Princeton
college
Harvard
MIT
number theory

Hi guys,

I am currently an undergraduate student, I am studying math and computer science (i am from germany). I'm interested especially in number theory, theory of information and cryptography. Could you please tell me which university in USA (ex. Princeton, Harvard, etc.) would be the best for me. Maybe Harvard? Maybe Princeton? I don't have any knowledge about graduate program on american universities. I was thinking about MIT, but I thing that math+com. sciense would be the too 'technical' for me.

Thanks

Don't limit yourself to one "best" choice. That's a popular enough combination that there are many good choices.
If you think you already know what you want to specialize in, the overall strength of the graduate program isn't as important as the professors in your chosen area. Look for those professors (and their web pages) to get some idea of what it will be like. If there are several at the same university, that's a very good sign.


Check out the official results of the 2004-2005 Mu Alpha Theta "Log1" Contest at http://www.natassessment.com and take the opportunity to congratulate your peers.  The first-place winners are:

Region 1  (Western US)

School:  Homestead High School, Cupertino, CA
Mu:  Zach Bunting, Thomas Jefferson High School, Auburn, WA
Alpha:  Danny Rorabaugh, Highline High School, Burien, WA
Theta:  Drew Haven, Homestead High School, Cupertino, CA

Region 2 (TX, OK, AR, LA)

School:  A&M Consolidated High School, College Station, TX
Mu:  Jessica Wang, A&M Consolidated High School, College Station, TX
Alpha:  Nicholas Schluterman, Southside High School, Ft. Smith, AR
Theta:  Jeffrey Chen, A&M Consolidated High School, College Station, TX

Region 3 (MS, AL, GA, FL)

School:  Vestavia Hills High School, Vestavia Hills, AL
Mu:  Benjamin Jedlovec, Grissom High School, Huntsville, AL
Alpha:  David Harris, Vestavia Hills High School, Vestavia Hills, AL
Theta:  Evan Zhao, Vestavia Hills High School, Vestavia Hills, AL

Region 4 (Northeastern US)

School:  Lakota West High School, West Chester, OH
Mu:  Muzhou Wang, Lakota West High School, West Chester, OH
Alpha:  Neal Jean, Beavercreek High School, Beavercreek, OH
Theta:  Joel Jean, Beavercreek High School, Beavercreek, OH

13th in my division...would have been 4th if my teacher didn't misspell my name of the first round. Is there any way to correct that?

Meh... 20th. Didn't get to do the last round, though, because my teacher forgot to give us the mail :(

only if i wasnt sick.... looks like iwas some top 50  because i missed thel ast round!!! (BTW, I could have won easily)

Eryaman,

     The results are official now, but if your coach e-mails me (clymer@natassessment.com) with the information on what score is wrong, we can make up an "honorary" certificate for the place you would have received.  We send coaches the scores of their students the week before we publish the results, so that data-entry and name-confusion issues can be corrected.

Fourth in Mu in Region III with a 127.  I'm happy with that.


$ a;b;c>0;abc\equal{}1$.Prove that:
$ a^3\plus{}b^3\plus{}c^3 \ge \sum \frac{a}{b\plus{}c}\plus{}\frac{3}{2}$

Using Chebyshev's inequality, [url=http://www.mathlinks.ro/viewtopic.php?p=966147#966147]this[/url] inequality and Nesbitt's inequality we conclude that
\begin{align*}\frac{a^3+b^3+c^3}{abc}&=\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}\\
&\ge\frac{3(a^2+b^2+c^2)}{ab+bc+ca}\\
&\ge2\sum\frac{a}{b+c}\\
&\ge\sum\frac{a}{b+c}+\frac 32.\end{align*}

another proof,
it's easy to Schow that $ \sum \frac{a}{b\plus{}c} \leq \frac{1}{4}\times \frac{a\plus{}b}{c}$
then it's suffice to prove that $ 4(a^3\plus{}b^3\plus{}c^3)\geq \sum \frac{a\plus{}b}{c}\plus{}6 \equal{}\sum ab(a\plus{}b)  \plus{}6$
$ \Leftrightarrow 4(a^3\plus{}b^3\plus{}c^3)\plus{}3abc \geq \sum ab(a\plus{}b)\plus{}9$
by Schur we have $ a^3\plus{}b^3\plus{}c^3\plus{}3abc\geq \sum ab(a\plus{}b)$ then it's suffice to prove that $ 3(a^3\plus{}b^3\plus{}c^3)\geq 9$
$ \Leftrightarrow a^3\plus{}b^3\plus{}c^3\geq 3$ wish obviouly true (use $ AM\minus{}GM$)

[quote="peine"]another proof,
it's easy to Schow that $ \sum \frac {a}{b \plus{} c} \leq \frac {1}{4}\times \frac {a \plus{} b}{c}$
then it's suffice to prove that $ 4(a^3 \plus{} b^3 \plus{} c^3)\geq \sum \frac {a \plus{} b}{c} \plus{} 6 \equal{} \sum ab(a \plus{} b) \plus{} 6$
$ \Leftrightarrow 4(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3abc \geq \sum ab(a \plus{} b) \plus{} 9$
by Schur we have $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\geq \sum ab(a \plus{} b)$ then it's suffice to prove that $ 3(a^3 \plus{} b^3 \plus{} c^3)\geq 9$
$ \Leftrightarrow a^3 \plus{} b^3 \plus{} c^3\geq 3$ wish obviouly true (use $ AM \minus{} GM$)[/quote]

It's same my solution  :lol:


[quote="Ramanujan"]Να δείξετε ότι για κάθε ακέραιο $n \, \geq 1$ ισχύει η ανισότητα

$\ \frac{1}{2}\left( F_{n}\sqrt[3]{\frac{F_{n+1}}{F_{n}}}+F_{n+1}\sqrt[3]{\frac{F_{n+2}}{F_{n+1}}}+F_{n+2}\sqrt[3]{\frac{F_{n}}{F_{n+2}}}\, \right) < F_{n+2}$

όπου $F_{n}$ η ακολουθία Fibonacci .   :)[/quote]

Ειναι η ανισοτητα ΑΜ-ΓΜ....... ετσι ραμανουζαν...?

