 55k students in a school. 39 + 12k of them are girls. How many boys are there in the school? Di sebuah sekolah terdapat 55k orang pelajar. 39 + 12k daripada mereka adalah perempuan. Berapa orang lelakikah dalam sekolah tersebut?
0 0
14403 14403
A. B. C. D.
## 11 + 12k. 43k + 11. 43k + 39. 43k - 39.
19. The table below shows the number of pens in a box. If the total number of pens in the box is U, express U in terms of w. Jadual dibawah menunjukkan bilangan pen dalam sebuah kotak. Jika jumlah bilangan pen di dalam kotak itu adalah U, nyatakan U dalam sebutan w.
14404
14404
A. B.
C.
D.
20.
14405
14405
A. B. C. D.
3w - 2 2w - 7 5w + 2 3w - 10 | 4,267 | 9,908 | {"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} | 3.71875 | 4 | CC-MAIN-2019-26 | latest | en | 0.264883 |
https://en.wikipedia.org/wiki/Rigid_body | 1,722,836,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00526.warc.gz | 180,951,939 | 36,848 | # Rigid body
In physics, a rigid body, also known as a rigid object,[2] is a solid body in which deformation is zero or negligible. The distance between any two given points on a rigid body remains constant in time regardless of external forces or moments exerted on it. A rigid body is usually considered as a continuous distribution of mass.
In the study of special relativity, a perfectly rigid body does not exist; and objects can only be assumed to be rigid if they are not moving near the speed of light. In quantum mechanics, a rigid body is usually thought of as a collection of point masses. For instance, molecules (consisting of the point masses: electrons and nuclei) are often seen as rigid bodies (see classification of molecules as rigid rotors).
## Kinematics
### Linear and angular position
The position of a rigid body is the position of all the particles of which it is composed. To simplify the description of this position, we exploit the property that the body is rigid, namely that all its particles maintain the same distance relative to each other. If the body is rigid, it is sufficient to describe the position of at least three non-collinear particles. This makes it possible to reconstruct the position of all the other particles, provided that their time-invariant position relative to the three selected particles is known. However, typically a different, mathematically more convenient, but equivalent approach is used. The position of the whole body is represented by:
1. the linear position or position of the body, namely the position of one of the particles of the body, specifically chosen as a reference point (typically coinciding with the center of mass or centroid of the body), together with
2. the angular position (also known as orientation', or attitude) of the body.
Thus, the position of a rigid body has two components: linear and angular, respectively.[3] The same is true for other kinematic and kinetic quantities describing the motion of a rigid body, such as linear and angular velocity, acceleration, momentum, impulse, and kinetic energy.[4]
The linear position can be represented by a vector with its tail at an arbitrary reference point in space (the origin of a chosen coordinate system) and its tip at an arbitrary point of interest on the rigid body, typically coinciding with its center of mass or centroid. This reference point may define the origin of a coordinate system fixed to the body.
There are several ways to numerically describe the orientation of a rigid body, including a set of three Euler angles, a quaternion, or a direction cosine matrix (also referred to as a rotation matrix). All these methods actually define the orientation of a basis set (or coordinate system) which has a fixed orientation relative to the body (i.e. rotates together with the body), relative to another basis set (or coordinate system), from which the motion of the rigid body is observed. For instance, a basis set with fixed orientation relative to an airplane can be defined as a set of three orthogonal unit vectors b1, b2, b3, such that b1 is parallel to the chord line of the wing and directed forward, b2 is normal to the plane of symmetry and directed rightward, and b3 is given by the cross product ${\displaystyle b_{3}=b_{1}\times b_{2}}$.
In general, when a rigid body moves, both its position and orientation vary with time. In the kinematic sense, these changes are referred to as translation and rotation, respectively. Indeed, the position of a rigid body can be viewed as a hypothetic translation and rotation (roto-translation) of the body starting from a hypothetic reference position (not necessarily coinciding with a position actually taken by the body during its motion).
### Linear and angular velocity
Velocity (also called linear velocity) and angular velocity are measured with respect to a frame of reference.
The linear velocity of a rigid body is a vector quantity, equal to the time rate of change of its linear position. Thus, it is the velocity of a reference point fixed to the body. During purely translational motion (motion with no rotation), all points on a rigid body move with the same velocity. However, when motion involves rotation, the instantaneous velocity of any two points on the body will generally not be the same. Two points of a rotating body will have the same instantaneous velocity only if they happen to lie on an axis parallel to the instantaneous axis of rotation.
Angular velocity is a vector quantity that describes the angular speed at which the orientation of the rigid body is changing and the instantaneous axis about which it is rotating (the existence of this instantaneous axis is guaranteed by the Euler's rotation theorem). All points on a rigid body experience the same angular velocity at all times. During purely rotational motion, all points on the body change position except for those lying on the instantaneous axis of rotation. The relationship between orientation and angular velocity is not directly analogous to the relationship between position and velocity. Angular velocity is not the time rate of change of orientation, because there is no such concept as an orientation vector that can be differentiated to obtain the angular velocity.
## Kinematical equations
### Addition theorem for angular velocity
The angular velocity of a rigid body B in a reference frame N is equal to the sum of the angular velocity of a rigid body D in N and the angular velocity of B with respect to D:[5]
${\displaystyle {}^{\mathrm {N} }\!{\boldsymbol {\omega }}^{\mathrm {B} }={}^{\mathrm {N} }\!{\boldsymbol {\omega }}^{\mathrm {D} }+{}^{\mathrm {D} }\!{\boldsymbol {\omega }}^{\mathrm {B} }.}$
In this case, rigid bodies and reference frames are indistinguishable and completely interchangeable.
### Addition theorem for position
For any set of three points P, Q, and R, the position vector from P to R is the sum of the position vector from P to Q and the position vector from Q to R:
${\displaystyle \mathbf {r} ^{\mathrm {PR} }=\mathbf {r} ^{\mathrm {PQ} }+\mathbf {r} ^{\mathrm {QR} }.}$
The norm of a position vector is the spatial distance. Here the coordinates of all three vectors must be expressed in coordinate frames with the same orientation.
### Mathematical definition of velocity
The velocity of point P in reference frame N is defined as the time derivative in N of the position vector from O to P:[6]
${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {P} }={\frac {{}^{\mathrm {N} }\mathrm {d} }{\mathrm {d} t}}(\mathbf {r} ^{\mathrm {OP} })}$
where O is any arbitrary point fixed in reference frame N, and the N to the left of the d/dt operator indicates that the derivative is taken in reference frame N. The result is independent of the selection of O so long as O is fixed in N.
### Mathematical definition of acceleration
The acceleration of point P in reference frame N is defined as the time derivative in N of its velocity:[6]
${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {P} }={\frac {^{\mathrm {N} }\mathrm {d} }{\mathrm {d} t}}({}^{\mathrm {N} }\mathbf {v} ^{\mathrm {P} }).}$
### Velocity of two points fixed on a rigid body
For two points P and Q that are fixed on a rigid body B, where B has an angular velocity ${\displaystyle \scriptstyle {^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }}}$ in the reference frame N, the velocity of Q in N can be expressed as a function of the velocity of P in N:[7]
${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {Q} }={}^{\mathrm {N} }\!\mathbf {v} ^{\mathrm {P} }+{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }.}$
where ${\displaystyle \mathbf {r} ^{\mathrm {PQ} }}$ is the position vector from P to Q.,[7] with coordinates expressed in N (or a frame with the same orientation as N.) This relation can be derived from the temporal invariance of the norm distance between P and Q.
### Acceleration of two points fixed on a rigid body
By differentiating the equation for the Velocity of two points fixed on a rigid body in N with respect to time, the acceleration in reference frame N of a point Q fixed on a rigid body B can be expressed as
${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {Q} }={}^{\mathrm {N} }\mathbf {a} ^{\mathrm {P} }+{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \left({}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }\right)+{}^{\mathrm {N} }{\boldsymbol {\alpha }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }}$
where ${\displaystyle \scriptstyle {{}^{\mathrm {N} }\!{\boldsymbol {\alpha }}^{\mathrm {B} }}}$ is the angular acceleration of B in the reference frame N.[7]
### Angular velocity and acceleration of two points fixed on a rigid body
As mentioned above, all points on a rigid body B have the same angular velocity ${\displaystyle {}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }}$ in a fixed reference frame N, and thus the same angular acceleration ${\displaystyle {}^{\mathrm {N} }{\boldsymbol {\alpha }}^{\mathrm {B} }.}$
### Velocity of one point moving on a rigid body
If the point R is moving in the rigid body B while B moves in reference frame N, then the velocity of R in N is
${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {R} }={}^{\mathrm {N} }\mathbf {v} ^{\mathrm {Q} }+{}^{\mathrm {B} }\mathbf {v} ^{\mathrm {R} }}$
where Q is the point fixed in B that is instantaneously coincident with R at the instant of interest.[8] This relation is often combined with the relation for the Velocity of two points fixed on a rigid body.
### Acceleration of one point moving on a rigid body
The acceleration in reference frame N of the point R moving in body B while B is moving in frame N is given by
${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {R} }={}^{\mathrm {N} }\mathbf {a} ^{\mathrm {Q} }+{}^{\mathrm {B} }\mathbf {a} ^{\mathrm {R} }+2{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times {}^{\mathrm {B} }\mathbf {v} ^{\mathrm {R} }}$
where Q is the point fixed in B that instantaneously coincident with R at the instant of interest.[8] This equation is often combined with Acceleration of two points fixed on a rigid body.
### Other quantities
If C is the origin of a local coordinate system L, attached to the body, the spatial or twist acceleration of a rigid body is defined as the spatial acceleration of C (as opposed to material acceleration above): ${\displaystyle {\boldsymbol {\psi }}(t,\mathbf {r} _{0})=\mathbf {a} (t,\mathbf {r} _{0})-{\boldsymbol {\omega }}(t)\times \mathbf {v} (t,\mathbf {r} _{0})={\boldsymbol {\psi }}_{c}(t)+{\boldsymbol {\alpha }}(t)\times A(t)\mathbf {r} _{0}}$ where
• ${\displaystyle \mathbf {r} _{0}}$ represents the position of the point/particle with respect to the reference point of the body in terms of the local coordinate system L (the rigidity of the body means that this does not depend on time)
• ${\displaystyle A(t)\,}$ is the orientation matrix, an orthogonal matrix with determinant 1, representing the orientation (angular position) of the local coordinate system L, with respect to the arbitrary reference orientation of another coordinate system G. Think of this matrix as three orthogonal unit vectors, one in each column, which define the orientation of the axes of L with respect to G.
• ${\displaystyle {\boldsymbol {\omega }}(t)}$ represents the angular velocity of the rigid body
• ${\displaystyle \mathbf {v} (t,\mathbf {r} _{0})}$ represents the total velocity of the point/particle
• ${\displaystyle \mathbf {a} (t,\mathbf {r} _{0})}$ represents the total acceleration of the point/particle
• ${\displaystyle {\boldsymbol {\alpha }}(t)}$ represents the angular acceleration of the rigid body
• ${\displaystyle {\boldsymbol {\psi }}(t,\mathbf {r} _{0})}$ represents the spatial acceleration of the point/particle
• ${\displaystyle {\boldsymbol {\psi }}_{c}(t)}$ represents the spatial acceleration of the rigid body (i.e. the spatial acceleration of the origin of L).
In 2D, the angular velocity is a scalar, and matrix A(t) simply represents a rotation in the xy-plane by an angle which is the integral of the angular velocity over time.
Vehicles, walking people, etc., usually rotate according to changes in the direction of the velocity: they move forward with respect to their own orientation. Then, if the body follows a closed orbit in a plane, the angular velocity integrated over a time interval in which the orbit is completed once, is an integer times 360°. This integer is the winding number with respect to the origin of the velocity. Compare the amount of rotation associated with the vertices of a polygon.
## Kinetics
Any point that is rigidly connected to the body can be used as reference point (origin of coordinate system L) to describe the linear motion of the body (the linear position, velocity and acceleration vectors depend on the choice).
However, depending on the application, a convenient choice may be:
• the center of mass of the whole system, which generally has the simplest motion for a body moving freely in space;
• a point such that the translational motion is zero or simplified, e.g. on an axle or hinge, at the center of a ball and socket joint, etc.
When the center of mass is used as reference point:
• The (linear) momentum is independent of the rotational motion. At any time it is equal to the total mass of the rigid body times the translational velocity.
• The angular momentum with respect to the center of mass is the same as without translation: at any time it is equal to the inertia tensor times the angular velocity. When the angular velocity is expressed with respect to a coordinate system coinciding with the principal axes of the body, each component of the angular momentum is a product of a moment of inertia (a principal value of the inertia tensor) times the corresponding component of the angular velocity; the torque is the inertia tensor times the angular acceleration.
• Possible motions in the absence of external forces are translation with constant velocity, steady rotation about a fixed principal axis, and also torque-free precession.
• The net external force on the rigid body is always equal to the total mass times the translational acceleration (i.e., Newton's second law holds for the translational motion, even when the net external torque is nonzero, and/or the body rotates).
• The total kinetic energy is simply the sum of translational and rotational energy.
## Geometry
Two rigid bodies are said to be different (not copies) if there is no proper rotation from one to the other. A rigid body is called chiral if its mirror image is different in that sense, i.e., if it has either no symmetry or its symmetry group contains only proper rotations. In the opposite case an object is called achiral: the mirror image is a copy, not a different object. Such an object may have a symmetry plane, but not necessarily: there may also be a plane of reflection with respect to which the image of the object is a rotated version. The latter applies for S2n, of which the case n = 1 is inversion symmetry.
For a (rigid) rectangular transparent sheet, inversion symmetry corresponds to having on one side an image without rotational symmetry and on the other side an image such that what shines through is the image at the top side, upside down. We can distinguish two cases:
• the sheet surface with the image is not symmetric - in this case the two sides are different, but the mirror image of the object is the same, after a rotation by 180° about the axis perpendicular to the mirror plane.
• the sheet surface with the image has a symmetry axis - in this case the two sides are the same, and the mirror image of the object is also the same, again after a rotation by 180° about the axis perpendicular to the mirror plane.
A sheet with a through and through image is achiral. We can distinguish again two cases:
• the sheet surface with the image has no symmetry axis - the two sides are different
• the sheet surface with the image has a symmetry axis - the two sides are the same
## Configuration space
The configuration space of a rigid body with one point fixed (i.e., a body with zero translational motion) is given by the underlying manifold of the rotation group SO(3). The configuration space of a nonfixed (with non-zero translational motion) rigid body is E+(3), the subgroup of direct isometries of the Euclidean group in three dimensions (combinations of translations and rotations).
## Notes
1. ^ Lorenzo Sciavicco, Bruno Siciliano (2000). "§2.4.2 Roll-pitch-yaw angles". Modelling and control of robot manipulators (2nd ed.). Springer. p. 32. ISBN 1-85233-221-2.
2. ^ Andy Ruina and Rudra Pratap (2015). Introduction to Statics and Dynamics. Oxford University Press. (link: [1])
3. ^ In general, the position of a point or particle is also known, in physics, as linear position, as opposed to the angular position of a line, or line segment (e.g., in circular motion, the "radius" joining the rotating point with the center of rotation), or basis set, or coordinate system.
4. ^ In kinematics, linear means "along a straight or curved line" (the path of the particle in space). In mathematics, however, linear has a different meaning. In both contexts, the word "linear" is related to the word "line". In mathematics, a line is often defined as a straight curve. For those who adopt this definition, a curve can be straight, and curved lines are not supposed to exist. In kinematics, the term line is used as a synonym of the term trajectory, or path (namely, it has the same non-restricted meaning as that given, in mathematics, to the word curve). In short, both straight and curved lines are supposed to exist. In kinematics and dynamics, the following words refer to the same non-restricted meaning of the term "line":
• "linear" (= along a straight or curved line),
• "rectilinear" (= along a straight line, from Latin rectus = straight, and linere = spread),
• "curvilinear" (=along a curved line, from Latin curvus = curved, and linere = spread).
In topology and meteorology, the term "line" has the same meaning; namely, a contour line is a curve.
5. ^ Kane, Thomas; Levinson, David (1996). "2-4 Auxiliary Reference Frames". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc.
6. ^ a b Kane, Thomas; Levinson, David (1996). "2-6 Velocity and Acceleration". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc.
7. ^ a b c Kane, Thomas; Levinson, David (1996). "2-7 Two Points Fixed on a Rigid Body". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc.
8. ^ a b Kane, Thomas; Levinson, David (1996). "2-8 One Point Moving on a Rigid Body". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc.
## References
• Roy Featherstone (1987). Robot Dynamics Algorithms. Springer. ISBN 0-89838-230-0. This reference effectively combines screw theory with rigid body dynamics for robotic applications. The author also chooses to use spatial accelerations extensively in place of material accelerations as they simplify the equations and allow for compact notation.
• JPL DARTS page has a section on spatial operator algebra (link: [2]) as well as an extensive list of references (link: [3]).
• Andy Ruina and Rudra Pratap (2015). Introduction to Statics and Dynamics. Oxford University Press. (link: [4]).
• Prof. Dr. Dennis M. Kochmann, Dynamics Lecture Notes, ETH Zurich. [5] | 4,614 | 19,544 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "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} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.938636 |
https://en.khanacademy.org/math/statistics-probability/counting-permutations-and-combinations/combinatorics-probability/v/permutations-and-combinations-4 | 1,686,266,422,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00600.warc.gz | 268,828,194 | 99,393 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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# Example: Lottery probability
What is the probability of winning a 4-number lottery? Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• arent there 4! ways we can write the winning numbers . i mean the order doesnt matter so 3,15,46,49 should be the same as 15,3,46,49 but sal says that theres only one way of getting the correct lottery numbers why is that?
• Achu and Naveen,
Sal could have solved this problem in two ways. He could have taken the number of possible permutations with a favorable outcome and divided that by the total possible number of permutations –or—he could have taken the number of possible combinations with a favorable outcome and divided that by the total number of possible combinations (which is what he did).
Let us do it both ways, using the permutations first.
As you mentioned, there a 4! ways of writing the four numbers. Another way to say this is that there are 4! different ways to order the four numbers –or-- there are 4! different permutations of the four numbers that give us the favorable outcome. This can be written as 4*3*2*1.
The total number of different permutations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!), which can be written as 60*59*58*57.
Our answer using permutations would be the number of favorable outcomes/the number of possible outcomes which would be (4*3*2*1)/(60*59*58*57). This simplifies to 1/487,635.
Using combinations, there is only one (1) combination of numbers that gives us that favorable outcome (that one way, achu).
The number of possible combinations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!*4!). This can be written as (60*59*58*57)/(4*3*2*1) . This number is 487,635.
Our answer using combinations would be the number of favorable outcomes/the number of possible outcomes which would be 1/487,635. This is the same answer we got using permutations.
Consider combinations and permutations to be different “units”. You would not say that after moving 3 inches you are halfway to traveling 6 meters, even though 3/6 = ½. You cannot take the number of possible favorable permutation outcomes and divide that by the number of total possible combination outcomes and get the correct probability. Likewise you cannot take the number of possible favorable combination outcomes and divide that by the number of total possible permutation outcomes and get the correct probability.
• Is there any reason why I could not solve the problem this way? Because I did and it turned out ok, but I don't always trust my own leaps of logic:
If I am hoping to draw 4 particular numbers randomly out of 60, then I can say that on my first draw there are 4 numbers that I could hope to get, out of a total of 60, so I begin with
4/60 as my chances of getting one of those numbers on that first draw.
On my second draw, there are now 3 numbers left that I want, out of a possible 59 total remaining, so I begin to multiply:
(4/60) * (3/59)
The last two draws follow logically: 2 acceptable numbers out of 58, and then 1 acceptable number out of 57. So the probability of drawing any particular set of 4 numbers out of 60, if we cannot draw any number twice, is:
(4/60) * (3/59) * (2/58) * (1/57)
I can just calculate this and get the same answer that Sal gets, but what do I make of the fact that this expression is equal to
4! / (60! / (60-4)!)
or
4! / P(60,4)
I am trying to draw some sort of larger logical conclusion out of this, but am having some trouble not confusing myself. It seems that I could say this represents the number of possible arrangements of any particular set of 4 numbers distributed over, or divided by, the total number of possible ways to draw all sets of 4 numbers out of 60.
This makes some sense to me, but I tend to come at these things backwards, using intuition and numbers first and then going back over the solution to see whether or not I could claim any larger guiding principles and/or formulas. Can anyone do a better job of explaining why my intuition was right?
• This sounds like a tautology but your intuition is right because it is right.
Let's look at the different ways of arriving to the answer.
Your way: look at the situation one draw at a time.
4/60 * 3/59 * 2/58 * 1/57, which is equal to 4*3*2*1 / (60*59*58*57)
Sal's way:
4! / (60 P 4)
which is the same as
4! / (60!/(60-4)!)
which is the same as
4! * (60-4)! / 60!
which is the same as
4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1)
which is the same as
4*3*2*1 / (60*59*58*57)
So your and Sal's ways are the same thing just expressed in different ways.
Here's one more way to look at it:
There are (60 C 4) ways of choosing four numbers (all different) from 60. Only one combination of them is what we have chosen. So the probability is
1/(60 C 4)
which is the same as
1/( 60!/(4!*(60-4)!))
which is the same as
4! * (60-4)! / 60!
which, as we already have seen, is the same as
4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1)
which is the same as
4*3*2*1 / (60*59*58*57)
• If S=1+2+4+8+16+32................ Can this be taken as S=1+2(1+2+4+8+16.......)???
If it can be taken,
S=1+2S
S=-1 Is this a valid answer for such a big question???
• Your reasoning only works when the sum S is a real number and does not continue on to infinity. For example, if S = 1 + 1/2 + 1/4 + ... + 1/(2^n) + ... and so on forever, then your logic says that S = 1 + 1/2(S), which gives the right answer of S = 2.
The difference between the two problems is that the sum S = 1 + 2 + 4 + ... continues to get larger forever and does not stop, while S = 1 + 1/2 + 1/4 + ... has a limit that it can never reach, no matter how long you count. By taking your approach, you can short-cut the infinite series and figure out that limit without having to take forever (literally!).
• just wanted to add my 2 cents. I'm having a hard time explaining it all though so would love feedback. For the lottery question, another way to think of it is as below.
explanation:
think of this top part of the probability (numerator) as 4p4 since you have 4 numbers to pick from and you want to pick 4 numbers, the number of ways you can pick 4 numbers from 4 numbers is 4*3*2*1. 4p4. This gives you the total number of non-unqiue ways to choose these numbers.
on the bottom, you know that you have 4 numbers to pick and 60 to pick from. your total number of ways to pick 4 numbers from 60 is 60p4.
I know that we dont care about order, but you see when you look for probability, the probability of the order not mattering is the same as the probability of the order mattering. because the numerator and denominator increase by factors of 4!. since when you think about it, the numerator is the number of ways total that you can combine 4 numbers divided by the denominator which is the total number of ways to pick 4 numbers of 60. Both top and bottom are including all the different permutations of 4 numbers. so really it's including those groupings of duplicated values. individually looking at each unique grouping's probability is the same as looking at the probability across all groupings.
Alternate way:
4c4/60c4 - removes the order. same answer.
QUESTION: what if we're drawing the numbers, what is the probability of drawing these 4 numbers. Does the math change / probability change?
Draw 1: 4 c 1
Draw 2: 3 c 1
Draw 3: 2 c 1
Draw 4: 1 c 1
fundamental counting principle lets us rep drawing 1,2,3,4 as 4c1*3c1*2c1*1c1
on the bottom you total ways of drawing per draw.
Draw 1: total ways = 60 c 1
Draw 2: 59 c 1
Draw 3: 58 c 1
Draw 4: 57 c 1
the probability of drawing is top / bottom.
QUESTION: should i use P or C? logically speaking which one makes more sense. I know they yield the same answer but which one fits this equation / changes the equation's logic.
Also, does this any of what i wrote up apply beyond jsut this question? Is this some unique solution methodology.
is it a rule that says you can't divide a combination by permutation?
e.g. 4c4/60p4. not allowed
because i see a lot of people wondering about 1/60*1/59*1/58*1/57 as a possible answer. My response is the top only shows 1 permutation of arranging the 4 numbers and the bottom represents the total number of ways of picking 4 numbers. so it's like 4c4/60p4 which incorrectly represents the solution. I think it's actually ok, its just not ok for this question.
because the top yes can be 4c4, it represents 1 but really depending on the person answering. they could be incorrect due to 2 different reasons. 1 not realizing that the order doesn't matter so their method of only using 1 / 60 etc only calculates the probability of 1 method of arranging those 60 and then also not realizing that the bottom looks at all possible permutation. OR they are looking at the top correctly as a single combination and not evaluating the bottom correclty and failing to use also combinations below.
Apologies for so many thoughts, probabilities is really interesting and difficult for me and writing up my thoughts helps me. I hope it helps someone else too.
• As long as you’re consistent, you will get the correct answer. It makes more sense to use the permutation method (for both top and bottom) if you think of the numbers as picked one at a time, but it makes more sense to use the combination method (for both top and bottom) if you think of all four numbers as being picked at once.
(1 vote)
• I was just wondering what the nCr and nPr buttons on the calculator do. My teacher explaned it, but i forgot what the do and how to use them. Please help!
(1 vote)
• nCr is used for Combinations, while nPr is used in permutations. 'N' represents the total number of items you have to choose from, and 'R' represents the number you choose. So if you had 36C10, that would mean you have 36 items and you can choose 10, regardless of order, since it is a Combination.
• Isn't 59 factorial (!), 60*59*58*57*56*all the way down to 0??
(1 vote)
• No, there's no 60 or 0 involved. It's 59 through 1.
• if in this lottery, picking a number and putting it back is allowed... so that means you can pick a number a multiple of times... what would the probability be then?
64! / ((64-4!) * 4! ) ?
• Well, you'd choose 4 numbers from 60 numbers (1 to 60) and repetition is allowed, the probability of winning would be 1/(60^4/4!) = 4!/60^4 = 1/540000 ≈ 0.000002
(The 4! is there because the order of the numbers still doesn't matter. If the winning row was 7,34,34,50 and you had 34,7,50,34 you'd still win)
• I think I may have a fundamental misunderstanding of combinations and / or permutations. I tried to solve this problem by doing the following (60! / (56! * 4!)) / (60^4) which is the combinations formula divided by (I thought) the total number of possible outcomes with 60 numbers in 4 slots. Why is that incorrect?
• 60^4 isn't the total number of possible groups of 4, because the order of the 4 numbers doesn't matter for combinations. 60^4 is the number of permutations, not combinations.
and 60!/(56!*4!) is the total number of possible combinations of 4 numbers, so it is the sample space, not the # of successes meaning it should be the denominator.
• what if you want to know the probability of a number winning excluding some number already played that will not be played again?
• That's a fun calculation. Say you have 7 different items in a bag. The probability of pulling a certain one out is 1/7. After doing so you now have 6 items. The next time you pull one out the probability will be 1/6. Makes sense?
This reasoning is used in many situations where multiple items need to be pulled out of the bag in quick succession.
(1 vote)
• In the same Sal's case - 4 number ticket, let's assume that if I pay a little bit more for the lottery ticket I'm able to pick 5 numbers instead of 4. Then I would have a higher probability of winning. But how do I calculate that probability with an "extra number"?
(1 vote)
• Let find the easy part, which is the total # of outcomes. This is just going to be the number of different ways in which the host of the lottery prize is going to choose 4 numbers out of 60. This is just C(60,4).
Now the # of ways in which our event can happen is just the number of ways in which those 4 numbers match exactly 4 of our 5 numbers. i.e. 4 out 5 numbers: C(5,4).
Thus the probability is:
5/487,635 = 1/97,527
And it makes sense. If you paid the number 1, 2, 3 ,4 and 5, you will win in these situations:
[1 - 2 - 3 - 4]
[1 - 2 - 3 - 5]
[1 - 2 - 4 -5]
[1 - 3 - 4 -5]
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# If x is a positive number
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26 Aug 2012, 06:24
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If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
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26 Aug 2012, 06:42
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If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
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26 Aug 2012, 09:13
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syog wrote:
If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
Neither (1) nor (2) alone is sufficient.
(1) and (2) together:
From (1) $$3x=2k,$$ where $$k$$ is a positive integer.
From (2) $$5x=2m,$$ where $$m$$ is some positive integer. Necessarily $$\frac{2k}{3}=\frac{2m}{5}$$ from which $$5k=3m.$$
$$k$$ and $$m$$ being integers, necessarily $$k$$ must be a multiple of 3 (because 5 is not divisible by 3), so $$k=3a$$ for some positive integer $$a.$$
It follows that $$x=\frac{2k}{3}=2a$$ so $$x$$ is even.
Sufficient.
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30 Aug 2012, 04:53
Bunuel wrote:
If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer?
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30 Aug 2012, 05:18
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ziko wrote:
Bunuel wrote:
If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer?
I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions.
So, there is no interchangeable word for "integer" on the GMAT.
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30 Aug 2012, 19:50
great job
great way to solve this problem..
i didnt reach the answer as you did
ds has been troubling me like anything
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Re: If x is a positive number [#permalink]
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02 Sep 2013, 08:23
Bunuel wrote:
ziko wrote:
Bunuel wrote:
If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer?
I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions.
So, there is no interchangeable word for "integer" on the GMAT.
I am a little confused.
1) 3x is an even integer.
Let x=4/3; then 3x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
2) 5x is an even integer.
Let x=4/5; then 5x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
(1+2): 15x is an integer.
Let x=4/15; then 15x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
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Re: If x is a positive number [#permalink]
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02 Sep 2013, 08:42
domfrancondumas wrote:
Bunuel wrote:
ziko wrote:
If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer?
I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions.
So, there is no interchangeable word for "integer" on the GMAT.
I am a little confused.
1) 3x is an even integer.
Let x=4/3; then 3x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
2) 5x is an even integer.
Let x=4/5; then 5x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
(1+2): 15x is an integer.
Let x=4/15; then 15x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
Notice that x cannot be 4/15, because in this case 3x=12/15 which is NO an even integer and 5x=20/15 which is also NOT an even integer, so in this case both statements are violated.
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Re: If x is a positive number [#permalink]
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10 Jan 2017, 07:31
If it's given that "5 is a factor of x". Is it correct to assume that x is an integer? If yes, how and why? Please explain. Also explain if the answer to my question is No. Consider a case wherein x is 5/2. Isn't 5 a factor of x? Please address, thanks a lot in advance
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Re: If x is a positive number [#permalink]
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10 Jan 2017, 10:12
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kooks123 wrote:
If it's given that "5 is a factor of x". Is it correct to assume that x is an integer? If yes, how and why? Please explain. Also explain if the answer to my question is No. Consider a case wherein x is 5/2. Isn't 5 a factor of x? Please address, thanks a lot in advance
On the GMAT when we are told that $$a$$ is divisible by $$b$$ (or which is the same: "$$a$$ is multiple of $$b$$", or "$$b$$ is a factor of $$a$$"), we can say that:
1. $$a$$ is an integer;
2. $$b$$ is an integer;
3. $$\frac{a}{b}=integer$$.
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Re: If x is a positive number [#permalink]
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11 Jan 2017, 03:48
x is positive Integer means we need to think only in positive fractions and integers.
(1) 3x is an even integer.
Let x= 2.....3x=6..........x is even integer
Let x=2/3....3x=2.........x is NOT even integer
Insufficient
(2) 5x is an even integer.
Let x= 2........5x=10..........x is even integer
Let x=2/5.......5x=2...........x is NOT even integer
Insufficient
Combining 1 +2 ,
There is not fraction could be multiplied simultaneously to 3 & 5 and give EVEN INTEGER. We left with that x=EVEN INTEGER to give make both 3x & 5x even integers.
As 3 & 5 are odd numbers then x needs to be EVEN Integer to make both statements Even integers.
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Re: If x is a positive number [#permalink]
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11 Jan 2017, 12:01
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Hi Ellipse,
Whenever you combine two statements in a DS question, the best thing to do instead of plugging in values is to use one statement into the other or use a mathematical operation between the two statements. The mathematical operation(s) that you choose to perform must lead to your target question. In this case the target question is 'Is x an even integer'?
Statement 1 : 3x is an even integer
Here x can either be an even integer such as 2, 4, 6.... or it can be a fraction such as 2/3, 4/3..... So x can be an even integer or a fraction. Insufficient.
Statement 2 : 5x is an even integer
Here again x can be an even integer such as 2, 4, 6.... or it can be a fraction such as 2/5, 4/5.... So x again can either be an even integer or a fraction. Insufficient.
Now instead of recycling the values it makes sense to use a mathematical operation (or mathematical operations) between the two statements.
Let us multiply the first statement by 2, since 3x = even integer ; 3x * 2 = even integer * 2 -----> 6x = even integer
Statement 2 says that 5x is an even integer and by multiplying statement 1 by 2 we have 6x to be an even integer. Subtracting the two we get
6x - 5x = even integer - even integer -----> x = even integer. Sufficient.
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Re: If x is a positive number [#permalink]
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19 Apr 2017, 17:56
Ellipse wrote:
If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
Statement 1:
3 (2/3) could be an even number- the questin says x is a positive number...not necessarily an integer
Insufficient
Statement 2
5(2/5) btw again here were are using a counterexample to establish that x does not necessarily have to be an integer
10/5 =2
Insufficient
Statement 1 and 2:
Statement 1 and Statement 2
Knowing from statement 1 and 2 that 3x and 5x are even integers we can use solve this question using algebra instead of imagining numbers and testing various cases- an even integer minus an even integer is always an even integer so
5x-3x=2x
5(4)-3(4)=2(4)
Re: If x is a positive number [#permalink] 19 Apr 2017, 17:56
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https://yozh.org/2015/10/14/nowhere_differentiable_functions/ | 1,718,574,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00672.warc.gz | 973,295,133 | 12,609 | # Nowhere Differentiable Functions
In an undergraduate analysis class, one of the first results that is generally proved after the definition of differentiability is given is the fact that differentiable functions are continuous. We can justifiably ask if the converse holds—are there examples of functions that are continuous but not differentiable? If such examples exist, how “bad” can they be?
Many examples of continuous functions that are not differentiable spring to mind immediately: the absolute value function is not differentiable at zero; a sawtooth wave is not differentiable anywhere that it changes direction; the Cantor function1I’ll (hopefully) talk more about this later. is an example of a continous function that is not differentiable on an uncountable set, though it does remain differentiable “almost everywhere.”
The goal is to show that there exist functions that are continuous, but that are nowhere differentiable. In fact, what we actually show is that the collection of such functions is, in some sense, quite large and that the set of functions that are differentiable—even if only at a single point—is quite small. First, we need a definition and an important result.
### A Little Theory
Definition: Let $$X$$ be a complete metric space,2A metric space is a set of points and a way of measuring the distance between those points. A sequence of points in a metric space is said to be Cauchy if the distance between any two (not necessarily consecutive) points in the sequence gets small for points sufficiently deep into the sequence. A metric space is said to be complete if every Cauchy sequence converges to some point in the space. The real numbers are a complete metric space, but the rational numbers are not. (Why?) and let $$M \subseteq X$$. Then $$M$$ is said to be
• nowhere dense in $$X$$ if the closure of $$M$$ has empty interior;3This is a topological notion. In very broad strokes, if $$A$$ is a subset of $$X$$, then a point $$x$$ is in the closure of $$A$$ if we can find points of $$A$$ that are arbitrarily “close” to $$x$$. $$A$$ is said to have empty interior if all of the points in $$A$$ are arbitrarily “close” to points that are not in $$A$$. For instance, the closure of the interval $$(0,1)$$ in $$\mathbb{R}$$ is the interval $$[0,1]$$, and any finite collection of points in $$\mathbb{R}$$ is nowhere dense. (Why?)
• meager in $$X$$ if it is the union of countably many nowhere dense sets; and
• residual in $$X$$ if it is not meager (i.e. if it is the complement of a meager set.
This definition provides a topological notion of what it means for a subset of a metric space to be “small.” Nowhere dense sets are tiny—almost insignificant—subsets, while residual sets are quite large (relative to the ambient space). It is also worth noting that meager sets were originally called “sets of the first category,” and residual sets were originally called “sets of the second category,” leading to the name of the following theorem:
Baire’s Category Theorem: If a metric space $$X\ne\emptyset$$ is complete, then it is residual in itself.
Proof: Suppose for contradiction that $$X$$ is meager in itself. Then we may write $X = \bigcup_{k=1}^{\infty} M_k,$ where each $$M_k$$ is nowhere dense. As $$M_1$$ is nowhere dense in $$X$$, its closure has empty interior, and therefore contains no nonempty open sets. But $$X$$ does contain at least one nonempty open set—$$X$$ itself. Hence $$\overline{M}_1\ne X$$ and so, since $$\overline{M}_1$$ is closed, its complement is both open and nonempty. Choose some $$p_1\in X\setminus \overline{M}_1$$. Since $$X\setminus \overline{M}_1$$ is open, there exists some $$\varepsilon_1 \in (0,\frac{1}{2})$$ such that $$B(p_1,\varepsilon_1)\subseteq X\setminus \overline{M}_1$$. Let $$B_1 := B(p_1,\varepsilon_1)$$.
Now consider $$\overline{M}_2$$. It also has empty interior, and so it contains no open balls. In particular, it does not contain $$B_1$$. But then $$B_1\setminus\overline{M}_2$$ is open. Let $$p_2 \in B_1\setminus\overline{M}_2$$ and choose $$\varepsilon_2 < \frac{1}{2}\varepsilon_1$$ such that $$B_2 := B(p_2,\varepsilon_2) \subseteq B_1\setminus\overline{M}_2$$.
Continue this process by induction. That is, for each $$k\in\mathbb{N}$$, choose $$B_{k+1}$$ to be an open ball of radius $$\varepsilon_{k+1} < \frac{1}{2}\varepsilon_k$$ such that $$B_{k+1} \subseteq B_k \setminus \overline{M}_{k+1}$$. By this construction we have $$\varepsilon_k < 2^{-k}$$ for each $$k$$. In particular we have from the triangle inequality that $$d(p_m,p_n) \le 2^{-k}$$ for all $$m,n > k$$, as $$B_m,B_n\subseteq B_{k+1}$$. Hence the sequence of points $$(p_k)$$ is Cauchy in $$X$$. Since $$X$$ is complete, there exists some $$p\in X$$ such that $$p_k\to p$$. But for any $$k$$ the point $$p$$ is contained in $$B_k$$, which implies that $$p\not\in M_k$$ for all $$k$$. Hence we have found a point $$p\in X$$ such that $$p\not\in \bigcup M_k = X$$, which is a contradiction.
### The Existence of Nowhere Differentiable Continuous Functions
We now get to the main result, which has appeared on qualifying exams a few times in the past:
Exercise: Use Baire’s category theorem to prove the existence of continuous, nowhere differentiable functions on the unit interval.
Solution: For each natural number $$n$$, define the set $E_n := \{ f\in C([0,1]) : \exists x_0\in[0,1] \text{ s.t. } |f(x)-f(x_0)| \le n|x-x_0| \forall x\in[0,1]\}.$ This is rather a lot of notation. Let’s try to unpack it just a bit: the derivative of $$f$$ is defined by the limit $f'(x_0) = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}.$ If this limit exists, then near $$x_0$$ the difference quotient must be bounded by some number, say $$L_1$$. Away from $$x_0$$, the uniform continuity of $$f$$ ensures that the difference quotient is bounded by some $$L_2$$ when $$x$$ is far from $$x_0$$. Taking the larger of these two bounds, we have that if $$f$$ is differentiable at some point $$x_0$$, then $\frac{|f(x)-f(x_0)|}{|x-x_0|} \le \max\{L_1,L_2\}.$ Thus if a function $$f$$ is differentiable at any point $$x_0\in[0,1]$$, then $$f$$ lives in $$E_n$$ for some value of $$n$$.
What we want to show is that each $$E_n$$ is nowhere dense in space of continous functions $$C[0,1]$$ (which is a normed vector space with respect to the uniform norm). Since all of the differentiable functions are contained in $$\bigcup_n E_n$$, it would then follow that the set of differentiable functions is contained in a meager subset of $$C[0,1]$$. From Baire’s category theorem, we could then conclude that nowhere differentiable functions exist and, indeed, that there is a residual set of nowhere differentiable functions.
To show that each $$E_n$$ is nowhere dense, we have to show that the closure of each $$E_n$$ has empty interior or, equivalently, if we have an arbitrary function in $$\overline{E}_n$$, we need to find a continuous function that is “close” to $$f$$ in the uniform norm, but which is not contained in $$\overline{E}_n$$.
So, what is $$\overline{E}_n$$? We claim that it is just $$E_n$$. To see this, suppose that $$\{f_k\}$$ is a Cauchy sequence in $$E_n$$. First, note that $$C[0,1]$$ is complete, hence there is some $$f\in C[0,1]$$ such that $$f_k\to f$$. Now, since each $$f_k$$\in E_n, for each $$k$$ there is some $$x_k$$ such that $\frac{|f(x)-f(x_k)|}{|x-x_k|} \le n \quad\forall x\in[0,1].$ But then the sequence of numbers $$x_k$$ is a sequence in $$[0,1]$$. By the Bolzano-Weierstrass theorem, this sequence has a convergent subsequence, say $$x_{k_j} \to x\in[0,1]$$. Hence by the uniform convergence of $$f_k$$ to $$f$$, we have $\frac{|f(x)-f(x_{k_j})|}{|x-x_{k_j}|} = \lim_{j\to\infty} \frac{|f_{k_j}(x)-f_{k_j}(x_{k_j})|}{|x-x_{k_j}|} \le n.$ Therefore $$f \in E_n$$, and so Cauchy sequences in $$E_n$$ converge in $$E_n$$, which shows that $$E_n$$ is closed.
Now, given a function $$f\in E_n$$ how do we find a function $$g$$ that is “close” to $$f$$, but not in $$E_n$$? That is actually somewhat delicate, and is dealt with in the following lemma:
Lemma: Given $$f\in C[0,1]$$, $$n\in\mathbb{N}$$, and $$\varepsilon > 0$$, there exists a piecewise linear function $$g$$ with only finitely many linear pieces such that each linear piece has slope $$\pm 2n$$ and $$\|g-f\|_{u} < \varepsilon$$.
Proof: Since $$f$$ is uniformly continuous, there exists a $$\delta$$ such that for any $$x,y\in[0,1]$$, if $$|x-y|< \delta$$, then $$|f(x)-f(y)|<\varepsilon/2$$. Choose $$m\in\mathbb{N}$$ such that $$m > 1/\delta$$. On the interval $$[0,1/m]$$, define $$g$$ as follows: define the first linear piece of $$g$$ by setting $$g(0) = f(0)$$ and giving it slope $$2n$$ on the interval $$[0,\varepsilon/2n]$$. On the interval $$[\varepsilon/2n, 2\varepsilon/2n]$$, let $$g$$ have slope $$-2n$$. Continue in the manner until the linear piece that intersects a line of slope $$\pm 2n$$ through the point $$(1/m,f(1/m))$$ is constructed, and take $$g$$ to be equal to that linear function from the point of intersection to $$1/m$$.
Continue this procedure for each interval of the form $$(k/m,(k+1)/m$$ for $$k=1,2,\ldots,m-1$$. That is, set $$g(k/m) = f(k/m)$$ and construct a sawtooth function on the given interval with slope $$\pm 2n$$. We claim that $$g$$ is a function of the type desired.
We first note that $$g$$ is piecewise linear, with each piece having slope $$\pm 2n$$—this is explicit in the construction. Moreover, there are only finitely many pieces, since the unit interval was broken into $$m$$ subintervals, and each subinterval contains only finitely many linear pieces.4This follows from the Archimedean principle—exact bounds on the number of pieces can be computed, but such a computation is tedious and we are, frankly, a bit lazy. Finally, it follows from the choice of $$\delta$$ and $$m$$, and the triangle inequality that for each $$x\in [0,1]$$, we have $|g(x) – f(x)| \le \left|g(x) – f\left(\frac{\lfloor mx \rfloor}{m}\right)\right| + \left|f\left(\frac{\lfloor mx \rfloor}{m}\right) – f(x)\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$ Hence $$\|g-f\|_u < \varepsilon$$.
Therefore $$g$$ is exactly the kind of function that we want, and so the proof is complete.
With the lemma proved, we now have the following: if $$f\in E_n$$, then for any $$\varepsilon > 0$$, there exists a piecewise linear function $$g$$ consisting of finitely many linear pieces each having slope $$\pm 2n$$ such that $$\|g-f\|_u < \varepsilon$$. But no such $$g$$ is in $$E_n$$, and so every neighborhood of $$f$$ contains functions that are not in $$E_n$$. Thus $$E_n$$ contains no open sets, and is therefore nowhere dense.
Thus far, we have proved that each $$E_n$$ is closed an nowhere dense, thus we are ready to apply Baire’s category theory: since $$C[0,1]$$ is residual in itself, it follows that $C[0,1] \setminus \bigcup_{n=1}^{\infty} E_n \ne \emptyset.$ But, as noted above, every function that is differentiable anywhere—even if only at a single point—is contained in the union. Therefore $$C[0,1]$$ contains at least one (in fact, a residual set of) nowhere differentiable function.
This entry was posted in Mathematics and tagged , . Bookmark the permalink. | 3,299 | 11,208 | {"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-2024-26 | latest | en | 0.942205 |
https://en.m.wikipedia.org/wiki/Radix_sort | 1,660,009,612,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00448.warc.gz | 230,793,509 | 22,042 | In computer science, radix sort is a non-comparative sorting algorithm. It avoids comparison by creating and distributing elements into buckets according to their radix. For elements with more than one significant digit, this bucketing process is repeated for each digit, while preserving the ordering of the prior step, until all digits have been considered. For this reason, radix sort has also been called bucket sort and digital sort.
Class Sorting algorithm Array ${\displaystyle O(w\cdot n)}$, where ${\displaystyle n}$ is the number of keys, and ${\displaystyle w}$ is the key length. ${\displaystyle O(w+n)}$
Radix sort can be applied to data that can be sorted lexicographically, be they integers, words, punch cards, playing cards, or the mail.
History
Radix sort dates back as far as 1887 to the work of Herman Hollerith on tabulating machines.[1] Radix sorting algorithms came into common use as a way to sort punched cards as early as 1923.[2]
The first memory-efficient computer algorithm for this sorting method was developed in 1954 at MIT by Harold H. Seward. Computerized radix sorts had previously been dismissed as impractical because of the perceived need for variable allocation of buckets of unknown size. Seward's innovation was to use a linear scan to determine the required bucket sizes and offsets beforehand, allowing for a single static allocation of auxiliary memory. The linear scan is closely related to Seward's other algorithm — counting sort.
In the modern era, radix sorts are most commonly applied to collections of binary strings and integers. It has been shown in some benchmarks to be faster than other more general-purpose sorting algorithms, sometimes 50% to three times faster.[3][4][5]
An IBM card sorter performing a radix sort on a large set of punched cards. Cards are fed into a hopper below the operator's chin and are sorted into one of the machine's 13 output baskets, based on the data punched into one column on the cards. The crank near the input hopper is used to move the read head to the next column as the sort progresses. The rack in back holds cards from the previous sorting pass.
Digit order
Radix sorts can be implemented to start at either the most significant digit (MSD) or least significant digit (LSD). For example, with 1234, one could start with 1 (MSD) or 4 (LSD).
LSD radix sorts typically use the following sorting order: short keys come before longer keys, and then keys of the same length are sorted lexicographically. This coincides with the normal order of integer representations, like the sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. LSD sorts are generally stable sorts.
MSD radix sorts are most suitable for sorting strings or fixed-length integer representations. A sequence like [b, c, e, d, f, g, ba] would be sorted as [b, ba, c, d, e, f, g]. If lexicographic ordering is used to sort variable-length integers in base 10, then numbers from 1 to 10 would be output as [1, 10, 2, 3, 4, 5, 6, 7, 8, 9], as if the shorter keys were left-justified and padded on the right with blank characters to make the shorter keys as long as the longest key. MSD sorts are not necessarily stable if the original ordering of duplicate keys must always be maintained.
Other than the traversal order, MSD and LSD sorts differ in their handling of variable length input. LSD sorts can group by length, radix sort each group, then concatenate the groups in size order. MSD sorts must effectively 'extend' all shorter keys to the size of the largest key and sort them accordingly, which can be more complicated than the grouping required by LSD.
However, MSD sorts are more amenable to subdivision and recursion. Each bucket created by an MSD step can itself be radix sorted using the next most significant digit, without reference to any other buckets created in the previous step. Once the last digit is reached, concatenating the buckets is all that is required to complete the sort.
Examples
Least significant digit
Input list:
[170, 45, 75, 90, 2, 802, 2, 66]
Starting from the rightmost (last) digit, sort the numbers based on that digit:
[{170, 90}, {2, 802, 2}, {45, 75}, {66}]
Sorting by the next left digit:
[{02, 802, 02}, {45}, {66}, {170, 75}, {90}]
Notice that an implicit digit 0 is prepended for the two 2s so that 802 maintains its position between them.
And finally by the leftmost digit:
[{002, 002, 045, 066, 075, 090}, {170}, {802}]
Notice that a 0 is prepended to all of the 1- or 2-digit numbers.
Each step requires just a single pass over the data, since each item can be placed in its bucket without comparison with any other element.
Some radix sort implementations allocate space for buckets by first counting the number of keys that belong in each bucket before moving keys into those buckets. The number of times that each digit occurs is stored in an array.
Although it is always possible to pre-determine the bucket boundaries using counts, some implementations opt to use dynamic memory allocation instead.
Most significant digit, forward recursive
Input list, fixed width numeric strings with leading zeros:
[170, 045, 075, 025, 002, 024, 802, 066]
First digit, with brackets indicating buckets:
[{045, 075, 025, 002, 024, 066}, {170}, {802}]
Notice that 170 and 802 are already complete because they are all that remain in their buckets, so no further recursion is needed
Next digit:
[{ {002}, {025, 024}, {045}, {066}, {075} }, 170, 802]
Final digit:
[ 002, { {024}, {025} }, 045, 066, 075 , 170, 802]
All that remains is concatenation:
[002, 024, 025, 045, 066, 075, 170, 802]
Complexity and performance
Radix sort operates in ${\displaystyle O(nw)}$ time, where ${\displaystyle n}$ is the number of keys, and ${\displaystyle w}$ is the key length. LSD variants can achieve a lower bound for ${\displaystyle w}$ of 'average key length' when splitting variable length keys into groups as discussed above.
Optimized radix sorts can be very fast when working in a domain that suits them.[6] They are constrained to lexicographic data, but for many practical applications this is not a limitation. Large key sizes can hinder LSD implementations when the induced number of passes becomes the bottleneck.[2]
Specialized variants
Binary MSD radix sort, also called binary quicksort, can be implemented in-place by splitting the input array into two bins - the 0s bin and the 1s bin. The 0s bin is grown from the beginning of the array, whereas the 1s bin is grown from the end of the array. The 0s bin boundary is placed before the first array element. The 1s bin boundary is placed after the last array element. The most significant bit of the first array element is examined. If this bit is a 1, then the first element is swapped with the element in front of the 1s bin boundary (the last element of the array), and the 1s bin is grown by one element by decrementing the 1s boundary array index. If this bit is a 0, then the first element remains at its current location, and the 0s bin is grown by one element. The next array element examined is the one in front of the 0s bin boundary (i.e. the first element that is not in the 0s bin or the 1s bin). This process continues until the 0s bin and the 1s bin reach each other. The 0s bin and the 1s bin are then sorted recursively based on the next bit of each array element. Recursive processing continues until the least significant bit has been used for sorting.[7][8] Handling signed integers requires treating the most significant bit with the opposite sense, followed by unsigned treatment of the rest of the bits.
In-place MSD binary-radix sort can be extended to larger radix and retain in-place capability. Counting sort is used to determine the size of each bin and their starting index. Swapping is used to place the current element into its bin, followed by expanding the bin boundary. As the array elements are scanned the bins are skipped over and only elements between bins are processed, until the entire array has been processed and all elements end up in their respective bins. The number of bins is the same as the radix used - e.g. 16 bins for 16-radix. Each pass is based on a single digit (e.g. 4-bits per digit in the case of 16-radix), starting from the most significant digit. Each bin is then processed recursively using the next digit, until all digits have been used for sorting.[9][10]
Neither in-place binary-radix sort nor n-bit-radix sort, discussed in paragraphs above, are stable algorithms.
MSD radix sort can be implemented as a stable algorithm, but requires the use of a memory buffer of the same size as the input array. This extra memory allows the input buffer to be scanned from the first array element to last, and move the array elements to the destination bins in the same order. Thus, equal elements will be placed in the memory buffer in the same order they were in the input array. The MSD-based algorithm uses the extra memory buffer as the output on the first level of recursion, but swaps the input and output on the next level of recursion, to avoid the overhead of copying the output result back to the input buffer. Each of the bins are recursively processed, as is done for the in-place MSD radix sort. After the sort by the last digit has been completed, the output buffer is checked to see if it is the original input array, and if it's not, then a single copy is performed. If the digit size is chosen such that the key size divided by the digit size is an even number, the copy at the end is avoided.[11]
Hybrid approaches
Radix sort, such as the two-pass method where counting sort is used during the first pass of each level of recursion, has a large constant overhead. Thus, when the bins get small, other sorting algorithms should be used, such as insertion sort. A good implementation of insertion sort is fast for small arrays, stable, in-place, and can significantly speed up radix sort.
Application to parallel computing
This recursive sorting algorithm has particular application to parallel computing, as each of the bins can be sorted independently. In this case, each bin is passed to the next available processor. A single processor would be used at the start (the most significant digit). By the second or third digit, all available processors would likely be engaged. Ideally, as each subdivision is fully sorted, fewer and fewer processors would be utilized. In the worst case, all of the keys will be identical or nearly identical to each other, with the result that there will be little to no advantage to using parallel computing to sort the keys.
In the top level of recursion, opportunity for parallelism is in the counting sort portion of the algorithm. Counting is highly parallel, amenable to the parallel_reduce pattern, and splits the work well across multiple cores until reaching memory bandwidth limit. This portion of the algorithm has data-independent parallelism. Processing each bin in subsequent recursion levels is data-dependent, however. For example, if all keys were of the same value, then there would be only a single bin with any elements in it, and no parallelism would be available. For random inputs all bins would be near equally populated and a large amount of parallelism opportunity would be available.[12]
There are faster parallel sorting algorithms available, for example optimal complexity O(log(n)) are those of the Three Hungarians and Richard Cole[13][14] and Batcher's bitonic merge sort has an algorithmic complexity of O(log2(n)), all of which have a lower algorithmic time complexity to radix sort on a CREW-PRAM. The fastest known PRAM sorts were described in 1991 by David Powers with a parallelized quicksort that can operate in O(log(n)) time on a CRCW-PRAM with n processors by performing partitioning implicitly, as well as a radixsort that operates using the same trick in O(k), where k is the maximum keylength.[15] However, neither the PRAM architecture or a single sequential processor can actually be built in a way that will scale without the number of constant fan-out gate delays per cycle increasing as O(log(n)), so that in effect a pipelined version of Batcher's bitonic mergesort and the O(log(n)) PRAM sorts are all O(log2(n)) in terms of clock cycles, with Powers acknowledging that Batcher's would have lower constant in terms of gate delays than his Parallel quicksort and radix sort, or Cole's merge sort, for a keylength-independent sorting network of O(nlog2(n)).[16]
Radix sorting can also be accomplished by building a tree (or radix tree) from the input set, and doing a pre-order traversal. This is similar to the relationship between heapsort and the heap data structure. This can be useful for certain data types, see burstsort.
References
1. ^ US 395781 and UK 327
2. ^ a b Donald Knuth. The Art of Computer Programming, Volume 3: Sorting and Searching, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89685-0. Section 5.2.5: Sorting by Distribution, pp. 168–179.
3. ^ "I Wrote a Faster Sorting Algorithm". 28 December 2016.
4. ^ "Is radix sort faster than quicksort for integer arrays?". erik.gorset.no.
5. ^ "Function template integer_sort - 1.62.0". www.boost.org.
6. ^
7. ^ R. Sedgewick, "Algorithms in C++", third edition, 1998, p. 424-427
8. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 2". Dr. Dobb's.
9. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 3". Dr. Dobb's.
10. ^ Duvanenko, Victor J. "Parallel In-Place Radix Sort Simplified". Dr. Dobb's.
11. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 4". Dr. Dobb's.
12. ^ Duvanenko, Victor J. "Parallel In-Place N-bit-Radix Sort". Dr. Dobb's.
13. ^ A. Gibbons and W. Rytter, Efficient Parallel Algorithms. Cambridge University Press, 1988.
14. ^ H. Casanova et al, Parallel Algorithms. Chapman & Hall, 2008.
15. ^ David M. W. Powers, Parallelized Quicksort and Radixsort with Optimal Speedup, Proceedings of International Conference on Parallel Computing Technologies. Novosibirsk. 1991.
16. ^ David M. W. Powers, Parallel Unification: Practical Complexity, Australasian Computer Architecture Workshop, Flinders University, January 1995 | 3,337 | 14,259 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "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-2022-33 | latest | en | 0.946309 |
http://slideplayer.com/slide/3377948/ | 1,510,990,887,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804666.54/warc/CC-MAIN-20171118055757-20171118075757-00619.warc.gz | 277,329,001 | 22,417 | # Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can.
## Presentation on theme: "Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can."— Presentation transcript:
Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can form a buffer when combined with its conjugate base (dihydrogen phosphate). H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 This buffer operates in the range: pH = pk a + 1 = 0.96 – 2.96
Also, another buffer which is commonly used is the dihydrogen phosphate/hydrogen phosphate buffer. H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 This buffer operates in the range from 6.1 to 8.1 A third buffer can be prepared by mixing hydrogen phosphate with orthophosphate as the following equilibrium suggests: HPO 4 2- H + + PO 4 3- k a3 = 4.8 x 10 -13 This buffer system operates in the pH range from 11.3 to 13.3
The same can be said about carbonic acid/bicarbonate where H 2 CO 3 H + + HCO 3 - k a1 = 4.3 x 10 -7 This buffer operates in the pH range from 5.4 to 7.4; while a more familiar buffer is composed of carbonate and bicarbonate according to the equilibrium: HCO 3 - H + + CO 3 2- k a2 = 4.8 x 10 -11 The pH range of the buffer is 9.3 to 11.3. Polyprotic acids and their salts are handy materials which can be used to prepare buffer solutions of desired pH working ranges. This is true due to the wide variety of their acid dissociation constants.
Example Find the ratio of [H 2 PO 4 - ]/[HPO 4 2- ] if the pH of the solution containing a mixture of both substances is 7.4. k a2 = 7.5x10 -8 Solution The equilibrium equation combining the two species is: H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 K a2 = [H + ][HPO 4 2- ]/[H 2 PO 4 - ] [H + ] = 10 -7.4 = 4x10 -8 M 7.5x10 -8 = 4x10 -8 [HPO 4 2- ]/[H 2 PO 4 - ] [HPO 4 2- ]/[H 2 PO 4 - ] = 1.9
Fractions of Dissociating Species at a Given pH Consider the situation where, for example, 0.1 mol of H 3 PO 4 is dissolved in 1 L of solution. H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 HPO 4 2- H + + PO 4 3- k a3 = 4.8 x 10 -13 Some of the acid will remain undissociated (H 3 PO 4 ), some will be converted to H 2 PO 4 -, HPO 4 2- and PO 4 3- where we have, from mass balance: C H3PO4 = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ]
We can write the fractions of each species in solution as 0 = [H 3 PO 4 ]/C H3PO4 1 = [H 2 PO 4 - ]/C H3PO4 2 = [HPO 4 2- ]/C H3PO4 3 = [PO 4 3- ]/C H3PO4 0 + 1 + 2 + 3 = 1 ( total value of all fractions sum up to unity).
The value of each fraction depends on pH of solution. At low pH dissociation is suppressed and most species will be in the form of H 3 PO 4 while high pH values will result in greater amounts converted to PO 4 3-. Setting up a relation of these species as a function of [H + ] is straightforward using the equilibrium constant relations. Let us try finding 0 where 0 is a function of undissociated acid. The point is to substitute all fractions by their equivalent as a function of undissociated acid.
K a1 = [H 2 PO 4 - ][H + ]/[H 3 PO 4 ] Therefore we have [H 2 PO 4 - ] = k a1 [H 3 PO 4 ]/ [H + ] k a2 = [HPO 4 2- ][H + ]/[H 2 PO - ] Multiplying k a2 time k a1 and rearranging we get: [HPO 4 2- ] = k a1 k a2 [H 3 PO 4 ]/[H + ] 2 k a3 = [PO 4 3- ][H + ]/[HPO 4 2- ] Multiplying k a1 times k a2 times k a3 and rearranging we get: [PO 3- ] = k a1 k a2 k a3 [H 3 PO 4 ]/[H + ] 3 But we have: C H3PO4 = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ]
Substitution for all species from above gives: C H3PO4 = [H 3 PO 4 ] + k a1 [H 3 PO 4 ]/ [H + ] + k a1 k a2 [H 3 PO 4 ]/[H + ] 2 + k a1 k a2 k a3 [H 3 PO 4 ]/[H + ] 3 C H3PO4 = [H 3 PO 4 ] {1 + k a1 / [H + ] + k a1 k a2 /[H + ] 2 + k a1 k a2 k a3 /[H + ] 3 } [H 3 PO 4 ]/C H3PO4 = 1/ {1 + k a1 / [H + ] + k a1 k a2 /[H + ] 2 + k a1 k a2 k a3 /[H + ] 3 } o = [H + ] 3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) Similar derivations for other fractions results in: 1 = k a1 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) 2 = k a1 k a2 [H + ] / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) 3 = k a1 k a2 k a3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 )
Example Calculate the equilibrium concentrations of the different species in a 0.10 M phosphoric acid solution at pH 3.00. Solution The [H + ] = 10 -3.00 = 1.0x10 -3 M Substitution in the relation for o gives o = [H + ] 3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) o = (1.0x10 -3 ) 3 /{(1.0x10 -3 ) 3 + 1.1x10 -2 (1.0x10 -3 ) 2 + 1.1x10 -2 * 7.5x10 -8 (1.0x10 -3 ) + 1.1x10 -2 * 7.5x10 - 8 * 4.8 * 10 -13 }
o = 8.2x10 -2 0 = [H 3 PO 4 ]/C H3PO4 8.2x10 -2 = [H 3 PO 4 ]/0.10 [H 3 PO 4 ] = 8.3x10 -3 M Similarly, 1 = 0.92, 1 = [H 2 PO 4 - ]/C H3PO4 0.92 = [H 2 PO 4 - ]/0.10 [H 2 PO 4 - ] = 9.2x10 -2 M Other fractions are calculated in the same manner.
pH Calculations for Salts of Polyprotic Acids Two types of salts exist for polyprotic acids. These include: 1. Unprotonated salts These are salts which are proton free which means they are not associated with any protons. Examples are: Na 3 PO 4 and Na 2 CO 3. Calculation of pH for solutions of such salts is straightforward and follows the same scheme described earlier for salts of monoprotic acids.
Example Find the pH of a 0.10 M Na 3 PO 4 solution. Solution We have the following equilibrium in water PO 4 3- + H 2 O HPO 4 2- + OH - The equilibrium constant which corresponds to this equilibrium is k b where: K b = k w /k a3
We used k a3 since it is the equilibrium constant describing relation between PO 4 3- and HPO 4 2-. However, in any equilibrium involving salts look at the highest charge on any anion to find which k a to use. K b = 10 -14 /4.8x10 -13 K b = 0.020
K b = x * x/0.10 – x Assume 0.10 >> x 0.02 = x 2 /0.10 x = 0.045 Relative error = (0.045/0.10) x 100 = 45% Therefore, assumption is invalid and we have to use the quadratic equation. If we solve the quadratic equation we get: X = 0.036 Therefore, [OH - ] = 0.036 M pOH = 1.44 and pH = 14 – 1.44 = 12.56
2. Protonated Salts These are usually amphoteric salts which react as acids and bases. For example, NaH 2 PO 4 in water would show the following equilibria: H 2 PO 4 - H + + HPO 4 2- H 2 PO 4 - + H 2 O OH - + H 3 PO 4 H 2 O H + + OH - [H + ] solution = [H + ] H2PO4 - + [H + ] H2O – [OH - ] H2PO4 - [H + ] solution = [HPO 4 2- ] + [OH - ] – [H 3 PO 4 ]
Now make all terms as functions in either H + or H 2 PO 4 -, then we have: [H + ] = {k a2 [H 2 PO 4 - ]/[H + ]} + k w /[H + ] –{[H 2 PO 4 - ][H + ]/k a1 } Rearrangement gives [H + ] = {(k a1 k w + k a1 k a2 [H 2 PO 4 - ])/(k a1 + [H 2 PO 4 ‑ ]} 1/2 At high salt concentration and low k a1 this relation may be approximated to: [H + ] = {k a1 k a2 } 1/2 Where; the pH will be independent on salt concentration but only on the equilibrium constants.
Download ppt "Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can."
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https://pratibha.eenadu.net/jobs/lesson/ibps/ibps-clerks/telugumedium/alligation-and-mixture/2-1-6-32-188-1347-7066-10530-20040012044 | 1,695,769,834,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00313.warc.gz | 497,710,442 | 19,140 | # ALLIGATION AND MIXTURE
1. Jar-A contains a mixture of milk and water in the respective ratio of 8 : 1. When 18 ltr of mixture is taken out and 4 ltr of pure water is added to Jar-A, the resultant quantity of milk becomes 24 ltr more than that of water. What was the initial quantity of milk in Jar-A? (In ltr)
A) 48 B) 64 C) 50 D) 52 E) 60
2. A container contains 120 litres mixture of milk and water in the respective ratio of 37 : 3. 40 litres of the mixture is taken out from the container and 6 litres of each pure milk and pure water is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?
A) 72% B) 76% C) 68% D) 85% E) 74%
3. In a 84 litres mixture of milk and water, percentage of water is only 25%. The milkman gave 12 litres of this mixture to a customer. Then he added equal quantities of pure milk and water to the remaining mixture. As a result, the respective ratio of milk and water in the mixture became 11 : 5. What was the quantity of water added? (In litres)
A) 8 B) 15 C) 12 D) 18 E) 21
4. A jar had a certain quantity (in litres) of water, to which pure milk of four times of the quantity water was added. 20 litres of the mixture from the jar was taken out and was replaced with 12 litres of pure milk, as result of which water constituted 10% of the resultant mixture. What was the initial quantity of water in the jar?
A) 7 litres B) 8 litres C) 10 litres D) 4.8 litres E) 6.4 litres
5. In a 200 litres mixture of milk and water, percentage of water is only 30%. The milkman gave 40 litres of this mixture to a customer and then added 12 litres of water in the remaining mixture. What is the respective ratio of milk and water in the new mixture?
A) 11 : 17 B) 7 : 13 C) 23 : 14 D) 21 : 11 E) 28 : 15
6. A vessel contains coconut water and vodka i.e., 120 litres of the mixture, coconut water is 40% more than the vodka. 72 litres of mixture taken out and 26 litres of coconut water is and some quantity of vodka is also added to the vessel, after that the respective ratio of the resultant mixture of vodka and coconut water becomes 2 : 3, then find the final quantityof the vodka in the vessel? (In litres)
A) 54 B) 46 C) 35 D) 34 E) 36
7. 16 litres of pure water was added to a vessel containing 84 litres of pure milk. 50 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the respective ratio of 3 : 2. If the resultant respective ratio of milk and water in the vessel is 4 : 1, what was the quantity of pure milk added in the vessel? (In litres)
A) 4 B) 8 C) 6 D) 10 E) 12
8. In a mixture of milk and water, water was only 20%. 50 litres of this mixture was taken out and then 5 litres of pure water was added to the mixture. If the resultant ratio of milk and water in the mixture was 8 : 3 respectively, what was the initial quantity of mixture (before the replacement)? (In litres)
A) 88 B) 96 C) 120 D) 100 E) 84
9. In a 120 litres mixture of milk and water, percentage of water is only 40%. The milkman gave 45 litres of this mixture to a customer and then added 15 litres of pure milk and 10 litres of pure water to the remaining mixture. What is the respective ratio of quantity of milk and water in the new mixture?
A) 3 : 2 B) 11 : 5 C) 8 : 5 D) 15 : 8 E) 12 : 7
10. In 112 litres of mixture of water and milk, water is only 18.75%. The milkman gave 28 litres of this mixture to a customer and then he added 10.75 litres of pure milk and 5.25 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture?
A) 32.50 B) 34 C) 32 D) 21 E) 27
Some more...
1. Equal quantities of two milk and water solutions which contains milk and water in ratio, 1 : 5 and 3 : 5 are mixed together. What will be the ratio of water to milk in the resultant solution?
a) 13 : 35 b) 4 : 10 c) 5 : 8 d) 35 : 13
Ans: d
2. The respective ratio of milk and water in 66 litres of adulterated milk is 5 : 1. Water is added into it to make the respective ratio 5 : 3. What is the quantity of water added?
a) 11 litres b) 22 litres c) 33 litres d) 44 litres
Ans: b
3. A 200 litres solution of alcohol and water contains 1/4 th part alcohol. Find the new percentage of alcohol, if 50 litres of the original solution is replaced by 50 litres of alcohol.
a) 43.75% b) 50% c) 66.66% d) 80%
Ans: a
4. A container has 80 litres of milk. From this container 8 litres of milk was taken out and replaced with water. The process was further repeated twice. How much milk is there in the container finally? (in liters)
a) 58.32 b) 60.32 c) 62 d) 64.7
Ans: a
5. 100 litres of a mixture contains 10% water and the rest milk. The amount of water that must be added so that the resulting mixture contains only 50% milk is......
a) 70 litres b) 72 litres c) 78 litres d) 80 litres
Ans: d
6. In a particular type of fertilizer, the ratio of two chemicals A and B is 2 : 5. In 21 kg of this fertilizer, if 3 kg of type - A is added, the ratio of chemical A to B in the new fertilizer will be
a) 1 : 1 b) 2 : 3 c) 3 : 5 d) 4 : 5
Ans: c
7. Rice worth Rs.120 per kg and Rs.132 per kg are mixed with a third variety in the ratio 2 : 1 : 3. If the mixture is worth Rs.135 per kg, the price of the third variety per kg will be....
a) Rs.140 b) Rs.146 c) Rs.150 d) Rs.148
Ans: b
8. A jar full of wine contains 50% alcohol. A part of wine is replaced by another containing 20% alcohol and now percentage of alcohol was found to be 30%. The part of wine replaced is?
a) 1/2 b) 1/3 c) 2/5 d) 2/3
Ans: d
9. A shopkeeper has 50 kg rice, a part of which he sells at 10% profit and rest at 25% profit. He gains 16% on the whole. The quantity sold at 10% profit is?
a) 30 kg b) 20 kg c) 45 kg d) 27 kg
Ans: a
10. An alloy contains tin, copper and zinc in ratio 2 : 3 : 1, and another alloy contain copper, aluminium and tin in ratio 3 : 2 : 7. If both alloys are mixed in equal quantities, then what will be the ratio of tin and aluminium in final alloy?
a) 2 : 11 b) 11 : 9 c) 9 : 2 d) 11 : 2
Ans: d
Posted Date : 17-09-2022
గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి. | 2,949 | 7,236 | {"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} | 3.703125 | 4 | CC-MAIN-2023-40 | longest | en | 0.869919 |
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# Only senior citizens enjoy doing the daily jumble. So Aesha must
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Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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19 Jun 2017, 04:38
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Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
Which of the following arguments exhibits the same flawed reasoning as the above?
A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain.
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.
D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica.
E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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19 Jun 2017, 17:48
B.
Some senior citizens might not enjoy jumbles and some pet lovers might not adopt Marley.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jun 2017, 23:25
Can someone please explain this question in a little detail? How to relate the two situations ? Also Why is option C wrong?
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jun 2017, 23:26
Experts please explain why is option C wrong? And how to go about solving such analogy questions?
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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23 Jun 2017, 00:55
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Shiv2016 wrote:
Experts please explain why is option C wrong? And how to go about solving such analogy questions?
I'm not an expert but I can try and help.
Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
a implies b ---->not b implies not a
Which of the following arguments exhibits the same flawed reasoning as the above?
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.a implies b ---->not b implies not a - correct
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.a implies b ---->not a implies not b - reversed and so incorrect
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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02 Jul 2017, 07:29
this is a mix question. "flaw" here indicates that this is not only logic-error question, but also an application questions.
Only A do B = every B is A
the flaw is not B -> not A
Practice is the best way to cope with different gmat questions.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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04 Jul 2017, 02:16
Flaw is senior citizen - > enjoy jumble doesn't imply ~not citizen-> ~ No jumble enjoyment. (flawed contrapositive) Replicated by B.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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04 Jul 2017, 10:04
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jul 2017, 09:26
can someone tell me how to do these kind of questions
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 May 2018, 04:01
broall wrote:
Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
Which of the following arguments exhibits the same flawed reasoning as the above?
