probability
videos
Alcumus
expected value
geometric series

Joe flipped a coin until he flipped heads 10 times in a row.
What's the expected number of flips Joe did?

Is the answer $ 2^{10}$ ?

The probability of getting 10 heads in a row is $ \frac{1}{2^{10}}$, so I guess $ 2^{10}$ must be the answer.

No, it's not. :)

Let the expected number of flips needed be $ E_n$ (for $ n$ heads in a row). We can flip T, giving us an expected value of $ E_n\plus{}1$ flips to finish, flip HT, giving us an expected value of $ E_n\plus{}2$ flips to finish, etc. all the way to HHH...HHHT with $ n\minus{}1$ H's, giving us an expected value of $ E_n\plus{}n$ flips to finish, or get all heads, taking $ n$.

Our equation, by the definition $ E \equal{} p_1e_1 \plus{} p_2e_2 \plus{} \cdots \plus{} p_ke_k$ where $ p_1 \plus{} p_2 \plus{} \cdots \plus{} p_k \equal{} 1$, is

$ E_n\equal{}\frac12(1\plus{}E_n)\plus{}\frac14(2\plus{}E_n)\plus{}\cdots\plus{}\frac1{2^n}(n\plus{}E_n)\plus{}\frac1{2^n}(n)$

$ \equal{}E_n\left(\frac12\plus{}\frac14\plus{}\cdots\plus{}\frac1{2^n}\right)\plus{}\left(\frac12\plus{}\frac24\plus{}\frac38\plus{}\cdots\plus{}\frac{n\minus{}1}{2^{n\minus{}1}}\plus{}\frac n{2^n}\right)\plus{}\frac n{2^n}$

$ \equal{}E_n\left(1\minus{}\frac1{2^n}\right)\plus{}2\minus{}\frac1{2^{n\minus{}1}}$ (first parentheses is a geometric series and the second an [url=http://www.artofproblemsolving.com/Wiki/index.php/Arithmetico-geometric_series]arithmetico-geometric[/url] series)

$ \implies\frac{E_n}{2^n}\equal{}2\minus{}\frac1{2^{n\minus{}1}}\implies E_n\equal{}2^{n\plus{}1}\minus{}2$.

Now, plugging in $ n \equal{} 10$, we have $ E_{10} \equal{} 2^{11} \minus{} 2 \equal{} \boxed{2046}$.

One interesting relation that arises is the recurrence equation $ E_n\equal{}2E_{n\minus{}1}\plus{}2$.

[quote="math154"]No, it's not. :)

Let the expected number of flips needed be $ E_n$ (for $ n$ heads in a row). We can flip T, giving us an expected value of $ E_n \plus{} 1$ flips to finish, flip HT, giving us an expected value of $ E_n \plus{} 2$ flips to finish, etc. all the way to HHH...HHHT with $ n \minus{} 1$ H's, giving us an expected value of $ E_n \plus{} n$ flips to finish, or get all heads, taking $ n$.

Our equation, by the definition $ E \equal{} p_1e_1 \plus{} p_2e_2 \plus{} \cdots \plus{} p_ke_k$ where $ p_1 \plus{} p_2 \plus{} \cdots \plus{} p_k \equal{} 1$, is

$ E_n \equal{} \frac12(1 \plus{} E_n) \plus{} \frac14(2 \plus{} E_n) \plus{} \cdots \plus{} \frac1{2^n}(n \plus{} E_n) \plus{} \frac1{2^n}(n)$

$ \equal{} E_n\left(\frac12 \plus{} \frac14 \plus{} \cdots \plus{} \frac1{2^n}\right) \plus{} \left(\frac12 \plus{} \frac24 \plus{} \frac38 \plus{} \cdots \plus{} \frac {n \minus{} 1}{2^{n \minus{} 1}} \plus{} \frac n{2^n}\right) \plus{} \frac n{2^n}$

$ \equal{} E_n\left(1 \minus{} \frac1{2^n}\right) \plus{} 2 \minus{} \frac1{2^{n \minus{} 1}}$ (first parentheses is a geometric series and the second an [url=http://www.artofproblemsolving.com/Wiki/index.php/Arithmetico-geometric_series]arithmetico-geometric[/url] series)

$ \implies\frac {E_n}{2^n} \equal{} 2 \minus{} \frac1{2^{n \minus{} 1}}\implies E_n \equal{} 2^{n \plus{} 1} \minus{} 2$.

Now, plugging in $ n \equal{} 10$, we have $ E_{10} \equal{} 2^{11} \minus{} 2 \equal{} \boxed{2046}$.

One interesting relation that arises is the recurrence equation $ E_n \equal{} 2E_{n \minus{} 1} \plus{} 2$.[/quote]

Oh, I'm not such advanced to understand your solution math154.
Where can I read more about expected numbers?  :maybe:

To learn about expected value more in depth, you could watch the (three? I think) expected value videos in Alcumus (I think you might have to answer a few/lots of questions first, so I'm not sure), get Introduction to/Intermediate Counting and Probability (expected value is only a small part of these, although well treated), etc. Or you can look around on Google. :)

But I think if you have a solid understanding of what expected value is, you should just do many expected value problems (like the problems [url=http://www.artofproblemsolving.com/Wiki/index.php/Expected_value#Problems]here[/url]).

[hide="Expected Value"]
If there are $ n$ possibilities that occur with probabilities $ p_1,p_2,p_3,\ldots,p_n$ that are all-encompassing (so $ p_1\plus{}p_2\plus{}\cdots\plus{}p_n\equal{}1$) that have values $ e_1,e_2,e_3,\ldots,e_n$ if they occur, then the expected value is defined as $ E\equal{}p_1e_1\plus{}p_2e_2\plus{}\cdots\plus{}p_ke_k$. The expected value is what is expected to be obtained on average if you were to repeat something many times; however, it doesn't have to be attainable.

So an example might help. Say we want to find the expected value of the sum of the numbers of two fair dice that are randomly rolled. The probability that we get a sum of $ 2$ is $ \frac1{36}$, the probability that we get a sum of $ 3$ is $ \frac2{36}$, and so on so that if we let $ p_i$ denote the probability that we roll a sum of $ i$ (where $ e_i\equal{}i$, because the value of each probability $ p_i$ if it occurs is $ i\implies e_i\equal{}i$), then we have $ p_2\equal{}\frac1{36}$, $ p_3\equal{}\frac2{36}$, $ p_4\equal{}\frac3{36}$, $ p_5\equal{}\frac4{36}$, $ p_6\equal{}\frac5{36}$, $ p_7\equal{}\frac6{36}$, $ p_8\equal{}\frac5{36}$, $ p_9\equal{}\frac4{36}$, $ p_{10}\equal{}\frac3{36}$, $ p_{11}\equal{}\frac2{36}$, $ p_{12}\equal{}\frac1{36}$ ($ p_1\plus{}p_2\plus{}\cdots\plus{}p_{12}\equal{}1$, so we have included all cases) is

$ E\equal{}p_2e_2\plus{}p_3e_3\plus{}\cdots\plus{}p_{12}e_{12}\equal{}2p_2\plus{}3p_3\plus{}\cdots\plus{}12p_{12}$

$ \equal{}\frac{2\cdot1\plus{}3\cdot2\plus{}4\cdot3\plus{}5\cdot4\plus{}6\cdot5\plus{}7\cdot6\plus{}8\cdot5\plus{}9\cdot4\plus{}10\cdot3\plus{}11\cdot2\plus{}12\cdot1}{36}$

$ \equal{}\frac{252}{36}\equal{}7$, so we are expected to get a $ 7$.

Note that the expected value doesn't have to be the most likely possibility, even though it is in this problem.

For example, if there was a machine that, with probability $ \frac25$ outputted $ 2$, with probability $ \frac25$ outputted $ 0$, and with probability $ \frac15$ outputted $ 1$, the expected output would be $ 1$, but $ 1$ is not the most likely output!
[/hide]
[hide="A Try at a Better Explanation of My Solution"]
So in my solution, we let $ E_n$ denote the expected number of flips needed to get $ n$ heads in a row.

If we get a tails on our first flip, then this process basically "repeats" itself and we are "starting over," so it is expected to take $ E_n$ flips after the tails to get to $ n$ heads. However, we've already "wasted" one flip, so this whole thing is expected to take $ 1\plus{}E_n$ flips, if the first flip is tails. This occurs with probability $ \frac12$.

If we get HT (a heads then a tails), then like above, it repeats itself, but this time, we've wasted two flips, so this whole thing is expected to take $ 2\plus{}E_n$ flips, and occurs with probability $ \frac14$.

If we get HHT, then, as above, it will take $ 3\plus{}E_n$ flips, and occurs with probability $ \frac18$.

We keep going until....HHH...HHHT (where there are $ n\minus{}1$ heads and one tail), which occurs with probability $ \frac1{2^n}$ and is expected to take $ n\plus{}E_n$ flips in all.

Finally, if we get HHHH...HHHH ($ n$ heads, no tails), then we are done. This takes $ n$ flips (so its value is $ n$), and occurs with probability $ \frac1{2^n}$.

Now, by our definition of expected value, we put this all together to get $ E_n\equal{}\frac12(1\plus{}E_n)\plus{}\frac14(2\plus{}E_n)\plus{}\cdots\plus{}\frac1{2^n}(n\plus{}E_n)\plus{}\frac1{2^n}(n)$, and the rest follows.
[/hide]

Thank you math154, you are very helpful!  :)


Simplify and express the following as a common fraction:

\[ \dfrac{2\bullet1\frac{1}{4}}{2\minus{}\dfrac{4 \minus{} \frac{7}{8}}{4 \minus{} \frac{1}{4}}}\]

$ \dfrac{2\times{1\frac14}}{2 \minus{} \dfrac{4 \minus{} \frac78}{4 \minus{} \frac14}}$

$ \implies \dfrac{\frac52}{2 \minus{} \dfrac{\frac {25}{8}}{\frac {15}{4}}}$

$ \implies \frac {\frac52}{\frac16} \equal{} \boxed{15}$.

gaussintraining, you made a slight mistake.  $ \frac{\frac{5}2}{2\minus{}\frac{\frac{25}{8}}{\frac{15}{4}}}\equal{}\frac{\frac{5}2}{\frac{7}6}\ne\frac{\frac{5}2}{\frac{1}6}$.  Therefore, the actual answer is $ \frac{\frac{5}2}{\frac{7}6}\equal{}\boxed{\frac{15}{7}}$

[quote="abhinand"]man alll the people I was with got 3/7.... was their a glitch?[/quote]

Yes there is a glitch. The problem that is displayed on FTW is $\frac{2\cdot \frac{1}{4}}{2-\frac{4-\frac{7}{8}}{4-\frac{1}{4}}}$ (which gives an answer of $\frac{3}{7}$) instead of $\frac{2\bullet1\frac{1}{4}}{2-\frac{4-\frac{7}{8}}{4-\frac{1}{4}}}$.

[BTW, I reported the problem]


Here is a Problem from the Team selection team of Bangladesh.

Let $a,b,c$ the sides of a triangle.
Prove that,
$\sum{\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}}\leq3$

I think this problem is taken from IMO 2006 short list.

Welcome you to Vietnam, my friend.


induction
graph theory
algebra
system of equations
combinatorics unsolved
combinatorics

Let $C$ be some ideal electrical circuit composed of only conducting wires and resistors. Let $C'$ be the same ciruit, except for the fact that the resistance of every resistor is doubled. Let $R$ and $R'$ be the respective resistance of $C$ and $C'$ across some two chosen points. Show, with mathematical justification, that $R' = 2R$.

you would probably use some tree or graph theory in this and of course
induction...i can't really go any deeper because i'm not skilled enough

[quote="cm"]first prove this is true for parallel and series, then because any circuit is a composition of parallel and series, by math induction, hence the result.[/quote]

Not necessarily true. Consider a circuit as follows: Take the edges of a cube. On every edge, attach a resistor. Then in this circuit (it doesn't matter which two points you take), no two resistors are in parallel or in series.

we will first attatch a battery of arbitrary power  :varepsilon: to the two points and solve the system.
There is always going to be nodes at which the total current is zero and also the sum of potential across any loop is zero. these will give you m equations and m unknowns.
The Total resistnace is then  :varepsilon: / I, where I is the current flowing through the battery.
note that if all resistances are doubled, then one trivial solution for the currents is just the solutioon of the previous system divided by 2, which satisfies all the equations. Then clearly
R' =  :varepsilon: / I' =2( :varepsilon: /I) =2R.
-Ali

What consists of a proof depends on how you define resistence. I assume you apply voltage V, measure the total current I and define R = V/I. The current at each wire is defined as a solution to a system of equations, and these are such that doubling all resistences and halving all currents gives a new solution.

If you have a different definition for the resistence of a network, please specify it.

On the topic of resistors, here's something that I've been wondering about:

Suppose that we are given a particular circuit consisting of only resistors and wires and we can measure the resistance $R$ across 2 points in this circuit.  Now, we are allowed to add additional wires to the circuit such that the original circuit still remains intact (that is, we do not modify the original circuit, just add connections across various points in the circuit).  Is it true that the new resistance $R'$ of the circuit is always less than or equal to $R$?

([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5853]This thread[/url] gives an example of the idea that I'm trying to convey.)

I think what you are asking for is called Rayleigh's Monotonicity Law.  You can find a discussion of it in Doyle and Snell's book on [url=http://front.math.ucdavis.edu/math.PR/0001057]Random Walks and Electrical Networks[/url], which is available on line.  See Section 1.4.