 :D

Γράφω τη λύση .
Από ΑΜ-ΓΜ είναι $\sqrt[3]{F_{n}^{2}F_{n+1}}\leq\frac{F_{n}+F_{n}+F_{n+1}}$
Όμοια και για τα άλλα....
Άρα $LHS\leq F_{n}+F_{n+1}+F_{n+2}=2F_{n+2}$ .

H ισότητα όμως δεν ισχύει  :D
Ωραία άσκηση Ramanajuan


trigonometry
calculus
calculus computations

Define the sequence $\{x_{n}\}$ by $x_{0}=a\in (0,1)$ and $x_{n+1}=\frac{4}{\pi^{2}}(\cos^{-1}x_{n}+\frac{\pi}{2})\sin^{-1}x_{n}(n=0,1,2,...)$. Show that the sequence converges and find its limit.

If we set $f(x)=\frac{4}{\pi^{2}}(\cos^{-1}x+\frac{\pi}2)\sin^{-1}x= 1-\left(\frac{2\cos^{-1}x}{\pi}\right)^{2}$, $f$ is increasing and maps $[0,1]$ to itself. In addition, $f(x)>x$ for all $x\in (0,1)$. By general principles of this sort of iteration, the sequence must converge to 1; it increases, and it can't stop except at a fixed point.

I don't really like Vietnamese problems most of the time- they're too complicated, with ugly details.

[quote="jmerry"]$\frac{4}{\pi^{2}}(\cos^{-1}x+\frac{\pi}2)\sin^{-1}x= 1-\left(\frac{2\cos^{-1}x}{\pi}\right)^{2}$[/quote]
Why?  :maybe:
[quote] they're too complicated, with ugly details.[/quote]
You are right! :D

Consider the relation of $\arcsin$ and $\arccos$ in $[0,1]$. Once we apply that, it's just a matter of expanding the product.

I didn't go through a proof that $f(x)>x$; I just graphed it on my calculator. That's equivalent to showing that $\cos t<1-\frac{4}{\pi^{2}}t^{2}$ for $x\in (0,\frac{\pi}{2})$, which shouldn't be too hard.


Let p and q be consecutive odd primes (i.e., no prime
numbers are between them). Show that p + q is a product of at least three primes.

[hide]
Since p and q are odd, the sum of them is even, so we knew $ 2$ is a factor. We just need 2 more factors.

Taking out the factor $ 2$ in the sum of p and q, we get the expression $ \frac{p+q}{2}$ which is the average of p and q. Because p and q are both odd, we know this average is an integer that is between p and q. And because p and q are consecutive prime, this average ineteger is composite so by definition it has at least 2 more factors.

Thus there are at least 3 factors to $ p+q$[/hide]


algebra
polynomial
function
complex numbers
complex analysis
complex analysis solved

here is an easy one:
find all complex polynomial $P$ such that $P(S) = S$ where $S$ is the unit circle.

Polynomials of the form $x^n$, I guess :).

Edit:

Actually, $ax^n$ with $|a|=1$ obviously work, so what I said up there is wrong :)

Actually, I was only thinking about monic polynomials when I said that. I'll see whether it can be extended to the general case.

Let $P(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$. Assume $P(z)\in S,\ \forall z\in S$. This means that $Q(x)=xP(x)$ has the same property. Now compute $Q(z_i),\ i\in\overline{0,n}$, where $z_i$ are the $n+1$'th roots of unity. We sum up the $Q(z_i)$, and we get $n+1$, but that is the sum of $n+1$ complex numbers with modulus $1$, so each one of them must be $1$.

From this we easily get $P(z)=z^n$: we can use Lagrange Interpolation (just the theorem, not the ugly formula :)). Since we know the falue of $P$ in $n+1$ points, it means that $P$ is determined uniquely. Since $P(z)=z^n$ satisfies our conditions, it means that this is it.

Let $P(x)=a_nx^n+...+a_0$. Consider $Q(x)=\overline{a_n}\frac{1}{x^n}+...+\overline{a_0}$. Then the analytic function $P(x)Q(x)$ satisfies $P(x)Q(x)=1$ on $S$, so $P(x)Q(x)=1$ on $\mathbb{C}\setminus\{0\}$. Do we need further explanations? :)


function
real analysis
real analysis unsolved

Find all the functions $f: \mathbb{N}^{*}\to\mathbb{R}$ such that: $f(1)+2\cdot f(2)+...+n\cdot f(n)=f(n+1)-f(1)$ for all  $n\in\mathbb{N}^{n}$.

I got $f(n)=0$, but my book does not agree with me :). Could the text be wrong?

That should be $\sum_{k=1}^n k\cdot k!$, which solves your problem.

