Prove that if $n$ divides any Fibonacci number then it divides infinitely many Fibonacci numbers

[hide]Look for a pattern in the congruences $\mod n$.
[/hide]

[quote="paladin8"][hide="general"] $F_{m}| F_{km}$ for positive integers $m, k$. [/hide][/quote]
[hide="More general"] $\gcd\left(F_{n},F_{m}\right)= F_{\gcd(n,m)}$ for any $m,n\in\mathbb{N}^{+}$. [/hide]


PROBLEM (Lokman GÖKÇE):If $a,b,c$ are sides of a triangle then, prove that: $a^{4}+b^{4}+c^{4}<2a^{2}b^{2}+2a^{2}c^{2}+2c^{2}b^{2}$


function
inequalities
combinatorics proposed
combinatorics

Find all functions $ f: \mathbb{N}\to mathbb{N}$ such that:
a) $ f(n\plus{}m) | f(n)\plus{}f(m)$ for all $ n,m \in \mathbb{N}$
b) integers $ n,m$ and $ k$ form the sides of a triangle if and only if integers $ f(n),f(m)$ and $ f(k)$ do so.

[hide]Assume $ f(a)\equal{}f(b)$ where $ a<b$. Then $ f(a),f(x),f(x\plus{}a)$ ($ x>b$) do not form sides of triangle, but $ f(b),f(x),f(x\plus{}a)$ form sides of triangle (because $ (b)\plus{}(x)>(x\plus{}a)$), a contradiction. Thus $ f$ is injective. Let $ f(1)\equal{}a$. We claim that $ f(n)\equal{}an$ for all $ n$, which will be proved by induction. Assume that this is true for $ n\equal{}1,2,\ldots,k\minus{}1$.

(i) $ k$ is even.
$ f(k),f(k),f(2k)$ do not form sides of triangle, hence one of the following inequalities must hold: $ f(k)\plus{}f(k)\le f(2k),f(k)\plus{}f(2k)\le f(k)$. The latter one is obviously wrong, hence $ f(2k)\ge2f(k)\equal{}2ak$. But $ f(2k)|f(k)\plus{}f(k)\equal{}2ak$, so $ f(2k)\le 2ak$. Therefore $ f(2k)\equal{}2ak$, as desired.

(ii) $ k$ is odd.
For $ k\equal{}3$, $ f(1),f(2),f(3)$ do not form sides of triangle, so $ a,2a,f(3)$ do not form sides of triangle. Therefore $ 2a\ge f(3)\plus{}a$ or $ f(3)\ge a\plus{}2a\equal{}3a$. If $ 2a\ge f(3)\plus{}a$, or $ f(3)\le a$, then $ f(1),f(1),f(3)$ form sides of triangle, which is wrong. Thus the latter inequality holds, that is $ f(3)\ge3a$. But $ f(3)|f(1)\plus{}f(2)\equal{}3a$, so $ f(3)\le 3a$, and hence $ f(3)\equal{}3a$.

For $ k>3$, $ f(k\minus{}2),f(k\minus{}1),f(k)$ form sides of triangle. So $ f(k\minus{}2)\plus{}f(k)>f(k\minus{}1)$, which gives $ f(k)>a$. Also note that $ f(\frac{k\minus{}1}2),f(\frac{k\plus{}1}2),f(k)$ do not form sides of triangle, hence $ f(k)\plus{}f(\frac{k\minus{}1}2)\le f(\frac{k\plus{}1}2)$ or $ f(\frac{k\minus{}1}2)\plus{}f(\frac{k\plus{}1}2)\le f(k)$. The first one implies $ f(k)\le a$, which is wrong. So the latter inequality holds, which gives $ f(k)\ge ak$. But $ f(k)|f(1)\plus{}f(k\minus{}1)\equal{}ak$, so $ f(k)\le ak$. Consequently, $ f(k)\equal{}ak$. This completes our proof.

Answer: $ f(n)\equal{}an$ for all $ n$, where $ a$ is a constant natural number.[/hide]

Remark: The first condition can be replaced by $ f(n\plus{}m)\le f(n)\plus{}f(m)$.

Perfect! ... and certainly we should replace the first condition to $ f(n\plus{}m)\leq f(n)\plus{}f(m)$ in this case. In the original problem I posed, the second condition was more open: integers $ f(n),f(m)$ and $ f(k)$ form the sides of a triangle if (NOT iff) $ n,m,k$ do so. I found it a bit hard and so I turned it to "if and only if", but still I believe the original problem would have some implications.


A 6-letter car plaque is to be made using the letters $A,...,Z$ such that the letters are in alphabetical order.  How many car plaques are possible?

Masoud Zargar

[hide]I get $\binom{26}{6}$. Choose any 6 letters and there exists a unique alphabetical arrangement.[/hide]

I assume here that you cannot have license plate AABCDE, since A does not come after A. For your question, should we count AABCDE and all other plates with duplicate letters?

[quote="boxedexe"]No, repetition of letters is not permitted.

EDIT: Also solve the problem when repetition is permitted. :)[/quote]
[hide]The answer is $\boxed{{31 \choose 6}}$.
This is the same as picking 6 |'s out of 25 _'s and 6 |'s, since we can represent every combination of letters that way. Each | represents a letter you pick, and each _ means you are going to the next letter.
For example, AAABBC would be
| | | _ | | _ | _ _ _ _ _ _ _ ...
Since each combination of letters has only one possible alphabetical arrangement, our answer is the number of ways to order $6$ |'s in $25+6=31$ objects, so our answer is ${31 \choose 6}$.[/hide]

[quote="boxedexe"]Now, what if I asked you to find the number of ways where repitition is allowed and the letters do not necessarily have to be in order.

Masoud Zargar[/quote]
[hide]This means that there are no restrictions on the letters, so the total is simply $\boxed{26^{6}}$[/hide]


ABC is a triangle and D is a point on the side BC, such that $BD\cdot AB=DC\cdot AC$. M is the midpoint of AD and N is the midpoint of BC. It is also known that <MBC+<MCA=<MBA+<MCB.
Prove that the circumcircle centre of the triangle BMC lies on MN.

Hi my dears, this is not the return of Ashegh!

Let me add something to this problem,we should also assume that in triangle$ABC$,$AB\geq <AC$.

Because the problem is wrong if$AB=AC$(I will explain why!)

This problem is a little tricky! it wants us to prove that circumcenter of triangle $BMC$, lies on$MN$

But it is equal to prove that $\angle BMC=90$(why?)

Suppose $AQ$ is bisector of $\angle BAC$. we know that$\frac{AB}{AC}=\frac{BQ}{QC}$

But as the problem said,we know:$\frac{AB}{AC}=\frac{CD}{DB}$

Then: $\frac{BQ}{QC}=\frac{CD}{DB}$

Then: $\frac{BQ}{QC+QB}=\frac{CD}{DB+CD}$.but because$QC+QB=DB+CD$,we conclude that:

$BQ=CD$,it is easy to find that:$QN=ND$

in triangle $ADQ$, $M$and$N$are the midpoints of segments $AD$and$DQ$,respectively.

Then:$MN$is parallel to $AQ$.

Assume $T,S$,on $BC$,in which $MT,MS$ are parallel to$AB,AC$,respectively.it is clear that

$MN$should be the bisector of $\angle TMS$.

Lets name the angles to continue the solution easier.

$\angle ABM=c$,$\angle MBC=d$,$\angle ACM=a$,$\angle MCB=b$,$\angle TMN=\angle SMN$

,$\angle BMN=w$,$\angle CMN=w'$

ok,according to above,$\angle TMB=c$,$\angle SMC=a$

$w-c=w'-a$then:w-w'=c-a

according to assumption of problem,$c+b=a+d$

then:$w-w'=d-b$, and:$w+b=d+w'$

we know that:$w+w'+d+b=180$,then:$w+b=r+w'=90$

the conclusion is that:$sinw=cosb$ and $sinw'=cosd$

$MN$, is the median of triangle $MBC$.

We know that:$\frac{sinw}{sinw'}=\frac{sind}{sinb}$ then:$\frac{sinw}{sinw'}=\frac{cosb}{cosd}$

Then:$\frac{sind}{sinb}=\frac{cosb}{cosd}$

$2sind.cosd=2sinb.cosb$and :$sin2b=sin2d$

according to it we should have:

$2d+2b=180$, or $2d=2b$

if  $2d+2b=180$,we find that:$d+b=90$ and it is proved that:$\angle BMC=90$

but if$2d=2b$,then:$d=b$and$c=a$ and the:$c+d=a+b$, and$\angle B=\angle C$,then it is proved that:

$AB=AC$,then in this case ,the problem is not necessarily true.then it is better to add the problem that:

$AB\geq <AC$

and every thing will be ok!

I will come back after $8$months!

Ashegh, I haven't read your solution (which seems interesting), but I just wanted to say that the problem [b]is[/b] true for AB = AC, because everything is symmetric.

In fact, the circumcenter of triangle BMC lies on MN not only if triangle BMC is right-angled at M, but also if BM = CM.

  Darij

[b]PS.[/b] This is not the return of Darij Grinberg ;)

oh,yes...

i forgot the main problem,which the circum center lies on $MN$!

but im sure that there exists a day which its name will be:

the returns of $Darij$ and $Ashegh$ :D


How do I do the degree sign?

I tried [code]$30^{degrees}$[/code] But didn't work. Also $ 30^{degree}$

[code]$30^\circ$[/code]

You willl see $ 30^\circ$

$ \sin{30^\circ}\equal{}0.5\equal{}\frac{1}{2}\equal{}\sin{\frac{\pi}{6}}$

Thanks for the help:)


Use 2, 3, 9, and 11 get the number 24 by using each number once.

[hide](11 - 2)sqrt(9) - 3[/hide]

OR

[hide](11 - 2)3 - sqrt(9)[/hide]

OR

[hide]((11 - 9)2)3![/hide]

OR

[hide]11 + 9 + 3! - 2
(any order of the first three, for example) [/hide]

In the traditional game of 24, you're restricted to addition, multiplication, subtraction, and division.

That's FINE...for THAT game.

You have to state the parameters.

You can stay with your "traditional game of 24" which was never mentioned, and the problem was thus open.

Unfortunately, the OP stated no parameters.  Personally, I don't like the use of unary operations in problems of this type because they essentially give you more numbers to work with than you actually have.

I prefer just addition, multiplication, subtraction, and division to do more with fewer operations.

However, are you wanting to include parentheses with your preference, or not?

Parentheses aren't an operation; they just represent the ability to apply the operations sequentially rather than in one expression, which I think is fine.

HEY  :mad:   :mad:   :mad:
I've proved that there are NO solutions by computer!
You should really specify which of the operations are allowed.

I typed...
"...are you wanting to include parentheses..."

I never stated that they were.  My question was akin to asking if someone wanted extras with their car.  You failed to pick up on that .
The use of parentheses with the four main operations is not automatic, either.
It is to be stated that they will or will not be used as part of the problem.

modeler,

thank you for confirming my high suspicions.

Well if you don't want parentheses, just type everything in prefix notation so that they're unnecessary.

For example: $ (3\plus{}5)(13\minus{}10)$ would be written as $ * \plus{} 3\ 5 \minus{} 13\ 10$ (I used an asterisk instead of \cdot because \cdot was too small)

In my opinion, nobody should disallow parentheses.

[quote="Hamster1800"]Well if you don't want parentheses, just type everything in prefix notation so that they're unnecessary.

For example: $ (3 \plus{} 5)(13 \minus{} 10)$ would be written as $ * \plus{} 3\ 5 \minus{} 13\ 10$ (I used an asterisk instead of \cdot because \cdot was too small)

In my opinion, nobody should disallow parentheses.[/quote]

Agree  :)
although i like to use postfix notation more as it is more natural.
My program uses postfix notation :lol:


superior algebra
superior algebra unsolved

Let $a,b$ element of a field $F$ where $b$ is not a square in $F$. Prove that
\[
\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}\quad\text{for some } m,n\in F
\]
if and only if $a^2-b$ is a square in $F$.

$(\sqrt m+\sqrt n)^2=m+n+\sqrt{4mn}$. Then $(m+n)^2-4mn=(m-n)^2$, which takes care of one side.

For the other side, $a+\sqrt{a^2-k^2}=\left(\sqrt{\frac{a+k}2}+\sqrt{\frac{a-k}2}\right)^2$.

I hope this helps as a detailed solution. I've attached a PDF file.


Are you sure about the problem?
If we take: $b, c \to - \infty$ and $a \to \infty$,
then: $b^3c, c^3b, bc \to \infty$.
Hence the sum also goes to infinity, so... :?

:roll:  The maximum is 3072 which is taken when a=b=c=4.


quadratics
calculus
derivative
function
algebra
quadratic formula
combinatorics unsolved

dts is half responsible for this problem.

Let $\displaystyle f(n) = \sum_{i=0}^{n} \sum_{j=i+1}^{n+1} \frac{{n \choose i} \cdot {n+1 \choose j}}{2^{i+j}}$.  Find a closed-form expression for $f(n)$.

(I hope there are no typos there!)

Ugh. This is brutal. All I've been able to do is to make the denominators uniform:

\begin{eqnarray}
f(n)&=&\sum_{i=0}^{n}\sum_{j=i+1}^{n+1}{{{n \choose i}\cdot{{n+1} \choose j}\over{2^{i+j}} \nonumber\\
&=&{{1}\over{2^{2n+1}}}\sum_{i=0}^{n}\sum_{j=i+1}^{n+1}\left(2^{n-i}{n \choose i}\right)\left(2^{n+1-j}{{n+1} \choose j}\right) \nonumber
}}\end{eqnarray}

Now, $2^{n-r}$ is the number of ways of choosing $0, 1, 2, \ldots, n-r$ objects from $n-r$ objects; ${n \choose r} = {n \choose {n-r}}$ is the number of ways of choosing $n-r$ objects from $n$ objects. Multiplying them would seem to yield the number of choosing $0, 1, 2, \ldots, n-r$ objects from $n$ objects, repeated a few times. That's just weird...

An observation...

