calculus
integration
trigonometry
function
limit
complex analysis
complex analysis unsolved

Calculate

    1) \[\int_{|z|=2}\frac{1}{z^{2}+1}\cdot\cos\frac{1}{z}dz\] 2) \[\int_{|z|=2}\frac{1}{(z+i)^{2}}\cdot e^{\frac{1}{z}}dz\] 3) \[\int_{0}^{\infty}\left(\frac{1}{x^{4}+1}\right)^{2}dx\] using Rezidue theorem.

     4) \[\int_{0}^{2\pi}\frac{dx}{(\sin x+2)^{2}}\] using Rezidue theorem

     5) \[\int_{0}^{2\pi}\frac{\cos x}{5-4\cos x}dx\] using Rezidue theorem.

[quote="Cezar Lupu"]Calculate

    1) \[\int_{|z|=2}\frac{1}{z^{2}+1}\cdot\cos\frac{1}{z}dz\] [/quote]

1) $\int_{|z|=2}\frac{1}{z^{2}+1}\cdot\cos\frac{1}{z}dz= \sum_{k=0}^{\infty}(-1)^{k}\int_{|z|=2}z^{2k}\cos\frac{1}{z}dz$
$=\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\frac{(-1)^{k+j}}{(2j)!}\int_{|z|=2}z^{2(k-j)}dz=0$

This really doesn't seem right... I'll come back to this later.

EDIT: No, it's right. Since $f(z)=\frac{1}{z^{2}+1}\cdot\cos\frac{1}{z}$ is an even function the Laurent series for $f(z)$ contains only even powers of z, and hence $a_{-1}=0,$ another way to put it is that the residues of $f(z)$ at $z=\pm i$ are negatives of each other so that their sum cancels.

[quote="Cezar Lupu"]Calculate

2) \[\int_{|z|=2}\frac{1}{(z+i)^{2}}\cdot e^{\frac{1}{z}}dz\] [/quote]

 2) Let $J: =\int_{|z|=2}\frac{1}{(z+i)^{2}}\cdot e^{\frac{1}{z}}dz = \sum_{k=0}^{\infty}\frac{1}{k!}\int_{|z|=2}\frac{1}{z^{k}(z+i)^{2}}dz$
$= 2\pi i \sum_{k=0}^{\infty}\frac{1}{k!}\left[ \mbox{Res}_{z=-i}\left(\frac{1}{z^{k}(z+i)^{2}}\right)+\mbox{Res}_{z=0}\left(\frac{1}{z^{k}(z+i)^{2}}\right) \right]$
$= 2\pi i \sum_{k=0}^{\infty}\frac{1}{k!}\left[ \mbox{Res}_{z=-i}\left(\frac{1}{z^{k}(z+i)^{2}}\right)+\mbox{Res}_{z=0}\left(\frac{1}{z^{k}(z+i)^{2}}\right) \right]$

using the [b]theorem[/b]: if $z_{0}$ is a pole of order $m\geq 2$ of $g(z),$ then \[\mbox{Res}_{z=z_{0}}g(z) = \lim_{z\rightarrow z_{0}}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}\left[ (z-z_{0})^{m}g(z)\right] .\] we can calculate the residues to be \[\mbox{Res}_{z=0}\left(\frac{1}{z^{k}(z+i)^{2}}\right)= \lim_{z\rightarrow 0}\frac{1}{(k-1)!}\frac{d^{k-1}}{dz^{k-1}}\left(\frac{1}{(z+i)^{2}}\right)\]  \[= \lim_{z\rightarrow 0}\frac{1}{(k-1)!}\left(\frac{(-1)^{k-1}k!}{(z+i)^{k+1}}\right) = ki^{k+1}\] and the other residue calculation is \[\mbox{Res}_{z=-i}\left(\frac{1}{z^{k}(z+i)^{2}}\right)= \lim_{z\rightarrow-i}\frac{d}{dz}\left(\frac{1}{z^{2}}\right) =-\frac{2}{i^{3}}=-2i\] and so the integral becomes

$J= 2\pi i \sum_{k=0}^{\infty}\frac{1}{k!}\left(-2i+ki^{k+1}\right) = 2\pi i \left(-2i\sum_{k=0}^{\infty}\frac{1}{k!}+\sum_{k=1}^{\infty}\frac{i^{k+1}}{(k-1)!}\right)$
$= 2\pi i \left(-2i e-\sum_{k=0}^{\infty}\frac{i^{k}}{(k)!}\right) = 2\pi i \left(-2i e-e^{i}\right)$
$= 2\pi \left(2e-ie^{i}\right)$

somebody check this!

TIA,
   --Ben

[quote="Cezar Lupu"]
     5) \[I=\int_{0}^{2\pi}\frac{\cos x}{5-4\cos x}dx\] [/quote]

Let $z= e^{i x}$ . Then $\cos x = \frac{ e^{i x}+e^{-i x}}{2}= \frac{ z^{2}+1}{2z}$ and  $dx = \frac{dz}{iz}$ .

Then $I = \int_{C}\frac{i}{4}\cdot \frac{z^{2}+1}{z(z-2)(z-\frac{1}{2})}$ where $C$ is the unit circle

and the singularities of the integrand are $z=0, \, z=2, \, z=\frac{1}{2}$
therefore the function has simple poles at $z=0, \, z=\frac{1}{2}$ .

Let $f(z) =\frac{i}{4}\cdot \frac{z^{2}+1}{z(z-2)(z-\frac{1}{2})}$.

The residues are

$\ \mbox{Res}_{z=0}(f(z)) =\lim_{z \to 0 }z f(z) = \frac{i}{4}$

$\ \mbox{Res}_{z=\frac{1}{2}}(f(z)) =\lim_{z \to \frac{1}{2}}(z-\frac{1}{2}) f(z) =-\frac{5i}{12}$

Now by Residue theorem $I=2 \pi i \cdot \left( \frac{i}{4}-\frac{5i}{12}\right) = \frac{ \pi}{3}$

[quote="Cezar Lupu"]Calculate
 3) \[I=\int_{0}^{\infty}\left(\frac{1}{x^{4}+1}\right)^{2}dx\] [/quote]

let $C$ be the contour consisting of the real axis from $-R$ to $R$ and the upper half-plane semicircle of radius $R >1$ .

It's \[\int_{C}\frac{1}{(z^{4}+1)^{2}}\, dz = \int_{-R}^{R}\frac{1}{(x^{4}+1)^{2}}\, dx+\int_{0}^{ \pi}\frac{i R e^{i \theta }}{(1+R^{4}e^{4 i \theta })^{2}}\, d \theta .\] The singularities of $f(z)= \frac{1}{(z^{4}+1)^{2}}$ are $z=e^{\left( \frac{ \pi}{2}\cdot k+\frac{ \pi }{4}\right)i }$ for $k=0,1,2,3$ and only for $k=0,1$ they are inside $C$ .

The residues are $\ \mbox{Res}_{z=e^{ \frac{ \pi }{4}i }}(f(z))=-\frac{3+3i}{16 \sqrt{2}}$ and $\mbox{Res}_{z=e^{\left( \frac{ \pi }{4}+\frac{ \pi }{2}\right)i}}(f(z))= \frac{3-3i}{16 \sqrt{2}}$ and by Residue theorem $\int_{C}\frac{1}{(z^{4}+1)^{2}}\, dz = \frac{3 \pi }{4 \sqrt{2}}$

Also \[\left| \int_{0}^{ \pi}\frac{i R e^{i \theta }}{(1+R^{4}e^{4 i \theta })^{2}}\, d \theta \right| \leq \frac{ \pi R}{(R^{4}-1)^{2}}\stackrel{ R \to \infty }{ \longrightarrow}0\] therefore \[\int_{-\infty}^{\infty}\frac{1}{(x^{4}+1)^{2}}\, dx = \lim_{R \to \infty}\int_{-R}^{R}\frac{1}{(x^{4}+1)^{2}}\, dx = \lim_{R \to \infty}\int_{C}\frac{1}{(z^{4}+1)^{2}}\, dz = \frac{3 \pi }{4 \sqrt{2}}\] and \[\int_{0}^{\infty}\frac{1}{(x^{4}+1)^{2}}\, dx = \frac{1}{2}\[ \int_{-\infty}^{\infty}\frac{1}{(x^{4}+1)^{2}}\, dx = \frac{3 \pi }{8 \sqrt{2}}\]

$4$ can be solved in the same way as $5$.

We have that $\sin z = \frac{z-z^{-1}}{2i}$, so it suffices to compute $\oint_{\mathcal C}\frac{4iz}{\left( z^{2}+4iz-1 \right)^{2}}\, dz$.

The only residue in the unit circle is $\left( \sqrt 3-2 \right) i$. We have that \[\textrm{Res}\left( \frac{4iz}{\left( z^{2}+4iz-1 \right)^{2}}, \left( \sqrt 3-2 \right) i \right) = \left. \frac{d}{dz}\frac{4iz}{\left( z+\left( \sqrt 3+2 \right) i \right)^{2}}\right|_{z = \left( \sqrt 3-2\right) i}=-\frac{2 \sqrt 3}9 i .\] The wanted integral is $2 \pi i \cdot \left(-\frac{2 \sqrt 3}9 i \right) = \frac{4 \pi \sqrt 3}9$.

I get the same result with Mathematica.


geometry
inequalities
trigonometry
geometry proposed

If $ x,y,z$ be real numbers, then

$ a^2(x\minus{}y)(x\minus{}z)\plus{}b^2(y\minus{}z)(y\minus{}x)\plus{}c^2(z\minus{}x)(z\minus{}y)\geq0$

[quote="aadil"]where is the geometry in the problem?[/quote]

Sorry
$ a,b,c$ side lenghts

[quote="Ligouras"]If $ x,y,z$ be real numbers, then

$ a^2(x \minus{} y)(x \minus{} z) \plus{} b^2(y \minus{} z)(y \minus{} x) \plus{} c^2(z \minus{} x)(z \minus{} y)\geq0$[/quote]
Let $ n\equal{}x\minus{}y,m\equal{}y\minus{}z,p\equal{}z\minus{}x$. Then $ n\plus{}m\plus{}p\equal{}0$. Since $ nm \cdot mp \cdot pn \equal{} (nmp)^2 \ge 0$ wlog assume $ nm \ge 0$, and we have to prove:
$ \minus{}(a^2np\plus{}b^2nm\plus{}c^2mp) \equal{} \minus{}b^2nm \plus{} (n\plus{}m)(a^2n\plus{}c^2m) \ge 0 \iff$
$ \minus{}b^2nm \plus{} (an)^2\plus{}(cn)^2\plus{}nm(a^2\plus{}c^2) \ge \minus{}b^2nm \plus{} 2nmac \plus{} nm(a^2\plus{}c^2) \equal{} nm( (a\plus{}c)^2\minus{}b^2) \ge 0$ which is obvious since $ a\plus{}c \ge b$. QED

[quote="Ligouras"]Prove that for any real numbers  $ x,y,z$ and for any triangle $ ABC$ there is the inequality $ \sum a^2(x \minus{} y)(x \minus{} z)\ge 0$ .[/quote]

[b]Lemma.[/b] For any real numbers $ u,v,w$ so that $ u\plus{}v\plus{}w\equal{}0$ and for any triangle $ ABC$ exists the inequality $ vwa^2\plus{}wub^2\plus{}uvc^2\le 0$ .

From upper lemma for $ u\equal{}y\minus{}z$ , $ v\equal{}z\minus{}x$ , $ w\equal{}x\minus{}y\ : \ \sum$ $ vwa^2\equal{}\sum a^2(x\minus{}y)(z\minus{}x)\le 0\Longleftrightarrow\sum a^2(x\minus{}y)(x\minus{}z)\ge 0$ .

[quote="Virgil Nicula"][quote="Ligouras"]Prove that for any real numbers  $ x,y,z$ and for any triangle $ ABC$ there is the inequality $ \sum a^2(x \minus{} y)(x \minus{} z)\ge 0$ .[/quote]

[b]Lemma.[/b] For any real numbers $ u,v,w$ so that $ u \plus{} v \plus{} w \equal{} 0$ and for any triangle $ ABC$ exists the inequality $ vwa^2 \plus{} wub^2 \plus{} uvc^2\le 0$ .

From upper lemma for $ u \equal{} y \minus{} z$ , $ v \equal{} z \minus{} x$ , $ w \equal{} x \minus{} y\ : \ \sum$ $ vwa^2 \equal{} \sum a^2(x \minus{} y)(z \minus{} x)\le 0\Longleftrightarrow\sum a^2(x \minus{} y)(x \minus{} z)\ge 0$ .[/quote]
The problem is clearly equivalent with the above lemma, but i fail to see an obvious proof of the fact that $ vwa^2 \plus{} wub^2 \plus{} uvc^2\le 0$..

[quote="Virgil Nicula"][b]Lemma.[/b] For any real numbers $ u,v,w$ so that $ u \plus{} v \plus{} w \equal{} 0$ and for any triangle $ ABC$ exists the inequality $ vwa^2 \plus{} wub^2 \plus{} uvc^2\le 0$ .[/quote]
[b]Proof.[/b] $ vwa^2 \plus{} u\left( wb^2 \plus{} vc^2\right)\le 0\Longleftrightarrow$ $ vwa^2\le (v \plus{} w)\left(wb^2 \plus{} vc^2\right)\Longleftrightarrow$ $ v^2c^2 \plus{} vw\left(b^2 \plus{} c^2 \minus{} a^2\right) \plus{} w^2b^2\ge 0\Longleftrightarrow$

$ v^2c^2 \plus{} 2bcvw\cos A \plus{} w^2b^2\ge 0\Longleftrightarrow$ $ \left(vc \plus{} wb\cos A\right)^2 \plus{} w^2b^2\sin ^2A\ge 0$ , what is truly. We have equality iff $ u \equal{} v \equal{} w \equal{} 0$ .

[b]Remark.[/b] This lemma is well-known and often usefully. For example, here is two particular cases :

$ \ \bullet\ u \equal{} b \minus{} c\ ,\ v \equal{} c \minus{} a\ ,\ w \equal{} a \minus{} b$ $ \Longrightarrow\ \sum (c \minus{} a)(a \minus{} b)a^2\le0\ \Longrightarrow$

$ \underline {\overline {\left\|\ a^2(a \minus{} b)(a \minus{} c) \plus{} b^2(b \minus{} c)(b \minus{} a) \plus{} c^2(c \minus{} a)(c \minus{} b)\ \ge\ 0\ \right\|}}$ ([u]inegalitatea [b]Schur[/b][/u] pentru $ r \equal{} 2$ ).