Let $ a,b,c,d$ be nonnegative real numbers, Prove that
\[{{{ \frac {/ab-cd/} {(a + b-c-d)^2}}+\frac {/ac-bd/} {(a +c-b-d)^2}}+\frac {/ad-bc/} {(a+d-b-c)^2}}  \geq 3
\]
When do we have equality?  :!:

[quote="taisaodunglai"]Let $ a,b,c,d$ be nonnegative real numbers, Prove that
\[{{{ \frac {/ab - cd/} {(a + b - c - d)^2}} + \frac {/ac - bd/} {(a + c - b - d)^2}} + \frac {/ad - bc/} {(a + d - b - c)^2}} \geq 3
\]
When do we have equality?  :!:[/quote]
This inequality does not hold. Try $ a=13^{12^{1988}},b=c=1,d=0$. ;)

yes canhang_2007 i wrong  :(
edit
Let $ a \ge b\ge c > d\ge 0$, no two of which are zero . Prove that
\[ \frac {\sqrt [4]{ab\minus{}cd}}{\sqrt {a\plus{}b\minus{}c\minus{}d}}\plus{} \frac {\sqrt [4]{ac\minus{}bd}}{\sqrt {a\plus{}c\minus{}b\minus{}d}}\plus{} \frac {\sqrt [4]{bc\minus{}ad}}{\sqrt {b\plus{}c\minus{}a\minus{}d}} \ge 3
\]

[quote="taisaodunglai"]yes canhang_2007 i wrong  :(
edit
Let $ a \ge b\ge c > d\ge 0$, no two of which are zero . Prove that
\[ \frac {\sqrt [4]{ab \minus{} cd}}{\sqrt {a \plus{} b \minus{} c \minus{} d}} \plus{} \frac {\sqrt [4]{ac \minus{} bd}}{\sqrt {a \plus{} c \minus{} b \minus{} d}} \plus{} \frac {\sqrt [4]{bc \minus{} ad}}{\sqrt {b \plus{} c \minus{} a \minus{} d}} \ge 3
\]
[/quote]
We need the condition $ b \plus{} c \ge a \plus{} d$ so that the last term has its sense. Now, to prove this inequality, after using AM-GM, we only need to show that
\[ (ab \minus{} cd)(ac \minus{} bd)(bc \minus{} ad) \ge (a \plus{} b \minus{} c \minus{} d)^2(a \plus{} c \minus{} b \minus{} d)^2(b \plus{} c \minus{} a \minus{} d)^2
\]
This inequality follows from the following inequalities
\[ (ab \minus{} cd)(ac \minus{} bd) \ge (bc \minus{} ad)(a \plus{} b \minus{} c \minus{} d)(a \plus{} c \minus{} b \minus{} d)
\]

\[ (ac \minus{} bd)(bc \minus{} ad) \ge (ab \minus{} cd)(a \plus{} c \minus{} b \minus{} d)(b \plus{} c \minus{} a \minus{} d)
\]

\[ (bc \minus{} ad)(ab \minus{} cd) \ge (ac \minus{} bd)(a \plus{} b \minus{} c \minus{} d)(b \plus{} c \minus{} a \minus{} d)
\]
These inequalities can be proved easily. ;) :)


Does there exists a base in which the numbers of the form:
\[ 10101, 101010101, 1010101010101,\cdots \]
are all prime numbers?


number theory
prime numbers
greatest common divisor
number theory unsolved

Can anybody give me some hint how to solve this problem?

Let's divide all natural numbers except 1 into n non-empty sets so that:
1) each natural number except 1 is element of exactly one of these n sets
2) if two different commensurable numbers a, b are elements of the same set then also their greatest common divisor is element of this set.

Prove that one of these n sets certainly contains all prime numbers besides maximally (n-1) exceptions (that signifies that "nearly" all prime numbers are in the same set - there can be maximally (n-1) prime numbers that may be in other sets).

Thanks.

Let the subsets be $A_1, \cdots , A_n$.
Since there are a finite numbers of parts and an infinite number of primes, we may assume that $A_n$ contains infinitely many primes.

Now, suppose for a contradiction, that $p_1, \cdots , p_n$ are pairwise distinct primes which do not belong to $A_n$.
Let $q_1, q_2, q_3, \cdots $ be a sufficientely long (hmm... of length $n+1$ should be enough) sequence of pairwise distinct primes which all belong to $A_n$.
For each $i <n$, among the numbers $q_1p_1, q_1p_2, \cdots, q_1p_n$ at most one belongs to $A_i$ otherwise we would have $q_1 \in A_i$ contradicting that $A_i$ and $A_n$ are disjoint. Thus, at least one of these numbers belongs to $A_n$.
The same for $q_jp_1, q_jp_2, \cdots, q_jp_n$ for all $j$.
Thus, there exist $i,j,k$ with $j \neq k$ such that both $q_jp_i$ and $q_kp_i$ belong to $A_n$. Thus $p_i \in A_n$, a contradiction.

Pierre.

Rrrr...
Pierre write it faster than I.  :(  :(  :(


I don't see why you would. Each problem is to be considered separately.

[quote="lena_siz"]Round 3 Criteria for problem 4 states:
"The solution can get full credit even if the degenerate points at the vertices of the ellipse are included in the locus. If the participant notes that those points must be removed, we give +1 point on any partial credit for some other error."

If I did omit these two points, but had everything right in the solution for Problem 4 anyways, can I get the extra credit on some other problem?[/quote]
No.

We allowed the degenerate points to simplify the grading, and we gave credit for equivalent mistakes to be fair. A mistake in another problem would not be equivalent, and transferring points from one problem to another would most certainly not simplify the grading.

I am already embarrassed about grading in haste on Round 3. Don't make me galled about it too by treating it as a free for all.

Erin Schram, USAMTS grader

I have never heard anyone ask about extra credit in USAMTS.

This ain't school, this is a math contest. This ain't a grade, this is a score.

I didn't mean to offend anyone with my question, because I was simply wondering about what the grading criteria was implying. :D


can you maybe take courses at university in high school at some point?  In which case skipping courses will behoove you, as it might allow you to dip your feet in earlier.  Otherwise, your next best alternative is self study.  however, that requires tremendous discipline on your part, as learning is a gradual process that happens step by step, day by day, rather than through five hours of cramming.

I know several people who have done it.

It's a good idea if and only if the benefit you get out of it outweighs everything you gave up (most importantly time which could have been spent doing other useful things, leisure/relaxation time, and also usually money). Note that these costs are not there when you are self studying because:

- If you think something is more beneficial at any point in time, you'd have the opportunity to put aside your self-study and do it
- If you really want to self study something years in advance, it's part of your leisure. Elements of stress, such as mandatory work and exams, are not there.
- You don't need to invest much money.

On the other hand, in a university course, you get the insight of the instructor, which can be great not only for understanding, but also for inspiration, especially if it's a subject you're interested in.


function
integration
geometry
3D geometry
linear algebra
matrix
induction

There were $ 7$ problems that i post only $ 4$ of them here because the other $ 3$ are really easy! :P

$ \textbf{1.}$ Does there exist a $ (1\minus{}1)$ function $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all $ x\in\mathbb{R}$ we have: $ f\left(x^{2}\right)\minus{}f^{2}(x)\geq\frac{1}{4}$

$ \textbf{2.}$ Let $ f: [0,1]\rightarrow\mathbb{R}$  be a continuous function and differentiable on $ (0,1)$. Also there exists a number $ a\in(0,1]$ such that $ \int^{a}_{0}f(x)dx\equal{}0$. Prove that:

$ \left|\int^{1}_{0}f(x)dx\right|\leq\frac{1\minus{}a}{2}Sup\left(\left\{\left|f^{\prime}(x)\right|\;: \;x\in(0,1)\right\}\right)$

$ \textbf{3.}$ In how many ways can we put the numbers of the set $ \left\{1,2,\cdots,12\right\}$ together such that we obtain a complete cube? (by using all the numbers)

$ \textbf{4.}$  Let $ A$ be a $ n\times n$ matrix with real entries. Also there exists a number like $ a\in[\minus{}2,2]$ exists such that $ A^{2}\minus{}aA\plus{}I_{n}\equal{}0_{n}$. Prove that for all positive integers $ m$ there exists a uniqe number $ a_{m}\in[\minus{}2,2]$ such that:

$ A^{2m}\minus{}a_{m}A^{m}\plus{}I_{n}\equal{}0_{n}$

 :)

[quote="sinajackson"]$ \textbf{1.}$ Does there exist a $ (1 \minus{} 1)$ function $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all $ x\in\mathbb{R}$ we have: $ f\left(x^{2}\right) \minus{} f^{2}(x)\geq\frac {1}{4}$[/quote]
[hide="Solution"]No: we have from the given equation that $ f(0) \minus{} f(0)^2 \geq \frac {1}{4}$ and $ f(1) \minus{} f(1)^2 \geq \frac {1}{4}$.  However, the only real $ x$ such that $ x \minus{} x^2 \geq \frac {1}{4}$ is $ \frac {1}{2}$, so for any such function $ f$ we have $ f(0) \equal{} f(1) \equal{} \frac {1}{2}$ and $ f$ is not one-to-one.[/hide]
[quote="sinajackson"]$ \textbf{3.}$ In how many ways can we put the numbers of the set $ \left\{1,2,\cdots,12\right\}$ together such that we obtain a complete cube? (by using all the numbers)[/quote]
What does this mean?

Problem 2 is [url=http://www.mathlinks.ro/viewtopic.php?t=1080]not exactly[/url] original :D

[quote="JBL"]
[quote="sinajackson"]$ \textbf{3.}$ In how many ways can we put the numbers of the set $ \left\{1,2,\cdots,12\right\}$ together such that we obtain a complete cube? (by using all the numbers)[/quote]
What does this mean?[/quote]

Sorry for my bad English! :oops:

putting these numbers together for example like : $ 231487659121011$

how many times this number will be a complete cube?!

sorry again:(

Ah, okay!  (In English, we say "perfect cube.")  Then the answer is [hide]none, since any such number is divisible by 3 but not by 9.[/hide]

[hide="4"]
Because $ I\equal{}A(aI\minus{}A)$, $ A$ is invertible. Then any $ a_m$ satisfying $ A^{2m}\minus{}a_mA^m\plus{}I\equal{}0$ must satisfy $ a_mI\equal{}A^m\plus{}A^{\minus{}m}$, so if $ a_m$ exists, it is unique.

It remains to show that for all positive integers $ m$, $ A^m\plus{}A^{\minus{}m}$ is a multiple $ a_m$ of the identity matrix and $ |a_m|\leq 2$. But we can show by induction that
\[ a_m\equal{}2\cos m\theta ,\ \theta\equal{}\arccos \frac{a}{2}\]
The base cases $ m\equal{}0,1$ are trivial (here $ A^0$ denotes $ I$). For inducting from $ m,m\plus{}1$ to $ m\plus{}2$,
\[ A^{m\plus{}2}\plus{}A^{\minus{}(m\plus{}2)}\equal{}(A^{m\plus{}1}\plus{}A^{\minus{}(m\plus{}1)})(A\plus{}A^{\minus{}1})\minus{}(A^m\plus{}A^{\minus{}m})\]
\[ \left( 4\cos (m\plus{}1)\theta \cos \theta \minus{}2\cos m\theta \right) I\]
The product-to-sum identity $ 2\cos (m\plus{}1)\theta \cos \theta \equal{}\cos m\theta \plus{}\cos (m\plus{}2)\theta$ completes the inductive step.
[/hide]


integration
calculus
calculus computations


1-If $ a\plus{}b\plus{}c\equal{}180\text{  In degree}$
prove that (and explore the solution as you can):
$ \csc \frac{a}{2}\plus{}\csc \frac{b}{2}\plus{}\csc \frac{c}{2}\ge 6$

2-what are the important Inequalites I must learn it to solve Inequalites? :help:

[hide="1"]
Jenson's on $ \frac{1}{3}\csc \frac{a}{2}\plus{}\frac{1}{3}\csc \frac{b}{2}\plus{}\frac{1}{3}\csc \frac{c}{2}$ should do it.[/hide]

2 - AM-GM, Cauchy-Schwarz, Schur, Jensen, Rearrangement and Chebychev are the basic ones. Everything else... oh well.


How many congruent 4-foot tall cylindrical pipes with an inside diameter of 2 inches are needed to hold the same amount of water as one pipe of the same height with an inside diameter of 12 inches?

The formula for the volume of a cylinder is $ \pi r^2h$. We want to know how many $ 1^248\pi$ are in $ 6^248\pi$ (I converted the 4 feet into inches). Dividing, we get $ \frac{36(48)\pi}{1(48)\pi}\equal{}\boxed{36}$

Notice how, at the end, the height does not change; therefore, the volume of the water is only affected by the base area of the cylinders, $\left(\frac{d}{2}\right)^2$. Since the areas are $1\pi$ and $36\pi$, respectively, we need $\frac{36\pi}{1\pi}=\boxed{36}$ pipes to fill one with diameter 12 inches.


Find the last decimal digit of the sum $ 1^1 \plus{} 2^2 \plus{} 3^3 \plus{} ... \plus{} 2001^{2001}$

Because last digit $ n^n$ had period 20 and $ 2000/20\equal{}100$ last digit these number is last digit $ 2001^{2001}$ and equal $ 1$.


integration
calculus
function
calculus computations

Solve for $x:$

$\int_{0}^{2[x+14]}{\{\frac{x}{2}\}dx=\int_{0}^{\{x\}}[x+14]dx}$
where $\{.\},[.]$ respectively denote the fractional and integral part of $x$.

[hide]
the answer to this was given as $-13>x\geq-14$ but I'm not getting it.plz help me with this.[/hide]

The notation is messed up. I will consider the equation
\[\int_{0}^{2[x+14]}{\{\frac{t}{2}\}dt=\int_{0}^{\{x\}}[t+14]dt }\]
The function $\{t/2\}$ has period $2$, which implies that its integral over any interval of length $2$ is the same, namely $\int_{0}^{2}\{t/2\}dt=\int_{0}^{2}(t/2)dt=1$. Hence the integral on the left is equal to $[x+14]$.
As for the other integral, $[t+14]=14$ when $t$ is between $0$ and $1$. So the second integral is $14\{x\}$. It remains to solve
\[[x+14]=14\{x\}\]


let Ai be seq,of nonnegative real nos. s.t.limAi =0 and summation Ai i going from1 to infinity is infinite.let S={Ai}
 then show for any 2 given POSITIVE real nos. p and q ,show there exist finite terms say B1,B2,...,Bk in S s.t.
 p<B1+B2+...+Bk<q.

