Is there any efficient, straightforward way to solve the Diophantine $ x^2 \plus{} y^2 \equal{} n$ (that is, without testing all the values, taking it various mods, etc.)? Things like the Bramagupta-Fibonacci identities seem like they could be able to reduce the problem a bit, though I don't see any way to apply them rigorously.

The Brahmagupta-Fibonacci identities solve the problem [b]completely[/b], at least in the sense that the set of solutions is describable in terms of the prime factorization of $ n$ over the Gaussian integers; see my blog post [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url].  In fact, since the problem is equivalent to finding the prime factorization of $ n$ (since from here one easily recovers the prime factorization over the Gaussian integers and knowing all the solutions also allows you to recover the prime factorization), the complexity in terms of $ n$ is exactly the complexity of integer factorization, which is sub-exponential and hence quite a lot better than the brute-force approach, which is $ O(2^{b/2})$.


search
linear algebra
linear algebra unsolved

given $ A \in M_n(R)$ such that $ trace(A)\equal{}0$.
a.there exists $ B \in M_n(R)$ such that $ A \sim B$ and $ [B]_{ii} \equal{}0,i\equal{}\overline{1,n}$
b.there exists $ C,D \in M_n(R)$ such that $ A\equal{}CD\minus{}DC$

blue_se7en,
If you search on this forum I think you'll find the answer.
for b, here is sharper result.
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1525734767&t=101774

im only an amateur :blush: ,thank for your advice

No worries, I'm also only an amateur :lol:

You'll find a solution here

http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=3979


calculus
integration
trigonometry
induction
inequalities
complex analysis
calculus computations

[color=white]............[/color]

take a [b][u]deep[/u] [/b]breath, then try and show the following:  :rotfl:

$\int_{0}^\infty\;\;( \;[ ( \prod_{i=1}^{p}\;\cos{(\alpha_{i}\;x)}) ( \prod_{j=1}^{q}\;\frac{\sin{(\beta_{j}\;x)}}{x}) ] \;\frac{\sin{(\gamma\;x)}}{x}\;) \;\;\textbf{d}x$

$\hspace*\;\;\;\quad=\;\boxed{\; \frac{\pi}{2}( \prod_{j=1}^{q}\;\beta_{j}) }$

where [u]each [/u]of $\alpha_{i}\;\;\;and\;\;\;\beta_{j}\;\in\mathbf R,$
[i]and [/i]$ \gamma\;>\;\sum_{i=1}^{p}\;|\;\alpha_{i}\;|\;+\;\sum_{j=1}^{q}\;|\;\beta_{j}\;|$

[ PS: i used complex variables to show it [i]first[/i], now i am attempting
to show it [i]using [/i]trigonometric integration..... ]  :maybe:

Here is a method of attack, avoiding complex analysis techniques:

The integral equality can be proved by induction on $q$.

First we check that, I won't do that, $P(1)$ is valid namely:

$\int_{0}^{\infty}\prod_{i=1}^{p}\cos(\alpha_{i}x)\frac{\sin \beta_{1}x }{x}\frac{\sin \gamma x}{x}dx=\frac{\pi}{2}\beta_{1}$

Then we show that if $P(q)$ holds true that $P(q+1)$ is also valid, of course the constants $\beta_{i}$ and $\alpha_{j}$, $\gamma$ should verify the proper inequality conditions.

So $P(q)$ holds true. Now $P(q+1)$ reads

$\int_{0}^{\infty}\prod_{i=1}^{p}\cos(\alpha_{i}x)\left(\prod_{j=1}^{q+1}\frac{\sin\beta_{j}x}{x}\right)\frac{\sin \gamma x}{x}dx=\frac{\pi}{2}\prod_{j=1}^{q+1}\beta_{j}$.

Now,

Let $f(\beta_{q+1})$ be the preceding integral. Then a calculation shows that

$f'(\beta_{q+1})=\int_{0}^{\infty}\prod_{i=1}^{p}\cos(\alpha_{i}x)\left(\cos \beta_{q+1}x\right)\left(\prod_{j=1}^{q}\frac{\sin \beta_{j}x}{x}\right)\frac{\sin \gamma x}{x}dx$

which by the hypotheses of induction, namely $P(q)$ holds true, is equal to

$f'(\beta_{q+1})=\frac{\pi}{2}\prod_{j=1}^{q}\beta_{j}$.

Integrate and  you get that

$f(\beta_{q+1})=\frac{\pi}{2}\prod_{j=1}^{q+1}\beta_{j}+C$ where $C$ is the constant of integration.
It is not hard to see that $C=0$ from which the desired result holds.

PS This is only a scketch of the proof.
PPS I hope I made not too many mistakes.


algebra
polynomial
linear algebra
linear algebra unsolved

I need help with this problem

A linear operator staisfies the equation
A^2-3A+2=0

I need to find the eigenvalues of A and represent A as a matrix. I also need to know the eigenstates of A

Well, obviously $f(x)=x^2-3x+2$ is the minimal polynomial for $A$, because $f(A)=0$ and has distinct linear factors $(x-1)$ and $(x-2)$.
So, the only eigenvalues of $A$ are $1,2$ and their geometric multiplicity is their multiplicity as roots so it's 1 for both. Therefore $A$ will be (in Jordan form, here diagonal) :
\[\left(\begin{array}{cc}I_k& 0\\0&2I_h\end{array}\right)\]
with $h,k$ nonzero natural numbers.
So, the eigenstates of $A$ are the multiples of $\{e_1,\ldots,e_n\}$ (obviously, in the Jordan basis).


abstract algebra
superior algebra
superior algebra unsolved

Determine which abelian groups A fit into a short exact sequence

$ 0 \rightarrow \mathbb{Z}_{p^m} \rightarrow A \rightarrow \mathbb{Z}_{p^n} \rightarrow 0$ with p prime. What about the case of short exact sequences $ 0 \rightarrow \mathbb{Z} \rightarrow A \rightarrow \mathbb{Z}_n \rightarrow 0$?

I guess that the order of A should be $ \mathbb{Z}_{p^{m\plus{}n}}$, but I don't know how to find A using mappings that fits into a short exact sequence.

Thanks.

For the second case, consider $ A\equal{}\mathbb{Z}$, and the mapping $ \mathbb{Z}\to A$ defined by $ a\mapsto na$, which is the multiplication by $ n$. The second map $ A\to \mathbb{Z}_n$ is just the canonical projection. Then u can easily check it is exact.

So, from that, we have the solution for the first case. I also guess $ A\equal{}\mathbb{Z}_{p^{m\plus{}n}}$ and the mappings defined as follows

$ f : \mathbb{Z}_{p^m}\to\mathbb{Z}_{p^{m\plus{}n}}$ is the multiplication by $ p^n$, i.e, $ f(z\plus{}p^m\mathbb{Z})\equal{}z.p^n\plus{}\mathbb{Z}_{p^{m\plus{}n}}$.

and $ g: \mathbb{Z}_{p^{m\plus{}n}}\to\mathbb{Z}_{p^n}$ is just the canonical projection.

Then the exactness is easily checked.

The map given in the previous post does not account for [i]all[/i] such groups:

Consider the short exact sequence $ 0\rightarrow\mathbf{Z}\stackrel{f}{\rightarrow}A\stackrel{g}{\rightarrow}\mathbf{Z}/n\rightarrow 0$.  Since $ f$ is injective and $ g$ is surjective, let $ a = f\left(1\right)$ and $ b$ be an element such that $ g\left(b\right) = 1\in\mathbf{Z}/n$.  Then, $ A$ is generated by $ a$ and $ b$.

Now, since $ nb\in\ker g$, we must have $ ma = nb$ for some $ m\in\mathbf{Z}$ by the exactness of the sequence (we may assume that $ \gcd\left(m,n\right) = 1$).  Consider the homomorphism defined by

\begin{align*}\psi &: \mathbf{Z}\times\mathbf{Z}\rightarrow A , \\ &: \left(1,0\right)\mapsto a , \\ &: \left(0,1\right)\mapsto b .\end{align*}
If $ \left(x,y\right)\in\ker\psi$, then $ ax + by = 0$.  Therefore,

\[ ax + by = 0 \implies nax + may = 0 \implies f\left(nx+my\right) = 0 ,\]
so $ nx + my = 0$ since $ f$ is injective.  Since $ m$ and $ n$ share no common factors, we have that $ n$ divides $ y$, so $ y = cn$ for some $ c\in\mathbf{Z}$.  This immediately implies that $ x = -cm$, so it follows that $ \ker\psi = \left(-m,n\right)\mathbf{Z}$ (the reverse inclusion is clear).

We also know that we can find $ s,t\in\mathbf{Z}$ such that $ ms+nt = \gcd\left(m,n\right)$.  Therefore, send the point $ \left(-m,n\right)\mapsto\left(0,\gcd\left(m,n\right)\right)$ and the First Isomorphism Theorem tells us that

\[ A \simeq \frac{\mathbf{Z}\times\mathbf{Z}}{\left\{0\right\}\times\gcd\left(m,n\right)\mathbf{Z}} \simeq \mathbf{Z}\times\mathbf{Z}/\gcd\left(m,n\right) .\]
It is now easily checked that $ \mathbf{Z}$ and $ \mathbf{Z}/d$ will result in an exact sequence for each $ d\in\mathbf{Z}$ such that $ d$ divides $ n$.  Therefore, we have [i]all[/i] groups $ A$ for which the stated sequence is exact.


calculus
integration
function
inequalities
real analysis
real analysis unsolved

find $f:[0,1]\to [0,\infty]$ a continuous function on $[0,1]$,derivable on $(0,1)$ with the following properties:

  i) $f(0)=0$
 ii) $0\leq f'(x) \leq 1$
 iii) $\int_{0}^{1} f(x)dx=1$ ;)

Suppose $f$ is continuous on $[0,1]$, differentiable on $(0,1)$ with $|f'(x)|\le1$, and $f(0)=0$.
By the Mean Value theorem, $|f(x)|\le x$ for all $x\in[0,1]$. Integrating this inequality, $\left|\int_0^1f(x)\,dx\right|\le\int_0^1|f(x)|\, dx\le\frac12$.

Now what, exactly, are you going to do about the obvious impossibility of your conditions?

Yes,you are right.I have made a mistake in my calculations. :oops:


Let BL bisector of an triangle ABC $ \angle BLA \equal{}60^\circ$   $ CL^2 \equal{}AL \cdot BL$.Prove that  $ BL\equal{}AC$

___________________________
Azerbaijan Land of the Fire    :rotfl:  :|

Take $A'\in (BL)|LA'=LA$; from the given relation $CL$ is tangent to the circle $\odot (BCA')$, i.e. $\angle ACA'=\angle CBA'$, hence $\triangle ACA'\equiv \triangle LBA$, thus implying $BL\equiv AC$. (This is [b]NOT[/b] my solution)

Best regards,
sunken rock

Another solution:
LK is bisector of ∠BLC .so we have ∆BLK≅∆BAL.take k´ϵ (BL) such that kk´=kL.
∆KLK´ is equilateral triangle and kk´∥LC we have to prove Bk´=LC.
We know BK/KC=(BK´)/(K´L)=BL/CL=CL/AL  and AB=BK→BK´=(BL×BK)/BC=(BL×AB)/BC
and AB/BC=AL/CL=CL/BL → (BL×AB)/BC =(CL×BL)/BL=CL


What is the smallest positive integer value of n such that 2^n+5^n+7^n is a multiple of 17?

I need the solution.

Thanks :]

[hide="Answer"]
Well, the answer is 3.
2^3+5^3+7^3=
8+125+343=476
476/17=28

I just plugged in answers--I'm sure there's some neat little trick as to how it works.[/hide]

I may come up with the solution :: I'll get back to you later.

yeah, when you have problems like that, where you don't know how to solve it, usually you just plug in numbers until you get the answer. Also, if you do it fast, sometimes it goes by faster than you would go had you known the formula.


Hello everyone! I have a few problem regarding Log.
First of all, it is a homework about Logarithm. I provide an image of my homework here:
[url=http://img443.imageshack.us/my.php?image=logarithmsbj4.jpg][img]http://img443.imageshack.us/img443/8108/logarithmsbj4.th.jpg[/img][/url]
(The ones squared with red are the ones I have problem with solving)

The main problem I have with Logs are the exponents. I know how to convert from logarithm form to exponent form and vice versa. But when solving for a variable exponent, it becomes difficult.

This is what I have done:
[url=http://img75.imageshack.us/my.php?image=log1kf8.jpg][img]http://img75.imageshack.us/img75/9156/log1kf8.th.jpg[/img][/url]
In problem 57.
I can solve a.
But b, c, and d are different.
In b. when solving on exponential form, how can I make a base with a variable exponent equal to a fraction (or decimal)?
In c. when solving on exponential form, how can I make a base with a variable exponent equal to a root?
The same goes with d.
In problem 58.
I could solve a. But it seems that b, c, and d, are solved differently. All of them contain negative exponents. How can I make a base with an exponent equal to another positive number if the exponent is negative?

Thanks in advance for the help!


[quote="hello"]I was wondering if there is any way to know if a website is safe to know that I can visit it? I mean that it does not produce viruses, spyware, etc?[/quote]If they don't try to sell you something, or they aren't an adult site and they aren't supplying warez then it's most likely safe :D

Also, if they have a security certificate they're probably secure, hence the name. ;)

It also helps to get an antivirus and anti-spyware program. There are free ones out there that work OK. Search Google

[quote="mathnerd314"]Also, if they have a security certificate they're probably secure, hence the name. ;)[/quote]Well anyone can make a secure certificate, that doesn't guarantee you that that site will be safe ;)

[quote="Valentin Vornicu"][quote="kezyxes"]If you use a secure browser (e.g. firefox, safari, konqurer), and dont' do anything stupid, you'll be safe, no matter what's on the website.[/quote]Haha, that's what most people who get infected say before they get infected "I use firefox, I am safe" :D No browser is safe, especially the popular ones (IE, Firefox!) ;)[/quote]

That's why I added "and don't do anything stupid" in :P

If one doesn't run untrusted code, installes security update as soon as they are released, and follows a good security policy, then there's no reason why he/she would get infected.

[quote="kezyxes"]That's why I added "and don't do anything stupid" in :P[/quote]Yes, most people however forget to mention that important detail :P. That's why I personally consider Firefox more dangerous for the regular users: it is so much advertised as being super safe, that the regular user will not think that he is in any danger at all, so the probability of him doing  something stupid increases :D

Ummmm, G-UNIT...