$ \ \bullet\ u \equal{} a(b \minus{} c)\ ,\ v \equal{} b(c \minus{} a)\ ,\ w \equal{} c(a \minus{} b)$ $ \Longrightarrow\ \sum b(c \minus{} a)c(a \minus{} b)a^2\le0\ \Longrightarrow$

$ \boxed { a(a \minus{} b)(a \minus{} c) \plus{} b(b \minus{} c)(b \minus{} a) \plus{} c(c \minus{} a)(c \minus{} b)\ \ge\ 0\ }$ ([u]inegalitatea [b]Schur[/b][/u] pentru $ r \equal{} 1$ ).

[b][u]Proposed problem (own)[/u].[/b] In a triangle $ ABC$ there is the inequality $ 0\le \sum a^2(a \minus{} b)(a \minus{} c)\le 2\cdot\sum a^2(b \minus{} c)^2$ .


Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this cryptarithmetic equation. Neither [b]N [/b]nor [b]V[/b] is zero.

([b]NET[/b])^3 = [b]VULTURINE[/b]

The solution which satisfies the above "cryptarithmetic equation" as well as conforms to all the rules as stated is given below:

[b] (682)^ 3 = 3 1 7 2 1 4 5 6 8 [/b] where  [b]  N =6, E=8, T=2, V=3, U=1, L=7, R=4, I=5. [/b]

It's not too hard to brute-force this, as there are few possibilities. (Note that VULTURINE ends in NE, while NET begins with NE).

First, let's think: how many cubes end in a different digit than their cube root? 6 possibilities are out:

last digit 0 -> 0^3 = 0
last digit 1 -> 1^3 = 1
last digit 4 -> 4^3 = 64
last digit 5 -> 5^3 = 125
last digit 6 -> 6^3 = 216
last digit 9 -> 9^3 = 729

Now you're down to 4 candidates.

last digit 2 -> 2^3 = 8 -> 82^3 = ...68 -> 682
last digit 3 -> 3^3 = 27 -> 73^3 = ...17 -> 173
last digit 7 -> 7^3 = 343 -> 37^3 = ...53 -> 537
last digit 8 -> 8^3 = 512 -> 28^3 = ...52 -> 528

682^3 = 317214568
173^3 = 5177717
537^3 = 154854153
528^3 = 147197952

It's easy to see only 682 works.

[quote] It's not too hard to brute-force this, as there are few possibilities. (Note that VULTURINE ends in NE, while NET begins with NE). It's easy to see only 682 works. [/quote]

Well, As far as I'm concerned, I acknowledge (to a certain extent) what you have said. But I personally felt that problem was never as easy as depicted by you and the underlying logic was not so obvious after-all.


search
number theory unsolved
number theory

$ p^{x}\equal{}y^4\plus{}4\equal{}(y^2\plus{}2)^2\minus{}(2y)^2\equal{}(y^2\plus{}2\minus{}2y)(y^2\plus{}2\plus{}2y)$ we have two cases
$ 1\minus{}\mbox{ y is odd} \implies \gcd(y^2\plus{}2\minus{}2y,y^2\plus{}2\plus{}2y)\equal{}1 \implies$
$ y^2\plus{}2\minus{}2y\equal{}1,and,y^2\plus{}2\plus{}2y\equal{}p^{x}$
$ \implies y\equal{}1 \implies p\equal{}5,x\equal{}1$
$ 2\minus{}\mbox{ y is even } \implies p\equal{}2,and,x<3 ,contradiction$.

[quote="makar"]Problem 2:
Find all triples $ (p, x, y)$ such that $ p^{x} \equal{} y^{4} \plus{} 4$, where $ p$ is a prime and $ x, y$ are natural numbers.[/quote]
We may rewrite the equation as $ p^x \equal{} (y^2 \minus{} 2y \plus{} 2)(y^2 \plus{} 2y \plus{} 2)$,which leads to $ y \equal{} 1 \Rightarrow x \equal{} 1$ and $ p \equal{} 5$;or $ y^2 \minus{} 2y \plus{} 2 \equal{} p^a$ and $ y^2 \plus{} 2y \plus{} 2 \equal{} p^b$ for some integers $ a$ and $ b$ such that $ a \plus{} b \equal{} x$.We have $ 4y \equal{} p^b \minus{} p^a \equal{} p^a(p^{b \minus{} a} \minus{} 1)$ from which we deduce that $ p|y$ (or $ p\equal{}2$).Back to the initial equation,looking at it $ (mod p)$ we conclude that $ p|4$,hence $ p \equal{} 2 \Rightarrow x \equal{} 2$ and $ y \equal{} 0$.

All the triplets that satisfy the equation are $ (2,2,0)$ and $ (5,1,1)$.

rachid:All the triplets that satisfy the equation are $ (2,2,0)$ and $ (5,1,1)$.
but $ (2,2,0)\notin{N^3}$....


prove that there exist two conbsecutive squares with at least 666 primes between them.

Are we allowed to use Tchebyscheff's bounds on the number of prime number in $[1;x]$ ($x \in \mathbb{N}$) ?
Or do you want a more elegant way ?

yeah if you mean $\frac{n}{ln n}$.(or maybe sth else.)

It is obviosly. Let N is big number. We have N intervales $(N^2,(N+1)^2),((N+1)^2,(N+2)^2)),\dots ((2N-1)^2,4N^2)$. In this N intervales we have at least $aN^2/ln(N)$ primes. Therefore exist n: $N\len<2N$, suth that in interval (n^2,(n+1)^2) containst at least $aN/ln(N)>666$ primes if N is large.


algebra
polynomial
number theory proposed
number theory

Let $\mathfrak F(x) \in \mathbb{Z} [x]$ and $p,q \in \mathbb{Z},p \neq q$ such that

$|\mathfrak F(q)|=|\mathfrak F(p)|=1$.

also let $\alpha \in \mathbb{Q},\mathfrak F(\alpha)=0$ :

Prove that :

$\alpha=\frac12(p+q)$     ,           and ,                 $|p-q|=1 \ or 2$

Let $\alpha=\frac rs$ where $r,s\in \mathbb Z$ and $(r,s)=1$.
Then $\mathfrak F(x)=(sx-r)g(x)$ for some $g(x)\in \mathbb Z[x]$ by gauss's lemma.
So we arrive at $|sp-r|=1$ and $|sq-r|=1$. If both of them are $1$ or $-1$ then $s=0$ which is contradiction.
So without loss of generality we have $sp-r=1$ and $sq-r=-1$.
By summing them we arrive at $s(p+q)-2r=0$ and so $\frac rs=\frac{(p+q)}{2}$.
By subtracting them we get $s(p-q)=2$ and so $p-q=\frac 2s$ so $|p-q|=\frac{2}{|s|}\leq 2$ which gives the second result.


superior algebra
superior algebra unsolved

We define the subfield $ E$ of $ \mathbb{R}$ as ${ E=\mathbb{Q}(\{2^{\frac{1}{n}}}| n \in \mathbb{N}\})$.
How can we prove that $ E$ is algebraic over $ \mathbb{Q}$ ?

Every $ x\in E$ is in some subfield $ \mathbb{Q}(2, \sqrt{2}, \ldots, \sqrt[k]{2})$ of $ E$ for some positive integer $ k$. This subfield of $ E$ is finite and thus algebraic over $ \mathbb{Q}$, so $ x$ is algebraic over $ \mathbb{Q}$.

[quote="Yimin Ge"]Every $ x\in E$ is in some subfield $ \mathbb{Q}(2, \sqrt {2}, \ldots, \sqrt [k]{2})$ of $ E$ for some positive integer $ k$.[/quote]

I can not make that rigorous  :blush:

Define $ E_k: \equal{} \mathbb{Q}(2,\sqrt{2},\ldots, \sqrt[k]{2})$ for positive integers $ k$. Then
\[ E \equal{} \bigcup_{k\equal{}1}^\infty E_k,
\]
so if $ x\in E$, it follows that $ x\in E_k$ for some positive integer $ k$.

And why is that equality true ?
My only problem is the inclusion of $ E$ in the right side.

Clearly,
\[ E': \equal{}\bigcup_{k\equal{}1}^\infty E_k \subset E.
\]
But $ \{ E_k \mid k\in \mathbb{Z}^\plus{}\}$ is a chain of fields wrt to set inclusion, so $ E'$ is a field. On the other hand, for each positive integer $ k$, $ \sqrt[k]{2}\in E'$. Thus, $ E'$ is a field containing $ \mathbb Q$ and $ \{ \sqrt[k]{2} \mid k\in\mathbb Z^\plus{}\}$ and since $ E$ is the smallest such field, it follows that $ E\subset E'$.


In the solid shown, all edges except edge $\overline{EF}$ have length $s.$ Edge $\overline{EF}$ has length $2s$. Given that $s=6\sqrt{2}$, what is the volume of the solid?
[img]6436[/img]

This has already been posted several times  :huh:

[hide="Hint"] Make two of the solid.  Try to piece them together to form a familiar shape. [/hide]

Or cut the irregular looking pieces off and sum the volumes separately as a prism and two pyramids.


Στο site της ΕΜΕ ανεβηκε μια πολυ ωραια ασκηση που προτεινε ο κ.Ψυχας σαν ασκηση του μηνα. Απο την ασκηση προκυπτει και ενα αλλο ενδιαφερον ερωτημα που φανταζομαι θα ενδιαφερει τον κ. Στεργιου. Εστω ημικυκλιο διαμετρου ΜΝ και τυχον σημειο Κ της ΜΝ. Οριζω επι του ημικυκλιου τα Α ,Β ,C ,D ωστε $ \angle{MKA}\equal{}\angle{DKN}$ και $ \angle{BKA}\equal{}\angle{CKD}$. ΝΔΟ οι BC, AD τεμνονται πανω στην ΜΝ.

Για την ωρα δεν το εχω ψαξει αρκετα το ερωτημα που προσθεσες και δεν εχω βρει κατι. Εχω ομως λυση στο ερωτημα της ΕΜΕ, αν θες να την γραψω πες μου.

[quote="Sakis"]Την εχω λυσει και εγω την ασκηση αλλα πιστευω πως, για λογους δεοντολογικους, δεν πρεπει να γραψουμε λυση στο internet πριν την ΕΜΕ.[/quote]

Δεν νομιζω να υπαρχει καποιο προβλημα, αφου η λυση της ασκησης μπορουσε να σταλει μεχρι τις 20 Νοεμβριου. Απλα η ΕΜΕ παντα αργει λιγο να ανεβασει τις λυσεις της. Βασικα ακομα δεν εχει ανεβασει τις λυσεις της προηγουμενης ασκησης και μιας αλλης περσινης

Αυτη ηταν περιπου και η δικια μου λυση, δηλαδη εδειξα πως τα Κ, Ε, Ζ ανηκουν στον ριζικο αξονα των ΑΚΕΒ και ΚΕCD.

[quote="NickNafplio"][quote="Sakis"]Δεν το ηξερα οτι η προθεσμια ειχε ληξει. Σε αυτην την περιπτωση θα ηθελα να δω την λυση που εδωσες.[/quote]

Οκ

[hide]Εστω $ A',B',C',D'$ τα συμμετρικα των $ A,B,C,D$ ως προς την ευθεια $ MN$ αντιστοιχα, τα οποια ανηκουν στο αλλο μισο ημικυκλιο του κυκλου με διαμετρο $ MN$

Απο τη συμμετρια εχω: $ KA \equal{} KA', KB \equal{} KB', KC \equal{} KC', KD \equal{} KD'$. Ακομα απο τη συμμετρια ειναι: $ \angle{MKA} \equal{} \angle{MKA'} \equal{} \angle{DKN}$, που σημαινει οτι οι γωνιες $ MKA'$ και $ DKN$ ειναι κατα-κορυφην και επομενος τα $ A',K,D$ ειναι συνευθειακα. Ομοιως ειναι και $ B,K,C':$ συνευθειακα (διοτι ειναι και $ \angle{BKM} \equal{} \angle{CKN}$). Απο τη δυναμη του $ K$ ως προς τον κυκλο εχω τωρα: $ A'K*KD \equal{} B'K*KC \Leftrightarrow AK*KD \equal{} BK*KC \Leftrightarrow \frac {AK}{KC} \equal{} \frac {BK}{KD}$, και επηδη και οι γωνιες $ AKC$ και $ BKD$ ειναι ισες, εχουμε οτι τα τριγωνα $ \triangle{AKC}$ και $ \triangle{BKD}$ ειναι ομοια.

Επομενος ειναι $ \angle{KAC} \equal{} \angle{KBD}$, και $ \angle{KBD} \equal{} \angle{KCA}$, αρα τα $ AKEB$ και $ KECD$ ειναι εγγραψημα, και επηδη απο τη δυναμη του $ Z$ ως προς το μεγαλο κυκλο ειναι: $ ZB*ZA \equal{} ZC*ZD$, εχουμε οτι το $ Z$ εχει ιδιες δυναμεις ως προς τους περιγεγραμενους κυκλους των 2 τετραπλευρων, αυτο ομως ειναι δυνατο μονο οταν η προεκταση της $ ZE$ διερχεται απο το $ K$ (αφου οι 2 κυκλοι τεμνονται στο $ E$ και στο $ K$), αρα τελικα εχουμε: $ Z,E,K:$ συνευθειακα και τελειωσαμε. [/hide][/quote]

Παιδιά σας ευχαριστώ πολύ. Θα την τυπώσω αύριο και θα δω αν την έχω στις σημειώσεις μου. Αν όχι , θα τη βάλω στο βιβλίο με τα ονοματά σας ως λύτες.
 Μπάμπης


superior algebra
superior algebra solved

Let $ G$ be a group with the property that there exists a prime $ p>2$ such that
$ p\plus{}\left|\text{Aut}\left(G\right)\right|\cdot\left|Z\left(G\right)\right|\equal{}\left|G\right|^2$, where $ \text{Aut}\left(G\right)$ is the group of all automorphisms of $ G$, and $ Z\left(G\right)$ is the center of $ G$. Find $ G$.