(Similar to osiris.)  When you factor out the powers of 2 to get a common denominator, the coeffecients of the binomial expansions of [tex](x+2y)^{n+1}[/tex] and [tex](x+2y)^n[/tex] are showing up.  For n=3, the coefficients of [tex](x+2y)^4[/tex] are 1, 8, 24, 32, and 16; and the coefficients of [tex](x+2y)^3[/tex] are 1, 6, 12, and 8.

Assuming I am performing the computations correctly, the value of f(3) is exactly:

[tex]f(3)=\frac{(1+8+24+32)\times 8+(1+8+24)\times 12+(1+8)\times 6+(1)\times 1}{2^7}=\frac{971}{128}[/tex]
[tex]f(3)=7.5859375[/tex]

Why is this in "Intermediate Topics"?

I don't have a closed form, but I have an interesting restatement. A closed form solution is too much to ask here.

Let $\displaystyle f(n) = \sum_{i=0}^{n} \sum_{j=i+1}^{n+1} {n \choose i} \cdot {n+1 \choose j}\cdot t^{i+j}$.
Then $\displaystyle f(n) = \sum_{i=0}^{n} \sum_{j=i+1}^{n+1} {n \choose i} \cdot ({n \choose j}+{n \choose j-1})\cdot t^{i+j}$
$\displaystyle f(n) = \sum_{i=0}^{n} \sum_{j=i+1}^{n} {n \choose i} \cdot {n \choose j}\cdot t^{i+j} + t\sum_{i=0}^{n} \sum_{k=i}^{n} {n \choose i} \cdot {n \choose k}\cdot t^{i+k}$
$\displaystyle f(n) = \sum_{i=0}^{n} \sum_{j=i+1}^{n} {n \choose i} \cdot {n \choose j}\cdot t^{i+j} + t\sum_{k=0}^{n} \sum_{i=0}^{k} {n \choose i} \cdot {n \choose k}\cdot t^{i+k}$
$\displaystyle f(n) = (\frac{t+1}{2})\sum_{i=0}^{n} \sum_{j=0}^{n} {n \choose i} \cdot {n \choose j}\cdot t^{i+j} + (\frac{t-1}{2})(\sum_{i=0}^{n} \sum_{k=i}^{n} {n \choose i} \cdot {n \choose k}\cdot t^{i+k}-\sum_{i=0}^{n} \sum_{k=i+1}^{n} {n \choose i} \cdot {n \choose k}\cdot t^{i+k})$
$\displaystyle f(n) = (\frac{t+1}{2})\sum_{i=0}^{n} \sum_{j=0}^{n} {n \choose i} \cdot {n \choose j}\cdot t^{i+j} + \frac{t-1}{2}\sum_{i=0}^{n}{n \choose i}^2 \cdot t^{2i}$
$\displaystyle f(n) = (\frac{t+1}{2})(1+t)^n(1+t)^n + \frac{t-1}{2}\sum_{i=0}^{n}{n \choose i}^2 \cdot t^{2i}$

For numerical values, this is where to stop- a sum of n+1 values is reasonable.

Now we want to evaluate $\displaystyle \sum_{i=0}^{n}{n \choose i}^2 \cdot t^{2i}$
Let $\displaystyle g_0(x) = \sum_{n=0}^{\infty}x^n \sum_{i=0}^{n}{n \choose i}^2 \cdot t^{2i}$
and $\displaystyle g_k(x) = \sum_{n=0}^{\infty}\sum_{i=0}^{n}x^n \cdot {n \choose i}\cdot {n \choose i-k} \cdot t^{2i-k}$
We have $\displaystyle g_0(x)+2g_1(x)+2g_2(x)+... = \sum_{n=0}^{\infty} \sum_{i=0}^{n} \sum_{j=0}^{n} {n \choose i} {n \choose j}\cdot x^n \cdot t^i \cdot t^j = \sum_{n=0}^{\infty} x^n \cdot (1+t)^{2n} = \frac{1}{1-x(1+t)^2}$
This works if y is small enough, and we have absolute convergence. This implies that for fixed x within the radius, $g_k(x)$ tends to 0 as $k$ tends to $\infty$.

Also, since ${n \choose i}\cdot {n \choose i-k} = {n-1 \choose i}\cdot {n-1 \choose i-k} + {n-1 \choose i-1}\cdot {n-1 \choose i-k} + {n-1 \choose i}\cdot {n-1 \choose i-k-1} +{n-1 \choose i-1}\cdot {n-1 \choose i-k-1}$ for $n > 0$,
$g_0(x) = 1+x((1+t^2)g_0(x)+2tg_1(x))$ and
$g_k(x) = x(tg_{k-1}(x)+(1+t^2)g_k(x)+tg_{k+1}(x))$ for all $k \ge 1$.

By the general theory of constant-coefficient linear recursions,
 $g_k(x)=a \cdot r^k + b \cdot (\frac{1}{r})^k$ where $r$ and $\frac{1}{r}$ are the two roots of $txr^2+(x+t^2x-1)r+tx=0$. Let $r$ be the root which is less than 1, so $\frac{1}{r} > 1$. Since $g_k(x)$ tends to 0, $b=0$.
Then $g_1(x)=rg_0(x)$, so $g_0(x) = 1+x((1+t^2)g_0(x)+2trg_0(x))$
and $\displaystyle g_0(x) = \frac{1}{1-x-t^2x-2rtx}$.
By the quadratic formula $\displaystyle 2rtx = 1-x-t^2x-\sqrt{(1-x-t^2x)^2-4t^2x^2}$, so
$\displaystyle g_0(x) = \frac{1}{\sqrt{(1-x-t^2x)^2-4t^2x^2}}= \frac{1}{\sqrt{(1-x(1-t)^2)(1-x(1+t)^2)}}$.

The desired sum $\displaystyle \sum_{i=0}^{n}{n \choose i}^2 \cdot t^{2i}$ is  the $n$th derivative of $g$ divided by $n!$, evaluated at $x=0$. The sum is a better form anyway- I suspect there is no closed form, except in the special cases $t=1$, $t=-1$ (The given case was $t=\frac{1}{2}$).

Does anyone mind if I move this to Advanced?

[quote="jmerry"]I have an interesting restatement.

$\displaystyle g_0(x) = \frac{1}{\sqrt{(1-x-t^2x)^2-4t^2x^2}}= \frac{1}{\sqrt{(1-x(1-t)^2)(1-x(1+t)^2)}}$.

The desired sum $\displaystyle \sum_{i=0}^{n}{n \choose i}^2 \cdot t^{2i}$ is  the $n$th derivative of $g$ divided by $n!$, evaluated at $x=0$. The sum is a better form anyway- I suspect there is no closed form, except in the special cases $t=1$, $t=-1$ (The given case was $t=\frac{1}{2}$).[/quote]

Very nice work, jmerry!  Although I haven't found the closed form, I'm more inclined to think there is a closed form, thanks to your work.

I'll make a couple more observations, which you undoubtedly noticed while deriving your summation formula for the nth derivative of g(0).  [Others might? find these observations helpful.]

Let $d(x)=\sqrt{1-\frac{x}{4}}$.  Let $e(x)=\sqrt{1-\frac{9x}{4}}$.  These are the factors in the denominator of your g_0 function.

Each of these functions has derivatives which are in closed form:

$\displaystyle d(x)|_{x=0}=1$
$\displaystyle d'(x)|_{x=0}=\frac{-1}{8}=-0.125$
$\displaystyle d''(x)|_{x=0}=\frac{-1}{8}\cdot\frac{1}{8}=-0.015625$
$\displaystyle d'''(x)|_{x=0}=\frac{-1}{8}\cdot\frac{1}{8}\cdot\frac{3}{8}=-0.005859375$
[tex]\displaystyle d''''(x)|_{x=0}=\frac{-1}{8}\cdot\frac{1}{8}\cdot\frac{3}{8}\cdot\frac{5}{8}=-0.003662109[/tex]

[tex]\displaystyle d^{(n)}(x)|_{x=0}=\frac{-(2n-3)!}{2^{n-2}(n-2)!\times 8^n}~~~,~n\ge 2[/tex]

$\displaystyle e(x)|_{x=0}=1$
$\displaystyle e'(x)|_{x=0}=\frac{-1\times 9}{8}=-1.125$
$\displaystyle e''(x)|_{x=0}=\frac{-1\times 9}{8}\cdot\frac{1\times 9}{8}=-1.265625$
$\displaystyle e'''(x)|_{x=0}=\frac{-1\times 9}{8}\cdot\frac{1\times 9}{8}\cdot\frac{3\times 9}{8}=-4.271484375$
$\displaystyle e''''(x)|_{x=0}=\frac{-1\times 9}{8}\cdot\frac{1\times 9}{8}\cdot\frac{3\times 9}{8}\cdot\frac{5\times 9}{8}=-24.027099609375$

[tex]\displaystyle e^{(n)}(x)|_{x=0}=\frac{-(2n-3)!\times 9^n}{2^{n-2}(n-2)!\times 8^n}~~~,~n\ge 2[/tex]

The Leibniz Identity gives us a general method (involving a summation) for finding the derivatives of the product [tex]h(x)=d(x)\times e(x)[/tex].

$\displaystyle h(x)|_{x=0}=d(0) e(0)=1$
$\displaystyle h'(x)|_{x=0}=d(0) e'(0) + d'(0) e(0)=-1.25$
$\displaystyle h''(x)|_{x=0}=d(0) e''(0) + 2d'(0) e'(0) +d''(0) e(0)=-1$
$\displaystyle h'''(x)|_{x=0}=d(0) e'''(0) + 3d'(0) e''(0)+ 3d''(0) e'(0) +d'''(0) e(0)=-3.75$
$\displaystyle h''''(x)|_{x=0}=-21.75$

$\displaystyle h^{(n)}(x)|_{x=0}=\sum_{i=0}^n{{n\choose i}\big(d^{(i)}(0)\big)\big(e^{(n-i)}(0)}\big)$

Can we turn this into closed form?  Better yet, can we turn the derivatives of this function's reciprocal into closed form?


function trigonometry integration calculus derivative vector algebra

Can you show me an example?
A function $ f \in C^{\infty}(\mathbb R)$,such that $ f(x) \equal{} 0,x \le 0$ and $ f(x) \equal{} 1,x \ge 1$
Thank you.

I think this works?

$ f(x)\equal{}\begin{cases}0&\text{if } x\leq 0\\
\sin{\left(\frac{x\pi}{2}\right)} & \text{if }0<x<1 \\
1&\text{if } x\geq1\end{cases}$

That's not $ C^{\infty}.$ It's not even $ C^1,$ since it's not differentiable at $ 0.$

No analytic function can satisfy these conditions. By analytic continuation, if it is zero on an interval, then it is zero everywhere.

Try this. Let $ \psi(x)\equal{}\begin{cases}e^{\minus{}\frac1x}&x>0\\0&x\le 0\end{cases}.$

Let $ C\equal{}\int_0^1\psi(x)\psi(1\minus{}x)\,dx$

Then $ f(x)\equal{}\int_{\minus{}\infty}^x\frac1C\,\psi(t)\psi(1\minus{}t)\,dt.$

$ f$ will have the required properties.

Thank you,Kent and AdrianP.

And does the more general theorem still true?
Let $ A$ and $ B$ be two closed subset with empty insection of $ \mathbb{R}$ and $ f_1$ and $ f_2$ be two $ C^{\infty}$ function on $ A$ and $ B$ respectively.
Then there exist a function $ f\in C^{\infty}(\mathbb{R})$ such that $ f|_A \equal{} f_1$ and  $ f|_B \equal{} f_2$

Or we can generalize it to any function of $ C^r$ where $ r$ can be $ 0,1,\cdots,\infty$,when $ r \equal{} 0$ it is the Tietze's theorem.

A question about the meaning of that. What exactly does "$ C^{\infty}$ on the closed set $ F$" mean? My inclination is to say that means that the function can be extended to a $ C^{\infty}$ function on some open set containing $ F.$ Is that how it should be interpreted?

according to Evans' [i]Partial Differential Equations[/i], pg. 618, if $ \Omega\subset\mathbb{R}^d$ is open, then we have the definitions

\[ C^k(\overline{\Omega}) \equal{} \{u\in C^k(U)|D^\alpha u \text{ unif. continuous on bdd subsets of }\Omega\text{ for }|\alpha|\leq k\}
\]

\[ C^\infty(\overline{\Omega}) \equal{} \bigcap_{k\geq 0} C^k(\overline{\Omega})
\]

Yes,because the derivate is limit.So $ C^{1}$ means the limit exist on $ F$.

That's fine- for [b]open[/b] sets. It's more complicated if we're looking at a boundary point. Does the directional derivative exist if following that vector one way immediately takes you outside the domain?
That's the point Kent was getting at- it's not obvious how to handle points on the boundary.


Evaluate the product: \product^{2^1999}_{k=0} (4*(sin(k*pi/2^2000))^2-3)

Answer : -3

Hint : Let x_0, x_1, \dots, x_{n-1} be the nth roots of unity and
       let P = \prod_{k=0}^{n-1} (x_k^2+x_k+1).
       Then P=0 if n=0 mod 3 and P=3 otherwise.

Hint to the hint : use resultants (see http://www.mathlinks.ro/viewtopic.php?t=3274&highlight=) to deduce that

\prod_{k=0}^{n-1} (x_k^2+x_k+1)
= Resultant(X^n-1,X^2+X+1)
= (z_1^n-1) (z_2^n-1),

where z_1 and z_2 are the roots of X^2+X+1.

Sorry, in the previous message, one should read

"Answer : 3".

Actually, the important thing is the hint on the use of resultants, which helps proving that

the product \product^{n}_{k=0} (4*(sin(k*pi/2*n))^2-3)  equals

  0, if n=0 mod 3
 -3, if n=1 mod 3
  3, if n=2 mod 3.


Three circles, radii $ 1,2,3$, are externally tangent to one another.
(a) Another circle is externally tangent to all three circle. Find its radius.
(b) Another circle is internally tangent to all three circle. Find its radius.

[quote="not_trig"]See [url]http://mathworld.wolfram.com/SoddyCircles.html[/url]. (Maybe I should memorize the formulae there)[/quote]

How to prove it?  :D

I think you use inversion to prove it, but I don't know.


calculus
integration
function
geometry
calculus computations

A cubic funtion $ y \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d\ (a\neq 0)$ intersects with the line $ y \equal{} px \plus{} q$ at $ x \equal{} \alpha ,\ \beta ,\ \gamma\ (\alpha < \beta < \gamma).$ Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta ,\ \gamma$.