I think the point was to make sure it's safe BEFORE you download anything. Because after you download something and you find that "oh no, it's a virus" it's a little too late to do anything about it. That's why antivirus programs help.

Mathnerd, I think G-UNIT was trying a little bit of irony ;)

[quote="G-UNIT"]how to determine if a website is safe?

  Simple.  Download/access whatever the website is offering, if you find you're computer is infected it means the website is not safe.  On the other hand, if you find your computer is functioning properly it means a) you have a virus that is undetectable/ takes time to kick in, or b) you don't have a virus and the website is safe.  I hope that helped.

  Best regards,

  G-UNIT[/quote]

LOL that would be so stupid.. i just use anti spyware/ antivirus stuff.


Let ABCD be a cyclic quadrilateral. C1, C2, C3, C4 are circles which have the sides AB, BC, CD, DA as chords, respectively. Ci intersects C(i+1) at Ai for 1<=i<=4 (C5=C1). Prove that the pts Ai lie on a circle.

Since ABA1A2 and so on are cyclic quadrilateral it is simple to prove
angle A1A4A3+angle A1A2A3 = \pi
The End.

Yeah, it's kind of easy, but I posted it because I had an inversion-based soln and I liked it. See ya :D

Dear Mathlinkers,
this result is known as the six circles theorem which comes from Auguste Miquel.
For  asynthetic proof
http://perso.orange.fr/jl.ayme  vol. 2 Du théorème de Reim au théorème des six cercles p. 8-12  with an interesting rule...
Sincerely
Jean-Louis


Look at picture..

I don't get how to do this problem, the answer is B but I don't get how it is.

How I did it:

If $ AD \equal{} 1$, then $ AP \equal{} 0.5$, so $ PD \equal{} 0.5\sqrt{3}$ by the 30-60-90 rule.  Then $ PB\equal{} 2\minus{} 0.5 \equal{} 1.5$.  $ BC\equal{}1$, so $ DB\equal{} \sqrt{3}$. So $ 0.5\sqrt{3} \plus{} \sqrt{3} \plus{} 1.5 \equal{} 1.5\sqrt{3} \plus{} 1.5$

[quote="Mewto55555"]Why is AP necessarily .5? We have a 30-60-90 triangle, so shouldn't AD=1, Ap=1/rt3, andPD=2/rt3??[/quote]

Wouldn't it be .5, since 1/2 is .5?  I am really confused... :|  :maybe:

$ APD$ is a 30-60-90 triangle with the altitude of the equilateral triangle it's in being $ AD$. (You can determine this by the fact that $ \angle ADP \equal{} 30^\circ$). Thus, we have $ \frac {\sqrt {3}s}{2} \equal{} 1 \implies \frac {1}{2}s \equal{} \frac {\sqrt {3}}{3}$. Remember that $ \frac {1}{2}s \equal{} AP$. Then we have $ DP \equal{} s$. So $ DP \equal{} \frac {2\sqrt {3}}{3}$. You can work from there.

The problem is all about 30-60-90 triangles.

We have AD = 1 (opposite 60 degree angle); so we have AP = $ 1/ \sqrt{3} \equal{} \sqrt{3}/3$ After this, we have DP = $ 2\sqrt{3}/3$

We also know that BC = 1 (opposite 30 degree angle); so we have BD = 2 and DC = $ \sqrt {3}$.

The only side length of BPD that we do yet know is PB which is simply $ \sqrt {3} \minus{} \sqrt{3}/3 \equal{} 2\sqrt{3}/3$

$ 2\sqrt{3}/3 \plus{} 2 \plus{} 2\sqrt{3}/3 \equal{} 2 \plus{} 4\sqrt {3}/3$

B


linear algebra
matrix
modular arithmetic
geometry
geometric transformation
rotation
group theory

I know the following may seem obvious for most of you, but it is not the case for me:
  How to prove using only elementary methods and no embeddings in algebraically closed fields that for any integer matrix A and any prime p we have $ Tr(A^p)-TrA$ divisible by p?

Since this was unknown to me and I like it, I'll post the obvious non-elementary approach :).

The eigenvalues of $A$ lie in some algebraically closed extension of $\mathbb Z_p$ over which we consider the matrix, and the trace of $A^p$ is the sum of the $p$'th powers of the eigenvalues $\lambda_1,\ldots,\lambda_n$ of $A$ (with the same multiplicities). The value we're looking at is, thus, $(\lambda_1^p+\ldots+\lambda_n^p)-(\lambda_1+\ldots+\lambda_n)$, which equals $(\sum\lambda_i)^p-\sum \lambda_i$, and this equals $0$, since $\sum\lambda_i=Tr(A)$ lies in $\mathbb Z_p$.

Grobber, I know this method, I want an elementary one.

An idea: how far can we get mimicing Euler's proof of Fermat's theorem?

Because the trace of a product does not depend on the order of that product,

$Tr\left((A+B)^p\right)=\sum_{k=0}^p{p \choose k}Tr(A^{p-k}B^k) \equiv Tr(A^p) + Tr(B^p)\pmod{p}.$

Is there a useful way of decomposing matrices additively to take advantage of this?

I think your idea works  :D  :
Take two matrices $ A_{1},A_{2} \in M_{n}(\mathbb{Z}) $.
Then $ (A_{1}+A_{2})^{p} = \sum_{\alpha_{1},\alpha_{2}, \ldots ,\alpha_{p}}
A_{\alpha_{1}} A_{\alpha_{2}} \ldots A_{\alpha_{p}} $ where
$ \alpha_{1}, \alpha_{2}, \ldots , \alpha_{p} $ go through $ 1,2 $.
Now the additive cyclic group $ \mathbb{Z}_{p} $ acts on such products
$ A_{\alpha_{1}} A_{\alpha_{2}} \ldots A_{\alpha_{p}} $ by rotations:
$ A_{\alpha_{1}} A_{\alpha_{2}} \ldots A_{\alpha_{p}} \rightarrow
A_{\alpha_{2}} A_{\alpha_{3}} \ldots A_{\alpha_{p}} A_{\alpha_{1}} \rightarrow
\ldots \rightarrow
A_{\alpha_{p}} A_{\alpha_{1}} \ldots A_{\alpha_{p-2}} A_{\alpha_{p-1}}
$
so element $ k \in \mathbb{Z}_{p} $ applied to
$ A_{\alpha_{1}} A_{\alpha_{2}} \ldots A_{\alpha_{p}} $
gives us $ A_{\alpha_{1+k}} A_{\alpha_{2+k}} \ldots A_{\alpha_{p+k}} $.
Now, because $ p $ is prime, every orbit of this action consists of
exactly $ p $ such products, except two orbits $ A_{1}^{p} $ (when
$ \alpha_{1} = \alpha_{2} = \ldots = \alpha_{p} = 1 $ ) and
$ A_{2}^{p} $ (when
$ \alpha_{1} = \alpha_{2} = \ldots = \alpha_{p} = 2 $ ) and
,which consist of one element.
Now, because $ tr(AB) = tr(BA) $ for every matrices $ A,B $, we get that
traces of all such products in given orbit is the same, so modulo $ p $ we get that
$ tr((A_{1} + A_{2})^{p}) \equiv tr(A_{1}^{p}) + tr(A_{2}^{p}) ( mod p ) $.
Therefore in order to prove that $ tr(A^{p}) \equiv tr(A) ( mod p ) $,
it is enough to check the case when $ A $ consists of $ 0 $'s , with the only
$ 1 $ in some place, and it is easy.


I have a request. I would like to copy $LaTeX$ source code here. But sometimes it is quite restricted. Please could you implement the following suggestions:

- displaymath structure to be replaced by $ structure
- textbf and others by the Italic/Fat options here
- etc.

Thanks for your efforts in implementing convenient matters.  :)

Orlando,

I would be happy to fulfill your request as long as you tell me to which packages these commands belong to, and what do they do exactly.

[b]FONTS[/b]

\textrm{...} roman \textsf{...} sans serif
\texttt{...} typewriter
\textmd{...} medium \textbf{...} bold face
\textup{...} upright \textit{...} italic
\textsl{...} slanted \textsc{...} small caps
\emph{...} emphasized \textnormal{...} document font

[b]FONT SIZE[/b]

\tiny tiny font
\scriptsize very small font
\footnotesize quite small font
\small small font
\normalsize normal font
\large large font
\Large larger font
\LARGE very large font
\huge huge
\Huge largest

[b]MATH FONTS[/b]

\mathrm{...} Roman Font
\mathbf{...} Boldface Font
\mathsf{...} Sans Serif Font
\mathtt{...} Typewriter Font
\mathit{...} Italic Font
\mathcal{...} CALLIGRAPHIC FONT
\mathnormal{...} Normal Font

EQUATIONS

\begin{displaymath} .... \end{displaymath} is the same as two $ signs
\begin{gather*} ... \end{gather*}

MISC

\itshape
...

I will tell you when I should miss other language items. Packagges are discussed here:

- http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf
- http://ctan.cict.fr/info/companion-rev/ch8.pdf

Thanks for your efforts, Valentin.  :)

Orl, you still haven't specified any package. Those are commands.

I think this should do for all I need (maybe some graphic stuff):

\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{latexsym}
\usepackage{wasysym}
\usepackage{graphics}


Don't think this is the appropriate section for your inequality.

You can google for some proofs.

Well it's obvious that $\sum_{n=1}^{j} \frac{1}{n^2}$ increases as $j$ increases, so all we need to show is that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.  There are [b]many[/b] ways to do this, but here is a simple one:

The fourier series for $f(x) = x^2$ on the interval $\left[-\pi, \pi \right]$ is $\frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos(nx)\Rightarrow f(\pi) = \pi^2 = \frac{\pi^2}{3} + 4 \sum_{n=1}^{\infty} \frac{1}{n^2} \Rightarrow \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$


trigonometry
geometry
trig identities
Law of Sines
geometry solved

Let $BCD, CAE, ABF$ be triangles outside of triangle $ABC$ such that $\angle BAF = \angle CAE, \angle CBD = \angle ABF, \angle ACE = \angle BCD$. Prove that lines $AD, BE$ and $CF$ are concurent.

Lets use law of sines, many many, but many times

By trigonometric Ceva, we know that they concur iff $\frac{sinEBA}{sinEBC}\frac{sinBCF}{sinFCA}\frac{sinCAD}{sinDAB}= 1$

From triangle$BEA$ we know that $\frac{EA}{BE}=\frac{sinEBA}{sinBAE}$
and from triangle $BEC$ we know that $\frac{EC}{BE}=\frac{sinEBC}{sinBCE}$

Dividing them we know that $\frac{sinEBA}{sinEBC}=\frac{EC}{EA}\frac{sinBCE}{sinBAE}$

And in the same way in the other two sides, multiplying them correctly we get in the right side of the equality what we want to prove and in the other side

$\frac{sinEAB}{sinCAF}\frac{sinCBF}{sinDBA}\frac{sinADC}{sinBCE}\frac{EC}{EA}\frac{FA}{BF}\frac{DB}{DC}$

Look $\angle EAB=\angle CAF$, and consecuently the other two pairs.

So we just have to prove that $\frac{EA}{EC}\frac{FA}{BF}\frac{DB}{DC}=1$

And again by law of sines in triangle $CEA$ we know that $\frac{EC}{EA}=\frac{sinEAC}{sinECA}$. Analog for the others

And the product we had turns to be $\frac{sinEAC}{sinBAF}\frac{sinDCB}{sinECA}\frac{sinABF}{sinDBC}=1$

And that is true, because the problem says it.

Jacobi's theorem is back...

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=186725#p186725]http://www.mathlinks.ro/Forum/viewtopic.php?t=30114 post #2[/url]

  Darij


induction
inequalities unsolved
inequalities

let a,b,c>0 prove that: (the sums are cyclic)
\[ \sum\frac{a^{2008}}{b^{2007}}\geq \sum\frac{a^{2007}}{a^{2006}}\]
(and you can replace 2008 with $ n$)

$ \frac{a^{2008}}{b^{2007}}\plus{}\frac{a^{2006}}{b^{2005}} \geq 2\frac{a^{2007}}{b^{2006}}$
applying this to the other cyclic terms:

$ \sum\frac{a^{2008}}{b^{2007}}\plus{}\sum\frac{a^{2006}}{b^{2005}}\geq 2\sum\frac{a^{2007}}{a^{2006}}$

applying it inductively we get the result.

Note: Our induction base will be: $ \sum\frac{a^2}{b}\geq \sum{a}$
which follows from:
$ \sum\frac{a^2}{b}\plus{}\sum{a}\geq 2\sum{a}$ (AM-GM) as in the beginning.
In the same way can be proved that:
$ \sum\frac{a^m}{b^n} \geq \sum\frac{a^p}{b^q}$
$ m\minus{}n\equal{}p\minus{}q\geq 0$

[quote="M.A"]let a,b,c>0 prove that: (the sums are cyclic)
\[ \sum\frac {a^{2008}}{b^{2007}}\geq \sum\frac {a^{2007}}{a^{2006}}
\]
(and you can replace 2008 with $ n$)[/quote]
  by AM-GM  ineq with 2008 real posotive numbers,  we have :
 $ \frac {a^{2008}}{b^{2007}} \plus{}  2007b  \geq 2008a$
 similar, we  done completly.


function
integration
logarithms
complex analysis
complex numbers
number theory

Can anyone explain to me the Riemann Hypothesis and the Riehmann Zeta Function?

The Riemann Zeta function is defined by

\[
\zeta (t) = \sum_{x = 1}^\infty \frac{1}{x^t}
\]

for real $t >1$.   This can be extended to the whole complex plane  via [url=http://mathworld.wolfram.com/AnalyticContinuation.html]analytic continuation[/url].

A root $r$ of $\zeta (r) = 0$ is called trivial if $r$ is a negative even number (all negative even numbers are roots of the Riemann Zeta function).  Riemann's Hypotheseis states that if $z$ is a non-trivial root of $\zeta (z) = 0$, then $\Re \{z\} = 1/2$.

That's the (very) shortened version.  The Riemann Hypothesis is a huge subject, and I cannot put it in one post.  If you want a whole lot more, see [url=http://mathworld.wolfram.com/RiemannHypothesis.html]MathWorld's page on the Riemann Hypothesis[/url]

Another fact:
Mathematicians have a formula that gives you roughly gives you the number of primes below a certain number, say $n$ If reimann was solved, it would have the effect that the error reading from this formula is less than $ \sqrt{n}$, but you can get as close to this as you like.

