function
induction
strong induction
Functional Equations

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $m,n\in \mathbb{N}$: \[f(f(m)+f(n))=m+n.\]

First, suppose $ f(a) \equal{} f(b) \equal{} c$. Then $ m \equal{} n \equal{} a$ and $ m \equal{} n \equal{} b$ gives $ f(2c) \equal{} 2a \equal{} 2b$, so $ a \equal{} b$ and $ f$ is injective.

Suppose we have $ f(a) \equal{} 0$. Then $ m \equal{} n \equal{} 0$ gives $ f(0) \equal{} 2a$. Now $ m \equal{} n \equal{} 0$ gives $ f(4a) \equal{} 0 \equal{} f(a)$. So $ a \equal{} 4a$ and $ a \equal{} 0$. Now plug $ m \equal{} n \equal{} 0$ in to get $ f(2f(0)) \equal{} 0$, so $ f(0) \equal{} 0$.

Plug $ m \equal{} 0$ in to get $ f(f(n)) \equal{} n$. Then let $ m \equal{} f(m)$. We get $ f(m\plus{}f(n)) \equal{} f(m)\plus{}n$. This equation is solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150759]here[/url], and both of its solutions also work for this one.

With $ \mathbb{N}$ we mean $ \{1,2,...\}$ so I'm afraid you can't use $ m \equal{} 0$ or $ n \equal{} 0$ here.  :noo:

There is a nice solution at http://www.kalva.demon.co.uk/short/soln/sh8819.html if you want.

Oops. Sorry about that. Injectivity still holds though, so I'll continue from that.

Consider any $ k\in\mathbb{N}$. Note that $ f(f(k)\plus{}f(k\plus{}2)) \equal{} f(2f(k\plus{}m))$, so $ f(k\plus{}2)\minus{}f(k\plus{}1) \equal{} f(k\plus{}1)\minus{}f(k)$. An easy induction allows us to conclude that $ f(2)\minus{}f(1) \equal{} f(k\plus{}1)\minus{}f(k)$ for all natural $ k$.

Now suppose that $ a < b$ and $ f(a) > f(b)$. Consider $ f(b)\minus{}f(b\minus{}1)\plus{}f(b\minus{}1)\minus{}f(b\minus{}2)\plus{}\ldots\plus{}f(a\plus{}1)\minus{}f(a)$. This allows us to conclude that $ f(b)\minus{}f(a) \equal{} (b\minus{}a)(f(2)\minus{}f(1))$. This also means that $ f(2) < f(1)$. But now we can conclude that $ f(k\plus{}1) < f(k)$, and this contradicts the well-ordering principle. Thus, $ f$ is nondecreasing.

Now we prove by strong induction that $ f(k) \equal{} k$. Start with the base case $ k \equal{} 1$. Notice that $ f(2f(1)) \equal{} 2$. If $ f(1) > 1$, then $ 2f(1) > 2$ contradicting $ f$ nondecreasing. So $ f(1) \equal{} 1$. Now suppose the hypothesis is true for $ k$. Let $ m \equal{} 1$ and $ n \equal{} k$, and we get $ f(k\plus{}1) \equal{} k\plus{}1$, completing the induction.

Hello! I'm a beginner at functional equations, so I'd like to ask you if the following solution is true.
First let $ a,b,c,d\in \mathbb{N}$ and we notice $ f\left(f\left(f\left(a\right) \plus{} f\left(b\right)\right) \plus{} f\left(f\left(c\right) \plus{} f\left(d\right)\right)\right) \equal{} f\left(a\right) \plus{} f\left(b\right) \plus{} f\left(c\right) \plus{} f\left(d\right)$, but $ f\left(f\left(a\right) \plus{} f\left(b\right)\right) \equal{} a \plus{} b$ and $ f\left(f\left(c\right) \plus{} f\left(d\right)\right) \equal{} c \plus{} d$ so $ f\left(a \plus{} b \plus{} c \plus{} d\right) \equal{} f\left(a\right) \plus{} f\left(b\right) \plus{} f\left(c\right) \plus{} f\left(d\right)$.
Setting $ a \equal{} b \equal{} c \equal{} d$ we get $ f\left(4a\right) \equal{} 4f\left(a\right)$.
Let's fix $ a \plus{} b \plus{} c \plus{} d$, $ a$ and $ b$. So $ f\left(a \plus{} b \plus{} c \plus{} d\right) \minus{} f\left(a\right) \minus{} f\left(b\right) \equal{} const$. Therefore $ f\left(c\right) \plus{} f\left(d\right) \equal{} const$ but also $ c \plus{} d \equal{} const$. So for any $ p,q,r,g\in \mathbb{N}$ from $ p \plus{} q \equal{} r \plus{} g$ follows $ f\left(p\right) \plus{} f\left(q\right) \equal{} f\left(r\right) \plus{} f\left(g\right)$.
Now we have $ f\left(1\right) \plus{} f\left(7\right) \equal{} f\left(4\right) \plus{} f\left(4\right)$, but $ f\left(4\right) \equal{} 4f\left(1\right)$ hence $ f\left(7\right) \equal{} 7f\left(1\right)$. We know also $ f\left(7\right) \equal{} f\left(2\right) \plus{} f\left(2\right) \plus{} f\left(2\right) \plus{} f\left(1\right)$ and since $ f\left(7\right) \equal{} 7f\left(1\right)$ $ f\left(2\right) \equal{} 2f\left(1\right)$.
Now we are going to prove by induction that $ f\left(n\right) \equal{} nf\left(1\right)$ for every $ n\in \mathbb{N}$.
For $ n \equal{} 1$, obvious!
Assume it's true for $ n \equal{} k$.
We are going to prove that it's true for $ n \equal{} k \plus{} 1$.
We have $ f\left(k\right) \plus{} f\left(3k \plus{} 1\right) \equal{} f\left(4k\right) \plus{} f\left(1\right)$. $ f\left(k\right) \equal{} kf\left(1\right)$, $ f\left(4k\right) \equal{} 4f\left(k\right) \equal{} 4kf\left(1\right)$, and we conclude $ f\left(3k \plus{} 1\right) \equal{} \left(3k \plus{} 1\right)f\left(1\right)$. Now $ f\left(3k \plus{} 1\right) \plus{} f\left(k \plus{} 1\right) \equal{} f\left(4k\right) \plus{} f\left(2\right)$ $ \Rightarrow$ $ f\left(k \plus{} 1\right) \equal{} \left(k \plus{} 1\right)f\left(1\right)$.
So the induction is finished and we proved $ f\left(n\right) \equal{} nf\left(1\right)$ for every $ n\in \mathbb{N}$.
Now we have $ f\left(f\left(m\right) \plus{} f\left(n\right)\right) \equal{} f\left(\left(m \plus{} n\right)f\left(1\right)\right) \equal{} \left(m \plus{} n\right)f^{2}\left(1\right)$. But $ f\left(f\left(m\right) \plus{} f\left(n\right)\right) \equal{} m \plus{} n$ therefore $ \left(m \plus{} n\right)f^{2}\left(1\right) \equal{} m \plus{} n$ hence $ f^{2}\left(1\right) \equal{} 1$ $ \Rightarrow$ $ f\left(1\right) \equal{} 1$. $ f\left(n\right) \equal{} nf\left(1\right)$ hence $ f\left(n\right) \equal{} n$ for every $ n\in \mathbb{N}$.

I think this problem can be solve shorter.By the equation,
we obtain f is monotone (which mean f(x)=f(y) if only if x=y)
let n=m+2 we have
f( f(m+2)+f(m) )=2m+2=f( 2 f(m+1))
because f is monotone we have
f(2) - f(1)=f(3) - f(2)=.......=f(n+1) - f(n) = d = constant for all n in N
and f(n) = (n-1)d + f(1)

[b]Lemma:[/b] $f$ is injective.
[i]Proof.[/i] If $f(m)=f(n),$ then $$m+k=f(f(m)+f(k))=f(f(n)+f(k))=n+k,$$ implying the claim. $\blacksquare$

Setting $m=n+2,$ yields that $f(n+2)+f(n)=2f(n+1).$ By induction, $f(n+1)-f(n)$ is constant. Clearly $f$ is linear. If $f(n)=an+b,$ then plugging in the original yields $a=1$ and $b=0,$ hence the identity function works.


Find the sum\[1+11+111+\cdots+\underbrace{111\ldots111}_{n\text{ digits}}.\]

Let $A=1+11+111+\cdots+\underbrace{111\ldots111}_{n\text{ digits}}$ \\
this expression can be written as $\sum_{i=0}^{n-1}(n-i)\cdot10^i = \frac{10^{n+1} -9n-10 }{81}$


I was just curious... How many people in the nation make the AIME annually?  I heard it was about 10,000, but I just wanted to check.

Like 13,300 or so last year.

http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2003-aa/03statsaime.html

BTW... How are things scored for the USAMO again? I forgot about that old post...

I'm pretty sure the USAMO index is AMC10 score+10*AIME score. So the max would be 300, if u got 150 on amc12 and 15 on aime. I'm also kinda sure u need about 7 or 8 aime to go on to the usamo if u took the amc 10.


If $ \sin x \equal{} 3 \cos x$, then what is $ \sin x \cos x$?

$ \textbf{(A)}\ \frac16 \qquad \textbf{(B)}\ \frac15 \qquad \textbf{(C)}\ \frac29 \qquad \textbf{(D)}\ \frac14 \qquad \textbf{(E)}\ \frac3{10}$

hello, from $ \sin(x)\equal{}3\cos(x)$ we get $ \tan(x)\equal{}3$ and we have $ \sin(\arctan(3))\cos(\arctan(3))\equal{}\frac{3}{10}$. The correct answer is $ \text{E}$.
Sonnhard.


If $ x,y$ are real then prove the following inequality
$ x^{2}\plus{}2xy\plus{}3y^{2}\minus{}6x\minus{}2y>\minus{}11$

The inequality must be $ x^2\plus{}2xy\plus{}3y^2\minus{}6x\minus{}2y\geq \minus{}11$

Let $ f(x)\equal{}x^2\plus{}2x(y\minus{}3)\plus{}3y^2\minus{}2y\plus{}11$

$ \Delta\equal{}b^2\minus{}4ac\equal{}\minus{}8(y\plus{}1)^2\leq 0, \ \forall y\in\mathbb{R}\Rightarrow f(x)\geq 0, \ \forall x\in\mathbb{R}$

[quote="RAGHAV GROVER"]If $ x,y$ are real then prove the following inequality
$ x^{2} \plus{} 2xy \plus{} 3y^{2} \minus{} 6x \minus{} 2y > \minus{} 11$[/quote]

$ x^{2} \plus{} 2xy \plus{} 3y^{2} \minus{} 6x \minus{} 2y\plus{}11 \equal{} (x\plus{}y\minus{}3)^2\plus{}2(y\plus{}1)^2$.


Where I can find some information about these inequalities:

$ \sqrt[k]{\frac{a_1\plus{}a_2\plus{}...\plus{}a_n}{n}}\ge\frac{\sqrt[k]{a_1}\plus{}\sqrt[k]{a_2}\plus{}...\plus{}\sqrt[k]{a_n}}{n}$

AND

$ \sqrt[k]{a_1a_2...a_n}\ge\cfrac{k}{\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n}}$ ?

 :)

There are entire books devoted to inequalities.  For information on some of the more well-known ones, see [url=http://en.wikipedia.org/wiki/Inequality]Inequality[/url]


geometry
3D geometry
tetrahedron
geometry unsolved

A tetrahedron is to be cut by three planes which form a parallelepiped whose three faces and all vertices lie on the surface of the tetrahedron.
(a) Can this be done so that the volume of the parallelepiped is at least $ \frac{9}{40}$ of the volume of the tetrahedron?
(b) Determine the common point of the three planes if the volume of the parallelepiped is $ \frac{11}{50}$ of the volume of the tetrahedron.

a) Answer is No.
b) ABCD is the tetrahedron with base ABC. M on DA, K on DB and N on KC and the plane passing through MKN is one of the three planes defined in the problem. Q is the intersection of the three planes. The answer is that Q is on MN such that MQ = 3AC/10.


geometry
ratio
3D geometry
tetrahedron
congruent triangles

Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$

[hide]This is hard to explain without a diagram.  After finding the correct tetrahedron, we can find the base which is an equilateral triangle with side $s\sqrt{2}$ where s is the side length of the cube.  That area is $\frac{s^2\sqrt{3}}{2}$.  The other three sides are congruent, so we only need to find one of the triangle.  One of these triangles has a base of $s\sqrt{2}$ and the height is $\sqrt{s^2+\frac{s^2}{2}}$ which is $\frac{s\sqrt{6}}2$.  So, after we multiply that by $s\sqrt{2}$, the area of one of the triangles is $\frac{s^2\sqrt{3}}{2}$.  After we multiply that by three (three congruent triangles) and add the base area, we get $2s^2\sqrt{3}$.  The surface area of a cube is simply $6s^2$, so $\frac{6s^2}{2s^2\sqrt{3}}$ simplifies to $\sqrt3$. [/hide]
As you can see, this is not a very good explanation.  I hope that I didn't make any mistakes during my explanation. :(

Reply: I think this is a very nice explanation. Just space things out a bit. :)
Hmm, is this abuse of moderation? If so, I'd better stop.  :D

[quote="breez"]Doesn't $\frac{6s^2}{2s^2\sqrt{3}}$ simplify to $\frac{3}{\sqrt{3}}$?[/quote]

Yes it does but $\frac{3}{\sqrt{3}}=\frac{3 \times \sqrt{3}}{3}=\sqrt{3}$.


I've always used:

$ \stackrel{\frown}{AB}$

to express arc AB.  There's got to be a nicer-looking way.  Help anyone?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 10[/url].


Now they hate me, Now they don't.
Sometimes they rush against me,
Sometimes they flow with me.
I am invisible.
I am everywhere at all times,
Nothing can touch me, even the wind and the ghosts.

Who or what is the "I" referred to in the riddle?

[quote="Ihatepie"][hide]time?[/hide][/quote]

Yes, that is correct.


I need to get more games for my TI-84. I currently have puzzpack and (with the Mirage OS) Centipede.

