quadratics
function
algebra
polynomial
Rational Root Theorem

Hey anyone able to get this problem from amoc 2001

Prove that the only integer pairs (x,y) s.t.

$x+x^{2}+x^{3}=y+y^{2}$

are (0,0) and (0,-1)

I made a bit of progress using mods but i think u'll need something else as well.

[quote="Seanie_G"]Prove that the only integer pairs (x,y) s.t.

$x+x^{2}+x^{3}=y+y^{2}$

are (0,0) and (0,-1)

[/quote]
Hi Seanie_G,
Just a quick hint from me:
Try to solve the equation in the form [b]f(x,y)=0[/b] in unknown x  & or in unknown y.
Apply the theorem of rational roots of algebraic equation to find possible integer roots of it.

hey sean which mod u work in?
this might help
$x^{3}=(x+y+1)(y-x)$ if only they made x a prime...

i worked mod 5

evaluate the LHS and RHS for x and y congruent to 0 through 4

and u get LHS congruent to 0, 3 or 4 and RHS congruent to 0 1 or 2

therefore LHS congruent to 0 and for this to be so x must be congruent to 0.

also x can be easily proven as even, so i have 10 divides x.

I think the rational root theorem idea is awesome for this though, just need to find a contradiction or sumthing between the 2 divisibility statements it gives you. (x divides RHS and y divides LHS).

$f(x,y)=y^{2}+y-(x^{3}+x^{2}+x)=0... (*)$

Treat this form like a quadratic in [b]y[/b]:
[b]x=-1[/b] doesn't satisfy [b](*)[/b],and according to diophantine terms, it must be $\sqrt{D}=\sqrt{1+4(x^{3}+x^{2}+x)}\geq 0$.
From the foregoing,note that it suffices to consider only the cases [b]y[/b]$\geq-1$.
Consequently, [b]x[/b] is a nonegative integer.
It follows that :$y\geq x$.Substitute in ([b]*[/b]) $y=x+n;n\in\mathbb{N}$ in order to get:
$x^{3}-2xn-n^{2}-n=0...(**)$
__________________________

Sum [b](*) [/b]and[b] (**)[/b] up:
$(*)+(**)=y^{2}-(x+n)^{2}=x+n-y\Rightarrow\boxed{(y-x-n)(y+x+n)=-(y-x-n)}\Longrightarrow$ arrive at the conclusion w\o congruences by yourself.

hey guys js an easy amoc question, if you wanted, or if yor bored.

Prove that there exists no function $f: R \implies R$ such that:

$f(r^{2})-(f(r))^{2}\geq \frac{1}{4}$ $\forall$ $r$
and
$f(x)=r$ has at most one solution.

lol anyway..
[hide]
$f(0)=f(1)=\frac{1}{2}$ contradiction.
[/hide]


is it just me or are some amc 12 questions like a 10 or 11 on aime.  Like the one about 2^2004 is a 604 digit number with its first digit as 1.  Find the number of numbers in set s (2^0,2^1,2^2...2^2003) whose first digit is four.

an easier question than that appeared on the 1990 AIME as #13.

Some of the later AMC12 questions (mostly #24's and #25's) can probably qualify as an AIME #5 or around that. I'd definitely agree that I have more trouble with AMC12 #24's and #25's than the first 5 problems on an AIME (in part due to time constraints, but some of them are just hard).

The same argument can be made about AIME vs. USAMO questions, though less often in my opinion because the USAMO has so much more time and so many less problems. Like I think this year's USAMO #1 and #4 were comparable to later AIME problems (#13-15) maybe not from this year's AIME1, but like, say, the 2003 AIME or something (just picking out a not too easy year from what I remember).

It could also be that the question came from one of the older AHSMEs, which had to select people for the USAMO and was therefore harder.

Edit--never mind, the 2004 means it was from the year 2004 :(.

[quote="chess64"]It's not that hard if you have enough time, I would say it would be a #1 or #2.[/quote]Really?  Most AIME #1 and #2 are quite straightforward.  This question (from AMC 2004B #25) requires a lot more cleverness, in my opinion.

[quote="qweretyq"]I think even one of this years IMO problems was comparable to an AIME problem. It was the only one I have ever solved!!
The one that used Fermat's Little Theorem. As long as you knew it you pretty much got the answer.[/quote]

IMO has some pretty easy problems every year. If you are willing to give it a shot, you can probably get a lot of the #1 and #4's on each IMO (though this year's #1 was not that easy for a #1)

Other notable "easy" problems are USAMO 2005 #4 (AIME level at best) and AIME2 2003 #13

[quote="white_horse_king88"]Is it just me, or are AHSME questions harder than AMC 12s? I mean, they always ask long sentency questions... really annoying. I do well on AMCs but horrendous on AHSMEs at home. :huh:[/quote]

Same here. I think there isn't significant difference concerning difficulty but since you have 30 problems, you get more pressure. Also, especially the ones in 80's (and I think early 90's problems are still quite difficult) were used to determine USAMO so they're much difficult (this stopped after 1982 since AIME was created 1983). I also find first 15 questions of AMC much easier.


hint: [hide]tan x= sin x / cos x, and cos(90 - x) = sin x[/hide]


Find the minimal $K_n$ so that if $a_1,a_2,...,a_n$ are positive real numbers satisfying \[ \left(\sum_{i=1}^n a_i\right)^n\geq\ K_n\cdot \sum_{i=1}^n a_i^n\] then for all $0\leq\ i,j,k\leq\ n$, $(i-j)(j-k)(k-i)\neq\ 0$, numbers $a_i$, $a_j$, $a_k$ are sides of a triangle.

[i]Edited by Myth[/i]

I have got a terrible value for $K_n$. Though, I am not sure it is correct.
So... $K_n=\frac{\big(4+(n-3)a\big)^n}{2+2^n+(n-3)a^n}$, where $a=\sqrt[n-1]{\frac{1+2^{n-1}}{2}}$.


[url=http://www.mathlinks.ro/viewtopic.php?t=5607]Problem A6 of the 1971 Putnam[/url] asks (essentially): Prove that if $ 1^x$, $ 2^x$, $ 3^x$, ... are all integers, then $ x$ is a positive integer.

I'm trying to recall a result that says something like if $ 2^x$, $ 3^x$, and $ 5^x$ are integers, then $ x$ is a positive integer, but $ 2^x$ and $ 3^x$ are not enough.  Does anyone else remember this result?

The Putnam problem, as stated in the weak form, is not difficult: assuming that $x \notin \mathbb{N}$, write $x=s+\alpha$ with $s \in \mathbb{N}_0$ and $0 < \alpha < 1$, and get a contradiction by verifying that, for $n$ large enough,
\[
0 < \sum_{0 \leq j \leq s} \binom{s}{j}(-1)^{s-j}(n+j)^x < 1.
\]

However, I could not show that $2^x,3^x, 5^x \in \mathbb{N}$ implies that $x$ is an integer  :( . Also, why is $2^x,3^x \in \mathbb{N}$ not sufficient? Is there an easy example showing this? I could not find such example  :blush: .

Another generalization: show that if $\alpha \in (0,1)$, then there exists a rational number $q$ such that $q^{\alpha} \notin \mathbb{Q}$.

--Vesselin

Can someone explain how vess arrived at his result? It is similar to Boole's formula...
Letting $ f(t) \equal{} t^x$, then $ \sum_{0 \leq j \leq s} \binom{s}{j}( \minus{} 1)^{s \minus{} j}(n \plus{} j)^x$ is exactly the s'th finite difference of $ f$ around $ x \equal{} n$, with $ h \equal{} 1$: $ \Delta^s_1[f](n)$.
From the relation $ \frac {d^n f}{d x^n}(x) \equal{} \frac {\Delta_h^n[f](x)}{h^n} \plus{} O(h)$, we obtain:
$ x(x \minus{} 1)\cdots(x \minus{} (s \minus{} 1)) n^{x \minus{} s} \sim \Delta^s_1[f](n) \plus{} C$ for some $ C$
For large $ n$, LHS gets closer to 0.
Still, can't prove vess's result.

[From Wikipedia:
$ \Delta^n_h[f](x) \equal{} \sum_{i \equal{} 0}^{n} ( \minus{} 1)^i \binom{n}{i} f(x \plus{} (n \minus{} i) h)$
$ \frac {d^n f}{d x^n}(x) \equal{} \frac {\Delta_h^n[f](x)}{h^n} \plus{} O(h)$
[url]http://en.wikipedia.org/wiki/Finite_difference[/url] ]

[quote="bambaman"] [...]
$ x(x \minus{} 1)\cdots(x \minus{} (s \minus{} 1)) n^{x \minus{} s} \sim \Delta^s_1[f](n) \plus{} C$ for some $ C$
For large $ n$, LHS gets closer to 0.
[...]
[/quote]
Actually, as $ n \to \infty,$ the LHS goes to $ \plus{} \infty.$ (remember that $ x \minus{} s \equal{} \alpha > 0$)
On the other hand, with your notations, $ \Delta_1^{s \plus{} 1}[f](n)$ does tend to $ 0$ as $ n \to \plus{} \infty,$ so that we may find some $ N \in \mathbb{N}$ such that, for any $ n \ge N, \ \ 0 \le \Delta_1^{s \plus{} 1}[f](n) < 1.$
But the bounded expression being an integer, this yields that $ \forall n \ge N, \ \ \Delta_1^{s \plus{} 1}[f](n) \equal{} 0,$ so that $ \Delta^s_1[f](n)$ should be constant from $ N$ on, which contradicts the fact that it goes to infinity at infinity.
I guess that is what [b]vess[/b] had in mind.


complex analysis
limit
ratio
algebra
polynomial
absolute value
complex analysis unsolved

Let $ r>1$, let $ D\equiv \left\{ z\in \mathbb{C}\parallel{}z|<r\right\}$, and let $ f$ be holomorphic on $ D\minus{}\{ 1\}$, where $ f$ has a pole at $ 1$.  Then, $ f$ can be written as a power series centered at $ 0$:  $ f(z)\equal{}\sum _{n\in \mathbb{N}}a_nz^n$, which converges for $ |z|<1$.  Prove that $ \lim \frac{a_n}{a_{n\plus{}1}}\equal{}1$.

Note that you cannot just apply the ratio test, because for the ratio test to work, you have to know [i]a priori[/i] that the limit of $ \frac{a_n}{a_{n\plus{}1}}$ exists, and this I cannot figure out how to prove.  Not to mention, the ratio test actually takes the limit of the [i]absolute value[/i] of the ratio of coefficients, not just the straight up ratio.

Any ideas/hints/solutions?  Anything at all would be much appreciated.

Write $ f(z)\equal{}g(z)\plus{}\frac{b_1}{z\minus{}1}\plus{}\frac{b_2}{(z\minus{}1)^2}\plus{}\cdots\plus{}\frac{b_n}{(z\minus{}1)^n}$, where $ g$ has no pole at $ 1$. Then $ g$ is analytic on a strictly larger disk, and the coefficients of its power series decay. The stuff coming from the $ b_i$ doesn't decay, and has some polynomial growth pattern, leading to a ratio of $ 1$.

I'm sorry, I don't quite understand this solution.  Could you possibly explain in a little more detail what you mean by this?

Thanks!

The essential point is that "analytic on a strictly larger disk"; that's what it means to be analytic on a closed disk, and this would fail if there were other poles on the circle. To make use of that, we subtract stuff off until what's left has a removable singularity at $ 1$ instead of a pole; that remainder's power series has a radius of convergence strictly larger than $ 1$.

[quote="GleasSpty"]So I agree with and understand everything you have said thus far, I'm just not seeing it:  what does all this have to do with the ratio of the coefficients of the power series of $ f$?

Sorry for being so slow and thanks for your help.[/quote]

Suppose $ f(z) \equal{} \sum b_nz^n \plus{} \frac {1}{(1 \minus{} z)}$, where $ \sum b_nz^n$ is analytic in the larger disc. What are the power series coefficients of $ f$ at 0?


induction
logarithms
modular arithmetic
number theory
number theory proposed

Let $ p$ be a prime number, $ p\not \equal{} 3$, and integers $ a,b$ such that $p\mid a+b$  and $ p^2\mid a^3 \plus{} b^3$. Prove that $ p^2\mid a \plus{} b$ or $ p^3\mid a^3 \plus{} b^3$.

[hide="My solution"]
[color=darkblue]We have $ p^2|a^3 \plus{} b^3 \equal{} (a \plus{} b)(a^2 \plus{} b^2 \minus{} ab)$
Because $ p$ is prime, there are 2 cases:
[b]1)[/b] $ p|a \plus{} b$, so $ \boxed{p^2|a \plus{} b}$.
[b]2)[/b] $ p|a^2 \plus{} b^2 \minus{} ab \equal{} (a \plus{} b)^2 \minus{} 3ab$. Because $ p\not \equal{} 3$, $ p|ab$. But $ p$ is prime, so it divides one of the numbers $ a,b$. Let $ p|a$, but $ p|a \plus{} b$, so $ p|a,b$, and $ \boxed{p^3|a^3 \plus{} b^3}$.[/color]
[/hide]

[quote="Ahiles"]
[color=darkblue]We have $ p^2|a^3 \plus{} b^3 \equal{} (a \plus{} b)(a^2 \plus{} b^2 \minus{} ab)$
Because $ p$ is prime, there are 2 cases:
[b]1)[/b] $ p|a \plus{} b$, so $ \boxed{p^2|a \plus{} b}$.
[/quote]
Why from $ p|a\plus{}b$ then $ p^2|(a\plus{}b)$
We can general problem as follow :
[b] Problem [/b]
$ p$ is a prime satisfy : $ p|a\plus{}b$ and $ p^k|a^n\plus{}b^n$ where $ n\equiv 1 (\mod 2)$
Prove that $ p^{k\minus{}v_p(n)}|a\plus{}b$ or $ p^n|a^n\plus{}b^n$
It is an consequence of this lemma :
 [b]Lemma[/b] If $ \gcd(a,b)\equal{}1$ and $ p|a\minus{}b$ then
$ v_p(a^n\minus{}b^n)\equal{}v_p(a\minus{}b)\plus{}v_p(n)$
 It is an well know lemma .