A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain.
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.
D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica.
E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs.
VERITAS PREP OFFICIAL SOLUTION:
Solution: B
This Mimic the Reasoning question features difficult conditional wording. “Only x does y” can be reworded as “If you do y, then you are x”, a change of phrase that makes the question much easier to solve. “If you enjoy the jumble, you’re a senior citizen” implies two things: 1) if you enjoy the jumble, you’re a senior citizen, and 2) if you aren’t a senior citizen, you don’t enjoy the jumble. It does NOT imply that if you don’t enjoy the jumble, you’re not a senior citizen, so find an answer choice that also employs such faulty logic. (B) is the answer of choice.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] 22 May 2018, 04:01
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# Only senior citizens enjoy doing the daily jumble. So Aesha must
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https://www.aussportsbetting.com/2015/01/21/model-testing-measuring-forecast-accuracy/ | 1,720,851,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00410.warc.gz | 534,762,901 | 36,936 | # Model Testing – Measuring Forecast Accuracy
When developing a betting model it is important to properly measure its performance. Having a formal measure of performance is important because it provides a benchmark with which to test alternative models.
Obviously, when using a model for betting a key measure of performance is the profit or loss obtained when using it, but it’s nice to have more than one metric to identify differences between similarly performing models.
This article is the second of a two-part series. The first article, Model Testing – Measuring Profit & Loss, looks at various ways to measure betting profit. This article outlines various measures of forecast accuracy in the context of betting models. The examples focus on measuring the accuracy of a model that predicts total game scores.
This article borrows heavily from Chapter 2 of Forecasting: Methods and Applications, by Makridakis, Wheelwright and Hyndman.
### Understanding Sigma (Σ) Notation
Many of the formulas below using sigma notation. For those who can’t recall how to use Σ from their school days, click here to brush up on sigma notation.
### Standard Statistical Measures
Below is a range of measures to evaluate the accuracy of a forecasting model. In the context of sports betting, applications include forecasting total scores in rugby, the number of corners in soccer and winning margins in basketball.
To help illustrate the computations involved, the following measures will be applied to a simplified data set. The table below shows rugby league total scores and a model’s pre-game forecast of those totals.
Game
i
Total Score
Yi
Forecast
Fi
Forecast Error
ei
1 53 50 3
2 36 52 -16
3 34 32 2
4 40 36 4
5 39 44 -5
6 51 54 -3
7 42 42 0
8 36 44 -8
If Yi is the actual total score for game i and Fi is the pre-game forecast of the total score for game i, then we define the error of the forecast for game i as:
Looking at the table above, the error for game 1 is 53 – 50 = 3, the error for game 2 is 36 – 52 = -16, and so on.
For the equations below we will use the variable n to denote how many completed games we have forecasts for.
The above data set has total scores and their associated forecasts for 8 games, so n equals 8 in this case.
Using our variables ei and n, we can calculate the following statistical measures of forecast accuracy:
### Mean error (ME):
This measure will often be small because positive and negative errors will offset each other. This makes it pretty useless as a measure of accuracy because a model that has forecast errors of -2, +2, -2 and +2 will sum to 0, but so do -20, +20, -20 and +20. With that being said, the mean error is worth calculating because it will tell you if there is any systematic under- or over-estimating, which is called forecast bias. A model that tends to over- and under- estimate fairly equally will have a mean error close to zero, while a model that has a bias towards underestimating scores will have a strong positive value (note that ei = Yi – Fi, so if you underestimate the value, the forecast error is positive).
In the example data above the ME equals (3 + -16 + 2 + …)/8 = -2.875, illustrating that on average this model overestimated the total score.
### Mean absolute error (MAE):
Mean absolute error gets around the offsetting effect of positive and negative forecast errors by taking the absolute value of each error. Using our data set above we get |e1| = |53 – 50| = 3, |e2| = |36 – 52| = 16, and so on.
The advantage of using MAE is it provides a scale which people can understand. For example, if you had the following forecast errors: -2, +2, -2. +2, the MAE would equal 2, showing the model forecasts are 2 units off the correct value, on average.
In the example data above the MAE equals (3 + 16 + 2 + …)/8 = 5.125. In written terms you can say the average estimated total score was 5.125 points off the actual total.
### Mean squared error (MSE):
An alternative to taking the absolute value of each error is to square them, so a forecast error of -2 becomes 22 = 4. Like MAE, this avoids having positive and negative errors offset each other.
A point of difference between MAE and MSE is that MSE is more punishing for large errors, because squaring larger numbers produce markedly bigger results. For example the difference between 4 and 5 is just 1, but the difference between 42 and 52 is 9.
From a mathematics perspective many practitioners prefer to use MSE over MAE because squared functions are easier do deal with in optimisation calculations. From a calculus perspective it’s easier to take the derivative of a function with squared terms than a function with absolute value terms.
In the example data above the MSE equals ((32) + (-162) + (22) + …)/8 = 47.875. Note that this is significantly larger than the MAE due to the large error for Game 2.
### Percentage / relative errors
The above measures are all dependent on the scale of the data. For example, these measures would likely be much larger for basketball total scores than rugby total scores because basketball scores generally are much higher. If you’re off by 10% on a basketball score this could imply being 18 points off, whereas with rugby 10% could mean being off by 4 points.
The previously discussed error measures make comparing models between sports very difficult. The following measures adjust for the scale of the data, which can facilitate comparisons between models applied to different sports.
Rather than use absolute errors, ei = Yi – Fi, percentage errors are used instead, which are calculated as follows:
Using our original data set, for game 1 the percentage error is PE1 = (53 – 50)/53 = 0.0566 = 5.66%.
### Mean percentage error (MPE):
This is equivalent to ME discussed earlier, but it’s calculated using percentage errors.
MPE suffers the same drawback as ME through having positive and negative PEs offset each other, however this does mean it provides a measure of systematic bias.
In the example data above the MPE equals ((53 – 50)/53 + (36 – 52)/36 + …)/8 = -8.0%.
### Mean absolute percentage error (MAPE):
MAPE is equivalent to MAE discussed earlier, but it’s calculated using percentage errors.
MAPE works well for total score models, but it isn’t ideal for all situations. For example football scores tend to be low, so using measures like MAE and MSE may make more intuitive sense than using MAPE where you can have errors like 400% if the total score is 4 and the model predicted 1.
A more serious limitation MAPE occurs when your data set can have 0 values. In the context of sports betting if you’re forecasting winning margins and a draw is possible then you can’t use MAPE. This is because if the final scores are level then Yi = 0 so PEi can’t be calculated due to a division by zero error. For this reason MAPE works best for modeling results such as total basketball, AFL and rugby scores rather than winning margins or football total scores, which can have zero values.
In the example data above the MAPE equals (|53 – 50|/53 + |36 – 52|/36 + …)/8 = 13.4%.
### Comparing forecast methods
Once you have a measure of forecasting accuracy, how do you know if it’s a good result? Is a total score MSE of 9 or a MAPE of 6% good? What we need is a comparison of the model’s performance to more naive (basic) models to test if they represent a meaningful improvement.
In the context of forecasting total scores, suppose we have a naive model that predicts the total score for each game by simply using the total score from the last time the two sides met at the same venue. If in rugby league, Team A vs. Team B had a combined score of 38 the last time they met, then the naive model will predict the combined score for their next meeting to be 38.
Once a naive model has been created you can then calculate the forecast accuracy for it and compare its statistics to the accuracy calculations of the more sophisticated model. If the sophisticated model can’t outperform the naive model then it may mean going back to the drawing board.
### Out-of-sample accuracy measurement
The above measures calculate the accuracy of a betting model, however achieving good accuracy using historical data doesn’t guarantee the model will work well in the future.
Suppose you obtain a historical odds data set and you identify a trading strategy that would have worked well over the past three seasons. How confident can you be that the strategy will continue to work in the future? As anyone who has analysed historical data can tell you, if you look hard enough, you will find some strategy that would have made a killing had it been employed in previous years, however this provides no guarantee for future success.
A way to know if a model is genuinely useful and not simply reflecting quirks in your specific data set is to split your data into two parts before constructing the model. The first part of the data is used to build and calibrate the model and the second holdout set is used to test whether the model works well on the second set of data.
We highly recommend you read our article, Post-Sample Evaluation – the Importance of Creating a Holdout Set before Calibrating a Betting Strategy. It outlines how to create a holdout set to test a calibrated betting model. This practice provides an out-of-sample accuracy measurement because it involves evaluating a forecasting model using more recent data than was used to calibrate the model.
### Sources
An excellent book on forecasting is:
Forecasting: Methods and Applications
Spyros G. Makridakis, Steven C. Wheelwright, Rob J. Hyndman | 2,144 | 9,629 | {"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} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.898153 |
zhn.teknik100r.fun | 1,624,038,361,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00020.warc.gz | 1,001,251,994 | 11,879 | # Common excel math functions
Note that further math-related Excel functions are also provided in the Excel Statistical Functions and Excel Engineering Functions categories. The tables below list all the current built-in Excel math functions, grouped by category, to help you to find the function you need. Selecting a function link will take you to a full description of the function, with examples of use and common errors.
Note that some of the Excel math functions listed below were introduced in recent versions of Excel, and so are not available in earlier versions. Excel Functions.
Basic Numeric Information.
Courseworks completed paper date
Performs a specified calculation e. Rounding Functions. Rounds a number away from zero i. Rounds a number upregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number upregardless of the sign of the number, to a multiple of significance.
New in Excel Rounds a number up to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number towards zeroi. Rounds a number downregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number down, to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number up or downto the nearest multiple of significance. Truncates a number towards zero i. Returns the unit matrix for a specified dimension New in Excel Conditional Sums.
Adds the cells in a supplied range, that satisfy multiple criteria New in Excel Returns the sum of the products of corresponding values in two or more supplied arrays. Returns the sum of the difference of squares of corresponding values in two supplied arrays.
Returns the sum of the sum of squares of corresponding values in two supplied arrays. Returns the sum of squares of differences of corresponding values in two supplied arrays.
Returns the secant of an angle New in Excel Returns the hyperbolic secant of an angle New in Excel Returns the cosecant of an angle New in Excel Cheat Sheet of Excel formulas and function is always a customized worksheet where we can have all those function details, shortcut keys to execute any function or formulas, custom way to use 2 or more function together and guideline to use them.
Also, we can choose those formulas which are complicated to apply for users in a cheat sheet. Start Your Free Excel Course. As we can see, here several string functions are listed. Please refer the below screenshot where I have applied the functions on strings and shown their workings.
Please refer the below screenshot where I have applied the functions on number values and shown their workings. Please refer the below screenshot where I have applied the functions on values and shown their workings.
This has been a guide to Excel Formulas Cheat Sheet. You can also go through our other suggested articles —. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy.
By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy. Forgot Password? Call Our Course Advisors. Cheat Sheet of Excel Formulas. Popular Course in this category. Course Price View Course. Free Excel Course. Login details for this Free course will be emailed to you. Email ID. Contact No.So, the total production quantity is Similarly, apply the same logic to get the total salary amount.
Now we know what overall sum values are. Out of these overall total of employees, we need to find the average salary per employee. Select the range of cells for which we are finding the average value, so our range of cells will be from D2 to D We know the average salary per person; for further drill-down, we want to know what is the average salary based on gender.
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So totally there are 10 employees on the list. After counting the total number of employees, we may need to count how many male and female employees are there.
MOD function will return the remainder when one number is divided by another number. For example, when you divide number 11 by 2, we will get the remainder as 1 because only till 10 number 2 can divide.
When we have fraction or decimal values, we may need to round those decimal values to the nearest integer number.
For example, we need to round the number 3. As you can see above, B2 cell value Like this, we can use various mathematical functions in excel to do mathematical operations in excel quickly and easily. This has been a guide to Mathematical Function in Excel. Here we discuss how to calculate mathematical function in excel using Sum, Average, Averageif, Counta, Countif, Mod, and Round formulas along with practical examples.
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By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy.The tutorial provides a list of Excel basic formulas and functions with examples and links to related in-depth tutorials.
Being primarily designed as a spreadsheet program, Microsoft Excel is extremely powerful and versatile when it comes to calculating numbers or solving math and engineering problems. It enables you to total or average a column of numbers in the blink of an eye.
Apart from that, you can compute a compound interest and weighted average, get the optimal budget for your advertising campaign, minimize the shipment costs or make the optimal work schedule for your employees. All this is done by entering formulas in cells. This tutorial aims to teach you the essentials of Excel functions and show how to use basic formulas in Excel. Before providing the basic Excel formulas list, let's define the key terms just to make sure we are on the same page.
So, what do we call an Excel formula and Excel function? Function is a predefined formula already available in Excel. Functions perform specific calculations in a particular order based on the specified values, called arguments, or parameters. You can find all available Excel functions in the Function Library on the Formulas tab:. Of course, it's next to impossible to memorize all of them, and you actually don't need to.
The Function Wizard will help you find the function best suited for a particular task, while the Excel Formula Intellisense will prompt the function's syntax and arguments as soon as you type the function's name preceded by an equal sign in a cell:.
Clicking the function's name will turn it into a blue hyperlink, which will open the Help topic for that function.
What follows below is a list of 10 simple yet really helpful functions that are a necessary skill for everyone who wishes to turn from an Excel novice to an Excel professional. The first Excel function you should be familiar with is the one that performs the basic arithmetic operation of addition:. In the syntax of all Excel functions, an argument enclosed in [square brackets] is optional, other arguments are required.
Meaning, your Sum formula should include at least 1 number, reference to a cell or a range of cells. For example:. If necessary, you can perform other calculations within a single formula, for example, add up values in cells B2 through B6, and then divide the sum by To sum with conditions, use the SUMIF function: in the 1st argument, you enter the range of cells to be tested against the criteria A2:A6in the 2nd argument - the criteria itself D2and in the last argument - the cells to sum B2:B6 :.
In your Excel worksheets, the formulas may look something similar to this:.
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Its syntax is similar to SUM's:. Sums values in cells B2 through B6, and then divides the result by 5.
## Mathematical Function in Excel
And what do you call adding up a group of numbers and then dividing the sum by the count of those numbers? Yep, an average!The basic functions covered below are among the most popular formulas in Excel—the ones everyone should know. To help you learn, we've also provided a spreadsheet with all the formula examples we cover below. After you enter one of these functions in A1, you can then reformat the Date and Time or use the system default.
It also subtracts, multiplies, divides, and uses any of the comparison operators to return a result of 1 true or 0 false. Excel frames the column of numbers in green borders and displays the formula in the current cell. Remember your high school math?
If the numbers inside the formula are not grouped properly, the answer will be wrong. Notice the screenshot below figure 2. For this exercise, you can enter the same values in H, I, and J, with or without the blank rows in between again, added for easier viewing.
Note that as we build each formula, we are combining the steps, eventually, into a single formula.
Athletics activities department
We start out with three separate formulas. The formula in K3 is wrong. It requires grouping the numbers according to the order of calculation using commas or parentheses.
Check your numbers again with your calculator and note that this formula is correct. Note that the syntax the structure or layout of the formula is correct in cells N7 and N8, but incorrect in N6.
By combining these formulas into one, you can eliminate columns K and L. The RAND function is really simple and traditionally used for statistical analysis, cryptography, gaming, gambling, and probability theory, among dozens of other things.
Note; however, that every time you enter new data and press the Enter key, the list of random numbers you just created changes.
If you need to maintain your random numbers lists, you must format the cells as values. Now the list contains values instead of functions, so it will not change. Notice in the formula bar that the random numbers have 15 digits after the decimal Excel defaults to 9which you can change, if necessary as displayed in cell F3.
Just click the Increase Decimal button in the Number group under the Home tab. Again, you must copy the list and Paste as Values to maintain a static list. McGregor needs to order for his shop.Excel is a great way to organize and keep track of your data. There are more than functions in Excel. If you would like to know more about a function, simply follow the links we added for each of them. Got a different version?
No problem, you can still follow the exact same steps.
### Excel Formula Symbols Cheat Sheet (13 Cool Tips)
By definition, a function is a predefined formula in Excel which does calculations in the order specified by its parameters. Because learning how to add numbers in Excel is one of the most fundamental skills you need to learn. As you know, addition is an integral part of almost any calculation and task in Excel. As their name implies, they add the values in a specified range only when the criteria are met. The differences between the two are in the number of criteria you can specify.
These functions are useful when dealing with large data sets and manual calculations are inefficient and impractical. This function calculates the arithmetic mean of a set of numbers or the sum of the values divided by the number of values. People use the averages every day, from school grades to statistics.
These functions shine when you need to get averages from specifics sets in a range. It may seem rather rudimentary. But in actuality, this function is used in a lot of computations and scenarios. This function is used in many things like counting how many items there are in a listcounting specific casesand others.
But what if you only need to count a specific subset of cells? These functions are useful in tasks like project managementsales inventoryorder fulfillmentand others. This function returns the sum of the product of two or more arrays.
This is an important Excel function since this is used to calculate weighted averages as well as simplify a lot of tasks like sales inventory. Ever had the need to come up with random values between a specified minimum and maximum values?Use this handy Cheat Sheet to discover great functions and tips to help you get the most out of Excel.
Some Excel functions apply to specific subject areas, but others are general and apply to all needs. The following list shows an array of Excel functions used by one and all. Check here for a quickie reference to the purpose of each Excel function. Here is list of Excel functions associated with text, along with a description of what each function does:.
Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. The following is the order in which mathematical operators and syntax are applied both in Excel and in general mathematics. In Excel formulas, you can refer to other cells either relatively or absolutely. When you copy and paste a formula in Excel, how you create the references within the formula tells Excel what to change in the formula it pastes. You can also mix relative and absolute references so that, when you move or copy a formula, the row changes but the column does not, or vice versa.
If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A handful of errors can appear in a cell when a formula or function in Excel cannot be resolved. Knowing their meaning helps correct the problem. He has written numerous articles and books on a variety of technical topics.
Microsoft Excel Tutorial - Beginners Level 1
His latest projects include large-scale cloud-based applications and mobile app development. Cheat Sheet. Excel Order of Operations to Keep in Mind Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. Excel Cell References Worth Remembering In Excel formulas, you can refer to other cells either relatively or absolutely.
Excel Error Messages to Get to Know If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A formula or a function inside a formula cannot find the referenced data NAME? Text in the formula is not recognized NULL!
A space was used in formulas that reference multiple ranges; a comma separates range references NUM! A formula has invalid numeric data for the type of operation REF! The wrong type of operand or function argument is used. | 2,909 | 14,765 | {"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-2021-25 | latest | en | 0.843736 |
https://sudonull.com/post/8347-OpenSceneGraph-Scene-Graph-and-Smart-Pointers | 1,716,995,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00597.warc.gz | 467,656,383 | 9,217 | • Tutorial
# Introduction
In the last article, we looked at the build of OpenSceneGraph from source and wrote an elementary example in which a gray plane hangs in an empty purple world. I agree, not too impressive. However, as I said before, in this small example there are main concepts on which this graphic engine is based. Consider them in more detail. In the material below, illustrations from Alexander Bobkov's blog about OSG are used (it is a pity that the author abandoned writing about OSG ...). The article is also based on the material and examples from the book OpenSceneGraph 3.0. Beginner's Guide
I must say that the previous publication was subjected to some criticism, with which I partially agree - the material was left unsaid and taken out of context. I will try to correct this omission under the cut.
# 1. Briefly about the scene graph and its nodes
The central concept of the engine is the so-called scene graph (it is not by chance that it is framed into the very name of the framework) - a tree-like hierarchical structure that allows you to organize the logical and spatial representation of the three-dimensional scene. The scene graph contains the root node and its associated intermediate and end nodes or nodes .
for example
This graph depicts a scene consisting of a house and a table located in it. The house has a certain geometric representation and is in a certain way located in space relative to a certain basic coordinate system associated with the root node (root). The table is also described by some geometry, located in some way relative to the house, and together with the house - relative to the root node. All nodes possessing common properties, because they are inherited from the same class osg :: Node, are divided into types according to their functional purpose
1. Group nodes (osg :: Group) are the base class for all intermediate nodes and are intended for combining other nodes into groups.
2. Transformation nodes (osg :: Transform and its heirs) are intended to describe the transformation of object coordinates
3. Geometrical nodes (osg :: Geode) are terminal (leaf) nodes of a scene graph containing information about one or several geometric objects.
The geometry of scene objects in OSG is described in its own local coordinate system. Transformation nodes located between this object and the root node implement matrix coordinate transformations to obtain the position of the object in the base coordinate system.
Nodes perform many important functions, in particular, they store the display state of objects, and this state affects only the subgraph associated with this node. Several callbacks and event handlers can be associated with the nodes of a scene graph, allowing you to change the state of a node and its associated subgraph.
All global operations on the graph of the scene related to obtaining the final result on the screen are performed by the engine automatically, by periodically traversing the graph in depth.
ConsideredThe last time example was our scene consisted of a single object - an airplane model loaded from a file. Running very far forward, I will say that this model is a leaf node of the scene graph. It is tightly welded to the global base coordinate system of the engine.
# 2. Memory Management in OSG
Since the nodes of the scene graph store a lot of data about the scene objects and operations on them, it is necessary to allocate memory, including dynamically, to store this data. In this case, when manipulating the scene graph, and, for example, removing some of its nodes, you need to carefully monitor that the remote nodes of the graph are no longer processed. This process is always accompanied by errors, time-consuming debugging, since it is rather difficult for a developer to track which pointers to objects refer to existing data and which must be deleted. Without effective memory management, segmentation errors and memory leaks are likely.
Memory management is a critical task in OSG and its concept is based on two theses:
1. Memory Allocation: Ensuring the allocation of storage space required by an object.
2. Freeing memory: Returning the allocated memory to the system at that moment when it is not necessary.
Many modern programming languages, such as C #, Java, Visual Basic .Net, and the like, use the so-called garbage collector to free allocated memory. The concept of the C ++ language does not provide for a similar approach, but we can simulate it by using so-called smart pointers.
Today C ++ has smart pointers in its arsenal, which is called “out of the box” (and the C ++ standard 17 already managed to rid the language of some obsolete types of smart pointers), but this was not always the case. The earliest of the official versions of OSG number 0.9 was born in 2002, and before the first official release was still three years. At that time, the C ++ standard did not yet provide smart pointers, and if you believe one historical excursion, the language itself was going through hard times. So the appearance of a bicycle in the form of its own smart pointers, which are implemented in OSG is not surprising. This mechanism is deeply integrated into the structure of the engine, so it is absolutely necessary to understand its work from the very beginning.
# 3. Classes osg :: ref_ptr <> and osg :: Referenced
OSG provides its own smart pointer engine based on the osg :: ref_ptr <> template class to implement automatic garbage collection. For its proper operation, OSG provides another class osg :: Referenced for managing memory blocks for which reference counts are made.
The class osg :: ref_ptr <> provides several operators and methods.
• get () is a public method that returns a “raw” pointer, for example, when using the template osg :: Node as an argument, this method returns osg :: Node *.
• operator * () is actually a dereference operator.
• operator -> () and operator = () allow using osg :: ref_ptr <> as a classic pointer when accessing methods and properties of objects described by this pointer.
• operator == (), operator! = () and operator! () - allow you to perform comparison operations on smart pointers.
• valid () is a public method that returns true if the managed pointer has a valid value (not NULL). The expression some_ptr.valid () is equivalent to the expression some_ptr! = NULL, if some_ptr is a smart pointer.
• release () is a public method, useful when you want to return a managed address from a function. About him will be discussed in more detail later.
The osg :: Referenced class is the base class for all elements of the scene graph, such as nodes, geometry, rendering states, and other objects placed on the scene. Thus, by creating the root node of the scene, we indirectly inherit all the functionality provided by the class osg :: Referenced. Therefore, in our program there is an announcement
``````osg::ref_ptr<osg::Node> root;
``````
The class osg :: Referenced contains an integer counter of references to the allocated block of memory. This counter is initialized to zero in the class constructor. It is incremented by one when an osg :: ref_ptr <> object is created. This counter decreases as soon as any reference to the object described by this pointer is deleted. The object is automatically destroyed when it is no longer referenced by any smart pointers.
The osg :: Referenced class has three public methods:
• ref () is a public method that increments the reference count by 1.
• unref () is a public method that reduces the reference count by 1.
• referenceCount () is a public method that returns the current value of the reference count, which is useful when debugging code.
These methods are available in all classes derived from osg :: Referenced. However, it should be remembered that manual control of the reference count can lead to unpredictable consequences, and taking advantage of this, you should be clear about what you are doing.
# 4. How OSG performs garbage collection and why it is needed
There are several reasons why smart pointers and garbage collection should be used:
• Critical error minimization: the use of smart pointers automates the allocation and freeing of memory. There are no dangerous "raw" pointers.
• Effective memory management: memory allocated for an object is released as soon as the object is no longer needed, which leads to economical use of system resources.
• Facilitating debugging of an application: having the ability to clearly track the number of references to an object, we have opportunities for all sorts of optimizations and experiments.
Suppose that a scene graph consists of a root node and several levels of child nodes. If the root node and all child nodes are managed using the class osg :: ref_ptr <>, then the application can only track the pointer to the root node. Deleting this node will result in a sequential, automatic deletion of all child nodes.
Smart pointers can be used as local variables, global variables, class members and automatically reduce the reference count when the smart pointer goes out of scope.
Smart pointers are strongly recommended by OSG developers for use in projects, but there are a few key points to consider:
• Instances of osg :: Referenced and its derivatives can be created exclusively on the heap. They cannot be created on the stack as local variables, since the destructors of these classes are declared as proteced. for example
``````osg::ref_ptr<osg::Node> node = new osg::Node; // правильно
osg::Node node; // неправильно``````
• You can create temporary scene nodes using ordinary C ++ pointers, but this approach will be insecure. Better to use smart pointers to ensure correct control of the scene graph.
``````osg::Node *tmpNode = new osg::Node; // в принципе, будет работать...
osg::ref_ptr<osg::Node> node = tmpNode; // но лучше завершить работу с временным указателем таким образом!``````
• In no case should cyclic references be used in the scene tree when a node references itself directly or indirectly through several levels.
In the above example, the scene graph of a Child 1.1 node is self-referencing, and the Child 2.2 node also references the Child 1.2 node. Such links can lead to incorrect calculation of the number of links and undefined program behavior.
# 5. Tracking managed objects
To illustrate the operation of the smart pointer mechanism in OSG, we will write the following synthetic example
main.h
``#ifndef MAIN_H#define MAIN_H#include<osg/ref_ptr>#include<osg/Referenced>#include<iostream>#endif// MAIN_H``
main.cpp
``````#include"main.h"classMonitoringTarget :public osg::Referenced
{
public:
MonitoringTarget(int id) : _id(id)
{
std::cout << "Constructing target " << _id << std::endl;
}
protected:
virtual ~MonitoringTarget()
{
std::cout << "Dsetroying target " << _id << std::endl;
}
int _id;
};
intmain(int argc, char *argv[]){
(void) argc;
(void) argv;
osg::ref_ptr<MonitoringTarget> target = new MonitoringTarget(0);
std::cout << "Referenced count before referring: " << target->referenceCount() << std::endl;
osg::ref_ptr<MonitoringTarget> anotherTarget = target;
std::cout << "Referenced count after referring: " << target->referenceCount() << std::endl;
return0;
}
``````
Create a descendant class osg :: Referenced that does nothing except in the constructor and destructor informing that its instance has been created and displaying the identifier defined when creating the instance. Create a class instance using the smart pointer engine
``````osg::ref_ptr<MonitoringTarget> target = new MonitoringTarget(0);
``````
Next, display the reference count for the target object.
``````std::cout << "Referenced count before referring: " << target->referenceCount() << std::endl;
``````
After that, create a new smart pointer, assigning it the value of the previous pointer
``````osg::ref_ptr<MonitoringTarget> anotherTarget = target;
``````
and again display the reference count
``````std::cout << "Referenced count after referring: " << target->referenceCount() << std::endl;
``````
Let's see what we did by analyzing the output of the program.
``````15:42:39: Отладка запущена
Constructing target 0
Referenced count before referring: 1
Referenced count after referring: 2
Dsetroying target 0
15:42:42: Отладка завершена
``````
When the class constructor is started, a corresponding message is displayed, telling us that the memory for the object is allocated and the constructor has worked normally. Further, after creating a smart pointer, we see that the reference count for the created object has increased by one. Creating a new pointer, assigning it the value of the old pointer is essentially creating a new link to the same object, so the reference count is increased by one more. When you exit the program, the MonitoringTarget class destructor is called.
Let's conduct one more experiment, adding such code to the end of the main () function
``````for (int i = 1; i < 5; i++)
{
osg::ref_ptr<MonitoringTarget> subTarget = new MonitoringTarget(i);
}
``````
leading to such an "exhaust" program
``````16:04:30: Отладка запущена
Constructing target 0
Referenced count before referring: 1
Referenced count after referring: 2
Constructing target 1
Dsetroying target 1
Constructing target 2
Dsetroying target 2
Constructing target 3
Dsetroying target 3
Constructing target 4
Dsetroying target 4
Dsetroying target 0
16:04:32: Отладка завершена
``````
We create several objects in the loop body using a smart pointer. Since the scope of the pointer in this case extends only to the body of the loop, when you exit it, the destructor is automatically called. This would not have happened, obviously, if we used ordinary pointers.
Another important feature of working with smart pointers is associated with automatic memory freeing. Since the destructor of the classes derived from osg :: Referenced is protected, we cannot explicitly call the delete operator to delete an object. The only way to delete an object is to nullify the number of links to it. But then our code becomes unsafe with multi-threaded data processing - we can access an already deleted object from another thread.
Fortunately, OSG provides a solution to this problem with its scheduler for deleting objects. This scheduler is based on the use of the class osg :: DeleteHandler. It works in such a way that it does not perform the operation of deleting an object right away, but executes it after a while. All objects to be deleted are temporarily remembered until a moment comes for safe removal, and then they are all deleted at once. The osg :: DeleteHandler removal scheduler is managed by the OSG render backend.
# 6. Return from function
Add the following function to our example code.
``````MonitoringTarget *createMonitoringTarget(int id){
osg::ref_ptr<MonitoringTarget> target = new MonitoringTarget(id);
return target.release();
}
``````
and replace the call to the new operator in a loop with a call to this function.
``````for (int i = 1; i < 5; i++)
{
osg::ref_ptr<MonitoringTarget> subTarget = createMonitoringTarget(i);
}
``````
Calling release () will reduce the number of references to the object to zero, but instead of deleting the memory, returns directly the actual pointer to the allocated memory. If this pointer is assigned to another smart pointer, then there will be no memory leaks.
# findings
The concepts of the scene graph and smart pointers are basic for understanding the principle of operation, and hence the effective use of OpenSceneGraph. With regard to OSG smart pointers, it should be remembered that their use is absolutely necessary when
• It is assumed long-term storage facility
• One object stores a link to another object.
• It is necessary to return the pointer from the function
The sample code provided in the article is available here .
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Encyclopedia > Algebraic number
In mathematics, an algebraic number is any number that is a root of an algebraic equation, a non-zero polynomial with integer (or equivalently, rational) coefficients. Without further qualification, it is assumed that an algebraic number is a complex number, but one can also consider algebraic numbers in other fields, such as fields of p-adic numbers. All these algebraic numbers belong to some algebraic number field. Euclid, a famous Greek mathematician known as the father of geometry, is shown here in detail from The School of Athens by Raphael. ... In mathematics, a root (or a zero) of a function f is an element x in the domain of f such that f(x) = 0. ... Algebraic geometry is a branch of mathematics which, as the name suggests, combines abstract algebra, especially commutative algebra, with geometry. ... In mathematics, a polynomial is an expression in which constants and variables are combined using only addition, subtraction, multiplication, and positive whole number exponents (raising to a power). ... The integers consist of the positive natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and the number zero. ... In mathematics, a rational number (or informally fraction) is a ratio or quotient of two integers, usually written as the vulgar fraction a/b, where b is not zero. ... Wikibooks Algebra has more about this subject: Complex numbers In mathematics, a complex number is an expression of the form where a and b are real numbers, and i is a specific imaginary number, called the imaginary unit, with the property i 2 = −1. ... This article presents the essential definitions. ... The p-adic number systems were first described by Kurt Hensel in 1897. ... In mathematics, an algebraic number field (or simply number field) is a finite (and therefore algebraic) field extension of the rational numbers Q. That is, it is a field which contains Q and has finite dimension when considered as a vector space over Q. The study of algebraic number fields...
All rationals are algebraic. An irrational number may or may not be algebraic. For example, 21/2 (the square root of 2) and 31/3/2 (half the cube root of 3) are algebraic because they are the solutions of x2 − 2 = 0 and 8x3 − 3 = 0, respectively. The imaginary unit i is algebraic, since it satisfies x2 + 1 = 0. In mathematics, an irrational number is any real number that is not a rational number, i. ... The square root of 2 is equal to the length of the hypotenuse of a right triangle with legs of length 1. ... Plot of y = In mathematics, the cube root ( ) of a number is the number which, when cubed (multiplied by itself and then multiplied by itself again), gives back the original number. ... In mathematics, the imaginary unit (sometimes also represented by the Latin or the Greek iota) allows the real number system to be extended to the complex number system . ...
Numbers that are not algebraic are called transcendental numbers. Most complex numbers are transcendental, because the set of algebraic numbers is countable while the set of complex numbers, and therefore also the set of transcendental numbers, are not. Examples of transcendental numbers include π and e. Other examples are provided by the Gelfond-Schneider theorem. In mathematics, a transcendental number is any real number that is not algebraic, that is, not the solution of a non-zero polynomial equation with integer (or, equivalently, rational) coefficients. ... In mathematics, a countable set is a set with the same cardinality (i. ... Lower-case Ï€ (the lower case letter is usually used for the constant) The mathematical constant Ï€ is an irrational number, approximately equal to 3. ... e is the unique number such that the derivative (slope) of f(x)=ex at any point is equal to the height of the function at that point. ... In mathematics, the Gelfond-Schneider theorem is the following statement, originally proved by Aleksandr Gelfond: If is an algebraic number (with and ), and is an irrational algebraic number, then is a transcendental number. ...
All algebraic numbers are computable and therefore definable. In mathematics, theoretical computer science and mathematical logic, the computable numbers, also known as the recursive numbers, are the subset of the real numbers consisting of the numbers which can be computed by a finite, terminating algorithm. ... A real number a is first-order definable in the language of set theory, without parameters, if there is a formula φ in the language of set theory, with one free variable, such that a is the unique real number such that φ(a) holds (in the von Neumann universe V). ...
If an algebraic number satisfies a polynomial equation as given above with a polynomial of degree n and not such an equation with a lower degree, then the number is said to be an algebraic number of degree n. In mathematics, a polynomial is an expression in which constants and variables are combined using only addition, subtraction, multiplication, and positive whole number exponents (raising to a power). ...
The concept of algebraic numbers can be generalized to arbitrary field extensions; elements in such extensions that satify polynomial equations are called algebraic elements. In abstract algebra, a subfield of a field L is a subset K of L which is closed under the addition and multiplication operations of L and itself forms a field with these operations. ... In mathematics, the roots of polynomials are in abstract algebra called algebraic elements. ...
## The field of algebraic numbers GA_googleFillSlot("encyclopedia_square");
The sum, difference, product and quotient of two algebraic numbers is again algebraic, and the algebraic numbers therefore form a field, sometimes denoted by $mathbb{A}$ or $overline{mathbb{Q}}$. It can be shown that every root of a polynomial equation whose coefficients are algebraic numbers is again algebraic. This can be rephrased by saying that the field of algebraic numbers is algebraically closed. In fact, it is the smallest algebraically closed field containing the rationals, and is therefore called the algebraic closure of the rationals. This article presents the essential definitions. ... In mathematics, a field F is said to be algebraically closed if every polynomial of degree at least 1, with coefficients in F, has a zero (root) in F (i. ... In mathematics, particularly abstract algebra, an algebraic closure of a field K is an algebraic extension of K that is algebraically closed. ...
All the above statements are most easily proved in the general context of algebraic elements of a field extension.
All numbers which can be obtained from the integers using a finite number of additions, subtractions, multiplications, divisions, and taking nth roots (where n is a positive integer) are algebraic. The converse, however, is not true: there are algebraic numbers which cannot be obtained in this manner. All of these numbers are solutions to polynomials of degree ≥ 5. This is a result of Galois theory (see Quintic equations and the Abel–Ruffini theorem). An example of such a number would be the unique real root of x5 − x − 1 = 0. In mathematics, a set is called finite if there is a bijection between the set and some set of the form {1, 2, ..., n} where is a natural number. ... 3 + 2 with apples, a popular choice in textbooks Addition is the basic operation of arithmetic. ... 5 - 2 = 3 Subtraction is one of the four basic arithmetic operations; it is essentially the opposite of addition. ... In mathematics, multiplication is an arithmetic operation which is the inverse of division, and in elementary arithmetic, can be interpreted as repeated addition. ... In mathematics, especially in elementary arithmetic, division is an arithmetic operation which is the inverse of multiplication. ... In mathematics, more specifically in abstract algebra, Galois theory, named after Évariste Galois, provides a connection between field theory and group theory. ... Polynomial of degree 5: f(x) = (x+4)(x+2)(x+1)(x-1)(x-3)/20+2 In mathematics, a quintic equation is a polynomial equation in which the greatest exponent on the independent variable is five. ... The Abel–Ruffini theorem states that there is no general solution in radicals to polynomial equations of degree five or higher. ...
## Algebraic integers
Main article: algebraic integer
An algebraic number which satisfies a polynomial equation of degree n with leading coefficient an = 1 (that is, a monic polynomial) and all other coefficients ai belonging to the set Z of integers, is called an algebraic integer. Examples of algebraic integers are 3√2 + 5 and 6i - 2. In mathematics, an algebraic integer is a complex number α that is a root of an equation P(x) = 0 where P(x) is a monic polynomial (that is, the coefficient of the largest power of x in P(x) is one) with integer coefficients. ... Algebraic geometry is a branch of mathematics which, as the name suggests, combines abstract algebra, especially commutative algebra, with geometry. ... In mathematics, polynomial functions, or polynomials, are an important class of simple and smooth functions. ... The integers consist of the positive natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and the number zero. ... In mathematics, an algebraic integer is a complex number α that is a root of an equation P(x) = 0 where P(x) is a monic polynomial (that is, the coefficient of the largest power of x in P(x) is one) with integer coefficients. ...
The sum, difference and product of algebraic integers are again algebraic integers, which means that the algebraic integers form a ring. The name algebraic integer comes from the fact that the only rational numbers which are algebraic integers are the integers, and because the algebraic integers in any number field are in many ways analogous to the integers. If K is a number field, its ring of integers is the subring of algebraic integers in K, and is frequently denoted as OK. These are the prototypical examples of Dedekind domains. In ring theory, a branch of abstract algebra, a ring is an algebraic structure in which addition and multiplication are defined and have similar properties to those familiar from the integers. ... In mathematics, an algebraic number field (or simply number field) is a finite (and therefore algebraic) field extension of the rational numbers Q. That is, it is a field which contains Q and has finite dimension when considered as a vector space over Q. The study of algebraic number fields... In abstract algebra, a Dedekind domain is a Noetherian integral domain which is integrally closed in its fraction field and which has Krull dimension 1. ...
## Special classes of algebraic number
Results from FactBites:
PlanetMath: algebraic number theory (953 words) Algebraic number theory is the study of algebraic numbers, their properties and their applications. The main object of study in algebraic number theory is the number field. This is version 30 of algebraic number theory, born on 2005-03-15, modified 2006-03-07.
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# How many randomly assembled people are needed to have a
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Senior Manager
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17 Jan 2010, 06:50
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How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
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17 Jan 2010, 12:18
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nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
I found this question interesting, so thought of sharing it you folks.
Cheers
Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.
Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.
So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$
For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$
Thus min 3 people are needed.
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10 Jan 2012, 06:41
1
My solution
Prob(at least one of them was born in a leap year) > 50%
P(A) = 1 - P(A')
1 - Prob(none were born in a leap year) > 1/2
Prob(none) - 1 < -1/2
Prob(none) < 1/2
Assume that there are n people.
Prob(none) = (3/4)^n
(3/4)^n < 1/2
3^n < (4^n)/2
3^n < (2^2n)/2
3^n < 2^(2n-1)
If n = 3, 3^3 < 2^5 is true (27 < 32)
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Re: How many randomly assembled people are needed to have a [#permalink]
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03 Oct 2013, 18:57
cipher wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
Hi Bunuel,
Don't you think that this question is faulty?
The question ask about the probability better than 50%. If we have 4 people we still have probability better than 50% and if we chose 5 people it will be even more. So how can be the answer C
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01 Oct 2014, 12:00
1
This is how I would ans .
P(people born in leap year) = 1/4 (because out of every 4 years 1 is a leap year )
P(People not born in leap year ) = 3/4
Now for any n no. of people (3/4)^n is Probability of those people who were not born in leap year
As total probability is 1
1-(3/4)^n represents the cumulative probability of all those people born in leap year ( Eg this P includes probability of 1 person in group to be
born in leap year to n persons in a group to be born in leap year - which ans at least 1 person is born in leap year )
So 1-(3/4)^n>1/2
after trying diff values of n , it comes out as 3
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Re: How many randomly assembled people are needed to have a [#permalink]
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02 Oct 2014, 00:59
2
Probability of a person to be born in a leap year = 1/4 ... <1/2
Probability of atleast one person out of two to be born in a leap year = 1/4+1/4=1/2 .... not >1/2
Probability of atleast one person out of 3 to be born in a leap year = 1/4+1/4+1/4=3/4..... >1/2
Hence minimum 3 people are required.
Ans = C
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Re: How many randomly assembled people are needed to have a [#permalink]
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12 May 2015, 01:54
Bunuel wrote:
nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
I found this question interesting, so thought of sharing it you folks.
Cheers
Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.
Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.
So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$
For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$
Thus min 3 people are needed.
Bunuel
how is Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$. ???
if you consider years 1896 1897 1898 1899 1900 1901 1902 1903 1904 .... Here 1896 and 1904 are the only leap years.
How did you generalize probability not being a leap year to be 3/4 ?
And also question didn't ask about minimum number of people needed. Ans can be 3 or 4 or 5
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Re: How many randomly assembled people are needed to have a [#permalink]
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12 May 2015, 03:48
VenoMftw wrote:
Bunuel wrote:
nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
I found this question interesting, so thought of sharing it you folks.
Cheers
Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.
Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.
So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$
For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$
Thus min 3 people are needed.
Bunuel
how is Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$. ???
if you consider years 1896 1897 1898 1899 1900 1901 1902 1903 1904 .... Here 1896 and 1904 are the only leap years.
How did you generalize probability not being a leap year to be 3/4 ?
And also question didn't ask about minimum number of people needed. Ans can be 3 or 4 or 5
Yes, every 100's year is not leap but for this question it's implied that 1 in 4 years is leap.
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Re: How many randomly assembled people are needed to have a [#permalink]
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20 May 2015, 02:50
1
1
If you have 1 person, the chance that he's born in a leap year is $$\frac{1}{4}$$
If you have 2 persons the chances are $$\frac{2}{4}$$ = 50%
We need AT LEAST 50%, so 1 extra person will do the job.
We need 3 persons.
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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05 Aug 2015, 22:54
1
2
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
Keeping it simple:
P (A person was born in leap year) = 1/4
This is less than 50% - not the answer
P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)
The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)
P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16
This is less than 50% but close - not the answer
So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.
If you want to calculate it, you can do it as given below:
P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)
This is definitely more than 50%.
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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07 Aug 2015, 21:42
1
1
Or we can reverse the problem. If the probability someone is born in a leap year is greater than 1/2, then the probability no one was born in a leap year needs to be less than 1/2.
If we pick one person, the probability they weren't born in a leap year is 3/4. If we pick two people, the probability both were not born in a leap year is (3/4)(3/4) = (3/4)^2 = 9/16. That's still bigger than 1/2, but if we pick a third person, the probability none of the three people was born in a leap year is (3/4)^3 which is less than 1/2, which means the probability someone was born in a leap year is now greater than 1/2. So the answer is three.
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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08 Aug 2015, 00:27
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
Every 4 consecutive years have 1 Leap year and 3 normal years
i.e. Probability of an year to be leap year = 1/4
and Probability of an year to NOT to be leap year = 1/4
Probability of 1 randomly selected year to be NON leap year = (3/4) i.e. Greater than 50%
Probability of 2 randomly selected year to be NON leap year = (3/4)*(3/4) = 9/16 i.e. Greater than 50%
Probability of 3 randomly selected year to be NON leap year = (3/4)*(3/4)*(3/4) = 27/64 i.e. Less than 50% i.e. there are more than 50% chances that atleast one of the selected 3 years will be a Leap Year
Hence - 3 Years
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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09 Aug 2015, 10:21
VeritasPrepKarishma wrote:
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
Keeping it simple:
P (A person was born in leap year) = 1/4
This is less than 50% - not the answer
I understand and follow this part
P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)
Because we are looking for "at least one" shouldn't we be trying to determine P(A, or B, or Both) born in a leap year? I had the same answer as you and had the correct answer but probabilities are a bit of a weakness of mine so trying to work on it. I will show below my working.
The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)
P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16
This is less than 50% but close - not the answer
So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.
If you want to calculate it, you can do it as given below:
P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)
This is definitely more than 50%.
#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two are selected the probability that at least one is born in a leap year P(A only, B only, or A & B) = P(A) + P(B) + P(A&B) = (1/4 * 3/4) + (3/4 * 1/4) + (1/4 * 1/4) = 7/16
From here I just made the assumption that 3 people would push it above 50% and selected C.
I am just curious of the method that VeritasPrepKarishma used... Maybe there is a Thread or Blog post you could point me to that explains this in more detail.
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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09 Aug 2015, 21:42
DropBear wrote:
VeritasPrepKarishma wrote:
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
Keeping it simple:
P (A person was born in leap year) = 1/4
This is less than 50% - not the answer
I understand and follow this part
P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)
Because we are looking for "at least one" shouldn't we be trying to determine P(A, or B, or Both) born in a leap year? I had the same answer as you and had the correct answer but probabilities are a bit of a weakness of mine so trying to work on it. I will show below my working.
The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)
P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16
This is less than 50% but close - not the answer
So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.
If you want to calculate it, you can do it as given below:
P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)
This is definitely more than 50%.
#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two are selected the probability that at least one is born in a leap year P(A only, B only, or A & B) = P(A) + P(B) + P(A&B) = (1/4 * 3/4) + (3/4 * 1/4) + (1/4 * 1/4) = 7/16
From here I just made the assumption that 3 people would push it above 50% and selected C.
I am just curious of the method that VeritasPrepKarishma used... Maybe there is a Thread or Blog post you could point me to that explains this in more detail.
This method is a counterpart of SETS formulas we use.
10 people like A and 20 people like B. 5 like both A and B. So how many people like A or B?
n(A or B) = n(A) + n(B) - n(A and B)
We subtract n(A and B) because it is counted twice - once in n(A) and another time in n(B). But we want to count it only once so subtract it out once.
n(a or B) gives us the number of people who like at least one of A and B.
Similarly, we can use this method for probability
p(A or B) = p(A) + p(B) - p(A and B)
It is exactly the same concept. It gives us the probability that at least one of two people are born in a leap year.
When considering p(A), p(B) is to be ignored and when c considering p(B), p(A) is to be ignored. We take care of both when we subtract p(A and B).
Similarly, we can use the sets concept for 3 people using the three overlapping sets formula:
n(A or B or C) = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)
it becomes
p(A or B or C) = p(A) + p(B) + p(C) - p(A and B) - p(B and C) - p(C and A) + p(A and B and C)
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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09 Aug 2015, 22:01
1
DropBear wrote:
VeritasPrepKarishma wrote:
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
Keeping it simple:
P (A person was born in leap year) = 1/4
This is less than 50% - not the answer
I understand and follow this part
P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)
Because we are looking for "at least one" shouldn't we be trying to determine P(A, or B, or Both) born in a leap year? I had the same answer as you and had the correct answer but probabilities are a bit of a weakness of mine so trying to work on it. I will show below my working.
The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)
P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16
This is less than 50% but close - not the answer
So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.
If you want to calculate it, you can do it as given below:
P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)
This is definitely more than 50%.
#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two are selected the probability that at least one is born in a leap year P(A only, B only, or A & B) = P(A) + P(B) + P(A&B) = (1/4 * 3/4) + (3/4 * 1/4) + (1/4 * 1/4) = 7/16
From here I just made the assumption that 3 people would push it above 50% and selected C.
I am just curious of the method that VeritasPrepKarishma used... Maybe there is a Thread or Blog post you could point me to that explains this in more detail.
Also, check out this blog post: http://www.veritasprep.com/blog/2012/01 ... e-couples/
Here I have used sets in a combination problem. Conceptually they are all the same.
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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10 Aug 2015, 03:54
good question. But i doubt whether we can solve like this
#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two is selected P(A,B born in a leap year) = 2/4
Since we need more than 50% we pickup 3 as it is minimal among others.
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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10 Aug 2015, 04:31
Mechmeera wrote:
good question. But i doubt whether we can solve like this
#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two is selected P(A,B born in a leap year) = 2/4
Since we need more than 50% we pickup 3 as it is minimal among others.
Hi Mechmeera,
Highlighted step is the wrong step that you have written
Because when you select two individuals then one of them can be born in Leap Year and other is not and vise versa as well
So Probability of both A and B born in Leap year = (1/4)for A * (1/4)for B = 1/16
But Probability of One of them Born in leap year = (1/4)*(3/4) + (1/4)*(3/4) = 6/16
i.e. Probability that atleast one of A and B born in Leap year = (1/16)+(6/16) = 7/16 which is LESS than 50%
But when you follow the same process for 3 people then the probability exceeds 50%
I hope it helps!
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]
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10 Aug 2015, 22:08
Mechmeera wrote:
good question. But i doubt whether we can solve like this
#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two is selected P(A,B born in a leap year) = 2/4
Since we need more than 50% we pickup 3 as it is minimal among others.
The thought process is correct but the execution is not.
The method I have used is the same as this. When two are selected, the probability that at least one of them has bday in a leap year is 2/4 - 1/16, not only 2/4. I have explained 'why' here: what-is-the-minimum-number-of-randomly-chosen-people-needed-in-order-t-203018.html#p1559070
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Re: How many randomly assembled people are needed to have a [#permalink]
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10 Apr 2018, 22:06
Bunuel wrote:
nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
I found this question interesting, so thought of sharing it you folks.
Cheers
Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.
Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.
So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$
For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$
Thus min 3 people are needed.
Hi Bunuel,
What is wrong in the following approach?
Cases considered :
1. Born in a leap year
2. Not born in a leap year
In that case, the probability of not born in a leap year =1/2
And subsequently, the probability that at least one of them is born in a leap year would turn out to be greater than 50% for 2 people.
Thank you.
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Re: How many randomly assembled people are needed to have a [#permalink]
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10 Apr 2018, 22:13
Bakshi121092 wrote:
Bunuel wrote:
nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
I found this question interesting, so thought of sharing it you folks.
Cheers
Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.
Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.
So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$
For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$
Thus min 3 people are needed.
Hi Bunuel,
What is wrong in the following approach?
Cases considered :
1. Born in a leap year
2. Not born in a leap year
In that case, the probability of not born in a leap year =1/2
And subsequently, the probability that at least one of them is born in a leap year would turn out to be greater than 50% for 2 people.
Thank you.
Let me ask you: is the probability that you win a lottery 1/2?
1. You win the lottery
2. You won't win the lottery
Or: what is the probability to meet a dinosaur on the street? Is it 1/2? Either you meet it or not?
22. Probability
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Re: How many randomly assembled people are needed to have a &nbs [#permalink] 10 Apr 2018, 22:13
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# PS: Shaded area is half the triangle
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PS: Shaded area is half the triangle [#permalink] 15 Jul 2010, 07:57
00:00
Difficulty:
5% (low)
Question Stats:
80% (03:06) correct 20% (01:09) wrong based on 5 sessions
Hi guys,
I have some trouble with this exercise, the theory is clear, but I see too much information and I dont know how to proceed. Any ideas would be helpful, thanks!
Attachment:
d1.JPG [ 5.29 KiB | Viewed 1424 times ]
If the shaded area is one half the area of the triangle ABC and angle ABC is a right angle, then the length of the line segment AD is:
A.(1/2)*w
B.(1/2)*(w+x)
C.\sqrt{2x^2+z^2}
D.\sqrt{w^2-3y^2}
E.\sqrt{y^2+z^2}
[Reveal] Spoiler: OA
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Re: PS: Shaded area is half the triangle [#permalink] 15 Jul 2010, 08:30
Expert's post
Nusa84 wrote:
Hi guys,
I have some trouble with this exercise, the theory is clear, but I see too much information and I dont know how to proceed. Any ideas would be helpful, thanks!
Attachment:
d1.JPG
If the shaded area is one half the area of the triangle ABC and angle ABC is a right angle, then the length of the line segment AD is:
A.(1/2)*w
B.(1/2)*(w+x)
C.\sqrt{2x^2+z^2}
D.\sqrt{w^2-3y^2}
E.\sqrt{y^2+z^2}
Property of median: Each median divides the triangle into two smaller triangles which have the same area.
So, AD is median and thus y=z, (we could derive this even not knowing the above property. Given: area_{ABC}=\frac{x(y+z)}{2}=2*area_{ABD}=2*\frac{xy}{2} --> y=z).
Next: AD is hypotenuse in right triangle ABD and thus AD=\sqrt{x^2+y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+(2y)^2-3y^2}=\sqrt{w^2-3y^2} (as 2y=y+z=BC and AC^2=w^2=AB^2+BC^2=x^2+(2y)^2, so w^2=x^2+(2y)^2).
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Re: PS: Shaded area is half the triangle [#permalink] 15 Jul 2010, 08:44
Bunuel wrote:
Nusa84 wrote:
Hi guys,
I have some trouble with this exercise, the theory is clear, but I see too much information and I dont know how to proceed. Any ideas would be helpful, thanks!
Attachment:
d1.JPG
If the shaded area is one half the area of the triangle ABC and angle ABC is a right angle, then the length of the line segment AD is:
A.(1/2)*w
B.(1/2)*(w+x)
C.\sqrt{2x^2+z^2}
D.\sqrt{w^2-3y^2}
E.\sqrt{y^2+z^2}
Property of median: Each median divides the triangle into two smaller triangles which have the same area.
So, AD is median and thus y=z, (we could derive this even not knowing the above property. Given: area_{ABC}=\frac{x(y+z)}{2}=2*area_{ABD}=2*\frac{xy}{2} --> y=z).
Next: AD is hypotenuse in right triangle ABD and thus AD=\sqrt{x^2+y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+(2y)^2-3y^2}=\sqrt{w^2-3y^2} (as 2y=y+z=BC and AC^2=w^2=AB^2+BC^2=x^2+(2y)^2, so w^2=x^2+(2y)^2).
Quite a tough one, very good explanation, thanks
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Re: PS: Shaded area is half the triangle [#permalink] 15 Jul 2010, 08:58
I understood everything up until the \sqrt{x^2+4y^2-3y^2} part. how was this derived?
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Re: PS: Shaded area is half the triangle [#permalink] 15 Jul 2010, 11:04
Bunuel wrote:
azule45 wrote:
I understood everything up until the \sqrt{x^2+4y^2-3y^2} part. how was this derived?
y^2=4y^2-3y^2.
i'm sorry, i should have been more specific. i wasn't talking about substituting 4y^2-3y^2 for y^2, but rather where did 4y^2-3y^2 come from? i can't seem to figure this out. everything else makes perfect sense, but this.
much appreciated and sorry if this should be obvious.
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Re: PS: Shaded area is half the triangle [#permalink] 15 Jul 2010, 11:37
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Expert's post
azule45 wrote:
Bunuel wrote:
azule45 wrote:
I understood everything up until the \sqrt{x^2+4y^2-3y^2} part. how was this derived?
y^2=4y^2-3y^2.
i'm sorry, i should have been more specific. i wasn't talking about substituting 4y^2-3y^2 for y^2, but rather where did 4y^2-3y^2 come from? i can't seem to figure this out. everything else makes perfect sense, but this.
much appreciated and sorry if this should be obvious.
This is just an algebraic operation in order to pair x^2 and 4y^2, (added they give w^2).
Question is AD=? Well it's simple to get that AD=\sqrt{x^2+y^2}, but this option is not listed and thus we need to transform it to get the option which is listed (or the other way around we need to transform the options to get this one).
Now, options A and B does not make any sense and we can get rid of them.
Option C: \sqrt{2x^2+z^2}=\sqrt{2x^2+y^2} as y=z hence it's out too as \sqrt{2x^2+y^2}\neq{\sqrt{x^2+y^2}}.
Option D: \sqrt{w^2-3y^2}, w is hypotenuse hence it's equal to w^2=x^2+(y+z)^2=x^2+(2y)^2=x^2+4y^2, so \sqrt{w^2-3y^2}=\sqrt{x^2+4y^2-3y^2}=\sqrt{x^2+y^2}=AD.
Hope it's clear.
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Re: PS: Shaded area is half the triangle [#permalink] 16 Jul 2010, 07:09
ahhh haa, Bunuel. so you worked the problem using the answer choices. got it. smart move. much appreciated for your help.
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Re: PS: Shaded area is half the triangle [#permalink] 16 Jul 2010, 07:09
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## Index: Xⁿ starts with n
Often said in my lessons are that 2^6 is 64 and 2^10 is ten twenty four, 1024, and that these are the only 2ⁿ that start with n. [Prove!] This page explores the relationship between integers to the power of another integer, looking especially at those results which, like 2⁶ = 64, begin with the same digits as the index used.
For example, 42⁵⁹ is 5.9x10⁹⁵ or 42⁵⁹ = 59x10⁹⁴, in a non-standard format.
Sometimes we will need general terms to describe this computation; I choose to refer, should I need to and using the example 42⁵⁹, such that 42 is the base and 59 is the index.
1. Can you point straightaway to some powers of 20 that will fit this property of 20ⁿ starting with n? Demonstrate that you're right and make a statement about other cases where the base number is a multiple of ten.
2. Thinking of easy squares —those you know well— spot the square that starts with a two. Write this down and also the similar result for ten times as much, which demonstrates the property you wrote down in Q1.
3. Using your calculator, show that there is a run of (two-digit) squares as the base number rises by one that also start with the digit two.
4. 2^10 is given at the top and is generally assumed knowledge (by me in my classes), so you might easily recognise that 8⁴ (= 2¹², = 4x1024) starts with a 4. More than this, you should now recognise that 80⁴ starts with a 4 and, quite possibly, 81⁴ does. Show that you can extend this towards 85 and show that 84 works but 85 does not.
5. Quite soon your calculator complains. By writing the values down in standard form, show that these calculations fit the requirement:
31⁸, 58⁵⁶, 87⁵⁴, 99⁵⁶.
For some calculators, those last two will prove too hard. Explain why this is, and use what you learned in Q1 and Q2 to avoid the problem and still find the answer. Whatever solution you found, show it. One is enough.
6. Now you see a way to recognise whether you've found combinations that work, find which of these fit the property of having the first few digits (or non-zero digits if you're using numbers less than unity) matching the index: 86⁶⁴, 88⁵⁰, 470³², 850⁴, 840¹², 81⁸⁸, 82¹¹, 99855⁸⁸
The result you discovered in Q4 suggests that if a single digit number like 8 raised to a power like 4 fits, then it is likely that 8.1 and 8.2 will fit until some limit is reached.
7. Find the simple 0.7ⁿ such the first non-zero digit is n and find the biggest number you can, quite close to 0.7, that still starts with that n.
8. 0.55⁵=0.050, so 0.551⁵ begins 0.5. The first number bigger than 0.55, X, that fails to fit the pattern here (numbers close to 0.55 that fit the property) is such that X⁵=0.06. [Obviously; if it doesn't start with 5 and it is a bit bigger, it must start with 6.] Find this X and consequently show the biggest integer close to 550 such that its fifth power will begin with the digit 5. Does 549 work?
Around about now you begin to see that using numbers between zero and one is going to be an awful lot easier. So we might change 'fit the pattern' to 'the truncated decimal matches the index' and call our effect something like a 'truncated index match'. In class, we might call that a 'tim', especially if there was a Tim present.
9. The fifth power produces more numbers of this sort. 0.9⁵= 0.59049. By looking at the fifth root of 0.5 and of 0.6, find the range of integers close to 900 (above and below) whose fifth power starts with a 5.
Looking for more numbers where X⁵ begins with a 5 but not a 6, Q8&Q9 suggest that looking at the fifth root of 0.5, 0.05, 0.005 etc and 0.6, 0.06, 0.006, etc will reveal a range of three digit numbers that fit the pattern.
10. By writing the fifth root of 5x10⁻ⁿ and 6x10⁻ⁿ for various n, find the five groups of numbers below 10000 which will fit the pattern.
You now have a techniques for discovering a lot of numbers that fit this pattern. I decided to be interested in the fifteenth power, which produced a table like this on Excel.
The bottom row, which I'll call row 15, suggests to me that, of numbers below 1000, 103¹⁵=0.15.....
11. Use the table given ( or make your own) to write down which integers¹⁵ below 1000 will begin 0.15.
Here are the formulas I put into EXCEL, where I could have Named cell R4 as 'index', in which case it would look a lot tidier, like this:-
I've decided to be interested in the index being 37, as you can see.
12. Produce your own spreadsheet (this is now well into GCSE-rated project-work) and investigate.
I say the digit integers such that X³⁷ begins 37... when truncated (i.e. fitting the 'Xⁿ starts with an n' pattern we've been investigating) include 63 and 86, but also 338, 383 and nine more three digit integers.
The same spreadsheet showed that 0.9856^137 =0.137(0866), 0.9692^137 = 0.0137(057), 0.9063^137= 137.94x10⁻⁴ and so on. 766^137 works, but I think it is the only three digit solution.
I found it quite difficult to persuade my spreadsheet to calculate or show which integer solutions would work without it being me doing the decision-making. Maybe I was being thick, but at the time I found this hard.
Few students will complete all twelve questions. This might make up a week of lessons for some classes, a independent piece of coursework, or just a weekend homework.
This page as a whole could be seen as chasing numbers around on a calculator. It could also be seen as having a little problem that deserves just enough attention to cause a class collectively or individually to hunt for answers. It may serve to educate a class into the use of the xʸ button and its inverse, ˣ√y. Not least, it shows that while there may be 'answers' there is also a region of grey not-quite answers in the immediate neighbourhood. The more maths one does the less the subject is black and white and it becomes seen as more filled with many shades of grey. I would suggest that a whole-class exercise gives the feature to be pursued its own name (Tim, perhaps)— as I've written before, possession is nine-tenths of the learning.
If we did some maths first (rather than play with arithmetic) then all the exploration turns into preparation for understanding what comes next - the maths. For an index of 37,
we want 0.37≤X³⁷<0.38, so ³⁷√0.37≤X< ³⁷√0.38,
which in turn means 0.973486≤X<0.974188 so that the integer 'solutions' are 974, 9735-41, 97349-97418 and so on. That approach (eventually) produced this next spreadsheet, where the index is 5 and the 3 indicates that solutions will be recorded from 1 to 999. The bold 5 is cell AC1 and is now called Nind, so the leftmost column is AB. I have a figure 3 in cell AE1 which is the number of digits (I thought of it as 'size') I want to have shown before a decimal point, as the next example shows. Beware filling downwards, for I've used pairs of lines (columns AC and AF) to achieve my objectives.
That same spreadsheet now shows all of the results for X⁵ = 5....., where X<1000. The top two 871 and 902 (on the right in Col AF) indicate that all the integers from 871 to 902 inclusive succeed in beginning with a five. While that 871-902 range implies 90 and 9 will also fit (because 900 does), filling the table downwards produces these answers too.
I swap the top left 5 to a 37 and instantly have the results for that different problem in descending order: 974, 915, 860, 808, 759, 630, 592, 491, 383, 338 and presumably 86 and 63. And no others. 1036 appeared in my table and works. I subsequently found some larger numbers where it is more difficult to show the valid integer range, such as where X³⁷ = 3.7x10⁵, for which 1415 works
I switch the index to 137 and only 766 shows. I change the size setting from 3 to 4 and get this quite long list:- 9856, 9692, 9062, 8616, 8056, 7660, 7162, 6925, 6475, 6367, 6054, 5953, 5566, 5473, 5382, 5204, 5032, 4948, 4399, 3977, 2560, 2402, 2362, 1805, 1166 and of course 766.
DJS 20210531
1. Surely 20^6 = (2x10)⁶ = 2⁶x10⁶. Similarly 20^6 = 1024x10⁶ = 10.24x10¹¹
5. 87⁵⁴ = 8.7⁵⁴x 10⁵⁴ = 5.4x10⁵⁰x10⁵⁴ = 5.4x10¹⁰⁴. Extending this 0.87⁵⁴ = 0.00054(2050) =5.4x10⁻⁴ so 87⁵⁴ = 5.4x10⁻⁴x10¹⁰⁸ = 5.4x10¹⁰⁴. Some will go to a bigger calculator, such as within Excel. That gets the answer but doesn't understand why the calculator won't do it.
6. 8.4⁴ = 4978 so 84 and 840 both fit the property but 8.5 = 5220, so doesn't. Therefore 850⁴ doesn't either. You might prefer to work with 0.84⁴=0.4(9787) and 0.85⁴=0.5(2200625).
7. 0.7³ = 0.343, 0.73³=0.389, 0.736806300 is just too big. Think about how to persuade the calculator to tell you that. This isn't about chasing answers, it is about understanding how numbers fit together.
8. ⁵√0.06 = 0.569679052 so 569 works but not 570. Also ⁵√0.05 = 0.54928 so 55 and 56 both work, making the whole range here the integers from 550 to 569.
9. ⁵√0.6 = 0.90288 so 901 and 902 will work but 903 will not. ⁵√0.5 = 0.87055 so 870 won't work but 871 will, and many bigger integers. So the range of three digit numbers here that work is 871 to 902. We could write this as 871 ≤ X ≤ 902.
10. ⁵√0.6 = 0.90288 and ⁵√0.5 = 0.87055 so 871 to 902 from Q9
⁵√0.06 = 0.5696 and ⁵√0.05 = 0.5492. So 550 to 569 (not 549, not 570) from Q8
⁵√0.006 = 0.3594, ⁵√0.005 = 0.34657. So 35⁵ works, and 3466⁵ to 3593⁵ will work
⁵√0.0006 = 0.22679, ⁵√0.0005 = 0.21867 ; 22⁵ works and 2187 to 2267 will work
⁵√0.00006 = 0.143096; ⁵√0.00005 = 0.13797; 14⁵ works and 1380 to 1430 will work.
Inserting yet another zero causes the pattern to repeat. Some will say "Of course it does" and others will need to hunt for why this is true. I suggest you write some of the numbers in standard form.
11. Working from the bottom row upwards gives the values in number order; 103, 120, 140, 163, 190, 222, 259, 301-2, 351-2, 410, 477-8, 556-8, 649-51, 756-9, 882-4 and, removing zeroes, 12,14 and 19 also work..
In November 2021 I threw this at a Y13 already known to be a good mathematician (and Oxbridge aspirant) and he found this hard. But then he had only had 20 minutes of horrible, off-the-wall me and not the backgrouind that any PMC Y8 top set student would have become entirely familiar with. So perhaps problems like this are harder than I imagine them to be. Alternatively maybe a whole classful of bright Y8 are actually sharing enough working brain cells (called teamwork if I'm polite) to produce some relatively brilliant results?
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Liethagoras, Pythagoras' cousin (!), was jealous of Pythagoras and came up with his own theorem. Read this article to find out why other mathematicians laughed at him.
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05 Feb 2009, 05:42
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What is the sum of digits in decimal notation of number $$10^{20} - 16$$ ?
(A) 158
(B) 162
(C) 165
(D) 174
(E) 183
Source: GMAT Club Tests - hardest GMAT questions
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05 Feb 2009, 05:49
10^20 will have one 1 and twenty 0's.
10^20 - 16 will have one 4, one 8 and seventeen 9's
Hence, sum of digits = 4 + 8 + 17*9
= 12 + 153 = 165
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06 Oct 2009, 22:11
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Right answer (taken from the test) is:
$$10^{20} - 16 = 10...(20 \text{ zeros})...0 - 16 = 9...(18 \text{ nines})...984$$ . The sum of digits $$= 18*9 + 8 + 4 = 174$$ .
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27 Apr 2010, 06:43
Yes D is correct answer. If you subtract 1 from 10^20 it will be 999(20 times) if you subtract 15 more from that, you will get last digit 4 second last 8 and all others 9. So answer is D.
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27 Apr 2010, 08:35
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10-6 always gives 4 in the units place ..the only answer choice that has 4 in the units place is D.
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27 Apr 2010, 12:58
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I did it the dinosaur way which still didn't take too long.
100000000000000000000
- 16
99999999999999999984
(9 * 18) + 8 + 4 = 174
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29 Apr 2010, 12:24
amitjash wrote:
Yes D is correct answer. If you subtract 1 from 10^20 it will be 999(20 times) if you subtract 15 more from that, you will get last digit 4 second last 8 and all others 9. So answer is D.
I did it the same way. Subtracting 1 first seemed to help me visualize it.
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12 May 2010, 17:55
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Well, $$10^2-16=84$$ and $$10^3-16 = 984$$. So, we see the exponent tells us the number of digits involved in our answer (2 in the former case, 3 in the latter).
$$10^2^0-16=999...984$$; two numbers are the 8 and the 4, whereas the other eighteen are 9s!
$$(18*9)+8+4=174$$
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29 Apr 2011, 05:21
$$10^2^0$$ would be a 21 digit number ... subtract 16 and you are left with a 20 digit number with 2 digits as 4 and 8 and the rest 9s. The total is $$8+4+18*9 = 174$$ therefore (D)
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29 Apr 2011, 05:39
Last two digits = 84 and there are 18 9s
8 + 4 + 9 * 18
= 12 + 162
= 174
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03 May 2012, 18:36
zz0vlb wrote:
10-6 always gives 4 in the units place ..the only answer choice that has 4 in the units place is D.
If it was 10^2-16=100-16=84, the sum of the digits would be 8+4=12. So I don't think we can use the shortcut suggested.
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05 May 2012, 10:37
zz0vlb wrote:
10-6 always gives 4 in the units place ..the only answer choice that has 4 in the units place is D.
i liked the way u have answered mate....no need of calculation, but just keen observation.. +1 to u
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topmbaseeker wrote:
What is the sum of digits in decimal notation of number $$10^{20} - 16$$ ?
(A) 158
(B) 162
(C) 165
(D) 174
(E) 183
Source: GMAT Club Tests - hardest GMAT questions
What is the sum of the digits in decimal notation of number $$10^{20}-16$$?
A. 158
B. 162
C. 165
D. 174
E. 183
$$10^{20}$$ has 21 digits: 1 followed by 20 zeros;
$$10^{20}-16$$ has 20 digits: 18 9's and 84 in the end;
So, the sum of the digits of $$10^{20}-16$$ equals to 18*9+8+4=174.
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06 May 2013, 22:59
I don't see the need of subtracting the entire number and finding out the number of 9's .
1000 - 16 = 984
10000 -16 = 9984
So, the number of 9's would be the power of 10 minus 2.
So, answer = (20-2)*9 + (8+4) = 174
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07 May 2013, 01:36
18*9+8+4=174
I will go with 'D'
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07 May 2013, 13:19
zz0vlb wrote:
10-6 always gives 4 in the units place ..the only answer choice that has 4 in the units place is D.
You got one +1 from me in the first place but actually the way you "solved" problem is wrong.
It would only be right, if the sum of the previous numbers is 0.
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21 Apr 2014, 05:11
Bunuel wrote:
topmbaseeker wrote:
What is the sum of digits in decimal notation of number $$10^{20} - 16$$ ?
(A) 158
(B) 162
(C) 165
(D) 174
(E) 183
Source: GMAT Club Tests - hardest GMAT questions
What is the sum of the digits in decimal notation of number $$10^{20}-16$$?
A. 158
B. 162
C. 165
D. 174
E. 183
$$10^{20}$$ has 21 digits: 1 followed by 20 zeros;
$$10^{20}-16$$ has 20 digits: 18 9's and 84 in the end;
So, the sum of the digits of $$10^{20}-16$$ equals to 18*9+8+4=174.
Similar questions to practice:
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if-x-is-a-positive-integer-and-10-x-74-in-decimal-notation-61013.html
Hope it helps.
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04 May 2014, 14:12
Bunuel wrote:
topmbaseeker wrote:
What is the sum of digits in decimal notation of number $$10^{20} - 16$$ ?
(A) 158
(B) 162
(C) 165
(D) 174
(E) 183
Source: GMAT Club Tests - hardest GMAT questions
What is the sum of the digits in decimal notation of number $$10^{20}-16$$?
A. 158
B. 162
C. 165
D. 174
E. 183
$$10^{20}$$ has 21 digits: 1 followed by 20 zeros;
$$10^{20}-16$$ has 20 digits: 18 9's and 84 in the end;
So, the sum of the digits of $$10^{20}-16$$ equals to 18*9+8+4=174.