Since |Z(G)| | |G|, it means that |Z(G)| | p, so |Z(G)| is p or 1.

For a g in G we define (f_g)(x)=gxg^(-1) (internal automprphisms). Let M be the set of all such automorphisms.

Case 1: |Z(G)|=1
It's obvious that |M|=|G| because all internal automorphisms al distinct because of the fact that the center of G contains only e (unit element of G).
Since |M| | |Aut(G)|, it means that |G| | |Aut(G)|, so |G| | p, and since |G|>1 we get |G|=p, which means that |Z(G)|=p (G is cyclic so it's also commutative), contradiction.

Case 2: |Z(G)|=p
|M|=|G/Z(G)|=|G|/|Z(G)| | |G|; |M| | |Aut(G)| => |M| is 1 or p. We assume it's p. This means |G|=p^2, and it's a known fact that a group with p^2 elements (p prime) is commutative. I won't prove it here because the proof isn't that short, but it isn't that difficult either. This means |Z(G)|=p^2>p, contradiction. We have found that |M| is 1, so |G|=|Z(G)|=p.

It's easy to check that if |G|=p then |Aut(G)|=p-1 and |Z(G)|=p, so the relationship p+|Aut(G)|*|Z(G)|=|G|^2 is verified.

So the answer is: |G|=p.

My solution was quite different. First of all, I considered a in G and the authomorphism f(x)=axa^-1. Since AutG is a group, we must have f...f (|AutG| times)= identity. From here we deduce that a^|AutG| belongs to Z(G) and since Z(G) is a group, we must have a^(|AutG|*|Z(G)|)=e. Since a^|G|=e, we must have a^p=e for any a in G. From Cauchy's theorem we deduce that |G|=p^k, with k>=1. Suppose k>1. It is easy to prove that Z(G) has at least 2 elements, so, since |Z(G)|||G|, we must have p||Z(G)|. If p^2||Z(G)|, then p^2|p, false. Then |Z(G)|=p and then (|Aut(G)|,p)=1. Since a^|AutG| is in Z(G) and a^p is in G for any a, we must have a=a^((|AutG|, p)) is in Z(G), which means that any element of G is in Z(G), which means |G|=p, false. Hence k=1 and |G|=p.

I still believe that the two solutions are quite different.


Is the maathematical ability a student possesses genetic, or is it rather more in a sense that the more experience you have, the better?

I donno. I'm just a little confused on that issue.

Sorry if this topic is not in its right place..  :blush:

I think a strong environment is one huge factor. Genetics may play a role, but anyone can be an IMO winner, as far as I think.

But some people just seem naturally inclined to math. I was always better at math than all my other subjects, but that's not saying I'm any better at math than someone else....

My parents stoped being able to help me with math passed 3rd grade, so I think genetics can be overcome.

i think a man is good in math when he has more experience!

Heres an example:

eventhough your father has a lot of experience in math it can only be passed to you if he teaches you not by passing his genetics to you.
just like my father he is good in math but he didnt taught me when i was i child. i only learn when my mom taught me, so now im the best in our class :first: .

thats why a man is good or even best in math because he has a lot of experience! :cool:

That's true. Even if genetics plays a factor, it can only take you so far.

Ohhhhhhh im sooooo gooood at this :first:

However, I think genetics are important. Your ability to comprehend and absorb new material, much like your IQ, is more predetermined than learned. You cannot learn IQ - it's how quick your mind is. Some might say that people who can learn faster have an edge at competition mathematics because their boundaries are limitless.

That's not entirely true, if it were, math would have never been discovered or created.

I don't think anybody really knows the answer at this point. Part of it, I think, is raw ability (not necessarily genetic) to compute quickly.

That is true. I personally, and don't take offense to this anyone, believe that genetics can help you possibly absorb information at an accelerated pace with makes learning the math easier. However, the learning part is dependable on the environment. Once you are past the learning stage, the creation stage is what usually requires an incredible amount of insight that [i]could[/i] and probably is at least partly, depends on genetics.

Considering that my little brother and I are the two "math geniuses" at my school, either environment/experience or genentics must have something to do with math ability. Not that we had any great math training or w/e... But I personally don't believe in genetics affecting anything other than physical traits / personal tastes. I'd say math ability is largely post-natal.

And thus is the old nature vs. nurture question--which should probably go up to debate in the Round Table section or somewhere like that :D

Yeah this should probably go to the Round Table.....

I think that part of it has to do with genetics....

Part of it to do with the environment and what opportunities you've been given to learn the material...

And a lot of it to do with how motivated you are. You could be the brightest kid in the world but if you're not willing to work, you won't learn anything. On the other hand, you could be a fairly "dumb" kid, have lots of motivation, work really hard, and become really good at math. It would just take you a little longer to understand things.

And then there's the people who just don't fit anywhere into the stuff I wrote above.... but.... they're probably not on this website because they're the kids who just plain don't care about math.  *sad*  :(

 :)  :)  :)

I want to thank all of you for making those replies, and I mostly agree with you.

Here's what I think:

  There are those people in the world that are better problem solvers and learnners than others. But I believe many of the smart people out there today grown up as a normal child, but he/she has started to solve problems and think logically earlier than others. This is one of the main reasons why. Also it can be just genetics, but if he/she who is gifted is willing to work harder, I suppose  it makes them even more brilliant. And finally there are those who are inspired by their environment. But, the main thing is willing to challenge yourself to solve problems.

But then, another question arises. If a person is quick at learning mathematical material, then is he/she/it necessarily good at solving problems?

[quote="jimhu"]
But then, another question arises. If a person is quick at learning mathematical material, then is he/she/it necessarily good at solving problems?[/quote]

Not really. I know more math than alot of the perople around me, even those who are of higher grade, but they still seem to score highter than me on contests. I don't really know why this is, I just know that I can must it, as I can learn pretty quickly, so I have to study alot. I think that how "smart" you are is defined on how fast you can learn material and how easilly you can understand material. That may be a gift, as not all people can achieve what they hope to. If a smart person and a less-smart person both work 5 hours each day, who do you think learns faster?

[quote="cindy28"]but still ur mind wont absord or think of something unless it is taught:

like a gifted child that can understand something easily but he wont understand it unless it is explained by the guardian

gets???? ;)[/quote]

So, if this was true... who invented the telephone, the TV, the microchip, the steamboat, the airplane, the radio, the car, the bifocals, the elevator... goes on and on. I'm sure nobody taught these geniouses how to invent these amazing applications. Plus, in math, every problem you do for the first time has not been taught to you. You teach yourself how to do it by experimenting and fooling around with the problem until some method works. Not everything is pre-taught.

I think there is definitely a correlation, but there are many exceptions.  But if you can truly *understand* material quickly, I think you would very likely be able to do well on math contests and such.  I know for a fact that it took me 2 years (maybe 1.5) to understand algebra very well though (if what I know can be considered "very well").

i think ur offending me!
on other hand this is what i can say

well now i realized that the two of them have something to do on how the child interact with math cause the two of them help each other.

genetics help the experience
                   genetics helps the child to answer the problems

experience help the genetics
                   experience give information on how to improve his/her ability

so the experience and genetics help the child on how to interact with math

i think none of them is enough for being successful in math.
you can be hard-working guy without matematical talent or you can be a talented one without working. both of them are not enough.
you have to work hard but a little bit of talent(actually much more talent to be in IMO)

Personally, I believe the previous post was quite right, because there are people who are just not fit for math. While there are people who try quite hard and are normal and are quite good. :)

but still there are people who are really good in math

right?

like the one who developed the e=mc^2

I don't think it's so much genetics as it is how your raised.  Genetics play some role, but I think if one toddler was to read say 9 hours a day and another was to watch tv 9 hours a day there is no question which would be smarter.

Well, that would depend on what channels the toddler watches and what books the other reads  :lol:

[quote="white_horse_king88"]However, I think genetics are important. . . . You cannot learn IQ - it's how quick your mind is.[/quote]

That's not the most considered view of most psychologists. The best informed psychologists agree that IQ tests are "samples of learned behavior," and that IQ is malleable.

http://learninfreedom.org/iqbooks.html

[quote="Jimmy"]I have to agree that genetics can be a large factor; some people can comprehend calculus by the second grade.[/quote]

Do you have specific examples in mind? What is the definition here of "comprehend calculus"?

so.........genetics gives larger factor than experience??? :?

I do not believe there is some genetic component that sits around in your brain and makes it impossible for you to understand calculus, or algebra, or w/e. I mean, if you work hard at it, then there is no reason that you should not be able to understand.

I think people that do not do well in math do not put in the appropriate amount of effort in order to understand the material. That said, some people are just fast. Examples: John Conway and Terrence Tao.

Anyone can learn math.


Solve the system:
$ y^3\equal{}x^3(9\minus{}x^3)$
$ x^2y\plus{}y^2\equal{}6x$

hello, i have found $ x\equal{}0,y\equal{}0$, $ x\equal{}1,y\equal{}2$ and $ x\equal{}2,y\equal{}2$.
Sonnhard.

[color=darkblue]Hi! This is my solution:

$ \bullet$ If $ x=0 \Longrightarrow y = 0    \Longrightarrow  \begin {cases} x=0 \\ y= 0 \end {cases}$ is one root of this system.

$ \bullet$ If $ x \neq 0$ system $ \Longleftrightarrow \begin {cases} x^6+y^3=9x^3 \\ 3x^4y+3y^2x^2=18x^3 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6+y^3=9x^3 \\ x^6+3x^4y+3y^2x^2+y^3 =27x^3 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6+y^3=9x^3 \\ (x^2+y)^3 =27x^3 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6+y^3=9x^3 \\ x^2+y =3x \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6+y^3=9x^3 \\ y =3x-x^2 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6+(3x-x^2)^3=9x^3 \\ y =3x-x^2 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^5-3x^4+2x^3=0 \\ y =3x-x^2 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^3(x^2-3x+2)=0 \\ y =3x-x^2 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^2-3x+2=0 \\ y =3x-x^2 \end {cases}$, because $ x\neq 0$.

$ \Longleftrightarrow \begin {cases} { [ \begin {array} {c}x=1\\x=2 \end {array}}  \\ y =3x-x^2 \end {cases}$

$ \Longleftrightarrow \begin {cases} x=1 \\ y =2 \end {cases}$ OR $ \begin {cases} x=2 \\ y =2 \end {cases}$

$ \bullet$ Hence this system have roots are: $ (0;0), (1;2), (2;2)$.[/color]

[quote="Dr Sonnhard Graubner"]hello, i have found $ x \equal{} 0,y \equal{} 0$, $ x \equal{} 1,y \equal{} 2$ and $ x \equal{} 2,y \equal{} 2$.
Sonnhard.[/quote]

Hi Dr Sonnhard Graubner. Can you post your solution? I think you shoud post it at here, if only give the roots then every One can do.

hello, solving your second equation for $ y$ we get
$ y_1\equal{}\minus{}\frac{x^2}{2}\plus{}\sqrt{\frac{x^4}{4}\plus{}6x}$
$ y_2\equal{}\minus{}\frac{x^2}{2}\minus{}\sqrt{\frac{x^4}{4}\plus{}6x}$
Substituting $ y_1$ in equation number one we get
$ \minus{}9x^3\minus{}\frac{x^6}{2}\minus{}9x^3\plus{}x^6\equal{}\sqrt{24\plus{}x^3}(\frac{1}{2}x^{\frac{9}{2}}\plus{}3x^{\frac{3}{2}})$
squaring bith sides and factoring we get
$ \minus{}27(\minus{}1\plus{}\sqrt{x})(1\plus{}\sqrt{x})(\minus{}2\plus{}x)x^3(1\minus{}\sqrt{x}\plus{}x)(1\plus{}\sqrt{x}\plus{}x)(4\plus{}2x\plus{}x^2)\equal{}0$
From here we get the solutions $ x_1\equal{}0$, $ x_2\equal{}1$ and $ x_3\equal{}2$.
The other case is analogously to solve.
Sonnhard.

[color=darkblue]I will give you another solution!

System $ \Longleftrightarrow \begin {cases} x^6 + y^3 = 9x^3 \\
y(x^2 + y) = 6x \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6 + y^3 = 9x^3 \\
y^3(x^2 + y)^3 = 216x^3 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6 + y^3 = 9x^3 \\
y^3(x^6 + y^3 + 3x^4y + 3x^2y^2) = 216x^3 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6 + y^3 = 9x^3 \\
y^3(9x^3 + 3x^4y + 3x^2y^2) = 216x^3 \end {cases}$   , (because $ x^6 + y^3 = 9x^3$).

$ \Longleftrightarrow \begin {cases} x^6 + y^3 = 9x^3 \\
y^3[9x^3 + 3x^2y(x^2 + y)] = 216x^3 \end {cases}$

But we have $ y(x^2 + y) = 6x$

So system $ \Longleftrightarrow \begin {cases} x^6 + y^3 = 9x^3 \\
y^3[9x^3 + 3x^2.6x] = 216x^3 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6 + y^3 = 9x^3 \\
y^3.27x^3 = 216x^3 \end {cases}$

$ \Longleftrightarrow \begin {cases} x^6 + y^3 = 9x^3 \\
 [ \begin {array} {c} x = 0 \\
y = 2 \end {array} \end {cases}$

$ \Longleftrightarrow { [ \begin {array} {c} {\begin {cases} x = 0 \\
y = 0 \end {array}} \\
{\begin {cases} y = 2 \\
x^6 - 9x^3 + 8 = 0 \end {cases}} \end {array}$

$ \Longleftrightarrow { [ \begin {array} {c} {\begin {cases} x = 0 \\
y = 0 \end {array}} \\
{\begin {cases} y = 2 \\
 [ \begin {array} {c} x = 1 \\
x = 2 \end {array} \end {cases}} \end {array}$

$ \Longleftrightarrow { [ \begin {array} {c} {\begin {cases} x = 0 \\
y = 0 \end {array}} \\
{\begin {cases} x = 1 \\
y = 2 \end {array}} \\
{\begin {cases} x = 2 \\
y = 2 \end {array}}\end {array}$

Hence this system have roots are: $ (0;0), (1;2), (2;2)$[/color]


calculus
integration
geometry
conics
parabola
calculus computations

3 points $ O(0,\ 0),\ P(a,\ a^2), Q( \minus{} b,\ b^2)\ (a > 0,\ b > 0)$ are on the parabpla $ y \equal{} x^2$.
Let $ S_1$ be the area bounded by the line $ PQ$ and the parabola and let $ S_2$ be the area of the triangle $ OPQ$.