[quote="Ahiles"][b]TTsphn[/b], we have $ p^2|(a \plus{} b)(a^2 \minus{} ab \plus{} b^2)$. Because $ p$ divides $ a \plus{} b$, it should divide or $ a \plus{} b$ one more time (so $ p^2|a \plus{} b$) or it divides $ a^2 \minus{} ab \plus{} b^2$[/quote]
If that you should type :
$ p|\frac{a\plus{}b}{p}$ :wink:

[quote="Ahiles"]Let $ p$ be a prime number, $ p\not \equal{} 3$, and integers $ a,b$ such that $ p|a \plus{} b$ and $ p^2|a^3 \plus{} b^3$. Prove that $ p^2\mid a \plus{} b$ or $ p^3\mid a^3 \plus{} b^3$.[/quote]
(Lemma) If $ p|\frac{a^3\plus{}b^3}{a\plus{}b}$, then $ p^2|\frac{a^3\plus{}b^3}{a\plus{}b}$.
[b]Proof: [/b] Suppose the "if" statement. Clearly $ p|p^2|(a\plus{}b)^2$, so $ p|3ab$, so at least one of $ a,b$ is a multiple of $ p$. But then the other must be, also, because of $ p|a\plus{}b$. Then we have the stronger statement $ p^2|3ab$. Reverting back to the stronger $ p^2|(a\plus{}b)^2$, we deduce that $ p^2|\frac{a^3\plus{}b^3}{a\plus{}b}$.

Using TTsphn's notation, we have two cases:

1. $ v_p(a\plus{}b)\equal{}1$
We compute: \[ v_p(a\plus{}b)\plus{}v_p\left(\frac{a^3\plus{}b^3}{a\plus{}b}\right) \equal{} v_p(a^3\plus{}b^3)\geq2\iff v_p\left(\frac{a^3\plus{}b^3}{a\plus{}b}\right)\geq1\]
But all the lemma says is that $ v_p\left(\frac{a^3\plus{}b^3}{a\plus{}b}\right)\not\equal{}1$. Thus it is $ \geq2$; thus $ \boxed{v_p(a^3\plus{}b^3)\geq3}$.
2. $ v_p(a\plus{}b)>1$
Clearly this is equivalent to the other "or" case, so the problem is solved.

---------------

TTsphn, where is a proof of that lemma? And i'm assuming it holds for negative $ b$?
Because then the solution to the general case is really just arithmetic :D

Yes, of course it holds also for negative $ b$. Try to prove it yourself, it's only induction and binomial theorem.

[quote="Ahiles"]Let $ p$ be a prime number, $ p\not \equal{} 3$, and integers $ a,b$ such that $ p|a \plus{} b$ and $ p^2|a^3 \plus{} b^3$. Prove that $ p^2\mid a \plus{} b$ or $ p^3\mid a^3 \plus{} b^3$.[/quote]
$p|a+b,p^2|(a+b)^3-3ab(a+b)\implies p^2|3ab(a+b)\implies p|3ab\implies p|a,b\implies p^3|a^3+b^3$
Else we have $p\not|a^2-ab+b^2$ and then from Lifting Exponent Lemma,since $p|a+b,p^2||a^3+b^3$

It has been almost three years since my last post here, and I've learned some math since then. I can see that I once viewed this as an isolated, number theory problem. In fact, this is a very trivial problem in:
http://en.wikipedia.org/wiki/P-adic_analysis

I don't see where you want to throw serious analysis in ;)
Well, one can do it using logarithms (I ignore $p=2$): If $p|b$, done. Otherwise, $p$-adically, you can reduce this to having some $\zeta = - \frac ab \in U^{(1)}$ such that $\zeta^3 \in U^{(2)}$. Thus if $\xi = \log_p(\zeta)$, you get $\xi \in \mathfrak p$ and $3 \xi \in  \mathfrak p^2$, which means that $\xi \in \mathfrak p^2$ or $\zeta \in U^{(2)}$.
Alternatively, you could directly involve the multiplicative structure of the $p$-adics, but that's just another use of $\log_p$.

$p^2|a^3+b^3$
if $p^2|(a+b)$ then done else $p|a^2-ab+b^2$
but, $p|(a+b)$
so, $p|(a+b)^2$
so, $p|3ab$
so, $p|ab$ since $p\neq 3$
let wlog $p|a$ then as $p|a+b$
so, $p|b$
so, $p^3|a^3+b^3$

[quote="kunny"]What's the source of the problem?[/quote]

This is [url=http://www.artofproblemsolving.com/resources.php?c=142&cid=78&year=2008]Romanian Junior Balkan Team Selection Test 2008[/url]. :)

$p$ divides $(a+b).$ so , $V_p{(a+b)}\ge1$
[by $V_p{(x)}$ , we mean the [i]highest power[/i] of $p$ that divides $x$].
now , if $V_p{(a+b)}\ge2$ , then $p^2$ divides $(a+b)$
otherwise , suppose $V_p{(a+b)}=1$. so,$a+b=pk$ , where $p$ doesn't divide $k$
so , $a^3+b^3=(a+b)(a^2-ab+b^2)=pk(a^2-ab+b^2)$
so , $p$ divides $(a^2-ab+b^2)=(a+b)^2-3ab$. so , $p$ divides $3ab$. as $p$ is not equal to $3$ , we have,
$p$ divides $ab$. so , $p$ divides $a$ or,$b$ or,both.
now , as $p$ divides $(a+b)$ also , so $p$ divides both $a$ & $b$.
so , $p^3$ divides $a^3+b^3$. hence done! :D

Why not this completely elementary approach? Write $d=\gcd(a,b)$, $a=dx$, $b=dy$, $\gcd(x,y)=1$. We are given $p\mid a+b = d(x+y)$ and $p^2 \mid a^3+b^3 = d^3(x^3+y^3)$ for some prime $p\neq 3$. Now, if $p\mid d$, it immediately follows $p^3 \mid d^3 \mid d^3(x^3+y^3) = a^3+b^3$ (it could  be that $p^2 \nmid d(x+y)=a+b$, if $p^2\nmid d$ and $p\nmid x+y$). If, on the other hand, $p\nmid d$, then $p\mid x+y$. Assuming $p^2 \nmid x+y$, notice that, since $p^2 \mid x^3+y^3 = (x+y)^3 - 3xy(x+y)$ and $p\neq 3$, it follows $p\mid xy$; but this is inconsistent with $p\mid x+y$. Therefore it must be $p^2\mid x+y \mid a+b$ (and it could be that $p^3 \nmid  d(x^3+y^3) = a^3+b^3$, if $p^3\nmid x+y$). Of course, a trivial case is for $0 = a+b = a^3+b^3$, when $p$ at any power divides both.

I was trying to write a one-line solution using LTE for this one, but I just realized this can't be done (because we finally show that $p | a$ and $p| b$)...

Why doesn't LTE work?
Case 1: $a \not\equiv 0 \bmod{p}$
We then have by LTE that $v_p(a^3+b^3)= v_p(a+b)$. This automatically implies that $p^2|a+b$ by the conditions.
Case 2: $a \equiv 0 \bmod{p}$
It is then easy to see that $p^3|a^3+b^3$ from the conditions.
This is basically $2$ one liners.

[quote="samaridin"]if a=5 b=5 p=5 =>p*p|a+b this is wrong[/quote]
[hide="Latexified:"][quote]If $a=b=5$ and $p=5$ then $p^2|a+b$, this is wrong.[/quote]
Just observe that $p^3|a^3+b^3$ holds good here and from quantumbyte's proof it is easily seen that here $a\equiv 0 \pmod {p}$ So $p^3|a^3+b^3$. See that the question says that we must prove that either of the two holds true. :) [/hide]

p|(a+b) we have (a+b)/p€N and used p2|(a+b)(a2-ab+b2) we deduce that p|   (a+b)/p(a2-ab+b2),  (p2=p*p,a2=a*a ,b2=b*b,(a+b)2=(a+b)*(a+b) )
Because p is a prime number ,we have:
a)p|   (a+b)/p →p2|(a+b)
b)p|(a2-ab+b2)→p|[(a+b)2-3*a*b]     and      p|(a+b)2  →p|3*a*b→p|a or p|b   because p is a prime number p≠3
p|a+b→p|a   and p|b→
p|a3 and p|b3→p|a3+b3 (a3=a*a*a and b3=b*b*b)
In conclusion p|a+b or p|a3+b3


Alright, with the second exam over the only problem left is figuring out the cutoff. We've been trying to collect information about other country's success on the campus, but this is kinda feasible and inexact. So I decided to open this topic for everyone to drop off the results she knows.

All following are just estimates of the number of solved problems, so don't get your hopes too high/low :)

Germany : $ 6 \minus{} \varepsilon/4/4/3.5/2.5/2.5$
Austria: $ 3/2/1.5/1/1/0$
China: $ 6/5/5/5/5/5$
Poland: $ 4/4/3/2/2/2$
Brazil: $ 5/4/4/4/3.5/3.5$
Turkey: $ 5/5/4/4/4/3.5$
Azerbaijan: $ 4/3/2.5/2/2/0.5$
India: $ 4\minus{}\varepsilon/4\minus{}\varepsilon/4\minus{}\varepsilon/3\minus{}\varepsilon/3\minus{}\varepsilon/3\minus{}\varepsilon$ Details on this would be nice!

A big thanks to everybody who contributed until now!

As you see, there are still 96 countries missing, so post what you know, and correct me if I'm wrong! I'll edit new information in here.

[quote="Agr_94_Math"]I think on Day 1 , India, I heard, 3 people solved the 1st 2 problems and mostly none of them solved the 3rd. On the 2nd day, probably all would not solved the last problem as it is very hard I think that the CHinese themselves couldnt solve it{5 of them}. So, I feel 4 would be the maximum possible by an Indian. I may be wrong also. But I shall post today if I get any news.[/quote]

Are you sure that three failed to solve the first two problems? :(
Can you also tell about the individual performance?

not that 3 failed to oslve the 1st 2. they might not have solved 1 out of the 2.
I think Gaurav patil has done well. Akashnil is sure ot have done well. I am not sure about htis.

firstday
süreyya +++
melih +++
umut +++
ufuk ++-
fehmi ++-
vefa ++-

[size=150]5/5/4/4/4/3,5
[/size]

[u][b]why don't the other countries write their results?
previously some of contestants were writing their results.[/b][/u]

Please do not say like that BG Yoda. There might have been health reasons also. EVen then, he has gone uptil the IMO and  probably he may even get a 10 out of 42 by collecting partial points.

[quote="BG Yoda"]:pilot:  :spider:  :play_ball:[/quote]

this seems super meaty post...dunno what you meant though

I expressed my opinion and surprise about the obtained results , provided that the difficulty of the problems was very disputable for such type of competition ( this is still IMO however , only the best students go there )... but in this way it sounded like I want to insult somebody and therefore I've edited the post with something senseless. There was already another post and because of that I couldn't delete mine.

If the moderator see this topic, let s/he delete my statements.

I have to say that IMO is merely elementary math contest.  Good or bad at such a contest has no effect on your future career. I know that there are some people, who could not do anything else than wining IMO medals. And for the whole life, they often live in the past.

My work has nothing to see with math, but i enjoy solving contest problems.

Well news from India.
Gaurav patil:3 + some partials{3-4}
Sameer Wagh- 4
Akashnil Dutta- 3
Akshay Mittal- 2.5
Ananth Shankar-3
Subhadip CHaudry- 3.5.

Wao, Japan is in the 2nd. :lol:

Congratulations to all of members of Japanese Teams! :)

Japanese Communities Moderator

kunny

[quote="April"]I don't know the results of the Vietnamese contestants, but Mr. Tran Nam Dung said in a Vietnamese Mathematical Forum that Vietnamese team will get 1 gold medal for sure, and another gold is still in doubt, 3 silvers and 1 bronze (or HM). Hopefully that what he said will come true  :)[/quote]

Final result: 2/2/2. It is good result for vietnamese team. Congratulations to contestants and leaders.

:D Congaratulations khashayar jan however the result was terrible for Iran. :(

[quote="sisioyus"][quote="kunny"]Wao, Japan is in the 2nd. :lol:

Congratulations to all of members of Japanese Teams!

Japanese Communities Moderator

kunny[/quote]
congratulations to your nation! :)[/quote]

Thank you!  :)

I am surprised at 5 Gold medals.

Congratulations to Chinese Team in the 1 st. :)

Chinese Team is very Powerful.

[quote="L_Euler"][quote="Fabiano"]VietNam
Hà Khương Duy: 7 7 7 7 7 4 =39
Phạm Đức Hùng 7 7 7 7 5 0 =33
Phạm Hy Hiếu:7 7 1 7 7 0 =29
Nguyễn Hoàng Hải: 7 3 1 7 7 0 =25
Tạ Đức Thành: 2 7 0 6 4 0 =19
Nguyễn Xuân Cương: 6 0 3 7 0 0 =16[/quote]

A bad result, VN was down!!!
[/quote]

It is good result for the team. During training camp Mr Khac Minh guessed for a lower result, but I believed that we will get 2 golds.

[quote="mto"]Changes in VietNam National Olympiad in recent years result in the down of Viet Nam Team 's rank at IMO. Many eligible guys get bad result in VN National Contest and could not attain the TST!!!![/quote]

[quote="L_Euler"]
That's true, we need a change in the VNMO and VNTST if wishing the high point and rank at IMO. In this year (and last year, 3,4 years ago,...), not few of hard students was miss in VMO and TST. :o[/quote]

I'm agreeing with you about this part. We should change something in VMO, TST and especially in motivation of students and teachers.

What about Ukraine? And is there any Internet page with full IMO results?

You can see the result here. :)

[url]http://www.imo-official.org/year_info.aspx?year=2009[/url]

[quote="L_Euler"][quote="Fabiano"]VietNam
Hà Khương Duy: 7 7 7 7 7 4 =39
Phạm Đức Hùng 7 7 7 7 5 0 =33
Phạm Hy Hiếu:7 7 1 7 7 0 =29
Nguyễn Hoàng Hải: 7 3 1 7 7 0 =25
Tạ Đức Thành: 2 7 0 6 4 0 =19
Nguyễn Xuân Cương: 6 0 3 7 0 0 =16[/quote][/quote]

Two gold medals; It is not bad result. My congratulations to all members of our country 's team.
Chinese Team is always success !. Many  congratulations to them.
However, I haven't seen The Result of Russian team yet. I guess they will be in the third.

