Hello everyone. I hope someone can help me on this problem.
I have a latex document and i am creating a list of figure in the beginning of the document by \listoffigures. I also have the pdf bookmarked. However, the link to the images does not go to the image the link describes. it goes about half a page further down. Does anyone know any possible reason? Thank a lot!!

Try deleting all the .aux, .toc etc files then compile twice again. If that doesn't work then have a look at http://tinyurl.com/3dvz34 which basically tells you to put \listoffigures after the other lists.

If that still doesn't work please post here a minimal document which is short but still exhibits the problem.

Here is a short version of my code that have the problem with.
Thanks for your input.

\documentclass[12pt]{report}

\usepackage{utepcsthesis}

\usepackage{amsmath, amssymb, amsfonts}

\usepackage[dvips]{graphicx}

\usepackage{graphics}

\usepackage[ps2pdf,a4paper=false,letterpaper=true,colorlinks=true,urlcolor=blue,bookmarks=true]{hyperref}

\begin{document}

\setcounter{tocdepth}{2}

\title{Title}

\maketitlepage

\include{myAbstract}

\include{acknowl}

\tableofcontents

\listoffigures

\StartBody

\include{myIntroduction}              % Introductory Chapter
\include{RelatedWork}               % Related Work
\include{amsm}                      % The protocol description

\bibliographystyle{plain}
\addcontentsline{toc}{chapter}{\bibname}
\bibliography{refs}

 \end{document}

That's going to be very difficult to debug from here as it uses proprietary files and includes lots of outside files.

What I meant by a minimal example was a short document with content in it which when run exhibits the problem. So take your document and start taking pieces out of it (both text and commands) until the problem stops and it will then be the last thing that is removed that is the cause.

There's help in doing this at [url=http://www.latex-einfuehrung.de/mini-en.html]What is a minimal working example?[/url]

This may help you solve it but if not then you can post your minimal example here.


Numbers 1, 3, 5, ..., 59 are written in the card, each number appearing on one card only
Arrange this cards so that the arrangement satisfies the following
If the topmost card is moved to the bottom of pile, then the card numbered 1 will be top most. Put this card numbered 1 on one side. Move the 3 topmost cards to the bottom of the pile. Then the card numbered 3 will be topmost. Put this card numbered 3 on one side. Repeat this procedure to move to topmost $ 2k\plus{}1$ (where $ 2\le k \le 28$) cards to the bottom of the pile in order so that the topmost card will be the one numbered $ 2k\plus{}1$ after the move. The last one is the card 59. Write down the original arrangement of these cards

What do you mean by "Put this card numbered 1 on one side."? :blush:

[hide="answer"]13 1 47 33 25 3 57 45 49 55 43 5 19 39 11 17 21 51 29 7 41 15 31 23 27 59 35 53 37 9[/hide]
[quote="gunes"]How can we move 57 or 59 cards to the bottom of the pile at the end of the game if we throw away 1,3,5... when we find'em?[/quote]
Say that we have left 2 cards. We move the first card to the end, and we repeat this procedure 57 times in total


calculus
integration
calculus computations

hello, setting $ x\equal{}1/t$ we get
$ \int\frac{\minus{}t}{\sqrt{\minus{}t^2\plus{}2t\plus{}1}}\,dt\equal{}\int\frac{\minus{}2t\plus{}2}{\sqrt{\minus{}t^2\plus{}2t\plus{}1}}\,dt\minus{}2\int\frac{1}{\sqrt{\minus{}t^2\plus{}2t\plus{}1}}\,dt$
Sonnhard.

Hi Sonnhard. I have been using CTRL C and CTRL V to get a backslash and now this isn't working. I was going to substitute u = x+ sqrt(x^2 + 2x-1) which would lead to a 1/(1+u^2)^2 integral and then substitute u = tan theta. I need to try to sort out my computer as it keeps crashing giving the message "cannot find server". Will get back when I can.

No nead of the factor $ 2$ to get

[hide] ${ \sqrt{-t^2+2t+1} - arc sin ({{t-1}\over \sqrt 2 }}) + C$ [/hide]


modular arithmetic
algebra unsolved
algebra

Give positive number x,y,z,n.
Investing n satisfy:
 x^2-y^2-z^2=n have exactly one solution.

If $n$ is even, then $(x,y,z)=(\frac{n}{2}+2k^{2}+2k+1,\frac{n}{2}+2k^{2}+2k,2k+1)$ is a solution for all $k\in \mathbb{N}$
and for $n$ odd we have $(x,y,z)=(\frac{n+1}{2}+2k^{2},\frac{n-1}{2}+2k^{2},2k)$ a solution for all $k\in \mathbb{N}$.

[quote="N7C"]Firstly,I've fixed the problem, the integers x, y, and z, are consecutive terms of an arithmetic progression.
Secondly,your Solution is not enough because this is a solution without matching your one. (x,y,z)=(13,10,7);[/quote]
Firstly,you've not say that x, y, and z, are consecutive terms of an arithmetic progression.
secondly,[b]olorin[/b] did not say that they are the only solutions

$x=y+t$,$z=y-t$ with $0<t<y$ leads to
$n+y^{2}=x^{2}-z^{2}=4yt$. So $y|n$ and $4|\frac{n}{y}+y$. (*)
If $2\nmid n$, then $y$ is odd and $0\equiv n+y^{2}\equiv n+1 \pmod{4}$.
If $2|n=y\frac{n}{y}$, then $y$ or $\frac{n}{y}$ is even.  (*) shows then, that both are even and $4|y\frac{n}{y}=n$.
And if $8|n$ we get similar $4|y$ and $4|\frac{n}{y}$ and $16|n$.

Now if $2\nmid n=ab$ with $1<a\leq b$, then $b$ is odd and then $b(a+b)=n+b^{2}\equiv n+1\equiv 0\pmod{4}$,
so that $4|a+b$ and
$(x,y,z)=(n+\frac{n+1}{4},n,n-\frac{n+1}{4})$ and $(b+\frac{a+b}{4},b,b-\frac{a+b}{4})$ are 2 different solutions.

If $8\nmid n=4ab$ with $1<a\leq b$, then $a,b$ are odd, and
$(x,y,z)=(2ab+\frac{ab+1}{2},2ab,2ab-\frac{ab+1}{2})$ and $(2b+\frac{a+b}{2},2b,2b-\frac{a+b}{2})$ are 2 different solutions.

And if $n=16ab$ with $1<a\leq b$, then
$(x,y,z)=(4ab+(ab+1),4ab,4ab-(ab+1))$ and $(4b+(a+b),4b,4b-(a+b))$ are 2 different solutions.
For $n=32$ we have 2 solutions $(x,y,z)=(7,4,1)$ and $(11,8,5)$.

So we must have $n=4$ or $n=16$ or $n=p\equiv-1 \pmod{4}$ or $n=4p$ or $n=16p$ for some odd prime $p$.
It's easy to check using (*), that in each of these cases there is a unique solution $(x,y,z)$.

My book gives me the value n=1155 which makes that equation has 10 solution.
But, with your solution when n is odd, I can find only 8 solution.
Please tell me why.


inequalities
trigonometry
function
inequalities proposed

Triangle ABC is acute-angled

Write x=(cosA)
         y=(cosB)
        z =(cosC)

Prove 2( x^2.y^2 +y^2.z^2 +z^2.x^2) >= 3xyz

There are so many inequalities of this type, that it's hard to tell whether it's original or not, but most probably not. However, the formulation might be original :).

Here's a solution: let $a=\frac {yz}x,\ b=\frac {zx}y,\ c=\frac{xy}z$. We must show that $S=a+b+c\ge \frac 32$. It's well-known that $x^2+y^2+z^2+2xyz=1\Rightarrow ab+bc+ca+2abc=1$. We now have $1=2abc+ab+bc+ca\le \frac 2{27}S^3+\frac 13S^2$, and from here the conclusion follows easily because $f(x)=\frac 2{27}x^3+\frac 13x^2$ is strictly increasing on nonnegative numbers.

I.  am sorry but this problem should have been submitted in the intermediate forum. I just realise that the result follows easily from the AM-Gm inequality together with the fact that cosAcosBcosC<=1/8.
Although it is easy this result might prove useful as a lemma.The proof submitted by grobber is very attractive and contains ideas that are useful in proving other inequalities.Thanks,Grobber!

[quote="Michael Niland"]I just realise that the result follows easily from the AM-Gm inequality together with the fact that cosAcosBcosC<=1/8.
[/quote]

Michael, can you show how to use AM-GM to prove this?

Grobber's proof is very nice! I have a proof which is different. I would like to see yours using AM-GM. Thanks.

Note that the inequality in question,

$2\cdot\left(y^2z^2+z^2x^2+x^2y^2\right) \geq 3xyz$

with x = cos A, y = cos B, z = cos C holds for [i]any arbitrary[/i] triangle ABC, not only for an acute triangle. In fact, while for an acute triangle, it was already proven above, for an obtuse or right triangle it is trivial, since the left hand side is nonnegative (as the sum of squares), while the right hand side is nonpositive (since one of the numbers x = cos A, y = cos B, z = cos C is either 0 or negative, while the other two are positive).

[b]PS.[/b] Michael, are you sure that the inequality is so easy using AM-GM? Looks like you have applied $\displaystyle \cos A \cos B \cos C \leq \frac18$ with reversed sign...

  Darij

Sorry my post of 18 August about using AM-GM was a false alarm.

Michael

A different approach:

[quote]If $A,B,C$ are angles of a triangle. Prove that

\[
2(\cos^2{A} \cos^2{B} +\cos^2{B} \cos^2{C} + \cos^2{C} \cos^2{A}) \geq 3\cos A \cos B \cos C.
\][/quote]

As Darij ponited out, it's true for any triangle. We may assume that the triangle is [b]acute[/b] since it's trivial for non-acute triangles. Making a substitution

$\displaystyle x=\frac{1}{2}\left( b^{2}+c^{2}-a^{2}\right), y=\frac{1}{2}\left( c^{2}+a^{2}-b^{2}\right), z=\frac{1}{2}\left( a^{2}+b^{2}-c^{2}\right)$,

then $x,y,z>0$ and $a^2=y+z, b^2=z+x, c^2=x+y$. The inequality is translated into:

$\displaystyle
\frac{2}{(x+y)(y+z)(z+x)}\left(\frac{x^2y^2}{x+y} + \frac{y^2z^2}{y+z} + \frac{z^2x^2}{z+x} \right) \geq \frac{3xyz}{(x+y)(y+z)(z+x)}$. ---------(*)

By Cauchy-Schwarz, $\displaystyle
\frac{x^2y^2}{x+y} + \frac{y^2z^2}{y+z} + \frac{z^2x^2}{z+x} \geq \frac{(xy+yz+zx)^2}{2(x+y+z)}$.

(*) is reduced to $(xy+yz+zx)^2 \geq 3xyz(x+y+z)$ which can be proved easily.

It is Nesbitte inequality by applying $\frac{cosAcosB}{cosC} = \frac{tgC}{tgA+tgB}$.

I didnt base this problem on the Nesbitt inequality.Here is my geometric proof

If K,L,M are points on the sides BC,CA,Ab of a triangle and the lines AK,BL,Cm cut at P then

PK/Ap +PL/BP+PM/CP >=3/2(this inequality was submitted in the intermediate section recently.I havent seen it anywhere but it must be already known)

Taking p at the orthocentre ,we get the inequality in this question.

Inequalities involving trig functions can by solved by interpreting them geometrically,or converting them to pieces of algebra or to the form of a well known inequality.It is good to see that all these methods have been used in this question .Thanks for all the good ideas.

[quote="Michael Niland"]PK/Ap +PL/BP+PM/CP >=3/2[/quote]

This can be viewed as Nesbitt too. If we let x=[PBC], y=[PCA], z= [PAB], where [] denotes the area. Then PK/AP = x/(y+z), etc. The inequality is translated into:

x/(y+z) + y/(z+x) + z/(x+y) >=3/2, which is exactly Nesbitt.


John Deere has 3 crates of pistons. Each crate has 1000 pistons. If it takes 150 pistons to make a tractor, how many tractors can he make with his current stock of pistons?

[Let's keep our problems constructive. Not the kind that encourage making more work for other people.]

[quote="Treething"]John Deere has 3 crates of pistons. Each crate has 1000 pistons. If it takes 150 pistons to make a tractor, how many tractors can he make with his current stock of pistons?[/quote]
[hide]
John Deere has a total of 3000 pistons.

So he can make a total of $\frac{3000}{150}=20$ tractors.[/hide]

Treething, does this really belong in the Mathcounts forum? Looks like Classroom Math at the very least, and more likely the Test Forum or Games and Fun.

And how does it have to do with banananananananananas?! :roll:

[quote="Treething"]Who edited it o.0 ?[/quote]
what did it say before?