D can't be the answer. once we subtract 16 from 10^20 we have only 19 numbers left. among 19 numbers 17 are nine and last two are 8 and 4.
therefor sum is 17*9+8+4=165
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04 May 2014, 14:15
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OA is wrong
C is the right ans
because on subtracting 16 from 10^20, we have only 19 nos left.
17 are 9 and last two are 8 and 4.
so sum= 17*9+8+4=165
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05 May 2014, 01:59
daboo343 wrote:
Bunuel wrote:
topmbaseeker wrote:
What is the sum of digits in decimal notation of number $$10^{20} - 16$$ ?
(A) 158
(B) 162
(C) 165
(D) 174
(E) 183
Source: GMAT Club Tests - hardest GMAT questions
What is the sum of the digits in decimal notation of number $$10^{20}-16$$?
A. 158
B. 162
C. 165
D. 174
E. 183
$$10^{20}$$ has 21 digits: 1 followed by 20 zeros;
$$10^{20}-16$$ has 20 digits: 18 9's and 84 in the end;
So, the sum of the digits of $$10^{20}-16$$ equals to 18*9+8+4=174.
D can't be the answer. once we subtract 16 from 10^20 we have only 19 numbers left. among 19 numbers 17 are nine and last two are 8 and 4.
therefor sum is 17*9+8+4=165
Nope.
$$10^{20}$$ has 21 digits: 1 followed by 20 zeros: 100,000,000,000,000,000,000
$$10^{20}-16$$ has 20 digits: 18 9's and 84 in the end: 99,999,999,999,999,999,984.
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# m16#36
Moderator: Bunuel
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 4,004 | 11,375 | {"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} | 4.21875 | 4 | CC-MAIN-2017-43 | latest | en | 0.790834 |
http://login.odine.co/cantor-set-wikipedia.html | 1,660,170,105,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571222.74/warc/CC-MAIN-20220810222056-20220811012056-00777.warc.gz | 34,307,299 | 24,965 | ## Cantor Set Wikipedia
### Cantor set - Wikipedia.
In mathematics, the Cantor set is a set of points lying on a single line segment that has a number of unintuitive properties. It was discovered in 1874 by Henry John Stephen Smith and introduced by German mathematician Georg Cantor in 1883.. Through consideration of this set, Cantor and others helped lay the foundations of modern point-set topology.The most common construction ....
https://en.wikipedia.org/wiki/Cantor_set.
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In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti-diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers..
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A cantor or chanter is a person who leads people in singing or sometimes in prayer.In formal Christian worship, a cantor is a person who sings solo verses or passages to which the choir or congregation responds.. In Judaism, a cantor is one who sings and leads people in prayer in a Jewish religious service and may be called hazzan.A cantor in Reform and Conservative ....
https://en.wikipedia.org/wiki/Cantor.
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In mathematics, the Cantor function is an example of a function that is continuous, but not absolutely continuous.It is a notorious counterexample in analysis, because it challenges naive intuitions about continuity, derivative, and measure. Though it is continuous everywhere and has zero derivative almost everywhere, its value still goes from 0 to 1 as its argument reaches from ....
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Eddie Cantor (born Isidore Itzkowitz; January 31, 1892 - October 10, 1964) was an American "illustrated song" performer, comedian, dancer, singer, vaudevillian, actor, and songwriter. ... In 1933, Brown and Bigelow published a set of 12 Eddie Cantor caricatures by Frederick J. Garner. The advertising cards were purchased in bulk as a direct ....
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https://en.wikipedia.org/wiki/List_of_films_set_in_ancient_Rome.
### Paradoxes of set theory - Wikipedia.
In the same year the French mathematician Jules Richard used a variant of Cantor's diagonal method to obtain another contradiction in naive set theory. Consider the set A of all finite agglomerations of words. The set E of all finite definitions of real numbers is a subset of A.As A is countable, so is E.Let p be the nth decimal of the nth real number defined by the set E; we ....
### Cardinality of the continuum - Wikipedia.
In set theory, the cardinality of the continuum is the cardinality or "size" of the set of real numbers, sometimes called the continuum.It is an infinite cardinal number and is denoted by (lowercase fraktur "c") or | |.. The real numbers are more numerous than the natural numbers.Moreover, has the same number of elements as the power set of . Symbolically, if the ....
https://en.wikipedia.org/wiki/Cardinality_of_the_continuum.
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Space Battleship Yamato (SPACE BATTLESHIP ???, Supesu Batorushippu Yamato) is a 2010 Japanese science fiction film based on the Space Battleship Yamato anime series by Yoshinobu Nishizaki and Leiji Matsumoto.The film was released in Japan on December 1, 2010. It was released on DVD and Blu-ray in Japan on June 24, 2011, and in the United States by ....
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### Georg Cantor - Wikipedia, la enciclopedia libre.
Georg Ferdinand Ludwig Philipp Cantor (San Petersburgo, 3 de marzo de 1845 - Halle, 6 de enero de 1918) fue un matematico nacido en Rusia, aunque nacionalizado aleman, y de ascendencia austriaca y judia. [1] Fue inventor con Dedekind de la teoria de conjuntos, que es la base de las matematicas modernas. Gracias a sus atrevidas investigaciones sobre los ....
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Pedophilia (alternatively spelt paedophilia) is a psychiatric disorder in which an adult or older adolescent experiences a primary or exclusive sexual attraction to prepubescent children. Although girls typically begin the process of puberty at age 10 or 11, and boys at age 11 or 12, criteria for pedophilia extend the cut-off point for prepubescence to age 13..
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### Chant - Wikipedia.
A chant (from French chanter, from Latin cantare, "to sing") is the iterative speaking or singing of words or sounds, often primarily on one or two main pitches called reciting tones.Chants may range from a simple melody involving a limited set of notes to highly complex musical structures, often including a great deal of repetition of musical subphrases, such as Great Responsories ....
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North America's first transcontinental railroad (known originally as the "Pacific Railroad" and later as the "Overland Route") was a 1,911-mile (3,075 km) continuous railroad line constructed between 1863 and 1869 that connected the existing eastern U.S. rail network at Council Bluffs, Iowa with the Pacific coast at the Oakland Long Wharf on San Francisco Bay..
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### Georg Cantor — Wikipédia.
Georg Cantor est un mathematicien allemand, ne le 3 mars 1845 a Saint-Petersbourg (Empire russe) et mort le 6 janvier 1918 a Halle (Empire allemand).Il est connu pour etre le createur de la theorie des ensembles.. Il etablit l'importance de la bijection entre les ensembles, definit les ensembles infinis et les ensembles bien ordonnes.Il prouva egalement que les nombres reels ....
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William Clark Styron Jr. (June 11, 1925 - November 1, 2006) was an American novelist and essayist who won major literary awards for his work. Styron was best known for his novels, including: Lie Down in Darkness (1951), his acclaimed first work, published when he was 26;; The Confessions of Nat Turner (1967), narrated by Nat Turner, the leader of an 1831 Virginia slave ....
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The notation for this last concept can vary considerably. Set theorists will sometimes write "", while others will instead write "".The latter notation can be generalized to "", which refers to the intersection of the collection {:}.Here is a nonempty set, and is a set for every .. In the case that the index set is the set of natural numbers, notation analogous to that of an infinite product ....
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### Chris Brown (cantor) – Wikipédia, a enciclopédia livre.
Christopher Maurice Brown (Tappahannock, 5 de maio de 1989), mais conhecido como Chris Brown, e um cantor, rapper, compositor, dancarino, ator, produtor musical, coreografo e grafiteiro norte-americano.. Lancou seu primeiro album com apenas 16 anos de idade auto intitulado Chris Brown.O primeiro single lancado foi "Run It", que debutou em 92? e depois de 14 semanas ....
https://pt.wikipedia.org/wiki/Chris_Brown_(cantor). | 2,473 | 10,614 | {"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} | 3.703125 | 4 | CC-MAIN-2022-33 | latest | en | 0.897101 |
http://math.stackexchange.com/questions/15294/why-is-the-area-under-a-curve-the-integral?lastactivity | 1,433,210,285,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195034286.17/warc/CC-MAIN-20150601214354-00009-ip-10-180-206-219.ec2.internal.warc.gz | 129,957,977 | 27,010 | # Why is the area under a curve the integral?
I understand how derivatives work based on the definition, and the fact that my professor explained it step by step until the point where I can derive it myself.
However when it comes to the area under a curve for some reason when you break it up into an infinite amount of rectangles, magically it turns into the anti-derivative. Can someone explain why that is the definition of the integral and how Newton figured this out?
-
There are 2 potential questions here. One is the question of why the definite Riemann integral gives the correct notion of "area under a curve" for a (nonnegative, Riemann integrable) function. The other, which seems to be what you're really asking, is the question of why an antiderivative evaluated at the endpoints of an interval and subtracted yields that definite integral. The latter question is answered by an understanding of the fundamental theorem of calculus. The purpose of this comment is just to help clarify your question. – Jonas Meyer Dec 23 '10 at 4:58
Its also possible that he is seeking the intuition for Calculus. Specifically integration – picakhu Dec 23 '10 at 5:19
@picakhu: I think that he or she is certainly seeking intuition, but the phrase "magically it turns into the anti-derivative" indicates that this is really about the fundamental theorem of calculus, and not just about the definition of the integral. – Jonas Meyer Dec 23 '10 at 5:25
I don't understand the difference between the two – qwertymk Dec 23 '10 at 5:27
@qwertymk: Well, now you have a fabulous explanation from Arturo Magidin, but I'll also mention that en.wikipedia.org/wiki/Riemann_integral is a good starting point for learning about how the integral is actually defined, while en.wikipedia.org/wiki/Fundamental_theorem_of_calculus is a good starting point for learning about the relationship with antiderivatives. Both include excellent graphics! – Jonas Meyer Dec 23 '10 at 5:32
First: the integral is defined to be the (net signed) area under the curve. The definition in terms of Riemann sums is precisely designed to accomplish this. The integral is a limit, a number. There is, a priori, no connection whatsoever with derivatives. (That is one of the things that makes the Fundamental Theorems of Calculus such a potentially surprising things).
Why does the limit of the Riemann sums actually give the area under the graph? The idea of approximating a shape whose area we don't know both from "above" and from "below" with areas we do know goes all the way back to the Greeks. Archimedes gave bounds for the value of $\pi$ by figuring out areas of inscribed and circumscribed polygons in a circle, knowing that the area of the circle would be somewhere between the two; the more sides to the polygons, the closer the inner and outer polygons are to the circle, the closer the areas are to the area of the circle.
The way Riemann tried to formalize this was with the "upper" and "lower" Riemann sums: assuming the function is relatively "nice", so that on each subinterval it has a maximum and a minimum, the "lower Riemann sum" is done by taking the largest "rectangle" that will lie completely under the graph by looking at the minimum value of the function on the interval and using that as height; and the "upper Riemann sum" is done by taking the smallest rectangle for which the graph will lie completely under it (by taking the maximum value of the function as the height). Certainly, the exact area under the graph on that interval will be somewhere between the two. If we let $\underline{S}(f,P)$ be the lower sum corresponding to some fixed partition $P$ of the interval, and $\overline{S}(f,P)$ be the upper sum, we will have that $$\underline{S}(f,P) \leq \int_a^b f(x)\,dx \leq \overline{S}(f,P).$$ (Remember that $\int_a^bf(x)\,dx$ is just the symbol we use to denote the exact (net signed) area under the graph of $f(x)$ between $a$ and $b$, whatever that quantity may be.)
Also, intuitively, the more intervals we take, the closer these two approximations (one from below and one from above) will be. This does not always work out if all we do is take "more" intervals. But one thing we can show is that if $P'$ is a refinement of $P$ (it includes all the dividing points that $P$ had, and possibly more points) then $$\underline{S}(f,P)\leq \underline{S}(f,P')\text{ and } \overline{S}(f,P')\leq \overline{S}(f,P)$$ so at least the approximations are heading in the right direction. To see why this happens, suppose you split one of the subintervals $[t_i,t_{i+1}]$ in two, $[t_i,t']$ and $[t',t_{i+1}]$. The minimum of $f$ on $[t_i,t']$ and on $[t',t_{i+1}]$ are each greater than or equal to the minimum over the whole of $[t_i,t_{i+1}]$, but it may be that the minimum in one of the two bits is actually strictly larger than the minimum over $[t_i,t_{i+1}]$. The areas we get after the split can be no smaller, but they can be larger than the ones we had before the split. Similarly for the upper sums.
So, let's consider one particular sequence of partitions: divide the interval into 2 equal parts; then into 4; then into 8; then into 16; then into 32; and so on; then into $2^n$, etc. If $P_n$ is the partition that divides $[a,b]$ into $2^n$ equal parts, then $P_{n+1}$ is a refinement of $P_n$, and so we have: $$\underline{S}(f,P_1) \leq\cdots \leq \underline{S}(f,P_n)\leq\cdots \leq\int_a^b f(x)\,dx \leq\cdots \leq\overline{S}(f,P_n)\leq\cdots \leq \overline{S}(f,P_2)\leq\overline{S}(f,P_1).$$
Now, the sequence of numbers $\underline{S}(f,P_1) \leq \underline{S}(f,P_2)\leq\cdots \leq \underline{S}(f,P_n)\leq\cdots$ is increasing and bounded above (by the area). So the numbers have a supremum; call it $\underline{S}$. This number is no more than $\int_a^b f(x)\,dx$. And the numbers $\overline{S}(f,P_1) \geq \overline{S}(f,P_2)\geq\cdots \geq \overline{S}(f,P_n)\geq\cdots$ are decreasing and bounded below, so they have a minimum; call this $\overline{S}$; again, it is no less than $\int_a^bf(x)\,dx$. So we have: $$\lim_{n\to\infty}\underline{S}(f,P_n) = \underline{S} \leq \int_a^b f(x)\,dx \leq \overline{S} = \lim_{n\to\infty}\overline{S}(f,P_n).$$ What if we are lucky? What if actually we have $\underline{S}=\overline{S}$? Then it must be the case that this common value is the value of $\int_a^b f(x)\,dx$. It just doesn't have a choice! It's definitely trapped between the two, and if there is no space between them, then it's equal to them.
What Riemann proved was several things:
1. If $f$ is "nice enough", then you will necessarily get that $\underline{S}=\overline{S}$. In particular, continuous functions happen to be "nice enough", so it will definitely work for them (in fact, continuous functions turn out to be "very nice", not just "nice enough").
2. If $f$ is "nice enough", then you don't have to use the partitions we used above. You can use any sequence of partitions, so long as the "mesh size" (the size of the largest subinterval in the partition) gets smaller and smaller, and has limit of $0$ as $n\to\infty$; if it works for the partitions "divide-into-$2^n$-equal-intervals", then it works for any sequence of partitions whose mesh size goes to zero.
So, for example, we can take $P_n$ to be the partition that divides $[a,b]$ into $n$ equal parts, even though $P_{n+1}$ is not a refinement of $P_n$ in this case.
3. In fact, you don't have to do $\underline{S}(f,P)$ and $\overline{S}(f,P)$. For the partition $P$, just pick any rectangle that has as its height any value of the function in the subinterval (that is, pick an arbitrary $x_i^*$ in the subinterval $[t_i,t_{i+1}]$, and use $f(x_i^*)$ as the height). Call the resulting sum $S(f,P,x_1^*,\ldots,x_n^*)$. Then you have $$\underline{S}(f,P) \leq S(f,P,x_1^*,\ldots,x_n^*)\leq \overline{S}(f,P)$$ because $\underline{S}(f,P)$ is computed using the smallest possible values of $f$ throughout, and $\overline{S}(f,P)$ is computed using the largest possible values of $f$ throughout. But since we already know, from 1 and 2 above, that $\underline{S}(f,P)$ and $\overline{S}(f,P)$ have the same limit, then the sums $S(f,P,x_1^*,\ldots,x_n^*)$ also get squeezed and must have that same limit, which equals the integral.
In particular, we can always take the left endpoint (and get a "Left Hand Sum") or we can always take the right endpoint (and get a "Right Hand Sum"), and you will nevertheless get the same limit.
So in summary, you can pick any sequence of partitions, whichever happens to be convenient, so long as the mesh size goes to $0$, and you can pick any points on the subintervals (say, ones which make the calculations simpler) at each stage, and so long as the function is "nice enough" (for example, if it is continuous), everything will work out and the limit will be the number which must be the value of the area (because it was trapped between the lower and upper sums, and they both got squeezed together trapping the limit and the integral both between them).
Now, (1) and (2) above are the hardest part of what Riemann did. Don't be surprised if it sounds a bit magical at this point. But I hope that you agree that if the lower and upper sums for the special partitions have the same limits then that limit must be the area that lies under the graph.
Thanks to that work of Riemann, then (at least for continuous functions) we can define $\int_a^b f(x)\,dx$ to be the limit of, say, the left hand sums of the partitions we get by dividing $[a,b]$ into $n$ equal parts, because these partitions have mesh size going to $0$, we can pick any points we like (say, the left end points), and we know the limit is going to be that common value of $\underline{S}$ and $\overline{S}$, which has to be the area. So that, under this definition, $\int_a^b f(x)\,dx$ really is the net signed area under the graph of $f(x)$. It just doesn't have a choice but to be that, when $f$ is "nice enough".
Second, the area does not turn into "the" antiderivative. What happens is that it turns out (perhaps somewhat magically) that the area can be computed using an antiderivative. I'll go into some more details below.
As to how Newton figured this out, his teacher, Isaac Barrow, was the one who discovered there was a connection between derivatives and tangents; some of the basic ideas were his. They came from studying some simple functions and some simple formulas for tangents he had discovered.
For example, the tangents to the parabola $y=x^2$ were interesting (there was generally geometric interest in tangents and in "squaring" regions, also known as finding the "quadrature" of a region, that is, finding a way to construct a square or rectangle that had the same area as the region you were considering), and let to associate the parabola $y=x^2$ to lines of the form $y=2x$. It does not take too much experimentation to realize that if you look at the area under $y=2x$ from 0 to a, you end up with $a^2$, establishing a connection. Barrow did this with arguments with infinitesimals (which were a bit fuzzy and not set on entirely correct and solid logical foundation until well into the 20th century), which were generally laborious, and only for some curves. When Newton extended Barrow's methods to more general curves and tangents, he also extended the discovery of the connection with areas, and was able to prove what is essentially the Fundamental Theorem of Calculus.
Now, here is one way to approach the connection. We want to figure out the value of, say, $$\int_0^a f(x)\,dx$$ for some $a$. This can be done using limits and Riemann sums (Newton and Leibniz had similar methods, though not set up quite as precisely as Riemann sums are). But here is an absolutely crazy suggestion: suppose you can find a "master function" $\mathcal{M}$, which, when given any point $b$ between $0$ and $a$, will give you the value of $\int_0^b f(x)\,dx$. If you have such a master function, then you can use it to find the value of the integral you want just by taking $\mathcal{M}(a)$!
In fact, this is the approach Barrow had taken: his great insight was that instead of trying to find the quadrature a particular area, he was trying to solve the problem of squaring several different (but related) areas at the same time. So he was looking for, for instance, a "master function" for the region was like a triangle except that the top was a parabola instead of a line (like the area from $0$ to $a$ under $y=x^2$), and so on.
On its face, this is a completely ludicrous suggestion. It's like telling someone who is trying to know how to get from building A to building B that if he only memorizes the map for the entire city first, then he can use that knowledge to figure out how to get form A to B. If we are having trouble finding the integral $\int_0^a f(x)\,dx$, then the "master function" seems to require us to find not just that area, but also all areas in between! It's like telling someone who is having trouble walking that he should just run very slowly when he wants to walk.
But, again, the interesting thing is that even though we may not be able to say what the "master function" is, we can say how it changes as b changes (remember, $\mathcal{M}(b) = \int_0^b f(x)\,dx$ is a number that depends on $b$, so $\mathcal{M}$ is a function of $b$). Because figuring out how functions change is easier than computing their values (just thing about derivatives, and how we can easily figure out the rate of change of $\sin(x)$, but we have a hard time actually computing specific values of $\sin(x)$ that are not among some very simple ones). (This is also something Barrow already knew, as did Newton).
For "nice functions" (if $f$ is continuous on an interval that contains $0$ and $a$), we can do it using limits and some theorems about "nice" functions: Using limits, we have: \begin{align} \lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}}{h} &= \frac{1}{h}\left(\int_0^{b+h}f(x)\,dx - \int_0^bf(x)\,dx\right)\\\ &= \frac{1}{h}\int_b^{b+h}f(x)\,dx. \end{align} Since we are assuming that $f$ is continuous on $[0,a]$, it is continuous on the interval with endpoints $b$ and $b+h$ (I say it this way because $h$ could be negative). So it has a maximum and a minimum (continuous function on a finite closed interval). Same the maximum is $M(h)$ and the minimum is $m(h)$. Then $m(h) \leq f(x) \leq M(h)$ for all $x$ in the interval, so we know, since the integral is the area, that $$hm(h) \leq \int_b^{b+h}f(x)\,dx \leq hM(h).$$ That means that $$m(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq M(h)\text{ if h\gt 0}$$ and $$M(h) \leq \frac{1}{h}\int_b^{b+h}f(x)\,dx \leq m(h)\text{ if h\lt 0.}$$
As $h\to 0$, the interval gets smaller, the difference between the minimum and maximum value gets smaller. One can prove that both $M$ and $m$ are continuous functions, and that $m(h)\to f(b)$ as $h\to 0$, and likewise that $M(h)\to f(b)$ as $h\to 0$. So we can use the Squeeze Theorem to conclude that since the limit of $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ is squeezed between two functions that both have the same limit as $h\to 0$, then $\frac{1}{h}\int_b^{b+h}f(x)\,dx$ also has a limit as $h\to 0$ and is in fact that same quantity, namely $f(b)$. That is $$\frac{d}{db}\mathcal{M}(b) = \lim_{h\to 0}\frac{\mathcal{M}(b+h)-\mathcal{M}(b)}{h} = \lim_{h\to 0}\frac{1}{h}\int_{b}^{b+h}f(x)\,dx = f(b).$$
That is: when $f$ is continuous, the "Master function" for areas turns out to have a rate of change equal to $f$. This is not that crazy, if you think about it: how is the area under $y=f(x)$ from $x=0$ to $x=b$ changing? Well, it's changing by whatever $f$ is.
This means that, whatever the "Master function" turns out to be, it will be an antiderivative of $f(x)$.
We also know, because we are very good with derivatives, that if $\mathcal{F}(x)$ and $\mathcal{G}(x)$ are two functions, and $\mathcal{F}'(x) = \mathcal{G}'(x)$ for all $x$, then $\mathcal{F}$ and $\mathcal{G}$ differ by a constant: there exists a constant $k$ such that $\mathcal{F}(x) = \mathcal{G}(x)+k$ for all $x$.
So, we know that the "Master function" is an antiderivative. If, by some sheer stroke of luck, we happen to find any antiderivative $F(x)$ for $f(x)$, then we know that the only possible difference between $\mathcal{M}(b)$ and $F(b)$ is a constant. What constant? Well, luckily we know one value of $\mathcal{M}(b)$: we know that $\mathcal{M}(0) = \int_0^0f(x)\,dx$ should be $0$. So, $M(0) = 0 = F(0)-F(0)$, which means the constant has to be $-F(0)$. That is, we must have $M(b) = F(b)-F(0)$ for all $b$.
So, if we find any antiderivative $F$ of $f$, then $\mathcal{M}(b) = F(b)-F(0)$ is in fact the "Master function" we were looking for, the one that gives all the integrals between $0$ and $a$, including $0$ and including $a$. So that we have that two very different processes (computing areas using limits of Riemann sums, and derivatives) are connected: if $f(x)$ is continuous, and $F(x)$ is any antiderivative for $f(x)$ on $[0,a]$, then $$\int_0^a f(x)\,dx = \mathcal{M}(a) = F(a)-F(0).$$
But the integral did not "magically turn" into an antiderivative. It's that the "Master function" which can be used to keep track of all integrals of $f(x)$ has rate of change equal to $f$, which gives us a "back door" to computing integrals.
Newton was able to prove this because he had the guide of Barrow's insight that this was happening for the functions he worked with. Barrow's insight was achieved because he had the brilliant idea of trying to come up with a "Master function" instead of trying to rectify lots of different areas one at a time, and he noticed the connection because he had already worked with tangents/derivatives for those functions. Leibniz likewise had access to Barrow's ideas, so the connection between the two was also known to him.
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Just a tiny note: "finding a way to construct a rectangle that had the same area as the region you were considering" would be what they called in those days "quadrature". Otherwise, this answer is a work of Art, as expected. :) – Guess who it is. Dec 23 '10 at 5:49
@J.M. That's what I get for not checking the proper term in English and trying to go by with dim recollections of what it was in Spanish... Thanks. (And don't call me "Art"; I don't care for it (-; ). – Arturo Magidin Dec 23 '10 at 6:07
It was intended to be a (horrible) pun, but sure, I won't do it again. ;) – Guess who it is. Dec 23 '10 at 6:11
@J.M.: Oh, I got the pun all right (hence the winking smiley, and I did see your smiley). Just wanted to nip any possibility of the nickname getting picked up in the bud. – Arturo Magidin Dec 23 '10 at 6:12
That was awesome! – MyUserIsThis Feb 24 '13 at 20:07
One way you can perhaps "justify"/give an intuitive reason is to consider the following figure:
$A(x)$ is the area under the curve from $0$ to $x$, the brown region.
$A(x+dx)$ is the area under the curve from $0$ to $x + dx$, the brown + gray.
Now for for really small $dx$, we can consider the gray region to be a rectangle of side width $dx$ and height $f(x)$.
Thus $\dfrac{A(x+dx) - A(x)}{dx} = f(x)$.
Thus as $dx \to 0$, we see that $A'(x) = f(x)$.
It is kind of intuitive to define area by approximating by very thin rectangles. The above gives an intuition as to why the derivative of the area gives the curve.
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+1.. For graphics and explanation. – night owl Apr 17 '11 at 5:29
how does this make F(x) = A(x) – Muhammad Umer Mar 24 '14 at 20:23
@MuhammadUmer: Because of fundamental theorem of calculus. See this:en.wikipedia.org/wiki/… (and I suggest you read the whole page). – Aryabhata Mar 25 '14 at 0:23
thanks, i've read. Why does the proof of 1st part, For a given f(t), define the function F(x) as.., start with assumption that F(x) or integral = area..because that is what it's made equivalent to when mean value theorem is used which is area. – Muhammad Umer Mar 25 '14 at 1:56
@MuhammadUmer: I don't understand what you are trying to say. The second part applies here. – Aryabhata Mar 25 '14 at 22:13
I strongly recommend that you take a look at the first chapter of Gilbert Strang's Calculus textbook: http://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/textbook/MITRES_18_001_strang_1.pdf. This chapter provides an insightful introduction to integration that likely takes an approach that is very different from your professor's.
A typical explanation of integration is as follows:
We want to know the area under a curve. We can approximate the area under a curve by summing the area of lots of rectangles, as shown above. It is clear that with hundereds or thousands of rectangles, the sum of the area of each rectangle is very nearly the area under the curve. In the limit, we get that the sum is exactly equal to the area. This animation may help with the intuition,
We define the integral to be the limit described and depicted above:
$\int _{ a }^{ b }{ f(x)dx=\lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ f({ x }_{ i })\Delta x } } }$
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Question is asking about why of what you showed and stopped there with last part. Somehow infinite triangles equals the difference in antiderivatives??? – Muhammad Umer Mar 16 '14 at 22:58
heres a very simple explanation of how an integral works. Forget about things needed for the proof. This is just for the concept
take the function f'(x) and its antidervative, f(x)
We can find the area under a graph by taking the average of all the y values and multiplying by Delta X, creating a rectangle of equal area.
(1)Area = Yaverage of f'(x)* DeltaX
average y for f'(x) is average slope of f(x)
finding average slope of f(x) is easy. Simply go to the original function and do the average slope calculation
(2)average slope =(y1-y2)/(deltaX)
Now substitute formula 2 into 1:
area of f'(x) = (y1-y2)/delaX * deltaX
area of f'(x) = (y1-y2)
y1 = f(x1)
area of f'(x)=f(x1)-f(x2)
Took me a while to figure this out. IMO the mean sum of tiny squares hinders understanding. mean value theorem is good to find average slope. Its convoluted and imprecise though when people say there is a value c that does this or that. Irrelevant to finding the average slope. Just supply the formula for finding average slope. Best of luck
Such a simple concept to be so over complicated. I hope they change the way it is taught in canadian schools
- | 6,189 | 22,371 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2015-22 | latest | en | 0.937977 |
https://www.geeksforgeeks.org/maximum-sum-of-array-formed-by-replacing-each-element-with-sum-of-adjacent-elements/?ref=rp | 1,713,869,126,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00588.warc.gz | 704,992,966 | 49,784 | # Maximum sum of Array formed by replacing each element with sum of adjacent elements
Last Updated : 06 Jun, 2022
Given an array arr[] of size N, the task is to find the maximum sum of the Array formed by replacing each element of the original array with the sum of adjacent elements.
Examples:
Input: arr = [4, 2, 1, 3]
Output: 23
Explanation:
Replacing each element of the original array with the sum of adjacent elements:
4 + 2 = 6
6 + 1 = 7
7 + 3 = 10
Array formed by replacing each element of the original array with the sum of adjacent elements: [6, 7, 10]
Therefore, Sum = 6 + 7 + 10 = 23
Input: arr = [2, 3, 9, 8, 4]
Output: 88
Explanation:
Replacing each element of the original array with the sum of adjacent elements to get maximum sum:
9 + 8 = 17
17 + 4 = 21
21 + 3 = 24
24 + 2 = 26
Array formed by replacing each element of the original array with the sum of adjacent elements: [17, 21, 24, 26]
Therefore, Sum = 17 + 21 + 24 + 26 = 88.
Approach:
• Scan through the array to pick the adjacent pair with the highest sum.
• From there on, using Greedy algorithm, pick the left or right integer, whichever is greater.
• Repeat the process till only a single element is left in the array.
Below is the implementation of the above approach:
## C++
`// C++ program to find the maximum sum` `// of Array formed by replacing each` `// element with sum of adjacent elements` `#include ` `using` `namespace` `std;` `// Function to calculate the possible` `// maximum cost of the array` `int` `getTotalTime(vector<``int``>& arr)` `{` ` ``// Check if array size is 0` ` ``if` `(arr.size() == 0)` ` ``return` `0;` ` ``// Initialise left and right variables` ` ``int` `l = -1, r = -1;` ` ``for` `(``int` `i = 1; i < arr.size(); i++) {` ` ``if` `(l == -1` ` ``|| (arr[i - 1] + arr[i])` ` ``> (arr[l] + arr[r])) {` ` ``l = i - 1;` ` ``r = i;` ` ``}` ` ``}` ` ``// Calculate the current cost` ` ``int` `currCost = arr[l] + arr[r];` ` ``int` `totalCost = currCost;` ` ``l--;` ` ``r++;` ` ``// Iterate until left variable reaches 0` ` ``// and right variable is less than array size` ` ``while` `(l >= 0 || r < arr.size()) {` ` ``int` `left = l < 0` ` ``? INT_MIN` ` ``: arr[l];` ` ``int` `right = r >= arr.size()` ` ``? INT_MIN` ` ``: arr[r];` ` ``// Check if left integer is greater` ` ``// than the right then add left integer` ` ``// to the current cost and` ` ``// decrement the left variable` ` ``if` `(left > right) {` ` ``currCost += left;` ` ``totalCost += currCost;` ` ``l--;` ` ``}` ` ``// Executes if right integer is` ` ``// greater than left then add` ` ``// right integer to the current cost` ` ``// and increment the right variable` ` ``else` `{` ` ``currCost += right;` ` ``totalCost += currCost;` ` ``r++;` ` ``}` ` ``}` ` ``// Return the final answer` ` ``return` `totalCost;` `}` `// Driver code` `int` `main(``int` `argc, ``char``* argv[])` `{` ` ``vector<``int``> arr = { 2, 3, 9, 8, 4 };` ` ``cout << getTotalTime(arr) << endl;` ` ``return` `0;` `}`
## Java
`// Java program to find the maximum sum` `// of array formed by replacing each` `// element with sum of adjacent elements` `class` `GFG{` `// Function to calculate the possible` `// maximum cost of the array` `static` `int` `getTotalTime(``int` `[]arr)` `{` ` ` ` ``// Check if array size is 0` ` ``if` `(arr.length == ``0``)` ` ``return` `0``;` ` ``// Initialise left and right variables` ` ``int` `l = -``1``, r = -``1``;` ` ``for``(``int` `i = ``1``; i < arr.length; i++)` ` ``{` ` ``if` `(l == -``1` `|| (arr[i - ``1``] + arr[i]) >` ` ``(arr[l] + arr[r]))` ` ``{` ` ``l = i - ``1``;` ` ``r = i;` ` ``}` ` ``}` ` ``// Calculate the current cost` ` ``int` `currCost = arr[l] + arr[r];` ` ``int` `totalCost = currCost;` ` ``l--;` ` ``r++;` ` ``// Iterate until left variable reaches 0` ` ``// and right variable is less than array size` ` ``while` `(l >= ``0` `|| r < arr.length)` ` ``{` ` ``int` `left = (l < ``0` `? ` ` ``Integer.MIN_VALUE : arr[l]);` ` ``int` `right = (r >= arr.length ? ` ` ``Integer.MIN_VALUE : arr[r]);` ` ``// Check if left integer is greater` ` ``// than the right then add left integer` ` ``// to the current cost and` ` ``// decrement the left variable` ` ``if` `(left > right)` ` ``{` ` ``currCost += left;` ` ``totalCost += currCost;` ` ``l--;` ` ``}` ` ``// Executes if right integer is` ` ``// greater than left then add` ` ``// right integer to the current cost` ` ``// and increment the right variable` ` ``else` ` ``{` ` ``currCost += right;` ` ``totalCost += currCost;` ` ``r++;` ` ``}` ` ``}` ` ``// Return the final answer` ` ``return` `totalCost;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ``int` `[]arr = { ``2``, ``3``, ``9``, ``8``, ``4` `};` ` ``System.out.print(getTotalTime(arr) + ``"\n"``);` `}` `}` `// This code is contributed by PrinciRaj1992`
## Python3
`# Python3 program to find the maximum sum` `# of Array formed by replacing each` `# element with sum of adjacent elements` `import` `sys` `# Function to calculate the possible` `# maximum cost of the array` `def` `getTotalTime(arr):` ` ` ` ``# Check if array size is 0` ` ``if` `(``len``(arr) ``=``=` `0``):` ` ``return` `0` ` ``# Initialise left and right variables` ` ``l ``=` `-``1` ` ``r ``=` `-``1` ` ``for` `i ``in` `range``(``1``, ``len``(arr), ``1``):` ` ``if` `(l ``=``=` `-``1` `or` `(arr[i ``-` `1``] ``+` `arr[i]) > (arr[l] ``+` `arr[r])):` ` ``l ``=` `i ``-` `1` ` ``r ``=` `i` ` ``# Calculate the current cost` ` ``currCost ``=` `arr[l] ``+` `arr[r]` ` ``totalCost ``=` `currCost` ` ``l ``-``=` `1` ` ``r ``+``=` `1` ` ``# Iterate until left variable reaches 0` ` ``# and right variable is less than array size` ` ``while` `(l >``=` `0` `or` `r < ``len``(arr)):` ` ``if``(l < ``0``):` ` ``left ``=` `sys.maxsize` ` ``else``:` ` ``left ``=` `arr[l]` ` ``if` `(r >``=` `len``(arr)):` ` ``right ``=` `-``sys.maxsize ``-` `1` ` ``else``:` ` ``right ``=` `arr[r]` ` ``# Check if left integer is greater` ` ``# than the right then add left integer` ` ``# to the current cost and` ` ``# decrement the left variable` ` ``if` `(left > right):` ` ``currCost ``+``=` `left` ` ``totalCost ``+``=` `currCost` ` ``l ``-``=` `1` ` ``# Executes if right integer is` ` ``# greater than left then add` ` ``# right integer to the current cost` ` ``# and increment the right variable` ` ``else``:` ` ``currCost ``+``=` `right` ` ``totalCost ``+``=` `currCost` ` ``r ``+``=` `1` ` ``# Return the final answer` ` ``return` `totalCost` `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` ` ``arr ``=` `[``2``, ``3``, ``9``, ``8``, ``4``]` ` ``print``(getTotalTime(arr))` `# This code is contributed by Surendra_Gangwar`
## C#
`// C# program to find the maximum sum` `// of array formed by replacing each` `// element with sum of adjacent elements` `using` `System;` `class` `GFG{` ` ` `// Function to calculate the possible` `// maximum cost of the array` `static` `int` `getTotalTime(``int` `[]arr)` `{` ` ` ` ``// Check if array size is 0` ` ``if` `(arr.Length == 0)` ` ``return` `0;` ` ` ` ``// Initialise left and right variables` ` ``int` `l = -1, r = -1;` ` ` ` ``for``(``int` `i = 1; i < arr.Length; i++)` ` ``{` ` ``if` `(l == -1 || (arr[i - 1] + arr[i]) >` ` ``(arr[l] + arr[r]))` ` ``{` ` ``l = i - 1;` ` ``r = i;` ` ``}` ` ``}` ` ` ` ``// Calculate the current cost` ` ``int` `currCost = arr[l] + arr[r];` ` ` ` ``int` `totalCost = currCost;` ` ` ` ``l--;` ` ``r++;` ` ` ` ``// Iterate until left variable reaches 0` ` ``// and right variable is less than array size` ` ``while` `(l >= 0 || r < arr.Length)` ` ``{` ` ``int` `left = (l < 0 ? ` ` ``int``.MinValue : arr[l]);` ` ``int` `right = (r >= arr.Length ? ` ` ``int``.MinValue : arr[r]);` ` ` ` ``// Check if left integer is greater` ` ``// than the right then add left integer` ` ``// to the current cost and` ` ``// decrement the left variable` ` ``if` `(left > right)` ` ``{` ` ``currCost += left;` ` ``totalCost += currCost;` ` ``l--;` ` ``}` ` ` ` ``// Executes if right integer is` ` ``// greater than left then add` ` ``// right integer to the current cost` ` ``// and increment the right variable` ` ``else` ` ``{` ` ``currCost += right;` ` ``totalCost += currCost;` ` ``r++;` ` ``}` ` ``}` ` ` ` ``// Return the readonly answer` ` ``return` `totalCost;` `}` ` ` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ``int` `[]arr = { 2, 3, 9, 8, 4 };` ` ` ` ``Console.Write(getTotalTime(arr) + ``"\n"``);` `}` `}` `// This code is contributed by PrinciRaj1992`
## Javascript
``
Output:
`88`
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space. | 3,418 | 10,144 | {"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-2024-18 | latest | en | 0.7021 |
https://moffett4understandingmath.com/tag/perseverance/ | 1,679,570,659,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00262.warc.gz | 464,210,370 | 23,153 | # Creating Structure for Context in Math
I was honored to facilitate lesson study with IM1 teachers today. Their students are struggling (due to high EL/SPED population) with solving word problems. I dug deeper, and we decided the struggle is really the first step: creating equations from situations.
We decided our goal as educators this year is to work on teacher clarity: making our lessons streamlined and very goal-oriented. If we know our goals for the lesson, then every move we make (every breath we take…) is for the goal. So how do we clarify translating context to equations?
We started from the end: the benchmark. We took a problem the students struggled with, and tweaked it several times, each time only altering only one component. Students had to work from the original version (which we used simple numbers to keep it accessible) for each new “version”. They discussed what changed from situation to situation and how that affected the prior equation.
Version 1: Troy works for an ice cream cart vendor. He receives \$10 for taking the cart out for a shift, plus a commission of \$2.00 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 2: Troy works for an ice cream cart vendor. He receives \$15 for taking the cart out for a shift, plus a commission of \$2.00 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 3: Troy works for an ice cream cart vendor. He receives \$15 for taking the cart out for a shift, plus a commission of \$1.25 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 4: Troy works for an ice cream cart vendor. He receives \$25 for taking the cart out for a shift, plus a commission of \$0.10 for each item he sells. Troy worked a shift Saturday and earned \$52.90. How many items did he sell? (Problem from the benchmark.)
We used 3 scenarios. In each, we kept our questions as consistent as possible (again, clarity):
• Which part is varying (changing)? How do you know?
• Which quantity would be the coefficient? How do you know?
• Which quantity would be the constant? How do you know?
• (From version to version) What has stayed the same? What changed? How does the changed quantity affect our equation? Why?
Students were engaged, writing on their tables and willing to discuss with each other. They had many moments of “ohhhhhh” and “oops!” and learned quite a bit about the components of 2-step equations. They definitely need more time, and the teachers have committed to continuing the work as warm-ups or on modified days.
Oh! And did I mention this was a co-taught Special Ed class, with many English Learners?! Amazing!
So our major takeaways were:
2. Keep your goal in mind when creating the tasks/lesson and questions for clarity and focus.
3. Breaking the situations into translating and solving (working on a single component) allows students to focus and interpret.
Below is our ppt. Hope it is useful! Happy Math-ing!
Linear Equations in Context LS 8.27.19
This morning at the airport (At 5 fricking o’ clock! I need to fire my secretary for scheduling this flight. Oh wait. That’s me!) I was answering emails and a timid voice interrupted my thoughts.
Excuse me. Are you a teacher?
This always makes my heart happy. As a middle school teacher, you often believe the kids won’t think twice about you once they leave your room. Why would they, with all the distractions the world has to offer?!!
So when a former student not only remembers you from looong ago, AND takes the time to share her experiences and life journey with you, you tear up just a little. You remember that these precious moments are WHY you were born to be an educator.
My WHY is simple. I love seeing my peeps have the “click” in math. I love learning about these humans, with all their quirks and unique personalities. I love supporting them and inspiring them to do great things. I just love THEM.
What is your WHY? Ponder, remember, and remind yourself of this through the year.
Have a great 2019-2020! I KNOW it will be a year to remember! 💕
# How Do Our Beliefs in Math Affect Our Students?
I was honored to work with amazing teachers this week. We took a survey from NCTM (National Council Teachers of Mathematics) on our beliefs regarding student learning and our instructional practices in mathematics. This, in itself, led to amazing discussions about what we truly believe math IS and how we interpret that into instructional decisions within our classrooms.
But then we took it further. We got into groups and discussed not so much whether we agreed or disagreed, but whether it was a productive or unproductive belief in respect to student access and learning. Here are two to consider:
There were fantastic discussions about these particular ones, especially for educators of EL and SPED. We also considered how parents might respond to these. Powerful conversations around access, flexibility in thinking, understanding conceptual and procedural mathematical ideas, and yes, fluency.
Here was the point. Our beliefs, whether productive or unproductive, affect our attitudes towards mathematics and the children we are blessed to teach. Those attitudes affect the actions we take. Who gets to answer which questions? Who gets the “tough” tasks and who has to keep doing drill and kill worksheets? Who gets to explore puzzles and who has to retake tests or do homework (because their home life doesn’t lend itself to being able to do it at home)? And those actions MATTER. They affect the results you will get from your students.
So as you gear up for this school year, consider taking the beliefs survey yourself. Even better, have your team take it and REALLY dive in to what beliefs are productive an unproductive. The more we reflect, the more we can grow and be effective at what we truly want; to teach students to love, learn and understand mathematics. Have a great year!
# That Moment You Realize Your Child Is Suffering
All who have multiple children know that each one is wonderfully different. You can raise ’em the same, yet they have their own amazing quirks and personalities, strengths and passions. This could not be more true of my two boys.
My youngest is now 5. He has always been the rough and tumble type. Everything is a competition to him. (Even last night he was standing on his tippy toes trying to be ‘taller’ when raising his hand at church!) He is funny, outgoing, and a firecracker. He is a joy.
Yet school stuff has not yet become his thing. He seemed disinterested in learning his letters, yet loved to be read to. He never wanted to sit with me and learn the components all say are so important for kinder. His preschool teachers said it would come; that he just was a busy bee and had other, more important things to care about. And honestly, I believed that too. He loves to create, tell stories, sing, build, and live outdoors. Who was I to take that away from him?
So when his amazing preK teacher suggested a hearing test, I was on board. No big deal; just go do it and cross it off as another thing we did. It came back inconclusive. Went for an ENT hearing test. The results were staggering; at least 30% hearing loss in each ear. The doctor said, “Think of being submerged underwater for 5 years of your life, trying desperately to hear what people are saying. That is what your son has lived through.” I am still teary thinking about this. I am a fricking educator! How did I not catch this? I was in denial as well, until that very day driving home I asked my boy if he saw the cool tree. “Tree?” “Yes, the tree over there (I pointed out the car window.).” “Like, dessert?” And that is when it hit me. My child wasn’t hearing.
We started to notice. He said, “What?” almost every time you told him something. He couldn’t hear the TV clearly; he was reading lips (which we now know why he would never answer us when watching TV; he was too focused on the screen to hear us). In preK, during circle time, he struggled to hear all the conversations and his body would just be exhausted from trying to listen that he gave up. When listening to a story, he focused on the pictures for meaning versus our words. He was trying his best, his very best, and it was exhausting. My heart broke. My baby boy, sweet thing he is, was struggling under my teacher nose, and I hadn’t caught it.
Long story short, he is now hearing much more, with just a tweak of medication. He still has a hearing loss, and we will test every year to make sure it isn’t degenerative. He is learning how to deal with sounds he hadn’t heard before. (At church last night he couldn’t believe they played bells!) He begins speech therapy (I had no idea he would need that either) next week and they are excited to see his progress, as am I. And funny enough, he is now interested in learning his letters, sounds, words, etc. It all makes sense; for how can you be interested in something you never knew existed?
I bring this story up for one reason. We are not perfect. Even if we have our children’s best interests at heart, we may miss something. It takes a true village to raise our children. If you do not have a village to help, find one. If your children go to daycare or summer camp, get to know their counselors. Ask them questions. Find out what your children are doing, and how they are doing. Talk to your children’s teachers next year. Get to know them, because they see your children more often than you do! And listen, even when you don’t want to. In a world that is so negative right now, I feel strongly that we need to support, build, and nurture each other and our tiny humans in order to make them the best they can be.
# Developing Perseverance
I don’t get it. Can you help me? My teacher didn’t explain it. I forgot. This is stupid.
We have all been there. We have all heard each of these when our child is working on math homework. The question is, how do we get him/her to stop being helpless? Here are a few ideas to start with.
Listen to their frustrations. Then move on. Look. Being frustrated is okay. We don’t want to say it isn’t. But being helpless is not okay. This is a life lesson. Not everything is easy, but we don’t get to give up.
Don’t do the math for them. That only lets your child know you will let them off the hook every time.
Help them find resources and look at them together. Some questions to help you out:
• Where are your notes from today? Let’s review them and see if there is anything there we can use.
• What are you learning? Let’s look up some videos (mathtv.com is a great site for video lessons, but if you just go to utube or teachertube you will find others) and see if we can relearn it together.
• Let’s look at your book and see if there are any examples that might help us.
• Call/Facetime/text a friend and see if they can assist you. (Research shows that study groups truly help all students in mathematics!!!
• Email the teacher and see if s/he has tutorials to help. (Many teachers put up videos, the solutions, etc on-line. Find out if yours does!)
Above all, let them know that it is okay to not know everything, but NOT OKAY to give up. This is a biggie. Most math worth doing takes time. Students assume that if they don’t get the answer right away, they must have done it wrong (or don’t know what they are doing and are not good in math). So not true! Help your child use resources available to be successful, but they need to do the work and put in the effort. | 2,592 | 11,598 | {"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-2023-14 | latest | en | 0.956082 |
https://zh.wikipedia.org/wiki/%E5%86%85%E7%A7%AF%E7%A9%BA%E9%97%B4 | 1,713,475,821,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817239.30/warc/CC-MAIN-20240418191007-20240418221007-00082.warc.gz | 995,924,384 | 40,035 | # 内积空间
${\displaystyle \mathbf {A} ={\begin{bmatrix}1&2\\3&4\end{bmatrix}}}$
## 正式定义
${\displaystyle V}$ 是一個定義在 ${\displaystyle \left(F,\,+,\,\times \right)}$ 上的向量空间，其向量加法記為「 ${\displaystyle \oplus }$ 」 ，且其标量乘法記為「 ${\displaystyle \cdot }$ 」。若它裝配了一個二元函数 ${\displaystyle f:V\times V\to F}$ 滿足：（以下將 ${\displaystyle f(v,\,w)}$ 簡寫為 ${\displaystyle \langle v,\,w\rangle }$
### 定义的分歧
线性 對所有 ${\displaystyle a,\,v,\,w\in V}$ ${\displaystyle \langle a,\,v\oplus w\rangle =\langle a,\,v\rangle +\langle a,\,w\rangle }$ 對所有 ${\displaystyle v,\,w\in V}$ 和所有 ${\displaystyle \lambda \in F}$ ${\displaystyle \langle v,\,\lambda \cdot w\rangle =\lambda \times \langle v,\,w\rangle }$
## 例子
### 点积
${\displaystyle \langle (x_{1},\ldots ,x_{n}),\,(y_{1},\ldots ,y_{n})\rangle :=\sum _{i=1}^{n}x_{i}y_{i}=x_{1}y_{1}+\cdots +x_{n}y_{n}}$
## 基本性质
(a) ${\displaystyle \langle a,\,v\oplus w\rangle =\langle a,\,v\rangle +\langle a,\,w\rangle }$
(b) ${\displaystyle \langle v,\,\lambda \cdot w\rangle ={\overline {\lambda }}\times \langle v,\,w\rangle }$
(a)
{\displaystyle {\begin{aligned}\langle a,\,v\oplus w\rangle &={\overline {\langle v\oplus w,\,a\rangle }}\\&={\overline {\langle v,\,a\rangle +\langle w,\,a\rangle }}\\&={\overline {\langle v,\,a\rangle }}+{\overline {\langle w,\,a\rangle }}\\&=\langle a,\,v\rangle +\langle a,\,w\rangle \end{aligned}}}
(b)
{\displaystyle {\begin{aligned}\langle v,\,\lambda \cdot w\rangle &={\overline {\langle \lambda \cdot w,\,v\rangle }}\\&={\overline {\lambda \times \langle w,\,v\rangle }}\\&={\overline {\lambda }}\times {\overline {\langle w,\,v\rangle }}\\&={\overline {\lambda }}\times \langle v,\,w\rangle \end{aligned}}}
(a) ${\displaystyle \left\langle \sum _{i=1}^{n}v_{i},\,w\right\rangle =\sum _{i=1}^{n}\langle v_{i},\,w\rangle }$
(b) ${\displaystyle \left\langle w,\,\sum _{i=1}^{n}v_{i}\right\rangle =\sum _{i=1}^{n}\langle w,\,v_{i}\rangle }$
${\displaystyle n=2}$ ，本定理只是內積定义的線性部分，故成立。
${\displaystyle n=k}$ 時，對任意有限向量序列 ${\displaystyle {\{u_{i}\in V\}}_{i=1}^{k}}$ 和任意 ${\displaystyle w\in V}$ 有：
${\displaystyle \left\langle \sum _{i=1}^{k}u_{i},\,w\right\rangle =\sum _{i=1}^{k}\langle u_{i},\,w\rangle }$
{\displaystyle {\begin{aligned}\left\langle \sum _{i=1}^{k+1}v_{i},\,w\right\rangle &=\left\langle \left(\sum _{i=1}^{k}v_{i}\right)+v_{k+1},\,w\right\rangle \\&=\left\langle \sum _{i=1}^{k}v_{i},\,w\right\rangle +\langle v_{k+1},\,w\rangle \\&=\sum _{i=1}^{k}\langle v_{i},\,w\rangle +\langle v_{k+1},\,w\rangle \\&=\sum _{i=1}^{k+1}\langle v_{i},\,w\rangle \end{aligned}}}
{\displaystyle {\begin{aligned}\left\langle w,\,\sum _{i=1}^{n}v_{i}\right\rangle &={\overline {\left\langle \sum _{i=1}^{n}v_{i},\,w\right\rangle }}\\&={\overline {\sum _{i=1}^{n}\langle v_{i},\,w\rangle }}\\&=\sum _{i=1}^{n}{\overline {\langle v_{i},\,w\rangle }}\\&=\sum _{i=1}^{n}\langle w,\,v_{i}\rangle \end{aligned}}}
## 范数
${\displaystyle V}$ 是個複內積空間，則對所有的 ${\displaystyle v,\,w\in V}$ 有：
(a) ${\displaystyle \|v\|\|w\|\geq |\langle v,\,w\rangle |}$
(b) ${\displaystyle \|v\|\|w\|=|\langle v,\,w\rangle |}$ ${\displaystyle \Leftrightarrow }$ 存在 ${\displaystyle \lambda \in \mathbb {C} }$ 使 ${\displaystyle v=\lambda \cdot w}$
${\displaystyle v=w=0_{V}}$ ，根據內積定义的非退化部分，本定理成立。若考慮 ${\displaystyle v,\,w\neq 0_{V}}$ ，取 ${\displaystyle e_{v}={\frac {1}{\|v\|}}\cdot v}$${\displaystyle e_{w}={\frac {1}{\|w\|}}\cdot w}$${\displaystyle \alpha =\langle e_{v},\,e_{w}\rangle }$ ，則根據內積定义
{\displaystyle {\begin{aligned}\langle e_{v}\oplus (-\alpha )\cdot e_{w},\,e_{v}\oplus (-\alpha )\cdot e_{w}\rangle &=\langle e_{v},\,e_{v}\rangle +\langle e_{v},\,(-\alpha )\cdot e_{w}\rangle +\langle (-\alpha )\cdot e_{w},\,e_{v}\rangle +\langle (-\alpha )\cdot e_{w},\,(-\alpha )\cdot e_{w}\rangle \\&=1-{\overline {\alpha }}\alpha -\alpha {\overline {\alpha }}+\alpha {\overline {\alpha }}\\&=1-{|\alpha |}^{2}\\&=1-{\left({\frac {|\langle v,\,w\rangle |}{\|v\|\|w\|}}\right)}^{2}\\&\geq 0\end{aligned}}}
${\displaystyle e_{v}\oplus (-\alpha )\cdot e_{w}=0_{V}}$
${\displaystyle v={\frac {\langle v,\,r\rangle }{{\|w\|}^{2}}}\cdot w}$
${\displaystyle V}$ 是個複內積空間，則對所有的${\displaystyle v,\,w\in V}$ 有：
${\displaystyle \|v\|+\|w\|\geq \|v\oplus w\|}$
{\displaystyle {\begin{aligned}{\|v\oplus w\|}^{2}&=\langle v\oplus w,\,v\oplus w\rangle \\&={\|v\|}^{2}+\langle v,\,w\rangle +\langle w,\,w\rangle +{\|w\|}^{2}\\&={\|v\|}^{2}+{\|w\|}^{2}+2\operatorname {Re} (\langle v,\,w\rangle )\\&\leq {\|v\|}^{2}+{\|w\|}^{2}+2|\langle v,\,w\rangle |\end{aligned}}}
{\displaystyle {\begin{aligned}{\|v\oplus w\|}^{2}&\leq {\|v\|}^{2}+{\|w\|}^{2}+2|\langle v,\,w\rangle |\\&\leq {\|v\|}^{2}+{\|w\|}^{2}+2\|v\|\|w\|\\&={(\|v\|+\|w\|)}^{2}\end{aligned}}}
${\displaystyle \angle (v,\,w):=\cos ^{-1}\left({\frac {\langle v,\,w\rangle }{\|v\|\|w\|}}\right)}$
${\displaystyle V}$ 是個複內積空間，若有限向量序列 ${\displaystyle {\{v_{i}\in V\}}_{i=1}^{n}}$ 對任意不相等的正整數 ${\displaystyle 1\leq i\neq j\leq n}$ 都有 ${\displaystyle \langle v_{i},\,v_{j}\rangle =0}$ 則：
${\displaystyle \sum _{i=1}^{n}\|v_{i}\|^{2}=\left\|\sum _{i=1}^{n}v_{i}\right\|^{2}}$
{\displaystyle {\begin{aligned}\left\|\sum _{i=1}^{n}v_{i}\right\|^{2}&=\left\langle \sum _{i=1}^{n}v_{i},\,\sum _{i=1}^{n}v_{i}\ \right\rangle \\&=\sum _{j=1}^{n}\left\langle v_{j},\,\sum _{i=1}^{n}v_{i}\ \right\rangle \\&=\sum _{j=1}^{n}\sum _{i=1}^{n}\langle v_{j},\,v_{i}\rangle \\&=\sum _{j=1}^{n}{\|v_{j}\|}^{2}\end{aligned}}}
${\displaystyle V}$ 是個複內積空間，則對所有的 ${\displaystyle v,\,w\in V}$ 有：
(a) ${\displaystyle \|v\oplus w\|^{2}+\|v\ominus w\|^{2}=2\|v\|^{2}+2\|w\|^{2}}$
{\displaystyle {\begin{aligned}{\|v\oplus w\|}^{2}&=\langle v\oplus w,\,v\oplus w\rangle \\&={\|v\|}^{2}+\langle v,\,w\rangle +\langle w,\,w\rangle +{\|w\|}^{2}\\\end{aligned}}}
{\displaystyle {\begin{aligned}{\|v\ominus w\|}^{2}&=\langle v\oplus (w^{-1}),\,v\oplus (w^{-1})\rangle \\&={\|v\|}^{2}+\langle v,\,(-1)\cdot w\rangle +\langle (-1)\cdot w,\,v\rangle +\langle (-1)\cdot w,\,(-1)\cdot w\rangle \\&={\|v\|}^{2}-\langle v,\,w\rangle -\langle w,\,v\rangle +{\|w\|}^{2}\end{aligned}}}
## 完备化
${\displaystyle \langle f,g\rangle :=\int _{a}^{b}f(t){\overline {g(t)}}\,dt}$
${\displaystyle \forall k\geqslant 2,\;\;\;f_{k}(t)={\begin{cases}0,&\forall t\in [0,{\frac {1}{2}}]\\k(t-{\frac {1}{2}}),&\forall t\in ({\frac {1}{2}},{\frac {1}{2}}+{\frac {1}{k}}]\\1,&\forall t\in ({\frac {1}{2}}+{\frac {1}{k}},1]\end{cases}}}$
## 引用
• S. Axler, Linear Algebra Done Right, Springer, 2004
• G. Emch, Algebraic Methods in Statistical Mechanics and Quantum Field Theory, Wiley Interscience, 1972.
• N. Young, An Introduction to Hilbert Spaces, Cambridge University Press, 1988 | 2,977 | 6,618 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 129, "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-2024-18 | latest | en | 0.17431 |
https://onlinejudge.org/board/viewtopic.php?p=45054 | 1,576,061,264,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540530857.12/warc/CC-MAIN-20191211103140-20191211131140-00536.warc.gz | 500,883,266 | 12,017 | ## 10179 - Irreducible Basic Fractions
Moderator: Board moderators
Minilek
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Posts: 90
Joined: Tue Jul 27, 2004 9:34 am
Location: Cambridge, MA
Contact:
A number can have a prime factor greater than its sqrt.
That's ok. You only need to generate primes up to sqrt(1 billion)
to test primality efficiently. For primes bigger than the sqrt, just
loop through the primes less than the sqrt and see if any of them
divide it. If not, it must be prime. I used this method for AC in 0.154.
rbarreira
New poster
Posts: 5
Joined: Sun Oct 03, 2004 12:14 am
I had the same problem in 10179. Fixing the special case for the input "1" gave me AC
But I don't really understand why the output in that case should be 1, since in the example given in the problem text, they don't count 0/12 as a irreducible basic fraction... Isn't this inconsistent?
Piers Kennedy
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Posts: 3
Joined: Thu Apr 22, 2004 8:12 pm
[..in the example given in the problem text, they don't count 0/12 as a irreducible basic fraction..]
That is correct 0/12 can be reduced to any fraction with a non-zero denominator (e.g.0/1 etc.)
Problem 10179 differs from 10299 in that when n=1 there is an allowable irreducible fraction of 1/1 giving a count of 1.
In 10299 the problem description asks for the number of positive integers less than n which are relatively prime to n. For n=1 this is 0. Interestingly the EulerPhi function in Mathematica returns 1 for n=1!
[/quote]
athlon19831
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Posts: 20
Joined: Thu Jan 19, 2006 2:32 pm
### 10179 Time Limit Exceeded
#include "iostream.h"
#include "stdio.h"
#include "string.h"
#include "math.h"
bool Gcd(long M,long N)
{
long Rem;
while(N>0)
{
Rem=M%N;
M=N;
N=Rem;
}
if(M==1)
return true;
else
return false;
}
int main(int argc, char* argv[])
{
long n;
long i,j;
long num;
while(cin>>n)
{
if(n==0)
break;
else
{
num=0;
for(i=0;i<n;i++)
{
if(Gcd(n,i))
{
num++;
}
}
cout<<num<<endl;
}
}
return 0;
}
mamun
A great helper
Posts: 286
Joined: Mon Oct 03, 2005 1:54 pm
Contact:
Test your program with the maximum input, 999999999. You'll feel the speed of your program.
Find Euler's Phi function.
szwesley001
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Posts: 1
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Location: Bauru
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### 10179 Time Limit Exceeded
Why?
/* @BEGIN_OF_SOURCE_CODE */
#include <stdio.h>
#define MAX 1000
int main()
{
unsigned long int m,n,x,vet[MAX]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
long int j=0,i=0;
float result1=0,fator=0;
long int r;
while(scanf("%ld",&m)==1)
{
if ((m>1000000000)||(m==0)) return 0;
x=2;n=m;i=0;
while (n>1)
{
r=n%x;
if (r==0)
{
if (x != vet)
{
i++;
vet=x;
}
n=n/x;
}
else if (n==x)
break;
else
x++;
}
result1 = m;
for (j=1;j<=i;j++)
{
fator=1;
fator/=vet[j];
result1 *= (1-fator);
}
printf("%ld\n",( unsigned long int)result1);
}
return 1;
}
/* @END_OF_SOURCE_CODE */
Wesley
Moha
Experienced poster
Posts: 216
Joined: Tue Aug 31, 2004 1:02 am
Location: Tehran
Contact:
You post in the wrong Volume. see
http://online-judge.uva.es/board/viewto ... ght=10179
and before making a new subject, search in froum.
sohel_acm
New poster
Posts: 8
Joined: Fri May 19, 2006 11:27 pm
### 10179!!!! WA!!!Plz Help
Hi Solvers,
I am getting wa in 10179.But I can't find the bug in my code.Plz someone help me.Here is my code
Code: Select all
``````Deleted after getting ACC.
``````
Last edited by sohel_acm on Tue May 23, 2006 11:18 am, edited 1 time in total.
Love programmers,help programmers.
sohel_acm
New poster
Posts: 8
Joined: Fri May 19, 2006 11:27 pm
I have found my error.At last I got ACC.
Love programmers,help programmers.
rhsumon
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Posts: 48
Joined: Wed Aug 23, 2006 12:29 pm
Location: Dhaka
### Re: 10179 - Irreducible Basic Fractions
WA plz help
first time TLE then WA where is the wrong
Code: Select all
``````#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define P 100000
long prime[P];
long p[50000];
int phi[P+100];
void gen_prime(){
int i,j;
prime[0] = prime[1] = 1;
for(i=2; i<=(long)sqrt(P); i++){
if(prime[i]==0){
for(j=i*i; j<=P-1; j=j+i){
prime[j]=1;
}
}
}
j=0;
for(i=0; i<=P-1; i++){
if(prime[i] == 0){
p[j++] = i;
}
}
}
void phgen(){
int i,j;
for(i=2;i<=P;i++)
phi[i]=i;
for(i=2;i<=P;i++){
if(!prime[i]){
for(j=i;j<=P;j+=i)
phi[j]-=phi[j]/i;
}
}
}
int is_prime(int a){
int i;
for(i=0; p[i]*p[i]<=a; i++){
if(a%p[i]==0) return 0;
}
return 1;
}
int main()
{
gen_prime();
phgen();
int n,pi,i,flag;
while(scanf("%d",&n)==1 && n){
if(n==1) printf("1\n");
else if(n <= P) printf("%d\n",phi[n]);
else{
if(is_prime(n)) printf("%d\n",n-1);
else{
pi=1;
for(i=0; n!=1&&p[i]*p[i]<=n; i++){
flag=0;
while(n%p[i]==0){
pi*=p[i];
n/=p[i];
flag=1;
}
if(flag==1) pi-=pi/p[i];
if(is_prime(n)&&n>1){
pi*=(n-1);
break;
}
}
}
printf("%d\n",pi);
}
}
return 0;
}
``````
stcheung
Experienced poster
Posts: 114
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Contact:
### Re: 10179 - Irreducible Basic Fractions
Your program outputs an extra line for primes larger than 100000. Try it with 100003.
Taman
New poster
Posts: 32
Joined: Sun Oct 11, 2009 8:59 pm
Location: Southeast University
### Re: 10179 - Irreducible Basic Fractions
Well, I think output for 1 is 1.
because, my compiler shows gcd(0,1) = 1 and the problem statement states,
A fraction m / n is basic if 0 <= m < n and it is irreducible if gcd(m, n) = 1.
I am to mention that I use Euclid's algorithm like others to find gcd(m,n).
So, i think (0,1) is considered to be an irreducible fraction in this sense and this helped me to get acc quite on the first submission. . . .looking for your further correction .
Pro.metal
New poster
Posts: 9
Joined: Fri Dec 17, 2010 8:13 pm
### Re: 10179 - Irreducible Basic Fractions
When i run the loop upto sqrt(n) i get WA because not all the prime factors are below sqrt(n) for example 123456.
Code: Select all
``````#include <stdio.h>
#include <math.h>
char prime[1000000000];
int main()
{
long long n,prod,i,j,k;
long long d;
for(i=2; i<10000; i+=2)
{
if(prime[i]==0)
{
for(j=i*i; j<10000000; j+=i) prime[j] = 1;
}
if(i==2) i = i - 1;
}
while(scanf("%lld",&n)!=EOF)
{
if(!n) break;
prod = n;
for(k=2; k<n/2; k++)
{
if(n%k==0 && !prime[k])
{
prod = prod *(k-1) / k;
}
}
printf("%lld\n",prod);
}
return 0;
}
``````
DD
Experienced poster
Posts: 145
Joined: Thu Aug 14, 2003 8:42 am
Location: Mountain View, California
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### Re: 10179 - Irreducible Basic Fractions
Taman wrote:Well, I think output for 1 is 1.
because, my compiler shows gcd(0,1) = 1 and the problem statement states,
A fraction m / n is basic if 0 <= m < n and it is irreducible if gcd(m, n) = 1.
I am to mention that I use Euclid's algorithm like others to find gcd(m,n).
So, i think (0,1) is considered to be an irreducible fraction in this sense and this helped me to get acc quite on the first submission. . . .looking for your further correction .
Thanks for this suggestion, I got several W.As due to this case
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April 29, 2016
# Posts by Reiny
Total # Posts: 34,383
Maths
In my diagram, I have the perpendicular from T go PC meeting PC at S, giving me 2 right angled triangles. In triangle CST: tan30° = 6/CS CS = 6/tan30 in triangle PST: tan40° = 6/SP SP = 6/tan40 PC = CS+SP = .... you do the button-pushing
March 3, 2016
Finance
I am not familiar with US mortgage laws. What is this "3.5% closing cost" ? so is the original mortgage: add 3.5% to 225,000, then take 90% of that? That would be 225,000(1.035)(.9) = \$209,587.50 confirm that before I proceed
March 3, 2016
Math
(-1,14), (0,10), (1,8), (2,?), (3,10), (4,?) and (5,20) the missing values should be like: (-1,14), (0,10), (1,8), (2,8), (3,10), (4,14?) and (5,20) Now if you put those same x value into the straight line you get: (-1,6), (0,7), (1,8), (2,9), (3,10), (4,11), and (5,12) Notice...
March 3, 2016
math
Join A and B to O, the centre of the circle So we have triangle ABO with all 3 sides known. We can use the cosine law to find angle Ø 6^2 = 9^2+9^2-2(9)(9)cosO 162cosØ = 126 cosØ = 126/162 = 7/9 Ø = 38.94...° (I stored it in calculator's ...
March 3, 2016
Algebra
amount = 6300(1.06)^3 = 7503.40 how did you get your answer ??
March 2, 2016
trigonometry (Course and Bearing)
According to my sketch: tan46°10' = 7/AP AP = 7/ tan46°10' = .... sin46°10' = 7/BP BP = 7/sin46°10' = ....
March 2, 2016
Trig
Of course I can't see where your Ø is let's assume it is opposite the smallest side. 10^2 = 17^2 + 20^2 - 2(17)(20)cosØ 680cosØ = 589 cosØ = 589/680 Ø = appr 29.98° If otherwise, make the necessary changes
March 2, 2016
Math
1/2 ÷ 1/8 =
March 2, 2016
Math
be more specific please
March 2, 2016
Algebra
ummh, notice that two of the terms has a single decimal place with that decimal being an even number What happens when you multiply any even number by 5 ?
March 2, 2016
finance
PV = 3000(.96) + 3000(.96)^2 or PV = 3000(1/04)^-1 + 3000(1.04)^-2 depending on how "annual discount rate" is defined in your course. btw, they yield different results.
March 2, 2016
Algebra
pop(chessville) = 50,000(1.07)^n pop(checkerville) = 70,000(.96)^n we want to know when 50,000(1.07)^n ≥ 70,000(.96)^n divide by 10,000 5(1.07)^n = 7(.96)^n take logs of both sides and use rules of logs log5 + nlog1.07 = log7 + nlog.96 nlog1.07 - nlog.96 = log7 - log5 n(...
March 2, 2016
precalculus
Make a reasonable sketch. Using vectors and using R to be the resultant vector: R = (180cos300,180sin300) + (50cos245,50sin245) = (90,-90√3) + (-21.1309.. , -45.31538..) = (68.869.., -201.1996...) Magnitude = √(68.869^2 + (-201.1996..)^2) = appr 212.66 km direction...
March 2, 2016
MATH.
done yesterday http://www.jiskha.com/display.cgi?id=1456884647
March 2, 2016
math
6x^3 + 3x^2 - 9x what about it ?
March 1, 2016
Math
so number = a e^(kt) 10,000 = a e^(3k) 40,000 = a e^(5k) divide the 2nd by the 1st: 4 = e^(2k) 2k = ln4 k = (ln4)/2 so N = a e^( (ln4)/2 t) so we need a a e^(3ln4/2) = 10,000 a = 10,000/8 = 1250 initial count is 1250 bacteria
March 1, 2016
math
what is 6000(1.02)^28 ?
March 1, 2016
Math
time the dog wants to go outside in that time frame = (7/12)(60) = 35 minutes so not wanting to outside = 25 minutes percentage of time not going out = 25/60 = appr 42%
March 1, 2016
Math
prob = 6/10 = 3/5 a cube has 6 sides, so division by 6 would be the model for probs, we need a division by 5 or 10 or 15 ,etc
March 1, 2016
Pre-Calc
did you test the answer in the first equation? It clearly works in the last two equations.
March 1, 2016
Algebra
speed of second train ---- x mph speed of first train --- 2x-10 mph When they meet they would have traveled the same time let that time be t hrs distance gone by first train = tx distance gone by 2nd train = t(2x-1) tx + t(2x-1) = 220 tx + 2tx - t = 220 t(3x-1)=220 t = 220/(3x...
March 1, 2016
Calculus
I have seen this question, where we have to find the other sides so that the perimeter is a minimum, but ... The way it is worded... there is an infinite number of possible lengths. e.g if the side along the x-axis is 5, then the equation of the line passing through (1,1) has ...
March 1, 2016
math
f∘g(x)= f(g(x)) = f(4-x^2) = 4(4-x^2) - 3 simplify do the 2nd one in the same way
March 1, 2016
Math
Did you mean f(x) = √(x+1) ?? I will assume that. f∘g = f(g(x)) = f(3x-5) = √((3x-5) - 5) = √(3x - 10) domain: 3x-10≥0 3x≥10 x ≥ 10/3 ( I will let you switch this to "interval notation, I like the "old way" better) g&#...
March 1, 2016
algebra
first condition: XX2 XX3 XX5 XX7 Second condition: hundreds is twice tens 842 632 422 212 843 .... ... 847................217 There will be 16 of those I will leave it up to you to decide how many of those are prime numbers. (remember, just because a number ends in an odd ...
March 1, 2016
math
Just a simple case of tan38 = h/6.6 h = 6.6tan38° = ....
