Does anyone know what scores the winners or top 10-12 got in previous years? After the format was changed.

Just curious to see if the score has fluctuated a lot or if earlier/later years seemed to have higher/lower scores.

[quote="frost13"]They stopped giving us score information the last couple years, but here is what I have from before that for 1st - 12th written:

1999  45 - 36
2000  42 - 35
2001  42 - 35
2002  43 - 38
2003  45 - 40[/quote]
I thought they gave the bar graphs this year...

Thanks frost,

does anyone have anything like around 1993-1994?

[quote="jhredsox"]it was 42 (neal) to 36 (i believe, but it might of been a tiebreaker for a few 35's)

wasn't there a perfect in '04

my coach knows almost nothing about mc (well..no one really does), but he kept on obsessing about it :D

-jorian[/quote]

No, Arjun had a 36...I had the highest score of 35 and got 14th Place...

[quote="perfect628"]How many times have you complained about that?[/quote]
Dude, he was correcting jhredsox with uncontradictable proof.

Ah, but consider the ...
It seems to me that he combined that with a complaint. Of course, it is rather tough to sense tone with typing...

Umm...I wasn't complaining...I was simplying correcting someone...

I always have ellipses at the end of my sentence... well almost always.

[quote="bpms"]
2005: 30 (I think)-39[/quote]

No, Noah Arbesfeld was 14th with a 31, the cutoff was 31.


calculus
integration
logarithms
calculus computations

Can anyone demostrate

$\int \frac{x^{2}dx}{\sqrt{ax^{6}+bx+c}}\, = \, \frac{1}{3 \sqrt{a}}\ln \left| x^{3}\sqrt{a}+\sqrt{ax^{6}+bx+c}\right|+C$

directly, without deriving. Thank you all.

[quote="jmerry"]It's false. The integral you propose almost certainly has no elementary antiderivative in general.

If that were $bx^{3}$ instead, it's a routine exercise. The answer you gave is still wrong (so differentating it won't work), but the integral is doable.[/quote]

I coudn´t solve it, then I went to Mathmatica 5.2 and it offerd this result

[url=http://imageshack.us][img]http://img305.imageshack.us/img305/5702/intpq8.png[/img][/url]

[quote="jmerry"]Mathematica is interpreting "$bx$" as a single object, not $b\cdot x$. Since we're not integrating with respect to $bx$, it is treated as a constant, just like $c$.[/quote]

Yes, tks


$ 2^5 \times 3^2 \times 5^3$ [i]is[/i] the correct prime factorization. Multiply it out on a calculator.

[quote="JokoMoko"]One of the problems in Volume 1: the Basics states that the prime factorization of 36000 is 2^5*3^2*5^3, but isn't it 2^4*3^2*5^2?

Because of this, I got a different answer than the book! Ugh! Please try and solve this math problem:
3 radical 36000/243 (the 3 in the beginning is an exponent)

 :yoda: [/b][/quote]

Do you mean symplify the radical?

Yes i put it into my calculator and the book is correct about the prime factorization so the problem:

${ \frac{\sqrt[3]{36000}}{243}}$

${ \frac{\sqrt[3]{2^5*3^2*5^3}}{243}}$

${ \frac{2\sqrt[3]{2^2*3^2*5^3}}{243}}$

${ \frac{2*5\sqrt[3]{2^2*3^2}}{243}}$

${ \frac{10\sqrt[3]{36}}{243}}$

I don't think that you can simplify it any more...

[quote="JokoMoko"]One of the problems in Volume 1: the Basics states that the prime factorization of 36000 is 2^5*3^2*5^3, but isn't it 2^4*3^2*5^2?
[/quote]
How can there be three 0s on the end if it has only two factors of 5?

He already figured out his mistake, if you look at the previous posts.


function
inequalities
number theory
relatively prime
number theory unsolved

Q_2, $f(m)=f(n)\Longrightarrow 2m=m+m=f(f(m)+f(m))=f(f(n)+f(n))=2n\Longrightarrow m=n$, so $f$ is injective.

$f(f(n)+f(n+2))=n+(n+2)=(n+1)+(n+1)=f(2f(n+1))\forall n\in\mathbb{N}\Longrightarrow f(n+2)-f(n+1)=f(n+1)-f(n)$. Now use telescopic sum!

I think that Q1 doesn't have any sense, if $m>n$ and $m^{n}=n^{m}$
then $f(m)=n<m=f(n)$, and it's the same for $m<n$.

Q3-> This is the ugliest solution that I have for now.

If we know that $f(n)=n$ and $f(n+k)=n+k$ then it stands that
$f(m)=m$ for all $n<m<n+k$, because this function is strictly
increasing. We will use Bertnard's theorem which says that for
every natural number $n$ there is a prime between $n$ and $2n$.

We know that $f(7)=7$ that's why $f(1)=1,...,f(6)=6$. Now take
$f(7)$. $f(14)=f(7)f(2)=14$, so $f(n)=n$ for all $7<n<14$. There
is some prime number between $7$ and $14$, and let's say that is
$p$. Then $f(2p)=f(p)f(2)=2p$, and $2p>14$ so $f(n)=n$ for $n<2p$.
Now there is some prime between $p$ and $2p$ let's say $q$, so
$f(2q)=f(2)f(q)=2q$ so $f(n)=n: n\leq 2q$, and so on...

[quote="SpongeBob"]I think that Q1 doesn't have any sense, if $m>n$ and $m^{n}=n^{m}$
then $f(m)=n<m=f(n)$, and it's the same for $m<n$.
[/quote]

But $f(m)=n$ OR $f(n)=m$ NOT both!

[quote="Goblin"]
This is easy, but is it $f(n)=n^{2}$ for all $n$?[/quote]
I know is it true! But now i can not remember prove.

[quote="antinoise007"]Q1>
f:N→N
f is strictly increasing
f(mn)=f(m)f(n)
If m≠n & m^n=n^m, then f(m)=n or f(n)=m

f(12)?

Q2>
f:N→N
f(f(m)+f(n))=m+n

show that f(n)=n

Q3>
f:N→N
f is strictly increasing
f(7)=7
If p,q are relatively prime,then f(pq)=f(p)f(q)

show that f(n)=n[/quote]
I think you really should read this->[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135914[/url]

$f(n)$ is $n^{2}$ for all $n$. This is also easy. Choose $r$ such that for $l,n$ we have $2^{r}<n^{l}<2^{r+1}$. If $f(n)<n^{2}$, i.e. $f(n)=n^{2}-k$, then $f(2^{r})<f(n^{l})<f(2^{r+1})$, and therefore \[4^{r}<(n^{2}-k)^{l}<4^{r+1}.\] From previous inequality we have $\frac{n^{2l}}{4}<4^{r}$ and therefore $\left(\frac{n^{2}}{n^{2}-k}\right)^{l}<4,$ for all $l$. This  is clearly impossible since $\frac{n^{2}}{n^{2}-k}>1$. Simularly we can get a contradiction if $f(n)>n^{2}$.

Like I said, I don't like that proof (for Q3), so while I was in school today, I thought of something else.

First paragraph still remains the same: if $f(n)=n$ and $f(n+k)=n+k$ then $f(m)=m$ for all $n\leq m \leq n+k$. And this is the main fact we use.
If we have that $f(n)=n$ and $f(n-1)=n-1$ then $f(n(n-1))=f(n)f(n-1)=n(n-1)$, so for every $m: n\leq m\leq n^{2}-n$ it stands that $f(m)=m$. Then take $f((n^{2}-n)(n^{2}-n-1))=f(n^{2}-n)f(n^{2}-n-1)=(n^{2}-n)(n^{2}-n-1)$ and so on, you can always find some number $x$ which is arbitarly big, and $f(n)=n$ for all $n\leq x$.

Bye


[quote="abacadaea"]is there a way to acess the game from the AoPS website instead of this link?[/quote]

Not currently, except for the link in the Game Information topic.

You should make one It would solve alot of trouble

[quote="minicon"]Could you put it in the "My Classes" Page just like Alcumus?[/quote]

We'll have a public link relatively soon.

I just get redirected to http://www.artofproblemsolving.com/Alcumus/invaliduser.php  :D

[quote="minicon"]Oh, did I spill the beans again?
 :blush:[/quote]
Wasn't the description already in the Focus article about AoPS?


algebra
polynomial
function
real analysis
real analysis unsolved

Let $P \in \mathbb{R}[X]$-fixed polynomial of odd degree, $degP$-odd. And $f: \mathbb{R} \to \mathbb{R}$, $f \in C^{\infty}$ such that $|f^{n}(x)| \leq |P(x)|$ for all $n \in \mathbb{N}$ and $x \in \mathbb{R}$. Prove that $f \equiv 0$.

Let $a$ be a zero of $p$. Then $f^{(n)}(a)=0$ for all $n$. Let $f_n(x)=\sum_{j=0}^n\frac{f^{(j)}(a)}{j!}(x-a)^j=0$.

For any $r$, there exists $M$ such that $|p(x)|\le M$ for all $x$ such that $|x-a|\le r$. Now by Taylor's theorem, $|f(x)-f_n(x)|\le \frac{M}{(n+1)!}x^{n+1}\to0$ whenever $|x-a|\le r$, and therefore $|f(x)|=0$ in this range. Since $r$ was arbitrary, $f$ is identically zero.

$P$ can be any continuous function with a zero. If $P$ has no zero, we merely get that the power series for $f$ converges to $f$ on the whole real line.

[quote="jmerry"]
$P$ can be any continuous function with a zero. [/quote]
Even continuity is not needed; it suffices that $P$ be finite at every point. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28646[/url]


Prove that for n positive integer, n fixed

\[x^{2n}+y^{2n}+z^{2n}=x^{2n+1}+y^{2n+1}+z^{2n+1}\]

has an infinity of solution for x, y, z integer...

I swear that this problem has a solution, my teacher haas a book with the solution, but he won't help me!!!!

After many tries (and fails  :( ) :

Trying with x=-z < 0 and y = -k*x gives the solutions :

x = -k*(2*k^(2n)+1), y = 2*k^(2n)+1 and z=k*(2*k^(2n)+1) are solutions  :D

Thank you very much! Could you please write it again in latex, because i think I didn't understand correctly...I ow you! :D


Any tips on how to improve my violin skills over the summer?

Starting from what level, and what's your goal?

It's mostly in the etudes. :surrender:  :weightlift: I know, i did that last summer.

Seriously, I'm buckling down this summer, but that's hardcore. Just normal would be lots of scales (30-60 min), etudes (40-60), and whatever pieces your teacher gives you.

Really good etude books;
Fiorillo, Dont (little), Kreutzer, Gavinies (once slow, once fast), Dont (big), then *gasp* Paganini, done in that order. If your done with Paganini, get Ernest 6 etudes. nobody ever gets to those, though. I dream of playing those..... :(

for scales, try carl flesch if you havent...
thats what im playing right now.

Try Mazas and Wolfhart etudes.
It's what I did, and I improved
A LOT!  :D


function
calculus
integration
absolute value
real analysis
real analysis unsolved

Please help me with this problem:

Find a sequence of functions fn(x) which converge to 0 in L^1 such that fn(x) doesn't converge to 0 for any value of x in R.

thanks

(by convergence in L^1 ,which is the space of integrable functions, i mean the integral of the absolute value of fn(x) converge to 0.)

Know your standard examples: this calls for the "rotating tooth". Take a thin bump of fixed height and translate it to cover the whole interval. Repeat with a thinner bump, and keep going infinitely. The widths and thus the integrals go to zero, but each point has a large value infinitely often.

This is what i ended up with myself too.
thanks


Let $z > 1$ be an integer and let $M$ be the set of all numbers of the form $\dsp
z_k = \sum \limits^n_{k=0} z^k,$ $k = 0,1, \ldots$ Determine the set $T$ of divisors of at least one of the numbers $z_k$ from $M.$

Isn't it obvious they are all number coprime to $z$?

yeah, if i'm not totally wrong...

Peter

Indeed. Assume $a,b\in T,\ (a,b)=1$. Then $a|z_n,b|z_m$. It's not hard to see that $z_m,\ z_n|z_{mn+1}\Rightarrow ab\in T$. Now assume $p$ is a prime which doesn't divide $z$. If $p^k|z-1$ and $p^{k+1}$ doesn't, then $\displaystyle p^u|z_{\phi(p^{k+u})-1}=\frac{z^{\phi(p^{k+u})}-1}{z-1}$. Combine these two facts and we get the result.

[TeX question: how do you make "does not divide"?]