Find the minimum value of $ \frac {S_1}{S_2}$.

[hide]
$ S_1 \equal{} \int^{a}_{ \minus{} b} x^2 dx \equal{} \frac {1}{3}(a^3 \plus{} b^3)$

denote by $ S_{PQ}$ de Area bounded by the line $ PQ$ and $ x \equal{} \minus{} b$ and $ x \equal{} a$

so :  $ S_2 \equal{} S_{PQ} \minus{} S_{OQ} \minus{} S_{OP}$

and $ (PQ): y \equal{} (a \minus{} b)x \plus{} ab$ ,  $ (OP): y \equal{} ax$ , $ (OQ): y \equal{} \minus{} bx$

then : $ S_2 \equal{} \int^{a}_{ \minus{} b} (a \minus{} b)x \plus{} ab. dx \minus{} \int^{a}_{0} ax. dx \minus{} \int^{0}_{ \minus{} b} \minus{} bx .dx$

         $ S_2 \equal{} \frac {1}{2} ab(a \plus{} b)$

so : $ \frac {S_1}{S_2} \equal{} \frac {2(a^3 \plus{} b^3)}{3ab(a \plus{} b)}\geq\frac {2ab(a \plus{} b)}{3ab(a \plus{} b)} \equal{} \frac {2}{3}$

:)[/hide]

the line PQ is on

$ (b^2\minus{}a^2)\equal{}m(\minus{}b\minus{}a) \to  m\equal{}a\minus{}b$
$ y\equal{}(a\minus{}b)x\plus{}ab$

so

$ S_1\equal{}\int_{\minus{}b}^a [(a\minus{}b)x\plus{}ab]\minus{}x^2dx\equal{}(a\minus{}b)\frac{a^2\minus{}b^2}{2}\plus{}ab(a\plus{}b)\minus{}\frac{a^3\plus{}b^3}{3}$

$ \equal{}\frac{a^3\plus{}3ab^2\plus{}3a^2b\plus{}b^3}{6}\equal{}\frac{(a\plus{}b)^3}{6}$

Yes, $ S_1\equal{}\frac{1}{6}(a\plus{}b)^3$.

Here is the standard solution of Japanese high school students who will have university entrance exam.

Since  the line $ PQ$ has intersection points with the parabpla at $ x\equal{}\minus{}b,\ a\ (a>0,\ b>0)$, let $ y\equal{}l(x)$ be the equation of the line $ PQ$, we can write
$ l(x)\minus{}x^2\equal{}\minus{}(x\plus{}b)(x\minus{}a)$. $ \therefore S_1\equal{}\minus{}\int_{\minus{}b}^a (x\plus{}b)(x\minus{}a)\ dx\equal{}\frac{1}{6}(a\plus{}b)^3$ by the formula $ \boxed{\int_{\alpha}^{\beta} (x\minus{}\alpha)(x\minus{}\beta)\ dx\equal{}\minus{}\frac{1}{6}(\beta \minus{}\alpha)^3\ (\alpha <\beta)}$


1. If $ \cos (x\minus{}y)\plus{}\cos (y\minus{}z)\plus{} \cos (z\minus{}x)\equal{}\frac {\minus{}3}{2}$

Prove that $ \cos x\plus{} \cos y\plus{} \cos z\equal{}0\equal{}\sin x\plus{} \sin y \plus{}\sin z$

2. Find the exact value of $ \tan ^2 \frac {\pi}{16}\plus{}\tan ^2 \frac {3\pi}{16}\plus{}\tan ^2 \frac {5\pi}{16}\plus{}\tan ^2 \frac {7\pi}{16}$

hints
[hide]
1. use complex numbers
2. refer M L Khanna for the sum (its a std sum)
[/hide]

1) implies
$ \sum (\cos^{2} x \plus{} \sin^{2}x) \plus{} 2\sum \cos x \cos y \plus{} 2\sum \sin x \sin y \equal{} 0 \Rightarrow (\sum \sin x)^{2} \plus{} (\sum \cos x)^{2} \equal{} 0$
so done


Ahha! I've been going through all my bits of paper and collected hundreds of problems from the last 4 years or so during my training. So I'll be starting to post them soon :)

where are these from? olympiads? training camps?


For $ n\equal{} 1,2,...,$ let $ x_n \equal{} \frac{(100\pi)^n}{n!}$. For which $ n$ is $ x_n$ largest.

[hide="Solution?"]
$ 100\pi\approx 314$.

$ f(n) \equal{} \frac {(100\pi)^n}{n!} \equal{} \frac {100\pi}{1}\cdot\frac {100\pi}{2}\cdots\frac {100\pi}{n}$

So, every time you multiply by $ \frac {100\pi}{x}$ where $ x\le 314$, the value of $ f(n)$ will increase.

However, when $ x\ge315$, then the value of $ f(n)$ will decrease because then you will be multiplying by a number less than one.

Hence, the maximum of $ f(n)$ occurs when $ n \equal{} 314$.

Answer: 314[/hide]


I'm planning to apply to the COSMOS ([url]http://www.ucop.edu/cosmos/welcome.html[/url]) program this year.  Has anyone here taken it?  What clusters would you recommend (I like most of the courses, in both math and science)?  I'm either gong to apply to UC Santa Cruz, or UC Davis.   Thanks for all your advice.

I'm also currently applying to COSMOS. From their website, it seems like the clusters teach many different topics and finish the program by allowing people to do a research project (kind of like RSI in a sense except not as intense). But that's all I know from just clicking around on their site.

As stated by star99, any advice from COSMOS alumni would be greatly appreciated.  :)

I can't decide between the Mathematics (Cluster 6) and Biomedical Sciences (Cluster 7) at Davis.  Since they are in high demand, they have to be put as first choices.  Has anyone attended them before?

Also, if I apply to the Biomedical course, would it be ok if I only use math teachers' recommendations?


[quote="nayel"]Quadrilateral $ ABCD$ is inscribed in a circle $ \omega$. $ AB$ and $ CD$ intersect at $ E$, the line through $ E$ parallel to $ AC$ meets $ BD$ at $ M$. $ MT$ is tangent to $ \omega$ at $ T$. Show that $ ME=MT$.[/quote]
Ok here is my solution.

Let $ AC \cap BD=X$
$ AC \parallel ME$ So,we can check that, $ \triangle XBA \sim \triangle MBE$
So,$ ME=\frac{AX.MB}{BX}$--(1)
again using the fact $ AC \parallel ME$
$ ME=\frac{DM.XC}{DX}$--(2)
and using power of point of $ X$ $ BX.XD=XC.XA$--(3)
multiplying (1) and (2).we get
$ ME^{2}=\frac{AX.MB.XC.DM}{BX.DX}$ using (3) we get $ ME^{2}=MB.DM=MT^{2}$ (using power of the point $ M$)
so,$ ME=MT$ as desired

$ QED$
Moon

[quote="K81o7"]I assume you intended this...

[hide]
Let $ AC, BD$ intersect at $ F$
\[ \frac{MB}{ME}=\frac{FB}{FA}\]
because $ \triangle{MBE}\sim \triangle{FBA}$
\[ \frac{MD}{ME}=\frac{FD}{FC}\]
because $ \triangle{MDE}\sim \triangle{FDC}$
Multiplying these,
\[ \frac{MB \cdot MD}{ME^{2}}=\frac{FB \cdot FD}{FA \cdot FC}\]
But the left hand side is equal to $ 1$, by Power of a Point.  Then $ MB \cdot MD=ME^{2}$.  But $ MB \cdot MD=MT^{2}$, by Power of a Point, and we are done.[/hide][/quote]

 I think it is better to see the figure , sent by moonmathpi 496 , with the solution. I put it right now.

[b]Babis[/b]

My solution:
[asy]size(200); pair A,B,C,D,planeE,M,T,X; A=origin; T=(50,50); M=(0,50); X=(100,0); draw(circumcircle(A,T,X)); B=(80,-40); draw(M--B); draw(M--T); D=intersectionpoints((circumcircle(A,T,X)),(M--B))[0]; C=(80,40); draw(A--C); draw(A--B--C--D--cycle); planeE=(-52,29); draw(A--planeE); draw(D--planeE); draw(M--planeE); dot(A); dot(C); dot(D); dot(B); dot(M); dot(T); dot(planeE); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,N); label("T",T,N); label("E",planeE,W); label("M",M,N);[/asy]

We have $ \angle EMD=\angle BME$ and $ \angle MED=\angle DCA=\angle DBA=\angle MBE$. Thus $ \triangle MED\sim \triangle MBE$. So $ \frac{ME}{MB}=\frac{MD}{ME}\Longrightarrow ME^{2}=MB\cdot MD=MT^{2}$. Therefore $ ME=MT$, as desired.

two lines, what do you say to that! :P

[quote="nayel"]
two lines, what do you say to that! :P[/quote]
very nice...but may be its bcz of ur experience or for being lucky  :P
and now i am thinking why i didnt see the fact :ninja:


Given triangle $ABC$ and an arbitrary point $P$ theree median $m_{a},m_{b},m_{c}$ prove that
$am_{a}PA++bm_{b}PB+cm_{c}PC\ge\frac{3}{2}abc$ equality hold iff $P\equiv L$ is Lemoine point of $ABC$.

You can actually prove the following inequality which I generalized from yours (my proof is simple)  :)

Given $M$ and $N$ two arbitrary poitns. Then $\frac{MA.NA}{bc}+\frac{MB.NB}{ca}+\frac{MC.NC}{ab}\geq 1$.

I guess you know when the inequality becomes equality  :wink: .

[hide="Proof"] Let $A', B', C'$ be the projections of $M$ on $BC, CA, AB$. Then $a.MA.NA = 2R.B'C'.NA \geq 2RS_{[AB_{1}NC_{1}]}$. Similarly we obtain $\sum a.MA.NA \geq 2R \sum S_{[AB_{1}NC_{1}]}\geq 2RS = abc$ [/hide].

That's righ treegoner you known source of that inequality  :D , I only remark it has another proof by complex number, and equality holds when $M,N$ are isogonal conjugate, but I think your proof is only valid for $M,N$ inside triangle actually with complex number we can prove for all $M,N$ on plane! :)


define a sequence $ u_n$ as : $ u_n\minus{}\frac{u_{3n}}{3}\equal{}1$.
prove that there exist $ n_0 \in N$ and a real number $ a$ such that for all $ n \ge n_0$ we have $ u_n\equal{}a$.

The statement is wrong. Let $ a_n\equal{}u_{3^n}$ and $ q_n\equal{}a_n\minus{}\frac32$. Then we get $ q_{n\plus{}1}\equal{}3q_n$.

[quote="Yustas"]The statement is wrong. Let $ a_n \equal{} u_{3^n}$ and $ q_n \equal{} a_n \minus{} \frac32$. Then we get $ q_{n \plus{} 1} \equal{} 3q_n$.[/quote]
Yes .
Let $ x_n\equal{}a_n\plus{}\frac{3}{2}$
Then $ 3x_n\equal{}x_{3n}$
If we chose $ x_n\equal{}n$ then it is not satisfy condition.

[quote="TTsphn"][quote="Yustas"]The statement is wrong. Let $ a_n \equal{} u_{3^n}$ and $ q_n \equal{} a_n \minus{} \frac32$. Then we get $ q_{n \plus{} 1} \equal{} 3q_n$.[/quote]
Yes .
Let $ x_n \equal{} a_n \plus{} \frac {3}{2}$
Then $ 3x_n \equal{} x_{3n}$
If we chose $ x_n \equal{} n$ then it is not satisfy condition.[/quote]
sorry , I had a mistake ,this  sequence is bounded . :oops:

$ x_{3n}\equal{}3x_n$
$ \Longrightarrow x_{3^m.k}\equal{}3^mx_k$
Because $ x_n$ is bound so $ x_k\equal{}0,\forall n$
This problem claim.


[quote="jli"]
if theres a ring and a ball on the inside edge of the ring, will the ball be attracted towards the middle of the ring?

-Johnny

[/quote]

If you mean due to gravitational forces between the two, I believe the ball will be attracted to the CM of the ring while the ring will be attracted to the CM of the ball.

[quote="Conjurer"]I guess they will try to be at the center of mass of the system, in which case YES. But wait and confirm on this.[/quote]

In a nut shell, the system will try to attain a configuration due to which the potential energy (here gravitational) will be minimum.

Well
...

let us assign the distance between the cm of the ring and the object as d;

Now we will try to finde the total potential energy asociated ..

let us take an element of the ring which subtends an angle da at the ob ject and at a distance of r from it..

then the potential energy dU=-(Gm*(dM)/r)

but dM=sr(da) where s is linear mass density ..

so the resultant dU=-(Gm*s(da))

integrating w.r.t a from 0 to 2pi we get an expression inependent of d ..

so my calculations point out that the object should remain where it is ...

[quote="Rohan"]
Let us take an element of the ring which subtends an angle $ da$ at the object and at a distance of $ r$ from it.

[/quote]

The ball is on the inside edge of the ring, which means $ r\equal{}0$.

$ \Delta U = - \int_{x_i}^{x_f} F(x)dx \\
 \\
U - U_i = - \int_{x = r}^{x = 0} \frac { - GMm}{x^2}dx \ldots\text{minus sign for attractive force along - ve Y axis} \\
 \\
U - U_i =  \frac { - GMm}{x} |^{x = 0}_{x = r} \\
 \\
U - U_i = \frac {GMm}{r} - \infty \\
 \\
\; \text{which means, } U_i > U, \text{or simply that the Centres of Masses coincide}$

But one question

How can u write the potential energy of the two mass system to be GMm/r^2 where r - distance between c.m.  ???

If that was the case then the points inside a hollow sphere wouldn't have remained at equal potential

Further in my solution the potential energy comes independent of d which means F=dU/dx=0 at all points inside the ring ..


Harvard
college
Support
geometry
geometric transformation
reflection
analytic geometry

Do you think (in high school) IB or AP is better?  Some say IB is better, but I am a math-first person, and I can get more math in with AP than IB.  This is because IB requires you to take a ToK course, which replaces some other AP course I could take.  Since I am more math/sciences than humanities, I think AP is better.  But does anyone have any firsthand advice?  Do colleges view IB higher?