The desired area is $ \left|\int_{\alpha} ^{\gamma } \{ax^3 \plus{} bx^2 \plus{} cx \plus{} d \minus{} (px \plus{} q)\}dx\right|\ dx$

$ \equal{} \left|\int_{\alpha }^\gamma a(x \minus{} \alpha )(x \minus{} \beta )(x \minus{} \gamma)\ dx\right|$

$ \equal{} \frac {|a|}{12}(2\beta \minus{} \gamma \minus{} \alpha )(\gamma \minus{} \alpha )^3$.

I think you should define the "area bounded by the graphs."  Does orientation matter?  If it does, your answer is "almost" right; if not, the answer is \[\frac{|a|}{12}\left((\beta-\alpha)^3(2\gamma-\alpha-\beta)+(\gamma-\beta)^3(\beta+\gamma-2\alpha)\right)\,.\]


function
inequalities
calculus
calculus computations

Find the range of $ a>0$ such that $ ax^3\minus{}3x^2\plus{}1\geq 0$ for all positive real numbers.

Let $ f(x) \equal{} ax^3\minus{}3x^2\plus{}1$.
$ f'(x) \equal{} 3ax^2\minus{}6x \equal{} 3x(ax\minus{}2)$
So one of the extremum is when $ x\equal{}0$, $ f(x) \equal{} 1$, independent to $ a$.
The other extremum is when $ x \equal{} \frac{2}{a}$, as $ a>0$, $ \frac{2}{a} >0$, so $ (0,1)$ is a local maximum while local minimum occur when $ x \equal{} \frac{2}{a}$.
So to achieve $ f(x) \geq 0 \forall x>0$, we just need to have $ f(\frac{2}{a}) \geq 0$,
i.e. $ 4 \geq a^2$ thus the range of $ a$ is $ 2\geq a >0$.

You almost have solved the problem. :(

Try again! :)

Oops really careless,

from $ f(\frac{2}{a}) \geq 0$ we should have $ 4 \leq a^2$ instead of $ 4 \geq a^2$
Thus the answer should be $ a \geq 2$ instead.

Thx kunny~~


Let $ n$ be an integer greater than $ 2$. Show that $ 2^{2^n \minus{} 1} \minus{} 2^n \minus{} 1$ is not prime.


calculus
integration
function
parameterization
derivative
trigonometry
real analysis

Is there some general methodology for computing higher order approximations of integrals of highly oscillatory functions of the form
\[ I(x) \equal{} \int_a^b e^{ix\psi (t)}dt
\]
where $ x$ is a large parameter, $ \psi$ is real analytic and has one stationary point $ \tau \in [a,b]$ with $ \psi'(\tau) \equal{} 0$, $ \psi''(\tau )\neq 0$, $ \psi (\tau )\neq 0$?  The method of [url=http://en.wikipedia.org/wiki/Stationary_phase_approximation]stationary phase[/url] works ok as a leading order approximation, but it seems it doesn't generally work for higher order approximations as the non-stationary part can yield a non-vanishing contribution.

Is there a fairly general method that's perhaps a little more specialized to this case than, say, steepest descent?

Consider a more general integral $ I(x)\equal{}\int_a^b fe^{ix\psi}$ with smooth $ f$. Assume that $ \psi$ is smooth, has only one critical point $ c$ on $ [a,b]$ and some (usually second) derivative doesn't vanish there. There are two big contributions to your integral as written: one from the critical point and one from the endpoints. Note that if $ f$ vanishes up to a high order at $ a,b,c$, then you can integrate by parts a few times introducing the factor $ \psi'$ to the denominator and show that $ I(x)$ decays faster than a big power of $ x$. Thus, what really matters is the Taylor expansion of $ f$ near $ a,b,c$. By making a smooth local change of variable near each of those points, we can reduce the investigation to that of the integral $ I_{m,k}(x)\equal{}\int_0^\infty f(t)e^{ixt^k}$ where $ f(t)$ is smooth, vanishes outside some compact interval, and equals $ t^m$ near the origin. Now compare $ I_{k,m}(x)$ to $ 2^{\frac {m\plus{}1}k}I_{k,m}(2x)$. You'll see that the difference is less than any power of $ x$. Hence, $ I_{k,m}(x)\equal{}A_{k,m}x^{\minus{}\frac{m\plus{}1}k}\plus{}O(x^{\minus{}d})$ with arbitrarily large $ d>0$. To find $ A_{k,m}$, consider $ f(t)\equal{}t^me^{\minus{}t^k}$ (the leading term in the asymptotics should be the same), make the usual change of variable and arrive at an expression involving the $ \Gamma$-function.

So, the stationary phase formula is almost as good and straightforward as the Laplace one in the sense that the derivation of the higher order terms is fairly much the same. To see if you can handle it properly, try to find the first few terms in $ \int_{0}^{\pi}e^{i x\cos t}$ (both endpoints will contribute).


inequalities
number theory proposed
number theory

Find the pairs of integers $(m,n)\ (m\geq 3,\ n\geq 4)$ for which $\frac{1}{x+y}\leq \frac{1}{mx}+\frac{1}{ny}$ holds for arbiterary positive real numbers $x,y.$

[quote="jensen"]are you sure?
because i get that :$\frac{1}{m}+\frac{1}{n} \geq \frac{2k}{(k+1)^2}$.and it is enough that $\frac{1}{m}+\frac{1}{n} \geq \frac{1}{2}$ and from that we get these answers:$(4,4),(3,4),(3,5)$.[/quote]
Yes I'm sure.And the result I get is nessasary and suffient.
And I think continue my proof,can get your result as well.
More,I think there is no use to have a limit that $m\geq 3,n\geq 4$
thus we can get more pairs of integers satisfy the condition.
P.S. I don't understand your solution,could you explain it clearly?

[quote="Hawk Tiger"]$(x+y)(\frac{1}{mx}+\frac{1}{ny})\geq (\frac{1}{\sqrt{m}}+\frac{1}{\sqrt{n}})^2$
so we should find the pair of $(m,n)$such that ${\frac{1}{\sqrt{m}}+\frac{1}{\sqrt{n}}}\geq 1$[/quote]

Hawk Tiger, your solution is also correct.


[quote="red_dog"]The equation can be written $ ([x]-\{x\})\left(1-\frac{2007}{[x]\{x\}}\right)=0$
If $ [x]-\{x\}=0$ then $ x=0$, but $ x\neq 0$
Then $ 2007=[x]\{x\}$ and we can take $ x=2007n+\frac{1}{n},n\in\mathbf{N},n\geq 2$[/quote]
Another solution is $ x=n+\frac{2007}{n}$

[quote="Rust"][quote="Yuriy Solovyov"]Another solution is $ x=n+\frac{2007}{n}$[/quote]
But you forget, that  n>2007. It is general solution. If n=2007m (m>1), we get red_dog's solution.[/quote]
 :blush:

[quote="snowyowl"]Er... sorry, what is {x}?[/quote]
$ \{x\}=x-[x]$, where $ [x]$ is the biggest integer that $ [x]\leq x$.
$ 0\leq\{x\}<1$ always. :roll:


parameterization
advanced fields
advanced fields unsolved

Given five  parameters of a design, can you say (at least in a special case)
how many different designs can be made?
(A good question for thinking)

In future I will give some information about BIBDs,OBDs,...and maybe the answer of this question! :) (added in edit)

[quote="ZetaX"]What do BIBD and OBD mean¿[/quote]
[b]BIBD[/b]:Balanced Incomplete Block Design
[b]OBD[/b]: Ordered Block Design


geometry
function
conics
parabola
analytic geometry
absolute value

If i asked what the area of the square formed by the points (1,i), (1,-i),

(-1,i) and (-1,-i) on the complex plane, would the answer be 4 or 4i?  Ignoring the absolute value of an imaginary, why do we measure the distance on the "imaginary" axis in real numbers?  If we did measure it in imaginaries the distance from the origin would be a^2-b^2, instead of a^2+b^2, if the point is in the form of a+bi, which can be negative, and makes some sense considering they are imaginary numbers.

Also if i gave the equation Y=abs(x)^2, where abs is absolute value, and asked for the area of the set of all points that are part of a tangent to this function, what is the answer?  I would assume the answer is [hide] 1/2[/hide], but i don't know the answer or how to solve it if there is an answer.

Keep in mind that i made both of these problems up off the top of my head.

[hide]On the first, I'm not sure about measurements in imaginary planes/space, so I'll just leave that alone.

The second problem can be explained logically.  I don't know if you realized this, but , which may make the problem much less difficult than you intended or thought.

Next, try drawing some tangent lines to the graph.  Notice where they go.  No matter how hard you try, any given tangent line to the graph will not go above the parabola at any point.

Or, another way to do it is to draw points randomly on the coordinate plane and try to draw tangent lines from them to the parabola.  You will soon notice that it is impossible to draw a tangent from any point above the parabola.

We have concluded that no tangent line can be drawn from any point above the parabola, and a tangent can be drawn from any point below the parabola.  Thus, the area of the set of points that lie on a tangent to  is , or, simply put, all of the points under the parabola, which is infinite.[/hide]

I knew the area of the tangents would be the same as the area of y is less than x^2, but doesn't the absolute value only include the values where x is positive?  In finding the area, is there any way you can find a percentage of the area of the whole coordinate plane, even though it is infinite?

I really need to learn latex.


inequalities
logarithms
inequalities proposed

Let $ x;y;z>0$and $ xyz \equal{}1$
Prove that
$ \frac{x^3}{x^4\plus{}1}\plus{}\frac{y^3}{y^4\plus{}1}\plus{}\frac{z^3}{z^4\plus{}1}\le \frac{3}{2}$

It is a very odd type of inequality, I mean the upper restriction is very crucial. I failed to hit it with all type of known results. If anyone has solved it please post the solution. I'm still trying.

Let $ f(x,y,z)\equal{}\frac{x^3}{x^4\plus{}1}\plus{}\frac{y^3}{y^4\plus{}1}\plus{}\frac{z^3}{z^4\plus{}1}\minus{}\frac{3}{2}$

 Then $ f(x,y,z)\minus{}f(x,\sqrt{yz},\sqrt{yz})\leq 0$

 But

$ f(x,\sqrt{yz},\sqrt{yz})\equal{}f(\frac{1}{t^2},t,t)\equal{}\minus{}\frac{3t^{12}\minus{}4t^9\plus{}3t^8\minus{}2t^6\plus{}3t^4\minus{}4t^3\minus{}2t^3\plus{}3}{2(t^8\plus{}1)(t^4\plus{}1)}\leq 0$

 and inequality is prove.

Is this method Mixing Variable? I can't understand the solution.

Can anyone solve without using it?

[quote="tranvanluan"]Let $ x;y;z > 0$and $ xyz \equal{} 1$
Prove that
$ \frac {x^3}{x^4 \plus{} 1} \plus{} \frac {y^3}{y^4 \plus{} 1} \plus{} \frac {z^3}{z^4 \plus{} 1}\le \frac {3}{2}$[/quote]
$ \sum_{cyc}\frac {x^3}{x^4 \plus{} 1}\leq\frac {3}{2}\Leftrightarrow\sum_{cyc}\left(\frac {1}{2} \minus{} \frac {x^3}{x^4 \plus{} 1} \plus{} \frac {\ln x}{2}\right)\geq0.$
Let $ f(x) \equal{} \frac {1}{2} \minus{} \frac {x^3}{x^4 \plus{} 1} \plus{} \frac {\ln x}{2}.$ then
$ f'(x) \equal{} \frac {(x \minus{} 1)(x^7 \plus{} 3x^6 \plus{} 3x^5 \plus{} 3x^4 \plus{} 5x^3 \minus{} x^2 \minus{} x \minus{} 1)}{2x(x^4 \plus{} 1)^2}.$
Hence, $ x_{min} \equal{} 1$ and $ x_{max} \equal{} x_1,$ where $ 0 < x_1 < 1.$
Not exist more root of $ x^7 \plus{} 3x^6 \plus{} 3x^5 \plus{} 3x^4 \plus{} 5x^3 \minus{} x^2 \minus{} x \minus{} 1$ on $ (0,1)$ since,
we see only one sign switching of the coefficients in this polynomial.
Thus, there is only one root of $ f$ on $ (0,x_1).$
Let $ x_0$ the root of $ f$ on $ (0,x_1).$
Easy to show that $ x_0 < 0.429.$
Hence, our inequality is proved if $ \min\{x,y,z\}\geq0.429.$
Let $ \min\{x,y,z\} < 0.429$ and $ g(x) \equal{} \frac {1}{2} \minus{} \frac {x^3}{x^4 \plus{} 1}.$
We obtain: $ g'(x) \equal{} \frac {x^2(x^4 \minus{} 3)}{(x^4 \plus{} 1)^2}.$
Hence, $ g(x)\geq g\left(\sqrt [4]3\right)$ for all $ x > 0$ and $ g\left(\min\{x,y,z\}\right) > g(0.429).$
Id est, $ \sum_{cyc}g(x)\geq g(0.429) \plus{} 2g\left(\sqrt [4]3\right) \equal{} 0.28...\geq0$ and our inequality is proved.

arqady I like the ln trick  :lol:
here is my solution:
setting : $ x\equal{}\frac{a}{b},y\equal{}...$
the inequality becomes : $ \sum_{cyc}{\frac{a^3b}{a^4\plus{}b^4}} \leq \frac{3}{2}$
we have:
$ 2LHS \leq  \sum_{cyc}{\frac{2a^3b}{a^2b^2\plus{}\sqrt{\frac{a^8\plus{}b^8}{2}}}} \leq \sum_{cyc}{\frac{a^3b}{ab\sqrt{\sqrt{\frac{a^8\plus{}b^8}{2}}}}} \equal{} \sum_{cyc}{ (\frac{2a^8}{a^8\plus{}b^8})^{1/4}} \leq \sum_{cyc}{\sqrt{\frac{2a^4}{a^4\plus{}b^4}}} \leq 3\equal{}2RHS$
the last inequality is true since it's Vasc's  :D ,and I saw somewhere on this forum , two nice proofs with cauchy and Jensen  :)

Nice :) .  Your solution has a bit celever :P

[quote="anas"]

......$ \sum_{cyc}{\sqrt {\frac {2a^4}{a^4 \plus{} b^4}}} \leq 3 \equal{} 2RHS$
the last inequality is true since it's Vasc's  :D ,and I saw somewhere on this forum , two nice proofs with cauchy and Jensen  :)[/quote]

I cannot find the topic you refer this inequality. :(

Can you please remember me the proof by cauchy?