[quote="grobber"][TeX question: how do you make "does not divide"?][/quote]

a \not |\ b results in $a \not |\ b$


For $ a,b,c,d$ positive real numbers, prove that

$ \frac{a^2}{b(b\plus{}c)}\plus{}\frac{b^2}{c(c\plus{}d)}\plus{}\frac{c^2}{d(d\plus{}a)}\plus{}\frac{d^2}{a(a\plus{}b)}\geq2$

[quote="marin.bancos"]For $ a,b,c,d$ positive real numbers, prove that

$ \frac {a^2}{b(b \plus{} c)} \plus{} \frac {b^2}{c(c \plus{} d)} \plus{} \frac {c^2}{d(d \plus{} a)} \plus{} \frac {d^2}{a(a \plus{} b)}\geq2$[/quote]

Hello [b]marin.bancos[/b]

Acording to  Am-Gm inequality we have :

$ \frac {a^2}{b(b \plus{} c)} \plus{} \frac {b}{b \plus{} c} \ge \frac {2a}{b \plus{} c}$

and we 'll prove the inequality :

$ \sum_{cyc} \frac {a}{b \plus{} c} \ge \sum_{cyc}\frac {b}{b \plus{} c}$

we have :

$ \sum_{cyc} \frac {a}{b \plus{} c} \ge \sum_{cyc}\frac {b}{b \plus{} c}$
$ \Leftrightarrow \frac {a \plus{} c}{b \plus{} c} \plus{} \frac {b \plus{} d}{c \plus{} d} \plus{} \frac {c \plus{} a}{d \plus{} a} \plus{} \frac {b \plus{} d}{a \plus{} b} \ge 4$

and we have :

$ \frac {a \plus{} c}{b \plus{} c} \plus{} \frac {b \plus{} d}{c \plus{} d} \plus{} \frac {c \plus{} a}{d \plus{} a} \plus{} \frac {b \plus{} d}{a \plus{} b} \equal{} \frac {(a \plus{} c)(a \plus{} b \plus{} c \plus{} d)}{(b \plus{} c)(a \plus{} d)} \plus{} \frac {(b \plus{} d)(a \plus{} b \plus{} c \plus{} d)}{(a \plus{} b)(c \plus{} d)} \ge \frac {4(b \plus{} d)}{a \plus{} b \plus{} c \plus{} d} \plus{} \frac {4(a \plus{} c)}{a \plus{} b \plus{} c \plus{} d} \equal{} 4$

Hence it 's enough to check that :

$ \sum_{cyc}\frac {a}{b \plus{} c} \ge 2$

Which is easy By $ CS$  :)

P/s:Thanks Mathias_Dk  :)

[quote="Abdek"]
By the rearangment inequality we have :

$ \sum_{cyc} \frac {a}{b \plus{} c} \ge \sum_{cyc}\frac {b}{b \plus{} c}$
[/quote]
Can you be more specific?

[quote="Mathias_DK"][quote="Abdek"]
By the rearangment inequality we have :

$ \sum_{cyc} \frac {a}{b + c} \ge \sum_{cyc}\frac {b}{b + c}$
[/quote]
Can you be more specific?[/quote]

W.L.O.G $ a\ge b\ge c \ged$
$ \sum_{cyc}\frac {b}{b + c} - \sum_{cyc} \frac {a}{b + c} = \sum b(\frac {1}{b + c} - \frac {1}{c + d})$
 ( + )developed Abel $ \sum b(\frac {1}{b + c} - \frac {1}{c + d}) \ge 0\sum_{cyc}$
--->$ \frac {b}{b + c}\ge\sum_{cyc} \frac {a}{b + c}$

Thank you, [b]Abdek[/b]! Your idea is really interesting. Your valuable remark is that
$ \left.\frac{a^{2}}{b(b\plus{}c)}\plus{}\frac{b^{2}}{c(c\plus{}d)}\plus{}\frac{c^{2}}{d(d\plus{}a)}\plus{}\frac{d^{2}}{a(a\plus{}b)}\geq\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}d}\plus{}\frac{c}{d\plus{}a}\plus{}\frac{d}{a\plus{}b}\geq2\right.$
In addition, I'd like to say that my proof is different than yours. I'm waiting for new solutions to the problem.

[quote="marin.bancos"]For $ a,b,c,d$ positive real numbers, prove that

$ \frac {a^2}{b(b \plus{} c)} \plus{} \frac {b^2}{c(c \plus{} d)} \plus{} \frac {c^2}{d(d \plus{} a)} \plus{} \frac {d^2}{a(a \plus{} b)}\geq2$[/quote]
By CS and AM-GM:
$ LHS \ge \frac{(\sum a^2)^2}{\sum a^2b^2\plus{} \sum a^2bc} \ge \frac{(\sum a^2)^2}{2\sum a^2b^2} \ge \frac{(\sum a^2)^2}{\frac{(\sum a^2)^2}{2}}\equal{}2$

[quote="marin.bancos"]For $ a,b,c,d$ positive real numbers, prove that

$ \frac {a^2}{b(b \plus{} c)} \plus{} \frac {b^2}{c(c \plus{} d)} \plus{} \frac {c^2}{d(d \plus{} a)} \plus{} \frac {d^2}{a(a \plus{} b)}\geq2$[/quote]
marin.bancos, what about the following proof?
$ \sum_{cyc}\frac {a^2}{b(b \plus{} c)}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2}{\sum(a^2b^2 \plus{} a^2bc)}.$
$ \frac {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2}{\sum(a^2b^2 \plus{} a^2bc)}\geq2\Leftrightarrow$
$ \Leftrightarrow a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2a^2c^2 \plus{} 2b^2d^2\geq2\sum_{cyc}a^2bc.$
But $ 2a^2c^2 \plus{} 2b^2d^2\geq4abcd.$
Hence, it remains to prove that
$ a^{4} \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 4abcd\geq2(a^2bc \plus{} b^2cd \plus{} c^2da \plus{} d^2ab),$ which see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56040

Dear [b]arqady[/b],
Thanks! Interesting idea... Hmmm... It's not very easy to approach the problem this way... Thanks again.

[quote="marin.bancos"]Dear [b]CSS-MU[/b],
You took into account that
$ \sum a^2b^2\geq\sum a^2bc$
Could you explain this inequality? Maybe I'm missing something...[/quote]

We have

$ a^2 b^2  \plus{} a^2 c^2  \ge 2a^2bc$. It's easy

oh

[quote="marin.bancos"]Dear [b]CSS-MU[/b],
You took into account that
$ \sum a^2b^2\geq\sum a^2bc$
Could you explain this inequality? Maybe I'm missing something...[/quote]
Sorry,[b]marin.bancos[/b]. I had a mistake.

We have from AMGM that $ \frac{a^2}{b(b\plus{}c)}\equal{}\frac{a^2}{b^2\plus{}bc}\geq\frac{a^2}{b^2\plus{}\frac{b^2\plus{}c^2}{2}}\equal{}\frac{2a^2}{3b^2\plus{}c^2}$
Therefore $ \frac{a^{2}}{b(b\plus{}c)}\plus{}\frac{b^{2}}{c(c\plus{}d)}\plus{}\frac{c^{2}}{d(d\plus{}a)}\plus{}\frac{d^{2}}{a(a\plus{}b)}\geq2\left(\sum_{cyc}\frac{a^2}{3b^2\plus{}c^2}\right)$, so the inequality we have to prove is $ \sum_{cyc}\frac{x}{3y\plus{}z}\geq1$, where $ a^2,b^2,c^2$, and $ d^2$ are replaced by $ x,y,z$ and $ w$ respectively.

But from Cauchy-Schwarz, $ \sum_{cyc}\frac{x}{3y\plus{}z}\equal{}\sum_{cyc}\frac{x^2}{3xy\plus{}xz}\geq\frac{(x\plus{}y\plus{}z\plus{}w)^2}{3\sum_{cyc}xy\plus{}2xz\plus{}2yw}$, so we have to prove $ (x\plus{}y\plus{}z\plus{}w)^2\geq3\sum_{cyc}xy\plus{}2xz\plus{}2yw$, which is equivalent to $ \sum_{cyc}(x\minus{}y)^2\geq0$.

Equality occurs if and only if $ a\equal{}b\equal{}c\equal{}d$.

YES!!! This is the simplest and nicest solution to my problem. Well done, [b]bakerbakura[/b]! I knew you are able to find it! Congrats!
The proofs given by [b]Abdek[/b] and [b]arqady[/b] are also nice. Thanks to all of you!


Hi,
This is Steven. I am new member to this site.The given article is really very nice.
=============
John
[url=http://www.trainwithmeonline.com]workout programs[/url]  [/url]

uhhhh... why are u posting workout things in a math forum??


What is the largest natural number that is a solution of $ 13x \plus{} 8 < 35$?

So Natural number $ x$ = $ x$ is Integer, $ x$>0

Now also $ 13x\plus{}8<35$   or $ 13x<27$    or $ x<\frac{27}{13}$ Now $ x$ is integer, so largest is $ 2$.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a Noetherian ring and x an indeterminate. Then $ dimR[x,x^{\minus{}1}]\equal{}dimR\plus{}1$ where dim means the Krull dimension.

The problem is the first exercise of chapter 10 of Eisenbud's Commutative Algebra with a View towards Algebraic Geometry. I've been thinking about it for a bit now and I'm not sure how to do this. My idea is that we can provide a lower bounds of the dimension because $ dimR[x]\equal{}dimR\plus{}1$ but I have no idea how to show that the dimension is no larger. Intuitively, I can see that if $ R$ is a field, then the variety $ xy\minus{}1$ is one dimensional so $ R[x,y]/(xy\minus{}1)$ should be 1 dimensional, but I'm not sure how to do it in the more general case. Any input is appreciated!

[quote="LkNsngth"]Let $ R$ be a Noetherian ring and x an indeterminate. Then $ dimR[x,x^{ \minus{} 1}] \equal{} dimR \plus{} 1$ where dim means the Krull dimension.

The problem is the first exercise of chapter 10 of Eisenbud's Commutative Algebra with a View towards Algebraic Geometry. I've been thinking about it for a bit now and I'm not sure how to do this. My idea is that we can provide a lower bounds of the dimension because $ dimR[x] \equal{} dimR \plus{} 1$ but I have no idea how to show that the dimension is no larger. Intuitively, I can see that if $ R$ is a field, then the variety $ xy \minus{} 1$ is one dimensional so $ R[x,y]/(xy \minus{} 1)$ should be 1 dimensional, but I'm not sure how to do it in the more general case. Any input is appreciated![/quote]

For the lower bound note that if $ P$ is a prime ideal of $ R,$ then $ P[x,x^{ \minus{} 1}]$ is a prime (but not maximal) ideal of $ R[x,x^{ \minus{} 1}].$

The upper bound quickly follows from $ \dim R[x] \equal{} \dim R \plus{} 1$ and this simple fact that $ R[x,x^{ \minus{} 1}]$ is the localization of $ R[x]$ at $ S \equal{} \{1,x,x^2, \cdots \}.$

Ahh thank you very much, but just to make sure, we know that $ dimR\plus{}1$ is an upper bound because whenever we localize, there is an injection from the localized ring into the original ring, in particular $ SpecR[S^\minus{}1]\subseteq SpecR$ correct?

[quote="LkNsngth"]Ahh thank you very much, but just to make sure, we know that $ dimR \plus{} 1$ is an upper bound because whenever we localize, there is an injection from the localized ring into the original ring, in particular $ SpecR[S^ \minus{} 1]\subseteq SpecR$ correct?[/quote]

First of all $ R_1$ is a subring of $ R_2$ doesn't imply that $ \dim R_1 \leq \dim R_2$. For example $ \mathbb{Z} \subset \mathbb{Q}$ but $ \dim \mathbb{Z}\equal{}1 > 0 \equal{} \dim \mathbb{Q}.$

Second of all, when you localize a ring at a multiplicative subset (consisting of regular elements of course!) of that ring, you get a ring which contains the original one not the other way

around. What I meant was to consider any finite chain of prime ideals of $ S^{\minus{}1}R[x]$, say: $ S^{\minus{}1}P_n \supset \cdots \supset S^{\minus{}1}P_0.$ This will give you a chain of prime ideals of $ R[x]$, i.e. $ P_n \supset \cdots \supset P_0$.

Thus $ \dim S^{\minus{}1}R[x] \leq \dim R[x]\equal{}\dim R  \plus{} 1.$

ah yes there was a typo above, I meant that there was an injection in the [i]ideals[/i] of the localized ring into the ideals of the original ring. So yes I believe we are thinking of the same thing. Thanks.


trigonometry
ratio
number theory
least common multiple
combinatorics unsolved
combinatorics

[quote="cancer"]Four lines are given in the plane such that no two intersect and no three are concurrent. The three intersection points on each line form two segments, giving eight segments in total. Can the lengths of these segments be

(a) $1,2,3,4,5,6,7,8$?
(b) different positive integers?[/quote]

(a) : No
(b) : Yes

(a) ==========================
Let AB the segment with length 1. Let (D1 and D2) the lines containing A and (D1 and D3) the lines containing B (D1 is the line AB). D2 and D3 intersect in C.
AB, AC and BC are all integers and AB=1. Hence AC=BC (else A,B and C would be on the same line).
Hence D4 intersects D2 in D between A and C and intersects D3 in E between B and C. Let F be the intersection between D1 and D4. WLOG consider A between F and B.
For easy writings, let $AB=1,AD=a,DC=b,BE=c, EC=d, FA=u,FD=v, DE=w$. Let $\alpha$ be the angle AB,AC.
$\cos(\alpha)=\frac{1}{2(a+b)}=\frac{1}{2(c+d)}$
In triangle ADF, we have $v^{2}=u^{2}+a^{2}+\frac{ua}{a+b}$ and $a+b|ua$ and hence $a+b|ub$
In triangle BEF, we have $(v+w)^{2}=(u+1)^{2}+c^{2}-\frac{c(u+1)}{a+b}$ and $a+b|c(u+1)$ and hence $a+b|d(u+1)$ (since $a+b=c+d$)

So we have $a+b\in\{7,8,9,10,11,12,13\}$ and $a+b|ub$, $a+b|ua$, $a+b|(u+1)c$, and $a+b|(u+1)d$, with $\{a,b,c,d,u,v,w\}=\{2,3,4,5,6,7,8\}$ and it is easy to check that these conditions are impossible.