1. Find the number of subsets of [b]N[/b] which contain NO 2 consecutive numbers in terms of n.

2. Find the number of expression as the sum of n positive integers of a fixed number m, order counted.

3. Number of 0-1 sequence of length n which contains exactly m "01"s

1.  Let $ S_n$ be the set of $ n$-element binary sequence such that no two $ 1$'s occur consecutively.  Clearly, $ S_1 \equal{} \{0,1\}$, $ S_2 \equal{} \{00,01,10\}$, and $ S_3\equal{}\{000,001,010,100,101\}$.  Let $ a_n : \equal{} \left|S_n\right|$.  Then, $ a_n$ is also the number of subsets of $ \{1,2,\dots,n\}$ which contain no two consecutive numbers (why?).

Since $ a_1\equal{}2$, $ a_2\equal{}3$, and $ a_3\equal{}5$, then it is promising that $ a_n\equal{}F_{n\plus{}2}$, where $ F_n$ is the $ n$-th Fibonacci number (starting from $ F_1\equal{}F_2\equal{}1$).   So, we want to show that $ a_{n} \equal{} a_{n\minus{}1}\plus{}a_{n\minus{}2}$ for $ n>2$ as well.

Write $ S_n \equal{} U_n \cup V_n$, where $ U_n$ is the set of elements of $ S_n$ ending with $ 0$ and $ V_n$ is the set of element of $ S_n$ ending with $ 1$.  It is clear that $ U_n \sim S_{n\minus{}1}$ since we can cut the final $ 0$ from each element in $ U_n$ to obtain an element in $ S_{n\minus{}1}$.  Since each element of $ V_n$ must end in $ 01$, so by similar argument, $ V_n \sim S_{n\minus{}2}$.  This proves the result.

2. Well known problem: $ \binom{m\minus{}1}{n\minus{}1}$.  (Suppose you have $ m$ stars and $ n\minus{}1$ bars.  There are $ m\minus{}1$ spaces between stars.  You choose $ n\minus{}1$ of the spaces to put the bars in.  Suppose for $ m\equal{}7$ and $ n\equal{}3$, an outcome is $ */**/****$, then it is equivalent to expressing $ 7\equal{}1\plus{}2\plus{}4$.)

3.  The are four types of such sequences: $ 1\minus{}0\minus{}1\minus{}0\minus{}\dots\minus{}1\minus{}0\minus{}1$,  $ 1\minus{}0\minus{}1\minus{}0\minus{}\dots\minus{}0\minus{}1\minus{}0$,  $ 0\minus{}1\minus{}0\minus{}1\minus{}\dots\minus{}1\minus{}0\minus{}1$, and $ 0\minus{}1\minus{}0\minus{}1\minus{}\dots\minus{}0\minus{}1\minus{}0$.  Using 2., these types have $ \binom{n\minus{}1}{2m}$, $ \binom{n\minus{}1}{2m\plus{}1}$, $ \binom{n\minus{}1}{2m\minus{}1}$, and $ \binom{n\minus{}1}{2m}$ sequences, respectively.
Thus, the answer is
\[ 2\binom{n\minus{}1}{2m}\plus{}\binom{n\minus{}1}{2m\minus{}1}\plus{}\binom{n\minus{}1}{2m\plus{}1}\,.\]

[quote="Batominovski"]1.  Let $ S_n$ be the set of $ n$-element binary sequence such that no two $ 1$'s occur consecutively.  Clearly, $ S_1 \equal{} \{0,1\}$, $ S_2 \equal{} \{00,01,10\}$, and $ S_3 \equal{} \{000,001,010,100,101\}$.  Let $ a_n : \equal{} \left|S_n\right|$.  Then, $ a_n$ is also the number of subsets of $ \{1,2,\dots,n\}$ which contain no two consecutive numbers (why?).

Since $ a_1 \equal{} 2$, $ a_2 \equal{} 3$, and $ a_3 \equal{} 5$, then it is promising that $ a_n \equal{} F_{n \plus{} 2}$, where $ F_n$ is the $ n$-th Fibonacci number (starting from $ F_1 \equal{} F_2 \equal{} 1$).   So, we want to show that $ a_{n} \equal{} a_{n \minus{} 1} \plus{} a_{n \minus{} 2}$ for $ n > 2$ as well.

Write $ S_n \equal{} U_n \cup V_n$, where $ U_n$ is the set of elements of $ S_n$ ending with $ 0$ and $ V_n$ is the set of element of $ S_n$ ending with $ 1$.  It is clear that $ U_n \sim S_{n \minus{} 1}$ since we can cut the final $ 0$ from each element in $ U_n$ to obtain an element in $ S_{n \minus{} 1}$.  Since each element of $ V_n$ must end in $ 01$, so by similar argument, $ V_n \sim S_{n \minus{} 2}$.  This proves the result.

2. Well known problem: $ \binom{m \minus{} 1}{n \minus{} 1}$.  (Suppose you have $ m$ stars and $ n \minus{} 1$ bars.  There are $ m \minus{} 1$ spaces between stars.  You choose $ n \minus{} 1$ of the spaces to put the bars in.  Suppose for $ m \equal{} 7$ and $ n \equal{} 3$, an outcome is $ */**/****$, then it is equivalent to expressing $ 7 \equal{} 1 \plus{} 2 \plus{} 4$.)

3.  The are four types of such sequences: $ 1 \minus{} 0 \minus{} 1 \minus{} 0 \minus{} \dots \minus{} 1 \minus{} 0 \minus{} 1$,  $ 1 \minus{} 0 \minus{} 1 \minus{} 0 \minus{} \dots \minus{} 0 \minus{} 1 \minus{} 0$,  $ 0 \minus{} 1 \minus{} 0 \minus{} 1 \minus{} \dots \minus{} 1 \minus{} 0 \minus{} 1$, and $ 0 \minus{} 1 \minus{} 0 \minus{} 1 \minus{} \dots \minus{} 0 \minus{} 1 \minus{} 0$.  Using 2., these types have $ \binom{n \minus{} 1}{2m}$, $ \binom{n \minus{} 1}{2m \plus{} 1}$, $ \binom{n \minus{} 1}{2m \minus{} 1}$, and $ \binom{n \minus{} 1}{2m}$ sequences, respectively.
Thus, the answer is
\[ 2\binom{n \minus{} 1}{2m} \plus{} \binom{n \minus{} 1}{2m \minus{} 1} \plus{} \binom{n \minus{} 1}{2m \plus{} 1}\,.
\]
[/quote]
In fact, I failed to express my meaning into words for question 3...

For instance,

when $ m \equal{} 1$, and $ n\equal{}3$,

There are

$ 001$, $ 010$, $ 011$, $ 101$ four possibilities

By type $ 1 \minus{} 0 \minus{} 1 \minus{} \dots \minus{} 0 \minus{} 1 \minus{} 0$, I meant the set of sequences beginning with some $ 1$'s, then some $ 0$'s, then $ \dots$, and then, finally, some $ 0$'s.  In your example, $ 011$ is of $ 0 \minus{} 1$ type and $ 010$ is of $ 0 \minus{} 1 \minus{} 0$ type.  Do you get it?  Try plugging in my formula and you will see that it gives the correct answer.


geometry
parallelogram
ratio
angle bisector
geometry proposed

Points $D$ and $E$ lie on on sides $BC$ and $CA$ triangle $ABC$, so that $BD=AE$. Lines $AD$ and $BE$ cross in point $P$. Angle bisector of $\angle ACB$ crosses segments $AD$ and $BE$ at points $Q$ and $R$. Prove that :
\[ \frac{PQ}{AD}=\frac{PR}{BE}  \]

A well-known problem:
Construct a paralellogram $PDAM\longrightarrow MA=BD=AE\longrightarrow \Delta EAM$ is isosceles at $A$ and $\widehat{EAM}=\widehat{ABC}+\widehat{BAC}\longrightarrow \widehat{AEM}=\frac{\widehat{ACB}}{2}\longrightarrow EM//RQ$
Two triangles:$EMB$ and $RQP$ have:$EM//RQ,PQ//BM,PR\equiv BE\longrightarrow$they are similar$\longrightarrow\frac{PQ}{BM}=\frac{PR}{BE}\longrightarrow\frac{PQ}{AD}=\frac{PR}{BE}$

By the way, [i]James Potter[/i], I don't know this nice (yet simple) problem.
But the novelty of this solution is the estimate  of the common ratio $\frac{PQ}{AD}=\frac{PR}{BE}$.[hide="Here is a longer proof."]
Suppose $BC\ne AC$, i.e. $a\ne b$. Denote $AE=BD=x$ and the intersection $F$
between the angle-bisector of the angle $\widehat {ACB}$ with the side $[AB]$, i.e. $\frac{FA}{FB}=\frac ba$.

[u]Menelaus' theorem[/u] to the transversals $\overline {BPE}$, $\overline {APD}$ for the $\triangle CAD,\ \triangle CBE\ :$

$\blacktriangleright\ \frac{BD}{BC}\cdot \frac{EC}{EA}\cdot \frac{PA}{PD}=1\Longrightarrow \frac{PA}{a}=\frac{PD}{b-x}=\frac{AD}{a+b-x}\Longrightarrow$ $\boxed {\ AP=\frac{a}{a+b-x}\cdot AD\ }\ \ (1)$
$\blacktriangleright\ \frac{AE}{AC}\cdot \frac{DC}{DB}\cdot \frac{PB}{PE}=1\Longrightarrow \frac{PB}{b}=\frac{PE}{a-x}=\frac{BE}{a+b-x}\Longrightarrow$ $\boxed {\ BP=\frac{b}{a+b-x}\cdot BE\ }\ \ (2)$

[u]Menelaus' theorem[/u] to the transversals $\overline {CQF}$, $\overline {CRF}$ for the $\triangle ABD,\ \triangle BAE\ :$

$\blacktriangleright\ \frac{CD}{CB}\cdot \frac{FB}{FA}\cdot \frac{QA}{QD}=1\Longrightarrow \frac{QA}{b}=\frac{QD}{a-x}=\frac{AD}{a+b-x}\Longrightarrow$ $\boxed {\ AQ=\frac{b}{a+b-x}\cdot AD\ }\ \ (3)$
$\blacktriangleright\ \frac{CE}{CA}\cdot \frac{FA}{FB}\cdot \frac{RB}{RE}=1\Longrightarrow \frac{RB}{a}=\frac{RE}{b-x}=\frac{BE}{a+b-x}\Longrightarrow$ $\boxed {\ BR=\frac{a}{a+b-x}\cdot BE\ }\ \ (4)$

$(1)\ \wedge\ (3)\Longrightarrow\boxed {\ PQ=|AP-AQ|=\frac{|a-b|}{a+b-x}\cdot AD\ }\ \ (5)$
$(2)\ \wedge\ (4)\Longrightarrow\boxed {\ PR=|BP-BR|=\frac{|a-b|}{a+b-x}\cdot BE\ }\ \ (6)$
$(5)\ \wedge\ (6)\Longrightarrow\boxed {\ \frac{PQ}{AD}=\frac{PR}{BE}=\frac{|a-b|}{a+b-x}\ }\ .$

[b]Remark.[/b] Let $E\in [AC$, $D\in [BC$ be two mobile points so that $AE=BD$. Then the geometrical locus of the middlepoint $L$ of the segment $[DE]$ is a line which is parallelly with the angle-bisector $CF$ of the angle $\widehat {ACB}$ and which passes through the middlepoint of the side $[BC].$[/hide]


$p=8k+1$ is a prime .Prove that  $X^6+x^4+x^2+1=0 mod p$  have exactly $6$ solution $modp$

[i]lemma[/i]:p is a prime ,d is divise p-1 then  $x^{d}-1=0 mod p$
have axactly d solution$mod p$
Then $x^8-1=0modp$ hás 8 solution and $x^2-1=0modp$ has 2 solution  :P

Rust please explain $(Z/pZ)=\ { s^k\ } \cdot t=s^{\frac{p-1}{s}}$
with examples if possible


calculus
integration
real analysis
real analysis unsolved

A particle of unit mass is projected from level ground with speed USqrt(2) at an elevation of PI/4 above the horizontal.It experiences a resisting force directly opposing the motion whose magnitude is k times the square of the particle's  speed where k(U^2)/g is a small number.

Show that after a lapse of time t during the flight the vertical component of the acceleration is approximately

                           -g - (U-gt)Sqrt[ (U-gT)^2 + U^2]

Deduce that the highest point of the trajectory is attained when t is approximately equal to

                          (U/g) -K(U^3)/(3g^2)[ Sqrt(8) -1]

I am aware that this problem can be solved using Picard'S method of successive approximations.But as the question was given to students of school level this question should be solved using more elementary methods.

[quote="Michael Niland"] Show that after a lapse of time t during the flight the vertical component of the acceleration is approximately

                           -g - (U-gt)Sqrt[ (U-gT)^2 + U^2]
[/quote]
I guess it should be $-g - %Error. "mathbold" is a bad command.
k (U-gt)\sqrt{ (U-gt)^2 + U^2}$. As to the "elementary" solution, my guess is that it was a physics exam and the expected reasoning was the following: since $k$ is small, physically it is obvious that the motion is approximately the same as the motion with $k=0$ in terms of position, speed, etc., in particular, the vertical velocity is $U-gt+o(1)$ and the horisontal one is $U+o(1)$ as $k\to 0$. Thus, we can just plug the motion with $k=0$ into the right hand part of the equation to get the acceleration with precision $o(k)$. The second part is obtained by integrating the vertical acceleration until the integral reaches $U$ with some approximations along the way. Really that is more or less equivalent to performing one iteration in Picard method but it avoids a reference to it. I doubt there is a simpler solution but I'll be happy to be proved wrong  :).


analytic geometry
polynomial
geometry proposed
geometry

Could anybody post any problem and solve with [b]triangle complex coordinates[/b]?