Thank you very much.

$ \sum \sqrt{ \frac{a}{a+b}} \le \sqrt{ 2(ab+bc+ca)[\sum \frac{1}{(a+b)(b+c)} ] } =\sqrt{\frac{4(ab+bc+ca)(a+b+c)}{(a+b+c)(ab+bc+ca)-abc} \le \frac{3}{\sqrt{2}}}$

The inequality becomes $ \sum_{cyc}x^3(y^4 \plus{} 1)(y^4 \plus{} 1)\leq\frac {2}{3}\prod_{cyc}(x^4 \plus{} 1)$
Expanding, we get $ \sum_{cyc}(x^3y^4z^4 \plus{} x^3y^4 \plus{} x^3z^4 \plus{} x^3)\leq\frac {3}{2}(x^4y^4z^4 \plus{} \sum_{cyc}x^4y^4 \plus{} \sum_{cyc}x^4 \plus{} 1)$
$ \sum_{sym}x^4y^4z^3 \plus{} \sum_{sym}x^4y^3 \plus{} \sum_{sym}x^3\leq\frac {3}{2}\sum_{sym}x^4y^4z^4 \plus{} \frac {3}{2}\sum_{sym}x^4y^4 \plus{} \frac {3}{2}\sum_{sym}x^4 \plus{} \frac {3}{2}$
By multiplying or dividing a suitable power of $ xyz$, we get
$ \sum_{sym}x^3y^3z^2 \plus{} \sum_{sym}x^{\frac {13}{3}}y^{\frac {10}{3}}z^{\frac {1}{3}} \plus{} \sum_{sym}x^{\frac {14}{3}}y^{\frac {5}{3}}z^{\frac {5}{3}}$
$ \leq\frac {3}{2}\sum_{sym}x^{\frac {8}{3}}y^{\frac {8}{3}}z^{\frac {8}{3}} \plus{} \frac {3}{2}\sum_{sym}x^4y^4 \plus{} \frac {3}{2}\sum_{sym}x^{\frac {16}{3}}y^{\frac {4}{3}}z^{\frac {4}{3}} \plus{} \frac {3}{2}x^{\frac {8}{3}}y^{\frac {8}{3}}z^{\frac {8}{3}}$
which can be easily proved with Muirhead's Inequality.

Although it is a low-level stuff, and pretty annoying, I like to use this for fractional inequality, with $ xyz \equal{} 1$

[color=red][Sorry, made some mistake here, but couldn't delete post][/color]

[quote="KenHungKK"]$ \sum_{sym}x^3y^3z^2 \plus{} \sum_{sym}x^{\frac {13}{3}}y^{\frac {10}{3}}z^{\frac {1}{3}} \plus{} \sum_{sym}x^{\frac {14}{3}}y^{\frac {5}{3}}z^{\frac {5}{3}}\leq\frac {3}{2}\sum_{sym}x^{\frac {8}{3}}y^{\frac {8}{3}}z^{\frac {8}{3}} \plus{} \frac {3}{2}\sum_{sym}x^4y^4 \plus{} \frac {3}{2}\sum_{sym}x^{\frac {16}{3}}y^{\frac {4}{3}}z^{\frac {4}{3}} \plus{} \frac {3}{2}x^{\frac {8}{3}}y^{\frac {8}{3}}z^{\frac {8}{3}}$[/quote]
I agree with arqady.... cause I could not prove it using $ M\ddot{u}irheads$ too. Will you detail this part of your proof please. :maybe:


How do I get the solutions for the problems in ACoPS?

You post problems on the forum and wait for people to respond.

You speak of the elusive ACoPS solutions manual.  :yoda:  :starwars:

 :rotfl:

And...how would that help?

Well I do know Paul Zeitz in person, I guess I could email him...

[quote="Fraxia"]Isn't there a PDF version of the solutions document? If you knew anyone that had that version, they could email it to you.[/quote]

That would basically have all the problems and then nobody would need to buy ACoPS

Well, ACoPS is more of a textbook than purely a problem book. So even if all the problems in it (with solutions) were available in a PDF document, it'd still be beneficial for a lot of kids like me to still get the actual book  :lol: Also, most of the problems in the book are taken from old contests, so they're already easily accessible to everyone.

EDIT: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=solving+problem+craft+art&t=8006&sid=2e2e550a2040d396b3ede27923de2945[/url]

This is where I read someone mention that PDF document - if that was the actual solutions manual.


Prove that the following equation had no solution in integers
$a^2-d^2-2c=c^2-b^2+1=c^2+a^2+2(b+c+bc-ad)+1$
edit- ok joml88
edit again- typo

[quote="Scrambled"]Prove that the following equation had no solution in rational numbers
a^2-d^2-2c=c^2-b^2+1=c^2+a^2+2(bc-ad)+1[/quote]

$a^2-d^2-2c=c^2-b^2+1=c^2+a^2+2(bc-ad)+1$

Scrambled, $\LaTeX$ is really, really easy to learn.  For this problem, all you have to do is place dollar signs around what you wrote!!!

What you wrote:

a^2-d^2-2c=c^2-b^2+1=c^2+a^2+2(bc-ad)+1

Place dollar signs around it:

$\$ a^2-d^2-2c=c^2-b^2+1=c^2+a^2+2(bc-ad)+1 $\$

and you get:

$a^2-d^2-2c=c^2-b^2+1=c^2+a^2+2(bc-ad)+1$

Is it so hard to type two little dollar signs?

Why won't you at least try to learn $\LaTeX$!? :mad:

anyone?

[hide="big hint"]this is a geometry problem, or at least, it can be nicely solved that way[/hide]


calculus
integration
geometry
calculus computations

Find the area of the region surrounded by the two curves $ y=\sqrt{x},\ \sqrt{x}+\sqrt{y}=1$ and the $ x$ axis.

The answer isn't going to be nice, but its simple from:

$ \int^{a}_{0}\sqrt{x}dx+\int_{a}^{1}(\sqrt{x}-1)^{2}dx, a = \frac{7-3\sqrt{5}}{2}$

Your answer is incorrect. What's the value of $ a$?

It said in my original post, after the integrals.

Sorry, I could find it, that is correct.

Sorry, just thought it would look really messy without that substitution. Thank you.


geometry
geometric transformation
rotation
Support

There are less than three weeks left until the Super Bowl! Who does everybody think is going to win?

I'm rooting for the Pats.

falcons vs patriots

big ben's time has come to an end

then falcons win in a blowout 35-9!!!

My expectation is that Pittsburg tramples Philly for the championship.  Being a Massachussets resident, I feel a bit sad that the Patriot's dominance is coming to an end, but one championship a year should be good enough for most people.  Atlanta?  The same Atlanta Falcons that got lost to Detroit and New Orleans and blown out by the Chiefs and Buccaneers?  I mean, Vick is pretty good, but he's not going to take them past the Eagles.

[quote="JBL"]The same Atlanta Falcons that got lost to Detroit [/quote]

Hey, don't insult the Lions. We're getting better, I tell you. We were so close to playoffs this year... I predict Detroit in the Superbowl in about ... 2008.

The closest Detroit will ever get to a superbowl is the one that is coming next year...  Other than that, Lions have no chance.

only reason falcons lost to the lions was cuz of mooch, cuz he knows the offensive style too well...
saints thing was cuz vick wasn't even playing....
the chiefs thing was embarassing but, none of the teams have such strong back rotations...
the bucs thing was cuz the picket just collasped all the time and Vick had some bad judgements...

Although I think the factor will be the defense and getting Warrick Dunn involved in the rushing and receiving, not just Vick.

Anyways... I was rooting for the Chargers, but now they are gone. I always like to pick a poor team from one year who I think will make the changes to improve greatly, so I'll pick the Clevland Browns for next year

[quote="blahblahblah"]Falcons versus Pats.

The Steelers didn't impress me at all on Sunday, and I think that the Steelers will need to throw for 250+ yards to beat the Pats, because I think that the Pats should do well against the big backs of the Steelers.

In the NFC, Atlanta has a way better defense than Minnesota did, and I just am not impressed by the Philly offense without T.O.  The reverse is true for Philly, but I'll take Michael Vick over Donovan Mcnabb in any playoff game, and that's why I like Atlanta.[/quote]

I agree with the first statement. Unless Big Ben can throw the ball well on Sunday, the Steelers have a very little chance of victory.

About the second statement, it is true Atlanta has a better defense than the Vikings, but taking Vick over McNabb in a playoff game? Vick didn't even throw for 100 yards against the Rams! I think the Eagles and Patriots are in similar situations. If the Eagles can stop the Falcons running game and contain Vick, they will win. Make Vick throw the ball, because the Falcons do not have good receivers and Vick is far from a good passer.

[quote="yif man12"][quote="blahblahblah"]Falcons versus Pats.

The Steelers didn't impress me at all on Sunday, and I think that the Steelers will need to throw for 250+ yards to beat the Pats, because I think that the Pats should do well against the big backs of the Steelers.

In the NFC, Atlanta has a way better defense than Minnesota did, and I just am not impressed by the Philly offense without T.O.  The reverse is true for Philly, but I'll take Michael Vick over Donovan Mcnabb in any playoff game, and that's why I like Atlanta.[/quote]

I agree with the first statement. Unless Big Ben can throw the ball well on Sunday, the Steelers have a very little chance of victory.

About the second statement, it is true Atlanta has a better defense than the Vikings, but taking Vick over McNabb in a playoff game? Vick didn't even throw for 100 yards against the Rams! I think the Eagles and Patriots are in similar situations. If the Eagles can stop the Falcons running game and contain Vick, they will win. Make Vick throw the ball, [b]because the Falcons do not have good receivers and Vick is far from a good passer[/b].[/quote]

Ever hear of Alge Crumpler? Plus like I said, running backs are one of the most underrated receivers on teams, so Dunn is a great receiver from what i've seen. Vick has the arm strengh, just not enough accuracy, the Falcons just need to move around their rotation of speed receivers and he'll do fine.

Also, rossum is turning into a big play maker as well.

Vick not needing to throw the ball is not the same as Vick having a bad passing game.  What we're finding is that Vick is rewriting the book on how quarterback stats need to be viewed.  Besides, Atlanta ran for about 3000 yards (I'm only off by one zero) and a whopping 8.2 yards per carry against the Rams - why would you want to throw in that situation?

As for McNabb, in the seasons before this one (aka, Pre T.O), he never had a passer rating of higher than 86, never completed more than 58.5% of his passes, and never threw for even 7 yards per attempt touchdowns.  So he's not exactly John Elway out there, and all those people jumping on the Donovan McNabb bandwagon need to remember that he's not a great passer, not a great scrambling quarterback, and his playoff performances have been historically lackluster (remember 2003? and 2002? and 2001?).

Vick is a difference maker.  Donovan McNabb is a good quarterback without T.O.  T.O. makes him a great quarterback.  I mean, that's what I'm basing my pick on.  I think these teams are a wash, except that Philly has a better secondary.  But I give Atlanta the edge at QB, and that's why I'm going with Atlanta.

Alge Crumpler is a tight end.  All of the Falcons receivers are average at best, otherwise, but surprise surprise, so are the Eagles' without T.O.

Crumpler isn't much of a blocking TE, so I consider him a receiver and most TE's ought to be as well because he is a main target for the QB to throw to.

I'm not a football fan so hopefully this won't seem like a stupid/odd/"don't you have a life?" question...Who's hosting the Super Bowl this year?

I'm making a distinction here because actually, Crumpler is a top-notch blocker, and obviously most of his routes are to the middle of the field.  Vick throws better there than he does to the sidelines.

Unfortunately, Vick does have tunnel vision when it comes to receivers; his reads go something like Price (who is just about never open), Crumpler, scramble.

NOOOOOOOOO!!!!!!!!!!!!! The eagles lost!!!

Not that's a surprise or nothing, but I'm in Philly, so my heart goes out to my hometeam.

Did anybody else vote that they thought that Patriots would win [i]after[/i] the Superbowl ended like I did...?

[quote="MithsApprentice"]Man...the eagles. *sniffs*

That touchdown with 2 minutes on the clock should've definitely been worth 11 points.

And was I the only one who noticed what horrible commercials we had this year?[/quote]

The commercials totally sucked this year. Big letdown.

I really don't get how the eagles weren't hurried at all when they were down 10 with less than 7 minutes to go. Makes you wonder if the result would've been different if they had hurried down the field.

Hehe, I knew the Patriots would win AND I knew the commercails would suck thanks to a list on http://www.howstuffworks.com.  :D

omg.... this year's Super Bowl was very disappointing. The game was so boring because it was a defensive struggle. It kind of blew open in the second half but it wasn't a high scoring game. I wanted to see more points on the scoreboard!!!

The commercials sucked as well. No memorable commericals like last year. Usually there is an abundance of funny commercials but they were lacking this year. It's probably due to last year's incident with Janet Jackson.... I think the producers of commercials were pressured to keep it appropriate.

On the plus side: The half-time show was free of incidents and it was pretty good. Although Sir Paul McCarntey isn't what he used to be... he still is a pleasure to listen to. All in all, this Super Bowl was definitely disappointing.

[quote="plokoon51"][quote="Jhyon8149"]omg.... this year's Super Bowl was very disappointing. The game was so boring because it was a defensive struggle. It kind of blew open in the second half but it wasn't a high scoring game. I wanted to see more points on the scoreboard!!! .[/quote]

as the saying goes, "offense wins games, defense wins championships"

...or something like that[/quote]

The people like offense, but the best way of winning is still with defense.

Say, that was one hell of a game T.O. had. I've gotta hand it to him.

how was the half time show? i only got to watch the end of the fourth quarter

well, paul mccartney was performing. he's a legend but times have changed

ooohhh.... anyone else?

(I've never even heard of that guy, but then again, i dont listen to that kind of music that much)

Paul McCartney was one of the Beatles, if that helps.


