function
trigonometry
real analysis
real analysis unsolved

Prove

$ \arcsin^{2}z\equal{}\frac{1}2\sum_{k\equal{}1}^{\infty}\frac{(2z)^{k}}{k^{2}{2k\choose k}}\qquad |z|\le 1$

Shouldn't the function be even?or is it Arcsin(arcsin(z))? :wink:

The function should be even.  :wink:

I committed a blunder again.  :rotfl:

Oh, it isn't possible to make corrections! :rotfl:

[quote="Dr. Doe"]Prove

$ \arcsin^{2}z \equal{}\frac{1}{2}\sum_{k \equal{} 1}^{\infty}\frac{(2z)^{2k}}{k^{2}{2k\choose k}}\qquad |z|\le 1$[/quote]
we have

$ \cos\;(n\theta)\equal{}1\;\minus{}\;\frac{n^{2}\theta^{2}}{2!}\plus{}\ldots\;\;\;\quad\ldots\;\;\;\quad\ldots\;\;\;(A)$

also,

$ \cos\;(n\theta)\equal{}1\;\minus{}\;\frac{n^{2}}{2!}\sin^{2}\theta\;\plus{}\;\frac{n^{2}\cdot(n^{2}\minus{}2^{2})}{4!}\sin^{4}\theta\;\minus{}\;\frac{n^{2}\cdot(n^{2}\minus{}2^{2})\cdot(n^{2}\minus{}4^{2})}{6!}\sin^{6}\theta\;\plus{}\;\frac{n^{2}\cdot(n^{2}\minus{}2^{2})\cdot(n^{2}\minus{}4^{2})\cdot(n^{2}\minus{}6^{2})}{8!}\sin^{8}\theta\;\minus{}\;\ldots\;\;\;\quad\ldots\;\;\;\quad\ldots\;\;\;(B)$

the identity in $ (B)$ easily follows from equating co-efficients of $ z^{n}$ on both sides of
$ \frac{1\minus{}z\cos\theta}{1\minus{}2z\cos\theta\plus{}z^{2}}\;\equal{}\;1\plus{}z\cos\theta\plus{}z^{2}\cos2\theta\plus{}z^{3}\cos3\theta\plus{}\ldots\plus{}z^{n}\cos n\theta$

equate co-efficients of $ n^{2}$ from both $ (A)$ and $ (B)$, and you get

$ \frac{\theta^{2}}{2!}\;\equal{}\;\frac{1}{2!}\sin^{2}\theta\;\plus{}\;\frac{2^{2}}{4!}\sin^{4}\theta\;\plus{}\;\frac{2^{2}\cdot 4^{2}}{6!}\sin^{6}\theta\;\plus{}\;\frac{2^{2}\cdot 4^{2}\cdot 6^{2}}{8!}\sin^{8}\theta\;\plus{}\;\ldots$

finally, put $ \theta\equal{}\sin^{\minus{}1}\;z$ to get what you want.

[b][u]alternate solution[/u][/b]: start with

$ \frac{1}{\sqrt{1\minus{}z^{2}}}\;\equal{}\;\sum_{k\equal{}0}^{\infty}\;\frac{(2k\minus{}1)!!}{(2k)!!}\;z^{2k}\;\;\;\quad\ldots\;\;\;\quad\ldots\;\;\;(C)$

and

$ \sin^{\minus{}1}\;z\;\equal{}\;z\plus{}\frac{1}{2}\cdot\frac{z^{3}}{3}\plus{}\frac{1\cdot 3}{2\cdot 4}\cdot\frac{z^{5}}{5}\plus{}\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\cdot\frac{z^{7}}{7}\plus{}\ldots\;\;\;\quad\ldots\;\;\;\quad\ldots\;\;\;(D)$

multiply $ (C)$ and $ (D)$ to get

$ \frac{\sin^{\minus{}1}\;z}{\sqrt{1\minus{}z^{2}}}\;\equal{}\;z\plus{}\frac{2}{3}\cdot z^{3}\plus{}\frac{2\cdot 4}{3\cdot 5}\cdot z^{5}\plus{}\frac{2\cdot 4\cdot 6}{3\cdot 5\cdot 7}\cdot z^{7}\plus{}\ldots$

now integrate both sides w.r.t. $ z$ to get the answer.

[u]note[/u]: $ (D)$ can be easily obtained by equating co-efficients of $ n$ on
both sides of $ (E)$ and $ (F)$

$ \sin\;(n\theta)\;\equal{}\;n\theta\minus{}\ldots\;\;\;\quad\ldots\;\;\;\quad\ldots\;\;\;(E)$

and

$ \sin\;(n\theta)\;\equal{}\;n\sin\theta\;\minus{}\;\frac{n\cdot (n^{2}\minus{}1^{2})}{3!}\sin^{3}\theta\;\plus{}\;\frac{n\cdot (n^{2}\minus{}1^{2})\cdot (n^{2}\minus{}3^{2})}{5!}\sin^{5}\theta\;\minus{}\;\frac{n\cdot (n^{2}\minus{}1^{2})\cdot (n^{2}\minus{}3^{2})\cdot (n^{2}\minus{}5^{2})}{7!}\sin^{7}\theta\;\plus{}\;\ldots\;\;\;\quad\ldots\;\;\;\quad\ldots\;\;\;(F)$

and finally putting $ \theta\equal{}\sin^{\minus{}1}\;z$

a follow-up Q... :D  can you show

$ (\sin^{\minus{}1}z)^{4}\; \equal{}\;\frac{3}{2}\;\sum_{k \equal{} 1}^{\infty}\;\left\{\;\left[\sum_{j \equal{} 1}^{k\minus{}1}\;\frac{1}{j^{2}}\right]\cdot\frac{(2z)^{2k}}{k^{2}\cdot\dbinom{2k}{k}}\right\}$

Can someone prove the formula

$ \cos\;(n\theta)\equal{}1\minus{}\frac{n^{2}}{2!}\sin^{2}\theta\plus{}\frac{n^{2} (n^{2}\minus{}2^{2})}{4!}\sin^{4}\theta\minus{}\frac{n^{2} (n^{2}\minus{}2^{2})(n^{2}\minus{}4^{2})}{6!}\sin^{6}\theta\plus{}...$

For me it is not obvious.


Let $ A$ be an arbitrary $ nxn$ matrix, $ det A\not\equal{}0$. Prove that $ AA^{T}$ is symmetric and positive definitive.

How about a $ nxk$ matrix?

For an $ n\times k$ matrix $ A,$ we will always have $ AA^T$ symmetric and positive semidefinite. Symmetric is obvious, and for positive semidefinite, we do this: let $ x$ be an arbitrary column vector in $ \mathbb{R}^n,$ so that $ A^Tx\in \mathbb{R}^k.$ The square of the norm of any vector is nonnegative, so:

$ x^T(AA^T)x=(x^TA)(A^Tx)=(A^Tx)^T(A^Tx)=\|A^Tx\|^2\ge 0.$

In order to have $ A^TA$ positive definite, we need to have this positive for $ x\ne0,$ meaning that we need $ A^Tx\ne 0$ for $ x\ne 0.$ This is true precisely when $ \text{rank}(A^T)=\text{rank}(A)=n.$ (In turn, this is only possible if $ n\le k$ and not guaranteed even then.)

The $ n \times n$ case is generalized by Sylvester's law of inertia; in particular, $ AA^T$ is congruent to the identity matrix.


algebra
polynomial
number theory proposed
number theory

Prove that for each prime $ p\equiv 1(mod6)$ there exists exactly one pair $ (a,b)\in\mathbb N\times\mathbb N$ such that $ p\equal{}\frac{a^2\plus{}3b^2}4$.

Moreover, for $ n\in\mathbb N$ such a representation $ n\equal{}\frac{a^2\plus{}3b^2}4$ exists iff all prime factors of the square-free part of $ n$ are $ \equiv 1(mod6)$. :)

[hide="concerning Gauss’ theorem"] The proof is a bit more tricky than for Gauss’ famous one (the one stating a similar result for $ p\equiv 1(mod4)$ about $ p\equal{}a^2\plus{}b^2$.)
I wonder if there are other theorems of this type – not sure at all :?:
 [/hide]

There is a vast generalisation of this:

Let $ n$ be a positive integer. Then there exists a monic irreducible integer polynomial $ f_n(x)$ such that for every odd prime $ p$ not dividing $ n$ or the discriminant of $ f_n(x)$ we have:

$ p \equal{} x^2 \plus{} ny^2$ (for integers $ x,y$)
$ \iff$
$ \left( \frac { \minus{} n}{p} \right) \equal{} 1$ and $ f_n(x) \equiv 0 \mod p$ has a solution.

Note that $ \left( \frac { \minus{} n}{p} \right)$ is the Legendre-symbol (thus $ \equal{} 1$ iff $ \minus{} n$ is a square $ \mod p$).

[hide="More exactly..."]we can take $ f_n(x)$ to be a minimal polynomial of an algebraic integer $ \alpha$ such that $ \mathbb Q[\alpha, \sqrt { \minus{} n}$ is the generalised class field corresponding to the order $ \mathbb Z [\sqrt { \minus{} n}]$.[/hide]

Now for $ n \equal{} 1,2,3,4$, we can take $ f_n(x) \equal{} x$. For $ n \equal{} 5$, we can take $ f_5(x) \equal{} x^2 \plus{} 1$ (giving that a prime is of type $ x^2 \plus{} 5y^2$ iff it is $ \equiv 1,9 \mod 20$). It gets more complicated as $ n$ increases.
For $ n \equal{} 27$, we have $ f_{27}(x) \equal{} x^3 \minus{} 2$, thus a prime is of type $ p \equal{} x^2 \plus{} 27y^2$ iff $ p \equiv 1 \mod 3$ and $ 2$ is a third power $ \mod p$.
For $ n \equal{} 64$, we have $ f_{64}(x) \equal{} x^4 \minus{} 2$, thus a prime is of type $ p \equal{} x^2 \plus{} 64y^2$ iff $ p \equiv 1 \mod 4$ and $ 2$ is a fourth power $ \mod p$.

[quote="spanferkel"]Prove that for each prime $ p\equiv 1(mod6)$ there exists exactly one pair $ (a,b)\in\mathbb N\times\mathbb N$ such that $ p \equal{} \frac {a^2 \plus{} 3b^2}4$.

[/quote]

For $ p\equal{}7$ I've found two pairs : $ (a,b)\equal{}(5,1)$ and $ (a,b)\equal{}(4,2)$ :huh: Am I wrong ?

The correct version would be to require the numbers to be odd.
By the way, see http://www.mathlinks.ro/viewtopic.php?t=86455 for how to tackle uniqueness. See http://www.mathlinks.ro/Forum/viewtopic.php?t=55714 for the existence part's idea. Both work with $ p$ instead of $ 4p$, but there is only one brick missing here; try to find it.

Sorry, I don’t know what happened the moment I wrote it up :blush: In my original problem it should read $ p=\frac{a^2+27b^2}4$. Then the solution is unique. :)
[hide] Note that the primes $ p\equiv 1(mod6)$ split up into 2 classes, because $ a$ und $ b$ are either both odd or both even. As ZetaX has already mentioned,
[quote="ZetaX"]For $ n = 27$, we have $ f_{27}(x) = x^3 - 2$, thus a prime is of type $ p = x^2 + 27y^2$ iff $ p \equiv 1 \mod 3$ and $ 2$ is a third power $ \mod p$.[/quote] the primes where $ a$ und $ b$ are both even are exactly those where $ 2$ is a third power $ \mod\ p$.[/hide]

And to make more precise what I mean by “other theorems of this type”:

For which other $ m \in\mathbb {N}$ is there a set $ K\subset \{1,...,m\}$ and $ s,t\in\mathbb Q$ [hide] apart from trivialities like dividing the above $ s,t$ by some squares, or replacing $ m$ by a multiple of it[/hide] such that the following holds:

For a prime $ p$, there exists a pair $ (a,b)\in\mathbb N\times\mathbb N$ such that $ p=sa^2+tb^2$ iff $ p\equiv k(\mod\ m)$ for a certain $ k\in K$ :D
(see ZetaX’ example for $ m=20, K=\{1,9\}, s=1,t=5$).

I don’t think there are other $ m$ such that $ \exists K$ with $ card(K)=1$. ZetaX, what do you think?

And: For a given cardinality of $ K$, are there only finitely many $ m$ ??


inequalities
calculus
derivative
logarithms
inequalities proposed

Given $ a, b, c \geq\ 0.$ Prove that:
$ \sqrt[]{a} \plus{} \sqrt[]{b} \plus{} \sqrt[]{c} \plus{} \sqrt[]{a\plus{}b\plus{}c} \geq\ \sqrt[]{a\plus{}b} \plus{} \sqrt[]{b\plus{}c} \plus{} \sqrt[]{c\plus{}a}$

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \sqrt []{a} \plus{} \sqrt []{b} \plus{} \sqrt []{c} \plus{} \sqrt []{a \plus{} b \plus{} c} \geq\ \sqrt []{a \plus{} b} \plus{} \sqrt []{b \plus{} c} \plus{} \sqrt []{c \plus{} a}$[/quote]
A squaring kills it.