Well, this is the same thing for semi-convergent series: you can get whatever you want with their subseries ...

hey please anybody gives explicit proof?my approach is totally inelegant.

Go along the sequence until you come to a term smaller than $q$. That's $B_{1}$. Now keep going in the sequence and pick as $B_{2}, \ldots$ each term such that the sum of all the stuff you've taken so far is still smaller than $q$. Stop when your sum is greater than $p$.

You know you'll find these terms because the terms go to 0 (so you can always find ones that don't exceed $q$), but if they go to 0 so fast that you can't get bigger than $p$, then obviously the series would converge.


modular arithmetic
number theory unsolved
number theory

It is divisible by $ 17$ for $ m\equal{}16 \cdot 17k \plus{} 1$.

[quote="bjh7790"]Prove that there are infinitely many odd number  m that makes

$ 8^m \plus{} 9m^2$ composite number.[/quote]
With $ m \equal{} 9$, we get $ 8^9 \plus{} 9^3 \equal{} 512^3 \plus{} 9^3 \equal{} 521u$

$ 521$ is a prime number. Then $ 8^{520} \equal{} 1\pmod{521}$ and so $ 8^{520\times 521} \equal{} 1\pmod{521}$ and so $ 8^{m \plus{} 520\times 521} \plus{} 9(m \plus{} 520\times 521)^2$ $ \equal{} 8^m \plus{} 9m^2\pmod {521}$

So $ 8^{9 \plus{} 520\times 521k} \plus{} 9(9 \plus{} 520\times 521k)^2 \equal{} 0\pmod{521}$ and so is a composite number.
Q.E.D.

[u][b]Edit [/b][/u]: Sorry for a so complex solution in front of elementary TomCiO's one  :blush:


fetos teleiwsa thn 3 lukeiou kai thelw na kanw proetoimasi gia tous foithtikous diagwnismous ( SEEMUS, IMC) alla d gnwrizw thn ulh pou apaiteitai kai poia vivlia tha me vohthhsoun. Borei kapoios na m pei thn ulh kai na m proteinei kapoia vivlia gia na 3ekinhsw.

Γεια σου φιλε και καλωσηρθες στο forum.
Θα προσπαθησω να δωσω μια ολοκληρωμενη απαντηση καθως πολλα παιδια μολις εισερχονται στο πανεπιστιμιο
θελουν να ασχοληθουν με διαγωνισμους που ισως να μην ασχοληθηκαν στο λυκειο(οπως και εγω :) ).
Λοιπον οι διαγωνισμοι οπως αναφερεις ειναι δυο ο Seemous και ο IMC.
Για τον Seemous:
Γινεται αν δεν κανω λαθος Μαρτιο και αφορα φοιτητες 1ου και 2ου ετους. Η υλη ειναι κατα κυριο λογο
Αναλυση και Γραμμικη Αλγεβρα. Εκει ριχνεται το βαρος μιας και καθε σχολη εχει αυτα τα δυο μαθηματα
στα πρωτα της ετη και ειναι πολυ βασικα. Επιπλεον ειναι Συνδιαστικη, Αναλυτικη Γεωμετρια, Θεωρια Αριθμων.
Το Σεπτεμβρη η ΕΜΕ κανει καποια μαθηματα τα οποια ειναι πολυ χρησιμα και μετα διοργανωνει εσωτερικο
διαγωνισμο για την επιλογη της ελληνικης ομαδας.
Για τον IMC:
Γινεται τελη Ιουλιου και αφορα φοιτητες μεχρι και 4ου ετους. Εξαιρουμενου του διαγωνισμου Putnam που
ειναι για τα Αμερικανικα πανεπιστιμια ειναι ο μεγαλυτερος μαθηματικος φοιτητικος διαγωνισμος. Και εκει
μεγαλο κομματι της υλης ειναι η Αναλυση και η Γραμμικη Αλγεβρα. Τωρα εκει χρειαζεται επιπλεον Συνδιαστικη,
Αναλυτικη Γεωμετρια, Θεωρια Αριθμων, Αλγεβρα, Μιγαδικη Αναλυση και Θεωρια Γραφηματων. Αυτο δεν σημαινει
πως ειναι απαραιτητο να τα γνωριζεις ολα αυτα μιας και μπαινουν λιγοτερα θεματα απο αυτα τα κεφαλαια(συνολο 10
με 12 θεματα).

Γενικα λοιπον οποιος θελει να ασχοληθει πρεπει απαραιτητα να διαβασει Αναλυση και Γραμμικη Αλγεβρα.
Επισης εμφανιζονται πολυωνυμα και ανισοτητες και γενικα οσο περισσοτερες γνωσεις εχει καποιος απο
μαθηματικα τοσο καλητερα θα παει.

Τα τρια παρακατω βιβλια πιστευω οτι ειναι τελεια για προετοιμασια αν και υπαρχουν πολλα ακομη:

1)Putnam and Beyond (στην ερωτηση πως θα ασχοληθω με φοιτητικους διαγωνισμους θα μπορουσα απλα να σου πω παρε αυτο το βιβλιο)
2)Berkley Problems in Mathematics
3)Problem Solving Strategies (αυτο ειναι για προετοιμασια για το λυκειο αλλα και για το πανεπιστιμιο ειναι πολυ χρησιμο)

Επισης χρησιμα ειναι τα παλια θεματα των διαγωνισμων αυτων καθως και του Putnam.
Τελος μην χασεις τα μαθηματα της ΕΜΕ τα Σαββατοκυριακα απο Σεπτεμβρη(αν γινουν βεβαια :) )

Αυτες ειναι οι ταπεινες μου αποψεις. Αν θελει καποιος ας με διορθωσει σε κατι η να προσθεσει κατι ακομα.

Καλησπέρα και καλωσήρθες στο forum. Απλώς, ήθελα να προσθέσω το βιβλίο του Paul Zeitz, "The Art and Craft of Problemsolving" στα προτεινόμενα βιβλία.
Επίσης, τα παλιά θέματα του Putnam, του κατά κοινή ομολογία δυσκολότερου μαθηματικού διαγωνισμού, υπάρχουν σε τρεις τόμους οι οποίοι μπορούν να  αγοραστούν μέσω του forum αλλά και μπορούν να βρεθούν σε ελληνικά βιβλιοπολεία (ο πρώτος δυσκολότερα).

για προτωετεις το putnam and beyond ισως ειναι βαρυ (αν και ειναι κορυφαιο βιβλιο ισως για μεγαλυτερους)...
οτι πρεπει ομως για την περιπτωση ειναι το art of problem solving απο το καλυτερα βιβλια του ειδους (απευθεινεται σε φοιτητες των πρωτων ετων που δεν εχουν εμπειρια απο διαγωνισμους και τους φτανει σε πολυ καλο επιεδο)

ενα αλλο πολυ καλο βιβλιο ειναι  του larson το problem solving through problems...

euxaristw polu gia ta vivlia pou proteinate. mou ekane entupwsh pws kaneis d anefere kapoio ellhniko vivlio. d uparxei kapoio kalo vivlio ellhna suggrafea h metafrasmeno sta ellhnika kuriws me gramikh algevra kai analush pou apo oti katalava einai ta vasikotera???

Για γραμμικη Αλγεβρα (και για προβληματα διαγωνισμων) το καλυτερο ειναι το Linear Algebra του Peter Lax. Ελληνικα βιβλια με προβληματα για προετοιμασια για τετοιους διαγωνισμους δεν νομιζω να υπαρχουν. Υπαρχουν καποια Ελληνικα βιβλια Αναλυσης που εχω ακουσει οτι ειναι παρα πολυ καλα (Νεγρεποντης πχ) αλλα δεν ξερω αν ειναι για προετοιμασια για διαγωνισμους. Επισεις για Ανωτερη Αλγεβρα που μπαινει παντα στον IMC και αν και δεν πολυμπαινει στον SEEMOUS μπορει να φανει χρησημη καμια φορα σε προβληματα αλλων κλαδων, τρομερο βιβλιο ειναι αυτο του Freileigh το οποιο (νομιζω) ειναι ξενο αλλα υπαρχει μεταφρασμενο στα Ελληνικα. Τελος οπως ειπαν και οι απο πανω, ενα απο τα καλυτερα βιβλια προετοιμασιας για αυτους τους διαγωνισμους ειναι το Putnam and Beyond.


Does anyone remember the question? The answer key says 4944, but i don't see how, if I remember the question correctly.

What is the smallest base-10 number composed only of 4s and 9s (at least one of each) such that the resultant number is divisible by both 4 and 9? Well we must have a number of 4s that is 0 mod 12 such that the number is divisible by 9 [sum-of-digits divisibility rule for 9], so since we have at least 1 4 we must have at least 12. Further, we have to have at least 1; we need only examine the last 2 digits to learn divisibility by 4. 49 and 94 aren't divisible by 4, so the last 2 digits must be 44, and the other digits can be arranged in any order. To minimize the value, though, we put the 9 in the hundreds place and put the remaining 4s to its left, for 4944 at the end.

i didnt exactly understand what SF said, but i did it like this.

[hide]the sum of the digits is going to be $4f+9n$ where f is the number of fours and n the number of nines.  For the number to be divisible by 9, the sum of the digits must be divisble by nine.  so $\frac{4f+9n}{9}=\frac{4f}{9}+n$.  So we need the smallest f such that $4f$ is divsible by 9.  Since 4 doesnt have any 3's in its factoriation, n=9.  Since we want the smallest number, we put as many 4's as we can first.  So we end up with 3 fours and 1 nine for the last 4 digits.  The smallest would be 4449, but 49 isnt divisible by 4, and same with 94 in 4494.  So the smallest one that is divisble by both ends in 4944.[/hide]

ah... must have been running out of time when i read it...

hey, can someone tell me the actually number 'n'? thanks

[quote="mathgeniuse^ln(x)"][quote="emilgouliev"]did it actually say at least one of each???[/quote]

Then it would be too easy.

9 4s.[/quote]
Nice try, but 4944 is a lot smaller.

[quote]Nice try, but 4944 is a lot smaller.[/quote]

Um i didnt take the amc 10b, but what do you mean 4944 is smaller, that makes no sense..

Wait a second.  4944 isn't even divisible by 9.  The number would have to be 4444444944.  My statement didn't make sense because I had assumed without checking that the answer was correct.  Maybe the test only asked for the last four digits.

I didn't take the 10b either.

4944 is the last 4 digits of the smallest possible number consisting of 4's AND 9's that is also divisible by 4 and 9

Yeah... Wasn't there a question that only had letter answers? I think the answer was S, but none of the answers is S. Is #12 supposed to be S, and they put 5?

No, they put s as choice D on that problem.

actually... it's not 12... WHAT HAPPENED TO THAT QUESTION???


:blush: I would like to have a better understanding of probability. Do you have any suggestions where to start? Maybe the question that I have is very simple and I apologize for posting it in this forum. Here is my question:

A rental truck agency services its vehicles on a regular basis, routinely checking for mechanical problems.  Suppose that the agency has 6 moving vans, two of which need to have new brakes.  During a routine check the vans are tested one at a time.
a) what is the probability that the last van with brake problems is the fourth tested?
b) what is the probability that no more than four vans need to be tested before both brake problems are detected?
c) given than one van with brakes is detected in the first two test, what is the probability that the remaining van is found on the third or fourth test?

Thanks a lot for your time and effort that you guys have put in this forum.  It is great!

For a problem like this, "the probability [event] has outcome [x]" is just the quotient of the number of ways [event] can have outcome [x] by the number of outcomes for [event].  So, for example, in (a) and (b), [event] is that all the vehicles are tested, one by one.  In part (a), outcome [x] is that the second of two busted vans is the fourth van total.  So you need to count the total number of orders the vans could be examined in, and compare this to the total number of ways that the vans can be tested in which the latter of the two busted vans in tested fourth.  Part (c) is slightly more complicated because the sample space (collection of possible outcomes of [event]) is limited -- there's a condition on it.  There are methods (e.g. Bayes Law) for dealing with this algebraically, but I think it's good for a beginner in probability to go through it manually.

As a beginner in probability this don't help much. Thank you anyway... can you be more specific please? I mean numbers involved so I can understand where to start..

I know that if the R van is in the fourth position, then the other R (the bad one) is in the first, second or third place...then from here I don't know how to count..

Thanks

So far so good.  Why not start with the easiest part, though: how many total ways are there to examine the vans, whether or not the second broken van is in the fourth position?  (This will be the denominator in our probabilities.)

Right-o, six choices for the last van, five for the van before it, and so on.  Then (because these choices are completely independent of each other) we get to multiply together the numbers 6, 5, 4, etc. and get 6!.  Now, apply the same thinking to the numerator in part (a): how many ways are there to order the vans so that the second broken one is in the fourth place?  (Trick of the trade: it didn't matter whether I counted 6! by starting at the back (as I did), starting at the front, or starting in some other position.  For the numerator, it does matter: some ways of counting are easier than others.  I would first start with the positions you know the most about.  In this case, that means counting the fourth, fifth and sixth positions first (i.e., start from the back).)

for the fourth one way, the fifth 5 ways and the sixth 4 ways since two vans are already placed... I think I am lost...

[quote="Isabella_"]I think I am lost...[/quote]

I agree ;-)

Maybe we should give the vans names: Broken van A, broken van B, not broken vans C, D, E and F.  Which vans can be in the 4th position?  We need a broken van there, so there are two choices (A or B).  Now, what do we know about the fifth and sixth vans?  Can they be broken?  Once we've taken care of those, are there any other restrictions on the first three vans?

On the fourth place can be broken van A or B, therefore there are 2!on the fifth and sixth place I have vans C, D, and E therefore I have 4! on the first three places I have a broken van which will be A or C then the good ones have only 3! choices

[quote="Isabella_"]On the fourth place can be broken van A or B, therefore there are 2![/quote]
Right, although writing it as $ 2!$ is misleading: you have two choices for the van in the fourth position (van A or B).

[quote="Isabella_"]on the fifth and sixth place I have vans C, D, and E therefore I have 4![/quote]
This doesn't make any sense.  You do have four choices for the vans 5 and 6 (any of C, D, E and F) but you are only taking two of them.  Thus, there are not $ 4!$ choices here but rather $ 4\cdot 3$ (four choices for the van in position 5 (any of C, D, E, F) and three choices for the van in position 6 (any of the three remaining from those four)).