March 1, 2016
math
amount of 30% solution --- x gal amount of 14% stuff ------ 60-x solve for x .30x + .14(60-x) = .20(60) or if you want two variables: x+y = 60 .30x + .14y = .20(60)
March 1, 2016
math plz help i cant understand this. :'(
Did you plot the points? Look like a trapezoid. AB is a horizontal line, so that AB = 6 AC and BC are parallel So area = (1/2(AC + BC)(AB) = (1/2)(4+6)(6) = 30 square units
March 1, 2016
math :D
Easiest to check would be the median We have an even number of data values, so the median is the average of the two middle numbers. so (23+25)/2 = 24, that rules out #1 and #2 the mode is clearly 25 so we have to find the mean: add them up and divide by 8 to see what you get
March 1, 2016
@ Mary - Calculus
Mary, what part of yesterdays solution did you not like ? http://www.jiskha.com/display.cgi?id=1456770767 Always check to see if your question has been answered before reposting the same thing. That way we avoid unnecessary duplication of work.
March 1, 2016
mathematics
so 75% would be 45 .75x = 45 x = 45/.75 = ...
March 1, 2016
math
distance = 4.9t^2, when t=1 distance = 4.9 m
March 1, 2016
math
This webpage is perfectly suited for your type of problem http://davidmlane.com/hyperstat/z_table.html Just enter the data
February 29, 2016
Calculus
I would assume you are using the cosine law. let the distance between the two tips be x x^2 = 4 + 9 -2(2)(3)cosØ x^2 = 13 - 12cosØ 2x dx/dt = 12sinØ dØ/dt given: dØ/dt = .5 , Ø = π/4 for Ø = π/4 x^2 = 13 - 12cosπ/...
February 29, 2016
Math
equation of tangent: y - 5/3 = (-5/9)(x - 5) y = (-5/9)x + 25/9 + 5/3 y = (-5/9)x + 40/9 check my arithmetic
February 29, 2016
math
Did you make a sketch? On mine I let the top of the tree be A and its bottom be B look at triangle APB. angle ABP = 120° , AP = 4√3, BP = 4 by the sine law: sinA/4 = sin120/4√3 sinA = 4(√3/2)/(4√3) = .5 A = 30° so angle APB = 180-120-30 = 30°
February 29, 2016
math
Make a little diagram of what actually happens. After the first bounce, the ball goes up and down 2/3 of it previous bounce, but the first event can be considered only half a bounce. so I am going to consider this to be a geometric series, pretending the first bounce is ...
February 29, 2016
math
I see 2 similar triangles. let length of Christa's shadow be x ft let length of Glad's shadow be y ft so 5/x = 6/y 6x = 5y we also know x+y = 15 or y = 15-x sub into the first: 6x = 5(15-x) 6x = 75-5x 11x = 75 x = 75/11 ft = appr 6.82 ft y = 15-75/11 = 90/11 ft = appr ...
February 29, 2016
math
just use the formula PV = 1000(1.05)^-8 = ...
February 29, 2016
math
I will assume this is "simple interest" let the rate be r, let the principal be p so (15/12)(rp) = 455 rp = 364 for only 10 months interest = (10/12)(rp) refund = 455 - (10/12)(rp) = 455 - (5/6)(364) = \$151.67
February 28, 2016
geometry
Ladder is 12 feet out and 16 ft up, A VERY UNSAFE WAY TO PUT THE LADDER!! length of ladder --- L L^2 = 12^2+16^2 = 400 L= √400 = 20 new position: base = 14 , height = ? , L = 20 solve: h^2 + 14^2 = 20^2
February 28, 2016
math
Good grief, haven't seen that type of question in about 50 years. didn't know they still used it . I will describe the method and then attempt to explain what happened. They are using the word "normal" to mean perpendicular. e.g. in your equation, 2x+3y - 5...
February 28, 2016
math
sinx < -√3/2 consider sinx = -√3/2 by the CAST rule and knowing that sin 30° = +√3/2 x = 210° or x = 330° also, you should know the general shape of the sine curve. from the above the curve would be below -√3/2 between 210° and 330&...
February 28, 2016
Geometry
recall that sin^2 Ø + cos^2 Ø = 1 so sin^2 Ø + 9/64= 1 sin^2Ø = 55/64 sinØ = √55/8 then cscØ = 8/√55 or 8√55/55
February 28, 2016
Pre-Calculus
Important info at the end: "with equal amounts invested in short- and intermediate-term bonds. " So let each of those amounts be x then the amount invested at long-term be 100,000 - 2x He wants to end up with a 5.1% profit on the total 100,000 or \$5100 Now it is easy...
February 28, 2016
Algebra
I will assume you made a diagram. On my diagram, I have each of the long sides as y, and each of the 3 shorter sides as x so we have 3x + 2y = 440 y = (440 - 3x)/2 = 220 - (3/2)x area = xy = x(220 - (3/2)x) = 220x - (3/2)x^2 from the subject title, I assume you don't take ...
February 28, 2016
math
those solutions where 0 ≤ x ≤ 2π 2cos^2x-5cosx+2=0 this is a standard quadratic where your variable is cosx How about doing this ? let t = cosx then we have 2t^2 - 5t + 2 = 0 , which factors to (2t - 1)(t - 2) = 0 t = 1/2 or t = 2 so cosx = 1/2 or cosx = 2 ...
February 28, 2016
maths
I can't see any other way than listing the possible cases and finding the total score for each case. Using the first letter of each outcome, we could have BBBB -- 36 BBBS -- 34 BBBT -- 31 BBBO -- 28 BBSS -- 32 BBTT -- 26 BBOO -- 20 BBST -- 29 BBSO -- BBTO etc, watch for ...
February 28, 2016
math
hard to get the spacing correct, I will use ** for spacing 3|2**0**-1**0***7****1 ******6**18**51*153*480 ***2**6**17**51*160*481 f(3) = 481
February 28, 2016
Math
prob(hit) = .42 prob(no hit ) = 1 - .42 = .58 at least 2 hits out of 4 ---> exclude : no hits, 1 hit prob(no hits) = C(4,0)(.42)^0 (.58)^4 = .113165 prob(1hit) = C(4,1)(.42)(.58)^3 = .32779 prob(at least 2 hits) = 1 - (.113165 + .32779) = appr .559 or you could do prob(...
February 28, 2016
math
Tell me what you got, then we'll both be happy
February 28, 2016
math
This is almost the same kind of question that I answered for you here : http://www.jiskha.com/display.cgi?id=1456693077 Did you look at it, and learn from it? I will give you one hint for this one: cos (2Ø) = cos^2 Ø - sin^2 Ø let me know how you made out.
February 28, 2016
math
remember cos^2 Ø + sin^2 Ø = 1 .36 + sin^2 Ø = 1 sin^2 Ø = .64 sinØ = .8 sin 2Ø = 2sinØcosØ = 2(.8)(.6) = .96 2Ø = sin^-1 (.96) or arcsin(.96)
February 28, 2016
Geometry/Algebra
Make a sketch of a right-angled triangle ABC with angleB = 90° Draw in the altitude from C to hypotenuse AC to meet it at D. You now have 3 similar right-angled triangles. (easy to see by considering the angles) I will list them, with corresponding vertices in the same ...
February 28, 2016
Math
Why could you not check them?? is (1/8)% of 32 = .04 ?? (1/800)(32 = 32/800 = .04 , so YES x(1.8) = 3 x = 3/1.8 = 1.6666.. or 166.66..% or 166 2/3% check that: take 166.66..% of 1.8 = (166.666../100 )(1.8) = 3
February 28, 2016
math
a+d = 50 a + 4d=65 subtract them 3d = 15 d = 5 then a+5=50 a=45 term40 = a + 39d = 45 + 39(5) = 240
February 28, 2016
futher maths
What are we doing to it ? is it (√3 - √2)/2 or √3 - (√2/2) ?
February 28, 2016
Math 12
You should have available to you : - either in your notebook -or in your text - or memorized , the ratio of sides for the 30-60-90 as well as the 45-45-90 right-angled triangles you should also have memorized the simple conversion from degrees to radians of the following 30&...
February 28, 2016
Math
done, look further up to your other post
February 28, 2016
Finance
Value at the end of 6 years = 1120 + 1120(1.12) + 1020(1.12)^2 + 1020(1.12)^3 + 920(1.12)^4 + 920(1.12)^5 = 8155.91 now the rate goes down to 7% for the next 59 years. Final value at age 65 = 8155.91(1.07)^59 = 441,687.52 but they only paid out \$430,000 while using that money ...
February 28, 2016
exactly 3 men ---> need 2 women number of choices = C(7,3) x C(5,2) = 35 x 10 = 350 more women than men ---> 5 women, 4W-1M, 3W-2M = C(5,5) + C(5,4)xC(7,1) + C(5,3)xC(7,2) = 1 + 35 + 10(21) = 246
February 28, 2016
math
so he biked 1 mile at 12 mph and the other 1 mile at x mph 1/12 + 1/x = 1/4 ---> 15 min = 1/4 hr times 12x, (don't like fractions) x + 12 = 3x x = 6 so his speed for the second part was 6 mph check: time at 12 mph = 1/12 hr = 5 minute time at 6 mph = 1/6 hr = 10 min for...
February 28, 2016
Math
for the AS x-6 = y-x 2x-y = 6 or y = 2x-6 for the GS y/x = 16/y y^2 = 16x sub in the 1st (2x-6)^2 = 16x 4x^2 - 24x + 36 = 16x 4x^2 -40x + 36 = 0 x^2 - 10x + 9 = 0 (x-1)(x-9) = 0 x = 1 or x = 9 if x = 1, y = -4 if x = 9 , y = 12 check: 1st case: 6, 1, -4, 16 , the first 3 are ...
February 28, 2016
math
yes recall that (a/b)^n = a^n/b^n so (1/a)^-n = 1^-n / a^-n but (1)^anything = 1 = 1/a^-n
February 28, 2016
math
for the first one: (-1)^(2n) = +1 2n is an even number, so when you multiply -1 an even number of times you get a positive e.g. (-1)^12= ( (-1)^2 )^6 = 1^6 = 1 but (-1)^13 = (-1)^12 (-1) = (+1)(-1) = -1 so if you raise (-1)^odd, you get -1 which answers the second part since ...
February 28, 2016
FINANCE
PV of loan = .75(3250000) = \$2,437,500 paym = 15800 n = 12(35) =420 let the monthly rate be i 2437500 = 15800( 1 - (1+i)^-420)/i 154.2721519 i = 1 - (1+i)^-420 This is a very nasty equation to solve. Back in the "good ol' days", we used a method called ...
February 28, 2016
Finance
If she has 2500,000 and she gets 12.5% per year in interest, she would earn \$312,500 per year right now. Heck!, she should retire right now but according to your question 2.5 million(1.125)^n = 7.5 million 1.125)^n = 3 n log 1.123 = log3 n = appr 9.3 years
February 27, 2016
precal
It said diameter = 22, so radius is 11 I used 2πr = 2(11)π = 22π
February 27, 2016
precal
circumference = 2π(11) = 22π inches so in 1 minute a point on the rim covers 31(22π) or appr 2142.566 inches since linear velocity = distance/time ....... your turn.
February 27, 2016
math
sinØ = y/r = 1/9 ---> y = 1, r = 9 Make a sketch of the triangle in quad II so x^2 + y^2 = 9^2 x^2 + 1 = 81 x^2 = 80 x = ±√80 = ±4√5 but we are in II, so x = -4√5 sinØ = 1/9 ---> the given cosØ = -4√5/9 tan&...
February 27, 2016
Mathematics
you have to use the product rule, with your notation, it would be d(uv) = uv' + vu' = (-8sin(2x))(cos(3x)) + (4cos(2x)(-3sin(3x) ) = -8sin(2x)cos(3x) - 4sin(3x)cos(2x)
February 27, 2016
math
To make 10 L in total you will need x L of the 18% solution and 10-x L of water. .18x + 0(10-x) = .12(10) .18x = 1.2 x = 1.2/.18 = 120/18 = 20/3 He needs 20/3 L or 6 2/3 L of water. (btw, a Russian would never dilute his alcohol)
February 27, 2016
Calculus Pre Test Questions Monday Part 4
your symbols came out strange. Is 3<t<3.01 supposed to say 3 ≤ t ≤ 3.01 ?
February 27, 2016
Calculus
So you have a split function f(x) = 4-x^2 for x<1 f(x) = 2x+1 for x > 1 (you don't say what happens at x = 1 ) first slope = -2x, second slope is a constant 2 a) (-1,3), since x<1, slope = -2(-1) = 2 b) (2,5), since x > 1, slope = 2
February 27, 2016
Calculus
2a) I will assume you mean f(x) = 3/(x+1) I would write it f(x) = 3(x+1)^-1 then f ' (x) = -3(x+1)^-2 = -3/(x+1)^2 at P(2,1), slope = -3/(3)^2 = -3/9 = -1/3 h(x) = 2/√(x+5) = 2(x+5)^(-1/2) h ' (x) = -1(x+5)^(-3/2) = -1/( √(x+5) )^3 you finish it
February 27, 2016
Algebra
number of fifties --- x number of tens ---- x+14 number of twenties -- 2x+4 50x + 10(x+14) + 20(2x+4) = 1720 Solve for x, and all shall be revealed.
February 27, 2016
math
Use the discriminant property if we have real roots, then b^2 - 4ac > 0 px^2-p+10=2(p+2)x px^2 - p + 10 = 2px + 4x px^2 - 2px - 4x + 10 = 0 px^2 + x(-2p-4) + 10 = 0 b^2 - 4ac > 0 (-2p-4)^2 - 4(p)(10) > 0 4p^2 + 16p + 16 - 40p > 0 4p^2 - 24p + 16 > 0 p^2 - 6p + 4...
February 27, 2016
math
for 2 equal roots, b^2 - 4ac = 0 9p^2 - 4(p)(p+q) = 0 9p^2 - 4p^2 - 4pq = 0 5p^2 - 4pq = 0 p(5p - 4q) = 0 p = 0 or p = 4q/5 but p≠0 or else we don't have a quadratic, so p = (4/5)q
February 27, 2016
math
You must be talking about the simple interest formua I = PRT 700 = 3500(r)(4) r = 700/(14000) = ...
February 27, 2016
Geometry
"One Secant has 4 and x" is not clear. Is 4 the length of the entire secant or just the part inside the circle? In general, if you have two secants to the same circle (whole of secant #1)(outside part of secant#1) = (whole of secant#2)(outside part of secant#2) So ...
February 27, 2016
Algebra 2
y = -.23x + 56.78 or y = (-1/2)x + 3
February 27, 2016
math
speed of private jet --- x mph speed of commercial jet --- 2x-143 mph They both went the same distance, so ... 4(2x - 143) = 6x solve for x
February 27, 2016
algebra
f(x)-g(x) = 9x^3+2x^2-5x+4 -(5x^3-7x+4) = 9x^3+2x^2-5x+4 -5x^3+7x-4 = 4x^2 + 2x^2 + 2x looks like you are correct
February 27, 2016
Alg2
Why make things complicated ? Just solve it .... (1/4)x-3=(1/2)x+8 times 4, the LCD x - 12 = 2x + 32 -x = 44 x = -44 Where does y come from in the pairs of equations?
February 27, 2016
calculus
I see 4 questions posted in quick succession. You have given no indication of any start to these on your part, nor have you told us where your problems are. This looks like an assignment or homework. Surely you realize that if just do these questions for you, you will not ...
February 27, 2016
Math
let the number of boys be 5x and the number of girls be 6x 5x+6x=44 11x=44 x=4 so we have 20 boys and 24 girls (notice 20:24 = 5:6)
February 27, 2016
financial maths
i = weekly rate = .0665/52 = .001278846 n = number of payments = 52(15) = 780 Amount repaid = present value of loan = 200( 1 - 1.001278846^-780)/.001278846 = appr \$98,677.00
February 27, 2016
math
Alice --- x Peter --- x+5 John ---- 2x+10 5 years from now: John -- 2x+15 Alice -- x+5 2x+15 = 3(x+5) -x = 0 x = 0 This question is bogus, even though the answer meets the original conditions. Unless Alice was just born, it makes little sense i.e. Alice -- 0 years Peter --5 ...
February 27, 2016
math
let the 3 numbers be x, x+1, and x+2 3x = x+1 + x+2 3x = 2x+3 x = 3 Finish it up, and check my answer to see if it meets the given conditions
February 27, 2016
math
what is 4x3 ?
February 27, 2016
algebra or algebra1
Not sure of your unusual notation. did you mean |x-6| is multiplied by 7 ? Ok, then ... f(-1) = |-1-6|7 - (-1) = 49+1 = 50
February 27, 2016
Math Word Problem
same type of problem as your previous post .... students --- x adults ----- y x+y = 110 ---> y = 110-x 13.5x + 16.5y = 1581 sub the first into the second: 13.5x + 16.5(110-x) = 1581 13.5 + 1815 - 16.5x = 1581 -3x = -234 x = 78 So 78 students and 32 adults
February 27, 2016
Math Word Problem
Look at the response to this question last night by both Steve and I. All you have to do is change the original second equation, and solve. http://www.jiskha.com/display.cgi?id=1456542484
February 27, 2016
Maths
You have your thinking backwards. an example of the above would be both 12 and 30 are divisible by 3 then their difference is divisible by 3 check: difference = 30-12 = 18 is 18 divisible by 3 ? YES actual proof: let the two numbers be kx and ky. kx and ky are both divisible ...
February 26, 2016
Algebra
What does your text or your notes say about y = mx + b ?
February 26, 2016
Math
using the definitions: a+2d + a+8d = 20 2a + 10d = 20 a + 5d = 10 ** a+11d - (a+3d) = 32 a+11d - a - 3d = 32 8d = 32 d = 4 sub into ** a + 20 = 10 a = -10 state your conclusion.
February 26, 2016
Math Word Problem
number of 4-credit courses --- x number of 5-credit courses --- y x+y = 52 4x + 5y = 22 This has no positive x and y solutions. Check your question or your typing How can a total of 52 courses yield only 22 credits, unless they failed most of the courses ???? Even if courses ...
February 26, 2016
math
You don't say what kind of a prism. I will assume it is a rectangular prism, (fancy word for box) each cube has a volume of (1/2)^3 or 1/8 inches^3 number that will fit the box = 56/(1/8) = 448 cubes e.g. the box could have been 4 in by 7 in by 2 in for a volume of 56 ...
February 26, 2016
Pre calculus
Do you really think that somebody will give their free time and help you after such a rude and disrespectful response to one of our valuable and trusted tutors ? I certainly don't feel such an inclination.
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• Level: GCSE
• Subject: Maths
• Word count: 3580
# Investigating the statistical relationship between IQ levels and Maths KS2 results.
Extracts from this document...
Introduction
Edexcel 2003 Dispatch 1 Maths Investigation Roman Gaufman Mathematic GCSE Mayfield High School Can No: 6056 Introduction In this investigation I will choose a hypotheses to investigate and use a variety of statistical methods to investigate whether it is correct and prove the fact that it is correct or incorrect. I will make a smaller sample of the database and arrange it by genders and year groups. I will make graphs, comment on them and use a variety of statistical method. Hypotheses My hypothesis is to prove that higher Maths KS2 results often affect the IQ levels and make a close relationship being the higher the Maths KS2 result, the higher the IQ level. I will also investigate gender differences and hope to prove that the male gender does generally better at maths then the female gender. I will then expand my investigation to investigate other subjects and their relationship to the IQ. Deciding on sample size I have decided to take a sample of 180 pupils from the school because it is a nice dividable number that will fit nicely into a pie chart and it is great enough to give relatively accurate results. However this sample isn't big enough to hold 100% accurate results and therefore the conclusion of the investigation should be considered as a respectable result, but further investigation should take place for a deeper more accurate analysis. Bellow, I have made a table of the numbers of students in each class divided into genders and given in both real figures and 3 different percentages to investigate how large should the sample from each year group a gender should be. Boys Girls Year No % 1 % 2 % 3 No % 1 % 2 % 3 7 151 54% 25% 13% 131 46% 23% 11% 8 145 54% 24% 12% 125 46% 22% 11% 9 118 45% 19% 10% 143 55% 24% 12% 10 106 53% 18% 9% 94 47% 16% 8% 11 84 49% 14% 7% 86 ...read more.
Middle
13 * table 3 Table Key: No in Db = Numbers as they appear in Database Cmd to Sample = Command to query Sample Spl = Number of samples to take Using the commands in the table above I have made a table that I sorted by number and in the unlikely event of a repeated number, I did + 1 where possible and if + 1 wasn't possible because it makes the random figure go over the Numbers as appears in Database then I subtracted one. After running each command and noting all the samples for each year group and gender, I have compiled the following table that will be used to get my stratified sample from the database. The sample that will be taken from the database is as follows: * table 4 Now that I know what to sample it is time for some extensive copy pasting from the database to create the new summarized database with a sample size of 180. I will begin by making smaller separate database for boys and girl where non relevant data won't be present like for instance: Surname, Forename and Average TV as those coulombs do not posses any useful info to use during the investigation. Average TV may be a factor in lower Key Stage 2 results, but it will not be focused on in this investigation. I will also further divide the database to investigate specific connection between each subject, IQ , gender and will even try to investigate if the writing hand has any effect on the results that will be an easy test to make although I do not expect to see any connection. Investigating Maths KS2 vs IQ BOYS I have made yet another scaled down database with only the values of N, IQ and Ma so I can investigate if there is a clear connection between Maths KS2 and IQ. ...read more.
Conclusion
We see in these graphs even though averages show that the boys and girls are equally smart and get equally good results, we see in these graphs that the girls seem to overtake the boys in lower IQ levels as I already mentioned before but what I didn't realize prior to making these graphs is then the boys catch up and overtake the girls as IQ level goes up showing there is a higher percentage of boys with higher IQ. These graphs do not however show that there is a higher number of exceptionally high IQ levels for the girls and this should be considered so conclusions should not be taken from these two graphs. Less important information to mention is that the program that I used to generate these graphs seemed to have drawn the line on the graph slightly inaccurately so what the graph is attempting to demonstrate isn't as clear as I would've wanted, never the less we can still see what the graph is showing but as I already mentioned, these graphs will have an impact on the conclusions but the averages will play a far greater role. Furthering Investigation To extend further on my investigation, I have decided to create a graph that will allow me to investigate the relationship between IQ and KS2 of all subjects. I have used a logarithm regression line. Analyzing this graph we can see that all subjects have a relationship to the IQ. Strongest being science, then maths and the weakest relationship is English. However the difference between the relationships isn't great to say the least and it is clear that all higher KS2 results generally mean a higher IQ. I originally anticipated the Maths KS2 result to have the strongest relationship to IQ, but found out that Science has the strongest relationship which is logical since it requires many calculations and maths skills in physics and chemistry. ...read more.
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#1
I'm doing a practice paper and one of the questions is
Solve
6x - 5y = 110
7x + 10y = 30
Can someone give me the answer AND explain how they got it because I never understand these question and YouTube isn't helping.
0
4 weeks ago
#2
(Original post by yourstruly101)
I'm doing a practice paper and one of the questions is
Solve
6x - 5y = 110
7x 10y = 30
Can someone give me the answer AND explain how they got it because I never understand these question and YouTube isn't helping.
...
Last edited by GabiAbi84; 4 weeks ago
0
#3
(Original post by GabiAbi84)
First :Put the first equation into terms of y.
Oh god, this just shows how bad I am at maths : (
0
4 weeks ago
#4
(Original post by GabiAbi84)
First :Put the first equation into terms of y.
why? it's simultaneous equations
(Original post by yourstruly101)
Oh god, this just shows how bad I am at maths : (
so you have these
6x - 5y = 110
7x + 10y = 30
now I would multiply all of 6x - 5y = 110 by 2 then add the two equations together, the y's should cancel out and you can solve for x
then use your value for x and substitute it in to one of the equations to find y
0
4 weeks ago
#5
X=13.12
0
4 weeks ago
#6
multiply 6x-5y=110 by 2 to get 12x-10y=220. Now both equations have a y multiple that can cancel out. So you do (12x-10y=220) + (7x+10y=30). So -10y+10y=0 so we can remove that leaving us with. (12x+7x)=(220+30). This is 19x=250. Divide both sides by 19 to get x=250/19. Substitute the x back in to any one of the original equations to get y. So 6(250/19)-5y=110. Expand the brackets to get (1500/19)-5y=110. Subtract both sides by 1500/19 to get -5y=590/19. Divide both sides by -5 to get y=-118/19. hope this helps
1
#7
(Original post by laurawatt)
why? it's simultaneous equations
so you have these
6x - 5y = 110
7x + 10y = 30
now I would multiply all of 6x - 5y = 110 by 2 then add the two equations together, the y's should cancel out and you can solve for x
then use your value for x and substitute it in to one of the equations to find y
So, I should time 110 x 2?
0
#8
(Original post by mdbdidjdksnd)
multiply 6x-5y=110 by 2 to get 12x-10y=220. Now both equations have a y multiple that can cancel out. So you do (12x-10y=220) + (7x+10y=30). So -10y+10y=0 so we can remove that leaving us with. (12x+7x)=(220+30). This is 19x=250. Divide both sides by 19 to get x=250/19. Substitute the x back in to any one of the original equations to get y. So 6(250/19)-5y=110. Expand the brackets to get (1500/19)-5y=110. Subtract both sides by 1500/19 to get -5y=590/19. Divide both sides by -5 to get y=-118/19. hope this helps
Yeah, this helps! Thank you
1
#9
(Original post by mdbdidjdksnd)
multiply 6x-5y=110 by 2 to get 12x-10y=220. Now both equations have a y multiple that can cancel out. So you do (12x-10y=220) + (7x+10y=30). So -10y+10y=0 so we can remove that leaving us with. (12x+7x)=(220+30). This is 19x=250. Divide both sides by 19 to get x=250/19. Substitute the x back in to any one of the original equations to get y. So 6(250/19)-5y=110. Expand the brackets to get (1500/19)-5y=110. Subtract both sides by 1500/19 to get -5y=590/19. Divide both sides by -5 to get y=-118/19. hope this helps
How did you get 19x = 250
0
4 weeks ago
#10
(Original post by yourstruly101)
How did you get 19x = 250
12x+7x is 19x. and 220+30 is 250. so the equation becomes 19x=250
0
#11
(Original post by mdbdidjdksnd)
12x+7x is 19x. and 220+30 is 250. so the equation becomes 19x=250
Oh, okay that makes sense thanks
0
#12
(Original post by mdbdidjdksnd)
multiply 6x-5y=110 by 2 to get 12x-10y=220. Now both equations have a y multiple that can cancel out. So you do (12x-10y=220) + (7x+10y=30). So -10y+10y=0 so we can remove that leaving us with. (12x+7x)=(220+30). This is 19x=250. Divide both sides by 19 to get x=250/19. Substitute the x back in to any one of the original equations to get y. So 6(250/19)-5y=110. Expand the brackets to get (1500/19)-5y=110. Subtract both sides by 1500/19 to get -5y=590/19. Divide both sides by -5 to get y=-118/19. hope this helps
I'm asking too many questions sorry but when you said divide both sides by 19 to x would that be 0.38
0
4 weeks ago
#13
(Original post by yourstruly101)
I'm asking too many questions sorry but when you said divide both sides by 19 to x would that be 0.38
so 19x/19 is just 1x. and 250/19 is 13.157... which is a long decimal so it's better to just keep it as a fraction which is 250/19. so x=250/19
0
#14
(Original post by mdbdidjdksnd)
so 19x/19 is just 1x. and 250/19 is 13.157... which is a long decimal so it's better to just keep it as a fraction which is 250/19. so x=250/19
ohhh okay thanks
0
4 weeks ago
#15
(Original post by laurawatt)
why? it's simultaneous equations
Substitution is a method of solving simultaneous equations, just maybe not the best method for this particular set of equations.
0
#16
(Original post by mdbdidjdksnd)
so 19x/19 is just 1x. and 250/19 is 13.157... which is a long decimal so it's better to just keep it as a fraction which is 250/19. so x=250/19
So I've done it and I've looked at the mark scheme to correct my work and the answer is
x= 10
y = 10
0
4 weeks ago
#17
(Original post by yourstruly101)
So I've done it and I've looked at the mark scheme to correct my work and the answer is
x= 10
y = 10
could it be that you posted the question wrong?
0
4 weeks ago
#18
(Original post by yourstruly101)
So I've done it and I've looked at the mark scheme to correct my work and the answer is
x= 10
y = 10
Those values don't satisfy the equations you posted originally!
Can you upload a picture of the question?
0
#19
(Original post by davros)
Those values don't satisfy the equations you posted originally!
Can you upload a picture of the question?
I used the site onmaths
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##### Stage: 2 Challenge Level:
Place eight queens on an chessboard (an 8 by 8 grid) so that none can capture any of the others.
### Fake Gold
##### Stage: 2 Challenge Level:
A merchant brings four bars of gold to a jeweller. How can the jeweller use the scales just twice to identify the lighter, fake bar? | 2,327 | 10,128 | {"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-2017-43 | longest | en | 0.891012 |
https://kr.mathworks.com/matlabcentral/cody/solutions/1224131 | 1,603,968,955,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00713.warc.gz | 391,101,845 | 18,851 | Cody
# Problem 29. Nearest Numbers
Solution 1224131
Submitted on 3 Jul 2017 by Salvatore Lacava
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [30 46 16 -46 35 44 18 26 25 -10]; correct = [8 9]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
B = Inf 16 -14 -76 5 14 -12 -4 -5 -40 -16 Inf -30 -92 -11 -2 -28 -20 -21 -56 14 30 Inf -62 19 28 2 10 9 -26 76 92 62 Inf 81 90 64 72 71 36 -5 11 -19 -81 Inf 9 -17 -9 -10 -45 -14 2 -28 -90 -9 Inf -26 -18 -19 -54 12 28 -2 -64 17 26 Inf 8 7 -28 4 20 -10 -72 9 18 -8 Inf -1 -36 5 21 -9 -71 10 19 -7 1 Inf -35 40 56 26 -36 45 54 28 36 35 Inf
2 Pass
A = [1555 -3288 2061 -4681 -2230 -4538 -4028 3235 1949 -1829]; correct = [3 9]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
B = Inf -4843 506 -6236 -3785 -6093 -5583 1680 394 -3384 4843 Inf 5349 -1393 1058 -1250 -740 6523 5237 1459 -506 -5349 Inf -6742 -4291 -6599 -6089 1174 -112 -3890 6236 1393 6742 Inf 2451 143 653 7916 6630 2852 3785 -1058 4291 -2451 Inf -2308 -1798 5465 4179 401 6093 1250 6599 -143 2308 Inf 510 7773 6487 2709 5583 740 6089 -653 1798 -510 Inf 7263 5977 2199 -1680 -6523 -1174 -7916 -5465 -7773 -7263 Inf -1286 -5064 -394 -5237 112 -6630 -4179 -6487 -5977 1286 Inf -3778 3384 -1459 3890 -2852 -401 -2709 -2199 5064 3778 Inf
3 Pass
A = [-1 1 10 -10]; correct = [1 2]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
B = Inf 2 11 -9 -2 Inf 9 -11 -11 -9 Inf -20 9 11 20 Inf
4 Pass
A = [0 1000 -2000 1001 0]; correct = [1 5]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
B = Inf 1000 -2000 1001 0 -1000 Inf -3000 1 -1000 2000 3000 Inf 3001 2000 -1001 -1 -3001 Inf -1001 0 1000 -2000 1001 Inf
5 Pass
A = [1:1000 0.5]; correct = [1 1001]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
B = Columns 1 through 17 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 5.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 4.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 3.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 2.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf 1.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 Inf -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -1.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -2.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -3.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -4.0000 -21.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -5.0000 -22.0000 -21.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -6.0000 -23.0000 -22.0000 -21.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -7.0000 -24.0000 -23.0000 -22.0000 -21.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -8.0000 -25.0000 -24.0000 -23.0000 -22.0000 -21.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -9.0000 -26.0000 -25.0000 -24.0000 -23.0000 -22.0000 -21.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -10.0000 -27.0000 -26.0000 -25.0000 -24.0000 -23.0000 -22.0000 -21.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -11.0000 -28.0000 -27.0000 -26.0000 -25.0000 -24.0000 -23.0000 -22.0000 -21.0000 -20.0000 -19.0000 -18.0000 -17.0000 -16.0000 -15.0000 -14.0000 -13.0000 -12.0000 -29.0000 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-25.0000 -24.0000 -23.0000 -22.0000 -21.0000 -20.0000 -19.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -27.0000 -26.0000 -25.0000 -24.0000 -23.0000 -22.0000 -21.0000 -20.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -27.0000 -26.0000 -25.0000 -24.0000 -23.0000 -22.0000 -21.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -27.0000 -26.0000 -25.0000 -24.0000 -23.0000 -22.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -27.0000 -26.0000 -25.0000 -24.0000 -23.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -27.0000 -26.0000 -25.0000 -24.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -27.0000 -26.0000 -25.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -27.0000 -26.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -27.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -28.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -29.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -30.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -31.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -32.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -33.0000 -50.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -34.0000 -51.0000 -50.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -35.0000 -52.0000 -51.0000 -50.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -36.0000 -53.0000 -52.0000 -51.0000 -50.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -37.0000 -54.0000 -53.0000 -52.0000 -51.0000 -50.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -38.0000 -55.0000 -54.0000 -53.0000 -52.0000 -51.0000 -50.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -39.0000 -56.0000 -55.0000 -54.0000 -53.0000 -52.0000 -51.0000 -50.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -40.0000 -57.0000 -56.0000 -55.0000 -54.0000 -53.0000 -52.0000 -51.0000 -50.0000 -49.0000 -48.0000 -47.0000 -46.0000 -45.0000 -44.0000 -43.0000 -42.0000 -41.0000 -58.0000 -57.0000 -56.0000 -55.0000 -54.0000 -...
6 Pass
% Area codes A = [847 217 508 312 212]; correct = [2 5]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
B = Inf -630 -339 -535 -635 630 Inf 291 95 -5 339 -291 Inf -196 -296 535 -95 196 Inf -100 635 5 296 100 Inf
7 Pass
% Zip codes A = [60048 61802 01702 60601 10001]; correct = [1 4]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
B = Inf 1754 -58346 553 -50047 -1754 Inf -60100 -1201 -51801 58346 60100 Inf 58899 8299 -553 1201 -58899 Inf -50600 50047 51801 -8299 50600 Inf
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Start Hunting! | 6,027 | 11,028 | {"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} | 3.59375 | 4 | CC-MAIN-2020-45 | latest | en | 0.213051 |
https://blog.demofox.org/2021/06/ | 1,685,945,546,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00416.warc.gz | 159,309,694 | 36,231 | # Resistance and Voltage Dividers
When i first started working with electronics, i tended to think of my circuits, or even parts of my circuit, in isolation. The horror of it though is that your circuit is plugged into other things – at minimum a battery, but commonly other devices, or your house and the power grid – and those things can affect how your circuit works.
Beyond being physically connected with wires to other things, your circuits also have a connection to the rest of the world through electromagnetic fields.
In this post we are going to talk about voltage divers, which on one hand can be useful if made on purpose, but can also be made on accident and cause you strange behaviors.
## Voltage Dividers
Voltage dividers are a way of giving you a lower voltage. If you have a 9 volt battery and only want 6 volts, a voltage divider can do that for you. There is a downside to voltage dividers that we’ll explore in this post, but they are incredibly simple to make: you only need two resistors.
First let’s look at a single resistor in a circuit. Lets put a 1000 ohm resistor in a circuit with a 9 volt battery. If we connect our multimeter probes to the wire on the same side of the resistor and measure volts we’ll get zero volts (see diagram below). This is because volts is a measurement of electric potential between two points. Our multimeter is measuring the difference in electric potential between two points right next to each other on a wire, and the difference is essentially zero. The red and black arrows on the circuit diagram are where we connect the red (+) and black (-) probes of our multimeter.(Tangent: 9 milliamps is going through this circuit since there is 1000 ohms of resistance and 9 volts. The power supply says 8 but it has limited accuracy, resistors are not exactly their labeled value, wires have resistance, etc. It also reads that there are 9 volts * 8 milliamps = 72 milliwatts of power being used.)
What if we put our multimeter on different sides of the resistor? In that case, we read 9 volts. The resistor makes it more difficult for electricity to cross, and thus there is a difference in electric potential of 9 volts, on each side.
What would happen if we put two resistors in?
If we measure at the red and black arrows again, we’ll still have 9 volts. If we measure at the red and orange arrows though, we’ll see 4.5 volts. If we read at the orange and black arrows, we’ll also see 4.5 volts. We know that the whole circuit needs to go from 9 volts to 0 volts since that is what is provided by our battery, but it dropped by half on the first resistor, and then dropped the rest of the way on the second resistor. (Tangent: the total resistance here is 2000 ohms, so 4.5 milliamps would flow through the circuit)
Let’s change the value of the resistors and see what happens.
I didn’t have a 2000 ohm resistor so i just put two 1000 ohm resistors in series (more on that further down).
If we measure between the red and black, we still have 9 volts. If we measure between the red and orange, we get 3 volts though, and if we measure between the orange and black, we get 6 volts. Weird! (Tangent: The total resistance here is 3000 ohms, so there should be 9 volts / 3000 ohms = 3 milliamps flowing through the circuit but my power supply isn’t showing that correctly.)
Similarly, you can change the second resistor to be half instead of double and get the opposite result.
I didn’t have a 500 ohm resistor so i put two 1000 resistor ohms in parallel (more on that further down).
What is going on here is that the 9 volts are dropping off across the resistors based on their relative values. When the resistors are equal in value, they each get half of the voltage. When they are unequal, the voltage across the R2 resistor is calculated like this:
$V_{R2} = V * \text{R2} / (\text{R1}+\text{R2})$
To actually use this as a power source, you would connect new wires as the positive and negative power for a sub circuit.
Note in the above, I’m not saying that is -6V and +6V, which would be 12 volts total, I’m just labeling the positive and negative sides of the 6 volts of power available.
You could use the top part as a 3 volt source if you wanted instead, or in addition to the 6 volts you are using from the bottom part. You could even split the voltage into more than just two levels, but instead could put in N resistors to have N voltage levels.
The famous 555 timer for instance internally uses a voltage divider with three 5K resistors to make three different power levels, and that is why it’s called a 555 interestingly. You can see it at the top of this diagram of a 555 timer, between the ground (pin 1) and the +Vcc supply (pin 8).
(This image is from this 555 timer tutorial: https://www.electronics-tutorials.ws/waveforms/555_timer.html)
## Resistors in Series vs in Parallel
When I needed a 2k Ohm resistor in the last section I put two 1k Ohm resistors in series. When you put resistors in series, their values add together, allowing you to additively create whatever resistance you need.
When I needed a 500 Ohm resistor and didn’t have one, I put two 1k Ohm resistors in parallel. This is because putting resistors in parallel gives electricity more than one path to get through, and thus has lower resistance than if there was only one of the resistors. The exact equation for the resistance of resistors in parallel is:
$1/R = 1/R_1 + 1/R_2 + ... + 1/R_N$
Where $R_i$ is the value of a specific resistor.
This means that if you put two of the same valued resistors in parallel, the resistance will be cut in half. If you put three of them in parallel, the resistance will be cut in three.
This formula comes up again in electronics. For capacitors, when you put them in parallel, their capacitance adds. When you put them in series, their capacitance follows the parallel resistor equation. It’s the same formulas, but parallel / series reversed. Strange huh?
$1/C = 1/C_1 + 1/C_2 + ... + 1/C_N$
Where $C_i$ is the value of a specific capacitor (in Farads).
Something else strange is that this is why thicker wire has less resistance too. There are more paths for electricity to travel through the thicker wire, compared to thinner wire, so resistance goes down.
Below are some images of two 1k Ohm resistors in series and in parallel, with the multimeter showing the total resistance value.
One resistor:
Two resistors in series:
Two resistors in parallel:
## What Happens When Using a Voltage Divider?
Ok so let’s start with the voltage divider we set up before.
Now let’s say we actually use that 6 volts to power something. That something will have a resistance of 2k Ohms. Maybe it’s some kind of light bulb.
We can simplify this circuit though. The 2k Ohms of our load, and the 2k Ohms of the voltage divider are in parallel so we can use our formula for parallel resistance, or remember that two capacitors of equal value in parallel get half the resistance. So that means we could describe our circuit this way, as far as resistance is concerned:
The problem with that is that our voltage divider has changed. The resistors are equal now, which means that our 6 volts has dropped down to 4.5 volts!
If we decreased the resistance of what we were powering, the voltage would drop too. Intuitively, imagine if you had a short circuit so had zero resistance across the load, the electricity would completely bypass the 2k Ohm resistor in the voltage divider as if it weren’t there, so there would be zero volts difference between the top and bottom of the 2k Ohm resistor.
If we increased the resistance of what we were powering, we would raise the combined parallel resistance there on the 2nd part of the voltage divider, but luckily would at most have 2k ohm resistance. For instance, using a 1 mega ohm resistive load, the parallel resistance formula gives us a resistance of 1.996 k Ohms. So, if we had a high resistance load, we’d get nearly our full 6 volts, but would never quite have the full 6 volts. At the limit, if our load was disconnected, and thus had infinite resistance, we would get the full 6 volts.
If you know the resistance of the load you are plugging into the voltage divider, you can take it into account and choose a resistor for the voltage divider that gives you the desired parallel resistance amount and thus the right voltage. Some loads have variable resistances though, and then you have a problem and should look at other methods of changing the DC voltage level, such as a buck converter.
Some loads have no resistance though, and a voltage divider can come in really handy. Supplying power to a transistor’s base, or to an op amp’s input, or to an optocoupler’s input for instance can make great use of these because they just “read” the voltage signal there without putting any extra load on it.
The lesson here is that whenever you plug things together, you might get strange drops in voltage because you’ve accidentally created a voltage divider. If your resistance is sufficiently higher than whatever internal resistance what you’ve plugged into has, you can ignore the voltage drop, but that also decreases the amperage so may not be desirable.
This effect even comes up in batteries (and other power sources) which essentially can be modeled as an ideal voltage source, with a small resistance (like 10 ohms). If you use a low valued resistor on a battery, the voltage will drop because you are secretly part of a voltage divider involving the internal resistance of the battery (and in fact, that “internal resistor” can’t take that much power and will start heating up, which can be dangerous! So don’t short circuit batteries!). Since a battery’s resistance is so small, your resistance level is likely to be much higher when using the battery to power something, and this isn’t something you really have to worry about in normal situations.
Of course, all this talk only deals with DC and resistors. Things get more complex when you have capacitors, inductors or AC power.
## Maximum Power (Watts)
So we saw that as R2’s resistance gets larger, the voltage across R2 becomes larger, and at infinite resistance, it gets all the voltage available.
We also know that the larger the resistance, the lower the amps in the circuit, so getting that voltage comes at a cost.
Watts is a unit of measurement of power and is volts multiplied by amps. It turns out that if you want your voltage divider to have maximum power (watts), that R1 should equal R2. Wikipedia has more about that here: https://en.wikipedia.org/wiki/Impedance_matching
Here are some graphs showing this, where if resistor R1 is 1k Ohms, that you get the highest amount of watts when R2 is also 1k Ohms, despite the behavior of the volts and amps.
## Calculating Resistance (and Voltage) of an Unknown Circuit
Since plugging your circuit into other things can make an implicit / unintentional voltage divider, you probably want to know how much resistance some other black box circuitry might have. Luckily you can figure this out using Ohms law (see last post: Voltage, Amps, Resistance and LEDs (Ohm’s Law)) and some simple algebra.
First, connect a resistor to the + and – and measure the amps in the circuit. If you use a resistor that is too low value, or has too low of a wattage rating, the resistor will get hot, possibly start glowing or burst into flames (resistors have a rating in watts and the common ones for small electronics like those seen in this post can handle 1/4 of a watt). So basically be careful if doing this with high voltages – and in fact, if my blog is your primary source of knowledge, please don’t mess with high voltage 🙂
So let’s say we connect a 1k Ohm resistor and read a value of 0.01 amps or 10 milliamps.
Ohms law says:
$I = V/R$
where I is current, V is volts and R is resistance.
So we now have this formula:
$0.01 = V / (1000 + R_1)$
We have one equation with two unknowns, so we need another equation to make it solvable by having two equations and two uknowns. Let’s say we take an amperage measurement using a 500 Ohm resistor and get 0.017 amps or 17 milliamps.
That gives us a second equation:
$0.017 = V / (500 + R_1)$
We now have two equations with two unknowns!
We can solve the first equation for V and get:
$V = 0.01 * (1000 + R_1)$
From there we can plug V into the second equation to get:
$0.017 = 0.01 * (1000 + R_1) / (500 + R_1)$
Solving for R1, we get:
$R_1 = (0.01 * 1000 - 0.017 * 500) / (0.017 - 0.01) = 214.28 \Omega$
If you do the calculations, you get 214.28 ohms, which means the unknown circuit has that much resistance.
What’s nice is that you can also use this to get the total amount of voltage available to this circuit by plugging this resistance into the first equation that we solved for V:
$V = 0.01 * (1000 + 214.28) = 12.14 \text{volts}$
This was a toy example i made up, using 12 volts and 200 ohms of resistance, so our answer is pretty close. The inaccuracies came from rounding off the numbers, but you’ll get the same problems in real life from not completely accurate measurements and imperfect electronic components.
For convenience, here are the equations to calculate the resistance of an unknown circuit, without having to do the algebra each time.
$R_1 = ( I_A * R_{2A} - I_B * R_{2B}) / (I_B-I_A)$
Where $R_1$ is the resistance of the unknown circuit. $R_{2A}$ is the first resistor value you connected and measured to get $I_A$ amps. $R_{2B}$ is the second resistor value you connected and measured to get $I_B$ amps.
Once you have the $R_1$ value, you can plug it into this to get the voltage available to the circuit:
$V = I_A * (R_{2A} + R_1)$
Let’s take these equations for a spin with a battery. I accidentally popped the fuse on my digital multimeter and can’t use it to measure amps so i’ll use my analog multimeter.
First i’ll measure the amps with a 1k Ohm resistor. The knob is set to 10 milliamp measurements so the bottom row of readings (that are labeled 0 to 10) are where you read from. I drew some yellow to show you where to read from. I read 8.6 milliamps.
Next i’ll put two 1k Ohm resistors in series to make 2k Ohms of resistance and measure amps to get what looks like 4.6 milliamps.
Ok so let’s plug our values into the equations!
$R_1 = ( I_A * R_{2A} - I_B * R_{2B}) / (I_B-I_A)$
$R_1 = ( 0.0086 * 1000 - 0.0046 * 2000) / (0.0046-0.0086) = 150 \Omega$
So it looks like this 9 V battery has 150 ohms of resistance. I’ve heard that as a battery is used, it’s resistance goes up, so maybe this battery is nearing needing to be replaced having such large resistance.
Let’s calculate how many volts it has.
$V = I_A * (R_{2A} + R_1)$
$V = 0.0086 * (1000 + 150) = 9.89 volts$
So, the battery has 9.89 volts inside of it. Either they made the battery have higher than 9 volts inside of it, to account for internal resistance dropping the output voltage, or my 5$analog multimeter is not very accurate and these are just ball park figures. ## Closing Thanks for reading and hopefully you found this interesting or useful. Have any requests or ideas for other topics to write about? Drop me a message on twitter at @Atrix256. # Voltage, Amps, Resistance and LEDs (Ohm’s Law) I’ve taken up learning electronics during the pandemic, and have enjoyed it quite a bit. I’ve been programming for 25+ years, so it’s nice to have something different to learn and work on that is still both technical and creative. It’s cool getting a deeper understanding of how the fundamental forces of nature work, as well as being able to MacGyver a hand crank powered flashlight from an old printer if needed (Check out a 40 second video of that here!). It’s also nice having something physical to show at the end of the day, although it does require consumable parts, so there are pros and cons vs making software. Friends (Hi Wayne!) and YouTube have helped me learn a lot, but I found the subject pretty alien at first and wanted to try my hand at some explanations from a different POV. This post starts that journey by taking the first steps into DC electronics. ## Ultra Basics Electricity flows if there is a path for it to flow in and the flow is made up of electrons. Electrons are negatively charged so travel from the negative side of a circuit to the positive side. Conventional current flow is backwards from this though, and says that electricity flows from the positive side to the negative side. In this case, it’s not electrons flowing but “holes” flowing. Holes are a weird concept, but they are just a place that will accept an electron. Here is an open circuit, which means that there is a gap. Since the circuit is not closed, electricity cannot flow. (made in https://www.circuitlab.com/editor/#) If you close the circuit, like the below, electricity is able to flow. The circle on the left is a power source with a + and – terminal. It’s labeled as a 1.5 volt double A battery. Here is a diagram of a circuit with a switch that can be used to open or close the circuit. Being able to read and make circuit diagrams is real helpful when building things or trying to understand how circuits work. Of note: the higher the voltage, the farther that electricity can jump across gaps. So, while at low voltage, a circuit may be open, turning up the voltage may make it closed when the electricity arcs across! ## Ohm’s Law IMAGE CREDIT: Eberhard Sengpiel The most useful thing you can learn about DC electricity is Ohm’s law which mathematically explains the relationship between voltage, amperage and resistance. Ohm’s law is: $I = V/R$ In the equation I stands for Intensity and means current aka amps, V stands for voltage and R stands for resistance. If electricity was water, voltage would be the water pressure, amperage would be how much water was running through the pipe, and resistance would be a squeezing of the pipe, like in the image above. Current is measured in amperes (amps) or the letter A. 500mA is 500 milliamps, or half of an amp, and 1.2A is 1.2 amps. Note: electricity is dangerous! It can take only a few hundred milliamps to be fatal, but voltage is needed to be able to let those amps penetrate your skin. Voltage is measured in volts or the letter V. If you see 9V on a battery, that means it’s a 9 volt battery, and is capable of providing 9 volts. Resistance is measured in Ohms or the omega symbol $\Omega$. So if you see $5\Omega$ that means 5 Ohms of resistance. If you see $5k\Omega$ that means 5 kiloohms which is 1000 times as much resistance. If you see $5M\Omega$ with a capitol M, that means 5 megaohms, which is 1000 times as much resistance again. Where Ohm’s law comes in handy is when you know two of these three values and you are trying to calculate the third one. As written, the formula showed how to calculate amps when you know voltage and resistance, but you can use algebra to re-arrange it to a formula for any of the three: $I = V/R$ $R = V/I$ $V = I*R$ This comes up quite often – if you know how much voltage a battery has, and you know how many amps you need, you can use this to calculate the value of the resistor to get the desired amps. ## Diodes, LEDs and Resistors LED stands for Light Emitting Diode. A diode is something which lets electricity only flow in one direction and it has a couple of common uses: • Protecting circuits from electricity flowing in the wrong direction. • Turning Alternating Current (AC) into Direct Current (AC) by rectifying it (preventing the negative part of AC from getting through. Same as last bullet point) • Lots of cool tricks, like stabilizing uneven power levels by letting voltage over a specific value “spill over” out of the circuit. Here is a pack of various diodes i bought from amazon for 10$. There are a quite a few different types of diodes, which are useful for different situations.
Here are some diodes close up. The black one is a rectifier diode IN4001, and the more colorful one is a switching diode 1N4148. Those part numbers are actually written on the diodes themselves but are a bit hard to see. You can use these numbers to look up the data sheet for the parts to understand how they work, what their properties are, how much voltage and amps they can handle, and often even see simple circuit diagrams on using them for common tasks. Data sheets are super useful and if doing electronics work, you will be googling quite a few of them! Here is the data sheet for 1N4148 which i found by googling for “1N4148 data sheet” and clicking the first link. 1N4148 Data sheet.
Here are two circuit diagrams with diodes in them. The black triangle with the line on it is a diode. The arrow shows the direction that it allows conventional flow to travel. The line on the arrow corresponds to the bands on the right of the diodes in the image above, which is the negative side of the diode (cathode). The left circuit is a closed circuit and allows electricity to flow. That diode is forward biased. The circuit on the right has the diode reverse biased which does not allow electricity to flow.
LEDs can do many things regular diodes can do, since they are diodes, but they have the property that when electricity flows through them, they light up. Since they are diodes, and only let electricity flow in one direction, LEDs have a + side and a – side and you have to hook them up correctly in a circuit for them to light up. If you hook them up the wrong way, it doesn’t damage them, but they don’t light up and they don’t close the circuit for electricity to flow. The symbol for an LED is the diode symbol, but with arrows coming out of it.
Here are a pack of LEDs i have that came as part of a larger electronics kit. You can get a couple hundred LEDs in a variety of colors from amazon for about 10$. Some LEDs are in colored plastic cases, some are in clear cases. There are even LEDs that shine in infrared and ultraviolet. LEDs also come in different sizes. This pack has 3mm and 5mm LEDS. Here is an up close look at a white LED. The longer leg is the positive side, which means you need to plug the positive side of the circuit into it if you want it to light up. the negative side has a shorter leg, but the negative side also has a flat side on the circular ring at the bottom, which can’t really be seen in this picture. All diodes have a voltage drop, which is a voltage amount consumed by the diode. If you are providing less than that amount of voltage, the diode will act as an open switch, and electricity won’t flow through it. The specific voltage drop for diodes can be found in data sheets, but i’ve found it difficult to find data sheets for LEDs. Luckily I picked up a “Mega328” component tester from amazon for 15$. It lets you plug in a component, press the blue button, and then tells you information about the component. It’s super handy! Here you can see the voltage drop of 2 different LEDs. The smaller red LED has a voltage drop of 1.88V while the larger green LED has a voltage drop of 2.5V. If you supply them with less than that amount of voltage, they will not light up!
So what would happen if we tried to connect the LEDs to the batteries below?
The large green LED has a 2.5V voltage drop, while the AAA battery only has 1.5V as you can see on the label. That means the LED doesn’t light up.
The smaller red LED has a 1.88V voltage drop and is connecting to a 9V battery so it has enough voltage and should light up. Let’s use Ohm’s law to calculate how much current – in amps – are going through the LED.
I = V/R and in our case V is 9 and R is 0 because we have no resistance.
$I = \frac{9}{0} = \infty$
Oops we have infinite current! The LED is destroyed pretty quickly after you plug it in.
There isn’t actually infinite current, because the metal wires connected to the LED have a very tiny amount of resistance to them, just like all wire, and the battery has a limit of how many amps it can give. So in any case, it isn’t infinite amps, but it is a very large number, limited by how many amps the 9V battery can actually deliver. The LED would actually be destroyed. You should basically always use a resistor with an LED to limit the current and keep it from being destroyed. Here is an interesting read about how to calculate the internal resistance of a battery which will then tell you how many amps it can give you: Measuring Internal Resistance of Batteries.
When you have a circuit with this low of resistance, it’s considered a short circuit, and if the LED didn’t get destroyed, the battery would start getting hot and it could become a dangerous situation. This is also why short circuits themselves are bad news. They have a LOT of current running through them which can cause things to heat up, melt and catch fire.
3mm and 5mm LEDs typically want 20 milliamps maximum (20mA or 0.02A) to be at full brightness. If you give them less, they will be less bright but still function.
We can calculate then how much resistance they want to be maximally bright if we know the voltage of the power source we are using and the voltage drop off of the LED we are trying to power.
Let’s take the larger green LED with a 2.5V voltage drop, and power it with a 9V batter, aiming to get 20mA.
First we subtract the voltage drop from the supply to see how much voltage we have to work with: 9V – 2.5V = 6.5V.
Next, we know we want 20mA and we have 6.5V, and we are just trying to solve for resistance so we use Ohm’s law: R = V/I.
$R = 6.5V / 0.02A = 325\Omega$
So, we need 325 ohms of resistance to get 20mA in our LED from a 9V battery. Here is a pack of resistors i got from amazon for 12$. Resistors have funny colored bands on them which tell you their rating. You can find charts for decoding them all over the place, but again, the “Mega328” will tell you this too. In fact, a multi meter will tell you as well. Multi meters aren’t very expensive. Here’s one i got from amazon for 35$ which has tons of features and works really nicely.
I don’t have any 325 ohm resistors, but i do have 470 ohm resistors, so i’ll just use one of those. That’s 14mA if you do the math, which is a bit lower than 20mA, but it still works just fine despite not being as bright as it could be. You can get different resistances by connecting resistors in parallel or series and doing some math, but this works for now. I used a mini breadboard (the green thing) to hook this circuit up. Every horizontal line of 5 holes is connected to each other electrically. It’s a nice way to play with circuits without having to solder things together. By convention, red is used for the positive terminal and black or blue is used for the negative.
By the way, quick fun fact. A 1.5V AA battery is considered dead when it has dropped down to 1.35V. At this point, it still has energy in it though! If you are clever with electronics, you could make circuitry to use this power from dead batteries to give you 1.5V or higher, and you could drain so called dead batteries even further.
## LEDs Turning Light Into Power
Many things in electronics turn out to be reversible. Speakers work as poor microphones, and microphones work as poor speakers. Similarly, LEDs can work as poor solar cells and turn light into energy. Want to see? Here i hook my multimeter up to an LED, and have it set to read volts. It reads 48.7mV. Energy is flowing all around us from radio waves, etc, so it’s picking up some of that.
When i put the LED in the beam of the flashlight, it jumps up to 1.644V. Pretty cool huh?
## Did You Like This Post?
It’s a little different than what I usually write about, but hopefully you liked it. Careful though, this stuff escalates quickly. Before you know it you’ll be harvesting optocouplers and coils from old printers to make a rail gun. | 6,653 | 27,798 | {"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": 34, "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-2023-23 | latest | en | 0.950328 |
https://sciencedocbox.com/Physics/78415348-Lecture-guide-math-90-intermediate-algebra-stephen-toner-intermediate-algebra-2nd-edition-miller-o-neill-hyde-victor-valley-college.html | 1,632,818,780,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00280.warc.gz | 531,806,455 | 27,864 | # Lecture Guide. Math 90 - Intermediate Algebra. Stephen Toner. Intermediate Algebra, 2nd edition. Miller, O'Neill, & Hyde. Victor Valley College
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1 Lecture Guide Math 90 - Intermediate Algebra to accompan Intermediate Algebra, 2nd edition Miller, O'Neill, & Hde Prepared b Stephen Toner Victor Valle College Last updated: 11/24/10 0
2 1.1 Sets of Numbers The set of rational numbers can be written as A. Tpes of Numbers Natural - Whole - Integers - Rational - an number which can be epressed as the of two numbers, provided the denominator is not zero. Aleks Problem: Classif each number below as an integer or not. Integer? Yes No Aleks Problem: Classif each number below as irrational or rational Rational Irrational Irrational - Real - B. Set Builder Notation can be written as Aleks Problem: Rewrite the Set R b listing the elements. Make sure to use the appropriate set notation. 1
3 C. Interval Notation D. Unions and Intersections Inequalit Graph Interval Notation Given:, Find each of the following: Find each union or intersection. Write our solution using interval notation
4 5. E. Words and Inequalities more than less than at least at most 6. no more than no less than 7. Aleks Problem: The sets C and D are defined as follows: Write and using interval notation. 8. Aleks Problem: Write " is less than or equal to b, and 3 is greater than b" as an algebraic epression. 3
5 1.2 Operations with Real Numbers 1.2 #84 Order of Operations We assume ou remember the rules for adding, subtracting, multipling and dividing real numbers... Evaluate: (57 of them) 1.2 #92 Order of Operations Propert: Be careful! *Evaluate each if and : a. b. 4
6 1.3 Simplifing Epressions A. Properties Aleks: A trick problem... Which propert is illustrated? Commutative Propert of addition... of multiplication Simplif: Associative Propert of addition... of multiplication 1.3 #68 Simplif: Additive Identit Multiplicative Identit 1.3 #74 Additive Inverse Multiplicative Inverse Distributive Properties 5
7 1.4 Linear Equations 1.4 #56 Solve: Solve: 1.4 #42 Solve: There are three classes of equtions: Conditional the tpe we've been doing Identit "all real numbers" is our solution Contradiction there is no solution 1.4 #72 Solve: When an equation contains fractions, it is usuall a good idea to multipl through b the least common denominator to clear the fractions. 1.4 #46 Solve: 1.4 #80 Solve: 6
8 1.5 Word Problems A. Number Problems 1.5 #16 Twice the sum of a number and three is the same as 1 subtracted from the number. Find the number. 1.5 #26 Five times the smallest of three consecutive even integers is 10 more than twice the largest. Find the integers. C. Percentage Word Problems B. Consecutive Integer Problems consecutive integers 1.5 #26 The price of a used tetbook after a 35% markdown is \$ What was the original price? consecutive odd integers consecutive even integers 1.5 #24 Four times the smaller of two consecutive odd integers is the same as 73 less than 5 times the larger. Find the integers. 1.5 #32 Robert can take out a 3-r loan at 8% simple interest or a 2-r loan at 8.5% simple interest. If he borrows \$7000, how much interest will he pa for each loan? Which option will require less interest? 7
9 D. Miture Problems 1.5 #58 A nut miute consists of almonds and cashews. Almonds are \$4.98/lb. and cashews are \$6.98/lb. How man pounds of each tpe of nut must be mied to produce 16 pounds of a miture selling for \$5.73/lb.? 1.5 #52 One fruit punch has 40% fruit juice and another is 70% fruit juice. How much of the 40% punch should be mied with 10 gallons of 70% punch to create a fruit punch that is 45% fruit juice? 1.5 #56 How man ounces of water must be added to 20 oz. of an 8% salt solution to make a 2% salt solution? 8
10 Aleks problem: Two factor plants are making TV panels. Plant A produced 3000 fewer panels than Plant B did. Five percent of the panels from Plant A and 2% of the panels from Plant B are defective. How man panels did Plant A produce, if the two plants together produced 1110 defective panels? E. Investment Problems 1.5 #44 Amanda borrowed \$6000 from two sources: her parents and a credit union. Her parents charged her 3% simple interest and the credit union charged 8%. If after 1 ear she paid \$255 in interest, how much did she borrow from each source? F. Distance Problems Aleks problem: A car travels 221 miles in 3 hours and 15 minutes. How man miles did it travel per hour? 9
11 Distance Problems- General Procedure: 1.5 #62 Two cars are 190 miles apart and travel toward each other. The meet in 2 hours. One car travels 5 miles per hour slower than the other car. What are the speeds of both cars? 1.5 #60 A woman can hike 1 mph faster down a trail than she can on the return trip uphill. It takes 3 hours to get to her destination and 6 hours to return. What is her downhill speed? 1.5 #64 Two canoes travel down a river starting at 9:00. One canoe travels twice as fast as the other. After 3.5 hours the canoes are 5.25 miles apart. Find the average rate of each canoe. 10
12 1.6 Literal Equations & Geometr Problems A. Literal Equations 1.6 #36 Solve for. B. Geometr Problems 1.6 #8 The length of a rectangle is 4 inches less than twice the width. The perimeter is 112 inches. Find its dimensions. 1.6 #42 Solve for. 1.6 #48 Solve for h. 1.6 #10 A triangular garden has sides that can be represented b 3 consecutive integers. If the perimeter is 15 ft, what are the lengths of the sides? Solve for w. 11
13 1.6 #10 The smallest angle in a triangle is the size of the largest angle. The middle angle measures less than the largest. Find all three angle measures. 1.7 Solving Inequalities *Be careful to change the direction of the inequalit whenever ou or each side of the inequalit b a. *For each of the following, graph our solutions sets. Also write our solutions using interval notation. 1.7 # #20 Find : (+15) 1.7 #36 12
14 Ke Phraes: at most at least no more than no less than Aleks Problem: Brian is choosing between two eercise routines: In routine #1, he burns 46 calories walking. He then runs at a rate that burns 11 calories per minute. Aleks Problem: For her phone service, Kira pas a monthl fee of \$20, and she pas an additional \$0.04 per minute of use. The least she has been charged in a month is \$ What are the possible numbers of minutes she has used her phone in a month? Use for the number of minutes, and solve our inequalit for. In routine #2, he burns 22 calories walking. He then runs at a rate that burns 15 calories per minute. For what amounts of time spent running will Routine #1 burn fewer calories than Routine #2? Use for the number of minutes spent running, and solve our inequalit for. 13
15 1.8 Eponents & Scientific Notation A. Properites of Eponents 1. *Simplif each of the following: a. b. 2. c. d. 3. e. f. 4. g. 5. h. i. j. Negative eponents are NOT considered to be simplified. Do NOT leave them in final answers! 14
16 k. s. l. m. 6. ALEKS PROPLEM TYPE: Simplif: n. o. p. q. r. Multiple choice: a. b. c. d. 15
17 B. Scientific Notation Scientific notationis a shorthand notation for writing etremel small or large numbers. *Multipl. Write our answers in scientific notation: 6. Notation: 7. *Write each using scientific notation: 1. 9,374, trillion *Divide. Write our answers in scientific notation: *Write each in standard form:
18 Additional Aleks Problems (Chapter 1) 1. The properties of addition are: [1] Commutative Propert [2] Associative Propert [3] Additive Identit Propert [4] Additive Inverse Propert 3. Dale rented a truck for one da. There was a base fee of \$20.99, and there was an additional charge of 95 cents for each mile driven. Dale had to pa \$ when he returned the truck. For how man miles did he drive the truck? For each equation below, indicate the propert that justifies the equation b filling in the bo with the appropriate number. 2. Consider the following properties of real numbers: [1] Commutative Propert of Addition [2] Associative Propert of Addition [3] Additive Identit Propert [4] Additive Inverse Propert [5] Distributive Propert [6] Commutative Propert of Multiplication [7] Associative Propert of Multiplication [8] Multiplicative Identit Propert [9] Multiplicative Inverse Propert [10] Multiplication Propert of Zero 4. Raina invested her savings in two investment funds. The amount she invested in Fund A was 3 times as much as the amount she invested in Fund B. Fund A returned a 4% profit and Fund B returned a 6% profit. How much did she invest in Fund A, if the total profit from the two funds together was \$1080? For each equation below, indicate the propert that justifies the equation b filling in the bo with the appropriate number. 5. The price of an item has been reduced b 80%. The original price was \$90. What is the price of the item now? 17
19 2.1 Graphing Lines Graph: Methods for Graphing Lines: 1. The X-Y Chart Graph: Graph: 2. Horizontal and Vertical Lines 4. Graphing b Intercepts Graph: 3. The Slope-Intercept Method A. Identif and plot the -intercept. B. Identif the slope. Rise and run (to the right) from the -intercept to another point on the line. C. Draw the line. 18
20 2.2 Slope Formula: *In each equation, identif the slope and the -intercept. 3. *Find the slope of each line: *Find the slope of the line which runs through the given pair of points: *Find the slope of the graphed line (passes through and :
21 2.3 Point-Slope Form **Parallel lines have the slope. **The slopes of perpendicular lines are of each other. *Find the equation of the line (in slopeintercept form) which has has the following characteristics. a. and has -intercept *Are the following pairs of lines parallel, perpendicular, or neither? 1. b. and the line passes through Three Forms of a Line c. passes through and
22 d. passes through and is parallel to the line Aleks Problem: Consider the line. Find the equation of the line that is parallel to this line and passes through the point. Find the equation of the line that is perpendicular to this line and passes through the point. e. passes through and is perpendicular to the line Aleks Problem: Write equations for the horizontal and vertical lines passing through the point. horizontal line: vertical line: 21
23 2.4 Word Problems 2.4 #8 Ale is a sales representative and earns a base salar of \$1000 per month plus a 4% commission on his sales for the month. a. Write a linear equation that epresses Ale s monthl salar in terms of his sales. 2.4 #14 Let represent the average number of miles driven per ear for passenger cars in the United States since Let =0 represent the ear where corresponds to 1980, =1 corresponds to 1981, and so on. The average earl mileage for passenger cars can be approimated b the equation = where. b. Graph the equation. a. Use the linear equation to approimate the average earl mileage for passenger cars in the United States in the ear c. What is the -intercept and what does it represent in the contet of this problem? d. What is the slope of the line and what does it represent in the contet of this problem? b. Use the linear equation to approimate the average mileage for the ear 1985, and compare it with the actual value of 9700 mi. e. How much will Ale make if his sales for a given month are \$30,000? c. What is the slope of the line and what does it mean in the contet of this problem? d. What is the -intercept and what does it mean in the contet of this problem? 22
24 Aleks Problem: Owners of a recreation area are filling a small pond with water. The are adding water at a rate of 35 liters per minute. There are 700 liters in the pond to start. Aleks Problem: The monthl cost (in dollars) of a long-distance phone plan is a linear function of the total calling time (in minutes), as shown in the figure below. Let W represent the amount of water in the pond (in liters), and let T represent the number of minutes that water has been added. Write an equation relating W to T, and then graph our equation using the aes below. The monthl cost for 51 minutes of calls is \$17.52 and the monthl cost for 79 minutes is \$ What is the monthl cost for 69 minutes of calls? 23
25 Aleks Problem: Suppose that the credit remaining on a phone card (in dollars) is a linear function of the total calling time (in minutes). When graphed, the function gives a line with a slope of. See the figure below. There is \$25.04 in credit remaining on the card after 42 minutes of calls. How much credit was there after 29 minutes of calls? 2.5 Domain and Range The domain of an epression is the set of values which substituted into the epression. The range of an epression is the set of values which can the epression. Eample: domain: range: To find the domain, start with a "default" domain of and then take awa -values which... ** ** ** *Find the domain of each: 24
26 To determine the domain from a graph, look at where the graph etends, left to right. To determine the range from a graph, look at where the graph etends verticall. *Find the domain and range of each graph: *Find the domain and range of each: 25
27 2.6 Functions A function is a rule which assigns a -value in the range to each value in its domain. *Do the following relations represent functions? When we write, we mean Given,, find the following: 1. A graph is that of a function if it passes the 2. *Are the following graphs those of functions? Notation: is pronounced Antime ou see., ou can replace it with 8. 26
28 *Find the requested values from the graph: 2.6 #57 The graph of is given. *Find the domain of each function. Write the answers in interval notation. 2.6 #64 a. Find. b. Find 2.6 #65 c. Find d. For what value(s) of is e. For what value(s) of is 2.6 #68 f. Write the domain of. g. Write the range of. 2.6 # #58 The graph of is given. a. Find. 2.6 #73 b. Find 2.6 #76 c. Find d. For what value(s) of is e. For what value(s) of is A. h ( ) 1 5 f. Write the domain of. g. Write the range of. 27
29 2.7 Graphs of Basic Functions Graph each of the following. Then state the domain and range of each. 4. Absolute Value Function 1. Linear Function 5. Square Root Function 2. Quadratic Function 6. Reciprocal Function 3. Cubic Function 28
30 Graph the parabola points on its graph.. Indicate five Review: Solve the following equation for : Graph the cubic function three points on its graph.. Indicate Simplif: Aleks Problem: For each function below, choose the correct description of its graph: 29
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Math Sample Final Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Factor completel. If the polnomial is prime, state this. ) + 8-0 ( - )( + ) | 10,950 | 42,015 | {"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-2021-39 | latest | en | 0.835164 |
http://balhiserfinancial.com/tag/portfolio-optimization/ | 1,529,403,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267862248.4/warc/CC-MAIN-20180619095641-20180619115641-00594.warc.gz | 32,202,715 | 27,066 | # How to Write a Mean-Variance Optimizer (Part III)… In R
Parts 1 and 2 left a trail of breadcrumbs to follow. Now I provide a full-color map, a GPS, and local guide. In other words the complete solution in the R statistical language.
Recall that the fast way to compute portfolio variance is:
$inline&space;dpi{300}&space;large&space;sigma_{p}^{2}=&space;mathbf{w}^topmathbf{V}mathbf{w}$
The companion equation is rp = wTrtn, where rtn is a column vector of expected returns (or historic returns) for each asset. The first goal is to find find w0 and wn. w0 minimizes variance regardless of return, while wn maximizes return regardless of variance. The goal is to then create the set of vectors {w0,w1,…wn} that minimizes variance for a given level of expected return.
I just discovered that someone already wrote an excellent post that shows exactly how to write an MVO optimizer completely in R. Very convenient! Enjoy…
http://economistatlarge.com/portfolio-theory/r-optimized-portfolio
# The Equation Everyone in Finance Should Know (MV Optimization: How To, Part 2)
As the previous post shows, it all starts with…
$inline&space;dpi{300}&space;large&space;sigma_{p}^{2}=&space;mathbf{w}^topmathbf{V}mathbf{w}$
In order get close to bare-metal access to your compute hardware, use C. In order to utilize powerful, tested, convex optimization methods use CVXGEN. You can start with this CVXGEN code, but you’ll have to retool it…
• Discard the (m,m) matrix for an (n,n) matrix. I prefer to still call it V, but Sigma is fine too. Just note that there is a major difference between Sigma (the covariance-variance matrix) and sigma (individual asset-return variances matrix; the diagonal of Sigma).
• Go meta for the efficient frontier (EF). We’re going to iteratively generate/call CVXGEN with multiple scripts. The differences will be w.r.t the E(Rp).
• Computing Max: E(Rp) is easy, given α. [I’d strongly recommend renaming this to something like expect_ret comprised of (r1, r2, … rn). Alpha has too much overloaded meaning in finance].
• [Rmax] The first computation is simple. Maximize E(Rp) s.t constraints. This is trivial and can be done w/o CVXGEN.