(b) ===========================
Let $\frac{\pi}{2}>\alpha>\beta>0$ two angles such that $\sin(\alpha),\cos(\alpha),\sin(\beta),\cos(\beta)$ are all rational.
On an line D1, let two points A,B such that $AB=1$
Let D2 the line intersecting D1 in A and such that $(D1,D2)=\alpha$
Let D3 the line intersecting D1 in B and such that $(D1,D3)=\pi-\alpha$
Let C the intersection of D2 and D3.
Let D0 the line such that $(D1,D0)=\beta$ and such that C belongs to D0. Let M the intersection between D1 and D0. A is between M and B.
Let F any point between M and A such that FA is a rational number.
Let D4 the line such that $(D1,D4)=\beta$ and such that F belongs to D4. Let D the intersection between D4 and D2 (D between A and C) and E the intersection between D4 and D3 (E between B and C).

In triangle ADF, all the angles have their sinus rational and FA is rational, so FA, AD an FD are rational.
In triangle ABC, all the angles have their sinus rational and AB=1 is rational. So AB, AC and BC are rational.
In triangle BEF, all the angles have their sinus rationak and FB=FA+1 is rational. So BE and EF are rational.
So $FA,AB,BE,EC=BC-BE,AD, DC=AC-AD,FD,DE=EF-FD$ are rational.

So we just have to increase this figure by a ratio which is the lcm of all the denominators of these rational numbers in order to have a construction in which the eight segments have lengths in natural numbers. And it is easy to choose $\alpha,\beta$ and FA such that all these distances are different.

Q.E.D


induction
combinatorics proposed
combinatorics

[b]Mader's Theorem. [/b] Every graph of average degree at least $4k$ has a $k$-connected subgraph.

It's very beautiful.  :D  Enjoy it!

Let's prove by inducton the following lemma:
If a graph $G$ with $m$ vertices contains no $k$-connected subgraph, than it has at most $(2m-1)k+1$ edges.
For $m\leq k$ it's obvious.
Now induction step:
Suppose for all $m<n$ this is already proven.
Let's prove it for $n$. Of course the whole graph is not $k$-connected, then we can choose $k-1$ edges such that their removal disconnects the graph into two pieces $A$, $B$ with no edges between $A$, $B$.
Let $|A|=a, |B|=b, a+b=n, a,b<n$. Then by induction hypothesis the total number of edges in the graph is at most
$(2a-1)k+1+(2b-1)k+1+k-1=(2n-1)k+1$.
The induction is finished and the lemma is proven.

The theorem is a direct consequence of the lemma.

[b]Note[/b] This problem resembles very much this one (especially their solutions are similar):

http://www.mathlinks.ro/Forum/viewtopic.php?t=20662

Very good!  :D This is a good example that general theorems are, more often than not, easier to prove than special cases. The trick in this problem is formulate the stronger assumption that a graph with more than $(2m-1)k+1$ vertices has a $k$-connected subgraph which, for inductive reasons, is easier to prove.

Here's something strange:

Trying to improve $4$ by a better constant by using the same method we must prove inductively that
a graph with $m$ vertices and no $k$-connected subgraph has at most $(m c-1)k+1$ edges. Let $k>1$.
To use induction, this must hold for $m\leq k$ and it suffices to look only at case $m=k$.
So we get $(ck-1)k+1=\frac{k(k-1)}2$ and so $ck\geq \frac{k+1}2$ so $c=\frac{k+1}{2k}$.
So we can replace $4$ by $\frac{k+1}k$.

This means if a graph has average degree at least $k+1$, it contains a $k$-connected subgraph. This cannot be true, since someone would point it out for sure and would improve it.

What is wrong? :?

Actually, here is the general result.

Given $k\geq 1$, denote by $f(n)$ the biggest number of edges a graph with $n$ vertices and no $k$-connected subgraph can have. Then $f(n)=g(n)$ where
$g(n)= \frac{n(n-1)}2$ for $n<k$ and $g(n)=\frac{(2n-k-1)k}2$ otherwise.

Proof. We can note that if $i+j>k$ then $g(i)+g(j)+k-1\leq g(i+j)=g(1)+g(i+j-1)+k-1$.
Now the proof is by induction:
For $n \leq k$ it's obvious.
For $n>k$ let $G$ be the graph with $n$ vertices, $f(n)$ the maximal number of edges and no $k$-connected subgraph.
Then there are at most $k-1$ edges whose removal disconnects $G$ into two pieces $A,B$ with
$|A|=i, |B|=j, i+j=f(n)$ and $1\leq i,j <n$.
Thus by induction hypothesis $f(n)\leq f(i)+f(j)+k-1=g(i)+g(j)+k-1\leq g(n)$.
Hence $f(n) \leq g(n)$.
From the other side we can achieve this number as follows:
If $n\leq k$ take $G=K_n$.
If not take $G$ be formed of numbers from $1$ to $n$ and any $j$ be connected to the smallest $k-1$ numbers different from $j$, giving us exactly $f(n)=\frac{k(2n-k-1)}2$
QED

I am sorry, but I have few stupid questions.
1) If graph has $n$ vertices than exists no triangle than it has at most $\frac{n^2}{4}$
edges and this value is achievable.

But the upper lemma states if there $(2n-1)*3 + 1$ then it has triangle, so false ?
2) Average degree of graph is $4k$ means it has at least $2nk$ edges ?

3) There exists Turan theorem that states if graph has n vertices and no $k$ – subgraph,
then it has maximum $\frac{(k-2)n^2}{2(k-1)}$ edges.
Or  I am understanding problem wrong ?

$k$-connected is not the same as complete subgraph of $k$ vertices.
It means that for any two vertices in this subgraph we have at least $k$ disjoint paths connecting them.


Anyone has the pdf of the J. Neukirch, Algebraic Number Theoem??? I just want to study the algebraic number theorem.
Thanks for sharing.

You want to study what? The theorem or the theory?

I want to know the method of doing some elementary problem.Beacuse I think algebraic number theory is very strong.It can help us to prove quite a lot of things.Can you tell me some theorem of this and some method ?

You can always ask questions in this forum. For the book, I recomend asking someone who has access to SpringerLink.


algebra
polynomial
Galois Theory
abstract algebra
superior algebra
superior algebra unsolved

$ K$ is a finite field with $ |K| \equal{} q$ and $ f \in K[x]$ is irreducible, then $ f$ divides $ x^{q^{n}} \minus{} x$ if and only if $ \deg f$ divides $ n$.

I was only able to prove the "only if" part:
If $ f$ divides $ x^{q^{n}} \minus{} x$,Let $ F$ be the splitting field of $ x^{q^{n}} \minus{} x$ over $ K$, then we have: $ [F: K] \equal{} n$, and choose any root $ u \in F$ of $ f$, we have $ [K(u): K] \equal{} \deg f$, then  $ \deg f$ divides $ n$.

Can somebody tell me the incompleted part :) ?

uhm.. if ${\rm deg}\,f$ divides $n$, then its splitting fields $L$ is a subfield of what you called $F$.
now consider a generator of $L$, say $\alpha$: it's a root of $x^{q^{n}}-x$, since it belongs to $F$. then $f$ divides $x^{q^{n}}-x$, because $f$ is irreducible.

[quote="ma_go"]uhm.. if ${\rm deg}\,f$ divides $n$, then its splitting fields $L$ is a subfield of what you called $F$.[/quote]
Hi,can you tell me how to show this?

it's a consequence of the fact that finite fields are uniquely determined by their cardinalities.

and not only they're isomorphic, but they have same embeddings in a fixed algebraic closure: galois group of a finite extension of a finite field is cyclic (and generated by frobenius automorphism), so you have one and only one subfield for any divisor of the degree of the extension, and that should suffice.

i hope i've been clear enough, even if i'm not that convinced about what i wrote ;)
(it's true, anyway, i'm just not convinced of my explanations :p)

Thank you for your patience. :)
But from what your wrote I think we also have to prove that $[L: K] \le [F: K]=n$...
can you show it?

uhm.. actually, since $\alpha$ is a root of $x^{q^{n}}-x$, automatically $[K(\alpha): K]\le n$.
did you mean this?

I still mean how can you prove the splitting field of $f$ is a subfield of $F$.
If this is true,for $deg f=n$,then we have $L=F$,and $[L: K]=n$
then we have $L=K(\alpha)$ where $\alpha$ is any root of $f$,will it be always true??? :?:

However,I think I find a better way to slove the problem:
Consider the field $K[x]/(f(x))$,we can easily have $x^{q^{n}}-x$ by $deg f|n$,then because $x$ is a root of the irreducible polynomial $f$,thus $f$ divides $x^{q^{n}}-x$.
Am I right?

[Edit:Sorry I forget that any finite extension of a finite field is normal and separable,then $K[x]/(f(x))$ is in fact the splitting field of $f$ over $K$.
Thank you again]

I have a more "elementary" way of doing this problem (forward), i.e. no splitting fields. Since I want to be as helpful and as clear as possible I would write out a lot of details some of which much seem redunant and obvious.
~~~
Before we begin there are three things I want to mention.
1)We will be working the in the mother of all extension fields (I mean the algebraic closure).
2)Let $K$ be the set of all zero to $x^{p^{n}}-x$ then $K$ is field under the same binary operations. (A theorem)
3)I will do it for $p$ instead of prime power $q=p^{a}$, I am sure it generalizes easily.
~~~
This is what we will show:
"If $f(x)$ is irreducible factor of $x^{p^{n}}-x$ of degree $d$ then $d|n$ in the field $\mathbb{Z}_{p}[x]$."

[b]Step 1:[/b]Given $f(x)$ irreducible over $\mathbb{Z}_{p}$ there exists $\alpha$ (in the mommy extension field) such that $f(\alpha) = 0$.

[b]Step 2:[/b]Algebraically adjoin $\alpha$ to $\mathbb{Z}_{p}$ to form the simple extension $\mathbb{Z}_{p}(\alpha)$. Note that $[\mathbb{Z}_{p}(\alpha) : \mathbb{Z}_{p}] = d$.

[b]Step 3:[/b]Construct the field $K$ to be the set of all zeros of the polynomial $x^{p^{n}}-x$. (See comment before this proof).

[b]Step 4:[/b]Note that $\alpha \in K$. Because $f(\alpha)=0$ and $f(x)$ is a factor of $x^{p^{n}}-x$ by hypothesis.

[b]Step 5:[/b]That means $\mathbb{Z}_{p}(\alpha) \subseteq K$. Because $\mathbb{Z}_{p}$ is the [i]minimial[/i] field that contains $\alpha$ (that is a basic property of simple extensions).

[b]Step 6:[/b]We can form the subfield relation $\mathbb{Z}_{p}\leq \mathbb{Z}_{p}(\alpha) \leq K$

[b]Step 7:[/b]Note that $[K: \mathbb{Z}_{p}] = n$ by Step 3.

[b]Step 8:[/b]By (Lagrange's theorem  :?: ) $[K: \mathbb{Z}(\alpha)][\mathbb{Z}(\alpha): \mathbb{Z}] = [K: \mathbb{Z}_{p}] \implies kd=n$

[b]Step 9:[/b]By the previous step, we see now that $d|n$.  :clap:

[quote="zhaobin"]

I was only able to prove the "only if" part:
If $f$ divides $x^{q^{n}}-x$,Let $F$ be the splitting field of $x^{q^{n}}-x$ over $K$,then we have:$[F: K]=n$,and choose any root $u \in F$ of $f$,we have $[K(u): K]=deg f$,then  $deg f$ divedes $n$.

 ?[/quote]
Hei,what you wrote is actully what I wrote,If I change the words "the splitting field of $x^{q^{n}}-x$ over $K$",to "the field of all roots of $x^{q^{n}}-x$ containing $K$"

But Thank you all the same :)


Problema IV

O teorema de Rolle afirma que dada uma função contínua f definida num intervalo fechado [a,b] e diferenciável em (a,b), se f(a) = f(b) então existe algum ponto c em (a,b) onde a tangente ao gráfico de f é horizontal, isto é:

\begin{align*}
  & f(a)=f(b)=0 \\
 &  \\
 & f'(c)=kf(c) \\
 &  \\
\end{align*}

Por outro lado:

$ g(x)=f(x){{e}^{-kx}}$

No caso pedido:

Calculate "k"

TKX
Antonio

1. Don't use <math></math> - that doesn't work. Instead, enclose your math expressions between $ \$$ signs.

2. I am having trouble deciphering exactly what the problem is here or exactly what you are asking.


MATHCOUNTS
AMC
AMC 10
AMC 8
geometry
AIME
USA(J)MO

Mathcounters, what are your AMC 8 scores?

[quote="Naonao"]Either i forgot mine or i don't no it but i no i didn't get a perfect and i did bad[/quote]

Please do capitalize and not use chatspeak. It makes it easier to read and makes us appear more professional.

:wow: :wow:
I got a perfect score!! (just found out from the site because my teacher hasnt gotten the scores back yet)

There was only one person from Delaware that got a perfect score... could that be you?  :P

Possible, there could be some other person with the same name from the same school in the same state  :D and if thats true, I didnt get a perfect score :noo: :noo:

Howd you know I was from Delaware??

I actually just found out I was a perfect scorer on the AMC 8 a few days ago (too).  You can tell from my username. :o

I just realized that your username is your real name  :o

Well, at least us perfect scorers are in good company (Neal Wu) :P

I'm not a Mathcount competitor, but I got 22/25.

question for everyone

Was anyone invinted to AMC10? (it was mentioned in AMC site that high scorers in AMC8 are invited to AMC10.)
But I don't think they actually invite people to AMC10...

tch, i'm taking the AMC10... aiming for a 150 :o ...or w/e the perfect score is :P

I got a perfect score on the AMC 8 (from Washington--there were five of us, two (including me) were on the national mathcoutns team last yeawr).
And no, you don't have to be invited to take the AMC 10. You can come no matter what your score was on the 8. The first one that you need to be invited to is the AIME.