At the Harvard info session, their admissions guy said that Harvard and many of the other colleges at their level (although that's not the wording he used) view AP and IBs at basically the same level.

[quote="henryyangrui"]whats is a ToK course and why is that my school only offers AP not IB?[/quote]

For more on IB, visit [url=http://www.ibo.org/ibo/index.cfm?page=/ibo/about&language=EN]their homepage.[/url]  But basically, IB is a diploma, along with programs, that is certified for use at any IB school throughout the world.  IB is the new "fashion" in education, with big public schools all over the US trying to get IB/MYP certified.  If I went to an IB school in the US, then went to an IB school in...Buenos Aires, then came back here, I wouldn't miss a beat.  That is what IB is in a nutshell, but for a more accurate description, visit their website.

As far as ToK, that stands for Theory of Knowledge, a philosophy-type course IB REQUIRES for their diploma.  But I have only heard of ToK from admissions officers and guidance counselors, and I have no firsthand experience, so my information may be less than accurate.

I personally loved my ToK course, we had a lot of fun and good debating.

Ib is worth it for me, the classes are smaller and you really get to know everyone better.  Plus, cuz its IB, we get the best teachers in the school.

[quote="mysmartmouth"]Do you think (in high school) IB or AP is better?  Some say IB is better, but I am a math-first person, and I can get more math in with AP than IB. [/quote]

A bit about the math side first. The teacher that teaches both IB HL math and AP Calc BC claims that HL is more challenging and goes beyond Calc BC.

That said, while HL math is so far the best math class I have taken in school (fantastic teacher), I still do not believe that it is offering me what I am looking for in math. The biggest problem with HL math (and this goes for many of the IB courses) is that they expect you to cover a lot of material on the syllabus, but at the end, you don't necessarily walk away with a lot more insight or as a better thinker. The material is covered at a very artificial level, and it's not enough to arouse interest for many people.

I wish I could say that I am not bored in my HL math class, but I can't. And while I enjoy the class very much (the teacher is excellent--without him, I probably would have never found my passion for math), I can't say that the class is providing an adequate challenge.

If you decide to go the IB route, do consider a math extended essay. At my school, this is a lonely road, but I've found a lot of support outside of school, and this has been both crucial and inspiring.

[quote="mysmartmouth"]IB requires you to take a ToK course, which replaces some other AP course I could take.  Since I am more math/sciences than humanities, I think AP is better.  But does anyone have any firsthand advice?  Do colleges view IB higher?[/quote]

The strength of IB (yes, sometimes IB can have its strengths here and there) is that it allows students to explore different fields. Personally, I was quite surprised to find this year that I am actually enjoying my SL classes (English and Economics) a lot--at times, even more so than my HL classes. So by all means pursue what you are interested in but don't limit yourself to fields too early, especially if you are not sure.

The strengths of the SL classes are different than those of HL classes. At first, I thought that I was wasting my time with the SL classes, especially since I had worked so hard in the HL feeder classes the year before. But with the right peers and the right teachers, these SL classes can actually be a blessing. Since we're not confined with a jam-packed and overly rigid syllabus, we have the freedom (read time) to pursue the aspects of the subject that the class finds most interesting at an intensity and in a direction that suits us most.

When you sign up for those HL classes, be careful. A lot of people think that 7 HLs is the way to go, but really, it's probably not. You may think that you're impressing colleges, but when you sit down in those classes, you may find that there is no reason why colleges should be impressed. Two years is a long time to endure if the class is pure and absolute torture. The general impression I get from IB is that it requires a lot of busywork. IB's not a programme that requires intelligence. Rather, it tests your ability to read checklists (i.e. the syllabus) and to meet those requirements. One other thing. If you happen to get a class with a REALLY bad teacher, be prepared to self-study for two years. (This can be both a curse and a blessing, depending on the subject matter and your interest level in it.)

As for TOK...let's just say I have some very mixed feelings towards that class. The PHILOSOPHY of the TOK course is excellent--it's just a shame that IB fails to practice what it preaches and fails to incorporate the philosophy of TOK into the rest of its curriculum and diploma. If taught well, TOK has the potential to instill reflection and genuine thinking. Otherwise, TOK simply sums to endless and fruitless frustration, especially for students who already live in the TOK-way, if you know what I mean.

[url=http://www.digitalbrain.com/roztru/TOK/Theory%20of%20Knowledge.db_psc]More information on TOK.[/url]

This topic has actually been discussed a lot but hasn't yet been beaten dead (in my opinion).

In all honesty, though, while each program has its merits, I doubt that either IB or AP can offer the intellectual intensity some seek.

[quote="adidasty"]from what ive heard, ap tests are a little harder than ib HL exams...but i dont know fur sure[/quote]

thats only on some cases....

we can see AP as making broad approaches towards a subject; thus their tests seem to be a little bit harder than the IB counterpart. however, IB emphasizes depth... so it is harder in tests sometimes.

if you like memorizing and studying strictly, then AP is good for you... For IB, you dont necessarily have to study alot, but make a lot of inputs and insights... when you study for IB, you should definitely take a lot of time thinkg about one thing.

Agreed. With the exception of the languages, IB is for the most part very conceptual. There's not a whole lot of memorization that needs to be done. Furthermore, for the math and sciences, IB provides Data Booklets (which consists of formulas, data tables, periodic tables, values of constants, etc., depending on the subject).

[quote="mel"]Furthermore, for the math and sciences, IB provides Data Booklets (which consists of formulas, data tables, periodic tables, values of constants, etc., depending on the subject).[/quote]

I think this approach is more in tune with "the real world", but unfortunately, I'm doing APs this year.

[quote="4everwise"]Are IB classes only offerred on the east coast? I haven't heard of a single high school here in California with those courses.[/quote]

actually a school iwas going to move to, Sunnyhills High School in fullerton(? or Garden Grove... i dunno its been a long time), has IB.

[quote="4everwise"]Are IB classes only offerred on the east coast? I haven't heard of a single high school here in California with those courses.[/quote]

Nope. I know of many schools in Washington and Oregon that offer IB.

yea i agree... in fact i am doing that my junior year... but only because i just found out that HL Math is worth nothing.... just bunch of waste of time

i think you can coordinate it with your coordinator in case your school dont offer one. I am starting on IB Korean Studies by myself, which is like a literature course, only in Korean...

[quote="blahblahblah"][quote="mel"][quote="blahblahblah"]Its worth noting that for math-y types, there's an additional IB course, Further Mathematics SL, which contains more advanced material on Statistics, Discrete Math, Analysis/Approximation, Abstract Algebra, and Geometry.  I can say that its well worth it.[/quote]

Too bad a lot of schools don't offer it.  :([/quote]

I would guess that no schools in North America do.  This means you'll have to talk to your IBC about it, and probably offer to self-study the material.[/quote]

exactly what i was talking about. by the way, blahblahblah, i live in North America, and we DO offer it.

Yes, that was a stupid comment of me, (I got my IB diploma last year) seeing as I know a high school in Vancouver that offers it.  But most schools won't offer a class with it - they'll say explicitly that its self study.  That was my route - self study, and get a teacher to mark portfolios.

The IB HL outline is really exciting. Lots of cool topics and such. Though it's really a shame that we don't get to look through all of those.

Don't know much about AP, but I think IB HL is definitily a good course.

[quote="blahblahblah"]If you do further math, you cover all the 'cool' topics ...[/quote]

Ooooh, explain (fascinated).

Also, to xxreddevilzxx:
Re: Your Signature

I am at 276, so I am only 39 away from Pi Day!

[quote="mysmartmouth"]
Also, to xxreddevilzxx:
Re: Your Signature

I am at 276, so I am only 39 away from Pi Day![/quote]

LOL. Also notice when you reach or approach 711. It is an important day. Maybe you can convince a Seven-11 storekeeper to give you a free big gulp :D

BTW, back to HL math: What exactly is covered?? And I tried to do HL, but i am giving up the idea of doing it because when I thought I could use the AoPS Olympiad Study Final Project paper to turn in for HL, i heard that IBO give you a specific topic instead of letting you choose one. Also, I cannot test in HL until Senior yr, which I hate to do. Is there any other good reason for doing HL other than that it covers "cool topics"?

Hmm, our school has Math HL.. and then this year was the only year they offered Further Math (iono if it's SL or HL.. prolly SL), but they're gonna stop cause there's not enough kids in it.  :(  There probably is such a thing as Further Math HL.. but very rare..  if it even exists. Or maybe it's just in our dreams.

[quote="Tunio"]Is there such thing as a Further Mathematics HL course?[/quote]
There is no Further Mathematics HL.


Is there an easy way to do this without visually planning the thing out?

Consider the front view.  Put a 1 or a 2 in each square, determined by the maximum possible number of cubes lined up behind that square.  It should look something like this:

[color=white]0     0     0[/color]     1
[color=white]0[/color]     1     [color=white]0[/color]     1
[color=white]0[/color]     1     1     1
1     2     2     1
1     2     2     1

As you examine the front view, you note that the first and forth columns have strictly 1's because of the top perspective. And because of the right side perspective, the third, forth, and fifth rows have 1's.


MATHCOUNTS
Alcumus
FTW
AMC
AMC 8
search
USA(J)MO

I have a big problem.

My mathcounts coach can't teach.
The Ph.D's in my school don't like mathcounts so they won't teach.

How do I learn.

learn it yourself  :roll:  :roll:

I learned from 6th grade up to 10th grade by myself, so you should be able to since I'm only a 6th grader :roll:

[quote="saszs"]sure thing[/quote]
was that sarcasm?

[quote="saszs"]I have a big problem.

My mathcounts coach can't teach.
The Ph.D's in my school don't like mathcounts so they won't teach.

[/quote]

ok. what do you want us to do about it? thats what aops, their forums, and their books are for

I would just study. There's plenty of free resources on this site. *cough alcumus cough FTW cough forums*
Also, you could buy books, like the AoPS books, or (promoting my coach here  :rotfl: ) the mc book by my coach(see the announcements) for some well-written mc tests. You could do those tests, see what you got wrong, and improve on that. good luck with mc!

Aops is where the beastly people rise from. Just get the books and if you don't know a question, ask it on aops.

[quote="remy1140"]AoPS was really helpful for me. For one year I studied the Intro textbooks and Volume 1, along with the Advanced MathCounts online course, and I really improved a lot since last year. I would recommend working on the Intro books. :P[/quote

ummm  I just joined the basic amc8/ mathcounts class.

:rotfl:

:yoda:

Well that's good.
I'm going to join the advanced.
So that should be enough, but if this is your first year, next year you should take the advanced. And then you should be beastly enough. The admins are kind and patient enough to teach those who want to be taught.

Buy the Introduction books from AoPS and study them thoroughly. Like, actually do the problems. If it's challenging, then thats good because you don't learn much from doing trivial problems. If it's way too challenging, then review the examples and the techniques used to solve them.

Good luck!

P.S. Remember to study math to study math, not just to win at Mathcounts, then you can learn better.

[i]Along with the AoPS books[/i], you might consider a book like "Hard Math for Middle School: IMLEM Edition".  It gives some problems similar to MATHCOUNTS and teaches how to do them.  Since you are not a first time competitor, this book may not help you much, but it would be great for someone at your school who is just starting out.  By searching online bookstores you can find it at a very reasonable price.

Another resource are the problem solutions (since about 2001 MATHCOUNTS has published solutions, and the handbooks have solutions that your coach might be willing to give you, it won't take much effort on the coach's part).  [b]Don't use them unless you absolutely have no clue[/b], or you have solved it and want to see if there is better way to do it.  Of course, you can search the MATHCOUNTS forum for help on solving many of the problems and finding other ways to solve problems.  If you can't find it, then post it!  I learned more from looking up problems on this forum than I did any other way.

Getting together with other mathletes for study sessions is also beneficial.  Other people find problems that you struggle with easy, and can share their insight.

Good luck preparing for next year!

You  should try:
a. books
b. classes
c. practice old tests off of aops forum or searching
d. find a friend who also wants to be in MC

[u]BEST OF ALL[/u]
 :o
 :wink:
g. Get mad and go homeschool or go to another school then win, and make fun of the people who didn't let your coach teach.

[quote="jonathanchou711"]I agree with the Find a friend part OMG. Math isn't half as fun when you're the only smart kid. I have a friend that is about my level in Mathcounts and it's actually a whole bunch more fun. You can compare answers and learn from them! When I had a practice in the summer all by myself... It was horrible.  :ninja:[/quote]

At our school we have some good at math people but no one else is USAMO-level yet.

One thing that I really find useful is do more AMC 10's for a while than Mathcounts. Then, when you start doing the MC tests again you'll find that they're really really easy compared to the AMC 10 and then you can get good at speed.

No, do USAMO problems before MC. (seriously, I did not take MC school before USAMO qualifying)

In reality though AMC 12's and easy AIME's are good for MC practice.

I remember in fourth grade being really excited at solving single AIME problems (#1-3) - doing MC probs would have bored me and not get me into USAMO.

[quote="james4l"]No, do USAMO problems before MC. (seriously, I did not take MC school before USAMO qualifying)
[/quote]

This.

My whole mc practice this year has been solving USAMO problems/Vietnam TST problems/red mop quizzes/Russia Olympiad/China TST.

Apparently that's helpful, since I won CA this year.  :maybe:

You guys need to spend more time on what you can do to improve instead of what you or someone else can't do.  You will be amazed what you can do if you really use your time constructively.

Practice on FTW and Alcumus.  And, yes, buy the books.  They're really helpful.

use Problem Solving Strategies by Arthur Engel

Of course, that book is better suited for one preparing for Olympiad exams, such as the USAMO :wink:

I doubt someone trying to get better at MATHCOUNTS would find much use with PSS.

When training, just do AMCs and AIMEs.

When its the week of the competition, do FTW.

Also, books are very good.

Going back to the first post...

First off, why would you need a Phd to teach some elementary~middle school stuff?
Second, the first time I learned math straight from an adult was this year.  Throughout elementary and middle school, I only ever "learned" math from a book and practiced problem solving by doing tons of problems.  Your teacher shouldn't have much of a negative influence on how you learn (although positive is certainly possible).
All it takes is some determination and inspiration.  Try looking for that instead.

Are you guys joking?

The USAMO is definitely not a good tool to get ready for Mathcounts.