Riemann Hypothesis says that $\pi(x)=\text{Li}(x)+O(\sqrt{x}\ln(x))$, where $\pi(x)$ is the number of primes $\le x$ and $\text{Li}(x)=\int_2^x \frac{1}{\ln t}\; dt$. In general, if there are no zeros of $\zeta(s)$ with $\Re s>\sigma$, then $\pi(x)=\text{Li}(x)+O(x^\sigma\ln(x))$.

Actually there is an exact formula for the number of primes less than or equal to $x$, but the third-order term is a sum over all nontrivial zeros of $\zeta(s)$.

[quote="ComplexZeta"]Actually there is an exact formula for the number of primes less than or equal to $x$, but the third-order term is a sum over all nontrivial zeros of $\zeta(s)$.[/quote]

I've never found a straight answer to this question: does the veracity of Riemann's exact formula depend on the Riemann hypothesis?  I've tried reading a little bit of Edwards' book, which is tough given that I know next to no complex analysis, and from what I could tell, it doesn't.  It didn't seem like Riemann actually ever used his hypothesis, only that he conjectured it.

[quote="ComplexZeta"]No, the exact formula is independent of the Riemann Hypothesis. Riemann never used the Riemann Hypothesis in his work: he was interested in the Prime Number Theorem ($\pi(x)\sim\text{Li}(x)$), which only requires that there no zeros with real part 1. (See my last post for the reason.) Incidentally, Riemann wrote exactly one paper on number theory.[/quote]

And it was his one paper that, from my understanding, essentially created analytic number theory!

True. The (mathematical) world was so shaken by Riemann's great 1859 paper that no one really published anything serious related to it until Hadamard and de la Vallee Poussin (independently) proved the Prime Number Theorem in 1896.


[hide="whoa hold it"]
Let's do some conversion first.  I like degrees so we convert to $ \frac{2\tan15}{1\minus{}\tan^2 15}$
Reverse Double Angle to get $ \tan 30$

[/hide]

[quote="jeez123"]Find the exact value of $ \frac {2\tan15^\circ}{1 \minus{} \tan^2\frac {\pi}{12}}$.[/quote]
[hide="Solution"]$ \frac{2\tan15^\circ}{1\minus{}\tan^2\frac{\pi}{12}}\equal{}\frac{2\tan{15^{\circ}}}{1\minus{}\tan^2{15^{\circ}}}\equal{}\tan{30}\equal{}\frac{\sin{30}}{\cos{30}}\equal{}\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\equal{}\frac{1}{\sqrt{3}}\equal{}\boxed{\frac{\sqrt{3}}{3}}$[/hide]


[b] [color=black][/color][color=white]After she had figured out the ages of Blythe's sons, Alexis herself, who has three daughter asked Blythe to attempt to figure out her daghters ages.  "yuor first clue is that the sum of their ages is 11' said Alexis.  That is not enough information" replied Blythe.  " the product of their ages is either 16 years less or 16 years older than your age. "still not enough information" said Blythe after careful considartion. " The daughter whose age, in years, is the greatest is learning how to play chess" said Alexis.  Blythe was then immediatly able to determine the ages of Alexis's three daughters.  Waht are their ages?[/color][/b]

[hide] There are 10 possible combinations of ages, given the first clue. Of these combinations, there must be 2 whose product differs by 32.
$\boxed{4\cdot4\cdot3=48}$.
$5\cdot5\cdot1=25$.
$5\cdot4\cdot2=40$.
$5\cdot3\cdot3=45$.
$6\cdot4\cdot1=24$.
$6\cdot3\cdot2=36$.
$7\cdot3\cdot1=21$.
$7\cdot2\cdot2=28$.
$\boxed{8\cdot2\cdot1=16}$.
$9\cdot1\cdot1=9$.
The boxed ones are the possible ages: $\boxed{4,4,3}$ or $\boxed{8,2,1}$.
A 4-year-old can learn chess.  :huh: [/hide]

and even if they were twins, one was probably be born 10 minutes bfore the other.

but still a nice problem

[quote="lotrgreengrapes7926"][quote="daine"][quote="lotrgreengrapes7926"][hide]
The boxed ones are the possible ages: $\boxed{4,4,3}$ or $\boxed{8,2,1}$.
A 4-year-old can learn chess.  :huh: [/hide][/quote]
[hide]But there are two four year olds and there is only one eldest child according to the riddle.[/hide][/quote]
One of the 4-year-olds could be older.  :|[/quote]

the problem said oldest in years....  :maybe:

[quote="junggi"][/quote][quote="lotrgreengrapes7926"][quote="daine"][quote="lotrgreengrapes7926"][hide]
The boxed ones are the possible ages: $\boxed{4,4,3}$ or $\boxed{8,2,1}$.
A 4-year-old can learn chess.  :huh: [/hide][/quote]
[hide]But there are two four year olds and there is only one eldest child according to the riddle.[/hide][/quote]
One of the 4-year-olds could be older.  :|[/quote][quote="junggi"]

the problem said oldest in years....  :maybe:[/quote]

Exactly.
Even if one 4 year old is older, she does not have a larger age in years.


Let $(a_n)$ be a sequence : $a_n =\sum_{k=n+1}^{\infty}a_k^2$ .
 Prove that : $\sum a_n$ is converge if and only if $a_n =0$ , for all $n \in N$

First, a_n >= 0 for all n.

Let R+ be the set of positive reals.
Define f : R+ -> R+ as f(x) = x^2 + x. Then f is a strictly increasing function.

Suppose a_1 > 0. we want to show \sum a_n diverges.
The given equation can be transformed to the following recurrence relation :
f( a_(n+1) ) = a_n

For each n, we have f(1/(n+1)) < 1/n. This means that a_m > 1/n implies a_(m+1) > 1/(n+1). But the harmonic series diverges, so is the series \sum a_n.


logarithms
inequalities
analytic geometry
graphing lines
slope
inequalities unsolved

If $ 0 < a\leq b\leq c \leq d$, and $ a\neq c, b\neq d$, prove that $ \frac {lna \minus{} lnc}{a \minus{} c}\geq\frac {lnb \minus{} lnd}{b \minus{} d}$.

This is obviosly true because $ f(x) \equal{} \ln{x}$ is concave.  :wink:

Obviously...? By the theorem of Rolle, there exists $ l$ in $ (a,c)$ such that the first one is equal to $ \frac{1}{l}$, and also $ y$ in $ (b,d)$ such that the second one equals $ \frac{1}{y}$, but we don't know the relation between $ l$ and $ y$.

Theorem of Rolle? Never heard of this one...

But, however, if you take the whole inequality geometrically, the LHS is the slope of the chord through $ (a, \ln{a})$ and $ (c, \ln{c})$, RHS is the slope of the chord through $ (b, \ln{b})$ and $ (d, \ln{d})$. Convexity now implies that the inequality is true.  :lol:

\[ \frac{{\ln a \minus{} \ln c}}{{a \minus{} c}} \ge \frac{{\ln a \minus{} \ln d}}{{a \minus{} d}} \ge \frac{{\ln b \minus{} \ln d}}{{b \minus{} d}}\]

[quote="Spribo"]The theorem of Rolle is a special case of M.V.T.[/quote]

Thanks  :) I'll have a look at it...

By the way, I don't know where you have the inequality from; I know it from A. Engel's Problem-Solving Strategies, where the original form is:
If $ 0<a\leq b\leq c\leq d$, then $ a^b b^c c^d d^a\geq b^a c^b d^c a^d$.

I know it from the proof of Karamata's inequality^^

@Martin N.:I also found it in that book!


Does anybody knows a good step by step tutorial website for PRECALCULUS and CALCULUS?

Thanks in advance.

Well, publishers of a calc book usually have great websites. When in doubt, try MathWorld. It has some great resources. There may be some other free online textbooks.  For PreCalc, there are many good websites when you search google.
Hope I helped.

http://archives.math.utk.edu/visual.calculus/index.html is a really good site.

[url]http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/tutindex.html[/url] is also an okay site which covers nearly all maths.

Where can I get tough calculus problems??


Find all functions $f: \mathbb R \to \mathbb R$ such that
\[ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\minus{}2f(xy)\]
for all reals $x$ and $y$.

[quote="aktyw19"]Find all  $ f : R \rightarrow R$
 $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \minus{} 2f(xy)$[/quote]

$ f(x \plus{} y \plus{} z) \equal{} f(x \plus{} (y \plus{} z)) \equal{} f(x) \plus{} f(y \plus{} z)$ $ \minus{} 2f(xy \plus{} xz) \equal{} f(x) \plus{} f(y) \plus{}$ $ f(z) \minus{} 2f(yz) \minus{} 2f(xy)$ $ \minus{} 2f(xz) \plus{} 4f(x^2yz)$

$ f(x \plus{} y \plus{} z) \equal{} f((x \plus{} y) \plus{} z) \equal{} f(x \plus{} y) \plus{} f(z)$ $ \minus{} 2f(xz \plus{} yz) \equal{} f(x) \plus{} f(y)$ $ \plus{} f(z) \minus{} 2f(xy) \minus{} 2f(xz)$ $ \minus{} 2f(yz) \plus{} 4f(xyz^2)$

And so $ f(x^2yz) \equal{} f(xyz^2)$ $ \forall x,y,z$

For $ uv\ne 0$, setting $ x \equal{} \sqrt [3]{\frac {u^2}v}$ and $ y \equal{} 1$ and $ z \equal{} \sqrt [3]{\frac {v^2}u}$, then $ f(x^2yz) \equal{} f(xyz^2)$ becomes $ f(u) \equal{} f(v) \equal{} c$ $ \forall u\ne 0,v\ne0$

Plugging this in the original equation, we get $ c \equal{}0$ and so $ f(x) \equal{} 0$ $ \forall x\ne 0$
Then, using $ x \equal{} 1$ and $ y \equal{} \minus{} 1$, we get $ f(0) \equal{} 0$

And so the unique solution $ f(x) \equal{} 0$ $ \forall x$


Just curious to see where everyone is headed and what you plan to major in.

I'm going to Cornell University on a research scholarship.
Majoring in applied physics and maybe a minor in mathematics.

Congratulations! Feel free to contact me if you have any questions about Cornell Engineering, or Cornell in general.

See also: [url]http://www.mathlinks.ro/viewtopic.php?t=274660[/url]

Mod might merge.

Cornell University
majoring in....not sure yet
i like chem, physics, math, and comp sci so i'll have to explore freshmen year
i'm thinking maybe AEP.


Do harmonic means have any relation to  harmonic sequences?   :blush:

In some special cases, yes. I'm not too sure, however, so I won't elaborate.

An arithmetic mean can be defined as the number exactly in between two numbers in an arithmetic sequence.
For example, take the sequence 3,7,11,15,19,...  The arithmetice mean of 11 and 19 is the number in between them, 15.  The arithmetic mean of 3 and 19 is the number in between them, 11.

A geometric mean can be defined as the number in between two numbers in a geometric sequence.
Take 3,6,12,24,48,...  The geometric mean of 3 and 48 is the number in between them, 12.  Likewise, the geometric mean of 3 and 12 is the number equidistant from them, 6.

Who wants to try harmonic sequences?


calculus
integration
function
real analysis
abstract algebra
real analysis unsolved

The function $f(x)$ satisfies the integral equation

$\int^{\infty}_{-\infty}\frac{f(t)}{(x-t)^{2}+a^{2}}dt = \frac{1}{x^{2}+b^{2}},$

where $a,b$ are real positive constants such that $0<a<b$. Determine the function $f(x)$

I know what i need to do, i'm just having difficulty getting started. I need to express the integral on the left-hand side as a convolution, then get the fourier transform of both sides, invert the transform and determine $f(x)$.

My problem is the convolution. I can't figure out how to do it. I'm basically working from this example:

${\int^{\infty}_{-\infty}f(x-t)f(t) dt = exp\{{-\frac{x^{2}}{2}}}\}$

In this instance, the integral is the convolution $(f*f)(x)$, so when i take the FT i get
${\hat{f}(k)\hat{f}(k) = F[exp\{{-\frac{x^{2}}{2}}}\}]$

In the question i have to do it's not so obvious...any ideas?

Define a function (it's known as the "Poisson kernel") by

$P_{a}(x)=\frac{a}{\pi(x^{2}+a^{2})}.$

I like to put $2\pi$ in the exponent of my definition of the Fourier transform, in which case we get $\widehat{P_{a}}(\xi)=e^{-2\pi a|\xi|}.$ If your $2\pi$ is in a different place, then you get a slightly different form, but it will still contain the exponential of an absolute value.

Your equation now reads

$\frac{\pi}{a}P_{a}*f(x)=\frac{\pi}{b}P_{b}(x).$

Take the Fourier transform:

$\frac{\pi }{a}e^{-2\pi a|\xi|}\widehat{f}(\xi)=\frac{\pi }{b}e^{-2\pi b|\xi|}.$

$\widehat{f}(\xi)=\frac{a}{b}e^{-2\pi(b-a)|\xi|}$

(You should be able to see why we had the restriction $b>a$.)

$f(x)=\frac{a}{b}P_{(b-a)}(x)=\frac{a}{b}\cdot\frac{b-a}{\pi((b-a)^{2}+x^{2})}.$

thanks a lot, this really helped, i would never have thought of using the poisson kernel


How many three-digit numbers are divisible by 13?

This is the arithmetic sequence of integers 104, 117, ..., 988.

This is a total of $ \frac{988 \minus{} 104}{13} \plus{} 1 \equal{} \boxed{69}$ numbers.


How many ordered triples of integers $ (a, b, c)$ satisfy
\[ |a + b| + c = 19 \ \text{and} \  ab + |c| = 97 \ \ \texttt{?}
\]
(A) $ 0$ (B) $ 4$ (C) $ 6$ (D) $ 10$ (E) $ 12$

Thanks a lot!