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \sqrt []{a} \plus{} \sqrt []{b} \plus{} \sqrt []{c} \plus{} \sqrt []{a \plus{} b \plus{} c} \geq\ \sqrt []{a \plus{} b} \plus{} \sqrt []{b \plus{} c} \plus{} \sqrt []{c \plus{} a}$[/quote]
Assume that $ c\equal{}\min\{a,b,c\}$, then
\[ \sqrt{b\plus{}c} \le \sqrt{b} \plus{}\frac{\sqrt{c}}{2}\]
\[ \sqrt{a\plus{}c} \le \sqrt{a} \plus{}\frac{\sqrt{c}}{2}\]
Adding up, we get
\[ \sqrt{a\plus{}c}\plus{}\sqrt{b\plus{}c} \le \sqrt{a} \plus{}\sqrt{b}\plus{}\sqrt{c}\]
Moreover, it is easy to see that
\[ \sqrt{a\plus{}b\plus{}c} \ge \sqrt{a\plus{}b}\]
Adding up, we get the result. :)

Square both sides of the original inequality and simplify, obtain

\[ \sum\left(\sqrt{ab}\plus{}\sqrt{c}\sqrt{a\plus{}b\plus{}c}\minus{}\sqrt{(b\plus{}c)(c\plus{}a)}\right)\geq0.\]

Each term is non-negative, since

\[ \left(\sqrt{ab}\plus{}\sqrt{c}\sqrt{a\plus{}b\plus{}c}\right)^2\equal{}(b\plus{}c)(c\plus{}a)\plus{}2\sqrt{abc}\sqrt{a\plus{}b\plus{}c}.\]

Equality holds iff $ abc(a\plus{}b\plus{}c)\equal{}0$, ie, iff at least one of $ a,b,c$ is zero.

[quote="arqady"]
A squaring kills it.[/quote]
Yes, I also use this way. But I like can_hang's proof  :)
The question:
Find all k satisfy:
$ \sqrt [k]{a} \plus{} \sqrt [k]{b} \plus{} \sqrt [k]{c} \plus{} \sqrt [k]{a \plus{} b \plus{} c} \geq\ \sqrt [k]{a \plus{} b} \plus{} \sqrt [k]{b \plus{} c} \plus{} \sqrt [k]{c \plus{} a}$
 :)

Should hold for any real $ k\geq1$.

Clearly for $ k\equal{}1$ the inequality becomes an equality always.  If $ a\equal{}b\equal{}c\equal{}0$, the inequality is trivially an equality.  If $ a\equal{}b\equal{}c>0$, the inequality becomes

\[ 3\plus{}\sqrt[k]3\geq3\sqrt[k]2.\]

This is true with equality for $ k\equal{}1$, and the respective derivatives with respect to $ k$ of the LHS and RHS are $ \minus{}3^\frac{1}{k}\frac{\ln3}{k^2}$ and $ \minus{}3\cdot2^\frac{1}{k}\frac{\ln2}{k^2}$, or it suffices to show that $ \frac{3}{2}^\frac{1}{k}\leq\frac{\ln 8}{\ln 3}$, clearly true for $ k\geq1$ since the LHS will be less than $ \frac{3}{2}$, and $ \frac{3}{2}\ln 3\equal{}\ln\sqrt{27}<\ln\sqrt{64}\equal{}\ln 8$.

Finally, assume that $ a,b,c$ are not all equal, $ c$ is the minimum, and $ k>1$.  Call $ \rho\equal{}\frac{1}{k}<1$.  The inequality is clearly true for $ c\equal{}0$.  Now, the derivatives of the LHS and RHS with respect to $ c$ are $ \rho c^{\rho\minus{}1}\plus{}\rho (a\plus{}b\plus{}c)^{\rho\minus{}1}$ and $ \rho(b\plus{}c)^{\rho\minus{}1}\plus{}\rho(c\plus{}a)^{\rho\minus{}1}$.  Calling $ x\equal{}\frac{a}{c}>1$ and $ y\equal{}\frac{b}{c}>1$, we find that the derivative of the LHS will be larger than the derivative of the RHS (and hence the inequality true) whenever $ 1\plus{}(x\plus{}y\plus{}1)^{\rho\minus{}1}\geq(x\plus{}1)^{\rho\minus{}1}\plus{}(y\plus{}1)^{\rho\minus{}1}$.  This is now clearly true by Jensen because $ 1\plus{}(x\plus{}y\plus{}1)\equal{}(x\plus{}1)\plus{}(y\plus{}1)$, but $ x\plus{}y\plus{}1\minus{}1\equal{}x\plus{}y>|x\minus{}y|\equal{}|(x\plus{}1)\minus{}(y\plus{}1)|$, the second derivative of $ x^{\rho\minus{}1}$ being $ (\rho\minus{}1)(\rho\minus{}2)x^{\rho\minus{}1}>0$.


[quote="santosh_blr"]4 [b]identical[/b] dice are tossed simultaneously. What is the probability that [b]at least[/b] 3 of the four nos. shown are different?[/quote]
casework. first we do it for just 3.
there are 6*5*4*3 ways for the first 3 to be different and hte next one to be the same. multiply by 4c3, tehn divide by 2 because we have one thats the same as another. thats 360*2=720. divide by 6^4=1296. Then add 6*5*4*3/1296 which is 360. so the total is 1080/1296=5/6


limit
real analysis
function
real analysis unsolved

for $ x\in (0,1)$ we define
$ A_n(x)$ is the number of appearance of digit 7 in decimal notation of x, at first n positions.
(for example when $ x\equal{}0,437874...$ then $ A_3(x)\equal{}1$ , $ A_5(x)\equal{}2$ etc...)
let $ B\equal{}\{ x \in (0,1) :$ for which exists $ \lim_{n \to \infty} \frac{A_n(x)}{n} \}$

is $ B$ Lebesgue - measureable?

Not only it is measurable, but also it has measure $ 1$. Moreover, $ A_n(x)/n\to 1/10$, $ n\to \infty$ almost everywhere.

Anyway, the measurability is quite straightforward.
We have $ x\in B$ iff (Cauchy criterion of convergence) $ \forall k > 0\ \exists N > 0\ \forall m,n > N\quad x\in B_{m,n,k}$, where $ B_{m,n,k} \equal{} \{x: |A_{m}(x)/m \minus{} A_n(x)/n| < 1/k\}$. Try to prove that $ B_{m,n,k}$ are measurable.

thank you.

now, it's easy, $ B_{m,n,k}$ is measureable because it's finit sum of intervals.

[quote="_el_doopa"]$ B_{m,n,k}$ is measureable because it's finit sum of intervals.[/quote]
Yes, exactly.

By the way, there was a typo in the definition of $ B_{m,n,k}$ ($ >$ instead of $ <$), I think, you had found it.

Another remark is that this argument works for any sequence of measurable functions and $ A_n$ are, naturally, measurable.


Two cars of masses m_1 and m_2 collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of v_1, and car 2 was traveling northward at a speed of v_2.   After the collision, the two cars stick together and travel off in a northeastern direction(with angle theta).

First, find the magnitude of the vector v, that is, the speed v of the two-car unit after the collision.

With the normal notations

$ P_i \equal{} m_1v_1\hat{i} \plus{} m_2v_2\hat{j}$
$ P_f \equal{} (m_1 \plus{} m_2)\vec{v}$

where $ \vec{v} \equal{} v_1'\hat{i} \plus{} v_2'\hat{j}$

Momentum conservation. Solve.

Edit:

$ \vec{v'} \equal{} (m_1v_1\hat{i} \plus{} m_2v_2\hat{j})/m_1 \plus{} m_2$

Its still a vector. You cannot add it like a scalar. So You are given $ \theta$. So Y component by x component = $ tan\theta$

$ \equal{}> (m_2v_2/m_1v_1) \equal{} tan\theta$

Now find the magnitude of the given vector and try to express it in terms of $ tan\theta$


If $2=p_1<p_2<p_3<...$ is a sequence of prime number .Prove that for $n>2$ the number $\frac {p_n!}{p_n(p_n+1)...(p_{n+1}-1)}\in Z$.

Let $p$ be an arbitrary prime that divides $(p_n)(p_n+1) \cdots (p_{n+1}-1)$, use Bertrand's postulate to limit $p_{n+1}$ and then compare the exponent of $p$ in $p_n!$ and $(p_n)(p_n+1) \cdots (p_{n+1}-1)$.


Let $ a_{i}$ is fibonacci sequence

for $ \forall m\in\mathbb{Z}$

$ \exists$ $ k\in\mathbb{N}$  $ \frac {a_{k}^{4} \minus{} a_{k} \minus{} 2}{m}\in\mathbb{Z}$  show that

If we can prove that $ \exists k \in\mathbb{N} : a_k \equiv \minus{}1 \ (mod \ m)$, then also $ a_k^4 \minus{} a_k \minus{} 2 \equiv (\minus{}1)^4 \minus{} (\minus{}1) \minus{} 2 \equiv 0 \ (mod \ m)$, which we need to prove.

If we look at the Fibonacci sequence (mod m), then we can derive the sequence backwards.  Thus the sequence is completely cyclic.  Also, $ (1, 1)$ was preceded by $ (\minus{}1, 1, 0)$, this means that there are infinitely many numbers that satisfy the condition.


group theory
superior algebra
superior algebra unsolved

Let $ G$ be a cyclic group or order $ n$ and $ K$ a local field of characteristic $ 0$. We consider $ K$ like a trivial $ G$-module. I ought to show that $ (O_K : nO_K)\equal{}1/|n|$ with $ O_K$ the ring of integers of $ K$ and $ |x|$ the normalized valuation over $ K$. Thanks in advance for hints.

Try to mimic the proof or the corresponding lemma in number theory.

BTW: Which role does $ G$ have in this statement? It appears to me to be redundant. ;)


My dear friend,
May be you have so many hard lessons cause the problem is not...easy.It made me busy all this morning the time which all competitors are doing the Vietnam2004 problems for the first day.Now i think i solve the problem,so i will have 7 points....at most!

Now back the problem i am wondering why you call the problem like " Normal Equation" but the key of my proof base on the proof of an old Vietnam MO(2002 I THINK).Hear goes,

 We will prove that the equation has none root!
 Assume that the equation has root in positive integers ,and so we can chose a root x1,...,x2004 such that the sum x1+....+x2004 attains minimum and WLG we can consider x1>= x2>= ....>= x2004.Let us consider the following equqtion

              f(x)= x 2  - 2002*a*x +  b + 4=0
where a=product{ xk : k>1} and b = sum{xk 2  : k>1}
So f has a positive integer root ,it is x1. This implies f has another root which we call  y1.Since the definition of the root (x1,....,x20004) so we have
   y1 \geq  x1 \geq  x2 \geq  .... \geq  x2004
.It means the number x2 lies outside the integval (x1; y1) .So we have  f(x2) >=0 it means
                  x2 2  - 2002ax2+b+4  \geq 0
so
             2005x2 2  -2002ax2  \geq 0
ie...
                    a/x2  \leq  2005/2002
This contracts the fact that x3,...,x2004 >=2 !!!!!!!!!!!!!!!!!
This completes the proof!
Hem, now i can drink a coffee to keep stop understading how my... poor English made you mad!  :D
With my best regards,

I thank you very much .
Maybe this problem is very easy for you.
Maybe this problem is bad.
when i take part in this Forum,i hope i post and solute many good problems.
I understand i'm very crazy and silly and not famous in Vietnam
But it 's a property that i see about the MarKov equation.
Thus it is not my unsolve problem.
I post it because i want that many people know about it.
The your solution is certainly right.
I like Numbertheory very much.
If you are very smart,please help me solute this problem:
"Find x,y,n in N/{0;1}satisfy:
1^n+2^n+.+x^n=1!+2!+...+y!.
Thank you very much.

Dear Thanhbinh12T

Hum i don't think it's easy problem!And of cause i love NT too[st in common? :D ] .About your new problem ,i have no....comment :(  .Indeed i think it long time and it's still open for me,if U know how to prove it ,i think it's time U should post the solution.Very glad to see your idea!

Ps: By the way ,plz tell me know what is the name of the contest which you said you put the new problem from this!


I like to mess around with avatars, but sometimes a website has a set limit of how big an avatar can be. I'd like someone to think up a formula with variables that allow me to change the size of an image to the maximum allowed size a website will take..

Example:

[quote]Change Avatar:
Leave blank to keep current avatar. Must be 125 by 125 pixels or smaller and no more than 50000 bytes, and one of these file types: jpg,jpeg,png[/quote]

Avatar to alter: http://www.hal-pc.org/~lindner/vogt/washu.gif
Current size: 613 x 479
Requested maximum size: 125 x 125

Question: By what mathematical formula can I know what percentages to shrink this image?

Example:

613 * 50% = 306.5
479 * 50% = 239.5

Although this is not 125 pixels, it is coming closer to it. However, this process of elimination by checking each percentage can be time consuming and tedious. Is there a faster way to keep the images proportions and bring them to or close to 125 pixels by algebra?

I'm looking for an equation that allows input variables that will come out with something equal to or less than 125x125. The input variables would be the current pixel size of the image. The output variables would be equal to or less than 125x125 pixels.

[quote="Cyberman"]Current size: 613 x 479
Requested maximum size: 125 x 125[/quote]

Let the current size be $(x, y)$ and the requested maximum be $(m, m)$ (I am going to assume that it is always square, since it tends to be.)  Then shrink by a percentage

$p = \frac{m}{ max(x, y) }\cdot 100$


I was just wondering which universities send teams to IMC, and if there was any chance if i could participate in IMC. Perhaps through an exchange program of some sort, a few years down the track? While I'm still learning quite a lot in the way of undergraduate mathematics, I'm still very interested in any competition in the same style as IMO.  (I'm a first year undergraduate at Sydney Uni, which unfortunately doesn't seem to be participate in these competitions, although there are quite a few ex-IMO people here).
I'm sorry if this post doesn't quite belong here.