[quote="Isabella_"]on the first three places I have a broken van which will be A or C then the good ones have only 3! choices[/quote]
For "A or C" put "A or B;" for the first three vans (one of which is broken, two of which are not) we have three choices and we can arrange them however we like, for $ 3!$ ways.

Thus, the total number of "good" arrangements (with the second broken van in the fourth position) is the product of these values, $ 2 \cdot (4 \cdot 3) \cdot 3!$; to calculate the associated probability, divide by $ 6!$ (the total number of arrangements).

wow, great explaining joel, i was in the same position as isabella but after reading your explanation im pretty sure i understand it, thanks


geometry
circumcircle
ratio
geometry proposed

[u][b]The author of this posting is : Huynh Anh Hao (edited by Darij Grinberg)[/b][/u]
____________________________________________________________________

Let two circles $\left(O_1\right)$ and $\left(O_2\right)$ cut each other at two points $A$ and $B$. Let a point $M$ move on the circle $\left(O_1\right)$. Denote by $K$ the point of intersection of the two tangents to the circle $\left(O_1\right)$ at the points $A$ and $B$. Let the line $MK$ cut the circle $\left(O_1\right)$ again at $C$. Let the line $AC$ cut the circle $\left(O_2\right)$ again at $Q$. Let the line $MA$ cut the circle $\left(O_2\right)$ again at $P$.
[b](a)[/b] Prove that the line $KM$ bisects the segment $PQ$.
[b](b)[/b] When the point $M$ moves on the circle $\left(O_1\right)$, prove that the line $PQ$ passes through a fixed point.

[u][b]The author of this posting is : grobber[/b][/u]
____________________________________________________________________

Point [b](b)[/b] follows easily from the fact that the mapping $P\mapsto Q$ is an involution on the circle $\left(O_2\right)$.

And here is a proof of [b](a)[/b] based on [b](b)[/b]:

By a simple angle chase, we can show that the angle between $MK,PQ$ is constant. Now, since $PQ$ always passes through a fixed point (call it $T$), it means that $X=MK\cap PQ$ lies on some circle passing through $K$ and $T$. Now, by looking at some particular positions of $X$ (for example, we can take the limiting case when $X=B$, and the case when $X$ is $O_2$), we can see that this circle is the one passing through $T$ and the midpoints of the chords $PQ$ of $\left(O_2\right)$ through $T$.

Here is an alternative solution to the problem:

Let S be the point of intersection of the line MB with the circle $\left(O_2\right)$ (different from B). Let Q' and S' be the midpoints of the segments PQ and PS, respectively. Then, of course, the line Q'S' is parallel to the line QS (as a midparallel in the triangle QPS).

On the other hand, using directed angles modulo 180, we have < BSQ = < BAQ (cyclic, since the points B, Q, S, A all lie on the circle $\left(O_2\right)$); in other words, < BSQ = < BAC. But on the other hand, < BAC = < BMC (cyclic, since the points B, C, A, M all lie on the circle $\left(O_1\right)$); hence, < BSQ = < BMC. This yields QS || MK. Together with Q'S' || QS, this implies Q'S' || MK.

Now, the circle $\left(O_1\right)$ is the circumcircle of triangle MAB, and the point K is the point of intersection of the tangents to this circle through the points A and B. In other words, the point K is the M-vertex of the tangential triangle of triangle MAB. Hence, by a well-known theorem, the line MK is a symmedian of triangle MAB. On the other hand, since the points A, B, S and P lie on one circle (namely, on the circle $\left(O_2\right)$), the segment PS is antiparallel to the side AB of triangle MAB. Since symmedians bisect antiparallels, it follows that the line MK bisects the segment PS. In other words, the line MK passes through the midpoint S' of the segment PS. But since we also have Q'S' || MK, we see that the lines Q'S' and MK are parallel and have a point in common (namely, the point S'). Thus, these two lines must coincide. Hence, the line MK passes through the point Q', i. e. through the midpoint of the sgement PQ. In other words, the line MK bisects the segment PQ. This proves problem [b](a)[/b].

For problem [b](b)[/b], let T be the point where the line AK meets the circle $\left(O_2\right)$ (apart from the point A). Then, I claim that the line PQ always passes through the pole of the line BT with respect to the circle $\left(O_2\right)$. Of course, this pole is a fixed point, and thus we just need to prove my claim and problem [b](b)[/b] is solved.

In fact, let the lines MK and AB intersect each other at a point W. Since the point K is the point of intersection of the tangents to the circle $\left(O_1\right)$ at the points A and B, the line AB is the polar of the point K with respect to the circle $\left(O_1\right)$. Hence, by a well-known property of polars, the points K and W divide the segment MC harmonically. In other words, the double ratio (MC; KW) = -1. Now, projecting from the point A onto the circle $\left(O_2\right)$, we transform this into (PQ; TB) = -1. In other words, the quadrilateral BPTQ is harmonic. But this immediately shows that the line PQ passes through the pole of the line BT with respect to the circle $\left(O_2\right)$. Thus, my claim is proven.

$\blacksquare$

  Darij

This is a simple problem !!
  I have a solution for this problem (just using similar triangles and cyclic quadrilateral )  :)

Since the tangents at $A$ and $B$ to $(O_1)$ meet at $K$,it readily follows that $MK$ is the $M$-symmedian of $\triangle{MAB}$.Let $MB$ cut $(O_2)$ at $Q_1$.Then $MK$ passes through the midpoint of $PQ_1$.Now we have $\angle{AQQ_1}=\angle{ABM}=\angle{ACM}=\angle{KCQ} \implies QQ_1 \parallel MK$.This immediately proves part(a)

For part(b) Darij can you please elaborate on your involution argument.

I have a argument for part (b) now

Let $R=AK \cap (O_2)$.It is well known that $AK,AB,AC,AM$ are harmonic pencils.Projecting onto $(O_2)$ we get $(R,B;Q,P)=-1$.Let the tangents to points $R$ and $B$ meet at $K_1$ and $PK_1$ meet $(O_2)$ at $Q'$.Then it is well known that $(R,B;Q',P)=-1$.This implies $Q \equiv Q'$ and $PQ$ passes through the pole of $BR$ which is fixed.


Let $\Gamma: 4x^{2}+9y^{2}=36$ be an ellipse with foci $F_{1}, F_{2}$. $P$ is a point on $\Gamma$ and $L$ is the tangent at $P$ to $\Gamma$. $E_{1}$ and $E_{2}$ are the foot of the perpendicular drawn from $F_{1}$ and $F_{2}$ to $L$ respectively. If $\angle F_{1}PF_{2}=60^{\circ}$, then find the area of quadrilateral $F_{1}E_{1}E_{2}F_{2}$.

Any ideas?

[hide="Hint"]A well-known result:

Let $\Gamma$ be an ellipse. $P$ is a point on $\Gamma$. $F_{1}, F_{2}$ are the two foci. $L$ is tangent to $\Gamma$ at $P$. The minor angle $PF_{1}$ makes with the tangent line is equal to the minor angle $PF_{2}$ makes with the tangent line.[/hide]


function
real analysis
real analysis unsolved

could you give an example of $ f: \mathbb{R}\to \mathbb{R}$ which is differentiable on $ \mathbb{R}$ but $ f'$ is nowhere continuous ?

thanks

That's impossible: the pointwise limit of any sequence of continuous functions, if exists at all, has plenty of points of continuity. The proof uses the Baire category theorem and runs, roughly speaking, as follows. First, fix an open interval $ I_0$ on which the pointwise convergence $ f_n\to f$ takes place. Fix $ \varepsilon\equal{}1$. Consider the closed sets $ F_n\equal{}\{x\in I_0: \forall m\ge n, |f_m(x)\minus{}f(x)|\le\varepsilon\}$. Since the union of $ F_n$ is $ I_0$, some $ F_n$ contains an open interval $ I_1$ compactly contained in $ I_0$. Note that $ \operatorname{osc}_{I_1} f\le 2$. Now fix $ \varepsilon\equal{}1/2$ and do the same construction on $ I_1$. You'll get an interval $ I_2\subset\!\subset I_1$ with $ \operatorname{osc}_{I_2} f\le 1$. Proceed all the way down and apply the nested interval lemma to the sequence of the intervals $ I_1,I_2,\dots$ to get a point of continuity.

The points of continuity are dense, but I don't remember the proof. It's in Oxtoby's Measure and Category

How do you conclude that one of the f_n's necessarily contains an interval?

$ \operatorname{osc}_I f\equal{}\sup_I f\minus{}\inf_I f$. The statement about $ F_n$ containing an interval is exactly the Baire category theorem for $ \mathbb R$.


every time when I take AMC I always make some stupid mistakes.  This year I made two, including the Josh and MIke problem and a stupid mistake on #15, which I forgot to divide the area of triangle by 2.  If i don't make those mistakes, my score can be greatly improved.  for AMC, the penalty for wrong answer is heavy.  and there are so many questions it is almost impossible to check each one of them.  so what should i do to cut down the mistakes.?

[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]This article[/url] from the articles section of AoPS might help  :)

1. Read article posted above.
2. Practice lots.  Work on speed.  It may help to be good at mental math.
3. Avoid calculator use.  Written work is sometimes faster, improves your accuracy in various areas, and makes your work easier to check.
4. Check your answer carefully right after you do a problem.  This includes virtually doing the problem all over again (starting from skimming the problem again) but having your computations already written out.  For very easy questions, I tend to check them over 2-3 times using different methods.  This will only take at most 20 seconds and will save you a lot of mistakes.


geometry
circumcircle
trigonometry
parallelogram
cyclic quadrilateral
trig identities
Law of Cosines

Let $ ABC$ be a triangle,and let $ M$ be the midpoint of $ BC$.The circumcircle of $ \triangle ACM$ meets side $ AB$ again at $ D$(other than $ A$).Points $ E$ and $ F$ lie on segments $ CB$ nad $ CA$,respectively,such that $ CE\equal{}EM$ and $ CF\equal{}3FA$.Suppose that $ EF \perp BC$.Prove that $ \angle ABC\equal{} \angle DEF$

[hide="Solution"]Denote $ J$ the intersection of $ MF$ and $ AB$.
Applying [b]Menelaus[/b]'s theorem, we can show that $ FM\equal{}FJ$
Then $ CJ//EF$ or $ CJ\perp BC$.
And use [b]Menelaus[/b]'s theorem again we get $ JA\equal{}\frac{1}{3}JB$
But $ BD.BA\equal{}BM.BC$ so $ BD.BJ\equal{}BC.BE$
Thus $ DECJ$ is cyclic quadrilateral $ \Rightarrow \angle BDE\equal{}90^o$
On the other side, $ \angle BDM\equal{}\angle ACB, \angle FEC\equal{}90^o$
Therefore $ \angle MDE\equal{}\angle EFC\equal{}\angle MFE$
Then $ DMEF$ is cyclic quadrilateral.
$ \Rightarrow \angle FDE\equal{}\angle FME\equal{}\angle ACB, \angle DMB\equal{}\angle DFE$
$ \Rightarrow \angle B\equal{}\angle DEF$.
We are done :D[/hide]

From Power of a Point, we get $BD\cdot BA=BM\cdot BC$, hence
\[BD=\frac{a/2\cdot a}{c}=\frac{a^2}{2c}=\frac{3a}4\cdot \frac{2a}{3c}=BE\cdot \frac{2a}{3c}.\quad (1)\]

In $\triangle CEF$, we get $\cos \gamma=\frac{CE}{CF}=\frac{a/4}{3b/4}=\frac{a}{3b}$. Here, by the Law of cosines, $2ab\cos\gamma=a^2+b^2-c^2$, hence
\[a^2+b^2-c^2=\frac23 a^2.\]
Rearrange this, you get
\[\frac43 a^2=a^2+c^2-b^2=2ac\cos\beta\implies \cos\beta=\frac{2a}{3c}.\quad (2)\]

Combining (1) and (2), we find that $D$ is the projection of $E$ on $AB$. Hence $\angle DEF=90^\circ-\angle DEB=\beta$. Done.

My solution:

let $ X\in AB $ such that $ \frac{AX}{XB}=\frac{1}{3} $ .

Easy to see $ XECF $ is a parallelogram .
From $ XF=EC=ME, XF \parallel ME $ and $ EF \perp BC \Longrightarrow XFEM $ is a rextangle . ... $ (\star) $
Since $ \angle DMB=\angle DAC=\angle DXE \Longrightarrow D, E, M, X $ are concyclic ,
so combine with $ ( \star) $ we get $ D, E, F, M, X $ are concyclic $ \Longrightarrow \angle DEF=\angle FXA=\angle CBA $ .

Q.E.D


Given the system of equations:

{
x - xy + y = 5
x^2 + 3xy + y^2 = -5
}

Solve for (x,y)

[hide="easier way"]You can set $a=x+y$ and $b=xy$, then the two equations become $a-b=5$ and $a^{2}+b=-5$, which avoids some messy algebra.[/hide]

[hide]Adding the first two equations,
$(x+y)^{2}+(x+y)=0$
$(x+y)(x+y+1)=0$
$x+y=0$ or $x+y=-1$

For the first case, we get:
$0+x^{2}=5 \implies (\pm \sqrt{5},\mp \sqrt{5})$

For the second case, we get:
$-1+x(x+1)=5 \implies x^{2}+x-6 = 0 \implies (-3,2)\text{ or }(2,-3)$[/hide]


Let $x$, $y$, $z$ be real positive numbers with $xyz=1$. Prove that:

$x^{2}+y^{2}+z^{2}+xy+xz+yz\ge2(\sqrt{x}+\sqrt{y}+\sqrt{z})$

$LHS \ge 2(xy+yz+zx) \ge 2 \sqrt{3xyz(x+y+z)}\ge 2 \sqrt{3(x+y+z)}\ge 2 \sqrt{( \sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}=RHS$


[hide]It looks like $ 2^x\cdot2^3=2^0 $, this $ x=-3 $. [/hide]


Consider a triangle with side lengths a b and c, such that a<b<c. The triangle is revolved about one of its sides to generate a solid. Which side will result in the greatest volume?