[hide="Initial Thoughts"]We have three variables and two equations.  This means that we are going to have to use the fact that a, b, and c are integers.  We will also probably do case work with the absolute values.[/hide]
[hide="Solution"]From the first equation
$ c\equal{}19\minus{}|a\plus{}b|$
Plugging this into the second equation we get
$ ab\plus{}|19\minus{}|a\plus{}b| |\equal{}97$
The two possibilities from this equation are
$ 19\minus{}|a\plus{}b|\equal{}97\minus{}ab$ and $ 19\minus{}|a\plus{}b|\equal{}ab\minus{}97$
And the four possibilities from these equations are
$ ab\minus{}78\equal{}a\plus{}b$, $ 78\minus{}ab\equal{}a\plus{}b$, $ 116\minus{}ab\equal{}a\plus{}b$, and $ ab\minus{}116\equal{}a\plus{}b$
Factoring each of these gives
$ (a\minus{}1)(b\minus{}1)\equal{}79$, $ (a\plus{}1)(b\plus{}1)\equal{}79$, $ (a\plus{}1)(b\plus{}1)\equal{}117$, and $ (a\minus{}1)(b\minus{}1)\equal{}117$
Because 79 is prime we have no solutions from the first two equations.  Then all we need to focus on are the second two equations.
Because $ 117\equal{}3*3*13$ we have 3 pairs of factors to consider and 6 pairs of a and b to consider because a and b are symmetric.
$ 1, 117 \implies a\equal{}0, b\equal{}116 \text{ or } a\equal{}2, b\equal{}118$
$ 3, 39 \implies a\equal{}2, b\equal{}38 \text{ or } a\equal{}4, b\equal{}40$
$ 9, 13 \implies a\equal{}8, b\equal{}12 \text{ or } a\equal{}10, b\equal{}14$
From the second equation we know that $ ab\leq97$ because $ |c|\geq0$.
This means that the only pairs of a and b that will work are $ a\equal{}0, b\equal{}116 \text{ and } a\equal{}8, b\equal{}12$
Because a and b are symmetric, we have 4 possibilities $ \implies \boxed{B}$[/hide]


function
calculus
derivative
inequalities
inequalities unsolved

hello mates i has eniquality veri nice good.
prove
$ xyz\geq x\plus{}y\plus{}z\minus{}2$

It's false. Try $ x \equal{} 3$, $ y \equal{} 0.01$, $ z \equal{} 0.01$. You will have: $ 0.0003\geq 1.02$, which obviously false.

you lier! dont try to make me rong! if you tri x=1y=1z=1 it is corect so dont spred ur lies to te comunitti

 :mad:  :mad:  :mad:

Welcome to MathLinks, Paboto. :lol:

Paboto, you forget to write some condition, didn't you.

Maybe the problem is the following.

[color=blue]Let $ x,\ y,\ z$ be real numbers such that $ |x| \leq 1,\ |y| \leq 1,\ |z| \leq 1$. Prove that $ xyz \geq x \plus{} y \plus{} z \minus{} 2$.[/color]

then it's obvious
$ xyz \ge \minus{}1 \ge x\plus{}y\plus{}z\minus{}2$

I think it's obvious but not that way. Why $ x\plus{}y\plus{}z\minus{}2 \leq \minus{}1$ ?

[quote="Lyons"]How is this ineq trivial? I can't get it. :([/quote]
We can use function $ F(x,y,z) \equal{} xyz \minus{} x \minus{} y \minus{} z \plus{} 2$ for $ \minus{} 1\leq x,y,z \leq 1$, and find extremums. In extremums: $ \frac {dF}{dx} \equal{} 0$, $ \frac {dF}{dy} \equal{} 0$, $ \frac {dF}{dz} \equal{} 0$. So, $ x \equal{} y \equal{} z \equal{} \minus{} 1$, or $ x \equal{} y \equal{} z \equal{} 1$, and $ F_{extr} \equal{} 0$. And easy to see that it's minimum.

Can someone solve the problem which I posted in #6 without differentiation?

Problem:$ |x|,|y|,|z|$≤$ 1$.Prove that $ xyz\geq x\plus{}y\plus{}z\minus{}2$
My solution:
$ \minus{}1$≤$ yz$≤$ 1$
=>$ yz\minus{}1$≤$ 0$
Consider $ f(x)\equal{}x(1\minus{}yz)\plus{}y\plus{}z\minus{}2$.Since $ 1\minus{}yz \geq 0$,$ f(x)$ will increase if $ x$ increase.
=>$ f(x)$≤$ f(1)\equal{}1\minus{}yz\plus{}y\plus{}z\minus{}2\equal{}y\plus{}z\minus{}yz\minus{}1\equal{}\minus{}(1\minus{}y)(1\minus{}z)$≤$ 0$.

That's right.

We can prove the inequality using the result $ ab>a\plus{}b\minus{}1$ with $ |a|\leq 1,\ |b|\leq 1$.


Find the largest k such that 3^11 is the sum of k consecutive positive integers.

[hide]m is the middle term, k is the number of terms. note that if k is odd, m will be an integer, and if k is even, m will be an integer plus a half. km=3^11, and since m can either be an integer or an integer divided by 2, kn=2*3^11, where n=2m, and k and n are integers. This means k is either of the form 3^i or 2*3^i. In the first case, m=3^(11-i), and in the second case, 2m=(3^(11-i)) any choices for k and m will give consecutive positive integers iff k/2<m => k<2m. Scanning the above cases, we find the largest k can be is 2*3^5=486.correct?[/hide]

erm, Mystic, where did the 2 come from in your first kn expression? I think it should be /2 instead of *2

i dont think so.  i think he's right.  What he did was take the sum of the arithmetic sequence, and to get rid of the /2 on the LHS, he multipled by 2 on the RHS.

Oh geeze. I'm such a moron. I know what I was thinking, but it was obviously wrong. Eh, that's what you get when you try to do math in your head when you're pressed for time, I guess.

[hide]2x3^5=486

ugh, im mad because i was writing a solution for some time but then i lost it. grrr. maybe ill write it later unless someone else comes and posts the same thing. but grrr

[/hide]

ill give an outlined solution to my answer:

[hide]it had to do with summing series of the form k(2n+k)/2=3^11, maximizing, and considering cases where k is odd and even. nuff said on my part[/hide]


What is angle chasing? From the context, I think it's trying to find an angle... but is that true (if that's the case, i've done a lot of angle chasing)?

There is a chapter about this in AOPS vol 1.  Angle chasing is simply the act of finding the measures of desired angles (as the names suggests), but it requires only the knowledge of fundamental tools in geomtry.  By fundamental tools I am reffering to complementry and supplementary angles, angles in a line, angles in a circle, angles in a triangle etc.  So in a way its sort of like simplifying, reducing things so they can be solved from established knowledge.

[quote="G-UNIT"]There is a chapter about this in AOPS vol 1.  .[/quote]

Yeah, I just ordered the book.

Oh, it seems like I was close- I've done so many of these problems  :D

What "skills" are there to angle chasing?

You need to be aquainted with most of the angle theorems (I mentioned a few above).  You also must have a keen eye for spotting angles that can be found, not necesseraily the one you want, but any angle that will put things in prospective.  Drawing parallel lines never hurts either, in fact its a good tool for creating extra opposite and interior angles.

well, knowing many properties of circles and triangles certainly helps. but, for harder problems, there is some creativity involved. you have to know how to attack the problem. that's where problem-solving comes along. :)

I really hate it when you have to construct things in the figure. How do you know when to do that (or is that in the art of angle chasing)?

by construct, do you mean add or extend a line somewhere? if so, that just takes experience. once again, that goes under creativity. but, usually, you don't have to do that for the easier problems.

[quote="h78reg"]by construct, do you mean add or extend a line somewhere? [/quote]

Yeah, everytime I get those problems, I'm stuck.  :blush:

[quote]Yeah, everytime I get those problems, I'm stuck. [/quote]

well, i'm still not too good at those problems, but, AOPS volume 1 helped a lot.

Angle chasing is essentially drawing lines, making geometrical inferations, finding some thing out, or a process of geometical (and sometimes algebraical!!!) steps in finding out an angle, or a relationship between to angles.


Let $ \{x_n\}_{n\geq 1}$ be a sequences, given by $ x_1 \equal{} 1$, $ x_2 \equal{} 2$ and
\[ x_{n \plus{} 2} \equal{} \frac { x_{n \plus{} 1}^2 \plus{} 3 }{x_n} .
\]
Prove that $ x_{2008}$ is the sum of two perfect squares.

From the recurrent relation, we have:
\begin{eqnarray*}x_{n+2}x_n&=&x_{n+1}^2+3\\x_{n}^2+3&=&x_{n+1}x_{n-1}\end{eqnarray*}
Adding the equations and factoring yields \[ \frac{x_{n}+x_{n+2}}{x_{n+1}+x_{n-1}}=\frac{x_{n+1}}{x_n}\qquad(1)\]
If we substitute $ n-1$ for $ n$ in $ (1)$, we find that \[ \frac{x_{n-1}+x_{n+1}}{x_{n}+x_{n-2}}=\frac{x_{n}}{x_{n-1}}\qquad(2)\]
In fact, we can generate $ n-1$ of these equalities, which will all telescope when multiplied together. Thus:\begin{eqnarray*}\left(\frac{x_{n}+x_{n+2}}{x_{n+1}+x_{n-1}}\right)\times\left(\frac{x_{n-1}+x_{n+1}}{x_{n}+x_{n-2}}\right)\times\cdots\times\left(\frac{x_{2}+x_{4}}{x_{3}+x_{1}}\right)&=&\frac{x_{n+1}}{x_n}\times\frac{x_{n}}{x_{n-1}}\times\cdots\times\frac{x_3}{x_2}\\&\iff &x_{n}+x_{n+2}=\left(\frac{x_3+x_1}{x_2}\right)\cdot x_{n+1}\\&\iff&x_{n+2}-4x_{n+1}+x_n=0\qquad(3)\end{eqnarray*}
Thus $ x_{2n+1}=2x_{n+1}^2-1$.

The two recurrent equations established thus far are enough to establish these two properties of $ \{x_n\}$:\[ x_{2n-1}=2x_n^2-1\qquad \text{ and } \qquad x_{2n}=\left(x_{n+1}-x_n\right)^2+1\qquad(3)\]
When $ n=1$, we check that $ x_1=2-1$ and $ x_2=(2-1)^2+1$ as claimed.

Suppose the results in $ (4)$ hold up until $ n=k$. Using the equation $ (3)$, we compute \begin{eqnarray*}\frac{x_{2n+1}-(2x_{n+1}^2-1)}{2}&&\qquad\text{(trying to show this is exactly zero)}\\&=&\frac{(4x_{n+1}^2+4x_{n}^2-8x_nx_{n+1}+4) - (2x_n^2-1) - (2x_{n+1}^2-1)}{2}\\&=& x_n^2+x_{n+1}^2+3-4x_{n+1}x_n\qquad\text{(by the original recurrent equation)}\\&=&x_n^2+x_nx_{n+2}-4x_{n+1}x_n\\&=&x_n(x_{n+2}-4x_{n+1}+x_n)=0\end{eqnarray*}

To complete the recursion, it remains to show the result for $ x_{2n+2}$:
\begin{eqnarray*}
x_{2n+2}+x_{2n}&=&4x_{2n+1}=8x_{n+1}^2-4\\
&=&8x_{n+1}^2+2(x_{n+1}^2-x_{n}x_{n+2})+2\qquad\text{(original recursion)}\\
&=&16x_{n+1}^2-2x_nx_{n+2}+2x_{n+1}^2-8x_{n+1}^2+2\\
&=&(x_n+x_{n+2})^2-2x_nx_{n+2}+2x_{n+1}^2-2x_{n+1}(x_n+x_{n+2})+2\qquad\text{(using derived linear recursion)}\\
&=&2x_{n+1}^2+x_n^2+x_{n+2}^2-2x_{n+1}(x_n+x_{n+2})+2\\
&=&(x_{n+1}-x_n)^2+1
+(x_{n+2}-x_{n+1})^2+1\\
&=&x_{2n}+(x_{n+2}-x_{n+1})^2+1\\
\end{eqnarray*}
This completes the recursion. Hence, $ \boxed{x_{2008}=(x_{1005}-x_{1004})^2+1}$ is precisely the sum of two squares$ \square$

I will prove by induction that $ x_{n+2}=4x_{n+1}-x_{n}$ for all positive integers $ n$.

Base Case: $ n=1$. Calculate $ x_3=\frac{x_2^2+3}{x_1}=\frac{2^2+3}{1}=7=4(2)-1=4x_2-x_1$.

Inductive Step: Assume that $ x_{n+1}=4x_{n}-x_{n-1}$. Then

\begin{align*}
x_{n+2}x_{n}-x_{n+1}^2&=3\\
x_{n+1}x_{n-1}-x_{n}^2&=3\\
x_{n+2}x_{n}-x_{n+1}^2&=x_{n+1}x_{n-1}-x_{n}^2\\
x_{n+2}x_{n}-x_{n+1}(4x_{n}-x_{n-1})&=x_{n+1}x_{n-1}-x_{n}^2\\
x_{n+2}x_{n}-4x_{n+1}x_{n}&=-x_{n}^2\\
x_{n+2}-4x_{n+1}&=-x_{n}\\
x_{n+2}&=4x_{n+1}-x_{n}\\
\end{align*}

(Dividing by $ x_n$ is legal, because it is not zero; if $ n=1$ or $ 2$, then $ x_n=1$ or $ 2$, and if $ n>2$, then $ x_{n}=\frac{x_{n-1}^2+3}{x_{n-2}}$ and $ x_{n-1}^2+3\ge 3$ cannot be zero.)

This completes the induction.

This shows that $ x_1,x_2,...$ is an integer sequence, since if $ x_{n}$ and $ x_{n+1}$ are integers, then $ x_{n+2}=4x_{n+1}-x_{n}$ is an integer. Now, I will now show by induction that $ x_n-1$ is a perfect square for even $ n$.

Base Case: $ n=2$. We have $ x_2-1=2-1=1^2$ is a perfect square.

Inductive Step: Assume that $ x_{n}-1$ is a perfect square; I will show that $ x_{n+2}-1$ is as well.

\begin{align*}
(x_{n}-1)(x_{n+2}-1)&=x_{n}x_{n+2}-(x_{n}+x_{n+2})+1\\
(x_{n}-1)(x_{n+2}-1)&=(x_{n+1}^2+3)-4x_{n+1}+1\\
(x_{n}-1)(x_{n+2}-1)&=x_{n+1}^2-4x_{n+1}+4\\
(x_{n}-1)(x_{n+2}-1)&=(x_{n+1}-2)^2\\
\end{align*}

Now, since $ (x_{n+1}-2)^2$ and $ x_{n}-1$ are perfect squares, and $ x_{n+2}-1$ is an integer it must also be a perfect square. This completes the induction.

Now, we have that $ x_{2008}-1$ is a perfect square, and since $ 1$ is a perfect square as well, $ x_{2008}$ is the sum of two squares.