I got a 25 on the AMC 8. Texas. I don't think you get invited to the AMC 10. I think you just take it. There were 9 perfects in Texas. The test was one of the harder ones. I think maybe it is just me.

just a question, how do you find out your score from the site? our teacher's been sick so we won't find out our scores for a while.

Actually, there's no rule about taking the AMC 8, which is rather separate from the AMC10 or 12 in the sense that it doesn't continue to AIME and USAMO. Anybody can take the 10 without taking the 8.

No offense to those who did well (not me) but AMC 8 is relatively useless in terms of qualifying for anything.

I took it last year too, and did horribly. I got somewhere around a 17.
[quote="binonunquineist"]The amc8 is superbly easy, even compared to MATHCOUNTS material. I got a 25/25 last year as a sixth grader and a 25/25 this year as a seventh grader. Now I just have to continue that.[/quote]
Thanks. :dry:

[quote="pgpatel"]Binonunquneist got a perfect score on the math SAT this year too...[/quote]

SAT II? or the math portion of the SAT I. I think the hardest part of the math portion of the SAT I only has very basic Algebra II and all the problems are very straightfoward, not many word ones like mathcounts, not sure though, that's what I've heard.

Math portion for duke tip. THat or on the math ACT.

25. Yup.

I thought the AMC 8 was a really long time ago, though, so why you're asking here and now is beyond me.

[quote="ehehheehee"][quote="Ignite168"][quote="lotrgreengrapes7926"]800 on the SAT I math isn't that hard. I mean even I did that. Qualifying for National MathCounts is a lot harder, in my opinion.[/quote]

Not if your from North Dakota or the Virgin Islands, and what does the SAT I cover? Just Algebra I and Geometry?[/quote]

I think it is supposed to have algebra II (but I aced it without knowing that), but yeah, it is easier to ace it than make it to nationals.  But just because you didn't ace it doesn't mean to can't make nats because 70 or so questions is a lot to not make stupid mistakes it.[/quote]

Yeah, the SAT is more about doing relatively easy problems accurately. Mathcounts is the real deal.

[quote="cowpi"]Yeah, the SAT is more about doing relatively easy problems accurately. Mathcounts is the real deal.[/quote]
Personally, I think that AMC 10, 12, and AIME (and USAMO) are a better gauge for mathematical ability, while MathCounts requires you to be fast and know lots of tricks for solving problems. Plus, they're harder.  :P Maybe I'm just saying this because I'm better at that stuff than MathCounts. :D

i got 23/25 on the AMC 8, 23/25 = 138 on the AMC 10B  ;)
So i would say im better at harder, longer problems than easier, faster problems.

[quote="ehehheehee"]AMC competitions don't fly you across the country to a nice hotel and vacation in addition to the competition.  They just mail you the forms.  It is harder, but probably because 1. you get more time 2. there is trig laws, polynomials, logarithms, imaginary numbers, e, and all that fun stuff...

My AOPS Book 2 that is supposed to help me will take another 1-2 weeks to come (I have waited one already)[/quote]
Unless you're REALLY REALLY smart and get to go to the IMO. :D


geometry
geometric transformation
reflection
geometry unsolved

$ ABCD$ is a quadrilateral. $ A'BCD'$ is the reflection of $ ABCD$ in $ BC,$ $ A''B'CD'$ is the reflection of $ A'BCD'$ in $ CD'$ and $ A''B''C'D'$ is the reflection of $ A''B'CD'$ in $ D'A''.$ Show that; if the lines $ AA''$ and $ BB''$ are parallel, then ABCD is a cyclic quadrilateral.

Denote $\alpha = \measuredangle DAB,\ \beta = \measuredangle ABC,\ \gamma = \measuredangle BCD,\ \delta = \measuredangle CDA$. In addition, denote $\alpha_1 = \measuredangle DAC,\ \alpha_2 = \measuredangle CAB$, $\beta_1 = \measuredangle ABD,\ \beta_2 = \measuredangle DBC$, $\gamma_1 = \measuredangle BCA,\ \gamma_2 = \measuredangle ACD$, $\delta_1 = \measuredangle CDB,\ \delta_2 = \measuredangle BDA$. Obviously, $\alpha = \alpha_1 + \alpha_2,\ \beta = \beta_1 + \beta_2,\ \gamma = \gamma_1 + \gamma_2,\ \delta = \delta_1 + \delta_2$.

The triangle $\triangle AA''C$ is isosceles with AC = CA'' and

$\measuredangle A''CA = \measuredangle A''CD' + \measuredangle D'CB + \measuredangle BCA = \gamma_2 + \gamma + \gamma_1 = 2 \gamma$

$\measuredangle CAA'' = \frac{180^\circ - \measuredangle A''CA}{2} = 90^\circ - \gamma$.

The triangle $\triangle BB''D'$ is isosceles with BD' = D'B'' and

$\measuredangle B''D'B = 360^\circ - (\measuredangle B''D'A'' + \measuredangle A''D'C + \measuredangle CD'B) =$ $360^\circ - (\delta_2 + \delta + \delta_2) = 360^\circ - 2 \delta$

$\measuredangle D'BB'' = \frac{180^\circ - \measuredangle B''D'B}{2} = \delta - 90^\circ$.

Let the lines BB'' and AA'' meet at a point X and we will calculate the angle $\measuredangle AXB \equiv \measuredangle A''XB''$. The lines AA, BB'' are parallel (and their intersection X at infinity) iff $\measuredangle AXB'' = 0$.

$\angle A''XB'' = \measuredangle D'BB'' + \measuredangle CBD' + \measuredangle ABC -180^\circ + \measuredangle CAB - \measuredangle CAA'' =$

$= \delta - 90^\circ + \beta_2 + \beta - 180^\circ + \alpha_2 - (90^\circ - \gamma) = \beta_2 + \alpha_2 - \alpha = \beta_2 - \alpha_1$

Thus we see that the lines AA'', BB'' are parallel iff $\alpha_1 = \beta_2$ or $\measuredangle DAC = \measuredangle DBC$, which is equivalent to the quadrilateral ABCD being cyclic.


inequalities
logarithms
search
inequalities unsolved

Let abc=1 ,a,b,c>o. Prove that:
$ 10(a^3\plus{}b^3\plus{}c^3)\minus{}9(a^5\plus{}b^5\plus{}c^5) \geq 3$

What are you going to do, then?

We can't subtract both sides directly in this case.

I'am sorry but this inequality is wrong, i said that in my first post in this topic. Am i wrong  :D

This problem should be $ 10(a^3\plus{}b^3\plus{}c^3)\minus{}9(a^5\plus{}b^5\plus{}c^5)\leq 3.$

Let $ f(a)\equal{}10a^3\minus{}9a^5\plus{}15\ln a\ (a>0)$, we have $ f'(a)\equal{}\frac{15}{a}(1\minus{}a)(3a^4\plus{}3a^3\plus{}a^2\plus{}a\plus{}1)$, yielding $ f(a)\leq f(1)\Longleftrightarrow f(a)\leq 1.\ \therefore f(a)\plus{}f(b)\plus{}f(c)\leq 3.$

I think he means $ 10(a^{3} \plus{} b^{3} \plus{} c^{3}) \minus{} 9(a^{5} \plus{} b^{5} \plus{} c^{5})\geq1$

where $ a\plus{}b\plus{}c\equal{}1$

which is here
http://www.mathlinks.ro/viewtopic.php?search_id=984446516&t=161399

 :)


algebra polynomial function limit calculus

How do we determine whether the degree of a polynomial $ h(x)$ is even or odd? [Use the below examples if necessary]

Also, suppose a certain root of the polynomial $ f(x)$ has multiplicity $ k$, where $ k > 1$. (i.e. double roots, triple roots etc.) How would this influence the graph of $ f(x)$?

This is a pretty detailed treatment:

http://en.wikipedia.org/wiki/Degree_of_a_polynomial

A polynomial is even if it is symmetric with respect to the y-axis.

f(x) is even if f(-x)=f(x)

A polynomial is odd if it is symmetric with respect to the origin.

f(x) is odd if -f(-x)=f(x)

If a root has odd multiplicity, then the graph crosses the x-axis at that root. If the multiplicity is odd, then the graph just touches the x-axis at that point. (Like the point (-3,0) in your 1st graph.

I think you had some typos in your last paragraph.

Do you mean that if the degree of the polynomial is even, then the polynomial is even? And if the degree is odd, then the polynomial is odd? I don't think they're related at all! [b]I'm asking about the degree![/b]

Hehe. Maybe I'm wrong.

A function is odd iff. it is point-symmetric about the origin. A polynomial function is odd iff. [b]all [/b]terms with non-zero coefficients are of odd degree. Constants are considered the coefficient of the $ x^0$ term.

A function is even iff. it is symmetric across the y-axis. A polynomial function is even iff. [b]all [/b]terms with non-zero coefficients are of even degree.

Just looking at your three graphs, none of those polynomials are even or odd.

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

Let  $ P(x)$ be a polynomial with a root $ r$ of multiplicity $ k$, and let $ Q(x) \equal{} \frac {P(x)}{(x \minus{} r)^k}$ be another polynomial* (with a removable discontinuity). Then in the neighborhood of $ x \equal{} k$, $ P(x) \approx Q(r) (x \minus{} r)^k$.

Since $ Q(r)$ does not vary with $ r$, we can write this as $ P(x) \approx c (x \minus{} r)^k$ for some constant $ c$. Now, what do you know about a function of this form -- what does it look like? Try varying $ c$ and $ r$ and $ k$ on Mathematica, and you'll figure it out  :)

The neighborhood I refer to can be very small, so you might have to zoom in. It's probably a good idea to pick other numbers that are far away from the root you are observing to create a larger neighborhood.

*- By $ Q(x) \equal{} \frac{P(x)}{(x\minus{}r)^k}$, I really mean that Q is the polynomial $ P(x)$ without the $ k$ factors of $ (x\minus{}r)$; I don't mean to multiply the factors in and then divide them out, since this would just make $ Q(r) \equal{} 0/0$. A better way of saying this is that $ Q(x) \equal{} \lim_{t \to x}\frac{P(t)}{(t\minus{}r)^k}$

Consider $ \lim_{x \rightarrow \pm \infty} h(x)$ for odd and even degree polynomials. (For odd-degree polynomials, these two sides will tend towards different vertical directions, namely one $ \rightarrow \infty$ and one $ \rightarrow \minus{}\infty$, while for even-degree polynomials, they will tend towards the same direction, namely both $ \rightarrow \infty$ or both $ \rightarrow \minus{}\infty$)  :wink:

[quote="The Zuton Force"]A function is odd iff. it is point-symmetric about the origin. A polynomial function is odd iff. [b]all [/b]terms with non-zero coefficients are of odd degree. Constants are considered the coefficient of the $ x^0$ term.

A function is even iff. it is symmetric across the y-axis. A polynomial function is even iff. [b]all [/b]terms with non-zero coefficients are of even degree.

Just looking at your three graphs, none of those polynomials are even or odd.

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

Let  $ P(x)$ be a polynomial with a root $ r$ of multiplicity $ k$, and let $ Q(x) \equal{} \frac {P(x)}{(x \minus{} r)^k}$ be another polynomial* (with a removable discontinuity). Then in the neighborhood of $ x \equal{} k$, $ P(x) \approx Q(r) (x \minus{} r)^k$.

Since $ Q(r)$ does not vary with $ r$, we can write this as $ P(x) \approx c (x \minus{} r)^k$ for some constant $ c$. Now, what do you know about a function of this form -- what does it look like? Try varying $ c$ and $ r$ and $ k$ on Mathematica, and you'll figure it out  :)

The neighborhood I refer to can be very small, so you might have to zoom in. It's probably a good idea to pick other numbers that are far away from the root you are observing to create a larger neighborhood.

*- By $ Q(x) \equal{} \frac {P(x)}{(x \minus{} r)^k}$, I really mean that Q is the polynomial $ P(x)$ without the $ k$ factors of $ (x \minus{} r)$; I don't mean to multiply the factors in and then divide them out, since this would just make $ Q(r) \equal{} 0/0$. A better way of saying this is that $ Q(x) \equal{} \lim_{t \to x}\frac {P(t)}{(t \minus{} r)^k}$[/quote]

Erm, ThetaPi is asking if the degree of the function is even or odd, not if the function is even or odd.  Count the amount of changes in the direction of the graph and add 1.  If this number is even, the degree of the function is even and vice versa.

Oh, I get what he's asking now.

The first thing I thought is that "the degree can easily be found just by looking at the equation, and taking the highest power. Then, is it really that hard to find whether it's even or odd?"

I assumed, therefore, that he meant whether the polynomial is even or odd.

I guess he meant finding the degree by the graph, but I don't see why he asked how to find whether the degree is even or odd, rather than just asking how to find the degree...

1)How do we know whether the degree of a polynomial  is even or odd if we only have its graph?
2)How would the multiplicity of a root influence the graph of the polynomial?
3)Where do the answers come from/derive from, without invoking something not yet learned? Please explain so I do not have to memorize yet more "magical" rules.

Given the above questions, I'll try to interpret/summarize (shown) and explain (hidden) the many prior good answers. Feel free to tell me (as a post or by private message) how and where I failed. I also apologize for not giving due credit to everyone.