[quote="tjhance"]Cave
Dragonscape
Falldown
Mario
RType
Rubik's Cube
Rushhour
Slime
Tetris
Uncleworm
PuzzPack
Nim (I made this program myself, nobody else can beat it :D)
Castle (Short text-based game I made)[/quote]

wow, what kind of calc do you use?

I have a TI-84+, but I don't have mirage OS. I've programmed a few games, though, including:
snake
tic-tac-toe
reaction time
something kinda like the impossible quiz
a maze that takes up 3 screens
blackjack for 1 person
guess a number between 1 and 1000 :P
press enter 10 times as fast as you can :rotfl:

I also have some other random programs like a timer that counts down.
Right now I'm trying to program pong, but it's not going so well.

Last year in math all my friends played calc games all day. They would be out of their seats playing some tournament or something, and my awesome teacher didn't care and just taught class as if people were listening.  :rotfl:

[quote="7h3.D3m0n.117"]Hmm. In 6th grade I filled my ARC memory to the brim with games. I deleted most of them in like 7th.

I currently have Ancient Phoenix (for kicks to waste time in Geometry), Bomberkids (tournament between my friend and I), and some custom games I made myself.[/quote]

What about that Bob game, where it was multiple choice and problem solving stuff.

[quote="D3m0n Shad0w"]What about that Bob game, where it was multiple choice and problem solving stuff.[/quote]

Nah, 12K worth of ARC space for a game I can play blindfolded?

I program my own games on my TI-89 when I'm bored.

My creations:
1. Blackjack
2. Video Poker
3. A bad version of Minesweeper
4. Deal or No Deal (with a lousy offer algorithm)

[quote="7h3.D3m0n.117"]TI-89? I thought you couldn't program on Titaniums.[/quote]
Yes you can.

On the TI-84, I've made this puzzle game, which was basically get to the start to the finish, but you can't change directions while you're moving because there's ice or something (kind of like those Pokemon puzzles...), and you can only change directions after you get stopped by a wall or a sticky space.

Since I was limited to a 8x16 grid, I spiced it up by adding teleporters and walls that only appear after you walk over them once.


vector
linear algebra
matrix
calculus
limit

A  problem involving dominoe tiles leads to the following recursion formulas
\[u_{n+1}= 2v_{n}+u_{n},\quad v_{n+1}= u_{n+1}+v_{n}, \]
for $n\geq 0$, with $u_{0}= v_{0}= 1$. Find a formula for $v_{n}$.

[quote="Kent Merryfield"]If we substitute the the first equation into the second, we get this system:
\begin{eqnarray}u_{n+1}&=&u_{n}+2v_{n}\\ v_{n+1}&=&u_{n}+3v_{n}\end{eqnarray}
Define the vector $w_{n}=\begin{bmatrix}u_{n}\\v_{n}\end{bmatrix}.$ We can write the recursion as
\[w_{n+1}=\begin{bmatrix}1&2\\ 1&3\end{bmatrix}w_{n}\]
And with that observation, I'll move the topic from calculus to linear algebra.
[hide="Hint."]Diagonalize[/hide][/quote]

Thank you. Is there any other way? And if using linear algebra, could you explain a bit more please?

We find the eigenvalues and eigenvectors of that $2\times 2$ matrix, which we call $A.$ The characteristic equation is $\lambda^{2}-4\lambda+1=0,$ which has roots $\lambda=2\pm\sqrt{3}.$

Let $\lambda_{1}=2+\sqrt{3}.$ The eigenvector corresponding to this may be written as $v_{1}=\begin{bmatrix}\sqrt{3}-1\\1\end{bmatrix}.$

That means that $Av_{1}=\lambda_{1}v_{1}.$

Let $\lambda_{2}=2-\sqrt{3}.$ The eigenvector corresponding to this may be written as $v_{2}=\begin{bmatrix}\sqrt{3}+1\\1\end{bmatrix}.$

That means that $Av_{2}=\lambda_{2}v_{2}.$

We have as the initial condition that $w_{0}=\begin{bmatrix}1\\1\end{bmatrix}.$ If we can find constants $c_{1}$ and $c_{2}$ such that $w_{0}=c_{1}v_{1}+c_{2}v_{2},$ then it will follow that $w_{n}=A^{n}w_{0}=c_{1}\lambda_{1}^{n}v_{1}+c_{2}\lambda_{2}^{n}v_{2}.$

Solving the appropriate linear equations for the coefficients $c_{1}$ and $c_{2},$ we arrive at

$\begin{bmatrix}u_{n}\\v_{n}\end{bmatrix}= \frac{\sqrt{3}}2\left(2+\sqrt{3}\right)^{2}\begin{bmatrix}\sqrt{3}-1\\1\end{bmatrix}+\left(1-\frac{\sqrt{3}}2\right) \left(2-\sqrt{3}\right)^{2}\begin{bmatrix}\sqrt{3}+1\\1\end{bmatrix}$

You can multiply all of that out to get specific formulas for $u_{n}$ and $v_{n}.$ Two things to note about the behavior of this for large $n.$ The first is that

$\lim_{n\to\infty}\frac{u_{n}}{v_{n}}=\sqrt{3}-1\approx .732.$

The second is that

$\lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}= \lim_{n\to\infty}\frac{v_{n+1}}{v_{n}}= 2+\sqrt{3}\approx 3.732.$


Solve for x where 0 <= x < 2pi:

tan(2x) * tan(4x) = 1

[quote="worthawholebean"][hide="possibly wrong solution"]So I can't actually solve for $x$ (exactly), but I can for $\tan x$. I did all the ugly algebra out and got that $\tan x\in \left\{2+\sqrt{3},2-\sqrt{3},-2+\sqrt{3},-2-\sqrt{3}\right\}$.[/hide][/quote]

[hide="actually u can solve for x exactly since"]
$\tan\left(\frac{5\pi}{12}\right)=2+\sqrt{3}$ and $\tan\left(\frac{\pi}{12}\right)=2-\sqrt{3}$[/hide]

[hide="alternatively..."]
It's equivalent to $\tan 2x=\cot 4x=\tan \left( \frac{\pi}{2}-4x\right)$. Then for some integer $n$, we get $2x=\frac{\pi}{2}-4x+\pi n$, so that $x=\frac{\pi}{12}+\frac{\pi}{6}n$. All such values of $n$ work [i]except[/i] those where $n\equiv 1\pmod{3}$ (that would make $\tan 2x$ undefined and $\tan 4x=0$). The solutions in the interval $[0,2\pi )$ are
\[\frac{\pi}{12},\frac{5\pi}{12},\frac{7\pi}{12},\frac{11\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12},\frac{19\pi}{12},\frac{23\pi}{12}\]
[/hide]


A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $ 25\%$ without altering the volume, by what percent must the height be decreased?

$ \textbf{(A)}\ 10 \qquad
\textbf{(B)}\ 25 \qquad
\textbf{(C)}\ 36 \qquad
\textbf{(D)}\ 50 \qquad
\textbf{(E)}\ 60$

[hide]
an increase of 25% is the same as multiplying by 5/4
that means the area of the base of the cylinder increases by
5/4*5/4 or 25/16
so the height needs to multiplied by 16/25, or a decrease of 36%
so C
[/hide]

[quote="#H34N1"]A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by 25% without altering the volume, by what percent must the height be decreased?
(A) 10 (B) 25 (C) 36 (D) 50 (E) 60[/quote]

[hide="solution"]Let the dimetions of the origional jar be $r$ and $h$.  The volume of the jar is $\pi r^{2}h$.

When the diameter is increased by 25%, the radius is also increased by 25% so $\pi r^{2}h=\pi (1.25 r)^{2}h_{1}$.

Solving for $h_{1}$, we get $h_{1}=\frac{16}{25}h$ which is 64% of th origional height.  100-64=[b]36[/b][/hide]


I WANNA JOIN!!!

FANTASYLOVER SIGNS UP!!!

but can u do unrated?? bc my rating is pretty high now...

I'll sign up but I would much rather play unrated. I think we should do unrated games.

[quote="mewto55555"]Then how will we do cds? Guest accounts?[/quote]

whats wrong about guest accounts?

i always play with super as guest account.

I'm not saying anything's wrong with guests. It was a suggestion.

some other ppl said they were interested in the other thread, like isabella and skald, i think.

Count me in, though I will fail immediately. :blush:

I am in.

Though I will fail against anyone but ************* (anynomous).

i linked the threads (or tried 2 :D)

meh.

i mite join.

idk.

WOAH WHAT THE . THERE ARE WAAAAAAAAAAAAAAAAAAAAAAY 2 MUCH  M3MM3RS. U SHUD MAKE IT A TOURNEY FOR PPLZ W/ 1350 AND UNDER. MAKE A ANOTHER TOURNAMENT FOR N00BS LIKE ME.

like who?

i dont mem  :P

and if u think i mem ash- thats just bad for u  :bomb:

but then again, Fantasy just might get lucky

I'm going to lose. Even if I was in "the groove", Fantasy pwns me.

Super's prediction:

FantasyLover 3, Superllamel 0

F(4-0)
F(4-0)
F(4-0)

FantasyLover 3, Superllamel 0

F(4-0)
F(4-0)
F(4-0)

Fantasy Wins!!!!

i highly doubt that, even though super predicted it..
unless super loses on purpose

lol

I don't have to try to lose, it will happen no matter what.

really, supers memory is undestructable...

SUPER DIDNT FORGET ANYTHING!!!

@vallon

yea, making a rule that neither one of us will like with one game left...

thats pretty good idea.

lol fantasy

btw, i haven't remembered everything

[hide="this message is only for fantasy"]ADMIT THAT UR BETTER THAN ME!!!!

and for those of you who looked and aren't fantasy:

it wasn't that important so  :harhar: [/hide]

=)
i no
i make wonderful rules don't i? :D
@super
i really believe u don't remember anything..... :roll:

[quote="vallon22"]let's make a rule that u have to play with ur actualy account (no using guests)
therefore, if super wants to lose, he'll have his rating die too :)[/quote] He's gonna play with his account anyways

[quote="guineapiggies"]*gasp* here goes joining my first tourney like izzy

COUNT MOI IN!!!  :D

(i'll die within 3 seconds)[/quote]

<sarcasm>

I know this is a really old post... but moi (if you watch the discovery channel, which is usually borimng but this time has a thing called [i]Deadliest Catch[/i]) is the guy on the crabbing ship who gets kicked out... i dont think he plays FTW. </sarcasm>

FantasyLover vs. superllamel

Game Number - Score - Winner

Game 1 - 4:3 - FantasyLover
Game 2 - 3:4 - superllamel
Game 3 - 1:4 - superllamel
Game 4 - 4:2 - FantasyLover
Game 5 - 3:4 - superllamel

superllamel wins.

meh expected result...

CONGRATULATIONS SUPER!!!

I didn't win yet Fantasy. I still need to beat you again.

You're going to win, so let's just get it over with.  :D

WITHOUT FURTHER ADO! THE FINAL RESULTS!!!!

We have a 4-way tie for 13th place between [b]Guineapiggies[/b] , [b]Thdanh90[/b] , [b]isabella2296[/b] ,and [b]kookamunga[/b]

We also have a 4-way tie for 9th place between [b]dannyhamtx[/b] , [b]ash-pokemon[/b] , [b]budi713[/b] , and [b]mathblitz[/b]

In a 3-way tie for 6th place is [b]skaldjfhrulez[/b] , [b]violin321[/b] , and [b]hurlder[/b]

In a 2-way tie for 4th place is [b]ernie[/b] and [b]brut3forc3[/b]

THE TOP 3!!!!

Coming in 3rd place: [b]Mewto55555!!!!!!![/b] :winner_third:

In second: After a neck to neck final that came down to the last question: [b]FantasyLover!!![/b]  :winner_second:

And, our grand champion: [b][size=200]SUPERLLAMEL!!![/size][/b] Congratulations super[size=59]-lame-o[/size] on your victory.

Note: This tournament is now officially over. Could someone please lock it to prevent spam from certain (AIME15) people?

For those who care: The first 8 or so pages of this thread are a relic of the time before tourney mania. When you could have 6 pages without any spam. When people would have games done [b]before[/b] the deadline. I miss those days. With CTC they can come again!


I only want to add a straight line and some of content in the first page, something like:

MathLinks for everyone © 2008
________________________________________________________________________________

Also, add some of color an increase its size.

Thanks

To draw a line use \rule[i][lift[/i]]{[i]width[/i]}{[i]heigh[/i]t} where
[i]lift[/i] is optional and moves the line up and down relative to the text (if omitted it will be on the text's baseline);
[i]width[/i] is the length of the line;
[i]height[/i] is how thick the line is.
To get colour add the color package \usepackage{color}
For example: $ \color{red}\rule{6cm}{0.3mm}\normalcolor$


The MATHCOUNTS International Airport had 738,114 departures in 2001. If these were distributed uniformly throughout each 24-hour period, what is the mean number of seconds between each departure time? Express your answer to the nearest whole number.

[hide]there are 31536000 seconds a year, or about .0234 departures each second. But we want seconds between departures, so we take 1/0.0234 or about 43 seconds. [/hide]

answer[hide]there are 31536000 seconds in a single year,  and the seconds ratio is bout .0234.  the question asks us about the seconds between departures, so we get 1/0.0234, or about 43 seconds.

PS u can get the original # by urself. [/hide]

Wait, I don't get the seconds ratio. Explain please.


what is the little phrase that appears right below your screename in a post?(such as "P versus NP" on mine or "hodge conjecture" on some others.

This are some of them...

New Member
P vs NP
Hodge Conjecture
Yang-Mills Theory
Poincaré Conjecture
Navier-Stokes Equations
Riemann Hypothesis
Birch and Swinnerton-Dyer Conjecture

I'm not sure if these are in order from least posts to greatest posts, but i'm sure until yang-mills theory.

Newmember
then 20 posts or something u get P vs. NP
50 posts Hodge Conjecture
100 posts yang-mills theory...

but don't get hanged up on this and try to post things that are not relevant to math and problem solving.