Here's how I did it:(and got from never done a math competition before to top 1/3rd in nats in 5 months time)

1. Find a friend or better yet, multiple ones. They don't have to be as good as you, just a decent enough level. You can work on team rounds together, correct each other, and work out hard problems.
2. Get the tests. Ask your teacher for all that he/she has. If he/she doesn't have any, bug people on these forums until you get them.  :lol:
3. Look around the wiki for certain tricks that help you simplify a problem.
4. Practice, Practice, Practice. Whether it be Mathcounts tests (best tool,) AMC 10s, 12s, or mock tests, doing math is what really helps improve.
5. Try nationals problems and find out your weaknesses. I would then either ask others to post questions on the topic, create a marathon on it, or buy a mathematics book on the subject.
6. Get a good nights sleep and have breakfast the day of the test. Try to stay cool, calm, and relaxed.

USAMO = doing math, so if it is new math, it is good math to practice with.

I prefer medium AIME problems to speedsolve, they're about Mathcounts level problems.

I endorse the last statement.

Bleh, learn calculus and prepare for the Putnam for next year. And the USAMO + TST.

[quote="pythag011"]Well, you shouldn't just train for mathcounts in my opinion; you should do a lot of non-mathcounts math, learn linear algebra, ergodic theory, do usamo problems, etc.

Don't just do mathcounts.[/quote]
I agree with you, but he was specifically asking for help in Mathcounts. Not math in general.

To learn Mathcounts just do Mathcounts level problems or higher. There's no real "requirement" of practicing math - you can do problems without a coach.

I never said that doing the USAMO wouldn't help. I said it does not help a lot.
Look at my quote. (especially the bolded part)

[quote="Ihatepie"]The USAMO is definitely not a [b]good[/b] tool to get ready for Mathcounts.[/quote]

apparently the genius's in my class/school hate math and the interested ones left my school.

I can't leave because the school I attend, it's supposed to be a college prep school.

and there is a slow person that likes math and I can't work with him.

Have you thought about studying on your own and not blaming others for everything?

Just because someone is slow doesn't mean you can't/shouldn't work with them. Teaching them is a good way of solidifying your grasp of the concepts.

Also, don't be condescending. Would you like it if you did some math thing with really super-smart people and they were like "I can't work with saszs he's too slow"?

@saszs: It's not his fault. Or yours. He could be a good chance for a teamworker. If you teach him patiently because he loves math, then he could become a smart kid too. Don't insult other people and think that you are the best because you aren't. There's always someone better than you.  :)

considering this rly has nothing to do with mc, lets lock this topic up nice and tight, ok.
saszs, maybe u should put ur nose in a book and teach urself something instead of complaining about it on the forum.  better yet, stop claiming its ur teachers fault when they have to teach more than 1 student at a time.  they obv dont have enough time to reach out and make the work harder just bc they have one gifted student in one particular class.  i can;t really say anything else that some1 else has established b4 me.  i gotta say tho that some of the kids on this forum are in for a huge surprise when they hit the real world...


Cossack's problem reminded me of a very nice problem I've encountered some time before: imagine that you break the Earth (uniform ball of mass M, radius R) into two hemispheres (for example, through the equator). What is the force F that those two hemispheres attract each other?

It is possible to solve by nasty calculus, but you don't need it!

could you do this with the Maxwell Stress Tensor? I think you can.

Yes you can use the analog of the electrostatic result that the electrostaatic pressure  ( force perunit area ) is $ \frac{1}{2} \epsilon _0 E^2 $ . Now you take the projection of the area of hemisphere along it's diametrical plane( $ \pi R^2 $ ) (since that will give us the effective force for when we multiply each pressure with a differential area and all such forces vectorially, knowing that $E$ is same at all points on surface, we simply get that Required force is Pressure $ \times $ net horizontal area .)

Thus , the force is $ \frac{1}{2} \epsilon _0 E^2 \times \pi R^2 $. Now you just have to form the gravitational analog!!!

I was thinking of foing this with energy when I remembered this result from electrostatics. Do you think energy method can be used???


[b]In the triangle ABC le AD, BE, CF be the altitudes of the triangle.
Prove that if the sum of the vectors AD + BE + CF = 0 then the triangle is equilateral.[/b]
This problem seems trivial but i could solve it only using boring analytical geometry. I would like a nice solution ...

[hide="hint"]Denote $m=BD/DC,n=CE/EA,p=AF/FB$, we have $mnp=1$ by Ceva's theorem. Now compute vectors $AD,BE,CF$ by vectors $AB,AC$ and numbers $m,n,p$. Final, use '' If $xAB+yAC=0$ then $x=y=0$''.[/hide]


Solve in integers the following equation
$(x+y+z)^3 = 4(x^3+y^3+z^3)+12xyz+9$

$(x+y+z)^3 -4(x^3+y^3+z^3)-12xyz = 3 (x+y-z)(y+z-x)(x+z-y)$

therefore the equation becomes:

$(x+y-z)(y+z-x)(x+z-y)=3$.

The solutions are $\{ (2,2,1) , (-1,-2,0) , (1,1,-1) \}$ and all the permutations.

 :)


search
logarithms
real analysis
real analysis theorems

let $ a_1\ge a_2\ge \dots >0$ and $ \sum a_n\equal{}\infty$. show that $ \sum\frac{a_n}{na_n\plus{}1}\equal{}\infty$

Is this possible to say?

Since it is equivalent to $ \sum \frac {1}{n\plus{}\frac{1}{a_n}}$ and $ \sum {a_n}$ diverges as well, the fractional term (\frac {1}{a_n}) will approach 0 and then it sort of becomes the harmonic series with $ \sum \frac{1}{n}$ which diverges... so it must diverge... ?

[quote="Euclid's Turtle"]Is this possible to say?
[...] $ \sum {a_n}$ diverges as well, the fractional term (\frac {1}{a_n}) will approach 0 [...][/quote]
I think this argument won't work

$ a_n\equal{}1\plus{}\frac{1}{n}$ is decreasing, $ \sum a_n$ diverges, but $ \frac{1}{a_n}\not\to 0$ .....

A classic problem on this site (it was several years ago) was this:

Suppose $ a_n$ is decreasing and positive. If $ \sum a_n \equal{} \infty,$ then $ \sum \min\left(a_n,\frac1n\right) \equal{} \infty.$

I'll have to look for the link, but the proof is an excellent place to use the condensation test for convergence. (That's how to exploit the monotonicity.)

This problem reduces to that old one, since $ \frac {a_n}{na_n \plus{} 1}\ge\frac12\min\left(a_n,\frac1n\right).$

Added in edit: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=137297315&t=17772]this[/url] is the link I was looking for.

[quote="knoppix"][quote="Euclid's Turtle"]Is this possible to say?
[...] $ \sum {a_n}$ diverges as well, the fractional term (\frac {1}{a_n}) will approach 0 [...][/quote]
I think this argument won't work

$ a_n \equal{} 1 \plus{} \frac {1}{n}$ is decreasing, $ \sum a_n$ diverges, but $ \frac {1}{a_n}\not\to 0$ .....[/quote]

Well I think I said that wrong, about the aproaching 0 thing. True it doesn't become 0, but $ \frac{1}{a}$ or
$ \frac{1}{999999*a}$ or $ \frac{1}{10^{100}*a}$ etc... diverge. And so the it still becomes somewhat like $ H_n$ with some denomiator-increasing factor.

Euclid's Turtle: I'm sorry, but you don't have a proof.

If you step back and look at the larger picture, you'll realize that "becomes somewhat like the harmonic series with some denominator-increasing factor" is simply not a statement you can use to constitute a proof.

If you're trying to argue that there is some positive constant $ C$ such that $ \frac{a_n}{na_n\plus{}1}>\frac{C}{n},$ then that statement isn't even true. (For a simple example, try letting $ a_n\equal{}\frac1{n\log n}.$)

But if you let your "some denominator-increasing factor" be unbounded, then that isn't anything you can use to argue divergence.

This is a very tricky problem. I suggest you follow my link and read about its parent.

Okay thanks :) Was trying for an easy and quick solution


assume $P$ is a polynomia.And $P(x)=0$ have $n-1$ roots which is larger than $1$
let $Q(x)=(x^2+1)P(x)P'(x)+x[(P(x))^2+(P'(x))^2]$
prove:$Q(x)=0$ have $2n-1$ roots at least

Hint :

Can be rewritten : $P^2 \cdot (X+P'/P) \cdot (XP'/P+1)$

I'll give further hints if needed  ;)

sorry,I need further hints :(
I even don't know why you can write in that form:
because $P$ may equal to 0

[quote="tµtµ"]Sorry I didn't follow this topic.

Use factorisation to prove that between two roots of P (we can presume that the roots of P are simple) there 2 roots of your weird polynomial.[/quote]
can you explain more detail,in fact your hint is the same to my book...so I still can't figure out,sorry.
and thanks :)

and if it is proved,there has only 2n-4 roots?how we come out that 2n-1?

Look for the sign of : $P^2 \cdot (X+P'/P) \cdot (XP'/P+1)$ as $X \rightarrow rootOfP^\pm$

i don't think OP statement is true.
when p(x)=(x-2)(x-3)
q(x) has 3 real roots: 0.78 , 2.55 , 2.75

related:
[url]https://artofproblemsolving.com/community/c7h1811022[/url]


function
arithmetic sequence
number theory proposed
number theory

Let m,n be positive integer ,prove that for some positive integer $a$ each of $\varphi(a),\varphi(a+1),\dots,\varphi(a+n)$ is multiple of $m$.

Consider arithmetic progression $1+m\cdot k$. Due to Dirichlet theorem, it contains infinitely many prime numbers. Let $p_0$, ..., $p_n$ be some of them. Using Chinese residue theorem, we can find $a$ s.t. $p_i\,|\,a+i$. Therefore, $\varphi(a+i)=(p_i-1)\cdot A_i$ and $p_i-1$ is divisible by $m$.

That's all.


limit
logarithms
function
calculus
calculus computations

$(1).$ Prove that for any $n\in N^*$, the equation $x^n=x+n$ has
one positive root $x_n,\ \lim_{n\to \infty} x_n=1,\ \lim_{n\to\infty} \frac{\ln n}{n(x_n-1)}=1$.

$(2).$ Prove that for any $n\in N,\ n\ge 3$, the equation $x^n=nx-1$ has
two positive roots $0<x_n<1<y_n<2,\ x_n\rightarrow 0,\ y_n\rightarrow 1.$

$(3).$ Let $x_n$ be the zero of the equation $\ln (n+x)+E=\sum\limits_{k=1}^n\frac 1k$, where $E$
is the Euler's constant, i.e.  ${\ \lim_{n\to\infty} (\sum_{k=1}^n \frac 1k}-\ln n)=E$. Prove that $x_n\rightarrow \frac 12.$

$(4).$ Prove that for any $n\in N^*$, the equation $x^n+\arctan x=1$
has one positive zero $x_n$ and $x_n\rightarrow 1.$

$(5).$ Prove that for any $n\in N^*$, the equation $x^n+\ln x=0$ has
one zero $x_n\in (0,1)$ and $x_n\nearrow 1.$

$(6).$ Prove that for any $n\in N,\ n>1$, the equation $\sqrt{x+1}-\sqrt x=\frac 1n$, $x\in [0,1]$ has one zero $x_n\in (0,1)$ and $x_n\rightarrow \frac{1}{e-1}.$

$(7).$ Prove that for any $n\in N^*$, the equation $x^n+nx=1$
has one positive root $x_n$, $x_n\searrow 0$ and $\lim_{n\to\infty} (1+x_n)^n=e.$

$(8).$ Prove that for the fixed $p\in N^*$ and for any $n\in N^*$, the equation
$x^n+px=1$ has one positive root $x_n$ and $x_n\nearrow \frac 1p.$
[u]A particular[/u] case: $p=1\Longrightarrow$ http://www.mathlinks.ro/Forum/viewtopic.php?t=70205

1) Define $f(x)=x-x^n-n$. $f(1)=-n<0$ and $f(n)=n^n>0$. From mean value theorem follows the existence of $x_n$.

We have that $1\le x_n\le n$ which implies that $1\le x_n=\sqrt[n]{x_n+n}\le \sqrt[n]{2n} \to 1$. Hence, $x_n \to 1$.
We have that : $\lim \frac{\ln n}{n (x_n-1)}=\lim \frac{\ln n}{n \ln x_n} \frac{\ln x_n}{x_n-1}=$
$\lim \frac{\ln n}{n \ln x_n}=\lim \frac{\ln n}{\ln x_n^n}=\lim \frac{\ln n}{ln (x_n+n)}=1$ since $x_n \to 1$.
I used the fact that $\frac{\ln x}{x-1}\to 1$ when $x \to 1$ by l'hopital rule.

Here, $\lim_{x\to 1} \frac{\ln x}{x-1}=1$ you can not apply the [u]l'Hospital's rules[/u] ! This is a remarkable function limit, this proving differently. What you made is a vicious circle.

$\lim_{x\to 1} \frac{\ln x}{1-x}=\lim_{y \to 0} \frac{y}{e^y-1}$ and now I can apply L'Hospital rules. No? What is wrong in my proof??

Please, you ask your teacher where is the mistake.

I think professor Nicula meant: when you apply l'Hospital rule for your limit you are using the fact that (e^x)' = e^x and this fact is demonstrated using your limit.
Can you see now the circle ? This limit is called a 'fundamental' limit and the proof of it is based on other arguments.
The same goes for (sin x / x) when x -> 0. You cannot apply here the l'Hospital rule.

Here are some hints [to most delicate points].

1. The function should be f(x) = x^n - x - n.

3. It is a classical proof that Bn = Sum(1,n)[1/k] - ln n decrease  to E, E in (0, 1).
   The ecuation Bn = ln(n + x) - ln n + E has the unique solution Xn = (e^(Bn - E) - 1) / (1/n).
   Using Stoltz Cesaro lemma (case 0/0) we need to find limit of [e^(1/(n+1) - ln(n + 1) - ln n ) - 1]/(1/n*(n + 1)).
   This can be reached using l'Hospital (case 0/0) for the same fraction, replacing n with x real.