[quote="Slizzel"]Yes and then we have
$1!\cdot f(1)+2!\cdot f(1)+...+ n!\cdot f(1)=(n+1)!\cdot f(1)-f(1)$

$\Longleftrightarrow f(1)\cdot \sum_{k=1}^{n}{k!}=f(1)\cdot[(n+1)!-1]$ which is true if $f(1)=0$  :huh:[/quote]
it should be like : $1f(1)1!+2f(1)2!+3f(1)3!+...+nf(1)n!$ which equals $(n+1)!-1$.. (you forgot to multiply the $k$ in front of $f(k)$... $kf(k)=kf(1)k!$ :P )


arg....i simply need to know how to find the area of 2 intersecting circles..c the image how would i find the red, yello ar the black part, PLEASE HELP  in the yelo w one the circles are semi cirlces and mthe sides of the square are 8

Um...if I read this right, the area of intersecting circles can be found by taking the area of the sector of the circle containing the intersection and subtracting the triangle that is not part of it.

The yellow part is part of the sector that is I think 90 degrees.

You take the area of a circle: 16pi, then find the sector: 4pi, and then subtract the triangle with is 8.  Then multiply by two to get the answer.

so 8pi-16 would be the yellow part

To get the black area...consider the square with vertices as the center of the four circles.  Then find the area of the 4 quarter circles and subtract the intersection.  Subtract that from the square and you get the answer...

I don't think the red and black one is possible with the information given. The size of the red areas could easily be drawn very small or very large.


Find the positive solution to $\frac 1{x^2-10x-29}+\frac 1{x^2-10x-45}=\frac 2{x^2-10x-69}.$

Hint: let y equal something in terms of x that may simplify the process of solving.

I assume the x2 is supposed to be x^2....

[hide]

the denominators of the fractions look really similar, so we'll let x^2-10x-29=y.  then the eqn. becomes 1/y+1/(y-16)=2/(y-40).  Then it's pretty easy to solve from there.[/hide]

sorry, copy and paste problems :P

4.  Find the positive solution to 1/(x^2- 10x- 29) + 1/(x^2- 10x- 45) = 2/(x^2- 10x- 69).

On my post, I could have let u = x:^2: - 10x - 29, and it would've been much easier.

yup, thats right... i also set y = x^2 - 10x but either way, its not too hard

its from the 1990 one... http://www.kalva.demon.co.uk/aime.html


I think this is a very hard problem.

Let $ a,b,c$ be given positive numbers and let $ E_{n}=a^{n}+b^{n}+c^{n}-\frac 1{a^{n}}-\frac 1{b^{n}}-\frac 1{c^{n}}$.
If $ E_{1}=0$, then $ E_{n}$ has the same sign for any positive integer $ n$.

I hope I am neither misunderstanding nor making mistake :) . Suppose $ E_{n}=a^{n}+b^{n}+c^{n}-\frac{1}{a^{n}}-\frac{1}{b^{n}}-\frac{1}{c^{n}}$. And suppose $ E_{1}=0$. Then $ E_{n}>0$ for $ n>1$ iff exactly one of $ a,b,c$ is greater than $ 1$.

Of course, $ E_{1}=0$. I corrected the typo.

Are you sure the signs are correct? Maybe the last two pluses should be minuses actually?  :maybe:

[quote="Valentin Vornicu"]Maybe the last two pluses should be minuses actually?  :maybe:[/quote]

Yes, Valentin. Now the statement is correct, according to Darkseer's remark. And his observation is valid. :lol:

Congradulate to your nice solution, hungkhtn :lol:

This is mine: Suppose $ a>1\ge b,c$. Then $ a-\frac{1}{a}=\frac{1}{b}-b+\frac{1}{c}-c$. We have $ a\ge\frac{1}{b},\frac{1}{c}\ge 1$. And $ a^{n}-\frac{1}{a^{n}}=(a-\frac{1}{a})f_{n}(a)$, where $ f_{n}(a)=a^{n-1}+a^{n-3}+...+a^{-n+1}$. Then $ E_{n}=(a-\frac{1}{a})f_{n}(a)+(b-\frac{1}{b})f_{n}(b)+(c-\frac{1}{c})f_{n}(c)$, but $ f_{n}(a)\ge f_{n}(b),f_{n}(c)>0$ since $ f$ is strictly increasing on $ [1,\infty)$ and $ f(b)=f(\frac{1}{b})$.

Very nice and ingenious your solution, Hungkthn and Darkseer. :lol:
I will posted a similar one. :D


geometry
3D geometry
algebra
polynomial
number theory unsolved
number theory

i cant seem to kill this problem. its driving me crazy¡¡¡¡¡¡¡¡¡¡

find all primes p such that 2p+6 is a perfect cube.

i think the only one is 29. i cant find a factorization to the equation¡¡¡¡ obviously no residues can be used. so how do i solve it????

so in light of these new primes and the fact that i haven´t been able to solve it it looks like there could be infinitly many primes with that property. i dont think there is a proof for only finitly many primes satisfying this property. what does everyone else think???? i´d really like to take this problem out of my mind :D

Are you absolutely positive that the problem was given in exactly this form at some contest? :? As far as I know, the only known results of the type "there are infinitely many primes of the form $P(n)$" where $P$ is some polynomial are for polynomials of degree $1$ (Dirichlet's theorem). In this case, we want the primes of the form $4k^3-3$. This looks a lot like an open problem to me.. :?

mmmmm if im not mistaken the problem said that there was only one prime such that 2p+6 is a perfect cube. but maybe there was a typo because for example the problem can be solved if instead of 6 we have 1 because in this case the problem is indeed trivial and our only prime is 13. so my guess is probably that there was a typo. yes when i was trying to solve the problem in the original form i arrived at the same thing you did grobber, fortunatly i was right in thinking that the problem was wrong. yes it does look like an open question.