What units digit $ a$ will make the addition problem correct?

\begin{tabular}{cccc}
&2 & 3 & $ a$\\
&5&2&4\\
+&3&6&$ a$\\\hline
1&2&3&4
\end{tabular}

[quote="GameBot"]What units digit $ a$ will make the addition problem correct?

$ \begin{tabular}{cccc}
&2 & 3 & a\\
&5&2&4\\
\plus{}&3&6&a\\\hline
1&2&3&4
\end{tabular}$[/quote]

$ 1234\minus{}230\minus{}524\minus{}360\equal{}2a \iff 2a\equal{}120 \iff a\equal{}\boxed{60}$.
Is not there the mistake in a problem sentence? :maybe:

Using the same logic as Kouichi, we get $2a=10$, so $a=5$.


Let $ a,b,c$satisfy: $ \left\{ \begin{array}{l} 2 \ge a \ge b \ge c \\  a \plus{} b \le 3 \\  a \plus{} b \plus{} c \le 3 \\  \end{array} \right.$
Prove that: $ S\equal{}2^a\plus{}2^b\plus{}2^c \le 7$
 :o

[quote="ML2008"]Let $ a,b,c$satisfy: $ \left\{\begin{array}{l} 2 \ge a \ge b \ge c \\
a \plus{} b \le 3 \\
a \plus{} b \plus{} c \le 3 \\
\end{array} \right.$
Prove that: $ S \equal{} 2^a \plus{} 2^b \plus{} 2^c \le 7$
 :o[/quote]
From system we get $ r < 0;$ $ (2,1,r)\succ(a,b,c)$,and $ f(x) \equal{} 2^x$ and $ f"(x)\ge0$ so by Karamata $ 7 \equal{} f(2) \plus{} f(1) \plus{} f(0)\ge f(2) \plus{} f(1) \plus{} f(r)\ge f(a) \plus{} f(b) \plus{} f(c)$

Hm... Actually you conclude the existence of a real $ r < 0$ such that $ (2, 1, r)$ majorizes $ (a, b, c)$. Of course this is also sufficient.

Nice solution!


inequalities
Cauchy Inequality
inequalities unsolved

Let a,b,c are positive real numbers, and abc=1
proof
$ {1}\plus{}\frac{3}{a\plus{}b\plus{}c}\geq\frac{1}{ab\plus{}bc\plus{}ac}$
If you have got a solution, please let me know,thank you.

I'm not sure whether {1} is a mistake, but if not...

$ 1+\frac{3}{a+b+c}\geq\frac{1}{ab+bc+ac}$

$ %Error. "Rightleftarrow" is a bad command.
1+\frac{3}{a+b+c}\geq\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

$ %Error. "Rightleftarrow" is a bad command.
(a+b+c+3)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq a+b+c$

Since $ \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 1$ and $ (a+b+c+3) \geq a+b+c$, the last inequality is true.

This problem is a typical example for subsitution method mentioned in titu book.
Here is the full solution
Set $ a\equal{} \frac{1}{x}$ . $ b\equal{} \frac{1}{y}$ and $ c\equal{}\frac{1}{z}$. From this, we still reserve the condition $ xyz\equal{}1$.
It follows to prove that
$ 1\plus{} \frac{3}{xy\plus{}yz\plus{}zx} \ge \frac{6}{x\plus{}y\plus{}z}$
Obviously, we have
$ 1\plus{} \frac{3}{xy\plus{}yz\plus{}zx} \ge 1\plus{} \frac{9}{(x\plus{}y\plus{}z)^2} \ge \frac{6}{xy\plus{}yz\plus{}zx}$ ( the second inequality is an implication of Cauchy inequality that is quite familiar with the majority of Mathlinkers)
We are done! :)


Does anyone know if MathLinks is still running? The website appears to be defunct, and most of the links don't work.

It was down for a while yesterday, but it seems to work now.

Wouldn't MathLinks be an international program (for placement on a particular forum here)?

Sorry-- I don't even know what MathLinks is, besides being a competition. There appears to be no information on the website, and the links still don't work...

MathLinks is a huge olympiad-problem solving forum. The contest is just part of all the stuff they have.

[quote="mathnovice"]Sorry-- I don't even know what MathLinks is, besides being a competition. There appears to be no information on the website, and the links still don't work.[/quote]

I just landed there,

http://www.mathlinks.ro

today, and everything looked okay.

The ".ro" at the end of the address indicates that it is based in Romania.

[quote="mathnovice"]I cannot believe that something like that even exists.

I also cannot believe that there are so many people who are smart enough to make a forum that focuses just on olympiad level problems.[/quote]

The world is a big place!

And just in the US there are several hundred students who participate in the USAMO each year. Not to mention the hundreds of college students and alumni of Olympiads who continue to engage in recreational problem solving even after high school.

What's impressive is that I heard that the Romanian National Olympiad has approximately 500 participants each year. That's more than twice the number of USAMO participants out of a country with only a fraction of the population. (Although I have no idea what the score distribution is like out of those 500 participants).


1) A bag contains 10 red balls, 10 green balls and 10 white balls. Five balls are drawn at random from the bag without replacement. What is the probability that there are balls of all three different colours among the five?

2) A poker hand is being dealt and the first three cards are of the same type (3 Queens for example). When the remaining two more cards are dealt, what is the probability of his having a full house? (Three of the same type and two of a different type-three Queens and two Kings for example)

2.
[hide]You have 49 cards left in the deck 48 of which are not of the same type that you alredy have. You then have 48 cards left 3 of which are the same type of the fourth card. You then have (48/49)*(3/48) which is 3/49.[/hide]

2) [hide]There are 49 cards remaining. Of them, 48 can be chosen if we are going to get a full house (you can't choose. That's 48/49. Then there are three more cards out of the 48 remaining that are the same as the previously chosen card. So (3/48)(48/49)=3/49.[/hide]

I'm not too sure on this but here it goes.

1.  I tried to find out the number of draws that would give 1 of each color.  This lead me to two combinations 1-1-3 and 1-2-2.  Each of these would repeat 3 times with the three different colors.  So now we need to know how many combinations of 5 draws would give us these two patterns.  Then I divided that by the total number of combinations possible to give the probability.

Here it goes:

3[(10 nCr 3)(10 nCr 1) (10 nCr 1) + (10 nCr 2)(10 nCr 2)(10 nCr 1)] / (30 nCr 5)

This gives the probability of .6789 or 67.89%.  I don't know if this is right I'm trying to teach myself probability and I'm not sure if it is working.

[quote="Euler1001"]
3[(10 nCr 3)(10 nCr 1) (10 nCr 1) + (10 nCr 2)(10 nCr 2)(10 nCr 1)] / (30 nCr 5)

This gives the probability of .6789 or 67.89%.  I don't know if this is right I'm trying to teach myself probability and I'm not sure if it is working.[/quote]

That's what I got   :)


topology
advanced fields
advanced fields unsolved

Terminology:
Let $ \Gamma$ be a group acting on a metric space $ X$.

The action is proper if each $ x\in X$ has a neighborhood $ U$ such that $ \gamma U \cap U=\emptyset$ for all but finitely many $ \gamma$

And a problem:

If the action is proper then for any compact set $ K\subseteq X$ for all but finitely many $ \gamma$ we have $ \gamma K \cap K=\emptyset$

I should have written that action is by isometries


calculus
algebra
function
domain
integration
complex analysis

The Laurent series expansion is given by $ \frac{1}{ex}\minus{}\frac{1}{e}\minus{}\frac{x}{2e}\minus{}\frac{x^{2}}{6e}\plus{}\frac{x^{3}}{24e}\plus{}\cdots$.  Try seeing what you get now (I haven't done the calculations, I'm just giving you a more explicit way to do the integral).

$ use  cauchy  integral  formula  sol \equal{} 2 \pi {i} f(0) \equal{} 2\pi {i}/e$[/quote]

[quote="car202"]$ \oint_{\left|z\right| \equal{} 3/2} \frac {e^{\frac {1}{z \minus{} 1}}}{z} dz$

I thought I have to evaluate residues at both of the singularities. But the answer is $ 2\pi i$ instead of $ 2\pi i \plus{} \frac {2\pi i}{e}$.
FYI, if the domain was $ |z| \equal{} 1/2$, the answer was $ \frac {2\pi i}{e}$.

Can you explain this to me? Thank you.[/quote]

You do.
\[ \oint_{|z|\equal{}3/2} \frac{e^{\frac{1}{z\minus{}1}}}{z}\equal{}2\pi i(r1\plus{}r2)\]
The residue at zero is easy.  It's a simple pole so $ r1\equal{}e^{\minus{}1}\equal{}1/e$.
The residue at z=1 is a little tricky.  We can expand the integrand about that point by considering:
\[ 1/z\equal{}\sum_{n\equal{}0}^{\infty} (\minus{}1)^n(z\minus{}1)^n\]
\[ e^{\frac{1}{z\minus{}1}}\equal{}\sum_{n\equal{}0}^{\infty}\frac{1}{(z\minus{}1)^n n!}\]
Then via Cauchy's product:
\[ \frac{e^{\frac{1}{z\minus{}1}}}{z}\equal{}\sum_{n\equal{}0}^{\infty}\sum_{k\equal{}0}^n \frac{(\minus{}1)^k (z\minus{}1)^k}{(z\minus{}1)^{n\minus{}k}(n\minus{}k)!}\]
Now, we want the $ 1/(z\minus{}1)$ terms so that we wish $ k\minus{}(n\minus{}k)\equal{}\minus{}1$ or $ n\equal{}2k\plus{}1$.  Therefore, the residue at z=1 is:
\[ \sum_{k\equal{}0}^{\infty}\frac{(\minus{}1)^k}{(k\plus{}1)!}\equal{}1\minus{}1/e\]
Thus:
\[ \oint_{|z|\equal{}3/2} \frac{e^{\frac{1}{z\minus{}1}}}{z}\equal{}2\pi i(r1\plus{}r2)\equal{}2\pi i\]


[quote="Fierytycoon"]For each arrange ment of the numbers 1, 4, 7, 10, 13, 16, 19, 22, 25, and 28 around a circle, let N be the largest of the ten sums obtained by adding each number to the sum of its neighbors. Minimize N.[/quote]
48

28 16 4 22 19 7 13 25 10 1 (pretend it wraps around)

if you assume all the sums are equal (which is obviously wrong), you get 10*N=3*sum of all the digits. Solving for N gives 43.5 . Since N must be a multiple of 3 (all the numbers are congruent mod 3), the smallest N can be is 45. Since at most half the sums can be the same (if you had more than half the sums the same, you would have two consecutive equal sums, which would mean two numbers were equal), we must have half the sums 45 (the maximum possible) and half the sums 42 (if we had anything else, the average would not be 43.5)

That means the pattern will look like this:

x y z x-3 y+3 z-3 x

since we have to alternate adding and subtracting 3. however, we run into the problem of having an equal number (notice the double x), so that's not possible. Trying the next smallest N, 48, we see that it works.

nice solution

So how would so solve a problem like this if the numbers weren't congruent in a mod?

One possible idea is that you can add or subtract or multiply the numbers by anything without changing the relative ordering of the sums. So add 2 to each term, divide by 3, and solve the same problem with the easier to work with numbers 1 through 10. At then end just multiply by 3 and subtract 2.


"There came three Indians of my accquaintance to see me,being lately married;they brought their wives with them.The men's names were Ram,Gopl and Amit;the women's Jaya,Uma and Sujata:but I forgot the name of each man's wife.[b]They told me they had been at market to but sheep;each person bought as many sheep as they gave rupees(a currency like dollars) for one sheep;Ram bought 23 sheep more than Uma;and Gopal bought 11 more than Jaya;likewise,each man laid out [/b]$63$[b] rupees more than his wife[/b].I desire to know the name of each man's wife."

[quote="Gyan"][hide]That's easy, every bollywood fan knows, Amit(ab) is married to Jaya. Sujata sounds like Sita, So Ram is Married to Sujata that leaves Gopal with Uma.
[/hide][/quote]

 :rotfl:  :rotfl:   Wow,nice thinking Mr.Gyan,though unfortunately it is a math question to be done in a mathematical way :D   but your answers are right :wow:

Of course even bolywood fans know that .[hide].. (Click only if you want to peek at the solution...)
[hide] $a^2-b^2=(a-b)(a+b)=63$,   and notice that possible value of  (a-b) can only be $(1,3,7)$ which would give only $(33,31),(12,9),(8,1)$ values for husband wife pair..[/hide][/hide]


Find all solutions in integers $ a$, $ b$, $ c$, $ d$

$ \frac{a}{b} \plus{} \frac{b}{a} \plus{} \frac{c}{d} \plus{} \frac{d}{c} \equal{} \frac{2009}{abcd}$

[hide]There is a neat solution to this. Look for factorizations.[/hide]

$ a^2dc\plus{}b^2dc\plus{}c^2ab\plus{}d^2ab\equal{}2009$

$ ad(ac\plus{}bd)\plus{}bc(bd\plus{}ac)\equal{}2009$

$ (ad\plus{}bc)(bc\plus{}ad)\equal{}2009$ Now it's just casework

There's a typo in last line, should be

$ (ad \plus{} bc)(bd \plus{} ac) \equal{} 2009$

You can go further.


Let $\mathbb{R}_{0}^{+}$ be the set of all nonnegative reals (including 0).

Find all functions $f(x)$ $\mathbb{R}_{0}^{+}\rightarrow \mathbb{R}_{0}^{+}$ verifying the two relations :

$f(x+1)=f(x)+1$ $\forall x\in \mathbb{R}_{0}^{+}$

$f(\frac{1}{f(x)})=\frac{1}{x}$ $\forall x\in \mathbb{R}_{0}^{+}$ such that $x\neq0$ and $f(x)\neq 0$

[quote="pohoatza"]Hello Patrick, see here: [url]http://www.mathlinks.ro/viewtopic.php?p=576234#576234[/url][/quote]

Ahhhh !  I'm sorry.

Thanks.