I'll thanks it a lot

Sincerely

Perhaps, he is referring to Eisenstein Integers.

Masoud Zargar

[quote="Virgil Nicula"]What are triangle complex coordinates ? Maybe barycentrical coordinates of a point w.r.t. a given triangle ?[/quote]

Look at [url]http://thejuniverse.org/geometry/LC/LCmath.html[/url]

Here is the explanation


[quote="FOURRIER"]Let $ a,b,c,d$ be positive reals such that

$ abcd > a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$

Then  prove that $ abcd > a \plus{} b \plus{} c \plus{} d \plus{} 8$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170261


πρέπει τα S-ττ να ικανοποιούν το 2ο όρο που έθεσα στο topic "sta logia" ,
2)Μία επιπλεόν ιδιότητα που πρέπει να έχουν είναι τα τρία τελευταία ψηφία τους να μην είναι όλα μηδέν
???

Θανο ξερεις την απαντηση? :wow:  :what?:


Consider the set $S=\{1,2,\cdots,p-1\}$,(where $p$ is a prime number),let
$a_{k}$ be the minimal number such that $p|k^{a_{k}}-1$,for $k \in S$
Prove:$\max_{k\in S}a_{k}=p-1$
Sorry if it has been posted,Good Luck :)


limit
logarithms
calculus
calculus computations

$ a_1\equal{}1$, $ a_{n\plus{}1}\equal{}a_n\plus{}\frac 1{S_n}$ where $ n\equal{}1,\,2,\ldots$ and $ S_n\equal{}a_1\plus{}a_2\plus{}\ldots \plus{}a_n$. Prove that $ \lim_{n\rightarrow\infty}a_n\equal{}\infty$. Take $ b_n\equal{}n(a_n^2\minus{}a_{n\minus{}1}^2)$; does it have a limit if yes what?

[size=75]$ \log n < a_n \le n$ by induction.
Hence $ a_n\to\infty$ and $ b_n$ cannot have finite limit.

Interesting, what is asymptotics of $ a_n$?[/size]

UPD: No, I'm wrong

I did not think about second part of the problem yet but the first one is simple. Our sequence is increasing, hence it has some limit $ a$. Suppose that $ 0 < a < \infty$. We can rewrite the condition of the problem as follows: $ n(a_{n \plus{} 1} \minus{} a_n) \equal{} \frac {n}{S_n}$. By Shtoltz theorem RHS converges to $ \frac {1}{a}$ thus $ n(a_{n \plus{} 1} \minus{} a_n)\to \frac {1}{a}$. But then for sufficiently large $ n$ we have  $ a_{n \plus{} 1}\ge a_n \plus{} \frac {1}{2na}$. The rest should be clear :)

You don't need Cesaro-Stolz. The sequence $ (a_n)_{n\geq 1}$ is increasing. Assume $ a_n \leq L < \infty$ for all $ n$, so $ S_n \leq nL$. But by iteration, $ a_{n\plus{}1} \equal{} \sum_{k\equal{}1}^n \frac {1} {S_k} \plus{} a_1 \geq a_1 \plus{} \frac {1} {L} \sum_{k\equal{}1}^n \frac {1} {k} \to \infty$ when $ n \to \infty$, contradiction.


logarithms
integration
real analysis
calculus
function
ratio
calculus computations

Having $ f(x)\equal{}\minus{}\ln(1\minus{}x),$ we all know we can find its series representation,

$ f(x) \equal{}  \minus{} \ln (1 \minus{} x) \equal{} \int_0^x {\frac{1}
{{1 \minus{} y}}\,dy}  \equal{} \int_0^x {\left\{ {\sum\limits_{k \equal{} 0}^\infty  {y^k } } \right\}\,dy}  \equal{} \sum\limits_{k \equal{} 0}^\infty  {\left\{ {\int_0^x {y^k \,dy} } \right\}}.$

How does one justify that $ \int_0^x {\left\{ {\sum\limits_{k \equal{} 0}^\infty  {y^k } } \right\}\,dy}  \equal{} \sum\limits_{k \equal{} 0}^\infty  {\left\{ {\int_0^x {y^k \,dy} } \right\}}$? Besides, how do I find the values for $ x$ so that the series converges?

As to the first question, there is a beautiful theorem in the Lebesgue integration theory that says that for every sequence of [b]non-negative[/b] functions $ f_n$ and every measurable set $ E$, one has
\[ \int_E\Bigl(\sum_n f_n\Bigr)\equal{}\sum_n\Bigl(\int_E f_n\Bigr)\]
so the justification reduces to noticing that $ y^k\ge 0$ when $ y\ge 0$.
As to the second question, the ratio test may be useful.

The name: [url=http://en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli.27s_theorem]Tonelli's theorem.[/url] It also works for any sum/integral variant, of course.

Tonelli is sort of overkill though. Tonelli is really hard to prove for general measure spaces, but this follows immediately from monotone convergence theorem or whatever it's called

For power series - and the original post did involve power series - the interchange is justified by uniform convergence as long as you confine yourself to a compact region of radius less than the radius of convergence.


Please try to show all work:
\[\frac{x^{5}+5x^{4}+0x^{3}+0x^{2}+3x+21}{x^{3}+0x^{2}+3x-1}\]
Edit: by the way, the problem originally read:
\[\frac{x^{5}+5x^{4}+3x+21}{x^{3}+3x-1}\]
***Please use division not factoring****

[hide]Both are equal to each other
I think you can factor it.

[/hide]

Division of polynomials is just like that of regular numbers, you multiply the divisor to get as close to the dividend as possible, without going over, and the multiple is part of the answer; then you subract like in regular numerical division.
You have the right idea with the $0x^{n}$ terms, it helps keep the orders of $x$ in place.


Let a,b, and c be non negative real number such that a^2+b^2+c^2+abc=4. Prove that 0 le ab+bc+ca-abc ge 2


inequalities
logarithms
trigonometry
inequalities proposed

I need some proof of this inequality:
Let be given a triangle $ ABC$. Prove that: :
$ sinA \cdot sinB \cdot sinC \le \frac{3 \sqrt{3}}{8}$.

There must be many proofs. Here is mine. The inequality is equivalent to

$ abc/8R^3 < \equal{} 3sqrt(3)/8$ or

$ S/R^2 < \equal{} 3sqrt(3)/4$ or

$ tr/R^2 < \equal{} 3sqrt(3)/4$ or

$ 2(a \plus{} b \plus{} c)r/R^2 < \equal{} 3sqrt(3)$ or

$ 2(2RsinA \plus{} 2RsinB \plus{} 2RsinC)r/R^2 < \equal{} 3sqrt(3)$ or

$ (sinA \plus{} sinB \plus{} sinC)r/R < \equal{} 3sqrt(3)/4$

Now use Jensen's inequality to prove $ sinA \plus{} sinB \plus{} sinC < \equal{} 3sqrt(3)/2$ and the well-known $ r < \equal{} R/2$ and you are done.

$ AM\minus{}GM$ and Jensen works here. :wink:

You can solve it by this following method: By setting x = tan(A/2), y = tan(B/2) and z = tan(C/2). We get the identity: xy + yz +zx = 1. After that, using the well - known identity: (x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) - xyz and AM - GM ineq.

how about taking just log on each side and use a Jensen about $ f(x) \equal{} \log {\sin{x}}$ ?


A chain of length 2d and mass 2m is placed over a frictionless table such that half of the rope is on the table; the other half dangling, and it is just starting to accelerate. Find its acceleration as a function of time.

Let x be the vertical displacement of the chain midpoint. As long as $-d \le x \le +d$, the gravitational force of the hanging part of the chain is

$F = \left(m + \frac x d\ m\right) g = \left(1 + \frac x d\right) mg$

and the mass of the entire chain being accelerated is 2m.

$2ma = F = \left(1 + \frac x d\right) mg$

$\ddot x = a = \left(1 + \frac x d\right) \frac g 2$

$\ddot x - \frac{g}{2d}\ x = \frac g 2$

This is a 2nd order differential equation with constant coefficients and with non-zero right side (non-homegenous). The general solution is the sum of the general solution of the corresponding homgenous equation plus any particular solution of the non-homgenous equation. The characteristic polynomial of the homogenous equation is

$\lambda^2 - \frac{g}{2d} = 0,\ \ \lambda = \pm \sqrt{\frac{g}{2d}}$

and the general solution of the homogenous equation is

$x_0(t) = C_1 \exp\left(+t \sqrt{\frac{g}{2d}}\right) + C_2 \exp\left(-t \sqrt{\frac{g}{2d}}\right)$

where $C_1, C_2$ are arbitrary constants. A particular solution can be found by the variation of the constants $C_1, C_2$ (i.e., by considering them as functions of t, differentiating x(t) twice under this assumption and substituting the result into the non-homogenous equation). But in this case, one particular solution $x_P(t)$ is easy to guess: a non-zero constant. Then $\ddot x_P = 0$ and $x_P(t) = -d$. (This particular solution corresponds to the entire chain lying on the table, with no portion hanging.) The general solution of the non-homogenous equation is then

$x(t) = x_0(t) + x_P(t) = C_1 \exp\left(+t \sqrt{\frac{g}{2d}}\right) + C_2 \exp\left(-t \sqrt{\frac{g}{2d}}\right) - d$

By the initial conditions, $x(0) = 0,\ \dot x(0) = 0$, hence

$C_1 + C_2 - d = 0,\ \ C_1 \sqrt{\frac{g}{2d}} - C_2 \sqrt{\frac{g}{2d}} = 0$

It follows that $C_1 = C_2 = \frac d 2$ and

$x(t) = d\ \frac{\exp\left(+t \sqrt{\frac{g}{2d}}\right) + \exp\left(-t \sqrt{\frac{g}{2d}}\right)}{2} - d = d \left[\cosh \left(t \sqrt{\frac{g}{2d}}\right) - 1\right]$

$a = \ddot x = \left(1 + \frac x d\right) \frac g 2 = \frac g 2 \cosh \left(t \sqrt{\frac{g}{2d}}\right)$

This is the chain acceleration as long as $x \le d$ or

$d \left[\cosh \left(t \sqrt{\frac{g}{2d}}\right) - 1\right] \le d,\ \ \ \cosh \left(t \sqrt{\frac{g}{2d}}\right) \le 2,\ \ \ t \le t_{\text{max}} = \sqrt{\frac{2d}{g}} \cosh^{-1} 2$

Since the inverse of $\cosh x$ is

$\cosh^{-1} x = \ln\left(x + \sqrt{x^2 - 1}\right)$,

the maximum time is

$t_{\text{max}} = \sqrt{\frac{2d}{g}} \ln \left(2 + \sqrt 3\right)$

For time $t > t_{\text{max}}$, the chain acceleration is simply equal to the gravitational acceleration g.


First use the rational root theorem and try to find some upper and lower bounds.

[quote="chess64"][hide](E), x = -18

You can just use a graphing calculator.   :P  I think that's the fastest way--it only takes about half a minute.

[/hide][/quote]

That's not the point.

[quote="Psycho"]The following function has only one real zero. Which of the answers contains that number?

\[f(x)=x^5+20x^4+44x^3+148x^2+90x+324\]

\[(A) \{-15, -1, 12, 18\}\]   \[(B) \{-324, -3, 6, 54\}\]   \[(C) \{-81, -24, 4, 36\}\]   \[(D) \{-72, -27, 3, 162\}\]   \[(E) \{-54, -18, 0, 5\}\]

off course, you can just check them all  :?  but i don't think that's the point, since you only have a limited amount of time :)
i'm curious about the methods you will come up with ;)[/quote]

This is just a seek for elimination. Positive numbers cannot be in, the numbers <-30 are trivially impossible, those >-5 as well, so that leaves us -15,-18,-27 and -24 to investigate. Or number theory fans like me will first see that it needs to be even and only then it needs to be negative :blush: so it's either -18 or -24, and I have to say I'd just fill them in at the contest, since it's faster than seeking anoher good modulo.

well, how can you solve it?? Can you give a detailed solution?

[quote="Peter VDD"]well I told you didn't I?

My full solution
* >0 means everything above 0 so never f(x)=0
* <-21 neither since 21>20,44*21>148,90*21>324 thus the polynome is strictly negative
* if x is integer (which all options are) then x must be even due to the coefficients.