Hi  Vinoth.

You can participate at Imc 2008 in Bulgaria independently.

Yeah the only problem would be getting there (from Sydney) :)

Thanks.
This year it's probably too late to participate, but next year I'll give it some serious thought and contact the organizers about the details.

do you have to be a mathematcian or math undergrad student to participate in the IMC, can engineers participate?

[quote="kurt.math"]What are the rules of the IMC?

Can a student who is technically in high school but is doing a mathematics degree compete?[/quote]

Well actually I have not seen any written rules except those short ones published at the web-site main page.

You should better ask J.J. but I think if you have strong desire this is not forbidden at least.

I know a precedent (actually J.J. mentioned this at the Closing Ceremony, at IMC-2008 I guess) when that summer IMO winner became also IMC winner (I guess gold medal and 1st prize respectively).

[quote="soumik"]Anyone with any trace of IMC 2011, its website and individual participatance details.....pls post here.....[/quote]
The only official web-site is http://www.imc-math.org.uk/
I suppose that the place of IMC 2011 has not been decided finally yet. I have heard different options: Baku, Krakow, again Budapest etc. It will become known in January 2011 from the web-site.
The organizing fee in 2010 was 280 Euroes for person. It covered accomodation, meals and social activities. So I can predict that in 2011 hardly it will be less than that. Maybe the same or a little more. So for participation in IMC 2011 your expencies will be tickets to the place where it will be plus organizing fee.
Good luck.


superior algebra
superior algebra unsolved

Let $ R$ be the ring of integers in a number field $ \mathbb{Q}(\alpha)$.  Let $ N(x)$ denote the norm of $ x$ in $ \mathbb{Q}(\alpha)$ over $ \mathbb{Q}$.
[u]Prove or disprove[/u]:  If $ a,b\in R$ such that the ideal $ (a,b)=(1)$, then $ \gcd(N(a),N(b))=1$ in $ \mathbb{Z}$.

I thought I tried that!  It must have been too late when I was looking at this.  Thanks ZetaX for your simplistic, yet elegant statements!!!


I've tried population-adjusting this year's final results using the method developed by Tasar in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41587]this post[/url].  Basically, we try to estimate what a country's total would be if its population were 10,000,000.

Why is Lichtenstein is top?  Well, it only has 34,000 people (about the size of the catchment area of my school, for example), so even 4 points is a pretty good total.  (Are there three people at the average school who could get nonzero IMO scores?)  Macau (450,000) and Iceland (294,000) are also very small.  Other than that, it's pretty much the usual suspects in the top ten.

Data from the two students from Israel who received incorrect papers has been left out of this analysis.

Go New Zealand!

Full ranking:
[hide] 214.93       Liechtenstein
 201.88       Macau
 198.3        Iceland
 188.48       Singapore
 187.09       Bulgaria
 182.82       Estonia
 180.84       Hungary
 173.62       Moldova
 163.5        Slovakia
 157.97       Hong Kong (10)
 152.46       Armenia
 149.31       Romania
 147.94       Latvia
 147.27       Taiwan
 140.83       Slovenia
 140.51       Macedonia
 137.7        Czech Republic
 135.82       Israel
 134.43       Belarus
 132.96       New Zealand (20)
 130.88       Croatia
 126.35       Georgia
 122.4        Cyprus
 117.9        Korea
 112.41       Lithuania
 109.31       Ireland
 107.01       Denmark
 101.19       Bosnia & Herzegovina
101.14       Albania
 98.73        Iran (30)
 97.5         Trinidad and Tobago
 93.59        Ukraine
 92.93        Uruguay
 88.43        Switzerland
 86.78        Austria
 86.65        Kazakhstan
 84.33        Finland
 81.32        Kyrgyzstan
 81.15        Costa Rica
 74.73        Norway (40)
 74.52        Australia
 71.91        Belgium
 71.73        Azerbaijan
 70.19        Serbia and Montenegro
 70.14        Russia
 61.19        Canada
 60.96        Luxembourg
 55.84        United Kingdom
 54.83        Greece
 52.16        Cuba (50)
 47.26        Japan
 46.93        Peru
 46.06        Sweden
 45.29        Germany
 40.57        Netherlands
 38.63        United States
 37.28        El Salvador
 37.01        Turkmenistan
 34.86        Poland
 31.82        Columbia (60)
 31.62        Vietnam
 30.51        Thailand
 29.87        Italy
 29.68        Turkey
 25.65        Portugal
 20.52        Puerto Rico
 19.38        Paraguay
 17.92        Argentina
 16.38        China
 16.07        Sri Lanka (70)
 15.69        France
 13.28        Kuwait
 12.86        Ecuador
 11.65        Spain
 10.21        Mexico
 9.12         South Africa
 9.02         Tunisia
 6.37         Malaysia
 6.09         Venezuela
 5.58         Morocco (80)
 5.07         Brazil
 4.18         Tajikistan
 4.09         Guatemala
 3.47         Philippines
 3.22         Indonesia
 1.16         Saudi Arabia
 1.06         Mozambique
 .86          India
 .67          Pakistan
 .2           Bangladesh (90)
 0            Bolivia[/hide]

Wow. At least we beat China. And Germany.

China has a large population, though.
But as I know, almost every there are some Chinese among USA, Canada, Austrilia, New Zealand, Malaysia,... Teams, they should be counted

Wow, 26th. I'd better make sure next year's Irish team knows about this. They can brandish it on most of the teams we usually get annihilated by - except for Iceland, of course!


I need your step-by-step here and need answer, of course. Thanks.

If 6a - 4 = - 2, what is the value of 12a - 4?

[quote="ckck"][quote="Anonymous52111"]I need your step-by-step here and need answer, of course. Thanks.

If 6a - 4 = 2-, what is the value of 12a - 2?[/quote]
[hide]Is that $-2$? If it is the answer is $a=\frac{1}{3}$. So $12\cdot\frac{1}{3}=4-2=2$.[/hide]
This should be in Classroom Math.[/quote]

um, the original post sats 12a-4, but in the quote its 12a-2. w/e.

i just say its -4, which is cooler because:

[hide]$6a-4=-2$
$6a-2=0$
$2(6a-2)=(0)(2)=12a-4=\boxed{0}$[/hide]

[quote="ckck"][quote="Anonymous52111"]I need your step-by-step here and need answer, of course. Thanks.

If 6a - 4 = 2-, what is the value of 12a - 2?[/quote]
[hide]Is that $-2$? If it is the answer is $a=\frac{1}{3}$. So $12\cdot\frac{1}{3}=4-2=2$.[/hide]
This should be in Classroom Math.[/quote]

Sorry, I have recently changed it. Thanks! Okay, I'm going to create the same question and put it in the Classroom Math.


induction
least common multiple
number theory solved
number theory

Given a positive integer n > 3, prove that the least common multiple of the products x_1x_2...x_k with k \geq 1, whose factors x_i are positive integers with x_1 + x_2 + ... + x_k \leq n, is less than n! .

[i]Mikls Schweitzer competition 1951 problem 8[/i]

I'll prove that x1*x2*...*xk|n!
I'prove this by induction on n.
If n=1 it is clear.

It is clear that I can suppose  \sum x_i=n or else add another xk+1 =n- \sum x_i.

If k=1 it is clear.
If k>1 we have n= \sum{i<k} (xi) +xk.
By the induction hypothesis we have x1*x2*...*x_(k-1)|(n-xk)!. To complete the proof it must be proven that xk|(n+1-x_k)(n+2-x_k)...n. There are xk consecutive terms in that product, so xk divides on of them and the whole product.
Finished.


linear algebra
matrix
calculus
derivative
vector
function

I cannot seem to find much information on this topic, and the wikipedia article doesn't seem too helpful, and I hope this is the right place to ask this question.

I need to take a derivative of a matrix with respect to a vector (or another matrix). More specifically, Let $ \vec{x} \in \Re^{n}, F: \Re^n \rightarrow \Re^{m\times p}$, then what might $ \frac{\partial{F(\vec{x})}}{\partial{\vec{x}}}$ look like, and what dimensions will it have?

What if I also have $ \vec{b} : \Re^n \rightarrow \Re^{p \times 1}$, how do I calculate ${ \frac{\partial{}}{\partial{\vec{x}}}(F(\vec{x})\vec{b}(\vec{x})})$ using the chain rule, and once again what are the dimensions of the output? If you apply the chain rule as one might expect, there seems to be a problem with dimensions, as the derivative will surely change the dimensions of whichever term it is acting on, creating an inconsistent multiplication.

Any references covering this would be very helpful to me.

Thanks,
Don

The derivative of $ F$ can't be written as a matrix; it's three-dimensional, living in a $ m\times n\times p$ tensor product.
Also, I don't see how $ F\cdot b$ is even defined.

Hi,

I understand that the derivative is going to be 3-dimensional, which is my problem.

As far as the $ F\cdot \vec{b}$, I don't see what you mean, $ F$ is an $ m \times p$ matrix, whose elements are functions of $ \vec{x}$, and $ \vec{b}$ is an $ p \times 1$ matrix, whose elements are functions of $ \vec{x}$. The product of $ F\cdot \vec{b}$ is an $ m \times 1$ matrix, and thus, perhaps the derivative of this quantity should be $ m \times n \times 1$ (and this in fact the result I want!), but I'd need to calculate it via the chain rule, which is where problems arise.

-Don

EDIT: Note I meant to say product rule rather than chain rule, as jmerry has pointed out.

I misread it as $ p\plus{}1$.

The chain rule derivative is a composition of linear operators. Then again, you're not differentiating a composition. This should be a version of the product rule, and if what you write down makes dimensional sense, it's very likely to be right.

Duh I meant to say product rule, not chain rule, oops  :blush:.

${{ \frac{\partial{}}{\partial{\vec{x}}}(F(\vec{x})\vec{b}(\vec{x})}) = \frac{\partial{F(\vec{x})}}{\partial{\vec{x}}}\cdot \vec{b}(\vec{x})} + F(\vec{x})\cdot \frac{\partial{\vec{b}(\vec{x})}}{\partial{\vec{x}}}$

This doesn't make dimensional sense, and I don't see how it could.

The second term is a $ (m \times p) \cdot (p \times n \times 1)$ which results in a $ m \times n$, this is correct.

The first term however is $ (m \times n \times p) \cdot (p \times 1)$ which I cannot figure out how to evaluate. I need to do some sort of dimension boosting on $ \vec{b}$ or dimension reduction on $ \frac{\partial{F(\vec{x})}}{\partial{\vec{x}}}$ in order to make it work. This is the part I cannot understand.

-Don

OK. Here's what I'd do: abandon the perspective that derivatives are matrices, and think in terms of (multi)linear operators. What is $ DF$ (the derivative of $ F$)? It takes $ n$-vectors to $ m\times p$ matrices. $ Db$ takes $ n$-vectors to $ p\times 1$ matrices. $ D(F\cdot b)(v) \equal{} DF(v)\cdot b \plus{} F\cdot Db(v)$

Why think in linear operators? That way, we can keep track of where to put the arguments.

Actually, looking at the $ (m \times n \times p) \cdot (p \times 1)$, I did the common sense thing and multiplied $ (m \times p) \cdot (p \times 1)$ $ n$ times, and then I have to transpose it because I actually want $ m \times 1 \times n$, but it does the trick!

So you're first post suggesting $ m \times n \times p$ ended up answering all of my question.

Thanks!
Don


linear algebra
matrix
linear algebra unsolved

Let T $\in$ $M_n (Z)$ s.t. $t_{ij}=1$ for each i,j.
Prove that exists a matrix A $\in$ $M_n (\{0,1\})$ s.t. $A^2=T$ if and only if n is a perfect sqare.

Nice problem, man :).

Suppose first that $n=k^2$. Consider the $n\times n$ matrix to be a block matrix, with $k^2$ blocks of size $k\times k$. We denote the $(i,j)$ block in thi block matrix by $B_{ij}$. Now, let $B_{ij}$ be the $k\times k$ block which has $1$ on the $j$'th line and zero everywhere else, for all $i,j\in\overline{1,k}$.

Conversely, suppose $A^2=T$, where $A$ is a binary matrix.This means that $A$ and $T$ commute. Now, the $(i,j)$ entry of $AT$ is the number of $1$'s on the $i$'th line of $A$, while the $(i,j)$ entry of $TA$ is the number of $1$'s on the $j$'th column of $A$, so all lines an all columns of $A$ have the same number $k$ of $1$'s. We now look at $T^2$. On the one hand, it's $nT$. On the other hand, it's $A^4=TA^2$, and since $TA=kT$, this is equal to $k^2T$, so $n=k^2$.


geometry
parallelogram
power of a point
radical axis
geometry unsolved

Let ABCD be a parrallelogram.  A circle, inside ABCD, is tangent to AB and AD and intersects diagonal BD in points E and F. Prove there exists a circle tangent to BC and CD and passing through E and F.

Dear,if we draw the related circle,its existence will be proved accordingly.

Suppose the tangency of the first circle to the lines$ AB,AD$,be points $ G,H$,respectively.