Well TSTs ought not to be trivial... but it certainly looks quite hard if there's no typo.


n is a positive integer  $n \geq 2$ Determine the maximum value of the sum of the natural numbers $k_1,k_2,...,k_n$ satisfying the condition:

$k_1^3+....+k_n^3 \leq 7n$

This problem and some solutions were previously posted at
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13801[/url].


(a) Two positive integers are chosen. The sum is revealed to logician A, and the sum of squares is revealed to logician B. Both A and B are given this information and the information contained in this sentence. The conversation between A and B goes as follows : B starts:
B : " I can't tell what the two numbers are."
A : " I can't tell what the two numbers are."
B : " I can't tell what the two numbers are."
A : " I can't tell what the two numbers are."
B : " I can't tell what the two numbers are."
A : " I can't tell what the two numbers are."
B : " Now I can tell what the two numbers are."
What are the two numbers ?

(b) When B first says that he cannot tell what the two numbers are, A
receives a large amount of information. But when A first says that he cannot
tell what the two numbers are, B already knows that A cannot tell what the
two numbers are. What good does it do B to listen to A ?

P.S. I'm quite sure that it was discussed before but I cannot find it.


geometry
number theory
national olympiad

I think that many people have the book "problems from around the world" of about last 10 years.

But I want to know that which problems are easier.I mean the contest of which countries are easier than other countries usually.

for example the contests of Canada,France,Austria,India and some specific parts of other countries are easier than some countries like China,Russia etc

So i want to know from the experienced people about it.

[color=indigo][disclaimer: this topic is not for telling that the questions of some countries are childlike etc. but just to know ur opinion about the difficulty level so that the who wants to start solving this books can get a hint][/color]

Are there books "Math Olympiads from around the world" on the Internet? I know there are two of them, I think 1996. and 1997., but are there the others?

umm of cause not all questions from Olympiads are good evenly IMO, but well I'm fond of: geometry problems from Iran, USA, China, Romania, Bulgaria or France. For combinatorics Russia's problems are lovely and attractive  :D, beside it USA's problems are hard and they contain something new ideas and many questions from China, Bulgary, ... are hard. What about algebra? I prefer Romania' problems but problems from USA, Iran, Vietnam or China are also nice. And finally Number Theory I like problems from Romania, USA, China and Russia.  :)

[quote="Moonmathpi496"]
for example the contests of Canada,France,Austria,India and some specific parts of other countries are easier than some countries like China,Russia etc
[/quote]
I don't think so.It depends from your location.It is just my opinion,but i think that for man,who live for example in Kazakhstan,Ukraine or in Russia easier to solve problems from Russian,China etc.But much harder to solve problems from Romania,Austrian-Polish,Bulgarian and other european national olympiads.That's why we can't talking about the level of olympiad problems from around the world.  :)
Where i can download all books from series "Math Olympiads from around the world"? :wink:

maa 1996-2001
http://www.ajorza.org/paper/andreescu/  :lol:

Thank you,apollo,for your help :) Do you know where i can find other books from that serie(2001-2006)? :wink:

It is written not in English so I can't read it.
Who can help me?

@TTsphn :Some book are writen by France- i think so :D .

[quote="Amazigh"]None of the above posts gives an answer of the real question of this topic :([/quote]Ya...I asked this question almost 1 year ago and still it is not properly answered :(

Whatever, what I have seen during the last year is that, it is not that easy to say which on is easy...(because it depends on many things, like if you are good at geometry you'll find them easier, even you may find some strong (in IMO) countries problems easier than the weaker ones]

But, obviously there are something that we can assume, like the problem of China is difficult than of (almost) any other country. :roll:

Ehh...I am also spamming :P

I can't open http://www.ajorza.org.

Anybody knows why?

May be their site is down (Why these sites...like kalva are down these days :!:  :?: )

Instead you can use this link: [url]http://web.archive.org/web/20070821052815/http://www.ajorza.org/[/url]


Solve for $ x$:

$ \lfloor x\rfloor\plus{}\lfloor 5x\rfloor\plus{}\lfloor 10x\rfloor\plus{}\lfloor 20x\rfloor\equal{}36k\plus{}35$

where $ 'k'$ is an integer and $ \lfloor x\rfloor$ denotes the floor function.

I don't have an answer, but what does $ 'k'$ denote? Thnx.

There is no solution.
\begin{align*}
5\lfloor x\rfloor &\leq \lfloor 5x \rfloor \leq 5\lfloor x\rfloor + 4 \\
10\lfloor x\rfloor &\leq \lfloor 10x\rfloor \leq 10\lfloor x\rfloor + 9 \\
20\lfloor x\rfloor &\leq\lfloor 20x\rfloor \leq 20\lfloor x\rfloor + 19
\end{align*}
If $ S=\lfloor x\rfloor + \lfloor 5x \rfloor + \lfloor 10x\rfloor + \lfloor 20x \rfloor$, then we have \[ 36\lfloor x\rfloor \leq S \leq 36\lfloor x\rfloor+32,\] thus $ 0\leq S\leq 32\pmod{36}$ and hence $ 35\pmod{36}$ is not attainable.