1) The sign (positive or negative) at both ends of the graph tells you about the degree of the polynomial $ h(x)=ax^n+bx{}^n{}^-{}^1+ ...$ (and the sign of the leading coefficient $ a$). If the degree, $ n$, is even, at the left and right ends, the graph should go up very sharply ("through the roof") for $ a>0$; down very sharply for $ a>0$. If the degree is odd, the graph will head (very sharply) in different directions at both ends.
The graph 9200.png shows that $ a>0$ and $ n$ is odd.
The other two graphs are not appropriate for this question because they do not show sharp changes at both ends, like all graphs of polynomials should.[hide]azjps drifts towards calculus and talks about limits, but even without knowing the definition of limit, we can see that as $ |x|$ grows large, at some point $ (|x|>K)$ the first term $ ax^n$ of a polynomial $ h(x)=ax^n+bx{}^n{}^-{}^1+ ...$ will "overwhelm" the rest of the polynomial:
$ |ax^n|>>|bx{}^n{}^-{}^1+ ...|$
As a consequence, for those values of $ x$ at the extremes $ ax^n$ is a good approximation of the polynomial, and the values of $ a$ and $ n$ determine the "behavior" of the graph at the "extremes".
You could also think of the factorization $ h(x)=a(x-r)^k(x-s)^l(x-t)^m...$, with real roots $ r$, $ s$, $ t$, and so on, and how the polynomial will change signs only at each root of odd multiplicity. The degree of the polynomial is the sum of all the root multiplicities, and it will be odd if and only if the sum of all the odd multiplicities is odd. You do not need to worry about non-real roots. If there are any, they will come in pairs, and will be reflected in your factorization by a polynomial of even degree that is positive for all values of x. For example, in the full factorization of $ x^3-1=(x-1)(x^2+x+1)$ the factor $ (x^2+x+1)$ is always positive for all real $ x$.[/hide]
2) A polynomial changes sign (positive to negative or the reverse) only at each odd multiplicity root. Even multiplicity roots will mean $ h(r)= 0$ is a maximum or minimum at root $ r$. There is an even multiplicity root in 9203.png at $ x=-3$. Maybe there's another one in 9201.png at $ x=-3$, but the graph is not very clear. The roots at $ x=1$ in your figures would be multiple roots of odd multiplicity, because of the horizontal tangent to the curve at that point. Other roots look like single roots, where the function changes sign.[hide]If you look at the factorization $ h(x)=a(x-r)^k(x-s)^l(x-t)^m...$, you will see that the polynomial will change signs only at each root of odd multiplicity. If $ r$ is a root of even multiplicity $ (k$ is even$ )$, then $ (x-r)^k$ will be positive on both sides of $ r$. As for $ x=1$ being a multiple root, the explanation I think of would go a bit too much into calculus and mostly repeat part of what The Zuton Force stated.[/hide]

[quote="ThetaPi"]It doesn't really answer my problem.[/quote]

It does read it carefully or tell me what textbook your using

[quote="KMST"]If the degree, $ n$, is even, at the left and right ends, the graph should go up very sharply ("through the roof") for $ a > 0$; down very sharply for $ a > 0$.[/quote]

I believe you meant that if it is even, then the graph goes either up [b]or[/b] down at both ends.

Either y approaches infinity on both sides, or it approaches negative infinity on both sides.  :wink:


inequalities
topology
triangle inequality
advanced fields
advanced fields unsolved

Let $ X \equal{} \{x \equal{} (x_n) | x_n \in \mathbb{R}\}$  be the set of all sequences of real numbers.

Let $ d: X\times X \rightarrow \mathbb{R}$ be defined as follows :

$ d(x,y) \equal{} \sum\limits_{n \equal{} 1}^{\infty} \frac {1}{2^n}. \frac {|x_n \minus{} y_n|}{|x_n \minus{} y_n| \plus{} 1},$ where

$ x \equal{} (x_n), y \equal{} (y_n)$

Show that $ d$ is a metric space on $ X$   :?:  :|

The only real problem is to show the triangle inequality, and this follows from the concavity of $ f(x)\equal{}\frac{x}{x\plus{}1}$.


Prove that the function $d(n)$, the number of divisors of $n$, and the function $\sigma(n)$, the sum of the divisors, are both multiplicative.

[hide="For the first one"]
Let $m$ and $n$ be coprime positive integers.
Let the divisors of of $m$ be $m_{1},m_{2},...,m_{d(m)}$ and the divisors of $n$ be $n_{1},n_{2},...,n_{d(n)}$, where all terms are distinct except that $m_{1}=n_{1}=1$.
$k$ is a divisor of $mn$ iff $k=m_{i}n_{j}$ for some $i,j$. But the number of ways this product can be formed is $d(n)\cdot d(m)$. Each of these divisors are distinct, so $d(mn)=d(m)\cdot d(n)$ [/hide]

[hide="Other solution for 1"]$d(mn)= d(m)d(n)$ for $m\bot n$ follows immediately from fact that $d\left(\prod_{1\leqslant i\leqslant k}p_{i}^{\alpha_{i}}\right)= \prod_{1\leqslant i\leqslant k}(\alpha_{i}+1)$ and all primes in the factorisations of $m$ and $n$ are different. [/hide]
[hide="Solution for 2"]As in my solution for $d(n)$, we just notice that $\sigma\left(\prod_{1\leqslant i\leqslant k}p_{i}^{\alpha_{i}}\right)= \prod_{1\leqslant i\leqslant k}\frac{p_{i}^{\alpha_{i}+1}-1}{p_{i}-1}$. [/hide]


The sum of the squares of three positive whole numbers is 165. Find the largest of the three squares.

Pretty much the only way to do this quickly and easily is to guess-and-check a little. We first see if 144 is one of them. 165 - 144 = 21, and 21 can't be written as the sum of two squares.

Next let's try 121. 165 - 121 = 44, but it doesn't work either.

Now we try 100. 165 - 100 = 65, and 65 is $ 8^2 \plus{} 1^2$! Thus, our answer is $ \boxed{100}$.


Και μετα να δειξετε οτι $ 300! > 100^{300}$  :P

Γνωρίζουμε ότι $ (1 \plus{} {\frac {1}{n}})^n < e < (1 \plus{} {\frac {1}{n}})^{n\plus{}1}$.Θέτοντας όπου $ n$ το $ 1,2,3,...,n$ διαδοχικά και πολλαπλασιάζοντας κατά μέλη έχουμε qed. :)
Το άλλο δεν προλαβαίνω να το κοιτάξω τώρα...

To άλλο προκύπτει άμεσα από το προηγούμενο αν βάλουμε όπου $ n$ το $ 300$ Αφού $ (\frac{300}{e})^{300} > 100^{300}$

[quote="mailo"]Νδο

$ n! > \left(\frac {n}{e}\right)^n$

ps. στο βιβλίο που βρήκα αυτή την ανισότητα έχει λύση με μαθηματική επαγωγή, αλλά εγώ σκέφτηκα μια πιο ωραία την οποία και θα γράψω αργότερα αν δεν την βρείτε εσείς.[/quote]

Απλώς μελέτησε το ολοκλήρωμα $ \int_1^n \log x dx$ :)

Από την άλλη, υπάρχει πάντα και ο Stirling που είναι πολύ πιο σφιχτός: $ n! > \sqrt {2\pi n} \left(\frac {n}{e}\right)^n$ :)

Cheerio,

Durandal 1707

Ωραιος Durandal! Αυτη ειναι ακριβως η λυση μου.

Απλα επρεπε να σκεφτει κανεις οτι

$ ln(2) \plus{} ln(3) \plus{} ... \plus{} ln(n) > \int_{1}^{n}ln(x) \ dx \Leftrightarrow$

$ ln(n!) > (xlnx \minus{} x)\Big|_1^n\Leftrightarrow$

$ ln(n!) > n\cdot ln(n) \minus{} (n \minus{} 1)\Leftrightarrow$

$ n! > \frac {n^n}{e^{n \minus{} 1}}\Leftrightarrow$

$ n! > \left(\frac {n}{e}\right)^n$  qed  :)


Let $ ABCD$ be  a quadrilateral has $ AB\parallel{} CD$.Denote $ M,N,P$ be midpoints of $ BC,AM,DM$
Now,Find the condition of two sides $ AB,BC$ to the intersection of $ BN$ and $ DP$ is inside of $ AMD$

[hide="result"]$ \frac {1}{3} < \frac {AB}{CD} < 3$[/hide]

[quote="Allnames"]Let $ ABCD$ be  a quadrilateral has $ AB\parallel{} CD$.Denote $ M,N,P$ be midpoints of $ BC,AM,DM$
Now,Find the condition of two sides $ AB,DC$ to the intersection of $ BN$ and $ DP$ is inside of $ AMD$

[hide="result"]$ \frac {1}{3} < \frac {AB}{CD} < 3$[/hide][/quote]
Any solution,see again my problem,it is my typo mistake and I have just fixed it,sorry


advanced fields
advanced fields unsolved

Let $\varphi: \pi_{1}(\mathbb{T}^{2})\to \pi_{1}(\mathbb{T}^{2})$ be any homomorphism. Prove that there exists $f: \mathbb{T}^{2}\to \mathbb{T}^{2}$ such that $f_{*}=\varphi$. Show also that if $\varphi$ is an isomorphism, then $f$ can be chosen to be a homomorphism.

Represent $\mathbb T^{2}$ as $\{(z,w)\colon |z|=|w|=1\}$, where $z$ and $w$ are complex numbers. The map $(z,w)\mapsto (z^{a}w^{b}, z^{c}w^{d})$ induces a homomorphism of $\pi_{1}(\mathbb T^{2})$ into itself. It remains to check that these are all homomorphisms of $\pi_{1}(\mathbb T^{2})=\mathbb Z^{2}$.


find all functions $f:\mathbb R\rightarrow\mathbb R$ such that:
            $f(x^n+f(y))=y+(f(x))^n$  for all x,y in real , n is the constant and n is natural
my solution does very long and boring
 I find $f(x)=x$ for all x in real and $f(x)=-x$ for all x in real are two functions satisfy the condition. :D  :D  :D  :D

May be this problem can be solved in the same way as IMO 1992 A(2).
What is AMM, N.T.Tuan?

to solve this pro, we have 2 case: n odd, n even
I think IMO92 very hard because how to find: $f(0)=0$
but i think similar pro very very hard: find all $f:\mathbb Q\to\mathbb Q$ such that: $f(x^2+f(y))=y+(f(x))^2$
@N.T.TUAN: are you NGUYEN TRONG TUAN in pleiku, iff it true- thanks for your new book :P


[quote="José"]Let $P$ be a point outside the circle $C$. Find two points $Q$ and $R$ on the circle, such that $P,Q$ and $R$ are collinear and $Q$ is the midpopint of the segmenet $PR$. (Discuss the number of solutions).[/quote]

$2*PQ^2=PQ*PR=PO^2-R^2=const$
So $PQ=\frac{\sqrt(PO^2-R^2)}{2}$
And $\le 2$ solutions

V:)


Your local hardwood store sells lawn mowers. Type A, costs 137 dollars and they make a profit of 40 dollars on each one. Type B, costs 120 dollars and they make a profit of 35 dollars on each one. They expect to sell at least 100 lawn mowers this month and they need to make a profit of at least 3,675 dollars on them. If they want to order at least one type of each lawn mower, how many should they order if they want to minimize their cost?

a) 70 -Type A
    30 -Type B

b) 35 - Type A
    65 - Type B

c) 65 - Type A
    35 - Type B

d) 30 - Type A
    70 - Type B

[hide="solution"] They earn 40 dollars for each Type A mower and 35 dollars for each Type B mower, and they want to make 3675 dollars.  So we have:
$ 40a\plus{}35b\equal{}3675$
where a is the number of Type A mowers and b is the number of Type B mowers.

We also know that they expect to sell 100 lawn mowers, which leads to:
$ a\plus{}b\equal{}100$
Solving the system of equations gives us a=35, b=65.  The answer is [b]B[/b].[/hide]


For each positive integer n, let k(n) demote the smallest number with the property

For any nxn lower triangular matrix A with nonzero diagonal entries, there exist k(n) elementary matrices Esub1, Esub2,... Esubk(n) such that

Esub1 Esub2 ... Esubk(n) A = Isubn

prove that k(n)< or = [n(n+1)]/2

hint: this question asks you to give an upper bound for how many elementary matrices are needed to invert an invertible lower triangular nxn matrix.

Describe the algorithm for solving a lower triangular system.  How many steps does it take?

can you please show me the beginning steps? i just couldn't figure out how to start it. :( And i have no idea the outcome of it will be k(n)< or = [n(n+1)]/2 even i tried very hard to think about the process of doing Lower triangular system.

Thank you very much.

whenever you see $ n(n\plus{}1)/2$ you should think of $ 1\plus{}2\plus{}\ldots\plus{}n$.

start with the $ n$th row.  how many elementary row operations does it take to make the $ n$th row of $ A$ agree with the $ n$th row of $ I_n$?  Call the resulting matrix $ A_1$

now how many elementary row operations, at most, does it take to make the $ n\minus{}1$th row of $ A_1$ agree with the $ n\minus{}1$th row of $ I_n$?

In general, how many nonzero entries are there, at most, in the $ n\minus{}k$th row of $ A$?


How many people have acne on AoPS, just curious.

You should've put a "kinda" catogory... With only 2 or 3 pimples on my face its kinda hard to say.

I do, but I still manage to look fabulous ;)

[quote="guineapiggies"]ummm occasionally. i mean have u every met someone that has NEVER had acne in their whole entire teenage life?[/quote]
Yes, I have.

Yes, and I HATE HATE HATE IT :mad:  :mad:  :mad:

[quote="Carnot"]yeah some hot girls just have [i]none at all[/i] (or maybe make-up covers up the minuscule amount but for practical purposes who cares)[/quote]

Actually, makeup works wonders. I'm just too manly to use it.

But like, if I were to, you would barely see anything. Have you ever seen those real-life pictures of some celebs and they have ridiculous acne but you don't see it in the movies?