So that leaves x=-18 as the only solution.[/quote]
i solved it almost the same way :)
first, get rid of all positive numbers
second, i knew it had to be even
and last, i knew that the biggest part of the value is determined by the \[x^5+20x^4\] so x had to lie as close as possible to 20... the first number i checked was -18 and it was correct immediatly :p

[quote="Peter VDD"]who are you actually? I see you got internet so you're not on IMO stage at the moment?[/quote]
me? i'm Thomas Decru... i highly doubt you know me :)
reached the FJMO-final two years ago (third place) and last year the FMO-final as well (no place  :lol: ). and if i didn't copy any answers wrong on my answering sheet, FMO-final this year again *little cheer :p *
(btw, the 101 boarder is really low this year, man  :?  it wasn't that hard at all imo... although all these things are totally off-topic ;) )

Indeed I don't know you... therefor I wondered who you would be (knowing mathlinks but not being on IMO training stage which is this weekend)

So how do you know about this site? Ah well you're right we're going off topic. Feel free to continue in PM :)

< and besides you should watch out saying 'imo' on this site, since it means the international olympiad ;) >


A circle has its center at P$ (0,\minus{}6)$. Given that the circle intersects the origin, what is the number of square units in its area? Express your answer in terms of $ \pi$.

The center of the circle is $ (0,\minus{}6)$, and a point that the circle intersects is $ (0,0)$.  Since the distance from $ (0,\minus{}6)$ and $ (0,0)$ is $ 6$ units, the radius of the circle is $ 6$ units.  Thus, the area of the circle is $ \pi 6^2\equal{}\boxed{36\pi}$


Hello,

here's something from a book called '100 Great Problems of Elementary mathematics' by Heinrich Dorrie, about how to derive a closed form for the expressions $ 1^k + 2^k + \cdots + n^k $. It uses a 'trick' which regards the values $ d^1, d^2, \ldots, d^k $ stemming from the binomial expansion of $ (n+d)^k $ not as powers of $ d $, but as variables subject to $ i $ constraints, where $ i $ is the respective exponent. I'm not fully convinced by it, and was hoping someone could shed light on this:

Binomial expansion tells us that

$ (v+d)^{k+1} - (v+d-1)^{k+1} = {k+1}v^{k} +
C^{k+1}_{2}v^{k-1}(d^2-(d-1)^2) +
C^{k+1}_{3}v^{k-2}(d^3-(d-1)^3) + \cdots $

We now define $ d^2 = (d-1)^2, d^3 = (d-1)^3 $, etc. The expression
simplifies to

$ (v+d)^{k+1} - (v+d-1)^{k+1} = {k+1}v^k $
$ \Rightarrow \sum^{n}_{v=1} v^k =
\frac{1}{k+1}\sum^{n}_{v=1}((v+d)^{k+1} - (v+d-1)^{k+1}) =
\frac{(n+d)^{k+1} - d^{k+1}}{k+1} $

The $ d^i $ are found as follows:

$ (d-1)^2 = d^2 \Rightarrow -2d^1 + 1 = 0 \Rightarrow d^1 = \frac{1}{2} $
$ (d-1)^3 = d^3 \Rightarrow -3d^2 + 3d^1 - 1 = 0 \Rightarrow -3d^2 + 3(\frac{1}{2}) - 1 = 0\Rightarrow d^2 = \frac{1}{6} $
$ (d-1)^4 = d^4 \Rightarrow -4d^3 + 6d^2 - 4d^1 + 1 = 0 \Rightarrow - 4d^3 + 6(\frac{1}{6}) - 4(\frac{1}{2}) + 1 = 0\Rightarrow d^3 = 0 $
etc etc. It turns out that the $ d^i $s are the Bernoulli numbers.

So say you want to find $ 1^2 + 2^2 + \cdots + n^2 $ We have

$ \sum^{n}_{v=1} v^2 = \frac{(n+d)^{3} - d^{3}}{3} = \frac{n^3 + 3n^{2}d^1 + 3nd^2}{3} $
and substituting $ d^1, d^2 $ and simplifying gives us $ \frac{1}{6}(n)(n+1)(2n+1) $, which turns out to be correct.

Any thoughts?

Welcome to mathlinks!

I'm not sure about your notation, but the idea is correct.  Maybe it will be clear if we do a simple example.

Suppose we want to find $\sum_{i = 1}^n i$.  For all $i$, we have $(i + 1)^2 - i^2 = 2i + 1$.  Summing this from $i = 1$ to $n$, the sum on the left telescopes, to get
\[(n + 1)^2 - 1 = 2 \sum_{i = 1}^n i + \sum_{i = 1}^n 1,\]
which becomes
\[n^2 + 2n = 2 \sum_{i = 1}^n i + n.\]

Therefore,
\[\sum_{i = 1}^n i = \frac{n^2 + 2n - n}{2} = \frac{n(n + 1)}{2}.\]

Similarly, to find $\sum_{i = 1}^n i^2$, take the identity $(i + 1)^3 - i^3 = 3i^2 + 3i + 1$ and again, sum from $i = 1$ to $n$ to get
\[(n + 1)^3 - 1 = 3 \sum_{i = 1}^n i^2 + 3 \sum_{i = 1}^n i + \sum_{i = 1}^n 1.\]
Re-arrange to isolate $\sum_{i = 1}^n i^2$.  Higher powers are done similarly.  Note that using this method, to find the sum of the $k$th powers, one must first find the sum of the 1st, 2nd, ..., $(k - 1)$st powers.

Hello, thanks.

I'm familiar with your derivation of the power sum, but it hadn't occured to me before that both methods might be essentially the same.

What I can't put my finger on tho is why

[quote]We now define $ d^2 = (d-1)^2, d^3 = (d-1)^3 $ [/quote]

works, if not in general, then at least in this context. (clearly this isn't consistent given our usual rules of arithmetic). What is the 'mechanism' that lies behind it? Maybe there is some way to 'map' my argument (or rather, Heinrich Dorrie's) into yours, or vice versa?

Sorry, not sure if what I'm trying to say is clear here....


post your highscores for this game
________________________
[url]http://www.ferryhalim.com/orisinal/g3/bells.htm[/url]
________________________
mine's 43200 at the moment

-jorian

3,840,000.

A friend of mine recently got 14 quadrillion.   :!:

Here is a video of 74 trillion:

http://www.youtube.com/watch?v=o6hvn2P_2VU&mode=related&search=

Edit:  Ooh, here's another of 16 quadrillion  :!:  :!: :

http://www.youtube.com/watch?v=7cN75aNiyuk&NR

That one is GREAT.  15760 atm, I just started 5 minutes ago ;P.

 :D  Edit:  117040.

Edit:  Above was on a laptop  :D

It's great how your score goes up exponentially. I got into 7 digits.

BTW: log_2(16,000,000,000,000,000/10) is approximately 51, which is a bit more than the number of birds you have to fly into in order to get 16 quadrillion points. Doesn't seem too impossible.

im on a laptop so it's not that hard to get a high score! :mad:
-jorian :blush:

210000

 :rotfl:

Dude how did all of you get into 7 digits. wow lol z

Bleh, my hand started cramping and it was hard to readjust so i messed up

48,028,228,060

OMG!!!

What is the trick to this game?  I know you just have to concentrate reallly hard and randomly twitch the mouse.  maybe i'm doing badly because i'm on a laptop -_-.

I dunno, take it easy?

adn get off a laptop if yo'ure using the touchpad thing

Haha, good idea.  I went on a PC for like, 5 minutes and got something in the 150,000's.

This game is too addicting, love it.

...Valentin posted it in his blog...that's where I discovered it.

-jorian

I just got 35 quadrillion...I dont know the exact #'s since it doesn't fit in the box..

God does my hand hurt

edit> it might have been 350 quad since I was getting 11,500 per bell and the guy in the 16quad video was only getting 9500ish per bell

edit2> yea probably 350 quad since 11500/bell was around the 200-300quad point level in another video..

so yea i'm done with this game

ya, but one miss of a bird could mess u up

-jorian

:clap2:

Ah, 136000....000 on the first try...miraculous if not impossible.

In fact, it is easy to prove that a number with that many zeros can not be formed by a sum of perfect numbers to powers of two times ten (if you had played the game, you would have realized this is how the score is taken).
Look at my screenshot to see what it actually looks like...

you mean a triangular number not a perfect number.

1695340 - still a n00b.

Anirudh, you just got your photoshop "pwndizzled" by me@home's awesome math skills.

[quote="me@home"]Ia sum of perfect numbers to powers of two times ten (if you had played the game, you would have realized this is how the score is taken).[/quote]

Unless you intentionally skip either bells or birds.

The score is still obviously fake, though.  Unless you can explain the incredibly low Previous High away.

lol. It was photoshoped, merged, and al other stuff so anti-forge couldn't detect it...but I was foolish in teh numbers :rotfl:  :blush:

The bells get soo small once you're up high.

Birds double scores, so scores increase effectively exponentially the higher up you go.  To make a more reasonable comparison betweek scores, take a logarithm.  :D

250,530. I played this game about 50 times

[quote="anirudh"][img]http://theani.isgreat.org/php/uploads/ani/winterbell.gif[/img]

I used printkey to capture the high score.

And on a desktop[/quote]

you know what that score looks remarkably like 1360, which is a possible score.  You can just capture the 0s in paint and add them to the score if done right it looks right.  And also, the "New high score" seems chopped at the "g" as if... you know what.  You could have stolen that one from an old "new high score" thing and then saved it in paint.
it's all a fake!!!!!!   :o

awesome game  :D

unfortunately my high score is only 21120

edit- managed to get up to 58770


induction
modular arithmetic
geometry
3D geometry
algebra
difference of squares
special factorizations

[hide="Or..."] Recall that $a^{n}-b^{n}= (a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$. [/hide]
[hide="Or..."] [b]Lemma:[/b]  $a \equiv b \bmod m, c \equiv d \bmod m \implies ac \equiv bd \bmod m$

[b]Lemma:[/b]  $a \equiv b \bmod m \implies a^{n}\equiv b^{n}\bmod m$ [/hide]

without looking at the hints,

we have p(1) is true, so the induction is anchored.

Note that 8^(k+1)-2^(k+1)=8(8^k-2^k)+8(2^k)-2^(k+1)

=8(8^k-2^k)+8(2^k)-2^(k)=8(8^k-2^k)+6(2^k).

Now 8^k-2^k is divisible by 6 and we've shown that 8^(k+1)-2^(k+1) is also divisible by 6, but this is p(k+1).

[quote="mdk"]Prove that all numbers of the form 8^n-2^n are divisible by 6.[/quote]

Extension: Prove that all numbers of the form $x^{n}-y^{n}$ are divisible by $x-y$ (unless $x=y$).

Easy.
If n is even, then we do a difference of squares reduction until we have the term $(x-y)(bleh)$
If n is odd, then we factor as a difference of cubes, and squares if we must, until we have the term $(x-y)(bleh)$

[hide="proof through induction"]
Base case, for $n=1$:
$x-y \mid x-y$
Assumption, for $n=k$:
$x^{k}-y^{k}\mid x-y$
or
$\frac{x^{k}-y^{k}}{x-y}= r$ for some integer r.
Let $n = k+1$:
$x^{k+1}-y^{k+1}= (x^{k})x-(y^{k})y$.
Rearranging our inductive hypothesis yields:
$x^{k}=r(x-y)+y^{k}$
Therefore, $x^{k+1}-y^{k+1}= (r(x-y)+y^{k})x-(y^{k})y$
$=xr(x-y)+(y^{k})x+(y^{k})(-y)$
$=xr(x-y)+y^{k}(x-y)$
$=(xr+y^{k})(x-y)$
Q.E.D.
[/hide]

[quote="DiscreetFourierTransform"]If n is odd, then we factor as a difference of cubes[/quote]

What if $3 \not | n$?  And what, exactly, is so bad about my solution in the second post?


Show that every Hilbert space is reflexive

Show that the Lp is reflexive for 1<p<inf

This follows from the fact that $\ell_{p}^{*}= \ell_{q}$ where $1/p+1/q = 1$ ($1< p, q < \infty$).


I'm really having confusing on the trig.  I get all these things like sine, cosine, tangent, cosecant, secant, and cotangent but I'm confusing on following terms which stuck me pretty bad.

* Reference angle *
* Periodic *
* Period *
* Amplitude *

Plus, I have some questions to ask!  :lol:

P.S. Directions are copied too.

1. Use a calculator to approximate two values of  :theta: (0  <=  :theta: < 2 :pi: ) that satsify the equation.  Round to three decimal places.

* cos  :theta: = 0.9848 --> Now, I got 0.175, which is one value. But I just can't other value.

2. Find the point  (x, y) on the unit circle that corresponds to the real number t.  Use the result to evaluate sin t, cos t, and tan t.

Now, I got sin, cos, tan pretty easily.  But I just can't get (x, y).

3. Approximate the solution of the equation where 0  <= t < 2 :pi:.

They actually gave two values for each answer but I can only do one.

a. sin t = 0.25 --> 0.25
b. cos t = -0.25 --> 1.82

Really big help here!! ;)

Thanks a lot  :P

Hint for #1: Remember [tex]\cos\theta=\cos(2\pi-\theta)[/tex]

2.  My guess is that polar coordinates would help: our point in polar coordinates is (1, theta).  Just put it in rectangular coordinates.

[quote="Silverfalcon"]Huh? Polar coordinate? I don't think I know it, even I heard it.

And beta, I don't really get what you mean because I did what you told me to do, but it doesn't give the correct answer. I got 0.55something like that but another answer's way bigger than that.[/quote]

If you did it right that number should also give the correct answer. Remember most trig functions are many to one mapping because of the nature of the triangles.