We write the condition in the form $ x_nx_{n\plus{}2}\equal{}x_{n\plus{}1}^2\plus{}3$ and if we write it for $ n$ the
$ n\minus{}1$ we get $ x_{n\minus{}1}x_{n\plus{}1}\equal{}x_n^2\plus{}3$ and substracting these two we have that
$ \frac{x_n}{x_{n\plus{}1}\plus{}x_{n\minus{}1}}\equal{}\frac{x_{n\plus{}1}}{x_{n\plus{}2}\plus{}x_n}\equal{}\frac{x_2}{x_3\plus{}x_1}\equal{}\frac{1}{4}$
so $ x_{n\plus{}1}\equal{}4x_n\minus{}x_{n\minus{}1}$ and from here easily we obtain that $ x_{n\plus{}2}\equal{}14x_n\minus{}x_{n\minus{}2}$
so if $ y_n\equal{}x_{2n}$ then we have that $ y_{n}\equal{}14y_{n\minus{}1}\minus{}y_{n\minus{}2}$
Now we will prove by induction that there exist a sequence $ b_n$ with $ b_{n\plus{}1}\equal{}4b_n\minus{}b_{n\minus{}1}$ and $ b_1\equal{}1$ and $ b_2\equal{}5$ such that $ y_n\equal{}b_n^2\plus{}1$ which will be the desired .
The result is obvious for $ n\equal{}1$ and $ n\equal{}2$ so by strong induction we suppose that it is true for all $ k\leq n$ and we will prove it for $ n\plus{}1$ .
We have that $ y_{n\plus{}1}\equal{}15y_n\minus{}15y_{n\minus{}1}\minus{}y_{n\minus{}2}$ or
$ y_{n\plus{}1}\minus{}1\equal{}15(y_n\minus{}1)\minus{}15(y_{n\minus{}1}\minus{}1)\plus{}y_{n\minus{}2}\minus{}1$ or $ y_{n\plus{}1}\minus{}1\equal{}15b_n^2\minus{}15b_{n\minus{}1}^2\plus{}b_{n\minus{}2}^2$

or $ y_{n\plus{}1}\minus{}1\equal{}15b_n^2\minus{}15b_{n\minus{}1}^2\plus{}(4b_{n\minus{}1}\minus{}b_n)^2$ which means that
$ y_{n\plus{}1}\minus{}1\equal{}(4b_n\minus{}b_{n\minus{}1})^2\equal{}b_{n\plus{}1}^2$ which is what we want and we are done .


If $ B\equal{}\left[\begin {array}{cc} \minus{}12&14\\\minus{}27&27 \end {array} \right]$ then find $ B^n\equal{}?$ when $ n\in N$

Diagonalize it! (It is diagonalizable.) First step: what are its eigenvalues and its eigenvectors? (Hint: the eigenvalues are integers, which makes the arithmetic easier.)


MIT
college
Stanford
Princeton
number theory

Ok, since it was mentioned in the other thread, but buried below some other comments, I decided to start a new thread for it.

Of thise who got accepted to MIT, who is planning on going to Campus Preview Weekend?

I am, and am curious to see who else is.

i'll be up there =P

maybe some of the aops people should meet at some point?

I'm going, and yes I think that'd be a good idea.

(that was kinda one of the eventual goals of this thread)

i'll most likely be there.

now let's start planning this shindig.

we should meet RIGHT NOW.

but we live far away from each other.

Let's not all be too anxious -.-. But anyway, if you're going kindly put your name on our facebook MIT CPW event. I don't have a link on me so i'll have to see about getting one later, but do sign up.

I will of course be around organizing activities and schedules for those hosted at my fraternity.

I'll be there as well.  We can all go wake Samson up at 3 pm on Saturday :D (jk)

Let's see, I know for sure that I'll be at the Battle of the Bands event as will towersfreak2006 (Samson).  Though, we'll be helping out with it since it's sponsored by our frat

To shamelessly plug my frat again :D you are always welcome to meet up at our frat.  There's plenty to do e.g. movies, tv, computer games, Wii, possibly xbox, ps, pool, foosball, air hockey, math, hanging out on the rooftop deck, touring Boston, and so on.

Now for my dorm:  Simmons Hall is having an arcade night.  Pool, ping pong, air hockey, foosball, skee ball, cards, ddr...yeah it should be awesome.

Or there will probably be some fun events at Random Hall where you already have like 50 USAMOers, e.g. I think they usually make liquid nitrogen ice cream.

Anyways, those are just a few options.  I'll try to show up at whatever event you guys decide to meet at.

Sorry for killing the thread :P

One more week!  Wonder what the weather will be like...

Yeah, so where's the meeting gonna be?  How about before or after the "President's Welcome & Faculty Keynote Lecture" on Friday which takes place from 9-10:30 am?  I would suggest beforehand at 8:30 am.  Kresge's the building that looks like part of a sphere.

Or if anyone has another idea...

Just let me know where you'll be.

Wasn't much to kill. -.- Anyway, I'm going to arrive on Wednesday night (advantage of being homeschooled eh?) but apparently there isn't much to do on Wednesday night. :-/ We should probably meet sooner than later, Thurs. perhaps?

Yeah, I'll be there.

I guess a Thursday meeting makes sense since there are fewer activities built into the cpw schedule that day. Maybe Thursday afternoon?

I also will be there- initials DLR - i'm a prospective student, and there is a pretty good chance you can find me wearing jeans and a red T-shirt that says "Deerfoot Lodge" in small white letters on the front corner. If that doesn't work... brown hair, brown eyes, 5'11'', 152lb, lol. I look forward to meeting some of you!

[quote="lotrfan"]If that doesn't work... brown hair, brown eyes, 5'11'', 152lb, lol. I look forward to meeting some of you![/quote]

That basically describes me as well :P (except I'm probably more like 5'10'').

If you really wanted to know what I looked like there are definitely pictures of me online that you should be able to find...

Um, so any planned meeting?  Or just randomness?

Just out of curiosity, where are people hosted?

[quote="joml88"][That basically describes me as well :P (except I'm probably more like 5'10'').[/quote]

haha, im probably more like 5'10'' too.

but anyway, what is the first place where everyone is going to be together, like for an introductory speech or something, if there is one? because we could just try to show up for that a few minutes early and come up with some landmark at which to congregate.

[quote="joml88"]There's a number theory talk from 5:30-7:00 on Thursday with a math mixer where you can talk to current math majors at MIT right after that in the same room from 7:00-8:00.  Free pizza will be served at the beginning of the number theory talk.  I say we meet there (in room 2-102) at 5:00 pm.  So count on me being there for sure at 5. :)[/quote]

good idea   :)

Sorry if this is posted before...

:) Enjoy::)

[url=http://tech.caltech.edu/TECH/mit/index.html]MIT Tech[/url]

haha, yeah.  Props to them.  That was a pretty good one in my opinion.

Oh man, CPW's nearing an end.  It's been sooooo awesome.  I met a few AoPSer including drunner who stayed with me.  It's been so much fun.  Not quite over yet though :D  Battle of the Bands tonight and then about 10 parties are taking place. :) :) :)

ya... so when the pizza ran out at the Number Theory thing I just left, lol. I hadn't had lunch and I was REALLY hungry. Of course the large calzone I ordered at the student center turned out to be a whole pizza folded in half, so that filled me up at least! So i didn't meet any of you, but CPW was still awesome.

I have the feeling that some people with several admission offers by top universities decided in favour for MIT based on MIT CPW. How is that? Is it more fun or whatever than Stanford Admit Weekend, Princeton Preview etc.?

four words: liquid nitrogen ice cream. :D

Did ny one give you the direction from MIT campus to Paris:

Go to maps.google.com
Type "MIT to paris"

[url=http://maps.google.com/maps?f=q&hl=en&q=MIT++to+++Paris&layer=&sll=44.715514,-40.957031&sspn=58.796864,164.53125&ie=UTF8&z=4&om=1]Enjoy[/url]

[quote="MithsApprentice"][quote="orl"]I have the feeling that some people with several admission offers by top universities decided in favour for MIT based on MIT CPW.[/quote]

I think almost all the kids here (definitely all the ones I talked to) were pretty much set on MIT after CPW. I was leaning away before I got here actually...but they definitely left a great impression (gj Joe - btw tell towers not to eat too much pork loin tonight -.- ) despite the terrible weather.[/quote]

Anymore reasons beside ice cream why MIT CPW has a major impact on yield rate compared to other short-time college visits?

It's probably the people here.  That was the biggest attraction for me and it's what makes me love MIT so much now.  Prestige and academics can only take you so far.  I think that the stereotype for MIT is something along the lines of a bunch of masochistic nerds who never sleep because we're always tooling.  This isn't quite true, and when you come to MIT people find out.  It's more like we never sleep because we always put fun before sleep...and we're not really masochists :D

Then one might wonder if everyone just puts on a show while prefrosh are here for CPW.  And this certainly doesn't happen.  I wouldn't sacrifice a little bit of my academics to have fun and try to get people to come to MIT.  Everything that happens at CPW happens during the regular year.  It's just condensed into a few days for CPW.  Furthermore, when you come to MIT you can tell how passionate we are.  There's no show...we just love this place (no matter how much we HTFP :P ).

Finally I should mention the diversity at MIT.  Whatever you're into, there's someone else here also into it.  At CPW the activities range all over the place.  You have all the usual tours of campus, labs, classes, and dorms.  Then you have the stuff like liquid nitrogen ice cream, mooninite LED construction, hacking tours, gaming, sports, and parties.

Btw, this was my first CPW but as you could probably tell, I was won over by MIT just during the regular school year (and orientation).

[quote="orl"]Joseph, thanks for reply why MIT CPW seems to be successful. Is there anybody who is willing to compare it with those events of similar colleges, e.g. Stanford Admit Weekend, Princeton April Hosting? What are they missing out, fail to do during these events? Maybe my observation is biased towards mathy kids who took math/science/cs competitions during high school but were living in remote states of the US and now see the chance to study together at a common place and MIT seems to have built a reputation for that.[/quote]
Unfortunately you're not going to find anyone who can compare MIT CPW to Princeton April Hosting as they happened at the same time this year.

Chances are that people who were more interested in MIT went to the MIT one and people more interested in Princeton went to the Princeton one; thus, it's not too surprising if people choose the university whose visitation weekend they attended.


Let a,b,c>0 such that   abc=1                                Prove that
$ 4(\sum\frac{a}{b\plus{}c})\plus{}3  \leq (a\plus{}b\plus{}c)^2$

$ 4(\sum \frac {a}{b \plus{} c}) \le \frac {\sum a^2 \plus{} 3\sum ab}{2}$
~> Done
[quote]
:( .Wrong again :((
[/quote]

[quote="tranquoc"]Let a,b,c>0 such that   abc=1                                Prove that
$ 4(\sum\frac {a}{b \plus{} c}) \plus{} 3 \leq (a \plus{} b \plus{} c)^2$[/quote]

$ 4(\sum\frac {a}{b \plus{} c})\plus{}3\leq 2\sum\frac{a}{\sqrt{bc}}\plus{}3\equal{}2\sum a\sqrt{a}\plus{}3\leq (a\plus{}b\plus{}c)^2$  :)

[quote="tranquoc"]Let a,b,c>0 such that   abc=1                                Prove that
$ 4(\sum\frac {a}{b \plus{} c}) \plus{} 3 \leq (a \plus{} b \plus{} c)^2$[/quote]
$ 5\sum_{cyc}\frac {a}{b \plus{} c} \plus{} \frac {3}{2}\leq(a \plus{} b \plus{} c)^2$ is true too,
But $ 6\sum_{cyc}\frac {a}{b \plus{} c}\leq(a \plus{} b \plus{} c)^2$ is wrong already
and $ 108\left(\sum_{cyc}\frac {a}{b \plus{} c}\right)^2\leq(a \plus{} b \plus{} c)^5$ is true already.   :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201500

Let a,b,c>o such that abc=1 prove that
$ \sum\frac{\sqrt{2(a^2\plus{}b^2)}}{c}\geq (a\plus{}b\plus{}c)^2\minus{}3$


Let $ABC$ be an acute triangle.We constructs the squares inscribed the triangle $ABC$ on $BC,CA,AB$ with center $A_{1},B_{1},C_{1}$,respectively.
Show that $AA_{1},BB_{1},CC_{1}$ are concurrent.
(Note that the square $MNPQ$ is constructed on $BC$ and inscribed the triangle $ABC$ iff $M,N\in BC$,$P\in CA$ and $Q\in AB$).

If the Squares are constructed out the triangle then we can solve it by using transformations
 Abdurashid


inequalities
function
quadratics
geometric series
algebra
difference of squares
special factorizations

[quote="ArcticMonkey"]You can factor the expresion $x^n+y^n$ only when n is an odd number. Otherwise you can try to use $x^2+y^2=(x+y)^2-2xy$. It helps sometimes. ;-)[/quote]

And to go further, $(x+y)^2-(\sqrt{2xy})^2=(x+y+\sqrt{2xy})(x+y-\sqrt{2xy})$.

[quote="myc"]$x^3+y^3= (x+y)(x^2-xy+y^2)$,

but how would you factor $x^4 +y^4$ ?  Is there a formula?

Thanks  :)[/quote]

No, there is not a formula.  However, you can always substitute $a=x^2$ and $b=y^2$ into the equation to reduce the degree.

It could be difference of squares.  They both mean the same thing practically.

[quote="jclarkemathL314159"][quote="Nekruzjon_eko"]$x^n+y^n$ is factorable if and only if $n$ is odd.[/quote]

Is that provable by a middle schooler?[/quote]

The proof isn't difficult.  It uses summation of a finite geometric series - certainly a middle school topic when handled properly :)

Edit:  This applies to the proof that $x + y | x^{2k+1} + y^{2k+1}$.  The proof that $x^{2n} + y^{2n}$ isn't factorable is less easy.

[quote="lotrgreengrapes7926"][quote="ShoeFactory"][quote="t0rajir0u"]There isn't a way to factor it with integer real coefficients.

Here's a nice related formula, however.  It's called the Sophie Germaine Identity.

$x^4 + 4y^4 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$

:)[/quote]

Or you can not remember that and use difference of squares.[/quote]
Except it's sum of squares.[/quote]

You add something to it to create a difference of squares so that you can factor.
$x^4 + 4y^4$
$x^4 + 4y^4 + 4x^2y^2 - 4x^2y^2$
$(x^2+2y^2)^2 - (2xy)^2$
$(x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$

how do you factor $x^5+y^5$? I just guessed and check.

 I knew there has to be a $(x^2-y^2)(x^3+something -y^3)$

by guessing  I got $(x^2-y^2)(x^3+xy^2+yx^2-y^3)$
Is there an easier way?

  and how would you start off with the proof? ( $x^n+y^n$  is factorable only and only if n is odd).