[quote="RussianRocket"]Just use Benzoyl Peroxide whenever a pimple comes up[/quote]

benzoyl peroxide works wonders :D

[quote="crazychinesemaniac"]yea..my sister had this horrible acne that was impossible to treat. She eventually had to take this really strong prescription treatment that made her tired and pale. She said it was worth it though.[/quote]

Was it Acutane, or some other form of isotretinoin? If so, I'm interested in more details.

I have horrendous acne that nothing has succeeded in treating. For the last few years, it's really died down, so its quite sparse, but I can't kill it off. It just won't go away. Like an effing zombie. I'm wondering if I may need something like that, or if time will take care of it.

RussianRocket wrote: [quote]
Just use Benzoyl Peroxide whenever a pimple comes up

guineapiggies wrote: [quote]
benzoyl peroxide works wonders[/quote][/quote]

haha yay benzoyl peroxide! That's what I use. :)

benzoyl peroxide, huh? i've gotta try... SO MUCH ACNE!!!!! (what's worse, i have a habit of picking at it, too. :( )

Benzoyl peroxide is basically Proactiv... although Proactiv has some other stuff in it too


geometry
incenter
circumcircle
trapezoid
complex numbers
geometry proposed

Let $K$ and $L$ be two points on the arcs $AB$ and $BC$ of the circumcircle of a triangle $ABC$, respectively, such that $KL\parallel AC$. Show that the incenters of triangles $ABK$ and $CBL$ are equidistant from the midpoint of the arc $ABC$ of the circumcircle of triangle $ABC$.

Denote by $I_1$ and $I_2$ the inceters of $\triangle{AKB}$ and $\triangle{BLC}$, respectively.
Extend $BI_1$ and $BI_2$ to meet the circumcircle(for the second time) of $\triangle{ABC}$ at $X$ and $Y$, then obviously $XI_1=YI_2$.
Since $M$- the midpoint of $\widehat{ABC}$ is found on the perpendicular bisectors of $AC$, $XY$ and $KL$
we can conclude that $\triangle{MI_1X}=\triangle{MI_2Y}$ and consequently $MI_1=MI_2$.

I think that the two segments are equal to each other iff B is midpoint of arc ABC. If I am wrong, I'm sorry.

I too think that this equality holds only when $AB=BC$
Let us assume $YI_2=XI_1$
It is easy to observe that $YI_2=YB=YC$ and $XI_1=XA=XB \Longrightarrow XB=YC  \Longrightarrow BC=XY$
and $AX=YB \Longrightarrow AB=XY \Longrightarrow AB=BC$

I'll post 2 solutions.One elementary and one by complex bashing:
Let $I_1$ and $I_2$ be the incenters of triangles $BKA$ and $BLC$. Now we intersect $BI_1$ and $BI_2$ with the circumcircle and we denote those points as $X$ and $Y$ respectively.Note that $XACY$ is a trapezoid so $XA=YC$.Now, it is well-known that $X$ is the circumcenter of a triangle $AKI_1$ so $AX=XI_1$.By the same reasoning $YC=YI_2$ so we conclude that $XI_1=YI_2$. Now because $MX=MY$ (trivial) and $\angle BXM= \angle BYM$ we have that $\triangle MI_1X= \triangle MI_2Y$ so $MI_1=MI_2$.

Complex bash:
Asign points $A , B , C, K$ complex numbers: $a^2, b^2, c^2, k^2$. Because $KL \parallel AC$ we get $l^2=\frac{a^2c^2}{k^2}$.
$m=ac$.
$x=\frac{m-i_1}{m-i_2}= \frac{ac+ab+ak+bk}{ac+cb+cl+bl}=\frac{ac+ab+ak+bk}{ac+cb+\frac{ac}{k}(b+c)} =\frac{ac+ab+ak+bk}{\frac{c}{k}(ak+bk+ab+ac)}=\frac{k}{c}$.
Now because $\frac{m-i_1}{m-i_2}= \overline{(\frac{m-i_2}{m-i_1})}$ it follows $MI_1=MI_2$.

[b]S[/b]olved with [b]mathscrazy, everythingpi3141592[/b]

[hide = solution]
Let $X$ be the midpoint of arc $ABC$. Let $I_1, I_2$ be the incentres of triangles $BAK$ and $BCL$ respectively. Further, let $M_1, M_2$ be the midpoint of minor arcs $AK, CL$. Consider the spiral similarity mapping segment $M_1I_1$ to $M_2I_2$. Because $M_1I_1 = M_1B = CM_2 = M_2I_2$, it follows that this spiral similarity is in fact a rotation. Moreover the centre of this rotation lies on $\odot(AM_1M_2)$. As $XM_1 = XM_2$, it follows that the centre is $X$ and then we have $XI_1 = XI_2$ as well, as desired.
[/hide]


inequalities
algebra
polynomial
inequalities unsolved

Let $a,b,c>0$ and $ab+bc+ca=3$.
Prove that: $\sum \frac{a}{\sqrt{\frac{5}{4}a+b}}\ge 2$.
It's true or false :) ?

can anyone provide some cleaner solution without such calculations??

[quote="Ji Chen"]If $ x,y,z$ be nonnegative numbers such that $ yz \plus{} zx \plus{} xy \equal{} 3,$ then

$ \frac {x}{\sqrt {5x \plus{} 4 y}} \plus{} \frac {y}{\sqrt {5y \plus{} 4z}} \plus{} \frac {z}{\sqrt {5z \plus{} 4x}}\geq 1,$

with equality if and only if $ x \equal{} y \equal{} z \equal{} 1.$
[/quote]
We can use also the following Holder.
$\left(\sum_{cyc} \frac {x}{\sqrt {5x \plus{} 4 y}}\right)^2\sum_{cyc}x(5x+4y)(x+2y+4z)^3\geq\left(\sum_{cyc}(x^2+6xy)\right)^3$.
Thus, it remains to prove that $3\left(\sum_{cyc}(x^2+6xy)\right)^6\geq(xy+xz+yz)\left(\sum_{cyc}x(5x+4y)(x+2y+4z)^3\right)^2$,
which is true.

[quote=xzlbq]
xzlbq ,Automatic inequality discovery [/quote]

:lol:  yes!  you have the greatest number of inequalities.

[quote=xzlbq][quote=mudok]whose computer program is the best?

Ji Chen, szl6208, xzlbq or Grotex?[/quote]

都科：
Ji Chen is very famous。

 szl6208 be good at sos,as you see.

xzlbq ,Automatic inequality discovery,

and Grotex is The bright younger generation。[/quote]

how can we get them?

[quote=mudok]whose computer program is the best?

Ji Chen, szl6208, xzlbq or Grotex?[/quote]

we needn't their tools at all

[quote=xlzq][quote=xzlbq][quote=mudok]whose computer program is the best?

Ji Chen, szl6208, xzlbq or Grotex?[/quote]

都科：
Ji Chen is very famous。

 szl6208 be good at sos,as you see.

xzlbq ,Automatic inequality discovery,

and Grotex is The bright younger generation。[/quote]

Their software can only solve trivial polynomial inequalities. xlzq doesn't even  need software and can beat all of these guys with sw easily.[/quote]
I believe that，

I remember your problems,are very very hard!
as

[quote=xlzq]$a,b,c >0$ and $a+b+c=1$,  prove that
\[ 2\geqslant a^{k_ob^2}+b^{k_oc^2}+c^{k_oa^2}\]
where $k_o=9 \left( \frac{ln3-ln2}{ln3} \right) \approx 3.32163$

It is a little bit more challenge than a trivial one posted [url=http://math.stackexchange.com/questions/2176571/if-ab-1-so-a4b2b4a2-leq1]here[/url][/quote]

是骡子是马拉出来溜,你做好准备，
（You're ready for the test....）

[quote=xlzq][quote=xzlbq][quote=mudok]whose computer program is the best?

Ji Chen, szl6208, xzlbq or Grotex?[/quote]

都科：
Ji Chen is very famous。

 szl6208 be good at sos,as you see.

xzlbq ,Automatic inequality discovery,

and Grotex is The bright younger generation。[/quote]

Their software can only solve trivial polynomial inequalities. xlzq doesn't even  need software and can beat all of these guys with sw easily.[/quote]

Are you crazy?


geometry
3D geometry
ratio
geometry unsolved

Consider a cylinder and a cone with a common base such that the volume of the
part of the cylinder enclosed in the cone equals the volume of the part of the cylinder outside the cone. Find the ratio of the height of the cone to the height of the cylinder.

Coz the small cone is similar to the large one, so the ratio of the height of these two cones is $ 1: \sqrt[3]{2}$.
So the required ratio is $ \sqrt[3]{2}: \sqrt[3]{2}\minus{}1$.


$ a,b,c>0$

prove that !

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

[quote="bigbang195"]$ a,b,c > 0$

prove that !

$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ac}$[/quote]

$ \Longleftrightarrow \sum_{cyc} (a^4c^2 \minus{} a^3b^2c)\geq 0$.

$ \Longleftrightarrow (c^2a^2 \plus{} ( \minus{} cb^2 \plus{} c^3 \plus{} bc^2)a \plus{} b^4 \plus{} c^4 \minus{} b^3c)(a \minus{} b)(a \minus{} c)$ $ \plus{} (b^2 \plus{} bc \plus{} c^2)(ab \plus{} ca \minus{} bc)(b \minus{} c)^2\geq 0$

assume $ a \equal{} max{(a,b,c)}$

we can prove

$ ab \plus{} ca \minus{} bc\geq 0$

$ f(a) \equal{} c^2a^2 \plus{} ( \minus{} cb^2 \plus{} c^3 \plus{} bc^2)a \plus{} b^4 \plus{} c^4 \minus{} b^3c\geq 0$.

$ (\Delta \equal{} \minus{} c^2(3c^2 \minus{} 5bc \plus{} 3b^2)(b^2 \plus{} bc \plus{} c^2)\le 0)$

so.........

 :lol:

[quote="frankvista"][quote="hedeng123"][quote="bigbang195"]$ a,b,c > 0$

prove that !

$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ac}$[/quote]

$ \Longleftrightarrow \sum_{cyc} (a^4c^2 \minus{} a^3b^2c)\geq 0$.

[/quote]

By AM-GM inequality,

$ \sum{a^4c^2} \equal{} \frac {1}{2} \sum{(a^4c^2 \plus{} b^4a^2)} \ge \sum{a^3b^2c}$.

We've done. #[/quote]
 :blush:

[quote="hedeng123"][/quote][quote="frankvista"][quote="hedeng123"][quote="bigbang195"]$ a,b,c > 0$

prove that !

$ \frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{a} \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ac}$[/quote]

$ \Longleftrightarrow \sum_{cyc} (a^4c^2 - a^3b^2c)\geq 0$.

[/quote]

By AM-GM inequality,

$ \sum{a^4c^2} = \frac {1}{2} \sum{(a^4c^2 + b^4a^2)} \ge \sum{a^3b^2c}$.

We've done. #[/quote][quote="hedeng123"][/quote]

:blush:

[hide="点"]我知道均值不等式可以做，没想到这么简单[/hide]

[quote="bigbang195"]$ a,b,c > 0$

prove that !

$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ac}$[/quote]

you are very good!

please see nice and strong:

$ \frac{a^3}{b}\plus{}\frac{b^3}{c}\plus{}\frac{c^3}{a}\geq (a^2\plus{}b^2\plus{}c^2)\frac{a^3\plus{}b^3\plus{}c^3}{a^2b\plus{}b^2c\plus{}c^2a}.$
BQ

[quote="xzlbq"][quote="bigbang195"]$ a,b,c > 0$

prove that !

$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ac}$[/quote]

you are very good!

please see nice and strong:

$ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq (a^2 \plus{} b^2 \plus{} c^2)\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2b \plus{} b^2c \plus{} c^2a}.$
BQ[/quote]

try $ a \equal{} \frac {1}{11},b \equal{} \frac {1}{71},c \equal{} 1$

 :lol:


lim(2arctanx/pi)^x    x tends to +inf
lim(e^n-(1+1/n)^n^2) n tends to inf

is this what you meant?
$ \lim_{x \to \infty}(2\frac {arctan(x)}{\pi})^x$
$ \lim_{n \to \infty}(e^n \minus{} (1 \plus{} \frac {1}{n})^{n^2})$

Sorry
[(2arctanx)/pi]^x
 sorry not (1-1/n),but (1+1/n),i typed wrongly.

The first:
the limit equals
limxtoinf e^((ln((2arctanx)/pi))/1/x)=
e^limxtoinf ((ln((2arctanx)/pi))/1/x),

Using L'HOPITAL yields

e^limxtoinf (pi/(2arctanx))(2/pi)(1/(1+x^2))/(-1/x^2)=
e^(-2/pi)

$ L \equal{} \lim_{x \to \infty}\left(\frac{2 \arctan x}{\pi}\right)^x \equal{} e^A$, where

$ A \equal{} \lim_{x \to \infty} x\left(\frac{2 \arctan x}{\pi} \minus{} 1\right) \equal{} \frac{1}{\pi} \lim_{x \to \infty} \frac{2 \arctan x \minus{} \pi}{\frac{1}{x}} \equal{} \frac{1}{\pi} \underbrace{\lim_{x \to \infty} \minus{} \frac{2x^2}{x^2 \plus{} 1}}_{using L'Hopital}$

$ \equal{} \minus{} \frac{2}{\pi}$, and so $ L \equal{} e^{\minus{}2/\pi}$.

is this correct for the second one?