Yale college Harvard MIT Princeton Stanford vector

I am interested in a rigorous math program as an undergrad, and I was admitted to Yale early action.  I've heard that their math is not quite as tough as at Harvard, MIT, Princeton, or Stanford, but that it is still a top 10 or so program.  I have also applied to Princeton, partially because of the reputation for better math.  Has anyone here gone through Yale's math program?  Does anyone know how difficult it is?  Yale lists a class "Math 230" which is a proof-based vector calc+ linear algebra class.  Is this as difficult as either Princeton's 200 series, Stanford's 51H-53H series, or Harvard's Math 25?

This is somewhat of a FAQ in this forum. I will be interested to see what knowledgeable students and alumni say in reply.

I don't know.  But you should look at more than just their multivariable calculus options.  For example, one reason why I personally wouldn't go to Yale for math is that the majority of Yale students, at least from the impression I get, are more interested in liberal arts/humanities than math.  There's probably, at least, a small group of math enthusiasts there that you can become friends with but the environment will surely be different from what you'll get at Harvard, MIT, or Princeton.  Harvard and Princeton also have a large emphasis on humanities, but there are plenty of people who love math as well.

In response to joml, I guess the diversity of students' interests is part of the reason I like Yale.  Although I'm very passionate about math, I'm not as single minded about it as many people are on this forum.  I am certainly looking for a broad education (and might end up majoring in anything from music to econ to math to comp sci), although I would prefer that there be a challenging math program, in case I decide to push my life in that direction.

[quote="musicheck"]In response to joml, I guess the diversity of students' interests is part of the reason I like Yale.[/quote]

Hi, musicheck, in your opinion is Yale stronger in "diversity of students' interests" than some of the likely "peer" colleges? What other colleges are you comparing Yale to in that regard?


To write the half-reaction equation for the following redox reaction:

$ Mg(s) \plus{} Br_{2}(l) \rightarrow MgBr_{2}(s)$

I find that the oxidation numbers are:

[code]  0       0       +2  -1[/code]
$ Mg(s) \plus{} Br_{2}(l) \rightarrow MgBr_{2}(s)$

Which means that Mg is oxidized and Br is reduced.

To write this in a half-reaction equation:

Oxidation: $ Mg \rightarrow Mg^{2 \plus{} }$

Reduction: Why is $ Br_{2} \rightarrow Br_{2}^{ \minus{} }$ wrong and $ Br_{2} \rightarrow 2Br^{ \minus{} }$ correct?

Thanks in advance!

$ Br_2 \plus{} e^\minus{} \longrightarrow Br_2^\minus{}$

$ Br_2 \plus{}2\, e^\minus{} \longrightarrow 2\,Br^\minus{}$

$ Mg \longrightarrow Mg^{2\plus{}} \plus{} 2\,e^\minus{}$

[quote="Carcul"]$ Br_2 \plus{} e^ \minus{} \longrightarrow Br_2^ \minus{}$

$ Br_2 \plus{} 2\, e^ \minus{} \longrightarrow 2\,Br^ \minus{}$

$ Mg \longrightarrow Mg^{2 \plus{} } \plus{} 2\,e^ \minus{}$[/quote]

Do you mean that both $ Br_{2} \rightarrow Br_{2}^{ \minus{} }$ and $ Br_{2} \rightarrow 2Br^{ \minus{} }$ are wrong?

Thanks!

[quote="elcric"][quote="Carcul"]Wrong. What about the electrons? Did you eat them?[/quote]

 :D  No - I didn't eat them.

How do $ Br_{2} \rightarrow Br_{2}^{ \minus{} }$ and $ Br_{2} \rightarrow 2Br^{ \minus{} }$ differ in terms of electrons?[/quote]

because the first equation only has 1 electron on the reactant side, while the second equation needs 2 electrons to correctly balance them.


Fie $A \in M_n(C)$ cu $detA=0$.Sa se arate ca $\exists a \in C$ astfel incat :$(A^* )^2  = aA^* $

$rank(A)=dim(Im A) \leq n-1$

If $rank(A)\leq n-2$ then $A^{*}=0$ any $a$ is solution

If $rank(A)=n-1$ then $rank(A^{*})=1$ it is known that any matrix M of rank 1

satisfy $M^2=tr(M)M$


Am I wrong or isn't it obvious that $(PQR)^{2}=R^{2}$?

[quote="yuyang"]Hello!

Let ${\cal D}=\{P,Q,R\in{\cal M}_{2}({\mathbb C})\,: \, PQRPQ=R\}$. Prove that for all $P,Q,R\in{\cal D}$, for all integers $n\geq 1$ we have : $(PQR)^{n}=P^{n}Q^{n}R^{n}$. :P[/quote]

$P^{-1}=Q$ ?? :blush:

Then, $(PQR)^{n}=R^{n}=P^{n}Q^{n}R^{n}$

sorry if i'm wrong :oops:


A Vandal and a Moderator are editing a Wikipedia article.  The article originally is error-free.  Each day, the Vandal introduces one new error into the Wikipedia article.  At the end of the day, the moderator checks the article and has a $ 2/3$ chance of catching each individual error still in the article.  After $ 3$ days, what is the probability that the article is error-free?

[hide]
The possible paths to a error free page are:
1. moderator corrects one error each day which has a probability of (2/3)(2/3)(2/3)=8/27.

2. moderator corrects one error first day, misses one second day, and corrects 2 on the third day.  Probability (2/3)(1/3)(2/3)(2/3) = 8/81.

3. moderator misses  the error on the first day, corrects two the second day and one the last day.  Prob: (1/3)(2/3)(2/3)(2/3) = 8/81.

4. moderator misses first day, corrects one and misses one second day, and corrects two last day.  Prob: (1/3)(2/3)(1/3)(2/3)(2/3) = 8/243.

5. moderator misses all errors first two days and corrects all three on the last day. Prob: (1/3)(1/3)(1/3)(2/3)(2/3)(2/3) = 8/729.

Add it all up and the probability is 392/729.
[/hide]

[hide="I got something different"]
Consider the error that was introduced on day $ 1$.  The probability that the Moderator misses this error on all three checks is $ \frac{1}{3^3}$, so the probability that the error is removed is $ 1\minus{}\frac{1}{3^3}$.  Similarly, the probability that the moderator misses the other two errors are $ 1\minus{}\frac{1}{3^2}$ and $ 1\minus{}\frac{1}{3}$.  So the probability that the article has no errors is:

$ \left (1\minus{}\frac{1}{3^3} \right )\left(1\minus{}\frac{1}{3^2} \right )\left (1\minus{}\frac{1}{3} \right )\equal{}\boxed{\frac{416}{729}}$.
[/hide]

bananaslug is missing one case:

[hide]
+5. moderator misses first day, misses one and corrects one the second day, and corrects two last day. Prob: 8/243

Note the (subtle!) difference between this and "moderator misses first day, corrects one and misses one second day, and corrects two last day."
[/hide]

On a side note, wikipedia doesn't have moderators. An administrator, or a sysop, perhaps?

oops  :blush:
[hide] I calculated the case of miss one on first day correct one second day as (1/3)(1/3)(2/3)(2/3)(2/3) when it should be (1/3)(4/9)(2/3)(2/3) since he has 2/9 chance of correcting the first mistake and 2/9 chance of correcting the second mistake.[/hide]


Can someone tell me what this book is about ? Does it disscus problem solving or is it just a collection of AMC problems. If it disscuses problem solving on what level ? Is it more like AoPS Vol.1 or AoPS Vol.2?

I have this book, it's more like a small textbook with proofs for the equations they give you and it gives you like 3 examples and then around 20 practice problems taken from AMC10/12 tests. It's an ok book but it only covers the topics in a superficial way and if the subject is new to you, it doesn't teach too well. I haven't tried the AOPS books yet so i can't say how it relates. I would recommend it if you know the topics already because personally i don't think it teaches too well. It has solutions in the back with explanations as well.


For every integer $ n>1$ prove that:

$ {\sum_{k\equal{}1}}^{n}\frac{1}{k^{2}}>\frac{3n}{2n\plus{}1}$

[quote="sinajackson"]see: http://www.mathlinks.ro/viewtopic.php?t=155177   :)[/quote]
Thank's  :)


Let $ a, b, c > 0$ such as $ a^2\plus{}b^2\plus{}c^2\equal{}1$.
Prove that !
 $ \left(a^2b^2\plus{}b^2c^2\plus{}c^2a^2\right)\left(\frac{1}{(1\minus{}ab)^2}\plus{}\frac{1}{(1\minus{}bc)^2}\plus{}\frac{1}{(1\minus{}ca)^2}\right) \le \frac{9}{4}$

[quote="bigbang195"]Let $ a, b, c > 0$ such as $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.
Prove that !
 $ \left(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\right)\left(\frac {1}{(1 \minus{} ab)^2} \plus{} \frac {1}{(1 \minus{} bc)^2} \plus{} \frac {1}{(1 \minus{} ca)^2}\right) \le \frac {9}{4}$[/quote]

The inequality is equivalent to

$ \left(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\right)$ $ \left(\frac {1}{(\sum a^2 \minus{} ab)^2} \plus{} \frac {1}{(\sum a^2 \minus{} bc)^2} \plus{} \frac {1}{(\sum a^2 \minus{} ca)^2}\right) \le \frac {9}{4}$

by AM-GM Inequality

$ \left(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\right)$ $ \left(\frac {1}{(\sum a^2 \minus{} ab)^2} \plus{} \frac {1}{(\sum a^2 \minus{} bc)^2} \plus{} \frac {1}{(\sum a^2 \minus{} ca)^2}\right)$

$ \le \left(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\right)$ $ \left(\frac {1}{(c^2 \plus{} \frac {a^2 \plus{} b^2}{2})^2} \plus{} \frac {1}{(a^2 \plus{} \frac {b^2 \plus{} c^2}{2})^2} \plus{} \frac {1}{(b^2 \plus{} \frac {a^2 \plus{} c^2}{2})^2}\right)$

substitution $ x \equal{} a^2,y \equal{} b^2,z \equal{} c^2$

we just need to prove

$ \left(xy \plus{} yz \plus{} zx\right)$ $ \left(\frac {1}{(z \plus{} \frac {x \plus{} y}{2})^2} \plus{} \frac {1}{(x \plus{} \frac {y \plus{} z}{2})^2} \plus{} \frac {1}{(y \plus{} \frac {z \plus{} x}{2})^2}\right)\le \frac {9}{4}$

$ \Longleftrightarrow 36\sum(a^4(a \minus{} b)(a \minus{} c)) \plus{} 144\sum(a^3(b \plus{} c)(a \minus{} b)(a \minus{} c)) \plus{}$ $ 229(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$

$ \plus{} 468\sum(b^2c^2(a \minus{} b)(a \minus{} c)) \plus{} 808abc\sum(a(a \minus{} b)(a \minus{} c))$ $ \plus{} 480abc\sum((b \plus{} c)(a \minus{} b)(a \minus{} c))\geq 0$

[i]If my memory is not too bad . There is a beautiful lemma as follows :

          Lemma : If $ a ;b;c$ are positive real numbers such that : $ a^2 \plus{} b^2 \plus{} c^2 \ \equal{} \ 1$ then :

    $ \frac{1}{(1\minus{}ab)^2} \plus{} \frac{1}{(1\minus{}bc)^2} \plus{} \frac{1}{(1\minus{}ca)^2} \le \frac{27}{4}$ :wink:

    Apply this lemma , everything else seems to be very easy[/i]


function
search
irrational number
algebra unsolved
algebra

I expect a solution to this interesting problem. Thanks in advance.

[quote="thaithuan_GC"]Find all continous functions $ f: \mathbb R\rightarrow \mathbb R$ such that $ f(x) - f(y) \in\mathbb Q$ for every pairs of real numbers $ (x,y)$ satisfying $ x - y\in\mathbb Q$[/quote]
This problem posted many times before ,you can search in our forum .

Lemma: let $ g$ be a continuous function satisfying $ g(x) \in \mathbb{Q} \forall x\in \mathbb{R}$. $ g$ is constant.

Proof: suppose by absurd $ g(a) \not \equal{} g(b)$ for some real numbers $ a, b$. Let $ c$ be an irrational number between $ g(a)$ and $ g(b)$. By Bolzano´s theorem, we must have $ c \equal{} f(d)$ for some real number $ d$ in the interval $ ]a, b[$. Contradiction (the image of $ g$ is contained in $ \mathbb{Q}$).

Now let $ g_r(x) \equal{} f(x \plus{} r) \minus{} f(x)$ for each rational $ r$. We must have $ f(x \plus{} r) \minus{} f(x)$ constant; then every rational is "almost a period" of $ f$. If we have $ f(x \plus{} 1) \equal{} f(x) \plus{} a$ for some real $ a$, necessarily $ f(x \plus{} \frac {1}{n}) \equal{} f(x) \plus{} \frac {a}{n} \forall n \in \mathbb{N} \Longrightarrow f(x \plus{} r) \equal{} f(x) \plus{} ar \forall r \in \mathbb{Q}$. In particular, $ f(r) \equal{} ar \plus{} f(0)$. From the continuity of $ f$, we conclude $ f$ has to be linear. The rest is trivial.


Τι μπορεί να γίνει σε αυτή  την άσκηση ;

 http://www.mathlinks.ro/viewtopic.php?t=183410

Se kati tetoia h Taylor sxedon panta voh8aei...:P
An den kanw la8os einai h Taylor expansion tou arcsin.5 = 30 degrees = π/6
Elpizw na einai swsto!

[quote="bilot4"]Δεν ξέρω αν πρέπει να το βάλω εδώ ή στο αγγλικό forum αλλά θα μπορούσατε να εξηγήσετε λίγο κ.Στεργιου ποιος είναι ο γενικός τύπος κάθε κλάσματος... εννοώ ποιός είναι ο επόμενος όρος :?:[/quote]

Δυστυχώς το πρόβλημα μου το έστειλαν δύο φυσικοί μέσω ενός τρίτου ( Σαβάλλας κλπ) και δεν ξέρω αν το βρήκαν από φυσσικό πρόβλημα ή το έδωσαν στα παιδιά τους από το σχολείο για τον Ευκλείδη.
 Θα το ρωτήσω το απόγευμα.