[quote="brad sue"]Hi ,
Please can I have some help with this demonstration:
Let a#0   ( #= different)
Show that if[b] a x b = a x c[/b] and [b] a.b=a.c[/b], then [b]b=c[/b] ( Hint: Cross both sides of the first equation with a)

thank you[/quote]

$ba=ac$
$a: b=a: c$
so,$b=c$


[quote="vo thanh van"]Let a,b,c>0 such that a+b+c=1.Prove that:
$\frac{4}{81(ab+bc+ca)}+abc \geq \frac{5}{27}$[/quote]
Simple SOS or Mixing Variables  :blush:  or after homogenize and assume that $a \geq\ b \geq\ c$ put $a=c+x$ and $b=c+y$ we are done ! :D


function
calculus
combinatorics unsolved
combinatorics

The partition problems are really easy to solve wth generating functions.
But I will put up (soon) an algebric sol.
Anyway, here's another one easy proved with GF.
x+2*y=n-1
2*x+3*y=n-2
.
.
.
n*x+(n+1)*y = 0.
Prove that the number sum of the number of solutions of each equation equals n+1.
:) :D

[quote]
The partition problems are really easy to solve wth generating functions
[/quote]

heh, you guys make it sound so easy.  any tips on how to get generating functions for some special paritions?

You can find this in Polya and Szego Problems and theorems in calculus Part I , just the first chapter.


[quote="myc"]how do you get from $ 12\sqrt{2-\sqrt{3}}$ to $ 6\sqrt{6}-6\sqrt{2}$?

thanks!!![/quote]
[hide="Solution"]We want to get rid of that radical inside of a radical.
Here's what kunny wrote:
$ 12\sqrt{2-\sqrt{3}}=6\cdot 2\sqrt{2-\sqrt{3}}=6\sqrt{4(2-\sqrt{3})}$

To get rid of the radical inside the radical, set:
$ \sqrt{a}-\sqrt{b}= \sqrt{4(2-\sqrt{3})}$
(for rational $ a,b$ with $ a>b$)
Square both sides
$ a+b-2\sqrt{ab}= 8-4\sqrt{3}$

Now, we can match up the rational parts together and the irrational parts together.

Hence, our system is:
$ a+b=8$
$ ab=12$

Inspection tells us that one solution is $ (a,b) = (6,2)$, and this is the only solution, since we assumed that $ a>b$.

So we know that:
$ \sqrt{6}-\sqrt{2}= \sqrt{4(2-\sqrt{3})}$

Therefore,

$ 6(\sqrt{6}-\sqrt{2}) = 6\sqrt{4(2-\sqrt{3})}= 12\sqrt{2-\sqrt{3}}$,

as desired.[/hide]

EDIT: Fixed error.

"Inspection tells us that one solution is $ (a,b) = (8,2)$, and this is the only solution, since we assumed that a>b"

Should be $ (a,b) = (6,2)$


Let$(x_n);(y_n);(z_n): x_0,y_0,z_0>0$and
$x_{n+1}=x_n+\frac{1}{y_n}$
$y_{n+1}=y_n+\frac{1}{z_n}$
$z_{n+1}=z_n+\frac{1}{x_n}$
Prove that $\exists lim_{n\rightarrow\infty}\frac{x_n}{\sqrt{n}}$
[u][b]a hard question[/b][/u]Find it. :)

You have "Prove that" and "Find it."

Because you don't have the limit set equal to anything, you must
only mean "find it."

Besides the fact that I don't want subjective words such as "hard," it wouldn't be "an hard."  It would be "a hard."

Anyway, regarding the former point, edit your post so it is clear what is to be done for the problem.


For any positive integer $x$, let $S(x)$ be the sum of the digits of $x$, and let $T(x)$ be $|S(x+2)-S(x)|.$  For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$  How many values $T(x)$ do not exceed 1999?

[hide]
For all $x$ with a unit's digit between 0 and 7, $T(x)$ should just be $2$.

For numbers that have the form $x=k\cdot 10^p -1$,
\begin{eqnarray*} T(x)&=&|(k+1)-(k-1+9p)|\\ &=&|2-9p| \end{eqnarray*}
Example: $k=3, p=4: x=29999, x+2=30001$
$T(x)=|(3+1)-(3-1+9(4))|=34$

This also applies to numbers of the form $x=k\cdot 10^p -2$:
\begin{eqnarray*} T(x)&=&|(k)-(k-1+9(p-1)+8)|\\ &=&|9p-2| \end{eqnarray*}
Also it shouldn't matter if $k$ is of n-digits..

So finally the problem is reduced down to finding the number of $a=|9p-2|$ that satisfy:
$0 < |9p-2| < 1999$

So there are $222+1=\boxed{223}$ values (counting "2" in the initial case) of $T(x)$ (?)...
[/hide]

[quote="Karth"]Sorry for reviving this thread, but can someone explain to me why the answer wouldn't be:

[hide]Infinity, because if $x = 99999999999997$, $T(99999999999997) = S(99999999999999) - S(99999999999997) = 2$. Likewise, this would work for $x = 99999999999996, 99999999999995, 99999999999994, etc.$. By adding an infinite amount of $9$s to the beginning, we can generate an infinite number of solutions. Please help, thanks  :) [/hide][/quote]

because we are counting how many "distinct" solutions less than 1999 there are, not how many possible solutions (not necessarily distinct).

[quote=vatatmaja]Sorry for the revive, but I don't understand xxreddevilzxx's explanation on why there aren't an infinite number of solutions...[/quote]

The question asks for the number of distinct values of $T(x)$ that does not exceed $1998$ not the number of distinct values of $x$. By the definition of the problem, the answer must be less than or equal to $1998$.

p11: If the number has unit digit 1-7, the value of $T(x)$ is just $2$. If the number has unit digit $8$, then upon testing values we can see the only possibilities are numbers $7 \pmod 9$, and the same thing for numbers with units digit $9$. The amount of numbers that are $7 \pmod 9$ not exceeding $1999$ is $\frac{1989}{9}+1 = 222$, and adding $1$ (because of $2$) gives $\boxed{223}$.

For this I just realized that we only needed to check the edge cases for powers of $10$:

$$|S(9)-S(7)| = |9-7| = 2$$

$$|S(10)-S(8)| = |1-8| = 7$$

$$|S(100)-S(98) = |1-17| = 16$$

$$|S(1000) - S(998)| = |1-26| = 25$$

We can see that the distinct values are determined when $x+2$ is a power of $10$ ($S(x+2) = 1$) and that $x$ is two less than a power of $2$ (say there are n $9$'s and one $8$)

Here, we can easily see that the distinct values are generated by $|1-(8+9n)| \le 1999$, which yields $222$ positive solutions and $n=0$. Our answer is $\boxed{223}$ solutions.


Hmm....  polls show Jay Nixon has a >10% lead over Hulshof...

In other news, Exit Polls will almost be released...

Not me. It was just a spam post... :rotfl:

The reason our governor will be Democrat is because the farmers obviously care EVEN LESS about the governor. :P

Farmers as n00bs...

But I am a farmer... :o :fool:

[b]Ferma[/b]t [b]Farme[/b]r. Rearrange bold part and add a letter.

Dude that's where he got his theorom. He thought of pythag's theorom but expanded it to hyper-cornfields.


function
real analysis
real analysis solved

f, g:[a,b]--> R  is derivable s.t. f' , g' are increasing positive
Prove that there is c \in )a,b(

((f(a)-f(b))/(b-a)).((g(a)-g(b))/(b-a)) = f '(c).g '(c)

Source Romania I don't the author

I think the author is Cristinel Mortici. I don't know exactly where I saw it before.

I think Rolle theorem but what is the function ?

It's obvious from the conditions that f' and g' are continuous. Let t(x) be the linear function whose graph passes through the points (a,f(a)) and (b,f(b)).

Let's assume f'(a)>=(f(b)-f(a))/(b-a). Then f'(x)>=f'(a)>=(f(b)-f(a))/(b-a) for all x>=a, so (f(x)-t(x))'>=0, so f(x)-t(x) is increasing, and since f(a)-t(a)=f(b)-t(b)=0, we get f(x)=t(x) for all x in [a,b]. We repeat this for g and we also do it for b (the other extremity of the interval [a,b]).

In the end we find that there are 2 possible cases:

(1)
f(x)=g(x)=t(x) and the conclusion follows quickly because f'(x)=(f(b)-f(a))/(b-a) and g'(x)=(g(b)-g(a))/(b-a) for all x.

(2)
f'(a)g'(a)<((f(a)-f(b))/(b-a)).((g(a)-g(b))/(b-a)) and
f'(b)g'(b)>((f(a)-f(b))/(b-a)).((g(a)-g(b))/(b-a)), so there must be a c in (a,b) s.t. f'(c)g'(c)=((f(a)-f(b))/(b-a)).((g(a)-g(b))/(b-a)) because the function f'(x)g'(x) is the product of 2 continuous functions, so it's continuous.

Hope I made myself understood.

[quote="grobber"]It's obvious from the conditions that f' and g' are continuous.[/quote]

Just a precision in Romania when  f is derivable does this imply
that f ' is continuous ?

f(a)g(a)<((f(a)-f(b))/(b-a)).((g(a)-g(b))/(b-a))

grobber is it f(a)g(a) or f'(a)g'(a) ?

Can you invente a problem for 3 functions

Let f,g,h :[a,b]-> R  derivable, s.t. f',g',h' increasing, positive

(f(a)-f(b))/(b-a).(g(a)-g(b))/(b-a).(h(a)-h(b))/(b-a)= something depending
f',g',h'

Yes Moubinool, I did mean f'(a)g'(a) and f'(b)g'(b), I'll edit the post in a jippy :D

I don't think that f derivable => f' is continuous. There is Darboux's thm which states that f' has Darboux's property, but that is weaker than continuity. When we also have f' continuous, we say that f is of class
C 1 (I think it's the same in English, because I've seen many members use this notation; isn't it the same in French?).

The problem is obviously true for n functions instead of 2.


Let $ ABC$ be a right angled triangle with right angle at $ A$. Let $ P$ be an arbitrary point on $ BC$ and $ Q,R$ be the feet of the perpendiculars from $ P$ to $ AB$ and $ AC$ respectively. Prove that \[ BP\cdot PC \equal{} BQ\cdot QA \plus{} AR\cdot RC\]

[asy]size(120);
pair A,B,C,P,Q,R;
A=origin;
B=(0,40);
C=(30,0);
P=(20,13.333);
Q=(0,13.333);
R=(20,0);
draw(A--B--C--cycle);
draw(P--Q);
draw(P--R);
label("A",A,SW);
label("B",B,NW);
label("C",C,SE);
label("P",P,NE);
label("Q",Q,W);
label("R",R,S);[/asy]

Note that this becomes the Pythagorean theorem when $ P$ is the midpoint of $ BC$.

Well if the proof didn't involve the pythagorean theorem then it would be a generalization. By the way, nice find  :coolspeak:


function
geometry
geometric transformation
real analysis
real analysis unsolved

Prove that function $ f : R^{2} \rightarrow R$ is continous function if we know that :
$ \forall_{x_{0}\in R} f(x_{0},y)$ is continous function and
$ \forall_{y_{0}\in R} f(x,y_{0})$ is continous and monotonic function.

By translation, etc., it is sufficient to prove continuity at (0,0), and we can assume that $ f(0,0)\equal{}0$.

Let $ \epsilon>0$. By continuity in x, there exists $ \delta_1>0$ such that $ |f(x,0)|<\epsilon/2$ whenever $ |x|\leqslant\delta_1$. By continuity in y, there exists $ \delta_2>0$ such that $ |f(\pm\delta_1,y)\minus{} f(\pm\delta_1,0)|<\epsilon/2$ whenever $ |y|\leqslant\delta_2$. Therefore $ |f(\pm\delta_1,y)|<\epsilon$ whenever $ |y|\leqslant\delta_2$. It follows from the monotonicity in x that $ |f(x,y)|<\epsilon$ whenever $ |x|\leqslant\delta_1$ and $ |y|\leqslant\delta_2$. Therefore $ f$ is continuous at (0,0).


[hide]
assume WLOG that $x\geq y\geq z$
lets prove that $\frac{x^{k+1}}{x^{k+1}+y^{k}+z^{k}}\geq \frac{y^{k+1}}{y^{k+1}+z^{k}+x^{k}}\geq \frac{z^{k+1}}{z^{k+1}+x^{k}+y^{k}}$and $z^{k+1}+y^{k}+x^{k}\geq y^{k+1}+x^{k}+z^{k}\geq x^{k+1}+z^{k}+y^{k}$
the first set of inequalities is easy and for the second one it's enough to prove one of them so
$y^{k+1}+x^{k}\geq x^{k+1}+y^{k}$
$(\frac{x}{y})^{k}\geq \frac{1-y}{1-x}$
from $z(x-y)\geq 0$ we have
$(x-y)(1-(x+y))\geq 0$
$x(1-x)\geq y(1-y)$
$\frac{1-y}{1-x}\leq \frac{x}{y}\leq (\frac{x}{y})^{k}$
so now we can chebyshev the initial inequality a couple of times.
$\sum_{cyc}\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}\geq \frac{x+y+z}{3}\sum_{cyc}\frac{x^{k+1}}{x^{k+1}+y^{k}+z^{k}}=\frac{1}{3}\sum_{cyc}\frac{x^{k+1}}{x^{k+1}+y^{k}+z^{k}}\frac{\sum_{cyc}(x^{k+1}+y^{k}+z^{k})}{\sum_{cyc}(x^{k+1}+y^{k}+z^{k})}\geq \frac{1}{3}*(3*\sum_{cyc}x^{k+1})\frac{1}{\sum_{cyc}x^{k+1}+2\sum_{cyc}x^{k}}\geq \sum_{cyc}x^{k+1}\frac{1}{\sum_{cyc}x^{k+1}+6\sum_{cyc}x^{k+1}}=\frac{1}{7}$
 :lol:
[/hide]

So from your proof, equality when \[\frac{x^{k+2}}{x^{k+1}+2x^{k}}=\frac{1}{21}\] at $x=\frac{1}{3}$
so \[=\frac{x^{k+2}}{\frac{7}{3}x^{k}}=\frac{1}{21}\] which is an identity is for all $k$, there is equality at $x=y=z=\frac{1}{3}$

:blush:
boy am i an idiot...
just glad i never forgot to finish a solution in a competition :)
thanks...