$ ln\left(\left(1\plus{}\frac{1}{n}\right)^n\right)\equal{}\sum_{k\equal{}1}^{\infty}\frac{\minus{}1^k}{\minus{}kn^{k\minus{}1}}$
$ \left(1\plus{}\frac{1}{n}\right)^n\approx e^{1\minus{}\frac{1}{2n}}$
$ \left(1\plus{}\frac{1}{n}\right)^{n^2}\approx e^{n\minus{}.5}$
$ e^n\minus{}\left(1\plus{}\frac{1}{n}\right)^{n^2}\approx e^{n\minus{}.5}(\sqrt{e}\minus{}1)$
which diverges


The orthocentre of the triangle $ ABC$ is $ H$, and its incircle, centred at $ I$, touches the sides $ AC$ and $ BC$ at the points $ P$ and $ Q$. Prove that if $ H$ lies on the line $ PQ$ then the line $ HI$ passes through the midpoint $ K$ of the side $ AB$.

let $ BH \cap AC =E$ and $ AH \cap BC=D$
by simple agle chasing we observe that $ PQ$ bisects $ \angle DHB$ and $ \angle EHA$ also quad $ ABDE$ is cyclic
therefore, $ \frac{EP}{PA}= \frac{EH}{HA}= \frac{DH}{HB}= \frac{DQ}{QB} =\gamma$
now extend $ HI$ to a point K such that  $ \frac{HI}{IK}= \gamma$ but, as $ IP \parallel HE$ and $ IQ \parallel HD$ we get (by basic proportionality theorem) $ KA$ is perpendicular to $ AC$ and $ KB$ is perpendicular to $ AB$
now drop perpendiculars to sides $ AC$ and $ BC$ from midpoint of $ AB=M$ at $ S$ and $ T$ respectively
by midpoint theorem
as $ SM \parallel BE$ we get, $ ES=SB$
as $ TM \parallel AD$ we get, $ DT=TB$
join $ HM$ and extend to $ K'$ such that $ HM=MK'$
therefore by midpoint theorem
 $ K'A$ is perpendicular to $ AC$ and $ K'B$ is perpendicular to $ AB$
$ \Longrightarrow K=K'$
$ \Longrightarrow H ,I ,M , K$ are colinear
QED

gauravpatil , could u please tell me what
$ XY\cap AB \equal{}L$  means??
Sorry for such a stupid question.

[quote="Aravind Srinivas L"]gauravpatil , could u please tell me what
$ XY\cap AB \equal{} L$  means??
Sorry for such a stupid question.[/quote]

$ XY$ and $ AB$ intersect at $ L$  :)

thanks Indybar.
Nice soln Gaurav INMO topper :P


geometry Princeton college MIT Harvard Stanford Yale

Being an undergrad, choosing graduate program is very hard.

I am into algebraic geometry, category theory, k-theory, and algebraic topology... sort of thing.

Could you name some top programs in these areas?
Could you name some schools that I should never go to if I have these interests? (for example, NYU)?

I haven't met my profs for a while (currently on one-year break), so hopefully someone here will guide me.

So far I am looking to apply to only top programs  :maybe: , and if that doesn't work out I will probably get master's and try again.

Schools that I am looking at now are Princeton, MIT, Harvard, Chicago, Berkeley, Stanford, Yale, Columbia, Michigan, and Toronto. I did most of the research by going to the websites, looking at faculty list and funding, and also USNEWS rankings (are these respected? I especially looked at algebra/number theory/algebraic geometry sub-ranking and also topology sub-ranking). If you want these rankings for yourself, you can PM me.

I got this answer from another website. I will put it here so other people can have a look.

[quote]Right now...

Harvard/MIT are best all around choices.

U Penn has some really good people (but I've never liked the enviornment).

Toronto recently acquired some really good people so its definitely worth looking at.

Now, if you are absolutely positive that you are looking into the Algebraic Geometry/Topology route, consider (some of these will seem surprising):

Columbia, SUNY-Stony Brook, University of Maryland-College Park, Northwestern, University of Chicago, UC San Diego (more Algebraic Geometry), Berkeley, UC Davis, Princeton, Michigan State University, UIUC (more Category Theory). Ohio state is developing a pretty strong geometry program.

Harvard, Suny-SB, Maryland, Toronto, Berkeley and Princeton each have at least one excellent potential advisor (i.e., very active in the field, producing good work, wide knowledge base, well respected and still accepting "average" students) in the field.[/quote]


inequalities
rearrangement inequality
inequalities proposed

Let $ a,b,c\geq 0$, s.t. $ a\plus{}b\plus{}c\equal{}3$. Prove that:
$ (a^2b\plus{}b^2c\plus{}c^2a)(ab\plus{}bc\plus{}ca)\le 9$.

P.S.: could someone explain me the solution here (see post #4):
http://www.mathlinks.ro/viewtopic.php?t=228534
Thank you very much.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ (a^2b \plus{} b^2c \plus{} c^2a)(ab \plus{} bc \plus{} ca)\le 9$.
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ t\geq1$ and $ (a^2b \plus{} b^2c \plus{} c^2a)(ab \plus{} bc \plus{} ca)\le 9\Leftrightarrow$
$ \Leftrightarrow3u^5\geq(a^2b \plus{} b^2c \plus{} c^2a)v^2\Leftrightarrow$
$ \Leftrightarrow6u^5 \minus{} v^2\sum_{cyc}(a^2b \plus{} a^2c)\geq v^2\sum_{cyc}(a^2b \minus{} a^2c)\Leftrightarrow$
$ \Leftrightarrow6u^5 \minus{} 9uv^4 \plus{} 3v^2w^3\geq(a \minus{} b)(a \minus{} c)(b \minus{} c)v^2.$
$ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\geq3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Hence, $ 2u^5 \minus{} 3uv^4 \plus{} v^2w^3\geq2u^5 \minus{} 3uv^4 \plus{} v^2\left(3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\geq0$ because
$ 2u^5 \minus{} 3uv^4 \plus{} v^2\left(3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow u^5 \minus{} u^3v^2\geq v^2\sqrt {(u^2 \minus{} v^2)^3}\Leftrightarrow t^3 \minus{} t \plus{} 1\geq0,$ which is true.
Hence, $ 6u^5 \minus{} 9uv^4 \plus{} 3v^2w^3\geq0$ and enough to prove that
$ (6u^5 \minus{} 9uv^4 \plus{} 3v^2w^3)^2\geq v^4(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2.$
Since, $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2 \equal{} 27(3u^2v^4 \minus{} 4v^6 \plus{} 6uv^2w^3 \minus{} 4u^3w^3 \minus{} w^6)$
we obtain $ (6u^5 \minus{} 9uv^4 \plus{} 3v^2w^3)^2\geq v^4(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\Leftrightarrow$
$ \Leftrightarrow v^4w^6 \plus{} uv^2(u^4 \plus{} 3u^2 \minus{} 6v^4)w^3 \plus{} u^{10} \minus{} 3u^6v^4 \plus{} 3v^{10}\geq0.$
Id est, it remains to prove that $ u^2v^4(u^4 \plus{} 3u^2 \minus{} 6v^4)^2 \minus{} 4v^4(u^{10} \minus{} 3u^6v^4 \plus{} 3v^{10})\leq0.$
But $ u^2v^4(u^4 \plus{} 3u^2 \minus{} 6v^4)^2 \minus{} 4v^4(u^{10} \minus{} 3u^6v^4 \plus{} 3v^{10})\leq0\Leftrightarrow$
$ \Leftrightarrow t(t^2 \plus{} 3t \minus{} 6)^2 \minus{} 4(t^5 \minus{} 3t^3 \plus{} 3)\leq0\Leftrightarrow(t \minus{} 1)^2(t^3 \minus{} 4t \plus{} 4)\geq0,$ which is true.
Done! :)

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ (a^2b \plus{} b^2c \plus{} c^2a)(ab \plus{} bc \plus{} ca)\le 9$.
[/quote]

Let $ \{x,y,z\}\equal{}\{a,b,c\}$ such that $ x\ge y\ge z$. By the Rearrangement Inequality we have
$ a^2b \plus{} b^2c \plus{} c^2a\equal{} a(ab)\plus{}b(bc)\plus{}c(ca)\le x(xy)\plus{}y(zx)\plus{}z(yz)\equal{}y(x^2\plus{}xz\plus{}z^2)$
Using AM-GM Inequality we get
$ (xy\plus{}yz\plus{}zx)y(x^2\plus{}xz\plus{}z^2)\le y\frac{(xy\plus{}yz\plus{}zx\plus{}x^2\plus{}xz\plus{}z^2)^2}{4}\equal{}$
$ \equal{}\frac1{4}y(x\plus{}z)^2(x\plus{}y\plus{}z)^2\equal{}\frac9{8}.2y.(z\plus{}x).(z\plus{}x)\le\frac9{8}.[\frac{2(x\plus{}y\plus{}z)}{3}]^3\equal{}9$
we have done !

@arqady: Your proof is long but I like it. My congratulations for your patience ! :lol:

[quote="arqady"]Very nice proof, dduclam!  :lol:
But it does not always exist.
Try to find a nice proof for the following inequality ( my gift for you ). I am sure that it does not exist although, this inequality  seems nice.

Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
:P[/quote]
I can prove easily the weaker by Am-Gm:
$ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [11]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$
My friend, I think the nice proofs always exist. I'll try to find with $ k\equal{}10$
 :)

[quote="arqady"]Very nice proof, dduclam!  :lol:
But it does not always exist.
Try to find a nice proof for the following inequality ( my gift for you ). I am sure that it does not exist although, this inequality  seems nice.

Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
:P[/quote]
I also killed $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}$
And I will try your problem ,arqady

[quote="arqady"][quote="Allnames"]
I also killed $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}$
[/quote]
The maximal result here is $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [5]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}.$ It's Vasc's inequality.[/quote]
But $ a^3\plus{}b^3\plus{}c^3\ge a^2\plus{}b^2\plus{}c^2$ :maybe:
I have just found a proof for your problem in pvthuan's book,It í also used pqr technique,and of course,not nice

[quote="Allnames"][quote="arqady"][quote="Allnames"]
I also killed $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}$
[/quote]
The maximal result here is $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [5]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}.$ It's Vasc's inequality.[/quote]
But $ a^3 \plus{} b^3 \plus{} c^3\ge a^2 \plus{} b^2 \plus{} c^2$ :maybe:
[/quote]
Of cause! But $ 5<10.$ :P

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ (a^2b \plus{} b^2c \plus{} c^2a)(ab \plus{} bc \plus{} ca)\le 9$.

P.S.: could someone explain me the solution here (see post #4):
http://www.mathlinks.ro/viewtopic.php?t=228534
Thank you very much.[/quote]
You can see
http://www.mathlinks.ro/Forum/viewtopic.php?t=146030

[quote="arqady"]Very nice proof, dduclam!  :lol:
But it does not always exist.
Try to find a nice proof for the following inequality ( my gift for you ). I am sure that it does not exist although, this inequality  seems nice.

Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
:P[/quote]
I think it exists, but don't find out yet, that's all  :P

Because in fact, the maximal result which you show also very easy with AM-GM:

[quote="arqady"][quote="Allnames"]
I also killed $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}$
[/quote]
The maximal result here is $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [5]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}.$ It's Vasc's inequality.[/quote]

$ (\sum a)^6\equal{}(\sum a^2\plus{}\sum bc\plus{}\sum bc)^3\ge 27(\sum a^2)(\sum bc)^2\ge 81(\sum a^2)abc(\sum a)$
$ \rightarrow \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [5]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}.$

[quote="nguoivn"]
I can prove easily the weaker by Am-Gm:
$ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [11]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$
[/quote]

I think somebody want to try  :)

For the collection.  :wink:  :)
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} 8.$ Prove that:
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [11]{\frac {a^2 \plus{} b^2 \plus{} c^2}{3}}
\]

[quote="arqady"]
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
:P[/quote]
The proof see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3107205#p3107205

[quote=arqady]Very nice proof, dduclam!  :lol:
But it does not always exist.
Try to find a nice proof for the following inequality ( my gift for you ). I am sure that it does not exist although, this inequality  seems nice.

Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
:P[/quote]

See here
[url]https://www.facebook.com/photo?fbid=2201540009983422&set=gm.2953521171602591[/url]


Let be $ a,b,c\ge 0$ such that $ ab\plus{}bc\plus{}ca\equal{}1$ . Prove that :

$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\ge \frac{5}{2}$

[quote="alex2008"]Let be $ a,b,c\ge 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$ . Prove that :

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\ge \frac {5}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31377
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101476


Prove that if a,b,c>0,then:
$ \sum \frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}$≥2

[quote="duongptnkvn"]Chebyshev inequality[/quote]
Can you explain more clearly?I don't think we can use Chebyshev to show this ineq.

By multiplying out it equivalents to $ \sum a^5b \geq \sum a^4b^2$ which is obviously true by Muirhead :wink:


How do i do these questions?
1)How can the word DECISIONS be arranged so that the "N" is somewhere to the RIGHT of "D" (my problem is the two I's in particular)
ANSWER: 45360

2)Bob is about to hang 8 shirts in the wardrobe. He has 4 different styles of shirt, 2 identical ones of each particular style. How many different arrangements are possible if no 2 identical shirts are next to one another?
ANSWER: 864

3) A motorist travels through 8 sets of traffic lights, each of which is red or green. He is forced to stop at 3 sets of lights
a) In how many ways could this happen? ANSWER: 56
b) What other number of red lights would give an identical answer to part a)? ANSWER: 5

4) Find how many 5 digit numbers are possible from the digits 9,8,7,6,5,4,3,2,1,0 if digits are to be in
a) ascending order? ANSWER: 126
b) descending order? ANSWER: 252

Any help on any of these questions really appreciated !!