[hide]If the triangle is rotated around side a, it will create two cones back to back, both with radius $h_a$, one with height x and one with height a-x. The volume is then $\frac{\pi h_a^2x}{3}+\frac{\pi h_a^2(a-x)}{3}=\frac{\pi h_a^2a}{3}$. $h_a$ can be written as $\frac{\sqrt{(s-a)(s-b)(s-c)}}{2a}$, by Heron's formula. Squaring that and multiplying by a, we get the only part of the volume that changes with the side, $\frac{(s-a)(s-b)(s-c)}{4a}$. We can see that the longer the side, the smaller this fraction will be since the numerator doesn't change and the denominator increases. So we want the shortest side, or a.[/hide]
Bad explanation as usual. Oh well.


Let $m,n$ be non zero integers and $a,b$ be odd numbers.For the point $X(2^ma,\ 2^nb)$ which lie in the first quadrant, denote the rule which make correspond to the integer $m-n$ by $(R)$.
If the two points $X(2^ma,\ 2^nb),\ X'(2^{m'}a',\ 2^{n'}b')$ which lie in the first quadrant lie in a line passing through the origin, then prove that  the integer $m-n$ equal to $m'-n'$ by coresponding by the rule $(R)$ respectively.

When $(a,b)$ and $(c,d)$ lie on a line through the orifin, there is a $r$ such that $a=rb , c=rd$ and this $r$ will be rational.
Now your desired number equals for $(x,y)$ to the $2$-adic valuation $v_2\left(\frac{x}{y}\right)$.
But $v_2\left(\frac{a}{b}\right) = v_2\left(r\right) = v_2\left(\frac{c}{d}\right)$ by definition.

Sorry for that I looked up but, I can't understand the word "ardic" and "valuation".Please give me an explanation, ZetaX.

kunny

The $2$-adic valuation $v_2 (n)$ of an integer is defined  as the number of factors of $2$ that are contained in $n \in\mathbb{Z}$.
It can be extended to rationals by defining $v_2 \left( \frac{m}{n} \right) = v_2(m)-v_2(n)$.


Does anyone know why auto-login may not work? On this computer, I click the checkbox to save my user name and screen name but it doesn't do anything. I'm thinking that my computer perhaps cannot store the cookie that saves this information, because it works when I use another computer.

And the following might be related:
The problem used to be worse. I would have to login multiple times in one visit (once to view my class forums on the forum menu, once again to enter the forum, once again to post something on the forum, and sometimes even once again when I click the submit post button). Thankfully, this seems to have fixed itself, as it has not occurred recently.

Does anyone know why this occurs? Thanks,

zfzf21

If you do not post your OS and Browsers' version it's hard to say anything but "delete your cookies". [url=http://www.mathlinks.ro/Forum/remove_cookies.php]Click here[/url] to delete them and see if it works ;)


calculus
integration
trigonometry
calculus computations

[quote="vector1"]$\int\frac{1}{(x^{2}+a)^\frac{3}{2}}dx$

Mathematica has a solution but I would like the details please!![/quote]
$\int\frac{1}{(x^{2}+a)^\frac{3}{2}}dx$
$y=\frac{x}{a^{1/2}}$,  $tanz=y$.
then $\int\frac{1}{(x^{2}+a)^\frac{3}{2}}dx$=$\frac{1}{a}\int\frac{1}{(y^{2}+1)^\frac{3}{2}}dy$=$\frac{1}{a}\int\frac{cos^{3}z}{cos^{2}z}dz$=$\frac{1}{a}\int{cosz}dz$=$\frac{1}{a}sinz+C$=$\frac{1}{a}sin(arctan(\frac{x}{a^{1/2}})+C$

[quote="Kent Merryfield"]When I teach trigonometric substitution, I always tell students to draw a triangle. (The triangle assumes first quadrant angles - outside the first quadrant, a little extra sign-checking thought may be needed.)

This right triangle has angle $z$ for which $\tan z=\frac{x}{\sqrt{a}}.$ So we mark the side opposite as $x,$ the side adjacent as $\sqrt{a},$ and the hypotenuse as $\sqrt{x^{2}+a}.$

Your answer came out as $\frac1a\sin z+C.$ We examine the triangle to see that $\sin z=\frac{x}{\sqrt{x^{2}+a}}.$ So we give the answer to the integration problem as

$\frac{x}{a\sqrt{x^{2}+a}}+C.$

Now we check:

$\frac{d}{dx}\left(\frac{x}{a\sqrt{x^{2}+a}}\right) =\frac1a\frac{d}{dx}(x(x^{2}+a)^{-1/2})$

$=\frac1a\left((x^{2}+a)^{-1/2}+x\left(-\frac12\right) (x^{2}+a)^{-3/2}(2x)\right)$

$=\frac1a\cdot\frac{(x^{2}+a)-x^{2}}{(x^{2}+a)^{3/2}}=\frac1{(x^{2}+a)^{3/2}}$

It checks. Theoretically, we could have solved it just by reversing the steps of the check, but in practice that's not an easy method to find.

Extremal's answer is correct but $\frac1a\sin\left(\arctan\left(\frac{x}{\sqrt{a}}\right)\right)$ is not the most useful form of that answer.[/quote]
I know other method but ( i think it is eqvivalent). I would be solve by this way. $sin^{2}z+cos^{2}z=1$ => $sinz=\frac{1}{1+\frac{1}{tan^{2}z}}$ we know $tanz=\frac{x}{a^{1/2}}$=> $sinz=\frac{x}{\sqrt{a+x^{2}}}$
thank you Kent Merryfield for this  This right triangle has angle $z$ for which $\tan z=\frac{x}{\sqrt{a}}.$ So we mark the side opposite as $x,$ the side adjacent as $\sqrt{a},$ and the hypotenuse as $\sqrt{x^{2}+a}.$

Your answer came out as $\frac1a\sin z+C.$ We examine the triangle to see that $\sin z=\frac{x}{\sqrt{x^{2}+a}}.$. I don't try to find good form for the answer because it's easy and can do every body + author of this problem in this forum.

Don't quote entire posts, especially the post immediately before yours. It's assumed you're responding to that unless you indicate otherwise.

Thanks all!

you just solved a magnetic flux density problem

BTW check out mathematicas solution for this, it is somewhat different.

I never thought of trig substitution!

[quote="vector1"]$\int\frac{1}{(x^{2}+a)^\frac{3}{2}}dx$

Mathematica has a solution but I would like the details please!![/quote]
Solution without trigonometry.
$t=\frac{a}{x^{2}}$
$\int\frac{1}{(x^{2}+a)^\frac{3}{2}}dx$=$-\frac{1}{2a}\int(t+1)^{-3/2}dt$=$\frac{1}{a\sqrt{t+1}}+C$=$\frac{x}{a\sqrt{a+x^{2}}}+C$


Prove that for any positive integer $ n$, $ 3 | \sigma(3n\plus{}2)$, where $ \sigma(n)$ is the summation of all divisors of $ n$.

[hide]Let $ k\equal{}3n\plus{}2$. Since $ k \equiv 2 \pmod 3$, we know that it is never a perfect square. Thus, all divisors of $ k$ can be paired such that each pair has a product of $ k$. Let $ r$ and $ s$ be two divisors that can be paired as such, and have product $ k$. We have $ rs\equal{}k\equiv 2 \pmod 3$. Examining all cases mod 3 gives $ r \equiv 1$, $ s \equiv 2$ or permutations. Thus, every pair of divisors of $ k$ has a sum divisible by 3, and the sum of all divisors together will have a sum divisible by 3.[/hide]

[quote="Brut3Forc3"]... we know that it is never a perfect square. Thus, all divisors of $ k$ can be paired ....[/quote]
Can not that pairing be done with numbers that are not perfect squares?  :huh:

EDIT - Nevermind. Thanks, [b]t0rajir0u[/b].

That's what he's saying.  Pairing cannot be done with perfect squares because perfect squares have an odd number of factors (in other words, the factor $ \sqrt{n}$ is unpaired).


Let's say I want to say HELLO in 8-bit ASCII.  That would be 0100100001000101010011000100110001001111.  Now would I write that in base-10 as 310400273487 or 072069076076079 something else?  I'm learning about RSA encryption and for some reason everyone only encrypts one-character messages :D

Heh, I've done stuff like that. I had a simplified version of you 8-bit hello where you used number other than 1 and 0 to show how many times a number appeard in a row.

So hello would be: 012014013010101202130120213012041.


Find all integers $ a$, $ b$ and $ n$ such that $ bn+a$ divides $ an+b$

[quote="asdrojas"]Find all integers $ a$, $ b$ and $ n$ such that $ bn\plus{}a$ divides $ an\plus{}b$[/quote]

There are a lot of ways for giving such solutions.
here is one :

Let $ b$ and $ n$ any integers.
Let $ p$ any divisor of $ n\minus{}1$
Let $ q$ any divisor of $ n\plus{}1$
Let $ r$ any divisor of $ b$

Then $ (a,b,n)\equal{}(pqr\minus{}bn,b,n)$ is solution and any solution is of this form.

proof :

$ 1).$ $ (a,b,n)\equal{}(pqr\minus{}bn,b,n)$ is solution :
$ bn\plus{}a\equal{}pqr$
$ an\plus{}b\equal{}pqr(n\minus{}\frac{n\minus{}1}{p}\frac{n\plus{}1}{q}\frac{b}{r})$ $ \equal{}(bn\plus{}a)(n\minus{}\frac{n\minus{}1}{p}\frac{n\plus{}1}{q}\frac{b}{r})$
Q.E.D

$ 2).$ Any solution is of this form :
$ bn\plus{}a|an\plus{}b$ $ \implies$ $ bn\plus{}a|an\plus{}b\plus{}bn\plus{}a$ $ \implies$ $ bn\plus{}a|(a\plus{}b)(n\plus{}1)$
Then it exists a divisor $ q$ of $ (n\plus{}1)$ and a divisor $ t$ of $ (a\plus{}b)$ such that :
$ n\plus{}1\equal{}qu$, $ a\plus{}b\equal{}tv$ and $ bn\plus{}a\equal{}tq$
Then $ (bn\plus{}a)\minus{}(a\plus{}b)\equal{}b(n\minus{}1)\equal{}t(q\minus{}v)$
Then $ t|b(n\minus{}1)$ and it exists a divisor $ p$ of $ (n\minus{}1)$ and a divisor $ r$ of $ b$ such that  $ t\equal{}pr$
So $ a\equal{}tq\minus{}bn\equal{}pqr\minus{}bn$
Q.E.D.


[quote="[url=http://www.spaceweather.com/]SpaceWeather.com[/url]"]
MOON & MARS:  When the sun sets tonight, go outside and look east.  The Moon and Mars are having a beautiful close encounter all weekend long.  The best night to look is Saturday when the distance between the pair shrinks to less than 2 degrees.  Campfire-red Mars so close to the silvery Moon is a sight wonderful to behold.

BLUE MOON ALERT:  This is an alert for Central and South Americans. The ongoing eruption of the Galeras volcano in southwestern Columbia is spewing fine ash miles high into the atmosphere.  Airborne volcanic ash can act as a color-filter, turning  moonlight blue. Although it is often said that Blue Moons are mythical, they can appear during volcanic eruptions, so this is a good time to look for a bright "Blue Moon" over your part of the world.
[/quote]

This should be nice. :)
Remember, pictures are encouraged.  :wink:

The board has been quite quiet lately.  :|

Yes, that it has. :(

The clear weather in much of the United States should make for some nice stargazing, though. :)


acceleration of block A is 10g/13 towards right and of block B is 5g/13 downwards

plz reply and tell whether it is correct or not............. :wink:

hey anurag plz see all the posts of accelerations part 1,2,3

and plz reply and tell whether my answers are correct or not

by the way these problems were damn simple may of class 9th level........ :P

 :)


Good luck to everyone who is taking the exam! My exam is set for tomorrow morning. Hopefully, all goes well, haha!   :surf:  :nhl:  :ddr:  :starwars:

I like these smilies. I like these smilies so I need this to be able to have 4 in the post.

I would also like to add my best wishes.  Good luck!


Solve for the inequality:

$ \frac{x\plus{}2}{x\minus{}3} > 2$

So I multiplied both sides by $ x\minus{}3$ and then solve and get $ x<8$. But the answer was $ 3<x<8$. How do you work this out algebraically?

[quote="jeez123"]Solve for the inequality:

$ \frac {x \plus{} 2}{x \minus{} 3} > 2$

So I multiplied both sides by $ x \minus{} 3$ and then solve and get $ x < 8$. But the answer was $ 3 < x < 8$. How do you work this out algebraically?[/quote]

Its never a good idea to divide through with variables in inequalities, since important bounds usually get lost. The best way is to transposition everything to one side as such:

$ \frac{x\plus{}2}{x\minus{}3}>2$
$ \Rightarrow \frac{x\plus{}2}{x\minus{}3}\minus{}2>0$
$ \Rightarrow \frac{\minus{}x\plus{}8}{x\minus{}3}>0$
So either the numerator and the denominator are both positive or both negative.

[u]Both positive[/u]
We have
$ \minus{}x\plus{}8>0\Rightarrow x<8$
$ x\minus{}3>0\Rightarrow x>3$
Which implies $ 3>x>8$.

[u]Both negative[/u]
We have
$ \minus{}x\plus{}8<0\Rightarrow x>8$
$ x\minus{}3<0\Rightarrow x<3$
This is clearly impossible to satisfy, since a number cannot be both greater than 8 and less than 3.

So the solutions are $ 3>x>8$.

[quote="jeez123"]Solve for the inequality:

$ \frac {x \plus{} 2}{x \minus{} 3} > 2$

So I multiplied both sides by $ x \minus{} 3$ and then solve and get $ x < 8$. But the answer was $ 3 < x < 8$. How do you work this out algebraically?[/quote]
The problem was that you assumed $ x\minus{}3$ was positive. If it isn't, then the sign gets switched to $ <$.