You can see the result in my post #61. :)

[quote="bertram"]
Germany : $ 6 \minus{} \varepsilon/4/4/3.5/2.5/2.5$
Austria: $ 3/2/1.5/1/1/0$
China: $ 6/5/5/5/5/5$
Poland: $ 4/4/3/2/2/2$
Brazil: $ 5/4/4/4/3.5/3.5$
Turkey: $ 5/5/4/4/4/3.5$
Azerbaijan: $ 4/3/2.5/2/2/0.5$
India: $ 4 \minus{} \varepsilon/4 \minus{} \varepsilon/4 \minus{} \varepsilon/3 \minus{} \varepsilon/3 \minus{} \varepsilon/3 \minus{} \varepsilon$ Details on this would be nice!
[/quote]

out of curiosity, did you ask this yourself? :D

[quote="Fabiano"]VietNam
Hà Khương Duy: 7 7 7 7 7 4 =39
Phạm Đức Hùng 7 7 7 7 5 0 =33
Phạm Hy Hiếu:7 7 1 7 7 0 =29
Nguyễn Hoàng Hải: 7 3 1 7 7 0 =25
Tạ Đức Thành: 2 7 0 6 4 0 =19
Nguyễn Xuân Cương: 6 0 3 7 0 0 =16[/quote]

What ??? The points are lower than usual ... :D
Hope Vietnam will be the CHAMPION in the next year :)

[quote="ZHANGWENZHONGKK"]:?:  What about USA and Russia?[/quote]
and Vietnam?[/quote]


Let x, y, z> 0, $ x ^ 2 \plus{} y ^ 2 \plus{} Z ^ 2 \equal{} 3$ Find min:
  $ \frac {x}{y^2\plus{}z^2} \plus{} \frac {y}{z^2\plus{}x^2} \plus{} \frac{z}{x^2 \plus{} y^2}$

$ \sum_{cyc} \frac {x}{y^2 \plus{} z^2} \equal{} \sum_{cyc} \frac {x}{3 \minus{} x^2} \ge 3 \times \frac {\frac {\sum_{cyc} x}{3}}{3 \minus{} (\frac {\sum_{cyc} x}{3})^2} \ge 3\times\frac {1}{3 \minus{} 1} \equal{} \frac {3}{2}$

Because $ 0 < \frac {1}{3}\sum_{cyc} x \le \frac {1}{3} \sqrt {3 \sum_{cyc} x^2} \equal{} \frac {1}{3} \sqrt {3 \times 3} \equal{} 1$

Sorry $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$
Let x, y, z> 0, $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$ Find min:
  $ \frac {x}{y^2 \plus{} z^2} \plus{} \frac {y}{z^2 \plus{} x^2} \plus{} \frac {z}{x^2 \plus{} y^2}$

[quote="aadil"]i think we can also try using caushy shwarz for the reals $ x,y,z$ and $ 1/(x^2 \plus{} y^2), 1/(y^2 \plus{} z^2),1(z^2 \plus{} x^2)$[/quote]
how does that give the minimum?


six people participated in a checkers tournament. Each participant played each of the other players exactly three times. How many games took place in all?

[hide]6 players, and each person plays each of the other 5 3 times.

Thats 15 matches per person. Multiply by 6, divide by 2 because we're repeating each game twice, and we get 45.[/hide]

[hide]The number of games will just be the fifth triangular number times 3. 3*(1+2+3+4+5)=45.[/hide]

[hide]Since each player faces the other 5 players exactly 3 times;
3 * (5 + 4 + 3 + 2 + 1) :arrow: 3 * ( 5 - 1 ) * 5; acording to (n+1)n
[b]=45[/b]
[i]45 games took place in all[/i][/hide]

[hide]
Each of the 6 people play each of the other 5 people 3 times each.
That is $6\times5\times3=90$ matches so far.
Yet, we double counted, because two people are in each match.
Thus the answer is $\boxed{45}$[/hide]

pianoforte, what does (6) over (2) mean? In beginner algebra language, please

By definition, $\binom {n}{m}$ is the number of ways to pick $m$ objects from $n$ objects ignoring order.

The formula for this is $\frac{n!}{m!(n-m)!}$.  (the !'s are factorial, ex. 5!=5*4*3*2*1, 4!=4*3*2*1.)

[hide]there are 5 choose 2 total games when played once. but multiply 3 to get 3 times each. That is 45 games total[/hide]


real analysis
vector
geometry
3D geometry
sphere
topology
Functional Analysis

Let $ D\equal{}(0,2\pi)$.
$ S\equal{}\{u\in L^2(D) : \parallel{}u\parallel{}_2 \equal{} 1\}$
$ B\equal{}\{u\in L^2(D) : \parallel{}u\parallel{}_2\leq 1\}$.
Prove that $ S$ and $ D$ are closed in $ L^2(D)$, $ B$ is weakly closed in $ L^2(D)$, and $ S$ is not weakly closed in $ L^2(D)$.

"Closed" works the same as in $ \mathbb R^2$ or any metric space. The second part is actually easier in an abstract Hilbert space with orthonormal basis $ (e_n)$: you show that $ e_n\to 0$ weakly. If you want a concrete example in $ L^2$, the sequence $ e^{inx}$ (normalized) will work, but the statement is equivalent to the Riemann-Lebesgue lemma.

$ \{u_n\}$  weakly converges to $ u$ if $ T(u_n)$ converges to $ T(u)$ for any $ T$ is bounded linear mapping from $ D$ to $ \mathbb{R}$.
$ W$ is weakly closed if $ u_n$ in $ W$ weakly converges to $ u$, then $ u\in W$.

$ B$ is weakly closed because it is a convex strongly closed subset of the normed vector space (in a locally convex space, the weak closure of a convex set and its strong closure coincide).

$ S$ is not weakly closed because its weak closure coincides with $ B$ (in any normed vector space, the weak closure of the unit sphere is the (closed) unit ball).

Yes, I know . But here, I dont mention any notion about topology and convex. I want to find a counter-example to show that : $ \{u_n\}\in S$ weakly converges to $ u$, but $ u\notin S$, that is $ \parallel{}u\parallel{}_2\ne 1$.


function
combinatorics proposed
combinatorics

Find the number of subsets $B$ of $\{1,2,3,...,2005\}$ such that the sum of the elements of $B$ is congruent to $2006 \mod 2048$.

I think i can solved it with generating functions and roots of unity.

here is something that you can do :
consider the product $P(x)=(1+x)(1+x^2)\cdots (1+x^{2005})$.every term in the product considers a subset of the given set.now note that $(1+x)(1+x^2)(1+x^4)\cdots (1+x^{1024})=\frac{1-x^{2048}}{1-x}=\sum_{k=0}^{2047} x^k$.now let $f(x)=\Pi_{k\neq 2^l}\  x^k=\sum_t a_t x^t$.so that  $P(x)=f(x)(1+x+x^2+\cdots+x^{2047})$. what we want is sum of coeffs of $P(x)$ of all of the terms $x^{s+k\cdot 2048},k=0,1,\ldots$ etc. where $s=2006$.
now for each $t$ the product $a_tx^t(1+x+x^2+\cdots+x^{2047})$ gives exactly one term of degree congruent to $s$ modulo $2048$. so every term $a_tx^t$ occuring in $f(x)$ contributes a total of $a_t$ to the sum that we wish to compute.so the required sum is simply the sum of the $a_t$'s. but that is simply $f(1)=2^{2005-11}=2^{1994}$.
hope everything is alright.

yup that's the basic gist of the official solution too, just that it wasn't in generating function form!


geometry
rotation
logarithms
integration
trapezoid
3D geometry
calculus

Consider the curve $y=\frac{1}{1+x^{2}}$ (you may like to sketch this).  Find the equation of the line which is tangent to the curve and passes through $(0,1)$.

By considering the area under the curve and this line, prove $\pi>3$.  Also, by considering the volume produced when the curve and the line are rotated about the $y$ axis, prove $\ln{2}>\frac{2}{3}$

[Sorry this question is so long, but I think you may appreciate the results]

The equation of the tangent line in question is $y=-\frac{1}{2}|x|+1,\ y=1$ and the tangent points are $(0,\ 1),\ \left(\pm 1,\ \frac{1}{2}\right).$
Now for the former part of the problem, $\int_{0}^{1}\frac{1}{1+x^{2}}\ dx>\left(1+\frac{1}{2}\right)\cdot 1\cdot \frac{1}{2}\cdots \text{the area of a trapezoid}.\Longleftrightarrow \frac{\pi}{4}>\frac{3}{4},$ yielding $\pi >3.$ For the latter part,

$\int_{\frac{1}{2}}^{1}\pi \left(\frac{1}{y}-1\right) dy>\frac{1}{3}\pi \cdot 1^{2}\cdot \left(1-\frac{1}{2}\right)\cdots \text{the volume of the cone}$

$\Longleftrightarrow \ln 2-\frac{1}{2}>\frac{1}{6},$ yielding $\ln 2 >\frac{2}{3}.$


geometry
3D geometry
number theory unsolved
number theory

Prove that the equation:
$x^3+y^3=z^3+4$ can not be solved in natural numbers $x,y,z$.

Cubes are -1, 0 or 1 (mod 9). That's it.


linear algebra
linear algebra unsolved
identity

$A,B\in M_n(C)$ such that $A,B$ commute with $[A,B]=AB-BA$

Prove that $e^Ae^B=e^{A+B+\frac{1}{2}[A,B]}$

[quote=loup blanc]Tom Calc.  To give one or two references is not a proof.
Do you know how to prove the BCH formula ?
On the other hand, the BCH formula is valid (that is, it is convergent)  only when the considered matrices are "small".[/quote]
A proof is given in Theorem 5.1 of link 1,a proof is given in the linked Wikipedia article.Sorry,but I don't find a reason to rewrite the proof of a well-known result. I think the case in #1 does not require a "smallness" condition on the matrices.But for $\ln(e^Xe^Y)=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]]-\frac{1}{12}[Y,[X,Y]]+...$ iff $||x||+||y||<2$ for $x,y\in \mathcal{B}$,since the mercator and BCH series both converges absolutely and the sums become equal.

the qn says "$A,B$ commute with $[A,B]=AB-BA$"
but $A,B$ commute $\implies [A,B]=AB-BA=O$

@below: ambiguous wording.. so [A,[A,B]]=[B,[A,B]]=0.

TomCalc , that you write is true but that is not the question.
Firstly, I once read the BCH's proof and it gave me stomach cramps; so I hurried to forget this proof.
Secondly, in the case we consider, there is practically nothing left of the formula; so we can think that there is a direct (and elementary) solution to our problem; and we are right to think that...

We use the standard theory of diff. equ. Let $C=[A,B]$.

Step 1. Let $X=e^{tA}e^{tB}e^{-1/2t^2C},Y=e^{t(A+B)}$. We want to prove that $X=Y$. Note that $X(0)=Y(0)$ and $YC=CY$.
$X'=AX+XB-tXC$; $Y'=(A+B)Y$. Then it remains to prove that $BY=YB-tCY$, that is,
$e^{t(A+B)}Be^{-t(A+B)}=B+tC$.
Step 2. Let $U=e^{t(A+B)}Be^{-t(A+B)},V=B+tC$. Note that $U(0)=V(0)$.
$U'=(A+B)U-U(A+B),V'=C$; we conclude with the fact that $C=(A+B)V-V(A+B)$.

[quote=loup blanc]Firstly, I once read the BCH's proof and it gave me stomach cramps; so I hurried to forget this proof. [/quote]
I don't remember any cramps,but the proof I read was from Hall's lie algebra for this
$\ln(e^Xe^Y)=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]]-\frac{1}{12}[Y,[X,Y]]+...$.I admit that the main BCH formula was confusing and time-consuming to learn at first.
Of course #1 has a elementary solution.Here is mine,very similar to yours:
Consider the function $f(x)=e^{xA}e^{xB}e^{x(A+B)}$.Differentiate with respect to $x$,use the identinty,$C=[A,B]\implies [C,A]=0=[C,B]$ and $e^ABe^-A=B+[A,B]$ we get a differential equation with initial condition $f(0)=1$ integrating it and setting $x=1$ gives the identity in #1.

Yours,is also nice @loup blanc.


$m$ an integer such that $m$ divides $a^2+1$for a certain$a>0$.Prove that $m$ is sum of two perfect square.

All the prime divisors are of the form $4k+1$, so they're sums of $2$ squares, and it's easy to prove that if two numbers are sums of $2$ squares, then so is their product.

of course it is very easy. (but in this exam, lots of student do not know these results and have to prove all these theorems ... here in france, there is not a lot of NT  :D  )


inequalities
calculus
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3\plus{}b^3\plus{}c^3\plus{}3abc \ge \frac{(a^2b\plus{}b^2c\plus{}c^2a)^2}{ab^2\plus{}bc^2\plus{}ca^2}\plus{}\frac{(ab^2\plus{}bc^2\plus{}ca)^2}{a^2b\plus{}b^2c\plus{}c^2a}\]
:)

[quote="mitdac123"]<=> $ abc\frac {\sum a^6 \minus{} \sum a^3b^3}{(\sum a^2b)(\sum ab^2)} \ge 0$
:)[/quote]
Yes, but mitdac123 also told me that he didn't find this by hand. So I am waiting for another approach (there exists a very nice and non-standard way to prove it :) )

So can you post your solution?
I am looking for it. :roll:
Nice ineq!

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {(ab^2 \plus{} bc^2 \plus{} ca)^2}{a^2b \plus{} b^2c \plus{} c^2a}
\]
:)[/quote]

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {(ab^2 \plus{} bc^2 \plus{} ca)^2}{a^2b \plus{} b^2c \plus{} c^2a}$

$ \leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \plus{} \sum_{sym} ab(a \plus{} b) \ge \frac {[\sum_{sym} ab(a \plus{} b)][\sum_{sym} a^2b^2(a^2 \plus{} b^2) \plus{} 2xyz\sum_{sym} ab(a \plus{} b)]}{(ab^2 \plus{} bc^2 \plus{} ca)(a^2b \plus{} b^2c \plus{} c^2a)}$

Assum $ a \plus{} b \plus{} c \equal{} 1$ .