[tex]2\pi=360\degrees[/tex] in radian measure.

For #3, remember what beta says.

cos :theta:  = cos ( 2pi - :theta: )

also, sin :theta:  = sin (pi -  :theta: )

Hope this helps.

Or in general,

cos x = cos (2a.pi +/- x), where a is an integer and x is measures in radians.

as the period is 2pi and the reference angle I guess is 0. (since cos x is an even function of x).

Sorry this may be wrong, it is more me using common sense to make formulas I've never seen before which rarely works!

[quote]3. Approximate the solution of the equation where 0  t < 2 .

They actually gave two values for each answer but I can only do one.

a. sin t = 0.25 --> 0.25
b. cos t = -0.25 --> 1.82 [/quote]

You can think about it in terms of a formula, but it's easier to imagine which quadrant sine :theta: , cos :theta: , and tan :theta:  are positive.

You have the equation [tex]\sin{t}=0.25[/tex].

Sine is positive in quadrant I (all positive there), and quadrant II (only sine is positive there).

So, you would have
[tex]t=\arcsin{0.25}[/tex] and [tex]t=\pi-\arcsin{0.25}[/tex].

Do you see why?  The arcsine will solve the equation for you between [tex][\frac{-\pi}{2},\frac{\pi}{2}][/tex], and then you need to use the magnitude of the angle and make it and angle [tex]\arcsin{0.25}[/tex] from the x-axis.

Since the trig functions are periodic, the solutions are:
[tex]t=\arcsin{0.25}+2k\pi[/tex]
and
[tex]t=\pi-\arcsin{0.25}+2k\pi[/tex]

Hope this helps.


trigonometry
algebra
function
domain
geometry
trig identities
Law of Sines

You're given three lengths, x, y, and z, and two angles A and B. They satisfy the following conditions:

0:^o: < A, B < 90:^o:
x2 = y2 + z2
x = y cos A + z cos B
y sin A = z sin B

Prove that A + B = 90:^o:.

Well, clearly those conditions are going to give us a right triangle.  "Clearly," of course, being my way of saying "I know it's true but not how to prove it."  So lets see if I can do better than that:

First, we know sin(A), cos(A), sin(B), and cos(B) are all positive numbers.  This is nice.
Take x = y*cos(A) + z*cos(B) and square it.
x:^2: = y:^2:*cos:^2:(A) + 2yz*cos(A)cos(B) + z:^2:cos:^2:(B)
y:^2: + z:^2: = y:^2:*cos:^2:(A) + 2yz*cos(A)cos(B) + z:^2:cos:^2:(B)
y:^2:*sin:^2:(A) + z:^2:*sin:^2:(B) = 2yz*cos(A)cos(B)
2y:^2:*sin:^2:(A) = 2yz*cos(A)cos(B)  and  2z:^2:*sin:^2:(B) = 2yz*cos(A)cos(B)
Taking y, z as positive (the degenerate cases could be taken care of seperately, if anyone cared -- I'm just going to ignore them because they aren't very important), we have that
y*sin:^2:(A) = z*cos(A)cos(B)  and  z*sin:^2:(B) = z*cos(A)cos(B)
Multiplying these two equations together, we get
yz*sin:^2:(A)sin:^2:(B) = yz*cos:^2:(A)cos:^2:(B)
sin:^2:(A)sin:^2:(B) = cos:^2:(A)cos:^2:(B)
Since everything is positve, just drop the squares.
sin(A)sin(B) = cos(A)cos(B)
0 = cos(A)cos(B) - sin(A)sin(B)
0 = cos(A + B)
A + B = 90:^o: as desired (because of the domain limitation).

I am afraid my solution is a bit simpler than that...

Okey, seems like noone's interested in this problem any more...

Solution:

[hide]Rearrange y sin A = z sin B to get y / sin B = z / sin A. This is the sine law. Thus it's possible to make a triangle with one side y, another z, the third some positive length a, the angle between a and y A, the angle between a and z B. Then by cosine law:

z2 = a2 + y2 - 2ay cos A,

y2 = a2 + z2 - 2az cos B.

Rearrange:

y cos A = (a2 + y2 - z2) / 2a,

z cos B = (a2 + z2 - y2) / 2a.

Thus y cos A + z cos B = a. However, we're given that y cos A + z cos B = x; thus x = a. Then, the triangle really has sides x, y, and z; we're given x2 = y2 + z2, so x must be the hypotenuse and y, z right sides. Thus A + B = 90o as required.[/hide]

That isn't that much nicer -- it is neat that you can build up the triangle itself, however.  I didn't think of using the law of sines to show it -- very clever.  Where's this problem from?


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\pi} x|\sin nx|\ dx\ (n\equal{}1,\ 2,\ \cdots)$.

$ \int x\sin nx\ dx \equal{} \frac{1}{n^2}\sin(nx) \minus{} \frac{x}{n}\cos(nx)$
$ \int_0^{\pi} x|\sin nx|\ dx\ \equal{}$
$ \sum_{k\equal{}0}^{n\minus{}1} |\frac{1}{n^2}\sin((k\plus{}1)\pi) \minus{} \frac{(k\plus{}1)\pi}{n^2}\cos((k\plus{}1)\pi) \minus{}
\frac{1}{n^2}\sin(k\pi) \plus{} \frac{k\pi}{n^2}\cos(k\pi)| \equal{}$
$ \sum_{k\equal{}0}^{n\minus{}1}\frac{(2k\plus{}1)\pi}{n^2} \equal{} \pi$.


inequalities
triangle inequality
complex analysis
complex analysis unsolved

A)
If f(z) has a power series expansion (sum) an z^n,  from n=0 to infinity, with an  [u]>[/u] 0, prove that
|f(w)| [u]<[/u] f(|w|) for |w| < R.

B) SHow that |e^(w) -1 | [u]<[/u] e^(|w|) - 1

Thank you for your input.

I will try to prove it for problem number (1) by using triangle inequality
[hide="Hint"]
Note that for any $ w<R$, $ |f(w)|\equal{}|a_0\plus{}a_1w\plus{}a_2w^2\plus{}a_3w^3\plus{}\cdots\plus{}a_n.w^n|\leq |a_0|\plus{}|a_1\parallel{}w|\plus{}|a_2\parallel{}w^2|\plus{}|a_3\parallel{}w^3|\plus{}\cdots\plus{}|a_n\parallel{}w^n|\equal{}a_0\plus{}a_1|w|\plus{}a_2|w|^2\plus{}a_3|w|^3\plus{}\cdots\plus{}a_n|w|^n\equal{}f(|w|)$
For part (B), you can use part (A) by setting $ f(z)\equal{}e^z\minus{}1$
[/hide]


geometry
rectangle
number theory
relatively prime
combinatorics unsolved
combinatorics

A rectangle $ABCD$ has side lengths $AB = m$, $AD = n$, with $m$ and $n$ relatively prime and both odd. It is divided into unit squares and the diagonal AC intersects the sides of the unit squares at the points $A_1 = A, A_2, A_3, \ldots , A_k = C$. Show that $A_1A_2 - A_2A_3 + A_3A_4 - ... + A_{k-1}A_k = {\sqrt{m^2+n^2}\over mn}. $

Usually, if you have questions about a problem which has recently been used in such a prestigious contest as the Balkan Olympiad, you'll find it on the forum if you look for it :).

No, China TSTs have no this problem.

In Mathlinks, a topic has been posted about this problem, but with only outlines of idea, and Myth had proved a relative problem.Can someone post clear solutions?

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Balkan&t=5332


1.  Using the digits 1,2,3,4,5 and 6 once and only once, find the 6 digit-numbers in which the numbers created from the first two digits is divisble by 2, the number created from the first three digits is divisible by 3, the number created from the first four digits is divisible by 4, the number created from the first five digits is divisible by 5, and the number itself is divisible by 6.

  2.  Using the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 once and only once, find 9-digit numbers in which the number created from the first two digits is divisible by 2, the number created from the first three digits is divisible by 3, ..., the number created from the first eight digits is divisible by 8, and the number itself is divisible by 9.

  P.S:  Is it just me, or has this forum been very inactive for awhile, by "inactive" i mean no one seems to be posting problems, in comparison to how many problems there are being posted in the basics and classroom forums.

[hide="1"]The fifth digit has to be 5, the 2nd, 4th, and 6th digits must be from the set {2,4,6}. Let's start with a two digit number that is even. Let's pick 12. Then, find a third digit that makes the three digit number divisible by three. Our choices are 3 and 6. We sort of reserved the 6 for either the 4th and 6th digits, so pick 3. The we must pick 6 for our next digit, since that's the only number that will make the 4 digit number divisible by 4. THen the 5 goes in the fifth digit. That leaves 4 to be put in the 6th spot, and we see that [b]123654[/b] is divisible by 6 also.

More can find other numbers if they want.[/hide]

[hide="2"]This is so nasty that I won't explain it fully, but we still know that the 5 belongs in the 5th digit, even digits belong in even places, etc.
Actually, I haven't found such a number yet. [/hide]


function
probability
calculus
integration
complex analysis
probability and stats

Suppose that $ X$ and $ Y$ are two independant random variables and that $ X\plus{}Y$ has a normal distribution.
Show that $ X$ and $ Y$ are two normal distributions ?

There's no reason this should be true.

Let's look on the characteristic function side.

Let $ X$ be a r.v. with characteristic function $ \chi_X(\omega)\equal{}e^{\minus{}\omega^2/4}e^{|\omega|}.$

Let $ Y$ be a r.v. with characteristic function $ \chi_Y(\omega)\equal{}e^{\minus{}\omega^2/4}e^{\minus{}|\omega|}.$

We have no trouble taking inverse Fourier transforms of that; these are probability distributions, and they're not normal.

Then if $ X$ and $ Y$ are independent and $ U\equal{}X\plus{}Y,$ we have that $ \chi_U(\omega)\equal{}e^{\minus{}\omega^2/2},$ and that makes $ U$ normal.

Or make it $ \chi_X(\omega)\equal{}e^{\minus{}\omega^2/4}(1\plus{}\omega^2)$ and $ \chi_Y(\omega)\equal{}e^{\minus{}\omega^2/4}\frac{1}{1\plus{}\omega^2}.$ That way, $ X$ and $ Y$ even have finite moments.

but $ X$ and $ Y$ have to be independant. In your examples, I think that they are not. ( I was very surprised too by this result)

Of course they're independent! $ X$ has a certain density function $ f(x)$ which we can find by taking the inverse Fourier transform of the formula I gave, and $ Y$ has a certain density $ g(y),$ and then $ (X,Y)$ has joint density $ f(x)g(y).$ That's independent.

The characteristic function of the sum of independent random variables is the product of the two characteristic functions. As you can see, this gives us great freedom to construct things.

Now if you want $ X$ and $ Y$ to be independent [b]and to have the same distribution[/b], that would be a completely different question. That takes away all of that freedom. Then, on the characteristic function side, if $ X\plus{}Y$ had characteristic function $ e^{\minus{}c|\omega|},$ then the characteristic functions of $ X$ and $ Y$ would have to be the same, and would have to then be the square root of this, namely $ e^{\minus{}c|\omega|/2}.$ Then $ X$ and $ Y$ would each be normal (with half the variance of $ X\plus{}Y.$)

I am very sorry  :blush:  I forgot to say that these random variables are real. (this is why the examples with the inverse Fourier transform do not work). Sorry to have read your post too quickly.

My examples are real-valued. The Fourier transform (or inverse Fourier transform) of a real, even function is real and even.

-----

However: Now that I step back, I see that there is one key element that I haven't proved, which is that my examples are positive. I know the inverse transforms are real, and I know they have integral $ 1,$ since that is the value of the transform when $ \omega\equal{}0.$ But I admit that I haven't checked that they are positive, which is the last key element needed for them to be probability measures.

There's a theorem that an even function which is convex on $ [0,\infty)$ must have a positive Fourier transform, but these examples don't quite meet that criterion.

I still think there's plenty of freedom to have non-normal RV's add up to be normal, but I haven't yet quite proved that I have counterexamples.

mlok, fedja - what do you think?

[quote="Kent Merryfield"]Or make it $ \chi_X(\omega) \equal{} e^{ \minus{} \omega^2/4}(1 \plus{} \omega^2)$ [/quote]Not bounded by 1 in absolute magnitude.

I could fix that by making it something like $ e^{\minus{}\omega^2/4}\left(1\plus{}\frac{\omega^2}{400}\right).$

But I still don't know whether that's positive, and perhaps it isn't.