Just to give an example of your proof, can you show the steps when $k=1$?

not really sure what you mean because the steps are identical...
DON'T look at this unless you wan't a really big hint...
ok, the inequality for k=1 is this
$\sum_{cyc}\frac{x^{3}}{x^{2}+y+z}\geq \frac{1}{7}$ where $x+y+z=1$
[hide] first we have $3\sum_{cyc}x^{2}\geq (\sum_{cyc}x)^{2}=\sum_{cyc}x$
then WLOG assume $x\geq y\geq z$ then prove that $\frac{x^{2}}{x^{2}+y+z}\geq \frac{y^{2}}{y^{2}+z+x}\geq \frac{z^{2}}{z^{2}+x+y}$ and $\frac{1}{x^{2}+y+z}\geq \frac{1}{y^{2}+z+x}\geq \frac{1}{z^{2}+x+y}$
both are really easy to prove.now we have
$\sum_{cyc}\frac{x^{3}}{x^{2}+y+z}\geq \frac{x+y+z}{3}\sum_{cyc}\frac{x^{2}}{x^{2}+y+z}=\frac{1}{3}(\sum_{cyc}\frac{x^{2}}{x^{2}+y+z}\sum_{cyc}(x^{2}+y+z)) \frac{1}{\sum_{cyc}(x^{2}+y+z)}\geq \frac{1}{3}(3\sum_{cyc}x^{2})\frac{1}{\sum_{cyc}(x^{2}+y+z)}= \frac{\sum_{cyc}x^{2}}{\sum_{cyc}x^{2}+2\sum_{cyc}x}\geq \frac{\sum_{cyc}x^{2}}{\sum_{cyc}x^{2}+6\sum_{cyc}x^{2}}=\frac{1}{7}$
equality holds iff $x=y=z=\frac{1}{3}$ which is easiest to obtain from
$3\sum_{cyc}x^{2}\geq (\sum_{cyc}x)^{2}$
[/hide]

[quote=N.T.TUAN]Let $k$ be a given natural number. Prove that for any positive numbers $x; y; z$ with
the sum $1$ the following inequality holds:
\[\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}}\geq \frac{1}{7}.\]
When does equality occur?[/quote]
See here:
http://www.artofproblemsolving.com/community/c6h1142219p5392189

There's a wonderful solution in the book [i]An Introduction to the Proving of Elementary Inequalities[/i] written by Jingjun Han.

Here is a bit of a different solution  :)

First of all notice that $\sum_{cyc}\frac{x^{k+2}}{x^{k+1}+y^k+z^k}\overset{\text{Generalized T2'S}}{\ge}\frac{\left(\sum_{cyc}x\right)^{k+2}}{3^k\left(\sum_{cyc}x^{k+1}+2\sum_{cyc}x^k\right)}$
Thus the inequality transforms into $7\left(\sum_{cyc}x\right)^{k+2}\ge3^k\left(\sum_{cyc}x^{k+1}+2\sum_{cyc}x^k\right)\Longleftrightarrow\frac{7}{3^k}\left(\sum_{cyc}x\right)^{k+2}\ge\sum_{cyc}x^{k+1}+2\sum_{cyc}x^k$
Moreover notice that from the problem condition we can furthermore rewrite the inequality as $\frac{7}{3^k}\ge\sum_{cyc}x^{k+1}+2\sum_{cyc}x^k$

Now let $f(x,y,z)=x^{k+1}+y^{k+1}+z^{k+1}+2x^k+2y^k+2z^k\text{ and }g(x,y,z)=x+y+z-1$ and also define $L=f(x,y,z)+\lambda g(x,y,z)$
$$\frac{\partial L}{\partial x}=kx^k+x^k+2kx^{k-1}+\lambda=0\Longrightarrow kx^k+x^k+2kx^{k-1}=-\lambda\color{red}\text{ (1) }$$
$$\frac{\partial L}{\partial y}=ky^k+y^k+2ky^{k-1}+\lambda=0\Longrightarrow ky^k+y^k+2ky^{k-1}=-\lambda\color{blue}\text{ (2) }$$
$$\frac{\partial L}{\partial z}=kz^k+z^k+2kz^{k-1}+\lambda=0\Longrightarrow kz^k+z^k+2kz^{k-1}=-\lambda\color{green}\text{ (3) }$$
Furthermore from $\color{red}\text{ (1) }$ and $\color{blue}\text{ (2) }$ we obtain $kx^k+x^k+2kx^{k-1}=ky^k+y^k+2ky^{k-1}\Longrightarrow k(x^k-y^k)+2k(x^{k-1}-y^{k-1})=-(x^k-y^k)$
Now $\text{FTSOC}$ assume that $x\neq y$
With this assumption we can get $$k+\frac{2k(x^{k-1}+y^{k-1})}{x^k-y^k}=-1\Longrightarrow 1+\frac{2(x^{k-1}-y^{k-1})}{x^k-y^k}=-\frac{1}{k}\Longrightarrow\frac{2(x^{k-1}-y^{k-1})}{x^k-y^k}=-\frac{1}{k}-1\text{ we can furthermore simplify it to }\frac{x^{k-1}-y^{k-1}}{x^k-y^k}=-\frac{1+k}{2k}$$
Thus $\frac{x^{k-1}-y^{k-1}}{x^k-y^k}<0$ which implies that either $x^{k-1}<y^{k-1}\text{ or }x^k<y^k$
[b]Case 1:[/b] $x^{k-1}<y^{k-1}$
From this we obtain $x<y\Longrightarrow x^k<y^k$ which forces  $\frac{\text{negative}}{\text{negative}}=\text{negative}$ which is clearly a contradiction.
[b]Case 2:[/b] $x^k<y^k$
From this we obtain $x<y\Longrightarrow x^{k-1}<y^{k-1}$ which furthermore forces $\frac{\text{negative}}{\text{negative}}=\text{negative}$ which is a contradiction.

Therefore $x$ must be equal to $y$

Moreover from $\color{blue}\text{ (2) }$ and $\color{green}\text{ (3) }$ we obtain $ky^k+y^k+2ky^{k-1}=kz^k+z^k+2kz^{k-1}\Longrightarrow k(y^k-z^k)+2k(y^{k-1}-z^{k-1})=-(y^k-z^k)$
Now $\text{FTSOC}$ assume that $y\neq z$
Using this assumption we can rewrite the equation as $$k+\frac{2k(y^{k-1}-z^{k-1})}{y^k-z^k}=-1\Longrightarrow1+\frac{2(y^{k-1}-z^{k-1})}{y^k-z^k}=-\frac{1}{k}\Longrightarrow\frac{2(y^{k-1}-z^{k-1})}{y^k-z^k}=-\frac{1}{k}-1\text{ continuing on we can simplify it down to }\frac{y^{k-1}-z^{k-1}}{y^k-z^k}=-\frac{1+k}{2k}$$
Thus $\frac{y^{k-1}-z^{k-1}}{y^k-z^k}<0$ which forces $y^{k-1}<z^{k-1}\text{ or }y^k<z^k$
[b]Case 1:[/b] $y^{k-1}<z^{k-1}$
From this we can obtain $y<z\Longrightarrow y^k<z^k$ which implies that $\frac{\text{negative}}{\text{negative}}=\text{negative}$ which is clearly a contradiction.
[b]Case 2:[/b] $y^k<z^k$
Here we obtain $y<z\Longrightarrow y{k-1}<z^{k-1}$ which forces $\frac{\text{negative}}{\text{negative}}=\text{negative}$ which is a contradiction.

Therefore $y$ must be equal to $z$ which in turn implies that $x=y=z$
Plugging this in we obtain $g(x,x,x)=3x-1=0\Longrightarrow x=y=z=\frac{1}{3}$
Therefore maximum is attained at $f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=3\cdot\frac{1}{3^{k+1}}+2\cdot3\cdot\frac{1}{3^k}=\frac{1}{3^k}+\frac{2}{3^{k-1}}=\frac{7}{3^k}$ $\blacksquare$.


function
algebra
polynomial
trigonometry
complex analysis

i have calculated
isolated essential singularity.
is it right?
1/(cos(1/z))=n pi
n approaches to infinity
cos(1/z)=0
1/z=2npi+pi/2
n approaches to infinity
z=0 is limit point of zeroes of f(z)
so z=0 is isolated essential singularity.

[quote="JAS"]what kind of singularity of function f(z)=sin(1/cos(1/z))?[/quote]

This is what it looks like to me:

By Picard, any non-polynomial entire function at a singular point of another function has an essential singularity at that point.  So the functions $ \sin(1/z)$, $ \cos(1/z + 2/(z - 2)^2)$,$ e^{1/z}$, $ \sin(\Gamma(z))$, $ \cos(\zeta(z))$ have isolated essential singular points at each pole of their arguments.

$ \cos(1/z)$ has an isolated essential singularity at the origin then, but also has a non-isolated set of zeros about the origin, one at each $ z = \frac {2}{(2n + 1)\pi},\; n = 0,\pm 1, \pm 2\cdots$.  Therefore, the function $ \frac {1}{\cos(1/z)}$, has a non-isolated essential singular point at the origin, surrounded by isolated poles. Finally, the function ${ \sin(\frac {1}{\cos(1/z)})}$  has a non-isolated essential singular point at the origin surrounded by isolated essential singular points.

[quote="JAS"]YOU have said $ \frac {1}{\cos(1/z)}$ has non isolated essential singularity ok but how isolated poles about origin because zeroes of $ cos(\frac {1}{z})$ are non isolated at each n therefore poles of $ \frac {1}{\cos(1/z)}$ should be non isolated and if $ \frac {1}{\cos(1/z)}$  has non isolated essential singularity can we say ${ sin(\frac {1}{\cos(1/z)})}$ will also have nonisolated essential singularity[/quote]

Hey Jas.

The zeros of $ \cos(1/z)$ are isolated.  Take for example the zero at $ z_0=2/\pi$ and it's deleted neighborhood $ D(z_0,0.1)$. There is no other zero in that neighborhood right?  How about take any of those zeros no matter how small, say the one at $ \frac{2}{(2n+1)\pi}$ for n very large.  The nearest zero to that one would be $ \frac{2}{(2(n+1)+1)\pi}$.  And therefore there exists a deleted neighborhood $ D(z_n,\delta)$ with $ \delta$ equal to say one-half the distance to $ z_{n+1}$ such that no other zero is contained in that neighborhood, that is, the zero is isolated.  This goes for all the zeros no matter how small their deleted neighborhood needs to be to make it so. Now, when we consider $ \frac{1}{\cos(1/z)}$, the zeros become poles and the singularity at the origin remains and becomes non-isolated since no matter how small it's deleted neighborhood, $ D(0,\delta)$, there will always exists (an infinite number of) points in that neighborhood that are singular, that is, when $ \frac{2}{(2n+1)\pi}<\delta$.

[quote="JAS"]and if $ \frac {1}{\cos(1/z)}$  has non isolated essential singularity can we say ${ sin(\frac {1}{\cos(1/z)})}$ will also have nonisolated essential singularity[/quote]

That's what I'm saying:  I think in general, if $ g(z)$ has a non-isolated singular point, then it's composition with an entire function $ f(z)$  will cause $ f(g(z))$ to have a non-isolated singular point as well.


Hi all,

I'm having some problems with these questions. Can anybody help me?

1.  A permutation of the intergers 1, 2, ..., 10 is a sequence a_1, a_2, ..., a_10 such that each of those integers appears once. Given such a permutation, we form the sequence of the partial sums

s_1 = a_1,
s_2 = a_1 + a_2,
s3 = a_1 + a_2 + a_3,
             .
             .
s_10 = a_1 + a_2 + ... + a_10.

Find the number of such permutations which have the property that the partial sums s_1, s_2, ..., s_10 are all not divisible by 3.

2. Evaluate

{1/2! + 1/3! + ... + 1/100!}

(Here |x| denotes the largest integer which is smaller than or equal to x)

3.  Determine the number of real solutions of
         {x/2} + {2x/3} = x
(here {x} denotes the largest integer which is smaller than or equal to x)

Thanks!

$e=\sum_{k=0}^{\infty}\frac{1}{k!}$
so {1/2! + 1/3! + ... + 1/100!}  is approximatly equal to e-2
so {1/2! + 1/3! + ... + 1/100!}  =.718281828459... (The error is less than 2/101! so you can basically forget it)
if you mean the greatest integer not greater than it, then the answer is 0.

[quote="lotrgreengrapes7926"][quote="IncS"]
2. Evaluate

{1/2! + 1/3! + ... + 1/100!}

(Here |x| denotes the largest integer which is smaller than or equal to x)

3.  Determine the number of real solutions of
         {x/2} + {2x/3} = x
(here {x} denotes the largest integer which is smaller than or equal to x)
[/quote]

$|\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}|$?? Is that it?

Do you mean "Determine the real solutions", as there are an infinite number, I believe.[/quote]

Yes, that's exactly what I mean.

NNever mind about the "determine real solutions" thing. I was wrong.

[quote]1. A permutation of the intergers 1, 2, ..., 10 is a sequence a_1, a_2, ..., a_10 such that each of those integers appears once. Given such a permutation, we form the sequence of the partial sums

s_1 = a_1,
s_2 = a_1 + a_2,
s3 = a_1 + a_2 + a_3,
.
.
s_10 = a_1 + a_2 + ... + a_10.

Find the number of such permutations which have the property that the partial sums s_1, s_2, ..., s_10 are all not divisible by 3. [/quote]

First let's put the numbers 1 through 10 into their groups mod 3:

{1, 4, 7, 10} {2, 5, 8} {3, 6, 9}

So we have  4 1s, 3 2s, and 3 3s. Now, let's only look at the numbers mod 3, as that's what matters if we want them to not be divisible by 3. Afterward we can multiply the answer by $3!3!4! = 864$, as the different values of a particular residue can be permutated over the times that that residue appears.