[quote="jeez123"]

$ \frac {x + 2}{x - 3} > 2$

So I multiplied both sides by $ x - 3$ and then solve and get $ x < 8$. But the answer was $ 3 < x < 8$. How do you work this out algebraically?[/quote]
In that case, multiply both sides by $ (x - 3)^2$ 'cause it's always positive, then

\begin{eqnarray*} \frac {{x + 2}} {{x - 3}} & > & 2 \\
(x - 3)(x + 2) & > & 2(x - 3)^2 \\
x^2 - x - 6 & > & 2x^2 - 12x + 18 \\
x^2 - 11x + 24 & < & 0 \\
(x - 3)(x - 8) & < & 0 \end{eqnarray*}

We have to possibilities

$ (x - 3) > 0\,\wedge\,(x - 8) < 0\implies 3 < x < 8$ or $ (x - 3) < 0\,\wedge\,(x - 8) > 0\implies\varnothing$


[quote="dina013"]Unde as putea gasi programa de olimpiada pe 2007?[/quote]

Da-mi o adresa de mail sau messenger si iti trimit.


So is it safe to say from the deafening silence, that no one got an email today?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147621[/url]

I heard that Harrison made Blue MOP (not MOSP... he used inversion and placed the O on the circle. Yea okay that was really bad... I was trying to make fun of a joke and use something we learned the other day from Steve's lecture).

Dum spiro [et dum Harriso[b]n[/b] intends to g[b]o[/b] to RSI], spero.


Can some mod delete this? (I'm not a moderator in this forum.)


We can define a "cool point" to be a point on a coordinate plane which has integral coordinates. For example, (7,8) and (5,9) are cool points. How many cool points exist within or on the graph of the circle $ (x\minus{}5)^2\plus{}(y\minus{}2)^2\equal{}100$?

$ \textbf{(A)} 317\qquad\qquad\qquad\textbf{(B)}
420\qquad\qquad\qquad\textbf{(C)}
100\qquad\qquad\textbf{(D)}
25\qquad\qquad\textbf{(E)}
10,000$

We only have to find the number of lattice points in the circle:
\[ x^2\plus{}y^2 \equal{} 10^2
\]

We see that each cool point in the circle has a "pair"...all except the point in the middle.  Then there must be an odd number of cool points, so the answer mus be $ A$ or $ D$.  Answer choice $ D$ is ridiculous so it has to be $ \boxed{A}$.

I don't really agree with your terminology. I know what you mean by "within" the graph, but strictly speaking, the circle is breadthless.

Can anyone provide a better term?

Also, jli's solution is not very mathematically rigorous. What if you had options 25, 37, 323, 317, 1000?

Any proof? Also, define what you mean by a "pair".

Basically what jli meant was that the arrangement of the lattice points is symmetrical, therefore each point has a mirror image - except the point at the center of the circle.

ehh we don't necessarily have to speak "strictly" here, but jli already provided that: any point that satisfies $ x^2\plus{}y^2\leq 10^2$.

I skipped this problem cause for some reason I thought it would be hard (wth circles? what was I thinking?). bleh I like jli's way for AMCs, but the only other way I can think of is plug in values for $ x$ and try to find $ y$ to find lattice points in one quarter of the circle, multiply by four, and add one to include the center point.


The area of the triangle bounded by the lines $y = x, y = -x$ and $y = 6$ is

$
\mathbf{(A)}\;  12\qquad
\mathbf{(B)}\;  12\sqrt2\qquad
\mathbf{(C)}\;  24\qquad
\mathbf{(D)}\;  24\sqrt2\qquad
\mathbf{(E)}\;  36$

[hide="Answer"]The triangle is formed by the three intersections of two lines at a time, so we have $(0,0)$, $(-6,6)$, and $(6,6)$. The triangle has base $12$ and height $6$, so the area is $\frac{1}{2}\cdot 6\cdot 12=36\Rightarrow \boxed{E}$.[/hide]


[quote="krnboy32"]simple but forgot how to do it...

Logbase16(32)[/quote]

that would be:

$\text{log}_{16}32$

[hide="hint"]

Basically you would have

$16^{x}= 32$.
[/hide]

[quote="krnboy32"]simple but forgot how to do it...

Logbase16(32)[/quote]

[hide]$\log_{16}32=\log_{2^{4}}2^{5}=x$
$2^{4(x)}=2^{5}\implies x=\frac54$

Of course, you could just realize that the fourth root of 16 is 2 and $2^{5}=32$ to get the same answer.[/hide]


does anyone know about where the amc 10/12 A will be administered? (preferably near winston-salem...) all i can find is B locations.

your school? WFU maybe? or go to mt. tabor lol. ask b-flat he should know... individual schools administer it

WFU offers B forms, not A. A tests are offerred by individual schools. Ask your math teacher for information. If your school is not offerring A form contests, ask another school whether they will let you take it. A test is next week - so hurry.

Mt. Tabor should have some extra tests available (for both the 10 and the 12) for the A, so you are welcome to take one.

PM me or my dad (gt59) if you are interested.

I know you had to sign the AMC 10A at Hanes, but it may be too late. It's tomorrow.

ok i took the AMC10A. i'll be taking another one at WFU. so thanks everyone who replied!


A log is cut into 4 pieces in 12 seconds using parallel slices. At the same rate, how many seconds will it take to cut the log into 5 pieces?

3 cuts, 12 seconds, so 4 cuts,[b] 16 seconds.[/b]

(that is, if the log is cylindrical and we cut parallel to the base... but that's aside the point)


MATHCOUNTS geometry trapezoid rectangle Mafia

Good luck. ("target #2 and 3" should be "target #3 and 4")

Edit: Darnit, I'm not allowed more than 3 attachments or consecutive posts, could someone just post something really quick?

ERRORS:
#13 should be $ 2x^2 \plus{} \frac {1}{8}$, not $ 2x^2 \minus{} \frac {1}{8}$.
In #26, it should be quadrilateral ABCD, not trapezoid ABCD.
Also, in #26, it should be "If the length of side BC [i]squared[/i] can be expressed in the form". Sorry for really messing this up. :(
In #10, it should say "all possible obtuse triangles [i]with integer side lengths[/i].

In the target round, #3 should be a 10X24 rectangle, not a 10X26.

Is this like for fun, or is it for like a real competition among everyone?

Edit, never mind, read the other post.

I can't put in more than 3 attachments in one post, and I'm not allowed to post consecutively for less than 10 minutes. :P

It's for fun.

Edit: But I will be scoring them and putting up some statistics.

About that... Why couldn't you just put all the pdfs into a zip file? That would only be one attachment then.

13 has two answers... maybe he means +1/8

Can you PM me the problems? I don't want to download anything.

Yes, sorry, I meant +1/8 on number 13.

And in number 26, it should be a quadrilateral, not a trapezoid. I'll edit that.

BOGTRO, sending them to you now.

Shoot.. I already sent my answers and I didn't answer 26. Can I still send it to you?

Sure, go ahead.

And yeah, I also forgot to put that side lengths were integers on #10. :blush:

are you sure problem 23 is fine?

i'll pm you my question?

This hasn't been going so great, as only 3 people have submitted answers so far.
I will probably be extending the deadline to sometime during the week after Nats, since many people are probably studying hard now.

wow, i briefly looked at the problems, and they look "beastly hard"

i hope 2008 nats isnt that hard... :|

oh, that reminds me. i thought the entire test was very creative, especially the doctor-mafia problem XD

[quote="Lawrence Wu"]can the doctor protect the mafia even thought it's pointless?
and since the doctor can protect himself, does that mean the mafia can kill the doctor?[/quote]

Yes, because he doesn't know who the mafia is.
Yes.

10 year bump..........?
And andersonw hasn't been active since 2015

How do people even find 10 year old threads?


Let $a,b,$ and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$. Then $a^{2}-b^{2}+c^{2}$ is:

Where did you get this problem?  It was #8 on the 2006 MMPC (michigan mathematics prize competition) and you don't live in michigan.

I'm using these topic quizzes I found online as 2 week prep for the AMC A coming up.

I googled them. I'm having trouble with some of them.  :(

Nice solution!! :clap2:  :clap:  :thumbup:


function
analytic geometry
calculus
integration
logarithms
trigonometry
linear algebra

Sorry if this problem was already solved here, but I did not find it.

I want generate random number with normal distribution (= Gaussian distribution) with non zero mean and  variance non equal to one (so Box Muller is not a case).

What is most effective way to do it?

I see possibility use equation (59) from http://mathworld.wolfram.com/NormalDistribution.html   - with evaluation of erf by equation (7) from http://mathworld.wolfram.com/Erf.html

- BUT this can be slow and due to using not long (sure not infinite) expansion of Maclaurin series for erf function also not precise.

What is more effective way ?

Thakns for any advice-hint-link.

There's a beautifully simple way, that harks back to the polar coordinates trick for evaluating the integral of the Gaussian in the first place.

Suppose $ U_1,U_2,\dots$ are uniform $ [0,1]$ random variables, independent each time called. (This is the usual output of pseudo-random number generators, or can be scaled to this.)

Let $ R\equal{}\sqrt{\minus{}2\ln U_1}$

Let $ \Theta\equal{}2\pi U_2$

That gives us the polar coordinates of a point. Now convert it to rectangular coordinates:

Let $ X\equal{}R\cos\Theta$

Let $ Y\equal{}R\sin\Theta$

$ X$ and $ Y$ will each have a standard normal distribution (mean zero, variance one) and will be independent of each other. To get a normal distribution of some other mean and variance will simply require a linear transformation applied to $ X$ and $ Y.$

To show that this does what I claim, suppose $ X$ and $ Y$ are independent standard normal distributions, let $ R\equal{}\sqrt{X^2\plus{}Y^2},$ and compute $ P(R>R_0)$ for any positive $ R_0.$

[quote="Kent Merryfield"] To get a normal distribution of some other mean and variance will simply require a linear transformation applied to $ X$ and $ Y.$[/quote]

Please, how to do this linear transformation?

Yes, change mean, it is simple
$ X \equal{} mean \plus{} R\cos\Theta$
and
$ Y \equal{} mean \plus{} R\sin\Theta$
but how to get new variance?

[quote="JBL"]How do you write $ \text{Var}(cX)$ in terms of $ \text{Var}(X)$ for a constant $ c$?[/quote]

Like that ?   ->

$ \text{Var}(cX)$ = c^2 *  $ \text{Var}(X)$

So if $ X$ is standard normal (mean zero and variance 1) and we want $ W$ to have mean $ \mu$ and variance $ \sigma^2$ (standard deviation $ \sigma),$ we let $ W\equal{}\mu\plus{}\sigma X.$

If you want a multivariate normal with a specified center of mass and a specified (not necessarily diagonal) covariance matrix - well, that's going to take knowing a little bit of linear algebra.


I wonder whether $ C_{6}O_{6}$(something has a similar structure as benzene ring) possesses cyclic delocalisation.

Something about the overlapping electron orbitals:
http://en.wikipedia.org/wiki/Delocalized_electron

yes that is indeed resonance .well one thing i never have heard of $ \text{C}_{6}\text{O}_{6}$

Assuming that Math^_^, by C6O6 is referring to a cyclohexane ring in which each carbon is double bonded to an oxygen, then there is no ressonance (electron delocalisation) because thereb are no electrons to delocalise: each carbon is sp2, but the lone p orbital, that in benzene is combined with the other five p orbitals from the other carbons to give the electron delocalisation, is in this case used to make a pi-bond with each oxygen. It is also not reasonable to consider positive charges on the oxygens because oxygen is quite more electronegative than carbon. However, if you substitute the 1,3, and 5 carbonyl groups by CH2 groups then you indeed get an aromatic compound (can you find it?).

However, if you, Math^_^, are referring to another compound then show us its structure.


linear algebra
matrix
vector
algebra proposed
algebra

From a group of $2n+1$ weights, if we remove any weight, the remaining $2n$, can be divided in two groups of $n$ elements, such that they have the same total weight. Prove all weights are equal.

Let $a_1, \cdots , a_{2n+1}$ be the weights. For each $i \in \{1, \cdots, 2n+1 \}$, when we exclude $a_i$, we may write an equation of the form
$\sum_{j \in A} a_j - \sum_{j \in B} a_j = 0$, (1)
where $A,B$ are two disjoints $n$-subsets from $\{1, \cdots , 2n+1 \} \setminus \{i \}$.
Now consider the $(2n+1) \times (2n+1)$ matrix $M$ whose entries in row $i$ are the coefficients of $a_1, \cdots, a_{2n+1}$ in (1).
Thus, the main diagonal of $M$ contains only $0$ and the off-diagonal entries of $M$ are $1$ or $-1$, with exactly $n$ entries equal to $1$ in each row.
Let $V$ be the $(2n+1) \times 1$ vector (sorry I don't know how to write it as a column) $V = (a_1, \cdots a_{2n+1})$ and $U = (1, \cdots , 1)$ (same problem  :?  ).
By construction, we have $M \cdot V = 0$, and it is clear that $M \cdot U = 0.$ Then $M$ has rank at most $2n$ and the desired conclusion will follow from the fact that $M$ has rank $2n$.
Let us consider the matrix $M$ modulo $2$ (that is each entry of $M$ is regarded as an integer modulo $2$). Then all the off-diagonal entries are $1$.
We perform row reduction by adding the first row of $M$, modulo $2$, to each other rown thereby obtaining a matrix $M'$ (draw it...).
Then $M$ and $M'$ have the same rank over the field of integers modulo $2$, but it is here clear (if you have drawn it) that $M'$ has rank at least $2n$.
Thus $M$ modulo $2$ has rank at least $2n$, which means that there is a minor of $M$ with size $2n \times 2n$ which is non-zero modulo $2$, and hence non-zero. Thus, $M$ has rank at least $2n$, and the conclusion follows.

Pierre.

It sounds similar, but i dont know linear algebra like you do, so i had to use more basic arguments concerning parity. to show what you also showed.

It depends...If you assume that the weights are commensurables (i.e. rationals), you can proceed by infinite descend.
In the general case, I'm quite sure that you are forced to use linear algebra, even if it is a bit masked.

