calculus
trigonometry
function
derivative
limit
real analysis
real analysis theorems

How does one prove that $ |\sin(x)|\leq |x| \quad \forall x \in \mathbb{R}$ if $ |x| < 1$ algebraically ?

Use the Taylor series.  But there is a nice geometric argument as well as the extremely straightforward calculus argument.

Alternatively, consider $ f(x)\equal{}\sin x \minus{}x$. Since sin and x are odd, we can drop the absolute values and just consider $ x\geq 0$ and thus $ f$. We have $ f(0)\equal{}0$. Also, we have $ f'(x)<0$ for $ x<1$, so $ f$ is decreasing and thus always $ \leq 0$. This gives what we want.

In general: the Mean Value Theorem shows that $ \sin x\minus{}\sin y\equal{}(x\minus{}y)\cos\xi$ for some $ \xi$ in between $ x$ and $ y.$ Since $ |\cos\xi|\le1,$ we get that $ |\sin x\minus{}\sin y|\le|x\minus{}y|$ for all $ x$ and $ y.$ Now just particularize this to the case in which $ y\equal{}0.$

This is sort of obvious, but if you use the usual "unit circle" interpretation of sin, and $ \theta\in[0,\pi/2]$, you can easily see that the straight line segment from $ (\cos\theta,0)$ to $ (\cos\theta,\sin\theta)$ is shorter than the circular arc from $ (1,0)$ to $ (\cos\theta,\sin\theta)$, which has length $ \theta$.  (Then it's easy to extend to all $ \theta$.)  That's sort of a synthetic "proof."

I'm guessing that's not what you wanted since you wanted to prove it "algebraically," but you probably meant to say "analytically" or "using calculus" since sin is not an algebraic function.

Perhaps my synthetic proof looks stupid, but it's worth noting that any calculus proof ultimately depends on how you decide to define these functions to begin with, which is sort of arbitrary.  My point is that it's almost as "obvious" that $ |\sin x|\leq |x|$ as it is "obvious" that $ |\sin x|\leq 1$, so then you're left asking things like, "How do you prove that $ |\sin x|\leq 1$"?  Imho, it's a useful exercise to see how to prove *all* standard properties of trig functions from first principles, but doing it piecemeal is a slightly weird thing to do because you never know whether you're going in circles.

Btw, Professor Merryfield's answer is the most appealing to me because it uses the most low-powered facts and is still elegant.

Well, my proof did require knowledge of the derivative of the sine. And to know that, I need to know that $ \lim_{x\to0}\frac{\sin x}{x}\equal{}1,$ and how am I going to prove that? That may very well bring me back to the arguments in the first paragraph of yenlee's post.

So here's a complete answer:  By Professor Merryfield's argument, the only properties of sin that you need are: (1) differentiability (so that MVT applies), and (2) the fact that $ |\sin'(x)|\leq 1$.  My point was that how you prove these things depends a bit on how you define sin.  Prof. Merryfield pointed out that it's not quite obvious that fact (2) is any easier than the theorem to be proved.

My preferred definition of sin is that it is the unique (smooth) solution to the ODE $ y''\plus{}y\equal{}0$ with initial values $ y(0)\equal{}0$ and $ y'(0)\equal{}1$.  Unfortunately, you need the existence and uniquess theorem for (constant coefficient) linear ODEs in order for this definition to make sense.  From the defining ODE and the initial conditions, one can then show that $ (\sin x)^2\plus{}(\sin' x)^2\equal{}1$ for all real $ x$.

Another standard definition of sin is to use power series.  Of course, this definition is equivalent to the one I just gave (as it must be), but it's less obvious to me how to prove the theorem directly from the power series definition (as suggested by t0rajir0u).

[quote="Kent Merryfield"]Well, my proof did require knowledge of the derivative of the sine. And to know that, I need to know that $ \lim_{x\to0}\frac {\sin x}{x} \equal{} 1,$ and how am I going to prove that? That may very well bring me back to the arguments in the first paragraph of yenlee's post.[/quote]

taylor explansion for sinx is the alternating sum over odds (1/k!)x^k (for cosine is the alternating evens!)

rewriting sinx, we see we can factor out x from the infinite expansion, leaving us with the same sum, just one less power on each x. nevertheless, as x-> 0, these all drop, leaving us with 1.[/code]

also, you could use the knowledge that the limit you are evaluating is merely the definition of derivative of sinx at x=0

Which is circular, because you can't get the derivative without calculating that limit in the first place.  Ditto the Taylor series.


function
geometry
circumcircle
3D geometry
calculus
integration
analytic geometry

CGMO 2006, Day 1.

Problems taken from here: http://www.aoshoo.com/bbs1/dispbbs.asp?boardID=15&ID=5693&page=1

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?p=596307#596307
Let $a>0$, the function $f: (0,+\infty) \to R$ satisfies $f(a)=1$, if for any positive reals $x$ and $y$, there is \[f(x)f(y)+f \left( \frac{a}{x}\right)f \left( \frac{a}{y}\right) =2f(xy)\] then prove that $f(x)$ is a constant.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?p=596309#596309
Let $O$ be the intersection of the diagonals of convex quadrilateral $ABCD$. The circumcircles of $\triangle{OAD}$ and $\triangle{OBC}$ meet at $O$ and $M$. Line $OM$ meets the circumcircles of $\triangle{OAB}$ and $\triangle{OCD}$ at $T$ and $S$ respectively.

Prove that $M$ is the midpoint of $ST$.

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?p=596312#596312
Show that for any $i=1,2,3$, there exist infinity many positive integer $n$, such that among $n$, $n+2$ and $n+28$, there are exactly $i$ terms that can be expressed as the sum of the cubes of three positive integers.

Edit: Thanks to the correction by ychjae.  :)

Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?p=596313#596313
$8$ people participate in a party.

(1) Among any $5$ people there are $3$ who pairwise know each other.  Prove that there are $4$ people who paiwise know each other.

(2) If Among any $6$ people there are $3$ who pairwise know each other, then can we find $4$ people who pairwise know each other?

Here is a PDF file of the problems of Day 1 Contest.

By the way, I hope someone will help me translating problem 4 better.

I think
"such that among $n$,$n+2$and$n+28$, there are [b]exactly [/b]$i$ terms that can be expressed as the sum of the cubes of three positive integers. "
is better

Day 2 Problems:

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?p=597353#597353
The set $S = \{ (a,b) \mid 1 \leq a, b \leq 5, a,b \in \mathbb{Z}\}$ be a set of points in the plane with integeral coordinates. $T$ is another set of points with integeral coordinates in the plane. If for any point $P \in S$, there is always another point $Q \in T$, $P \neq Q$, such that there is no other integeral points on segment $PQ$.  Find the least value of the number of elements of $T$.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?p=597356#597356
Let $M= \{ 1, 2, \cdots, 19 \}$ and $A = \{ a_{1}, a_{2}, \cdots, a_{k}\}\subseteq M$. Find the least $k$ so that for any $b \in M$, there exist $a_{i}, a_{j}\in A$, satisfying $b=a_{i}$ or $b=a_{i}\pm a_{i}$ ($a_{i}$ and $a_{j}$ do not have to be different) .

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?p=597358#597358
Given that $x_{i}>0$, $i = 1, 2, \cdots, n$, $k \geq 1$. Show that: \[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\] Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?p=597360#597360
Let $p$ be a prime number that is greater than $3$. Show that there exist some integers $a_{1}, a_{2}, \cdots a_{k}$ that satisfy: \[-\frac{p}{2}< a_{1}< a_{2}< \cdots <a_{k}< \frac{p}{2}\] making the product: \[\frac{p-a_{1}}{|a_{1}|}\cdot \frac{p-a_{2}}{|a_{2}|}\cdots \frac{p-a_{k}}{|a_{k}|}\] equals to $3^{m}$ where $m$ is a positive integer.

Here is a PDF file of the complete CGMO 2006 problems.

[quote="yptsoi"]Do you know the cutting scores of awards, shobber?[/quote]
I don't know......Maybe you can ask the HK teachers 陸美蓮, 黃雲霞, or the contestants 潘瑛, 葉翎, 陳芳妮, 黎芷均, 林曉宜, 李家意, 梁穎斌, 魏瀟瑤.

i know the cutting score of gold is 81. :P  :D


function
limit
calculus
real analysis
real analysis unsolved

Is it true that when we have uniformly convergent series of functions $g_{k}: N\to R$,
$\sum_{i=1}^\infty g_{k}(n)$ then :

$\lim_{n\to\infty}\sum_{i=1}^\infty g_{k}(n)=\sum_{i=1}^\infty\lim_{n\to\infty}g_{k}(n)$

($\lim_{n\to\infty}g_{k}(n)$ exists)

//could you move this msg to computations and tutorials?

huh, whatever, i think it's true and that it is a part of more general theorem, am i wrong ?

First, regarding moving this message: I think it belongs right here. In terms of American college courses, the "computations and tutorials" should mostly concern calculus and differential equations. Material from "Real Analysis" or "Advanced Calculus" should mostly be here.

Second: it's false.

Let $g_{k}(n)=\begin{cases}\frac1k&1\le n\le k^{2}\\0&n>k^{2}\end{cases}$

Then as $k\to\infty,\ g_{k}(n)\to0$ uniformly, but

$\lim_{k\to\infty}\sum_{n=1}^{\infty}g_{k}(n)=\infty.$

Those sums should all be $\sum_{k=1}^\infty$, right? Summing over $i$ makes no sense.

That's not quite the right example; the series $\sum g_{k}$ does not converge uniformly.

With uniform convergence of the series, the limits are equal whenever both exist (by the same argument that tells us a uniform limit of continuous functions is continuous).
There are however situations in which one exists and the other doesn't: try $g_{k}(n)=\begin{cases}(-1)^{k+n}\cdot\frac1k& n\ge k\\0&n<k\end{cases}$.

Ok, so because i've added that $\lim_{n\to\infty}g_{k}(n)$ exists then i was right :)


Could someone give me a few basic induction proof problems to practice (or a link)?  Thanks!

Here's a pretty easy one that's pretty common, I think.

Prove that

$ 1\plus{}3\plus{}5\plus{}\cdots(2n\minus{}1)\equal{}n^2$

There's also:

Prove that

$ (1\plus{}2\plus{}\cdots n)^2\equal{}1^3\plus{}2^3\plus{}\cdots n^3$

The second chapter of this book has several introductory induction problems.
[url=http://shell.cas.usf.edu/~eclark/elem_num_th_book.pdf]Elementary Number Theory[/url]

($ n\in\mathbb{N}$ for all problems)
Summations:
$ \frac{1}{1\cdot2}\plus{}\frac{1}{2\cdot3}\plus{}\frac{1}{3\cdot4}\plus{}\cdots\plus{}\frac{1}{n(n\plus{}1)}\equal{}\frac{n}{n\plus{}1}\\
2\plus{}5\plus{}8\plus{}\cdots\plus{}(3n\minus{}1)\equal{}\frac{n(3n\plus{}1)}{2}$

Inequality:
$ 2^n\geq2n\\
3^n\geq3n$

Divisibility:
$ n(n\plus{}1)$ is even.
$ 4^n\minus{}1$ is divisible by 3.

[quote="10000th User"]Everything looks good enough.  :)[/quote]

Thanks a lot!


arithmetic sequence
geometric sequence
algebra unsolved
algebra

Prove that for an integer n >= 3 there exists an arithmetic progression a_1; a_2; ... ; a+n
and a geometric progression b_1; b_2; .... ; b_n so that b_1 < a_1 < b_2 < a_2 < : : : < b_n < a_n.

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . But unlock because you are a new member.

Can you post this solution ?.I am new member so i can't open the page you have posted .


algebra
polynomial
trigonometry
complex numbers
algebra unsolved

Let $a_1,a_2,...a_n \in R$ satisfy that:
$f(x)= \sum \frac{cos(x+a_i)}{2^i}$and $f(x_1)=f(x_2)=0$.
Prove that:$\frac{x_2-x_1}{\pi} \in \mathbb{Z}$

[color=red]Edited by Megus[/color]

I guess that solution starts like this:

$f(x_1)-f(x_2)= 0 = \sum {cos (x_1+a_i) - cos (x_2+a_i)\over 2^i} = -sin {x_1-x_2\over 2} \sum {sin {x_1+x_2+2a_i\over2} \over 2^{i-1}}$
$f(x_1)+f(x_2)= 0 = \sum {cos (x_1+a_i) + cos (x_2+a_i)\over 2^i} = cos {x_1-x_2\over 2} \sum {cos {x_1+x_2+2a_i\over2} \over 2^{i-1}}$

If second part of each equation is different from 0 then from $-sin {x_1-x_2\over 2} = cos {x_1-x_2\over 2} = 0$ follows that ${{x_2-x_1}\over {\pi}} \in \mathbb{Z}$ but I can't prove that second part of each equation is different from 0...

Since $\cos(x+a)=\cos x\cos a-\sin x\sin a,$ we have
\[ f\left( x\right) =A\cos x-B\sin x,  \]
where $A=\sum\frac{\cos a_{i}}{2^{i}}$ and $B=\sum\frac{\sin a_{i}}{2^{i}}.$

Now, if $A\neq0$ or $B\neq0$ the conclusion follows easily. Suppose $A=B=0$
and consider the complex numbers $z_{k}=\cos a_{k}+i\sin a_{k}.$ We have
$\left\vert z_{k}\right\vert =1$ for all $k$ and
\[ \sum\frac{z_{k}}{2^{k}}=0.  \]
Write the latter as
\[ z_{1}=-\frac{z_{2}}{2}-\frac{z_{3}}{2^{2}}-\ldots-\frac{z_{n}}{2^{n-1}},  \]
and take absolute values in both sides. This yields

\[ 1=\left\vert \frac{z_{2}}{2}+\frac{z_{3}}{2^{2}}+\ldots+\frac{z_{n} }{2^{n-1}}\right\vert \leq\frac{\left\vert z_{2}\right\vert }{2}+\frac{\left\vert z_{3}\right\vert }{2^{2}}+\ldots\frac{\left\vert z_{n}\right\vert }{2^{n-1}}= \]
\[ =\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{n-1}}=1-\frac{1}{2^{n-1}}, \]

a contradiction.


Source: ACoPS
Prove that $n!<\left(\frac{n+1}{2}\right)^{n}$, for $n=2,3,4,....$

Here's how i solved:
[hide]
$n!=n\cdot (n-1)\cdot (n-2)\cdot\cdot\cdot2 \cdot 1$
Then i paired up the first and the last ,etc.
So $n\cdot 1<\left(\frac{n+1}{2}\right)^{2}$ by AM-GM (the equality happens if $n=1$, but since $n=2,3,4,....$ RHS is always greater than LHS.
$(n-1)\cdot 2<\left(\frac{n+1}{2}\right)^{2}$ and so on..
now we multiply out all the inequalities and we are done.
[/hide]

Oops, I posted this post twice...
I thought I could erase my posts........