Thanks for all the help  :P

It can very easily be the difference of squares try it there is no difference what you end up with is (x+y)(x-y)(x+y)(x-y)

There's a much nicer proof.  Using the Factor Theorem, we're looking for roots of \[x^{n}=-y^{n}.\] When $n$ is even, then by the Trivial Inequality, the RHS and LHS are only equal when $x = y = 0$, not yielding any linear functions for which the equation is identically true.

However, when $n$ is odd, we have the solution $x =-y$, which is a linear function, so $x+y = 0$, hence $x+y$ is a factor.

In regards to $x^{4}+4y^{4}$, we find the Sophie Germaine factors from $x^{4}=-4y^{4}$ and solving for $x$ in terms of $y$ using De Moivre and complex numbers.  The four solutions make up two pairs of complex conjugates.  Each pair represents the roots of a quadratic equation with real coefficients, hence the quadratic factors.

When looking for factors, think Factor Theorem.

[quote="t0rajir0u"][quote="t0rajir0u"]The proof that $x^{2n}+y^{2n}$ isn't factorable is less easy.[/quote]

I don't know why I said this, considering it's not true.  For example,

$x^{6}+y^{6}= (x^{2}+y^{2})(x^{4}-x^{2}y^{2}+y^{4})$[/quote]

Oops, yeah, reading over this I bit too, despite the fact that I just a section about it in the Intermediate Algebra book.  There are necessarily no linear factors with real coefficients, but substituting to take out all even exponents (like $a = x^{2}$ and $b = y^{2}$) can help draw out linear factors with real coefficients in the new variables.

[quote="t0rajir0u"][quote="t0rajir0u"]The proof that $x^{2n}+y^{2n}$ isn't factorable is less easy.[/quote]

I don't know why I said this, considering it's not true.  For example,

$x^{6}+y^{6}= (x^{2}+y^{2})(x^{4}-x^{2}y^{2}+y^{4})$[/quote]

Did you mean to say $x^{ 2^{n}}+y^{2^{n}}$ ?


limit
LaTeX
integration
logarithms
function
calculus
algebra unsolved

\lim_n { ( (1+\frac{1}{n})*(1+\frac{2}{n})^{2}*...*(1+\frac{n}{n})^{n} )^{\frac{1}{n^2}}  }

I hope you get it from the latex code (i m  new here so I dont get it, how to post something in latex : use [latex] [/latex] switch??)

help

If $L=\lim_{n\to\infty}\prod_{k=1}^{n}\left(1+{k\over n}\right)^{k/n^{2}}$, then

$\ln L=\lim_{n\to\infty}\sum_{k=1}^{n}{k\over n^{2}}\ln\left(1+{k\over n}\right)=\lim_{n\to\infty}\sum_{k=1}^{n}\left({1\over n}\cdot{k\over n^{}}\ln\left(1+{k\over n}\right)\right)$

which is Riemann's sum for the function $f(x)=x\ln(1+x)$ on the segment $[0,1]$. Hence

$\ln L=\int_{0}^{1}x\ln(1+x)\mathrm{d}x=\mathrm{[integration\ by\ parts]}={1\over 4}$

Hence $L=e^{1/4}$

EDIT: Beaten coz of the wordiness :P


[quote="ashrafmod"]find the pair of positive integers (m,n)for which
$m^{2}-n^{2}=1995$[/quote]
[hide]$(m+n)(m-n)=1995$
(998,997)
(334,331)
(202,197)
(146,139)
(74,59)
You can find the rest of pairs on your own...[/hide]

Pair? There are many pairs.

[hide="solution"]

We see that

(m-n)(m+n)=1995=3*5*7*19

We can proceed with casework.

There is the case thar m-n=1, m=998 and n=997.

The next case is that m-n=3 and m+n=665. That way, m=334 and n=331.

m-n=5, m+n=399, m=202, n=197.

m-n=7, m+n=285, m=146, n=139.

m-n=19, m+n=105, m=62, n=43.

m-n=15, m+n=133, m=74, n=59.

m-n=21, m+n=95, m=58, n=37.

m-n=57, m+n=35, m=46, n=-11. uh-oh, n is negative. we can't count that.

m-n=105, m+n=19  n is negative again.

If m-n is greater than m+n, then n is ngative. The remaining cases make n negative. So here are all of the pairs:

(998,997)
(334,331)
(202,197)
(146,139)
(74,59)
(62,43)
(58,37)[/hide]


[quote="Kle"]Verify that y^4=2e^(-x^2)+Ce^(-2x^2) is a solution ti the differential equation y'+xy=xe^(-x)y^(-3)
thankx[/quote]
Hm. It's very easy , isn't it ? Do it yourself !

thank you for ur help or merely mocking statements..... anh_steiner.. I hope you understand English too.... thankx

Well, to "verify", you just have to replace the given "y" in the diff. equation..  :roll:

I think's it good to give little help  :D . I'll let you do the verification. ;)
Well Kle,  perhaps you know implicit differentiation.
if we have $f(y)$ the function in $y$ and $y$ is the function in $x$, then by chain rule $\frac{df}{dx} = \frac{df}{dy} \frac{dy}{dx}$.
Now in your problem let $f(y)=y^4$. Then $\frac{df}{dx} = 4y^3 \frac{dy}{x} = 4y^3 y'$
Since $y^4 =2e^{(-x^2)}+C e^{(-2x^2)}$ differentiatied with respect to $x$ give $\frac{d(y^4)}{dx}=\frac{df}{dx}=-4xe^{(-x^2)}-4xCe^{(-2x^2)}$.
So We have $4y^3 y' = -4xe^{(-x^2)}-4xCe^{(-2x^2)} \Rightarrow y' = \frac{-4xe^{(-x^2)}-4xCe^{(-2x^2)}}{4y^3 }$

now Plug $y'$ to $y'+xy=xe^{(-x)}y^{(-3)}$. And tell me what is you get? :)


ratio
algorithm
floor function
geometric sequence
continued fraction
irrational number

We all know what division is, right?

I came across a topic that is new to me.

What is [color=red]CONTINUED DIVISION?[/color]
I searched the internet to no avail.

If it's not on the internet, it can't really exist..

Where did you see this?

[quote="pianoforte"]Could it be continued fraction?[/quote]

Do you mean infinite geometric sequence? or like the one you find the golden ratio wtih? (${1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}}$)?

What you're looking for is probably cotinued fractions, as pianoforte said.

Given any positive integers $m,n$ it can be shown (with the division algorithm) the rational number $\frac{m}n$ has a unique representation as a continued fraction of the form \[a_0+\frac1{a_1+\frac1{a_2+\ddots_{\frac1{a_n}}}},\] where the $a_i$ are positive integers.

There are also infinite continued fractions that look like the finite one above but $n$ is taken to infinity: \[a_0+\frac1{a_1+\frac1{a_2+\frac1{a_3+\ldots}}}\]I don't know much about infinite continued fractions but I'm assuming there is a unique one equal to each positive irrational number. Try searching for "infinite continued fractions" and you'll find some very pretty equations.

[quote="mathnerd314"]Every positive irrational number? Can you prove that? I know there's one for $\pi$ and e and all square roots of integers, but what about other irrationals?[/quote]I'm not trying to rigorously prove the existence or uniqueness, but this is what I came up with:

Let $\alpha$ be a positive irrational number, and defined the numbers $\{\alpha_n\}$ by the following:

$\alpha=\lfloor\alpha\rfloor+\alpha_1\\\\\frac1\alpha_1=\lfloor\frac1\alpha_1\rfloor+\alpha_2\\\\\frac1\alpha_2=\lfloor\frac1\alpha_2\rfloor+\alpha_3\\\\\cdots\\\\\frac1\alpha_n=\lfloor\frac1\alpha_n\rfloor+\alpha_{n+1}\\\\\cdots$

Letting $\lfloor\alpha\rfloor=a_0$ and $\lfloor\frac1\alpha_n\rfloor=a_n$ for $n\ge1$, we can combine these equations to get \[\alpha=a_0+\frac1{a_1+\frac1{a_2+\ldots}}.\]

Of course for each $n$, $\alpha_n$ is irrational and by definition $0<\alpha_n<1\Longrightarrow\frac1\alpha_n>1$. This means that for every $n\ge1$, $a_n$ is at least $1$. Thus we have an infinite continued fraction.


Is it true that for $ a,b,c > 0$, $ \sum_{cyc} a^5 \plus{} \sum_{cyc} a^2b^2c \ge \sum_{sym}a^4b$? It is stronger than Schur, and I think it's false, but I have no way to check (because I am rather terrible at finding counterexamples...)

[quote="not_trig"]Is it true that for $ a,b,c > 0$, $ \sum_{cyc} a^5 \plus{} \sum_{cyc} a^2b^2c \ge \sum_{sym}a^4b$? [/quote]
It's wrong. Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.9.$


induction
combinatorics proposed
combinatorics

Let $ N$ be postive integer. Some integers are written in a black board and those satisfy the following conditions.

1. Any numbers written are integers which are from 1 to $ N$.

2. More than one integer which is from 1 to $ N$ is written.

3. The sum of numbers written is even.

If we mark $ X$ to some numbers written and mark $ Y$  to all remaining numbers, then prove that we can set the sum of numbers marked $ X$ are equal to that of numbers marked $ Y$.

:wink:  What about $ N\equal{}10$ and written numbers are $ 2,3,9$?

Pierre.

Sorry for my poor English. How about this?

Let $ N$ be postive integer. Some integers are written in a blackboard. Suppose that :

1. The written number is all  belong to $ 1,\ 2,\ \cdots N.$

2. Each of integer of $ 1,\ 2,\ \cdots N$ is written at least one.

3. The sum of numbers written in the black board is even.

If we mark $ X$ to some numbers written and mark $ Y$  to all remaining numbers, then prove that we can set the sum of numbers marked $ X$ are equal to that of numbers marked $ Y$.

Suppose the number $i$ appears $a_i$ times. If $a_i>2$, we can mark one $i$ with $X$ and the other with $Y$, this does not affect the equality of the sums. So we can assume that $a_i\le2$ for all $i$. The sum of numbers is at most $2+4+\ldots+2N$. Since each integer $1,2,\ldots,N$ appears at least once, it is enough to prove that we can mark some of $1,2,\ldots,N$ with $X$ such that the sum is any integer from not exceeding $1+2+\ldots+N$. The following sets of numbers have the desired properties:

$\{1\},\{2\},\ldots,\{N\}$
$\{N+1\},\{N+2\},\ldots,\{N+(N-1)\}$
$\{N+(N-1)+1\},\{N+(N-1)+2\},\ldots,\{N+(N-1)+(N-2)\}$
...
$\{N+(N-1)+(N-2)+\ldots+1\}$

The problem just says that given a [b]multiset[/b] of integers, such that each of its elements is between $1$ and $N$, and conversely, each integer between $1$ and $N$ belongs to the multiset, and moreover such that the sum of its elements is even, one can partition the multiset into two sub-multisets having the same sum of elements.

The condition on parity is obviously necessary, given the result asked.

I will post my solution, basically the idea is to take any sum less than the half of the total and start adding new numbers. Following this we see that the if $A$ is the set with the numbers such that the sum of the elements of $A$ is the greatest possible less than the half of the total, let's say $t$, this may work.

Solution.
As above we consider $A$ a set which elements have the greatest possible sum less than $t$, if the sum is $t$ we finish, and let $B$ be the set of the remaining numbers. Let's assume that the sum of $A$ is $t-x$ for some integer $x$, now all the numbers from $1$ to $x$ belong to $A$. Indeed, if some number $y$, with $1 \le y \le x$, belong to $B$ we change it to $A$, but now $A \cup \{y\}$ contradicts the maximality, and if $y=x$ we are done, so all the numbers from $1$ to $x$ belong to $A$. Now if $x+1$ is in $B$, we change it to $A$ and we change $x$ to $B$, and again the new $A$ contradicts the maximality, we proceed again doing this induction and all the numbers belong to $A$ but this is a contradiction so if $A$ exists then, it's sum is $t$.

order the integers  $a_1\leq ...\leq a_n$

let the total sum of all integers be $2\sigma$

clearly $a_n\leq \sigma$

step 1:

if $a_n=\sigma$ then label $a_n$ by $X$ and the rest by $Y$.

if $\sigma -a_n>0$, define $\sigma_1=\sigma-a_n$ ,

let $\sigma_k$ denotes the number obtained after the subtraction in step $k$.

in step $k$ we choose the maximal $a_{j_{k}}$  from the remaining integers of $(a_1,a_2...a_{n})$ i.e the biggest interger not used yet in any subtraction to get $\sigma_k=\sigma-a_{j_{k}}$

keep applying the above algorithm. eventually we get : $\sigma_i\geq0 $ & $\sigma_{i+1}<0$ for a big enough $i\in(1...n)$,

if $\sigma_i=0$ we are done by step (1), i.e label $a_{j_i}$ and the other integers used in the algorithm by $X$ and the rest $Y$.

if $\sigma_i>0$ then we must have that $0<\sigma_i<\min(a_{j_1},a_{j_2}...a_{j_{i+1}}) $ .

therfore $\sigma_i\in(a_1,a_2...a_n)\backslash(a_{j_1},a_{j_2}...a_{j_{i+1}}) $,then label $\sigma_i,a_{j_1},a_{j_2}...a_{j_{i}}$ by $X$ and the rest by $Y$

[hide=Induction]For $N=1$ the statement is true. If the statement is true until $N-1$. We will prove that it is true with $N$. Indeed, denote $A,B$ be the sums of numbers marked $X,Y$ for $N-1$, respectively then $A=B$. We consider the two cases:

[b]Case 1.[/b] If $2 \mid N$ then remove $\tfrac 12 N$ from $A$ to $B$ (if $A$ doesn't contain $\tfrac 12 N$ then remove from $B$ to $A$) and then add $N$ to $A$. Hence, each sum is added $\tfrac 12 N$ so they are equal.

[b]Case 2.[/b] If $2 \nmid N$ then there exist two sums $C,D$ obtained from numbers from $1$ to $N-3$ such that $C=D$ according to the induction hypothesis. Now we need to put $N-2,N-1,N$ in $C,D$ in order to make a new two equal sums.

We can remove $\tfrac 12 (N-1)$ from $C$ to $D$ and then add $N,N-1$ to $C$ and add $N-2$ to $D$. Hence, the new sums are equal to $$C- \tfrac 12 (N+1)+N+N-1= D+ \tfrac 12 (N+1)+N-2.$$
We are done.
[/hide]

Here is a simpler solution.