[quote="Yustas"]The statement is true under the hypothesis of independence. This is very famous result and it is not easy at all, you can find the proof in Feller.[/quote]
Yes, the result is due to Cramer (1936).  The proof in Feller (Volume II, Chapter 15) is interesting and pretty short.  The idea is that the characteristic function of a normal distribution is never 0.  Thus the characteristic functions of $ X$ and $ Y$ are never 0.  Also, we can show that the characteristic functions of $ X$ and $ Y$ don't grow much faster than $ e^{|\omega|^2}$.  Using a result from complex analysis, we can then show that the characteristic functions of $ X$ and $ Y$ are of the form an exponential of a quadratic.  From there, we can quickly show that $ X$ and $ Y$ are normal.


calculus
integration
function
limit
real analysis
real analysis unsolved

Consider the continuous functions $ f: [0,\infty) \rightarrow R$ and $ g: [0,1] \rightarrow R.$ If $ \mathop {\lim }\limits_{x\rightarrow\infty}f(x) \equal{} L \in R,$ prove that
\[ \mathop {\lim }\limits_{n\rightarrow\infty}\frac {1}{n}\int_{0}^{n}f(x)g(\frac {x}{n})\,dx \equal{} L\int_{0}^{1}g(x)\,dx\]
I think I found a proof of this. When I took a peek at the solutions however, this problem was done differently I think. Besides a change of variables in an integral, all I did was use epsilon-delta arguments and basic ways to estimate integrals. Can anyone else confirm this approach or did I mistakenly think this problem was a bit simple (which is quite possible)? Thank you in advance.

This may help you.

We have
\[ \frac{1}{n}\int_{0}^{n}f(x)\cdot g\left (\frac{x}{n}\right )\,\mathrm{d}x\equal{}\int_{0}^{1}f(nt)\cdot g(t)\,\mathrm{d}t.\]
Now, the original problem is equivalent to
\[ \lim_{n\to\infty}\left |\int_{0}^{1}(L\minus{}f(nt))\cdot g(t)\,\mathrm{d}t\right |\equal{}0.\]

Let $ u \equal{} \frac {x}{n}$. Then $ dx \equal{} ndu$.

So $ \lim_{n \to \infty} \frac {1}{n} \int_{0}^{n} f(x)g(\frac {x}{n}) dx \equal{} \lim_{n \to \infty} \frac {1}{n} \int_{0}^{1} f(nu)g(u) ndu \equal{} \lim_{n \to \infty} \int_{0}^{1} f(nu)g(u)du$

So we want to show that $ \int_{0}^{1} (f(nu) \minus{} L)g(u)du \to 0$ as $ n \to \infty$.

Let $ 0 < \epsilon < 1$

Now since $ f(x) \to L$ then there exists $ X \in \Re$ such that $ \forall x \geq X$ we have $ |f(x) \minus{} L| < \epsilon$. So we choose $ n$ in the above such that $ X/n < \epsilon$. Then for $ u \in [\epsilon,1]$ we have $ |f(nu) \minus{} L| < \epsilon$. Let $ \sigma \equal{} sup\{|f(x) \minus{} L|: x \in [0, \infty)\}$. This certinaly exists since on the interval $ [0,X]$ the function is bounded above (from the theorem: a continuous function on a closed bounded interval is bounded, and attains its bounds) and in the interval $ [X,\infty)$, the function is bounded above by $ \epsilon$. $ \eta \equal{} \sup\{|g(x)|: x \in [0,1]\}$ also exists. So we can therefore say that for this $ n$, and all greater values of $ n$:

$ \int_{0}^{1} (f(nu) \minus{} L)g(u)du \equal{} \int_{0}^{\epsilon} (f(nu) \minus{} L)g(u)du \plus{} \int_{\epsilon}^{1} (f(nu) \minus{} L)g(u)du$ $ < \epsilon \sigma \eta \plus{} (1 \minus{} \epsilon)\epsilon\eta$

This can be made as small as we like (by taking $ \epsilon$ sufficiently small, and $ n$ sufficiently large) since $ \sigma$ and $ \eta$ are independent of $ \epsilon$.

Thanks guys, this is the exactly the solution I had in mind. The official solution does not make a change of variables, but rather defines a new function. It also applies l'Hopital's rule.

I believe you can also do a Stolz-Cesaro type proof

why not using Lebesgue Control convergence theorem.


Find all funtion $ f : R \mapsto R$:
              $ f(f(x)+y)=f(f(x)-y)+4 f(x)y$   for all x,y \in{R}

$ y=f(x),f(0)=a$ gives $ f(2f(x))=a+(2f(x))^{2}$
then $ \forall x\in \{2f(x)|\ x\in \mathbb{R}\}: \ f(x)=a+x^{2}$
NOW prove that  $ \{2f(x)|\ x\in \mathbb{R}\}=\mathbb{R}$  :wink: and find $ a$ by the relation $ f(f(x)+y)=f(f(x)-y)+4f(x)y$  :wink:

It's easy to see that,but when we solve,it's not as simple as we feel.Can anyone please post it's solution ?

[quote="aviateurpilot"]NOW prove that  $ \{2f(x)|\ x\in \mathbb{R}\}=\mathbb{R}$  :wink: [/quote]
for this you have to prove that $ f$ is surjective,but I think its not that simple,or maybe Im missing something stupid here...  :oops:
anyway @Nbach:as red_dog said,we prove that $ f(x)=x^{2}+a$ now put this in the original equation,and it becomes:
$ f(x)^{2}+y^{2}+2f(x)y+a=f(x)^{2}+y^{2}-2f(x)y+a+4f(x)y$
which is true for every $ a$ so we get that:
$ f(x)=x^{2}+a$ where $ a \in \mathbb{R}$

[quote=pco][quote="red_dog"]Let $ f(x)+y=u, \ f(x)-y=v$
Then the equation becomes $ f(u)=f(v)+u^{2}-v^{2}$
For $ v=0$ let $ f(0)=a$
Then $ f(x)=x^{2}+a$[/quote]

Yes, but you did not show that the equation $ f(u)=f(v)+u^{2}-v^{2}$ was true for any real $ u,v$.
You just showed that it was true for $ u,v$ such that it exist $ x,y$ such that $ f(x)+y=u$ and $ f(x)-y=v$.

For example, the solution $ f(x)=0$ $ \forall x$ can not be found by this way.[/quote]

but is there a way to make his solution i correct....i mean can we show by proving surjectivity that $ f(u)=f(v)+u^{2}-v^{2}$ is true for any real $ u,v$. .i already read your solution but i want to understand  this also...


integration
limit
calculus
calculus computations

The antiderivative is on the standard list, and those are standard values. These should take no more than five seconds each.

[hide="Click here"]$ \int \frac{1}{1\plus{}t^2}\,dt \equal{} \arctan t \plus{} C$, therefore,

$ \int_0^1 \frac{1}{1\plus{}t^2}\,dt \equal{} [\arctan t]_0^1 \equal{} \frac{\pi}{4}$

$ \int_0^{\infty} \frac{1}{1\plus{}t^2}\,dt \equal{} \lim_{a \to \infty} \int_0^a \frac{1}{1\plus{}t^2}\,dt \equal{} \lim_{a \to \infty} [\arctan t]^a_0 \equal{} \lim_{a \to \infty} \arctan a \minus{} 0 \equal{} \frac{\pi}{2}$[/hide]


how do you solve this problem?

On each tile in a set of dominoes, each integer 0 through 12 is paried exactly once with each integer 0 through 12. One number of the pair is written on one half of the domino, and the other number is written on the other half. If one domino is randomly selected from all dominoes in this set, what is the probability that the same number will be on both halves? Express your answer as a common fraction.

[hide="answer"]$ \frac{1}{7}$[/hide]

[quote="FantasyLover"]how do you solve this problem?

On each tile in a set of dominoes, each integer 0 through 12 is paried exactly once with each integer 0 through 12. One number of the pair is written on one half of the domino, and the other number is written on the other half. If one domino is randomly selected from all dominoes in this set, what is the probability that the same number will be on both halves? Express your answer as a common fraction.

[hide="answer"]$ \frac {1}{7}$[/hide][/quote]

theres 13+12+11+...+2+1 dominoes total...=91

13 of them have same number (0,0) (1,1) (2,2) etc.
so 13/91=1/7

blah... misinterpreted problem...

thought $ \frac{13}{169}$...

thanks aime

I did too...I hate problems with weird wordings and verbiage. :wink:

Another way to think about this problem:

If we pair all of them: (1,1),(1,2),...(2,1),(2,2)...
then there is:
1/13 chance of getting same numbers
6/13 chance of A>B
6/13 chance that A<B

Since we doublecounted A>B and A<B, we subtract 6/13.

Hence our answer is 1/(1+6) = 1/7

I seirously wanna say that james solution was way better....
thats what you get from usamo qualifiers :D

[quote="james4l"]Er, how did you know I was a USAMO qualifier? My solution works if you know what you're doing...[/quote]

pythag011 told me.

Yes i know your way works...thats whyi was commending it.


Hello,

I understand Serbia is sending a team of six people.  Now what is interesting is that the official IMO site [url]http://imo2006.dmfa.si/participants/SER.html[/url]  calls it "Serbia".

Until recently the country was "Serbia-Montenegro" but now Montenegro has become independent.  Does anyone know of the implications of this?
Are the members of this IMO 2006 team all from Serbia itself, or also from Montenegro?  Was a selection organized in Montenegro for 2006?

And will there be joint participation next year, or simply no participation from one of those countries or even both?

Serbia and Montenegro are two independent countries now and they have separate teams. The team that contested on this IMO was just from Serbia. And in future they will have separate teams.

You are certain?  When did selection for this year's Serbian team start and end then?
And you are quite certain that Montenegro wants to participate in the future?

Yes. I dunno exactly when the competition was held. I dunno for Montenegro, but Serbia will paricipate in the future on their own.

[quote="José"][url=http://imo2006.dmfa.si/participants.html]Here[/url] it says that Montenegro took part.[/quote]
It also says that Montenegro sent nobody to the IMO. It's just their flag and name of country that's there, and that is it.

Montenegro couldn't participate because it niether got the official invitation nor it had a selected team at time. (that's my opinion). It became officialy independent after all the IMO 2006 deadlines...
But how could Serbia participate?
1. Serbia is the legal descendent of Serbia & Montenegro (I guess...: Montenegro separated from the federation: Serbia didn't proclaim independence...)
2. All 6 contestants in the ex-Serbia & Montenegro team were Serbs from Serbia (this sentence sounds clumsy, but my friends from the Balcans will probably understand, and the rest may try...:)), so they could go as representatives of Serbia.

There... I think I got it right:)

Selection tst was held on 15.04. when Serbia and Montenegro  were in sam state, and there were students from Montenegro. But none of them qualified for IMO.


geometry
vector
linear algebra
matrix
trigonometry

i can't do this problem in a precalc book...

FInd the area of triangle ABC.

A( 1, -1, 2)  B (2, 0, -1)  C (0, 2, 1)

Um...you could use vectors...by finding the angle between 2 vectors, and then finding their magnitudes, then using the formula for areas using sines...

[hide]I used the distance formula to find that the sides were $ \sqrt{11}, \sqrt{11}, \sqrt{12}$. Then you can determine that the altitude to the side $ \sqrt{12}$ is $ 2\sqrt{2}$. Thus, the area is $ 2\sqrt{6}$.[/hide]

The easiest way I know is using vectors, but there are a lot of ways to do this.

[hide="For one..."]$ \vec{AB} \equal{} (1,1, \minus{} 3)$ and $ \vec{AC} \equal{} ( \minus{} 1, 3, \minus{} 1)$.

Then the area of the triangle is $ \frac {|\vec{AB} \times \vec{AC}|}{2}$

$ \vec{AB} \times \vec{AC} \equal{} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\
1 & 1 & \minus{} 3 \\
\minus{} 1 & 3 & \minus{} 1\end{vmatrix} \equal{} (8,4,4)$

$ |(8,4,4)| \equal{} \sqrt {8^2 \plus{} 4^2 \plus{} 4^2} \equal{} \sqrt {96}$

Area of the triangle is $ \frac {\sqrt {96}}{2} \equal{} 2\sqrt {6}$.[/hide]

we see that the sides are $ \sqrt{11}, \sqrt{11}$ and $ 2\sqrt{3}$
so its isosceles.  the height from the long side is $ 2\sqrt{2}$ so the area is $ 2\sqrt{6}$

$ S \equal{} \frac{1}{2}\sqrt{|\overrightarrow{AB}|^2|\overrightarrow{AC}|^2\minus{}(\overrightarrow{AB} \cdot \overrightarrow{AC})^2}$

use heron's formula. put: $ AB=a ; AC=b ; BC=c$ and $ p=\frac{a+b+c}{2}$. with heron's formula.
$ S=\sqrt{p(p-a)(p-b)(p-c)}$

Use Heron's formula or any of the trig relationships dear! Its a piece of cake!


[quote="ckck"][quote="Genecks"]x^2 = 20

1) $x^{2}= 20$
2) $x = \sqrt{20}$
3) $x = 2\sqrt{5}$

Is that correct?

or do I also need this:

$x =-2\sqrt{5}$

Thus making...

{$2\sqrt{5}$,$-2\sqrt{5}$}[/quote]
Yea thats right but usually you would only take the positive one.[/quote]
Actually, for an equation you should probably give all solutions.

On the other hand, if the problem had been to simplify $\sqrt{20}$, you would usually only give the positive answer.

I'm suppose to give all solutions.

Is the $-2\sqrt{5}$ also a correct solution?

The original question is to find all $x$ such that $x^{2}= 20$.  Since $-2 \sqrt{5}$ is such an $x$ it also needs to be included in your answer.