[color=darkred]Edit: Added a lower bound to make solution more clear.[/color]

But k could be smaller than $ \lfloor{x}\rfloor$, couldn't it?

It doesn't matter though. Because $ k$ and $ \lfloor x \rfloor$ are both integers, $ 36\lfloor x\rfloor\leq S\leq 36 \lfloor x\rfloor \plus{} 32$ implies that $ S$ cannot be $ \equiv 33,34,35\pmod {36}$. But since $ 36k \plus{} 35$ is always $ 35\pmod {36}$, we have a contradiction.


Determine all positive integers $ n$, for  which $ 2^{n\minus{}1}n\plus{}1$ is a perfect square.

Let $ 2^{n\minus{}1}n\plus{}1\equal{}N^{2}$, then if n>2, $ 2^{n\minus{}3}n\equal{}(N\minus{}1)/2.(N\plus{}1)/2\equal{}k(k\plus{}1)$. Now, we've got to make the LHS a product of two consecutive integers and since both of these integers should have the opposite pairity, we should make $ 2^{n\minus{}3}d$ and $ n/d$ consecutive integers, where d divides n
But since $ 2^{n\minus{}3}>n\plus{}1$ holds for $ n>5$, we should have $ n<6$
Substituting for n values 1 through 5, we get that 5 is the only solution

the answer can only be 5..........lajanugen did it in a nice way

Let $2^{n-1}n+1=x^2$.Then $n\ge 3$ and $2^{n-1}n=(x+1)(x-1)$.Since $x+1,x-1$ are same parity, we can suppose that $x+1=2^as,x-1=2^bt$($a,b\ge 1$, s,t:odd, $a+b=n-1, st=n$).Since $x-1,x+1$ are distinct in mod 4,either $a$ or $b$ is $1$.

[b]Case1[/b]:a=1
$2=2(s-2^{n-3}t)\Leftrightarrow s-2^{n-3}t=1$.
Then $t(2^{n-3}t+1)=n$, and $n\ge 2^{n-3}$.So $n\le 5$.

[b]Case2[/b]:b=1
$2=2(2^{n-3}s-t)\Leftrightarrow 2^{n-3}s-t=1$.
Then $s(2^{n-3}s-1)=n$, and $n\ge 2^{n-3}-1$.So $n\le 5$.

From case1,2, $n\le 5$.We can easily examine all $n\le 5$ cases, and we find only solution $\boxed{n=5}$.$\blacksquare$

[color=#f00][b]Ans:[/b][/color] $n=5, 2^{5-1}\cdot 5+1=81=9^2$.
[i][color=#00f]Proof:[/color][/i] Suppose that we have $m\in \mathbb{N}$ such that \[2^{n-1}\cdot n+1=m^2 \iff 2^{n-1}\cdot n=(m-1)(m+1).\] Since we can easily see that $n=1$ yields no solution, we can just consider $n\geqslant 2,$ then the LHS is even which implies that $m$ must be odd. We now rewrite $m=2k+1$ for some $k\in \mathbb{N} \cup \{0\}.$ Then our equation rewrites into \[2^{n-3}\cdot n = k(k+1).\]
We now split into two cases:
[i][color=#ff9a00]Case 1:[/color][/i] $k$ is even and $k+1$ is odd.
This means that we can write $k=2^{n-3}\cdot a$ for some positive integer $a$ which is a divisor of $n$. So we have the following condition, \[2^{n-3}\cdot n=2^{n-3}\cdot a(2^{n-3}\cdot a+1)\geqslant 2^{n-3}(2^{n-3}+1)\] or $n\geqslant 2^{n-3}+1.$ However, it can be easily proven that $\forall n\geqslant 6,$ we have $2^{n-3}+1> n$ by a simple induction. So, we just have to check the cases when $n\leqslant 5.$
[i][color=#ff9a00]Case 2:[/color][/i] $k+1$ is even and $k$ is odd.
Similarly, we can write $k+1=2^{n-3}\cdot a$ for some positive integer $a$ which is a divisor of $n$. So we have the following condition, \[2^{n-3}\cdot n=2^{n-3}\cdot a(2^{n-3}\cdot a-1)\geqslant 2^{n-3}(2^{n-3}-1)\] or $n\geqslant 2^{n-3}-1.$ Again, we can also show that  $\forall n\geqslant 6,$ we have $2^{n-3}-1> n$ by a simple induction. So, we just have to check the cases when $n\leqslant 5.$
Checking all the possible values for $n\leqslant 5,$ we find that $n=5$ is indeed the only solution.


vector
linear algebra
matrix
function
linear algebra unsolved

Let $A_i (x_i,y_i)$ be four collinear points in the plane s.t. there are invertible matrices $M \in \mathcal M_4 \left( \mathbb C \right)$ in which the first two lines are $(x_1,x_2,x_3,x_4)$ and $(y_1,y_2,y_3,y_4)$. Prove that the sum of the elements of $M^{-1}$ does not depend on what $M$ we choose.