Let $a_1\leq a_2\leq \cdots \leq a_n$ be the elements of this sequence, in ascending order (in particular, $n\geq N$, $a_1=1$, $a_n=N$). The idea is to look at the running sum, $S_\ell=\sum_{j=1}^\ell a_j$. Now, suppose that no such coloring is possible, and $S=\sum_{j=1}^n a_j$. Let $k$ be the first time such that $S_k$ exceeds $S/2$. Namely, $a_1+\cdots+a_{k-1}=S/2-j$ for some $j\geq 1$, and. $a_1+\cdots+a_k = S/2+i $, for some $i \geq 1$. In particular, $a_k=i +j$. But from the construction, $i \leq a_k$, and therefore, there is a $k'$ such that $a_{k'}=i$. With this, $S_k-a_{k'}=\sum_{1\leq j\leq k,j\neq k'}=S/2$, and thus, we get the desired contradiction.


calculus
integration
function
analytic geometry
calculus computations

$ \int_{\minus{}2}^{2} \int_{\minus{}\sqrt{4\minus{}y^2}}^{\sqrt{4\minus{}y^2}}  \int_{\sqrt{x^2\plus{}y^2}}^{2} xz  dz dx dy$

Well doing it the traditional way, it's obviously 0 (the "middle" integral becomes 0 by propertes of an odd function integral), but I don't know how to convert it into cylindrical. Should I draw it out first? When I did try to draw it I feel like I need to separate the integral, but I'm not sure. Can someone please help guide me through this problem?

Cylindrical coordinates is really just polar coordinates in three dimensions.  So your $ x$ and $ y$ are transformed just as if they would be in polar coordinates.  And $ z$ remains $ z$.  However, you need to account for the change in variables and add in an $ r$ term.  For example, if you were going to integrate a circle, $ \int_{\minus{}1}^{1}\int_{\minus{}\sqrt{R^{2}\minus{}x^{2}}}^{\sqrt{R^{2}\minus{}x^{2}}}\,dy\,dx$, the integral would turn into $ \int_{0}^{2\pi}\int_{0}^{R}r\,dr\,d\theta$ (note the $ r$ in the integrand that seems to come from nowhere [it doesn't]).

Yeah the $ r$ comes from the Jacobian
But how should I approach this particular problem?

As Jrav has mentioned, $ z$ is fixed anyway, so

1. What does the boundary of $ x$ and $ y$ tell you?
$ \minus{} \sqrt{4 \minus{} y^2} \leq x \leq \sqrt{4 \minus{} y^2}$ and $ \minus{} 2 \leq y \leq 2$?
2. With that piece of information, what substitution should you make for $ x$ and $ y$, and $ dxdy$?


calculus
integration
trigonometry
logarithms
derivative
limit
calculus computations

With $|a|<1,$ of course?

There's a recent topic in the Complex Analysis section which is related, although not identical.

Hi,
I think this is a method which might work.

for $a=1$ the integral is $\pi$.
Use the integral

$F(a)=\int\limits_{0}^{\pi}\ln(1+a^{2}-2a\cos(x))dx$.
It is known that $F(a)=0$ for $a^{2}<1$ and $F(a)=\pi\ln a^{2}$ for $a^{2}\geq1$.
Now take the derivative of $F(a)$ with respect to $a$ and you get that your integral is $0$ when $a^{2}<1$ and is equal to $\frac{2\pi}{a}$ for $a^{2}> 1$.
Didi

I'm trying to evaluate your $F(a)$ by using Leibniz Integral Rule, and I believe that requires finding the integral in my original post.

This is the technique:

Use the following Fourier expansion

$\ln(1-2a\cos(x)+a^{2})=-2(a\cos(x)+\frac{a^{2}\cos(2x)}{2}+\frac{a^{3}\cos(3x)}{3}+\cdots)$, valid for $|a|<1$.

Now, if you integrate this equality you definitely get that $F(a)=0$.
If $|a|>1$, use the substitution $a=\frac{1}{b}$ and you should get it.
Didi

Here's very simple solution for $|a|\ne1$.
First, $1+a^{2}-2a\cos x=(a-\cos x)^{2}+\sin^{2}x=|a-e^{ix}|^{2}$.
Hense, \[F(a)=\lim_{n\to\infty}\frac{\pi}n\sum_{k=1}^{n}\ln|a-e^{i\frac{k}{n}}|^{2}=\pi\lim_{n\to\infty}\frac1n\ln|a^{n}-1|^{2}=...\]


Let the sequence $ \left \{ a_n \right \}_{n\geq1}$ be defined by $ a_1 \equal{} 2$, $ a_2 \equal{} 8$ and $ a_{n \plus{} 2} \equal{} 3a_{n \plus{} 1} \minus{} a_n \plus{} 5( \minus{} 1)^n$. Prove that if $ a_n$ is prime, then $ n$ must be a power of 3.

i think nobody bothers to do this...anyone?

i'll post my solution tomorrow.


Hi everyone. This is a 24 game marathon. We will start today, and the rules will be posted shortly. :D MathCount123 and I are running this.

This is not my joining conformation or anything but I'm just curious. Is this like math or what?

In my Introduction to Algebra class, there's a 24 marathon. It was quite successful, and I thought since no one besides the people who signed up for it could use it, why not make an everybody-excessible one, in classroom math? So I talked with my friend Mfranzs at school, and we decided to try it out. Here are the rules:

[color=darkred]1)You may only add, subtract, multiply, and divide. NOTHING else, INcluding exponents, just like the original 24 game. [/color]

[color=red]2)When you answer a problem, you must remember to make a new one up. It doesn't take long to make one up, but make sure it works. If the person doesn't make one up, then someone else can make a
new pr[/color]oblem

[color=orange]3)You may only make problems with one or two didit numbers. Otherwise, it gets too confusing
ex of ok:1,1,8,3 ex of ok: 10,10,40,10 NOT ok: 123,144,333,775 NOT ok:1111,1111,2222,3333
EVEN MORE NOT OK:123,55543,13,6443 [/color]

[color=brown]4)Try not to repeat problems; people will get bored with them. Me (MathCount123) and Mfranzs will look through the marathon, and every once in a while, if a problem is repeated at least 3 times, either one of us may say this problem: n, y, e, m, has been used too many times. Try to stay away from using it again from now on. [/color]

[color=green]5)If rules are broken too many times in a row by a couple (5+) different people, I will post the rules again (this post). If only one or two people are accidently breaking rules, its ok. Mfranzs or I will PM whoever is breaking the rules, and remind them about what rule they are breaking, and how to stop it, et cetera.[/color]

[color=olive]NOTE: this rule is just to keep the marathon users not confused.NOTE: the rules aren't really rules, just guidelines [/color]

[color=cyan]6)Any suggestioons for new rules are welcome. Ask questions if you have any by PM'ing me (MathCount123) or mfranzs [/color]

[color=darkblue]7)Have fun!!! Lets try to keep this marathon unlocked for a month, and get at least 75 problems in[/color]

No bad words; please do not fill in the blanks about what this smiley is saying :cursing:

[size=200]Ignoring mfranzs time starting date, here is the first problem

5 6 5 1[/size]

If you don't know what the 24 game is here it is. You take 4 numbers, and try and add, subtract, multiply, or divide them to make 24.

Here is an example problem.

     4
2         3
     1

You could do (4+2)*(3+1)=24

_____________________________________________________________________
The answer to MathCount123s question is 6*5-5-1=24

here is another one

[size=200]5   5   5   1[/size]

[color=darkblue][size=200]5 6 5 1[/size][/color]

Opening ceremony... and BEGIN!

mfranzs, dont forget to post a new problem once you did one heres the new one:

1,1,1,22

(please do not answer this one mfranza)

I would like to add a new rule: If you are a magic 24 problem maker, and no one can get it right, please post the answer after a couple days after your problem has been posted. If you made a mistake making the problem, just post that you made a mistake, and we'll continue

$ (2 \plus{} 3)\times5 \minus{} 1$

Edit: didn't see mfranzs's post

$ \frac{(25\plus{}23)}{24}\plus{}22$
[size=150]3,3,3,3[/size]

$ (11 \minus{} 9)\times4\times3$

$ 42$, $ 68$, $ 95$, $ 24$

(Edited because I made a mistake and don't think the original problem worked)

Do you got to scroll down the whole way every time

push the down arrow button on the scroller on the left

or the same down arrow on the sidebar on the right


Find integers which satisfy:
$3^{a}+7^{b}=3^{c}+7^{d}$, where $a\neq c, b\neq d$ and $a,b,c,d \geq 1$

what does $c_{3}(x)$ mean? I don't think you prove it, Rust
I just wanted a proof of non-solvability.

$c_{p}(n)$ is minimal T, suth that $n|(p^{T}-1)$.


prove the following system of equations does not have any non-trivial integer solutions

$6x^2+y^2=t^2$
$x^2+6y^2=z^2$

[hide]

Solution of the system is a solution of the sum of both equations, so say there is a solution to . Consider it mod 8; the left side is 0 mod 8, and all squares are 0,1,4 mod 8, so both t and zero must be divisible by 8, so we can replace  and , so then plugging back in, we get . Do it again here, x and y must be both divisible by , then replace  and , and we get , lalalala infinite descent lalalala.

[/hide]

Good idea, but $4 + 4 \equiv 0 \mod 8$, so it doesn't hold up.  (In fact, the sum equation has lots of non-trivial solutions: $8(3^2 + 4^2) = 200 = 2^2 + 14^2$, for instance.)

oops, sorry question editted

Elemennop, has the idea :)

wouldn't adding the system yield a left side of $7x^2 + 7y^2$ ?  But i guess infinite descent would still work...

well you can start of the proof by dividing all sides by gcd(x,y,z,t) then by the same argument contradiction follows

quantum leap wrote:wouldn't adding the system yield a left side of  ?  But i guess infinite descent would still work...

haha, he edited it.

and yea, my bad, the new question has an identical approach; [hide]

add them, left side is 0 mod 7 obviously, and squares are 0,1,2,4 mod 7,  and in THIS case the only way we can have is both of the squares on the right are 0mod 7, and then we have the exact same approach and we're done[/hide]


Could somebody write out common mathematical symbols used in proofs mainly that most people would be unfamiliar with?

For example, what is "$ \in"$?

I'm sorry I couldn't just google them, because I don't know what to search.  I don't know what the symbol is called.

"is an element of", usually of a set.
You can say that $ 3\in \{1,2,3,4,5\}$ or $ 3\in\mathbb{R}$
Like saying $ x\in \mathbb{N}$ means that x is part of the set of natural numbers.

Also, for the rest of mathematical symbols that you might see, try [url=http://en.wikipedia.org/wiki/Table_of_mathematical_symbols]Table of mathematical symbols[/url].

A couple that you might be interested (but in general should avoid, except for the last one, as it clutters solutions):

$ \therefore$ means therefore

$ \forall$ means for all

$ \exists$ means there exists

$ \propto$ means proportional to

[quote="grn_trtle"]There's a lot of symbols, but if you specify ones in particular that you don't know, we can tell you what they mean.

(Also, considering you had to type \in for the $ \text{\LaTeX}$ for that symbol, wasn't it kind of obvious? :D)

You'll find information about other notations used in set theory here: http://en.wikipedia.org/wiki/Set-builder_notation[/quote]

I quoted someone who used that symbol, but I didn't know exactly what the "in" meant.  :)

And thanks b-flat, that was very helpful.


Prove that there exist a permutation of {$ {1;2;3;...;n}$} is {$ {a_1;a_2;...;a_n}$} such that $ a_1\plus{}a_2\plus{}...\plus{}a_k$ is divide by $ a_{k\plus{}1}$ where $ k\equal{}1;...;n\minus{}1$.

the permutation $ k\plus{}1, 1, k\plus{}2, 2, ..., 2k, k$ suffices for $ n\equal{}2k$, and $ k\plus{}1, 1, k\plus{}2, 2, ..., k, 2k\plus{}1$ suffices for $ n\equal{}2k\plus{}1$.


Ello..
Can you guys please kindly help me out with this question?? :lol:

The sum of two numbers is 32, and the product of these two numbers is 48. What is the sum of the reciprocals of the two numbers??  Express your answer as a common fraction...

Thanks... :blush:

[hide="hint"]What is $ \dfrac{1}{x}\plus{}\dfrac{1}{y}$, in terms of x and y?[/hide]

[hide]$ \frac{1}{x} \plus{} \frac{1}{y} \equal{} \frac{y}{yx} \plus{} \frac{x}{xy} \equal{} \frac{y\plus{}x}{xy} \equal{} \frac{32}{48} \equal{} \boxed{\frac{2}{3}}$[/hide]

[quote="minicon"]That looks right lokesh.  :lol:
 :arrow:  jogaman323, please title your topics something other then [color=red]...........[/color].  Its hard to tell what it is.
An appropriate topic would be Weird Question , not periods. :D[/quote]

 I'm very sorry... I didn't realize that.
           Sorry....  :oops_sign:


Let $ 0<a<1$ and $ I\equal{}(0,a)$. Find all functions $ f: I\to\mathbb{R}$ satisfying at least one of the conditions below:
a)$ f$ is continuous on $ I$ and $ f(xy)=xf(y)+yf(x)$ for all $x,y\in I$.
b)$ f(xy)=xf(x)+yf(y)$ for all $x,y\in I$.