Let $ a \plus{} b \plus{} c \equal{} p \equal{} 1;ab \plus{} bc \plus{} ca \equal{} q;abc \equal{} r$. We have :

$ Ineq \leftrightarrow r[3q^3 \minus{} 9q^2 \plus{} 6q \minus{} 1 \plus{} 3r(3q \minus{} 2)] \le\ 0$

. Follow schur we have $ r \ge\ \frac {4q \minus{} 1}{9}$

. Let $ f(r) \equal{} 3r(3q \minus{} 2) \plus{} 3q^3 \minus{} 9q^2 \plus{} 6q \minus{} 1 \rightarrow f '(r) \equal{} 3q \minus{} 2 \le\ 0$ ( because $ q \le\ \frac {1}{3}$)

$ \rightarrow f(r) \le\ f(\frac {4q \minus{} 1}{9}) \equal{} (3q \minus{} 1)^2(q \minus{} 1) \le\ 0$

Equation hold on $ a\equal{}b\equal{}c$ or $ abc\equal{}0$

We have done! :)

[quote="Allnames"][quote="onlylove_math"]
$ \rightarrow f(r) \le\ f(\frac {4q \minus{} 1}{9}) \equal{} (3q \minus{} 1)^2(q \minus{} 1) \le\ 0$

We have done! :)[/quote]
I think your proof is wrong because with yours,The equation holds only when $ p \equal{} \frac { 1}{3}$ ,It is not enough,Try $ p \equal{} \frac {1}{4}$,It means $ c \equal{} 0,a \equal{} b$ :lol: ,But I think you are  very hard ,my friend
Please check again and fix your mistake :)  :)[/quote]
Thank my friend . I had mistake. I have fixed  :)

[quote="shaaam"][quote="shaaam"][quote="shaaam"]after hours of expanding, i got this:

(all sums are cyclic)

$ \sum a^7bc \plus{} \sum a^4b^4c \plus{} 2\sum a^5b^2c^2 \geq 2\sum a^4bc^4 \plus{} \sum a^4b^2c^3 \plus{} \sum a^3b^4c^2$

can someone please tell me if that is right?[/quote]

i am sorry.. that is wrong above.......

but i did the expanding again......
and i got this :

$ \sum a^7bc \plus{} \sum a^4b^4c \geq 2\sum a^4c^4b$

Help me to prove this , please..... thanks!!!![/quote]

Sorry for my stupidity......

$ \sum a^7bc \plus{} \sum a^4b^4c \geq 2\sum a^4c^4b$

is equivalent to

$ \sum a^7bc \geq \sum a^4c^4b$

Dividing both sides by abc will yield

$ \sum a^6 \geq \sum a^3b^3$

and that is clearly right, since it is equivalent to :

$ (a^3 \minus{} b^3)^2 \plus{} (b^3 \minus{} c^3)^2 \plus{} (c^3 \minus{} a^3)^2 \geq 0$

believe it or not ,  i spent 3 hours on expanding!!!  :blush:[/quote]
It is okay, shaam. Thank you very for you patience. :)

Here is another way you can think of this problem:

Subtract $ \sum_{sym}a^2b$ to both sides and rewrite as
$ a(a\minus{}b)(a\minus{}c)\plus{}b(b\minus{}a)(b\minus{}c)\plus{}c(c\minus{}a)(c\minus{}b)\geq \left[(a\minus{}b)(b\minus{}c)(c\minus{}a)\right]^2\left(\frac{1}{a^2b\plus{}b^2c\plus{}c^2a}\plus{}\frac{1}{a^2c\plus{}c^2b\plus{}b^2a}\right)$
notice that if we subtract $ 0\le t\le \min{a,b,c}$ from each of $ a,b,c$ then LHS decreases and RHS increases.
So we only need to show that the inequality holds when one of the variables is set to $ 0$

So we need to prove
$ a^3\plus{}b^3\geq \frac{a^4b^2}{ab^2}\plus{}\frac{a^2b^4}{a^2b}\equal{}a^3\plus{}b^3$
which is obviously true.

So the inequality is true and equality is achieved only when one of the variables is 0 or when all three are equal.

how about this ?

Let $ a,b,c$ be positve real numbers then [b]prove or disprove[/b] that

$ a^3\plus{}b^3\plus{}c^3\plus{}3abc \ge \frac{(a^2b\plus{}b^2c\plus{}c^2a)^3}{(ab^2\plus{}bc^2\plus{}ca^2)^2}\plus{}\frac{(ab^2\plus{}bc^2\plus{}ca)^3}{(a^2b\plus{}b^2c\plus{}c^2a)^2}$

This inequality is sharper, but i think it is wrong.
Can someone please help me?

[quote="shaaam"]how about this ?

Let $ a,b,c$ be positve real numbers then [b]prove or disprove[/b] that

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^3}{(ab^2 \plus{} bc^2 \plus{} ca^2)^2} \plus{} \frac {(ab^2 \plus{} bc^2 \plus{} ca)^3}{(a^2b \plus{} b^2c \plus{} c^2a)^2}$

This inequality is sharper, but i think it is wrong.
Can someone please help me?[/quote]
Yes it's wrong $ a\equal{}b\equal{}0,9$ and $ c\equal{}1,2$.

But this sharper one is true:

$ \sum a(a \minus{} b)(a \minus{} c)\ge \left[(a \minus{} b)(b \minus{} c)(c \minus{} a)\right]^2 \left(\frac {1}{a^2b \plus{} b^2c \plus{} c^2a \minus{} 3abc} \plus{} \frac {1}{a^2c \plus{} c^2b \plus{} b^2a \minus{} 3abc}\right)$
Same proof as Albanian Eagle’s one. :) ;)

For an even sharper one see here towards the end:
http://www.mathlinks.ro/Forum/viewtopic.php?p=1321626 :D

[quote]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {(ab^2 \plus{} bc^2 \plus{} ca)^2}{a^2b \plus{} b^2c \plus{} c^2a}
\][/quote]
I have a solution by using AM GM inequality.Can you find it?  :)

[quote="shaaam"][quote="PHONGVAN"][quote]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {(ab^2 \plus{} bc^2 \plus{} ca)^2}{a^2b \plus{} b^2c \plus{} c^2a}
\]
[/quote]
I have a solution by using AM GM inequality.Can you find it?  :)[/quote]

With only and only AM-GM? i think it is just impossible.
if you have done it, then well done :first:[/quote]
There is a very short solution (about 8lines) with even no AM-GM (of course no EMV since I dont like this method) ;) :)

[quote="can_hang2007"][/quote][quote="shaaam"][quote="PHONGVAN"][quote]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 + b^3 + c^3 + 3abc \ge \frac {(a^2b + b^2c + c^2a)^2}{ab^2 + bc^2 + ca^2} + \frac {(ab^2 + bc^2 + ca)^2}{a^2b + b^2c + c^2a}
\]
[/quote]
I have a solution by using AM GM inequality.Can you find it?  :)[/quote]

With only and only AM-GM? i think it is just impossible.
if you have done it, then well done :first:[/quote][quote="can_hang2007"]
There is a very short solution (about 8lines) with even no AM-GM (of course no EMV since I dont like this method) ;) :)[/quote]
 :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {(ab^2 \plus{} bc^2 \plus{} ca)^2}{a^2b \plus{} b^2c \plus{} c^2a}
\]
:)[/quote]
this is my solution.
If $ a \equal{} 0$ or $ b \equal{} 0$ or $ c \equal{} 0$ ,it's true.

if $ abc > 0$

Put $ x \equal{} \frac {a}{b} , y \equal{} \frac {b}{c},z \equal{} \frac {c}{a}$.We need prove
 $ \frac {x}{z} \plus{} \frac {y}{x} \plus{} \frac {z}{y} \plus{} 3\geq\frac {(xy \plus{} yz \plus{} zx)^2}{xyz(x \plus{} y \plus{} z)} \plus{} \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx}$

$ \frac {x}{z} \plus{} \frac {y}{x} \plus{} \frac {z}{y} \geq\frac {x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2}{xyz(x \plus{} y \plus{} z)} \plus{} \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx}$

$ \frac {x^2}{z} \plus{} \frac {z^2}{y} \plus{} \frac {y^2}{x}\geq\frac {(x^2 \plus{} y^2 \plus{} z^2)(x \plus{} y \plus{} z)}{xy \plus{} yz \plus{} zx}$

$ \frac {x^3y}{z} \plus{} \frac {y^3z}{x} \plus{} \frac {z^3x}{y}\geq\ x^2y \plus{} y^2z \plus{} z^2x$

By using AM GM'inequality, we have:
 $ \frac {x^3y}{z} \plus{} xyz\geq\ 2x^2y$

$ \frac {y^3z}{x} \plus{} xyz\geq\ 2y^2z$

$ \frac {z^3x}{y} \plus{} xyz\geq\ 2z^2y$

$ \ x^2y \plus{} y^2z \plus{} z^2x\geq\ 3xyz$

We have done! :)

This is basically, midtac123's solution at post number 2.

Also you have a typo.

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {(ab^2 \plus{} bc^2 \plus{} ca)^2}{a^2b \plus{} b^2c \plus{} c^2a}
\]
:)[/quote]
Actually, this inequality is equivalent to the interesting and nice inequality of VasC:
\[ \left( x\plus{}\frac{1}{y} \minus{}1\right) \left( y\plus{}\frac{1}{z}\minus{}1\right) \plus{}\left( y\plus{}\frac{1}{z} \minus{}1\right) \left( z\plus{}\frac{1}{x} \minus{}1\right) \plus{}\left( z\plus{}\frac{1}{x} \minus{}1\right) \left( x\plus{}\frac{1}{y}\minus{}1\right) \ge 3\]
for any $ x,y,z>0.$ ;) :)

[quote="manlio"][quote="can_hang2007"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {(ab^2 \plus{} bc^2 \plus{} ca)^2}{a^2b \plus{} b^2c \plus{} c^2a}
\]
:)[/quote]
Actually, this inequality is equivalent to the interesting and nice inequality of VasC:
\[ \left( x \plus{} \frac {1}{y} \minus{} 1\right) \left( y \plus{} \frac {1}{z} \minus{} 1\right) \plus{} \left( y \plus{} \frac {1}{z} \minus{} 1\right) \left( z \plus{} \frac {1}{x} \minus{} 1\right) \plus{} \left( z \plus{} \frac {1}{x} \minus{} 1\right) \left( x \plus{} \frac {1}{y} \minus{} 1\right) \ge 3
\]
for any $ x,y,z > 0.$ ;) :)[/quote]
Sorry, but I don't see how to make the two inequalities equivalent. :(
Can anyone help me?
Thank you very much.[/quote]
Indeed, Manlio, there is a problem here, since an inequality is symmetric and the other is only cyclic.

[quote="Vasc"][/quote][quote="manlio"][quote="can_hang2007"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 + b^3 + c^3 + 3abc \ge \frac {(a^2b + b^2c + c^2a)^2}{ab^2 + bc^2 + ca^2} + \frac {(ab^2 + bc^2 + ca)^2}{a^2b + b^2c + c^2a}
\]
:)[/quote]
Actually, this inequality is equivalent to the interesting and nice inequality of VasC:
\[ \left( x + \frac {1}{y} - 1\right) \left( y + \frac {1}{z} - 1\right) + \left( y + \frac {1}{z} - 1\right) \left( z + \frac {1}{x} - 1\right) + \left( z + \frac {1}{x} - 1\right) \left( x + \frac {1}{y} - 1\right) \ge 3
\]
for any $ x,y,z > 0.$ ;) :)[/quote]
Sorry, but I don't see how to make the two inequalities equivalent. :(
Can anyone help me?
Thank you very much.[/quote][quote="Vasc"]
Indeed, Manlio, there is a problem here, since an inequality is symmetric and the other is only cyclic.[/quote]
My statement did not get any problem, VasC. Let me explain how could I make they "equivalent":
[b]Problem. [/b][i]Let $ x,y,z$ be positive real numbers. Prove that[/i]
\[ \sum \left( x + \frac {1}{y} - 1\right) \left( y + \frac {1}{z} - 1\right) \ge 3.
\]
[b][i]Proof. [/i][/b]We rewrite the inequality as
\[ \sum \frac {y}{x} + \left( \frac {1}{xyz} - 2\right) \sum x + \left( 1 - \frac {2}{xyz}\right) \sum xy + 3 \ge 0.
\]
Putting $ xyz = k^3$, then there exist $ a,b,c > 0$ such that $ x = \frac {ka}{b},y = \frac {kb}{c},z = \frac {kc}{a}.$ The inequality becomes
\[ \sum \frac {a^2}{bc} + \left( \frac {1}{k^2} - 2k\right) \sum \frac {a}{b} + \left( k^2 - \frac {2}{k}\right) \sum \frac {b}{a} + 3 \ge 0,
\]

\[ f(k) = \sum a^3 + \left( k^2 - \frac {2}{k}\right) \sum a^2b + \left( \frac {1}{k^2} - 2k\right) \sum ab^2 + 3abc \ge 0
\]
We have that
\[ f'(k) = \frac {2(k^3 + 1)}{k^3} \left( k\sum a^2b - \sum ab^2 \right)
\]

\[ f'(k) = 0 \Leftrightarrow k = \frac {\sum ab^2}{\sum a^2b}.
\]
From now, according to the Variation Board, we can deduce our inequality to show that
\[ f\left( \frac {\sum ab^2}{\sum a^2b}\right) \ge 0
\]
or equivalently,
\[ a^3 + b^3 + c^3 + 3abc \ge \frac {(a^2b + b^2c + c^2a)^2}{ab^2 + bc^2 + ca^2} + \frac {(ab^2 + bc^2 + ca^2)^2}{a^2b + b^2c + c^2a}.
\]
That's all. Hope I dont get any stupid mistakes.  :blush: :)