6. I smell a rat here :)

7. f(x) = x^n + n*x - 1, on [0, inf). f'(x) = n*x^(n-1) + n > 0. f is strict increasing. f(0) = -1. f(1/n) = (1/n)^n.
   So there exist a unique Xn in (0, 1/n) s.t. f(Xn) = 0. It follows that Xn -> 0.
   (1) Xn^n + n * Xn = 1.
   Then, 0 < Xn^n < Xn -> 0, so from (1), n*Xn -> 1.
   It follows that (1 + Xn)^n = [(1 + Xn)^(1/Xn)]^n*Xn -> e^1 = e.

   The delicate part is to show that Xn is decreasing.
   If g(x) = (1 - x^n)/x for x > 0, g'(x) < 0, so g is strictly decreasing.

   We show that (2) g(Xn) < g(Xn+1) and from here Xn is decreasing.
   (2) <=> n < (1 - Xn+1^n)/Xn+1 = ((n+2)Xn+1 - 1)/Xn+1^2 <=> n*Xn+1^2 - (n+2)Xn+1 + 1 < 0.
   The roots are: R1 = (n + 2 - rad(n^2 + 4))/2n, R2 = (n + 2 + rad(n^2 + 4))/2n.
   The condition is that Xn+1 lies in (R1, R2).
   Because R2 > 1 > Xn+1 it remains to show R1 < Xn+1 <=> 2/(n + 2 + rad(n^2 + 4)) < Xn+1 <=>
   2/(n + 1 + rad((n-1)^2 + 4)) < Xn. If a = rad((n-1)^2 + 4), 2/(n + 1 + a) < Xn <=> f(2/(n + 1 + a)) < 0 <=>
   2^(n-2) < (n + a + 1)^(n-2) * (n + a + 1)/ (n + a - 1) which is true because (n + a + 1)/ (n + a - 1) > 1.

[I am not sure why my html is deactivated, when in my options window it says the opposite, it was a good option parallel to latex...]

Thanks, MareleG, for your explanation of the vicious circle !

The vicious circle depends only on definitions. Using the "high-school" definitions we get vicious circle. Using "series" definitions we get a correct proof by xirti.

What rat was smellt at 6?

[quote="perfect_radio"]What rat was smellt at 6?[/quote]
[color=darkblue][b]Sorry, Perfect_Radio ![/b] The correct enunciation of the exercise $ 6$ is following :[/color]
[quote][color=darkred]Prove that for any $ n\in N,\ n > 1$, the equation $ \sqrt [n]{x\plus{}1}\minus{}\sqrt [n] x \equal{}\frac{1}{n}$, $ x\in [0,1]$ has one zero $ x_{n}\in (0,1)$ and $ x_{n}\rightarrow\frac{1}{e\minus{}1}.$[/color][/quote]
[color=darkblue]I proposed "[b]Problem[/b] $ 6$" in [b]G.M.B.[/b] many years ago.
It is appears at [b]ex.[/b] $ 75.9$ (the page $ 454$) in the Romanian book
"[b]Analiza matematica pentru clasa a XI - a, Ed. TEORA, Bucuresti 1999[/b]".[/color]

2)$ f(x) \equal{} X^{n}\minus{}nx\plus{}1$
$ f'(x) \equal{} n(x^{n\minus{}1}\minus{}1)$
Consider sign of $ f'(x)$ we have
$ f(0) \equal{} 1$
$ f(1) \equal{} 2\minus{}n < 0$
$ f(2) \equal{} 2^{n}\minus{}2n\plus{}1 > 0$
So have $ x_{n},y_{n}$ root
a)$ 0 < x_{n}$
$ x_{n}<\frac{2}{n}$
$ lim_{n\to\infty}\frac{2}{n}\equal{} 0$
So $ lim_{x_{n}}\equal{} 0$
Same method for $ y_{n}$

5)$ f(x)\equal{}x^{n}\plus{}lnx (x\in (0,\plus{}\infty)$
$ f(x)$ monotone($ f'(x)\equal{}1\plus{}\frac{1}{x}>0$)
$ f(1)\equal{}1$
$ f(x)_{x\to}<0$
So$ f(x)$ has unique solution on $ (0,\plus{}\infty)$
$ x_{n}\to 1$ has same method as above solution.

6)$ \frac{1}{\sqrt{x}\plus{}\sqrt{x\plus{}1}}\equal{}\frac{1}{n}$
So $ \sqrt{x}\plus{}\sqrt{x\plus{}1}\equal{}n$
So $ \sqrt{x}\equal{}n\plus{}\frac{1}{n}$
$ x_{n}\to\plus{}\infty$

Prove that there exist a sequence $ \{a_{n}\}\in [0;\frac{\pi}{2}$] such that
$ \text{cos}{a_{n}}\equal{} a_{n}^{n}$
Prove that $ a_{n}$ converge and find it limit.
This problem I take from Diendantoanhoc.org.

4)$ f(x) = x^{n}+\text{arctan}{x}-1$
$ f(x)$ monotone on $ [0,1]$
Easy to check $ x_{n}$ converge then it has limit>Call it is $ l$
Then $ l^{n}+\text{arctan }{l}-1 = 0$
$ \leftrightarrow l = (1-\text{arctan}{l})^{\frac{1}{n}}\to\1$
So $ \lim_{n\to\infty}x_{n}= 1$

[color=darkblue]Please, [b]TTsphn,[/b] see and read the last my reply. The [b]exercise[/b] $ 6$ was wrote wrongly (see down the box) . The its correct enunciation is following :[/color]

[color=darkred]$ (6).$ Prove that for any $ n\in N,\ n > 1$, the equation $ \boxed{\sqrt [n]{x\plus{}1}\minus{}\sqrt [n] x \equal{}\frac{1}{n}}$, $ x\in [0,1]$ has one zero $ x_{n}\in (0,1)$ and $ x_{n}\rightarrow\frac{1}{e\minus{}1}.$[/color]

$ p>1$
$ f(x) \equal{} x^{n}\plus{}px\minus{}1$
$ f'(x) \equal{} nx^{n\minus{}1}\plus{}p > 0$ so $ f(x)$ monotone on $ R^{\plus{}}$
$ f(\frac{1}{p}) \equal{} (\frac{1}{p}) > 0$
$ \lim_{x\to 0}f(x) < 0$
So $ f(x)$ has unique solution on $ (0,\plus{}\infty)$
More than $ x <\frac{1}{p}$ so
$ 1 > px_{n}> 1\minus{}(\frac{1}{p})^{n}$
So $ \lim_{n\to\infty}x_{n}\equal{}\frac{1}{p}$

$ n>1$
Prove that following equation has unique  solution on $ (0,\plus{}\infty)$
$ \sum_{i\equal{}1}^{n}\frac{1}{\sqrt{nx\plus{}i}}\equal{}\sqrt{n}$
b)Prove that $ x_{n}$ converge and find it limit.

Consider equation
$ \sum_{i\equal{}1}^{n}\frac{1}{{i^{2}}x\minus{}1}\equal{}\frac{1}{2}$
a)Prove that equation has unique solution on $ (1,\plus{}\infty)$
b)Prove that $ x_{n}$ converge and find it limit.

These problem is quite good so can some body solve it?

[quote="TTsphn"]Prove that exist unique $ f(x): [0,\plus{}\infty)\to\text{R}$ such that
$ f(x)^{e^{f(x)}}\equal{} x$
b)Find $ \lim_{x\to\infty}\frac{f(x)}{\text{ln}x}$[/quote]

for the existence of $ f(x)$ study the function $ h(t)\equal{}t.e^{t}$

b) Lambert function http://mathworld.wolfram.com/LambertW-Function.html

the limit is 1


James and Steve were busily emptying their book shelves from one room to another. James on his own could have cleared the room in 18 hours. Steve also would take 18 hours. Unfortunately, John was moving the books back into the first room from the second. He could fill the first room in 36 hours. How long will it take the three of them to complete the

The first two together would fill the second room in 9 hours. However, in 9 hours, John had moved $\frac{1}{4}$ of the books back to the first room, so they take another $\frac{9}{4}$ hours to move those books back, in which time John moves $\frac{1}{16}$ of the books to the first room, etc. Basically, we come up with an infinite series which we cannot solve exactly until we know how many books there are, because then we know when the fraction becomes equal to one last book that James and Steve are moving to the second room, in which case it is done.


Factorize $x^{3}-(a+2)x+\sqrt{a+1}$ in $\mathbb{R}$

Put $\sqrt{a+1}=p\implies a=p^2-1.$ Then

\begin{eqnarray*}x^3-(a+2)x+\sqrt{a+1} &=& x^3-(p^2+1)x+p\\ &=& x^3-p^2x-x+p\\ &=& x(x-p)(x+p)-(x-p)\\ &=& (x-p)(x^2+px-1)\\ &=& (x-\sqrt{a+1})(x^2+x\sqrt{a+1}-1)\end{eqnarray*}

The second factor can be split using the quadratic formula, giving

$x^3-(a+2)x+\sqrt{a+1}=(x-\sqrt{a+1})(x-\frac{\sqrt{a+1}+\sqrt{a+5}}{2})(x-\frac{\sqrt{a+1}-\sqrt{a+5}}{2})$


induction
number theory proposed
number theory

Which one is bigger $100^{300}$ or $300!$

It's very easy to show by induction that $n!>(\frac{n}{e})^{n}$


There is a cool site with flash animated desktops. Everybody take a look at  http://fantasyartdesign.com/a-animated-wallpapers/01flash-animation-desktop.htm The site has 29 pages of different Macromedia flash, interactive desktop backgrounds for standard and windscreen monitors, interactive animated desktop wallpapers, and desktop mind and puzzle games. It has the best modern surrealism art gallery and free digital 3d art software as well.
[img]http://www.fantasyartdesign.com/3d-art-surrealism/02artfantasy3d/artworks/3d-fantasy-art/18image3d-twilight-s.jpg[/img]
[url]http://fantasyartdesign.com[/url]

yeah....looks pretty interesting.....But if you're interested in these flash based desktops, try this stuff as well.... http://www.3dna.net/  It's an amazing way of looking at your desktop...

I had [url]http://www.hdinstallers.com[/url] come out and hook up my 40" Sony 1080P Television to my computer and home theater system.

If you really want to see some great animation you have to get hooked up with HD large screen tv.


How many different questions are in the Alcumus database?  :huh:  :?:
Also, is there any way to tell how many problems I've truly done?
I know there are different degrees of repeat counting when you look in different places. For example, on the home page, I've done 1843, and when adding up correct+incorrect+giveup on the report, I get 1765.

Thanks!

There should be a lot of problems.
Since they obviously don't have $ \infty$ problems, they probably add more that are Intro to C/P level all the time, just in case people run out.

I have had alot of the problems given to me twice. :rotfl:  :maybe: [/img]

once you have done all the problems in the Alcumus database once, it gives you problems that you got wrong the first time around


For how many positive integers $m$ is \[\frac{2002}{m^{2}-2}\] a positive integer?

[hide]
I disagree with the answer key, which says the answer is $3$. [/hide]

[quote="Arne"][quote="lotrgreengrapes7926"]
[hide]
I disagree with the answer key, which says the answer is $3$. [/hide][/quote]

Well, you're wrong then :)[/quote]
Oh, yeah. Never mind.  :blush:

Yeah, on the prime factorization of $2002$, I thought it was $2\cdot7\cdot11\cdot17$ instead of $2\cdot7\cdot11\cdot13$. Which gives an extra value of $m=6$. :roll:


function
induction
limit
inequalities
algebra unsolved
algebra

find all the functions $f: \mathbb{R^{+}}\longrightarrow\mathbb{R^{+}}$ such that :

$f(xf(y))+f(yf(x)) = 2xy ,$ $\forall x , y \in\mathbb{R^{+}}$

proves that $f(x)=x$  $\forall x \in\mathbb{R^{+}}$

EDIT:let's suppose x=y; we have
f(xf(x))=x^2
always for induction we have
f(x^(2^n)f(x)/x)=x^(2^n)
from this we get
if a>b then f(a)>f(b) and if a<b then f(a)<f(b)   (1)
if it' is not true that f(x)=x then
if t=x^(2^n)f(x)/x
or f(x)>x so x^(2^n)f(x)/x>x^(2^n) so f(x^(2^n))>x^(2^n) instead we have f(x^(2^n))=x^(2^n)
or f(x)<x so x^(2^n)f(x)/x<x^(2^n) so f(x^(2^n))<x^(2^n) instead we have f(x^(2^n))=x^(2^n)
so f(x)=x and f(t)=f(x^(2^n)*x/x)=x^(2^n)x/x=t and f(t)=t for every real number t
all right?

Let me check. First I'll type up for you:

let's suppose $x=y;$ we have
$f(xf(x))=x^{2}$
always for induction we have
$f(x^{2^{n}}f(x)/x)=x^{2^{n}}$
from this we get
if $a>b$ then $f(a)>f(b)$ and if $a<b$ then $f(a)<f(b)$ (1)
if it' is not true that $f(t)=t$ then
if $t=x^{2^{n}}f(x)/x$
or $f(x)>t$ so $x^{2^{n}}f(x)/x>x^{(}2^{n})$ so $f(x^{2^{n}})>x^{2^{n}}$ instead we have $f(x^{2^{n}})=x^{2^{n}}$
or $f(x)>t$ so $x^{2^{n}}f(x)/x>x^{2^{n}}$ so $f(x^{2^{n}})>x^{2^{n}}$ instead we have $f(x^{2^{n}})=x^{2^{n}}$
so $f(x)=x$ and $f(t)=f(x^{2^{n}}*x/x)=x^{2^{n}}x/x=t$ and $f(t)=t$ for every real number $t$

I'm not sure, could you pleases elaborate some steps?