I'll answer your last question.In descending order , the answer is calculated by a ${10 \choose 5 }  = 126 $ .
In the ascending order , however , it is calculated by a $ {9 \choose 5 } = 252 $.
Here's the reasoning for that -- In the descending order , if a zero comes first , it can just be put as the last digit , whereas , you cant do that in the ascending order.....

I'll take care of number 3.  Since we go through 8 lights and 3 are red we have R, R, R, G, G, G, G, G in some order.  The way to arrange repeated items in sets is the total number factorial divided by how many times it repeats factorial.  In this case we get $8!/(3!5!)$ which equals 56.  Note for part b that this answer would be the same if we had 3 green lights and 3 red lights.

for number 1
since there are nine numbers, lets start off w/ 9P9, or 9!

since there are 2 S and 2 I (I think this is where you said your problem comes?) divide the expression by 2! (for the 2 S) and another 2! (for the 2 I)
now we have 9!/(2!2!)

half the time, N is to the right of D, beacause the other half of the time, N is to the left of D. so divide the expression by half again

9!/(2!2!) * 1/2 = 45360

guys, is there a link to permutations problems or some practice sets we can download?

Thanks

Why dont you search this forum ? I remember seeing a huge number of permutations and combinations here .....

Can someone please answer no.2, im not getting the answer

[hide="Solution Outline"]Find the total number of ways to order the shirts without the restriction that identical shirts are not adjacent, then use a massive amount of PIE bash to find the number of ways to arrange the shirts so that at least one pair of identical shirts are adjacent, then subtract the second value from the first. [/hide]

(Probably not the best solution. )


After noticing that several people could not solve this question, i decided to post it on the intermediate section of this forum

Here goes the question: what do you think the next 2 numbers in the following arrangement will be? 125, 135, 145, 155, _, _. to find the solution use your inductive reasoning skills. you base your response on what you think will happen. (the answer is not 165 and 175. it is 163 and 168. why?)

I have tried several things and am not going anywhere, any ideas?

There is, in fact, no reason, given only the first few terms of the sequence, why the next few should not be $165, 175$, as that is by far the simplest pattern that can be drawn from the sample data.  Any other answer is only a matter of preference, and I think it should be made clearer by assigning the next term $= 163$ and asking to find the next two terms after that, which is much less ambiguous  :D

However, given the parentheses, then the problem should really be "Find a rule for the sequence $125, 135, 145, 155, 163, 168$," since this is what the problem is really asking for.

Are you sure the rule for this sequence is math related?

yea pretty sure it is math related, any thoughts

simply put, your answer key is wrong  :lol:


MATHCOUNTS has posted the School and Chapter competitions.  You may now discuss any and all aspects of the competition, including scores, questions, solutions, and answers.

Thank you for your patience, as we awaited this moment!

The contests are posted [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=29]here[/url].

Is the MathCounts website down for everyone, or just me?

It should be working fine, is it something with your browser?

You should try Mozilla Firefox, gets less errors.

It seems to be always down the Saturday of states competitions.


number theory
least common multiple
modular arithmetic

[quote="mgao"]What is the smallest number that is divisible by 13 but when divided by the whole nmbers 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 yields a remainder of 1?

$?r1$
$13)\overline{abcdef...}$[/quote]
[hide]The lcm of 2-12 is 27720. So the number is 1 + a factor 27720. 27720 has a remainder of 4. Triple it to get one with a remainder of twelve and add one to that to get [u][b]83161[/b][/u][/hide]

[hide]LCM(2,3,...11,12)=27720
which implies that 27721 is the smallest number giving a remaider of 1 when divided by 1-12.

$27721\equiv 5\equiv -8\pmod{13}$
$27720\equiv 4 \pmod{13}$

we want to add multiples of 27720 to 27721 (since adding 27720 won't change the fact that it is $\equiv 1 \pmod{1-12}$) until it is divisible by 13. $-8+2*4\equiv 0\pmod{13}$, so we add two of them. 27721+2*27720=$\boxed{83161}$.[/hide]


$\text{Abel Summation: }$

Let $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ be two finite sequences of numbers.  Then \[ a_1b_1+a_2b_2+\cdots+a_nb_n = (a_1-a_2)b_1 + (a_2 - a_3)(b_1+b_2)+\cdots + (a_{n-1} - a_n)(b_1+b_2 + \cdots+ b_n)+a_n(b_1+b_2+\cdots+b_n). \]

[hide="1"]
square both sides, nested roots cancel, isolate root, square, squared terms cancel, $\boxed{x=26}$. Kinda easy... I'll get to the others later..[/hide]

[quote="chess64"][quote="PenguinIntegral"][quote="RiemannZetaFunction"][hide="2"]$\frac{n}{3^{n}}$[/hide][/quote]
Proof please.[/quote]

Just multiply it by 3 and do what you usually do to find the sum of geometric series.[/quote]
Multiply the sequence by 3? Then you get 1, 2/3, 1/3, which isn't geometric (I think).

I have no clue where $n$ got in his answer...

[hide]
Define $S=\frac 13 +\frac 29+\frac 3{27}+\frac 4{81}\cdots$.  So then $\frac 13 S = \frac 19 + \frac 2{27}+\frac 3{81}.$  Thus \[ S-\frac 13 S=\frac 13 + \frac 19 + \frac 1{27}+\cdots= \frac 13\cdot \frac 1{1-\frac 13}=12=\frac 23S \Rightarrow S= \frac 34. \][/hide]

[quote="PenguinIntegral"][quote="chess64"]
Just multiply it by 3 and do what you usually do to find the sum of geometric series.[/quote]
Multiply the sequence by 3? Then you get 1, 2/3, 1/3, which isn't geometric (I think).[/quote]

He means calculate $3S - S$.  :)

Edit:  [hide="4"] This problem and others like it have been posted many, many times before.  Let $\tau$ be a non-unity fifth root of unity.

$f(\tau) = 0$

Hence the remainder is $f( \tau^5 ) = f(1) = \boxed{5}$. [/hide]

Edit:  Incidentally, RiemannZetaFunction's answer to this problem is nonsense.  [hide="Generalization of 2"] Let $G(x) = \sum_{k=1}^{\infty} kx^k$.  Then

$x G(x) = \sum_{k=1}^{\infty} (k-1) x^{k}$

$G(x) - x G(x) = \sum_{k=1}^{\infty} x^k = \frac{1}{1 - x} - 1$

$\boxed{ G(x) = \frac{1 - (1 - x)}{ (1 - x)^2 } = \frac{x}{ (1 - x)^2 } }$. [/hide]

Edit:  Abel Summation is not cool.  :P [hide="3"] Let $G(x) = \sum_{k=1}^{n} kx^{k-1}$.  Then

$\int G(x) \, dx = \sum_{k=1}^{n} x^k = \frac{x^{n+1} - x}{x - 1}$

So that

$G(x) = \boxed{ \frac{ nx^{n+1} - (n+1)x^n + 1}{ (x-1)^2 } }$.  (I don't know if this has a neater form or not.) [/hide]

[quote="t0rajir0u"]

Edit:  [hide="4"] This problem and others like it have been posted many, many times before.  Let $\tau$ be a non-unity fifth root of unity.

$f(\tau) = 0$

Hence the remainder is $f( \tau^5 ) = f(1) = \boxed{5}$. [/hide]

[/quote]

the idea

[quote="t0rajir0u"][quote="SnowStorm"][b]Problem 253[/b]
Let $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$.  What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?[/quote]

[hide="Solution"]
$(x-1) g(x) = x^6 - 1$

So $g(x)$ has as its roots the sixth roots of unity.  That is, $g(x^{12}) = g(1)$ for all of these roots.  But $g(1) = \boxed{6}$.
[/hide]

[hide="Longer explanation"]
$g(x^{12}) = g(x) Q(x) + R(x)$, where $Q(x)$ is the quotient polynomial and $R(x)$ is the remainder polynomial which must have a degree of $< 5$.

We have established that $g(x)$ has as its roots the five non-unity sixth roots of unity.  Call them $\omega_1, \omega_2, ... \omega_5$.  For each of these roots we have $\omega_k^6 = 1$.  Because they are the roots of $g(x)$, obviously $g(\omega_k) = 0$.  Then

$g(\omega_k^{12}) = g(\omega_k) Q(\omega_k) + R(\omega_k)$
$g(1) = R(\omega_k)$
$6 = R(\omega_k)$

For five different values of $k$.  Because $R(x)$ must be $4^{th}$-degree or lower, this means

$R(x) = 6$

For all $x$.  Hence the remainder is $6$.  QED.
[/hide][/quote]

[quote="PenguinIntegral"][quote="kimby_102"][hide="#5"]
First let $\alpha=4, 16, 36, and 64$. Then our equ. becomes $\frac{x^2}{\alpha-1}+\frac{y^2}{\alpha-9}+\frac{z^2}{\alpha-25}+\frac{w^2}{\alpha-49}=1$.
Now, multiply $(\alpha-1)(\alpha-9)(\alpha-25)(\alpha-49)$ on both sides. Then u would get a fourth degree polynomial in terms of $\alpha$. Also the roots of this poly are $4,16,36,64$ or $(\alpha-4)(\alpha-16)(\alpha-36)(\alpha-64)=0$. The coefficients of the two polynomials should be equal. Thus $1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64$. The answer is $36$. Whew.....
[/hide][/quote]
I don't think you can set a variable to more than one value.[/quote]

She described the process she carried out very poorly.  She made the equation $\frac{x^2}{\alpha-1}+\frac{y^2}{\alpha-9}+\frac{z^2}{\alpha-25}+\frac{w^2}{\alpha-49}=1$.  From the given, this equation has roots at $\alpha = 4, 16, 36, 64$.  So it is the polynomial she claims, and then she equates coefficients.  This question is from an old AIME, if I recall -- a very tough one!

[quote="kimby_102"]Who is she?  :mad:

BTW, I'm sorry if my solution was bad. I was writing it on 1am.........[/quote]

Sorry!  It wasn't bad (actually quite nice) just your first step was poorly described.


My entire class of dedicated nerds (including our teacher) has been trying to solve this for weeks:

13, 14, 19, 5, 1

Target: 8

For those unfamiliar with Krypto, the idea is to add, subtract, multiply, and/or divide the first five numbers to reach the target number.

[quote="Hokkage"]Did you use a program? And if so, could you post the code? And if not, nice going, crazy-smart. How long did it take?[/quote]

I did use a program, but I didn't write it(I could have though, since I wrote a Maplet that solves 24 problem).
__________________________________________________________________________
EDITED: For the code for 24 problem, go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38886&postorder=asc&start=14]here[/url]. It's the first complex program I wrote, so it might be inefficient. It's in Maple(only language I know until I learnt java in September) so you might not be able to understand it.


induction
inequalities solved
inequalities

The sequence is defined as a recurrence relation:

(1) a_0 = 0, a_1 = 1 and a_2 = 1
(2) a_n+2 + a_n+1 = 2(a_n+1 + a_n)

a.) Prove that all elements of the sequence are perfect squares !
b.) Derive an explicit formula and prove it to be correct !

I guess the sequence is :
(2) a_(n+2) + a_(n+1) = 2(a_(n+1) + a_n)  isn't it?

cheers!

hmm.. I thought that [b] siuhochung [/b] thought that the sequence was like so:
 a_n+2 + a_n+1 = 2(a_n+1 + a_n)  but now that I also computed a_3 it is 3...  something is missing :D

There's some problem, with the recurrence given,
the general term is (1/3)(2^n-(-1)^n)

The recurrence becomes : a(n + 2) = a(n + 1) + 2a(n) and since a(1) = a(2) = 1, a(3) = 1 + 2.1 = 3. As far as I know 3 is no perfect square not yet :D So there seems to be something wrong with your table orl ! Are you sure this is the right recurrence ? In your table the squares of the Fibonacci numbers appear, but they don't satisfy your recurrence.

sorry for the typo:

(2) a_{n+2} + a_{n-1} = 2(a_{n+1} + a_{n})

Do you have a good site where one can get a description of the all the math symbols available in the forum ....

guess nobody took a look at this one after the correction was made! :(

here is what I saw

a_n=F_(n+1)^2,n>=1 where:

F_0=1, F_1=1 and F_(n+1)=F_n+F_(n-1)

and then induction to prove that we are right! :D

cheers!

part a.) is not too difficult and it can be solved independently from part b.) or try to derive an explicit formula and deduce the fact from it...

Well, the problem IS called "square of Fibonacci", so I guess it's not hard to see that the terms are squares of the Fibonacci terms! :D

Hi grobber,

I said derive !!! That's the square of the Binet formula is evident from my table and the problem's subject !!!  :D


http://www.cybernations.net/

Join Cyber Nations today!

And join the "Federated Allies for Cultural Exchange" alliance!
(our forums is http://www.face-cn.com/, just register there and post an application to join)

Cybernations is a very enjoyable text-based nation building game. On average at least one login per day is made.

Please join!
Also, my ruler name on Cybernations is "XRCatD"

On a side note, xscapezaer also plays Cyber Nations.

it seems interesting...

I might join.

I [i]might[/i].

yeah, most people who play LW start first with Cn.
i play cn too. i'm spaghetti and my nation is lanayru
unfortuantely im already in an alliance (random insanity alliance)
EDIT: would you like to tech trade? :)

Tip #1: Join an alliance QUICK if you haven't because there are alliance fishers out there.

I join and 6 or 7 hours later I have 8 offers to join 8 different clans.

darn, i didn't get you
hint: i'm an alliance fisher too.
and i got a lot more spam. random insanity alliance sent me a message in the first 20 seconds and by the first 5 minutes i had 17 invitations