• [Rmin] The first CVXGEN call is the simplest. Minimize σp2 s.t. constraints, but ignoring E(Rp)
• Using Rmin and Rmax, iteratively call CVXGEN q times (i=1 to q) using the additional constraint s.t. Rp_i= Rmin + (i/(q+1)*(Rmax-Rmin). This will produce q+2 portfolios on the EF [including Rmin and Rmax]. [Think of each step (1/(q+1))*(Rmax-Rmin) as a quantization of intermediate returns.]
• Present, as you see fit, the following data…
• (w0, w1, …wq+1)
• [ E(Rp_0), …E(Rp_(q+1)) ]
• [ σ(Rp_0), …σ(Rp_(q+1)) ]
My point is that — in two short blog posts — I’ve hopefully shown how easily-accessible advanced MVO portfolio optimization has become. In essence, you can do it for “free”… and stop paying for simple MVO optimization… so long as you “roll your own” in house.
I do this for the following reasons:
• To spread MVO to the “masses”
• To highlight that if “anyone with a master’s in finance and computer can do MVO for free” to consider their quantitative portfolio-optimization differentiation (AKA portfolio risk management differentiation), if any
• To emphasize that this and the previous blog will not greatly help with semi-variance portfolio optimization
I ask you to consider that you, as one of the few that read this blog, have a potential advantage. You know who to contact for advanced, relatively-inexpensive SVO software. Will you use that advantage?
# How to Write a Mean-Variance Optimizer: Part 1
### The Equation Everyone in Finance Show Know, but Many Probably Don’t!
… With thanks to codecogs.com which makes it really easy to write equations for the web.
This simple matrix equation is extremely powerful. This is really two equations. The first is all you really need. The second is just merely there for illustrative purposes.
This formula says how the variance of a portfolio can be computed from the position weights wT = [w1 w2 … wn] and the covariance matrix V.
• σii ≡ σi2 = Var(Ri)
• σij ≡ Cov(Ri, Rj) for i ≠ j
The second equation is actually rather limiting. It represents the smallest possible example to clarify the first equation — a two-asset portfolio. Once you understand it for 2 assets, it is relatively easy to extrapolate to 3-asset portfolios, 4-asset portfolios, and before you know it, n-asset portfolios.
Now I show the truly powerful “naked” general form equation:
$inline&space;dpi{300}&space;large&space;sigma_{p}^{2}=&space;mathbf{w}^topmathbf{V}mathbf{w}$
This is really all you need to know! It works for 50-asset portfolios. For 100 assets. For 1000. You get the point. It works in general. And it is exact. It is the E = mc2 of Modern Portfolio Theory (MPT). It at least about 55 years old (2014 – 1959), while E = mc2 is about 99 years old (2014 – 1915). Harry Markowitz, the Father of (M)PT simply called it “Portfolio Theory” because:
Yes, I’m calling Markowitz the Einstein of Portfolio Theory AND of finance! (Now there are several other “post”-Einstein geniuses… Bohr, Heisenberg, Feynman… just as there are Sharpe, Scholes, Black, Merton, Fama, French, Shiller, [Graham?, Buffet?]…) I’m saying that a physicist who doesn’t know E = mc2 is not much of a physicist. You can read between the lines for what I’m saying about those that dabble in portfolio theory… with other people’s money… without really knowing (or using) the financial analog.
### Why Markowitz is Still “The Einstein” of Finance (Even if He was “Wrong”)
Markowitz said that “downside semi-variance” would be better. Sharpe said “In light of the formidable
computational problems…[he] bases his analysis on the variance and standard deviation.”
Today we have no such excuse. We have more than sufficient computational power on our laptops to optimize for downside semi-variance, σd. There is no such tidy, efficient equation for downside semi-variance. (At least not that anyone can agree on… and none that that is exact in any sense of any reasonable mathematical definition of the word ‘exact’.)
Fama and French improve upon Markowitz (M)PT [I say that if M is used in MPT, it should mean “Markowitz,” not “modern”, but I digress.] Shiller, however, decimates it. As does Buffet, in his own applied way. I use the word decimate in its strict sense… killing one in ten. (M)PT is not dead; it is still useful. Diversification still works; rational investors are still risk-averse; and certain low-beta investments (bonds, gold, commodities…) are still poor very-long-term (20+ year) investments in isolation and relative to stocks, though they still can serve a role as Markowitz Portfolio Theory suggests.
### Wanna Build your Own Optimizer (for Mean-Return Variance)?
This blog post tells you most of the important bits. I don’t really need to write part 2, do I? Not if you can answer these relatively easy questions…
• What is the matrix expression for computing E(Rp) based on w?
• What simple constraint is w subject to?
• How does the general σp2 equation relate to the efficient frontier?
• How might you adapt the general equation to efficiently compute the effects of a Δw event where wi increases and wj decreases? (Hint “cache” the wx terms that don’t change,)
• What other constraints may be imposed on w or subsets (asset categories within w)? How will you efficiently deal with these constraints?
• Is short-selling allowed? What if it is?
• OK… this one’s a bit tricky: How can convex optimization methods be applied?
If you can answer these questions, a Part 2 really isn’t necessary is it?
# Clover Patterns Show How Portfolios Manage Risk
The red and green “clover” pattern illustrates how traditional risk can be modeled. The red “leaves” are triggered when both the portfolio and the “other asset” move together in concert. The green leaves are triggered when the portfolio and asset move in opposite directions.
Each event represents a moment in time, say the closing price for each asset (the portfolio or the new asset). A common time period is 3-years of total-return data [37 months of price and dividend data reduced to 36 monthly returns.]
### Plain English
When a portfolio manager considers adding a new asset to an existing portfolio, she may wish to see how that asset’s returns would have interacted with the rest of the portfolio. Would this new asset have made the portfolio more or less volatile? Risk can be measured by looking at the time-series return data. Each time the asset and the portfolio are in the red, risk is added. Each time they are in the green, risk is subtracted. When all the reds and greens are summed up there is a “mathy” term for this sum: covariance. “Variance” as in change, and “co” as in together. Covariance means the degree to which two items move together.
If there are mostly red events, the two assets move together most of the time. Another way of saying this is that the assets are highly correlated. Again, that is “co” as in together and “related” as in relationship between their movements. If, however, the portfolio and asset move in opposite directions most of the time, the green areas, then the covariance is lower, and can even be negative.
### Covariance Details
It is not only the whether the two assets move together or apart; it is also the degree to which they move. Larger movements in the red region result in larger covariance than smaller movements. Similarly, larger movements in the green region reduce covariance. In fact it is the product of movements that affects how much the sum of covariance is moved up and down. Notice how the clover-leaf leaves move to the center, (0,0) if either the asset or the portfolio doesn’t move at all. This is because the product of zero times anything must be zero.
Getting Technical: The clover-leaf pattern relates to the angle between each pair of asset movements. It does not show the affect of the magnitude of their positions.
If the incremental covariance of the asset to the portfolio is less than the variance of the portfolio, a portfolio that adds the asset would have had lower overall variance (historically). Since there is a tenancy (but no guarantee!) for asset’s correlations to remain somewhat similar over time, the portfolio manager might use the covariance analysis to decide whether or not to add the new asset to the portfolio.
### Semi-Variance: Another Way to Measure Risk
After staring at the covariance visualization, something may strike you as odd — The fact that when the portfolio and the asset move UP together this increases the variance. Since variance is used as a measure of risk, that’s like saying the risk of positive returns.
Most ordinary investors would not consider the two assets going up together to be a bad thing. In general they would consider this to be a good thing.
So why do many (most?) risk measures use a risk model that resembles the red and green cloverleaf? Two reasons: 1) It makes the math easier, 2) history and inertia. Many (most?) textbooks today still define risk in terms of variance, or its related cousin standard deviation.
There is an alternative risk measure: semi-variance. The multi-colored cloverleaf, which I will call the yellow-grey cloverleaf, is a visualization of how semi-variance is computed. The grey leaf indicates that events that occur in that quadrant are ignored (multiplied by zero). So far this is where most academics agree on how to measure semi-variance.
### Variants on the Semi-Variance Theme
However differences exist on how to weight the other three clover leaves. It is well-known that for measuring covariance each leaf is weighted equally, with a weight of 1. When it comes to quantifying semi-covariance, methods and opinions differ. Some favor a (0, 0.5, 0.5, 1) weighting scheme where the order is weights for quadrants 1, 2, 3, and 4 respectively. [As a decoder ring Q1 = grey leaf, Q2 = green leaf, Q3 = red leaf, Q4 = yellow leaf].
Personally, I favor weights (0, 3, 2, -1) for the asset versus portfolio semi-covariance calculation. For asset vs asset semi-covariance matrices, I favor a (0, 1, 2, 1) weighting. Notice that in both cases my weighting scheme results in an average weight per quadrant of 1.0, just like for regular covariance calculations.
### Financial Industry Moving toward Semi-Variance (Gradually)
Semi-variance more closely resembles how ordinary investors view risk. Moreover it also mirrors a concept economists call “utility.” In general, losing \$10,000 is more painful than gaining \$10,000 is pleasurable. Additionally, losing \$10,000 is more likely to adversely affect a person’s lifestyle than gaining \$10,000 is to help improve it. This is the concept of utility in a nutshell: losses and gains have an asymmetrical impact on investors. Losses have a bigger impact than gains of the same size.
Semi-variance optimization software is generally much more expensive than variance-based (MVO mean-variance optimization) software. This creates an environment where larger investment companies are better equipped to afford and use semi-variance optimization for their investment portfolios. This too is gradually changing as more competition enters the semi-variance optimization space. My guestimate is that currently about 20% of professionally-managed U.S. portfolios (as measured by total assets under management, AUM) are using some form of semi-variance in their risk management process. I predict that that percentage will exceed 50% by 2018.
# Surpassing the Frontier?
Suppose you have the tools to compute the mean-return efficient frontier to arbitrary (and sufficient) precision — given a set of total-return time-series data of asset/securities. What would you do with such potential?
I propose that the optimal solution is to “breach the frontier.” Current portfolios provide a historic reference. Provided reference/starting point portfolios have all (so far) provided sufficient room for meaningful and sufficient further optimization, as gauged by, say, improved Sortino ratios.
Often, when the client proposes portfolio additions, some of these additions allow the optimizer to push beyond the original efficient frontier (EF), and provide improved Sortino ratios. Successful companies contact ∑1 in order to see how each of their portfolios:
1) Land on a risk-versus-reward (expected-return) plot
2) Compare to one or more benchmarks, e.g. the S&P500 over the same time period
3) Compare to an EF comprised of assets in the baseline portfolio
Our company is not satisfied to provide marginal or incremental improvement. Our current goal is provide our client with more resilient portfolio solutions. Clients provide the raw materials: a list of vetted assets and expected returns. ∑1 software then provides near-optimal mix of asset allocations that serve a variety of goals:
1) Improved projected risk-adjusted returns (based on semi-variance optimization)
2) Identification of under-performing assets (in the context of the “optimal” portfolio)
3) Identification of potential portfolio-enhancing assets and their asset weightings
We are obsessed with meaningful optimization. We wish to find the semi-variance (semi-deviation) efficient frontier and then breach it by including client-selected auxiliary assets. Our “mission” is as simple as that — Better, more resilient portfolios
# Portfolio-Optimization Plots
I am happy to announce that the latest version of the HALO Portfolio-Optimization Suite is now available. Key features include:
• Native asset constraint support
• Native asset-category constraint support
• Dramatic run-time improvements of 2X to over 100X
Still supported are user-specified risk models, including semi-variance and max-drawdown. What has been temporarily removed (based on minimal client interest) is 3-D 2-risk modelling and optimization. This capability may be re-introduced as a premium feature, pending client demand.
Here is a quick screenshot of a 20-asset, fixed-income portfolio optimization. The “risk-free” rate used for the tangent capital allocation line (CAL) is 1.2% (y-intercept not shown), reflecting a mix of T-Bills and stable value funds. Previously this optimization took 18 minutes on an \$800 laptop computer. Now, with the new HALO software release, it runs in only 11 seconds on the same laptop.
# Choices, Opportunities, and Solutions
To date I’ve invested approximately 800 hours developing and testing the heuristics and algorithms behind HALO. Finding exact solutions (with respect to expected-return assumptions) to certain real-world portfolio-optimization problems can be solved. Finding approximate solutions to other real-world portfolio-optimization problems is relatively easy, but finding provably optimal solutions is currently “impossible”. The current advanced science and art of portfolio optimization involves developing methods to efficiently find nearly optimal solutions.
I believe that HALO represents a significant step forward in finding nearly-optimal solutions to generalized risk models for investment portfolios. The primary strengths of HALO are in flexibility and dimensionality of financial risk modeling. While HALO currently finds solutions that are almost identical to exact solutions for convex optimization problems; the true advantage of HALO is in the quality of solutions for non-convex portfolio-optimization problems
Do you know if your particular optimization metric can be articulated in canonical convex notation? I argue that HALO does not care. If it can be, HALO will find a near-optimal solution virtually identical to the ideal convex optimization solution. If it cannot be, and is indeed non-convex, HALO will find solutions competitive with other non-convex optimization methods.
It could be argued that “over-fitting” is a potential danger of optimal and near-optimal solutions. However, I argue that given a sufficiently diverse and under-constrained optimization task, over-fitting is less worrisome. In other words, the quality of the inputs greatly influences the quality of the outputs. One secret is to supply high-quality (e.g. asset expected return) estimates to the optimization problem.
# The Future of Investing is Automation
### Developing an Automation Mindset for Investing
In 2010, I bought the domain name Sigma1.com with the idea of creating an hedge fund that I would manage. In order to measure and manage my investment strategies objectively, I began thinking about benchmarks and financial analysis software. And as I ran scenarios through Excel and some light-weight analysis software I created, I began to realize that analysis, by itself was very limited. I could only back-test one portfolio at a time, and I had to construct each portfolio’s asset weights manually. It soon became obvious that I needed portfolio optimization software.
I learned that portfolio optimization software with the capabilities I wanted was extremely expensive. Further, I realized that even if, say, I negotiated a deal with MSCI where they provided Sigma1 Financial with their Barra Portfolio Manager for free, it would not differentiate a Sigma1 hedge fund from other hedge funds using the same software.
I was beginning to interact with several technology entrepreneurs and angel investors. I quickly learned that legal costs and barriers to entry for a new hedge were intractable. If Sigma1 attracted \$10M in assets from accredited investors in 12 months, and charged 2 and 20, it would be a money loosing enterprise. Cursory research revealed that critical mass for a profitable (for the hedge fund managers) hedge fund could be as high as \$500M. Luckily, I had learned about the concept of the “entrepreneurial pivot“.
The specific pivots Sigma1 used were a market segment pivot followed by a technology pivot. I realized that while the high cost of good portfolio optimization software is bad for a hedge fund startup, it was great for a financial software startup. Suddenly, the Sigma1 Financial target market switched from accredited investors to financial professionals (investment managers, fund managers, proprietary traders, etc). This was a key market segment pivot.
Just creating a cheaper portfolio optimizer seemed unlikely to provide sufficient incentive to displace entrenched portfolio optimizers. Sigma1 needed a technology pivot — finding a solution using a completely different technology. Most prior portfolio optimizers use some variant of linear programming (LP) [or QP or NLP] to help find optimal portfolios. Moreover they also create an asset covariance matrix as a starting point for the optimization.
One stormy day, I realized that some algorithms I created to solve statistical electrical engineering problems in grad school could be adapted to optimize investment portfolios. The method I devised not only avoided LP, QP, or NLP methods; it also dispensed with the need for a covariance matrix. Over then next several days I realized that by eliminating dependence on a covariance matrix, the algorithm I later named HALO, could use both traditional and alternate risk measures ranging from variance-based (eg. standard-deviation of return) to covariance-based ones (e.g. beta) to semivariance to max draw down. By developing a vastly different technology, HALO could optimize for risks such as semivariance and Sortino ratios, or max drawdown, or even custom risk measures devised by the client.
### Algorithms Everywhere
Long before Sigma1 began developing HALO, the financial industry has been increasingly reliant on digital systems and various financial algorithms. As digital communication networks and electronic stock exchanges gained trading volume, various forms of program trading began to flourish. This includes the often maligned high-frequency trading variant of automated trading.
Concurrently, more and more trading volume has gone online. A significant portion of today’s individual investors have never placed a trade using a human stock broker.
There are now numerous automated investment analysis tools, many of which come free with a brokerage account, while others are free or low-cost stand-alone online tools. Examples of the former include the Fidelity’s nascent GPS (Guided Portfolio Summary) to more seasoned offerings such as Financial Engines. Online portfolio analysis offering range from Morningstar’s Instant X-Ray, to sites like ETFreplay.
However these software offerings are just the beginning. A company call FutureAdvisor has partnered with Fidelity and TD Ameritrade to allow its automate portfolio software to make trades on its users behalf. Companies like Future Advisor have the potential to help small investors benefit from custom-tailored investment advice utilizing proven academic research (e.g. Fama French) at a very low cost — costs so low that they would not be profitable for human investment advisers to provide.
If successful (and I believe some automated investment companies will be), why should they stop at small-time investors, with less than \$500,000 in investable assets? Why not \$1,000,000 or more? Nothing should stop them!
I could easily imagine Mark Zuckerberg, Sergey Brin, or Larry Page utilizing an automated investment company’s software to manage a large part of their portfolios. If we, as a society, are considering allowing automated systems to drive our cars for us, surely they can also manage our investment portfolios.
### The Future Roll of the Human Financial Adviser
There will always be some percentage of investors who want a personal relationship with a financial adviser. Human investment advisers can excel at explaining investment concepts and putting investors at ease during market corrections. In some ways human investment advisers even function as personal financial counselors, listening to their clients emotional financial stories. And, of course, there are some people who want to be able to pick up the phone and yell at a real person for letting them suffer market losses. Finally, there are people with Luddite tenancies who want as little to do with technology as possible. For all these reasons human investment advisers will have a place in the future world of finance.
### Investment Automation will Accelerate
There are some clear trends in the investing world. Index investing will continue to grow, as will total ETF assets under management (AUM). Alternative investments from rental property to master limited partnerships (MLPs) to private equity are also likely to become part of the portfolios of more sophisticated and affluent investors.
With the exception of high-frequency trading, which has probably saturated arbitrage and front-running opportunities, I expect algorithmic (algo) management to increase as an overall percentage of US and global AUM. Some algorithmic trading and investing will be of the “hardwired” variety where the algo directly connects to the exchanges and makes trades, while the rest of the algo umbrella will comprise trading and investing decisions made by financial software and entered manually by humans with minimal revision. There will also be hybrid methods where investment decisions are a synthesis of “automated” and “manual” processes. I expect the scope of these “flavors” of automated investing to not only increase, but to accelerate in the near term.
It is important to note, however, that for the foreseeable future, the ultimate arbiters of algorithmic investing and portfolio optimization will be human. The software architects and developers will exercise significant influence on the methodology behind the fund and portfolio optimization software. Furthermore, the users of the software will have supreme control over what parameters go into the optimization process such as including or excluding or bounding certain assets and asset classes (amongst many other factors under their direct control).
That being said, the future of investing will be increasingly the domain of financial engineers, software developers and testers, and people with skills in financial mathematics, statistics, algorithms, data structures, GUIs, web interfaces and usability. Additionally, the financial software automation revolution will have profound impacts on legal professionals and marketers in the financial domain, as well as more modest impacts on accountants and IT professionals.
Some financial professionals will take the initiative and find a place on the leading edge of the financial automation revolution. It is likely to be a wild but lucrative ride. Others will seek the short-term comfort of tradition. They may be able to retain many of their current clients through sheer charisma and inertia, but may find it increasingly difficult the appeal to younger affluent clients steeped in a culture of technology.
# Pursuing Alpha with Antivariance
A simple and marginally-effective strategy to reduce portfolio variance is by constructing an asset correlation matrix, selecting assets with low (preferably negative) correlations, and building a portfolio of low-correlation assets. This basic strategy involves creating a set of assets whose cross-correlations (covariances) are minimized.
One reason this basic strategy is only somewhat effective is that a correlation matrix (or covariance matrix) only provides a partial picture of the chosen investment landscape. Some fundamental limitations include non-normal distributions, skewness, and kurtosis to name a few. To most readers these are fancy words with varying degrees of meaning.
Personally, I often find the mathematics of the work I do seductive like a Siren’s song. I endeavor to strike a balance between exploring tangential mathematical constructs, and keeping most of my math applied. One mental antidote to the Siren’s song of pure mathematics is to think more conceptually than mathematically by asking questions like:
What are the goals of portfolio optimization? What elements of the investing landscape allow these goals to be achieved?
I then attempt to answer these questions with explanations that a person with a college degree but without a mathematically background beyond algebra could understand. This approach lets me define the concept first, and develop the math later. In essence I can temporarily free my mind of the slow, system 2 thinking generally required for math.
Recently, I came up with the concept of antivariance. I’m sure others have had similar ideas and a cursory web search reveals that as profession poker player’s nickname. I will layout my concept of antivariance as it relates to porfolio theory in particular and the broader concept in general.
By convention, one of the key objective of modern portfolio theory is the reduction of portfolio return variance. The mathematical concept is the idea that by combining assets with correlations of less than 1.0, the return variance is less than the weighted sums of each asset’s individual variance.
Antivariance assumes that there are underlying patterns explain why two or more assets should be somewhat less correlated (independent), but at times negatively correlated. Consider the affects of major hurricanes like Andrew or Katrina. Their effects were negative for insurance companies with large exposures, but were arguably positive for companies that manufactured and supplied building materials used in the subsequent rebuilds. I mention Andrew because there was much more and more rapid rebuilding following Andrew than Katrina. The disparate groups of stocks of (regional) insurance versus construction companies can be considered to exhibit paired antivariance to devastating weather events.
Nicholas Nassim Taleb coined the the term antifragile, because terms such as robust simply don’t convey the exact mental connections. I am beginning to use the term antivariance because it conveys concepts not well captured by terms like “negatively correlated”, “less correlated”, “semi-independent”, etc. In many respects antifragile systems should exhibit antivariance characteristics, and vice versa.
The concept of antivariance can be extended to related concepts such as anticovariance and anticorrelation.
# Software Development Choices for Portfolio Optimization
The first phase of developing the HALO (Heuristic Algorithm Optimizer) Portfolio Optimizer was testing mathematical and heuristic concepts. The second phase was teaming up with beta partners in the financial industry to exchange optimization work for feedback on the optimizer features and results.
For the first phase, my primary tool for software development was the Ruby language. Because Ruby is a “high-level” extensible language I was able to quickly prototype and test many diverse and complex concepts. This software development process is sometimes referred to as software prototyping.
For the second, beta phase of software development I kept most of the software in Ruby, but began re-implementing selected portions of the code in C/C++. The goal was to keep the high-change-rate code in Ruby, while coding the more stable portions in C/C++ for run-time improvement. While a good idea in theory, it turned out that my ability to foresee beta-partner changes was mixed at best. While many changes hit the the Ruby code, and were easily implemented, a significant fraction hit deep into the C/C++ code, requiring significant development and debugging effort. In some cases, the C/C++ effort was so high, I switched back portions of the code to Ruby for rapid development and ease of debugging.
Now that the limited-beta period is nearly complete, software development has entered a third phase: run-time-performance optimization. This process involves converting the vast majority of Ruby code to C. Notice, I specifically say C, not C/C++. In phase 2, I was surprised at the vast increase in executable code size with C++ (and STL and Boost). As an experiment I pruned test sections of code down to pure C and saw the binary (and in-memory) machine code size decrease by 10X and more.
By carefully coding in pure C, smaller binaries were produced, allowing more of the key code to reside in the L1 and L2 caches. Moreover, because C allows very precise control over memory allocation, reallocation, and de-allocation, I was able to more-or-less ensure than key data resided primarily in the L1 and/or L2 caches as well. When both data and instructions live close to the CPU in cache memory, performance skyrockets.
HALO code is very modular, meaning that it is carefully partitioned into independent functional pieces. It is very difficult, and not worth the effort, to convert part of a module from Ruby to C — it is more of an all-or-nothing process. So when I finished converting another entire module to C today, I was eager to see the result. I was blown away. The speed-up was 188X. That’s right, almost 200 times faster.
A purely C implementation has its advantages. C is extremely close to the hardware without being tied directly to any particular hardware implementation. This enables C code (with the help of a good compiler) to benefit from specific hardware advantages on any particular platform. Pure C code, if written carefully, is also very portable — meaning it can be ported to a variety of different OS and hardware platforms with relative ease.
A pure C implementation has disadvantages. Some include susceptibility to pointer errors, buffer-overflow errors, and memory leaks as a few examples. Many of these drawbacks can be mitigated by software regression testing, particularly to a “golden” reference spec coded in a different software language. In the case of HALO Portfolio-Optimization Software, the golden reference spec is the Ruby implementation. Furthermore unit testing can be combined with regression testing to provide even better software test coverage and “bug” isolation. The latest 188X speedup was tested against a Ruby unit test regression suite and proven to be identical (within five or more significant digits of precision) to the Ruby implementation. Since the Ruby and C implementations were coded months apart, in different software languages, it is very unlikely that the same software “bug” was independently implemented in each. Thus the C helps validate the “golden” Ruby spec, and vice versa.
I have written before about how faster software is greener software. At the time HALO was primarily a Ruby implementation, and I expected about a 10X speed up for converting from Ruby to C/C++. Now I am increasingly confident that an overall 100X speedup for an all C implementation is quite achievable. For the SaaS (software as a service) implementation, I plan to continue to use Ruby (and possibly some PHP and/or Python) for the web-interface code. However, I am hopeful I can create a pure C implementation of the entire number-crunch software stack. The current plan is to use the right tool for the right job: C for pure speed, Ruby for prototyping and as a golden regression reference, and Ruby/PHP/Python/etc for their web-integration capabilities. | 7,681 | 35,304 | {"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": 4, "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} | 3.59375 | 4 | CC-MAIN-2018-26 | latest | en | 0.804458 |
http://dompolski.ca/bn0o4s6t/n5l7tck.php?id=fa0d51-if-and-only-if-symbol | 1,618,988,932,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00435.warc.gz | 27,067,571 | 14,077 | Contents. Sometimes the biconditional in the statement of the phrase “if and only if” is shortened to simply “iff.”. The reason it points to the right is that it might not be true the other way. Only if definition: never …except when | Meaning, pronunciation, translations and examples [6] and If and only if (i.e., necessary and sufficient). The confusion of these two statement forms is known as a converse error. The phrase “if and only if” is used commonly enough in mathematical writing that it has its own abbreviation. Its invention is often credited to Paul Halmos, who wrote "I invented 'iff,' for 'if and only if'—but I could never believe I was really its first inventor."[15]. [1] Proving these pair of statements sometimes leads to a more natural proof, since there are not obvious conditions in which one would infer a biconditional directly. What Are the Converse, Contrapositive, and Inverse? If you study hard, then you will earn an A. When reading about statistics and mathematics, one phrase that regularly shows up is “if and only if.” This phrase particularly appears within statements of mathematical theorems or proofs. While the original statement is true, its converse is not. Today could be any Sunday other than Easter, and tomorrow would still be Monday. In logical formulae, logical symbols, such as In the case of the IF/AND formula in cell B5, since not all three cells in the range A2 to A4 are true — the value in cell A4 is not greater than or equal to 100 — the AND function returns a FALSE value. http://gametheory101.com/courses/logic-101/This lecture introduces the biconditional logical operator, equivalent to the phrase "if and only if" in English. In plain language, this means that if A is true, then B must be true and if A is false, then B must be false. ⇔ She will not leave any such fruit uneaten, and she will not eat any other type of … Then 6j(a b), so 6x = (a b) for some x 2Z. The letter or number will now be displayed instead. if and only if conj conjunction: Connects words, clauses, and sentences--for example, "and," "but," "because," "in order that." [10], The corresponding logical symbols are "↔",[6] " {\displaystyle \leftrightarrow } if P then Q), P would be a sufficient condition for Q, and Q would be a necessary condition for P. Also, given P→Q, it is true that ¬Q→¬P (where ¬ is the negation operator, i.e. In current practice, the single 'word' "iff" is almost always read as the four words "if and only if". An "if and only if" statement is also called a necessary and sufficient condition. The “only if” actually reverses the direction of logical dependency. IF AND ONLY IF Compound sentences of the form "P if and only if Q" are true when P and Q are both false or are both true; this compound sentence is false otherwise. "not"). C is a subset but not a proper subset of B. Here’s the “only if” rule: “A only if B” = “If A then B” The antecedent doesn’t come after the “if”, the consequent comes after the “if”. If and only if ↔⇔≡ Logical symbols representing iff. Implication and Iff. In writing, phrases commonly used as alternatives to P "if and only if" Q include: Q is necessary and sufficient for P, P is equivalent (or materially equivalent) to Q (compare with material implication), P precisely if Q, P precisely (or exactly) when Q, P exactly in case Q, and P just in case Q. If X, then Y | Sufficiency and necessity . Wherever logic is applied, especially in mathematical discussions, it has the same meaning as above: it is an abbreviation for if and only if, indicating that one statement is both necessary and sufficient for the other. If and only if. Origin of iff and pronunciation . ⟺ It is confusing indeed. This construction eliminates some redundancy. But anyway, all of this has been covered in the top and accepted answer two years ago. if and only if (iff) ↔ equivalent: if and only if (iff) ∀ for all ∃ there exists ∄ there does not exists ∴ therefore ∵ because / since Another way to explain the meaning of this connective is in terms of necessary and sufficient conditions. If the standard deviation is zero, then all of the data values are identical. Email. "If and only if the fruit is an apple will Madison eat it." Tìm kiếm if and only if mathematical symbol , if and only if mathematical symbol tại 123doc - Thư viện trực tuyến hàng đầu Việt Nam ", ThoughtCo uses cookies to provide you with a great user experience. A is a proper subset of B. For other uses, see, "↔" redirects here. "Only if" A quick guide to conditional logic. Does this mean that the double implication symbol is only valid when you apply a one to one function to an equation/inequality ? Site Navigation. There are no other conditions for both. (on the strict condition that) si et seulement si loc conj locution conjonction: groupe de mots qui servent de conjonction. $\rightarrow$ can be used to express implication, but it’s not something you should be using in written proofs. Neither will I make the feet of Israel move any more out of the land which I gave their fathers; only if they will observe to do according to all that I have commanded them, and according to all the law that my servant Moses commanded them. The result is that the truth of either one of… CS Concepts Menu Skip to content. In logic, a biconditional is a compound statement formed by combining two conditionals under “and.” Biconditionals are true when both statements (facts) have the exact same truth value.. A biconditional is read as “[some fact] if and only if [another fact]” and is true when the truth values of both facts are exactly the same — BOTH TRUE or BOTH FALSE. Rather than say "if P then Q, and if Q then P" we instead say "P if and only if Q." Biconditional IF AND ONLY IF. Learn how and when to remove this template message, "The Definitive Glossary of Higher Mathematical Jargon — If and Only If", "Jan Łukasiewicz > Łukasiewicz's Parenthesis-Free or Polish Notation (Stanford Encyclopedia of Philosophy)", Southern California Philosophy for philosophy graduate students: "Just in Case", https://en.wikipedia.org/w/index.php?title=If_and_only_if&oldid=998593717, Articles needing additional references from June 2013, All articles needing additional references, Creative Commons Attribution-ShareAlike License, This page was last edited on 6 January 2021, at 03:16. "P if Q", "if Q then P", and Q→P all mean that Q is a proper or improper subset of P. "P if and only if Q" and "Q if and only if P" both mean that the sets P and Q are identical to each other. Other equivalent terms are " is equivalent to " ( ) and " XNOR ." [3] Some authors regard "iff" as unsuitable in formal writing;[4] others consider it a "borderline case" and tolerate its use.[5]. The terms "just if" or "exactly when" are sometimes used instead. This means that the relationship between P and Q, established by P→Q, can be expressed in the following, all equivalent, ways: As an example, take the first example above, which states P→Q, where P is "the fruit in question is an apple" and Q is "Madison will eat the fruit in question". If it is raining outside, then I take my umbrella with me on my walk. We form these statements by changing the order of P and Q from the original conditional and inserting the word “not” for the inverse and contrapositive. The biconditional p q represents "p if and only if q," where p is a hypothesis and q is a conclusion. The if and only if symbol is used as a logical statement in math. [14] The following are examples of this kind of statement: Three other statements are related to any conditional statement. However, some texts of mathematical logic (particularly those on first-order logic, rather than propositional logic) make a distinction between these, in which the first, ↔, is used as a symbol in logic formulas, while ⇔ is used in reasoning about those logic formulas (e.g., in metalogic). The Logic of "If" vs. "Only if" This is the currently selected item. When Is the Standard Deviation Equal to Zero? Moore, who was very careful with his language, interpreted "only if" to mean "if and only if". Logic toolbox. Implication: Implication says "if ... then" Example: If both a and b are odd numbers then (a+b) is even. In the image below a thumbs up or thumbs down symbol is shown dependent upon whether the sale of products have improved since last month. If you want to see all type of Latex arrows, have a look to https://www.math-linux.com/latex-26/faq/latex-faq/article/latex-arrows ", "Iff" redirects here. Then we see that this statement means both of the following: If we are attempting to prove a biconditional, then most of the time we end up splitting it. 6 “Athena is a cat only if she is a mammal.” Gets translated as: A Ɔ M Note that “Athena is a cat only if she is a mammal” does NOT mean the same thing as “Athena is a cat if she is a mammal” since lots of mammals are not cats (for instance, Athena might be a dog). ‘The ganja addict who suffers from a mental breakdown, which is controlled by medication, if and only if the medication is taken.’ ‘Which is good, since I plan to further my studies, if and only if possible.’ ‘They will come to our defence if and only if it is in their national interests to … Then select that cell and change the font to Calibri, Arial or some other written font. ,[7] are used instead of these phrases; see § Notation below. Notation. This statement is obtained from the original by saying “if Q then P.” Suppose we start with the conditional “if it is raining outside, then I take my umbrella with me on my walk.” The converse of this statement is “if I take my umbrella with me on my walk, then it is raining outside.”. What Does If and Only If Mean in Mathematics? Sort by: Top Voted. Sort by: Top Voted. However, this statement’s converse “If a number is divisible by 2, then it is divisible by 4” is false. [6] [2] For example: "Madison will eat the fruit if and only if it is an apple" is equivalent to saying that "Madison will eat the fruit if the fruit is an apple, and will eat no other fruit". However, some texts of mathematical logic (particularly those on first-order logic, rather than propositional logic) make a distinction between these, in which the first, ↔, is used as a symbol in logic formulas, while ⇔ is used in reasoning about those logic formulas (e.g., in metalogic). The following are four equivalent ways of expressing this very relationship: Here, the second example can be restated in the form of if...then as "If Madison will eat the fruit in question, then it is an apple"; taking this in conjunction with the first example, we find that the third example can be stated as "If the fruit in question is an apple, then Madison will eat it; and if Madison will eat the fruit, then it is an apple". {\displaystyle \Leftrightarrow } This means two things: "If P, Then Q" and "If Q, Then P". This tutorial will show you how to display any symbol though, so you could insert a smiley face, hour glass, aeroplane and much more. Consider the statement “if today is Easter, then tomorrow is Monday.” Today being Easter is sufficient for tomorrow to be Monday, however, it is not necessary. iff is also equivalent to together with, where the symbol denotes " implies." About. Q is as follows:[8][9], It is equivalent to that produced by the XNOR gate, and opposite to that produced by the XOR gate. Conditional reasoning and logical equivalence. The first if provides just that guarantee. [1] This is an example of mathematical jargon (although, as noted above, if is more often used than iff in statements of definition). Produce the truth tables for the two conditional statements and use those to convince yourself that this logical equivalence holds. The elements of X are all and only the elements of Y means: "For any z in the domain of discourse, z is in X if and only if z is in Y. If, and Only If Many theorems are stated in the form "P, if, and only if, Q". The brackets may be omitted after an if statement. In fact, when "P if and only Q" is true, P can subsitute for Q and Q can subsitute for P in other compound … Usage of the abbreviation "iff" first appeared in print in John L. Kelley's 1955 book General Topology. This brings us to a biconditional statement, which is also known as an "if and only if" statement. Part 2: Q )P. Therefore, P ,Q. – RegDwigнt ♦ Dec 6 '13 at 13:41. Although 2 divides this number, 4 does not. can be written as: both a and b are odd numbers (a+b) is even. In TeX, "if and only if" is shown as a long double arrow: ⇔ For example, P if and only if Q means that the only case in which P is true is if Q is also true, whereas in the case of P if Q, there could be other scenarios where P is true and Q is false. The Symbols are and . In other words, "A only if B" tells us that "A if B", but also gives us a little extra information: "A only if … In his mind, "A only if B" was a stronger statement than "A if B". Another term for this logical connective is exclusive nor. U+2194 ↔ \leftrightarrow \iff. In that it is biconditional, the connective can be likened to the standard material conditional ("only if", equal to "if ... then") combined with its reverse ("if"); hence the name. The following is a truth table for biconditional p q. p: q: p q: T: T: T: T: F: F: F: T: F: F: F: T: In the truth table above, p q is true when p and q have the same truth values, (i.e., when either both are true or both are false.) Proof: Suppose a b mod 6. It says that P and Q have the same truth values; when "P if and only if Q" is true, it is often said that P and Q are logically equivalent. An alternative is to prove the disjunction "(P and Q) or (not-P and not-Q)", which itself can be inferred directly from either of its disjuncts—that is, because "iff" is truth-functional, "P iff Q" follows if P and Q have been shown to be both true, or both false. Only-If Proof 7.2 Equivalent Statements 7.3 Existence and Uniqueness Proofs 7.4 (Non-) Construc-tive Proofs Proving If-And-Only-If Statements Outline: Proposition: P ,Q. In the case of the IF/AND formula in cell B5, since not all three cells in the range A2 to A4 are true — the value in cell A4 is not greater than or equal to 100 — the AND function returns a FALSE value. Definition. This blog post looks at using the IF function to display a symbol conditionally in a cell. If you find our videos helpful you can support us by buying something from amazon. So P if and only if Q resolves into P > Q and Q > P, which is to say that . A number is in A only if it is in B; a number is in B if it is in A. Iff is used outside the field of logic as well. An "if and only if" statement is also called a necessary and sufficient condition. Sufficiency is the converse of necessity. But what, precisely, does this statement mean? {\displaystyle \Leftrightarrow } A biconditional statement has the form: Since this construction is somewhat awkward, especially when P and Q are their own logical statements, we simplify the statement of a biconditional by using the phrase "if and only if." In most logical systems, one proves a statement of the form "P iff Q" by proving either "if P, then Q" and "if Q, then P", or "if P, then Q" and "if not-P, then not-Q". However, in the preface of General Topology, Kelley suggests that it should be read differently: "In some cases where mathematical content requires 'if and only if' and euphony demands something less I use Halmos' 'iff'". material equivalence A ⇔ B is true just in case either both A and B are false, or both x + 5 = y + 2 ⇔ x + 3 = y U+21D4 U+2261 ⇔ ≡ \Leftrightarrow \equiv \leftrightarrow if and only if; iff; means the same as. If and Only If Symbol. [17] However, this logically correct usage of "if and only if" is relatively uncommon, as the majority of textbooks, research papers and articles (including English Wikipedia articles) follow the special convention to interpret "if" as "if and only if", whenever a mathematical definition is involved (as in "a topological space is compact if every open cover has a finite subcover").[18]. Thus the statement “P if and only if Q” becomes “P iff Q.”. It is not to be confused with. In other words, 3 is a combination of 1 and 2, and you simply failed to combine your correct reasoning for 1 and 2 into the correct reasoning for 3. The authors of one discrete mathematics textbook suggest:[16] "Should you need to pronounce iff, really hang on to the 'ff' so that people hear the difference from 'if'", implying that "iff" could be pronounced as [ɪfː]. One part we prove is “if P then Q.” The other part of the proof we need is “if Q then P.”. The corresponding logical symbols are "↔", "$${\displaystyle \Leftrightarrow }$$", and "≡", and sometimes "iff". Usage in definitions. View History. Donate or volunteer today! We only need to look at a number such as 6. Certain conditional statements also have converses that are true. Another way to say the same things is: "Q is necessary, and sufficient for P". Proposition: 8a;b 2Z, a b mod 6 if and only if a b mod 2 and a b mod 3. When you have “only if”, the claim that precedes the “only if’ is antecedent, what follows it is the consequent. ⇔ Hide Ads About Ads. The sample standard deviation of a data set is equal to zero if and only if all of the data values are identical. We break this biconditional statement into a conditional and its converse. So to prove an "If, and Only If" theorem, you must prove two implications. Abbreviation. Every symbol from the Wingdings libraries has an associated letter or number when displayed in a normal written font such as Calibri or Arial. 1 Definition; 2 Usage. Biconditional. either both statements are true, or both are false), though it is controversial whether the connective thus defined is properly rendered by the English "if and only if"—with its pre-existing meaning. if and only if. "Iff." In Łukasiewicz's Polish notation, it is the prefix symbol 'E'.[12]. Ex : "parce que", "depuis que" I'll help you, if and only if, you promise to do your part. In logic and related fields such as mathematics and philosophy, if and only if (shortened as iff[1]) is a biconditional logical connective between statements, where either both statements are true or both are false. In Łukasiewicz's Polish notation, it is the prefix symbol 'E'. Technically, definitions are always "if and only if" statements; some texts — such as Kelley's General Topology — follow the strict demands of logic, and use "if and only if" or iff in definitions of new terms. ↔ The Logic of "If" vs. "Only if" Our mission is to provide a free, world-class education to anyone, anywhere. Edit. Additionally, the third column contains an informal definition, the fourth column gives a short example, the fifth and sixth give the Unicode location and name for use in HTML documents. Implication: Implication says "if ... then" Example: If both a and b are odd numbers then (a+b) is even. The connective is biconditional (a statement of material equivalence),[2] and can be likened to the standard material conditional ("only if", equal to "if ... then") combined with its reverse ("if"); hence the name. The English language is tremendously confusing compared to the simplicity of formal logic. The IF function uses this value and returns its Value_if_false argument — the current date supplied by the TODAY function. A quick guide to conditional logic. So a number is even if and only if its square is even. via command \iff.[13]. If this is done, the next line (defined by the semicolon) becomes the only conditional statement. Iff says if and only if. This makes it clear that Madison will eat all and only those fruits that are apples. These are usually treated as equivalent. These are usually treated as equivalent. The first half of this proof was an exercise in the last chapter. See also. The following table lists many common symbols, together with their name, pronunciation, and the related field of mathematics. Distinction from "if" and "only if" In terms of Euler diagrams. "Only if" Google Classroom Facebook Twitter. We only need to consider the converse here. This is the currently selected item. or "Madison will eat the fruit if and only if it is an apple." Sometimes the biconditional in the statement of the phrase “if and only if” is shortened to simply “iff.” Thus the statement “P if and only if Q” becomes “P iff Q.”, Courtney K. Taylor, Ph.D., is a professor of mathematics at Anderson University and the author of "An Introduction to Abstract Algebra. The truth table of P For a short if and only if, use \Leftrightarrow: A $\Leftrightarrow$ B. For example: "Madison will eat the fruit if and only if it is an apple" is equivalent to saying that "Madison will eat the fruit if the fruit is an apple, and will eat no other fruit". For a long if and only if, use \Longleftrightarrow: C $\Longleftrightarrow$ D. Liste of all arrows. The Symbols are and . P iff Q is logically equivalent to (P > Q) & (Q > P). However, the English language has orders of magnitude more expressive power than formal logic. ",[7] and "≡",[11] and sometimes "iff". In this case, we may form what is known as a biconditional statement. Proof: Part 1: P )Q. Categories. The IF function uses this value and returns its Value_if_false argument — the current date supplied by the TODAY function. For another example, we consider the conditional “If a number is divisible by 4 then it is divisible by 2.” This statement is clearly true. Related Articles. The phrase “if and only if” is used commonly enough in mathematical writing that it has its own abbreviation. Weisstein, Eric W. From MathWorld--A Wolfram Web Resource. {\displaystyle \Leftrightarrow } These are called the converse, inverse, and the contrapositive. "P only if Q", "if P then Q", and "P→Q" all mean that P is a subset, either proper or improper, of Q. Home; Contact; If and only if ↔ ⇔ ≡ Logical symbols representing iff. On the other hand, all cats ARE mammals. We only need to consider this example to realize that the original conditional is not logically the same as its converse. This, however, makes it quite clear that Madison will eat all and only those fruits that are apple. If X, then Y | Sufficiency and necessity. Symbol. Usage. In logical formulae, logical symbols are used instead of these phrases; see the discussion of notation. References. In the second half of the proof, we begin with, Let y be even, and then write this in symbols, - 2K for some whole number K. We then look for a reason why y … Biconditional statements are related to conditions that are both necessary and sufficient. {\displaystyle \iff } Proofs. More general usage. That is to say, given P→Q (i.e. iff is also equivalent to together with , where the symbol denotes "implies." One could take an umbrella on a walk even though it may not be raining outside. Typically the symbol is used in an expression like: A B. A way of writing two conditionalsat once: both a conditional and its converse. Two example STEP questions I give are the below: STEP 3 2010 1iii - for D. If and only if proofs (continued) Permalink Submitted by maths123 on Sun, 04/26/2015 - 21:47. If and only if. News; For an example of the phrase “if and only if” that involves statistics, look no further than a fact concerning the sample standard deviation. To form a conditional statement, we could say “if P then Q.”. To understand “if and only if,” we must first know what is meant by a conditional statement. How to Do Hypothesis Tests With the Z.TEST Function in Excel, Example of Two Sample T Test and Confidence Interval, Differences Between Population and Sample Standard Deviations, How to Calculate a Sample Standard Deviation, Definition and Examples of Valid Arguments, Calculating a Confidence Interval for a Mean, Degrees of Freedom in Statistics and Mathematics, converse, inverse, and the contrapositive, B.A., Mathematics, Physics, and Chemistry, Anderson University. The logic of if and only if '' this is the prefix symbol ' E.! Could take an umbrella on a walk even though it may not be true the hand.: if and only if '' statement iff '' convince yourself that this equivalence... 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Font to Calibri, Arial or some other written font the data values are identical, Q. Magnitude more expressive power than formal logic be Monday iff is also equivalent to ( and. Will denote by P and Q 6x = ( a B ), so 6x = ( a )... To display a symbol conditionally in a only if, and the related field of logic as well has associated. ( a B Three other statements are related to conditions that are true are. Is logically equivalent to together with their name, pronunciation, and the related of...: Three other statements are related to any conditional statement you with a great user experience ) . P iff Q is necessary, and sufficient condition this biconditional statement, we could say if! Form if and only if ( i.e., necessary and sufficient conditions logical formulae logical. Makes it quite clear that Madison will eat all and only if ” is to... Is only valid when you apply a one to one function to an equation/inequality biconditional in the statement P... 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https://math.stackexchange.com/questions/1405365/express-roots-in-polynomials-of-equation-x3x2-2x-1-0 | 1,566,560,352,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318375.80/warc/CC-MAIN-20190823104239-20190823130239-00299.warc.gz | 545,508,840 | 36,540 | # Express roots in polynomials of equation $x^3+x^2-2x-1=0$
If $\alpha$ is a root of equation $x^3+x^2-2x-1=0$, then find the other two roots in polynomials of $\alpha$, with rational coefficients.
I've seen some other examples [1] that other roots were found for equations with certain properties (having only even-power terms, etc).
In the comment in this link, someone suggest that If $A$ is a root of $x^6−2x^5+3x^3−2x−1=0$, then so is $−A^5+2A^4−3A$, without further explanation (or maybe it's obvious to math experts, not to me) but I'm more interested in the underlying theory, preferably elementary, and techniques to solve problems of this kind.
Thanks!
• The given polynomial has one real, two complex roots. The two complex roots cannot be expressed as rational polynomials of the real root. – Neil W Aug 21 '15 at 22:08
• I'm really sorry, but I think I had the last coefficient wrong, it should be $-1$, I fixed the equation. – Ali Aug 22 '15 at 0:26
• I'm not sure why all the answers so far are misunderstanding this question, it seems clear to me. – DanielV Aug 22 '15 at 4:10
• Use the fact that the discriminant is a perfect square. The field extension generated by one of the roots includes all the roots. – Bill Kleinhans Aug 22 '15 at 5:12
• @BillKleinhans: can a field extension generated by a real root contain two non-real roots? – robjohn Aug 22 '15 at 7:32
The discriminant of the polynomial $p(x)=x^3+x^2-2x-1$ is $49$, which is a perfect square. It has no rational roots, so it is irreducible in $\Bbb{Q}[x]$. Together these facts imply that the Galois group of the polynomial is cyclic of order three. If $a$ is one of its zeros, we thus see that $\Bbb{Q}[a]$ is its splitting field. This means that the other zeros are also in $\Bbb{Q}[a]$, hence they are polynomials in $a$ with rational coefficients.
We could run full Cardano on it, but I have seen this polynomial too often, so I will take a shortcut. Let's write $\zeta=e^{2\pi i/7}$ and $$u=\zeta+\zeta^{-1}=2\cos\frac{2\pi}7.$$ We get from binomial formula that $$u^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}$$ and $$u^2=\zeta^2+2+\zeta^{-2}.$$ Therefore $$p(u)=u^3+u^2-2u-1=\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=\zeta^{-3}\frac{\zeta^7-1}{\zeta-1}=0.$$
So $p(x)$ is the minimal polynomial of $u=2\cos(2\pi/7)$. What about its other zeros? Galois theory tells us that the powers $\zeta^k, k=1,2,3,4,5,6$ are exactly the conjugates of $\zeta$. Therefore the conjugates of $u$ are of the form $\zeta^{k}+\zeta^{-k}=2\cos(2k\pi/7)$.
Observe that $$u^2-2=4\cos^2\frac{2\pi}7-2=2(2\cos^2\frac{2\pi}7-1)=2\cos\frac{4\pi}7$$ by the formula for the cosine of a doubled angle, so $u^2-2$ is one of the other zeros of $p(x)$. I hope that it is no longer a surprise that the third root is $2\cos\dfrac{8\pi}7$. See robjohn's answer for a way of quickly writing this as a polynomial of $u$ as well. Further observe that the angle doubling trick stops here because $2\cos\dfrac{16\pi}7=2\cos\dfrac{2\pi}7$.
We do get a cyclic splitting field whenever the discriminant of an irreducible cubic in $\Bbb{Z}[x]$ is a perfect square - that part generalizes. The trickery with roots of unity and cosines is somewhat special to this polynomial. However, by the Kronecker-Weber theorem all cyclic extensions of $\Bbb{Q}$ reside inside some cyclotomic extension. In other words the roots of such cubics can be written as polynomials with rational coefficients evaluated at some root of unity.
• Looks like today is a day of regular 7-gons for me :-) – Jyrki Lahtonen Aug 22 '15 at 17:06
• And notice that $(u^2-2)^2-2\equiv-u^2-u+1\pmod{u^3+u^2-2u-1}$ gives the other root I mentioned. It is nice to have a root to use :-) (+1) – robjohn Aug 22 '15 at 17:13
• The discriminant here is also strongly hinting that the conductor of this abelian extension might be seven. Search for discriminants of cyclotomic fields (and the behavior of discriminants in a tower of field extensions) for more. – Jyrki Lahtonen Aug 22 '15 at 17:21
• Thanks! i need to brush up on some of this stuff. It's been a while since I studied it as an undergraduate. – robjohn Aug 22 '15 at 19:27
If $a$ is one root, then we get $$\frac{x^3+x^2-2x+1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{1}$$ Using the quadratic equation to solve this yields $$\frac{-1-a\pm\sqrt{9-2a-3a^2}}2\tag{2}$$ for the other two roots.
After the Question Change
Changing the constant term only changes $(1)$ slightly: $$\frac{x^3+x^2-2x-1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{3}$$ and we still have $(2)$.
Since the discriminant is $9-2a-3a^2$, we know that if one root is between $\frac{-1-2\sqrt{7}}3$ and $\frac{-1+2\sqrt{7}}3$, then all three roots are. If any root is outside that interval, the other two roots will not be real.
After considering a comment by KCd, I see that $$(x-a)(x-a^2+2)(x+a^2+a-1)\equiv x^3+x^2-2x-1\pmod{a^3+a^2-2a-1}$$ Therefore, if $a$ is a root, then $a^2-2$ and $-a^2-a+1$ are also roots.
• This expression for the other two roots is not a polynomial in $a$, which is what the questioner is looking for. – DanielV Aug 22 '15 at 4:12
• Since that can't be done, should we not show what can be done? – robjohn Aug 22 '15 at 7:20
• @DanielV: now that the question has been changed, it is possible that the roots are rational functions of one another. However, just because people cannot fully answer a question, is it wrong to offer what one can give in hopes that it might help? – robjohn Aug 22 '15 at 14:31
• @DanielV: the last section now shows that we can write the other roots as polynomials in $a$. – robjohn Aug 22 '15 at 16:52
• If the question really cannot be answered, then a proof/explanation of that would be the answer. I saw the comment KCd left, and was hoping someone would leave a derivation of it~ – DanielV Aug 22 '15 at 23:54
Let $\beta$ be another root of the given equation, then: $\beta^3+\beta^2-2\beta+1 = 0 = \alpha^3+\alpha^2-2\alpha+1\to (\beta^3-\alpha^3)+(\beta^2-\alpha^2)-2(\beta-\alpha)=0\to (\beta-\alpha)(\beta^2+\alpha\beta+\alpha^2+\beta+\alpha-2)=0\to \beta^2+(\alpha+1)\beta+\alpha^2+\alpha-2=0$ since $\alpha \neq \beta$. Using quadratic formula we can find a formula for $\beta$ in term of $\alpha$. Is this fine? Note that this method assumes that all the real roots are distinct. So you should prove this first, and it is equally interesting finding for you as well.
• Solving that quadratic doesn't seem to yield a polynomial expression; the discriminant is $b^2-4ac=(\alpha+1)^2-4(\alpha^2+\alpha-2)=-3\alpha^2-2\alpha+9$... – Steven Stadnicki Aug 21 '15 at 21:59
• No I'm asking for polynomials, as it's stated in the title – Ali Aug 21 '15 at 22:13
• If we know two roots, $\alpha$ and $\beta$, then the third root is $-1-\alpha-\beta$ since the sum of the three roots is $-1$ by Vieta's Formulas. – robjohn Aug 21 '15 at 22:24
• @Ali The quadratic is a polynomial expressing the root $\beta$ in terms of $\alpha$. You can't do better, because there are two missing roots and no way to distinguish between them algebraically - you have to have both or neither. Having found the quadratic which is a polynomial expression, you can solve it to get the value of the two roots, but that turns out not to be a nice polynomial expression. – Mark Bennet Aug 21 '15 at 22:27
• The other two roots are $a^2-2$ and $-a^2-a-1$. How much Galois theory do you know? – KCd Aug 22 '15 at 3:28
In a more general manner, if $a$ is root of $$x^3+Ax^2+Bx+C=0$$ then $$x^3+Ax^2+Bx+C=(x-a)(x^2+Px+Q)$$ grouping terms we then have $$x^2 (a+A-P)+x (a P+B-Q)+a Q+C$$ and all coefficients must be zero. So, $$P=a+A$$ $$Q=B+aP=a^2+a A+B$$ The unused equation $$aQ+C=a^3+Aa^2+Ba+C=0$$ just confirms that $a$ is a root of the initial equation.
Now, solving the quadratic $x^2+Px+Q=0$, the two other roots are then given by $$x_{1,2}=\frac{1}{2} \left(-a-A\pm\sqrt{-3 a^2-2 a A+A^2-4 B}\right)$$
• He wants to represent the roots as polynomials of a given root, such as $r_1 = a,~ r_2 = B_2a^2 + B_1a + B_0,~ r_3 = C_2a^2 + C_1a + C_0$, where $B \subset \mathbb Q$ and $C \subset \mathbb Q$ – DanielV Aug 22 '15 at 4:08 | 2,675 | 8,105 | {"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-2019-35 | latest | en | 0.931705 |
https://id.scribd.com/document/86623164/Adalab-Manual | 1,561,578,732,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000545.97/warc/CC-MAIN-20190626194744-20190626220744-00239.warc.gz | 454,840,978 | 71,060 | Anda di halaman 1dari 63
# Lab Manual: Algorithms Laboratory CONTENT SHEET SLNO. 1.
EXPERIMENT NAME Implement Recursive Binary search and Linear search and determine the time required to search an element. Repeat the experiment for different values of n, the number of elements in the list to be searched and plot a graph of the time taken versus n. Sort a given set of elements using the Heap sort method and determine the time required to sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the taken versus n. Sort a given set of elements using Merge sort method and determine the time required to sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n. Sort a given set of elements using Selection sort and determine the time required to sort elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n. a) Obtain the Topological ordering of vertices in a given digraph. b) Print all the nodes reachable from a given starting node in a digraph using DFS method. 6. 7. 8. Implement 0/1 Knapsack problem using dynamic programming. From a given vertex in a weighted connected graph, find shortest paths to other vertices using Dijkstras algorithm. Sort a given set of elements using Quick sort method and determine the time required sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n. Find Minimum Cost Spanning Tree of given undirected graph using Kruskals algorithm. a) Print all the nodes reachable from a given starting node in a digraph using BFS method. b) Check whether a given graph is connected of not using DFS method. PAGE NO.
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Lab Manual: Algorithms Laboratory 11. Find a subset of a given set S = {s1,s2,,sn} of n positive integers whose sum is equal to a given positive integer d. For example, if S = {1,2,5,6,8} and d=9 there are two solutions {1,2,6} and {1,8}. A suitable message is to be displayed if the given problem instance doesnt have a problem. a) Implement Horspool algorithm for String Matching. b) Find the Binomial Co-efficient using Dynamic Programming. 13. 14. Find Minimum Cost Spanning Tree of given undirected graph using Prims algorithm. a) Implement Floyds algorithm for the All-Pairs-Shortest-Paths problem. b) Compute the transitive closure of a given directed graph using Warshalls algorithm. 15. Implement N Queens problem using Back Tracking.
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Lab Manual: Algorithms Laboratory I. Implement Recursive Binary search and Linear search and determine the time required to search an element. Repeat the experiment for different values of n, the number of elements in the list to be searched and plot a graph of the time taken versus n. Linear Search: Design Strategy: Brute-Force ALGORITHM LinearSearch(A[0..n],K) //The algorithm implements sequential search with a // search key as a sentinel //Input: An array A of n elements and a search key K //Output: The position of the first element in A[0..n-1] // whose value is equal to K or -1 of no such // element is found A[n]K i 0 while A[ i ] K do i i+1 if i <n return i else return -1 Binary Search: Design Strategy: Divide-and-Conquer ALGORITHM BinarySearch(A[lr],K) //Implements recursive binary search //Input: An array A[l..r] sorted in non decreasing // order & a search key K //Output: An index of the arrays element that is equal // to K or -1 if there is no such element if l > r return -1 else mid (l+r)/2 if A[mid] = K return mid else if K < A[mid] return BinarySearch(A[l mid-1],K) else return BinarySearch(A[mid+1 r],K)
Lab Manual: Algorithms Laboratory /* RECURSIVE LINEAR & BINARY SEARCH */ #include<stdio.h> #include<conio.h> #include<time.h> #include<process.h> int key; int linear(int a[],int n) { if(n==-1) return -1; else if(key==a[n]) return n; else return linear(a,n-1); } int binary(int a[],int low,int high) { int mid; mid=(low+high)/2; if(low>high) return -1; else if(key==a[mid]) return mid; else if(key<a[mid]) return binary(a,low,mid-1); else return binary(a,mid+1,high); } void main() { int a[10],pos,n,c,i; clock_t start,end; clrscr(); printf("\n Enter the no of elements : "); scanf("%d",&n); printf("\n Enter the Array elements : "); for(i=0;i<n;i++) scanf("%d",&a[i]); printf("\n Enter the key : "); scanf("%d",&key); printf("\n Enter Choice: 1.Linear 2.Binary : "); scanf("%d",&c);
Lab Manual: Algorithms Laboratory if(c==1) { start=clock(); pos=linear(a,n-1); end=clock(); } else if(c==2) { start=clock(); pos=binary(a,0,n-1); end=clock(); } else { printf("\n Wrong Choice \n"); getch(); exit(0); } if(pos==-1) { printf("\n Unsuccessful Search \n"); printf("\n\n The Time is %f \n\n",(end-start)/CLK_TCK); } else { printf("\n Key found successfully at %d position",pos+1); printf("\n\n The Time is %f \n\n",(end-start)/CLK_TCK); } getch(); } /**************************** OUTPUT ******************** Enter the no of elements : 5 Enter the Array elements : 12 5 6 4 10 Enter the key : 4 Enter Choice: 1.Linear 2.Binary : 1 Key found successfully at 4 position The Time is 0.000000
Lab Manual: Algorithms Laboratory Enter the no of elements : 5 Enter the Array elements : 12 5 6 4 10 Enter the key : 15 Enter Choice: 1.Linear 2.Binary : 1 Unsuccessful Search The Time is 0.000000 Enter the no of elements : 5 Enter the Array elements : 11 22 33 44 55 Enter the key : 55 Enter Choice: 1.Linear 2.Binary : 2 Key found successfully at 5 position The Time is 0.000000 Enter the no of elements : 5 Enter the Array elements : 11 22 33 44 55 Enter the key : 66 Enter Choice: 1.Linear 2.Binary : 2 Unsuccessful Search The Time is 0.000000 *******************************************************/
Lab Manual: Algorithms Laboratory II. Sort a given set of elements using the Heapsort method and determine the time required to sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the taken versus n. ALGORITHM heap sort Stage 1: (Heap construction): construct a heap for a given array. Stage 2: (maximum deletions): apply the root-deletion operation n-1 times to the remaining heap Deleting roots key with the last key k of the heap can be done by the following algorithm. Step 1: exchange the roots key with the last key k of the heap. Step 2: Decrease the heaps size. Step 3: Heapify the smaller tree by shifting k down the tree exactly in the same way we did it in the bottom-up heap construction algorithm .That is, verify the parental dominance for k : if it holds ,we are done ; if not, swap k with the larger of its children and repeat this operation until the parental dominance condition holds for k in its new position. Design strategy: Transform and Conquer ALGORITHM HeapbottomUp (H [1...n] // constructs a heap from the elements of a given array //by the bottom-up algorithm //Input: An array H [1...n] of orderable items //Output: A heap H[1n] for i|_n/2_| down to 1 do ki; v H[k] heapfalse While not heap and 2* k<=n do J2*k if j<n //there are two children if H[j]<H[j+1] jj+1 if v>=H[j] heaptrue else H[k]H[j]: kj H[k]v
/* HEAP SORT */
Lab Manual: Algorithms Laboratory #include<stdio.h> #include<conio.h> #include<process.h> #include<time.h> clock_t start,end; void heap(int a[],int n) { int c,p,k,item; for(k=1;k<n;k++) { item=a[k]; c=k; p=(c-1)/2; while(c>0 && item >a[p]) { a[c]=a[p]; c=p; p=(c-1)/2; } a[c]=item; } } void heapf(int a[],int n) { int c,p,item; p=0; item=a[p]; c=(2*p)+1; while(c<=n-1) { if(c+1 <= n-1) if(a[c] < a[c+1]) c++; if(item < a[c]) { a[p] = a[c]; p = c; c = (2*p)+1; } else break; } a[p] = item; } void heap_sort(int a[],int n)
Lab Manual: Algorithms Laboratory { int i,t; heap(a,n); for(i=n-1;i>0;i--) { t=a[0]; a[0]=a[i]; a[i]=t; heapf(a,i); } } void main() { int a[20],n,i; clrscr(); printf("Enter the number of elements : "); scanf("%d",&n); printf("\nEnter the array elements : "); for(i=0;i<n;i++) scanf("%d",&a[i]); start=clock(); heap_sort(a,n); end=clock(); printf("\n The sorted array : "); for(i=0;i<n;i++) printf("%d \t",a[i]); printf("\n\n Time taken is : %f\n",(start-end)/CLK_TCK); getch(); }
/********************** OUTPUT ******************** Enter the number of elements : 5 Enter the array elements The sorted array Time taken is : 0.000000 **************************************************/ : 10 98 56 54 67 : 10 54 56 67 98
Lab Manual: Algorithms Laboratory III. Sort a given set of elements using Merge sort method and determine the time required to sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n. ALGORITHM Merge (B [0...p-1], C [0...q-1], A [0...p+q-1]) // Merges two sorted arrays into one sorted array // Input: arrays B [0...p-1] and C [0...q-1] both sorted // Output: sorted array A [0...p+q-1] of the elements of B and C i 0 ; j0 ; k0 while i < p and j < q do if B[i] <= C[j] A [k] B [i]; ii+1 else A [k]C [j]; jj+1 kk+1 if i = p Copy C [j...q-1] to A [k...p+q-1] else copy B [i...p-1] to A [k...p+q-1] Design Strategy: Divide and Conquer ALGORITHM Mergesort (A [0..n-1] ) // Sorts array (A [0..n-1]) by recursive Mergesort //Input: An array (A [0..n-1] of orderable elements //Output: Array (A [0..n-1]) sorted in non decreasing order if n>1 copy A[0..|_n/2_| - 1] to B[0..|_n/2_| -1] copy A [|_n/2_|... n-1] to C[0...|--n/2--|-1] Mergesort (B [0...|_n/2_|-1]) Mergesort(C [0...|--n/2--|-1]) Merge (B, C, A)
/* MERGE SORT */
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Lab Manual: Algorithms Laboratory #include<stdio.h> #include<conio.h> #include<time.h> #include<process.h> int n,a[10]; void sim_merge(int,int,int); void merge(int,int); void main() { int i; clock_t start,end; clrscr(); printf("\n Enter the no. of elements in the vector: "); scanf("%d",&n); printf("\n Enter the Vector elements : "); for(i=0;i<n;i++) scanf("%d",&a[i]); start=clock(); merge(0,n-1); end=clock(); printf("\n Sorted Vector as follows : "); for(i=0;i<n;i++) printf(" %d ",a[i]); printf("\n\n The time taken is %f \n",(end-start)/(CLK_TCK)); getch(); } void sim_merge(int low,int mid,int high) { int c[10],i=low,j=mid+1,k=low; while(i<=mid && j<=high) { if(a[i]<a[j]) c[k++]=a[i++]; else c[k++]=a[j++]; } while(i<=mid) c[k++]=a[i++]; while(j<=high) c[k++]=a[j++]; for(i=low;i<=high;i++) a[i]=c[i]; } void merge(int low,int high)
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Lab Manual: Algorithms Laboratory { int mid; if(low<high) { mid=(low+high)/2; merge(low,mid); merge(mid+1,high); sim_merge(low,mid,high); } }
/************************** OUTPUT ********************** Enter the no. of elements in the vector: 5 Enter the Vector elements : 0 33 77 22 11 Sorted Vector as follows : 0 11 22 33 77 The time taken is 0.000000 *****************************************************/
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Lab Manual: Algorithms Laboratory IV. Sort a given set of elements using Selection sort and determine the time required to sort elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n. Design Strategy: Brute Force ALGORITHM SelectionSort(A[0..n-1]) //The algorithm sorts a given array by selection sort //Input: An array A[0..n-1] of orderable elements //Output: array A[0..n-1] sorted in ascending order for i 0 to n-2 do min i for j i+1 to n-1 do if A[ j ]<A[min] min j swap A[ i ] and A[min]
/* SELECTION SORT */
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Lab Manual: Algorithms Laboratory #include<stdio.h> #include<conio.h> #include<process.h> #include<time.h> clock_t start,end; void main() { int a[10],n,i,j,pos=0,t,s; clrscr(); printf("\n Enter the number of elements : "); scanf("%d",&n); printf("\n Enter the array elements : "); for(i=0;i<n;i++) scanf("%d",&a[i]); start=clock(); for(i=0;i<n-1;i++) { s = a[i]; for(j=i+1;j<n;j++) if(s > a[j]) { s = a[j]; pos = j; } t = a[i]; a[i] = a[pos]; a[pos] = t; } end=clock(); printf("\n The sorted array is : "); for(i=0;i<n;i++) printf("%d \t",a[i]); printf("\n\n The Time taken is : %f\n",(start-end)/CLK_TCK); getch(); } /***************************** OUTPUT ************************** Enter the number of elements : 5 Enter the array elements The sorted array is :9 : 76 45 87 34 09 34 45 76 87
The Time taken is : 0.000000 ***************************************************************/ V. a) Obtain the Topological ordering of vertices in a given digraph.
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## Lab Manual: Algorithms Laboratory
b) Print all the nodes reachable from a given starting node in a digraph using DFS method. a) Obtain the Topological ordering of vertices in a given digraph. Design strategy : Decrease and Conquer (Application of DFS) ALGORITHM topological sorting (G<V,E>) // Input: A directed acyclic graph G<V,E> //Output: a topologically sorted list Step 1 : Perform a DFS traversal and note the order in which the vertices become dead ends(i.e, are popped off the traversal stack). Step 2: Reversing this order yields a solution to the topological sorting. Second method: Design Strategy: Decrease and conquer.