We should work with this topic, as you have posted your full solution here.

[quote="kimby_102"]Source: ACoPS
Prove that $n!<\left(\frac{n+1}{2}\right)^{2}$, for $n=2,3,4,....$

Here's how i solved:
[hide]
$n!=n\cdot (n-1)\cdot (n-2)\cdot\cdot\cdot2 \cdot 1$
Then i paired up the first and the last ,etc.
So $n\cdot 1<\left(\frac{n+1}{2}\right)^{2}$ by AM-GM (the equality happens if $n=1$, but since $n=2,3,4,....$ RHS is always greater than LHS.
$(n-1)\cdot 2<\left(\frac{n+1}{2}\right)^{2}$ and so on..
now we multiply out all the inequalities and we are done.
[/hide]

Oops, I posted this post twice...
I thought I could erase my posts........[/quote]
$n!>\left(\frac{n+1}{2}\right)^{2}$, for $n=3,4,5....$.  :maybe:

[quote="nebula42"]I think you meant $n!<(\frac{n+1}{2})^{n}$. Try plugging in $n=3$ into the inequality you have, for example. Then we get $6<4$, which is obviously incorrect.[/quote]
Yes, you are right.. I was in a hurry...
edited.

that's what i was thinking. induction is so powerful  :)


calculus
integration
trigonometry
logarithms
limit
real analysis
real analysis unsolved

Let  $f,\; f\in C[0,\pi] ,$ be such that   $f(\pi-x)=-f(x)\; ,\; \forall x \in [0,\pi] ,$   and $I[f]: =\int_{0}^{ \frac{\pi}{2}}\left(f(x)-f(2x)\right) \cot{x}\; dx \;$ exists (as proper or improper integral).[b]
Find  the  value of  $I[f]$ . [/b]

[quote="Carcul"][hide="Answer"]Use, for example, $f(x) = \cos x$, and then the answer is $I[f] = \ln 2$.[/hide][/quote]
Hi "Carcul",
 the answer, i.e. [b][u] the  explicit (analytic)  form of [/u] [/b] $I[f]$ depends (in my proposed question) on $f$ .

Ok, I thought I[f] was supposed to be the same for every f(x) verifying the condition you gave. Sorry about that.

Let $f(x)=\sum_{n=1}^{\infty}a_{2n-1}\cos(2n-1)x$ be the cosine Fourier expansion of $f$.

The initial condition on $f$ implies that the Fourier coefficients $a_{2n}=0$ where

$a_{n}=\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos nx dx$.

In this case, if my calculations are correct, I got that

$I(f)=\ln(2)f(0)-$

$-\sum_{n=1}^{\infty}a_{2n-1}(2n-1)\sum_{m=1}^{\infty}\frac{1}{m}[\frac{1}{2}\left(\frac{1}{2n-1+2m}+\frac{1}{2n-2m-1}\right)-\left(\frac{1-(-1)^{m-1}}{2(2n+m-1)}+\frac{1-(-1)^{m-1}}{2(2n-1-m)}\right)]$.

We have

$I(f)=\sum_{n=1}^{\infty}a_{2n-1}\int_{0}^{\pi/2}\left(\cos(2n-1)x-\cos2(2n-1)x\right)\cot x dx$.

Integrate by parts and we get that

$I(f)=\sum_{n=1}^{\infty}a_{2n-1}(2n-1)\int_{0}^{\pi/2}\ln \sin x\left(\sin (2n-1)x-2\sin(2(2n-1)x\right)dx$

Using the following Fourier expansion

$\ln \sin x=-\ln 2-\sum_{m=1}^{\infty}\frac{\cos 2mx}{m}$, $0<x<\pi$,

and after some calculations we get the result above.

If we use the sine Fourier expansion of $f$ then note that $b_{2n-1}=0$. A similar calculation can be used.

Observe that

if $f(x)=\cos x$ then $I(f)=\ln 2$ since $f(0)=1$ and the double sum reduces to 0.
There may be some computational mistakes, however I think this is the idea.

It is not hard to check that the double sum reduces to 0 and hence the integral equals $f(0)\ln 2$.

[quote="didilica"]..... hence the integral equals $f(0)\ln 2$.[/quote]
[b]Indeed, this is the  answer ! [/b]

[quote="flip2004"]Let  $ f,\; f\in C[0,\pi] ,$ be such that   $ f(\pi-x)=-f(x)\; ,\; \forall x \in [0,\pi] ,$   and $ I[f]: =\int_{0}^{ \frac{\pi}{2}}\left(f(x)-f(2x)\right) \cot{x}\; dx \;$ exists (as proper or improper integral).[b]
Find  the  value of  $ I[f]$ . [/b][/quote]
Since that integral converges, it also equals the Cauchy Principal Value:
\[ \lim_{\epsilon \to 0}\int_\epsilon^{\frac{\pi}2-\epsilon}\left( f(x)-f(2x) \right) \cot x \, dx . \]
Substitute $ y = 2x$:
\[ I_\epsilon = \int_\epsilon^{\frac{\pi}2-\epsilon}f(x) \cot x \, dx-\frac12 \int_{2 \epsilon}^{\pi-2 \epsilon}f(y) \cot \frac{y}2 \, dy . \]
By the well-known identity $ \cot \frac{y}2 = \cot y+\csc y$ we get that
\[ I_\epsilon = \int_\epsilon^{\frac{\pi}2-\epsilon}f(x) \cot x \, dx-\frac12 \int_{2 \epsilon}^{\pi-2 \epsilon}f(y) \cot y \, dy-\frac12 \int_{2 \epsilon}^{\pi-2 \epsilon}f(y) \csc y \, dy . \]
If $ g(y) = f(y) \csc y$, then $ g \left( \pi-y \right) =-g(y)$, so the last integral is $ 0$. Thus,
\begin{eqnarray*}I_\epsilon &=& \int_\epsilon^{\frac{\pi}2-\epsilon}f(x) \cot x \, dx-\frac12 \int_{2 \epsilon}^{\pi-2 \epsilon}f(x) \cot x \, dx \\ \ &=& \int_\epsilon^{2 \epsilon}f(x) \cot x \, dx+\int_{2 \epsilon}^{\pi-2 \epsilon}f(x) \cot x \, dx+\int_{\pi-2 \epsilon}^{\frac{\pi}2-\epsilon}f(x) \cot x \, dx-\frac12 \int_{2 \epsilon}^{\pi-2 \epsilon}f(x) \cot x \, dx \\ \ &=& \int_\epsilon^{2 \epsilon}f(x) \cot x \, dx+\frac12 \int_{2 \epsilon}^{\pi-2 \epsilon}f(x) \cot x \, dx-\int^{\pi-2 \epsilon}_{\frac{\pi}2-\epsilon}f(x) \cot x \, dx .\end{eqnarray*}
Now look at the last integral:
\[ \int^{\pi-2 \epsilon}_{\frac{\pi}2-\epsilon}f(x) \cot x \, dx = \int_{\frac{\pi}2-\epsilon}^\frac{\pi}2 f(x) \cot x \, dx+\int_\frac{\pi}2^{\pi-2 \epsilon}f(x) \cot x \, dx . \]
Using the fact that $ h(x) = f(x) \cot x$ satisfies $ h \left( \pi-x \right) = h(x)$, we get that the latter integral is $ \frac12 \int_{2 \epsilon}^{\pi-2 \epsilon}f(x) \cot x \, dx$.
Returning, we see that
\[ I_\epsilon = \int_\epsilon^{2 \epsilon}f(x) \cot x \, dx+\int_{\frac{\pi}2-\epsilon}^\frac{\pi}2 f(x) \cot x \, dx . \]
Now we can finally let $ \epsilon \to 0$. The second integral goes to $ 0$, there's no doubt about that.
It remains to establish
\[ \lim_{\epsilon \to 0}\int_\epsilon^{2 \epsilon}f(x) \cot x \, dx . \]
A side computation needed to finish the problem is
\[ \int_\epsilon^{2 \epsilon}\cot x \, dx = \ln \left( \frac{\sin 2 \epsilon}{\sin \epsilon}\right) = \ln \left( 2 \cos \epsilon \right) . \]
So we break $ \int_\epsilon^{2 \epsilon}f(x) \cot x \, dx$ into
\[ \int_\epsilon^{2 \epsilon}\left( f(x)-f(0) \right) \cot x \, dx+\int_\epsilon^{2 \epsilon}f(0) \cot x \, dx . \]
The second part goes to $ f(0) \ln 2$, whilst the first goes to $ 0$, by some easy epsilonistics.

So our integral is $ f(0) \ln 2$...

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=288711&highlight=specific#288711]Here's[/url] my inspiration source, in case you were wondering.
I hope that the vast majority of you are still awake, although I feel I may have exaggerated with the details :blush:

That's all, folks! :bye:

[quote="perfect_radio"][quote="flip2004"]Let  $ f,\; f\in C[0,\pi] ,$ be such that   $ f(\pi-x)=-f(x)\; ,\; \forall x \in [0,\pi] ,$   and $ I[f]: =\int_{0}^{ \frac{\pi}{2}}\left(f(x)-f(2x)\right) \cot{x}\; dx \;$ exists (as proper or improper integral).[b]
Find  the  value of  $ I[f]$ . [/b][/quote].......That's all, folks! :bye:.....[/quote]
Very nice solution,thanks !


Given are $a>0\ ,\ b\ge 0$. Ascertain the limit of the sequence
$x_{n}=\sqrt{a+\sqrt{a+\sqrt{a+\ldots+\sqrt{a+\sqrt{a+\sqrt b}}}}}$,
where are $n\in N^{*}$ radicals.

I assume ascertain means "compute".

$x_{n}=\sqrt{a+\sqrt{a+\sqrt{a+\ldots+\sqrt{a+\sqrt{a+\sqrt b}}}}}$

$(x_{n})^{2}={a+\sqrt{a+\sqrt{a+\ldots+\sqrt{a+\sqrt{a+\sqrt b}}}}}$

$(x_{n})^{2}-a={\sqrt{a+\sqrt{a+\ldots+\sqrt{a+\sqrt{a+\sqrt b}}}}}$

Here's the ingenius crux move! :P

${(x_{n})^{2}-a=x_{n}}$

$x_{n}=\frac{1+\sqrt{1+4a}}{2}$
(Ignore the negative root)

P.S. In my method, I find the value of b to be irrelevant. Am I doing anything wrong? :blush:

i am not convinced that it converges...(does it?)

[b]vishalarul[/b], what you did was not wrong but you have to show two things first before you do what you did:
(1) Show that the sequence is an increasing sequence, and
(2) Show that it has an upper bound, say, L, i.e. each $x_{n}\le L.$

Then, there is a theorem that guarantees us that the sequence converges to a finite value, which could be different from $L$. And, now we can do what you did to get our limit.

Showing the first part is true isn't hard, but I am unable to get an upper bound though I am pretty sure there is one.


If you highlight the MATHCOUNTS Problem Series homepage website (or whatever ur class is) and then save it, then wait like a year later and try it on the toolbar, then will it still be there (even though it was taken down)?

Example:

http://www.blahblahwhateveryeahugotthepointwhatimean.com/

will it still work if i put it on the toolbar (even though it was taken down)?

Also, if u go to My Classes and click MATHCOUNTS Problem Series and highlight the transcripts like:

Transcripts
Class Transcript 06/11
Class Transcript 06/18
Class Transcript 06/25
Class Transcript 07/02
Class Transcript 07/09
Class Transcript 07/16
Class Transcript 07/23
Class Transcript 07/30
Class Transcript 08/06
Class Transcript 08/13
Class Transcript 08/20
Class Transcript 08/27

and save into MS will it work if u click on one (literally)?

(just asking)

 :D

I don't think that it will work. To be on the safe side, just do it the traditional way.

What would be the traditional way?

I just clicked "print" and then saved the web page on to my computer. Would that work?

I think it will work as long as it doesn't require an internet access to open it. My traditional way would be: click and drag to highlight all the materials, copy and paste it into word, where I saved the copy. I also saved it on my jump drive and another computer just in case things broke down or get lost.

I did the same thing as toadoncart in case they leave it up, which they probably won't.  I saved mine on a computer, my jump drive and a CD.  If you have copy and paste abilities on your '03 windows, toadoncart, you could click the "Print" Button, then Alt-A, Alt-C and open a text editor and then Alt-V.  That should work.  Good Luck! :D

i'm using a more recent made computer that isn't '03, but as i pressed those first two thingies u told me to do, it just opened the favorites.

OK, well I don't use windows so I don't know about favorites, but click $ \boxed{\text{Print}}$ and then highlight the text you want to save, then at the top under "Edit", click Copy.  Then open a text editor and go back to "Edit" and click Paste.  That should work.  :D

Nope. Tried it. For my text editor, it doesn't show any of the images or $ \mathfrak{LaTex}$

I REALLY Don't know then.  Oh, I messed up the code.  Instead of "Alt" use "Ctrl".
[hide="Do not open if you LOVE windows"][img]http://www.synath.com/pics/windows_sucks.jpg[/img]
I use mac, but  those kind of images are funny :rotfl:[/hide]


rotation
calculus
integration
geometry
geometric transformation

A thin, circular plate of negligible thickness and radius $ R$ has a charge of $ \plus{}Q$ (where $ Q>0$) evenly distributed over it. The plate rotates in a clockwise direction at an angular velocity of $ \omega$ . What is the magnitude and direction of the magnetic field experiences at the centre of the circular plate?

Is this the [hide="answer?"]$ \vec{B}\equal{}\frac{\mu_{0}Q\omega}{2\pi R}$, downwards[/hide]
Is there any other way to do this than brute force integration?

That's what I got. What do you mean brute force integration?

[hide="My Solution"]

Take a thin disk of radius $ r$ and thickness $ \Delta r$. So $ \Delta Q \equal{}\frac{2\pi r\Delta r}{\pi R^{2}}Q$ and if $ f$ is the frequency of rotation we have that $ I \equal{} f\Delta Q$ where $ I$ is the current generated. Now we can see that $ \Delta B \equal{}\frac{\mu_{0}I}{2r}\equal{}\frac{\mu_{0}\omega Q}{2\pi R^{2}}\Delta r$

Hence $ B \equal{}\int .dB \equal{}\int^{R}_{0}\frac{\mu_{0}\omega Q}{2\pi R^{2}}. dr \equal{}\frac{\mu_{0}\omega Q}{2\pi R}$.  [/hide]

I meant a more elegant method. But I don't think there is one, mainly because of not enough symmetry.

[quote="shreyas_patankar"]mainly because of not enough symmetry.[/quote]
 :wink:  well that statement somewhat sounds gr8


My PM has been disabled so I'm posting it here, so maybe someone can tell me what happened. My brother, THECHAMP, claims that he was on AoPS and logged off. His password and e-mail were changed within five minutes after he left and an imposter used his account briefly, using rather aggravating comments. Finally, he left by saying, "BBL." My brother, THECHAMP, would like his account back and hopes the administrators can find out exactly who's done this. Anyone who has any suspicions or news that can help catch the person responsible would be helpful.-------------------Thanks  :)

This would belong in FTW Questions & Support.