[quote="can_hang2007"][/quote][quote="Vasc"][/quote][quote="manlio"][quote="can_hang2007"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ a^3 + b^3 + c^3 + 3abc \ge \frac {(a^2b + b^2c + c^2a)^2}{ab^2 + bc^2 + ca^2} + \frac {(ab^2 + bc^2 + ca)^2}{a^2b + b^2c + c^2a}
\]
:)[/quote]
Actually, this inequality is equivalent to the interesting and nice inequality of VasC:
\[ \left( x + \frac {1}{y} - 1\right) \left( y + \frac {1}{z} - 1\right) + \left( y + \frac {1}{z} - 1\right) \left( z + \frac {1}{x} - 1\right) + \left( z + \frac {1}{x} - 1\right) \left( x + \frac {1}{y} - 1\right) \ge 3
\]
for any $ x,y,z > 0.$ ;) :)[/quote]
Sorry, but I don't see how to make the two inequalities equivalent. :(
Can anyone help me?
Thank you very much.[/quote][quote="Vasc"]
Indeed, Manlio, there is a problem here, since an inequality is symmetric and the other is only cyclic.[/quote][quote="can_hang2007"]
My statement did not get any problem, VasC. Let me explain how could I make they "equivalent":
[b]Problem. [/b][i]Let $ x,y,z$ be positive real numbers. Prove that[/i]
\[ \sum \left( x + \frac {1}{y} - 1\right) \left( y + \frac {1}{z} - 1\right) \ge 3.
\]
[b][i]Proof. [/i][/b]We rewrite the inequality as
\[ \sum \frac {y}{x} + \left( \frac {1}{xyz} - 2\right) \sum x + \left( 1 - \frac {2}{xyz}\right) \sum xy + 3 \ge 0.
\]
Putting $ xyz = k^3$, then there exist $ a,b,c > 0$ such that $ x = \frac {ka}{b},y = \frac {kb}{c},z = \frac {kc}{a}.$ The inequality becomes
\[ \sum \frac {a^2}{bc} + \left( \frac {1}{k^2} - 2k\right) \sum \frac {a}{b} + \left( k^2 - \frac {2}{k}\right) \sum \frac {b}{a} + 3 \ge 0,
\]

\[ f(k) = \sum a^3 + \left( k^2 - \frac {2}{k}\right) \sum a^2b + \left( \frac {1}{k^2} - 2k\right) \sum ab^2 + 3abc \ge 0
\]
We have that
\[ f'(k) = \frac {2(k^3 + 1)}{k^3} \left( k\sum a^2b - \sum ab^2 \right)
\]

\[ f'(k) = 0 \Leftrightarrow k = \frac {\sum ab^2}{\sum a^2b}.
\]
From now, according to the Variation Board, we can deduce our inequality to show that
\[ f\left( \frac {\sum ab^2}{\sum a^2b}\right) \ge 0
\]
or equivalently,
\[ a^3 + b^3 + c^3 + 3abc \ge \frac {(a^2b + b^2c + c^2a)^2}{ab^2 + bc^2 + ca^2} + \frac {(ab^2 + bc^2 + ca^2)^2}{a^2b + b^2c + c^2a}.
\]
That's all. Hope I dont get any stupid mistakes.  :blush: :)[/quote]
Very nice. That means that you either obtained a strong symmetric inequality from a cyclic inequality, or solved the cyclic inequality using a symmetric inequality. But the inequalities are not equivalent, they are only "equivalent". :P

[quote="shaaam"]I am sorry for my stupidity , but what does "Variation Board" mean?
Can you please explain it? Is it something to do with calculus?

Thanks

 :)

P.S: I didn't find it in Wikipedia.[/quote]
Sorry, my English is bad. Maybe I have mistakes in writing it in English. Let me explain more:
Let $ k_0 \equal{}\frac{\sum ab^2}{\sum a^2b}$, we can see that $ f'(k) \le 0 \forall k \in (0,k_0]$ and $ f'(k) \ge 0 \forall k \in [k_0, \plus{}\infty)$, which shows that $ f(k)$ is decreasing for all $ k \in (0,k_0]$, and $ f(k)$ is increasing for all $ k \in [k_0, \plus{}\infty)$. Therefore $ f(k)$ attains minimum when $ k\equal{}k_0$ and to prove the inequality $ f(k) \ge 0$, we just need to prove $ k(k_0) \ge 0$. That's what I want to say. :)

[quote="Vasc"]Very nice. That means that you either obtained a strong symmetric inequality from a cyclic inequality, or solved the cyclic inequality using a symmetric inequality. But the inequalities are not equivalent, they are only "equivalent". :P[/quote]
Yes.  :lol:

[quote="sfsreporter"]hey Can-Hang

cant we prove by simple applying Cauchy to the RHS??

if i am correct we have a 3 line solution...[/quote]

Can you please post your solution?

I am very interested in it as I also tried to solve it by BCS, but I failed. :( .

Thank you very much. :)

[quote="sfsreporter"]ps: i am new to inequalities... my solution can be dangerously irritating to ppl passionate about inequalities...plz forgive me if i'm wrong. :blush:  :blush:

 By Cauchy it can be said

$ \frac {({a^2b \plus{} b^2c \plus{} c^2a})^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {({ab^2 \plus{} bc^2 \plus{} ca^2})^2}{a^2b \plus{} b^2c \plus{} c^2a} \le \sum \frac {a^4b^2}{ab^2} \plus{} \sum \frac {a^2b^4}{a^2b}$

$ \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} a^3 \plus{} b^3 \plus{} c^3$

and $ a^3 \plus{} b^3 \plus{} c^3\ge 3abc$ by am-gm

thus

$ \frac {({a^2b \plus{} b^2c \plus{} c^2a})^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {({ab^2 \plus{} bc^2 \plus{} ca^2})^2}{a^2b \plus{} b^2c \plus{} c^2a} \le a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc$ :maybe:[/quote]

Sorry, but you used

$ a^3 \plus{} b^3 \plus{} c^3\ge 3abc$

not correctly. :wink:

ok manlio then can it be said that $ min(a^3\plus{}b^3\plus{}c^3)\equal{}3abc$??

then it is all done

No sfs. It is true that $ a^3\plus{}b^3\plus{}c^3\ge 3abc$ in all cases (if they are positive).
The problem is you are again falling into the trap:
$ A\le B\ge C\implies A\le C$ which is not true.

(In here
$ A\equal{}\frac {({a^2b \plus{} b^2c \plus{} c^2a})^2}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} \frac {({ab^2 \plus{} bc^2 \plus{} ca^2})^2}{a^2b \plus{} b^2c \plus{} c^2a}$

$ B\equal{}\sum \frac {a^4b^2}{ab^2} \plus{} \sum \frac {a^2b^4}{a^2b}\equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} a^3 \plus{} b^3 \plus{} c^3$

$ C\equal{}a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc$)

yes Akash

i hav had a talk regarding this with can hang..

i'm seeing how to get out of this ..

i understand the case but if u see my last post i have argued that $ min {a^3\plus{}b^3\plus{}c^3}\equal{}3abc$

where i feel that substitution of the minimum of B (in ur case ) should be greater than A and thus would hold the inequality.

i am lookin for a better arguement and i'll come back to this soon...maybe...

I'm sorry to tell you this sfs. But I think that this approach can't be fixed. And I don't understand what you are trying to do with $ \min (a^3\plus{}b^3\plus{}c^3)$.

[quote="can_hang2007"]
\[ f(k) \equal{} \sum a^3 \plus{} \left( k^2 \minus{} \frac {2}{k}\right) \sum a^2b \plus{} \left( \frac {1}{k^2} \minus{} 2k\right) \sum ab^2 \plus{} 3abc \ge 0
\]
We have that
\[ f'(k) \equal{} \frac {2(k^3 \plus{} 1)}{k^3} \left( k\sum a^2b \minus{} \sum ab^2 \right)
\]

\[ f'(k) \equal{} 0 \Leftrightarrow k \equal{} \frac {\sum ab^2}{\sum a^2b}.
\]
[/quote]
Dear Can_Hang,

that is just such a nice idea to take the Schur-like expression as $ f(k)$ and to look at this four-dimensional thing from an entirely different angle by examining $ f'(k)$. Cheers! :thumbup:

[quote="Vasc"] the inequalities are not equivalent, they are only "equivalent". :P[/quote]
It's always funny when semantics enter into mathematics. :P The big question now: can we say that yours,
\[ \sum \left( x \plus{} \frac {1}{y} \minus{} 1\right) \left( y \plus{} \frac {1}{z} \minus{} 1\right) \ge 3.
\]
is definitely stronger than Iran96, or is it only "stronger" than it (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1317354]here[/url]) ??  :lol:


hehe I was just able to install latexrender on a test phpbb3 forum (thanks for [url=http://www.phpbb.com/community/viewtopic.php?f=72&t=653165&start=0#p3667815]this[/url], stevem :wink: ) but I happened to discover by chance that when I input

[code][tex]\text{include}[/tex][/code]
it came back with the "[unparseable or potentially dangerous latex formula]" error. It doesn't work with any sequence of letters including "include" lol.. for example, "includ" will display properly, but "includeaaaaa," "aaaabbbincludecccddd," etc. will return the same error.

is it some kind of system variable?

$ \text{i}\text{nclude}$

That seems to be a pretty cheesy trick, but it did seem to work here.

yeah it does work here, but for some reason, not on my phpbb3/latexrender installation..

EDIT: oops sorry I didn't see your LaTeX code :oops: yeah it works on my installation that way too. hmm.. but just doing \text{include} works on AoPS, and I wondered how they fixed it (if it didn't work in the first place)

okay um I don't think it's the word "include" that's the problem..

[code][tex]inserttexthere[/tex][/code]

does NOT return:

$ inserttexthere$

it returns _nothing_.

[code][tex]'[/tex][/code]

returns "39;"

am I missing a package or something?

This might not fix everything, but you should be doing it anyway: use $ and $$ instead of [tex].

huh? I thought the dollar signs worked by default only on AoPS and not on normal installations of latexrender.. doesn't seem to work on mine, at least.. :huh:

The default on other forums may be different, but the standard in a normal $ \text{\LaTeX}$ document is dollar signs.

oh I see.. eh well apparently it's a problem with phpbb3 again: [url]http://www.shatters.net/forum/viewtopic.php?f=2&t=12397&p=103507#p103506[/url]

so it doesn't like "loop" either. :(

I guess I'll just have to include the extra brackets like in{}clude for now..

undefined117: it's not a phpBB3 problem but a deliberate security measure by LatexRender. In class.latexrender.php have a look at the lines following
[code]var $_latex_tags_blacklist = array([/code]and you will see all the words that are excluded. You can change them if you wish but they are there for a very good reason and I would leave them there unless your forum is completely private.

ohh hehe I never noticed that part. Thanks! How come it's disabled on AoPS, though? Is it because AoPS has customized code?

AoPS has a specially customised version of LatexRender (eg the use of dollar signs). You are free to make any changes you like as well on your forum but always keep security in mind.


Find (a, b) integers such that: [a[bn]]=n-1, where n is a integer.

[x]=t / t is integer, t<=x<t+1  :wink:

Look this links:

http://www.mathlinks.ro/Forum/viewtopic.php?p=516506&search_author=carlosbr&show_results=posts&sort_dir=desc&return_chars=200#516506

http://www.mathlinks.ro/Forum/viewtopic.php?t=88510


function
real analysis
real analysis theorems

Show that the sum and product of two simple functions are simple.
Show that
$ \chi_{A\cap B} \equal{} \chi_A\cdot \chi_B$
$ \chi_{A\cup B} \equal{} \chi_A \plus{} \chi_B \minus{} \chi_A\cdot \chi_B$
$ \chi_{A'} \equal{} 1 \minus{} \chi_A$.

$ \chi_{A}\chi_{B}=\{
\begin{array}{c l}
  1\cdot 0 & x\in A,x\notin B\\
  0\cdot 1 & x\notin A,x\in B\\
  0\cdot 0 & x\notin A,x\notin B\\
  1\cdot 1 & x\in A,x\in B
\end{array}

=\{
\begin{array}{c l}
  0 & elsewhere\\
  1 & x\in A\cap B
\end{array}
$

$ \chi_{A^c}+\chi_{A}=
\{
\begin{array}{c l}
  1 & x\in A^c, x\notin A \\
  1 & x\notin A^c, x\in A \\
  0 & x\notin A^c, x\notin A \\
  2 & x\in A^c, x\in A \\
\end{array}
=
\{
\begin{array}{c l}
  1 & x\in A^c \\
  1 & x\in A \\
  0 & x\in\emptyset \\
  2 & x\in\emptyset\\
\end{array}
=1$

$ \chi_{A}+\chi_{B}-\chi_{A}\chi_{B}=
\{
\begin{array}{c l}
  1 & x\in A, x\notin B \\
  1 & x\notin A, x\in B \\
  0 & x\notin A, x\notin B \\
  2 & x\in A, x\in B\\
\end{array}
-\chi_{A}\chi_{B}=
\{
\begin{array}{c l}
  1-0 & x\in A, x\notin B \\
  1-0 & x\notin A, x\in B \\
  0-0 & x\notin A, x\notin B \\
  2-1 & x\in A, x\in B\\
\end{array}=
\{
\begin{array}{c l}
  0 & x\in (A\cup B)^c \\
  1 & elsewhere \\
\end{array}$


topology
trigonometry
vector
advanced fields
advanced fields unsolved

$ M,N$ are $ C^{\infty}$ manifolds, $ f: M \minus{} > N$ is $ C^r$.

There is a well known approximation thm: $ C^{\infty}$ maps are dense in $ C^r(M,N)$ with the Whitney topology?

Now how do we conclude: since by this thm we can approximate $ f$ by $ C^{\infty}$ maps,
then $ f$ is homotopic to a $ C^{\infty}$ map $ g$?

For compact $ M$ , I found a proof using tubular nbh in Dieudonne 's "treatise of analysis" Thm 16.26.4.

Lack of compactness creates some complications here. Consider $ M \equal{} \mathbb R$ and $ N \equal{} \mathbb R^2\setminus (\mathbb Z\times \{0\})$. The map $ f(t) \equal{} (t,(t^2\plus{}1)^{ \minus{} 1}\cos \pi t)$ is a smooth embedding of $ M$ into $ N$. It's easy to find an approximating sequence $ f_n$ converging (in whatever sense) to $ f$ such that none of $ f_n$ is homotopic to $ f$.

[quote="mlok"]Lack of compactness creates some complications here. Consider $ M \equal{} \mathbb R$ and $ N \equal{} \mathbb R^2\setminus (\mathbb Z\times \{0\})$. The map $ f(t) \equal{} (t,(t^2 \plus{} 1)^{ \minus{} 1}\cos \pi t)$ is a smooth embedding of $ M$ into $ N$. It's easy to find an approximating sequence $ f_n$ converging (in whatever sense) to $ f$ such that none of $ f_n$ is homotopic to $ f$.[/quote]

It seems that Hirch "differential topology" thm 4.3.5's proof used this to prove every vector bundle is $ C^{\infty}$.

Can I see why your $ fn$ is not homotopoic to $ f$?