Show that if a and b are relatively prime integers, then (a+2b,2a+b) = 1 or 3

[hide="Solution"]
Let $ a\ge b$
$ (a\plus{}2b,b\plus{}2a)\equal{}(a\minus{}b,a\plus{}2b)\equal{}(3b,a\minus{}b)$, so that gcd is the divisor of $ 3b$
It can't have any part of $ b$ because then would be $ b\mid a\minus{}b$ so $ (a,b)>1$-contradiction, so the gcd is divisor of 3.
[/hide]


Show  for every $ \alpha > 0$:
$ \int^1_0 \frac{sin\frac{1}{x}}{1\plus{}x^\alpha} dx > 0$

Simillar problem

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=31180&search_id=1868721776[/url]

similar but not the same, because there is $ sinx$

Just make the change of variable $ x\equal{}\frac1y$ to get

$ \int_1^{\infty}\frac{\sin y}{y^2\plus{}y^{2\minus{}\alpha}}\,dy.$

Now try integrating that by parts.


trigonometry
probability
geometry
AMC
AIME
AMC 12
USA(J)MO

How often do trig problems pop up on the AMC 12s?  More or less than probablility problems?  More or less than number theory problems?  Is it more advantageous to have a solid number theory/probablity foundation or a better trignometry foundation?

Thanks!!!

They're all about the same. But if I wanted to have more of a solid foundation of one thing more than the others, it would be geometry (which encompasses trigonometry). You have to know geometry well not only on the AMC, but also very well to solve the hard problems on the AIME (and certainly some easy/medium ones), as geometry problems are almost guaranteed these days to be one of the harder problems.

Probability is also useful to be good at, but past the AIME, it's almost nonexistent (correct me if I'm wrong... but I mean particularly on Olympiads). When I say Probability is almost nonexistent, however, this definitely does not mean that combinatorics is. Combinatorics is used all over the place.

Number Theory is useful as well. It helps on the AMC and on at least 3 or 4 problems on an average on the AIME (on problems 1 - 10). On problems 11 - 15, maybe about 1 or 2 problems would be number theory related. When I say this, I don't mean that the entire problem is based on number theory. When you get to higher levels, the problems become harder, one of the reasons being that different types of math are mixed together i.e. geometry + probability, number theory + algebra, etc.

All in all, I'd be good at geometry, but that's just my personal preference.  :)

is it regular geometry that is being tested , eg. clever ways to manipulate diagrams, triangles, circles, arcs, ect?  I just scanned through some of the AMC 12 problems and I don't see a lot of trig in there.  thanks!

Well, truthfully, all problems can be solved without trigonometry. However, the hardest of problems require ingenious constructions, which take time to learn how to make. If you want to better yourself at geometry in general, however, you will want to learn trigonometry. It is a tool that helps almost trivialize problems sometimes. You don't have to use trigonometry on geometry problems on the AMC 12, but you can.

[quote="Karth"]Well, truthfully, all problems can be solved without trigonometry. [/quote]
Actually thats not true.  For example look at problem 17 in 2007 AMC 12 A.  There are no problems that require trig on the AMC 10, but on AMC 12 there are some problems that deal only with trig like the sum to products or product to sum trig formulas,  but yes there are not a lot of them.

But also, alot of geometry problems can be solved with trig but next year, without caculators, that will be a little harder (before I can just use like random laws of sines and find sines of abnormal angles to find side lengths), but its always good to know some basic trig like sin and cosine of all 15 degree or multiples of them angles and like sin 36 or something..

do you suggest I should strengthen my number theory/ probablility foundation or take intro to trig/complex numbers this summer?  I've taken intro to proability and intro to number theory, but my foundation isn't that good, especially with number theory.  Should I buy the number theory book and go over that?  since there aren't that many trig problems on the amc and calculators aren't allowed anymore (really? ).  I want to score a 110 or above on the AMC12 next year.

thanks

Oh, yes. I did overlook that one... unless a problem explicitly mentions trigonometry, you don't have to know it. But, not knowing it is a bad idea. In other words, unless you bump up your level of knowledge to high USAMO or IMO standards (minus the trigonometry), you won't be able to solve the medium and hard (and even some easy) AMC problems. Also, taking the AoPS classes doesn't help if you don't do problems on your own. In fact, the only way to get better at problem solving is by doing problems (direct quote from Arthur Engel, IMO Coach of Germany). So, if you keep doing problems to strengthen number theory and probability, that'd be good. But, don't keep focused on these two only. Try to work on trig/complex numbers as well so as to keep varied so that you don't get bored of only one or two topics.

[quote="Karth"]Oh, yes. I did overlook that one... unless a problem explicitly mentions trigonometry, you don't have to know it.[/quote]
This is not true.  Once, a class (at MOP, I think) was given a problem set.  The topic of the lecture had not been announced.  Several problems had been solved before anybody realized that the subject of the lecture was trig.  The AMC's only exams which do not assume knowledge of trig are the AMC 8 and AMC 10.

Don't worry about subject distribution.  It's pretty even among the four subjects algebra, combinatorics, geometry, and number theory.  Just do a mix of problems.  In reality, there is probably not as much difference as you perceive in your problem-solving ability from one subject to another.

[quote]unless you bump up your level of knowledge to high USAMO or IMO standards (minus the trigonometry), you won't be able to solve the medium and hard (and even some easy) AMC problems[/quote]

I disagree.  First, olympiads do not use knowledge significantly more advanced than the knowledge the AMCs use, so "IMO standards" is not a meaningful phrase here.  It is the application of knowledge which is more sophisticated.  Second, as the abundance of zeros each year on the USAMO show, it is quite possible to pass the AMCs and AIME with flying colors without any competence in olympiad material.

Also, it is true that almost every geometry problem in the AMCs can be solved without trig.  I cannot remember ever having used a special value of a trigonometric function (other than 0, 30, 45, 60, and 90 degrees) on an AMC contest.  In fact, most nontrivial "trig" problems are more algebraic than geometric.

Finally, I think that "foundations" are overrated.  We like to labor under the delusion that were it not for insufficient "foundations", we could solve more sophisticated problems easily.  More often, our knowledge of basic things and our ability to solve easy problems are fine, and we cannot easily solve harder problems because they are harder than those we are used to solving.  It is good to solve problems at the limit of your ability.

That said, you have probably learned almost no number theory in school, in which case you want to learn more, such as the material of AoPS II.  However, you can learn most of the prerequisite material by solving problems and then reading [i]good[/i] solutions, and looking up any theorems you don't know; but don't fall into the trap of always thinking that you can't solve a problem because it uses a sophisticated theorem you don't know :  it is probably a sophisticated application of a theorem you know well that is trapping you.

[quote=Boy Soprano II]Also, it is true that almost every geometry problem in the AMCs can be solved without trig.  I cannot remember ever having used a special value of a trigonometric function (other than 0, 30, 45, 60, and 90 degrees) on an AMC contest.  In fact, most nontrivial "trig" problems are more algebraic than geometric.[/quote]

Not yet, but without caculator next year, I bet theyll have some problem like find cos36-cos72 or something...

[quote="turtlecloud"]Is cos 36 - cos 72 = 1/2 ? wasn't that an old AHSME problem?[/quote]

yea thats why I used it as an example.  They used to have those problems before the scientific caculator but now obviously they wont have that because you can just plug it into your caculator.

But without caculators next year, they might do those "types" of problems again.

In the AMC 10/12 Math Club Package, we have a chart on Page 25 of problem type frequency as classified here at the AMC by the constructors of the contests.
I have reproduced the relevant part of that chart.
                             2002             2003            2004         2005       2006
                             12 A   12 B    12 A   12 B  12A  12B  12A 12B  12 A   12 B
Number Theory     3        4         2         2       2      3      3      2      2        2
Triangle Geom      3        5          3         4       1      3      1      1      0       2

Note 1:  Probability does not occur as a problem type classification in this chart, indicating that problems involving probability are likely classified under "Problem Solving" or "Other".

Note 2:  AMC 12 problems often involve several topics.  What we have labeled as Number Theory, you might consider as something else.  Likewise for probability and trigonometry.

Note 3:  Many problems can be worked in several ways,  e.g. with trig functions and also without.

Note 4:  What are classified as Triangle Geometry problems may or may not involve "trigonometry" in the whatever sense you use the term.

No, the chart in the Math Club Package is not available on-line.  Teachers and interested students may purchase it for classroom  and personal use.  Use the registration form, last item in the right hand column.  You can also order it via the usual means of communication with the AMC.

-- Steve Dunbar
MAA Director, American Mathematics Competitions


function
limit
ratio
algebra
rational function
algebra unsolved

find all polinomials with real coefficients $P(x)$ satisfying the condition :
 $P(x)P(x+1) =P(x^{2}+2)$ for all $x \in R$

How can you be sure that the degree will be zero?

[quote="Rust"]Only P(x)=1 (degree is 0).[/quote]
Can you solve detail ? Thanks. :)

i don t agree with you because P(x)=x²-x+2 is a solution

[quote="Rust"]If $P(x)=0$ then $P(x^{2}+2)=0$ and $P((x-1)^{2}+2)=0$. It give $degP(x)=\infty$ if $degP(x)>0$.[/quote]

the problem with that is you don't know that $\exists x$ such that $P(x)=0$.  So what you've discovered is that $P(x) \neq 0 \forall x$, which is true in the counterexample that beljad5 gave.

Exist at least one complex roote x0, if deg(P(x))>0. It give another roote $x^{2}+2\not =x$ or $(x-1)^{2}+2\not =x$. It give infinetely many roots, because we can choose $|x_{n}| \to \infty$.

[b]Solution:[/b] $(x^{2}-x+2)^{k}$ where $k\geq 0$, and a trivial solution $0$.

First, it is easy to see that $P_{k}(x) = (x^{2}-x+2)^{k}$ is a solution, since
1) $x^{2}-x+2$ is a solution;
2) the product of two solutions is a solution.

Let us show that there are no other solutions. The argument of Rust shows that $P(x)$ does not have real roots. Therefore, $\deg P(x)$ is even; denote $2d = \deg P(x)$. It is easy to see that the leading coefficient of $P$ must be equal to $1$.

Consider the function \[f(x) = \frac{P(x)}{P_{d}(x)}.\] It verifies the equation \[f(x) f(x+1) =f(x^{2}+2).\] Note that \[{\lim_{t\to+\infty}f(x) = (\text{ratio of leading coefficients of }P \text{ and }P_{d})}=1.\] If $f\equiv 1$, we are done. So assume to the contrary that $f$ is not a constant. Since $f(x)$ is a rational function either it should tend to $1$ from above as $x\to+\infty$, or it should tend to $1$ from below as $x\to+\infty$.

In the former case, for sufficiently large $x$, $f(x)$ is decreasing, and $f(x) > 1$. We have $f(x^{2}+2) = f(x)f(x+1) > f(x)$. We get contradiction.

In the latter case,  for sufficiently large $x$, $f(x)$ is increasing, and $f(x) < 1$. We have $f(x^{2}+2) = f(x)f(x+1) < f(x)$. We get contradiction.

And that problem?
Find all polinomials P(x) with real coefficients  satisfying the condition :
$P(x).P(2x^{2})=P(x^{3}+x), x /in R$


geometry
trigonometry
trig identities
Law of Cosines
power of a point

$ \triangle ABC$, with $ AB \equal{} BC \equal{} CA \equal{} 11$, is on the interior of a circle.  $ AB$ is extended through $ B$ to meet the circle at $ P$, $ BC$ is extended through $ C$ to meet the circle at $ Q$, and $ CA$ is extended through $ A$ to meet the circle at $ R$.  If $ BP \equal{} 11$, $ CQ \equal{} 33$, $ AR \equal{} 55$, and the area of the circle is $ 3\pi K$, compute $ K$.

[hide="Sketch"]
Extend CB past B to meet the circle at S, similarly define T and U. Let BS=a, CT=b, AU=c. By power of a point on A,B, and C, we get a system for a, b and c, which is easily solvable. At this point I think there is an easier way to finish the problem than what I did, but here is my method. By law of cosines on $ \triangle PAT$, we can solve for PT, similarly, we can solve PR. We can use the formula Area= $ (1/2)ab sin C$ on both $ \triangle PAT$ and $ \triangle PAR$ to find the area of PRT, then we can use the fact that area$ \equal{} \frac {abc}{4R}$ to find the radius of the circle.

EDIT: Actually, we know $ PC \equal{} 11 \sqrt {3}$ and $ PC \perp RT$ since PAC is a 30-60-90 triangle, this makes computing the area of PRT and PR and PT easier.
[/hide]

[hide="Answer"]637[/hide]

The solution is correct but quite time-consuming.  There is a much faster way.
[hide="Hint"]Power of a point many many times.  Also remember that right inscribed angles intercept diameters.[/hide]


Ok, this is just a test.

But you, as the "mob", will have to guess my 4 number combination correctly.  I will not cheat.  When I reply, I will give you 2 numbers, the former will be how many are correct and in the right spot, and the latter will be how many numbers are correct, but in the wrong spot.  You may start now.

You will have 15 guesses.

* Note: This number may change in the next game.

Next time we hsould guess different digits.

dude ernie you totally stole this from me.