And I wouldn't think of anyone who would do that.

Are you sure that happened?  Did he type in the wrong pass?  Was he banned?

this is quite odd and unusual. i would go ahead with ernie's beliefs, but i dont know. this is probably a first for me seeing something like this

EDIT: maybe pm a mod? admin?

Yeah, I'm sure a mod knows your password, and as long as you can get verification that the username is yours, an admin would probably "remind" you of your password.

We do not have a way to retrieve your password. You can reset your password at:

http://www.artofproblemsolving.com/Forum/profile.php?mode=sendpassword

by providing your username and email address.


Does (lambda^2)>0 mean that lambda is real?

[quote="kouboy"]yes, there is no notion of greater and lesser in the complex field.[/quote]

Thanks kouboy  :)


You spin me right round baby right round..

So anyways, Gary Brolsma has a New Numa Numa video out.. It sucks, and he's just milking his 15 minutes.

Yeah I saw it. And I totally agree but he offers others to make a new numa numa so I guess we might find a good fan made one.

For those who haven't seen it:
[url]http://www.newnuma.com[/url]

XD when will the shady internet references stop?


1)  let x;y;z be positive.prove that:
    $ \sum\frac {x \plus{} y}{z \plus{} \sqrt [3]{4(x^3 \plus{} y^3)}}\le2$

Yes, we can do it!

The following estimation is crucial
$ \sqrt[3]{4(a^3\plus{}b^3)} \ge a\plus{}b$

We are done!
Surely we know how to finish it off in an elegant way!

wow,it is very simple.I also don't understand why I think many complex solutions
  thank u


Se consideră un grup G  şi funcţia $f: G\to G$, $f(x)=x^4$. Să se arate că, dacă funcţia f este morfism de grupuri, atunci funcţia $g: G\to G$, $g(x)=x^{n^{2}}$ , este morfism de grupuri pt. orice n natural nenul.

Am avut-o eu in concurs :D
Iti zic ideile:
1) $(xy)^3=y^3x^3$ si deci $(xy)^9=x^9y^9$
3) Demonstrezi prin inductie dupa $n$. Daca $n$ e par,$n^2$ se scrie ca $4k$ si deci
$(xy)^{4k}=((xy)^4)^k=(x^4y^4)^k=x^{4k}y^{4k}$ din ipoteza de inductie.
Daca $n$ impar se scrie ca $4k+9$ si avem $(xy)^{4k+9}=(xy)^{4k}(xy)^{9}=x^{4k}y^{4k}x^9y^9$

Mai demonstezi ca $x^4$ comuta cu toate elementele lui $G$ de forma $y^3$ si se termina problema.

Pentru $n$ impar nu vad de ce functioneaza. Ridicarea la puterea $k$ nu reprezinta un morfism, astfel ca egalitatea
\[ (xy)^{4k}=x^{4k}y^{4k} \]
nu este corecta(pentru $n$ par nu era o problema, caci atunci $k$ era patrat perfect si se putea folosi o ipoteza inductiva).


Sa se afle toate numerele reale care verifica ecuatzia :

4xyz - x^4 - y^4 - z^4 = 1

Ceva mai greu??Rezolvati ecuatia $\ x^6+5y^6=6y^5$(own)


algebra
polynomial
calculus
derivative
real analysis
real analysis unsolved

prove that fopr evey n\in N* the polynomial P_n=(X+n^2)(2X+n^2)...(nX+n^2)-2n^(2n) has a unique real positive root x=x_n and find lim{n->oo} x_n.

Remember : if x(k) >=0 and a(k) >= 0  then (x(1)+a(1))*...*(x(n)+a(n)) >= ((x(1)*...*x(n))1/n + (a(1)*...*a(n))1/n)n

We can write for all x>=0 : Pn(x) >= (x*(n!)1/n + n2)n - 2n2n

We have : (x*(n!)1/n + n2)n - 2n2n >=0 <====> (x*(n!)1/n + n2) - 2n2 <====> x >= n2(21/n-1)/((n!)1/n)

So for a(n) = n2(21/n-1)/((n!)1/n) we have P(a(n)) >=0

But Pn(0) = n2n - 2n2n < 0 ,  so there is at least a positive root  x(n) , 0 < x(n) = 0 , so Pn(x) is increasing for x>=0, so there is an unique positive root x(n) .

It's easy to see that lim(n--> \infty){a(n)}=0 , so x(n) is convergent and lim(n--> \infty){x(n)}=0  .

.........

i found your proof pretty confused.. and the result is not good...


Anyone? Here's a hint.

[hide]add a $1$ at the start.[/hide]


Let $a\in\mathbb N$ and $m=a^2+a+1$. Find the number of $0\leq x\leq m$ that:\[x^3\equiv1(\mbox{mod}\ m)\]

[quote="ZetaX"]All prime divisors $p \neq 3$ of $a^2+a+1$ are $\equiv 1 \mod 3$.
So the desired number is (by chinese remainder theorem) $3^k$ where $k$ is the number of prime divisors of $m$ different from $3$.[/quote]

How to use the chinese remainder theorem in this case? Can you show?

Let $f(a)$ be the desired number of solutions $\mod a$.
We have to show for coprime integers $a,b$: $f(ab)=f(a)f(b)$.

Now let $a_1,a_2,...,a_{f(a)}$ be the solutions $\mod a$, $b_1,b_2,...,b_{f(b)}$ be the solutions $\mod b$.
Then for every pair $(a_i,b_j)$ there is a $x , 0\leq x <ab$ with $x \equiv a_i \mod a$ and $x \equiv b_j \mod b$ (that's the point where the chinese remainder theorem is used). But then $x^3 \equiv 1 \mod a$ and $x^3 \equiv 1 \mod b$, so $x^3 \equiv 1 \mod ab$. This gives immediatly $f(ab)  \geq f(a) f(b)$. But if $x$ is a solution $\mod ab$, it is a solution $\mod a$ and $\mod b$, so we reach finally $f(ab)=f(a)f(b)$.


Ugly solution:

We can obtain four possible systems: \[ \begin{cases}x+2y=4\\-2x+4y=4\end{cases} \]
Cramer gives \[ \displaystyle(x,y)=\left(1,\frac{3}{2}\right) \]

\[ \begin{cases}2x-3y=4\\-2x+4y=4\end{cases} \]
Giving \[ (x,y)=(14,8) \]
But this can not be a solution because $max\{x+2y,2x-3y\}=x+2y=30$

\[ \begin{cases}x+2y=4\\10y-3x=4\end{cases} \]
Giving \[ (x,y)=(2,1) \]
But this is not valid since $min\{-2x+4y,10y-3x\}=-2x+4y=0$

\[ \begin{cases}2x-3y=4\\10y-3x=4\end{cases} \]
Giving \[ \displaystyle(x,y)=\left(\frac{52}{11},\frac{20}{11}\right) \]
But this is not valid since $min\{-2x+4y,10y-3x\}=-2x+4y<0$

So this gives us only one valid solution: \[ \displaystyle(x,y)=\left(1,\frac{3}{2}\right) \]


Let $ x,y,z\geq0,|t|\le 2$,prove that:

$ \left( \sqrt {{y}^{2}\plus{}{z}^{2}\plus{}kyz}\minus{}\sqrt {{z}^{2}\plus{}{x}^{2}\plus{}kzx}
 \right) ^{2}\plus{} \left( \sqrt {{z}^{2}\plus{}{x}^{2}\plus{}kzx}\minus{}\sqrt {{x}^{2}\plus{}{y}^{
2}\plus{}kxy} \right) ^{2}\plus{} \left( \sqrt {{x}^{2}\plus{}{y}^{2}\plus{}kxy}\minus{}\sqrt {{y}^{2
}\plus{}{z}^{2}\plus{}kyz} \right) ^{2}\leq  \left( y\minus{}z \right) ^{2}\plus{} \left( z\minus{}x
 \right) ^{2}\plus{} \left( x\minus{}y \right) ^{2}$

t=1,I have good proof.
BQ

[quote="xzlbq"]Let $ x,y,z\geq0,|t|\le 2$,prove that:

$ \left( \sqrt {{y}^{2} \plus{} {z}^{2} \plus{} kyz} \minus{} \sqrt {{z}^{2} \plus{} {x}^{2} \plus{} kzx} \right) ^{2} \plus{} \left( \sqrt {{z}^{2} \plus{} {x}^{2} \plus{} kzx} \minus{} \sqrt {{x}^{2} \plus{} {y}^{ 2} \plus{} kxy} \right) ^{2} \plus{} \left( \sqrt {{x}^{2} \plus{} {y}^{2} \plus{} kxy} \minus{} \sqrt {{y}^{2 } \plus{} {z}^{2} \plus{} kyz} \right) ^{2}\leq \left( y \minus{} z \right) ^{2} \plus{} \left( z \minus{} x \right) ^{2} \plus{} \left( x \minus{} y \right) ^{2}$

t=1,I have good proof.
BQ[/quote]

WLOG, assume $ x\equal{} \min \{x,y,z \}$, fix $ x\minus{}y$ and $ x\minus{}z$, then RHS is constant, but LHS is decreasing of $ x$ (importance, to be prove...), so we just neet to prove that when $ x\equal{}0$, that's obvious...


Se considera ecuatia: a1^2/(x-b1)+a2^2/(x-b2)+.....+an^2/(x-bn)=c ,unde ak, bk sunt din R,k=1..n, n>=11 si c din R* .Aratati ca ecuatia are toate radacinile rationale.

Ecuatia este:  $ \frac {{a_1}^2}{x \minus{} b_1} \plus{} \frac {{a_2}^2}{x \minus{} b_2} \plus{} \dots \plus{} \frac {{a_n}^2}{x \minus{} b_n} \equal{} c$, unde $ a_k;b_k\in\mathbb{R}$ pentru $ k\in\{1;2;\dots ;n\}$ si $ c\in\mathbb{R}^*$, iar $ n\ge 11$.
Am impresia ca totusi enuntul corect ar fi: [b]Aratati ca ecuatia are toate radacinile reale![/b]


algebra
polynomial
calculus
integration
function
domain
Ring Theory

Prove that the polynomial ring R[x] over commutative integral domain R with identity  is "the ring of general ideals" iff
R-field   (Maybe i used wrong notation:"the ring of general ideals" means that every ideal in R is of the form $aR$,
where $a\in{R})$

The standard term there is "principal ideal domain". This is a standard question.

What you call "the ring of general ideals" is usually called a PID, short for principal ideal domain.

If $R$ is a field, consider a polynomial $P$ of minimal degree in the ideal. It's easy to show that this polynomial generates the ideal (take a polynomial $F$ in the ideal, write $F(x)=Q(x)P(x)+r(x)$ with $deg r<deg P$; it follows that $r$ is null, so $P|F$).

Conversely, if $R$ is not a field, let $a$ be a non-invertible, non-null element. Look at the ideal generated by $ax$ and $(a-1)x^2+1$. If it were generated by some polynomial $P$, then $P$ would have degree at most one. If it's $1$, then $P(x)=\alpha x+\beta|ax\Rightarrow \beta=0\Rightarrow \alpha x|(a-1)x^2+1$, which is clearly false. This means that $P$ is a constant $t$. Since $t|(a-1)x^2+1$, we find $t$ to be invertible, so our ideal is the entire ring $R[x]$. In particular, we have polynomials $Q,F$ s.t. $Q(x)[(a-1)x^2+1]+F(x)ax=1\ (*)$. Now give the value $1$ to $x$ in $(*)$. We get $a[Q(1)+F(1)]=1$, meaning that $a$ is invertible, contradiction.

[quote="grobber"]
Conversely, if $R$ is not a field, let $a$ be a non-invertible, non-null element.[/quote]
Consider the ideal generated by $x$ and $a$. I think it works.


During the winter, the ratio of deer to foxes was 7 to 2. If there were 210 foxes, what was the number of deer?

[quote="math92"][hide]
if the ratio was 7:2 then for every 7 foxes there was 2 deer so 210/7=30 so 30*2=$\boxed{60}$[/hide][/quote]

You have the ratio wrong, I think. It's for every 2 foxes, there are 7 deer.

So the correct answer is:

[hide]
$\frac{7}{2}=\frac{x}{210}$
$2x=1470$
$x=735$
[/hide]


Find the smallest number $a$ such that a square of side $a$ can contain five disks of radius $1$, so that no two of the disks have a common interior point.

its obvious that a=1 , 2, 3 doesn't work

i show that a=4 doesn't work and we have an obvious example for a=5 but i cant show the sketch of my proof

because i cant work with geogebra but i draw my picture in geometer' sketchpad and i cant upload it here

can anyone help me?

Is it clear we should find the least such [b]integer[/b] $a$ ? There is no indication we are not in fact asked to find the least [b]real number[/b] $a$ with that property.

sorry.i made a mistake :blush:

i don't pay attention to this fact but i believe that it must be natural because i think that if $a$ works then there

exist a real positive $e$ such that e is very small and $a-e$ works too and it shows that we dont have minimum in

the reals and another reason is that its balticway competition and they wanna tricky idea in their problems not

troublesome problems :D

No, this argument invoked by you seems to imply "least integer" (which always exists), since "least real" may not exist; but your argument is flawed. If $a$ works, then there does not necessarily exist a small $e$, such that $a-e$ also works - since for the minimal $a$, some disks will touch the border of the square, so we cannot shrink it. This can be formally argued by Weierstrass' theorem, the case be. Second-guessing the composers is also not that good an idea :)

[quote]This can be formally argued by Weierstrass' theorem[/quote]

can you explain more about this theorem and its statement ?

meanwhile if the problem rearrangement such that any two disks dont even touch each other then my argument

works and my example for a=5 works too

I think the idea is to reason something like this:

Consider the smalles square that can contain 2 disks of radius 1.  It is intuitive and not hard to prove that the square would contain the 2 disks in a diagonal arrangement.  The length of the diagonal of the square would be $2 + 2\sqrt{2}$, so its side would be $2 + \sqrt{2}$.

Now let s be the length of some square that contains 5 disks of radius 1.  Divide the square into 4 squares by adding a vertical and horizontal line.  By the pigeon hole principal, at least 1 of these 4 smaller squares contains the centers of 2 of the disks.  Let's say it is the square in the lower left.  If we enlarge this smaller square by 1 unit upward and to the right, we now have a square of side $\frac{s}{2} + 1$ that completely contains 2 unit disks.

So we have the formula $\frac{s}{2} + 1 \geq 2 + \sqrt{2}$, or $s \geq 2 + 2\sqrt{2}$.

But we can achieve equality by placing 1 disk in the exact center of a square and placing the other 4 disks each tangent to the center disk and 2 of the square's sides.  So we conclude a = $2 + 2\sqrt{2}$.