[quote="mlok"]I was thinking of $ f_n$ that follows $ f$ in the strip $ |x|\le n$ and then goes horizontal.[/quote]
But those $ f_n$ are not $ C^{\infty}$. Note the appromation is by $ C^{\infty}$ maps.

[quote="mlok"]Argh. Make them $ C^{\infty}$, then.[/quote]

I don't have a homotopy at hand: but it is also difficult to see why they are not homotopic to $ f$.
i.e: $ H: M \times I \minus{}>N$ with $ H(M,0)\equal{}f$ and $ H(M,1)\equal{}f_1$, how to show $ H$ can't be continuous?

(BTW, I don't see why $ N$ has to be a horizontal stripe,  $ N\equal{}R^2$ would be same thing?)

Ok, I found the proof.
For any manifold $ N$, this is true by Hirsch's book, Thm 5.1.5.


A sequence of complex numbers $ z_0,z_1,z_2,....$ is defined by the rule
\[ z_{n \plus{} 1} \equal{} \frac {i z_n}{\overline{z_n}}
\]where $ \overline{z_n}$ is the complex conjugate of $ z_n$ and $ i^2 \equal{} \minus{} 1$. Suppose that $ |z_0| \equal{} 1$ and $ z_{2005} \equal{} 1$. How many possible values are there for $ z_0$?

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 2005\qquad
\textbf{(E)}\ 2^{2005}$

I don't found an answer, but i think with 30% that answer is 2
that mean B .
i suppose so...

[quote="scorpius119"][hide]From the recursive definition, multiply top and bottom by $z_n$ to get $z_{n+1}=\frac{iz_n^2}{|z_n|^2}$. Since $|z_0|=1$ and from the recursion, $|z_n|=\left| \frac{iz_{n-1}^2}{|z_{n-1}|^2} \right|=1$ for $n>0$, $|z_n|^2=1$. Then $z_{n+1}=iz_n^2$. From $i^4=1$, $i^2=1$, use the recursion 2005 times to get $z_{2005}=-iz_0^{2^{2005}}$ or $z_0^{2^{2005}}=i$ (from $z_{2005}=1$). This equation has $2^{2005}$ solutions. [b](E)[/b][/hide][/quote]

Why is $|z_n|^2=1$?

I had shown that $|z_n|$ equals 1 using induction. The base case is given, and inductive step uses $|i|=1$ and $|xy|=|x||y|$ (as well as the division equivalent).

[hide="solution"]since $|z_0|=1, \frac{|i|*|z_0|}{|\overline{z}|}=|z_1|=1$, same for all terms of the sequence

say $a_{2004}=a+bi$, $a_{2005}=1=\frac{i(a+ib}{a-ib}=\frac{i(a^2-b^2+2iab}{a^2+b^2}$ and $a^2+b^2=1$

so $a_{2005}=1=-2iab$, and we have a system, and there will be 2 sets of $\{a,b\}$ for going $a_{2005} \rightarrow a_{2004}$, and each time, we will double our solution set, so $2^{2005} \implies \boxed{E}$[/hide]

To Altheman,
     Your solution is an elegant approach, but why would there be 2 sets of $ \{a,b\} $ possible every time you move from one complex number to another that is lower in sequence?

[quote="scorpius119"][hide] $|z_n|=\left| \frac{iz_{n-1}^2}{|z_{n-1}|^2} \right|=1$ for $n>0$[b](E)[/b][/hide][/quote]

Some one please explain how does he get those?

[quote="scorpius119"][hide]From the recursive definition, multiply top and bottom by $z_n$ to get $z_{n+1}=\frac{iz_n^2}{|z_n|^2}$. Since $|z_0|=1$ and from the recursion, $|z_n|=\left| \frac{iz_{n-1}^2}{|z_{n-1}|^2} \right|=1$ for $n>0$, $|z_n|^2=1$. Then $z_{n+1}=iz_n^2$. From $i^4=1$, $i^2=-1$, use the recursion 2005 times to get $z_{2005}=-iz_0^{2^{2005}}$ or $z_0^{2^{2005}}=i$ (from $z_{2005}=1$). This equation has $2^{2005}$ solutions. [b](E)[/b][/hide][/quote]

Also, i don't know how he got the answer from the last step,please help!

http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_12B_Problems/Problem_22

I don't understand this whole e thing. How does that have any relevance to the solution itself?

The whole e thing is just another way to write complex numbers. This article explains is very well.
http://www.math.hmc.edu/calculus/tutorials/complex/

Just start reading from the section marked "Polar Form".

In the case of this problem since the magnitude of $z_0$ is 1 you know that r is also 1.


Prove that for each positive integer n there exist n consecutive positive integers none of which is an integral power of a prime number.

Simple application of the Chinese remainder theorem: consider the first 2n primes, and then you have the simultaneous solution of x = 0 mod (2*3), x = 0 (mod 5*7), etc. which is obviously never a perfect power.

1989 IMO Problem 5

I just went to the IMO section, and randomly picked a year, and then picked what looked like the most attemptable problem.

i was wondering the same question... but thanks for asking.. :) :D


Hello

Does AoPs ship out to P.O Boxes? I have a mailox. Would I just put the mailbox # in the shipping information?

Thanks

Yes, we ship to PO Boxes.

Put "PO Box 2185" (or whatever it is) in the address line.


Let $ a$, $ b$, $ c$ be positive reals. Prove that
$ \frac {c\plus{}a}{a\plus{}b} \plus{} \frac {a\plus{}b}{b\plus{}c} \plus{} \frac {b\plus{}c}{c\plus{}a} \leq \frac bc \plus{} \frac ca \plus{} \frac ab$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=5263 problem [b](2)[/b].

  darij


geometry
rectangle
combinatorics proposed
combinatorics

Let $ n $ and $ r $ be positive integers and $ A $ be a set of lattice points in the plane such that any open disc of radius $ r $ contains a point of $ A $. Show that
for any coloring of the points of $ A $ in $ n $ colors there exists four points of the same color which are the vertices of a rectangle.

Here, a row will be a horizontal line passing through a lattice point, while a column will be a vertical line passing through a lattice point.

We can rephrase the given condition as follows: there is some $k$ s.t. for every $k$ rows and every $k$ columns, there is at least one point of $A$ among their intersections.

Partition the lattice into $k\times k$ squares, and regard each such square as a lattice point in another, coarser lattice. To each such square associate a point of $A$ among the intersections of the $k$ rows and $k$ columns, and give the square the color $(u,i),\ u\in\overline{1,k^2},\ i\in\overline{1,n}$, where $u$ represents the position of the chosen point of $A$, while $i$ represents the color that we assign to this point. Clearly, a monochromatic rectangle ("rectangle" means "rectangle with sides parallel to the axes") in this new lattice corresponds to a monochromatic rectangle in the old lattice, so all we have to prove is this:

Given a coloring with $n$ colors of the points of the lattice, we can find a monochromatic rectangle.

Consider an $(n+1)\times\left(n\cdot\dfrac{n(n+1)}2+1\right)$ rectangle made up of points in this lattice. Each column contains a pair of points with the same color, so select such a pair for each column. There must be $\dfrac{n(n+1)}2+1$ columns for which the selected pairs have the same color, and since a column with $n+1$ points has $\dfrac{n(n+1)}2$ pairs of points, there must be two columns having the same selected pair with the same color. The four points in the two mentioned pairs are the vertices of a monochromatic rectangle.


induction
arithmetic sequence
algebra proposed
algebra

Find all f:N->N s.t.

i) f is strictly increasing;
ii) f(2n)=f(n)+n;
iii) f(n) prime => n prime.

Nice one. Is there a more elementary solution than mine (thgat is one with no use of Dirichlet's theorem)?

Let f(1) = a.
Then f(2) = a + 1, and an easy induction leads to f(2 p ) = a - 1 + 2 p for each nonnegative integer p.

Since f in strictly increasing, it follows that f(n) = n + a - 1 for each positive integer n.

Let p b a prime with p \geq  a. Then p = f(n) for some n > 0 which has to be a prime number too. That is n = p - (a-1) is a prime.
Repeating the reasoning, we deduce that for each integer i such that p - i(a-1) > a, the number p - i(a-1) is a prime.

Now, let's suppose that a > 1.
From Dirichlet's theorem about prime in arithmetic progression, and since gcd(a,a-1) = 1 and a-1 > 0, it is knwon that there exist an infinite number of primes of the form a + i(a-1).
So, we may choose a prime p  \geq  a 2  such that p = a + q(a-1) for some q  \geq  a.
From above, we then have p - i(a-1) is a prime for i = 0,1,... such that p-i(a-1) > a, and then in particular for i = q-a. But p - (q-a)(a-1) = a 2  which is not a prime. A contradiction.
Thus a = 1  and f(n) = n which, conversely, is a solution of the problem.

Pierre.

well my solution also uses dirchlet's theorem(i'm confused about whether this is proved for all arithmetic progressions or the one of the form ak+1)
what's ur solution ,grobber?

Dirichlet's theorem is valid for each arithmetic progression of the form {a+ib / i = 0,1,....} where a,b > and gcd(a,b) = 1.

Pierre.

[quote="pbornsztein"]Nice one. Is there a more elementary solution than mine (that is one with no use of Dirichlet's theorem)?

Let $ f(1) \equal{} a$.
Then $ f(2) \equal{} a \plus{} 1$, and an easy induction leads to $ f(2^p) \equal{} a \minus{} 1 \plus{} 2 ^p$ for each nonnegative integer p.

Since f in strictly increasing, it follows that f(n) = n + a - 1 for each positive integer n.

Let p be a prime with $ p \geq  a$. Then $ p \equal{} f(n)$ for some $ n > 0$ which has to be a prime number too. That is $ n \equal{} p \minus{} (a\minus{}1)$ is a prime.
Repeating the reasoning, we deduce that for each integer $ i$ such that $ p \minus{} i(a\minus{}1) > a$, the number $ p \minus{} i(a\minus{}1)$ is a prime.

Now, let's suppose that $ a > 1$.
From Dirichlet's theorem about prime in arithmetic progression, and since $ gcd(a,a\minus{}1) \equal{} 1$ and $ a\minus{}1 > 0$, it is known that there exist an infinite number of primes of the form $ a \plus{} i(a\minus{}1)$.
So, we may choose a prime $ p  \geq  a^2$  such that $ p \equal{} a \plus{} q(a\minus{}1)$ for some $ q  \geq  a$.
From above, we then have $ p \minus{} i(a\minus{}1)$ is a prime for $ i \equal{} 0,1,...$ such that $ p\minus{}i(a\minus{}1) > a$, and then in particular for $ i \equal{} q\minus{}a$. But $ p \minus{} (q\minus{}a)(a\minus{}1) \equal{} a^2$  which is not a prime. A contradiction.
Thus $ a \equal{} 1$ and $ f(n) \equal{} n$which, conversely, is a solution of the problem.

Pierre.[/quote]


geometry
geometric transformation
reflection
algebra
polynomial
inequalities
analytic geometry

Since so many topics in the forum have become jobless and are of no use to the new batch , lets start the olympiad discussion early this year .
Please post anything related to olympiads here until the NSE'S and RMO are completed.
Books/Papers/Problems etc.
It would be better to form a single topic than everything spread out here and there.

:roll:
all the best...remember to do well

Guess it's to early!
No one's sending any posts yet! :(
@franky
You forgot NSEB!:D

[quote="siddharth2010"]cut off is scratchin heads all day all nights long......
....anybody predict waez going to be cutoff dis tym for nsep :oops:[/quote]
Huh? The exam hasn't even been opened for registration!!

The cutoff varies every year depending upon the difficulty level of the paper.Just give your best and don't worry about cutoffs.

I mean,it's difficult to execute it(for me atleast) :(

ya ...maths unpredictable. :lol: ..my frnd shakti prasad went all da way from rmo to inmo this yr  just by solving 2 problems fully and 2 halfhazardly..and from inmo to ocsc just by solving 3-4 problems in inmo..his luck wasnt that good in ocsc though but its becoming a sort of trend that more of combinatorics masters are having an edge over number theory masters. :!: ..its strange but its happening as my mate said me after ocsc experience.

What bothers me more is that I am very much prone to tension,anxiety and finally messing up in the exam.
  :(  :(

same with me scary :(
[hide]Similarly messed up JEE this year :( [/hide]

Never imagined u guys might be facing the same problem. :(
But ya got a decent air,right ?:)
[hide]Mine is of extreme form.
We had hist and math together,I had to stay up till late the previous night and ended up losing 30 marks in math in this term :( What can be more tragic?[/hide]

You call AIR 4932 decent? :o  :o Before the exam,I was expecting atleast TOP 1000.But messed chem as usual :mad:

Dat's not an infinite series sum.
[hide]I am worse than u in these type of sums :blush: [/hide]

Good news
[hide]at least for delhi region
hose in other regions please check for your coordinator[/hide]
Here's what amitabh tripathi has to say

Amitabha Tripathi to mathematicsoly.
show details 2:16 PM (1 hour ago) Reply

I have just been informed that [u][b]RMO 2009 is now going to be held on Sunday, 29 November[/b][/u]. This will enable all of you to take both NSE & RMO, if you so wish. I am happy that prompt action was taken by all concerned in this regard.

A. Tripathi

It was bound to happen

but seems that IAPT has an upperhand on Maths authorities :rotfl:  :rotfl:

Yup
Date has been changed over here in WB as well.
I guess the change applies for all the states. :)

Thank you guys for the information about RMO.
All the best.

could someone tell a good problem book for organic chemistry?

@ i was thinking of doing(and doing) mechanics from HC Verma for NSEA(SR)....will it be enough for clearing it(or @ least getting a good rank) considering that i wud (most probably) be able to do most of the Maths Q's of NSEA owing to my RMO preparations...and can some1 suggest a good G.K book for NSEA questions like where was aryabhatta launched from??....n such factual stuff..??

do any of u know that y r the Short Ans q's given in H.C.???one of my freineds was telling me that they were(and perhaps are) asked in the interviews!!!..k so i didntget this unique question

a car is going towards North and another is going towards East......tell which1 has higher velocity????? strange naa...i think i have some sort of reasoning..but i'mm not quite sure about the Ans

[quote]Whats the confusion there? They're obviously the same. Unless you're getting practical. And thats the first time I've heard about interview ques being asked directly from HCV....[/quote]
the fact i've ASKED a Q..means they are not same....i mean just look at HCV ...1 of the short ANS Q's...from kinematics i guess...or phy n math....this Q directly says Which 1 is more......n as far as i've read HC..if he feels there is a "both are same"option..he wud have given it so plz try to think of something,...oh n btw when 1 of my friends asked this Q to his PHy teacher in FIITJEE, that guy said th althou he thin that it is =..some guys say that it isnt..some say east is more..some say north is more..so just wanted some views...i wud like to sahre my reasoning after some responses>... :)

My apologies, they are not the same, however they ARE equal in magnitude. Unless you are defining an 'imaginary' Cartesian plane where north will become greater than east. Nothing in the manner HCV speaks changes the facts.  :wink: He's mortal for heaven's sake. He's not even Feynman or some international author with high English language proficiency. There are still a few mistakes in HCV, however rare.