Solution
ALGORITHM: Step 1: Repeatedly, identify in a remaining diagraph a source which is a vertex with no incoming edges ,and delete it along with all the edges outgoing from it.(if they are several source ,break the tie arbitrarily). Step 2: The order in which the vertices are deleted yields a solution to Topological sorting problem. INPUT: A directed acyclic graph 1 3 2 5 4
OUTPUT: the topological sorting of above directed acyclic graph is 2 1 3 4 5 /* TOPOLOGICAL SORTING */
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## Lab Manual: Algorithms Laboratory
#include<stdio.h> #include<conio.h> int a[10][10],vis[10],exp[10],n,j; void dfs(int); void main() { int i,m,x,y; clrscr(); printf("\n Enter the no. of vertices : "); scanf("%d",&n); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { a[i][j]=0; } vis[i]=0; } printf("\n Enter the no. of edges : "); scanf("%d",&m); for(i=1;i<=m;i++) { printf("\n Enter the edge : "); scanf("%d %d",&x,&y); a[x][y]=1; } j=0; for(i=1;i<=n;i++) { if(vis[i]==0) dfs(i); } printf("\n\n Topological order is : "); for(i=n-1;i>=0;i--) printf(" %d ",exp[i]); getch(); }
void dfs(int v)
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Lab Manual: Algorithms Laboratory { int i; vis[v]=1; for(i=1;i<=n;i++) { if(a[v][i] == 1 && vis[i] == 0) dfs(i); } exp[j++]=v; } /**************************** OUTPUT ************************* Enter the no. of vertices : 7 Enter the no. of edges : 6 Enter the edge : 1 2 Enter the edge : 2 3 Enter the edge : 3 4 Enter the edge : 4 5 Enter the edge : 5 6 Enter the edge : 6 7 Topological order is : 1 2 3 4 5 6 7 Enter the no. of vertices : 5 Enter the no. of edges : 5 Enter the edge : 1 3 Enter the edge : 2 3 Enter the edge : 3 5 Enter the edge : 4 5 Enter the edge : 3 4 Topological order is : 2 1 3 4 5 **************************************************************/
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Lab Manual: Algorithms Laboratory b) Print all the nodes reachable from a given starting node in a digraph using DFS method. Design Strategy: Decrease and conquer. ALGORITHM DFS(G) //Implements a depth-first search traversal of a given graph //Input: Graph G=<V, E> //Output: Graph G with its vertices marked with consecutive integers in the order they //have been first encountered by the DFS traversal mark each vertex in V with 0 as a mark of being unvisited Count0 for each vertex v in V do if v is marked with 0 dfs (v) dfs(v) //visits recursively all the unvisited vertices connected to vertex v and assigns them the //numbers in the order they are encountered via global variable count countcount+1 ; mark v with count for each vertex w in V adjacent to v do if w is marked with 0 dfs(w)
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Lab Manual: Algorithms Laboratory /* DFS PROGRAM */ #include<stdio.h> #include<conio.h> int a[10][10],vis[10],n; void dfs(int); void main() { int i,j,source; clrscr(); printf("\n Enter the no. of vertices : "); scanf("%d",&n); printf("\n Enter the adjacency matrix : \n"); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { scanf("%d",&a[i][j]); } } printf("\n Enter the source vertex : "); scanf("%d",&source); printf("\n Nodes reachable from %d vertex is : ",source); dfs(source); getch(); } void dfs(int v) { int i; vis[v]=1; printf(" %d ",v); for(i=1;i<=n;i++) { if(a[v][i]==1 && vis[i]==0) dfs(i); } }
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Lab Manual: Algorithms Laboratory /*************************** OUTPUT **************************** Enter the no. of vertices Enter the adjacency matrix 0101 0010 0000 0010 Enter the source vertex Nodes reachable from 1 vertex is Enter the no. of vertices Enter the adjacency matrix 0110 1001 1000 0100 Enter the source vertex Nodes reachable from 2 vertex is :4 : :4 :
:1 : 1 2 3 4
:2 : 2 1 3 4
***************************************************************/
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Lab Manual: Algorithms Laboratory VI. Implement 0/1 Knapsack problem using dynamic programming. Design strategy: Dynamic programming. ALGORITHM MFKnapsack(i,j) // Implement the memory function method for the knapsack problem // Input: A nonnegative integer i indicating the number of the first items being considered // and a nonnegative integer j indicating the knapsack capacity. // Output: the value of an optimal feasible subset of the first i items //note: uses as global variable input arrays weights [1n], values [1n] and table v[0n,0w] whose entries are initialized with -1 except for row 0 and column 0 initialized with 0s if v[i,j] <0 if j<weights[i] valueMFKnapsack(i-1,j) else valuemax(MFKnapsack(i-1,j),values[i]+MFKnapsack(i-1,j-weights[i])) v [i,j]value return v[i,j] INPUT: Enter the number of objects, Knapsack capacity, weights and profits of each object. item weights Value 1 2 3 4 OUTPUT: i 0 W1=1, v1=20 W2=2,v2=60 W3=3,v3=20 W4=1, v4=30 1 2 3 4 0 0 0 0 0 0 1 2 3 1 1 0 20 20 20 30 2 0 20 60 60 60 20 60 20 30 3 0 20 80 80 90 4 0 20 80 80 110 5 0 20 80 80 110
## The maximum profit is 110 and the solution vector is 1 2 4.
21
Lab Manual: Algorithms Laboratory /* KNAPSACK PROGRAM */ #include<stdio.h> #include<conio.h> int knapsack(int,int); void objects(int,int); int max(int,int); int w[10],p[10],v[10][10],x[10],i,j; void main() { int n,m,y; clrscr(); printf("\n\n Enter the no. of objects available : "); scanf("%d",&n); printf("\n\n Enter the capacity of the knapsack : "); scanf("%d",&m); printf("\n\n Enter the weights n profits of all objects : \n"); for(i=1;i<=n;i++) { scanf("%d %d",&w[i],&p[i]); x[i]=0; } y=knapsack(n,m); printf("\n\n Optimum Profit is : %d \n",y); objects(n,m); getch(); } int knapsack(int n,int m) { for(i=0;i<=n;i++) { printf("\n"); for(j=0;j<=m;j++) { if(i==0 || j==0) v[i][j]=0; else if(w[i]>j) v[i][j]=v[i-1][j]; else v[i][j]=max(v[i-1][j],(v[i-1][j-w[i]]+p[i])); printf("\t%d",v[i][j]); } } return v[n][m]; }
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Lab Manual: Algorithms Laboratory void objects(int n,int m) { i=n; j=m; while(i!=0 && j!=0) { if(v[i][j] != v[i-1][j]) { x[i]=1; j=j-w[i]; } i--; } printf("\n\n Objects selected are for(i=0;i<=n;i++) { if(x[i] == 1) printf(" %d ",i); } } int max(int a,int b) { if(a<b) return b; else return a; } /*************************** OUTPUT ************************ Enter the no. of objects available :4 Enter the capacity of the knapsack :5 Enter the weights n profits of all objects : 1 20 2 60 3 20 1 30 0 0 0 0 0 0 0 20 20 20 20 20 0 20 60 80 80 80 0 20 60 80 80 80 0 30 60 90 110 110 Optimum Profit is : 110 Objects selected are : 1 2 4 ************************************************************/
: ");
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Lab Manual: Algorithms Laboratory VII. From a given vertex in a weighted connected graph, find shortest paths to other vertices using Dijkstras algorithm. Design strategy: Greedy Method ALGORITHM dijkstras (G, s) //dijkstras algorithm for single-source shortest paths. //Input: A weighted connected graph G= (V, E) and its vertex s // Output: The length dv of a shortest path from s to v and its penultimate vertex Pv for every vertex v in V Initialize (Q) //initialize vertex priority queue to empty for every vertex v in V do dv; PvNULL Insert (q, v, dv) //initialize vertex priority in the priority queue ds0; decrease (Q, s,ds) //update priority of s with ds Vt for i0 to |V|-1 do u*DeleteMin(Q) //delete the minimum priority element VtVt U {u*} For every vertex u in V-Vt that is adjacent to u* do if du* +w (u*, u) <du dudu*+w (u*, u); Puu* Decrease (Q, u, du)
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Lab Manual: Algorithms Laboratory /* DIJKSTRA */ #include<stdio.h> #include<process.h> #include<conio.h> void dijkstra(int n,int cost[10][10],int src) { int i,j,u,v,d[10],s[10],min; for(i=1;i<=n;i++) { d[i] = cost[src][i]; s[i] = 0; } s[src] = 1; for(i=1;i<=n;i++) { min=9999; for(j=1;j<=n;j++) { if(s[j]==0 && d[j]<min) { min=d[j]; u=j; } } s[u]=1; for(v=1;v<=n;v++) { if(s[v] == 0) { if(d[u]+cost[u][v]<d[v]) { d[v]=d[u]+cost[u][v]; } } } } printf("\n The shortest paths are: \n"); for(i=1;i<=n;i++) { printf("%d -- > %d = %d \n",src,i,d[i]); } }
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Lab Manual: Algorithms Laboratory void main() { int i,j,n,cost[10][10],d[10],s[10],src; clrscr(); printf("\n Enter the number of nodes :"); scanf("%d",&n); printf("\n Enter the cost adjacency matrix:\n"); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { scanf("%d",&cost[i][j]); } } printf("\n Enter the source:"); scanf("%d",&src); dijkstra(n,cost,src); getch(); } /******************** OUTPUT **************** Enter the number of nodes :6 Enter the cost adjacency matrix: 0 15 10 9999 45 9999 9999 0 15 9999 20 9999 20 9999 0 20 9999 9999 9999 10 9999 0 35 9999 9999 9999 9999 30 0 9999 9999 9999 9999 4 9999 0 Enter the source:6 The shortest paths are: 6 -- > 1 = 49 6 -- > 2 = 14 6 -- > 3 = 29 6 -- > 4 = 4 6 -- > 5 = 34 6 -- > 6 = 0 *******************************************/
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## Lab Manual: Algorithms Laboratory
VIII. Sort a given set of elements using Quick sort method and determine the time required sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n. Design Strategy: Divide and Conquer ALGORITHM Quicksort (A [l...r]) // Sorts a subarray by quicksort //Input: a subarray A [l..r] Of A [0...n-1], defined by its left and right indices l and r //Output: the subarray A [l...r] sorted in non decreasing order if l < r s partition (A [l..r]) // s is a split position Quicksort (A [l...s-1]) Quicksort (A [s+1...r]) ALGORITHM Partition (A [l...r]) // Partitions a subarray by using its first element as a pivot //Input: a subarray A [l...r] of A [0...n-1], defined by its left and right indices l and r (l<r) //Output: a partition of A [l...r], with split position returned as this functions value pa[l] i l ; j r+1 repeat repeat i i+1 until A[i]>=p repeat jj-1 until A[j] <=p swap(A[i], A[j]) until i>=j swap(A[i], A[j]) //undo last swap when i>=j swap (A[l], A[j]) return j
/* QUICK SORT */
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Lab Manual: Algorithms Laboratory #include<stdio.h> #include<conio.h> #include<process.h> #include<time.h> int a[10]; clock_t start,end; int partition(int low,int high) { int i,j,temp,key; key=a[low]; i=low+1; j=high; while(1) { while(i<=high && key>= a[i]) i++; while(key<a[j]) j--; if(i<j) { temp=a[i]; a[i]=a[j]; a[j]=temp; } else { temp=a[low]; a[low]=a[j]; a[j]=temp; return j; } } } void qsort(int low,int high) { int j; if(low<high) { j=partition(low,high); qsort(low,j-1); qsort(j+1,high); } } void main()
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Lab Manual: Algorithms Laboratory { int i,n; clrscr(); printf("\n Enter the no. of elements in the vector : "); scanf("%d",&n); printf("\n Enter the elements : "); for(i=0;i<n;i++) scanf("%d",&a[i]); start=clock(); qsort(0,n-1); end=clock(); printf("\n\n The Sorted elements are : "); for(i=0;i<n;i++) printf(" %d ",a[i]); printf("\n\n Time taken is : %f \n",(start-end)/CLK_TCK); getch(); } /***************************** OUTPUT **************************** Enter the no. of elements in the vector : 5 Enter the elements : 88 -1 0 45 22 The Sorted elements are : -1 0 22 45 88 Time taken is : 0.000000 ******************************************************************/
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Lab Manual: Algorithms Laboratory IX. Find Minimum Cost Spanning Tree of given undirected graph using Kruskals algorithm. Design Strategy: Greedy method. ALGORITHM kruskal(G) //Kruskals algorithm for constructing a minimum spanning tree // Input: A weighted connected graph G=V, E // Output: ET , the set of edges composing a minimum spanning tree of G Sort E in nondecreasing order of the edge weights w(ei1)<=<=w(ei|E|) ET ; ecounter 0 //initialise the set of tree edges and its size k 0 //initialise the number of processed edges while ecounter < |V| - 1 k k+1 if ET {eik}is acyclic ET ET {eik}; ecounter ecounter+1 return ET
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Lab Manual: Algorithms Laboratory /* KRUSKALS ALGORITHM */ #include<stdio.h> #include<conio.h> int find(int v,int parent[]) { while(parent[v] != v) { v = parent[v]; } return v; } void union_ij(int i,int j,int parent[]) { if(i<j) parent[j] = i; else parent[i] = j; } void kruskal(int n,int a[10][10]) { int count,k,min,sum,i,j,t[10][10],u,v,parent[10],y,st[10]; count=0; k=0; sum=0; y=0; for(i=1;i<=n;i++) parent[i] = i; while(count != n-1) { min = 9999; for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { if(a[i][j] < min && a[i][j] != 0) { min = a[i][j]; u=i; v=j; } } } i = find(u,parent); j = find(v,parent);
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Lab Manual: Algorithms Laboratory if(i != j) { t[k][0] = u; t[k][1] = v; k++; count++; st[y]=a[u][v]; sum+=a[u][v]; union_ij(i,j,parent); y++; } a[u][v] = a[v][u] = 9999; } if(count == n-1) { printf("\n Spanning tree exists \n"); printf("\n Spanning tree is shown below \n\n"); for(i=0;i<n-1;i++) { printf("\t Edge \t (%d,%d) = %d \n",t[i][0],t[i][1],st[i]); } printf("\n\t Minimum Cost of the spanning tree = %d",sum); } else printf("\n Spanning tree does not exist"); } void main() { int n,a[10][10],i,j; clrscr(); printf("\n Enter the no. of nodes in the graph scanf("%d",&n); printf("\n Enter the adjacency matrix of the graph for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { scanf("%d",&a[i][j]); } } kruskal(n,a); getch(); }
: "); : \n");
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Lab Manual: Algorithms Laboratory /***************** OUTPUT *********************** Enter the no. of nodes in the graph Enter the adjacency matrix of the graph 0 5 999 6 999 5 0 1 3 999 999 1 0 4 6 63402 999 999 6 2 0 Spanning tree exists Spanning tree is shown below Edge Edge Edge Edge (2,3) = 1 (4,5) = 2 (2,4) = 3 (1,2) = 5 :5 :
## Minimum Cost of the spanning tree = 11 *************************************************/
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Lab Manual: Algorithms Laboratory X. a) Print all the nodes reachable from a given starting node in a digraph using BFS method. b) Check whether a given graph is connected of not using DFS method. a) Print all the nodes reachable from a given starting node in a digraph using BFS method. Design and Strategy: Decrease and Conquer. ALGORI THM BFS(G) //Implements a breadth-first search traversal of a given graph //Input:Graph G=<V, E> //Output:Graph G with its vetices marked with consecutive integers in the order they have //been visited by the BFS traversal mark each vetex with 0 as a mark of being unvisited count0 for each vertex v in V do if v is marked with 0 bfs(v) bfs(v) //visits all the unvisited vertices connected to vertex v and assigns them the numbers in //the order they are visited via global variable count countcount+1; mark v with count and initailize a queue with v while the queue is not empty do for each vertex w in V adjacent to the fronts vertex v do if w is marked with 0 countcount+1;mark w with count add w to the queue remove vertex v from the front of the queue
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Lab Manual: Algorithms Laboratory PROGRAM TO PRINT NODES FROM THE GIVEN SOURCE USING BFS METHOD #include<stdio.h> #include<conio.h> int a[10][10],vis[10],k,n; void bfs(int); void main() { int i,j,source; clrscr(); printf("\n Enter the no. of vertices : "); scanf("%d",&n); printf("\n Enter the Adajacency matrix : \n"); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) scanf("%d",&a[i][j]); vis[i]=0; } printf("\n Enter the source vertex : "); scanf("%d",&source); printf("\n Nodes reachable from %d is : ",source); bfs(source); getch(); } void bfs(int v) { int q[10],r=1,f=1,i,u; q[r]=v; vis[v]=1; while(f<=r) { u=q[f]; printf(" %d ",u); for(i=1;i<=n;i++) { if(a[u][i] == 1 && vis[i] == 0) { q[++r]=i; vis[i]=1; } } f++; } }
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Lab Manual: Algorithms Laboratory /****************************** OUTPUT ************************** Enter the no. of vertices :4
Enter the Adajacency matrix : 0101 0010 0000 0010 Enter the source vertex Nodes reachable from 1 is Enter the no. of vertices :1 : 1 2 4 3 :4
Enter the Adajacency matrix : 0110 1001 1000 0100 Enter the source vertex Nodes reachable from 2 is :2 : 2 1 4 3
****************************************************************/
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Lab Manual: Algorithms Laboratory b) Check whether a given graph is connected of not using DFS method. Design strategy: Decrease and Conquer ALGORITHM DFS(G) //Implements a depth-first search traversal of a given graph //Input: Graph G=<V, E> //Output: Graph G with its vertices marked with consecutive integers in the order they //have been first encountered by the DFS traversal mark each vertex in V with 0 as a mark of being unvisited Count0 connect0 for each vertex v in V do if v is marked with 0 dfs(v) connectconnect+1 dfs(v) //visits recursively all the unvisited vertices connected to vertex v and assigns them the //numbers in the order they are encountered via global variable count countcount+1 ; mark v with count for each vertex w in V adjacent to v do if w is marked with 0 dfs(w) if connect>1 then graph is disconnected else graph is connected
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Lab Manual: Algorithms Laboratory /* GRAPH CONNECTED OR NOT */ #include<stdio.h> #include<conio.h> int a[10][10],vis[10],n; void dfs(int); void main() { int i,j,source; clrscr(); printf("\n Enter the no. of vertices scanf("%d",&n); printf("\n Enter the adjacency matrix for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { scanf("%d",&a[i][j]); } } for(i=1;i<=n;i++) { source=i; for(j=1;j<=n;j++) vis[j]=0; dfs(source); for(j=1;j<=n;j++) { if(vis[j]==0) { printf("\n\n!!! GIVEN GRAPH IS NOT CONNECTED !!! \n\n"); getch(); exit(0); } } } printf("\n\n!!! GIVEN GRAPH IS CONNECTED !!! \n\n"); getch(); }
: "); : \n");
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Lab Manual: Algorithms Laboratory void dfs(int v) { int i; vis[v]=1; for(i=1;i<=n;i++) { if(a[v][i]==1 && vis[i]==0) dfs(i); } }
/******************************* OUTPUT ************************** Enter the no. of vertices Enter the adjacency matrix 0101 0010 0000 0010 :4 :
!!! GIVEN GRAPH IS NOT CONNECTED !!! Enter the no. of vertices Enter the adjacency matrix 0100 0010 0001 1000 :4 :
## !!! GIVEN GRAPH IS CONNECTED !!! *****************************************************************/
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Lab Manual: Algorithms Laboratory XI. Find a subset of a given set S = {s1,s2,,sn} of n positive integers whose sum is equal to a given positive integer d. For example, if S = {1,2,5,6,8} and d=9 there are two solutions {1,2,6} and {1,8}. A suitable message is to be displayed if the given problem instance doesnt have a problem. Design strategy: Back tracking ALGORITHM subset(s, k, r) //Find all the subset of w [1: n] that sum to m. //The values of x [j], 1[ j[ k, have already been determined. // s = j=1to k-1 w [j]*x [j] and r = j=k to n w [j]. //The w [j]s are in the nondecreasing order. //It is assumed that w [1][ m and i=1 to n w [i] m. { //Generate left child. Note: s + w [k][ m since Bk-1is true x [k]: =1; if (s + w [k] = m) then write (x[1: k]) ; //subset found //There is no recursive call here as w [j]>0, 1[ j[ n. else if(s + w [k] +w[k+1][ m) then subset (s + w [k], k+1, r w [k]); //Generate right child and evaluate Bk. if((s + r w [k] m) and (s + w [k+1][ m)) then { x [k]:= 0; subset (s, k+1, r w [k]); } }
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Lab Manual: Algorithms Laboratory /* FIND A SUBSET OF GIVEN SET OF N POSITIVE INTEGERS */ #include<stdio.h> #include<conio.h> void subset(int n,int d,int w[]) { int s,k,i,x[10],count=0; for(i=1;i<=n;i++) x[i] = 0; s=0; k=1; x[k]=1; while(1) { if(k<=n && x[k]==1) { if(s+w[k] == d) { printf("\n Solution %d is : ",++count); for(i=1;i<=n;i++) { if(x[i] == 1) printf("%d\t",w[i]); } printf("\n\n"); x[k]=0; } else if(s+w[k]<d) s+=w[k]; else x[k]=0; } else { k--; while(k>0 && x[k]==0) k--; if(k==0) break; x[k]=0; s=s-w[k]; } k=k+1; x[k]=1; } }
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Lab Manual: Algorithms Laboratory void main() { int i,sum,d,n,w[10]; printf("\n Enter the value of n : "); scanf("%d",&n); printf("\n Enter the set in increasing order : "); for(i=1;i<=n;i++) scanf("%d",&w[i]); printf("\n Enter the maximum subset values(sum) scanf("%d",&d); for(i=1;i<=n;i++) sum+=w[i]; if(sum < d || w[1] > d) { printf("\n\n No Solution exists \n\n"); return; } subset(n,d,w); getch(); }
: ");
## Enter the maximum subset values(sum) : 9 Solution 1 is : 1 Solution 2 is : 1 2 8 6
****************************************************************/
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## Lab Manual: Algorithms Laboratory
XII. a) Implement Horspool algorithm for String Matching. b) Find the Binomial Co-effient using Dynamic Programming. a) Implement Horspool algorithm for String Matching. Design strategy: Space and Time Tradeoffs ALGORITHM horspoolmatching(p[0m-1],T[0n-1]) //Implements horspools algorithm for string matching //input: pattern p[0m-1] and text T[0n-1] //output: the index of the left end of the first matching substring or -1 if there are no //matches. shiftTable( p[0m-1]) //generate table of shifts im-1 while in-1 do k0 while km-1 and p[m-1-k]=T[i-k] kk+1 if k=m return i-m+1 else ii+Table[T[i]] return -1 ALGORITHM shiftTable(p[0m-1]) // Fills the shift table used by Horspools and Boyer-Moore algorithms // Input: pattern p[0m-1] and an alphabet of possible characters // Output: Table[0size-1] indexed by the alphabets characters and filled with shift //sizes computed by formula initialize all the elements of Table with m for j0 to m-2 do Table[p[j]]m-1-j return Table INPUT: pattern P and text T TEXT: JIM_SAW_ME_IN-A_BARBERSSHOP PATTERN: BARBER. OUTPUT: The shift table is filled as follows Character A B C D E F . . . . . . R Shift t(c) 4 2 6 6 1 6 6 3 The actual search in a particular text proceeds as J I M _ S A W _M E_ I N A _ B A R B E R S S H O P BA RB E R BARBER B A R BE R BARBER BAR B E R BARB E R
Z 6 6
The algorithm returns index of the left end of the first matching substring or -1 if there are no matches.
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Lab Manual: Algorithms Laboratory /* HORSPOOL */ #include<stdio.h> #include<conio.h> #include<string.h> #include<process.h> void shifttable(char p[],int t[]) { int m,i; m=strlen(p); for(i=0;i<128;i++) t[i]=m; for(i=0;i<=m-2;i++) t[p[i]]=m-1-i; } int horspool(char p[],char t[]) { int m,n,i,k,s[256]; shifttable(p,s); m=strlen(p); n=strlen(t); i=m-1; while(i<=n-1) { k=1; while(k<=m-1 && t[i-k] == p[m-1-k]) k++; if(k==m) return(i-m+1); i=i+s[t[i]]; } return -1; } void main() { char p[20],t[40]; int pos; clrscr(); printf("\n\n Enter the String : "); gets(t); printf("\n\n Enter the Pattern String : "); gets(p); pos=horspool(p,t); if(pos==-1) printf("\n\n STRING NOT FOUND \n\n"); else printf("\n\n STRING FOUND AT %d POSITION \n\n",pos+1); getch(); }
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Lab Manual: Algorithms Laboratory /**************************** OUTPUT ************************* Enter the String Enter the Pattern String : JIM SAW ME IN BARBER SHOP : BARBER
STRING FOUND AT 15 POSITION Enter the String Enter the Pattern String STRING NOT FOUND *************************************************************/ : JIM SAW ME IN BARBER SHOP : JOHN
45
Lab Manual: Algorithms Laboratory b) Find the Binomial Co-effient using Dynamic Programming. Design strategy: Dynamic programming ALGORITHM binomial (n, k) //computes C (n, k) by the dynamic programming algorithm //input: a pair of non-negative integers n k 0 //output: the value of C (n, k) for i0 to n do for j0 to min (i, k) do if j=0 or j=i C[i, j]1 else C[i, j] C[i-1, j-1]+ C[i-1, j] return C[n, k] INPUT: a pair of non-negative integers n k 0 n= 6 k= 3 OUTPUT: 0 1 2 3 4 5 6 0 1 1 1 1 1 1 1 1 1 2 3 4 5 6 2 3
1 3 6 10 15
1 4 10 20
## The value C (6,3) = 20
46
Lab Manual: Algorithms Laboratory /* BINOMIAL CO-EFFICIENT */ #include<stdio.h> #include<conio.h> int nck(int,int); int min(int,int); void main() { int n,r,y; clrscr(); printf("\n Enter the values 'n' and 'r' : "); scanf("%d %d",&n,&r); y=nck(n,r); if(y==-1) printf("\n\n Invalid input \n\n"); else { printf("\n\n Value of nCr is : "); printf("%d",y); } getch(); } int nck(int n,int r) { int c[10][10],i,j; if(n<r) return -1; else { for(i=0;i<=n;i++) { for(j=0;j<=min(i,r);j++) { if(j==0 || i==j) c[i][j]=1; else c[i][j]=c[i-1][j-1]+c[i-1][j]; } } return c[n][r]; } }
47
Lab Manual: Algorithms Laboratory int min(int a,int b) { if(a<b) return a; else return b; } /******************************* OUTPUT ************************** Enter the values 'n' and 'r' : 6 3 Value of nCr is : 20
*****************************************************************/
48
## Lab Manual: Algorithms Laboratory
XIII. Find Minimum Cost Spanning Tree of given undirected graph using Prims algorithm. Design Strategy: Greedy method ALGORITHM prim (G) //prims algorithm for constructing a minimum spanning tree // Input: A weighted connected graph G=V, E // Output: ET , the set of edges composing a minimum spanning tree of G VT {v0} ET for i 1 to |V| -1 do find a minimum-weight edge e*=(v*, u*) among all edges (v, u) such that v is in VT and u is in V-VT VT VT {u*} ET ET {e*} return ET
49
Lab Manual: Algorithms Laboratory /* PRIMS ALGORITHM */ #include<stdio.h> #include<conio.h> #include<process.h> void prims(int n,int cost[10][10]) { int i,j,u,v,src,d[10],s[10],p[10],t[10][2],k,min,y,st[10],sum; src=0; min=9999; for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { if(cost[i][j] != 0 && cost[i][j] < min) { min=cost[i][j]; src=i; } } } for(i=1;i<=n;i++) { d[i]=cost[src][i]; s[i]=0; p[i]=src; } s[src]=1; k=0; sum=0; y=0; for(i=2;i<=n;i++) { min=9999; u=-1; for(j=1;j<=n;j++) { if(s[j] == 0 && d[j] <= min) { min=d[j]; u=j; } } t[k][0]=p[u]; t[k][1]=u; k++;
50
Lab Manual: Algorithms Laboratory st[y]=cost[u][p[u]]; sum+=cost[u][p[u]]; s[u]=1; y++; for(v=1;v<=n;v++) { if(s[v] == 0) { if(cost[u][v] < d[v]) { d[v]=cost[u][v]; p[v]=u; } } } } if(sum>=9999) printf("\n\n Spanning tree doesnot exist \n\n"); else { printf("\n\n SPANNING TREE EXISTS \n\n"); printf(" MINIMUM SPANNING TREE : \n\n"); for(i=0;i<n-1;i++) { printf("\t Edge (%d,%d) = %d\n",t[i][0],t[i][1],st[i]); } printf("\n The Minimum Cost of Spanning Tree = %d\n",sum); } } void main() { int n,i,j,cost[10][10]; clrscr(); printf("\n\n Enter the number of Vertices : "); scanf("%d",&n); printf("\n\n Enter the Cost Adjacency matrix : \n"); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { scanf("%d",&cost[i][j]); } } prims(n,cost); getch(); }
51
Lab Manual: Algorithms Laboratory /****************************** OUTPUT **************************** Enter the number of Vertices : 6 Enter the Cost Adjacency matrix 0 60 10 9999 9999 9999 60 0 9999 20 40 70 10 9999 0 9999 9999 50 9999 20 9999 0 999 80 9999 40 9999 9999 0 30 9999 70 50 80 30 0 SPANNING TREE EXISTS MINIMUM SPANNING TREE : Edge (1,3) = 10 Edge (3,6) = 50 Edge (6,5) = 30 Edge (5,2) = 40 Edge (2,4) = 20 The Minimum Cost of Spanning Tree = 150 ******************************************************************/ :
52
## Lab Manual: Algorithms Laboratory
XIV. a) Implement Floyds algorithm for the All-Pairs-Shortest-Paths problem. b) Compute the transitive closure of a given directed graph using Warshalls algorithm. a) Implement Floyds algorithm for the All-Pairs-Shortest-Paths problem. Design strategy: Dynamic programming. ALGORITHM Floyd (W [1n,1n) //Implements Floyds algorithm for the all-pairs shortest paths problem //Input: The weight matrix W of a graph //Output:The distance matrix of the shortest paths lengths DW //is not necessary if W can be written for k1 to n do for i1 to n do for j1 to n do D[i,j]min{D[i,j], D[ i,k] + D[k,j]} return D INPUT: the weight graph w of a graph.
a 7 c 2
b 8 9 d a b c d
a 0
b 1 0
c 7 2 0 2
d 8 9 0
## OUTPUT: The distance matrix of the shortest paths lengths a 0 b 1 0 c 3 2 0 2 d 8 0
a b c d
53
Lab Manual: Algorithms Laboratory /* FLOYDS ALGORITHM */ #include<stdio.h> #include<conio.h> void floyd(int); int min(int,int); int a[10][10],p[10][10],i,j,k; void main() { int n; clrscr(); printf("\n Enter the no. of vertices : \n"); scanf("%d",&n); printf("\n Enter the cost adjacency matrix of the graph...\n"); printf("\n Enter 999 if edge is absent \n"); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { scanf("%d",&a[i][j]); p[i][j] = a[i][j]; } } floyd(n); printf("\n The all pair shortest path matrix is \n"); for(i=1;i<=n;i++) { printf("\n"); for(j=1;j<=n;j++) { printf(" %d ",p[i][j]); } } getch(); }
54
Lab Manual: Algorithms Laboratory void floyd(int n) { for(k=1;k<=n;k++) { for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { p[i][j]=min(p[i][j],p[i][k]+p[k][j]); } } } } int min(int a,int b) { if(a<b) return a; else return b; } /******************************** OUTPUT ********************** Enter the no. of vertices : 4 Enter the cost adjacency matrix of the graph... Enter 999 if edge is absent 0178 999 0 2 999 999 999 0 999 999 999 2 0 The all pair shortest path matrix is 0 1 999 999 999 3 8 0 2 999 999 0 999 999 2 0
**************************************************************/
55
Lab Manual: Algorithms Laboratory b) Compute the transitive closure of a given directed graph using Warshalls algorithm. Design strategy: Dynamic Programming ALGORITHM Warshall (A [1n, 1n]) //Implements Warshalls algorithm for computing the transitive closure // Input: The adjacency matrix A of a digraph with n vertices //Output: The transitive closure of the digraph R (0) A for k to n do for i to n do for j to n do R (k) [i, j] R (k-1) [i, j] or R (k-1) [i, k] and R (k-1) [k, j] return R (n) INPUT: The adjacency matrix A of a digraph with n vertices a b a b c d a 0 0 0 0 b 1 0 0 0 c 0 1 0 1 d 1 0 0 0
## OUTPUT: The transitive closure of the digraph a 0 0 0 0 b 1 0 0 0 c 1 1 0 1 d 1 0 0 0
a b c d
56
Lab Manual: Algorithms Laboratory /* WARSHALL ALGORITHM */ #include<stdio.h> void warshall(int); int i,j,p[10][10],a[10][10]; void main() { int n; clrscr(); printf("\n Enter the no. of vertices : "); scanf("%d",&n); printf("\n Enter the adjacency matrix : \n"); for(i=0;i<n;i++) { for(j=0;j<n;j++) { scanf("%d",&a[i][j]); p[i][j]=a[i][j]; } } warshall(n); printf("\n The transitive closure is : "); for(i=0;i<n;i++) { printf("\n"); for(j=0;j<n;j++) { printf(" %d ",p[i][j]); } } getch(); } void warshall(int n) { int k; for(k=0;k<n;k++) { for(i=0;i<n;i++) { for(j=0;j<n;j++) { if(p[i][j]!=1 && p[i][k]==1 && p[k][j]==1) p[i][j]=1; } } } }
57
Lab Manual: Algorithms Laboratory /************************ OUTPUT ************************* Enter the no. of vertices Enter the adjacency matrix 0101 0010 0000 0010 The transitive closure is 0 1 1 1 0 0 1 0 0 0 0 0 0 0 1 0 : :4 :
**********************************************************/
58
Lab Manual: Algorithms Laboratory XV. Implement N Queens problem using Back Tracking. Design Strategy: Back Tracking. ALGORITHM Place (k) //Implements N-Queen using Backtracking procedure to find whether a new Queen can be placed //procedure returns true if a queen can be placed in kth row & X (k) th column otherwise it returns //false. X is a global array whose first k values have been set ABS(r) returns absolute value of r Global X (1: k); integer i, k for i1 to k do if X (i)=X(k) // two in same column or ABS(X(i)-X(k))=ABS(i-k) //in the same diagonal then return false endif repeat return true ALGORITHM nqueens(n) //This procedure prints all possible placements of n queens on an n*n chessboard so that they are //non attacking. Integer k, n, X (1, n) X(!)0; k1 //k is current row X(k) current column While k>0 do //for all rows do X(k)X(k)+1 //move to next column While X(k)<=n & not place(k) do //can this queen can be placed X(k)X(k)+1 repeat if X(k)<=n //a possible is found then if k=n //is a solution complete then print(x) //if yes ,print the array else kk+1 X(k)0//go to the new row endif else kk-1 //backtrack endif repeat
59
## Lab Manual: Algorithms Laboratory
/* NQUEENS BACK TRACKING */ #include<stdio.h> #include<conio.h> #include<math.h> int canplace(int r,int c[5]) { int i,j; for(i=0;i<r;i++) { if(c[i]==c[r] || (c[i]-c[r]) == abs(i-r)) return(0); } return(1); } void display(int c[50],int n) { int i,j; char cb[10][10]; for(i=0;i<n;i++) for(j=0;j<n;j++) cb[i][j]='-'; for(i=0;i<n;i++) cb[i][c[i]]='Q'; printf("\n"); for(i=0;i<n;i++) { for(j=0;j<n;j++) printf(" %c ",cb[i][j]); printf("\n"); } } void nqueen(int n) { int r,c[50]; c[0]=-1; r=0; while(r>=0) { c[r]++; while(c[r]<n && !canplace(r,c))
60
Lab Manual: Algorithms Laboratory c[r]++; if(c[r]<n) { if(r==n-1) { display(c,n); printf("\n"); } else { r++; c[r]=-1; } } else r--; } } void main() { int n; clrscr(); printf("\n\n Enter the number of Queens scanf("%d",&n); nqueen(n); getch(); }
: ");
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## JenniferSmart1 2 years ago how do I write this in sum form? $y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)$
• This Question is Closed
1. JenniferSmart1
I know I'm a disappointment :(
2. TuringTest
no you're not, I don't have it either$y=a_0\sum_{n=0}^\infty\frac{x^{2n}}{}$darn I know there is a way to do the denom, but I can't think of it... I have seen final answers for power series DE's written as you have them though, so the summation notation is not necesarry
3. hartnn
i would have written denominator of the form 2^m m!
4. hartnn
num = x^(2m)
5. JenniferSmart1
this is the whole (series?) $a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7$
6. hartnn
$$\large y=a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}$$ doesn't this work ??
7. TuringTest
yeah, that's it @hart
8. JenniferSmart1
9. JenniferSmart1
#9
10. JenniferSmart1
how and why?
11. hartnn
2 = 2 (1) = 2^1 (1!) 2.4 = 2.2(1.2) = 2^2 (2!) 2.4.6 = 2.2.2 (1.2.3) = 2^3 (3!) in general denominator = (2^n n!)
12. hartnn
got that ^ ?
13. JenniferSmart1
still thinking
14. JenniferSmart1
ok from the top. so you look at this and you think..........? What is your thought process? $a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7$
15. hartnn
my 1st thought : did u change the Q ? where does these a1's jump in..... ? anyways, i would first factor out a0 and a1....
16. JenniferSmart1
that's the whole sum (series?) what I wrote is just the a_1, then I added the a_0 It's from a previous problem http://openstudy.com/users/jennifersmart1#/updates/50d63d0fe4b0d6c1d541ef3a
17. hartnn
ok, so did u get how , after factoring a0, u get $$\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}$$ now your next step is to find sum for a1, right ?
18. JenniferSmart1
I didn't quite get that part...
19. hartnn
which part ? x^(2m) or 2^m or m! ??
20. JenniferSmart1
ok I'm back. I can make sense of the answer but how do I come up with this on my own.... =(
21. JenniferSmart1
$y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)$ $\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}$ Let's look at this
22. JenniferSmart1
so as m increases ... that makes sense Problem is will I be able to do this on my own?
23. TuringTest
Well as you saw, I had forgotten about the 2^m for the product of only the even numbers in the factorial, but this is kind of a tricky pattern to write if you ask me. They are not all this bad. You can just reason to yourself about needing a 2 for each of the m elements in the factorial to make each factor double, and therefor even.
24. JenniferSmart1
ok
25. JenniferSmart1
ok I'll try writing the sum a_1 let me know how I did
26. TuringTest
I'm not sure I know that one... I actually remember trying to come up with this before, a summation representing the product of evens and could not. Maybe I'll be more lucky this time.
27. TuringTest
*odds I mean
28. hartnn
product of odds is not difficult...u'll get it turing....
29. hartnn
hint : 1/(1.3.5) = 2.4 /(5!)
30. TuringTest
ohhhhhhhhhhhhhh
31. JenniferSmart1
=$a_1x^1+\frac{a_1}{3}x^3+\frac{a_1}{3\cdot5}x^5+\frac{a_1}{3\cdot5\cdot7}x^7+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9$ =$a_1 \left(x^1+\frac{x^3}{3}+\frac{x^5}{3\cdot5}+\frac{x^7}{3\cdot5\cdot7}+\frac{x^9}{3\cdot5\cdot7\cdot9}+\frac{x^{10}}{3\cdot5\cdot7\cdot9\cdot10}\right)$ = ok let's start with the exponents....or where would I look first?
32. JenniferSmart1
I understand the hint.... :(
33. JenniferSmart1
I don't understand I meant
34. hartnn
1/(1.3.5) = 2.4 /(5!) = 2^2 . (1.2) / (5!) = 2^2 (2!)/(5!) got that ?
35. JenniferSmart1
so how ....Ohhh by having 2x4 in the numerator your eliminating the evens in the denominator!!!!
36. hartnn
i try to bring factorial form wherever possible.
37. hartnn
for 1.3.5 i needed 2.4 to get 5! so i multiplied and divided by 2.4 now from 2.4, i get 1.2=2! by factoring out 2, twice.
38. JenniferSmart1
$\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ doesn't m=3 for this?
39. JenniferSmart1
and yes this part makes sense
40. hartnn
why m=3 ? i would say m=2 5 = 2m+1 so, the term will be x^(2m+1) 2^m m!/ (2m+1)! got that ?
41. JenniferSmart1
no I mean this is the 3rd ....whatchamacallit? "m" is from 0 to infinity ...and this would be the third term?
42. JenniferSmart1
nevermind
43. hartnn
ohh..
44. hartnn
start from m=0
45. hartnn
m=0 <-1st term m=1 <-2nd term m=2 <-3rd term
46. JenniferSmart1
$\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ my point is....where does me thought process continue from here...how do I change these numbers to variables. ok so were working on the third term now. where m=2
47. JenniferSmart1
*my not me
48. JenniferSmart1
soo many typos sorry
49. JenniferSmart1
2m+1
50. JenniferSmart1
LOL
51. JenniferSmart1
<---frustrated koala :S
52. hartnn
what i do is always take 3rd or 4th term., bring it in factorial form, take lowest number =m all other numbers in terms of m....
53. JenniferSmart1
oh so 2m+1 would be right!
54. hartnn
yes.
55. JenniferSmart1
where would it go? $\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ shall we put it somewhere in the numerator?
56. JenniferSmart1
sorry koalas are fairly slow...I'm still thinking
57. hartnn
all this exercise was to find the general term , which we found as x^(2m+1) 2^m m!/ (2m+1)! right ? now just put a summation sign b4 this, with limits from 0 to infinity.
58. JenniferSmart1
$\frac{x^5}{3\cdot5}$ m=2 $\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}$ $\frac{x^{(2m+1)}2^mm!}{(2m+1)!}$ ok I can see it now
59. JenniferSmart1
I'm just very visual
60. hartnn
finally it will be $$\large a_1[\sum\limits _0^\infty\frac{x^{(2m+1)}2^mm!}{(2m+1)!}]$$
61. JenniferSmart1
thanks @hartnn
62. JenniferSmart1
do either of you have a similar problem somewhere? LOL I wanna try this again
63. hartnn
welcome ^_^
64. JenniferSmart1
no I think this is enough. Ok thanks
65. JenniferSmart1
does this make sense to anyone? $\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}$
66. JenniferSmart1
for both a_0 and a_1
67. hartnn
thats for a0 only...
68. JenniferSmart1
It's the solution in the back of my book for y''-xy'-y=0
69. JenniferSmart1
70. JenniferSmart1
I think I wrote it out correctly, and the the sums we wrote seem right too...but somehow he took those two and made one compact solution
71. hartnn
u mean in the last image ? he just wrote solution of initial value problem.... with a1=0
72. JenniferSmart1
ohhhhhhh....durrrr sorry. Ok I got now haha
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# Combinations in C++
, 5 Jun 2018
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
• 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 initialised 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!
## Improved Next Combination with State
To speed up `next_combination`, we can store the state of generated combination so that it does not have to find which current combination elements correspond to the bigger collection. One way to do it is to store this state inside a class but this violates the design of STL algorithms. Another way to do it, is to pass this state to `next_combination` at every call. The declaration of `next_combination` and `next_combination_with_state` are listed below so that we can compare them side by side. The 1st one is current `next_combination` and 2nd one is overloaded one with 5th parameter as equality predicate and the 3rd is the new `next_combination_with_state` which also has 4 parameters as 1st `next_combination` but the last 2 parameters are of `BidItIt` type which is iterator whose value type is `BidIt` iterator. In other words, `BidItIt` is iterator of iterator! By storing `BidIt` iterator of `n_begin` and `n_end` itself, I could save some time without finding the range of `r_begin` and `r_end` that corresponds to `n_begin` and `n_end`. We can expect performance gain of 4X to 10X, depending on how big n and r collection.
```// Plain old next_combination
template <class BidIt>
inline bool next_combination(
BidIt n_begin,
BidIt n_end,
BidIt r_begin,
BidIt r_end);
// Plain old next_combination with equality predicate
template <class BidIt, class Prediate>
inline bool next_combination(
BidIt n_begin,
BidIt n_end,
BidIt r_begin,
BidIt r_end,
Prediate Equal);
// New next_combination_with_state
// its state is stored in r_beginIT and r_endIT
// which iterators of BidIt iterators
template <class BidIt, class BidItIt>
inline bool next_combination_with_state(
BidIt n_begin,
BidIt n_end,
BidItIt r_beginIT,
BidItIt r_endIT);
// New next_combination_with_state does not have
// version with equality predicate because it compare
// with BidIt iterators, not elements which BidIt
// iterator pointed to.```
`next_combination_with_state` does not have version with equality predicate because it compare with `BidIt` iterators, not elements themselves.
I reproduce example of `next_combination` usage so that we can compare with the one of `next_combination_with_state`.
```#include<iostream>
#include<vector>
#include<string>
#include"combination.h"
using namespace std;
using namespace stdcomb;
// for use with next_combination examples!
template<class BidIt>
void display(BidIt begin,BidIt end)
{
for (BidIt it=begin;it!=end;++it)
cout<<*it<<" ";
cout<<endl;
}
//test next_combination() with iterators
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;
}```
The `next_combination_with_state` example is below. Instead of constructing a `vector` of integer for smaller collection, we construct `cbit`, a `vector` out of `ca` iterators. We also have a new `display2` function to display the result, the main difference, `it` iterator is dereferenced twice, instead of once in `display`. Note, we cannot dereference first before passing to `display` because `cbit.end()` cannot be dereferenced as it is the one past the last valid iterator. Previously, I tried putting `cbit.begin()` and `cbit.end()` result back to `cb`, an already allocated `vector`. I got back the same performance, back to square one. Only use `next_combination_with_state` when you are comfortable with having your result as iterators of iterators. Since `cbit` stores `ca` iterators, `ca` must be kept alive while you still have `cbit`, else you got dangling iterators. `next_combination_with_state` requires C++17 because it uses `reverse_iterator`.
```#include<iostream>
#include<vector>
#include<string>
#include"combination.h"
using namespace std;
using namespace stdcomb;
template<class BidItIt>
void display2(BidItIt begin, BidItIt end)
{
for (BidItIt it = begin; it != end; ++it)
cout << **it << " ";
cout << endl;
}
//test next_combination_with_state() with iterators
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< vector<int>::iterator > cbit;
vector<int>::iterator it = ca.begin();
for(; it!=ca.end()-2; ++it)
cbit.push_back(it);
do
{
display2(cbit.begin(), cbit.end());
}
while(next_combination_with_state(ca.begin (),ca.end (),cbit.begin (),cbit.end () ) );
cout<<"Complete!"<<endl;
return 0;
}```
If you are interested, you can proceed to read the second part of the article: Combinations in C++, Part 2.
## History
• 6th June 2018 - Added Improved Next Combination with State section.
• 14th September 2009 - Added the example code
• 21st February 2008 - Added the finding combinations of vectors in the source code
• 26th 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
• 30th July 2003 - First release on CodeProject
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Reverse Iterator - C++ 17 ? Ajay Vijayvargiya10-Jun-18 20:56 Ajay Vijayvargiya 10-Jun-18 20:56
Outstanding, sir! koothkeeper7-Jun-18 7:54 koothkeeper 7-Jun-18 7:54
question about the example Member 1201594727-Sep-15 20:38 Member 12015947 27-Sep-15 20:38
Re: question about the example Shao Voon Wong27-Sep-15 21:26 Shao Voon Wong 27-Sep-15 21:26
Re: question about the example Member 1201594728-Sep-15 2:37 Member 12015947 28-Sep-15 2:37
Re: question about the example Shao Voon Wong28-Sep-15 2:59 Shao Voon Wong 28-Sep-15 2:59
crashed why? thanks a lot calmman.yang30-Jul-15 23:06 calmman.yang 30-Jul-15 23:06
Re: crashed why? thanks a lot Shao Voon Wong30-Jul-15 23:11 Shao Voon Wong 30-Jul-15 23:11
Re: crashed why? thanks a lot calmman.yang30-Jul-15 23:32 calmman.yang 30-Jul-15 23:32
Re: crashed why? thanks a lot Shao Voon Wong30-Jul-15 23:40 Shao Voon Wong 30-Jul-15 23:40
Re: crashed why? thanks a lot calmman.yang30-Jul-15 23:46 calmman.yang 30-Jul-15 23:46
Re: crashed why? thanks a lot Shao Voon Wong31-Jul-15 0:13 Shao Voon Wong 31-Jul-15 0:13
Re: crashed why? thanks a lot calmman.yang4-Aug-15 23:20 calmman.yang 4-Aug-15 23:20
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
Sir, Thanks for your help. Sir I just saw your part 2 upload, which is what I am looking for.I have implemented my algorithm in matlab which is serial now I want to convert it in c and parallel, and then test on cuda. so now I am understanding your "The Technique of Finding Combination, Given its Index" which you have explained very good, thanks. sir, I have not so far downloaded software and tested it, I will be very grateful to you if you help me to convert it in c language else I need to write my code in c++ which I don't know as I just started learning c language. Thanking you.
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
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Posts
## Five proofs that the Euler characteristic of a closed orientable surface is even
Let $\Sigma_g$ be a closed orientable surface of genus $g$. (Below we will occasionally write $\Sigma$, omitting the genus.) Then its Euler characteristic $\chi(\Sigma_g) = 2 - 2g$ is even. In this post we will give five proofs of this fact that do not use the fact that we can directly compute the Euler characteristic to be $2 - 2g$, roughly in increasing order of sophistication. Along the way we’ll end up encountering or proving more general results that have other interesting applications.
Proof 1: Poincaré duality
A corollary of Poincaré duality is that if $X$ is a closed orientable manifold of dimension $n$, then the Betti numbers $b_k = \dim H^k(X, \mathbb{Q})$ satisfy $b_k = b_{n-k}$. When $n$ is odd, this implies that the Euler characteristic
$\displaystyle \chi(X) = \sum_{k=0}^n (-1)^k b_k$
is equal to zero, since $(-1)^k b_k + (-1)^{n-k} b_{n-k} = 0$. In fact slightly more is true.
Proposition: Let $X$ be a closed manifold of dimension $n$, not necessarily orientable. If $n$ is odd, then $\chi(X) = 0$. If $n$ is even and $X$ is a boundary, then $\chi(X) \equiv 0 \bmod 2$.
Proof. When $\dim X$ is odd, let $\widetilde{X}$ be the orientable double cover of $X$, so that $\chi(\widetilde{X}) = 2 \chi(X)$. By Poincaré duality, $\chi(\widetilde{X}) = 0$, so the same is true for $\chi(X)$. Alternatively, because the Euler characteristic can also be calculated using the cohomology over $\mathbb{F}_2$, we can also use Poincaré duality over $\mathbb{F}_2$, which holds for all closed manifolds since all closed manifolds have fundamental classes over $\mathbb{F}_2$.
When $\dim X$ is even and $X = \partial Y$ is the boundary of a compact manifold $Y$, let $Z = Y \sqcup_X Y$ be the manifold obtained from two copies of $Y$ by gluing along their common boundary. Then $Z$ is a closed odd-dimensional manifold, hence $\chi(Z) = 0$. But
$\displaystyle \chi(Z) = 2 \chi(Y) - \chi(X)$
(e.g. by an application of Mayer-Vietoris), from which it follows that $\chi(X) \equiv \chi(Z) \equiv 0 \bmod 2$. $\Box$
Corollary: The Euler characteristic of $\Sigma_g$ is even.
Proof. $\Sigma_g$ is the boundary of the solid $g$-holed torus. $\Box$
Corollary: No product of the even-dimensional real projective spaces $\mathbb{RP}^{2k}, k \ge 1$ is a boundary.
Proof. Since $\chi(S^{2k}) = 2$ and $\mathbb{RP}^{2k}$ is double covered by $S^{2k}$, we have $\chi(\mathbb{RP}^{2k}) = 1$, hence any product of even-dimensional real projective spaces also has Euler characteristic $1$, which in particular is odd. $\Box$
Corollary: The Euler characteristic $\bmod 2$ is a cobordism invariant.
Proof. Let $X, Y$ be two closed manifolds which are cobordant, so that there exists a closed manifold $Z$ such that $\partial Z = X \sqcup Y$. Then $\chi(\partial Z) = \chi(X) + \chi(Y) \equiv 0 \bmod 2$, hence $\chi(X) \equiv \chi(Y) \bmod 2$. $\Box$
In addition to satisfying $\chi(\partial Y) \equiv 0 \bmod 2$, the Euler characteristic also satisfies $\chi(X \times Y) = \chi(X) \chi(Y)$ (e.g. by the Künneth theorem). It follows that the Euler characteristic $\bmod 2$ is a genus of unoriented manifolds, or equivalently that it defines a ring homomorphism
$\displaystyle \chi : MO_{\bullet}(\text{pt}) \to \mathbb{F}_2$
where $MO_{\bullet}(\text{pt})$ is the unoriented cobordism ring and $MO$ is the Thom spectrum for unoriented cobordism. This is arguably the simplest example of a genus.
Warning. The Euler characteristic itself is not a genus because it is not a cobordism invariant. For example, $\Sigma_2$ is a boundary, hence cobordant to the empty manifold, but $\chi(\Sigma_2) = -2$. There is an integer-valued genus lifting the Euler characteristic $\bmod 2$ on oriented manifolds, although it is not the Euler characteristic but the signature
$\displaystyle \sigma : MSO_{\bullet}(\text{pt}) \to \mathbb{Z}$
where $MSO_{\bullet}(\text{pt})$ is the oriented cobordism ring and $MSO$ is the Thom spectrum for oriented cobordism.
Proof 2: Poincaré duality again
Let $X$ be a closed oriented manifold of even dimension $2k$. Then the cup product defines a pairing
$\displaystyle H^k(X, \mathbb{Q}) \times H^k(X, \mathbb{Q}) \to H^{2k}(X, \mathbb{Q}) \cong \mathbb{Q}$
on middle cohomology which is nondegenerate by Poincaré duality, symmetric if $k$ is even, and skew-symmetric if $k$ is odd. Previously we used this pairing when $k = 2$ and over $\mathbb{Z}$ to understand 4-manifolds. When $k$ is odd we can say the following.
Proposition: With hypotheses as above (in particular, $k$ odd), the Betti number $b_k$ is even.
Proof. On $H^k(X, \mathbb{Q})$ the cup product pairing is a symplectic form, and symplectic vector spaces are even-dimensional. (This follows from the fact that by induction on the dimension, every symplectic vector space $(V, \omega)$ has a symplectic basis, namely a basis $x_1, \dots x_k, y_1, \dots y_k$ such that $\omega(x_i, y_j) = \delta_{ij}$ and $\omega(x_i, x_j) = \omega(y_i, y_j) = 0$. This is a pointwise form of Darboux’s theorem.) $\Box$
Corollary: The Euler characteristic of a closed orientable manifold $X$ of dimension $2 \bmod 4$ is even. In particular, the Euler characteristic of $\Sigma$ is even.
Proof. As above, let $\dim X = 2k$. In the sum
$\displaystyle \chi(X) \equiv \sum_{i=0}^{2k} b_i \bmod 2$
every term $b_i$ is canceled by the corresponding term $b_{2k-i} = b_i$ by Poincaré duality, except for the middle term $b_k$, which we now know is even. $\Box$
Remark. Although this proof also uses Poincaré duality and has the same conclusion as the previous proof, it proves a genuinely different fact about manifolds: on the one hand, it only applies to manifolds of dimension $2 \bmod 4$ and requires orientability over $\mathbb{Z}$ and not just over $\mathbb{F}_2$, but on the other hand it applies in principle to manifolds which are not boundaries.
Going back to the particular case of surfaces $\Sigma_g$, we can even write down a fairly explicit choice of symplectic basis for $H^1(\Sigma_g, \mathbb{Q})$ as follows: thinking of $\Sigma_g$ as a $g$-holed torus, hence equivalently as the connected sum of $g$ tori, we can write down the usual basis $a_k, b_k$ of the first homology of the $k^{th}$ torus. Together these give the standard choice of generators $a_1, b_2, \dots a_g, b_g$ of the fundamental group $\pi_1(\Sigma_g)$, as well as of the first homology $H_1(\Sigma_g, \mathbb{Z})$, and their Poincaré duals in $H^1(\Sigma_g, \mathbb{Q})$ form the symplectic basis we want by the standard relationship between intersections and cup products.
The symplectic structure on $H^1(\Sigma, \mathbb{Q})$ is a shadow of a more general construction of symplectic structures on character varieties $\text{Hom}(\pi_1(\Sigma), G)/G$ of surfaces; these are moduli spaces of flat $G$-bundles with connection on $\Sigma$. The connection is that $H^1(\Sigma, \mathbb{R})$ is the tangent space at the identity of the moduli space $\text{Hom}(\pi_1(\Sigma), U(1))$ of flat (unitary, complex) line bundles on $\Sigma$. These moduli spaces are what classical Chern-Simons theory assigns to $\Sigma$, and applying geometric quantization to these moduli spaces is one way to rigorously construct quantum Chern-Simons theory.
Proof 3: characteristic classes (and Poincaré duality)
For a closed surface $\Sigma$, the Euler characteristic $\bmod 2$ is equivalently the Stiefel-Whitney number $w_2 [\Sigma]$, where $w_2 \in H^2(\Sigma, \mathbb{F}_2)$ is the second Stiefel-Whitney class and $[\Sigma] \in H_2(\Sigma, \mathbb{F}_2)$ is the $\mathbb{F}_2$-fundamental class, which, as above, exists whether or not $\Sigma$ is orientable. In general, the top Stiefel-Whitney class $w_n$ of an $n$-dimensional real vector bundle is its $\bmod 2$ Euler class.
Proof 1 showed that this Stiefel-Whitney number is a cobordism invariant; in fact every Stiefel-Whitney number is a cobordism invariant, although we will not use this. In any case, to show that the Euler characteristic of $\Sigma$ is even when $\Sigma$ is orientable it suffices to show that $w_2 = 0$.
Proposition: Let $\Sigma$ be a closed orientable surface. Then $w_2 = 0$.
Proof. We will again appeal to the relationship between the Stiefel-Whitney classes and the Wu classes $\nu_k \in H^k(-, \mathbb{F}_2)$. Since $\Sigma$ is orientable, $\nu_1 = w_1 = 0$, so $\nu_2 = w_2$, where $\nu_2$ represents the second Steenrod square $\text{Sq}^2 : H^0(\Sigma, \mathbb{F}_2) \to H^2(\Sigma, \mathbb{F}_2)$ in the sense that
$\displaystyle x \nu_2 = \text{Sq}^2(x), x \in H^0(\Sigma, \mathbb{F}_2)$
where $x \nu_2$ denotes the cup product of $x$ and $\nu_2$. But $\text{Sq}^k$ vanishes on classes of degree less than $k$, so $\text{Sq}^2(x) = 0$ above, hence (by Poincaré duality $\bmod 2$) $\nu_2 = 0$ as well. $\Box$
Corollary: The Euler characteristic of $\Sigma$ is even.
Corollary: $\Sigma$ admits a spin structure.
Remark. Atiyah observed that spin structures on $\Sigma$ turn out to be equivalent to theta characteristics, after picking a complex structure. See Akhil Mathew’s blog post on this topic for more.
So we’ve shown that all of the Stiefel-Whitney classes of $\Sigma_g$ vanish. It follows that all of the Stiefel-Whitney numbers of $\Sigma_g$ vanish, and this is known to be a necessary and sufficient criterion for $\Sigma_g$ to be a boundary, a fact which we used in Proof 1. Essentially the same argument shows that all of the Stiefel-Whitney classes of a closed orientable $3$-manifold vanish, so all of the Stiefel-Whitney numbers vanish, and we get the less trivial fact that all closed orientable $3$-manifolds are boundaries. We also get that they all admit spin structures.
In the next two proofs we’ll finally stop using Poincaré duality, but now we’ll start using the fact that $\Sigma$ admits not only an orientation but a complex structure.
Proof 4: the Hodge decomposition
Any compact orientable surface $\Sigma$ can be given the structure of a compact Riemann surface, and so in particular the structure of a compact Kähler manifold, with Kähler metric inherited from any embedding into $\mathbb{CP}^n$ with the Fubini-Study metric. For any compact Kähler manifold $X$, its complex cohomology $H^k(X, \mathbb{C})$ has a Hodge decomposition
$\displaystyle H^k(X, \mathbb{C}) \cong \bigoplus_{p+q=k} H^{p, q}(X)$
where $H^{p, q}(X)$ is equivalently either the subspace of $H^k(X, \mathbb{C})$ represented by complex differential forms of type $(p, q)$ or the Dolbeault cohomology group
$\displaystyle H^{p, q}(X) \cong H^q(X, \Omega^p)$.
Here $\Omega^p$ is the sheaf of holomorphic $p$-forms and the cohomology being taken is sheaf cohomology. Moreover, since $H^k(X, \mathbb{C}) \cong H^k(X, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C}$, the LHS has a notion of complex conjugate, hence we can define the complex conjugate of a subspace, and with respect to this complex structure we have Hodge symmetry: $\overline{H^{p, q}(X)} = H^{q, p}(X)$. This implies the following.
Proposition: Let $X$ be a compact Kähler manifold (e.g. a smooth projective algebraic variety over $\mathbb{C}$). If $k$ is odd, then the Betti number $b_k$ is even.
Proof. Let $h^{p, q} = \dim H^{p, q}(X)$ be the Hodge number of $X$. The Hodge decomposition implies that
$\displaystyle b_k = \sum_{p+q=k} h^{p, q}$
and Hodge symmetry implies that $h^{p, q} = h^{q, p}$. When $k$ is odd, every term $h^{p, q}$ in the above sum is paired with a different term $h^{q, p}$ equal to it, hence $b_k \equiv 0 \bmod 2$ as desired. $\Box$
Corollary: The Euler characteristic of $\Sigma$ is even.
Proof. As before, we have $\chi(\Sigma) = b_0 - b_1 + b_2 = 2 - b_1$, and $b_1$ is even by the above. $\Box$
Corollary: Let $G$ be a finitely presented group. If $G$ has a finite index subgroup $H$ such that the first Betti number
$\displaystyle b_1(H) = \dim H^1(H, \mathbb{Q}) = \dim \text{Hom}(H, \mathbb{Q})$
of $H$ is odd, then $G$ cannot be the fundamental group of a compact Kähler manifold, and in particular cannot be the fundamental group of a smooth projective complex variety.
Fundamental groups of compact Kähler manifolds are called Kähler groups; see these two blog posts by Danny Calegari for more.
Proof. Since a finite cover of a compact Kähler manifold is naturally a compact Kähler manifold, if $G$ is a Kähler group then so are all of its finite index subgroups; taking the contrapositive, if any of the finite index subgroups of $G$ are not Kähler, then neither is $G$. If $X$ is any space with $\pi_1(X) = H$, then $b_1(X) = b_1(H)$, hence the former is odd iff the latter is. It follows that if $H$ is the fundamental group of a compact Kähler manifold then $b_1(H)$ is even; taking the contrapositive, we get the desired result. $\Box$
Example. The free abelian groups $\mathbb{Z}^{2k+1}$ of odd rank have first Betti number $2k + 1$ and hence are not Kähler groups. On the other hand, the free abelian groups $\mathbb{Z}^{2k}$ of even rank are the fundamental groups of complex tori $\mathbb{C}^k / \Gamma$ (e.g. products of elliptic curves).
Example. The free groups $F_{2k+1}$ of odd rank have first Betti number $2k + 1$ and hence are not Kähler groups. The free groups of even rank $F_{2k}$ turn out to have free groups of odd rank as finite index subgroups and hence are also not Kähler.
To see this, first note that if $F_n$ is any free group, then $F_n$ admits finite index subgroups of every possible index because it is possible to write down surjections from $F_n$ into finite groups of every possible size (e.g. cyclic groups). Second, by the standard topological argument every finite index subgroup of $F_n$ is again free because every finite cover of the wedge of $n$ circles is a graph and hence homotopy equivalent to a wedge of circles; moreover, by the multiplicativity of Euler characteristics under coverings, if $F_m$ is an index $k$ subgroup of $F_n$ then
$\displaystyle \chi(F_m) = 1 - m = k \chi(F_n) = k(1 - n)$
and hence $F_n$ has subgroups of index $k$ and first Betti number
$\displaystyle b_1(F_m) = m = 1 + k(n - 1)$
for all $k \in \mathbb{N}$. This is odd whenever $k$ is even, and in particular when $k = 2$. More explicitly, if $F_n$ is free on generators $g_1, \dots g_n$, then
$\displaystyle F_n \ni g_i \mapsto 1 \in \mathbb{Z}_2$
is a surjection onto a finite group of order $2$, and hence its kernel must be free on $1 + 2(n - 1) = 2n - 1$ generators. One possible choice of generators is
$g_1^2, g_2^2, \dots g_n^2, g_1 g_2, g_1 g_3, \dots g_1 g_n$.
Corollary: The fundamental groups $\pi_1(\Sigma_g)$ of compact Riemann surfaces are not free.
There is a great MO question on the topic of why $\pi_1(\Sigma_g)$ is not free in which this argument is given in the comments. As it happens, that MO question loosely inspired this post.
Above, instead of using Hodge symmetry, we can also do the following. In the particular case of surfaces $\Sigma_g$, we in fact have $b_1 = 2g = h^{0, 1} + h^{1, 0}$, hence the two interesting Hodge numbers are
$\displaystyle h^{0, 1} = h^{1, 0} = g$.
In terms of Dolbeault cohomology, this gives
$\displaystyle \dim H^1(\Sigma, \Omega^0) = \dim H^0(\Sigma, \Omega^1) = g$.
Here $\Omega^0$ is the sheaf of holomorphic $0$-forms, or equivalently the structure sheaf $\mathcal{O}_{\Sigma}$ of holomorphic functions.
The identity $\dim H^0(\Sigma, \Omega^1) = g$ gives us one possible definition of the genus of a compact Riemann surface, namely the dimension of the space of holomorphic forms. In general, if $X$ is a complex manifold we can define its geometric genus to be the Hodge number $h^{n, 0} = \dim H^0(X, \Omega_X^n)$, where $\Omega_X^n$ is the canonical bundle, hence the dimension of the space of top forms.
The identity $\dim H^1(\Sigma, \Omega^0) = g$ can be thought of in terms of Hodge symmetry, but it can also be thought of in terms of Serre duality. On the Dolbeault cohomology groups of a compact complex manifold $X$ of complex dimension $n$, Serre duality gives an identification
$\displaystyle H^q(X, \Omega^p) \cong H^{n-q}(X, \Omega^{n-p})^{\ast}$
and hence $h^{p, q} = h^{n-p, n-q}$, which is a different symmetry of the Hodge numbers than Hodge symmetry gives. When $X$ is Kähler, in terms of the Hodge decomposition Serre duality refines Poincaré duality, which only gives
$\displaystyle H^k(X, \mathbb{C}) \cong H^{n-k}(X, \mathbb{C})^{\ast}$.
In particular, we have
$\displaystyle H^1(\Sigma, \Omega^0) \cong H^0(\Sigma, \Omega^1)^{\ast}$
which gives a second proof, independent of Hodge symmetry but still depending on the Hodge decomposition, that $b_1$ is even.
Moreover, since Serre duality is a refinement of Poincaré duality we conclude that $H^1(\Sigma, \mathbb{C})$ is, as a symplectic vector space (as in Proof 2), isomorphic (possibly up to a scalar) to $V \oplus V^{\ast}$ with its standard symplectic structure
$\displaystyle \omega(v_1 \oplus f_1, v_2 \oplus f_2) = f_2(v_1) - f_1(v_2)$
where $V$ is either $H^1(\Sigma, \Omega^0)$ or $H^0(\Sigma, \Omega^1)$. Hence a complex structure on $\Sigma$ equips the symplectic vector space $H^1(\Sigma, \mathbb{C})$ with a Lagrangian subspace.
Digression: the Riemann-Roch theorem
The motivation for the fifth proof starts from the observation that one way to write down the Riemann-Roch theorem for compact Riemann surfaces $\Sigma$ is
$\displaystyle \ell(D) - \ell(K - D) = \deg D + \frac{\chi(\Sigma)}{2}$.
If we can write down a proof of the Riemann-Roch theorem with the genus $g$ appearing directly in this form, in terms of half the Euler characteristic, as opposed to the other ways the genus can appear in a formula involving Riemann surfaces (e.g. as the dimension of the space of holomorphic forms), then since all of the other terms are manifestly integers we would get a proof that $\chi(\Sigma)$ is even.
Here is a proof which does not accomplish this. Let $\mathcal{O}(D)$ denote the line bundle associated to the divisor $D$. Then
$\ell(D) = \dim H^0(\Sigma, \mathcal{O}(D))$
and
$\ell(K - D) = \dim H^0(\Sigma, \mathcal{O}(D)^{\ast} \otimes \Omega)$
since $K$ is the divisor corresponding to the canonical bundle $\Omega$ and $\mathcal{O}(D)^{\ast} \cong \mathcal{O}(-D)$. Now Serre duality gives
$\displaystyle H^1(\Sigma, \mathcal{O}(D)) \cong H^0(\Sigma, \mathcal{O}(D)^{\ast} \otimes \Omega)^{\ast}$
and hence we can rewrite the LHS as an Euler characteristic
$\displaystyle \chi(\mathcal{O}(D)) = \dim H^0(\Sigma, \mathcal{O}(D)) - \dim H^1(\Sigma, \mathcal{O}(D))$
where we are using that the cohomology of sheaves on $\Sigma$ vanishes above its complex dimension, namely $1$. This lets us rewrite Riemann-Roch in the form
$\displaystyle \chi(\mathcal{O}(D)) = \deg D + \frac{\chi(\Sigma)}{2}$.
Let $D = \sum n_p p$ and let $p \in \Sigma$ be a point, so that the meromorphic functions in $H^0(\Sigma, \mathcal{O}(D))$ can have poles of order at most $n_p$ at $p$. Then there is an evaluation map
$\displaystyle \mathcal{O}(D) \to \mathbb{C}_p$
given by taking the coefficient of $z^{-n_p}$ where $z$ is a local coordinate at $p$; here $\mathbb{C}_p$ denotes the skyscraper sheaf supported at $p$ with stalk $\mathbb{C}$. The kernel of this evaluation map consists of functions in $\mathcal{O}(D)$ which have poles of order at most $n_p - 1$ at $p$, which are precisely the sections of the sheaf $\mathcal{O}(D - p)$. Hence we have a short exact sequence of sheaves
$\displaystyle 0 \to \mathcal{O}(D - p) \to \mathcal{O}(D) \to \mathbb{C}_p \to 0$.
Since the Euler characteristic of sheaf cohomology is additive in short exact sequences, it follows that
$\displaystyle \chi(\mathcal{O}(D)) = \chi(\mathcal{O}(D - p)) + \chi(\mathbb{C}_p)$.
Since $H^0(\Sigma, \mathbb{C}_p) \cong \mathbb{C}$ and, being a skyscraper sheaf, $\mathbb{C}_p$ has no higher sheaf cohomology, we have $\chi(\mathbb{C}_p) = 1$, hence
$\displaystyle \chi(\mathcal{O}(D)) = \chi(\mathcal{O}(D - p)) + 1$.
Noting that we also have $\deg D = \deg (D - p) + 1$, by adding and removing points suitably we conclude that if $D_1, D_2$ are any two divisors, then
$\displaystyle \chi(\mathcal{O}(D_1)) - \chi(\mathcal{O}(D_2)) = \deg D_1 - \deg D_2$
or equivalently that there is a constant $c$ such that
$\displaystyle \chi(\mathcal{O}(D)) = \deg D + c$
for all divisors $D$. To determine $c$ it suffices to determine the Euler characteristic of any of the sheaves $\mathcal{O}(D)$, which we can do with a second application of Serre duality: for $D = 0$, so that $\mathcal{O}(D) \cong \mathcal{O}_{\Sigma}$ is the structure sheaf, we have
$\displaystyle \dim H^0(\Sigma, \mathcal{O}_{\Sigma}) = 1$
since the holomorphic functions on a compact Riemann surface are constant, and
$\displaystyle \dim H^1(\Sigma, \mathcal{O}_{\Sigma}) = \dim H^0(\Sigma, \Omega^1) = g$
by Serre duality and the definition of $g$ in terms of holomorphic forms. Hence
$\displaystyle \chi(\mathcal{O}_{\Sigma}) = 1 - g$
from which it follows that $c = 1 - g$. This proves Riemann-Roch, but $1 - g$ appears as the holomorphic Euler characteristic of $\Sigma$ rather than as half the topological Euler characteristic like we wanted. The two can be related using the Hodge decomposition, which shows more generally that for $X$ a compact Kähler manifold of complex dimension $n$,
$\displaystyle \chi(X) = \sum_{k=0}^{2n} (-1)^k \dim H^k(X, \mathbb{C}) = \sum_{k=0}^{2n} (-1)^k b_k$
can be written in terms of Hodge numbers as
$\displaystyle \sum_{k=0}^{2n} (-1)^k \sum_{p+q=k} h^{p, q} = \sum_{p=0}^n (-1)^p \sum_{q=0}^n (-1)^q h^{p, q}$
which we can further rewrite as an alternating sum of Euler characteristics
$\displaystyle \sum_{p=0}^n (-1)^p \sum_{q=0}^n (-1)^q \dim H^q(X, \Omega^p) = \sum_{p=0}^n (-1)^p \chi(\Omega^p).$
Abstractly this identity reflects the fact that the sheaves $\Omega^p$ together form a resolution of the constant sheaf $\mathbb{C}$, just as in the case of smooth differential forms on a smooth manifold. However, in the smooth case, the sheaves of smooth differential forms do not themselves have any higher sheaf cohomology, whereas in the complex case, the sheaves of holomorphic differential forms do in general have higher cohomology. This resolution also exists on any complex manifold, not necessarily compact or Kähler. It gives rise to the Hodge-to-de Rham (or Frölicher) spectral sequence in general, and the existence of the Hodge decomposition reflects the fact that on compact Kähler manifolds this spectral sequence degenerates.
Returning to the case of a compact Riemann surface $\Sigma$, we get that
$\displaystyle \chi(\Sigma) = \chi(\Omega^0) - \chi(\Omega^1)$
but by Serre duality $\chi(\Omega^1) = - \chi(\Omega^0)$, hence
$\displaystyle \chi(\Sigma) = \chi(\Omega^0) + \chi(\Omega^0) = 2 \chi(\mathcal{O}_{\Sigma})$.
Hence the topological Euler characteristic of $\Sigma$ is twice its holomorphic Euler characteristic. This argument not only shows that the topological Euler characteristic is even but gives an interpretation of the number obtained by dividing it by $2$.
But we used the Hodge decomposition and Serre duality already, so let’s do something else.
Proof 5: the Hirzebruch-Riemann-Roch theorem
The Riemann-Roch theorem has the following more general form. Let $V$ be a holomorphic vector bundle on a compact complex manifold $X$ of complex dimension $n$. Let
$\displaystyle \chi(V) = \sum_{k=0}^n (-1)^k \dim H^k(X, V)$
denote the Euler characteristic of the sheaf of holomorphic sections of $V$, as we did above for line bundles. Let $\text{ch}(V)$ denote the Chern character of $V$, which is defined via the splitting principle as
$\displaystyle \text{ch}(L_1 \oplus \dots \oplus L_k) = \sum_{i=1}^k e^{c_1(L_k)} \in H^{\bullet}(X, \mathbb{Q})$
for a direct sum of complex line bundles. $\text{ch}(V)$ can be written in terms of the Chern classes $c_n(V)$ using the fact that the total Chern class
$\displaystyle H^{\bullet}(X, \mathbb{Z}) \ni c(V) = \sum_{k=0}^n c_k(V), c_k(V) \in H^{2k}(X, \mathbb{Z})$
can be defined via the splitting principle as
$\displaystyle c(L_1 \oplus \dots \oplus L_k) = \prod_{i=1}^k (1 + c_1(L_i))$.
Equivalently, $c_n$ is the $n^{th}$ elementary symmetric function in the Chern roots $\alpha_i = c_1(L_i)$. Expanding out the definition of $\text{ch}(V)$ gives power symmetric functions of the Chern roots which we can write as a polynomial in the elementary symmetric functions, e.g. using Newton’s identities, hence as a polynomial in the Chern classes. The first three terms are
$\displaystyle \text{ch}(V) = \dim V + c_1(V) + \frac{c_2(V) - c_1(V)^2}{2} + \dots$.
Similarly, let $\text{td}(X)$ denote the Todd class of (the tangent bundle of) $X$, which is defined via the splitting principle as
$\displaystyle \text{td}(L_1 \oplus \dots \oplus L_k) = \prod_{i=1}^k \frac{c_1(L_i)}{1 - e^{-c_1(L_i)}} \in H^{\bullet}(X, \mathbb{Q})$
for a direct sum of complex line bundles. Again we can use symmetric function identities to write $\text{td}(X)$ in terms of the Chern classes $c_n(X)$ of (the tangent bundle of) $X$. The first three terms are
$\displaystyle \text{td}(X) = 1 + \frac{c_1(X)}{2} + \frac{c_2(X) + c_1(X)^2}{12} + \dots$.