Find $ [2\sqrt{a_n}]$ if:
$ a_n\equal{}10^{2n}\minus{}10^n\plus{}1$
for natural $ n$.


inequalities
quadratics
linear algebra
matrix
inequalities proposed

If $ a$, $ b$,$ c$ and $ d$ are arbitrary real numbers prove that following inequality holds:

$ 11a^{2} \plus{} 11b^{2} \plus{} 221c^{2} \plus{} 131d^{2} \plus{} 22ab \plus{} 202cd \plus{} 48c \plus{} 6 \geq$ $ 98ac \plus{} 98bc \plus{} 38ad \plus{} 38bd \plus{} 12a \plus{} 12b \plus{} 12d$

In other words, bring the quadratic expression to the standard form (eliminating the linear terms) and check that the matrix is nonnegative definite and that the free term is also nonnegative? Nah, too boring. :noo:


Find function $f: \mathbb{R}\to\mathbb{R}$ satisfy the following conditions:
$i)$  There exists a positive real $M$ such that: $-M\le f(x) \le M$
$ii)$ $f(1)=1$
$iii)$$f(x+\frac{1}{x^2})=f(x)+(f(\frac{1}{x}))^2$

[quote="mathmanman"]Such a function does not exist. ;)
A proof by contradiction works well.
First of all, if we take $x=1$, we can get that : $f(2) = 2$.

And... I'll continue my solution later :)[/quote]

Yes, you're right. And today is Sunday!

Ok, thanks for the reminder. ;)

So, since we have $f(2) = 2$, we can say that $M \ge 2$, hence, there exists $x_0 \in \mathbb{R}$ such that : $M \ge f(x_0) > M - \frac 14$.
So : $M \ge f(x_0 + \frac 1{x_0^2}) = f(x_0) + (f(\frac 1{x_0}))^2 \ge M - \frac 14 + (f(\frac 1{x_0}))^2$. So, we deduce that :$(f(\frac 1{x_0}))^2 \le \frac 14$, so that :$f(\frac 1{x_0}) \ge - \frac 12$. Then, if we take $x_0 : = \frac 1{x_0}$, we get :$M \ge f(\frac 1{x_0} + x_0^2) = f(\frac 1{x_0}) + (f(x_0))^2 \ge -\frac 12 + (M - \frac 14)^2$.
Hence, we have that :
$\frac 12 > \frac 12 - \frac 1{16} \ge c(c-1-\frac 12) \ge 2 \cdot \frac 12 = 1$.
So, we have reached the desired contradiction that yields the conclusion. :)


13.  ABCD is a quadrilateral with AB = 8, BC = 6, BD = 10, ∠A = ∠D and ∠ABD = ∠C. Find CD.

This problem seems to be very easy for a #13 on AIME, but it's a good problem.

[quote="beta"]∠A = ∠D [/quote]
that refers to the whole angle D right?

[hide]label point E on CD such that DE = 8
ABED is an isosceles trapezoid
BD = AE = 10; angle ABD = angle DEA = angle C

then extend AB to F such that BF = CE
BFCE is another isosceles trapezoid similar to ABED
triangle BEC is similar to ADE by angle angle
CE/ED = 6/10
CE/8 = 6/10
CE = 4.8
CD = CE+ED = 4.8 + 8 = 12.8[/hide]

Extend AB, CD to meet at E. Then look for two similar triangles.


[quote="E^(pi*i)=-1"]Let $a*b=\displaystyle\frac{a}{a-b}$.

For what values of x is $x*(x*3)$ undefined?[/quote]

[hide]$\boxed{x\neq 3}$, $\boxed{x\neq 4}$, $\boxed{x\neq 0}$

$x*3=\frac{x}{x-3}$, $x\neq 3$

$x*(\frac{x}{x-3}=x-\frac{x}{x-\frac{x}{x-3}}$
$x-\frac{x}{x-\frac{x}{x-3}}\neq 0$
$x=\frac{x}{x-\frac{x}{x-3}}$
$x(x-3)=x$
$x^2-4x=0$
$x\neq 4$
$x\neq 0$[/hide]

[hide]The first step is $x*3=\frac{x}{x-3}$. Thus x=3 is impossible.

Then $x*(x*3)=\frac{x}{x-\frac{x}{x-3}}=\frac{x^2-3x}{x^2-4x}$. If x=0 or x=4, that would be undefined.

x can't equal 0, 3, or 4. [/hide]


In a $ N\times N$ array of numbers, all rows are different (two rows are said to be different even if they differ only in one entry). Prove that there is a column which can be deleted in such a way that the resulting rows will still be different.

First prove it's true for 2 by 2. Then work via induciton. Suppose for all (n-1) x (n-1) matrix with that property that all rows are different we may remove a column, we prove the same for n x n. For a n x n matrix, take the entries from (1,1) to (n-1,n-1). There is a column i that may be removed satisfying the conditions, since we assumed it was true for (n-1) x (n-1). Now add the nth row.
The same column i that was removed for the (1,1) to (n-1,n-1) may not be removed if and only if, the nth row is exactly the same to some other row m except for column i. Now it can be shown that we can remove the nth column. Since, we have shown that the nth row is different to row m because the ith column is still there. Also row m must be different from any other row at a column other than the ith or the nth column (via the original hypothesis). Since row n msut be equal to row m at any column other than the ith one, row n is also different to any other column other than the ith or the nth column. Thus we may remove the nth column. So via the inductin hypothesis, it is true for all n x n matrices.


geometry
combinatorics proposed
combinatorics

Consider a triplet $(\mathcal P,\mathcal L, \sigma)$ in which $\mathcal P$ a set of points, $\mathcal L$ is a set of subsets of $\mathcal P$ which we call lines, $\sigma: \mathcal P\rightarrow \mathcal P$ is a mapping which maps 3 collinear points to 3 collinear points and 3 noncollinear points on 3 noncollinear points.

We have 3 axioms: [those of an affine geometry]
- there's exactly 1 line through ever pair of (different) points
- for any line L and any point p not on L, there's exactly 1 line L' through P for which $L\cap L'=\emptyset$ (this we call parallel -- this is euclides' parallel postulate)
- there are 4 points of which no 3 are collinear (I think this condition is too strong - I think I proved it for $\mathcal P$ nonempty)

Show that $\sigma$ either fixes a point, or fixes a line, or maps a line to a parallel line. (at least one of these, not exactly one)

[note to the mods: this is combinatorics and [b]not[/b] by any means olympiad geometry..]


geometry
circumcircle
geometric transformation
reflection
trigonometry
projective geometry
angle bisector

Given is a triangle $ ABC$. Let $ D$, $ E$, $ F$ be the tangency point of its incircle $ \rho (I)$ with the sides $ BC$, $ CA$, $ AB$, respectively. Let $ IN$ the internal bisector of the angle $ BIC$, where $ N\in BC$, and $ T$ the intersection of $ AN$ and $ EF$. Prove that $ DT$ is the internal bisector of the angle $ EDF$.

Is $ N$ an arbitrary point on the angle bisector of $ \angle BIC$?

$ (O)$ is circumcircle of the $ \triangle ABC,$ $ M \in (O) \cap AI$ is circumcircle of the $ \triangle IBC.$ Reflect $ D \in (I)$ in $ IN$ into $ U \in (I).$ $ IM, ID \perp BC$ are isogonals with respect to the angle $ \angle CIB$ $ \Longrightarrow$ $ U \in IM \equiv AI.$ Let $ AI$ cut $ (I)$ again at $ V,$ the midpoint of the arc $ EF$ of opposite to $ D,$ so that $ DV$ bisects the angle $ \angle FDE.$ $ A$ is the pole of $ EF$ with respect to $ (I).$ Tangent $ v$ of $ (I)$ at $ V$ cuts $ BC$ at $ X,$ the pole of $ DV.$ $ EF$ is a polar of $ A$ and $ DU$ a polar of $ N,$ they intersect at the pole $ Y$ of $ AN.$ We have to show that $ EF, DV, AN$ are concurrent $ \Longleftrightarrow$ their poles $ A, X, Y$ are collinear.

Project the circle $ (I)$ to a circle $ (I^*)$ and a line $ l$ through $ A$ perpendicular to $ AI$ to infinity (adding another one to the forest out there). $ A^*$ at infinity is the image of $ A.$ Image $ U^*V^*$ of $ UV$ remains a diameter of $ (I^*)$ and image $ E^*F^*$ of $ EF$ becomes a diameter of $ (I^*)$ perpendicular to $ U^*V^*.$ The right $ \triangle I^*V^*X^* \sim \triangle U^*D^*V^* \sim \triangle U^*I^*Y^*$ are similar, having equal angles $ \Longrightarrow$ the right $ \triangle I^*V^*X^* \cong \triangle U^*I^*Y^*$ are congruent, because $ U^*I^* \equal{} I^*V^*,$ therefore $ I^*Y^* \equal{} V^*X^*$ and $ X^*Y^* \parallel U^*V^*$ $ \Longrightarrow$ $ A^*, X^*, Y^*$ are collinear $ \Longrightarrow$ $ A, X, Y$ are also collinear $ \Longrightarrow$ $ EF, DV, AN$ (where $ DV$ bisects the angle $ \angle FDE$) are concurrent at $ T.$

[quote="Virgil Nicula"][color=darkred][b][u]Lemma.[/u][/b] Let $ ABC$ be a triangle. For $ X\in (BC)$ , $ Y\in (CA)$ , $ Z\in (AB)$ denote $ P\in AX\cap YZ$ . Then $ \boxed {\ \frac {PZ}{PY} \equal{} \frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot\frac {AC}{AB}\ }$ .[/color]
[hide="Proof."]
$ \left\{\begin{array}{c} \frac {XB}{XC} \equal{} \frac {AB}{AC}\cdot\frac {\sin\widehat {XAB}}{\sin\widehat {XAC}} \\
 \\
\frac {AZ}{AY}\cdot\frac {\sin\widehat {PAZ}}{\sin\widehat {PAY}} \equal{} \frac {PZ}{PY}\end{array}\right\|\ \bigodot\ \implies\ \frac {PZ}{PY} \equal{} \frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot\frac {AC}{AB}$ .[/hide][/quote]

[quote="April"] [color=darkred]Given is a triangle $ ABC$. Let $ D$, $ E$, $ F$ be the tangency point of its incircle $ (I)$ with the sides $ BC$ , $ CA$ , $ AB$  respectively.

Let $ [IN$ the bisector of $ \widehat {BIC}$, where $ N\in BC$, and $ T\in AN\cap EF$ . Prove that $ [DT$ is the bisector of $ \widehat {EDF}$ .[/color] [/quote]

[color=darkblue][b][u]Proof (metric).[/u][/b] This old and well-known [b]April[/b]'s problem is a nice application of the upper lemma. Indeed, $ \frac {TF}{TE} \equal{} \frac {NB}{NC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}$ ,

i.e. $ \frac {TF}{TE} \equal{} \frac bc\cdot\frac {NB}{NC}$ . Since $ \frac {NB}{NC} \equal{} \frac {IB}{IC} \equal{} \sqrt {\frac {ac(p \minus{} b)}{p}\cdot\frac {p}{ab(p \minus{} c)}}$ , i.e. $ \frac {NB}{NC} \equal{} \sqrt {\frac {c(p \minus{} b)}{b(p \minus{} c)}}$ $ \implies$ $ \boxed {\ \frac {TF}{TE} \equal{} \sqrt {\frac {b(p \minus{} b)}{c(p \minus{} c)}}\ }$ . But

$ \frac {DF}{DE} \equal{} \frac {2\cdot DB\cdot\sin\frac B2}{2\cdot DC\cdot\sin\frac C2} \equal{} \frac {p \minus{} b}{p \minus{} c}\cdot\sqrt {\frac {(p \minus{} a)(p \minus{} c)}{ac}\cdot\frac {ab}{(p \minus{} a)(p \minus{} b)}}$ , i.e. $ \boxed {\ \frac {DF}{DE} \equal{} \sqrt {\frac {b(p \minus{} b)}{c(p \minus{} c)}}\ }$ . In conclusion,

$ \frac {TF}{TE} \equal{} \frac {DF}{DE} \equal{} \sqrt {\frac {b(p \minus{} b)}{c(p \minus{} c)}}$ , i.e. the ray $ [DT$ is the bisector of the angle $ \widehat {EDF}$ .[/color]

Here is an alternative solution for those interested. The first part is very similar to yetti's solution, but the methods to finish the proof are less sophisticated. I adjusted some names of points to make the figure more compatible with yetti's solution, I hope I didn't miss some points during the adjustment.

[u][b]Proof:[/b] [/u]

First we notice that $ NI$ and $ DV$ are parallel.
We can prove this using the fact that $ IM$ and $ ID$ are bisected by $ IN$, and chasing some angles.

Now, we will prove that the polar lines of $ V,T$ and $ D$ are concurrent, from which the conclusion will follow.
Call $ R$ the intersection of the perpendicular from $ D$ to $ IN$ and $ EF$. It is now clear that $ AR$ is the polar line of $ T$ and $ BC$ is the polar line of $ D$, both with respect to the incircle. We want to prove that the intersection of both polar lines, which we will call $ S$, lies on the polar line of $ V$ wrt the incircle.

To do this, it suffices to note that $ AR,AC,AN,AB$ is a harmonic bundle, which follows from the fact that $ R$ and $ T$ divide $ E$ and $ F$ harmonically. We deduce that $ S$ and $ N$ divide $ B$ and $ C$ harmonically, and since $ IN$ is the bisector of $ \angle BIC$, we have that $ IS \perp IN \parallel DV$. It now follows that $ IS \perp DV$, and hence $ SV$ is tangent to the incircle.