[quote="sahilsharma94"][quote]My apologies, they are not the same, however they ARE equal in magnitude. Unless you are defining an 'imaginary' Cartesian plane where north will become greater than east. Nothing in the manner HCV speaks changes the facts[/quote]
should i take that to mean u think they are =, coz i have arugments that suggest the V north and V east are gr8er(separate arguments obviously)[/quote]
Assuming the same speed(Question is invalid without the logical assumption) the magnitudes of velocity MUST be equal because they depend ONLY on the instantaneous speed and not the direction. The direction however, disqualify the 'velocities' from being EQUAL
So we have
1 An argument over the validity of the question, which can only be answered under a given set of assumptions set on the basis of convention.
2. A result with a proof given the assumptions are true.

What more do I need to explain here?  :o

hey nuclear...plz try to undertsand what the Q says...my interpretation is that you know although the odometers show the same speed...the speeds ARE not equal..its like the weighing scale show = weight when we weigh 1 kg of cotton and 1 kg of iron..but thats not actually =...so i think HC measn that although the ODO shows us that the cars have = speed...they dont and we have to reson out why one has more speed than other...............

Taking a non-ideal scenario and breaking the assumption I stated just because its an HCV question isn't fair in my opinion. But for the purpose of questioning, your line of practical thinking is fine. It will have to be something to do with the trade winds/westerlies etc. and relative distribution of populations in the wind deflection directions.


[color=blue][b]Solve this equation:
 $ 6^{x \minus{} 1} \equal{} 6log(6x \minus{} 5) \plus{} 1$[/b][/color]

hello, defining $ f(x)=6^{x-1}-6\ln(6x-5)-1$ we get $ \lim_{x \to \pm\infty}f(x)=+\infty$ further we get $ f^'(x)=6^{x-1}\ln(6)-\frac{36}{6x-5}$ and $ f^'(x)=0 \Leftrightarrow x^{\star}=\frac{1}{6}\frac{5\ln(6)+6\text{LambertW}(6\cdot6^{\frac{1}{6}})}{\ln(6)}$ with $ f(x^{\star})<0$, so we have one local minimum and only two solutions of our equation $ x_1=1$ or $ x_2 \approx 2.505186494$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, defining $ f(x) = 6^{x - 1} - 6\ln(6x - 5) - 1$ we get $ \lim_{x \to \pm\infty}f(x) = + \infty$ further we get $ f^'(x) = 6^{x - 1}\ln(6) - \frac {36}{6x - 5}$ and $ f^'(x) = 0 \Leftrightarrow x^{\star} = \frac {1}{6}\frac {5\ln(6) + 6\text{LambertW}(6\cdot6^{\frac {1}{6}})}{\ln(6)}$ with $ f(x^{\star}) < 0$, so we have one local minimum and only two solutions of our equation $ x_1 = 1$ or $ x_2 \approx 2.505186494$.
Sonnhard.[/quote]
[color=blue][b]May u explain more clealy? I am still understand why you use "Lanbert W" here .for?[/b][/color]


For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$?


algebra
polynomial
superior algebra
superior algebra unsolved

let f(x) be an irredueible polynomial over F of degree n,let K be a field extention of F with [K:F]=m. if gcd(n,m)=1,show that f is irredueible over K. :oops:

Let $f=gh$ in $F$. If $\alpha$ is zero of $g$ then we have $[F[\alpha]: F]=\deg g=k$, $[K[\alpha]: K]=n$ nad $[F: K]=m$. Therefore $[F[\alpha]: K[\alpha]]=\frac{mk}{n}$ and since $(m,n)=1$ we have $n\mid k$, which is a contradiction.


When I search something, I get an error like this:

General Error

SQL requests not achieved

DEBUG MODE

SQL Error: 1062 Duplicate entry '1228898744' for key 1

SQL Request: UPDATE phpbb_search_results SET search_id = 1228898744, search_time = 1263182430, search_array = 'a:6:{s:2:"hl";s:8:"toss egg";s:2:"by";s:0:"";s:3:"dir";s:4:"DESC";s:4:"show";s:6:"topics";s:5:"chars";i:200;s:3:"ids";s:34:"323832,315421,308002,305058,230035";}' WHERE session_id = '57d51ff8635c439da1bc8be4c2c4b879'

Line : 1331
File : class_search.php

What happened?

Actually it works now. I tried it ~5 times yesterday and it never worked, but it works now.


function
integration
calculus
inequalities
analytic geometry
graphing lines
slope

In this problem, I'd like to state a result about the existence of equilibria in certain types of auctions, and then we can try to figure out what's going on in the general case. It would also be nice to have an economic interpretation for the result.

In the auctions we consider, there'll be $ n$ bidders competing for a single item. Each bidder will rate the item to have a value $ v_i$; the $ v_i$ are independent, identically distributed random variables on [0, 1] according to a CDF $ F(t)$. The bidders don't know each other's valuations; they only know their own.

In the auction, each bidder submits a bid $ b_i$ without knowing anyone else's bid, and the high bid wins. The bidder then pays $ L(y_1, \ldots, y_n)$, a linear function of the bids in decreasing order (note: the bids are $ b_i$ when unordered and $ y_i$ when in decreasing order). In particular, the standard Vickrey auction is $ L(\textbf{y}) \equal{} y_2$ and the standard first price auction (the sealed-price version of English auction) is $ L(\textbf{y}) \equal{} y_1$.

The question is this: \textit{what conditions on $ L$ are necessary to ensure the existence of an equilibrium for all continuous $ F$?}

It's very easy to see that if $ L \equal{} y_2$, the second-price auction, $ b(v) \equal{} v$ is the equilibrium bidding function no matter what is $ F$. Also, if $ L \equal{} y_1$, there will always be an equilibrium functions. However, if $ L \equal{} y_3$, my prof says that he can construct a function $ F$ for which there is no equilibrium, and I have constructed a function $ F$ for which there is no equilibrium when $ L \equal{} \frac{1}{n\minus{}1} \sum_{k\equal{}2}^n y_k$.

Now, let me briefly review the necessary steps for finding equilibria in this kind of situation:

(1) Show that if an equilibrium bidding function $ b(v)$ exists, it must be increasing in $ v$.

(2) Assuming that the function exists, find the function by maximizing the expected surplus of a bidder assuming that everyone else is bidding according to the equilibrium function already. (This amounts to solving a linear first order differential equation, so you always get a function.)

(3) Check that the resulting function is actually increasing. If it is, we're done. If it isn't, there can be no equilibrium. (The above method is guaranteed to find the function if it exists.)

So, leaving (1) as an exercise, the first major step to approach the general problem would be to find the purported equilibrium function for general $ F$ and $ L$. I am fairly sure this is possible without unreasonable amounts of effort. Then we need to figure out why sometimes there can be $ F$ for which the function is not increasing and sometimes no $ F$ can cause any harm.

To help get this started, I will (soon) post the derivation of the bidding function and the proof of non-increasingness for the $ L$ I mentioned above (the "average-of-others" auction).

Proof of no equilibrium in case $ L(y_1, \ldots, y_n) \equal{} \frac{1}{n\minus{}1} \sum_{k\equal{}2}^n y_k$.

Assume that there is an equilibrium bidding function $ b(v)$. Clearly, $ b(0) \equal{} 0$ and $ b(1) \equal{} 1$. Suppose that bidder #1 bids [i]as if her value was $ z$[/i], that is, she bids $ b(z)$. This is a valid assumption, as the range of $ b$ must be [0, 1]. Suppose further that all other bidders bid at equilibrium.

The expected surplus of bidder 1 is thus

$ ES(z) \equal{} \mathbb{E}[S(z)] \equal{} \mathbb{E}\left[v_1 \minus{} \frac{1}{n\minus{}1}\sum_{k\equal{}2}^n b(v_k) | z > v_k, k \equal{} 2, 3, \ldots, n\right].$

Note that $ L$ doesn't distinguish between the order of the bids, so we may consider them as unordered. Additionally, as the $ v_k$ are all independent, we may split up the expectation into

$ ES(z) \equal{} v_1 F(z)^{n\minus{}1} \minus{} F(z)^{n\minus{}2}\int_0^z b(t) dF(t)$

If $ b$ is the equilibrium bidding function, then $ ES'(z)$ should be identically 0 (as a function of $ v_1$) when $ z \equal{} v_1$. This will give us an equation involving just $ v_1$ and $ b(v_1)$, which we can then use to find $ b$. We compute:

$ ES'(z) \equal{} (n\minus{}1)f(z)F(z)^{n\minus{}2} v_1 \minus{} (n\minus{}2)f(z)F(z)^{n\minus{}2} \int_0^z b(t) dF(t) \minus{} f(z)F(z)^{n\minus{}2} b(z).$

Setting $ v \equal{} v_1 \equal{} z$ and making that equal to 0, we have

$ (n\minus{}1)vf(v)F(v)^{n\minus{}2} \minus{} (n\minus{}2)f(v)F(v)^{n\minus{}2} \int_0^v b(t) dF(t) \minus{} f(v)F(v)^{n\minus{}2} b(v) \equal{} 0$

Setting $ H(v) \equal{} \int_0^v b(t) dF(t)$ turns this into a differential equation:

$ (n\minus{}1)vF(v)^{n\minus{}2}f(v) \minus{} (n\minus{}2)H(v)F(v)^{n\minus{}3}f(v) \minus{} H'(v)F(v)^{n\minus{}2} \equal{} 0$
$ (H(v)F(v)^{n\minus{}2})' \equal{} (n\minus{}1)vF(v)^{n\minus{}2}f(v)$
$ H(v)F(v)^{n\minus{}2} \equal{} C \plus{} \int_0^v (n\minus{}1) tF(t)^{n\minus{}2}f(t)dt$

Obviously $ C$ must be 0. (Look at $ v\equal{}0$.) Integration by parts of the LHS gives

$ H(v)F(v)^{n\minus{}2} \equal{} vF(v)^{n\minus{}1} \minus{} \int_0^v F(t)^{n\minus{}1} dt$
$ H(v) \equal{} vF(v) \minus{} \frac{1}{F(v)^{n\minus{}2}}\int_0^v F(t)^{n\minus{}1} dt$

We know $ H'(v) \equal{} f(v)b(v)$, so differentiating gives

$ b(v) \equal{} v \plus{} \frac{n\minus{}2}{F(v)^{n\minus{}1}} \int_0^v F(t)^{n\minus{}1} dt$

This, we hope, is the equilibrium bidding function. So far, in our probabilistic argument, we have used the fact that $ b$ is increasing and continuous. (I also assumed that $ F$ was continuous and differentiable, but that's not the thing at issue here.) What we need to do is show that there is a continuous, differentiable $ F$ for which $ b$ is not increasing. This violates our assumption about the existence of the equilibrium solution (for that $ F$).

So, if $ b$ were increasing, then $ b'$ would be positive. We compute:

$ b'(v) \equal{} 1 \plus{} \frac{(n\minus{}1)(n\minus{}2)}{F(v)^n}f(v)\int_0^v F(t)^{n\minus{}1} dt \plus{} (n\minus{}2)$

So we'd like to show that there exists $ F$ for which

$ n\minus{}1 \minus{} \frac{(n\minus{}1)(n\minus{}2)}{F(v)^n} f(v) \int_0^v F(t)^{n\minus{}1}dt < 0$,

or

$ (n\minus{}2)f(v)\int_0^v F(t)^{n\minus{}1} dt > F(v)^n$.

However, $ f(v)$ can be any positive function that integrates to 1. In particular, it can be unbounded. So of course we will be able to satisfy that inequality, as long as $ n > 2$. This shows that $ b$ need not be increasing.

An example is to let $ n\equal{}3$ and $ F(v) \equal{} \frac{1}{2}(1 \minus{} \sqrt{1 \minus{} 2v})$ for $ v < 1/2$ and whatever we want afterwards. You can compute $ b$ (I advise that you don't try it by hand) and show that $ b$ has a local max at $ v \equal{} \frac{40}{81}$.

Conjecture:
(1) The "only way to screw up the bidding function" is to have a PDF that gets very large somewhere.
(2) If $ n > 2$, any auction in which $ L$ has nonzero coefficients on $ y_k$ for any $ k > 2$ will not always have an equilibrium bidding function.
(3) If $ L(y_1, \ldots, y_n) \equal{} \lambda_1y_1 \plus{} \lambda_2y_2$, there is always an equilibrium.

I am vaguely confident of 1, pretty confident of 2, and not very confident of 3, but I thought I should at least state some things to aim for. Also, these results all seem highly counter-intuitive in terms of the economic interpretation. Why should a steep slope in the CDF of bidders' valuations make equilibrium impossible? Why should it matter if the third highest bidder (or the other bidders) affect the price?

Anyway, I'm working on this stuff right now, sort of. If anyone else is interested, I'd be delighted to talk about it. I don't know how much auction theorists care about this, so I don't know if it'd be publishable if we got a full characterization of when there's equilibrium in this class of auctions. I also don't actually know if it's still open (although my prof didn't already know, so it might be). Some game theory or econ journal might take it.

[quote="Xevarion"](3) If $ L(y_1, \ldots, y_n) \equal{} \lambda_1y_1 \plus{} \lambda_2y_2$, there is always an equilibrium. [/quote]
First, I should point out that in order for these auctions to "make sense" as auctions, we will need some requirement on $ L$. For now, let's assume that $ \lambda_i \ge 0$ for all $ i$; it's possible that we could relax this condition, but let's worry about that later.