Hint 1
[hide]
Since points $A_i$ are collinear, and the vectors $(x_1, x_2, x_3, x_4)$ and $(y_1, y_2, y_3, y_4)$ are linearly independent, the vector $(1, 1, 1, 1)$ is a linear combination of vectors $(x_1, x_2, x_3, x_4)$ and $(y_1, y_2, y_3, y_4)$.
[/hide]

Hint 1 solution:
[hide]
Let
$X: = (x_1, x_2, x_3, x_4) = (a, a+ \alpha t_1, a+ \alpha t_2, a+ \alpha t_3)$,
$Y: =(y_1, y_2, y_3, y_4) = (b, b+ \beta t_1, b+ \beta t_2, b+ \beta t_3)$.
Since $X$ and $Y$ form two first rows of a non-singular matrix, they are linearly independet.

Then
\[ (\beta X - \alpha Y) = (\beta a - \alpha b, \beta a - \alpha b, \beta a - \alpha b, \beta a - \alpha b).  \]
This vector is not equal to $0$, because $X$ and $Y$ are linearly independent.
[/hide]

Hint 2
[hide]
If $M(a,b,c,d)^T = (k,l,m,n)^T$, then $a+b+c+d$ is a function of $k$ and $l$, which does not depend of the third and fourth rows of $M$.
[/hide]

Hint 3
[hide]
Consider the equation $M(a,b,c,d)^T = (1,1,1,1)^T$. The sum of all elements of $M^{-1}$ equals $a+b+c+d$.
[/hide]

The official solution is much like Yury's solution.

[hide="The official solution"]
Let $\displaystyle A_1, A_2, A_3, A_4$ be on the line $\displaystyle ax+by+c=0$. If $\displaystyle c=0$, then $\displaystyle M$ isn't invertible, contradiction. So $\displaystyle c \neq 0$.

Let $\displaystyle M^{-1} = \left( m_{ij} \right)_{1 \leq i,j \leq 4}$ and $\displaystyle I_4 = \left( a_{ij} \right)_{1 \leq i,j \leq n}$.

Then \[ \sum_{i=1}^4 x_k m_{ij} = a_{1j} , \, \sum_{i=1}^4 y_k m_{ij} = a_{2j} ,  \]
so \[ \sum_{i=1}^4 \left( a x_i + b y_i \right) m_{ij} = a \cdot a_{1j} + b \cdot a_{2j} , \]
i.e. \[ \sum_{i=1}^4 m_{ij} = - \frac{a}{c} a_{1j} - \frac{b}{c} a_{2j} .  \]
Finally, \[ \sum_{i,j} m_{ij} = - \frac{a}{c} \sum_{j=1}^4 a_{1j} - \frac{b}{c} \sum_{j=1}^4 a_{2j} = - \frac{a+b}{c} ,  \]
which indeed does not depend on $\displaystyle M$.
[/hide]


[quote="Vinxenz"]Prove that $ a^{in}+a^{jn}=c^{n}$ $ \forall a,c,i,j \in{N}$ with $ n$ even has no solution.[/quote]
What about $ a=c=0$!? Maybe did you mean "no [i]trivial[/i] solutions"? :maybe:

Obviously, it can't be $ n = 0$. So $ n \ge 2$. If $ i = j$, then $ 2a^{nj}= c^{n}$ and $ 1+nj \cdot v_{2}(a) = n\;\! v_{2}(c)$, viz $ 1 \equiv 0 \bmod n$ (absurd). Here as usual, $ v_{2}(\cdot)$ is a dyadic valuation. Then we lose no generality assuming $ j > i \ge 1$, thus getting
\[ a^{in}\cdot (1+a^{n(j-i)}) = c^{n}\quad\qquad (1) \]
But actually $ \gcd(a^{ni}, 1+a^{n(j-i)}) = 1$. Hence $ (1)$ is possible iff $ c = a^{i}\cdot b$, for some $ b \in \mathbb{N}= \{1, 2, \ldots, \}$. So we're reduced to prove there don't exists $ x,y \in \mathbb{N}$ and $ n,k \in \mathbb{N}$ such that $ n \ge 2$ and $ 1+x^{n}= y^{n}$, which is pretty obvious, because that $ y \ge x+1$ and consequently $ y^{n}-x^{n}= (y-x) \cdot\sum_{k=0}^{n-1}y^{n-1-k}\;\! x^{k}\ge y^{n-1}+x^{n-1}\ge 2$.

As a margin note, observe we don't really need $ n$ to be even, it's enough to suppose $ n \ne 1$. Moreover, we don't even need $ a, c \in \mathbb{N}$: by a minor effort, we can prove it's in fact sufficient $ a,c\in\mathbb{Z}$ and $ ac \ne 0$. :maybe:


What is the first moment between 4 and 4:30 that the minute hand and the hour hand will form a perfect right angle. (to the fractions of the minute)

Let $ x$, $ y$ be the position of hr hand and min hand, respectively, then
$ y \equal{} 12x mod12$
To satisfy the given condition, there must be
$ y \equal{} x \plus{} 3 mod12$ or $ y \equal{} x \minus{} 3 mod12$,
$ 11x \equal{} 3 mod12$ or $ 11x \equal{} 9 mod12$,
thus, $ x \equal{} (12k \plus{} 3)/11$ (1), or $ (12k \plus{} 9)/11$ (2), where $ k$ is a positive integer.
Since $ x$ must be in the interval $ (4, 9/2)$,
solving for the inequalities,
for (1), there's no integer $ k$.
for (2), $ k$ can only be $ 3$; substitute it into (2), $ x \equal{} 45/11$.
The answer is: $ 4: 60/11$.

Say $ t$ is the number of minutes after 4:00...

The hour hand at time $ t$ is located at $ 20 \plus{} \frac {t}{12}$

The minute hand at time $ t$ is located at...$ t$.