Let $ GH$ intersect the rays $ CB,CD$,at $ I,J$,respectively,$ AB,CD$ are parallel,then triangles

$ AGH,DJH,BJI$,are similar to each other,and because of the equality of $ AG$and$ GH$,

we have: $ CI\equal{}CJ$ and $ DJ\equal{}DH$ and $ BG\equal{}BI$(1)

we know that as the power of point,$ BF.BE\equal{}BG^2$

according to(1),we have:$ BF.BE\equal{}BI^2$,and it shows that the circle$ (C)$ which goes through $ I,F,E$

is tangent to $ CB$,at point $ I$.

the similar wat about point $ D$,concludes that circle$ (C')$ which goes through $ J,E,F$,

is tangent to $ CD$,at point $ J$.

but if $ (C)$and $ (C')$,be two seprate circle,accorcing to (1),$ C$ lies on the radical axis of these two circles.

And $ E,F,C$ will be collinear.

But we know that $ E,F$ are on the diagonal,and will never be collinear with point$ C$.

It concludes that our hypothesis that $ (C)$and$ (C')$ are two seprate circle, is wrong.

And we have $ I,F,E,J$ lie on a circle which is tangent to $ CB,CD$at points$ I,J$respectively.

And the problem is solved :wink:


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 10 A
AMC 8

I want to start a topic on the amc 10

1. how are you guys going to prepare

2. what is the difference between amc 10 a and b

thanks
 :lol:  :rotfl:

1. Actually, I'm preparing more for the AIME, since I already made AIME last year, and I believe I can do decently on the 10. Last year, I mainly read Volume 1 (during the summer) and then did lots and lots of problems. I also started Volume 2.

2. There really is no difference. The main thing is that you take them on two different days. They're supposedly supposed to be the exact same difficulty level.

1. i might study for AMC 10 (by taking previous year's tests) not because i am afraid of not passing, but because i wanna get a perfect and i want some buffer in case i fail AIME.

2. no intended diff.

still can't get over the fact that i found the square of the sum instead of the sum of the squares on #3 last year. and the #9 fiasco, where many people (including myself) didn't read the question and so i put 919 like many other people..... so yeah could've gotten a 7, maybe more.....

[quote="#H34N1"][quote="cognos599"]i want some buffer in case i fail AIME[/quote]

lol haha same.
I need to get cover for teh 138057235087345087345083745083745 stupid mistakes I make on AIMEs[/quote]

That's over 9000!

Anyways, I suggest that you focus on studying for the AIMEs, cuz the 10 is unusually easy to pass.  Read AoPS and do past problems from the 10 and 12.

There is no difference, I think.  I wouldn't know, cuz my school only took the A.

Just gonna practice old tests and stuff. Just making sure- You can take both the 10 and the 12, right? Will they take your better score when seeing if you make AIME? What about for the USAMO Index(not even close to USAMO yet, but just asking)?

[quote="stupididiot1337"]1. By doing the AMC-12s. Yes, very stupid, but my mom thinks that that will work. I think otherwise.

2. The B gives you more time to prepare. So, if you desperately need more time and the A has not start yet, I opt that you go and take the B instead. PS- that also pushes the date of the AIME forward.[/quote]
But isn't the B a little harder than A? That way, those who have more time to prepare don't have an unfair advantage?

Nope, no difference. But B costs more. I am using AIME's to prepare since once I do AIME's, the AMC's should seems easy. This worked well for the AMC 8 (except for the stupid mistakes part which got me on the fail boat with a 16 :wallbash_red:


Let $ q$ be an odd prime.
Are the $ \frac{q\minus{}3}{2}$ units $ (1\plus{}\zeta_q), \cdots ,(1\plus{}\zeta_q^{\frac{q\minus{}3}{2}})$ mutually independent? In other words, if $ V$ is the group generated by these units and $ U$ that of all units, is $ U/V$ a finite group?

No, it isn't. The rank of $ U$ is $ \frac {q \minus{} 1}{2}$, by Dirichlet's unit theorem. The rank of $ V$ is less or equal to $ \frac {q \minus{} 3}{2}$ which implies that the rank of $ U/V$ is >0.

[color=red]EDIT[/color] Flawed proof, since $ rk(U)\equal{}\frac {q \minus{} 1}{2}\minus{}1$.

No the rank is really $ \frac{q\minus{}3}{2}$ ;)

Sigh, it was certainly too late to deduce the needed 1 here.  :blush: Thanks for pointing it out.

EDIT: Not "deduce", but "subtract" to let the sentence have more sense.  :huh:


Let be $ a,b\in \mathbb{R}$ such that $ 16a^4\plus{}12a^2\plus{}9b^2\plus{}8a\le 3$ . Prove that : $ a(8a^2\plus{}b)\le 1$

We are given $ \frac {16}{3}a^4 \plus{} 4a^2 \plus{} 3b^2 \plus{} \frac {8}{3}a \le 1$.
If a is positive, then $ 3b^2 \plus{} \frac {a^2}{12} \ge ab$, so we have to show $ \frac {16}{3}a^4 \plus{} \frac {47}{12}a^2 \plus{} \frac {8}{3}a \ge 8a^3$, but by AM-GM $ \frac {16}{3}a^4 \plus{} \frac {47}{12}a^2 \ge 9.14 a^3 \ge 8 a^3$
If a is negative then assume what we have to prove is false. We would have $ b^2 \ge (8a^2 \minus{} \frac {1}{a})^2$, so from the given we would have $ 16a^4 \plus{} 12a^2 \plus{} 9(8a^2 \minus{} \frac {1}{a})^2 \plus{} 8a \le 3$, which is false (the minimum is a=0.48949228, f(a)=7.852859). If you really need to show this without a calculator you could let a=-k, where k is positive, expand, and use AM-GM a couple of times.

$ 16a^{4} \plus{} 12a^{2} \plus{} 9b^{2} \plus{} 8a\leq 3 \Rightarrow 16a^{4} \plus{} 12a^{2} \plus{} 9b^{2} \plus{} 8a \plus{} 1\leq 4$
 The inequality $ a(8a^{2} \plus{} b)\leq 1$ is equivalent with $ 32a^{2} \plus{} 4ab\leq 4$
Now we need to prove that $ 16x^{4} \plus{} 12a^{2} \plus{} 9b^{2} \plus{} 8a \plus{} 1\geq 32a^{3} \plus{} 4ab$
$ 16a^{2}(a \minus{} 1)^{2} \plus{} (2b \minus{} a)^{2} \plus{} 5b^{2} \minus{} 5a^{2} \plus{} 8a\geq 0$
Since $ 16a^{2}(a \minus{} 1)^{2},(2b \minus{} a)^{2}$ and $ 5b^{2}$ are always positive we need to prove that $ 8a \minus{} 5a^{2}\geq 0$
This is true because $ \frac {8}{5} > 1 > a$

From the given inequality $ 16a^{4} \plus{} 12a^{2} \plus{} 9b^{2} \plus{} 8a\leq 3$ we see that the maximal value of $ a$ is when $ b \equal{} 0$ since $ 9b^{2} \geq 0$
Put $ b \equal{} 0$ and see that $ a < 1$.


A particle of mass m is at the centre of a rough (friction coefficient $\mu$)  horizontal table radius 2a and height greater than 4a.

An inelastic light string length 4a is attached to the particle and passes over a small smooth pulley at the edge of the table.

A particle of mass M is attached to the other end of the string and is  dropped from rest at the pulley.

Find the least value for $\mu$ if the system finally comes to rest with the particle of mass m still on the table.
(Neglect the weight of the string and the inertia of the pulley)

In order for the system to be at rest, the change of $M$'s potential energy must be compensated by the work of the friction force. This force is $\mu mg$ and its work is $\mu mgs$, where $s \leq 2a$ is the distance covered by $m$ on the table.

The body $M$ will drop by $4a-(2a-s) = 2a+s$, so the change in $M$'s potential energy is $Mg(2a+s)$. This is equal to $\mu mgs$, so we get
\[\mu = \frac{M}{m}\left(\frac{2a}{s}+1\right). \]
This is clearly minimal when $s$ is as large as possible, and since $s \leq 2a$ the largest value is $s = 2a$. So the least friction coefficient is
\[\mu = \frac{2M}{m}. \]


$ (x \plus{} y \plus{} z)(u \plus{} v \plus{} w)( \frac {1}{xu} \plus{} \frac {1}{yv} \plus{} \frac {1}{zw}) \geq 21 \plus{} \frac{x}{y} \plus{} \frac{y}{z} \plus{} \frac{z}{x} \plus{} \frac{u}{v} \plus{} \frac{v}{w} \plus{} \frac{w}{u}$

LHS$ \equal{}3\plus{}\frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}\plus{}\frac{x}{z}\plus{}\frac{z}{y}\plus{}\frac{y}{x}\plus{}\frac{u}{v}\plus{}\frac{v}{w}\plus{}\frac{w}{u}\plus{}\frac{u}{w}\plus{}\frac{w}{v}\plus{}\frac{v}{u}$
$ \plus{}\frac{xu}{yv}\plus{}\frac{xu}{zw}\plus{}\frac{xv}{zw}\plus{}\frac{xw}{yv}\plus{}\frac{yu}{zw}\plus{}\frac{yv}{xu}\plus{}\frac{yv}{zw}\plus{}\frac{yw}{xu}\plus{}\frac{zu}{yv}\plus{}\frac{zv}{xu}\plus{}\frac{zw}{xu}\plus{}\frac{zw}{yv}$
$ \geq15\plus{}\frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}\plus{}\frac{x}{z}\plus{}\frac{z}{y}\plus{}\frac{y}{x}\plus{}\frac{u}{v}\plus{}\frac{v}{w}\plus{}\frac{w}{u}\plus{}\frac{u}{w}\plus{}\frac{w}{v}\plus{}\frac{v}{u}$ by AMGM
$ \geq21\plus{}\frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}\plus{}\frac{u}{v}\plus{}\frac{v}{w}\plus{}\frac{w}{u}$ by AMGM

Equality occurs if and only if $ x\equal{}y\equal{}z$ and $ u\equal{}v\equal{}w$.

[quote="xzlbq"]$ (x \plus{} y \plus{} z)(u \plus{} v \plus{} w)( \frac {1}{xu} \plus{} \frac {1}{yv} \plus{} \frac {1}{zw}) \geq 21 \plus{} \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \plus{} \frac {u}{v} \plus{} \frac {v}{w} \plus{} \frac {w}{u}$[/quote]
See here: http://can-hang2007.blogspot.com/2009/12/inequality-13-b-q-liu.html


Small Inequality :
Let $a ,b ,c$ be positive reals that \[\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 1\] Prove \[\frac{c}{ab}+\frac{a}{bc}+\frac{b}{ca}\geq 1\]

From the given condition, $a+b+c\geq abc$. We have to prove now $a^{2}+b^{2}+c^{2}\geq abc$.
Obviously the inequality $(a^{2}+b^{2}+c^{2})^{2}\geq 3abc(a+b+c)$ is ture.
Hence $a^{2}+b^{2}+c^{2}\geq \sqrt3 abc \geq abc$.


With calculus it can of course be done easily.
I don't know if my without-calculus solution is correct.

$\frac{1 + \cos 2x + \sin ^2x}{\sin 2x} = \frac{5 - 3 \cos 2x}{\sin2x}$
That becomes $5 \mathrm{cosec} 2x - 3 \cot 2x$.
Setting $r \mathrm{cosec} \theta = 5$ and $r \cot \theta = 3$, we get r = 4.
Then the equation becomes $4(\mathrm{cosec}\theta \mathrm{cosec} 2x - \cot \theta \cot 2x)$.
From there, we get $\frac{4 - 4 \cos \theta \cos 2x}{\sin \theta \sin 2x}$, For minimality ,denominator should be maximum, I suppose, and so $\sin 2x$ and $\sin \theta$ are $1$ and the answer is $4$.

[quote="ifai"]Let $f(x)=\frac{1+\cos 2x +8\sin^2 x}{\sin 2x}$.
Find $\min_{0<x<\frac{\pi}2} f(x)$[/quote]
And do it without using calculus.


I've found $ \tan (a)=\frac{\sin (2 a)}{\cos (2 a)+2 \cos (6 a)}$

But i think i went to  the wrong place... since you said the solution is simple.

A simple solution to this equation is zero or pi.

but this is non-sense.

so i think your problem have some mistake, or mine solution, don't know yet

Here is my solution
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=892909#892909


Hi, I am a 2nd year student in Economics. I have some difficulties with probability. Here is a problem that I cannot solve and I need some help.

It is supposed that only 50% of the students of a sample will go up to the level master. One chooses by chance 18 students.
Which is the probability so that at least 10 go up to the level master.

I want just to know in this exercise, which is the law of probability that one must apply.
Thank you for your assistance, and I'm so sorry for my bad English.

Look up Binomial distribution.

The answer you are looking for is
$\left( \frac{1}{2^{18}}\right) \left( \binom{18}{10}+\binom{18}{11}+\binom{18}{12}+\dots \binom{18}{18}\right)$

BTW, the calculations can be simplified a little,
[hide]Due to symmetry, this is equal to $\frac{1}{2}-\left( \frac{1}{2^{18}}\right) \binom{18}{9}$[/hide]

Thanks you very much for the answer. It helps me a lot !
 :)


If $E_{1},E_{2}$ are projections onto independent subspaces, then $E_{1}+E_{2}$ is a projection. True or false?

(#3 from Hoffman & Kunze 6.7)

False. I assume that the criterion for $E$ to be considered a projection is that $E$ be idempotent.