Ο Ηλίας έχει δίκαιο ... :)   (καλά πώς το έπιασες ξέρεις το κόλπο ? :wink: )

Γιατί γενικά $ \arcsin x \equal{} x \plus{} \frac {x^3}{6} \plus{} \frac {3x^5}{40} \plus{} \frac {1\cdot 3\cdot 5\cdot x^7}{2\cdot 4\cdot 6\cdot 7} \plus{} ...$

kai gia $ x \equal{} \frac {1}{2}$ παίρνουμε το ζητούμενο άθροισμα

Wow! Sharp eyes salonikios... :!:

Eλπίζω μόνο να μην πέσει θέμα τέτοιου είδους στον Ευκλείδη ή σε παρόμοιο διαγωνισμό (το λέω εξαιτίας της πιθανής πηγής προέλευσης στην οποία αναφέρεται ο stergiu). Αλλιώς... καλύτερα να του αλλάξουμε όνομα του διαγωνισμού σε Trivial Pursuit ;)

Και πάλι, nice catch salonikios :) Cheerio,

Durandal 1707

Παιδιά ευχαριστώ πολύ ! Πρόκειται για εξαιρετική παρατήρηση. Θα το πω σ'αυτούς τους .... τύπους να μας λένε τουλάχιστον την πηγή και το επίπεδο των ερωτήσεων. Σκεφτείτε ότι παιδευόμουνα μια ώρα με τηλεσκοπικά αθροίσματα :( .

Μπάμπης


(Could not think of a better title -sorry if it was not nice)

a) Find two least composite numbers $ n$ such that $ n | 2^n \minus{}2$ and $ n|3^n \minus{} 3$.

b) Find the least positive integer $ n$ cuh that $ n|2^n \minus{}2$ but $ n$ does not $ | 3^n \minus{}3$.

c)Find the least positive integer $ n$ such that $ n$ does not divide $ 2^n \minus{}2$ but $ n| 3^n \minus{}3$.

[quote="Agr_94_Math"](Could not think of a better title -sorry if it was not nice)

a) Find two least composite numbers $ n$ such that $ n | 2^n \minus{} 2$ and $ n|3^n \minus{} 3$.

b) Find the least positive integer $ n$ cuh that $ n|2^n \minus{} 2$ but $ n$ does not $ | 3^n \minus{} 3$.

c)Find the least positive integer $ n$ such that $ n$ does not divide $ 2^n \minus{} 2$ but $ n| 3^n \minus{} 3$.[/quote]
May an apply of Fecma? But I have no answer  :P  ? Post any idea  :D


Let $ S$ be the set of lattice points in the region defined by $ 0 \leq x \leq 3$ and $ 0 \leq y \leq 3$. Triangles are formed by choosing three non-collinear members of $ S$ as vertices. Compute the number of distinct triangles with an area of $ \frac {3}{2}$.

Area of triangle = $ \frac 12\times {base}\times {height}$.
The various distances possible are $ 1,\sqrt 2, 2,\sqrt 5, \sqrt 8, 3, \sqrt {10}, \sqrt {13}, \sqrt {18}$ We can only hve $ \frac 12 \times 1 \times 3$ or $ \frac 12 \times 3 \times 1$. There are only $ \boxed {12}$ such triangles.

How about the triangles whose bases aren't parallel to the axes? For example, the one formed by connecting $ (0,0)$, $ (3,3)$ and $ (2,1)$ has area $ \frac {3}{2}$.


Find all positive integers n such that the polynomial with n+1 terms :
  $ P(x) = x^{4n} + x^{4(n-1)} + ... + x^8 + x^4 + 1 $
 is divisible by the polynomial with n+1 terms :
  $ Q(x) = x^{2n} + x^{2(n-1)} + ... + x^4 +x^2 + 1 $

This is not too difficult: [hide]it is only true for n even.  Proof: assume Q(x) | P(x).  Then there exists some polynomial R(x) such that R(x)*Q(x) = P(x).  Multiply through by (x^4 - 1).  Then

 and dividing, we see that we must have a polynomial , so that i and -i must be roots of .  This only happens when n is even.[/hide]

Using expression for sum of term of geometrical progression we obtain $P(x)=\frac{x^{4n+4}-1}{x^4-1}$ and $Q(x)=\frac{x^{2n+2}}{x^2-1}$, hence
\[\frac{P(x)}{Q(x)}=\frac{(x^{4n+4}-1)(x^2-1)}{(x^{2n+2}-1)(x^4-1)}=\frac{x^{2n+2}+1}{x^2+1}\].
Thus $x^{2n+2}+1$ must be divisible by $x^2+1$, i.e. $x^{2n+2}+1=(x^2+1)R(x)$ for some polynomial $R(x)$. We have $0=(i^2+1)R(i)=i^{2n+2}+1$ $\Leftrightarrow$ $n$ is even. Moreover, if $n$ is even we obtain $(-i)^{2n+2}+1=0$. It implies that $i$ and $-i$ are roots of $x^{2n+2}+1$. It follows that $x^{2n+2}+1$ is divisible by $x-i$ and $x+i$ $\Rightarrow$ $x^{2n+2}+1$ is divisible by $(x-i)(x+i)=x^2+1$.


If in applying the quadratic formula to a quadratic equation
\[ f(x)\equiv ax^2 \plus{} bx \plus{} c \equal{} 0,
\]
it happens that $ c \equal{} \frac {b^2}{4a}$, then the graph of $ y \equal{} f(x)$ will certainly:

$ \textbf{(A)}\ \text{have a maximum} \qquad\textbf{(B)}\ \text{have a minimum} \qquad\textbf{(C)}\ \text{be tangent to the x \minus{} axis} \\
\qquad\textbf{(D)}\ \text{be tangent to the y \minus{} axis} \qquad\textbf{(E)}\ \text{lie in one quadrant only}$

[hide]Taking the discriminant, we have $ b^2-4(\frac{b^2}{4a^})(a)=b^2-b^2=0$. If we apply the quadratic formula, we have $ x=\frac{-b\pm0}{2a}$. Obviously, there is only one value of x for this, so the graph would be tangent to the x-axis, or $ \textbf{ (C) }$ [/hide]


perfect cube
number theory
Perfect power
Diophantine equation

Let $ x,y,z$ be positive integer such that $ \frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x} \in \mathbb{N}$ ,prove that $ xyz$ is perfect cube

The equation  $ x{y}^{2}\plus{}z{x}^{2}\plus{}y{z}^{2}\equal{}kxyz$  is homogenous, so we can assume that  $ gcd(x,y,z)\equal{}1$.

    Now, let   $ {p}^{a}$   be the highest power of p which divides any of the variables, say x.

    Consider that   $ y{z}^{2}\equal{}x(kyz\minus{}{y}^{2}\minus{}xz)$  , and it is obvious that  $ {p}^{a}$ either divides $ y$ (1) or $ {z}^{2}$ (2).

    1; It is evident that  $ {p}^{a}$must also divide  $ y$. But this implies that $ {p}^{2a}$ must divide $ x(kyz\minus{}{y}^{2}\minus{}xz)$; contradiction.

     2; Note that p doesn´t divide $ (kyz\minus{}{y}^{2}\minus{}xz)$ . Let b be the highest number for which $ {p}^{b}$ divides $ z$. It is true that $ 2b\equal{}a$ and hence that p has the power 3b in xyz, which is a cube.

[quote="fredrik1232009"]The equation  $ x{y}^{2} \plus{} z{x}^{2} \plus{} y{z}^{2} \equal{} kxyz$  is homogenous, so we can assume that  $ gcd(x,y,z) \equal{} 1$.

    Now, let   $ {p}^{a}$   be the highest power of p which divides any of the variables, say x.

    Consider that   $ y{z}^{2} \equal{} x(kyz \minus{} {y}^{2} \minus{} xz)$  , and it is obvious that  $ {p}^{a}$ either divides $ y$ (1) or $ {z}^{2}$ (2).

    1; It is evident that  $ {p}^{a}$must also divide  $ y$. But this implies that $ {p}^{2a}$ must divide $ x(kyz \minus{} {y}^{2} \minus{} xz)$; contradiction.

     2; Note that p doesn´t divide $ (kyz \minus{} {y}^{2} \minus{} xz)$ . Let b be the highest number for which $ {p}^{b}$ divides $ z$. It is true that $ 2b \equal{} a$ and hence that p has the power 3b in xyz, which is a cube.[/quote]
In Case 1;
Why $ p^{2a} | x(kyz\minus{}y^2 \minus{}xz)$ then it's contradiction.   :?:
Thank you.

The contradiction follows from the assumption that $ p^{a}$ is the highest power of p which divides any of the variables.

Since$ p^{2a}| x(kyz\minus{}y^{2}\minus{}xz)$, which is a concequence of (1), would imply that $ p^{2a}| y$, the assumption would be contradicted.


trigonometry
calculus
calculus computations

Hi everyone,

Could you please help me with this problem?

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that Rn(x) goes to 0] Also, find the associated radius of convergence.

F(x)=sin2x

f(x)=sin2x          f(0)=0                  co=0
f'(x)=2cos2x       f'(0)=2                 c1=2
f''(x)=-4sin2x      f''(0)=0                 c2=0
f'''(x)=-8cos2x+xcosx  f'''(x)=-8       c3=-8

f^(n)(x)=-1^n               f^(n)(0)=-1^n  cn=(-1)^n??????????????????

sin2x=sum. n=0 to infinity (-1)^n/???????????

lim as n goes to infinity |(an+1)/(an)|
=?????????

I'm not sure how to find cn. Could someone please check what I've done and help me find cn, please? Also, a Talor series is another word for a Maclaurin series, right?

Thank you very much

[quote="chocolatelover"]f'''(x)=-8cos2x+xcosx[/quote]

This is incorrect; there is no second term.  The pattern should emerge quickly.  Alternately, you could find the Maclaurin series of $ f(u) \equal{} \sin u$ and substitute $ u \equal{} 2x$ into the power series later.

A Maclaurin series is a Taylor series centered at $ 0$.  Taylor series are more general, and can be centered at any constant $ c$.  See [url=http://en.wikipedia.org/wiki/Taylor_series]Wikipedia[/url].


It is easy to prove, that $ a\le 4$ and chek, that $ a \equal{} 4$ give not solution.
$ a \equal{} 2$ give $ (2,2,2),(2,2,4),(2,4,8)$.
$ a \equal{} 3$ give $ (3,5,15)$.

ModEdit: luckily there's a solution on http://www.kalva.demon.co.uk/imo/isoln/isoln921.html

[quote="Peter"]Find all integers $\,a,b,c\,$ with $\,1<a<b<c\,$ such that \[(a-1)(b-1)(c-1)\hspace{0.2in}\text{is a divisor of}\hspace{0.2in}abc-1.\][/quote]
$x=a-1 \; , \; y=b-1 \; , \; z=c-1 \; \Longrightarrow \; xyz \mid (xy+yz+zx+x+y+z)$

$x \geq 3 \; \Longrightarrow \; y \geq 4 \; , \; z \geq 5 \; \Longrightarrow$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} \leq \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{12}+\frac{1}{20}+\frac{1}{15}=\frac{59}{60}<1$

$\Longrightarrow \; xyz>xy+yz+zx+x+y+z$. It is a contradiction.

$\Longrightarrow \; x \leq 2$

The rest is easy and all solutions are:

$(x,y,z)=\{(1,3,7),(2,4,14)\}$

$(a,b,c)=\{(2,4,8),(3,5,15)\}$

The problem is IMO 1992 Probrem 1.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=154338&#p154338[/url]


What is the largest even integer that cannot be written as the sum of two odd composite positive integers?

The solution can be found at the[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1984&p=392634]contest section[/url]


vector
linear algebra
linear algebra unsolved

The story: [hide]I was in the class and the professor was teaching the Linear Functionals. Meanwhile, this came to my mind:[/hide]

Suppose that $ V$ is a vector space over a field $ \mathbb{F}$, and $ \mathbb{E}$ is a subfield of $ \mathbb{F}$. Define $ W^*$ be the set of all linear functionals from $ V$ to $ \mathbb{E}$. i.e. $ W^*\equal{}\{ L(V,\mathbb{E}) \}\equal{}\{ T: V \rightarrow \mathbb{E} | T \mbox{is a linear transformation} \}$. One can simply prove that this is a vector space over $ \mathbb{E}$.

Now I want to find the dimension of this vector space. I tried to find it in terms of dimension of $ \mathbb{F}$ as a vector space over $ \mathbb{E}$, and dimension of $ V$ over $ \mathbb{F}$. But it is just about the finite dimensional case. I don't have any idea how to precise it, and extend it for infinite dimensional case.

The only way for this to make sense is as maps linear over $ E$. The only structure on $ V$ that matters is as a vector space over $ E$. The dimension $ n$ of that is the product of the degree of the extension $ F/E$ and the dimension of $ V$ over $ F$.


function
real analysis
topology
Functional Analysis
real analysis unsolved

$U$ is a relatively compact open set in $R^{n}$, define a function on $R^{n}$
by $f(x)=d(x, R^{n}-U)$, i.e. distance from $x$ to complement of $U$.
This function is continuous, of course.

(1) Is it differentiable? is it inifinitely differentiable?

(2) If it is, do we have same result for any metric space?

(my question is inspired by the following setup: In a compact metric space $X$, the algebra generated by such functions is dense in $C(X)$ by Stone-Weirstrass. (Easy to see the countable family of such functions (indexed by $U$) seperates points).

So if they are also $C^{\infty}$, then it would be very nice. For example, then they would be dense in the Frechet space of all distributions.

I don't know what "relatively compact" means, but consider $U = (0, 1)^{2}\subset{\bf R}^{2}$.  This function is clearly not $C^\infty$ along the lines $y = x$ and $x+y = 1$, and something similar will be true for any polyhedron in ${\bf R}^{n}$.  Probably in general that function can only be as nice as the boundary of $U$, or something like that.

It can go bad even if the boundary is smooth- consider a ball, and look at what happens at the center.
Also, the distance is never differentiable on the boundary of $U$.

The distance function is Lipschitz (obviously), and this is about as much as you can say. Re: 2)  how do you define differentiability on a metric space?

Actually, the distance function can be differentiable at some boundary points of $U$. Indeed, if $\partial U$ has positive Lebesgue measure, then $d(x,\mathbb R^{n}\setminus U)$ is differentiable a.e. on $\partial U$ by Rademacher's theorem.


inequalities
trigonometry
inequalities proposed

The following inequalities, proposed by Pham Van Thuan, Hanoi University of Science, are posted as a dedication to Harazi and Vasc, our "Mathlinkers".