Hmm...I think I remember a solution from Myth, using $\mathbb {R}$ as a vector space over $\mathbb {Q}$...I will try to find it.

Pierre.

Bingo!  :D

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=basis+equal&t=18134

Pierre.

well Pierre, it seems like a very nice technic, I would be very happy if you could explain a little bit how to use that basis change. My solution also uses linear algebra but not so advanced like yours.


In an acute-angled triangle $ ABC$, $ D$ is the foot of the altitude from $ A$, and $ P$ a point on segment $ AD$. The lines $ BP$ and $ CP$ meet $ AC$ and $ AB$ at $ E$ and $ F$ respectively. Prove that $ AD$ bisects the angle $ EDF$.

This is a particular case of the following fact:

$P$ is a point on the plane of $\triangle ABC.$ Lines $PA,PB,PC$ cut $BC,CA,AB$ at $D,E,F.$ Define $U \equiv EF \cap AD$ and let $V$ be the orthogonal projection of $U$ onto $BC.$ Then lines $VU,BC$ bisect $\angle FVE.$

Let $ H_a\equiv EF\cap BC$. It is well- known that $ (H_aDBC)\equal{}\minus{}1$. Let $ G_a\equiv EF\cap AD$ $ \Longrightarrow (H_aG_aEF)\equal{}\minus{}1$. Hence $ D(H_aG_aEF)\equal{}\minus{}1$, but $ DH_a\perp DG_a$, thus $ DG_a$ is the internal bisector of $ \angle EDF$ $ \square$

Let $DF$ and $DE$ intersect the parallel through $A$ to $BC$ at $X$ and $Y$ respectively and, as $XY\parallel BC\perp AD$, we need to prove $AX=AY$.
But $\frac{AX}{BD}=\frac{AF}{BF}$ and $\frac{AY}{DC}=\frac{AE}{CE}$, or, dividing side by side the two equalities: $\frac{AX}{AY}=\frac{AF\cdot BD}{BF}\cdot\frac{CE}{AE\cdot DC}$, but, as from Ceva for $\triangle ABC$ with $P$, the right side of the last equality is $1$, hence our goal has been achieved.

Best regards,
sunken rock

Solution with Complex Bash:

Let $D = 0$, $A = ai$, $B = b$, $C = c$ and $P = pi$.

$E$ lies on $AC$ and $BP$, so for some $\lambda, \gamma \in \mathbb{R}$,
$$
E = c + \lambda(ai - c) = b + \gamma(pi - b)
$$
Comparing real and imaginary coefficients, we have $\gamma = \frac{\lambda a}{p}$ and $\lambda = \frac{cp - pb}{cp - ab}$, so
$$
E = c + \left(\frac{cp - pb}{cp - ab}\right) \cdot (ai - c).
$$
Similarly,
$$
F = b + \left(\frac{bp - pc}{bp - ac}\right)\cdot (ai - b).
$$

We wish to show $\angle EDA = \angle ADF$, that is (with directed angles)
\begin{align*}
\operatorname{Arg}\left(\frac{D - E}{D - A}\right) &= \operatorname{Arg}\left(\frac{D - A}{D - F}\right)\\
\iff \operatorname{Arg}\left(\frac{E}{ai}\right) &= \operatorname{Arg}\left(\frac{ai}{F}\right)\\
\iff \operatorname{Arg}\left(E\right) - \frac{\pi}{2} &= -\operatorname{Arg}\left(F\right)+ \frac{\pi}{2}\\
\iff \operatorname{Arg}\left(E\cdot F\right) &=\pi,
\end{align*}
so it suffices to show that $E\cdot F$ is real, and expanding,
\begin{align*}
E\cdot F &= \left(c + \left(\frac{cp - pb}{cp - ab}\right) \cdot (ai - c)\right) \cdot \left( b + \left(\frac{bp - pc}{bp - ac}\right)\cdot (ai - b)\right) \\
&= \frac{(-a b c-i a b p+i a c p+b c p)(-a b c+i a b p-i a c p+b c p)}{(c p-a b)(b p-a c)}\\
&=\frac{a^{2} b^{2} c^{2}+a^{2} b^{2} p^{2}-2 a^{2} b c p^{2}+a^{2} c^{2} p^{2}-2 a b^{2} c^{2} p+b^{2} c^{2} p^{2}}{a^{2} b c-a b^{2} p-a c^{2} p+b c p^{2}},
\end{align*}
which is real, so this holds.

(Am I correct in saying that we can say $\operatorname{Arg}(E\cdot F) = \pi$ is equivalent to saying $E \cdot F$ is real as we use directed angles and won't have configuration issues because of the acute $ABC$? If not, you'd probably have to prove that it's negative too)


If $ a_1,a_2,....a_{101}$ is an arrangemetn of numbers 1,2,...101 then porve that the product of $ (1\minus{}a_1)(1\minus{}a_2)..(1\minus{}a_{101})$ is an even integer

as your question there exists a $ n\in[1,101]$ such that $ a_n \equal{} 1$ so
the product $ (1 \minus{} a_1)(1 \minus{} a_2)....(1 \minus{} a_(101)) \equal{} Q_t(1 \minus{} a_n) \equal{} Q_t(1 \minus{} 1) \equal{} 0$ and zero is an even integer.(Q.E.D.)
here $ Q_t$ is an integer.  :)


Vmo2001
Day1
1.Two circles $(O_1)$and $(O_2)$ intersect at $A$ and $B$ and their common tangent touches $(O_1)$ at $P_1$ and $(O_2)$ at $P_2$.Let $M_1$ and $M_2$ be  the orthogonal  projections of $P_1$ and $P_2$ on the line $O_1O_2$.The line  $AM_1$ cuts $(O_1)$ again at $N_1$ and  $AM_2$ cuts $(O_2)$ again  at $N_2$.Prove that the points $B,N_1,N_2$ are collinear.
2.Let be given a positive integer $n$ and two coprime integers $a,b$ greater than $1$. Let  $p$ and $q$ be two odd divisors of $a^{6^n}+b^{6^n}$ different from $1$.Find the remainder of $p^{6^n}+q^{6^n}$ when divided by $6.12^n$.
3.Given real numbers $a,b$ the sequence $(x_n)_{n=0}^{\infty}$ is defined by $x_0=a$ and  $x_{n+1}=x_n+bsinx_n$ for all $n\geq 0$.
(a)If b=1 .prove that for every $a$ , the sequence $(x_n)$ has a finite limit  when $n\to \infty$ and find this limit.
(b)Prove that for every $b>2$ there exists a real number  $a$ for which the sequence $(x_n)$ does not have a finite limit when $n\to \infty$  .
Day2
1.Let $x,y,z$ be positive real numbers that satisfyng :
                 $\frac{1}{\sqrt2}\leq z<\frac{1}{2}.\min\{x\sqrt2,y\sqrt3\}$;
                     $x+z\sqrt3\geq \sqrt6$;
                       $y\sqrt3+z\sqrt10\geq 2\sqrt5$.
      Find the maximum value of $P(x,y,z)=\frac{1}{x^2}+\frac{2}{y^2}+\frac{3}{z^2}$.
2.Consider the fuction $g(x)=\frac{2x}{x^+1}$.Find all continuous fuctions $f(-1,1)\to R$ that satisfying :  $(1-x^2)f(g(x))=(1+x^2)^2.f(x)$ for all $x\in\(-1,1)$.
3.Let $n\geq 1$ be a given integer .Consider a permutation $(a_1,a_2,...,a_{2n})$ of the fisrt $2n$ positive integers such that th numbers $|a_{i+1}-a_i|$ are distinct for $i=1,2,...2n-1$.Prove that $a_1-a_{2n}=n$ if and only if $1\leq a_{2k}\leq n$ for every $k=1,2,...n$.

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=788932#788932
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=788935#788935
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=788936#788936
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=788939#788939
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=788940#788940
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=788942#788942


I thought of some question (not a problem) I don't know, please help me.
    let ABCD inscribed in a circle with radius r, can we calculate the maximum of AB*BC+CD*DA? or AB*BC+BC*CD+CD*DA+DA*AB? or AB*CD+BC*AD?:maybe:

For the first and the third i think we use Ptolemy's Theorem for inscribed quadrilateral(ABCD).

I see the third one using Ptolemy's theorem, but how do you do the first one?

@kritikos: He's asking for the maximum not the minimum. Ptolemy's theorem gives a lower bound. If that's not what you mean, could you please elucidate?

We have the hypothesis: [b] ABCD inscribes a circle of radius r [/b]

Thus $ AB.CD \plus{} BC.AD \equal{} AC.BD$

We know that $ AC, BD \leq 2r$

Thus, the maximum occurs when ABCD is  a square.

For the first one, notice that

$ S_{ABC} \equal{} \frac {1}{2} BA.BC sin{\angle{BAC}}$

From this, we deduce: $ BA.BC \plus{} CD.DA \equal{} 1/2 \frac{S_{ABCD}}{ sin{\hat{B}}}$
We now can find the maximum.

Hope this help.

So for the first one, what is the maximum of Sabcd/sinB?


The numbers from 1 to 7 are witten on a chalkboard

1 2 3 4 5 6 7

You may choose any two numbers and replace them by their positive difference. (Thus you could erase the numbers 3 and 7 and write a 4 in their place.) Continue this process until there is only one number left.

What is the smallest possible number that can be obtained in this manner ?

What is the largest possible number ?

How do the answers change if we instead use the numbers from 1 to 8, or the numbers from 1 to 9 ?

Why is this so ?

First consider the min case,

Clearly, every four consecutive integers can be converted to one $0$.
eg $5-4=1$, $7-6=1$, $1-1=0$

Case 1: the numbers are $1,2,3,...,4n$
By eliminating the numbers 4 by 4, a $0$ is left

Case 2, the numbers are $1,2,3,...,4n+3$
By eliminating the numbers 4 by 4, starting from the large number, 1, 2, 3 are left.
$3-2=1$, $1-1=0$

Case 3, the numbers are $1,2,3,...,4n+1$
The process is equivalent to placing + or - signs before numbers, the parity is unchanged. Therefore the final number is odd.
By eliminating the numbers 4 by 4, a $1$ is left, it is the smallest

Case 4, the numbers are $1,2,3,...,4n+2$
Similar to Case 3, the final number is odd.
By eliminating the numbers 4 by 4, $1, 2$ is left
$2-1=1$, it is the smallest

The max case can be done by converting the numbers except the last one to 0 or 1, then take the difference between it and the largest number.


Solve the system:
$\begin{cases}\displaystyle x+y+z=1\\ \displaystyle\frac{1}{x}+\frac{1}{y}+\frac{1}{x}=1\\x^n+y^n+z^n=1.\end{cases}$
where $n\in\mathbb{Z}.$

[quote="nickolas"]Solve the system:
$x+y+z=1$
$x^n+y^n+z^n=1$
where $n\in\mathbb{Z}.$[/quote]
Um... so $n=1$? :?

Let
$x+y+z=1$
$xy+yz+xz=xyz=\beta$, so $x,y,z$ are roots of polynomial:
$W(t)=t^3-t^2+\beta t -\beta$, but $W(1)=0$, so one of numbers $x, y,z$ is equal to $1$. WLOG $x=1$ If we put this to out equations we get $y=-z$ and $n\equiv{1}\pmod{2}$.

The second equation gives $xy+yz+zx=xyz$ (where obviously $xyz\neq 0$). Put $p=xy+yz+zx=xyz.$ Then the desired numbers are roots of the polynomial

$P(t)=t^3-t^2+pt-p=(t-1)(t^2+p)\iff\{x,y,z\}=\{1, \sqrt{-p},-\sqrt{-p}\}$

It is assumed that the solution is sought for in $\mathbb{C}$

Let $n=2k$. Then $1=x^n+y^n+z^n=1+(-p)^k+(-p)^k\iff p=0$, but $p=xyz\neq 0$. Hence in this case the system has no solution.

Let $n=2k+1.$ Then $1=x^n+y^n+z^n=1+\sqrt{-p}\cdot(-p)^k-\sqrt{-p}\cdot(-p)^k$, hence any $p$ satisfies the conditions.

Conclusion: For even $n$ the system has no solution and for odd $n$ it has infinitely many solutions of the form $\{x,y,z\}=\{1,\sqrt{-p},-\sqrt{-p}\}$ where $p$ is an arbitrary real number.

Note: For a real solution it's enough to take $p<0$, giving $\{x,y,z\}=\{1,w,-w\}$ where $w\in\mathbb{R}$


probability
parameterization
conditional probability

[quote="fizsle"]1.The probability of a bomb hitting a bridge is ½ and 2 direct hits are required to destroy it. The least number of bombs required, so that the probability of the bridge being destroyed is greater than 0.9 is ?

  a.what is the probability of destruction of the bridge if only 5
     bombs are dropped?

2. In a bag there are 10 black, 8 white and 5 red balls. Three balls are chosen at random and 1 is found to be black. What is the probability that the rest 2 are white?[/quote]that is so easy question. :oops:

#2 Solved...

I need help with the #1.... anybody there???

[quote="fizsle"]1.The probability of a bomb hitting a bridge is ½ and 2 direct hits are required to destroy it. The least number of bombs required, so that the probability of the bridge being destroyed is greater than 0.9 is ?

  a.what is the probability of destruction of the bridge if only 5
     bombs are dropped?

2. In a bag there are 10 black, 8 white and 5 red balls. Three balls are chosen at random and 1 is found to be black. What is the probability that the rest 2 are white?[/quote]

[hide]
Problem 1 is a binomial random variable with $p = q = \frac12$ and $n = 2$.

$Pr(\text{destruction}) = \binom{N}{2}(\frac12)^{2}(\frac12)^{N-2}= \frac{(N^{2}-N)}{2}\cdot (\frac12)^{N}$

$\Rightarrow Pr(\text{destruction}) = (N^{2}-N)\cdot(\frac12)^{N+1}$

I trust you can solve the first part. As for the 'a' part, if 5 bombs are dropped then $Pr(\text{destruction}) = (5^{2}-5)\cdot (\frac12)^{6}= \frac{20}{32}= \frac{5}{8}$.