Let $E_{1}=\begin{bmatrix}1&1\\0&0\end{bmatrix}$ and $E_{2}=\begin{bmatrix}0&0\\0&1\end{bmatrix}.$ Then $E_{1}$ and $E_{2}$ are both idempotent, and the range of $E_{1}$ and the range of $E_{2}$ are independent subspaces. However, $E_{1}+E_{2}$ is not idempotent.

Even if you require projections to be self-adjoint, it's still false.

Let $E_{1}=\frac1{2}\,\begin{bmatrix}1&1\\1&1\end{bmatrix}$ and $E_{2}=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$

These are self-adjoint idempotents. Geometrically, they are orthogonal projections on to the $x$ axis and the line $x=y.$ But $E_{1}+E_{2}$ is not an idempotent; not a projection.

[Make it self-adjoint projections onto orthogonal (rather than merely independent) subspaces, and the situation changes.]


[hide]The number of ways to create $ 1987$ groups of $ 1987$ people each from a pool of $ 1987^2$ people is given by $ \frac{(1987^2)!}{(1987)!^{1987}}$. There are $ 1987!$ ways of permuting the group numbers while keeping the group compositions the same, so that big number must be divisible by $ 1987!$[/hide]


Find all $ f: R\rightarrow R$ such that:
$ f(x\minus{}f(y))\equal{}4f(x)\minus{}f(y)\minus{}4x$

$ x\equal{}f(y) \Longrightarrow f(f(y))\equal{}\frac{5f(y)\plus{}f(0)}{4}$
So $ \forall t \in Im(f)$ we have $ f(t)\equal{}\frac{5t\plus{}f(0)}{4}$ (*)
$ x: \equal{}f(x)$ and (*) $ \Longrightarrow f(f(x)\minus{}f(y))\equal{}f(x)\minus{}f(y)\plus{}f(0)$
So $ \forall x,y$ $ f(x)\minus{}f(y)\plus{}f(0) \in Im(f).$
$ x: \equal{}f(x)\plus{}f(0) \Longrightarrow f(f(x)\minus{}f(y)\minus{}f(0))\equal{}4f(f(x)\plus{}f(0))\minus{}f(y)\minus{}4f(x)\minus{}4f(0)$
Since $ f(x)\minus{}f(y)\plus{}f(0) \in Im(f)$ we have by (*)
$ \frac{5(f(x)\minus{}f(y)\plus{}f(0)))\plus{}f(0)}{4}\equal{}4f(f(x)\plus{}f(0))\minus{}f(y)\minus{}4f(x)\minus{}4f(0)$
Which gives $ f(f(x)\plus{}f(0))\equal{}\frac{21f(x)\minus{}f(y)\plus{}22f(0)}{16}$        (1)
$ x: \equal{}f(x)\plus{}f(y) \Longrightarrow$ $ f(f(x)\plus{}f(y))\equal{}\frac{21f(x)\plus{}20f(y)\plus{}f(0)}{16}$
Replacing in that equation $ y\equal{}0$ we get
$ f(f(x)\plus{}f(0))\equal{}\frac{21f(x)\plus{}21f(0)}{16}$                               (2)
Combining (1) and (2) we get that $ f(y)\equal{}f(0).$
Which doesn't satisfy the original equation.
There is no such function.

[quote="ozgurkircak"]$ x \equal{} f(y) \Longrightarrow f(f(y)) \equal{} \frac {5f(y) \plus{} f(0)}{4}$
So $ \forall t \in Im(f)$ we have $ f(t) \equal{} \frac {5t \plus{} f(0)}{4}$ (*)
$ x: \equal{} f(x)$ and (*) $ \Longrightarrow f(f(x) \minus{} f(y)) \equal{} f(x) \minus{} f(y) \plus{} f(0)$
So $ \forall x,y$ $ f(x) \minus{} f(y) \plus{} f(0) \in Im(f).$
$ x: \equal{} f(x) \plus{} f(0) \Longrightarrow f(f(x) \minus{} f(y) \minus{} f(0)) \equal{} 4f(f(x) \plus{} f(0)) \minus{} f(y) \minus{} 4f(x) \minus{} 4f(0)$
Since $ f(x) \minus{} f(y) \plus{} f(0) \in Im(f)$ we have by (*)
$ \frac {5(f(x) \minus{} f(y) \plus{} f(0))) \plus{} f(0)}{4} \equal{} 4f(f(x) \plus{} f(0)) \minus{} f(y) \minus{} 4f(x) \minus{} 4f(0)$
Which gives $ f(f(x) \plus{} f(0)) \equal{} \frac {21f(x) \minus{} f(y) \plus{} 22f(0)}{16}$        (1)
$ x: \equal{} f(x) \plus{} f(y) \Longrightarrow$ $ f(f(x) \plus{} f(y)) \equal{} \frac {21f(x) \plus{} 20f(y) \plus{} f(0)}{16}$
Replacing in that equation $ y \equal{} 0$ we get
$ f(f(x) \plus{} f(0)) \equal{} \frac {21f(x) \plus{} 21f(0)}{16}$                               (2)
Combining (1) and (2) we get that $ f(y) \equal{} f(0).$
Which doesn't satisfy the original equation.
There is no such function.[/quote]
Yes,all right  :P


geometry
parallelogram
geometric transformation
rotation
congruent triangles
perpendicular bisector
complex numbers

[b] I send a problem. Unfortunately I lost the source.[/b]

This can be proven by a lot of angle chasing but is there a more elegant way of solving it?

Try

[hide="Congruent Triangles"]

Extend $ PB$ and $ QB$ to meet $ CR$ and $ AR$ at $ Q'$ and $ P'$, respectively.

We have $ PP'$ is the perpendicular bisector of $ AB$, because $ PAP'B$ is a rhombus.  Clearly, $ \triangle{PP'B}$ is equilateral, so $ PP'\equal{}PB$.  Also, we have $ PARQ'$ is a parallelogram, so $ RQ'\equal{}PA\equal{}PB$.  Thus, $ PB\equal{}PP'\equal{}RQ'$.

We similarly have $ QB\equal{}QQ'\equal{}RP'$.

Finally, we have $ m\angle{PBQ}\equal{}m\angle{RQ'Q}\equal{}m\angle{PP'R}\equal{}120^{\circ}$.  Then $ \triangle{PBQ}\cong\triangle{PP'R}\cong\triangle{RQQ'}$, as desired.
[/hide]

[quote="K81o7"]Try

[hide="Congruent Triangles"]

Extend $ PB$ and $ QB$ to meet $ CR$ and $ AR$ at $ Q'$ and $ P'$, respectively.

We have $ PP'$ is the perpendicular bisector of $ AB$, because $ PAP'B$ is a rhombus.  Clearly, $ \triangle{PP'B}$ is equilateral, so $ PP' \equal{} PB$.  Also, we have $ PARQ'$ is a parallelogram, so $ RQ' \equal{} PA \equal{} PB$.  Thus, $ PB \equal{} PP' \equal{} RQ'$.

We similarly have $ QB \equal{} QQ' \equal{} RP'$.

Finally, we have $ m\angle{PBQ}\equal{} m\angle{RQ'Q}\equal{} m\angle{PP'R}\equal{} 120^{\circ}$.  Then $ \triangle{PBQ}\cong\triangle{PP'R}\cong\triangle{RQQ'}$, as desired.
[/hide][/quote]

Ah ok thats a good solution, mine was similar but included a little more angle chasing.

uhh complex numbers anybody?

it takes like 5 seconds with that

i think you can also do a solution with some rotations or something...

[color=darkblue]It is [b]a particular case[/b] (when $ B\in AC$) of the problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=45678

Go to [b]a "richer" problem[/b] from the my topic http://www.mathlinks.ro/Forum/viewtopic.php?t=166351 [/color]

Now that I think about it, is this not just a special case of Napoleon's Theorem?

good point. Then obviously we can apply any of our other proofs for napoleon's theorem with a degenerate triangle

this is still very trivial with complex numbers


function
limit
trigonometry
real analysis
topology
real analysis solved

(4) Let f be a real uniformly continuous function on the bounded set E in R (reals). Prove that f is bounded on E. Show that the conclusion is false if boundedness of E is omitted from the hypothesis.

f:E --> R  f is unif cont,  E is bdd $E \subset R$.   Show $\forall x \in E, \exists M \in R \ni M \geq |f(x)|$.

Def unif cont:
\[\forall \epsilon >0, \exists \delta >0 \ni x,y \in E, |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon \]

so I think if i make $\epsilon $ in terms of M then i can show f is bounded but i don't know how to contruct it and then it looks like it doesn't matter if E is bdd.

here is what i think is an answer:

pick $ \delta \ni \forall x,y, |x-y|<\delta $ then there will be some finite $ \epsilon $ such that $|f(x)-f(y)| < \epsilon $ .  This is clear because if it were to blow for some x and y then f is not unif cont.  now i'm unclear on how to construct $ \epsilon $ .  I was thinking choose y such that $ |f(y)| = \inf (|f(E)|) $ and choose M large enough such that $\epsilon = M - |f(y)| >0 $ so that

\[|f(x)| - |f(y)| \leq |f(x)-f(y)|<\epsilon = M - |f(y)| \Rightarrow |f(x)| < M \]

Is this right/valid?  is it a good way to show this?

oh and i guess if E is not bdd then no such $\delta $ exists for example $E=R, f(x)=x$

Luther, you're picking $\epsilon$ and $\delta$ in the wrong order. You start with $\epsilon$, then $\delta$ depends on $\epsilon.$ You don't "construct $\epsilon.$"

Here's an argument (which comes from the insight that there's something related to compactness beneath this): The bounded set $E$ is contained in some bounded interval $[a,b]$ Let $\epsilon=1,$ and choose $\delta$ so that if $|x-y|<\delta$, then  $|f(x)-f(y)|<1.$ Partition $[a,b]$ into $N$ nonoverlapping intervals $J_k$, each of length $\le \delta.$ Clearly we must have $N\delta>b-a,$ or $N>\frac{b-a}{\delta}.$ That's still a finite number.

For each $k,\,1\le k\le N,$ choose $x_k\in J_k\cap E$ if that set is nonempty.

Now suppose $x\in E.$ There is some $x_k$ such that $|x-x_k|<\delta.$ Thus, $|f(x)-f(x_k)|<1,$ or $|f(x)|<|f(x_k)|+1.$

But then, for any $x\in E,\,|f(x)|<\max_k|f(x_k)|+1.$ Since a finite set has a maximum, we are done.

The generalization of this result to general metric spaces: a uniformly continuous function on a totally bounded set is bounded. (I'll let you look up the words "totally bounded" to figure out what they have to do with this proof.)

There's a few ways to do this.  One of the ways is via Cauchy sequences - any bounded set contains a convergent subsequence (Bolzano-Weierstrass), any convergent sequence in $\mathbb{R}$ is Cauchy, for any Cauchy sequence $\{x_n\}$ it follows that $\{f(x_n)\}$ is Cauchy (this requires unifom continuity) and hence is convergent.  But any unbounded sequence can be made arbitrarily large; use this to derive a contradiction.

I think that we can also extend $f$ to a continuous function on $\overline{E}$, because if we have $f(x)\to \infty$ as $x\to a$ for some $a\in \overline{E}\setminus E$, then it cannot be uniformly continuous, because by definition we would have that $\forall N>0$, there exists $\delta>0$ and $x\in E$ s.t $d(x,a)<\delta$ implies $\vert f(x)\vert >N$.  From here, we can invoke the theorem that says a continuous real function on a compact metric space (closed and bounded will do, here) attains its maximum and minimum.

One of the things I find helpful in approaching problems like this is asking: Why is uniform continuity important?  We can pick functions on bounded intervals (say, the reciprocal function) which are continuous (just get rid of $x=0$) but that obviously are not bounded.  So uniform continuity just means that it doesn't blow up too fast.

[quote="blahblahblah"]I think that we can also extend $f$ to a continuous function on $\overline{E},$ because if we have $f(x)\to \infty$ as $x\to a$ for some $a\in \overline{E}\setminus E,$ then ...[/quote]
What you're trying to prove here is true, but that's not the argument. Remember that there are other ways for a sequence or function to diverge than by going to infinity.

The steps of that continuous extension to the closure go something like this. Better in this case to prove it directly rather than by contradiction.

Take a point $a\in \overline{E}\setminus E.$

There must be a sequence $(x_n)$ in $E$ such that $x_n\to a.$ (That's what closure does for you.)

Since $(x_n)$ is convergent, it must be Cauchy.

Since $(x_n)$ is Cauchy, $(f(x_n))$ is Cauchy. THIS is where you need $f$ to be uniformly continuous.

Tentatively define $f(a)=\lim f(x_n).$

Prove that what you have is well-defined.

Prove that what you have is continuous on $\overline{E}.$

Here's the thought exercise for you, blahblahblah: consider the process of attempting to extend $f(x)=\sin\left(\frac1x\right)$ on $(0,1)$ to the closure, and see where the process breaks down.

As for the original problem, there were always two ways to do it. One method (yours) utilizes the continuous extension to the (compact) closure. The other method (mine) does not mention the extension of the function and does not at first seem to be about a compact set. However, I needed the set to be totally bounded, and the closure of a set is compact if and only if the set is totally bounded.