[u][b]Figure:[/b][/u]

[img]http://img244.imageshack.us/img244/2209/aprilinternalbisectoret9.png[/img]

Nice to meet you on MathLinks Forum, Jan!  :P In general, my solution is as same as yours. I use a remark that $ IN$ is the external angle bisector of $ \measuredangle AID\equiv\measuredangle VID$, and I call $ S$ the intersection of $ BC$ with the line passes through $ I$ and perpendicular to $ IN$. But your solution seems more natural and easier than mine  :D

[b]P.S.[/b] An elementary solution can be found at http://www.mathlinks.ro/viewtopic.php?t=24699

let $ DK$ be the angle bisector of $ \angle EDF$ where  $ K \in p(I)$
by simple angle chasing $ ($ $ \angle AKD \equal{} \angle AIN$ $ )$ we can show that  $ DK$ is parallel to $ IN$
so the $ polar(N)$ is perpendicular to $ DK$
$ polar(A) \cap polar(N) \equal{}$ $ pole$ of $ DT \equal{} R$
therefore $ REFT$ are harmonic
let $ DK \cap EF \equal{} T'$
also $ (REFT')$ are harmonic(internal external angle bisectors)
so $ T' \equal{} T$


Prove that if value of $ P(x)\equal{}ax^2\plus{}bx\plus{}c$  for arbitrary x being an integer is $ K^4$ for some K ( K-natural number) then $ a\equal{}b\equal{}0$

By a problem u also posted in this forum, we obtain $ P(x) \equal{} (dx \plus{} e)^{2}$
From now $ dx \plus{} e$ must be square for large enough $ x$, which is then $ d\equal{}0$
The conclusion follows


Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 4\left(\frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3}\right) \plus{} 48\geq7(a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\]

we must prove that:
\[ \frac {a \plus{} b \plus{} c}{2}\sum (a \minus{} b)^2 \plus{} 2(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(\sum \frac {(a \minus{} b)^2}{ab}) \geq 7 (\sum \frac {(a \minus{} b)^2}{ab})\]
But we have this ineq:
\[ \frac {a \plus{} b \plus{} c}{2}\sum (a \minus{} b)^2 \plus{} 2(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(\sum \frac {(a \minus{} b)^2}{ab}) \geq \frac {a \plus{} b \plus{} c}{2}\sum (a \minus{} b)^2 \plus{} 6(\sum \frac {(a \minus{} b)^2}{ab})\]
=> we 'll prove:
\[ \sum (a \minus{} b)^2(a \plus{} b \minus{} c) \geq 0\]
That is very easy

[quote="DangChienbn"]we must prove that:
\[ \frac {a \plus{} b \plus{} c}{2}\sum (a \minus{} b)^2 \plus{} 2(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(\sum \frac {(a \minus{} b)^2}{ab}) \geq 7 (\sum \frac {(a \minus{} b)^2}{ab})\]
[/quote]
It should be
\[ \frac {a \plus{} b \plus{} c}{2}\sum (a \minus{} b)^2 \plus{} 2(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(\sum \frac {(a \minus{} b)^2}{a^2b^2}) \geq 7 (\sum \frac {(a \minus{} b)^2}{ab})\]
:wink:

OK. Thanks. i was stupid  :blush:
Can you post your solution, auqady :D ?

[quote="DangChienbn"]
Can you post your solution, auqady :D ?[/quote]
I'll wait a little.
The equality holds also when $ a\equal{}\sqrt[3]4$ and $ b\equal{}c\equal{}\frac{1}{\sqrt[3]2}.$ :wink:

[quote="arqady"][quote="DangChienbn"]
Can you post your solution, auqady :D ?[/quote]
I'll wait a little.
The equality holds also when $ a \equal{} \sqrt [3]4$ and $ b \equal{} c \equal{} \frac {1}{\sqrt [3]2}.$ :wink:[/quote]
Oh, It is the reason why I couldn't use SOS to prove this ineq  :blush:

Nice, arquady!

The inequality is equivalent to
$ \left( t^3 \plus{} \frac {2}{t^3} \minus{} 3 \right)^2 \ge 0$ which is obviously true.

You know what I mean by t here, don't you arquady! (MV works effectively) :)

[hide="to dangchienbn"]Con nho anh la ai o VUT chu?[/hide]


function
integration
calculus
real analysis
real analysis unsolved

let $ f,g\in L^{2}( \minus{} \infty, \plus{} \infty)$ and $ f$ vanishes outside a bonded closed set. Show that
\[ \sum_{k \equal{} \minus{} \infty}^{ \plus{} \infty}2^{ \minus{} n}[\int_{ \minus{} \infty}^{ \plus{} \infty}f(x)g(2^{ \minus{} n}x \minus{} k)dx]^{2}\to 0\] as $ \boxed{n\to \infty}$
any idea is appriciated. thanks

We can assume $ f,g\ge 0$ - else replace them with their absolute values. $ f$ is zero outside the interval $ [a,b],$ so the integral can be restricted to that interval. Use Cauchy-Schwarz:

$ \sum_{k \equal{} \minus{} \infty}^{\infty}2^{ \minus{} n}\left[\int_a^bf(x)g(2^{ \minus{} n}x \minus{} k)\,dx\right]^2$

$ \le\sum_{k \equal{} \minus{} \infty}^{\infty}2^{ \minus{} n}\|f\|_2^2\int_a^bg(2^{ \minus{} n}x \minus{} k)^2\,dx$

$ \equal{}\|f\|_2^2\sum_{k \equal{} \minus{} \infty}^{\infty}\int_{2^{ \minus{} n}a}^{2^{ \minus{} n}b}g(u \minus{} k)^2\,du$

Now let $ h(u) \equal{} \sum_{k \equal{} \minus{} \infty}^{\infty}g(u \minus{} k)^2.$ $ h(u)$ is periodic of period $ 1,$ and the integral of $ h$ over any interval of length $ 1$ is the same as the integral of $ g^2$ over all of $ \mathbb{R}.$ Our original sum is now bounded by

$ \|f\|_2^2\int_{2^{ \minus{} n}a}^{2^{ \minus{} n}b}h(u)\,du.$

But that is the integral of an $ L^1$ function over a set whose measure tends to zero as $ n\to\infty.$ Measures induced by integrable functions are absolutely continuous, meaning that the integral of the function over a set of arbitrarily small measure must be arbitrarily small. Hence the limit is zero.

Thanks Prof. Kent,
 I understand  that $ h(u)$ is periodic of period $ 1$ but I am not quite sure what you wrote  :

[quote="Kent Merryfield"]

Now let $ h(u) \equal{} \sum_{k \equal{} \minus{} \infty}^{\infty}g(u \minus{} k)^2.$  the integral of $ h$ over any interval of length $ 1$ is the same as the integral of $ g^2$ over all of $ \mathbb{R}.$.[/quote]

if possible, could you please explain more? thanks

Let $ f\in L^1(\mathbb{R}).$ (In our case, $ f\equal{}g^2.$) Let $ h(x)\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}f(x\minus{}k).$

As you said, $ h$ is periodic of period $ 1,$ and for that reason, $ \int_a^{a\plus{}1}|h(x)|\,dx$ doesn't depend on $ a.$

$ \int_0^1|h(x)|\,dx\le\int_0^1\sum_{k\equal{}\minus{}\infty}^{\infty}|f(x\minus{}k)|\,dx\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}\int_0^1|f(x\minus{}k)|\,dx$

(That was justified by the MCT.)

$ \equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}\int_k^{k\plus{}1}|f(u)|\,du\equal{}\int_{\minus{}\infty}^{\infty}|f(u)|\,du.$

The same argument (with a DCT justification) without the absolute values shows that

$ \int_0^1h(x)\,dx\equal{}\int_{\minus{}\infty}^{\infty}f(x)\,dx$ and indeed,

$ \widehat{h}(k)\equal{}\int_0^1e^{\minus{}2\pi ikx}h(x)\,dx\equal{}\int_{\minus{}\infty}^{\infty}e^{\minus{}2\pi ikx}f(x)\,dx\equal{}\widehat{f}(k)$ for $ k\in\mathbb{Z}.$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group such that $ \forall H,K\leq G \implies H\cup K\leq G$ show that $ G$ is cyclic.

I can prove it for a finite groups. Is it true for infinite groups as well?

What do $ \le$ and $ \cup$ mean here? It can't be the standard meanings (inclusion for $ \le$ in particular).

@jmerry: $ \cup$ mean union. $ \leq$ means subgroup.
@-oo-: You have a counterexample?
-------------------------------------------------------
I mean to say.....

Given a group $ G$ so that the [b]union[/b] of any two subgroups is again a subgroup show that it must be a cyclic group. (I can prove it for finite groups).

I think if $ G$ is finitely generated, you could try something like this.  Suppose that $ G = <a,b>$.  This can be generalized pretty easily to finitely many generators.  Consider the subgroups $ <a>$ and $ <b>$.  If $ <a> \cup <b>$ is a group, it must be $ G$, since it contains $ a$ and $ b$.  So without loss of generality, $ ab = b^{n}$ for some $ n \in \mathbb{Z}$, so $ a = b^{n-1}$, or $ <a> \subset <b>$, so $ G = <b>$.

This seems like the obvious thing to do, is that what you did?
If you don't assume that $ G$ is f.g, then you might need Zorn's Lemma to do chains of cyclic subgroups or something.

It makes sense now- as originally stated, it was something about the union of subgroups still being a subset, which was trivially true for any group.

Incidentally, here's a counterexample: Let $ G$ be the group of dyadic rationals mod 1, under addition.

The complete list of all groups with this property:
For each prime $ p$, we have the cyclic groups $ \mathbb{Z}/p^{n}\mathbb{Z}$ for each $ n$, and also the limit of these groups, which can be represented as the rationals with power-of-$ p$ denominators.
That's it.


find m to satisfy $2^{sin^{2}x}+3^{cos^{2}x}=(4-3m)3^{sin^{2}x}$ has root

[quote="Rust"]for all m, suth that $0\le m\le 1$ exist x, satisfyed your equation.[/quote]

can you explain?

[quote="Rust"]for all m, suth that $0\le m\le 1$ exist x, satisfyed your equation.[/quote]
i think you should detail :(

[quote="Rust"]Let $y=sin^{2}x, \ 0\le y\le 1$, then equation is $2^{y}+3^{1-y}=(4-3m)3^{y}$.
Therefore $m=f(y)=\frac{4*3^{y}-3^{1-y}-2^{y}}{3^{y+1}}$.
$f(0)=0,f(1)=1$ and $0\le f(y)\le 1$, when $0\le y\le 1.$[/quote]
ok,i think we should complete that f(y) increases on [0,1]
and f(0)=0 and f(1)=1 .Hence 0<=m<=1.
thanks !!! :)

[quote="Rust"]I don't chek, that f(y) increase. It is easy to chek, that $0\le f(y)\le 1$.
1. $f(y)\ge 0$ if $g(y)=4*9^{y}-6^{y}-3\ge )$. It is easy to chek, that $g(y)$ increase, when y>=0.
2. $f(y)\le 1$ if $t(y)=9^{y}-6^{y}-3\le 0$. Because t(y) increase too and $t(1)=1$ we get $f(y)\le 1$.[/quote]
it's not really obvious!!!


[quote="abacadaea"]i dont understand the question
[quote]How often buses go around the men[/quote]
whats that supposed to mean?[/quote]
How often buses outstrips the men?

[quote="Lampetija"]Buses are going down the street two ways 25 km/h. Buses arrive to bus stops every 16 min. How often buses go around the men, who is going 5 km/h? How often he is going to see bus going towards him?
Answers: after 20 min., after 13 1/3 min. This is right answers, but I don't understand how to get them. PLease help me. :maybe:[/quote]

I am surprised that this question actually have answers.. :o

i don't quite understand the question so i don't know how to help you. is the man walking? and if so, does he start at the same time as the buses start? is that even part of the question? sorry i am just really confused.

[quote="Lampetija"]Buses are going down the street two ways 25 km/h. Buses arrive to bus stops every 16 min. How often buses go around the men, who is going 5 km/h? How often he is going to see bus going towards him?
Answers: after 20 min., after 13 1/3 min. This is right answers, but I don't understand how to get them. PLease help me. :maybe:[/quote]
I think...
That maybe you should clean up some grammar and such.
I can't understand the problem.

I think there are an endless evenly spaced chain of buses going both ways, and you have to determine how often a bus on either direction meets with the person.

[quote="Lampetija"][quote="abacadaea"]do they come in circles or what. im still lost?[/quote]
Maybe, I don't know, this is not written in the book.[/quote]

I think he means that this is how it was written in the book.


pigeonhole principle
geometry
combinatorics unsolved
combinatorics

Can someone provide a hint for this question? I tried dividing the square into smaller squares and use the pigeonhole principle but it doesn't work out nicely.

A square has side 20. S is a set of 1999 points inside the square and the 4 vertices. Show that we can find three points in S which form a triangle with area $\le$ 1/10.

Assume you add points one at a time and see how many triangles you get in a triangulation of the square (i.e. a partition of the square in triangles having vertices in $S$).

At first you have the $4$ vertices of the square, so no triangles. Then you add another point to get a triangulation of the square with four triangles. After that, at each step, you add a point inside an existing triangle, so one triangle dissappears (the large one) and three new triangles appear, meaning that your triangulation has two more triangles than in the previous step (assume no three points are collinear, because then you would have a triangle of area $0$). This mmeans that after $1999$ steps you have $4+2\cdot 1998=4000$ triangles in your triangulation. This means that at least one of them must have area $\le\frac{\hbox{area of the square}}{4000}$, and since the area of the square is $400$, we get the result.

But can you assume no three points are collinear?

If three points are collinear, then the triangle formed by them has area $0<\frac 1{10}$.

Start from one edge of the square and form a triangle with (one of) the nearest point(s). And go on doing this with new edges. The square will be divided into triangles and areas of the triangles will be greater than 0. Using v-e+f=2 since v=2003 e=(3f+1)/2 we found f=4001. This means there are 4000 triangles inside the square. And it follows as grobber done... (am I wrong?)

I don't understand what you're doing.....are you building a 3D polyhedron with the 1999 points? and how did you get e=(3f+1)/2?

There are f-1 triangles inside. every side is used by two triangles except the sides of the square. So there are [3(f-1)+4]/2 edges.
I'm trying to find a way to divide the square into triangles.

Ok I guess I don't know how V-E+F=2 works on two dimensional figures. If I take a square and cut it into 4 triangles by its diagonals, what are the values for V, E, F?

v=5 (4 vertex and the intersection of diagonals)
e=8
f=5 (4 triangles and the outside of square)


Sa se determine valorile lui a din R pentru care sirul cu termenul general: \[a_{n}=\frac{\ln{2}}{2}+\frac{\ln{3}}{3}+\ldots+\frac{\ln{n}}{n}-a\ln^{2}{n}\] este convergent.

Folosim niste estimari cu integrale, nu?

$\int \frac{\ln x}{x}\, dx = \frac12 \cdot \ln^{2}(x)+\mathcal C$, dupa care facem aceeasi pasi pe care-i facem cand demonstram ca $\sum_{k=1}^{n}\frac1{k}-\ln n$ e convergent.

[quote="Paul V."]s-ar putea sa mearga si asa cum zici,dar am mai stat de capul exercitiului si cred k s-ar putea sa mearga asa: \[a_{n}=\sum_{k=2}^{n}\frac{\ln{k}}{k}-a\ln^{2}{n}\] Aleg \[x_{k}=\ln{k},\;k\ge2\]  \[y_{k}=k,\;k\ge2\] Observam k se poate aplica Stolz-Cesaro si avem \[\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}\left(\sum_{k=2}^{n}\frac{x_{k}}{y_{k}}-a\ln^{2}{n}\right)= \lim_{n\to\infty}\sum_{k=2}^{n}\frac{x_{k+1}-x_{k}}{y_{k+1}-y_{k}}-\lim_{n\to\infty}a\ln^{2}{n}=\lim_{n\to\infty}\sum_{k=2}^{n}\ln{\frac{k+1}{k}}-\lim_{n\to\infty}a\ln^{2}{n}=-\lim_{n\to\infty}a\ln^{2}{n}\] Deoarece $a_{n}$ trebuie sa fie convergent valoare lui $\fbox{a=0}$[/quote]

Te inseli Paul. perfect_radio are dreptate; folosind metoda amintita de el, se arata ca sirul dat este convergent daca si numai daca $a=\frac{1}{2}$.

da, aveti dreptate asa e, m-am lasat dus de val,sorry,oricum metoda ce am dat-o eu este gresita  :blush:
Numai bine la toti!!!
tks for you're help again!!!