As well, notice that multiples of 3 don't change the number mod 3, so we can put the multiples of 3 anywhere we like, ie we must only look at how the residues 1 and 2 mod 3 are distributed. There are $_10C_3 = 100800$ ways to put multiples of 3 in between the other numbers, so we'll multiply the answer by that amount afterward.

Now for the residues 1 and 2. Let (m,n) represent a case where m 1 values and n 2 values have been added to our sum (by case, we're only worried when the number of 1s or 2s mod 3 change, ie we only go to a new case when we add 1 or 2 mod 3).
Notice that the sum will be a multiple of 3 iff $m \equiv n \mod 3$.

Each step of the way, we'll either add 1 to m or add 1 to n. Because m can't equal n, one of them will have to always be greater than the other or else they'll pass each other, and since we end up at (4,3), that'll have to be m. So the very first step will be to add 1, putting the sum at 1 mod 3.

Now, a particular case will be 1 or 2 mod 3 if it isn't a multiple of 3. If this is true, then adding either 1 or 2 (depending on what the sum is) will make our number a multiple of 3, which means that we must add the number that does not make it a multiple of 3, ie there's only one choice at each step. So, if there does exist an order that works, it's unique. We check that 1121212 works, so there's one possiblity.

Combining everything together we get $864 \times 100800 = 87091200$.


Let $\{a_{n}\}_{n \geq 0}$ and $\{b_{n}\}_{n \geq 0}$ be two sequences of integer numbers such that:

i. $a_{0}=0$, $b_{0}=8$.

ii. For every $n \geq 0$, $a_{n+2}=2a_{n+1}-a_{n}+2$, $b_{n+2}=2b_{n+1}-b_{n}$.

iii. $a_{n}^{2}+b_{n}^{2}$ is a perfect square for every $n \geq 0$.

Find at least two values of the pair $(a_{1992},\, b_{1992})$.

${\left\{\alpha_{n+2}=2 \alpha_{n+1}-\alpha_{n}+2,\alpha_{0}=0\right\}}$
${\left\{\beta_{n+2}=2 \beta_{n+1}-\beta_{n},\beta_{0}=8\right\}}$

We can solve it easily, since it's linear

${\alpha_{n}=n^{2}+c_{1}n-n}$
${\beta_{n}=n c_{2}+8}$

${\alpha_{n}^{2}+\beta_{n}^{2}=\left(n c_{2}+8\right)^{2}+\left(n^{2}+c_{1}n-n\right)^{2}}$

Seting ${n=1}$

Making a Pythagorean Triple

${c_{1}=a^{2}-b^{2}}$, ${c_{2}+8=2 a b}$

${c_{1}=(a-b) (a+b)}$ ,${c_{2}=2 (a b-4)}$

Substituing on the previosly equation

${\alpha_{n}=n \left(a^{2}-b^{2}+n-1\right)}$
${\beta_{n}=2 (a b-4) n+8}$

$a,b$ are just integers numbers.

just set $n=1992$ and you have infinitely many solutions

Hello Yagaron,

[quote="yagaron"]${\left\{\alpha_{n+2}=2 \alpha_{n+1}-\alpha_{n}+2,\alpha_{0}=0\right\}}$
${\left\{\beta_{n+2}=2 \beta_{n+1}-\beta_{n},\beta_{0}=8\right\}}$

We can solve it easily, since it's linear

${\alpha_{n}=n^{2}+c_{1}n-n}$
${\beta_{n}=n c_{2}+8}$[/quote]

I'm ok till this point (although, in a previous post - now deleted -, I said it was wrong  :oops:  :oops: ).
But I don't agree with the reminder.

I rewrite the expressions :
${\alpha_{n}=n^{2}+a n}$ where $a=a_{1}-1$
${\beta_{n}=b n+8}$ where $b=b_{1}+8$

${\alpha_{n}^{2}+\beta_{n}^{2}=n^{4}+2an^{3}+(a^{2}+b^{2})n^{2}+16b n+64}$

We can easily identify this with ${\alpha_{n}^{2}+\beta_{n}^{2}=(n^{2}+a n+8)^{2}}$ which gives $a^{2}+b^{2}=a^{2}+16$ and $16 a n=16 b n$

And two immediate solutions : $a=b=4$ and $a=b=-4$ and :

$\alpha_{1992}=1992^{2}+4*1992$ and $\beta_{1992}= 4*1992+8$
and
$\alpha_{1992}=1992^{2}-4*1992$ and $\beta_{1992}=-4*1992+8$

--
Patrick

i made the following

$a_{n}^{2}+b_{n}^{2}$ is a perfect square for every $n \geq 0$

but in the expressions
${\alpha_{n}=n^{2}+c_{1}n-n}$
${\beta_{n}=n c_{2}+8}$

i have 2 unknow constantes. how to find then!?

to simmplify i just set $n=1$ and made a pythagorean triple with the squares of the 2 sequences. and make a substituition for the 2 constantes to a,b.

but i forget that this is valid only for $n=1$

i just changed the constants. i did not find then...

i forget to review the function. sorry.

but i still think that there is more solutions for the constants.

c ya

[quote="yagaron"]${\left\{\alpha_{n+2}=2 \alpha_{n+1}-\alpha_{n}+2,\alpha_{0}=0\right\}}$
${\left\{\beta_{n+2}=2 \beta_{n+1}-\beta_{n},\beta_{0}=8\right\}}$

We can solve it easily, since it's linear

${\alpha_{n}=n^{2}+c_{1}n-n}$
${\beta_{n}=n c_{2}+8}$

${\alpha_{n}^{2}+\beta_{n}^{2}=\left(n c_{2}+8\right)^{2}+\left(n^{2}+c_{1}n-n\right)^{2}}$

...

just set $n=1992$ and you have infinitely many solutions[/quote]

OK, the first part is right, but not the second one.   It is true that $\alpha_n^2+\beta_n^2=n^4+2(c_1-1)n^3+c_2^2n^2+(c_1-1)^2n^2+16c_2 n+64$ must be a perfect square, but for all $n$.  Let $k_1>\frac{c_2^2}{2}>k_2$, where $k_1,k_2$ are integers.  Then,
$\alpha_n^2+\beta_n^2-(n^2+(c_1-1)n+k_1)^2=(c_2^2-2k_1)^2n^2+(16c_2-2k_1(c_1-1))n+64-k_1^2$, where for sufficiently large $n$, the first term dominates and is negative, hence $\alpha_n^2+\beta_n^2<(n^2+(c_1-1)n+k_1)^2$ for infinitely many values of $n$.  Similarly, $\alpha_n^2+\beta_n^2>(n^2+(c_1-1)n+k_2)^2$ for infinitely many values of $n$, or $k=\frac{c_2^2}{2}$ must be an integer, and $\alpha_n^2+\beta_n^2=(n^2+(c_1-1)n+k)^2$ for infinitely many values of $n$, or for infinitely many values of $n$,

\[0=\alpha_n^2+\beta_n^2-\left(n^2+(c_1-1)n+\frac{c_2^2}{2}\right)^2=c_2(16-c_2(c_1-1))n+\frac{256-c_2^2}{4}\]

It suffices with two of them, so that from one side, $c_2^2=256$ for $c_2=\pm16$, yielding $c_1-1=1$ or $c_1-1=-1$, for $c_1=2$ or $c_1=0$ respectively.  There are therefore exactly two pairs of values $(\alpha_1,\beta_1)$, and the corresponding values of $(a_{1992},b_{1992})$ can easily be found.  The problem might as well have read "find all possible values of..." instead of "find at least two values of...".


MATHCOUNTS
analytic geometry
geometry
search
function
AoPSwiki
algorithm

Triangle $ ABC$ has $ A(0,0)$, $ B(14,2007)$, $ C(11,1337)$. How many lattice points lie inside the triangle?

[hide="hint"]When I was tying my Shoelaces this morning, I saw a penny on the ground and Picked it up.[/hide]

1679..

The solution is based soley on the fact if you know Picks or not..so I dont know if I should post it..because someone will be like how did you get that..

but is that a unique solution or are there any better ways to do it because I dont think you learn Pick in mathcounts

Oh yea also note the coordinates are relatively prime (importatnfor boundaries)

[quote="13375P34K43V312"]Now that I think about it, this problem is pretty bad...it requires two formulas that require pretty sophisticated analytic geometry to prove. I mean, shoelace isn't completely necessary, but it trivialiazes the first part of the problem.[/quote]

just making sure was that right?

I dont have a caculator in front of me so i actually multiplied it out..

What's the Shoelace Theorem, and what are "Picks"?  This actually sounds pretty interesting, so could anyone explain it?  I tried the search function for Shoelace, but it didn't give direct answers.  I tried AoPSWiki, but it didn't work either. :blush:

Here's a link that explains shoelace, but I have no idea what picks are.
http://www.mathreference.com/la-det,shoe.html

You didn't post the link.

Shoelace is an algorithm for finding the area of a polygon using only the coordinates of its vertices. It involves putting the coordinates in order (either clockwise or counterclockwise) in a matrix and doing some straightforward calculations.

Pick's Theorem states that the area of a polygon with vertices at lattice points is given by $ I+\frac{B}{2}-1$, where $ B$ is the number of lattice points on the perimeter, $ I$ is the number of lattice points in the interior.

We should add those to the Wiki...
They're very useful.  MATHCOUNTS will occasionally have problems which are trivialized by these theorems.  Those problems are basically testing you to see "do you know this formula?"

[quote="13375P34K43V312"]
They've (hopefully) realized that those types of problems don't really teach you anything except how to memorize random things, since the proofs of many of these theorems involve pretty sophisticated stuff.[/quote]
let's celebrate if they have realized
 :clap:
MATHCOUNTS, I thought, was always lacking good geometry problems.  Maybe they were better back in the day.

[quote="ordinary boy"][quote="13375P34K43V312"]I think there are less and less of those now. There's also less and less of good geometry problems, which is a shame, but that's a different story. But I don't think there are any more Chicken McNugget problems, or problems that are like "find the area of the polygon with vertices at [10 different points]," etc.

They've (hopefully) realized that those types of problems don't really teach you anything except how to memorize random things, since the proofs of many of these theorems involve pretty sophisticated stuff.[/quote]

well doesnt mathcounts have like problems with a triangle and 10 more traingles in it and say
"how many triangles are in this figure?"[/quote]

What's your point?

[quote="ordinary boy"]well it teaches you nothing except to count triangles..which 2 year olds can do.[/quote]

That's not true. I myself am not a huge fan of those problems, but it teaches you to systematically count overlaps, etc., and in a good problem, find symmetries that make the calculation/counting less heavy.

And by the way, in those problems, if there are ten small triangles, the answer is never 10.


Determine all continuous function $ R\to R$ such that :

$ xf(x)\minus{}yf(y)\equal{}(x\minus{}y)f(x\plus{}y)$ , $ \forall x,y \in R$

$ x\equal{}ny,n\ge 2$ give $ f((n\plus{}1)y)\equal{}\frac{n}{n\minus{}1}f(ny)\minus{}f(y)\to f(ny)\equal{}a(y)ny\plus{}b(y)$. And $ a(y/m)\equal{}a(y),b(y/m)\equal{}b(y)$, therefore for continuosly function $ a(y)\equal{}a,b(y)\equal{}b$ constants and $ f(x)\equal{}ax\plus{}b.$


just a tricky question i came across and found interesting when reviewing for the physics C ap exam...

The acceleration of a certain mass undergoing simple harmonic motion is given by $x''(t) = -9x$.  What is the period, $T$, of this mass's oscillation if $x$ is its displacement and $t$ is the time elapsed?

[hide]$T = 2\pi/3$[/hide]

Yeah although that ultimately comes from differential equations.  It's not that lengthy by the way.

Or you can achieve that equation by looking at the correlation between a oscillating pendulum and an object undergoing uniform circular motion.

[quote="ffdbzathf"]this problem [i]can[/i] be done with differential equations, but that would be very lengthy and unnecessary.  there is a much quicker way to do it.  i've added a hint below.

[hide]hint: $x = Acos(\omega t + \phi_0)$[/hide][/quote]

Umm, firstly, anyone with competence who has taken a diff eq class should be able to solve that in his/her head. Secondly, if for not solving a differential equation, where did you arrive at that equation given in your hint?

this problem wasn't aimed at students who had taken a differential equation class...this was simply a problem from AP Physics C.  that equation is part of the curriculum in simple harmonic motion.  as others have mentioned, this forum really needs some sub-forums.

from the definition of SHM, if $x''=-w^2x$ then, $T=\frac{2\pi}{w}$


find $a,b,c\in\mathbb{R}$ such that

$2a=b+c$
$2b=a+c$
$2c=a+b$

1. This should definitely NOT be in Intermediate. Classroom Math would be better.

2. Not enough information to solve; it appears the original poster forgot to type the left half of the last equation, and if he meant to say $2c = a+b$, then it is just a dependent system, as that can be easily derived from the first two.

The three planes intersect at the line $\vec{r}=s(1,1,1),\ s\in\mathbb{R}$, that is, $(a,b,c)=(x,x,x)\in\mathbb{R}^{3}$.  You could either use cross product or set a parameter and solve linear systems in terms of that.

You easily get that a=b=c and these numbers can be any real numbers.  :)

just subtract any pair of equations to get a=b, etc


limit
real analysis
real analysis unsolved

Let    $ L_n\equal{}\sqrt[n\plus{}1]{(n\plus{}1)!} \minus{}\sqrt[n]{n!}$

  prove  that   \[ \lim_{n \to \infty} L_n\equal{}e^{\minus{}1}\]

It's Lalescu's sequence, right?

Discussed here: http://www.mathlinks.ro/viewtopic.php?t=52803

and here: http://www.mathlinks.ro/viewtopic.php?t=233953

and a nice text about its generalization: http://pefmath2.etf.bg.ac.yu/files/118/868.pdf


inequalities
logarithms
inequalities proposed

If $ a$ and $ b$ are positive real numbers such that $ a\plus{}b\equal{}2$, then

$ a^b\plus{}b^a \le 2$.