Well, it doesn't work because $f(x)$ doesn't have a limit - from either side (which is what I was tacitly assuming).  Anyways, I am disappointed because I thought I had two solutions, and now I have one.  :(

But $\sin \left(\frac {1}{x}\right)$ isn't uniformly continuous, as a small point.

[quote="Kent Merryfield"]
Prove that what you have is continuous on $ \overline{E}.$
[/quote]

What would be a better way to prove the extension is continuous on $ \overline{E}$ ?  I tried to prove it by sequence argument. Please some one help.


algebra
polynomial
function
domain
induction
algebra open

Consider a polynomial P(x) such that

P(n)=2^n

for n=1,2,3,....,n,n+1.

Find P(n+2).

Hello there modeler.Please tell in detail.Is it necessary that domain of   y=P(x) is set of natural numbers.The question only says that P(x) is a polynomial of degree n.

$\binom{x-1}{n}: ={(x-1)(x-2)\cdots (x-n)\over n!}$ is defined for all $x\in\mathbb{C}$ and $n\in\mathbb{N}_{0}$.
And it's a polynom of degree $n$.

Nice solution [b]modeler[/b] btw.


geometry
3D geometry
sphere
geometric transformation
rotation

Hi,... i came across this problem: 10 pesons are to be sitted in a circular table whose chairs are numbered....

My concern is this: is the (n-1)! still applicable? Im suspecting not cause the chairs are numbered. help me... is it n! or (n-1)! ? thanks!!!:)

Yep.

Because if the chairs are numbered, who cares if it's in a circle or in a line or in a sphere or whatever?

The (n-1)! for a circle comes about when rotations are considered to be the same. Here they are not because the chairs are numbered. It doesn't matter where the chairs are, just what number they are.

So then it's like, you have a 10 people seated in a row of chairs, how many orders are there and the answer is 10!

yeah, i thought of that too, i just wanted to make sure. thanks a lot!!!:D


Is there a latex command on this forum that enables the counterclockwise line integral.  Basically I am looking for the equivalent of \varointctrclockwise.  Can anyone help me out.

I'm not exactly sure what you're asking, since I've not taken multivariable calculus yet, but is \oint what you are looking for?

[tex]\displaystyle \oint[/tex]

But what if you want a little arrow on the circle pointing in the counterclockwise (positive) direction?


Hello
I just got a old newspaper where I saw this problem.
Below is a figure made up of  $ 36$ unit squares.We have to divide it into four and three equal parts.
I have divided into four parts but still could not divide it into three parts.
Thank u.

[quote="33286"]I'm assuming that they have to be the same shape as well, as otherwise it is very easy.
Could you simply divide each column into three parts? Then the parts of the pieces would be connected only by corners; I'm not sure if that's allowed.[/quote]
hello
Yes they should be divided into three parts of same shape.
Thank u.

Just say they must be congruent.  The only thing that I can tell is they must be 12 units...

If the squares marked A and B are in the same piece, then there is no way that both of the other pieces can be long enough to be congruent to that piece.
The same applies for C and D.
A and D clearly cannot be in the same piece. The other pieces could not be the same shape.
B and C cannot because they are too far apart vertically.
A and C cannot. (Try it!) Neither can B and D.
Therefore there must be at least 4 pieces. (One each for A, B, C, and D)


Well, 2 extra weeks will help, I think. Does anyone know if I can get my school to the AIME II instead of the AIME I, or can I only take the AIME II if my school has some valid scheduling conflict?

The rules say nothing about scheduling conflicts.
However the AIME is free and the AIME II is not ($25 for up to 10 students,$50 for 11+), so if your school is anything like mine, they'll blow off the proposal just for that.

The rules are quite explicit (quoting from page 7 of the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AMC1012/TM06hs.html]2006 AMC Teachers' Manual[/url]).

[quote][b]Second AIME Testing Date[/b]
Situations in which a student may take a second version of the AIME to be held on WEDNESDAY, March 22, 2006, keeping their USAMO eligibility open are:
1. School is closed on March 7th (i.e. spring break, weather).
2. Student is out of school the entire day due to attendance at an academic/school related event.
3. Student is ill and can not attend school on March 7th.[/quote]
I believe, however, that the AMC is somewhat laissez-faire about enforcing the strict interpretation of this rule, allowing schools to schedule the Alternate AIME for any reasonable reason.  There is, as mentioned above, a fee for administering the Alternate AIME, which tends to deter most schools from administering it frivolously.

I have a question.  Is there some type of form or like checkbox that the supervising teacher must fill out in order to select either the AIME I.  Or is it assumed that you are taking the AIME I unless they hear otherwise from you, and they will just send you the necessary materials before the AIME test date.

I believe that they send you the regular 3/7/06 AIME materials by default.  You then have to contact the AMC if you want the Alternate AIME (you have to pay them too).

Out of curiosity, are you allowed to take both AIMEs, like you can with the AMC?

An update: my school was planning to do AIME II, anyways, because of a science fair.

I actually feel kinda sad now. I actually want to take the AIME I.  :lol:

Why?
1. If the AIME I is easy, I'll kick myself for not doing it.
2. If the AIME I is hard, I'll lose confidence going into test.
3. AMC 12B hard $\Rightarrow$ AIME II hard?
4. Plus the two week wait will be excruciating. I'll see all these AoPSers with scores of 11-15 on the AIME I, and I'll be saying,"Aaah! The index will be 250!"

I'm kinda scared since I need a 10, most likely, to qualify for the USAMO.  So I'm really borderline in getting 10 on practice tests now. It would be cruel if the AIME I had like no geometry questions, and the AIME II had 8. :D

[quote]3. AMC 12B hard  AIME II hard? [/quote]

So the B index will be smaller, like last year. Not necessarily a bad thing.

Bad for people who only took the B and qualify by index.

aime ii is usually harder. No real reason. Probably just coincidence. But the average score has been harder on the aime ii for most years. Look at last year as an example. The index differed by about 10 (?) points. It really sucked if you were going for the floor on the aime ii, since by the index, getting a 9 on the aime ii was like getting a 10 on the aime i.

2 weeks aren't gonna help that much. The only case when 2 weeks helps a lot is if you've got a break from school (spring break). But then again, if you've got spring break during that time, you're generally taking the aime ii anyways.

[quote="yif man12"]aime ii is usually harder. No real reason. Probably just coincidence. But the average score has been harder on the aime ii for most years. Look at last year as an example. The index differed by about 10 (?) points. It really sucked if you were going for the floor on the aime ii, since by the index, getting a 9 on the aime ii was like getting a 10 on the aime i.

[/quote]

I'm a 11th grader, so the index doesn't matter. Plus wasn't it 233 vs. 220.5? I haven't taken the AIME II 2005 as a practice yet, but it did seem easier that the AIME I (like #14, which I got in a few minutes).


A bowl contains only red marbles, blue marbles and green marbles. The probability of selecting a red marble from the bowl is $ \frac{3}{13}$. The probability of selecting a blue marble from the bowl is $ \frac{2}{5}$. There are fewer than 100 marbles in the bowl. What is the probability of selecting, at random and without replacement, a green marble and then a red marble from the bowl on the first two selections? Express your answer as a common fraction.

For our given probabilities for red and blue to occur, the total number of marbles must be a multiple of $ 13 \times 5 \equal{} 65$. $ 65$ is the only multiple of 65 which is less than 100, so this is our total. From this, we find all the rest of the infromation.

$ \frac{3}{13} \equal{} \frac{15}{65}$ so 15 red.

$ \frac{2}{5} \equal{} \frac{26}{65}$ so 26 blue.

$ 65 \minus{} 15 \minus{} 26 \equal{} 24$ green.

Our probability is $ \frac{24}{65} \times \frac{15}{64} \equal{} \frac{9}{104}$.


[quote="amcavoy"][quote="MrWise"]$3\sqrt{tg2x+1}(sin2x+2cos2x)=5(sin2x+3cos2x)$[/quote]
What is that first part under the radical?  Also, for trigonometric functions, put a "" before the command.  For example, $\$$ \sin x $\$$ produces $\sin x$ while $\$$ sin x $\$$ produces $sin x$.[/quote]

$tg 2x$ I think is the same as $\tan 2x$. In some countries it is written in one way, in others, second is used.

Convert everything into tangent.

$\sqrt{\tan{2x}+1}=\frac{5}{3} \cdot \frac{\sin{2x}+3\cos{2x}}{\sin{2x}+2\cos{2x}}$, divide both the numerator and denominator by $\cos{2x}$ and we get that $\sqrt{\tan{2x}+1}=\frac{5}{3} \cdot \frac{\tan{2x}+3}{\tan{2x}+2} \Longrightarrow \tan{2x}+1=\frac{25}{9} \cdot (\frac{\tan{2x}+3}{\tan{2x}+2})^2$. Hmmm...perhaps this way might get a little messy (i.e. a cubic equation in solving for $\tan{2x}$)

[quote="frt"][quote="Iversonfan2005"]we get that $\sqrt{\tan{2x}+1}=\frac{5}{3} \cdot \frac{\tan{2x}+3}{\tan{2x}+2} \Longrightarrow \tan{2x}+1=\frac{25}{9} \cdot (\frac{\tan{2x}+1}{\tan{2x}+2})^2$.[/quote]
There's a typo. It should say $\tan{2x}+1=\frac{25}{9} \left(\frac{\tan{2x}+3}{\tan{2x}+2}\right)$ ;)[/quote]

There's a typo. It should say $\tan{2x}+1=\frac{25}{9} \left(\frac{\tan{2x}+3}{\tan{2x}+2}\right)^2$ ;)

[quote="frt"]Oh oops :oops:  :P

Did anyone finish the solution?[/quote]

i don't know...doing it my way does not give any rational value for $x$ so either i'm missing something or.....$x$ is not supposed to be rational (perhaps the calculator is needed to do this for solving the cubic and taking the inverse $\tan$) :(

I think so is correctly : $3\sqrt{tg2x+1}(sin2x+\boxed {5}cos2x)=5(sin2x+3cos2x)$,
i.e. $\boxed {\ 1+\tan 2x=\frac{25}{9}\cdot \left(\frac{3+\tan 2x}{5+\tan 2x}\right)^2\ }\Longrightarrow \tan 2x=0$ a.s.o.

[quote="Iversonfan2005"]hah, now that's much better :D, so $x=0^\circ, 90^\circ, 180^\circ$, or anything coterminal[/quote]

You edited it and yet it's still wrong: It should be $x=90k^\circ, k\in\mathbb{Z}$ (or $x = 0^\circ, 90^\circ, 180^\circ, 270^\circ$ or anything coterminal)

no, i did not edit anything ;) , Virgil Nicula proposed a different problem

and agreed, i forgot $270^\circ$

 ;)


I think I've caught a little case of senioritis and I know I'm going to have to do some self-study. There are a few books out (Princeton Review, Cliffs, Barrons, etc...); I'm looking for the best.

Also if you've taken the AP Language and Ap Government tests, I'd appreciate any recommendations for those too.

I had barrons. I think the exams there, especially the multiple choice is more difficult then the actual ap exam.

I wasn't impressed with the Princeton Review's, but maybe the AP Calc exam is extremely easy for us AoPS folks.

I used the Barron's book as well.  It was pretty good - there were a few annoying errors, but mostly it was accurate.  And everything on last year's test was in the book; the main problem will be having the focus to read everything (especially logistic differential equations, which came up last year as a FRQ, ugh).

The new one, written by the most experienced people is "Be Prepared for the AP Calculus Exam" by Mark Howell and Martha Montgomery from skylight publishing.

I highly recommend Barron's.... it gives a concise review/outline of all the essentials, and it provides thorough explanations of all problem solutions. Plenty of problems in there too... I usually just pick it up once or twice a day, look at a few random problems, see if I can do them, then go on my merry way  :)


Steve writes a list of consecutive integers, starting with 1, at a speed of 5 digits a second. What is the sum of the last two digits he writes in 50 seconds?

Please help. I don't know how to solve this.  :maybe:

Here's a pretty quick way:

12345
67891

01112
13141
51617
18192
...
98991

00101
10210
31041
05106
10710
81091
...
You could have noticed this without writing it up, if you can calculate efficiently how much time you need to write a "decade" (ten numbers). For a one digit decade you need 2 secs (note that in the last 1/5 sec you begin with the next decade). For two digit decade you need 4 secs (note same as in the first case, the next decade begins in the last 1/5 sec), and for three digit decade you need 6 secs.

2+9x4=38. 50-38=12. In twelve seconds you can write exactly two decades of three digit numbers, and begin the third, so the last two digits were (812)[b]91[/b]. Hence, their sum is 10. Is there a faster algorithm?

:maybe: Probably I made some dumb mistake, but I still don't get it...

Basically, what are digits #249 and #250. 9 for single digits, so we have a remaining $ 241$ for 2 digit, that won't be enough, so we write $ 2\times{90}$ digts in 2-digit numbers (can you see why?). So now we've used 189 digits out of 250 $ 250\minus{}189\equal{}61$ With 61 digits, we can write the first 20 numbers, 100-119, and then the first digit of 120, so $ 9\plus{}1\equal{}10$ I get the same as hsiljak....

Hmm...perhaps I copied the problem wrong? When I click on "Ask on Forum" nothing happens(no tab or anything, it's probably just my computer), so I have to copy it out manually. But thanks guys, I know how to solve it now next time I see it.  :)

wow i so confused.... :blush:  :maybe:  :blush:


Find the sum of the recipricols of all the integers whose only prime divisors are 3 and 5.  That is to say:
1+1/3+1/5+1/9+1/15+...