There are 5 degrees in one minute, so 90 degrees represents an 18 minute difference...

So, since the angle is formed between 4 and 4:30, the hour hand is 18 minutes ahead of the minute hand, giving us this equation:

$ t \plus{} 18 \equal{} 20 \plus{} \frac {t}{12}\Leftrightarrow$
$ \frac {11t}{12} \equal{} 2\Leftrightarrow$
$ t \equal{} \frac {24}{11}$

At the time of $ \frac {24}{11}$ minutes after 4, the minute and hour hand form a right angle. This answer is off though. Why?

And...could you explain what you're doing timwu? I don't see it...Specifically...
isn't $ 12x(\mod{12})\equiv0$ How does that help you?

[quote]There are 5 degrees in one minute, so 90 degrees represents an 18 minute difference... [/quote]
mm...here is where you go astray...60 minutes is equivalent to 360 degrees, so 90 is 15 minutes, not 18...

The method I used is some basic modulo, if you don't know what it is, keep in mind: if a=b mod n, then a and b has the same remainder when divided by n.
It's better to use the scale than degrees, 15 min is "3" on a clock. Because min hand is 12 times faster than hr hand, $ y \equal{} 12x mod12$ (when it turns 360 degrees, it goes back, that's why modulo is used). Similarly, to get a 90- degree angle, y should be either 3 greater or less than x modulo 12. Then the following is easy to see...

EDIT: for integer, your statement is true; but x, which means "hours" here, may not be integer, so it doesn't always be 0.

Ah, there are 6 degrees in a minute, not 5. My mistake.

Oh! I understand what you're doing now. I didn't know where you got the $ 3\mod12$ from, but now I realize it's just 15 (the right angle) expressed in terms of mod. The rest makes sense now :D

That's an interesting way of doing it.

Thank you :blush: This method is not my original, in fact, most clock problems in high school contests use this as a standard solution...


Could someone explain why: 1,1,1-trideuterio-2-methylcyclohex-1-ene doesn't show optical isomerism?
In general would a compound like: $ CH_{3}CXYCD_{3}$ show optical isomerism?
(The compound above isn't too complex, its (1,1,1-trideuterio) just a methyl group with the hydrogen atoms replaced by deuterium).

yes if $ \text{x}$ $ \text{y}$ are differnt

A molecule is chiral (which means it exhibits optical isomerism) if it cannot be superimposed on its mirror image, in some sense it means that the chiral centre (in this case the carbon atom of $ CH_{3}CXYCD_{3}$) must be attached to four different types of species.
So in general if $ X$ and $ Y$ are not the same species, then $ CH_{3}CXYCD_{3}$ is chiral.

Yes of course $ X$ and $ Y$ are different. But my question was really about $ CH_{3}$ and $ CD_{3}$ being different. The compound which (nobody seems to have noticed) has a plane of symmetry, even though a $ CH_{3}$ (methyl group) isn't exactly the same thing as a $ CD_{3}$ group. So my question was:  why aren't they considered different when attached to a carbon atom in general, and specifically in the doubt about the hexene ring in question (when determining the optical activity that is.)

are u sure about the name of the compound u posted :|

I think a figure should make it much more clear.

Will this compound show optical isomerism or not?

well the name is wrong it should be
$ \text{1\minus{}methyl(1',1',1' trideuterio)\minus{}2\minus{}methyl cyclohex\minus{}1\minus{}ene}$
also there is no chiral carbon atom.

[quote="IrinaSharova"] in some sense it means that the chiral centre [/quote]

Just a remark: for a molecule to be chiral it is not necessary for it to have a chiral center.

[quote="Cyprusad"]Could someone explain why: 1,1,1-trideuterio-2-methylcyclohex-1-ene doesn't show optical isomerism?[/quote]

That molecule doesn't show optical isomerism simply because its trans isomer would be too unstable. In general, six membered ring alkenes don't show cis-trans isomerism precisely because of the very high energy the trans isomer would have. Eight membered rings are the smallest ones for which trans isomers are stable enough to be isolated. However, if your compound were an open chain one then it could (theoretically) exist as cis and trans isomers.

Thanks again, [b]Carcul[/b], I just looked it up, I get it now  :)


I didn't want to post this in the "Announcements" section, so here it is: this whole thing about keeping flags doesn't work well at all. Some of the flags refuse to go away even after you've viewed the post several times, and that doesn't seem right. It wouldn't be bad at all if it worked as it should, but as it is, I think we were beter off without it.

Anyway, I think you can tell what my vote was on that poll :)..

yes grobber, unfortuantely, it doesn't work as on my local version of the site, most probably due to the different speed of the updates on the database, so either I will find a way around this, or just quit using unread flags.

but first I need the opinion of more people :?

well the vote is pretty clear

and i never saw a lot of people asking for it either :?

Just a quick remark and then I'm done about this: frankly, I don't think I remember the last time when a flag DID disappear, even though I've read some of the messages more than 10 times :). They just keep piling up..

ok, I got your point. I will remove the flag thiny untill I can test it more properly (I have to install locally the database, but I must do it in Linux, cuz in windows it sux)

btw, for Dust (in another post, which I can't remember at this moment) the server in obviously hosted on Linux :P ;)

Well, I don't really care about the flags, but can we not remove the "View posts since last visit" function?

hmmmz, didn't thought about that ... but I remember some people (Moubi) explicitely asking for that ... :?