[b]Correction[/b]: The result above should be $\frac{20}{64}=\frac{5}{16}$.  (There a 3 kinds of mathematicians: those who can count, and those who cannot.  :blush: )

Problem #2: The answer that is required is the conditional probability of choosing white balls second and third given that a black ball was chosen first, $P(W_{2}W_{3}|B_{1})$

$P(W_{2}W_{3}|B_{1}) = P(W_{2}|B_{1})\cdot P(W_{3}|B_{1}W_{2})$

$P(W_{2}|B_{1}) = \frac{8}{22}= \frac{4}{11}$

$P(W_{3}|B_{1}W_{2}) = \frac{7}{21}= \frac{1}{3}$

$\therefore P(W_{2}W_{3}|B_{1}) = \frac{4}{11}\cdot\frac{1}{3}$
$=\frac{4}{33}$[/hide]

[hide]Suppose $n$ bombs are dropped.  We find the probability of the bridge being destroyed by considering the complement, the probability of the bridge not being destroyed.  There are $\binom{n}{0}+\binom{n}{1}=n+1$ ways for the bridge to not be destroyed, each with probability $\frac{1}{2^{n}}$.  Therefore, we have $1-\frac{n+1}{2^{n}}>\frac{9}{10}\Rightarrow 2^{n}>10(n+1)$.  The smallest integer solution is $n=7$.[/hide]

[quote="xsquaredster"][quote="matt276eagles"]Therefore, we have $1-\frac{n+1}{2^{n}}>\frac{9}{10}\Rightarrow 2^{n}>10(n+1)$.  The smallest integer solution is $n=6$.[/quote]

Not so. $2^{6}= 64 < 10(6+1) = 70$. $n=7$ works just fine though.[/quote]

Oops, I'll edit my solution.

[hide="number 2"]

The probability that two white balls are drawn from three total given that 1 is black ($P(W|B)$, the | is the "given that" symbol) is

$\frac{P(W\cap{B})}{P(B)}$, where $P(W\cap{B})$ is the probability that two white balls and a black ball are drawn if one draws three balls.

This value is $\frac{10}{23}\cdot\frac{8}{22}\cdot\frac{7}{21}\cdot{3\choose2}=\frac{40}{253}$

$P(B)=\frac{10}{23}$.

So $P(W|B)=\frac{\frac{40}{253}}{\frac{10}{23}}=\frac{4}{11}$.

okay... did I do something wrong?  :maybe: [/hide]

[quote="236factorial"]...

This value is $\frac{10}{23}\cdot\frac{8}{22}\cdot\frac{7}{21}\cdot{3\choose2}=\frac{40}{253}$

$P(B)=\frac{10}{23}$.

So $P(W|B)=\frac{\frac{40}{253}}{\frac{10}{23}}=\frac{4}{11}$.

okay... did I do something wrong?  :maybe: [/quote]

Yes, the factor $\binom{3}{2}$ is unnecessary. You've already accounted for each probability, $\frac{10}{23}\cdot\frac{8}{22}\cdot\frac{7}{21}$. Multiplying by $\binom{3}{2}$ increases the answer by a factor of $3$.

[quote="fizsle"]1.The probability of a bomb hitting a bridge is ½ and 2 direct hits are required to destroy it. The least number of bombs required, so that the probability of the bridge being destroyed is greater than 0.9 is ?

  a.what is the probability of destruction of the bridge if only 5
     bombs are dropped?

2. In a bag there are 10 black, 8 white and 5 red balls. Three balls are chosen at random and 1 is found to be black. What is the probability that the rest 2 are white?[/quote]

[hide="Number 1"]  Let's say that we need at least x bombs.  We want to find all the ways to have at least two of the bombs hit and divide that by the total number of ways, since each one has equal probability.  There is 1 way for 0 bombs to hit and x ways for one bomb to hit.  Since there are a total of x bombs and each has a $\frac{1}{2}$, there are a total of $2^{x}$ arrangements.  Then there is a probability of $\frac{2^{x}-x-1}{2^{x}}$ of the bridge being destroyed.  We know that this is greater than 0.9 but it is not greater than 0.9 for x-1 bombs.  In other words, $\frac{2^{x-1}-x+1-1}{2^{x-1}}\le 0.9<\frac{2^{x}-x-1}{2^{x}}$ or $\frac{2^{x}-2x}{2^{x}}\le 0.9<\frac{2^{x}-x-1}{2^{x}}$  This is also equivalent to $0.1-\frac{2x}{2^{x}}\le 0<0.1-\frac{x-1}{2^{x}}$ and $20x\ge 2^{x}>10x+10$.  It is easy to see from this that x=7 because 64<60+10 but 128>70+10.[/hide]

[hide="Number 1a"]
Like before, there are 1+5=6 ways to choose hit or miss for each of the 5 bombs such that less than two hit.  This gives a probability of $\frac{2^{5}-6}{2^{5}}=\frac{13}{16}$[/hide]

[hide="Number 2"]
This is simply the probability of choosing 2 white from 9 black, 8 white, and 5 brown, which is $\frac{8}{22}\times \frac{7}{21}=\frac{4}{33}$[/hide]

However, if problem 2 means EXACTLY one is black, then it is a probability of $\frac{8}{13}\times \frac{7}{12}=\frac{14}{39}$

[quote="fizsle"]Thanks Dr. .... can you help me understand the logic...I don't get it...(i'm no mathematician) :( ....[/quote]

Sure, but first check my edit of part 1a. I made an arithmetic mistake.

The notation $\binom{N}{n}$ means choose 'n' things out of a collection of 'N' things. The N things in this case are bombs dropped. The n things are direct hits ($n=2$ required for destruction). $\binom{N}{2}= \frac{N!}{2!(N-2)!}=\frac{N(N-1)}{2!}$.

The parameter $p =\frac12$ is the probability of achieving a hit for any single bomb. (The parameter $q = 1-p =\frac12$ is the probability that any single bomb will miss.)

Therefore the overall probability that you will achieve 2 hits out of N bombs dropped is $\binom{N}{2}p^{2}q^{N-2}$.


I am given the roots of 3 + root2    and 3 - root 2. I need to get the quadratic equation from this. Can someone help me out

First, note that any quadratic function (a function of the form $ax^2+bx+c$) can be factored into the form

\[a(x-r_1)(x-r_2)\]

where $r_1$ and $r_2$ are the roots of the quadratic.  The roots of this quadratic are $3+\sqrt2$ and $3-\sqrt2$, so this quadratic must be of the form

\[a(x-(3+\sqrt2))(x-(3-\sqrt2))\]

Using the FOIL method, we can expand the above to get

\[a(x^2-6x+7)\]

No matter what $a$ is used in the above, the roots will always be $3+\sqrt2$ and $3-\sqrt2$, so this is all that can be determined, unless more information is given.

Thank you very, very much man. Its all making sense now!


function
inequalities
complex analysis
real analysis
real analysis unsolved

My teacher claimed that a positive harmonic function on the unit disc is bounded. I don't see how this follows. Is it true? It does not seem to conform with Harnack's inequality either.

In my notes it's scribbled:
$ u \geq 0$, $ u \equal{} \Re f$, $ e^{ \minus{} f}$ ... bounded $ u$.

Yeah this is simply not true. $ z \to \frac {1 \minus{} |z|^2}{|1 \minus{} z|^2}$ is harmonic in the unit disc, and positive. If we approach 1 radially it blows up.

What's written in your notes makes sense, though: if $ u\ge 0$ is harmonic, and $ f$ is a holomorphic function such that $ \mathrm{Re}\,f\equal{}u$, then $ e^{\minus{}f}$ is bounded.

Yeah, I think maybe what he said is not that $ u$ is bounded, but that since $ e^{ \minus{} f}$ is bounded it has nontangential limits, and so does $ u$. I got confused by poor notes.


What are $ \sinh$ and $ \cosh$, and how are they used?

How could I prove that $ \cosh^2x\minus{}\sinh^2x\equal{}1$?

Thanks!


vector
linear algebra
linear algebra unsolved

Prove or disprove:Every subspace [b]W[/b] of a vector space[b] V[/b] is spanned by a finite set [b]S[/b].

If you take $W$ to be an infinite subspace of some infinite vector space $V$, then $W$ can obviously not be spanned by a finite set... Or am I missing something?

You're totally right. Now one can pose a question like this:
Let $V$ be an infinite dimentional vector space with $a$ as its cardinal. What can we say about the cardinality of the basis of this vector space?
Let me put it this way:What is the sufficient condition for a  given independent  infinite subset of an infinite dimensional vector space to be a basis?
PS Zorn's lemma guarantees that a basis exists.


algebra
polynomial
function
inequalities
limit
calculus
calculus computations

Thank you.

What if you wanted to find a certain amount of accuracy (to a certain number of decimal points)?

You might want to look at [url=http://clem.mscd.edu/~talmanl/PDFs/APCalculus/Lagrange.pdf]this[/url].

[quote="Shivar"]How would you use Taylor polynomials to approximate $e$?[/quote]

As pointed out $e^{x}=1+x+x^{2}/2!+...$

So let $x=1$ to get, $e = 2+\frac{1}{2!}+...$

The question is how accurate it is. We will use the Lagrange form of the Remainder to answer this question. Which I state for reference. (Here Taylor polynomials are centered at zero).

[b]Taylor's Theorem:[/b] Let $f$ be a differentiable function on $(a,b)$ with $a<0<b$ is $n$ times differentiable. And let $0\not = x\in (a,b)$. If $T_{n-1}(x)$ represents the $n-1$ degree Taylor polynomial and if $R_{n}(x)$ is the remainder, that is, $R_{n}(x) = f(x)-T_{n-1}(x)$ then there exists a number $y$ between $0$ and $x$ so that, $R_{n}(x) = \frac{f^{(n)}(y)}{n!}\cdot x^{n}$.

So consider $e^{x}$ with $a=-\infty$ and $b=+\infty$ which is certainly differentiable for any $n$. So if we choose $x=1$ then there exists a number $0<y<1$ so that $R_{n}(1) = \frac{f^{(n)}(y)}{n!}\cdot (1)^{n}$.
But then,
$|R_{n}(1)| = | \frac{e^{y}}{n!}| \leq \frac{3}{n!}|$
You can use the rightmost inequality to tell you the error.

[quote="sylow_theory"][b]Taylor's Theorem:[/b] Let $f$ be a differentiable function on $(a,b)$ with $a<0<b$ is $n$ times differentiable. And let $0\not = x\in (a,b)$. If $T_{n-1}(x)$ represents the $n-1$ degree Taylor polynomial and if $R_{n}(x)$ is the remainder, that is, $R_{n}(x) = f(x)-T_{n-1}(x)$ then there exists a number $y$ between $0$ and $x$ so that, $R_{n}(x) = \frac{f^{(n)}(y)}{n!}\cdot x^{n}$.
But then,
$|R_{n}(1)| = | \frac{e^{y}}{n!}| \leq \frac{3}{n!}|$
You can use the rightmost inequality to tell you the error.[/quote]
Lagrange form of the remainder states that $R_{n}(x)=\frac{f^{(n+1)}(y)}{(n+1)!}\cdot x^{n+1}$.  I noticed you made that mistake on another problem as well.

And on the last step instead of $3$, obviously $3$ works, but more strongly we could just say $e$ for the numerator.

He's using a different $R_{n}(x)$, shifted by one term. His $R_{n}(x)$ is clearly stated as the remainder $f(x)-T_{n-1}(x)$ of the Taylor polynomial of degree $n-1$.

Oh.  I see.. It was just that I had a question on Lagrange Form of the Remainder and eventually figured it out but he used it in that manner but it didn't produce the correct answer.. I eventually got it and see why it works that way though.

[quote="Shivar"]How would you use Taylor polynomials to approximate $e$?[/quote]
[Proposal] Us the following facts: If $T_{n-1}(X): =\sum_{k=0}^{n-1}\frac{X^{k}}{k!}$, then
\[\begin{array}{|c|}\hline \\ T_{n-1}(X)+\frac{X^{n}}{n!}e^{\frac{ X}{n+1}}\le e^{X}\le T_{n-1}(X)+\frac{X^{n}}{n!}\frac{n+e^{X}}{n+1}\; ,\; \; (X\ge 0) .\\ \\ \hline \end{array}\]
Further consider $X=1 .$

[size=75]
[b]Above inequalities give us supplementary informations for remainder in Taylor's Theorem.[/b][/size]

[quote="skimnc"]
And on the last step instead of $3$, obviously $3$ works, but more strongly we could just say $e$ for the numerator.[/quote]
But then we need to know exactly how much $e$ is, if we really know how much it is why approximate  :wink:

[quote="sylow_theory"]But then we need to know exactly how much $e$ is, if we really know how much it is why approximate  :wink:[/quote]
My opinion: Let $a \in{\mathbb Z}$ and $x\in (a,a+1)$ . It's a curiosity (because we are mathematicians) to find
first  decimals $a_{1},a_{2},...a_{n},...$  from the decimal representation  $x=a.a_{1}a_{2}...a_{n}...$ . In this sense,please see
an interesting book by Steven Finch , [i]Mathematical constants[/i]  ,available on-line at
[color=red][URL deleted by a moderator][/color]

By the way, which is your favourite definition of Napier's constant $e$ ?
Using your definition, try to prove, in a simple way,  that  $e\in (2,3) .$
Perhaps, some definitions are:
1) $e: =\sum\limits_{k=0}^{\infty}\frac{1}{k!}$ ;
2) $e: =\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}$ ;
3) $e$ is the unique positive number  $a$ such that $a^{x}\ge x+1 \; ,\; \forall x \in{\mathbb R},$ ;
4) $e$ is the unique positive number  $a$ such that $a^{x}\ge x^{a}\; \; ,\; \forall x \in (0,\infty)$ ;
5)  $e$ is the unique positive number  $a$ such that $\frac{x^{2}+6x+12}{x^{2}-6x+12}\le a^{x}\le \frac{2+x}{2-x}\; \; ,\; \forall x \in [0, 2)$ ;
...