[quote="M4RI0"]Determine all $\theta$ such that $0 \leq 0 \leq \frac{\pi}{2}$and $sin^{5}\theta+cos^{5}\theta=1$[/quote]
[hide="add"]It is possible to change here one of the digit 5 or both ones to any numbers greater than 2. [/hide]
[hide="key"]$0<x<1$ implies $x^{2}> x^{5}$[/hide]
[hide="answer"]$\theta = 0$ or $\theta = \frac{\pi}{2}$[/hide]


Let $ a_1,a_2,...,a_n$ be positive real numbers such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 1$. Prove that

$ a_1\cdot a_2^{\frac {2}{3}} \plus{} a_2\cdot a_3^{\frac {2}{3}} \plus{} ... \plus{} a_{n \minus{} 1}\cdot a_n^{\frac {2}{3}} < \frac {3}{7}$

[quote="hollandman"]Let $ a_1,a_2,...,a_n$ be positive integers such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 1$. Prove that

$ a_1\cdot a_2^{\frac {2}{3}} \plus{} a_2\cdot a_3^{\frac {2}{3}} \plus{} ... \plus{} a_n\cdot a_1^{\frac {2}{3}} < \frac {3}{7}$[/quote]
It's obviously wrong. :wink:

I think he was referring to the $ a_i$ being positive integers... :roll:


Find the number of degrees in the acute angle formed by the [b]hour [/b]and [b]second[/b] hands of a clock at 20 seconds after 1:24. Express your answer as an improper fraction reduced to lowest terms.


Find the smallest sum of three consecutive natural numbers which are relatively prime.

Notice that there must be only one even number. Thus, the first sets are (1,2,3) and (3,4,5) but 1 and 3 are not relatively prime, so 3+4+5=12.

The only problem is AoPS says that the answer is six.

[quote="james4l"]but 1 and 3 are not relatively prime[/quote]
This is a false statement, since $ 1$ is relatively prime to every other natural number. Also, obviously $ 2$ and $ 3$ are relatively prime. Hence, $ 1\plus{}2\plus{}3\equal{}\boxed{6}$.

The definition of relatively prime is that $ (a,b) \equal{} 1$, where $ (a,b)$ is the greatest common factor of $ a$ and $ b$.


trigonometry
absolute value
complex numbers

For any complex number $w = a + bi$, $|w|$ is defined to be the real number $\sqrt{a^2 + b^2}$.  If $w = \cos{40^\circ} + i\sin{40^\circ}$, then
\[ |w + 2w^2 + 3w^3 + \cdots + 9w^9|^{-1}  \]
equals

$\textbf{(A)}\ \frac{1}{9}\sin{40^\circ} \qquad \textbf{(B)}\ \frac{2}{9}\sin{20^\circ} \qquad \textbf{(C)}\ \frac{1}{9}\cos{40^\circ} \qquad \textbf{(D)}\ \frac{1}{18}\cos{20^\circ} \qquad \textbf{(E)}\text{ none of these}$

[hide="Partial answer"]$w=a+bi=\cos 40+i\sin 40=e^\frac{2\pi i}{9}$
$|w+2w^2+3w^3+...+8w^8+9w^9|=|e^\frac{2\pi i}{9}+2e^\frac{4\pi i}{9}+...+8e^\frac{16\pi i}{9}+9e^{2\pi i}|\Rightarrow$
$|\cos 40+i\sin 40+2(\cos 80+i\sin 80)+...+9(\cos 360+i\sin 360)|$

The imaginary parts will cancel out to $i(\sin 200+3\sin 240+5\sin 280+7\sin 320)$, while the real parts become $10(\cos 40+\cos 80+\cos 120+\cos 160+1)$.

Okay... I just don't know how to simplify either of these. :huh:[/hide]

[hide]

Notice the nine 9th roots of unity = 1. Also, .

[/hide]

Now I'm stuck. :/

[hide="Solution"]
Let $S=w+2w^2+3w^3+\cdots+9w^9.$  Thus

\begin{eqnarray*} S &=&w+2w^2+3w^3+\cdots+9w^9\\ wS &=& w^2+2w^3+\cdots+9w^{10}\\ S-wS &=& w+w^2+w^3+\cdots+w^9-9w^{10} \end{eqnarray*}

Now, using the fact that $1+w+w^2+\cdots+w^8=0$ (since $w$ is a 9th root of unity), we have that $w+w^2+w^3+\cdots+w^9=0.$  We also note that $w^9=1$.  Therefore, $S(1-w)=-9w\Rightarrow S=\frac{-9w}{1-w}.$ Furthermore, $|S|^{-1}=\frac{|1-w|}{|-9w|}=\frac{|1-w|}9.$  So we now compute

\begin{eqnarray*} |1-w| &=& |(1-\cos 40^\circ)-\sin 40^\circ|\\ &=& \sqrt{(1-\cos 40^\circ)^2+(-\sin 40^\circ)^2}\\ &=& \sqrt{(\cos^2 40^\circ+\sin^2 40^\circ)+1-2\cos 40^\circ}\\ &=& \sqrt{2-2\cos 40^\circ}\\ &=& 2\sin 20^\circ\\ &\Rightarrow& |S|^{-1}=\frac 29\sin 20^\circ \end{eqnarray*}[/hide]

Congratulations joml88! Beautiful solution!  ;)

Wow, I can't believe I made such a careless error.

that is beautiful, when I first saw the problem I went pale, but you brought color to my cheeks...

[quote="joml88"][hide="Solution"]
Furthermore, $|S|^{-1}=\frac{|1-w|}{|-9w|}=\frac{|1-w|}9.$  So we now compute

\begin{eqnarray*} |1-w| &=& |(1-\cos 40^\circ)-\sin 40^\circ|\\ &=& \sqrt{(1-\cos 40^\circ)^2+(-\sin 40^\circ)^2}\\ &=& \sqrt{(\cos^2 40^\circ+\sin^2 40^\circ)+1-2\cos 40^\circ}\\ &=& \sqrt{2-2\cos 40^\circ}\\ &=& 2\sin 20^\circ\\ &\Rightarrow& |S|^{-1}=\frac 29\sin 20^\circ \end{eqnarray*}[/hide][/quote]

wow i got lost right there.... heh

can anyone explain that a lil better? the $| |$ are magnitude signs right...? not absolute value?

If what you mean by magnitude is what I think you mean, it's essentially the same thing. Magnitude is always a positive value, and absolute value is always positive.

[quote="gotztahbeazn"]
can anyone explain that a lil better? the $| |$ are magnitude signs right...? not absolute value?[/quote]

They define them in the problem statement...  Also, can you take the absolute value of a complex number?  The answer is no...but you can find the magnitude of complex numbers

wow very nice solution.
i was able to notice that they were the 9 9th roots of one but was not able to find the answer w/ plugging it into my calculator.

o and i was thought that the absolute value meant the distance a number is from the origin. so it seems that they mean the same thing.  or was i thought incorrectly?  :?

No, you were taught correctly :)

In fact, when working with the complexs, it's also the distance to the origin, but, in the complex plane..
So, we can still say it's the absolute value (or, if you want to use the right vocabulary : modulus ;))

[quote="protoss3.14"]yay I got B :lol:[/quote]

Thanks for sharing :roll:. But take a look at the date of the last post before posting.


quadratics
function
algebra unsolved
algebra

Let $ f(x)\equal{}ax^{2}\plus{}bx\plus{}c, a,b,c \in \mathbb{R}$ and $ |f(x)| \le 1$ for $ |x| \le 1$.

Prove that $ |f'(x)| \le 4$ for $ |x| \le 1$.

For $ x \in [ \minus{} 1,1]$, we have:

$ 1 \geq |f(x)| \equal{} \left|ax^2 \plus{} bx \plus{} c\right| \equal{} \left|\frac {(2ax \plus{} b)^2}{4a} \minus{} \frac {b^2 \minus{} 4ac}{4a}\right| \equal{} \left|\frac {f'(x)^2}{4a} \minus{} \frac {b^2 \minus{} 4ac}{4a}\right|$

replacing $ f(x)$ with $ \minus{} f(x)$ if need be, we may assume that $ a > 0$. we consider two cases.

1. $ |\frac { \minus{} b}{2a}| \geq 1$. in this case, for $ |x|\leq 1$, $ |f'(x)| \equal{} |2ax \plus{} b|\leq |2ax| \plus{} |b|\leq 2a \plus{} |b|\leq 2|b|$. so it suffices to prove that $ |b|\leq 2$.

since $ |f(\pm 1)|\leq 1$, we have $ |a \plus{} b \plus{} c| \leq 1$ and $ |a \minus{} b \plus{} c| \leq 1$, so $ |2b| \equal{} |a \plus{} b \plus{} c \minus{} (a \minus{} b \plus{} c)|\leq |a \plus{} b \plus{} c| \plus{} |a \minus{} b \plus{} c|\leq 2$, implying that $ |b|\leq 1$. (so in this case we actually get $ |f'(x)|\leq 2$.)

2. $ |\frac { \minus{} b}{2a}| \leq 1$. write $ f(x) \equal{} a(x \minus{} h)^2 \plus{} k$; then $ h \equal{} \frac { \minus{} b}{2a}$. by replacing $ f(x)$ with $ f( \minus{} x)$ if need be, we may assume that $ 0\leq h\leq 1$. then $ f$ assumes its maximum on $ [ \minus{} 1,1]$ at $ \minus{} 1$. it assumes its minimum at $ x \equal{} h$. so we get $ f(h) \equal{} k\geq \minus{} 1$ and $ f( \minus{} 1) \equal{} a(1 \plus{} h)^2 \plus{} k\leq 1$. then $ a(h \plus{} 1)\leq a(h \plus{} 1)^2\leq 2$. now, for $ |x|\leq 1$, $ |f'(x)| \equal{} |2ax \minus{} 2ah| \equal{} 2a|x \minus{} h|\leq 2a| \minus{} 1 \minus{} h| \equal{} 2a|1 \plus{} h| \equal{} 2a(1 \plus{} h)\leq 2\cdot 2 \equal{} 4$.


Prove
$ \frac{d_{max}}{d_{min}}>\equal{} \frac{\sqrt{3}}{2}(\sqrt{n}\minus{}1)$

[quote="tranquoc"]Prove
$ \frac {d_{max}}{d_{min}} > \equal{} \frac {\sqrt {3}}{2}(\sqrt {n} \minus{} 1)$[/quote]

Please define your notations ($ d_{max}$, $ d_{min}$, $ n$). Anyway, I guess you mean this one?
http://www.mathlinks.ro/Forum/viewtopic.php?t=3416

  darij


I'm not going to college yet, but I figure that it's not to early to start asking some questions.

My oldest cousin went to MIT and majored in mathematics and computer science.  He went to a prep school and is very bright, although he was not really involved with the AMCs or math outside of school.

I asked someone close to him about what it was like to major in math at MIT; she said that he was well prepared for the workforce, but that he disliked the competition at MIT.  She mentioned that MIT had "two math tracks" and that my cousin was not able to take the upper track due to "math geniuses" and "Russians who had gone to math camp every summer since they were seven."

How true is this?  And what level, in terms of success on the AIME/USAMO/IMO, as well as math learning ability, should a student be at in order to be prepared to take high-level math classes at MIT or another top school?  How can one prepare for a university mathematics education [i]at the highest level[/i]?

We have two math tracks?  That's news to me.

What we do have is three different math major options, but I've been told that it honestly doesn't matter too much which option you pick.  Taking classes is mostly a matter of convincing your advisor you can handle it; prerequisites don't matter too much.

Clearly Olympiad preparation is a big help; however, it's not at all necessary for success here.  Just do a lot of interesting mathematics.  If you want an idea of the kind of stuff we do here, look no further than [url=http://ocw.mit.edu/OcwWeb/web/home/home/index.htm]MIT OCW[/url].

[quote="t0rajir0u"]We have two math tracks?  That's news to me.[/quote]

I think that the person who explained this to me was probably a little confused about what my cousin had said: he had probably meant that the "higher track" consisted of taking tougher courses like 18.100C starting off, and then more rigorous theoretical math courses as an undergraduate, whereas the "lower track" consisted of taking of taking a basic (altough MIT-hard version) of single variable calculus to begin, and not taking all the desired theoretical math courses because of issues with difficulty.

Maybe, but then I don't see what that has to do with Russian math geniuses, unless your cousin was complaining about those classes being full.  Then I can't comment, since I haven't had that problem yet.  (There are plenty of classes to take.)

On the topic of MIT, is MIT worth all the tough courses and rigorous workload?

I hear all the time that MIT has a very heavy workload and everyone who goes there dies. So is it worth it or more of a waste?

Look, it's not as if the Ivies have light workloads compared to MIT, and the size of your workload here is completely dependent on which classes you take and how many classes you take.  You can stop at 4 or 5 a semester or overload on 6 to 10 like some people I know; it's completely up to you.

For deciding between top schools, the choice almost always comes down to culture rather than workload (assuming it didn't come down to financial aid; that's one place MIT could improve on), and MIT has a unique culture that I really appreciate.


A track meet awards 6, 5, 4, 3, 2 and 1 points for $ 1^\text{st}$, $ 2^\text{nd}$, $ 3\text{rd}$, $ 4^\text{th}$, $ 5^\text{th}$ and $ 6^\text{th}$ places, respectively. Team $ A$'s only two runners originally earned 6 points in the first race with both runners finishing in the top 6, but then the winner of the first race was disqualified. How many points did Team $ A$ have after the ranks were adjusted up?

Both runner's ranks and scores go up by 1, so their adjusted total score is 8.

Since both finished in the top 6, Team A can not have the winner, or else his/her partner would get 0. When the first place is disqualified, 2nd gets 6 points, 3rd gets 5 points, and so on. Both of them advance by 1 point each, so the total is 8.


geometry
inequalities unsolved
inequalities

Let $ABCD$ be a circumscribable quadrilateral with parallel sides $AD$ and $BC$ and area $F$. Prove

$ AB + CD \geq 2\sqrt F $

In any quadriater we have :
S :le:  :sqrt: (p-a)(p-b)(p-c)(p-d) :le: (2p-a-b)(2p-c-d)/4=(a+b)(c+d)/4 :le: p 2 /4
We have in our case AB+CD=AD+BC=p
2 :sqrt: S :le: p=AB+CD


trigonometry
geometry
geometric transformation
reflection
parallelogram
perpendicular bisector
geometry proposed

Let $ ABC$ be a triangle with $ AB \equal{} AC$. $ M$ is the midpoint of $ BC$. Take $ N$ on $ AC$ such that $ MN \perp AC$. Point $ P$ is chosen on segment $ MC$ such that $ MP \equal{} PC\sin^2C$. Prove that $ AP\perp BN$.