Can we use Jensen on $ f(x) \equal{} x^{2 \minus{} x}$? We have $ a^{2\minus{}a}\plus{}(2\minus{}a)^a \le 2$ by Jensen.

[quote="dgreenb801"]Can we use Jensen on $ f(x) \equal{} x^{2 \minus{} x}$? [/quote]
Check it! :wink:

it looks very simple. I might have a mistake somewhere.

Let $ a\geq 1\geq b$

we have to prove

$ a(1\minus{}\frac{1}{a^{1\minus{}b}})\plus{}b(1\minus{}b^{a\minus{}1})\geq 0$

but this is true as $ a,b$ are positive and

$ a^{1\minus{}b}\geq 1 \geq b^{a\minus{}1}$

Nice, Eagle.

But, what you think about these inequalities below (in the same conditions):

(b) $ a^{2b}\plus{}b^{2a} \le 2$;

(c) $ a^{3b}\plus{}b^{3a} \le 2$.

[quote="Vasc"]Nice, Eagle.

But, what you think about these inequalities below (in the same conditions):

(b) $ a^{2b} \plus{} b^{2a} \le 2$;

(c) $ a^{3b} \plus{} b^{3a} \le 2$.[/quote]
Very nice, VasC. $ k \equal{} 3$ is also the best constant for the inequality
\[ a^{kb} \plus{} b^{ka} \le 2
\]
:)
But my solution for it is quite long and ugly. :( However, I got a nice solution by Bernoulli for $ k \equal{} 2$. :)

[quote="Vasc"]Nice, Eagle.

But, what you think about these inequalities below (in the same conditions):

(b) $ a^{2b} \plus{} b^{2a} \le 2$;
[/quote]
Ok, I will post my solution for this
Assume that $ a \ge b$, then $ a \ge 1 \ge b$. Then the Bernoulli Inequality gives us
\[ a^b \le 1\plus{}b(a\minus{}1) \equal{}1\plus{}b\minus{}b^2\]
\[ b^a \equal{}b\cdot b^{a\minus{}1} \le b [1\plus{}(a\minus{}1)(b\minus{}1)] \equal{}b[1\minus{}(b\minus{}1)^2] \equal{}b^2(2\minus{}b)\]
Hence
\[ a^{2b}\plus{}b^{2a} \le (1\plus{}b\minus{}b^2)^2\plus{}b^4(2\minus{}b)^2\]
It suffices to show that
\[ (1\plus{}b\minus{}b^2)^2\plus{}b^4(2\minus{}b)^2 \le 2\]
\[ \Leftrightarrow b^6 \minus{}4b^5\plus{}5b^4\minus{}2b^3\minus{}b^2\plus{}2b\minus{}1 \le 0\]
\[ \Leftrightarrow (b\minus{}1)^2(b^4\minus{}2b^3 \minus{}1 ) \le 0\]
which is obviously true since $ 0 <b \le 1$.
:)

[quote="Vasc"]Nice, Can.
Actually, there are 6 ways to prove this inequality by Bernoulli's Inequality, since

$ a^{2b} \equal{} (1 \plus{} a^2 \minus{} 1)^b\le 1 \plus{} b(a^2 \minus{} 1)$,

$ a^{2b} \equal{} ((1 \plus{} a \minus{} 1)^b)^2\le (1 \plus{} b(a \minus{} 1))^2$

and

$ b^{2a} \equal{} (1 \plus{} b^4 \minus{} 1)^\frac a{2}\le 1 \plus{} \frac a{2}(b^4 \minus{} 1)$,

$ b^{2a} \equal{} b^2(1 \plus{} b^2 \minus{} 1)^{a \minus{} 1}\le b^2(1 \plus{} (a \minus{} 1)(b^2 \minus{} 1))$,

$ b^{2a} \equal{} b^2(b^{a \minus{} 1})^2\le b^2(1 \plus{} (a \minus{} 1)(b \minus{} 1))^2$.

I have no nice solution to this very special inequality

$ a^{3b} \plus{} b^{3a} \le 2$.[/quote]
Another Bernoulli way is
With $ a \ge 1 \ge b$, then
\[ a^{2b} \le [1 \plus{} b(a \minus{} 1)]^2
\]
and
\[ \frac {1}{b^a} \ge 1 \plus{} a(\frac {1}{b} \minus{} 1) \equal{} \frac {b \plus{} a(1 \minus{} b)}{b}
\]

\[ \Rightarrow b^{2a} \le \frac {b^2}{[b \plus{} a(1 \minus{} b)]^2}
\]
:)

why is $ k\equal{}3$ the best constant for this inequality?

[quote="Vasc"]If $ a$ and $ b$ are positive real numbers such that $ a \plus{} b \equal{} 2$, then

$ a^b \plus{} b^a \le 2$.[/quote]

I am interested in a solution with Bernoulli inequalty for the following:

What is the maximum value of $ a^b\plus{}b^a$, given that $ a,b$ are positive real numbers such that $ a\plus{}b\equal{}1$.

[quote="Albanian Eagle"]why is $ k \equal{} 3$ the best constant for this inequality?[/quote]
We can write the desired inequality as $ f(a)\le 0$ for $ 0<a<2$, where
$ f(a)\equal{}a^{kb}\plus{}b^{ka}\minus{}2$ for $ b\equal{}2\minus{}a$.
Since $ f(1)\equal{}0$ and $ f'(1)\equal{}0$, it is necessary to have $ f''(1) \le 0$. Since $ f''(1)\equal{}2k(k\minus{}3)$, we get the necessary condition $ k\le 3$.

[quote="hitek"] What is the maximum value of $ a^b \plus{} b^a$, given that $ a,b$ are positive real numbers such that $ a \plus{} b \equal{} 1$.[/quote]
The maximum value is $ \sqrt 2$.
But the following inequality is stronger and very nice:

[i]If $ a$ and $ b$ are positive real numbers such that $ a\plus{}b\equal{}1$, then[/i]

$ a^{2b}\plus{}b^{2a} \le 1$.

[quote="hitek"][quote="Vasc"]
The maximum value is $ \sqrt 2$.
.[/quote]
Can you solve it with Bernoulli, Vasc?[/quote]
I have a solution to your inequality, but it is not with Bernoulli, and it is not very nice.
I hope somebody can find and post a nice (or ugly) solution.

On the other hand, there are two open inequalities in this topic:

(1) $ \ \ \ \ \ a^{3b} \plus{} b^{3a} \le 2$, for $ a \plus{} b \equal{} 2$;

(2) $ \ \ \ \ \ a^{2b} \plus{} b^{2a} \le 1$, for $ a \plus{} b \equal{} 1$.

Nice solution, pvthuan. :lol:
My solution uses the inequalities
$ b^a\le ka\plus{}(\sqrt 2 \minus{}k)b$,
$ a^b\le 1\minus{}ka\plus{}(k\minus{}\sqrt 2)b$,
where $ k\equal{}\frac {1\minus{}ln 2}{2\sqrt 2}$.

[quote="Vasc"]Nice solution, pvthuan. :lol:
My solution uses the inequalities
$ b^a\le ka \plus{} (\sqrt 2 \minus{} k)b$,
$ a^b\le 1 \minus{} ka \plus{} (k \minus{} \sqrt 2)b$,
where $ k \equal{} \frac {1 \minus{} ln 2}{2\sqrt 2}$.[/quote]
We can prove it by Bernoulli Inequality like this
\[ b^{a} \equal{} \frac {(2b)^a}{2^a} \le \frac {1 \plus{} a(2b \minus{} 1)}{2^a} \equal{} \frac {1 \plus{} a \minus{} 2a^2}{2^a}
\]
It suffices to prove that
\[ \frac {1 \plus{} a \minus{} 2a^2}{2^a} \plus{} \frac {1 \plus{} b \minus{} 2b^2}{2^b} \le 1
\]
It is easy to show that
\[ 2^x \ge 1 \plus{} x\ln 2
\]
For all $ x \in R$.
Hence
\[ 2^{a \minus{} 1/2} \ge 1 \plus{} (a \minus{} \frac {1}{2}) \ln 2
\]

\[ \Rightarrow 2^a \ge \frac {1}{\sqrt {2}}(2 \plus{} (2a \minus{} 1) \ln 2)
\]
We have to show
\[ \frac {1 \plus{} a \minus{} 2a^2}{2 \plus{} (2a \minus{} 1) \ln 2} \plus{} \frac {1 \plus{} b \minus{} 2b^2}{2 \plus{} (2b \minus{} 1) \ln 2} \le 1
\]

\[ \Leftrightarrow \frac {1 \plus{} a \minus{} 2a^2}{2 \plus{} (2a \minus{} 1) \ln 2} \plus{} \frac {3a \minus{} 2a^2}{2 \minus{} (2a \minus{} 1) \ln 2} \le 1
\]

\[ \Leftrightarrow (2 \plus{} \ln 2)(1 \minus{} \ln 2) (2a \minus{} 1)^2(2 \minus{} \ln 2 \plus{} 2a \ln 2)\ge 0
\]
which is true. :)
P/s: The solution of pvthuan is nice and I like it. :)

For $ a\plus{}b\equal{}2$, the inequalities hold

$ a^{2b}\plus{}b^{2a}\plus{}(\frac {a\minus{}b}{2})^2 \le 2$;

$ a^{3b}\plus{}b^{3a}\plus{}(\frac {a\minus{}b}{2})^4 \le 2$,

with equality for $ a\equal{}b\equal{}1$, or $ a\equal{}0$ and $ b\equal{}2$, or $ a\equal{}2$ and $ b\equal{}0$.

The last inequality is a refining of a conjecture above. :D

[quote="Vasc"]Nice solution, pvthuan. :lol:[/quote]
But it is a little wrong, because $ 2\minus{}2q\ge 1$. :(

[quote="Albanian Eagle"]it looks very simple. I might have a mistake somewhere.

Let $ a\geq 1\geq b$

we have to prove

$ a(1 \minus{} \frac {1}{a^{1 \minus{} b}}) \plus{} b(1 \minus{} b^{a \minus{} 1})\geq 0$

but this is true as $ a,b$ are positive and

$ a^{1 \minus{} b}\geq 1 \geq b^{a \minus{} 1}$[/quote]
This problem is similar a problem from toan hoc tuoi tre magazine!
your solution seem to like my solution:
 set $ a\equal{}1\minus{}x,b\equal{}1\plus{}x$
and $ check f(x)$ :wink:

[quote="Vasc"][quote="Vasc"]Nice solution, pvthuan. :lol:[/quote]
But it is a little wrong, because $ 2 \minus{} 2q\ge 1$. :([/quote]
But my proof still works :)


function
calculus
derivative
probability and stats

I need som help please with the following exercise:

Let X be a random variable. The cumulants $ k_n$ of X are defined by
the cumulant-generating function
g(t) = log(E(exp(tX))).
The nth cumulant is given by $ k_n$ = $ g^{(n)}(0)$ (the nth derivative of g at
0).
Find the first five cumulants of the normal one dimensional distribution.

Thanks

Hi georgech

The Laplace Transform of Normal Law with mean $ \mu$ and Variance $ \sigma^2$ is

$ e^{\mu.t\plus{}(\sigma.t)^2/2}$

you have then :

$ \kappa_1\equal{}\mu$, $ \kappa_2\equal{}\sigma^2$, $ \kappa_i\equal{}0$, $ \forall i>2$


algebra
polynomial
Cubic
number theory
Iteration
IMO Shortlist

Let $ p(x)$ be a cubic polynomial with rational coefficients. $ q_1$, $ q_2$, $ q_3$, ... is a sequence of rationals such that $ q_n \equal{} p(q_{n \plus{} 1})$ for all positive $ n$. Show that for some $ k$, we have $ q_{n \plus{} k} \equal{} q_n$ for all positive $ n$.

Suppose that $ p(x) = 1/e(ax^{3} + bx^{2} + cx + d) $ where $ a,b,c,d,e \in \mathbb{Z} $,
$ e > 0 $. Then suppose that $ m,M \in \mathbb{Z} $, $ n,N \in \mathbb{N} $, the fractions
$ \frac{m}{n} $ , $ \frac{M}{N} $ are reduced and $ p(\frac{m}{n}) = \frac{M}{N} $.
But $ p(\frac{m}{n}) = \frac{am^{3}  +  bm^{2}n + cmn^{2} + dn^{3}}{en^{3}} $.
Suppose $ k = gcd(am^{3}  +  bm^{2}n + cmn^{2} + dn^{3},en^{3}) $. Denote also
$ x = am^{3} $, $ y = bm^{2}n + cmn^{2} + dn^{3} $. Then $ k | x+y $ and $ k | en^{3} $.
Therefore $ k | e(x^{3} + y^{3}) $ ( because $ x+y | x^{3} + y^{3} ) $ , but
$ k | en^{3} | ey^{3} $, so $ k | ex^{3} = ea^{3}m^{9} $ , and as before $ k | en^{3} $.
So, at the end, $ k | ea^{3}m^{9} , ea^{3}n^{3} $. But $ gcd(m^{9},n^{3}) = 1 $, therefore
$ k | ea^{3} $. So we get that $ gcd(am^{3}  +  bm^{2}n + cmn^{2} + dn^{3},en^{3}) | ea^{3} $,
therefore $ Mea^{3} \geqslant en^{3} $ , so $ n \leqslant a \sqrt[3]{M} $ .