[hide]

Wait, I didn't get why you multiplied the two fractions together at the last part; don't you add them?[/hide]

Pakagawa:

[hide]As an example, look at (1+1/3+1/9)(1+1/5+1/25). If you multiply this out, you get nine terms: 1+1/3+1/9+1/5+1/(3*5)+1/(9*5)+1/25+1/(3*25)+1/(9*25). Notice that all of these terms are different, and that we have listed every term whose denominator is 3^i*5*j for 0 <= i, j<= 2. If we then take the appropriate limits and make it the product of one series by another series, we get every term whose denominator is 3^i*5^j for all i and j. That's the whole original series. Furthermore, each such term appears exactly once.

(An analysis technicality that's probably out of place in intermediate level: for a positive term series, which this is, convergence is the same as absolute convergence, and with absolute convergence, we can rearrange the series - add all its terms up in a different order - without changing the result. This proof surely rearranges the order of the original series.)[/hide]

And oh, yeah, mathkid is right. I shouldn't put words in Magnara's mouth, but I would assume that was exactly the argument he intended.

I saw what I was missing. He only had the squares of 1/3 and 1/5.

[quote="Kent Merryfield"]And oh, yeah, mathkid is right. I shouldn't put words in Magnara's mouth, but I would assume that was exactly the argument he intended.[/quote]

The words in mouth was entirely justified.


algebra
polynomial
modular arithmetic
number theory proposed
number theory

[quote="warut_suk"]Find all primes $p,q$ such that $a^{3pq} = a (mod3pq)$ for all integers $a$[/quote]
The key is using Lagrage theorem. Just consider some cases as following
1) Consider $3<p<q$
Let us consider the polynomial $P(x)=x^{3pq}-x$. From the fact that $P(a)\equiv0\pmod{3pq}$ for all $a\in\mathbb Z$ so $P(x)=(x^{2(p-1)(q-1)}-1)R(x)$ where $R(x)$ is a polynomial with interger coefficcients! So $3pq-1$ is divisible by $2(p-1)(q-1)$. Now we have to find $p,q$ are primes such that $2(p-1)(q-1)|3pq-1$. So
\[3pq-1=a(pq-p-q+1) ==> (a-3)pq+a+1=a(p+q)\]
Note $a>3$ and if $a>6$ then $a-3>a/2$ and from $pq>2(p+q)$ we get the contraction!. So $a=4,5$ for each $a$ it's easy to deduce $p,q$.

2) $3<p=q$, $p=2<3<q$ the same as above.
3) $p=2,q=3$, $p=q=3$
4) $p=3<q$

Hmm... I think there should be an easier solution!  :D


Source: BMO 2001, Round 1, Problem 1

Find all positive integers $m, n$, where $n$ is odd, that satisfy
$\frac{1}{m} + \frac{4}{n} = \frac{1}{12}$.

[quote="Andreas"]Source: BMO 2001, Round 1, Problem 1

Find all positive integers $m, n$, where $n$ is odd, that satisfy
$\frac{1}{m} + \frac{4}{m} = \frac{1}{12}$.[/quote]

which is $n$ ?

$\frac{1}{m}+\frac{4}{n} = \frac{1}{12}$ <=>
$\frac{n+4m}{nm} = \frac{1}{12}$
suppose m = 1
 $\frac{n+4}{n} \neq \frac{1}{12} \forall n$

suppose m > 1
 then $\frac{n+4m}{nm} = \frac{1}{12}$ <=> $n+4m \geq nm$

 $\forall$  odd $n \geq 9$  $n+4m \geq nm$ only for m = 1

 let n = 1:  $\frac{1+4m}{m} \neq \frac{1}{12}$ $\forall$ positive m
 let n = 3:  $\frac{3+4m}{3m} \neq \frac{1}{12}$ $\forall$  positive m
 let n = 5:  $\frac{5+4m}{5m} \neq \frac{1}{12}$ $\forall$  positive m
 let n = 7:  $\frac{7+4m}{7m} \neq \frac{1}{12}$ $\forall$  positive m

Suppose $n \leq m$.
Then $\frac{1}{m} + \frac{4}{n} = \frac{1}{12} \leq \frac{5}{n}$ $\Longrightarrow$ $n \leq 60$.

We also have $\frac{1}{m} + \frac{4}{n} = \frac{1}{12} \geq \frac{4}{n}$ $\Longrightarrow$ $n \geq 48$.

So solutions are $(588, 49)$, $(204, 51)$, $(76, 57)$.
We do the same for $m < n$, that gives no solutions.


calculus
integration
geometry
geometric transformation
rotation
LaTeX
real analysis

Hi Guys,

Working my way through some integration questions and i've come across another question which i'm finding a little tricky.

Is there anybody that could explain the solution to this problem:-

[url]http://putfile.com/pic.php?pic=main/8/21713203764.jpg&s=f10[/url]

Thank you

Anybody have any ideas? I know it's a little tricky...

Use the rotation formula, $V=\pi \int_{a}^{b}y^{2}dx$..

Could somebody check if this looks right or not?

MY answer:

[url]http://putfile.com/pic.php?pic=main/8/21717492325.jpg&s=f10[/url]

That looks right yes.. :)

btw.. latex makes it easier to print formulas on this forum hehe, you might wanna look into
http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php
 ;)

Also, write equal signs instead of "Thus,".


search
inequalities proposed
inequalities

Let $x,y,z$ be positive real numbers. Prove that
\[\sum\sqrt{\frac{x}{8y+z}}\ge 1 \]
 :)

[quote="The soul of rock"]I remember that I was posted here twice! Please search before you post![/quote]
Yeah, but do you have a solution for it? :-/


The earth's metallic core extends out to 3000 km, about half the earth's radius.  Imagine that the field we observe at the earth's surface, which has a strength of roughly 0.5 gauss at the north magnetic pole, is caused by a current flow around the "equator" of the core.  How big would that current be, in amperes?

[hide] I = 3*10^9 amps [/hide]

Use the Biot-Savart Law.  Since the current is uniform at the equator, you have \[\mathbf{B}=\frac{\mu_{0}I}{4\pi}\int\frac{d\mathbf{L}\times\hat{\mathbf{r}}}{r^{2}}.\] Then you know that $d\mathbf{L}=r d\theta$ for $\theta\in\left[0,2\pi\right]$.  Since the current travels in a circle around the center of the earth, $d\mathbf{L}\cdot\hat{\mathbf{r}}=0$ so you can get rid of the cross product in the integrand.  The final step is to evaluate this integral and find $\left|\mathbf{B}\right|$ which will give you the answer you already posted.


in Square ABCD circle O is tangent to side AB and side BC and diagonal AC.  Prove that the shortest distance from a point on the circumference of circle O (point P) to angle B (point B) is half of the radius of the inscribed circle.

I found cases where this worked by cannot prove it myself.  I also dont know if this belongs in High School Basics.

EDIT: changed a to p,  sorry, and i found the inradius to be $ \frac{2x \minus{} x\sqrt{2}}{2}$ because in a right trianle is $ \frac{a \plus{} b \minus{} c}{2}$ where a and b are legs and c is hypotenuse

are you sure?  im getting (sqrt2)-1 * inradius

and i dont think think you meant point A for the point on the circumference of the circle?

Well...
Draw the altitude from $ B$ to $ AC$.  The in radius is found using $ rs\equal{}A$, where $ s$ is the semi-perimeter.  Then you find the length of the altitude and subtract twice the radius for the shortest distance.  =)

[quote="mathcrazed"]Well...
Draw the altitude from $ B$ to $ AC$.  The in radius is found using $ rs \equal{} A$, where $ s$ is the semi-perimeter.  Then you find the length of the altitude and subtract twice the radius for the shortest distance.  =)[/quote]

that gets you sqrt2-1 i think


Let $ (a_n)_{n\geq1}$ be a sequence such that $ a_n>0 \forall n\geq1$.
Let $ S_n\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n$.
Prove  that if $ a_{n\plus{}1}\leq \frac{1}{S_{n\plus{}1}}((S_n\minus{}1)a_n\plus{}a_{n\minus{}1}) \forall n \geq 2$, then $ lim a_n\equal{}0$

$ S_{n\plus{}1}a_{n\plus{}1}\leq S_na_n\minus{}a_n\plus{}a_{n\minus{}1},\  S_{n}a_{n}\leq S_{n\minus{}1}a_{n\minus{}1}\minus{}a_{n\minus{}1}\plus{}a_{n\minus{}2},$$ ..., S_{3}a_{3}\leq S_2a_2 \minus{}a_2\plus{}a_1$
You add and obtain : $ S_{n\plus{}1}a_{n\plus{}1}\plus{}S_3a_3\leq  \minus{}a_n\plus{}S_2a_2\plus{}a_1\ \Longrightarrow \ 0<a_{n\plus{}1}\leq \frac {S_2a_2\plus{}a_1}{S_{n\plus{}1}}$
Manifestly $ S_{n\plus{}1}$ is an increasing sequence so it converges to a  finite or infinite limit. .
If $ \lim S_{n\plus{}1}\equal{}L$ (finite) then $ a_{n\plus{}1}\equal{}S_{n\plus{}1}\minus{}S_n \Longrightarrow  \lim a_{n\plus{}1}\equal{} L\minus{}L\equal{}0$
If $ \lim S_{n\plus{}1}\equal{}\plus{}\infty$ then $ 0<a_{n\plus{}1}\leq \frac {S_2a_2\plus{}a_1}{S_{n\plus{}1}} \Longrightarrow   \lim a_{n\plus{}1}\equal{}0$
 :cool:


geometry
geometric transformation
reflection
rotation
congruent triangles

Let ABC be a triangle. Let A' be the reflected image with axis BC, B' the reflected image of B with axis CA and C' the reflected image of C with axis AB.

[b]Determine the possible sets of angles of triangle ABC for which A'B'C' is EQUILATERAL?[/b]

Does anyone have any idea about it??? My brain is empty~~ :(

[quote="Quevvy"]I think it's this:  :P
[hide]There's only one because angle A' will be the same as angle A and angle B' will be the same as angle B and so on. So then, for triangle A'B'C' to be equilateral, triangle ABC would also need to be equilateral. The angles would be 60, 60 and 60.[/hide][/quote]

I agree that the reflections from an equilateral triangle will create an equilateral triangle, though you might need to explain your reasoning a bit more. Why will the angles remain the same after reflection?

A reflection is a way to make congruent triangles (along with rotation and translation). [hide]So angle ACB corresponds (and is congruent) to angle ACB' and then angle CAB corresponds (and is congruent) to angle CAB'. Then, angle ABC corresponds (and is congruent) to angle AB'C. You can just apply that to all the other angles so then the original triangle [i]must[/i] be an equilateral triangle.[/hide]
Also, completely off topic, how do you make something hidden with a title?

[quote="Quevvy"]
Also, completely off topic, how do you make something hidden with a title?[/quote]
 You put an "=blahblahblah" between the word hide and the bracket.
i.e. [code] [hide=something] Woo, party [/hide] [/code] becomes [hide="something"]Woo, party[/hide]

[quote="Quevvy"][hide="thanks"] :omighty: [/hide][/quote]
[hide=" :ninja:  Rofl,"]
no prob.  :rotfl: [/hide]


let a;b;c be three positive real numbers , prove that
$ \sum \frac{a(a^{3}+b^{3})}{2a^{3}+b^{3}}\geq \frac{2}{3}(\sum a)$


Capitalization?

Being that Harvard came out with the anarchist e.e. cummings, one could suggest that capitalization isn't a serious thing. It's all up to the eye of the beholder.

What is up with all the Harvard and MIT stuff? They don't bust out with great biological feats these days. The majority of research is bound by ethical concerns. Stupid..

[quote="flying_watermelons"]ok, i'm not joking around. seriously, what do ivy league schools look for? (i understand that this is a very common topic)

thank you guys so much! :D[/quote]
Ivy League schools could fill their classes 2-3 times (if not more) with outstanding applicants.  I think that unless you have a major hook, it's fairly random seeing as how they receive the top applicants in the country.

According to data found as part of the [url=http://knowledge.wharton.upenn.edu/papers/1287.pdf]Revealed Preferences Study[/url], Harvard and MIT both are increasingly likely to admit an applicant as that applicant's SAT score increases, but this is a probabilistic rather than deterministic tendency. See this forum's sticky thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16376]Books on College Admission[/url] for more information. And, yes, if any onlooker has suggestions for revisions of that thread, I would be glad to consider them.


Here is a hint for those who are stuck: equality holds at $ (t,t,t), (2t,t,t), (t,2t,t), (t,t,2t), (2t,2t,t), (2t,t,2t), (t,2t,2t)$, where $ t$ is a real number. Also, this inequality holds for any REAL numbers $ x,y,z$.

[quote="Rofler"]$ 4\sum_{sym} x^6 \plus{} 77\sum_{sym} x^4y^2 \ge 28 \sum_{sym} x^5y \plus{} 53 \sum_{sym} x^3z^3$[/quote]

$ \Longleftrightarrow (2y \minus{} x)^2(y \minus{} 2x)^2(x \minus{} y)^2 \plus{} (2z \minus{} y)^2(z \minus{} 2y)^2(y \minus{} z)^2 \plus{}$ $ (2x \minus{} z)^2(x \minus{} 2z)^2(z \minus{} x)^2\geq 0$


hallo,
find all solutions of the equation \[(1+x!)(1+y!)=(x+y)!\] where $x,y \in N$.  :)

It is equavalent $(1+\frac{1}{x!})(1+\frac{1}{y!})=C_{x+y}^{x}$.
If $x\ge2, y\ge 2$ it give contradition. Therefore x=1 or y=1. It give x=1,y=2 and x=2,y=1.