Here's another approach. Put $f(x)=xg(x)$. Then
\[g(y)g(xyg(y))+g(x)g(xyg(x))=2\]
We have to prove that $g(x)=1$.
First, we have $g(x)>m_{1}$ for all $x>0$, where $m_{1}=0$.
Let us have $g(x)>m_{n}$.
Then for all $x,y$
\[g(x)g(xyg(x))=2-g(y)g(xyg(y))<2-m_{n}^{2}\]
Take $y=\frac1{g(x)}$. We'll get $g(x)<M_{n}$ for all $x>0$, where $M_{n}=\sqrt{2-m_{n}^{2}}$.
There's $y>0$ such that $g(y)\leqslant1$ (otherwise we 'd always had $g(y)g(xyg(y))+g(x)g(xyg(x))>2$). Take this $y$. Then for all $x>0$
\[g(x)=\frac{2-g(y)g(xyg(y))}{g(xyg(x))}>\frac{2-M_{n}}{M_{n}}=m_{n+1}\]
It's easy to prove that for all $n$
\[m_{n}<m_{n+1}<1<M_{n+1}<M_{n}\]
and from here one can deduce that
\[\lim_{n\to\infty}m_{n}=\lim_{n\to\infty}M_{n}=1\]
But for all $x>0$
\[m_{n}<g(x)<M_{n}\]
Hence $g(x)=1$

Very nice method, I like it :P
I am a beginner at functional inequalities, there was a beginners problem (well it is in a current contest actually but a rather easy one) that I tried to solve using limits and  bounds like you did but to no avail, then I said to myself: this is an easy problem so look for an easy solution.
And that day I saw the solution and it was so trivial I was angry
but the results I had established and related functions were nice also.
[b]
[color=red]The main thing I want to ask is, can you point me to a good resource of functional inequalities?[/color][/b]

I posted some two years ago on the forum the following problem: find all functions $f$ from $R$ to $R$ such that $f(xf(y))+f(yf(x))=2xy$. I don't think this is simple.  :wink:


search
function
induction
modular arithmetic
number theory
prime numbers

Hello all...

here is the problem it has been bothering me for days!

There are infinitely many primes of the form 6k - 1 where k s a positive
integer.

prove!

anyone??
and help would be greatly appreciated

thanks!

You should try using the Search button. You should realize a problem like this is quite popular.

You know what the best thing is about the search function?  Only finding posts where somebody answers by yelling about using the search function.

If there were only finitely many, let their product be $ Q$.  Look at the factors of $ 6Q \minus{} 1$.

Welcome to the forums!

proof by induction maybe?

and as for using the search button,,, i couldnt find much at all

There is plenty on $ 4n\plus{}5, 12n\plus{}5, 6n\plus{}5$. Sure they are not EXACTLY the same but you can figure out how to do it from them.

[quote="f1mpladed"]thanks LRain... i see what your saying ..what i know is that except for 2... all other primes are odd (duh!) but i dont know how to approach what your saying exactly[/quote]
Show that a number congruent to $ 5 \bmod 6$ must have a prime factor congruent to $ 5 \bmod 6$.

[quote="Fermat's Little Turtle"]There is plenty on $ 4n \plus{} 5, 12n \plus{} 5, 6n \plus{} 5$. Sure they are not EXACTLY the same but you can figure out how to do it from them.[/quote]
The first two cases are much less trivial than this one.  The trivial cases are $ 4n \plus{} 3$ and $ 6n \plus{} 5$.

@FLT: First, the $ 6n \minus{} 1$ case is exactly the same as the $ 6n \plus{} 5$ case.  Second, I've found that it's much more constructive to actually do the search yourself, post a link to one or two past threads, [i]and post the words you searched for to find them[/i] -- this replaces the relatively unhelpful comment "do a search" with a comment that's much more useful in two ways: it provides immediate satisfaction for the question at hand, and it provides a practical lesson on how to behave in the future.

thanks to everyone that has been helping... slight problem.. in class a proofs class, we havent done modulus or anything like that

If you post homework for a class, you should be very clear that that's what you're doing, so people know to give hints rather than post full solutions.

That said, if you know what modulo means, nobody's forcing you to use *that word* to describe the concept.  A good place to start in writing a proof is making sure that *you* understand the ideas, then as a second step, prepare the presentation so that you fulfill the requirement of the assignment, or your audience, or whatever.

The problem inherently involves remainders (and that's all modulo is) so there's no way getting around this idea.  You need to be able to deduce what t0rajir0u said.  I'll rephrase it without modulo so you get the idea: Any number of the form $ 6k \plus{} 5$ must have a prime factor of the form $ 6k \plus{} 5$.

There are lots of articles establishing that there are arbitrarily long arithmetic progressions of primes. Terence Tao did a fansinating work.

Search button or no search button, but it clearly belongs to Number theory, not to Abstract algebra. :P. Moved.

ok.. thanks i guess ill keep looking at it and try to figure it out.. its a bonus problem that im trying to get credit for... so any solutions would be greatly appreciated

gracias

[quote="pigming"]There are lots of articles establishing that there are arbitrarily long arithmetic progressions of primes.[/quote]
Yes, but this has very little to do with primes in arithmetic progressions.

[quote="f1mpladed"]its a bonus problem that im trying to get credit for... so any solutions would be greatly appreciated[/quote]
I don't know if this is explicitly against the rules of the forum, but it's unethical for you to ask us to write solutions for your extra credit, unless your teacher/professor wrote a note on the syllabus which states, "If you don't feel like turning in your own solutions, you may turn in solutions written by somebody else who is not a student here, and who you have never met."

We have already told you how to do the problem.  The rest is up to you.

let's the say,we have n prime numbers  which apply to this form.
$ p_1,p_2,..,p_n\equiv \minus{} 1\pmod{6}$
$ 6.(p_1.p_2.....p_n) \minus{} 1 \equal{} A$
if A is a prime Contradiction!
if A is a coprime ; this number has a prime divisitor which applies to this form.
But this prime number is different to n prime number
Contradiction!


Assume that n is a non-negative integer. Determine all non-negative integers g, so that [tex]g^4+g^3+g^2+g+1=n^2[/tex].

It seems that this one appears many times before here...
Anyway, I agree that it uses idea that has to be seen.

Pierre.

its not too hard to show for g even and greater than 0:

$(g^2+\frac{g}{2})^2 < g^4+g^3+g^2+g+1 < (g^2+\frac{g}{2}+1)^2$

and for g odd, and greater than 3,

$(g^2+\frac{g}{2}-\frac{1}{2})^2 < g^4+g^3+g^2+g+1 < (g^2+\frac{g}{2}+\frac{1}{2})^2$

so the only pairs (g,n) that work are (0,1), (3,11).

solmething like that...

[quote="pbornsztein"]It seems that this one appears many times before here...
Anyway, I agree that it uses idea that has to be seen.[/quote]
It is Pre-Olympiad...

That's why I added a second sentence  :D

Pierre.

[quote="pbornsztein"]That's why I added a second sentence  :D [/quote]
I guessed it.  :D


Help me solve this problem.
People of country T-land lived on the big plain many years ago. It is happened so that they started to quarrel, so they began to build walls to separate from each other.
One day they realized that walls surround some part of the country. Your task is to determine which wall was build first to surround a part of the T-land. M (1<=M<=200000) walls are given.
 Example:
   4
   0 0 1 0
   0 1 0 0
   1 0 0 1
   2 2 5 7
The answer for this example is 3 because the third segment constructs a triangle consisting of segments 1, 2 and 3, and the triangle separates plane from its interior.

The official version of this says walls cannot intersect, and only meet each other at vertices, etc. So it looks like its just a simple set/union problem (you want the first time that a wall connects two points in the same set).

TripleM thanks for all those soltions  :omighty:


The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite arc $BC$ is a semicircle while arc $AB$ and arc $AC$ are supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is midpoint of that portion of the tangent intercepted by the extended lines $AB$ and $AC$. More precisely, the point $D$ on arc $BC$ is the midpoint of the segment joining the points $D'$ and $D''$ where the tangent at $D$ intersects the extended lines $AB$ and $AC$. Similarly for $E$ on arc $AC$ and $F$ on arc $AB$.
Prove that triangle $DEF$ is equilateral.

[Edit: I finished this question 6 minutes be4 the contest ended, but I made a stupid mistake after all. And I didn't get to finish the last part of question $1$, I got a pattern like $\sum f(2006, n) = \sum g(x) (^{2006}_{x})$ where    $g(x) = 2^x-g(x-1)$ if $x$ is even and $g(x) = 2^x +g(x-1)$ if $x$ is odd. ]
[Further Edit: The sequence $g(x)$ looks like the following:
$g(1) = 1, g(2) = 1, g(3) = 3, g(4) = 5, g(5) = 11, g(6)= 21, ...$ where $g(k)+g(k-1) = 2^{k-1}$

angle chasing owns it.
i think problem 1 is solved differently. let number of children who got 0,1,2 candies be a,b,c. then the function value equals number of int solutions to:
a + b + c = n
b + 2c = k.
i probably made a stupid mistake, too. How many marks do u think they will take off?

I did angle chasing with 5 minutes left, and showed that angle D is 60 degrees. Then I said, "similarly, angles E and F are 60 degrees."  :?

I did that too, but certainly, I was trying to mess up my paper, coz I'm sure, if that's written in LaTeX, the examiner will give me a big $0$ on question 5.

lol. the fact angles E and F are 60 degrees can be proved in 5 sec, and that's better than proving angle D, because i don't think it's the same argument, while it's the same for E and F. i might be wrong, i haven't checked.

ok. Obviously all angles at pt A are 90, F, E, D are midpts so F'F = F"F = FA, E'E = E"E = EA, AD = D'D = DD". that's very important
let F"FA and E'EA be 2k ,2l respectively. By angle chord   theorem and some isosceles triangles, get angle FAB = k, CAE = l, ADE = 2l, FAD = 2k.Let FDD' be 2m. Then DAD" is k + m, FED = 2m.But A is 90, so BAD is 90 - k - m. FAB is k, so FAD is 90 -m.
90 - m = 2m.
2m = 60. Do the same for pt F this time it is by analogy.QED

[quote="Singular"]Can you please show the proof that Angle E, Angle F is 60 deg? "in 5 seconds"?  That is the entire problem[/quote]
Let angle CDD'' be $\alpha$ and CEE'' be $\beta$.
angle CAD is $\alpha$ since line CD'' is tangent line.
angle CD''D' is also $\alpha$ since D is the circumcentre of triangle AD'D''.
By similiar reason, $\beta$ = angle EAC= angle EE''A.

angle EAD = angle EAE'' + angle DAE'' = $\alpha+\beta$
angle ECD = angle ECA + angle DCA = $2(\alpha+\beta)$

angle EAD + angle ECD = 180, so $\alpha+\beta = 60$. Therefore, angle F = angle EAD = 60.


function
integration
symmetry
calculus
calculus computations

Is there a simple answer for
\[\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}\]
? So far I, in duplication the classic solution for $\frac{1}{n^{2}}$, I managed to calculate this to
\[\sum_{n=1}^{\infty}\frac{n*\pi^{n}}{(2n+1)!}\]
But this does not seem to be easily computed.

I don't have time to work this right now, but this is a classic use of the Poisson summation formula.

[quote="JBL"]It has appeared on the forum before.  One can replace "1" with "$a$" and the answer is in the form of a trigonometric function of $a$.  From the point in your solution that you're at, substitute $\pi = x$, divide by $x$ and integrate.  Then you should have something you can work with.  (Actually, maybe $\pi = x^{2}$ is better.  Either way will work, though.)[/quote]

Can you show me or post a link to the other topics?

I tried searching but was unsuccessful -- I just know that I've posted in such a thread.  (My guess is that this happened in 2007, but I'm not entirely confident of that.)

I don't really see it as a variation on the zeta function at all. It's something else.

Start from this:

$\frac1{1+\xi^{2}}=\pi\int_{-\infty}^{\infty}e^{-2\pi|x|}e^{2\pi ix\xi}\,dx$

Now define the function

$f(x)=\pi\sum_{k=-\infty}^{\infty}e^{-2\pi|x-k|}$

This gives us a function that is periodic of period 1. We can calculate its Fourier coefficients:

$\widehat{f}(n)=\int_{0}^{1}f(x)e^{-2\pi inx}\,dx=\frac1{1+n^{2}}.$

Now, by Fourier inversion,

$\sum_{n=-\infty}^{\infty}\frac1{1+n^{2}}=f(0)= \pi\sum_{k=-\infty}^{\infty}e^{-2\pi |k|}$

$=\pi\left(\frac2{1-e^{-2\pi}}-1\right)= \pi\,\frac{1+e^{-2\pi}}{1-e^{-2\pi}}$

So, by symmetry, the sum we want is

$\sum_{n=1}^{\infty}\frac1{1+n^{2}}= \frac12\left(\pi\,\frac{1+e^{-2\pi}}{1-e^{-2\pi}}-1\right).$

I guess I was hoping for a solution based on the power series of sin x, which is used to solve the $\zeta{2}$ (see the combinatorial identity in the op).

One could also study fourier series of $f(t)=e^{t}$
\[c_{n}=(1+in)\frac{(-1)^{n}}{2(1+n^{2})}\frac{e^\pi-e^{-\pi}}{\pi}\]
and apply Parseval
\[\sum_{n\in\mathbb{Z}}|c_{n}|^{2}=\frac{1}{2\pi}\int_{-\pi}^\pi|f(t)|^{2}dt \]
to get the answer.
Or we could evaluate $\frac{1}{2}(f(\pi+)+f(\pi-))$ at the jump and
equal this to the series at $t=\pi$


calculus
derivative
inequalities
inequalities unsolved

let $x$ be a real number such that $x>1$ and $n$ an integer such that $n>1$
prove that : $1+\frac{x-1}{nx}<\root{n}\of{x}<1+\frac{x-1}{n}$

I think they are both easily handled by a derivative check.  Were you looking for a specific method?

We have:

1 + (x-1)/n = ((n - 1) + x)/n and by AM-GM inequality:

(1 + 1 + ... + 1 + x)/n > x^(1/n)

We can´t have equality because x>1
That was the second part.
Now, by GM-HM inequality:

x^(1/n) > n/(1 + 1 + ... + 1 + 1/x) = n/(n-1 + 1/x) = nx/(nx - x + 1)

Let´s see now that nx/(nx - x + 1) >= (nx + x - 1)/nx = 1 + (x-1)/nx which is clearly true because (nx)^2  >= (nx)^2 - (x-1)^2 .
This means x^(1/n) > 1 + (x-1)/nx as wanted.

I think we can use Bernuirli's ineq:
I have:[tex]x^\frac{1}{n}=(1+x-1)^\frac{1}{n}\leq\ 1+\frac{x-1}{n}[/tex]


Giving the nonegetive integer sequence $u_{1},u_{2},..u_{n},...$ such that $u_{n+1}>u_{u_{n}}$ for all $n \in N^{+}$.
Prove that $2^{u_{n}-1}+1$ doesn't divisible to $n$   for all $n \in N^{+}$   :)  :)

Nice combination of an IMO problem and Sierpinski theorem. ;)  ;)