Now suppose we have a sequence of rational numbers $ q_{1},q_{2}, \ldots $ such that
$ q_{n} = p(q_{n+1}) $. Write $ q_{n} = \frac{a_{n}}{b_{n}} $ as a reduced fraction
where $ a_{n},b_{n} \in \mathbb{Z} $, $ b_{n} > 0 $. Then, as we saw before,
$ b_{n+1} \leqslant a \sqrt[3]{b_{n}} $. Therefore the sequence $ \{ b_{n} \} $ is
bounded. Also note that the sequence $ \{ q_{n} \} $ is bounded. Indeed, there exists
some $ C > 0 $ such that for every $ x \in \mathbb{R} $, $ |x| > C $ we have
$ |p(x)| > 2|x| $. Therefore it is impossible that $ |q_{n}| > C $ for infinitely many
$ n $'s, because $ q_{1} $ is bounded.
So $ \{ b_{n} \} $ is bounded and $ \{ q_{n} \} $ is bounded, therefore $ \{ a_{n} \} $
is bounded. Therefore the sequence $ q_{n} $ takes only a finite number of values.
Therefore there exists a rational number $ q $, which appears in the sequence $ \{ q_{n} \} $
infinitely many times. Because the number $ q $ appears at least two times, there is
$ k \in \mathbb{N} $, such that
$ \underbrace{p \circ p \circ \ldots \circ p}_{k times} (q) = q $. Therefore
$ k $ is a period for $ q_{n} $, because there exists $ N $ as large as we want,
such that $ q_{N} = q $, so because of $ q_{n-1} = p(q_{n}) $, we get that
$ q_{n-k} = q_{n} $ for every $ k+1 \leqslant n \leqslant N $. So because we can take $ N $
to be as large as we want, we get that $ q_{n-k} = q_{n} $ for every
$ k+1 \leqslant n $.
Therefore $ q_{n} $ is periodic.


An archer shoots an arrow into the air such that its height at any
time, t, is given by the function h(t) = -16t^2 + kt + 3. If the maximum height of the arrow occurs at time t = 4, what is the value of k?

(1) 128     (3) 8
(2) 64       (4) 4

I'm kind of busy at the moment but here's a [hide="hint"]Think about the vertex form of a quadratic equation. Find k so that the term that is the y-value of the vertex is 4.[/hide]


Compute the sum $S=3^{2}+2\cdot 3^{3}+3\cdot 3^{4}+....+(n-1)3^{n}$  as a function of $n$ where $n$ is a positive integer .


i know many of you have been playing and trying to beat your scores on the arithmetic game....this is just a thought i had....wouldn't it be cool if there was a base numbers arithmetic game?...where for each problem you are asked to change a number in one particular base to another base...and then your score would be how many numbers you could convert in one minute...i don't know....just a thought

Could you please give me a link to the game you are talking about?

It would probably be fun, but I don't think it would be as useful as the original game.

why not?...learning to convert between bases fast is just as important as learning how to add, subtract, multiply, and divide fast, right?

not many base problems do appear in MC, as far as I know.

so it's not as important as adding, subtracting, etc. btw, you can't convert between bases without basic arithmetic skills

Well, it'd be great practice, but how exactly would we do this?

well....u get a number in some particular base and then ur told to convert it into another base...it worked out fine....in my head.. :blush:

No, I meant like how would we make such a game or put it on the internet, or was that not the idea...

Hmm... I don't think it would help too much in Mathcounts, but it would be fun...

it helps us? oh wait... i think he was saying that someone could make that game... idk

archimedes1 has a cool avatar... i took his and added e dancing with pi

Hey I know Java - but I ain't a bright programmer -  :blush: . Isn't there alreayd a game thogh?

I checked out the game source- he uses XML, CSS, and JavaScript (and of course HTML). This is more complicated than it seems.

oh well then anyone who is good with this stuff.. try!

[quote="ZhangPeijin"][quote="archimedes1"]Some bright programmer would donate his/her time, or one of us could just learn JAVA.  Easy as that.[/quote]
How does programming in JAVA help us in math?  :huh:[/quote]
alkjash asked how would make such a game.


calculus
derivative
algebra open
algebra


Let f: R to R be a continuous function such that f(x+sqrt(2))+f(x+sqrt(3))=f(x) for all x in R. Show that f is a constant function.

[quote="Lazarus"]I might be misunderstanding what you're saying, but if f(x) = k, then doesn't the given turn to 2k = k?[/quote]
Which is true if $ k=0$, so maybe the problem should say that if that functional equation is satisfied then $ f(x)=0$.

EDIT: What about the Dirac Delta function $ \delta (x)$  :rotfl:

Maybe he means the zero function? Though it seems that a suitable exponential function might work? Not sure.

[quote="JBL"]My bet is that it's supposed to read $ f(x+\sqrt{2}) = f(x+\sqrt{3}) = f(x)$.[/quote]

that is what it's supposed to read.

[quote="sylow_theory"]Here is an incomplete solution.
The condition,
$ f(x+\sqrt{2}) = f(x)$
Says the function is [i]periodic[/i].

So, $ f(x+\sqrt{2}n)=f(x)$ for all $ n\in \mathbb{Z}$.

But the problem is that,
$ f(x+\sqrt{3}n)=f(x)$ also.

So the [i]fundamental period[/i] is $ 0$ (and henceforth it must be constant), it cant be related to $ \sqrt{2}$ because then we cant obtain $ \sqrt{3}$.

But, I am not so satisfied with that solution cuz were do we use the fact that $ f$ is continous?[/quote]

You did use the continuity.  In that argument, there is implicitly the statement that "We can always find integers $ a, b$ s.t. $ a\sqrt{3}$ and $ b\sqrt{2}$ are arbitrarily close together (for however "arbitrarily close is defined").  And hence continuity is needed.

Given the information in the problem, we deduce that the function repeats every $ a\sqrt{3}+b\sqrt{2}$, where $ a, b$ are any integers.  We can make that arbitrarily small, thus showing that the function repeats in arbitrarily small intervals, but we can't make [b]any[/b] real number as a linear combination of those two.

[quote="K81o7"]Given the information in the problem, we deduce that the function repeats every $ a\sqrt{3}+b\sqrt{2}$, where $ a, b$ are any integers.  We can make that arbitrarily small, thus showing that the function repeats in arbitrarily small intervals, but we can't make [b]any[/b] real number as a [i]rational[/i] linear combination of those two.[/quote](Slightly modified by me.)

An obvious example is the function $ \chi$ given by $ \chi(x) = 1$ if $ x = a\sqrt{2}+b\sqrt{3}$ for $ a, b \in \bf Z$ and $ \chi(x) = 0$ otherwise.  This function satisfies the given relation but is not constant.  The reason we can do this is that it's not continuous.


Hello,

Let $ n \in \mathbb{N}^{*}$, Prove that :\[ i)\; \min_{k \in \mathbb{N}^{*}} \lfloor k \plus{} \frac {n}{k}\rfloor \equal{} \lfloor\sqrt {4n \plus{} 1}\rfloor.\] \[ ii)\;\Big| \big\{n\sqrt2\,\big\} \minus{} \big\{n\sqrt3\,\big\}\Big| > \frac1{20n^3}\]

[quote="modularmarc101"]Whats min?[/quote]
min = minimum valeur

[hide="Problem (i)"]The minimum of the given expression is obtained for ${ k_{\min}}=\left\lfloor\sqrt n+{1\over 2}\right\rfloor$ (i.e. for the integer closest to $ \sqrt{n}$).

Put $ s=\left\lfloor\sqrt n+{1\over 2}\right\rfloor$. Since the expression under the floor sign can never be integer, we have $ s<\sqrt n+{1\over 2}<s+1\iff s^2-s+{1\over 4}<n<s^2+s+{1\over 4}\iff s^2-s+1\leqslant n\leqslant s^2+s$

First we'll discuss $ s=1$. Then $ 1\leqslant n\leqslant 2$. For those values, the initial identity is easily checked to hold - $ \left\lfloor 1+{1\over 1}\right\rfloor=\lfloor\sqrt{4\cdot 1+1}\rfloor$ and $ \left\lfloor 1+{2\over 1}\right\rfloor=\left\lfloor 2+{2\over 2}\right\rfloor=\lfloor\sqrt{4\cdot 2+1}\rfloor$. So now we assume $ s\geqslant 2$.

[b]Case 1.[/b] $ s^2-s+1\leqslant n\leqslant s^2-1$.

Then $ 2s-1+{1\over s}\leqslant s+{n\over s}\leqslant 2s-{1\over s}\implies \left\lfloor s+{n\over s}\right\rfloor = 2s-1$

Also, $ 2s-1<\sqrt{4s^2-4s+5}\leqslant\sqrt{4n+1}\leqslant\sqrt{4s^2-3}<2s$, hence $ \lfloor\sqrt{4n+1}\rfloor=2s-1=\left\lfloor s+{n\over s}\right\rfloor$

[b]Case 2.[/b] $ s^2\leqslant n\leqslant s^2+s-1$

Then $ 2s\leqslant s+{n\over s}\leqslant 2s+1-{1\over s}\implies\left\lfloor s+{n\over s}\right\rfloor=2s$

Also, $ 2s<\sqrt{4s^2+1}\leqslant\sqrt{4n+1}\leqslant\sqrt{4s^2+4s-3}<2s+1$, hence $ \lfloor\sqrt{4n+1}\rfloor=2s=\left\lfloor s+{n\over s}\right\rfloor$

[b]Case 3.[/b] $ n=s^2+s$

Then $ s+{n\over s}=2s+1$ and $ \sqrt{4n+1}=2s+1$, hence $ \lfloor\sqrt{4n+1}\rfloor=2s+1=\left\lfloor s+{n\over s}\right\rfloor$

This completes the proof.[/hide]

The second problem had already been posted here, and I have written a solution in that topic. You may check it at: http://www.mathlinks.ro/Forum/viewtopic.php?t=172272.


I think they did that to gain more money.  Besides that, I don't see any alterior motives.

-interesting_move

[quote="Syntax Error"]how come the mathcounts site releaes the state and chapter tests from each year, but they dont post the national test.  is this so people will buy the booklet thingie?

P.S.
i would just like to add that it just turned 7 am here. and that ive been up for about 2 hours. well, in bed i was up for 2 hours, i couldnt sleep. and i have to leave for my school enrollment in 45 minutes. aargh[/quote]

No wonder, I was thinking on going later, so I would meet you people at the enrollment, but oh well, if I did, I would've met Prashant.

yeah, i kinda got sidetracked talking to rep123max, so i left at 815. but i still woke up at the same time

yeah I think so. I think they just don't want to lose the money for the comp. tests

Moogra, this is OLD.

Please stop posting in ond threads

[quote="xxm0rg07hxx"]Might want to edit that.  It's old, not ond.[/quote]

I think everybody knows what he means.


Groudon and Kyogre have awaken once again, and this time, a PKMN trainer has caught Rayquaza, so Rayquaza isn't there to stop them from fighting. The PKMN trainer is too lazy and scared to stop both of these terrifying lvl 100 monsters. Now you used to be from Team Magma but switched to Aqua later, and they kicked you out, but you still want to help one side. (Magma helps Groudon, Aqua helps Kyogre; you want to be on the winning side) You have to decide which one will win. Niether PKMN knows 1-hit KO moves, and these are the moves: (the parenthesis in this order is strength, accuracy, PP)

Kyogre: Effect: Drizzle (rains)
Ice Beam       (95, 100, 10)  10% of freezing opponent
Water Spout    (150, 90, 5)   strength is less as HP is higher
Hydro Pump     (120, 80, 5)   -none-
Surf           (95, 100, 15)  -none-

Groudon: Effect: Drought (sunlight)
Earthquake     (100, 100, 10)  -none-
Fire Blast     (120, 80, 5)    10% of burning opponent
Eruption       (150, 90, 5)    the lower the HP, higher the str
Solar Beam     (120, 100, 10)  see below (long description)

Solar beam: 1st turn absorbs light, second turn attack on normal conditions. If it is sunny, 1 turn move and more effective. If it is raining, strength is halved.

So what do you think? Don't forget to say why.

Hmm... I have trouble making polls

For Kyogre, I would replace hydro with aqua tail (physical move to counter blissey and better accuracy) and surf with thunder (against other water types).

For groudon, and all the moves are crap  except for earthquake; solar beam is a lame idea on groudon (low pp, only 120 as compared to 150 for earthquake, etc.). Eruption is lame, replace it with swords dance or bulk up. And replace with probably dragon claw. Hm, what should be the last move, maybe stone edge for the fliers. So yeah, that would make these guys so much better.

But anyway, Kyogre. Drizzle + STAB + GODLY special attack = PWNGE. And he has an obvious attribute advantage of Groudon, so he'd win anyway.  :P

[quote="dragon96"]Which one would win [b]given[/b] these move then? And yes, its Hoenn region, so there is no aqua tail. If I did have a kyogre, then I would teach it thunder, but that would be with TMs. Kyogre doesn't learn Thunder.[/quote]
Oh fine. Still kyogre.

kyogre uses surf/hydro pump
game over (groudon's x4 weak to water)

Actually its only 2 times. Does any one know how much hp each has at lvl 100 and how much damage Groudon takes at lvl 100 from Kyogre's surf?

[quote="dragon96"][quote="theprodigy"]groudon is fire/ground so its 4x weak to water
now stfu

and if EVs and IVs are same they will have the same hp and speed[/quote]
Groudon is only ground type, its just the Groudon has a lot of fire moves because they can't think of a ground equivalent of water spout.  I think personally that Groudon would win because Groudon is slower, so Kyogre's effect goes first, then Groudon's effect takes place (drought). The strongest move Kyogre should do would be Water spout, but the HP is full. drought halve's Water spout's power, and since Kyogre is water type, you multiply 1.5x, resulting with strength 112.5, which probably won't kill. The Groudon uses solar beam, which only takes 1 turn, since it is sunny, and Kyogre also deals with 2 times damage (super effective!) so the strength is 240.[/quote]
If I'm not wrong, Kyogre is Water/Ice type. So, damage is 4x (I think :|)

No. Both Kyogre and Groudon are single types. And grass is 1x damage, so it would still be 2 times.

[quote="theprodigy"]groudon is fire/ground so its 4x weak to water
now stfu

and if EVs and IVs are same they will have the same hp and speed[/quote]
Groudon is ground only, so stop being lame and randomly dissing people around the boards. It's not cool, you know.

And pacman, kyogre is only water. But still, Kyogre has the advantage. While drought typically cancels out the water weakness, kyogre also has drizzle, so it doesn't work. And either way, groundon's attacks aren't good against kyogre's. Except solar beam. And kyogre's special attack is higher, so it's moveset it more suited for it thn gruodon's is.

sure groudon's only one type BUT I KNOW FOR A FACT THAT KYOGRE AND GROUDON HAVE THE SAME FCKING SPEED!!!

kyogre's sp atk and sp def>groudon's sp and sp def

therefore if u kept using special attacks kyogre would win
simple stuff :roll:

[quote="james4l"]Kyogre: Groudon does not have any moves that do double-damage.[/quote]

Read carefully. Solarbeam.

but it takes 2 turns to recharge. All other moves are 1/2 damage.