One can also judge by divisibility over 10:if $x\geq5,y\geq 5$,there are clearily no solutions..because $(x+y)!$ is divisible by 10, but $1+x!$ and $1+y!$ aren`t.So we are left with a finite number of cases,from which we get the solutions x=1,y=2 and x=2,y=1.


How many integers less than 10000 can be made with the eight digits 0, 1, 2, 3, 4, 5, 6, 7?

[hide]Since 7777 is the greatest integer, everything in base 8 up to 7777 can be made with those eight digits.
$ 7+7\cdot8+7\cdot8^{2}+7\cdot8^{3}=4095$[/hide]

[hide] 0's an integer too, so I got $ 4095+1=\boxed{4096}$ [/hide]

[quote="Lazarus"]Why 4095? Why not $ 7\cdot7! = 35280$?

The first digit can be one of seven different numbers, and the remaining seven digits can be arranged 7! different ways.[/quote]
since the integers are less than 10000, you can't have that many arrangements.

[quote="abacadaea"][quote="Lazarus"]Why 4095? Why not $ 7\cdot7! = 35280$?

The first digit can be one of seven different numbers, and the remaining seven digits can be arranged 7! different ways.[/quote]

because an eight digit number is greater than 10000. instead you would have $ 7\sum_{k=0}^{k=3}{\binom{7}{k}}$ because we have $ 7*\binom{7}{3}$ for 4 digit numbers, $ 7*\binom{7}{2}$ for 3 digit numbers, $ 7*\binom{7}{1}$ for 2 digit numbers, and $ 7*\binom{7}{0}$ for 1 digit numbers. that equals 64. how are you guys getting 4096[/quote]
The digits can permute, so you can't use combinations.

I love how stuff here are reated like 4+ for totally wrong answers..


i have a problem

has anyone used kaplan's sat subject test math level 2 05-06 edition? i'm on pg 27-30, it's 10 problem diagnostic test, i found like 5 errors.  :(

? anyone know anything about this book? am i right or wrong?

in the meantime, i will switch to princeton review :lol:

[quote="chilly_snow"]Ya... I was so angry when I was studying using that book. I found a lot of mistakes.  :what?:[/quote]

should i continue to use it?

After I did the test this Dec., I found that most of the topics covered in the book were not on the test. Also, the book didn't provide enough materials on Combinations and Probability. Since the Kaplan's book was the only book I used to study for the test (acutally I also used the Coles Notes), so I have no comment on whether you should continue to use it. sorry..

[quote="rain"]from my experience, barrons is the best hands down just because thier problems are like twice as hard as anyon else[/quote]
I agree.. I am using the Barrons to study for the Chemistry test, and it is much better compared to the Kaplan's book.  :P  Unfortunately, I didn't know that until the night before I wrote the Math test..  :dry:

Barron's SAT2 Math is better. No question about it.


For two real numbers $ a$,$ b$, with $ ab\neq 1$, define the $ \ast$ operation by
$ a\ast b\equal{}\frac{a\plus{}b\minus{}2ab}{1\minus{}ab}$. Start with a list of n\geq 2 real numbers whose entries x all satisfy $ 0<x<1$. Select any two numbers $ a$ and $ b$ in the list; remove them and put the number $ a\ast b$ at the end of the list, thereby reducing its length by one. Repeat this procedure until a single number remains.

a. Prove that this single number is the same regardless of the choice of pair at each stage.
b. Suppose that the condition on the numbers x is weakened to $ 0<x\leq 1$. What happens if the list contains exactly one $ 1$?

Part a) seems to ask to prove that the operation is commutative and associative, so that you can do your calculation by applying the operation repeatedly on a list of numbers, with any order or pairing scheme, just like with the usual multiplication and addition operations.

For part b), you'll have to calculate $ 1\ast x$ or $ y\ast 1$ at some point in your computation, where one of the real numbers involved in the $ \ast$ operation is $ 1$. What would your result be then? What would happen as you keep calculating?

Alright thans a lot, this question is from CMO the canadian mathematical olympiad, i will try it with your hints.


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Let $ f: [0;1]\rightarrow\R$ a concave and continuous function with $ f(0)=1$. Show the following inequality:
$ \frac{p+2}{2} \int_{0}^{1} x^pf(x)dx \le \int_{0}^{1}f(x)dx-\frac{p}{2(p+1)}$, for any $ p \ge 1$.

Ok, I think I got the answer though it took me a while  :)

By writing it in a nicer form, the inequality is equivalent to:
$ \int_{0}^{1} x^p (f(x) \minus{} 1) dx \le \frac {2}{p \plus{} 2} \int_{0}^{1}(f(x) \minus{} 1)dx$.
If we set $ g: [0;1] \rightarrow R$ and $ g(x) \equal{} f(x) \minus{} 1$ we get that g is concave and $ g(0) \equal{} 0$.
So the ineq becomes: $ \frac {2}{p \plus{} 2} \int_{0}^{1} g(x)dx \ge \int_{0}^{1}x^p g(x)dx$.
As $ g$ is concave we have that for any $ \alpha \in [0;1], g(\alpha x \plus{} (1 \minus{} \alpha)0) \ge \alpha g(x) \plus{} (1 \minus{} \alpha)g(0)$ but by using the fact that $ g(0) \equal{} 0$ we get that $ g(\alpha x) \ge \alpha g(x)$ for any $ \alpha \in [0;1]$. So, $ g(x^k)\ge x^{k \minus{} 1}g(x)$ for any $ x\in [0;1]$.
$ \int_{0}^{1}g(x)dx \equal{} \int_{0}^{1}g(x^{\frac {p}{2} \plus{} 1})x^{\frac {p}{2}} \cdot (\frac {p}{2} \plus{} 1)dx$, by making a change of variable.
And then we get that $ \int_{0}^{1}g(x)dx \ge \frac {p \plus{} 2}{2} \int_{0}^{1} x^p g(x) dx$ and hence the conclusion.


analytic geometry
geometry
area of a triangle

Points A(0,0), B(6,0) and C(3,5) are the vertices of a triangle in a coordinate plane. How many square units are in the area of the triangle?

Since the distance from $ A$ to $ B$ is easy to calculate, we will use that as our base.

Using the distance formula $ \sqrt{(x_{1}\minus{}x_{2})\plus{}(y_{1}\minus{}y_{2})}$, the distance from $ A$ to $ B$ is $ 6$ units.

To find the altitude of the triangle, you can find the perpendicular straight line distance from the $ X\minus{}axis$ to $ C$.

Since it is perpendicular, then the altitude is the distance from $ (3,0)$ to $ C$. We shall call $ (3,0)$: $ D$.

Using the distance formula, the distance from $ C$ to $ D$ is 5.

Then plugging these numbers into the formula for the area of a triangle $ A\equal{}\frac{bh}{2}$ it is easy to find the area: $ 15$ square units.


In terms of $\sqrt3$, what is $\sqrt{27}+\sqrt{12}$?

In terms of $2^x$, what is $2^3\cdot4^3\cdot8^2$?


How is the hat symbol made? Looks like "^" over a symbol.

There's \hat{a} which gives you $ \hat{a}.$

I happen not to like that as the symbol for Fourier transform, so I use \widehat{f} instead:

$ \widehat{f}(\xi)\equal{}\int_{\minus{}\infty}^{\infty}f(x)e^{\minus{}2\pi i\xi x}\,dx.$


function integration calculus algebra polynomial vector domain

Does there exist eigenfunctions for the Laplace or Fourier transform?

More specifically, a function $f$ such that $F(s) = \lambda f(s)$.

Where $F(s) = \int_{0}^{\infty}e^{-st}f(t) dt$.

Or $F(s) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}e^{-ist}f(t) dt$.

You mean the eigenvalues in the Fredholm Integral Equation?
I asked the same question [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141932.[/url]

My question was about the Laplace Integral Equation.

I can answer the question about the Fourier Integral Equation. The "Gaussian" is a classic example $f(x) = e^{-x^{2}}$, it is an eigenfunction to this equation.

The complete set of Fourier eigenfunctions is given by [url=http://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions]Hermite functions[/url]. Don't know about eigenfunctions for the Laplace transform.

[quote="mlok"] Don't know about eigenfunctions for the Laplace transform.[/quote]
Nobody does  :rotfl:

PTynan89: with that particular definition of the Fourier transform, you wanted $e^{-\frac{x^{2}}{2}}$ rather than $e^{-x^{2}}.$

The Fourier transform, written that way, has the property that its fourth power is the identity operator. That limits its possible eigenvalues to $\pm1,\pm i.$ All four of these numbers are eigenvalues, and all four eigenspaces are infinite dimensional. While the Hermite functions provide an interesting and useful orthogonal basis for these four eigenspaces, as with any basis of a (bigger than one-dimensional) vector space, there is nothing unique about this basis.

The problem with the Laplace transform is that as they are commonly used, the transforms are often functions with different domains than the functions they are transforms are. If the Laplace transform is not seen as acting from a space into itself, then the issue of eigenfunctions is moot.

The Hermite functions form an orthogonal basis for $L^{2}(\mathbb{R}).$ If you're willing to expand beyond square-integrable functions, you can find additional eigenfunctions (but no additional eigenvalues - still only $\pm1,\pm i.$) One interesting example: the "comb function" given by $\sum_{k=-\infty}^{\infty}\delta(x-k)$ where that is the Dirac delta.

[Caveat: that works if you define the transform as $\widehat{f}(\xi)=\int_{-\infty}^{\infty}e^{-2\pi i\xi x}f(x)\,dx.$ Using the transform as defined above, I would need to adjust the spacing of the "comb" to make it an eigen"function", which I'm too lazy to do right now.]

The problem for the Laplace transform is not that clear, indeed. On one hand, you can always consider some function like $f(t)=\frac 1{\sqrt t}$ (or, more generally, an appropriate linear combination of $t^{-\alpha}$ and $t^{-(1-\alpha)}$, $0<\alpha<1$) to get an "eigenfunction". On the other hand, if you want to get one in some "reasonable" function space like $L^{1}$ or $L^{2}$, the answer is, probably, "No", though I do not see how to prove it right away. Still, the Laplace transform can be considered as a bounded self-adjoint operator in $L^{2}(0,+\infty)$, so, in this sense, the question about its eigenfunctions is completely well posed.

[quote="fedja"]The problem for the Laplace transform is not that clear, indeed.[/quote]
I'd love to know myself. I asked already a few people in my university and on other forums.


How'd you guys do?

I missed almost every case on number one :( cause the number formatter I used added commas for some strange reason and I didn't even get to the second one...argh

What division did you try out for?

I tried Gold.  And I did horrible.  All I feel like saying is that I obviously won't make it to Gold.

EDIT:  Nvm, I didn't do [i]that[/i] horrible, cause Kolstad changed some stuff with the test cases.  But it's still bad, and I probably won't get into gold.

How much do you have to get to be invited to gold??

I tried for gold.  I was already in silver.

made gold!!

i didnt do the second one, but i got lucky on the first one and got all of them (i checked it with like 10 min left and found a couple of really bad bugs). :)

I guess I'm back in silver.  Not sure since I'm not listed there.  Oh well.  Does anyone know where I can find the test case data?  I want to see what my program did wrong.

Hi all.
I need help with the algorithms in silver. I have got half (!?) of the test cases with a very naive algo, and i have not  solved the second. I tried to figure out a greedy algo but failed to find one. Any help is welcome.

well... i could not participate at silver... i made bronze and gold.. i got 100% of bronze but made very bad at gold... later i checked my codes and found some very stupid code bugs (not algorithm bugs) so i am very happy, and after checking all the 6 codes at analysis mode they r 100% :D

here they r:

(i hope they will help)

could u explain your algos for cowski and flight?  Code is hard to follow especially in cpp since I use java.  Thanks

There are anylisis's on the website if anyone needs them (but only for bronze and gold).


Draw a circle and arrange six circles tangent to it such that they touch both the original circle and their two neighbors. Then the three lines joining opposite points of tangency concur.
Please~answer my question. :ewpu:

An exhaustive answer to your post is included, with many other interesting things, in the old book "The Seven circles theorem", written in about 1970's by C.J.A.Evelyn, G.B.Money-Coutts and J.A.Tyrrell. But I affraid that today this book is absolutely out of print.

I guess you want to see the proof of the theorem, and find out little bit more about it. Rabinowitz' paper on the subject looks more than satisfactory: [url]http://www.mathpropress.com/stan/bibliography/sevenCircles.pdf[/url].

There is also some discussion on http://www.mathlinks.ro/viewtopic.php?t=5190 (or, more precisely, there was some discussion 4 years ago).

  darij

Thank you,everyone.
I tried to find out something for seven circle theorem.but I couldn't search anything at geometry section in mathlinks.
So I posted it. :blush:
God bless all of you! :)


George wants to find out how many three letter words in the English language have an ``a'' as the middle letter. How many possible arrangements of letters must he check?

There are $ 26$ choices for the first letter, just 1 choice for the middle letter(that's "a"), and 26 choices for the last letter. The answer is then $ 26 \cdot 1 \cdot 26 \equal{} \boxed{676}$.