[quote="N.T.TUAN"]Let $ 0 < a < 1$ and $ I \equal{} (0,a)$. Find all functions $ f: I\to\mathbb{R}$ satisfying at least one of the conditions below:
a)$ f$ is continuous on $ I$ and $ f(xy) \equal{} xf(y)\plus{}yf(x)\forall x,y\in I$.
b)$ f(xy) \equal{} xf(x)\plus{}yf(y)\forall x,y\in I.$[/quote]

a) : Let $ g(x)\equal{}a^{\minus{}x}f(a^{x})$. We have $ g$ continuous,  defined on $ (1,\plus{}\infty)$ and $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ $ \forall x\in (1,\plus{}\infty)$
It's then easy, for $ b>1$ and $ p\geq q\geq 1\in\mathbb{N}$ to write :
$ g(q\frac{pb}{q})\equal{}qg(\frac{pb}{q})$ since $ \frac{pb}{q}\in(1,\plus{}\infty)$ and $ g(pb)\equal{}pg(b)$ since $ b\in(1,\plus{}\infty)$
And so $ g(\frac{p}{q}b)\equal{}\frac{p}{q}g(b)$ $ \forall b>1\in\mathbb{R}$ and $ \forall p\geq q\geq 1\in\mathbb{N}$
And so, since $ g$ is continuous, $ g(x)\equal{}cx$ $ \forall x>1$
So $ f(x)\equal{}\alpha x\ln(x)$ $ \forall x\in I$ and it is easy to check that this necessary condition is sufficient.

b) : $ f(xy)\equal{}xf(x)\plus{}yf(y)$ $ \implies$ $ f(x^{2})\equal{}2xf(x)$ and then also $ 2f(xy)\equal{}f(x^{2})\plus{}f(y^{2})$. So we can write $ f(x^{2}y^{2})$ in two ways :
$ f(x^{2}y^{2})\equal{}f(XY)\equal{}Xf(X)\plus{}Yf(Y)\equal{}x^{2}f(x^{2})\plus{}y^{2}f(y^{2})$
$ f(x^{2}y^{2})\equal{}f(Z^{2})\equal{}2Zf(Z)\equal{}2xyf(xy)\equal{}xy(f(x^{2})\plus{}f(y^{2}))$
Comparing these two expressions,we get, for $ x\neq y$, $ xf(x^{2})\equal{}yf(y^{2})$ and so $ f(x)\equal{}\frac{c}{\sqrt{x}}$
Putting back this expression in $ f(x^{2})\equal{}2xf(x)$, we get $ \frac{c}{x}\equal{}2c\sqrt{x}$ and so $ c\equal{}0$
And so $ f(x)\equal{}0$

So the requested functions are $ f(x)\equal{}\alpha x\ln(x)$ for any $ \alpha\in\mathbb{R}$


geometry
Euler
circumcircle
incenter
geometric transformation
reflection
trigonometry

Let $ ABC$ be a triangle with $ \angle A < 60^\circ$. Let $ X$ and $ Y$ be the points on the sides $ AB$ and $ AC$, respectively, such that $ CA \plus{} AX \equal{} CB \plus{} BX$ and $ BA \plus{} AY \equal{} BC \plus{} CY$ . Let $ P$ be the point in the plane such that the lines $ PX$ and $ PY$ are perpendicular to $ AB$ and $ AC$, respectively. Prove that $ \angle BPC < 120^\circ$.

Consider the triangle of the excenters $ O_1O_2O_3$ $ P$ is its circumcenter. $ ABC$ is the orthic triangle of $ O_1O_2O_3$. Then, $ P$ lies in the Euler line of $ O_1O_2O_3$ in fact the circumcenter $ O$ of the orthic triangle, $ ABC$ it's halfway from the orthocenter of $ O_1O_2O_3$ ($ I$) and $ P$.

By easy angle computations and using the fact that $ <A<60$° is easy to see that $ <BIC<<BOC$ then, because $ P$ is located in the same line as $ I$ and $ O$ (in direction to $ O$ from $ I$) its angle must be even less than $ <BOC\equal{}2<A<120$°

Sorry for bumping, but I can't see why P is the circumcenter of the excentral triangle. Could someone explain why to me?

Also, shouldn't P be located in the direction from I to O, not the other way, because the Euler line has the nine point center (circumcenter of ABC) halfway between the circumcenter (point P) and orthocenter (incenter of ABC)? Or am I just being stupid?

Does this solution work? Or do I have to prove some of the stuff in paragraph 2 as Lemmas?

Let $ I$ be the incenter of triangle $ ABC$ and let the incircle of $ ABC$ touch $ AB$ and $ AC$ at $ X'$ and $ Y'$ respectively. Let $ M$ and $ N$ denote the midpoints of $ AB$ and $ AC$ and $ O$ the circumcenter of $ ABC$. Note that by definition, $ BC \plus{} AX' \equal{} AC \plus{} BX'$ and therefore it follows that $ AX' \equal{} BX$. Therefore $ X$ is the reflection of $ X'$ in $ M$. Similarly, $ Y$ is the reflection of $ Y'$ in $ N$. Hence $ PX$ is the reflection of $ IX'$ in $ OM$ and $ PY$ is the reflection of $ IY'$ in $ ON$. Therefore $ P$ is the reflection of $ I$ in $ O$.

Now consider the circumcircle $ \Gamma$ of $ OBC$. Since $ \angle{A} < 60$, it follows that $ \angle{BOC}\equal{} 2\angle{A}<90\plus{}\angle{A}/2 \equal{} \angle{BIC}$. Hence point $ I$ is in the interior of $ \Gamma$. Therefore $ P$, the reflection of $ I$ in point $ O$ which lies on $ \Gamma$, must be in the exterior of $ \Gamma$.

Therefore it follows that since $ P$ is in the exterior of $ \Gamma$, that $ \angle{BPC}<\angle{BOC}\equal{}2\angle{A}<120$.

Call the excentres of the $\triangle ABC$ $I_A,I_B,I_C$
Then quite obviously, $P$ is the circumcentre of $\triangle I_AI_BI_C$
Call $I_BB \cap I_CC = H$ Then again obviously $H$ is the orthocentre of triangle $I_AI_BI_C$
[b]Claim : [/b]$\angle BPC \le \angle BHC$
If we prove $P$ lies outside $\odot BHC$, we are done.
We know $I_ABHC$ is cyclic with $I_AH$ as diameter.
But $I_AH = 2I_AP \cos(90-\frac{A}{2})=2I_AP \sin (\frac{A}{2}) < I_P$
$\implies P$ lies outside $\odot BHC$
$\implies \angle BPC < \angle BHC = 90+\frac{A}{2} < 120$
Thus we are done

Nice! :)

Let $I$ be the incenter of $\triangle ABC$. Note that $P$ is the circumcenter of the excentral triangle $\triangle J_AJ_BJ_C$ of $\triangle ABC$. Secondly, $$J_AI=2J_AP \cos \left(90^{\circ}-\frac{A}{2}\right)=2J_AP \sin \left(\frac{A}{2}\right)>J_AP,$$ so $P$ lies outside the circumcircle of $\triangle BIC$. Consequently, $\angle BPC<\angle BIC=\left(90^{\circ}+\frac{A}{2}\right)<120^{\circ}$ as desired.

[quote=MathPanda1]What is the motivation for looking at $I$ and $O$ and the circumcircle of $BOC$? Thanks![/quote]

The length condition suggests excenters.

[hide=Solution]Note that $X$ is the contact point of the $C$-excircle and $Y$ is the contact point of the $B$-excircle(since we get $AX = s-b$ and $AY = s-c$). Thus, it suffices to consider the excentral triangle $I_AI_BI_C$. Let $K$ be the point on $(I_AI_BI_C)$ such that $KI_BI_C$ is isosceles.

Lemma 1: $P$ is the circumcenter of $I_AI_BI_C$

Proof: $\angle PI_BI_C = 90-\angle I_BAC = 90-\angle I_BI_AI_C$, and similarly, $\angle PI_CI_B = 90-\angle I_BI_AI_C$, and we know that the circumcenter satisfies the $2$ angle conditions, and only $1$ point can simultaneously satisfy the $2$ angle conditions(since it is the intersection of $2$ lines).

Now, let $H_K$ denote the orthocenter of $KI_BI_C$. Let $H$ be the orthocenter of $I_AI_BI_C$.

Lemma 2: $KH_K = I_AH$

Proof: We use complex numbers, letting $(I_AI_BI_C)$ as the unit circle. Note that $h_K = k+b+c \implies h_K-k = b+c$. Furthermore, $h = a+b+c \implies h-a = b+c$.

Lemma 3: $\angle I_BI_AI_C > 60^{\circ}$

Proof: Note that $\angle I_BAC = \angle I_BI_AI_C$ and similarly, $\angle I_CAB = \angle I_BI_AI_C$, so $\angle BAC = 180-2\angle I_BI_AI_C$. Since $\angle BAC < 60^{\circ}$, this implies $\angle I_BI_AI_C > 60^{\circ}$.

Furthermore, we have $\triangle I_AI_BI_C$ is acute, since $\angle BAC = 180-2\angle I_BI_AI_C$, implying $\angle I_BI_AI_C$ is acute, and analogously for the other angles. Now, we claim that $P$ lies outside $(BCH)$. Note that since $\angle I_ABH = \angle I_ACH = 90^{\circ}$, this is just the circle with diameter $I_AH$.

Lemma 4: $P$ lies outside of the circle with diameter $KH_K$

Proof: Note that $\angle I_BPI_C = 2 \angle I_BI_AI_C$. Furthermore, $\angle I_BH_KI_C = 180-\angle I_BI_AI_C$ by cyclic quadrilaterals, so since $\angle I_BI_AI_C > 60^{\circ}$, this means that $\angle I_BPI_C > \angle I_BH_KI_C$, and thus $P$ lies closer to $BC$ than $H$, and thus cannot lie inside $(BCH)$.

Now, note that this means that $R > KH_K = I_AH$, where $R$ is the circumradius of $I_AI_BI_C$. Thus, $P$ lies outside of the circle with diameter $I_AH$, since $I_AP = R$. However, the locus of points $N$ where $\angle BNC > 120^{\circ}$ is clearly contained within the circle diameter $I_AH$, since $\angle BMC$, where $M$ is on the circle with diameter $I_AH$, is at most $120^{\circ}$, as it is either equal to $BI_AC$ or $BHC = 180-BI_AC$. Thus, we must have $\angle BPC < 120^{\circ}$.
[/hide]

You can find that $O$ is the midpoint of $I$ and $P$
So it can be solved easily.

P= the circumcenter of IaIbIc
 O circumcenter of ABC  =the center of 9point circle of IaIbIc
  I  the incenter of triangle ABC =   the orthocenter of triangle IaIbIc
     So  PI  is th euler line of IaIbIc so P,O,I are on the same line.
                               So point  p is out of triangle IBC
 so we have that the angle BPC≤the angle BIC=90+A/2≤120
and the problem is easelly solved


limit
logarithms
LaTeX
trigonometry
calculus
calculus computations

If  $ \lim_{x \to 0} \left(\frac{1\plus{}cx}{1\minus{}cx}\right)^{\frac{1}{x}} \equal{} 4$,then

$ \lim_{x \to 0} \left(\frac{1\plus{}2cx}{1\minus{}2cx}\right)^{\frac{1}{x}} \equal{} ?$

[quote="guru_bhai"]If  $ \lim_{x \to 0} \left(\frac {1 \plus{} cx}{1 \minus{} cx}\right)^{\frac {1}{x}} \equal{} 4$,then

$ \lim_{x \to 0} \left(\frac {1 \plus{} 2cx}{1 \minus{} 2cx}\right)^{\frac {1}{x}} \equal{} ?$[/quote]
 use $ \lim_{y \to 0}{(1 \plus{} y)^{\frac {1}{y}}} \equal{} e$ we have
$ \lim_{x \to 0}{(1 \plus{} cx)^{\frac {1}{x}}} \equal{} \lim_{x \to 0}{(1 \plus{} cx)^{\frac {1}{cx}c}} \equal{} e^c$
$ \lim_{x \to 0}{(1 \minus{} cx)^{\frac {1}{x}}} \equal{} \lim_{x \to 0}{(1 \minus{} cx)^{ \frac {1}{ \minus{} cx}\minus{} c}} \equal{} e^{ \minus{} c}$
then $ \lim_{x \to 0} \left(\frac {1 \plus{} cx}{1 \minus{} cx}\right)^{\frac {1}{x}} \equal{} \frac {e^c}{e^{ \minus{} c}} \equal{} e^{2c} \equal{} 4$
 same $ \lim_{x \to 0} \left(\frac {1 \plus{} 2cx}{1 \minus{} 2cx}\right)^{\frac {1}{x}} \equal{} \frac {e^{2c}}{e^{ \minus{} 2c}} \equal{} e^{4c} \equal{} (e^{2c})^2 \equal{} 16$

Just like \lim is a built-in command in LaTeX, so is \ln (and also \log, \sin, \cos, etc.).


I'm not sure if this is exactly the right place for this, but it's the closest I could find, and it seems that the moderators here are very efficient at moving things that get misplaced anyway...

Does anyone know of an International Physics Competition similar to the IMC, for undergraduate students?  I am aware of competitions such as the BAUPC that people sit in their own universities, but I'm looking for one that involves a gathering - it's more fun and more interesting!

[b]
So why don't we start a IPC forum at this site? :D
Lots of physics problems are connected with equations, trig and even calculus![/b]

Shober please stop writing in [b]bold[/b] text.

Alright。。。。。。。。。


Let $x_1, x_2, x_3, x_4$ be positive numbers with the product $1$. Prove that:
$\sum_{i=1}^4x_i^3\geq \max \{\sum_{i=1}^4x_i,\sum_{i=1}^4\frac{1}{x}\}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=32855

  darij

posted before, and the solution is that
$\sum x_i^3 \ge \frac{(\sum x_i)^3}{16} \ge \sum x_i$ (amgm, holder)
$\sum x_i^3 \ge \sum x_1x_2x_3 = \sum \frac{1}{x_1}$ (amgm)


If  $ z^{2008} \minus{} 2008z \plus{} 2008 \equal{} 0$. Proove that $ z$ doesn't belongs in $ \mathbb{R}$

[hide]Assume for the sake of contradiction that $ z$ is real.  Then graph $ f(z) \equal{} z^{2008}$ and $ g(z) \equal{} 2008z\minus{}2008$.  Clearly, $ f(z) > g(z)$ where $ z \le 1$.  And after that, $ f(z)$ grows faster than $ g(z)$, so $ f(z) > g(z)$ for all $ z$.  Hence, the graphs of $ f$ and $ g$ never intersect, meaning $ f(x)\minus{}g(x) \equal{} z^{2008} \minus{} 2008z \plus{} 2008 \equal{} 0$ has no real roots.[/hide]

[hide="Outline"]By the rule of signs, we can't have any negative roots.

Positive roots are the same as the roots of $ \minus{} 2008 \plus{} 2008z \minus{} z^{2008}$

Now you can use AM-GM in some form to find the appropriate relative maximum value of $ 2008z \minus{} z^{2008}$, and show that it's less than $ 2008$, which means that $ \minus{} 2008 \plus{} \text{max} (2008z \minus{} z^{2008} ) < 0$, which means no positive roots.[/hide]

[hide]Suppose $ z\in\mathbb{R}$. Obviously, if $ z\le 1$ then $ z^{2008}\plus{}2008(1\minus{}z)>0$. so $ z\ge 1$. Setting $ z\equal{}1\plus{}x$ and using Bernoulli's inequality, we have $ (1\plus{}x)^{2008}\minus{}2008(1\plus{}x)\plus{}2008\ge 1\plus{}2008x\minus{}2008x\minus{}2008\plus{}2008 \equal{} 1>0$.[/hide]

[b]bertram[/b], I don't think the use of Bernoulli's inequality (I just looked up) is necessary here since you set up that $ x\geq 0$. $ (1\plus{}x)^{2008}\geq 1\plus{}2008x$ directly follows from binomial theorem.