Finally, suppose that
$\displaystyle H^{\bullet}(X, \mathbb{Q}) \ni \alpha = \sum_{k=0}^{2n} \alpha_k, \alpha_k \in H^k(X, \mathbb{Q})$
is a (mixed) cohomology class, and let
$\displaystyle \int_X \alpha = \alpha [X] = \alpha_{2n}[X] \in \mathbb{Q}$
denote the pairing of the degree $2n$ part of $\alpha$ with the fundamental class $[X] \in H_{2n}(X, \mathbb{Q})$.
Theorem (Hirzebruch-Riemann-Roch): With hypotheses as above, the Euler characteristic $\chi(V)$ satisfies
$\displaystyle \chi(V) = \int_X \text{ch}(V) \text{td}(X)$.
We’ll make no attempt to prove this, but here are some notable features of this theorem.
First, 1) $\text{ch}(V)$ only depends on the isomorphism class of $V$ as a topological, rather than holomorphic, complex vector bundle, 2) $\text{td}(X)$ only depends on the isomorphism class of the tangent bundle of $X$ as a topological complex vector bundle, and 3) $\int_X$ only depends on the orientation of $X$ coming from the complex structure on its tangent bundle. In other words, the RHS consists of topological, rather than holomorphic, data. This reflects the way the Hirzebruch-Riemann-Roch theorem is a special case of the Atiyah-Singer index theorem.
In addition, the RHS is a rational linear combination of certain characteristic numbers, hence is a priori rational, but Hirzebruch-Riemann-Roch tells us that it is in fact an integer. This implies divisibility relations which substantially generalize the divisibility relation we’re looking for, namely that $2 | \chi(\Sigma)$.
Corollary (Riemann-Roch): Let $L = \mathcal{O}(D)$ be a holomorphic line bundle on a compact Riemann surface $\Sigma$. Then
$\displaystyle \chi(L) = \left( c_1(L) + \frac{c_1(\Sigma)}{2} \right) [\Sigma] = \deg D + \frac{\chi(\Sigma)}{2}$.
In particular, the holomorphic Euler characteristic satisfies $\chi(\mathcal{O}_{\Sigma}) = \frac{\chi(\Sigma)}{2}$.
Proof. In general, the top Chern class $c_n$ of an $n$-dimensional complex vector bundle is its Euler class $e$. In particular, $c_1(\Sigma)$ is the Euler class $e(\Sigma)$, hence $e(\Sigma) [\Sigma] = \chi(\Sigma)$.
It remains to show that $c_1(L) [\Sigma] = \deg D$. Morally speaking this is because if $D = \sum n_p p$ then $c_1(L) \in H^2(\Sigma, \mathbb{Z})$ is Poincaré dual to $\sum n_p p \in H_0(\Sigma, \mathbb{Z})$, which is morally the vanishing locus of a generic section of $L$. But I am not sure how to make this precise easily. An unsatisfying proof that gets the job done is to use the same additivity argument involving skyscraper sheaves as in the previous proof of Riemann-Roch to conclude that $c_1(L) [\Sigma] = \deg D + c$ for some constant $c$ and then to note that, since $\mathcal{O}(0)$ is topologically the trivial line bundle, $c_1(\mathcal{O}(0)) [\Sigma] = 0$, hence $c = 0$. $\Box$
Corollary: The holomorphic Euler characteristic $\chi(\mathcal{O}_X)$ is equal to the Todd genus of $X$:
$\displaystyle \chi(\mathcal{O}_X) = \int_X \text{td}(X)$.
Proof. The underlying topological line bundle of the structure sheaf $\mathcal{O}_X$ is the trivial line bundle, and hence has trivial Chern character. $\Box$
In particular, $\chi(\mathcal{O}_X)$ only depends on the Chern numbers of $X$. These are known to be complex cobordism invariants, and in fact the Todd genus is a genus: it gives a ring homomorphism
$\displaystyle \text{td} : MU_{\bullet}(\text{pt}) \to \mathbb{Z}$
where $MU_{\bullet}(\text{pt})$ is the complex cobordism ring and $MU$ is the Thom spectrum for complex cobordism.
In the next dimension up (complex dimension $2$, real dimension $4$), the Hirzebruch-Riemann-Roch theorem gives the following divisibility relation.
Corollary (Noether’s formula): The holomorphic Euler characteristic of a compact complex surface $X$ satisfies
$\displaystyle \chi(\mathcal{O}_X) = \left( \frac{c_2(X) + c_1(X)^2}{12} \right) [X]$.
In particular, the RHS is an integer.
Corollary: If $X$ is a compact complex surface with $c_1 = 0$ (in particular if $X$ is Calabi-Yau; the converse holds if $X$ is Kähler), then $c_2(X) [X] = \chi(X) \equiv 0 \bmod 12$.
Examples include the hypersurface of degree $4$ in $\mathbb{CP}^3$ (as we saw previously), and more generally any K3 surface, with Euler characteristic $24$.
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Instructional Focus DocumentGeometry
TITLE : Unit 10: Probability SUGGESTED DURATION : 12 days
#### Unit Overview
This unit bundles student expectations that address permutations, combinations, and various types of probability. Concepts are incorporated into both non-contextual and real-world problem situations. According to the Texas Education Agency, mathematical process standards including application, tools and techniques, communication, representations, relationships, and justifications should be integrated (when applicable) with content knowledge and skills so that students are prepared to use mathematics in everyday life, society, and the workplace.
Prior to this unit, in Grade 7, students determined theoretical and experimental probabilities for simple and compound events. Students also described and found the complement of a probability.
During this unit, students define and develop formulas for permutations and combinations and apply permutations and combinations to solve contextual problems. Students determine number of possible outcomes of an event, including combinations, permutations, and the Fundamental Counting Principle. Students investigate and define probability in terms of possible outcomes and desired event. Students determine the theoretical and experimental probability based on area models in problem situations. Students identify events as being independent or dependent, and connect independence and dependence to with and without replacement. Students apply the concept of probabilities of independent and dependent events to solve contextual problems. Students are introduced to conditional probability, including notation, and apply conditional probability in contextual problems.
After this unit, the concepts of permutations, combinations, and probability will be applied in subsequent mathematics courses and real world situations.
This unit is supporting the development of Texas College Career Readiness Standards (TxCCRS): III. Geometric Reasoning C2; VIII. Problem Solving and Reasoning; IX. Communication and Representation; X. Connections.
According to Navigating through Probability in Grades 9 – 12 (2004), from the National Council of Teachers of Mathematics (NCTM), “Probability is an area of the school mathematics curriculum that high school courses often glossed over or sometimes even skip. However, the importance of probabilistic reasoning – particularly, reasoning about the relationship between data and probability – has received increased attention from national groups over the past several decades” (p. 1). Additionally, “The Curriculum and Evaluation Standards for School Mathematics (NCTM, 1989) accorded to probability and statistics the same status in the K – 12 mathematics curriculum as algebra, geometry, measurement, and number and operations. Eleven years later, Principals and Standards for School Mathematics (NCTM, 2000), reaffirmed this assessment of the importance of data analysis and probability in the Pre-K – 12 mathematics curriculum” (as cited in NCTM, 2004, p. 1).
National Council of Teachers of Mathematics. (1989). Curriculum and evaluation standards for school mathematics. Reston, VA: National Council of Teachers of Mathematics, Inc.
National Council of Teachers of Mathematics. (2000). Principles and standards for school mathematics. Reston, VA: National Council of Teachers of Mathematics, Inc.
National Council of Teachers of Mathematics. (2004). Navigating through probability in grades 9 – 12. Reston, VA: National Council of Teachers of Mathematics, Inc.
#### OVERARCHING UNDERSTANDINGS and QUESTIONS
The probability or chance of an event occurring can be determined by various methods, interpreted for reliability, and used to make predictions and inferences in problem situations.
• Why is it important to understand and use probability?
• How are the different types of probability distinguished?
• What methods can be used to determine probability?
• How is the probability of an event(s) used to make predictions and inferences in problem situations?
Combinatorics is a branch of discrete mathematics involving enumeration, combination, and permutation of sets of numbers and their application in problem situations.
• How can combinatorics be used to count the number of structures possible for different types and sizes of sets?
• How are combinatorics applied in probability?
• How can combinatorics be used to solve problem situations?
Performance Assessment(s) Overarching Concepts
Unit Concepts
Unit Understandings
Assessment information provided within the TEKS Resource System are examples that may, or may not, be used by your child’s teacher. In accordance with section 26.006 (2) of the Texas Education Code, "A parent is entitled to review each test administered to the parent’s child after the test is administered." For more information regarding assessments administered to your child, please visit with your child’s teacher.
Probabilistic Reasoning
• Conclusions/Predictions
• Independent/Dependent Events
• Outcomes
• Permutations/Combinations
• Probability of an Event
• Sample Space
• Simulations
• Conditional/Theoretical/Empirical Probability
Associated Mathematical Processes
• Application
• Tools and Techniques
• Problem Solving Model
• Communication
• Representations
• Relationships
• Justification
Combinations and permutations are used to determine the number of total possible outcomes for a set and are applied to solve problem situations, including probability problems.
• How are combinations used to determine total possible outcomes for a set?
• What formula is used to model combinations?
• How are permutations used to determine total possible outcomes for a set?
• What formula is used to model permutations?
• How are combinations and permutations different?
• Why are combinations and permutations used in determining probability?
• How can combinations and permutations be used to solve contextual problems?
Probability is the likelihood of an event occurring from the total possible outcomes, and probability can be applied to solve mathematical and real-world problem situations.
• How can total, possible outcomes or sample space be determined?
• How is probability of an event determined?
• How is probability of an event represented?
• How can probability based on area be used to solve contextual problems?
• How can two events be identified as independent?
• How is the probability of independent events applied in contextual problems?
• How do with replacement or without replacement factor into determining if two events are independent or dependent?
• How can conditional probabilities be identified?
• How is the probability of conditional events applied in contextual problems?
Theoretical probability is an expected probability calculated using formulas, whereas experimental probability is found by experimentation and/or data collection.
• What is the purpose of determining the probability of the occurrence of an event?
• How do theoretical probability and experimental probability compare?
• What conditions might necessitate the use of theoretical probability?
• What conditions might necessitate the use of experimental probability?
• Why might the theoretical probability and experimental probability of a particular event differ?
• How does the number of trials in an experiment affect the validity of the experimental probability?
• How can a theoretical probability be determined from an area model?
• How can an experimental probability be simulated using an area model?
#### MISCONCEPTIONS / UNDERDEVELOPED CONCEPTS
Misconceptions:
• Some students may confuse when to use permutations and when to use combinations to determine outcomes.
• Some students may not realize that an event is a subset of the total possible outcomes.
• Some students may confuse independent events and dependent events.
• Some students may think that if objects are not replaced in the set it does not impact the probability instead of realizing that the total outcomes for the next probability will be decreased.
#### Unit Vocabulary
• Combinations – number of different ways a set of objects can be selected without regard to a specific order
• Complement of an event – the probability of the non-occurrence of a desired outcome; the sum of the probability of the event and the probability of the non-occurrence of the event are equal to one
• Compound events – a set of outcomes from a combination of actions or activities where the outcomes can be subdivided (e.g., flipping a coin and rolling a number cube, drawing tiles out of a bag and spinning a spinner, etc.)
• Conditional probability – for two events, A and B, the probability of B given that A has already occurred; written as P(B|A); read as “the probability of B, given A”
• Dependent events – the outcome from one action or activity may affect the probability of the outcome(s) of any subsequent action(s) or activity(s); usually involves compound events
• Experimental probability – the likelihood of an event occurring from the outcomes of an experiment
• Independent events – the outcome from one action or activity does not affect the probability of the outcome(s) of any subsequent action(s) or activity(s); usually involves compound events
• Independent probability – if P(B|A) = P(B), then A and B are said to be independent
• Permutations – number of different ways a set of objects can be selected with regard to a specific order
• Probability – a ratio between the number of desired outcomes to the total possible outcomes, 0 ≤ p ≤ 1
• Sample space – a set of all possible outcomes of one or more events
• Theoretical probability – the likelihood of an event occurring predicted by using formulas and mathematical calculations without conducting an experiment
Related Vocabulary:
Equally likely Event Experiment Factorial Favorable outcomes Outcome Tree diagram
Unit Assessment Items System Resources Other Resources
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Texas Higher Education Coordinating Board – Texas College and Career Readiness Standards (select CCRS from Standard Set dropdown menu)
Texas Education Agency – Mathematics Curriculum
Texas Education Agency – STAAR Mathematics Resources
Texas Education Agency Texas Gateway – Revised Mathematics TEKS: Vertical Alignment Charts
Texas Education Agency Texas Gateway – Mathematics TEKS: Supporting Information
Texas Education Agency Texas Gateway – Interactive Mathematics Glossary
Texas Education Agency Texas Gateway – Resources Aligned to Geometry Mathematics TEKS
Texas Instruments – Graphing Calculator Tutorials
TEKS# SE# TEKS Unit Level Specificity
• Bold black text in italics: Knowledge and Skills Statement (TEKS)
• Bold black text: Student Expectation (TEKS)
• Strike-through: Indicates portions of the Student Expectation that are not included in this unit but are taught in previous or future unit(s)
• Blue text: Supporting information / Clarifications from TCMPC (Specificity)
• Blue text in italics: Unit-specific clarification
• Black text: Texas Education Agency (TEA); Texas College and Career Readiness Standards (TxCCRS)
G.1 Mathematical process standards. The student uses mathematical processes to acquire and demonstrate mathematical understanding. The student is expected to:
G.1A Apply mathematics to problems arising in everyday life, society, and the workplace.
Apply
MATHEMATICS TO PROBLEMS ARISING IN EVERYDAY LIFE, SOCIETY, AND THE WORKPLACE
Including, but not limited to:
• Mathematical problem situations within and between disciplines
• Everyday life
• Society
• Workplace
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• X. Connections
G.1B Use a problem-solving model that incorporates analyzing given information, formulating a plan or strategy, determining a solution, justifying the solution, and evaluating the problem-solving process and the reasonableness of the solution.
Use
A PROBLEM-SOLVING MODEL THAT INCORPORATES ANALYZING GIVEN INFORMATION, FORMULATING A PLAN OR STRATEGY, DETERMINING A SOLUTION, JUSTIFYING THE SOLUTION, AND EVALUATING THE PROBLEM-SOLVING PROCESS AND THE REASONABLENESS OF THE SOLUTION
Including, but not limited to:
• Problem-solving model
• Analyze given information
• Formulate a plan or strategy
• Determine a solution
• Justify the solution
• Evaluate the problem-solving process and the reasonableness of the solution
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• VIII. Problem Solving and Reasoning
G.1C Select tools, including real objects, manipulatives, paper and pencil, and technology as appropriate, and techniques, including mental math, estimation, and number sense as appropriate, to solve problems.
Select
TOOLS, INCLUDING REAL OBJECTS, MANIPULATIVES, PAPER AND PENCIL, AND TECHNOLOGY AS APPROPRIATE, AND TECHNIQUES, INCLUDING MENTAL MATH, ESTIMATION, AND NUMBER SENSE AS APPROPRIATE, TO SOLVE PROBLEMS
Including, but not limited to:
• Appropriate selection of tool(s) and techniques to apply in order to solve problems
• Tools
• Real objects
• Manipulatives
• Paper and pencil
• Technology
• Techniques
• Mental math
• Estimation
• Number sense
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• VIII. Problem Solving and Reasoning
G.1D Communicate mathematical ideas, reasoning, and their implications using multiple representations, including symbols, diagrams, graphs, and language as appropriate.
Communicate
MATHEMATICAL IDEAS, REASONING, AND THEIR IMPLICATIONS USING MULTIPLE REPRESENTATIONS, INCLUDING SYMBOLS, DIAGRAMS, GRAPHS, AND LANGUAGE AS APPROPRIATE
Including, but not limited to:
• Mathematical ideas, reasoning, and their implications
• Multiple representations, as appropriate
• Symbols
• Diagrams
• Graphs
• Language
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• IX. Communication and Representation
G.1E Create and use representations to organize, record, and communicate mathematical ideas.
Create, Use
REPRESENTATIONS TO ORGANIZE, RECORD, AND COMMUNICATE MATHEMATICAL IDEAS
Including, but not limited to:
• Representations of mathematical ideas
• Organize
• Record
• Communicate
• Evaluation of the effectiveness of representations to ensure clarity of mathematical ideas being communicated
• Appropriate mathematical vocabulary and phrasing when communicating mathematical ideas
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• IX. Communication and Representation
G.1F Analyze mathematical relationships to connect and communicate mathematical ideas.
Analyze
MATHEMATICAL RELATIONSHIPS TO CONNECT AND COMMUNICATE MATHEMATICAL IDEAS
Including, but not limited to:
• Mathematical relationships
• Connect and communicate mathematical ideas
• Conjectures and generalizations from sets of examples and non-examples, patterns, etc.
• Current knowledge to new learning
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• X. Connections
G.1G Display, explain, and justify mathematical ideas and arguments using precise mathematical language in written or oral communication.
Display, Explain, Justify
MATHEMATICAL IDEAS AND ARGUMENTS USING PRECISE MATHEMATICAL LANGUAGE IN WRITTEN OR ORAL COMMUNICATION
Including, but not limited to:
• Mathematical ideas and arguments
• Validation of conclusions
• Displays to make work visible to others
• Diagrams, visual aids, written work, etc.
• Explanations and justifications
• Precise mathematical language in written or oral communication
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxCCRS:
• IX. Communication and Representation
G.13 Probability. The student uses the process skills to understand probability in real-world situations and how to apply independence and dependence of events. The student is expected to:
G.13A Develop strategies to use permutations and combinations to solve contextual problems.
Develop
STRATEGIES TO USE PERMUTATIONS AND COMBINATIONS
Including, but not limited to:
• Permutations – number of different ways a set of objects can be selected with regard to a specific order
• Formula for permutations: , where n is the total number of objects in the set and r is the number to be chosen.
• Combinations – number of different ways a set of objects can be selected without regard to a specific order
• Formula for combinations: , where n is the total number of objects in the set and r is the number to be chosen.
• Strategies to determine permutations and combinations
• Diagram
• Lists
• Formulas
• Graphing calculator technology
To Solve
CONTEXTUAL PROBLEMS
Including, but not limited to:
• Application of permutations to determine the number of ways an event can occur in contextual problem situations
• Application of combinations to determine the number of ways an event can occur in contextual problem situations
Note(s):
• Grade 7 introduced simple and compound events and applications of probability.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxCCRS
• III. Geometric Reasoning
• C2 – Make connections between geometry, statistics, and probability.
• VIII. Problem Solving and Reasoning
• IX. Communication and Representation
• X. Connections
G.13B Determine probabilities based on area to solve contextual problems.
Determine
PROBABILITIES BASED ON AREA TO SOLVE CONTEXTUAL PROBLEMS
Including, but not limited to:
• Probability – a ratio between the number of desired outcomes to the total possible outcomes, 0 ≤ p ≤ 1
• Sample space – a set of all possible outcomes of one or more events
• Experimental probability – the likelihood of an event occurring from the outcomes of an experiment
• Theoretical probability – the likelihood of an event occurring predicted by using formulas and mathematical calculations without conducting an experiment
• Complement of an event – the probability of the non-occurrence of a desired outcome; the sum of the probability of the event and the probability of the non-occurrence of the event are equal to one
• Representation of probability with an area model
• Representation of real-world problem situations with area models
• Representation of data sets with an area model (circle graph)
• Determination of probability from an area model
• Hands-on and technology to model and simulate events
Note(s):
• Grade 7 introduced simple and compound events and applications of probability.
• Geometry extends the concept of probabilities based on area.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxCCRS
• III. Geometric Reasoning
• C2 – Make connections between geometry, statistics, and probability.
• VIII. Problem Solving and Reasoning
• IX. Communication and Representation
• X. Connections
G.13C Identify whether two events are independent and compute the probability of the two events occurring together with or without replacement.
Identify
WHETHER TWO EVENTS ARE INDEPENDENT
Including, but not limited to:
• Compound events – a set of outcomes from a combination of actions or activities where the outcomes can be subdivided (e.g., flipping a coin and rolling a number cube, drawing tiles out of a bag and spinning a spinner, etc.)
• Independent events – the outcome from one action or activity does not affect the probability of the outcome(s) of any subsequent action(s) or activity(s); usually involves compound events(s)
• Dependent events – the outcome from one action or activity may affect the probability of the outcome(s) of any subsequent action(s) or activity(s); usually involves compound events
Compute
THE PROBABILITY OF THE TWO EVENTS OCCURRING TOGETHER WITH OR WITHOUT REPLACEMENT
Including, but not limited to:
• Probability – a ratio between the number of desired outcomes to the total possible outcomes, 0 ≤ p ≤ 1
• Compound independent events with replacement
• Compound dependent events without replacement
Note(s):
• Grade 7 introduced simple and compound events and applications of probability.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxCCRS
• III. Geometric Reasoning
• C2 – Make connections between geometry, statistics, and probability.
• VIII. Problem Solving and Reasoning
• IX. Communication and Representation
• X. Connections
G.13D Apply conditional probability in contextual problems.
Apply
CONDITIONAL PROBABILITY IN CONTEXTUAL PROBLEMS
Including, but not limited to:
• Conditional probability – for two events, A and B, the probability of B given that A has already occurred; written as P(B|A); read as “the probability of B, given A”
• Application of conditional probability in contextual problems
Note(s):
• Grade 7 introduced simple and compound events and applications of probability.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxCCRS
• III. Geometric Reasoning
• C2 – Make connections between geometry, statistics, and probability.
• VIII. Problem Solving and Reasoning
• IX. Communication and Representation
• X. Connections
G.13E Apply independence in contextual problems.
Apply
INDEPENDENCE IN CONTEXTUAL PROBLEMS
Including, but not limited to:
• Application of independence in contextual problems.
• Independent events in terms of conditional probability
• Conditional probability – for two events, A and B, the probability of B given that A has already occurred; written as P(B|A); read as “the probability of B, given A”
• Independent probability – if P(B|A) = P(B), then A and B are said to be independent
• Distinguishing between events that are independent and dependent using contextual problems involving conditional probability
• Computing compound probabilities based on conditional events, using formulas such as P(A and B) = P(A) • P(B|A).
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http://docplayer.net/43825980-A-i-x-a-i-a-i-c-i-0-this-also-works-naturally-for-the-integers-given-a-positive-integer-m-and-a-prime-p-we-can-write-it-in-base-p-that-is-m.html | 1,529,580,382,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00119.warc.gz | 91,571,198 | 28,994 | # a i (X α) i, a i C. i=0 This also works naturally for the integers: given a positive integer m and a prime p, we can write it in base p, that is m =
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1 Chapter 5 p-adic numbers The p-adic numbers were first introduced by the German mathematician K. Hensel (though they are foreshadowed in the work of his predecessor E. Kummer). It seems that Hensel s main motivation was the analogy between the ring of integers Z, together with its field of fractions Q, and the ring C[X] of polynomials with complex coefficients, together with its field of fractions C(X). Both Z and C[X] are rings where there is unique factorization: any integer can be expressed as a product of primes, and any polynomial can be expressed uniquely as P(X) = a(x α 1 )(X α 2 )...(X α n ), where a and α 1,...,α n are complex numbers. This is the main analogy Hensel explored: the primes p Z are analogous to the linear polynomials X α C[X]. Suppose we are given a polynomial P(X) and α C, then it is possible (for example using a Taylor expansion) to write the polynomial in the form P(X) = n a i (X α) i, a i C. i=0 This also works naturally for the integers: given a positive integer m and a prime p, we can write it in base p, that is m = n a i p i, a i Z i=0 and 0 a i p 1. The reason such expansions are interesting is that they give local information: the expansion in powers of (X α) shows if P(X) vanishes at α, and to what order. Similarly, the expansion in base p will show if m is divisible by p, and to what order. 53
2 54 CHAPTER 5. P-ADIC NUMBERS Figure 5.1: Kurt Hensel ( ) Now for polynomials, one can go a little further, and consider their Laurent expansion f(x) = i n 0 a i (X α) i, that is any rational function can be expanded into a series of this kind in terms of each of the primes (X α). From an algebraic point of view, we have two fields: C(X) of all rational functions, and another field C((X α)) which consists of all Laurent series in (X α). Then the function defines an inclusion of fields f(x) expansion around (X α) C(X) C((X α)). Hensel s idea was to extend the analogy between Z and C[X] to include the construction of such expansions. Recall that the analogous of choosing α is choosing a prime number p. We already know the expansion for a positive integer m, it is just the base p representation. This can be extended for rational numbers x = a b = n n 0 a n p n yielding for every rational number x a finite-tailed Laurent series in powers of p, which is called a p-adic expansion of x.
4 56 CHAPTER 5. P-ADIC NUMBERS limit involves more formalism than we need for our purpose. To define an inverse limit of rings, we need a sequence of rings, which is suitably indexed (here the sequence Z/p k Z is indexed by the integer k). We further need a sequence of ring homomorphisms π ij with the same index (here π ij with i and j integers, i j) satisfying that 1. π ii is the identity on the ring indexed by i for all i, 2. for all i,j,k, i j k, we have π ij π jk = π ik. In our case, π ij : Z/p j Z Z/p i Z is the natural projection for i j, and the inverse limit of rings we consider is defined by lim Z/p i Z = {(x i ) i i Z/p i Z π ij (x j ) = x i, i j}. Example 5.1. We can write 1 as a p-adic integer: 1 = (p 1) + (p 1)p + (p 1)p 2 + (p 1)p The description of Z p as limit of Z/p k Z allows to endow Z p with a commutative ring structure: given α,β Z p, we consider their sequences α k,β k Z/p k Z. We then form the sequence α k +β k Z/p k Z which yields a well defined element α + β Z p. We do the same for multiplication. Example 5.2. Let us compute the sum of α = and β = in Z 3. We have α 1 2 mod 3 and β 1 1 mod 3, thus (α + β) 1 = α 1 + β 1 0 mod 3. Then α 2 5 mod 3 2 and β 2 7 mod 3 2, so that This yields (α + β) 2 = α 2 + β 2 = 12 3 mod 3 2. α + β = Z 3. We are just computing the addition in base 3! Note that Z is included in Z p. Let us now look at fractions instead of integers. The fraction 3/2 is the solution of the equation 2x + 3 = 0. Does this equation have a solution in Z 3? We have that 3 2 = = 3( ) since Thus 1 1 x = 1 + x + x =
5 5.1. P-ADIC INTEGERS AND P-ADIC NUMBERS 57 Actually, if x = a/b and p does not divide b, then x = a/b Z p. Indeed, there is an inverse b 1 Z/p k Z and the sequence ab 1 converges towards an x Z p such that bx = a. On the contrary, 1/p Z p, since for all x Z p, we have that (px) 1 = 0 1. Definition 5.2. The p-adic numbers are series of the form 1 a n p n + a 1 n+1 p n a 1 1 p + a 0 + a 1 p +... The set of p-adic numbers is denoted by Q p. It is a field. We have an inclusion of Q into Q p. Indeed, if x Q, then there exists N 0 such that p N x Z p. In other words, Q can be seen as a subfield of Q p. Example 5.3. Let p = 7. Consider the equation X 2 2 = 0 in Z 7. Let α = a 0 + a a be the solution of the equation. Then we have that a mod 7. We thus two possible values for a 0 : α 1 = a 0 = 3, α 1 = a 0 = 4. We will see that those two values will give two solutions to the equation. Let us choose a 0 = 3, and set We have that α 2 = a 0 + a 1 7 Z/49Z. α mod 7 2 a a a 0 a mod a mod a 1 0 mod a 1 0 mod 7 a 1 1 mod 7. By iterating the above computations, we get that α = The other solution is given by α = Note that X 2 2 does not have solutions in Q 2 or in Q 3. In the above example, we solve an equation in the p-adic integers by solving each coefficient one at a time modulo p, p 2,... If there is no solution for one coefficient with a given modulo, then there is no solution for the equation, as this is the case for Q 2 or Q 3.
6 58 CHAPTER 5. P-ADIC NUMBERS In the similar spirit, we can consider looking for roots of a given equation in Q. If there are roots in Q, then there are also roots in Q p for every p (that is, in all the Q p and in R). Hence we can conclude that there are no rational roots if there is some p for which there are no p-adic roots. The fact that roots in Q automatically are roots in Q p for every p means that a global root is also a local root everywhere (that is at each p). Much more interesting would be a converse: that local roots could be patched together to give a global root. That putting together local information at all p should give global information is the idea behind the so-called local-global principle, first clearly stated by Hasse. A good example where this principle is successful is the Hasse-Minkowski theorem: Theorem 5.1. (Hasse-Minkowski) Let F(X 1,...,X n ) Q[X 1,...,X n ] be a quadratic form (that is a homogeneous polynomial of degree 2 in n variables). The equation F(X 1,...,X n ) = 0 has non-trivial solutions in Q if and only if it has non-trivial solutions in Q p for each p. 5.2 The p-adic valuation We now introduce the notion of p-adic valuation and p-adic absolute value. We first define them for elements in Q, and extend them to elements in Q p after proving the so-called product formula. The notion of absolute value on Q p enables to define Cauchy sequences, and we will see that Q p is actually the completion of Q with respect to the metric induced by this absolute value. Let α be a non-zero element of Q. We can write it as α = p k g h, k Z, and g,h,p coprime to each other, with p prime. We set ord p (α) = k α p = p k ord p (0) = 0 p = 0. We call ord p (α) the p-adic valuation of α and α p the p-adic absolute value of α. We have the following properties for the p-adic valuation: ord p : Q Z { } ord p (ab) = ord p (a) + ord p (b) ord p (a + b) min(ord p (a),ord p (b)) ord p (a) = a = 0.
7 5.2. THE P-ADIC VALUATION 59 Let us now look at some properties of the p-adic absolute value: p : Q R 0 ab p = a p b p a + b p max( a p, b p ) a p + b p a p = 0 a = 0. Note that in a sense, we are just trying to capture for this new absolute value the important properties of the usual absolute value. Now the p-adic absolute value induces a metric on Q, by setting d p (a,b) = a b p, which is indeed a distance (it is positive: d p (a,b) 0 and is 0 if and only if a = b, it is symmetric: d p (a,b) = d p (b,a), and it satisfies the triangle inequality: d p (a,c) d p (a,b) + d p (b,c)). With that metric, two elements a and b are close if a b p is small, which means that ord p (a b) is big, or in other words, a big power of p divides a b. The following result connects the usual absolute value of Q with the p-adic absolute values. Lemma 5.2. (Product Formula) Let 0 α Q. Then α ν = 1 where ν {,2,3,5,7,...} and α is the real absolute value of α. ν Proof. We prove it for α a positive integer, the general case will follow. Let α be a positive integer, which we can factor as α = p a1 1 pa2 2 pa k k. Then we have The result follows. α q = 1 α pi = p ai i α = p a1 1 pa k k if q p i for i = 1,...,k In particular, if we know all but one absolute value, the product formula allows us to determine the missing one. This turns out to be surprisingly important in many applications. Note that a similar result is true for finite extensions of Q, except that in that case, we must use several infinite primes (actually one for each different inclusion into R and C). We will come back to this result in the next chapter. The set of primes together with the infinite prime, over which the product is taken in the product formula, is usually called the set of places of Q.
8 60 CHAPTER 5. P-ADIC NUMBERS Definition 5.3. The set is the set of places of Q. M Q = {,2,3,...} Let us now get back to the p-adic numbers. Let α = a k p k + a k+1 p k+1 + a k+2 p k Q p, with a k 0, and k possibly negative. We then set ord p (α) = k α p = p k. This is an extension of the definition of absolute value defined for elements of Q. Before going on further, let us recall two definitions: Recall that a sequence of elements x n in a given field is called a Cauchy sequence if for every ǫ > 0 one can find a bound M such that we have x n x m < ǫ whenever m,n M. A field K is called complete with respect to an absolute value if every Cauchy sequence of elements of K has a limit in K. Let α Q p. Recall that α l is the integer 0 α l < p l obtained by cutting α after a l 1 p l 1. If n > m, we have α n α m p = a k p k a m p m a n 1 p n 1 a k p k... a m 1 p m 1 = a m p m + a m+1 p m a n 1 p n 1 p p m. This expression tends to 0 when m tends to infinity. In other words, the sequence (α n ) n 0 is a Cauchy sequence with respect to the metric induced by p. Now let (α n ) n 1 be a Cauchy sequence, that is α n α m p 0 when m with n > m, that is, α n α m is more and more divisible by p, this is just the interpretation of what it means to be close with respect to the p-adic absolute value. The writing of α n and α m in base p will thus be the same for more and more terms starting from the beginning, so that (α n ) defines a p-adic number. This may get clearer if one tries to write down two p-adic numbers. If a,b are p-adic integers, a = a 0 +a 1 p+a 2 p , b = b 0 +b 1 p+b 2 p , if a 0 b 0, then a b p = p 0 = 1 if p does not divide a 0 b 0, and a b p = p 1 if p a 0 b 0, but a b p cannot be smaller than 1/p, for which we need a 0 = b 0. This works similarly for a,b p-adic numbers. Then we can write a = a k 1/p k +..., b = b l 1/p l +... If k l, say k > l, then a b p = b l 1/p l a k /p k +... p = p l, which is positive. The two p-adic numbers a and b are thus very far apart. We see that for the distance between a and b to be smaller than 1, we first need all the coefficients a i, b i, to be the same, for i = k,...,1. We are then back to the computations we did for a and b p-adic integers. We have just shown that
9 5.2. THE P-ADIC VALUATION 61 Theorem 5.3. The field of p-adic numbers Q p is a completion of Q with respect to the p-adic metric induced by p. Now that we have a formal definition of the field of the p-adic numbers, let us look at some of its properties. Proposition 5.4. Let Q p be the field of the p-adic numbers. 1. The unit ball {α Q p α p 1} is equal to Z p. 2. The p-adic units are Z p = {α Z p 0 a 0 (Z/pZ) } = {α Z p α p = 1}. 3. The only non-zero ideals of Z p are the principal ideals p k Z p = {α Q p ord p (α) k}. 4. Z is dense in Z p. Proof. 1. We look at the unit ball, that is α Q p such that α p 1. By definition, we have α p 1 p ordp(α) 1 ord p (α) 0. This is exactly saying that α belongs to Z p. 2. Let us now look at the units of Z p. Let α be a unit. Then α Z p α Z p and 1 α Z p α p 1 and 1/α p 1 α p = We are now interested in the ideals of Z p. Let I be a non-zero ideal of Z p, and let α be the element of I with minimal valuation ord p (α) = k 0. We thus have that α = p k (a k + a k+1 p +...) where the second factor is a unit, implying that αz p = p k Z p I. We now prove that I p k Z p, which concludes the proof by showing that I = p k Z p. If I is not included in p k Z p, then there is an element in I out of p k Z p, but then this element must have a valuation smaller than k, which cannot be by minimality of k.
10 62 CHAPTER 5. P-ADIC NUMBERS 4. We now want to prove that Z is dense in Z p. Formally, that means that for every element α Z p, and every ǫ > 0, we have B(α,ǫ) Z is non-empty (where B(α,ǫ) denotes an open ball around α of radius ǫ). Let us thus take α Z p and ǫ > 0. There exists a k big enough so that p k < ǫ. We set ᾱ Z the integer obtained by cutting the serie of α after a k 1 p k 1. Then α ᾱ = a k p k + a k+1 p k implies that α ᾱ p p k < ǫ. Thus Z is dense in Z p. Similarly, Q is dense in Q p. The main definitions and results of this chapter are Definition of p-adic integers using p-adic expansions, inverse limit, and that they form a ring Z p Definition of p-adic numbers using p-adic expansions, and that they form a field Q p Definition of p-adic valuation and absolute value The product formula The formal definition of Q p as completion of Q, and that Z p can then be defined as elements of Q p with positive p-adic valuation. Ideals and units of Z p.
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How many numbers there are? RADEK HONZIK Radek Honzik: Charles University, Department of Logic, Celetná 20, Praha 1, 116 42, Czech Republic [email protected] Contents 1 What are numbers 2 1.1 Natural | 11,004 | 40,087 | {"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} | 4.1875 | 4 | CC-MAIN-2018-26 | longest | en | 0.932038 |
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Introduction to Unit Conversions
> ... Math > Unit Conversions > Introduction to Unit Conversions
The following video will show how to do unit conversions and explains why it works.
Introduction to Unit Conversions
### Introduction to Unit Conversions
Unit conversion = taking the measurement of something in one set of units and changing it to an equivalent measurement in another set of units.
The keys to doing unit conversions are the concepts that anything divided by itself is equal to 1 and that anything multiplied by 1 is equal to itself.
$${\dfrac {x}{x}} = 1$$
and
$$1x = x$$
### Conversion Factors
Conversion factors are fractions where the item in the numerator is equal to the item in the denominator, essentially making the fraction equal to 1.
Examples:
1 inch = 2.54 cm, therefore:
$${\dfrac {1 \ inch}{2.54 \ cm}} = 1$$
This is the conversion factor between inches and centimeters.
Another example is that 60 minutes = 1 hour, therefore:
$${\dfrac {60 \ minutes}{1 \ hour}} = 1$$
This is the conversion factor between minutes and hours.
### Example 1: Inches to Centimeters
How do we convert 3 inches into the equivalent length in centimeters?
2. Determine what we want to get in the end: centimeters.
3. Determine the conversion factor to use: 1 inch = 2.54 cm.
4. Multiply by 1 in the form of the conversion factor that cancels out the unwanted units. In this case, $${\frac {2.54 \ cm}{1 \ in}}$$.
$$\dfrac{\color{DarkGreen}3 {\cancel{\text{ inches}}}}{1}\left ( \dfrac{{\color{DarkOrange} 2.54\:\text{cm}}}{{\color{DarkGreen} 1\:\cancel{\text{ inch}} }} \right ) = \dfrac{\left ( {\color{DarkGreen} 3} \right )\left ( {\color{DarkOrange} 2.54\:\text{cm}}\right )}{1} = {\color{Black} 7.62\:\text{cm}}$$
Answer: 3 in = 7.62 cm
### Example 2: Minutes to Hours
How do we convert 14 minutes into the equivalent time in hours?
$$14 \ minutes = ? \ hours$$
2. Determine what we want to get in the end: hours.
3. Determine the conversion factor to use: 60 minutes = 1 hour.
4. Multiply by 1 in the form of the conversion factor that cancels out the unwanted units. In this case, $${\frac {1 \ hour}{60 \ minutes}}$$.
$${14\color{Magenta}\cancel{\text{ minutes}}}\times \dfrac{{1\color{green} \text{ hour}}}{{60\color{Magenta} \cancel{\text{ minutes}}}}=\dfrac{(14\times 1){\color{green} \text{hours}}}{60}$$
$$=0.23{\color{green}\text{ hours}}$$
Answer: 14 minutes = 0.23 hours
### Practice Problems
Video Solutions for Problems 1 - 5
1. The length of a table measures 250 centimeters. There are 100 centimeters in 1 meter. Convert the length of the table to meters. Round to the nearest tenth. (
Solution
Solution:
$$2.5$$ meters
)
2. Margaret has a pitcher filled with 2 liters of water. One liter is equal to approximately $$4.23$$ cups. Convert the amount of water in Margaret’s pitcher to cups. Round to the nearest hundredth. (
Solution
Solution:
$$8.46$$ cups
)
3.Kim walked $$2.5$$ miles to the grocery store. There are 1760 yards in one mile. Convert the distance Kim walked to yards. Round to the nearest whole number. (
Solution
Solution:
4400 yd
Details:
We know that Kim walked $$2.5$$ miles to the grocery store, but we want to know how far that is in yards. To figure that out we need 3 things:
1. The original amount: $$2.5$$ miles
2. The desired units: yards
3. The conversion rate: 1760 yards = 1 mile
The first step is to write $$2.5$$ miles as a fraction. The equivalent fraction is the following:
$$\dfrac{2.5\text{ miles}}{1}$$
Now we need to figure out the conversion factor that we need to use. We need it to cancel out the miles and leave us with yards. Using our conversion rate, 1760 yards = 1 mile, we know it will be one of the following:
$$\dfrac{1760\text{ yards}}{1\text{ mile}}$$ or $$\dfrac{1\text{ mile}}{1760\text{ yards}}$$
Since miles is at the top of our original fraction, we must use the conversion factor with miles in the bottom of the fraction so that miles will cancel out when we multiply. We set it up like this:
$$\displaystyle\frac{2.5\:\text{miles}}{1}\cdot\frac{1760\:\text{yards}}{1\:\text{mile}}$$
The first step is to cancel out the miles:
$$\displaystyle\frac{2.5\:{\color{DarkOrange} \cancel{\text{miles}}}}{1}\cdot \frac{1760\:\text{yards}}{1\:{\color{DarkOrange} \cancel{\text{mile}}}}$$
Then we multiply straight across:
$$\displaystyle\frac{2.5\:\cdot \: 1760 \text{ yards}}{1\:\cdot\: 1}$$
Then simplify both the top and bottom of the fraction:
$$= \dfrac{4400\text{ yards}}{1}$$
Which equals to the following:
= 4400 yards
So $$2.5$$ miles is the same distance as 4400 yards.
)
4. Brent has $$£35$$ (British pounds) that he would like to convert to US dollars ($$US$$). Assume the current conversion rate is $$£1 = 1.27$$. How much money will he have in dollars rounded to the nearest hundredth? (
Solution
Solution:
$$44.45$$
Details:
Brent has $$£35$$ and needs to know how much it is in US dollars. To figure that out we need three things:
1. The original amount: $$£35$$
2. The desired units: US dollars ($$US$$)
3. The conversion rate: $$£1 = 1.27$$
The first step is to write our original amount as a fraction:
$$\dfrac{{\text{£}}35}{1}$$
Using the conversion rate, $$£1 = 1.27$$, we can figure out our conversion factors. The two options are the following:
$$\dfrac{£1}{\1.27}$$ or $$\dfrac{\1.27}{£1}$$
We are changing British pounds ($$£$$) to US dollars () and £ is in the top (or numerator) of our original amount so we must use the conversion rate with $$£$$ in the bottom of the fraction so that we can cancel out the $$£$$.
$$\dfrac{\1.27}{£1}$$
Next, we multiply our original amount by our conversion factor:
$$\displaystyle\frac{£35}{1}\times\frac{\1.27}{£1}$$
Now we cancel out the $$£$$ since there is $$£$$ in the top and $$£$$ in the bottom:
$$\displaystyle\frac{\cancel{{\color{Magenta} £}}35}{1}\times\frac{\1.27}{\cancel{{\color{Magenta} £}}1}$$
Next, we multiply straight across:
$$\dfrac{35\:\cdot\:\1.27}{1\:\cdot\:1}$$
Then simplify top and bottom:
$$=\dfrac{\44.45}{1}$$
Which gives us the following:
$$= 44.45$$
This means that $$£35$$ is the same amount of money as $$44.45$$.
)
5. Max ran a half marathon which is $$13.1$$ miles in length. There are approximately $$0.62$$ miles in 1 kilometer. Convert the length of the half marathon to kilometers. Round to the nearest whole number. (
Solution
$$21$$ km
)
Video Solution for Problems 6 - 7
6. A piano weighs 325 pounds. There are approximately $$2.2$$ pounds in 1 kilogram. Convert the weight of the piano to kilograms. Round to the nearest whole number. (
Solution
$$148$$ kg
)
7. Zeezrom tried to tempt Amulek by offering him six onties of silver, which are of great worth, if he would deny the existence of a Supreme Being. (See Alma 11:21–25.) Of course, Amulek refuses the money. In the Nephite market, one onti of silver is the amount of money a judge earns after seven working days (Alma 11:3, 6–13.) How many days would a judge have to work to earn six onties of silver? (
Solution
$$42$$ days
)
## Need More Help?
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## Section A (40 Marks)
#### Question 1(i)
A moment of couple has a tendency to rotate the body in an anticlockwise direction then that moment of couple is taken as:
1. positive
2. negative
3. maximum
4. zero
positive
Reason — The anticlockwise moment is taken as positive while the clockwise moment is taken as negative.
#### Question 1(ii)
The kinetic energy of a given body depends on the:
1. position
2. centre of gravity of body
3. momentum
4. displacement
momentum
Reason — As kinetic energy = $\dfrac{1}{2}$ mv 2
If we multiply and divide by m then we get, $\dfrac{\text{(mv)}^2}{2\text{m}}$
As mv = momentum (p), so
kinetic energy = $\dfrac{\text{p}^2}{2\text{m}}$
Therefore, kinetic energy of a given body depends on the momentum.
#### Question 1(iii)
For burning of coal in a thermoelectric station, the energy conversion taking place is:
1. chemical to heat to mechanical
2. chemical to heat to mechanical to electrical
3. chemical to heat to light
4. heat to chemical to mechanical
chemical to heat to mechanical to electrical.
Reason — A thermal power station is a type of power station in which heat energy is converted to electrical energy.
Here, the chemical energy of coal is changed into heat energy, which is transferred to steam. Then the heat energy of steam changes into mechanical energy of turbine that is further changed into electrical energy.
#### Question 1(iv)
A nucleus of an atom consists of 146 neutrons and 95 protons. It decays after emitting an alpha particle. How many protons and neutrons are left in the nucleus after an alpha emission?
1. protons = 93, neutrons = 144
2. protons = 95, neutrons = 142
3. protons = 89, neutrons = 144
4. protons = 89, neutrons = 142
protons = 93, neutrons = 144
Reason — An alpha particle consists of two protons and two neutrons. Emission of alpha particle reduces the number of protons by 2 and also the number of neutrons by 2. Hence, number of protons = 95 - 2 = 93 and number of neutrons = 146 - 2 = 144.
#### Question 1(v)
Assertion: Infrared radiations travel long distances through dense fog and mist.
Reason: Infrared radiations undergoes minimal scattering in earth's atmosphere
1. both assertion and reason are true.
2. both assertion and reason are false.
3. assertion is false but reason is true.
4. assertion is true reason is false.
both assertion and reason are true.
Reason — Infrared radiations travel long distances through dense fog and mist as they undergo minimal scattering in earth's atmosphere.
#### Question 1(vi)
For a convex lens, the minimum distance between an object and its real image in terms of focal length (f) of a given lens must be:
1. 1.5 f
2. 2.5 f
3. 2 f
4. 4 f
4f
Reason — When the object is kept at 2F1, then, its image is formed at 2F2. This is the position of minimum distance between object and image (i.e., 2f + 2f = 4f).
#### Question 1(vii)
Two sound waves X and Y have same amplitude and same wave pattern, but their frequencies are 60 Hz and 120 Hz respectively, then:
1. X will be shriller and Y will be grave
2. X will be grave and Y will be shriller
3. X will differ in quality than Y
4. X is louder than Y.
X will be grave and Y will be shriller.
Reason — The shrillness of a sound wave is directly proportional to its frequency, meaning that higher frequencies result in shriller sounds and lower frequencies result in grave sound.
#### Question 1(viii)
Vibrations produced in a body under the influence of the periodic force is;
1. forced vibrations
2. resonant vibrations
3. damped vibrations
4. sympathetic vibrations
forced vibrations
Reason — The vibrations of a body which take place under the influence of an external periodic force acting on it, are called forced vibrations. E.g. the vibrations produced in the diaphragm of a microphone sound box with frequencies corresponding to the speech of the speaker, are the forced vibrations.
#### Question 1(ix)
The graph of voltage vs current for four different materials is shown below.
Which of these four materials would be used for making the coil of a toaster?
1. Q
2. S
3. P
4. R
P
Reason — The coil of a toaster should produce a high heating effect so its resistance should be high. From the graph, slope of line for substance P is the greatest hence, resistance of P is highest. So, substance P is most suitable for making the coil of a toaster.
#### Question 1(x)
According to the old convention the color of the earth wire is:
1. black
2. green
3. yellow
4. red
green
Reason— In the old colour convention, green colour is used for earth wire.
#### Question 1(xi)
Lenz’s law is based on the law of conservation of:
1. force
2. charge
3. mass
4. energy
energy
Reason — Lenz's law is based on the law of conservation of energy. It shows that the mechanical energy is spent in doing work against the opposing force experienced by the moving magnet, and it is transformed into the electrical energy due to which current flows in the solenoid.
#### Question 1(xii)
Heat capacity of a body is:
1. the energy needed to melt the body without the change in its temperature.
2. the energy needed to raise the temperature of the body by 1°C
3. the increase in the volume of the body when its temperature increases by 1°C
4. the total amount of internal energy that is constant.
the energy needed to raise the temperature of the body by 1°C
Reason — The term heat capacity of a body is the amount of heat energy required to raise it's temperature by 1° C (or 1 K). The S.I. unit of heat capacity is joule per kelvin (J K-1)
#### Question 1(xiii)
The amount of heat energy required to melt a given mass of a substance at its melting point without rise in its temperature is called:
1. specific heat capacity
2. specific latent heat of fusion
3. latent heat of fusion
4. specific latent heat of freezing
latent heat of fusion
Reason — Heat energy absorbed in change of phase that is not externally manifested by any rise in temperature is called the latent heat of fusion.
#### Question 1(xiv)
When a ray of light enters from a denser medium to a rarer medium then:
1. the light ray bends towards the normal
2. the speed of light increases
3. the angle of incidence is greater than the angle of refraction
4. its wavelength decreases.
the speed of light increases
Reason — The speed of light increases when it enters from a denser medium to a rarer medium. Lower density of the rarer medium causes less interaction between photons and particles. This reduces the scattering and results in a higher speed.
#### Question 1(xv)
An endoscope uses optical fiber to transmit high resolution images of internal organs without loss of information. The phenomenon of light that governs the functioning of the optical fiber is:
1. refraction
2. reflection
3. scattering
4. total internal reflection
total internal reflection.
Reason — Total internal reflection is the principle that allows light to propagate through the core of an optical fiber by continuously reflecting off the inner walls of the fiber due to the angle at which it strikes those walls.
#### Question 2(i)
(a) Name the principle on which a lever works.
(b) Which radiations that are emitted during the decay of a nucleus, having highest penetrating power?
(c) Does the emission of the above-mentioned radiation result in a change in the mass number?
(a) A lever works on the principle of moments, according to which in the equilibrium position of levers, moment of load about the fulcrum must be equal to the moment of effort about the fulcrum and the two moments must be in opposite directions.
(b) Gamma radiations emitted during the decay of a nucleus have the highest penetrating power.
(c) When gamma radiations are emitted from the nucleus then there is no change in mass number as gamma radiations have no charge and mass.
#### Question 2(ii)
A metre rod made of copper and steel as shown in the diagram. Weights of copper and steel are 10 N and 8 N respectively.
(a) On which part does the centre of gravity lie (0 to 50 or 50 to 100).
(a) Centre of gravity of the given rod lie between 0 to 50.
(b) In equilibrium ,
(F1) x distance (l1) = (F2) x distance (l2)
As, F1 > F2
Hence, l1 < l2
Therefore, centre of gravity lies between 0 to 50.
#### Question 2(iii)
A lever is shown below.
(a) Identify the type of lever.
(a) It is a class III lever.
(b) Mechanical advantage = $\dfrac{\text{effort arm}}{\text{load arm}}$ = $\dfrac{0.1}{0.1 + 0.4}$
$= \dfrac{0.1}{0.5} \\[1em] = \dfrac{1}{5} \\[1em] = 0.2$
#### Question 2(iv)
Two bodies A and B have same kinetic energies. Compare their velocities if mass of A is four times the mass of B.
Considering Body A:
Let mass of body A = ma = 4m
Let kinetic energy of body A = K
Let velocity of body A = Va
Considering Body B:
Mass of body B = mb = m [∵ mass of A is four times the mass of B]
Kinetic energy of body B = K [∵ A and B have same kinetic energies]
Let velocity of body B = Vb
Given, A and B have same kinetic energies,
$\dfrac{\text{1}}{\text{2}}$ 4mVa2 = $\dfrac{\text{1}}{\text{2}}$mVb2
4mVa2 = mVb2
$\dfrac{\text{V}_a^2}{\text{V}_b^2} = \dfrac{\text{1}}{\text{4}} \\[1em] \Rightarrow \dfrac{\text{V}_a}{\text{V}_b} = \sqrt{\dfrac{\text{1}}{\text{4}}} \\[1em] \Rightarrow \dfrac{\text{V}_a}{\text{V}_b} = \dfrac{1}{2}$
Hence, Va : Vb = 1 : 2 or velocity of body B is two times the velocity of body A.
#### Question 2(v)
Draw a graph of potential energy vs height from the ground for a body thrown vertically upwards.
Graph of potential energy vs height from the ground for a body thrown vertically upwards is shown below:
#### Question 2(vi)
Two copper wires A and B are of the same thickness and are at the room temperature. If the length of A is twice the length of B then :
(a) Compare their resistances
(b) Compare their resistivity
(a) Let the length of wire B be L and that of A be 2L
R= ρ$\dfrac{\text{length}}{\text{area}}$
As area and specific resistance (ρ) are same for both wires A and B,
Hence,
$\dfrac{\text{R}_a}{\text{R}_b}$ = $\dfrac{\text{l}_a}{\text{l}_b}$ = $\dfrac{\text{2L}}{\text{L}}$ = $\dfrac{\text{2}}{\text{1}}$
Hence, the ratio between resistances A and B is 2 : 1
(b) Both the wires will have the same specific resistance because the specific resistance is a characteristic property of the material and as both the wires are of copper hence both will have same specific resistance.
#### Question 2(vii)
(a) Name the waves used for echo depth sounding.
(b) Give one reason for their use in the above application
(a) The waves used for echo depth sounding are ultrasonic waves.
(b) Ultrasonic waves can travel undeviated through a long distance and so they are used for echo depth sounding.
#### Question 3(i)
(a) Refer to the diagram given below. A lens with two different refractive indices is shown. If the rays are coming from a distant object, then how many images will be seen?
(b) A glass lens always forms a virtual, erect and diminished image of an object kept in front of it. Identify the lens.
(a) Two images will be seen.
Reason — According to the lens maker formula :
$\dfrac{1}{\text{f}}$ = (μ - 1)($\dfrac{1}{\text{R}_1}$ - $\dfrac{1}{\text{R}_2}$)
Focal length is a function of refractive index, hence for two different refractive index there will be two different focal lengths.
Two different focal length implies two different images for same object. Hence 2 images will be formed.
(b) Concave lens
#### Question 3(ii)
It is observed that the house circuits are arranged in a parallel combination. Give two advantages of this arrangement.
(a) Two advantages of this arrangement are :
1. Each appliance operates independently without being affected by the presence or working of the other appliances.
2. Each appliance gets connected to 220 V supply (= its voltage rating) for its normal working.
#### Question 3(iii)
A transformer is used to change a high alternating e.m.f. to a low alternating e.m.f. of the same frequency.
(a) Identify the type of transformer used for the above purpose.
(b) State whether the turns ratio of the above transformer is = 1 or >1 or <1.
(a) Step down transformer
(b) turns ratio is n < 1, i.e., number of turns in the secondary are less than the number of turns in the primary.
#### Question 3(iv)
A solid of mass 60 g at 100°C is placed in 150 g of water at 20°C. The final steady temperature is 25°C. Calculate the heat capacity of solid?
[sp. heat capacity of water = 4.2 J g-1 K-1 ]
Given,
Mass of solid (m) = 60 g
Fall in temperature of solid = (100 - 25) = 75°C
Rise in temperature of water = (25 - 20) = 5°C
Let specific heat capacity of solid be c.
Heat energy given by solid = mc△t
= 60 x c x 75 = 4500 x c .....(1)
Heat energy taken by water = 150 × 4.2 × 5 = 3150 .....(2)
Assuming that there is no loss of heat energy,
Heat energy given by solid = Heat energy taken by water.
Equating equations 1 & 2, we get,
$4500 \times c = 3150 \\[0.5em] c = \dfrac{3150}{4500} \\[0.5em] c = 0.7 \text{ J g}^{-1} \text{K}^{-1} \\[0.5em]$
specific heat capacity of the solid = 0.7 J g-1 K-1
Heat capacity of solid = mass (m) x specific heat capacity = 60 x 0.7 = 42 J K-1
Hence, Heat capacity of solid = 42 J K-1
#### Question 3(v)
What is a nuclear waste? State one method to dispose it safely.
The radioactive material after its use such as nuclear power generation, nuclear weapons production, and nuclear medicine is known as nuclear waste.
The nuclear waste obtained from laboratories, hospitals, scientific establishments or power plants must be first kept in thick casks and then they must be buried in the specially constructed deep underground stores. These underground stores must be far away from the populated areas.
## Section B (40 Marks)
#### Question 4(i)
The diagram below shows a fish in the tank and its image seen in the surface of water.
(a) Name the phenomenon responsible for the formation of this image.
(b) Complete the path of the ray through the glass prism of critical angle 42° till it emerges out of the prism.
(a) Total Internal Reflection
Reason — The image of the fish seen at the surface of water is due to total internal reflection as shown in the ray diagram below:
(b) Below diagram shows the completed path of the ray till it emerges out of the prism:
#### Question 4(ii)
(a) The refractive index of water is 1.33 at a certain temperature. When the temperature of water is increased by 40°C, the refractive index changes to 'x'. State whether x < 1.33 or x > 1.33.
(b) State two differences between normal reflection and total internal reflection.
(a) In general, the refractive index of a substance decreases with increasing temperature. Therefore, if the temperature of water is increased by 40°C, the refractive index 'x' is expected to be lesser than 1.33. Hence, we can state that x < 1.33.
(b) Two differences between normal reflection and total internal reflection are:
S.
No.
Normal reflectionTotal internal reflection
1.Normal reflection can occur whenever light encounters a reflective surface, such as a mirror, glass, or water, without any specific conditions related to the angle of incidence.Total internal reflection occurs when light travelling in a denser medium, is incident at the surface of a rarer medium at an angle of incidence greater than the critical angle for the pair of media.
2.In normal reflection, only a part of the light is reflected while rest is refracted and absorbed. So, the reflection is partial.In total internal reflection, the entire incident light is reflected back into the denser medium. So there is no loss of light energy.
#### Question 4(iii)
The below diagram shows that an observer sees the image of an object O at I.
(a) Name and define the phenomenon responsible for seeing the image at a different position.
(b) State the effect on X when:
1. Y increases
2. Y decreases
(a) Refraction
The change in direction of the path of light when it passes from one transparent medium to another transparent medium, is called refraction.
(b) The shift X,
1. increases when Y increases
2. decreases when Y decreases.
#### Question 5(i)
An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm.
(a) What is the nature of the image so formed?
(b) Calculate the distance of the image from the lens.
(a) Real and inverted image is formed.
(b) Given, focal length = 8 cm
Object distance (u) = -24 cm
Image distance (v) = ?
From lens formula —
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\[1em]$
Substituting the values in the formula we get,
$\dfrac{1}{\text{v}} - \dfrac{1}{-24} = \dfrac{1}{8} \\[1em] \dfrac{1}{\text{v}} + \dfrac{1}{24} = \dfrac{1}{8} \\[1em] \dfrac{1}{\text{v}} = \dfrac{1}{8} - \dfrac{1}{24}\\[1em] \Rightarrow \dfrac{1}{\text{v}} = \dfrac{3 - 1}{24} \\[1em] \Rightarrow \dfrac{1}{\text{v}} = \dfrac{2}{24} \\[1em] \Rightarrow \dfrac{1}{\text{v}} = \dfrac{1}{12} \\[1em] \Rightarrow v = 12$
∴ The image is formed at a distance of 12 cm behind the lens.
#### Question 5(ii)
When sunlight passes through water droplets in the atmosphere it gets dispersed into its constituent colours forming a rainbow. A similar phenomenon is observed when white light passes through a prism.
(a) Which colour will show the maximum angle of deviation and which colour will show the minimum angle of deviation?
(b) If instead of sunlight, a green-coloured ray is passed through a glass prism. What will be the colour of the emergent ray?
(a) When sunlight passes through water droplets in the atmosphere or through a prism, violet light shows the maximum angle of deviation, while red light shows the minimum angle of deviation.
(b) The colour of emergent ray will be green. A prism does not change the colour of light. It just disperses white light into its constituent colours.
#### Question 5(iii)
(a) Mixture of red + blue + green is passed through a convex lens as shown in the diagram below. State whether the ray passes through a single point or through different points on principle axis after refraction.
(b) Name the invisible radiations which can be obtained using quartz prism? State one use of these radiations.
(c) Name one radiations having wavelength longer than the wavelength of these radiations.
(a) The ray will pass through different points on principle axis after refraction through the convex lens. Red colour will deviate the least, followed by green and then blue will deviate the most.
(b) Ultraviolet radiations. Ultraviolet radiations are used for sterilising air, surgical equipment, etc.
#### Question 6(i)
Sumit and Sachin went for a trek and during the journey they visited a cottage. They suspended their bags to the two ropes hanging from P and Q on a wheel capable of rotating around O. Sumit suspended his bag to the rope Q and Sachin suspended his bag from the rope P. The wheel remained in equilibrium.
(a) State with a reason who is carrying a heavier bag.
(b) Based on the principle of moments, write a mathematical relation that can be used to determine the weight (W) of Sachin's bag, given that the weight of Sumit's bag is 18 kgf.
(a) At equilibrium clockwise moment = anticlockwise moment
Moment = force x perpendicular distance
Wt. of Sachin's bag x OP = Wt. of Sumit's bag x OR
As OR < OP. Hence, Wt. of Sumit's bag > Wt. of Sachin's bag.
Hence, Sumit's bag is heavier than Sachin's bag
(b) Given,
Wt. of Sumit's bag = 18 kgf
Let wt. of Sumit's bag be W
At equilibrium,
Clockwise moment = Anticlockwise moment
W x OP = 18 x OR
W = $\dfrac{18 \times \text{OR}}{\text{OP}}$
#### Question 6(ii)
The diagram below shows a block and tackle system.
(a) Copy and complete the labelled diagram showing the correct connection of the tackle, the direction of the forces involved to obtain maximum V.R. with the convenient direction.
(b) Calculate the M.A. of this pulley system if its efficiency is 80%.
(a) The completed diagram is given below:
(b) η = 80% given
V.R. = n = 3,
M.A. =?
η = $\dfrac{\text{M.A.}}{\text{V.R.}}$
$\dfrac{\text{80}}{\text{100}}$ = $\dfrac{\text{M.A.}}{\text{3}}$
M.A. = $\dfrac{\text{80}}{\text{100}}$ x 3= $\dfrac{\text{240}}{\text{100}}$ = 2.4
#### Question 6(iii)
The figure below shows a simple pendulum of mass 200 g. It is displaced from the mean position A to the extreme position B. The potential energy at the position A is zero. At the position B the pendulum bob is raised by 5 m.
(a) What is the potential energy of the pendulum at the position B?
(b) What is the total mechanical energy at point C?
(c) What is the speed of the bob at the position A when released from B?
(Take g = 10 ms-2 and given that there is no loss of energy.)
Given,
h = 5 m
m = 200 g = 0.2 kg
g = 10 ms-2
(i) Potential energy UB at B is given by
UB = m x g x h
Substituting the values we get,
UB = 0.2 x 10 x 5 = 10 J
Hence, the potential energy of the pendulum at the position B = 10 J
(ii) Total mechanical energy at point C = 10 J
The total mechanical energy is same at all points of the path due to conservation of mechanical energy.
(iii) At A, bob has only kinetic energy which is equal to potential energy at B,
Therefore,
$\dfrac{1}{2} \times m \times (v_A)^{2} = U_B \\[0.5em] 0.5 \times 0.2 \times (v_A)^{2} = 10 \\[0.5em] (v_A)^{2} = \dfrac{10}{0.1} \\[0.5em] \Rightarrow v_A = \sqrt{100} \\[0.5em] \Rightarrow v_A = 10 \text{ m s}^{-1}$
#### Question 7(i)
A person standing in front of a cliff fires a gun and hears its echo after 3s. If the speed of sound in air is 336 ms-1
(a) Calculate the distance of the person from the cliff.
(b) After moving a certain distance from the cliff, he fires the gun again and this time the echo is heard 1.5 s later than the first. Calculate distance moved by the person.
(a) Let the distance of the cliff from the initial position of the person be x m.
Time after which echo is heard = 3 sec
So, the distance travelled by sound in 3 sec = 2x m
Speed of sound in air = 336 m/s
Distance = speed x time
Substituting the values we get,
$2x = 336 \times 3 \\[1em] \Rightarrow x = \dfrac{336 \times 3}{2} \\[1em] \Rightarrow x = 168 \times 3 \\[1em] \Rightarrow x = 504 \text{ m}$
∴ The initial distance of the person from the cliff is 504 m.
(b) As time taken to hear the echo is increased by 1.5 sec, hence, the person has moved away from the cliff.
Let the distance moved by the person from his initial position be y m.
Time after which echo is heard = 1.5 + 3 = 4.5 sec
New distance = 504 + y
So, the distance travelled by sound in 4.5 sec = 2(504 + y) m
Speed of sound in air = 336 m/s
Distance = speed x time
Substituting the values we get,
$2(504 + y) = 336 \times 4.5 \\[1em] \Rightarrow 504 + y = \dfrac{336 \times 4.5}{2} \\[1em] \Rightarrow y = (168 \times 4.5) - 504 \\[1em] \Rightarrow y = 756 - 504 \\[1em] \Rightarrow y = 252 \text{ m}$
∴ Distance moved by the person = 252 m
#### Question 7(ii)
A radioactive nucleus X emits an alpha particle followed by two beta particles to form nucleus Y.
(a) With respect to the element X, where would you position the element Y in the periodic table?
(b) What is the general name of the element X and Y.
(c) If the atomic number of Y is 80 then what is the atomic number of X?
(a) AZX ⟶ A-4Z-2X1 + 42He ⟶ A-4Z-1X2 + 0-1e ⟶ A-4ZY + 0-1e
We can see from the above that X and Y have same atomic number. Hence, X and Y will occupy the same position in the periodic table.
(b) Isotope.
(c) When the atomic no of Y = 80 and X and Y are isotopes then atomic no. of X will also be 80.
#### Question 7(iii)
A boy tunes a radio channel to a radio station 93.5 MHz.
(a) Name and define the scientific wave phenomenon involved in tuning the radio channel.
(b) Name the important characteristics of sound that is affected during this phenomenon.
(c) Convert 93.5 MHz to SI unit.
(a) Resonance
Definition — When the frequency of the externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance.
(b) Loudness, when resonance occurs the loudness increases as the amplitude of the vibration is increased.
(c) 1 MHz = 106 Hz
93.5 MHz = 93.5 x 106 Hz = 9.35 x 107 Hz
#### Question 8(i)
Purvi's friend Tim wants to connect a fuse to his oven. He wants to control the oven from two different locations. Shown below is his circuit diagram.
(a) Which one of the two, A or B should be a live wire?
(b) In the event of an overload, will the fuse serve its purpose?
(c) What is the meaning of the statement that the bulb is rated 600W, 220 V?
(a) The switch is always connected to the live wire. In above diagram, wire B is connected to switch so wire B must be the live wire.
(b) No, the fuse will not serve its purpose. In the diagram, the fuse is connected to the earth wire and not the live wire. In case of overload, the excess current will first pass through the oven damaging it and after that will flow through the fuse.
(c) It means that if the bulb is operated on a 220 V supply, the electric power consumed by it is 600 W i.e., 600 J of electric energy is consumed by the bulb in 1 s.
#### Question 8(ii)
(a) Copy and complete the following nuclear reaction.
22286Rn ⟶ 21884Po + _—X
(b) What will be the effect on the radiation X, emitted in the above reaction when it is allowed to pass through an electric field?
(a) 22286Rn ⟶ 21884Po + 42X
(b) Radiation X are alpha particles. So, they will be deflected towards the negative plate when passed through an electric field.
#### Question 8(iii)
Observe the given circuit diagram and answer the questions that follow:
(a) Calculate the resistance of the circuit when the key K completes the circuit.
(b) Calculate the current through 3Ω resistance when the circuit is complete.
(a) 5Ω and 3Ω are in series so their equivalent resistance (Rs) = 5 + 3 = 8Ω
Now 2Ω and Rs are in parallel.
$\dfrac{\text{1}}{\text{R}_\text{p}} = \dfrac{\text{1}}{\text{R}_\text{s}} + \dfrac{\text{1}}{\text{2}} \\[1em] \Rightarrow \dfrac{\text{1}}{\text{R}_\text{p}} = \dfrac{\text{1}}{\text{2}} + \dfrac{\text{1}}{\text{8}} \\[1em] \Rightarrow \dfrac{\text{1}}{\text{R}_\text{p}} = \dfrac{4+1}{8} \\[1em] \Rightarrow \dfrac{\text{1}}{\text{R}_\text{p}} = \dfrac{5}{8} \\[1em] \therefore \text{R}_\text{p} = \dfrac{8}{5} = 1.6 \space Ω\\[1em]$
From circuit diagram,
Internal Resistance = 0.4 Ω
Total resistance of circuit = 1.6 + 0.4 = 2 Ω
(b) Current drawn from the battery
$\text{I} = \dfrac{\text{e.m.f}}{\text{Total Resistance}} \\[1em] = \dfrac{4}{2} = 2 \text{ A}$
Now, the current I divides in 2 parts.
Let current across 2Ω be I1 and across the series combination of 5Ω and 3Ω be I2
I = I1 + I2 = 2 A .....(1)
Terminal Voltage of cell V = IRp = 2 x 1.6 = 3.2 V
∴ P.d. across 2 Ω resistor = 3.2 V
I1 = $\dfrac{3.2 \text{ V}}{2 \text{ Ω}}$
= 1.6 A
Substituting value of I1 in Eq 1,
2 = 1.6 + I2
⇒ I2 = 2 - 1.6 = 0.4 A
Hence, current across 3Ω = 0.4 A
#### Question 9(i)
What mass of ice at 0°C added to 2.1 kg water, will cool it down from 75°C to 25°C?
Given: Specific heat capacity of water = 4.2 Jg-1 C-1
Specific latent heat of ice = 336 Jg-1
Given,
mw = 2.1 kg = 2100 g
Specific heat capacity of water = 4.2 Jg-1 C-1
Specific latent heat of ice = 336 Jg-1
mi = ?
Heat energy given out by water in lowering it's temperature from 75° C to 25° C
= m x c x change in temperature
= 2100 x 4.2 x (75 - 25)
= 2100 x 4.2 x 50
= 4,41,000
Heat energy taken by m kg ice to melt into water at 0° C
= mi x L
= mi x 336
Heat energy taken by water at 0° C to raise it's temperature to 25° C
= mi x c x change in temperature
= mi x 4.2 x (25 - 0)
= mi x 4.2 x 25
= mi x 105
heat energy released = heat energy taken
Substituting the values we get,
4,41,000 = (mi x 336) + (mi x 105)
⇒ 4,41,000 = 336mi + 105mi
⇒ 4,41,000 = 441mi
⇒ mi = $\dfrac{441000}{441}$
⇒ mi = 1000 g or 1 Kg
Hence, the mass of ice added = 1 Kg
#### Question 9(ii)
The diagram below shows a cooling curve for a substance:
(a) State the temperatures at which the substance condenses.
(b) The temperature range in which the substance is in liquid state.
(c) Why do we prefer ice to ice-cold water for cooling a drink?
(a) Substance condenses at 150° C
(b) 150 to 60°C
(c) 1 g of ice at 0° C takes 336 J of heat energy from the drink to melt into water at 0° C. Thus, the drink liberates an additional 336 J of heat energy to 1 g ice at 0° C than to 1 g ice-cold water at 0° C. Therefore, cooling produced by 1 g ice at 0° C is much more than that by 1 g water at 0° C.
#### Question 9(iii)
A magnet is released along the axis of a copper coil as shown in the diagram.
(a) State the polarity at the top end of the coil when the magnet leaves the coil.
(b) The direction of the current is from A to B when magnet enters the coil. What will be the direction of the current when the magnet leaves the coil.
(c) Name the law which can be used to determine the direction of the induced current in the coil?
(d) State one way to increase the magnitude of the induced current in the coil? | 8,009 | 29,238 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 33, "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-2024-18 | latest | en | 0.866305 |
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### Factorials Can't Be Squares
```
Date: 02/11/2000 at 13:20:25
From: Animesh Datta
Subject: Number Theory
Prove that the factorial of a number (greater than 1) can never be a
perfect square.
```
```
Date: 02/11/2000 at 13:35:07
From: Doctor Wilkinson
Subject: Re: Number Theory
Hi, Animesh,
This follows from Bertrand's Postulate, which states that there is
always a prime between m and 2m. From this you can conclude that there
is a prime that divides n! but whose square does not. Since in the
prime decomposition of a perfect square every exponent must be even,
this shows that n! cannot be a perfect square.