[quote="yetti"]What is pont $ N$ :?:[/quote]

Sorry, I missed that. $ N$ is on $ AC$ such that $ MN\perp AC$

I edited the problem

Let $ K$ be a reflection of $ N$ in the perpendicular bisector $ AM$ of $ BC$ and $ L$ a reflection of $ N$ in the midpoint $ M$ of $ BC.$ $ BNCL$ is a parallelogram, because its diagonals cut each other in half at $ M$ $ \Longrightarrow$ $ LC \parallel BN.$ Let $ KL \perp BC$ cut $ AC$ at $ Q.$ From the right $ \triangle NKQ$ with $ AK \equal{} AN,$ $ A$ is the midpoint of $ QN$ and

$ \frac {\overline{QA}}{\overline{AC}} \equal{} \frac {\overline{AN}}{\overline{AC}} \equal{} \frac {\overline{AM}}{\overline {AC}}\ \sin C \equal{} \frac {\overline{AC}}{\overline {AC}}\ \sin^2 C \equal{} \frac {\overline{MP}}{\overline{PC}}.$

It follows that $ AP \parallel QM.$  From $ (QL \equiv KL) \perp (MC \equiv BC), (QN \equiv AC) \perp (LM \equiv MN)$  $ \Longrightarrow$ $ Q$ is orthocenter of $ \triangle LMC$ $ \Longrightarrow$ $ QM \perp LC$ its remaining altitude $ \Longrightarrow$ $ (AP \parallel QM) \perp (BN \parallel LC).$

Very uninteresting problem (just draw AP perpendicular to BN, evaluate MP and ask the converse). I ended up with tons of trigonometry.

Congratulations yetti, your solution is really really beautiful!

[b]Related problem[/b]: Take AP perpendicular to BN. If AP is the bisector of MÂC, find BÂC.

If $ \{K\}\equal{}AP\cap MN$, then it is easy to see that $ K$ is the midpoint of $ MN$.
Then the problem becomes quite straightforward.
For example one could notice that:
\[ \overrightarrow{AK}\cdot{\overrightarrow{BN}}\equal{}\frac{1}2 (\overrightarrow{AN}\plus{}\overrightarrow{AM})(\overrightarrow{BM}\plus{}\overrightarrow{MN})\equal{}\frac{1}2 (\overrightarrow{AN}\cdot \overrightarrow{BM}\plus{}\overrightarrow{AM}\cdot\overrightarrow{MN})\equal{}\]
\[ \equal{}\frac{1}2( \overrightarrow{AN}\cdot \overrightarrow{MC}\plus{}(\overrightarrow{AC}\plus{}\overrightarrow{CM})\cdot \overrightarrow{MN}))\equal{}\frac{1}2 \overrightarrow{MC}\cdot (\overrightarrow{AN}\plus{}\overrightarrow{NM})\equal{}\frac{1}2 \overrightarrow{MC}\cdot \overrightarrow{AM}\equal{}0,\]
thus $ AP\perp BN$.

Same as the above post,we could prove that:K is the midpoint of MN.
Then apply Ceva's Theorem to triangle MNC we have:
$ \frac{AC}{AN}.\frac{KN}{KM}.\frac{PM}{PC}\equal{}1$
Because $ \frac{AC}{AN}\equal{}\frac{AC}{AM}.\frac{AM}{AN}\equal{}\frac{1}{sin^2C}$,we conclude that:
$ \frac{PM}{PC}\equal{}sin^2C$
Q.E.D

[color=darkblue]It's not so pretty but it's a solution:

We wanna prove that:

$ AB^2\minus{}AN^2\equal{}PB^2\minus{}PN^2$   $ (*)$

But

$ NC\equal{}MC\cdot \cos{C}$

$ MN\equal{}MC\cdot \sin{C}$

$ MC\equal{}PC\cdot (1\plus{}\sin^2 C)$

$ AN\equal{}AC\minus{}NC\equal{}AB\minus{}NC\equal{}AB\minus{}MC\cdot \cos{C}$

By Stewart in $ \triangle{MNC}$ we get

$ PN^2\equal{}\dfrac{MN^2\cdot PC\plus{}NC^2 \cdot MP}{MC}\minus{}MP\cdot PC\equal{}$

$ \equal{}\dfrac{MC^2 \cdot PC \cdot \sin^2 C\plus{}MC^2\cdot PC \cdot \cos^2{C}\cdot \sin^2{C}}{MC}\minus{}PC^2\sin^2{C}\equal{}$

$ \equal{}MC \cdot PC \cdot \sin^2 C (1\plus{}\cos^2{C})\minus{}PC^2\cdot \sin^2{C}\equal{}$

$ \equal{}PC^2 \sin^2 C(1\plus{}\cos^2{C})(1\plus{}\sin^2{C})\minus{}PC^2 \cdot \sin^2{C}\equal{}$

$ \equal{}PC^2\sin^2 C(1\plus{}\cos^2{C}\cdot\sin^2{C})$

Also

$ PB\equal{}MB\plus{}PM\equal{}MC\plus{}MP\equal{}PC(2\sin^2{C}\plus{}1)$

Then, by putting in $ (*)$ we get

$ AB^2\minus{}(AB\minus{}MC\cdot \cos{C})^2\equal{}PC^2(2\sin^2{C}\plus{}1)^2\minus{}PC^2\sin^2 C(1\plus{}\cos^2{C}\cdot\sin^2{C})$

$ 2MC^2\minus{}MC^2\cdot \cos^2{C}\equal{}PC^2(2\sin^2{C}\plus{}1)^2\minus{}PC^2\sin^2 C(1\plus{}\cos^2{C}\cdot\sin^2{C})$

$ PC^2(1\plus{}\sin^2{C})^2(2\minus{}\cos^2{C})\equal{}PC^2(2\sin^2{C}\plus{}1)^2\minus{}PC^2\sin^2 C(1\plus{}\cos^2{C}\cdot\sin^2{C})$

$ (1\plus{}\sin^2{C})^2(2\minus{} \cos^2{C})\equal{}(2\sin^2{C}\plus{}1)^2\minus{}\sin^2 C(1\plus{}\cos^2{C}\cdot\sin^2{C})$

$ (2\sin^2{C}\plus{}1)(\cos^2{C}\plus{}\sin^2{C}\minus{}1)\equal{}0$

Which is clearly true.[/color]

[quote="No Reason"]Same as the above post,we could prove that:K is the midpoint of MN.[/quote] We could. Knowing this, you apply Menelaus theorem for $ \triangle MNC$:[quote="No Reason"]
Then apply Ceva's Theorem to triangle MNC we have:
$ \frac {AC}{AN}.\frac {KN}{KM}.\frac {PM}{PC} \equal{} 1$
Because $ \frac {AC}{AN} \equal{} \frac {AC}{AM}.\frac {AM}{AN} \equal{} \frac {1}{sin^2C}$,we conclude that:
$ \frac {PM}{PC} \equal{} sin^2C$
Q.E.D[/quote]
But that's given. Where is $ AP \perp BN$ ?

I chose the point P such that $ AP \perp BN$ not the point P in the problem.(but for more clearly,this point should be denoted as P')  :(


function
search
topology
advanced fields
advanced fields unsolved

does exist a non compact metric space E such that every continuous function f from E to E admit a fixed point?

http://www.mathlinks.ro/viewtopic.php?search_id=816219946&t=114137  is an interesting discussion, I think (actually, more like a monologue, but surely interesting  :D ). However, I'd like to suggest you posting this kind of problems in the more appropriate analysis forum.

[quote="harazi"] However, I'd like to suggest you posting this kind of problems in the more appropriate analysis forum. [/quote]
Well, the questions about topology can really be posted in either one (though the trend is to post "abstract topology" questions in Advanced Fields and those with words like "metric space", "fixed point", etc. in Analysis, indeed). Anyway, don't worry about such things too much now. By your 20th post you'll figure out which forum is for what and before that I or some other mod can move your posts if we decide that they are misplaced. Also, use the search button before posting. You do not need to get obsessed with it, but, since you are new here, you may not realize that many good problems have already been discussed. So, it is a good idea to make a search by the key words (in this case something like "fixed point" or "metric space") before posting a new question. It may well happen that somebody else has posted it already. :)


Find all positive integers $ k,m$ such that $ (1\plus{}2\plus{}\cdots \plus{}n)^k\equal{} 1^m\plus{}2^m\plus{}\cdots \plus{}n^m$ for every $ n\in \mathbb N$

Let $ X_n \equal{} (1 \plus{} 2 \plus{} ... \plus{} n)^k, Y_n \equal{} 1^m \plus{} 2^m \plus{} ... \plus{} n^m$.
For big n $ X_n \equal{} \frac {n^{2k}}{2^k}(1 \plus{} o(1))$, $ Y_n \equal{} \frac {n^{m \plus{} 1}}{m \plus{} 1}(1 \plus{} o(1)).$
Therefore $ 2k \equal{} m \plus{} 1 \equal{} 2^k$ or $ k \equal{} m \equal{} 1$ or $ k \equal{} 2,m \equal{} 3$, Both case is solution.

For $ n\equal{}2$, we get $ 3^k\equal{}1\plus{}2^m$. For $ n\equal{}3$, we get $ 6^k\equal{}1\plus{}2^m\plus{}3^m$. Subtract the first equation from the second equation, we get $ 6^k\minus{}3^k\equal{}3^m$, thus $ 2^k\minus{}1\equal{}3^{m\minus{}k}$. Considering modulo 3, $ k$ is even. So $ (2^{k/2}\plus{}1)(2^{k/2}\minus{}1)\equal{}3^{m\minus{}k}$. Therefore $ 2^{k/2}\equal{}2$ and $ k\equal{}2$. We also get $ m\equal{}3$. These values are clearly true.


Hello all, I came across the following proof by induction the other day:

Prove that for all integers n such that n is greater than or equal to two,

n!> n^n

(p.s does anyone know how to write equations on this forum)

[quote="avemuri57"]
n!> n^n

(p.s does anyone know how to write equations on this forum)[/quote]

Put your equation in dollar signs, like

[code]$n! > n^n$[/code]

which comes out as $ n! > n^n$.

By the way, $ n! \equal{} \underbrace{n \cdot (n \minus{} 1) \cdots 1}_{n\ \text{times,\ each} < n} < n^n \equal{} \underbrace{n \cdot n \cdots n}_{n\ \text{times}}$.

[quote="avemuri57"]Prove that for all integers n such that n is greater than or equal to two,

$ n!> n^n$[/quote]

Counter examples: 2>4, nope. 6>27, nope. etc.

Haha,

Sorry for mistype, i meant $ n!<n^n$

But still, $ n>2$  .....

That proof can be done directly.  A little more interesting is the inequality $ n! < \left( \frac{n\plus{}1}{2} \right)^n$.

...for which I have a proof.
\[ \left(\frac{\sum_{i\equal{}1}^n i}{n}\right)>\sqrt[n]{1\cdot 2\hdots n}\equal{}\sqrt[n]{n!}\iff \left(\frac{n\plus{}1}{2}\right)^n>n!\]

I also found that given $ n,k\in\{\mathbb{Z^\plus{}}: k,n\neq1\}$ and $ k<n!$ then $ k^n\neq n!$.

Cool.

What is more it is very easy to prove:
$ \forall_{k\geq e}n!*k^{n}>n^{n}$  :wink:


algorithm
algebra
polynomial
algebra open

HI.... i need to get sqrt(n) in a different way... some algorithm or something.... i tried taylor polynomials around 1 but is not too good.... and also this

sqrt(n)=1+(n-1)/(2+(n-1)/2........)))))))) i need something more like an algorithm... not infinite

[/quote]

You could try the iterative formula:

$ x_n \equal{} \frac {x_{n \minus{} 1}^2 \plus{} s}{2 x_{n \minus{} 1}}$

For a large enough $ n$, this would give you a good approximation of $ \sqrt {s}$. Plugging $ x_0 \equal{} \frac {s}{2}$ as a start value yields a good result.

wow nice............ thanks..... just what i need...

milin, in your numerator, I think you meant $ x_{n\minus{}1}^2$, not merely $ x_{n\minus{}1}$.


graph theory
combinatorics proposed
combinatorics

Prove that there exist graph $ G$ with $ n(2n+1)$ vertexes,such that max. size of the clique in $ G$ is $ 2n$ and for every subgraph $ A$ of $ G$:
$ \omega(A)\neq\omega(G\A)$,where $ \omega(A)$ IS max size of the clique in $ A$.

I do not understand the statement : a clique is a complete subgraph. So if the initial graph $ G$ is not complete, then you may remove any vertex which does not belong to a maximal clique without destroying the maximal clique of $ G$...

Maybe are you asking for the opposite, that is $ G$ is a subgraph of $ A$?

Pierre.

What is $ G/A$? The graph obtained by removing all the vertices in $ A$?
Thus, your statement seems contradictoring IMO 2007 pb 3
http://www.mathlinks.ro/Forum/viewtopic.php?t=159838

Pierre.

[quote="Erken"]I know that they are similar but not the same.[/quote]
They are clearly not the same, since their statements contradict each other.

The IMO problem says:
Let G be a graph with even w(G).  Prove that there exists an
induced subgraph B of G such that $ w(B)\equal{}w(G\minus{}B)$.

Your problem says:
Prove that there exists a graph on n(2n+1) vertices with even w(G)=2n,
such that for all subgraphs A of G, $ w(A)\ne w(G\minus{}A)$.

If you plug the induced subgraph B of the IMO problem into your
statement, you get a clean contradiction.

[quote="Erken"]Oops i mean that $ w(G / A)\neq w(A)$[/quote]

Doesn't G\A stand for edge contractions?
Is there a difference between G\A and G/A in graph theory?

I'm sorry,because i mean that $ A$ is a clique too,and i had problem with internet at my mobile phone last time.
Now i know its solution,and i am not interseted in this problem anymore,thank you.

[quote="pbornsztein"]So, post your solution. Then we would close the thread.

Pierre.[/quote]
But you didn't even try to solve it,maybe i must wait sometime,and maybe someone else post it?
OK,i'll post my solutions tomorrow :wink:


abstract algebra
induction
group theory
superior algebra
superior algebra solved

(a) Prove that if $x \mapsto x^3$ is an automorphism of a group $G$, then $G$ is abelian.
(b) Prove that it is sufficient for $G$ to be abelian that the map $x \mapsto x^n$ ($n > 1$) is a group homomorphism if $G$ is torsion-free.

I assume that the problem text is actually like this:.
Let $G$ be a torsion free abelian group. Prove that $G\to G: x\to x^n$ is a group homomorphism.

I don't see why do you need the group to be torsion free.
$x^ny^n=(xy)^n\Leftrightarrow x^{n-1}y^{n-1}=(yx)^{n-1}=(xy)^{n-1}$ and an easy induction proves $x^ny^n=(xy)^n\ \forall x,y\in G$.
Trivially $1^n=1$.