Prove that if $x,y,z>0$ then
\[\frac{x}{y^2+yz+z^2}+\frac{y}{x^2+xz+z^2}+\frac{z}{x^2+xy+y^2}\geq\frac4{x+y+z+\frac{x^3+y^3+z^3}{x^2+y^2+z^2}}.\]
Prove that if $x,y,z>0$ then
\[\frac{x}{y^2+yz+z^2}+\frac{y}{x^2+xz+z^2}+\frac{z}{x^2+xy+y^2}\geq\frac4{x+y+z+3\frac{x^3+y^3+z^3}{(x+y+z)^2}}.\]
Prove that if $x,y,z>0$ and $x+y+z=1$ then
\[\frac{x}{y^2-yz+z^2}+\frac{y}{x^2-xz+z^2}+\frac{z}{x^2-xy+y^2}\geq\frac4{x^2+y^2+z^2+x^3+y^3+z^3}.\]
Prove that if $a,b,c>0$ then
\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac b{c+a}}\geq \sqrt{\frac{a+b}c}.\]

Thanks, but you know, I think that other mathlinkers deserve these inequalities, I did nothing in the last year and probably I won't do from now on.  ;)   Probably when nice inequalities will be discussed on mathlinks, but that's a very far future...

Sorry, but the first three inequalities all follow directly using Cauchy+ (really)  :( small amount of computations.   :(

Yes, Harazi you are right. I thought these inequalities were quite strong, but I was wrong.  :( I find the last one very interresting.  ;)

Of course I'm right.  :D  :P  Then why don't you post a solution for the last one? I have one, straightforward, but ugly.

Well,I haven't finished it yet. I was thinking in the same way and then to use an identity.When, I'm threw with it, I'll post my solution. ;)

I am definitely too lazy now to post solutions to the first three inequalities (indeed, the first one is, as Harazi said, a simple Cauchy, the second one follows from the first, and the third one is a consequence of Chebyshev $ \sum\frac{x}{y^{2}\minus{}yz\plus{}z^{2}}\sum x^{2}\geq 3\sum\frac{x^{3}}{y^{2}\minus{}yz\plus{}z^{2}}$ (with cyclic sums), of the inequality $ \sum\frac{x^{3}}{y^{2}\minus{}yz\plus{}z^{2}}\geq\sum x$ (from http://www.mathlinks.ro/Forum/viewtopic.php?t=6551 ), of the given condition $ \sum x \equal{} x\plus{}y\plus{}z \equal{} 1$ and of the inequality $ \frac{3}{\sum x^{2}}\geq\frac{4}{x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}x^{3}\plus{}y^{3}\plus{}z^{3}}$ (which is straightforward) [[b]EDIT:[/b] this proof of the third inequality is wrong!]). The fourth one is quite neat, although it is missing a condition. Here is how I suppose it was meant (with different notations, the solution will show why):

[color=blue][b]Problem.[/b] Let x, y, z be three positive reals such that $ z\geq x$ and $ z\geq y$. Prove that $ \sqrt{\frac{x}{y\plus{}z}}\plus{}\sqrt{\frac{y}{z\plus{}x}}\geq\sqrt{\frac{x\plus{}y}{z}}$.[/color]

[hide="Solution"][i]Solution.[/i] Upon multiplication with $ \sqrt{\frac{z}{x\plus{}y}}$, the required inequality $ \sqrt{\frac{x}{y\plus{}z}}\plus{}\sqrt{\frac{y}{z\plus{}x}}\geq\sqrt{\frac{x\plus{}y}{z}}$ becomes

$ \sqrt{\frac{x}{y\plus{}z}}\cdot\sqrt{\frac{z}{x\plus{}y}}\plus{}\sqrt{\frac{y}{z\plus{}x}}\cdot\sqrt{\frac{z}{x\plus{}y}}\geq\sqrt{\frac{x\plus{}y}{z}}\cdot\sqrt{\frac{z}{x\plus{}y}}$,

or, equivalently,

$ \sqrt{\frac{zx}{\left(x\plus{}y\right)\left(y\plus{}z\right)}}\plus{}\sqrt{\frac{yz}{\left(z\plus{}x\right)\left(x\plus{}y\right)}}\geq 1$.

Now, let a = y + z, b = z + x, c = x + y. Then, since b + c - a = (z + x) + (x + y) - (y + z) = 2x > 0, so that b + c > a, and similarly, c + a > b and a + b > c, the numbers a, b, c are the sidelengths of a triangle. The semiperimeter $ s \equal{}\frac{a\plus{}b\plus{}c}{2}$ of this triangle equals $ s \equal{}\frac{a\plus{}b\plus{}c}{2}\equal{}\frac{\left(y\plus{}z\right)\plus{}\left(z\plus{}x\right)\plus{}\left(x\plus{}y\right)}{2}\equal{} x\plus{}y\plus{}z$; thus, s - a = (x + y + z) - (y + z) = x and similarly s - b = y and s - c = z. Hence, the inequality in question, $ \sqrt{\frac{zx}{\left(x\plus{}y\right)\left(y\plus{}z\right)}}\plus{}\sqrt{\frac{yz}{\left(z\plus{}x\right)\left(x\plus{}y\right)}}\geq 1$, becomes

$ \sqrt{\frac{\left(s\minus{}c\right)\left(s\minus{}a\right)}{ca}}\plus{}\sqrt{\frac{\left(s\minus{}b\right)\left(s\minus{}c\right)}{bc}}\geq 1$.

Now, by the half-angle formulas, we have $ \sqrt{\frac{\left(s\minus{}c\right)\left(s\minus{}a\right)}{ca}}\equal{}\sin\frac{B}{2}$ and $ \sqrt{\frac{\left(s\minus{}b\right)\left(s\minus{}c\right)}{bc}}\equal{}\sin\frac{A}{2}$, so it remains to prove $ \sin\frac{B}{2}\plus{}\sin\frac{A}{2}\geq 1$. This is equivalent to $ \sin\frac{A}{2}\plus{}\sin\frac{B}{2}\geq 1$.

Note that $ z\geq x$ yields $ a \equal{} y\plus{}z\geq x\plus{}y \equal{} c$; since in a triangle, the greater side lies opposite to the greater angle, this yields $ A\geq C$, and thus $ 90^{\circ}\minus{}\frac{A}{2}\leq 90^{\circ}\minus{}\frac{C}{2}$. Similarly, $ 90^{\circ}\minus{}\frac{B}{2}\leq 90^{\circ}\minus{}\frac{C}{2}$. Also, since the inequality in question is symmetric in A and in B, we can WLOG assume that $ 90^{\circ}\minus{}\frac{A}{2}\leq 90^{\circ}\minus{}\frac{B}{2}$, so that $ 90^{\circ}\minus{}\frac{A}{2}\leq 90^{\circ}\minus{}\frac{B}{2}\leq 90^{\circ}\minus{}\frac{C}{2}$.

But since the angles A, B, C are the angles of a triangle, we have A + B + C = 180°; hence,

$ \left(90^{\circ}\minus{}\frac{A}{2}\right)\plus{}\left(90^{\circ}\minus{}\frac{B}{2}\right)\plus{}\left(90^{\circ}\minus{}\frac{C}{2}\right) \equal{} 270^{\circ}\minus{}\frac{A\plus{}B\plus{}C}{2}$
$ \equal{} 270^{\circ}\minus{}\frac{180^{\circ}}{2}\equal{} 180^{\circ}$;

also, $ 90^{\circ}\minus{}\frac{A}{2}> 0^{\circ}$ (since $ 90^{\circ}>\frac{A}{2}$, because 180° > A), and similarly $ 90^{\circ}\minus{}\frac{B}{2}> 0^{\circ}$ and $ 90^{\circ}\minus{}\frac{C}{2}> 0^{\circ}$. Hence, $ 90^{\circ}\minus{}\frac{A}{2}$, $ 90^{\circ}\minus{}\frac{B}{2}$, $ 90^{\circ}\minus{}\frac{C}{2}$ are the angles of a triangle. Since we also have $ 90^{\circ}\minus{}\frac{A}{2}\leq 90^{\circ}\minus{}\frac{B}{2}\leq 90^{\circ}\minus{}\frac{C}{2}$, we can now apply http://www.mathlinks.ro/Forum/viewtopic.php?t=4379 and see that $ \cos\left(90^{\circ}\minus{}\frac{A}{2}\right)\plus{}\cos\left(90^{\circ}\minus{}\frac{B}{2}\right)\geq 1$. In other words, $ \sin\frac{A}{2}\plus{}\sin\frac{B}{2}\geq 1$. Proof complete.[/hide]

  Darij

You should try it out yourself first because the skill is standard.
[hide]
$(\sum \frac{x}{y^2+yz+z^2})(x+y+z)(xy+yz+zx) = (\sum \frac{x}{y^2+yz+z^2})(\sum x(y^2+yz+z^2)) \ge (x+y+z)^2$
Thus $L.H.S. \ge \frac{x+y+z}{xy+yz+zx}$
$R.H.S. \le \frac{3}{x+y+z}$
It remaisn to show $(x+y+z)^2 \ge 3(xy+yz+zx)$ which is trivial.[/hide]

[quote="siuhochung"][quote="darij grinberg"]the third one is a consequence of Chebyshev $\sum\frac{x}{y^2-yz+z^2}\sum x^2\geq 3\sum\frac{x^3}{y^2-yz+z^2}$[/quote]

Sign reversed?[/quote]

Indeed. This is what comes out if one doesn't do any inequalities for a month... Anyway, the third inequality is wrong: take $x=y=\frac12$ and z = 0 (or z arbitrary close to 0, before somebody posts another remark of this kind). Probably the proposer had something different in mind.

  darij

The third one is not true:

[quote]Prove that if $x,y,z>0$ and $x+y+z=1$ then
\[\frac{x}{y^2-yz+z^2}+\frac{y}{x^2-xz+z^2}+\frac{z}{x^2-xy+y^2}\geq\frac4{x^2+y^2+z^2+x^3+y^3+z^3}.\]
[/quote]This is not true, for counter-example,  $\,\,x=\frac{3}{7},\,\, y=\frac{3}{7}, \,\,z=\frac{1}{7}.$

I am curious about you, Jichen? You appears in Mathlinks just to "solve problem" (not "talk" or "chat" with other Mathlinkers) and almost your posts are to generalize the post of one person (pvthuan)? Who are you? Are you Jichen appearing in Crux? Sorry because of my question but I really curious. If you are not afraid, can you answer my question?

Sorry to other Mathlinkers because my question is not about the problem discussed here.

To GacKiem, all you need to know is that he is a Chinese friend of mine.

It's not enough  :huh: (for me)! Because he is so good with a lot of difficult (and sometimes complicated) inequalities. At least may I know that he is "Jichen" appearing in Crux or not?

Sorry to GacKiem, I am not take notice of your postsin in season . I am feazed by mathematical problems for a long time. Since 1988, I am appearing in Crux.


Nowadays, with the existent technology, it's "easy" to calculate the distance between to planets or stars... but how the first people used to calculate the distance between two stars or planets??? I mean, you can tell me the ancient greece people, or Kepler, or Newton, or anyone you want.... liks or explanations are thanks!!!

I think they could do it not only for their physics knowledge, but also for their outstanding math skills! They cound master basic geometry and conics and that ensure them to be able to do the calculation.

well, aristarh's theory assumed that the distance betwen stars is enormous (that ancient greeks couldn't accept, so they concluded that aristarh's wrong)

then I know that Newton combined the third kepler's law and the law for centripental force (I don' t know is "centripental" written like this), made some lemas and got the correct distance betwen earth and the moon. he assumed (right one this time too) that you can calculate the distance betwen every two objects in the space by the same way if they have a gravitation force betwen themselves.

that's the physics way of calculating the distance.

this is math way:

1752. frenchs astronomers La Lande and La Caille used trigonometry in calculating the distance betwen the earth and the moon. they took two points, A and B, on the same earth meridiem (this word isn't spelled corectly). they measured the angles (alpha') and (betha') which are determined by the moon and the vertical lines in A and B ("zenit directions"). the moon, A, B and the center of earth are in the same plain.

this isn't grammaticly correct text. sorry. I hoped I helped you.

if you need more information about La Landes and La Cailles method (all the calculations) just say it.

I'd need a graph for the math idea.... thanks! what about a link?

Can you explain me again and with more details the physics idea?


I'd say that $\mathrm{atan}\equiv\arctan\equiv\tan^{-1}$, and $d_{1}, d_{2}$ are the diagonals.

DB = d1; sorry... miss out one detail.. possible to solve?

I think you would need at least 1 length...

And what are $a,b$? (lower-case letters)

a and b are angle that are unknown.
solve in term of a,b,d1,d2;

I have this edited version. All information are in already. Is this question possible to solve?

How could AD = $\frac{d_{1}^{2}+d_{2}^{2}}{4}$?

The altitude to BD isn't 1/2 AC....


geometry
perimeter
inequalities
inequalities unsolved

Let $ a,b,c$ be sidelengths of a triangle which perimeter 3. Prove that
\[ \frac1{\sqrt[3]{b\plus{}c\minus{}a}}\plus{}\frac1{\sqrt[3]{c\plus{}a\minus{}b}}\plus{}\frac1{\sqrt[3]{a\plus{}b\minus{}c}}\ge\frac9{ab\plus{}bc\plus{}ca}
\]

The following Inequality is true or false for the same condition?
\[ \frac1{\sqrt[4]{b\plus{}c\minus{}a}}\plus{}\frac1{\sqrt[4]{c\plus{}a\minus{}b}}\plus{}\frac1{\sqrt[4]{a\plus{}b\minus{}c}}\ge\frac9{ab\plus{}bc\plus{}ca}
\]

The second inequality is valid and nice, Dduclam. In addition, the following is false with the same condition
$ \frac{1}{\sqrt[5]{b\plus{}c\minus{}a}}\plus{}\frac{1}{\sqrt[5]{c\plus{}a\minus{}b}}\plus{}\frac{1}{\sqrt[5]{a\plus{}b\minus{}c}} \ge \frac{9}{ab\plus{}bc\plus{}ca}.$
;) :)