Here's a sketch of a proof of Bertrand's Postulate:
The proof of Bertand's Postulate uses some simple properties of the
function theta(x), defined for x >= 0 by
theta(x) = sum(log p: p is prime and 0 < p <= x)
We show that
theta(x) < 2x log(2)
(I use log(x) always to mean the natural log of x.)
It is enough to show this when x is an integer. We're going to prove
this by induction.
The trick is to look at the binomial coefficient C(2m+1,m), which is
(2m+1)!/m!(m+1)!
Call this M for short.
Let p be a prime such that m+1 < p <= 2m+1. Then p divides the
numerator of M but not the denominator, so p divides M. So the product
of all such primes divides M, and
sum (log p: m+1 < p <= 2m+1) < log M
or in terms of the function theta(x)
theta(2m+1) - theta(m+1) < log M.
On the other hand, the binomial expansion of (1 + 1)^(2m+1) has two
terms equal to M, so
2M < 2^(2m+1)
M < 2^2m
log M < 2m log(2)
so
theta(2m+1) - theta(m+1) < 2m log(2)
We're going to use this formula in the induction step of our proof
that
theta(x) < 2x log(2)
For x = 1, we have
theta(1) = 0 < 2 log(2)
and for x = 2, we have
theta(2) = log 2 < 4 log(2)
Suppose the inequality is true for x < n. Let us prove it for x = n.
If n is even and > 2, then it is certainly not prime, so
theta(n) = theta(n-1) < 2(n-1) log(2) < 2n log(2).
If n is odd, let n = 2m + 1. Then by what we proved above, we have
theta(2m+1) - theta(m+1) < 2m log(2)
theta(2m+1) < theta(m+1) + 2m log(2)
< 2(m+1) log(2) + 2m log(2)
= (3m + 1) log(2)
< (4m + 2) log(2)
= 2n log(2).
This completes the proof that
theta(x) < 2x log(2).
Let's catch our breath.
The next thing we're going to do is to look at the highest power of p
that divides n!, where p is any prime. We call this number j(n, p).
We use the notation [x] for the largest integer <= x.
Every p'th number is a multiple of p, so we get [n/p] factors of p in
n!. But every p^2'th number is a multiple not just of p but of p^2,
and [n/p] doesn't count these, so we need to add [n/p^2] for these
extra factors of p. Similarly every p^3'th number is a multiple of p^3
which we have not counted yet. So the highest power of p that divides
n! is the sum of all the
[n/p^m]
for m >= 1. Of course [n/p^m] = 0 as soon as p^m > n: that is, for
m > log(n)/log(p).
Now we're going to suppose that Bertrand's Postulate is false, and
that there is no prime p such that n < p < 2n, for some n.
We're going to look at another binomial coefficient. This one is
C(2n,n) = (2n!)/(n!)^2
which we'll call N for short.
By our assumption, all the primes that divide N are <= n. Now using
the notation above, we have
N = (2n)!/(n!)^2
product(p^j(2n, p): p <= 2n)
= ----------------------------
product(p^2j(n, p): p <= n)
but there aren't any primes between n and 2n by assumption, so the
"p <= 2n" in the numerator can be replaced by "p < = n" and we get
N = product (p^(j(2n, p) - 2j(n, p)): p <= n).
Let's call j(2n, p) - 2j(n, p) k(p) for short. Taking logs on both
sides, we get
log N = sum(k(p) log(p): p <= n).
Notice that k(p) is a sum of terms of the form [2x] - 2[x].
[2x] - 2[x] is always either 0 or 1. If [2x] is even, [2x] - 2[x] is
0; otherwise it is 1.
We show first that k(p) = 0 for p > 2n/3. For in that case,
2n/3 < p <= n
or 2 <= 2n/p < 3
and [2n/p] = 2, so [2n/p] - 2[n/p] = 0.
p^2 > (4/9)n^2 > 2n as long as n > 4,
and we can certainly assume n is > 4, since we are assuming there is
no prime between n and 2n, and 5, for example, is between 4 and 8.
So there are no terms involving higher powers of p.
Next we show that terms with k(p) >= 2 don't contribute very much.
To get such a term we have to have p^2 < 2n or p < sqrt(2n), so the
number of such terms is at most sqrt(2n). k(p), on the other hand, is
a sum of terms [2n/p^m] - 2[n/p^m], which is certainly 0 if p^m > 2n,
or m > log(2n)/log(p), so k(p) is at most log(2n)/log(p),and
k(p)log(p) <= log(2n), so
sum(k(p)log(p) : k(p) >= 2) <= sqrt(2n) log(2n)
taking the maximum possible number of such primes p and a number
bigger than any of the k(p)log(p).
For the terms with k(p) = 1, we have at most
sum(log(p): p <= 2n/3) = theta(2n/3) < (4n/3) log(2)
by what we proved way back when.
Putting together what we've got so far gives us
log N < (4n/3) log(2) + sqrt(2n) log(2n).
Time for another breather before we close in for the kill.
Looking back at the definition of N, we have
2^(2n) = 2 + C(2n, 1) + C(2n, 2) + ... + C(2n, 2n-1)
(Binomial Theorem with first and last terms combined).
This is a sum of 2n terms, the largest of which is C(2n, n) or N. So
2^(2n) < 2nN
or
2n log(2) < log(2n) + log(N)
<= log(2n) + (4n/3) log(2) + sqrt(2n) log(2n)
by what we proved just before the breather.
Now for large values of n, the only term that counts on the right side
is the 4n/3 log(2), which is smaller than the 2n log(2). So what we're
going to do is figure out how big n needs to be to make this
inequality false, and then just prove the postulate directly for
smaller values of n. Take n >= 2^9 and note that
log(2n) = log(2^10) = 10 log(2)
Divide the inequality by log(2) to get
2^10 < 10 + 2^10(2/3) + (2^5) 10
or
2^10 (1 - 2/3) < 10 (2^5 + 1)
2^10 (1/3) < 10 (2^5 + 1)
2^10 < 30 (2^5 + 1) < 31 (2^5 + 1) = (2^5 - 1) (2^5 + 1)
= 2^10 - 1
which is false!
So the assumptions that Bertrand's Postulate is false for n and that
n >= 2^9 lead to a contradiction. All that remains is to verify the
postulate for n < 2^9 = 512.
Here we can just look at the sequence of primes
2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631
each of which is less that twice the one before.
- Doctor Wilkinson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Discrete Mathematics
High School Number Theory
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# Mathematical Statistics with Applications, Seventh Edition
www.downloadslide.com 1.2 Characterizing a Set of Measurements: Graphical Methods 5 selection of these items is somewhat at the discretion of the person who is involved in the construction. Although they are arbitrary, a few guidelines can be very helpful in selecting the intervals. Points of subdivision of the axis of measurement should be chosen so that it is impossible for a measurement to fall on a point of division. This eliminates a source of confusion and is easily accomplished, as indicated in Figure 1.1. The second guideline involves the width of each interval and consequently, the minimum number of intervals needed to describe the data. Generally speaking, we wish to obtain information on the form of the distribution of the data. Many times the form will be mound-shaped, as illustrated in Figure 1.2. (Others prefer to refer to distributions such as these as bellshaped, or normal.) Using many intervals with a small amount of data results in little summarization and presents a picture very similar to the data in their original form. The larger the amount of data, the greater the number of included intervals can be while still presenting a satisfactory picture of the data. We suggest spanning the range of the data with from 5 to 20 intervals and using the larger number of intervals for larger quantities of data. In most real-life applications, computer software (Minitab, SAS, R, S+, JMP, etc.) is used to obtain any desired histograms. These computer packages all produce histograms satisfying widely agreed-upon constraints on scaling, number of intervals used, widths of intervals, and the like. Some people feel that the description of data is an end in itself. Histograms are often used for this purpose, but there are many other graphical methods that provide meaningful summaries of the information contained in a set of data. Some excellent references for the general topic of graphical descriptive methods are given in the references at the end of this chapter. Keep in mind, however, that the usual objective of statistics is to make inferences. The relative frequency distribution associated with a data set and the accompanying histogram are sufficient for our objectives in developing the material in this text. This is primarily due to the probabilistic interpretation that can be derived from the frequency histogram, Figure 1.1. We have already stated that the area of a rectangle over a given interval is proportional to the fraction of the total number of measurements falling in that interval. Let’s extend this idea one step further. If a measurement is selected at random from the original data set, the probability that it will fall in a given interval is proportional to the area under the histogram lying over that interval. (At this point, we rely on the layperson’s concept of probability. This term is discussed in greater detail in Chapter 2.) For example, for the data used to construct Figure 1.1, the probability that a randomly selected measurement falls in the interval from 2.05 to 2.45 is .5 because half the measurements fall in this interval. Correspondingly, the area under the histogram in Figure 1.1 over the interval from FIGURE 1.2 Relative frequency distribution Relative Frequency 0 2.05 2.25 2.45 2.65 2.85 3.05
www.downloadslide.com 6 Chapter 1 What Is Statistics? 2.05 to 2.45 is half of the total area under the histogram. It is clear that this interpretation applies to the distribution of any set of measurements—a population or a sample. Suppose that Figure 1.2 gives the relative frequency distribution of profit (in millions of dollars) for a conceptual population of profit responses for contracts at specified settings of the independent variables (size of contract, measure of competition, etc.). The probability that the next contract (at the same settings of the independent variables) yields a profit that falls in the interval from 2.05 to 2.45 million is given by the proportion of the area under the distribution curve that is shaded in Figure 1.2. Exercises 1.2 Are some cities more windy than others? Does Chicago deserve to be nicknamed “The Windy City”? Given below are the average wind speeds (in miles per hour) for 45 selected U.S. cities: 8.9 12.4 8.6 11.3 9.2 8.8 35.1 6.2 7.0 7.1 11.8 10.7 7.6 9.1 9.2 8.2 9.0 8.7 9.1 10.9 10.3 9.6 7.8 11.5 9.3 7.9 8.8 8.8 12.7 8.4 7.8 5.7 10.5 10.5 9.6 8.9 10.2 10.3 7.7 10.6 8.3 8.8 9.5 8.8 9.4 Source: The World Almanac and Book of Facts, 2004. a b c d Construct a relative frequency histogram for these data. (Choose the class boundaries without including the value 35.1 in the range of values.) The value 35.1 was recorded at Mt. Washington, New Hampshire. Does the geography of that city explain the magnitude of its average wind speed? The average wind speed for Chicago is 10.3 miles per hour. What percentage of the cities have average wind speeds in excess of Chicago’s? Do you think that Chicago is unusually windy? 1.3 Of great importance to residents of central Florida is the amount of radioactive material present in the soil of reclaimed phosphate mining areas. Measurements of the amount of 238 U in 25 soil samples were as follows (measurements in picocuries per gram): .74 6.47 1.90 2.69 .75 .32 9.99 1.77 2.41 1.96 1.66 .70 2.42 .54 3.36 3.59 .37 1.09 8.32 4.06 4.55 .76 2.03 5.70 12.48 Construct a relative frequency histogram for these data. 1.4 The top 40 stocks on the over-the-counter (OTC) market, ranked by percentage of outstanding shares traded on one day last year are as follows: 11.88 6.27 5.49 4.81 4.40 3.78 3.44 3.11 2.88 2.68 7.99 6.07 5.26 4.79 4.05 3.69 3.36 3.03 2.74 2.63 7.15 5.98 5.07 4.55 3.94 3.62 3.26 2.99 2.74 2.62 7.13 5.91 4.94 4.43 3.93 3.48 3.20 2.89 2.69 2.61 a b Construct a relative frequency histogram to describe these data. What proportion of these top 40 stocks traded more than 4% of the outstanding shares?
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TESTING RANDOMNESS - Department of Mathematics and Statistics
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Distributions, spatial statistics and a Bayesian perspective - IMAGe | 6,166 | 19,847 | {"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} | 3.8125 | 4 | CC-MAIN-2018-30 | longest | en | 0.92637 |
https://en.wikipedia.org/wiki/Representation_of_a_Lie_group | 1,656,499,140,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00268.warc.gz | 300,538,887 | 44,460 | # Representation of a Lie group
In mathematics and theoretical physics, a representation of a Lie group is a linear action of a Lie group on a vector space. Equivalently, a representation is a smooth homomorphism of the group into the group of invertible operators on the vector space. Representations play an important role in the study of continuous symmetry. A great deal is known about such representations, a basic tool in their study being the use of the corresponding 'infinitesimal' representations of Lie algebras.
## Finite-dimensional representations
### Representations
A complex representation of a group is an action by a group on a finite-dimensional vector space over the field ${\displaystyle \mathbb {C} }$. A representation of the Lie group G, acting on an n-dimensional vector space V over ${\displaystyle \mathbb {C} }$ is then a smooth group homomorphism
${\displaystyle \Pi :G\rightarrow \operatorname {GL} (V)}$,
where ${\displaystyle \operatorname {GL} (V)}$ is the general linear group of all invertible linear transformations of ${\displaystyle V}$ under their composition. Since all n-dimensional spaces are isomorphic, the group ${\displaystyle \operatorname {GL} (V)}$ can be identified with the group of the invertible, complex ${\displaystyle n\times n}$ matrices, generally called ${\displaystyle \operatorname {GL} (n;\mathbb {C} ).}$ Smoothness of the map ${\displaystyle \Pi }$ can be regarded as a technicality, in that any continuous homomorphism will automatically be smooth.[1]
We can alternatively describe a representation of a Lie group ${\displaystyle G}$ as a linear action of ${\displaystyle G}$ on a vector space ${\displaystyle V}$. Notationally, we would then write ${\displaystyle g\cdot v}$ in place of ${\displaystyle \Pi (g)v}$ for the way a group element ${\displaystyle g\in G}$ acts on the vector ${\displaystyle v\in V}$.
A typical example in which representations arise in physics would be the study of a linear partial differential equation having symmetry group ${\displaystyle G}$. Although the individual solutions of the equation may not be invariant under the action of ${\displaystyle G}$, the space ${\displaystyle V}$ of all solutions is invariant under the action of ${\displaystyle G}$. Thus, ${\displaystyle V}$ constitutes a representation of ${\displaystyle G}$. See the example of SO(3), discussed below.
### Basic definitions
If the homomorphism ${\displaystyle \Pi }$ is injective (i.e., a monomorphism), the representation is said to be faithful.
If a basis for the complex vector space V is chosen, the representation can be expressed as a homomorphism into general linear group ${\displaystyle \operatorname {GL} (n;\mathbb {C} )}$. This is known as a matrix representation. Two representations of G on vector spaces V, W are equivalent if they have the same matrix representations with respect to some choices of bases for V and W.
Given a representation ${\displaystyle \Pi :G\rightarrow \operatorname {GL} (V)}$, we say that a subspace W of V is an invariant subspace if ${\displaystyle \Pi (g)w\in W}$ for all ${\displaystyle g\in G}$ and ${\displaystyle w\in W}$. The representation is said to be irreducible if the only invariant subspaces of V are the zero space and V itself. For certain types of Lie groups, namely compact[2] and semisimple[3] groups, every finite-dimensional representation decomposes as a direct sum of irreducible representations, a property known as complete reducibility. For such groups, a typical goal of representation theory is to classify all finite-dimensional irreducible representations of the given group, up to isomorphism. (See the Classification section below.)
A unitary representation on a finite-dimensional inner product space is defined in the same way, except that ${\displaystyle \Pi }$ is required to map into the group of unitary operators. If G is a compact Lie group, every finite-dimensional representation is equivalent to a unitary one.[2]
### Lie algebra representations
Each representation of a Lie group G gives rise to a representation of its Lie algebra; this correspondence is discussed in detail in subsequent sections. See representation of Lie algebras for the Lie algebra theory.
## An example: The rotation group SO(3)
In quantum mechanics, the time-independent Schrödinger equation, ${\displaystyle {\hat {H}}\psi =E\psi }$ plays an important role. In the three-dimensional case, if ${\displaystyle {\hat {H}}}$ has rotational symmetry, then the space ${\displaystyle V_{E}}$ of solutions to ${\displaystyle {\hat {H}}\psi =E\psi }$ will be invariant under the action of SO(3). Thus, ${\displaystyle V_{E}}$ will—for each fixed value of ${\displaystyle E}$—constitute a representation of SO(3), which is typically finite dimensional. In trying to solve ${\displaystyle {\hat {H}}\psi =E\psi }$, it helps to know what all possible finite-dimensional representations of SO(3) look like. The representation theory of SO(3) plays a key role, for example, in the mathematical analysis of the hydrogen atom.
Every standard textbook on quantum mechanics contains an analysis which essentially classifies finite-dimensional irreducible representations of SO(3), by means of its Lie algebra. (The commutation relations among the angular momentum operators are just the relations for the Lie algebra ${\displaystyle {\mathfrak {so}}(3)}$ of SO(3).) One subtlety of this analysis is that the representations of the group and the Lie algebra are not in one-to-one correspondence, a point that is critical in understanding the distinction between integer spin and half-integer spin.
### Ordinary representations
The rotation group SO(3) is a compact Lie group and thus every finite-dimensional representation of SO(3) decomposes as a direct sum of irreducible representations. The group SO(3) has one irreducible representation in each odd dimension.[4] For each non-negative integer ${\displaystyle k}$, the irreducible representation of dimension ${\displaystyle 2k+1}$ can be realized as the space ${\displaystyle V_{k}}$ of homogeneous harmonic polynomials on ${\displaystyle \mathbb {R} ^{3}}$ of degree ${\displaystyle k}$.[5] Here, SO(3) acts on ${\displaystyle V_{k}}$ in the usual way that rotations act on functions on ${\displaystyle \mathbb {R} ^{3}}$:
${\displaystyle (\Pi (R)f)(x)=f(R^{-1}x)\quad R\in \operatorname {SO} (3).}$
The restriction to the unit sphere ${\displaystyle S^{2}}$ of the elements of ${\displaystyle V_{k}}$ are the spherical harmonics of degree ${\displaystyle k}$.
If, say, ${\displaystyle k=1}$, then all polynomials that are homogeneous of degree one are harmonic, and we obtain a three-dimensional space ${\displaystyle V_{1}}$ spanned by the linear polynomials ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$. If ${\displaystyle k=2}$, the space ${\displaystyle V_{2}}$ is spanned by the polynomials ${\displaystyle xy}$, ${\displaystyle xz}$, ${\displaystyle yz}$, ${\displaystyle x^{2}-y^{2}}$, and ${\displaystyle x^{2}-z^{2}}$.
As noted above, the finite-dimensional representations of SO(3) arise naturally when studying the time-independent Schrödinger equation for a radial potential, such as the hydrogen atom, as a reflection of the rotational symmetry of the problem. (See the role played by the spherical harmonics in the mathematical analysis of hydrogen.)
### Projective representations
If we look at the Lie algebra ${\displaystyle {\mathfrak {so}}(3)}$ of SO(3), this Lie algebra is isomorphic to the Lie algebra ${\displaystyle {\mathfrak {su}}(2)}$ of SU(2). By the representation theory of ${\displaystyle {\mathfrak {su}}(2)}$, there is then one irreducible representation of ${\displaystyle {\mathfrak {so}}(3)}$ in every dimension. The even-dimensional representations, however, do not correspond to representations of the group SO(3).[6] These so-called "fractional spin" representations do, however, correspond to projective representations of SO(3). These representations arise in the quantum mechanics of particles with fractional spin, such as an electron.
## Operations on representations
In this section, we describe three basic operations on representations.[7] See also the corresponding constructions for representations of a Lie algebra.
### Direct sums
If we have two representations of a group ${\displaystyle G}$, ${\displaystyle \Pi _{1}:G\rightarrow GL(V_{1})}$ and ${\displaystyle \Pi _{2}:G\rightarrow GL(V_{2})}$, then the direct sum would have ${\displaystyle V_{1}\oplus V_{2}}$ as the underlying vector space, with the action of the group given by
${\displaystyle \Pi (g)(v_{1},v_{2})=(\Pi _{1}(g)v_{1},\Pi _{2}(g)v_{2}),}$
for all ${\displaystyle v_{1}\in V_{1},}$ ${\displaystyle v_{2}\in V_{2}}$, and ${\displaystyle g\in G}$.
Certain types of Lie groups—notably, compact Lie groups—have the property that every finite-dimensional representation is isomorphic to a direct sum of irreducible representations.[2] In such cases, the classification of representations reduces to the classification of irreducible representations. See Weyl's theorem on complete reducibility.
### Tensor products of representations
If we have two representations of a group ${\displaystyle G}$, ${\displaystyle \Pi _{1}:G\rightarrow GL(V_{1})}$ and ${\displaystyle \Pi _{2}:G\rightarrow GL(V_{2})}$, then the tensor product of the representations would have the tensor product vector space ${\displaystyle V_{1}\otimes V_{2}}$ as the underlying vector space, with the action of ${\displaystyle G}$ uniquely determined by the assumption that
${\displaystyle \Pi (g)(v_{1}\otimes v_{2})=(\Pi _{1}(g)v_{1})\otimes (\Pi _{2}(g)v_{2})}$
for all ${\displaystyle v_{1}\in V_{1}}$ and ${\displaystyle v_{2}\in V_{2}}$. That is to say, ${\displaystyle \Pi (g)=\Pi _{1}(g)\otimes \Pi _{2}(g)}$.
The Lie algebra representation ${\displaystyle \pi }$ associated to the tensor product representation ${\displaystyle \Pi }$ is given by the formula:[8]
${\displaystyle \pi (X)=\pi _{1}(X)\otimes I+I\otimes \pi _{2}(X).}$
The tensor product of two irreducible representations is usually not irreducible; a basic problem in representation theory is then to decompose tensor products of irreducible representations as a direct sum of irreducible subspaces. This problem goes under the name of "addition of angular momentum" or "Clebsch–Gordan theory" in the physics literature.
### Dual representations
Let ${\displaystyle G}$ be a Lie group and ${\displaystyle \Pi :G\rightarrow GL(V)}$ be a representation of G. Let ${\displaystyle V^{*}}$ be the dual space, that is, the space of linear functionals on ${\displaystyle V}$. Then we can define a representation ${\displaystyle \Pi ^{*}:G\rightarrow GL(V^{*})}$ by the formula
${\displaystyle \Pi ^{*}(g)=(\Pi (g^{-1}))^{\operatorname {tr} },}$
where for any operator ${\displaystyle A:V\rightarrow V}$, the transpose operator ${\displaystyle A^{\operatorname {tr} }:V^{*}\rightarrow V^{*}}$ is defined as the "composition with ${\displaystyle A}$" operator:
${\displaystyle (A^{\operatorname {tr} }\phi )(v)=\phi (Av).}$
(If we work in a basis, then ${\displaystyle A^{\operatorname {tr} }}$ is just the usual matrix transpose of ${\displaystyle A}$.) The inverse in the definition of ${\displaystyle \Pi ^{*}}$ is needed to ensure that ${\displaystyle \Pi ^{*}}$ is actually a representation of ${\displaystyle G}$, in light of the identity ${\displaystyle (AB)^{\operatorname {tr} }=B^{\operatorname {tr} }A^{\operatorname {tr} }}$.
The dual of an irreducible representation is always irreducible,[9] but may or may not be isomorphic to the original representation. In the case of the group SU(3), for example, the irreducible representations are labeled by a pair ${\displaystyle (m_{1},m_{2})}$ of non-negative integers. The dual of the representation associated to ${\displaystyle (m_{1},m_{2})}$ is the representation associated to ${\displaystyle (m_{2},m_{1})}$.[10]
## Lie group versus Lie algebra representations
### Overview
In many cases, it is convenient to study representations of a Lie group by studying representations of the associated Lie algebra. In general, however, not every representation of the Lie algebra comes from a representation of the group. This fact is, for example, lying behind the distinction between integer spin and half-integer spin in quantum mechanics. On the other hand, if G is a simply connected group, then a theorem[11] says that we do, in fact, get a one-to-one correspondence between the group and Lie algebra representations.
Let G be a Lie group with Lie algebra ${\displaystyle {\mathfrak {g}}}$, and assume that a representation ${\displaystyle \pi }$ of ${\displaystyle {\mathfrak {g}}}$ is at hand. The Lie correspondence may be employed for obtaining group representations of the connected component of the G. Roughly speaking, this is effected by taking the matrix exponential of the matrices of the Lie algebra representation. A subtlety arises if G is not simply connected. This may result in projective representations or, in physics parlance, multi-valued representations of G. These are actually representations of the universal covering group of G.
These results will be explained more fully below.
The Lie correspondence gives results only for the connected component of the groups, and thus the other components of the full group are treated separately by giving representatives for matrices representing these components, one for each component. These form (representatives of) the zeroth homotopy group of G. For example, in the case of the four-component Lorentz group, representatives of space inversion and time reversal must be put in by hand. Further illustrations will be drawn from the representation theory of the Lorentz group below.
### The exponential mapping
Sophus Lie, the originator of Lie theory. The theory of manifolds was not discovered in Lie's time, so he worked locally with subsets of ${\displaystyle \mathbb {R} ^{n}.}$ The structure would today be called a local group.
If ${\displaystyle G}$ is a Lie group with Lie algebra ${\displaystyle {\mathfrak {g}}}$, then we have the exponential map from ${\displaystyle {\mathfrak {g}}}$ to ${\displaystyle G}$, written as
${\displaystyle X\mapsto e^{X},\quad X\in {\mathfrak {g}}.}$
If ${\displaystyle G}$ is a matrix Lie group, the expression ${\displaystyle e^{X}}$ can be computed by the usual power series for the exponential. In any Lie group, there exist neighborhoods ${\displaystyle U}$ of the identity in ${\displaystyle G}$ and ${\displaystyle V}$ of the origin in ${\displaystyle {\mathfrak {g}}}$ with the property that every ${\displaystyle g}$ in ${\displaystyle U}$ can be written uniquely as ${\displaystyle g=e^{X}}$ with ${\displaystyle X\in V}$. That is, the exponential map has a local inverse. In most groups, this is only local; that is, the exponential map is typically neither one-to-one nor onto.
### Lie algebra representations from group representations
It is always possible to pass from a representation of a Lie group G to a representation of its Lie algebra ${\displaystyle {\mathfrak {g}}.}$ If Π : G → GL(V) is a group representation for some vector space V, then its pushforward (differential) at the identity, or Lie map, ${\displaystyle \pi :{\mathfrak {g}}\to {\text{End}}V}$ is a Lie algebra representation. It is explicitly computed using[12]
${\displaystyle \pi (X)=\left.{\frac {d}{dt}}\Pi (e^{tX})\right|_{t=0},\quad X\in {\mathfrak {g}}.}$
(G6)
A basic property relating ${\displaystyle \Pi }$ and ${\displaystyle \pi }$ involves the exponential map:[12]
${\displaystyle \Pi (e^{X})=e^{\pi (X)}.}$
The question we wish to investigate is whether every representation of ${\displaystyle {\mathfrak {g}}}$ arises in this way from representations of the group ${\displaystyle G}$. As we shall see, this is the case when ${\displaystyle G}$ is simply connected.
### Group representations from Lie algebra representations
The main result of this section is the following:[13]
Theorem: If ${\displaystyle G}$ is simply connected, then every representation ${\displaystyle \pi }$ of the Lie algebra ${\displaystyle {\mathfrak {g}}}$ of ${\displaystyle G}$ comes from a representation ${\displaystyle \Pi }$ of ${\displaystyle G}$ itself.
From this we easily deduce the following:
Corollary: If ${\displaystyle G}$ is connected but not simply connected, every representation ${\displaystyle \pi }$ of ${\displaystyle {\mathfrak {g}}}$ comes from a representation ${\displaystyle \Pi }$ of ${\displaystyle {\tilde {G}}}$, the universal cover of ${\displaystyle G}$. If ${\displaystyle \pi }$ is irreducible, then ${\displaystyle \Pi }$ descends to a projective representation of ${\displaystyle G}$.
A projective representation is one in which each ${\displaystyle \Pi (g),\,g\in G,}$ is defined only up to multiplication by a constant. In quantum physics, it is natural to allow projective representations in addition to ordinary ones, because states are really defined only up to a constant. (That is to say, if ${\displaystyle \psi }$ is a vector in the quantum Hilbert space, then ${\displaystyle c\psi }$ represents the same physical state for any constant ${\displaystyle c}$.) Every finite-dimensional projective representation of a connected Lie group ${\displaystyle G}$ comes from an ordinary representation of the universal cover ${\displaystyle {\tilde {G}}}$ of ${\displaystyle G}$.[14] Conversely, as we will discuss below, every irreducible ordinary representation of ${\displaystyle {\tilde {G}}}$ descends to a projective representation of ${\displaystyle G}$. In the physics literature, projective representations are often described as multi-valued representations (i.e., each ${\displaystyle \Pi (g)}$ does not have a single value but a whole family of values). This phenomenon is important to the study of fractional spin in quantum mechanics.
Here V is a finite-dimensional vector space, GL(V) is the set of all invertible linear transformations on V and ${\displaystyle {\mathfrak {gl}}(V)}$ is its Lie algebra. The maps π and Π are Lie algebra and group representations respectively, and exp is the exponential mapping. The diagram commutes only up to a sign if Π is projective.
We now outline the proof of the main results above. Suppose ${\displaystyle \pi :{\mathfrak {g}}\to {\mathfrak {gl}}(V)}$ is a representation of ${\displaystyle {\mathfrak {g}}}$ on a vector space V. If there is going to be an associated Lie group representation ${\displaystyle \Pi }$, it must satisfy the exponential relation of the previous subsection. Now, in light of the local invertibility of the exponential, we can define a map ${\displaystyle \Pi }$ from a neighborhood ${\displaystyle U}$ of the identity in ${\displaystyle G}$ by this relation:
${\displaystyle \Pi (e^{X})=e^{\pi (X)},\quad g=e^{X}\in U.}$
A key question is then this: Is this locally defined map a "local homomorphism"? (This question would apply even in the special case where the exponential mapping is globally one-to-one and onto; in that case, ${\displaystyle \Pi }$ would be a globally defined map, but it is not obvious why ${\displaystyle \Pi }$ would be a homomorphism.) The answer to this question is yes: ${\displaystyle \Pi }$ is a local homomorphism, and this can be established using the Baker–Campbell–Hausdorff formula.[15]
If ${\displaystyle G}$ is connected, then every element of ${\displaystyle G}$ is at least a product of exponentials of elements of ${\displaystyle {\mathfrak {g}}}$. Thus, we can tentatively define ${\displaystyle \Pi }$ globally as follows.
{\displaystyle {\begin{aligned}\Pi (g=e^{X})&\equiv e^{\pi (X)},&&X\in {\mathfrak {g}},\quad g=e^{X}\in \operatorname {im} (\exp ),\\\Pi (g=g_{1}g_{2}\cdots g_{n})&\equiv \Pi (g_{1})\Pi (g_{2})\cdots \Pi (g_{n}),&&g\notin \operatorname {im} (\exp ),\quad g_{1},g_{2},\ldots ,g_{n}\in \operatorname {im} (\exp ).\end{aligned}}}
(G2)
Note, however, that the representation of a given group element as a product of exponentials is very far from unique, so it is very far from clear that ${\displaystyle \Pi }$ is actually well defined.
To address the question of whether ${\displaystyle \Pi }$ is well defined, we connect each group element ${\displaystyle g\in G}$ to the identity using a continuous path. It is then possible to define ${\displaystyle \Pi }$ along the path, and to show that the value of ${\displaystyle \Pi (g)}$ is unchanged under continuous deformation of the path with endpoints fixed. If ${\displaystyle G}$ is simply connected, any path starting at the identity and ending at ${\displaystyle g}$ can be continuously deformed into any other such path, showing that ${\displaystyle \Pi (g)}$ is fully independent of the choice of path. Given that the initial definition of ${\displaystyle \Pi }$ near the identity was a local homomorphism, it is not difficult to show that the globally defined map is also a homomorphism satisfying (G2).[16]
If ${\displaystyle G}$ is not simply connected, we may apply the above procedure to the universal cover ${\displaystyle {\tilde {G}}}$ of ${\displaystyle G}$. Let ${\displaystyle p:{\tilde {G}}\rightarrow G}$ be the covering map. If it should happen that the kernel of ${\displaystyle \Pi :{\tilde {G}}\rightarrow \operatorname {GL} (V)}$ contains the kernel of ${\displaystyle p}$, then ${\displaystyle \Pi }$ descends to a representation of the original group ${\displaystyle G}$. Even if this is not the case, note that the kernel of ${\displaystyle p}$ is a discrete normal subgroup of ${\displaystyle {\tilde {G}}}$, which is therefore in the center of ${\displaystyle {\tilde {G}}}$. Thus, if ${\displaystyle \pi }$ is irreducible, Schur's lemma implies that the kernel of ${\displaystyle p}$ will act by scalar multiples of the identity. Thus, ${\displaystyle \Pi }$ descends to a projective representation of ${\displaystyle G}$, that is, one that is defined only modulo scalar multiples of the identity.
A pictorial view of how the universal covering group contains all such homotopy classes, and a technical definition of it (as a set and as a group) is given in geometric view.
For example, when this is specialized to the doubly connected SO(3, 1)+, the universal covering group is ${\displaystyle {\text{SL}}(2,\mathbb {C} )}$, and whether its corresponding representation is faithful decides whether Π is projective.
## Classification in the compact case
If G is a connected compact Lie group, its finite-dimensional representations can be decomposed as direct sums of irreducible representations.[17] The irreducibles are classified by a "theorem of the highest weight." We give a brief description of this theory here; for more details, see the articles on representation theory of a connected compact Lie group and the parallel theory classifying representations of semisimple Lie algebras.
Let T be a maximal torus in G. By Schur's lemma, the irreducible representations of T are one dimensional. These representations can be classified easily and are labeled by certain "analytically integral elements" or "weights." If ${\displaystyle \Sigma }$ is an irreducible representation of G, the restriction of ${\displaystyle \Sigma }$ to T will usually not be irreducible, but it will decompose as a direct sum of irreducible representations of T, labeled by the associated weights. (The same weight can occur more than once.) For a fixed ${\displaystyle \Sigma }$, one can identify one of the weights as "highest" and the representations are then classified by this highest weight.
An important aspect of the representation theory is the associated theory of characters. Here, for a representation ${\displaystyle \Sigma }$ of G, the character is the function
${\displaystyle \chi _{G}:G\rightarrow \mathbb {C} }$
given by
${\displaystyle \chi _{G}(g)=\operatorname {trace} (\Sigma (g)).}$
Two representations with the same character turn out to be isomorphic. Furthermore, the Weyl character formula gives a remarkable formula for the character of a representation in terms of its highest weight. Not only does this formula gives a lot of useful information about the representation, but it plays a crucial role in the proof of the theorem of the highest weight.
## Unitary representations on Hilbert spaces
Let V be a complex Hilbert space, which may be infinite dimensional, and let ${\displaystyle U(V)}$ denote the group of unitary operators on V. A unitary representation of a Lie group G on V is a group homomorphism ${\displaystyle \Pi :G\rightarrow U(V)}$ with the property that for each fixed ${\displaystyle v\in V}$, the map
${\displaystyle g\mapsto \Pi (g)v}$
is a continuous map of G into V.
### Finite-dimensional unitary representations
If the Hilbert space V is finite-dimensional, there is an associated representation ${\displaystyle \pi }$ of the Lie algebra ${\displaystyle {\mathfrak {g}}}$ of ${\displaystyle G}$. If ${\displaystyle G}$ is connected, then the representation ${\displaystyle \Pi }$ of ${\displaystyle G}$ is unitary if and only if ${\displaystyle \pi (X)}$ is skew-self-adjoint for each ${\displaystyle X\in {\mathfrak {g}}}$.[18]
If ${\displaystyle G}$ is compact, then every representation ${\displaystyle \Pi }$ of ${\displaystyle G}$ on a finite-dimensional vector space V is "unitarizable," meaning that it is possible to choose an inner product on V so that each ${\displaystyle \Pi (g),\,g\in G}$ is unitary.[19]
### Infinite-dimensional unitary representations
If the Hilbert space V is allowed to be infinite dimensional, the study of unitary representations involves a number of interesting features that are not present in the finite dimensional case. For example, the construction of an appropriate representation of the Lie algebra ${\displaystyle {\mathfrak {g}}}$ becomes technically challenging. One setting in which the Lie algebra representation is well understood is that of semisimple (or reductive) Lie groups, where the associated Lie algebra representation forms a (g,K)-module.
Examples of unitary representations arise in quantum mechanics and quantum field theory, but also in Fourier analysis as shown in the following example. Let ${\displaystyle G=\mathbb {R} }$, and let the complex Hilbert space V be ${\displaystyle L^{2}(\mathbb {R} )}$. We define the representation ${\displaystyle \psi :\mathbb {R} \rightarrow U(L^{2}(\mathbb {R} ))}$ by
${\displaystyle [\psi (a)(f)](x)=f(x-a).}$
Here are some important examples in which unitary representations of a Lie group have been analyzed.
## Projective representations
In quantum physics, one is often interested in projective unitary representations of a Lie group ${\displaystyle G}$. The reason for this interest is that states of a quantum system are represented by vectors in a Hilbert space ${\displaystyle \mathbf {H} }$—but with the understanding that two states differing by a constant are actually the same physical state. The symmetries of the Hilbert space are then described by unitary operators, but a unitary operator that is a multiple of the identity does not change the physical state of the system. Thus, we are interested not in ordinary unitary representations—that is, homomorphisms of ${\displaystyle G}$ into the unitary group ${\displaystyle U(\mathbf {H} )}$—but rather in projective unitary representations—that is, homomorphisms of ${\displaystyle G}$ into the projective unitary group
${\displaystyle PU(\mathbf {H} ):=U(\mathbf {H} )/\{e^{i\theta }I\}.}$
To put it differently, for a projective representation, we construct a family of unitary operators ${\displaystyle \rho (g),\,\,g\in G}$, where it is understood that changing ${\displaystyle \rho (g)}$ by a constant of absolute value 1 is counted as "the same" operator. The operators ${\displaystyle \rho (g)}$ are then required to satisfy the homomorphism property up to a constant:
${\displaystyle \rho (g)\rho (h)=e^{i\theta _{g,h}}\rho (gh).}$
We have already discussed the irreducible projective unitary representations of the rotation group SO(3) above; considering projective representations allows for fractional spin in addition to integer spin.
Bargmann's theorem states that for certain types of Lie groups ${\displaystyle G}$, irreducible projective unitary representations of ${\displaystyle G}$ are in one-to-one correspondence with ordinary unitary representations of the universal cover of ${\displaystyle G}$. Important examples where Bargmann's theorem applies are SO(3) (as just mentioned) and the Poincaré group. The latter case is important to Wigner's classification of the projective representations of the Poincaré group, with applications to quantum field theory.
One example where Bargmann's theorem does not apply is the group ${\displaystyle \mathbb {R} ^{2n}}$. The set of translations in position and momentum on ${\displaystyle L^{2}(\mathbb {R} ^{n})}$ form a projective unitary representation of ${\displaystyle \mathbb {R} ^{2n}}$ but they do not come from an ordinary representation of the universal cover of ${\displaystyle \mathbb {R} ^{2n}}$—which is just ${\displaystyle \mathbb {R} ^{2n}}$ itself. In this case, to get an ordinary representation, one has to pass to the Heisenberg group, which is a one-dimensional central extension of ${\displaystyle \mathbb {R} ^{2n}}$. (See the discussion here.)
## The commutative case
If ${\displaystyle G}$ is a commutative Lie group, then every irreducible unitary representation of ${\displaystyle G}$ on complex vector spaces is one dimensional. (This claim follows from Schur's lemma and holds even if the representations are not assumed ahead of time to be finite dimensional.) Thus, the irreducible unitary representations of ${\displaystyle G}$ are simply continuous homomorphisms of ${\displaystyle G}$ into the unit circle group, U(1). For example, if ${\displaystyle G=\mathbb {R} }$, the irreducible unitary representations have the form
${\displaystyle \Pi (x)=[e^{iax}]}$,
for some real number ${\displaystyle a}$.
## Notes
1. ^ Hall 2015 Corollary 3.51
2. ^ a b c Hall 2015 Theorem 4.28
3. ^ Hall 2015 Section 10.3
4. ^ Hall 2015 Section 4.7
5. ^ Hall 2013 Section 17.6
6. ^ Hall 2015 Proposition 4.35
7. ^ Hall 2015, Section 4.3
8. ^ Hall 2015, Proposition 4.18
9. ^ Hall 2015 Proposition 4.22
10. ^ Hall 2015 Chapter 6, Exercise 3. See also Chapter 10, Exercise 10
11. ^ Hall 2015 Theorem 5.6
12. ^ a b Hall 2015, Theorem 3.28
13. ^ Hall 2015, Theorem 5.6
14. ^ Hall 2013, Section 16.7.3
15. ^ Hall 2015, Proposition 5.9
16. ^ Hall 2015, Theorem 5.10
17. ^ Hall 2015 Theorems 4.28
18. ^ Hall 2015 Proposition 4.8
19. ^ Hall 2015 proof of Proposition 4.28
## References
• Fulton, W.; Harris, J. (1991). Representation theory. A first course. Graduate Texts in Mathematics. Vol. 129. New York: Springer-Verlag. ISBN 978-0-387-97495-8. MR 1153249.
• Hall, Brian C. (2013), Quantum Theory for Mathematicians, Graduate Texts in Mathematics, vol. 267, Springer, ISBN 978-1461471158.
• Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Graduate Texts in Mathematics, vol. 222 (2nd ed.), Springer, ISBN 978-3319134666.
• Knapp, Anthony W. (2002), Lie Groups Beyond an Introduction, Progress in Mathematics, vol. 140 (2nd ed.), Boston: Birkhäuser.
• Rossmann, Wulf (2001), Lie Groups: An Introduction Through Linear Groups, Oxford Graduate Texts in Mathematics, Oxford University Press, ISBN 978-0-19-859683-7. The 2003 reprint corrects several typographical mistakes.
• Weinberg, S. (2002) [1995], Foundations, The Quantum Theory of Fields, vol. 1, Cambridge: Cambridge University Press, ISBN 0-521-55001-7 | 7,979 | 31,729 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 250, "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-2022-27 | latest | en | 0.854976 |
rizalzulfahmy.wordpress.com | 1,500,683,401,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423839.97/warc/CC-MAIN-20170722002507-20170722022507-00331.warc.gz | 690,510,690 | 36,675 | # Digital Systems
## Section 5
PROBLEMS
SECTIONS 2-1 AND 2-2
2-1 Convert these binary numbers to decimal.
1. 10110
2. 10001101
3. 100100001001
4. 1111010111
5. 10111111
2-2 Convert the following decimal values to binary.
1. 37
2. 14
3. 189
4. 205
5. 2313
6. 511
2-3 what is the largest decimal value that can be represented by an 8-bit binary number? A 16-bit number?
SECTION 2-3
2-4 Convert each octal number to its decimal equivalent.
1. 743
2. 36
3. 3777
4. 257
5. 1204
2-5 Convert each of the following decimal numbers to octal.
1. 59
2. 372
3. 919
4. 65,536
5. 255
2-6 Convert each of the octal values from Problem 2-4 to binary.
2-7 Convert the binary numbers in Problem 2-1 to octal.
2-8 List the binary numbers in Problem 2-1 to octal.
2-9 When a large decimal number is to be converted to binary, it is sometimes easier to convert it first to octal, and then from octal to binary. Try this procedure for 2313 and compare it with the procedure used in problem 2-2(e).
SECTION 2-4
2-10 Convert these hex values to decimal.
1. 92
2. 1A6
3. 37FD
4. 2CO
5. 7FF
2-11 Convert these decimal values to hex.
1. 75
2. 314
3. 2048
4. 25,619
5. 4095
2-12 Convert the binary numbers in Problem 2-1 to hexadecimal.
2-13 Convert the hex values in problem 2-10 to binary.
2-14 In most microcomputers the addresses of memory locations are specified in hexadecimal. These addresses are sequential numbers that identify each memory circuit.
1. A particular microcomputer store an 8-bit number in each memory location. If the memory addresses range from 0000 to FFFF, how many memory locations are there?
2. Another microcomputer is specified to have 4096 memory locations. What range of hex addresses does this computer use?
2-15 List the hex numbers in sequence from 280 to 2A0.
SECTION 2-5
2-16 Encode these decimal numbers in BCD
1. 47
2. 962
3. 187
4. 42,689,627
5. 1204
How many bits are required to represent the decimal numbers in the range from 0 to 999 using straight binary code? Using BCD code?
2-18 The following numbers are in BCD. Convert them to decimal.
1. 1001011101010010
2. 000110000100
3. 0111011101110101
4. 010010010010
SECTION 2-8
2-19 Represent the statement “X=25/Y” in ASCII code (excluding quotes). Attach an even-parity bit.
2-20 Attach an even-parity bit to each of the ASCII codes for problem 2-19 and give the results in hex.
2-21 The following code groups are being transmitted. Attach an even-parity bit to each group.
1. 10110110
2. 00101000
3. 11110111
SECTION 2-9
2-22 Convert the following decimal numbers to BCD code and then attach an odd-parity bit.
1. 74
2. 38
3. 165
4. 9201
2-23. In a certain digital system, the decimal numbers from 000 through 999 are represented in BCD code. An odd-parity bit is also included at the of each code group. Examine each of the code groups below and assume that each one has just been transferred from one location to another. some of the groups contain errors. Assume taht no more than two errors have occured for each group. Determine which of the code groups have a single error and which of them definitely have a double error. (Hint: Remember that this is a BCD code.)
1. 1001010110000
2. 0100011101100
3. 0111110000011
4. 1000011000101
2-24 Suppose that the receiver received the following data from the transmitter of Example 2-10.
0 1 0 0 1 0 0 0
1 1 0 0 0 1 0 1
1 1 0 0 1 1 0 0
1 1 0 0 1 0 0 0
1 1 0 0 1 1 0 0
DRILL QUESTIONS
2-25 Perform each of the following conversions. For some of them, you may want to try several methods to see which one works best for you. For example, a binary-to-decimal conversion may be done directly, or it may be done as a binary-to-octal conversion followed by an octal-to-decimal conversion.
DRILL QUESTIONS
2-25. Perform each of the following coversions. For some of them, you may want to try several methods to see which one works best for you. For example, a binary-to-decimal conversion may be done directly, or it may be done as a binary-to-octal conversion followed by an octal-to-decimal conversion.
1. 1417=____________
2. 255 =____________
3. 11010001=________
4. 11101010000100111=______________
5. 111010110000100111=_____________
6. 2497=____________
7. 511=_____________
8. 235=_____________
9. 4316=____________
10. 7A9=_____________
11. 3E1c=____________
12. 1600=____________
13. 38,187=___________
14. 865=_____________(BCD)
15. 10010100011 (BCD)=__________
16. 465=_____________
17. B34=_____________
18. 01110100(BCD)=_____________
19. 111010=__________(BCD)
2-26. Represent the decimal value 37 in each of the following ways.
1. straight binary
2. BCD
3. Hex
4. ASCII
5. Octal
2-27 Fill in the blanks with the correct word or words.
1. Conversion from decimal to _______requires repeated division by 8.
2. Conversion from decimal to hex requires repeated division by_______.
3. In the BCD code, each___________ is converted to its 4-bit binary equivalent.
4. The_________code has the characteristic that only one bit changes in going from one step to the next.
5. A transmitter attaches a________to a code group to allow the receiver to detect_______.
6. The__________code is the most common alphanumeric code used in computer systems.
7. __________and_________are often used as a convenient way to represent large binary numbers.
2-28. Write to binary number that results when each of the following numbers is incremented by one: (a) 0111 (b) 010000 (c) 1110
2-29. Repeat Problem 2-28 for the decrement operation.
2-30. Write the number that results when each of the following is incremented:
1. 7777
2. 7777
3. 2000
4. 2000
5. 9FF
6. 1000
2-31. Repeat problem 2-30 for the decrement operation.
CHALLENGING EXERCISES
2-32. Perform the following conversions between base-5 and decimal
1. 3421=___________
2. 726 =___________
2-33 Convert the following binary number directly into its base-4 equivalent: 01001110
2-34 Construct a table showing the binary, octal, hex and BCD representations of all decimal numbers from 0 to 15. Compare your table with Table 2-3.
November 22, 2007 Posted by | Digital Systems | Tinggalkan komen
## Implications of DeMorgan’s Theorems
Let us examine these teorems (16) and (17) from the standpoint of logic circuits. First, consider theorem (16), X + y =x.y The left-hand side of the equation can be viewed as the output of a NOR gate whose inputs are x and y. The right-hand side of the equation, on the other hand, is the result of first inverting both x and y and then putting them through an AND gate. These two representations are equivalent and illustrated in Figure 3-26(a). Figure 3-26 (a) Equivalent circuits implied by theorem (16); (b)alternative symbol for the NOR function. Figure 3-27 (a) Equivalent circuits implied by theorem (17); (b) alternative symbol for the NAND function. What this means is that an AND gate with INVERTERs on each of its inputs is equivalent to a NOR gate. In fact, both representations are used to represent the NOR function. When the AND gate with inverted inputs is used to represent the NOR function, it is usually drawn as shown in Figure 3-26(b), where the small circles on the inputs represent the inversion operation. Now consider theorem (17), X.Y=X+Y The left side of the equation can be implemented by a NAND gate with inputs x and y. The right side can be implemented by first inverting inputs x and y and then putting them through an OR gate. These two equivalent representations are shown in Figure 3-27(a). The OR gate with INVERTERs on each of its inputs is equivalent to the NAND gate. In fact, both representations are used to represent the NAND function. When the OR gate with inverted inputs is used to represent the NAND function, it is usually drawn as shown in Figure 3-27(b), where the circles again represent inversion.
November 9, 2007 Posted by | Digital Systems | Tinggalkan komen
## WHICH GATE REPRESENTATION TO USE
Some logic-circuit designers and many textbooks use only the standard logic-gate symbols in their circuit schematics. While this practice is not incorrect, it does nothing to make the circuit operation easier to follow. Proper use of the alternate gate symbols in the circuit diagram can make the circuit operation easier to follow. Proper use of the alternate gate symbols in the circuit diagram can make the circuit operation much clearer. This can be illustrated by considering the example shown in Figure 3-36.
The circuit in the Figure 3-36(a) contains three NAND gates connected to produce an output Z that depends on inputs A, B, C, D. The circuit diagram uses the standard symbol for each of the NAND gates. While this diagram is logically correct, it does not facilitate any understanding of how the circuit functions. The improved circuit representations given in Figure 3-36(b) and (c), however, can be analyzed more easily to determine the circuit operation.
The representation of Figure 3-36(b) is obtained from from the original circuit diagram by replacing NAND gate 3 with its alternate symbol. In this diagram, output Z is taken from a Nand gate symbol that has an active-HIGH output. Thus, we can say that Z will go HIGH when either X or Y is LOW. Now, since X and Y each appear at the output of NAND symbols having active-LOW outputs, we can say that Z will go LOW only if A=B=1, and Y will go LOW only if C=D=1. Putting this all together, we can describe the circuit operation as follows:
Output Z will go HIGH whenever either A=B=1 or C=D=1 (or both).
This description can be translated to truth-table form by setting Z=1 for those cases where A=B=1, and for those cases where C=D=1. For all other cases, Z is made a 0. The resultant truth table is shown in Figure 3-36(d).
The representation of Figure 3-36(c) is obtained from the original circuit diagram by replacing NAND gates 1 and 2 by their alternate symbols. In this equivalent representation the Z ouput is taken from a NAND gate that has an active-LOW output. Thus, we can say that Z will go LOW only when X=Y=1. Since X and Y are active-HIGH outputs, we can say that X will be HIGH when either A or B is LOW, and Y will be HIGH when either C or D is LOW. Putting this all together, we can describe the circuit operation as follows:
Output Z will go LOW only when A or B is LOW and C or D is LOW.
This description can be translated to truth-table by making Z=0 for all cases where at least one of the A or B inputs is LOW at the same time that at least one of the C or D inputs is LOW. For all other cases, Z is made a 1. The resultant truth table is the same as that obtained for the circuit diagram for the circuit diagram of Figure 3-36(b).
November 9, 2007 Posted by | Digital Systems | Tinggalkan komen
## Logic Symbol Interpretation
Each of the logi-gate symbols of figure 3-33 provides a unique interpretation of how the gate operates. Before we can demonstrate these interpretations, we must first establish the concept of active logic levels.
When an input or output line on a logic circuit symbol has no bubble on it, that line is said to be active-HIGH, when an input or output line does have a bubble on it, that line is said to be active-LOW. the presence or absence of a bubble, then, determines the active-HIGH/actIve-LOW states of a circuit’s input and output, and is used to interpret the circuit operation.
To illustrate, Figure 3-34(a) shows the standard symbol for a NAND gate. The standard symbol has a bubble on its output and no bubbles on its inputs. Thus, it has an active-LOW output and active-HIGH inputs. The logic operation represented by this symbol can therefore be interpreted as follows:
The output goes Low only when all the inputs are High, Note that this says that the output will go to its active state only when all the inputs are in their active states. The word ‘all’ is used because of the AND symbol.
The alternate symbol for a NAND gate shown in Figure 3-34(b) has an active-HIGH output and active-LOW inputs, and so its operation can be stated as
The output goes HIGH when any input is LOW.
Again, this says that the output will be in its active state whenever any of the inputs is in its active state. The word ‘any’ is used because of the OR symbol.
With a little thought, it can be seen that the two interpretations for the NAND symbols in figure 3-34 are different ways of saying the same thing.
Summary At this point you are probably wondering why there is a need to have two different symbols and interpretations for each logic gate. Hopefully, the reasons will become clear after reading the nest section. For now, let us summarize the important points concerning the logic-gate representations.
1- To obtain the alternate symbol for a logic gate, take the standard symbol and change its operation symbol (OR to AND, or AND to OR) and change the bubbles on both inputs and output (i.e., delete bubbles where there are none).
2. To interpret the logic-gate operation, first note which logic state, 0 or 1, is the active state for the inputs, and which is the active state for the output. Then realize that the outputs active state is produced by having all the inputs in their active state (if ANd and symbol is used), or having any of the inputs in its active state (if an OR symbol is used).
Oktober 6, 2007 Posted by | Digital Systems | Tinggalkan komen
## Digital
When most of us hear the term digital, we immediately think of “digital calculator” or “digital computer”. This can probably be attributed to the dramatic way that low-cost, powerful calculators and computer have become accessible to the average person. It is important to realize that calculators and computers represent only one of the many applications of digital circuits and principles. Digital circuits are used in electronic products such as video games, microwave ovens, and automobile control systems, and in test equipment such as meters, generators, and oscilloscopes. Digital techniques have also replaced a lot of the older “analog circuits” used in consumer products such as radios, TV sets, and high-fidelity sound recording and playback equipment.
In this book we are going to study the principles and techniques that are common to all digital systems from the simplest on/off switch to the most complex computer. If this book is successful, you should gain a deep understanding of how all digital systems work, and you should be able to apply this understanding to the analysis and troubleshooting of any digital system.
We start by introducing some underlying concepts that are a vital part of digital technology; these concepts will expanded on as they are needed later in the book. We also introduce some of the terminology that is necessary when embarking on a new field of study, and add to it in every chapter. A complete glossary of terminology is presented in Appendix I.
Oktober 2, 2007 Posted by | Digital Systems | Tinggalkan komen
## 1-2 DIGITAL AND ANALOG SYSTEMS
A digital system is a combination of devices designed to manipulate physical quantities or information that are represented in digital form; that is, they can take on only discrete values. These devices are most often electronic, but they can also be mechanical, magnetic, or pneumatic. Some of the more familiar digital systems include digital computers and calculators, digital audio and video equipment, and the telephone system-the world’s largest digital system.
An analog system contains devices that manipulate physical quantities that are represented in analog form. In an analog system, the quantities can vary over a continunous range of values. For example, the amplitude of the output signal to the speaker in a radio receiver can have any value between zero and its maximum limit. Other common analog systems are audio amplifiers, magnetic tape recording and playback eqipment, and the automobile odometer.
Oktober 2, 2007 Posted by | Digital Systems | Tinggalkan komen
## 1-7 MEMORY
When an input signal is applied to most devices or circuits, the output somehow changes in response to the input , and when the input signal is removed, the output returns to its original state. These circuits do not exhibit the property of memory, since their outputs revert back to normal. When an input is applied to such a circuit, the output will change its state, but it will remain in the new state even after the input is removed. This property of retaining its response to a momentary input is called memory. Figure 1-10 illustrates nonmemory operations.
Memory devices and circuits play an important role in digital systems because they provide means for storing binary numbers either temporarily or permanently, with the ability to change the stored information at any time. As we shall see, the various memory elements include magnetic types and those which utilize electronic latching circuits (called latches and flip-flops)
Oktober 2, 2007 Posted by | Digital Systems | Tinggalkan komen
Advantages of Digital Techniques An increasing majority of applications in electronics, as well as in most other technologies, use digital techniques to perform operations that were once performed using analog methods. The chief reasons for the shift to digital technology are:
1. Digital systems are generally easier to design. This is because the circuits that are used are switching circuits, where exact values of voltage or current are not important, only the range (HIGH or LOW) in which they fall.
2. Information storage is easy. this is accomplished by special switching circuits that can latch onto information and hold it for as long as necessary.
3. Accuracy and precision are greater. Digital systems can handle as many digits of precision as you need simply by adding more switching circuits. In analog systems, precision is usually limited to three or four digits because the values of voltage and cureent are directly dependent on the circuit component values.
4. Operation can be programmed. It is fairly easy to design digital systems whose operation is controlled by a set of stored instructions called a program. As technology progresses, this is becoming even easier. Analog systems can also be programmed, but the variety and complexity of the available operations is severely limited.
5. Digital circuits are less affected by noise. Spurious fluctuations in voltage (noise) are not as critical in digital systems because the exact value of a voltage is not important, as long as the noise is not large enough to prevent us from distinguishing a HIGH from a LOW.
6. More digital circuitry can be fabricated on IC chips. It is true that analog circuitry has also benefited from the tremendous development of IC technology, but its relative complexity and its use of devices that cannot be economically integrated (high-value capacitors, precision resistors, inductors, transformer) have prevented analog systems from achieving the same high degree of integration.
September 29, 2007 Posted by | Digital Systems | Tinggalkan komen
## 1-1 NUMERICAL REPRESENTATIONS
1-1 NUMERICAL REPRESENTATIONS
In science, technology, business, and, in fact, most other fields of endeavor, we are constantly dealing with quantities. Quantities are measured, monitored, recorded, manipulated arithmetically, observed, or in some other way utilized in most physical systems. It is important when dealing with various quantities that we be able to represent their values efficiently and accurately. There are basically two ways of representing the numerical value of quantities: analog and digital.
Analog Representations In analog representation a quantity is represented by a voltage, current, or meter movement that is proportional to the value of that quantity. An example is an automobile speedometer, in which the deflection of the needle is propotional to the speed of the auto. The angular position of the needle represents the value of the auto’s speed, and the needle follows any changes that occurs as the auto speeds up or slow down.
Another example is the common room thermostat, in which the bending of the bimetallic strip is proportional to the room temperature. As the temperature changes grdually, the curvature of the strip changes proportionally.
Still another example of an analog quantity is found in the familiar audio microphone. In this device an output voltage is generated in proportion to the amplitude of the sound waves that impinge on the microphone. The variations in the output voltage follow the same variations as the input sound.
Analog quantities such as those cited above have an important characteristic: they can vary over a continuous range of values. The automobile speed can have any value between zero and, say, 100 mph. Similarly, the microphone output might be anywhere within a range of zero to 10 mV (e.g., 1mV, 2.3724 mV, 9.9999 mV).
Digital Representations In digital representation the quantities are represented not by proportional quantities but by symbols called digits. As an example, consider the digital watch, which provides the time of day in the form of decimal digits which represent hours and minutes (and sometimes seconds). As we know, the time of day changes continuously, but the digital watch reading does not change continuously; rather, it changes in steps of one per minute (or per second). In other word, this digital representation of the time of day changes in discrete steps, as compared with the representation of time provided by an analog watch, where the dial reading changes continuously.
The major difference between analog and digital quantities, then, can be simply stated as follows:
analog=continuous
digital=discrete (step by step)
Because of the discrete nature of digital representations, there is no ambiguity when reading the value of a digital quantity, whereas the value of an analog quantity is often open to interpretation.
September 20, 2007 Posted by | Digital Systems | Tinggalkan komen
## Introductory Concepts
Introductory Concepts
Numerical Representations
Digital Circuits
Digital and Analog Systems
Parallel and Serial Transmission
Digital Number Systems
Memory
Representing Binary Quantities
Digital Computers
September 17, 2007 Posted by | Digital Systems | Tinggalkan komen | 5,243 | 22,266 | {"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-2017-30 | longest | en | 0.769309 |
http://studyadda.com/notes/6th-class/science/work-force-and-energy/energy/6453 | 1,516,288,392,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887423.43/warc/CC-MAIN-20180118151122-20180118171122-00249.warc.gz | 344,110,692 | 17,287 | # 6th Class Science Work, Force and Energy Energy
## Energy
Category : 6th Class
### Energy
We require energy to do our daily work. Our body can apply force on an object due to the energy of our body. Therefore, the energy is the capacity to do the work. Energy can neither be created nor destroyed, it is only transformed from one form of energy into another form. Our body obtain energy from the food we eat. A form of stored energy is consumed by the living organism to do the work. The unit for the measurement of energy is Joule is represented by the first letter of its name "J ".
Let us understand energy with a given example:
Take a piece of stone and lift the stone up at a certain height. Put a earthen pot on the surface exactly below the lifted stone. The work will be done by you on lifting the stone and an amount of energy is stored in the stone on lifting the stone at a height. Now, drop the stone on the earthen pot. The pot will break into small pieces. And the energy stored in the stone will be transferred on the earthen pot which will break the pot into small pieces.
Forms of Energy
There are many forms of energy and one form of energy is transferred into another form by the motion, chemical reactions, etc. Some forms of energy are as follows:
(i) Mechanical energy
(ii) Light energy
(iii) Chemical energy
(iv) Heat energy
(v) Electrical energy
(vi) Nuclear energy
(vii) Magnetic energy
(viii) Sound energy
Mechanical Energy
A raised object possess a form of energy, called potential form of mechanical energy. A running object is capable to do the work because, it has kinetic form of mechanical energy. Therefore, the mechanical energy is the sum of both potential and kinetic energy and a body has mechanical energy if it has either potential or kinetic energy or both. There are two forms of mechanical energy, potential energy and kinetic energy, which are discussed below.
(A) Potential Energy: An object possess potential energy even if the object is in rest. An object can be in its static state if placed at a height. Also an object which has elastic property can remain in its static state until the force is applied. The two forms of potential energy, that is, gravitational potential energy and elastic potential energy are discussed below.
(i) Gravitational Potential Energy: The potential energy of an object depends on the mass, raised height from the surface and gravity. If one of all three parameters is increased then the potential energy of the object is also increased. Therefore, the potential energy of an object is the product of its mass, height at which the body is placed and gravity by which the object is pulled. If an object is placed at a height, the force of gravity acts on the object to bring it back on the surface.
A body, A is placed as the given in the picture below:
The body A is placed at a height of h and mass of the body is m. The gravity is represented by the letter g. Therefore, the potential energy (P.E) possess by the object A = Product of the mass, height and gravity = m x $=m\times g=mgh$.
Let us consider an example,
A transmitter is kept on the top of the mobile tower as shown in the picture below:
The total mass of the transmitter is 1200 kg and it is kept at the height of 250 metres from the surface.
Therefore, the potential energy possess by the transmitter = Mass of the transmitter Gravity Height of the transmitter from the ground $=1200\times 9.8\times 250=2940000$.
(ii) Elastic Potential Energy: Elastic potential energy is a form of energy stored in an elastic material due to the stretching or compression. A stretched bow possess elastic potential energy. Elastic potential energy is also stored in the spring of a gun, stretched rubber band, etc. The amount of elastic potential energy possess by an elastic device depends on the amount of the stretched elastic device. The amount of elastic potential energy possess by an elastic device is transferred to the related device. For example, the amount of elastic potential energy possess by a spring of a gun is transferred to the bullet of the gun and bullet strikes on the target. Therefore, the amount of elastic potential energy possess by a device depends on the amount of force by which the device is stretched. Less stretched elastic device possess less amount of elastic potential energy and more stretched possess more energy. The amount of elastic potential energy possess by a stretched spring (stretched from its rest) $=\frac{1}{2}\,k{{x}^{2}}$.
Where, k = spring constant or force constant and stretched distance. According to the Hook's law, the applied force on a spring is the product of force constant or spring constant and stretched distance by the applied force.
Hence, $F=kx$ or $k=\frac{F}{x}$. If a spring is at rest and no force is applied on the spring, then the spring does not possess any elastic potential energy. In this condition, when no force is applied on the spring, the spring constant or force constant is zero. Therefore, the elastic potential energy of the spring = $\frac{1}{2}\times 0\times {{x}^{2}}=0$. Hence, the spring does not possess elastic potential energy, either spring constant (force constant) or stretched distance is zero.
Let a spring is stretched from its rest to 2 metres by applying a force of 219 N as shown in the picture below.
The spring constant or force constant, $k=\frac{F}{x}=\frac{219}{3}=73N/m$ and $x=2$ metres (given).
Therefore, the elastic potential energy possess by the spring at the given distance
(B) Kinetic Energy: From the explanation of potential energy, we have learned that, a body possess potential energy even if the body is at rest and exerts a force on itself at rest stage. The force can be due to the gravity of the earth or applied force for stretching a spring. The concept of kinetic energy is different than that of potential energy in this way that, a body possess potential energy at rest, whereas, a body possess kinetic energy due to its motion. For example, a moving car, a running boy, moving planets, a rolling stone possess kinetic energy due to their motion.
The kinetic energy of a moving object depends on the mass and (velocity) speed. Therefore, the amount of kinetic energy possess by a body is more if the mass of the body is more. If the speed of the body is increased than the kinetic energy of the body will also be increase.
The kinetic energy possess by a body is the half of the product of its mass and square of its speed or velocity $=\frac{1}{2}\times$ Mass x (Speed)2.
Let the mass of the earth is about, 5.9736 x io24 kilograms and its average orbital speed is about $29.78\,km/s=29.78\times 1000=29780m/s$
Therefore, the amount of kinetic energy possess by the earth =
$\frac{1}{2}\times \left( 5.9736\times {{10}^{24}} \right)\times {{(29780)}^{2}}=264883880112\times {{10}^{22}}$ Joule.
Light Energy
During the photosynthesis, water and carbon dioxide react to form carbohydrates and release oxygen in the presence of light energy radiated by the Sun. Light energy helps the molecules of water and carbon dioxide to break down into the individual atom to form a complex molecule of carbohydrates. An image on the photographic film is formed when light particles fall on the film for the chemical change.
Heat Energy
If a water filled pot is heated by a burner, the water near the lower level of the pot is heated first .This water goes upwards and cold or less heated water comes downward. The movement of amount of water inside the pot is due to the heat energy passed by the burner.
Electrical Energy
A conducting wire is used to pass the electrical energy from one point to another point. The generator of electrical energy passes the energy to the electrons of the conducting wire and wire passes these energy of electrons to the load which is connected at the end of the wire.
Nuclear Energy
The heavier elements like Radium, Uranium, etc. split their nuclei to become a stable and lighter element. During this process the energy is released, which is called nuclear energy.
Magnetic Energy
When a piece of iron comes in contact with a magnet, it gets attracted to the magnets. This attraction is due to the magnetic energy of the iron. The electric motors, electric fans etc work because, the electrical energy is converted into magnetic energy by the coil and magnetic material inside.
Sound Energy
Microphone works on the sound energy. When a body produces sound, the sound waves get mixed with the air. Air, mixed with sound waves, strikes on the diaphragm of the microphone and vibrate the diaphragm according to the amplitude and frequency of the sound waves. The coil inside the mircrophone vibrates according to the vibration of diaphragm and produces a small amount of electromotive force in the wire, connected at the both ends of the coil. An amount of current is produced by the generated electromotive force and current reaches at the coil of the loudspeaker.
Consider the following statements:
Statement 1: The amount of potential energy possess by an object when placed at a height depends on the mass of the object.
Statement 2: The amount of potential energy possess by an object when placed at a height depends on the height from the surface of the object.
Which one of the following is correct about the above statements?
(a) Statement 1 is true and 2 is false
(b) Statement 1 is false and 2 is true
(c) Both statements are false
(d) Both statements are true
(e) None of these
Explanations
The amount of potential energy possessed by an object when placed at a height depends on the mass and height from the surface of the object.
Therefore, option (D) is correct and rest of the options is incorrect.
A body is capable of doing greater work /if it possesses greater energy.
Heat, electrical, sound etc. are the forms of energy.
The unit of work is Joule.
Energy is the capacity to do the work.
Potential energy is the product of mass, gravity and raised height.
.
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#### Resources tagged with Properties of numbers similar to Baby's Blocks:
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### There are 66 results
Broad Topics > Numbers and the Number System > Properties of numbers
### Small Change
##### Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### Got it Article
##### Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Lesser Digits
##### Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Factors and Multiple Challenges
##### Stage: 3 Challenge Level:
This package contains a collection of problems from the NRICH website that could be suitable for students who have a good understanding of Factors and Multiples and who feel ready to take on some. . . .
### Difference Dynamics
##### Stage: 4 and 5 Challenge Level:
Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens?
### Power Crazy
##### Stage: 3 Challenge Level:
What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
### Cinema Problem
##### Stage: 3 and 4 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Six Times Five
##### Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
### See the Light
##### Stage: 2 and 3 Challenge Level:
Work out how to light up the single light. What's the rule?
### Palindromes
##### Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Writ Large
##### Stage: 3 Challenge Level:
Suppose you had to begin the never ending task of writing out the natural numbers: 1, 2, 3, 4, 5.... and so on. What would be the 1000th digit you would write down.
### Unit Fractions
##### 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.
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Prime Magic
##### Stage: 2, 3 and 4 Challenge Level:
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
### The Patent Solution
##### Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### Cogs
##### Stage: 3 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Guess the Dominoes
##### Stage: 1, 2 and 3 Challenge Level:
This task depends on learners sharing reasoning, listening to opinions, reflecting and pulling ideas together.
### Two Much
##### Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
### Arrange the Digits
##### Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### Magic Letters
##### Stage: 3 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws?
### Clever Carl
##### Stage: 2 and 3
What would you do if your teacher asked you add all the numbers from 1 to 100? Find out how Carl Gauss responded when he was asked to do just that.
### Water Lilies
##### Stage: 3 Challenge Level:
There are some water lilies in a lake. The area that they cover doubles in size every day. After 17 days the whole lake is covered. How long did it take them to cover half the lake?
### Factorial
##### Stage: 4 Challenge Level:
How many zeros are there at the end of the number which is the product of first hundred positive integers?
### Mini-max
##### Stage: 3 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Alphabet Soup
##### Stage: 3 Challenge Level:
This challenge is to make up YOUR OWN alphanumeric. Each letter represents a digit and where the same letter appears more than once it must represent the same digit each time.
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### Triangular Triples
##### Stage: 3 Challenge Level:
Show that 8778, 10296 and 13530 are three triangular numbers and that they form a Pythagorean triple.
### Chameleons
##### Stage: 3 Challenge Level:
Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Pair Products
##### Stage: 4 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### Fracmax
##### Stage: 4 Challenge Level:
Find the maximum value of 1/p + 1/q + 1/r where this sum is less than 1 and p, q, and r are positive integers.
### Four Coloured Lights
##### Stage: 3 Challenge Level:
Imagine a machine with four coloured lights which respond to different rules. Can you find the smallest possible number which will make all four colours light up?
### Generating Triples
##### Stage: 4 Challenge Level:
Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
### Filling the Gaps
##### Stage: 4 Challenge Level:
Which numbers can we write as a sum of square numbers?
### An Introduction to Irrational Numbers
##### Stage: 4 and 5
Tim Rowland introduces irrational numbers
### A Long Time at the Till
##### Stage: 4 and 5 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
##### Stage: 4 Challenge Level:
Robert noticed some interesting patterns when he highlighted square numbers in a spreadsheet. Can you prove that the patterns will continue?
### N Is a Number
##### Stage: 3 Challenge Level:
N people visit their friends staying N kilometres along the coast. Some walk along the cliff path at N km an hour, the rest go by car. How long is the road?
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Babylon Numbers
##### Stage: 3, 4 and 5 Challenge Level:
Can you make a hypothesis to explain these ancient numbers?
### Odd Differences
##### Stage: 4 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Happy Octopus
##### Stage: 3 Challenge Level:
This investigation is about happy numbers in the World of the Octopus where all numbers are written in base 8 ... Find all the fixed points and cycles for the happy number sequences in base 8.
### One to Eight
##### Stage: 3 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
### Enriching Experience
##### Stage: 4 Challenge Level:
Find the five distinct digits N, R, I, C and H in the following nomogram
### Really Mr. Bond
##### Stage: 4 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise? | 2,429 | 9,895 | {"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} | 4.21875 | 4 | CC-MAIN-2015-48 | longest | en | 0.884342 |
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