Therefore $x\to x^n$ is a group homomorphism.

Come on ... the wording is: $G$ is a torsion free, and $x\rightarrow x^{n}$ is an homomorphism ($n>1$). Show that $G$ is abelian.

Let $f: G\to G$, $f(x)=x^n$, $n>1$. $f$ is a group homomorphism by hypothesis.
$f(1)=1\Leftrightarrow x^n=1$. Since $G$ is torsion free the last is equivalent to $x=1$, therefore $f$ is injective.

$x^ny^n=(xy)^n\Rightarrow x^{n-1}y^{n-1}=(yx)^{n-1}\ \forall x,y\in G$.
$x^{(n-1)^2}y^{(n-1)^2}=(y^{n-1}x^{n-1})^{n-1}=((xy)^{n-1})^{n-1}=(xy)^{(n-1)^2}\ \forall x,y\in G$.
$x^{(n-1)^2}y^{(n-1)^2}=x(x^{n(n-2)}y^{n(n-2)})y=x(x^{n-2}y^{n-2})^ny\Rightarrow$ $x(x^{n-2}y^{n-2})^ny=(xy)^{n^2-2n+1}\Rightarrow (x^{n-2}y^{n-2})^n=(yx)^{n(n-2)}\ \forall x,y\in G$.
Using the injectivity of $f$, the last gives $x^{n-2}y^{n-2}=(yx)^{n-2}\ \forall x,y\in G$.
$x^{n-2}y^{n-2}=(yx)^{n-2}\Rightarrow x^{n-1}y^{n-1}=x(yx)^{n-2}y=(xy)^{n-1}\Rightarrow g$ is an endomorphism, where $g: G\to G$, $g(x)=x^{n-1}$. Just like for $f$, $g$ is injective. But then $(yx)^{n-1}=x^{n-1}y^{n-1}=(xy)^{n-1}\Rightarrow xy=yx\ \forall x,y\in G$.

A proof by induction can also be given.


anyone have a scanned version I can look at? pm me

Dude same. I heard it is really challenging and that is why they use it at MIT for the advanced undergrad classes for mechanics. So if anyone has it, please pm me. I could not find it on the internet anywhere.


$(\tan x+\frac{1}{4}\cot x)^{n}=\cos^{n}x+\sin^{n}x$ with $n\in N$

[quote="N.T.TUAN"]We have $|\tan{x}+\frac{1}{4}\cot{x}|=|\tan{x}|+\frac{1}{4}|\cot{x}|\geq 1$ and $|\cos^{n}{x}+\sin^{n}{x}|\leq |\cos{x}|^{n}+|\sin{x}|^{n}\leq \sin^{2}{x}+\cos^{2}{x}=1$, for all $n>1$. Done! :D[/quote]
I think it should be
$|\tan{x}+\frac{1}{4}\cot{x}|\geq |\tan{x}|+\frac{1}{4}|\cot{x}|$
but
$|\cos{x}|^{n}+|\sin{x}|^{n}\leq \sin^{2}{x}+\cos^{2}{x}=1$ for all $n\ge 2$ but It isn't all now with $n=0,1$ what is it? and when does the equality hold to infer the solution of the equation,you should write clearly because it isn't $x+1=0$ it is a trigonometric equation, and moreover you only $\Rightarrow |\tan{x}+\frac{1}{4}\cot{x}|=|\cos^{n}{x}+\sin^{n}{x}|$ you lost the reverse direction!
There are so many problems here.  :D

[quote="skywalker"]I have recognized the mistake of you
$|\tan{x}+\frac{1}{4}\cot{x}|\le |\tan{x}|+\frac{1}{4}|\cot{x}|\geq 1$
 :D  :D
Really?[/quote]
He made no mistake.
$y: =2\tan x$, then $|\tan{x}+\frac{1}{4}\cot{x}|=\left |{y+{1\over y}\over 2}\right |={|y|+{1\over |y|}\over 2}=1+{(|y|-1)^{2}\over 2|y|}\geq 1$.

[quote="skywalker"]I don't understand anything here $x\in R$ why
$|\tan x+\frac{1}{4}\cot x|=|\tan x|+\frac{1}{4}|\cot x|$
here we don't have $x>0$ surely $|a+b|=|a|+|b|$forall $a,b\in R$ :rotfl:  :rotfl:[/quote]
If $ab\geq 0$ then $|a+b|=|a|+|b|.$

ok I see that's right, but can you show clearly the root :?:

If $n>2$ it hasn't got root!
If $n=2$ it has got roots are $x=\pm \frac{\alpha}{2}+k\pi(k\in\mathbb{Z})$, where $\cos{\alpha}=\frac{3}{5}(0<\alpha<\pi)$.
If $n=1$ , now I can not :D


invariant
combinatorics unsolved
combinatorics

The box contains 2000 white balls (W), 0 blue balls (B), and 0 red balls (R). I will denote $ W\plus{}W\to B\plus{}R$ the process where I take two white balls and replace them by one blue and one red ball.
Similarly one can do the following replacements:
$ W\plus{}W\to B\plus{}R$
$ B\plus{}B\to W$
$ R\plus{}R\to W$
$ B\plus{}W\to R$
$ R\plus{}W\to B$
1. After finitely many steps there are three balls. Show that exactly one of them is white.
2. Is it possible that there is exactly one ball in the box?

[hide="Solution"]Searching for an invariant, we note that balls R and B are closely related. The last two equations suggest that W somehow switches the parity of the balls ($ B\to R$ or $ R\to B$).
We see that we can assign a value of $ \minus{} 1$ to white balls, $ i$ to red balls, and $ \minus{} i$ to blue balls. The product of the values of the balls is invariant under multiplication when the described operations are applied. For the initial conditions, the product is $ ( \minus{} 1)^{2000} \equal{} 1$.

We also note that the combined number of blue and red balls is constant, modulo $ 2$, under the given operations. When there are $ 3$ balls left in total, there must either be $ 0$ red and blue balls ($ 3$ white) or $ 2$ red and blue balls($ 1$ white). However, the former case is impossible, because the product of the values of the three balls would be $ ( \minus{} 1)^3 \equal{} \minus{} 1$, not $ 1$. This completes the proof of part 1.

Part 2 follows trivially from the invariant that we found--the value of the hypothetical one ball is either $ \pm i$ or $ \minus{} 1$, never $ 1$ as it must be.[/hide]

err... can you please explain what is parity?  :blush:

Parity is congruence mod 2 (ie, the property of being odd/even).

[hide="Solution"]
1. If we do not distinguish between Red balls and Blue balls, the parity of these non-white balls remains constant. Hence there must be either one or three white balls.
Now suppose we end up with three balls.
It is easy to see, since the number of non-white balls increases by 2 in the first step, decreases by 2 in each of the next two steps, and remains unchanged in the others, there must have been as many first steps as second and third combined. Suppose there were $ r$ steps of type 2 and $ s$ steps of type 3.
Hence there were $ r \plus{} s$ steps of type 1, and $ 997 \minus{} (r \plus{} s)$ steps of types 4 and 5 together (as a count on the number of white balls shows).
Now, we count the blue balls.
Number of blue balls added by step 1 = $ r \plus{} s$.
Number of blue balls removed by step 2 = $ 2r$.
Hence, number of steps of type 5 = $ r \minus{} s$.
Similarly, number of steps of type 4 = $ s \minus{} r$.
Hence, $ r \equal{} s$ and $ 997 \minus{} (r \plus{} s)$ = 0, which is impossible for integers $ r$ and $ s$.
(This proof shows that an odd reduction in the number of whites without affecting the others is impossible. There may be a smaller solution though, and this one might have loopholes.)

2. Since at each step the number of balls decreases by at most 1, there must be three balls left at some stage before we reach one ball. We consider the two possible arrangements:

$ W, R, R$
Possible branches:
$ W \plus{} (R \plus{} R) \rightarrow W \plus{} W \rightarrow B \plus{} R$
$ (W \plus{} R) \plus{} R \rightarrow B \plus{} R$
Thus, we never reach a single ball from here.

$ W, B, R$
Since after the first move we reach $ 998W, B, R$, an argument as in 1 shows that this state is unreachable.
[/hide]


calculus
integration
function
limit
geometry
rectangle
real analysis

1. Find lim {x->Pi/2}  \int {0->x} 1/(1+(tgt)^a) dt where a>0.

2. Find lim {n->00}  \int {-a->a} dx/[(1+x^2+x^2n)*(1+e^x)] where a is in (0,1).

Hope you enjoy them (even if they are pretty easy... :D  :D )

[quote="grobber"]
Find lim {n->00}  \int {-a->a} dx/[(1+x^2+x^2n)*(1+e^x)] where a is in (0,1).
[/quote]
If 0<a<1 and |x| \leq a <1=> x^(2n)-->0 when n-->+oo
If x\in [-a;a] is fixed the sequence of continous functions
f_n(x)= 1/[(1+x^2+x^2n)*(1+e^x)]
converge to f(x)= 1/[(1+x^2)*(1+e^x)]
|f_n(x)|  \leq  1/[(1+x^2)*(1+e^x)] dominated function integrable
on the compact [-a;a]
Lebesgue's dominated convergence theorem gives
 \lim {+00}  \int {-a->a} dx/[(1+x^2+x^2n)*(1+e^x)] =
\int {-a->a} dx/[(1+x^2)*(1+e^x)]

oubinol

[quote="grobber"]1. Find lim {x->Pi/2}  \int {0->x} 1/(1+(tgt)^a) dt where a>0.
[/quote]

 \pi /4

cheese, heu cheers
oubi

Told you they were easy... The limit for the first one is easy to calculate, btw, it's arctg(a). Could you give your soln for the second one?

[quote="grobber"]1. Find lim {x->Pi/2}  \int {0->x} 1/(1+(tgt)^a) dt where a>0.
[/quote]

First existence of \int_{0 to  \pi /2}f(x)dx where f(t)=1/(1+(tgt)^a)

lim_{t->0} f(t) = 1
lim_{t-> pi/2} f(t)= 0
we can say f is continous on [0;pi/2] with f(pi/2)=0

change variable y=pi/2 -x
K=\int_{0 to  \pi /2}(tany)^a/(1+(tany)^a)dy
so 2K=\int_{0 to  \pi /2}dy=pi/2
K=pi/4
chers
oubinool

For wich one did find arctan(a) ?

It's arctan(a) for the SECOND problem (I said "first" in my previous post because it was the first one you solved). It's easily obtained by x->-x.

[quote="oubi"]
\int {-a->a} dx/[(1+x^2)*(1+e^x)]
[/quote]

K=\int {-a->a} dx/[(1+x^2)*(1+e^x)]
the change variable y=-x gives
K= \int {-a->a} e^ydy/[(1+y^2)*(1+e^y)]
2K=\int {-a->a} (1+e^x)/[(1+x^2)*(1+e^x)]dx=
\int {-a->a} dx/(1+x^2)= 2arctan(a)

You know grobber at the begining i doubted about the arctan(a) because
there is no primitive with usual function for 1/[(1+x^2)*(1+e^x)]
Which grade these two problems ?

These two problems are 12th grade problems! :D :D

Yeah, they're 12'th grade problems. They were proposed for some editions of a Romanian contest: Gheorghe Titeica. It's pretty famous in Romania. It's usually considered to be among the toughest highschool contests. These are some of the easiest problems I could find (the kind i can solve   :D  :D)

These integrals problems are nice for high school contests.
Moubinool OMARJEE
 :D  :D  :D  :D  :D :D  :D

[quote="grobber"]

2. Find lim {n->00}  \int {-a->a} dx/[(1+x^2+x^2n)*(1+e^x)] where a is in (0,1).
[/quote]

grobber,

The change variable y=-x in the begining is beter
G_n(a)=\int {-a->a} dx/[(1+x^2+x^2n)*(1+e^x)]  simplify
G_n(a)=\int {-a->a} e^y/[(1+y^2+y^2n)*(1+e^y)]dy

 2G_n(a)= \int {-a->a} dx/(1+x^2+x^2n) (*)
lim G_n(a)=1/2\int {-a->a} dx/(1+x^2)=\int {0->a} dx/(1+x^2)=Arctan(a)

In (*) factorise 1+x in the denominator  to get an asymptotic expansion G_n(a) when n->+oo (remind 1/(1+u) = 1-u+o(u))
G_n(a)= \int {0->a} dx/(1+x^2)(1+x^2n/(1+x))
=\int {0->a} dx/(1+x^2) - \int {0->a} x^(2n)/(1+x)dx +  t_n(a)
Integrate by part the second integral
G_n(a)=Arctan(a) -a^(2n+1)/(2n+1).1/(1+a)+
\int_{0 to a}x^(2n+1)/(2n+1).(2x/(1+x)) dx + t_n(a)

with  t _n(a) -->0 when n->+oo

That's what I did when I initially solved the problem: I used the substitution x->-x right at the start to get rid of e^x+1.

[quote="Moubinool"]
First existence of \int_{0 to  \pi /2}f(x)dx where f(t)=1/(1+(tgt)^a)

lim_{t->0} f(t) = 1
lim_{t-> pi/2} f(t)= 0
we can say f is continous on [0;pi/2] with f(pi/2)=0

change variable y=pi/2 -x
K=\int_{0 to  \pi /2}(tany)^a/(1+(tany)^a)dy
so 2K=\int_{0 to  \pi /2}dy=pi/2
K=pi/4
[/quote]

the essence of this solution is that:
f(t) is central symmetric of point (\pi/4, 1/2). it is easy to proof that f(\pi/4+x)+f(\pi/4-x)=1... :D
hence the area below f(t) on [0,1] is exactly half the area of rectangle [0,pi/2]x[0,1]...
the rectangle's area is \pi/2, therefore  \int f(t)= \pi/4.

cheers.


for downloads, I have MacTeX.mpkg.zip, baxic-miktex-2.7.3248.exe, texshop_1.43, Bibdesk, and TXC Setup_1StableRC1.exe

Are those the downloads needed for a Macintosh OS X (the older version) in order to type up LaTeX documents?

Thanks!

From the "[url=http://www.tug.org/mactex/downloading.html]downloading[/url]" page:
[quote="TeX Users Group"]Some browsers will automatically unzip the file and produce a package named MacTeX-2008.mpkg with Apple's familiar icon. Other browsers will produce a zip file; double click this file to unzip and produce the package. Then double click [b]MacTeX-2008.mpkg[/b] and follow the simple instructions to install. After installation, go to [b]/Applications/TeX[/b] and read the short document "READ ME FIRST." Follow the instructions in this document, and you'll be using TeX within a few minutes.[/quote]

That's all you need; the MacTeX download includes TeXShop and BibDesk, and the .exe downloads are Windows-only stuff anyway. Just trash the other files.

For some reason, it won't let me open LaTeXiT, why?

I dragged it to the dock, but no screen or anything popped up, why?