Anyway, I worked out the solution for the bidding function in the case where the amount paid depends only on the top two bids. In this case, the expected surplus (profit) of bidder 1, assuming she bids as if her value was $ z$, is

$ S(z) \equal{} \mathbb{E}[v_1 \minus{} \lambda_1 b(z) \minus{} \lambda_2 b(y_2) | z > y_2]$
$ S(z) \equal{} F(z)^{n\minus{}1}(v_1 \minus{} \lambda_1b(z)) \minus{} \lambda_2 \int_0^z b(t)(n\minus{}1)f(t)F(t)^{n\minus{}2}dt$

So, as before, we take the derivative and set $ z \equal{} v_1 \equal{} v$:

$ S'(z) \equal{} (n\minus{}1)f(z)F(z)^{n\minus{}2}(v_1 \minus{} \lambda_1 b(z)) \minus{} \lambda_1 F(z)^{n\minus{}1} b'(z) \minus{} \lambda_2 b(z) (n\minus{}1) f(z)F(z)^{n\minus{}2}$

So we want to find $ b$ to satisfy

$ (n\minus{}1)f(v)F(v)^{n\minus{}2}(v \minus{} \lambda_1 b(v)) \minus{} \lambda_1 F(v)^{n\minus{}1} b'(v) \minus{} \lambda_2 b(v) (n\minus{}1) f(v)F(v)^{n\minus{}2} \equal{} 0$

or

$ \lambda_1 F(v)^{n\minus{}1} b'(v) \plus{} (n\minus{}1)(\lambda_1 \plus{} \lambda_2)f(v)F(v)^{n\minus{}2}b(v) \equal{} (n\minus{}1)vf(v)F(v)^{n\minus{}2}$

This is still a pretty approachable, standard linear first order ODE. Using the method of integrating factor, or by observation if we're really clever, we see that we need to multiply both sides by $ F(v)^{(n\minus{}1)\lambda_2/\lambda_1}$, and integration gives

$ F(v)^{(n\minus{}1)(1 \plus{} \lambda_2/\lambda_1)}b(v) \equal{}  C \plus{} \frac{n\minus{}1}{\lambda_1}\int_0^v t f(t) F(t)^{(n\minus{}1)(1\plus{}\lambda_2/\lambda_1) \minus{} 1}dt$

Clearly $ C \equal{} 0$. (Consider $ v\equal{}0$.) Integration by parts on the RHS allows us to simplify to

$ b(v) \equal{} \frac{1}{\lambda_1 \plus{} \lambda_2}\left( v \minus{} \frac{1}{F(v)^{(n\minus{}1)(1 \plus{} \lambda_2/\lambda_1)}} \int_0^v F(t)^{(n\minus{}1)(1 \plus{} \lambda_2/\lambda_1)}dt\right)$

In the process of doing this, we divided by $ \lambda_1$, so we assumed it wasn't 0. But you can easily go back and check that if $ \lambda_1 \equal{} 0$, you get what you should from the standard Vickrey argument. Also, if you set $ \lambda_2 \equal{} 0$ here, you get the known result for the first-price auction. So it's believable. Also, this function is obviously increasing. I'll leave that as an exercise for the reader.

So the conjecture I quoted above is proven! Yay!

Also, we can see that this particular auction will make sense as long as $ \lambda_1 \plus{} \lambda_2 > 0$. However, unless $ \lambda_2/\lambda_1 > \minus{}1$, things are going crazy over on the right. As long as we impose those two conditions, we will have a good equilibrium. It's not clear to me how these conditions will generalize, though. ($ \sum \lambda_i > 0$ wouldn't be a surprise, but what about the other one!?)

Anyone made any progress with the general case yet? The [url=http://en.wikipedia.org/wiki/Order_statistics#Distribution_of_each_order_statistic_of_an_absolutely_continuous_distribution]pdfs of order statistics[/url] might help.


limit
real analysis
real analysis unsolved

Let sequence $ \{x_n\}$ of real numbers is such, that for every polynom $ P(x)$ of power $ 2$ with non-negative integer coefficients we have:
$ \lim_{n\to\infty}{(x_{n}\plus{}x_{P(n)})}\equal{}0$
Does this mean that $ \lim_{n\to\infty}{x_{n}}\equal{}0\;$?


Hi,
I just recently re-installed MikteX/WinEdt after formatting my hard drive. I can compile documents using just the Latex command, but the Texify command doesn't work - I get an error saying "latex.exe not found". I can convert the DVI file using dvi2pdf and then view it, but with PDFTexify, I get an error saying "pdflatex.exe" not found. I checked the paths and they seem to be correct. Also, I cannot run any macros - if I record and save a macro, I get the message saying that the macro couldn't be found when I try to run it. I tried reinstalling both MikteX and WinEdt but I have the same problem. I googled the problem but didn't find anything substantial. Can anyone tell me what the problem could be? Any and all help is appreciated.
SSN.

Hi,
I am not sure what changed since the installation, but I do not have the problem anymore; which, come to think of it, is even more frustrating.
SSN.


Determine all continuous functions $f: \mathbb R \to \mathbb R$ such that
\[ f(x + f(y)) = f(x) +y^n\]
for all $x,y \in \mathbb R$ and all positive integers $n$.

Are you sure there are solutions for any $ n$?  For even $ n$, for example, unless I've made some mistake we seem to get a quick contradiction:

Taking $ x\equal{}0$ implies that $ f$ is unbounded on $ \mathbb{R}$.  Taking $ y\equal{}0$ implies either $ f$ is periodic or $ f(0)\equal{}0$, but a continuous periodic function cannot be unbounded, so $ f(0)\equal{}0$ and $ f(f(y))\equal{}y^n$.  Take $ x\equal{}\minus{}f(y)$, then $ f(\minus{}f(y))\equal{}\minus{}y^n\equal{}\minus{}f(f(y))$, so $ f$ is surjective and $ \minus{}f(x)\equal{}f(\minus{}x)$ for all $ x\in \mathbb{R}$.  But if $ n$ is even, then $ f(\minus{}x\minus{}f(y))\equal{}f(\minus{}x)\plus{}(\minus{}y)^n\equal{}\minus{}f(x)\plus{}|y|^n\neq \minus{}f(x)\minus{}|y|^n\equal{}\minus{}f(x\plus{}f(y))$, contradiction if $ y\neq 0$.

[quote="Victory.US"]Determine all function $ f$ defined and continuous on $ R$ such that
$ f(x \plus{} f(y)) \equal{} f(x) \plus{} y^n$, $ \forall x,y \in R; n \in N^*$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} f(y)) \equal{} f(x) \plus{} y^n$

As Joeblow said : If $ f(0)\ne 0$, $ f(x)$ is continuous, unbounded (see $ \lim_{y\to\plus{}\infty}f(x\plus{}f(y))$) and periodic (see $ P(x,0)$), which is impossible. So $ f(0)\equal{}0$

Then $ P(0,x)$ $ \implies$ $ f(f(x))\equal{}x^n$ and so $ f(x^n)\equal{}f(x)^n$
$ P(x^n,f(y))$ $ \implies$ $ f(x^n\plus{}y^n)\equal{}f(x^n)\plus{}f(y^n)$ and so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y\ge 0$

This last equation implies immediately $ f(x)\equal{}f(1)x$ $ \forall$ non-negative rational $ x$ and continuity gives $ f(x)\equal{}ax$ $ \forall x\ge 0$

Using then $ y\ge 0$ and $ x\ge\max(0,\minus{}ay)$, $ P(x,y)$ implies $ a(x\plus{}ay)\equal{}ax\plus{}y^n$ and so $ a^2\equal{}1$ and $ n\equal{}1$ and it is easy to see that $ x$ and $ \minus{}x$ are indeed solutions.

So, there exist solutions only for $ n\equal{}1$ and then the only two solutions are $ f(x)\equal{}x$ $ \forall x$ and $ f(x)\equal{}\minus{}x$ $ \forall x$


invariant
algebra
polynomial
Galois Theory

[hide]our system is equivalent to
\[ \left\{ {\begin{array}{*{20}c} {\frac {1}{x^3} \plus{} y^3 \equal{} 19 } \\
{\frac {y^3}{x^3} \plus{} \frac {y^2}{x^2} \equal{} \minus{} 1 } \\
\end{array}} \right.
\]
now , denote that $ \frac {1}{x} \equal{} a ; y \equal{} b$ we obtain the equivalental system :
\[ \left\{ {\begin{array}{*{20}c} {a^3 \plus{} b^3 \equal{} 19 } \\
{(ab)^3 \plus{} (ab)^2 \equal{} \minus{} 1 } \\
\end{array}} \right.
\]
in other words
\[ \left\{ {\begin{array}{*{20}c} {a^3 \plus{} b^3 \equal{} 19 } \\
{(ab)^3 \plus{} (ab)^2 \equal{} \minus{} 1 } \\
\end{array}} \right.
\]
denote S=a+b and P=ab
hmm.. then I use Cardano's formula to calculate P , then use Viete when we obtain S and P [/hide]

Cardano's is kind of overkill - adding (I)+3(II) gives $ (a\plus{}b)^3 \equal{} 16$ or $ a\plus{}b \equal{} 2\sqrt[3]{2}$. Then, you get $ ab$ by dividing through $ \plus{}b$ in (II).

We find $ p(y)=0$ where degree(p)=18; with $ z=y^3$ we obtain $ q(z)=-1-1441z^2-57z+1079z^4-6707z^3+z^6-57z^5=0$. $ q$ is irreducible over $ \mathbb{Q}$ and the $ q$-Galois group: $ 2.S_4$ has $ 48=2^4.3$ elements and is solvable; then the equations $ q(z)=0$ and $ p(y)=0$ are solvable by the radicals $ \sqrt{}$ and $ ()^{1/3}$.
Maple gives the exact solutions; there exist 2 real solutions for $ z$: ${ 19+\alpha,-\alpha}$ with $ \alpha\approx{0.164}$. In fact $ q$ is invariant  if we change $ z$ into $ 19-z$. Thus solve  $ q$ has the complexity of "solve a polynomial of degree 3".


[color=darkred]For a triangle $ABC$ and the positive given numbers $m,n$ consider the convex quadrilateral $XYZT$

so that $\{\begin{array}{c}XY=ma\ ,\ ZY=mb\\\ XT=nc\ ,\ ZT=nb\\\ \widehat{YXT}\equiv\widehat{ABC}\end{array}$. Ascertain the length of the diagonal $XZ$ depending on $\{a,b,c\}$.[/color]

well I have computed $XZ$ but its not a simple answer and also I have used $m,n$ in my answer,it seems a stupid solution...  :maybe:
[hide]first assume that $TY=r$,now with $\cos$ law in $\triangle XTY$ we can compute $r$ as follow:

$r^{2}=m^{2}a^{2}+n^{2}c^{2}-2mnca.\cos B$ and also we know that $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ab}$

so we get that:

$r^{2}=m^{2}a^{2}+n^{2}c^{2}-\frac{mnc.(a^{2}+c^{2}-b^{2})}{b}$ so we have computed $r$ in terms of $a,b,c$...

now again by $\cos$ law in $\triangle XYT$ we get that $n^{2}c^{2}=m^{2}a^{2}+r^{2}-2mar.\cos y_{1}$ where $y_{1}=\angle XYT$

so we get that $\cos y_{1}=\frac{m^{2}a^{2}+r^{2}-n^{2}c^{2}}{2mar}$ so we have computed $\cos y_{1}$ in terms
of $a,b,c$ so we can compute $\sin y_{1}$ in terms of $a,b,c$ acording to $\sin^{2}y_{1}+\cos^{2}y_{1}=1$ so assume
that $\cos y_{1}=\alpha_{1},\sin y_{1}=\beta_{1}$

now with $\cos$ law in $\triangle YZT$ we get that $n^{2}b^{2}=r^{2}+m^{2}b^{2}-2rmb.\cos y_{2}$ where $y_{2}=\angle TYZ$ so we can compute $\cos y_{2}$ in terms of $a,b,c$ as foloow:

$\cos y_{2}=\frac{r^{2}+m^{2}b^{2}-n^{2}b^{2}}{2rmb}$ and also we write $\sin y_{2}$ in terms of $a,b,c$ now
assume that $\cos y_{2}=\alpha_{2},\sin y_{2}=\beta_{2}$ now we know that $\cos XYZ=\cos (y_{1}+y_{2})=\alpha_{1}\alpha_{2}-\beta_{1}\beta_{2}$ so we can write $\cos XYZ$ in terms of $a,b,c$ to,now finally with a $\cos$ law in $\triangle XYZ$ we get that:

$XZ^{2}=m^{2}b^{2}+m^{2}a^{2}-2m^{2}ab\cos XYZ=m^{2}(a^{2}+b^{2}-2ab.\cos XYZ)$ so we get that:

$XZ=m\sqrt{a^{2}+b^{2}-2ab.\cos XYZ}$ which is in terms of $a,b,c$

but I`m not in the mood to compute all of the huge equations and $\cos$laws above (probably I cant),so if any one could compute it,it would be great... :) [/hide]


Let $(a_1,a_2,...,a_n)$ be a permutation of $(1,2,...,n)$ for odd $n$. We call $n$ "good" if there is a permutation which satisfies the following:

$a_1-a_2+a_3-...-a_{n-1}+a_n>0$
$a_2-a_3+a_4-...-a_n+a_1>0$
$\vdots$
$a_n-a_1+a_2-...-a_{n-2}+a_{n-1}>0$

Find all "good" $n$'s.

The good $n$'s are those of the form $4k+1$, I think.

We may assume WLOG that the numbers are arranged on a circle in the order $a_1,\ldots,a_n$ going in a clockwise direction, and that $a_1=1$. The sum of the first two expressions listed there is $2a_1=2$, so there must be at least one such expression (at least two, actually), which is equal to $1$. However, if $n=4k+3$, those expressions are positive and even, so none of them can be $1$, giving a contradiction.

For $n=4k+1$ I haven't checked it properly, but it looks like arranging the numbers $2,4,6,\ldots,4k,4k+1,4k-1,\ldots,5,3,1$ in a clockwise direction on the circle works.

[quote="pbornsztein"]It comes from ISL 1991. ;)[/quote]
Now I see that it really is from ISL 1991, but in the moment of posting I didn't know that, my teacher showed it to me and I posted it here.

My solution is almast the same as grobber's one, the only differnce is that in the second case I arranged the numbers $1,\ 3,\ 5,\ ...,\ 4k+1,\ 2,\ 4,\ 6,\ ...,\ 4k$.


i want some elegant soln to this please......

$x-y=2$
$x^5-y^5=992$