AIME that was a lousy guess

really unhelpful

I guess 7788

I also guess 9922

7788- 1 correct: 0 in the right spot

9922- 2 correct: 1 in the right spot

Guesses: 5

9842- 3 correct: 1 in the right spot
9572- 3 correct: 2 in the right spot

Guesses: 9

0011
2233
2456
7788
9922
4455
0920
9842
9572

I shoulda done 10 guesses...

Guess 10:  9274

I'm pretty sure this is it...

correct!

i chose my lucky number :D

Ok, I have chosen my second number.  10 guesses.

cf249: 1 pt

Wth, most of the guesses are made by me and he gets the point?

I think we should pm you guesses secretly, and compete, or something.

Anyway,

Guess 1: 0011
Guess 2: 2233
Guess 3: 4455

Sorry chenhsi, that cannot be guaranteed.

0011- 1 correct, 1 in the right spot
2233- 0 correct
4455- 1 correct, 0 in the right spot
1212- 1 correct, 1 in the right spot
6677- 1 correct, 0 in the right spot
4117- 2 correct, 1 in the right spot

WHAT????

By 4455, there has to be either a 4 or a 5. By 6677, there has to be a 6 or a 7. By 2233, there can't be 2s or 3s, so by 1212, there has be 2 ones.

By __11 and 1_1_, either it is _11_ or 1__1. Haven't I covered all possible possibilities?

Because 7/5 can only be in the first 2 digits, and 6/4 can only be in the second 2 digits.

By 1212, you have only 1 one.  I don't repeat.

Does 1 correct, 1 in right spot mean 1 correct but in wrong spot, 1 in right spot?

Because in 1212, how can I only have 1 one? There are a total of 2 correct in there, and 2233 guarentees that the 2s cannot be correct.

No, 1 correct, 1 in right spot means that only one number you guessed is in the correct answer, and it is in the same spot in your guess as it is in the answer.

For example, if the correct code was 2490, and you guessed 4411, you would be told 1 correct and 1 in the right spot.
(Note:  The above numbers are completely random)

0011- 1 correct, 1 in the right spot
2233- 0 correct
4455- 1 correct, 0 in the right spot
1212- 1 correct, 1 in the right spot
6677- 1 correct, 0 in the right spot
4117- 2 correct, 1 in the right spot
1561- 2 correct, 0 in the right spot
7114- 2 correct, 1 in the right spot
5718- 3 correct, 3 in the right spot
5719- chenhsi wins!

Next number.  9 guesses. :P

ernie: 0
mob: 2

4334- 1 correct: 0 in the right spot
0123- 2 correct: 0 in the right spot
3278- 2 correct: 2 in the right spot
3295- 1 correct: 1 in the right spot
3259- 1 correct: 1 in the right spot
3716- 3 correct: 1 in the right spot
3677- 2 correct: 1 in the right spot


[quote="hungkhtn"]Let a,b,c,d be non-negative real numbers with sum 4. Prove that
\[\frac{1+ab}{1+b^{2}c^{2}}+\frac{1+bc}{1+c^{2}d^{2}}+\frac{1+cd}{1+d^{2}a^{2}}+\frac{1+da}{1+a^{2}b^{2}}\ge 4. \]
[/quote]
I have a solution for this problem by AM-GM but I can't solved it by AC or SMV. Who's help?


[b]Can someone help with this? :roll: [/b]

[i]Suppose you have a map with cities on them (think of them as simply circles) with roads connecting them (think of them as lines connecting the circles). Suppose you started in one sicty, say A, and you want to get to city B. Suppose you can only see the roads that are connected to the city you are currently in. Finally, suppose you can  leave and read any type of information at cities and roads. You can also read information in any adjoining city or road. Now, device a method by which you can get to city B from A, given the above constraints.

Hint: drop successive numbers in each new city you visited so you know how to backtrack. Also  drop a marker at each city that tells you whether you visited all of the surrounding cities (so this city becomes a dead-end).

Important: State exactly what information you need to store for the method you devise.[/i]

I suggest you search for 'depth first search', since that is precisely what you want here. In fact, any time of search will work, but thats probably the easiest to understand first off.

Depth-first seach will take some modification. What you get is known as Tremaux's algorithm

Tremaux's algorithm is a depth first search, and I don't see any reason why the conditions above prohibit a normal depth first search - in fact, that looks exactly like the definition of DFS to me.. if you are right, which condition in the problem am I missing?

Can you give me some link for Tremaux algorithm

Trevuxe is a DFS phrased as a maze algorithm. Basically a DFS where the nodes and paths are modified, not an internal data structure.


AMC
USA(J)MO
USAMO
inequalities proposed
inequalities

let a,b,c postisive numbers.prove that
$ \frac {(2a \plus{} b \plus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(2b \plus{} c \plus{} a)^2}{2b^2 \plus{} (a \plus{} c)^2} \plus{} \frac {(2c \plus{} b \plus{} a)^2}{2c^2 \plus{} (b \plus{} a)^2}\leq 8$  :wink:

[quote="youngtrai"]let a,b,c postisive numbers.prove that
$ \frac {(2a \plus{} b \plus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(2b \plus{} c \plus{} a)^2}{2b^2 \plus{} (a \plus{} c)^2} \plus{} \frac {(2c \plus{} b \plus{} a)^2}{2c^2 \plus{} (b \plus{} a)^2}$  :wink:[/quote]
Are you sure? :P

USAMO 2003.See this problem at
http://www.mathlinks.ro/Forum/viewtopic.php?t=46
http://www.mathlinks.ro/Forum/viewtopic.php?t=28678
http://www.mathlinks.ro/Forum/viewtopic.php?t=43278
http://www.mathlinks.ro/Forum/viewtopic.php?t=48989
http://www.mathlinks.ro/Forum/viewtopic.php?p=336208#p336208


algebra function domain trigonometry LaTeX

Hi, I use TexMaker, which I've installed because the filesize was much smaller to TexCenter. However, I've found that I can't write any math on it. In particular, I can't make any tables. This may be because it doesn't recognize the packages I try to put down, but I'm not sure. Here is the table I tried to include (which I am only writing here  so that I can come back and copy and paste it as an image on Word, since TexMaker doesn't work. Sorry for taking up space).

[hide]
\[ \begin{array}{|c|c|c|c|} \hline Equation & Domain & Range & Standard/Vertex Form \\
\hline a(x) \le \frac {1}{2} \cos x - 6 & x \in \Re & y \in [ - 6.5, - 5.5] & \\
\hline b(x) = - 4 & x \in [ - 8,8] & y \in { - 4} & \\
\hline c(x) = - 2 & x \in [ - 8,8] & y \in { - 2} & \\
\hline d(x) = - |x - 2| + 7 & x \in [0,4] & y \in [5,7] & \\
\hline e(x) = 6 & x \in [1,3] & y \in {6} & \\
\hline f(x) = 5 & x \in [0,4] & y \in {5} & \\
\hline g(x) = |4x + 15| - 6 & x \in [ - \frac {11}{2}, - \frac {9}{2}] \cap [ - \frac {7}{2}, - \frac {5}{2}] & y \in [ - 2,0] & \\
\hline h{x} = - x^2 - \frac {15}{2} x - \frac {189}{16} & x \in [ - \frac {11}{2}, - \frac {5}{2} & y \in [0,2.8] & \\
\hline i(x) = - 2x + 20 & x \in [ - 8, - 6] \cap [6,8] & y \in {2} & \\
\hline j(x) = 2 & x \in [4,8] & y \in [0,4] & \\
\hline k{x} = x - 4 & x \in [8, - \infty) & y \in ( - \infty, 4] & \\
\hline \#1: x = - 8 & x \in { - 8} & y \in ( - \infty,2] & \\
\hline \#2: x = - 7 & x \in { - 6} & y \in ( - \infty,2] & \\
\hline \#3: x = 7 & x \in {6} & y \in ( - \infty,2] & \\
\hline \#4: x = 8 & x \in {8} & y \in ( - \infty,2] & \\
\hline \#5: (x - 1)^2 + (y - 4)^2 \le \frac {1}{4} & x \in [\frac {1}{2}, \frac {3}{2}] & y \in [\frac {7}{2}, \frac {9}{2} & \\
\hline \#6: (x - 4)^2 + (y - 4)^2 \le \frac {1}{4} & x \in [\frac {5}{2}, \frac {7}{2}] & y \in [\frac {7}{2}, \frac {9}{2} & \\
\hline \#7: (x - 1)^2 + (y - 1)^2 \le 1 & x \in [0, 2] & y \in [ - 2, 0] & \\
\hline \#8: (x - 3)^2 + (y - 1)^2 \le 1 & x \in [2,4] & y \in [ - 2,0] & \\
\hline \#9: (x + 12)^2 + (y - 14)^2 = 4 & x \in [ - 14,10] & y \in [12,16] & \\
\hline \#10: x = y^2 + 6y - 1 & x \in [ - 9, - 8] & y \in [ - 4, - 2] & \\
\hline \#11: \alpha (x) = & x \in [ - \frac {1}{2}, \frac {11}{2}] & y \in [ - 1, 3] & \\
\hline \end{array}\]
[/hide]

I think that your links to the applications are wrong. Go under Options -> Configure Texmaker, and check if the links are correct. If not, just fix them.

Can you tell what exactly is supposed to be there? I don't know what's right or not


Given $ ab+bc+ca=1$ and $ a$, $ b$, and $ c$ are nonnegative
Prove $ a\sqrt{b^{2}+c^{2}+bc}+b\sqrt{a^{2}+c^{2}+ca}+c\sqrt{a^{2}+b^{2}+ab}\geq\sqrt{3}$

[quote="arqady"][quote="Allan Z"]Given $ ab+bc+ca=1$
Prove $ a\sqrt{b^{2}+c^{2}+bc}+b\sqrt{a^{2}+c^{2}+ca}+c\sqrt{a^{2}+b^{2}+ab}\geq\sqrt{3}$[/quote]
Are you  mean that $ a,$ $ b$ and $ c$ are non-negative numbers?
If so,the hint see here:
[hide]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156534
[/hide][/quote]

opps my bad.. edited

Very short.

[hide]
By AM-GM, $ b^{2}+c^{2}\ge 2bc$, so $ b^{2}+c^{2}+bc \ge \frac{3}{4}(b+c)^{2}$ (divide the previous inequality by $ 4$ and subtract it from this to get equality).

Hence,

\[ \sum_{\text{cyc}}a\sqrt{b^{2}+c^{2}+bc}\ge \sum_{\text{cyc}}\frac{\sqrt{3}}{2}a(b+c)=\sqrt{3}\]

[/hide]


integration
function
real analysis
real analysis solved

here is a generalization of this problem
http://www.mathlinks.ro/phpBB/viewtopic.php?t=330.

You should try it! :D :D

Compute lim(x-->0) [a1*a2*..*an*x^n-sh(a1x)*...*sh(an*x)]/x^(n+2) where a1,..,an are real numbers and n is natural.

cheers! :D :D :D :D

[quote="Lagrangia"]

Compute lim(x-->0) [a1*a2*..*an*x^n-sh(a1x)*...*sh(an*x)]/x^(n+2) where a1,..,an are real numbers and n is natural.

cheers! :D^4[/quote]
I use sh(u) = u +u^3/6+o(u^3) when u tend to 0
a1*a2*..*an*x^n-sh(a1x)*...*sh(an*x)
= a1.a2....anx^n( 1- ( 1+a1x/6+o(x))...(1+a2x/6+o(x))
=a1.a2....an.x^n( (a1+...+an).x/6+o(x))
the limit of Lagrangia problem is a1.a2...an(a1+...+an)/6
    Mo \cup bino \int

moubi you said that : sh(u) = u +u^3/6+o(u^3) when u tends to 0, what is o(u^3) I don't understand.. :(

and using the same thing to expand sh(u), how can you write ch(u),th(u), cth(u)....

thank you very much! :D :D

Hi Lagrangia,

Let f and g functions f = o(g) around x_0 means
lim_{x-->x_}(f(x)/g(x)) = 0
f=o(g) small o is the notation of Landau
for example x^3 = o(x^2) when x is near 0, lim_{x-->0}x^3/x^2=0
shu = u + u^3/6 +o(u^3) ideed when u is near u
lim_{u-->0}(shu - u -u^3/6)/u^3=0 apply l'Hospital several times.

Use the power serie of exp(u)=1+u/1!+u^2/2!+u^3/3!+u^4/4! +o(u^4)
chu=(exp(u)+exp(-u))/2= 1 + u^2/2! + u^4/4! +o(u^4)

when u near 0
thu = shu/chu = (u+u^3/3!+u^5/5!+...)(1+u^2/2!+u^4/4!+...) with
the division
th(u)= u -u^3/3 +2/15u^5 - 17/315u^7 +o(u^7)
for cth(u) = ch(u)/sh(u) = 1/u +u/3 -u^3/45 +o(u^3)

you can also said
shu = u +u^3/3! + u^3. \varepsilon (u) where
 \varepsilon (u) is a function such that
lim_{u-->0} \varepsilon (u)=0

http://mathworld.wolfram.com/LandauSymbols.html
 \cap  \cap  \emptyset  \cup \beta \int