Kyoger has 4 moves that do 2x dmg. (I think)

Solar beam takes no charge time if you have sunny day, but solar beam doesn't have STAB and it's a special attack (groudon's physical attack is better).

No need for that kinda languauge, btw. And anyway, since when did i say that there speed wasn't the same?

addressing to whom?

Anyways, i think kyogre is pretty much the winner here, with a type advantage and all that.

another thing
kyogre beats rayquaza super easy, groudon doesn't :D

arceus owns both

Really the only way Groudon stands a chance is if he knows Grass Knot, and that's next generation.

Sinnoh TM 86

100 acc
20 PP
strength depends on how heavy the opponent is (heavier, the stronger)

Kyogre's water spout is pretty good. Combine it with that one item that 1.5s your damage but skips a turn, and you win. Hmmm I haven' tested on level 100s though

kyogre and groudon have the same speed :wallbash_red:

i think the weather will be neutral...

[quote="mewto55555"][quote="herefishyfishy1"]I think it depends on their Individual Values for speed.[/quote]

I believe its decided on which trainer throw the ball first. IDK how thats decided though.[/quote]

I know my kyogre is so much faster than the groudons I have faced(link battle). And my kyogre always won against a groudon. :P

[quote="Nerd_of_the_Ages"][quote="dragon96"][quote="herefishyfishy1"]Kyogre obviously wins. Hydro Pump is super effective AND gets a 1.5x Drizzle bonus and a 1.5x Same Type Attack Bonus. Compare that with Solar Beam. Same power, 0.5x not-very-effective, no STAB. Earthquake gets no weather bonus AND has a lower power.

Oh, and Water Spout and Eruption actually have HIGHER power if HP is higher.[/quote]

1) Kyogre doesn't get a Drizzle bonus because Groudon is slower by a little bit, so Drizzle goes first, then Drought comes up immediatly after. In fact the Hydro Pump is halved because of Drought.

2) Solar beam would take one move, no STAB, but is super effective.[/quote]
I thought it would be like, rain, then sun, then rain, then sun, so on.[/quote]

The slower pokemon's effect comes to battle, not alternate.

[quote="mewto55555"][quote="theprodigy"]what's grass knot?

let me guess... physical grass-type :roll:[/quote]

Not only is it physical grass type, but it increass in power based on your opponents weight, which makes it beastly against Kyogre/Milotic.[/quote]

Grass knot is freaking special.

NOOO! The pokemon threads are dying!


calculus
integration
function
algorithm
calculus computations

hello
 please try calculate the integral $ \int {\frac{{{e^{2x}} \minus{} {e^x}}}{{{e^x} \minus{} x}}dx}$

hello, i think there is no known primitive function of your integral.
Sonnhard.

If you really want to check that it has no elementary antiderivative, you could either do the Risch algorithm by hand (and spend about a day on it) or looking for some tricky way to use Liouville's theorem.


How many integers between 200 and 300 are perfect squares?


Let $a,b,c$ be the dimensions of a right paralilepide and $d$ its diagonal.
If $a,b,c,d \in N$ prove that

$\frac{ab+bc+ca}{a+b+c+d} \in N$  .!!!

 :)  :)

What is that, is it just a rectangular prism?

The expression is

$\frac{(a+b+c)^2 - d^2}{2(a+b+c+d)} = \frac{a+b+c-d}{2}$

which is trivial to prove is an integer.

$(a+b+c+d)^2$ and $2(a+b+c+d)d$ are both divisible by $a+b+c+d$, so subtract them to get that $2(ab+bc+ca)$ is divisible by $a+b+c+d$.  Then WOP and WLOG one of a, b, c is odd so two must be even and d must be odd; then 2 | a+b+c+d but not 4.  Verify that 2 | ab+bc+ca and then we're done.


Let $ABC$ be a triangle and let $D$ and $E$ be the points where the incircle of $ABC$ touches the sides $BC$ and $AC$ respectively.
Prove: If $AD=BE$ then the triangle is isosceles.

As we know CE=CD using sine theorem in triangles $ADC, BEC$
 In $ADC \frac{sinC}{BE} EC=sinEBC$ and In BEC $\frac{sinC}{AD} DC=sinDAC$ so $sinDAC=sinEBC$
and $DAC=EBC$ it means AEDB cyclic chords $AD=BE$ so angles $<ABE=<BAE$ so it is obvious


This coming year I am planning on taking the multivariable sequence from EPGY.  Does anyone know how well these classes transfer to major institutions (MIT, Princeton, Northwestern, HMC...)?  Will I have to retake these classes or is it common to be able to move on?

I will also most likely be taking calculus C this summer at epgy (cause i taught myself and i think i'm lacking some fundamentals, but i'll be taking the bc exam anyways). For the people that have taken it, how is the class difficulty wise, was the material easy to learn, and how much contact did you make with your mentor? Also, what would you choose - Epgy calc c or CTY calc bc and why? Thanks people


Prove that $2\cdot \frac{2\cdot6\cdot \cdots \cdot (4n-6)\cdot(4n-2)}{n!}$ in a integer number.

Hint :

[hide]This is $(2n)! /n!^2= { 2n \choose n }$[/hide]

Are you sure that $2\cdot \frac{2\cdot6\cdot \cdots \cdot (4n-6)\cdot(4n-2)}{n!}=2\cdot\frac{(2n!)}{n!}$

[hide]\begin{eqnarray*} \frac{2\cdot6\cdot \cdots \cdot (4n-6)\cdot(4n-2)}{n!}&=&2^n\cdot \frac{1\cdot3\cdot \cdots \cdot (2n-3)\cdot(2n-1)}{n!}\\ &=&2^n\cdot \frac{(2n)!}{(n!)2\cdot 4\cdot \cdots \cdot (2n-2)\cdot(2n)}\\ &=&\frac{(2n)!}{(n!)^2}\\ &=&\binom{2n}{n}\\ \end{eqnarray*}

Since that expression can count things, it's a positive integer. Now this was half the given expression, and multiplication by 2 will still make it a positive integer.[/hide]

We all know that it's true.. but can you all prove that $\binom{2n}{n}$ is an integer?

$\binom{2n}{n}$ is the $n$th point in the $(2n+1)$th row of Pascal's triangle, and since the triangle is closed under the set of integers than the expression must be an integer.


geometry
circumcircle
conics
parabola
parallelogram
trapezoid
perpendicular bisector

Let a triangle $ABC$  with $AB=AC$  and a point $M$ which  varies on the segment $BC$ .Let a circle $(K,R_{1})$ which  passes through $M,\ B$ and it is tangent to $AB$  and let a circle $(L,R_{1})$  which passes through $C$ and $M$ and it is tangent to $AC$ .Prove that the perpendicular bisector of $KL$ passes through a fixed point .

This fixed point is $A$. The circles are tangent to the $AB$ and $AC$ respectively in $B$ and in $C$ so the powers of the point of point $A$ wrt. to these two circles are the same since $|AB|^{2}=|AC|^{2}$. $KL$ is a radical axis of these two circles, thus conclusion follows.

I do not see how the conclusion follows. KL is not the radical axis, but the center line of the 2 circles (K), (L) intersecting at $M \in BC$ and tangent to AB, AC at B, C, respectively.

Since the power of A to (K), (L) are equal, AM is their radical axis. Let N be the other intersection of (K), (L) different from M. $N \in AM$ is in the radical axis, $AM \cdot AN = AB^{2},$ $\frac{AM}{AB}= \frac{AB}{AN},$ triangles $\triangle ABM \sim \triangle ANB$ with the common angle at A are similar by SAS, $\angle ANB = \angle ABM \equiv \angle ABC,$ N lies on the circumcircle (O) of $\triangle ABC.$ Let D be the midpoint of BC and let the perpendicular bisector $AD \equiv AO$ of BC meet (O) again at Z. AZ is a diameter of (O), $\angle ANZ$ is right. Since the radical axis AN is perpendicular to the center line KL, $KL \parallel NZ.$ Let M', N' be feet of a common normal from O to the parallel lines $KL \parallel NZ,$ $OM' \equiv ON' \parallel AM \equiv AN,$ and let a parallel to BC through M' meet AZ at D'. The right triangles $\triangle ON'Z \sim \triangle ANZ$ are centrally similar with similarity center Z and coefficient 1/2. Since $M'N' = \frac{MN}{2},$ so are the parallel lines $M'D' \parallel MD \equiv BC,$ hence $ZD' = \frac{ZD}{2}.$ Thus the line M'D', the perpendicular bisector of ZD, is fixed and it is the locus of the feet of normals from O to KL. This means that KL is tangent to a parabola with the focus O and the vertex tangent identical with the perpendicular bisector of ZD. Obviously this parabola passes through B, C. Parabola tangents do not pass through a fixed point.

Sorry I made a mistake in the first post .The problem is
Let a triangle $ABC$  with $AB=AC$  and a point $M$ which  varies on the segment $BC$ .Let a circle $(K,R_{1})$ which  passes through $M,\ B$ and it is tangent to $AB$  and let a circle $(L,R_{1})$  which passes through $C$ and $M$ and it is tangent to $AC$ .Prove that the perpendicular bisector of  $KL$ passes through a fixed point

[quote="yetti"]Since the power of A to (K), (L) are equal, AM is their radical axis. Let N be the other intersection of (K), (L) different from M. $N \in AM$ is in the radical axis, $AM \cdot AN = AB^{2},$ $\frac{AM}{AB}= \frac{AB}{AN},$ triangles $\triangle ABM \sim \triangle ANB$ with the common angle at A are similar by SAS, $\angle ANB = \angle ABM \equiv \angle ABC,$ N lies on the circumcircle (O) of $\triangle ABC.$ ... let the perpendicular bisector AO of BC meet (O) again at Z. AZ is a diameter of (O), $\angle ANZ$ is right. Since the radical axis AN is perpendicular to the center line KL, $KL \parallel NZ.$[/quote]
The perpendicular bisector of NZ is a midline of the right triangle $\triangle ANZ$ parallel to AN, hence it passes through the the midpoint of AZ - circumcenter O, which is fixed. The isosceles triangles $\triangle BKM \sim \triangle MLC \sim \triangle BZC$ are all similar, having equal base angles $\angle MBK \equiv \angle CBZ = 90^\circ-\angle ABC = 90^\circ-\angle ACB = \angle BCZ \equiv \angle MCL$ Thus KZLM is a parallelogram, KZ = LM = LN. KLNZ is then an isosceles trapezoid, so that KL, NZ have a common perpendicular bisector passing through O.

[quote="silouan"][color=darkred]Let a triangle $ABC$  with $AB=AC$  and a point $M$ which  varies on the segment $BC$ .Let a circle $w_{1}=C(I_{1},r_{1})$ which  passes through $M,\ B$ and it is tangent to $AB$  and let a circle $w_{2}=C(I_{2},r_{2})$  which passes through $C$ and $M$ and it is tangent to $AC$ .Prove that the perpendicular bisector of $I_{1}I_{2}$ passes through a fixed point .[/color][/quote]
[color=darkblue]Denote the circle $w=C(I_{0},r_{0})$ which is tangent in the points $B$ , $C$ to the lines $AB$ , $AC$ respectively.
At the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=113718 you proved that $r_{1}+r_{2}=r_{0}$ (constant).
The your problem becomes a well-known property of an isosceles triangle (applying to the triangle $BI_{0}C$ and the mobile points $I_{1}$ , $I_{2}$) :[/color]

[color=darkred][b]Enunciation.[/b] Let $ABC$ be a isosceles triangle $(AB=AC)$ and let $M\in (AB)$ , $N\in (AC)$ be two mobile points so that $BM=AN\ .$
Then the perpendicular bisector of the segment $[MN]$ passes through a fixed point.

[b]Proof.[/b] Denote : the middlepoints $D,E,F,P$ of the sides $[BC],[CA],[AB]$ and the segment $[MN]$ respectively ; the point $L\in AD$ for which $LP\perp MN$ ; the projections $U,V$ of the point $L$ to the sides $[AB],[AC]$ respectively. Prove easily that $P\in [EF]\ .$ Thus, the quadrilaterals $MULP$ and $NVPL$ are cyclically $\implies$ $\widehat{PUL}\equiv\widehat{PML}\equiv\widehat{PNL}\equiv\widehat{PVL}$ $\implies$ $\widehat{PUL}\equiv\widehat{PVL}$ and $UV\parallel BC$ because $AU=AV$ $\implies$ $P\in UV\ .$ Therefore, $UV\equiv EF$, i.e. $U\equiv F$ and $V\equiv E$ $\implies$ the point $L$ is fixed and $LA=\frac{AB^{2}}{2\cdot AD}\ .$ [/color]


Find all integers $n$ such that:

$\frac{2^n+1}{n^2} \in \mathbb{Z}$

I would have thought this would have been posted before, alas i cannot find it, as i dont know what year it was from...

You can see the solution this
[url=http://www.kalva.demon.co.uk/imo/isoln/isoln903.html]www.kalva.demon.co.uk/imo/isoln/isoln903.html[/url] ;)

Find all$(a,n)$such that:$a,n\in Z^+$and
              $n^a|(a^n+1)$
 :)


function
inequalities
algebra
domain
inequalities unsolved

Prove, that if the inequality:

\[f\left(\frac{x+y}{2}\right)\geq\frac{f(x)+f(y)}{2}\]

holds for any $x,y$ from the domain of $f$, then:

\[f(\alpha x+(1-\alpha)y)\geq\alpha f(x)+(1-\alpha)f(y)\]

holds for any $x,y$ and $\alpha >0$

jensen's define convex by this way.  and next equation define convex function now. listen it's study in your class. you must know it . I can say: it's to analogisly defination of convex function.

See [i]Topics in Inequalities- Theorems and Techniques[/i] at [url=http://ultrametric.googlepages.com/]Hojoo Lee's website[/url]. The proof is at the beginning of chapter 4.