The flags seem to be working fine for me at the moment.  I think they're useful because I only want to check the posts which I haven't seen before (so it's a tremendous bother when all the flags disappear all of a sudden, which happens sometimes even though I don't close all the windows).  I would really hate to see them taken away.

I guess it would be good if they could be set in the Preferences.

unfortunately, setting them in the preferences (newly Options menu -> to be easier to spot by the new users) menu is relatively trouble some, and (I belive) takes more time to load (the page).

I will try to fix the code, and if I fail I will remove it.

It seems to be that the "View post since left" function is doing something different than what it supposed to - instead of showing new posts, it's showing unread posts. Right now, my resort to that is to click "Mark all forums read" button every time I finish with "View post since left", so that "View post since left" would actually be meaningful.

So Valentin, would it be possible for you to place a "Mark all forums read" on the bottom of the search page? So that I don't have to travel all the way down the main page to reach "Mark all forums read".  :lol:

It's not about the flags, but still, in some topic there was an idea about another style for the site :D
Just an opinion

I'd say no to that. This is the main style for most sites, and it's great. :)

Considering the flag, it's always been a problem-mod, on any site. I suggest you just remove it until it works enough :P

Sorry if I explain wrong , but I mean whean Alice puts a number in square x, Bob puts the number in the other x, so that the arrangement of the numbers remains the same :P
Hope no it's clear

I mean another choice, not a change.
It's democracy :), on my browser, one style, on your's another :)


xkcd.com

ignore the alt-text; that's useless

the comic is exactly what I've been saying for years though

My, arnav's power has controlled xkcd now.


What gift would you give a math teacher?

:? rubik's cube... math book... pi t shirt...?

woman.

why would it say this:
The message is too small. Please make the message longer before submitting.

It was too small.
They thought it was spam.

Why are you getting the teacher a gift? Is it just because you had a great year together, or is it because she wrote you a recommendation that got you into (your hard-to-get-into school of choice)? Or maybe it's a goodbye or retirement thing?

Personally, if I were a teacher, I wouldn't like to get a Rubik's cube. If you don't know how to solve it, then what's the point? If you do, you'd probably already have a half dozen cubes at home.
I think a t-shirt is a great idea. I know people who have gotten their class to design the shirt and donate some money to have a professional make the shirts.


algebra
polynomial
calculus
linear algebra
matrix
function
Euler

I could work out the first few terms of X! manually, but I was certain that there must be some general algorithm/formula that gives the nth term.

Please post any suggestions.

Thanks

There is no such formula.  Stirling's approximation is about as good as it gets.

A possible solution is to use the so-called [b]Stirling Numbers [/b]  $ s(n,k) $  of [i]the first kind[/i]. These are defined as the coefficients from the development
\[
 (1) \; \; (x)^{[n]} = \sum\limits_{k=0}^{n}s(n,k)x^k \; ,\]
where $ (x)^{[0]}:=1 $ and  $ (x)^{[k]}=x(x-1)\cdots (x-k-1) $ when $k\ge 1.$ Let us note that the coefficients  $ S(n,k) $  from identity
\[
(2)\; \; \;  x^n =\sum\limits_{k=0}^n S(n,k)(x)^{[k]}
\]
are known as [b]Stirling numbers[/b] of [i] the second kind[/i].

There is a lot of references on Internet about these ,,Stirling numbers". Try Google !

Stirling numbers are important in Combinatorics,Statistics and Numerical Analysis.
In a forthcoming message I try to send you some source of referencecs. However,the Stirling numbers are uinvestigated in many book devoted to ,,Calculus of Differences". Try the monograph by Ch.Jordan published many years ago in Chelsea Publ.House.

Now , if you select  $ n:=N\; ,\; x=N $ , you find  from (1) the identity
\[
(*)   \; \; N! =\sum\limits_{k=0}^N s(N,k) N^k
\]
and
\[
(**) \; \; \; N^N =\sum\limits_{k=0}^N S(N,k)k!{N\choose k}\; .
\]
Perhaps your intest is  to use (*) . Regards, Alex

Regarding the [b]Stirling numbers of first kind[/b] $s(n,k) $ defined by means of identity
\[
x(x-1)\cdots(x-n-1)=\sum\limits_{k=0}^{n}s(n,k)x^k \; ,
\]
I list following questions :

[b]1).[/b] Particular cases :
\[ s(0,0)=1\; ,\; s(n,0)=0 \; \mbox{\rm for}\; n \ge 1\; ,\]
\[
s(n,1)=(-1)^{n-1}(n-1)!\; \; \; ,\; \; s(n,2)=(-1)^{n-2}(n-1)!\sum\limits_{k=1}^{n-1}\frac{1}{k}\; .\]
\[|s(n,k)|=(-1)^{n-k}s(n,k)\; .\]
[b]2).[/b]  Recurrences  :
\[ s(n,k)=s(n-1,k-1) -(n-1)s(n-1,k) ,
\]
\[
s(n,k)=\frac{(-1)^{n-k}}{2}\sum\limits_{i=k+1}^{n}{i-1\choose k-1}s(n,i)n^{i-k}\; .
\]
[b]3)[/b] Connection with ,,$r$-order" harmonic numbers : Let $H_n^{(r)}:=\sum\limits_{k=1}^n\frac{1}{k^r}\; . $ Then
\[
s(n,k)=(-1)^{n-k}\frac{(n-1)!}{(k-1)!}w(n,k-1)\]
where the $ w $-matrix $||w(n,j)||$  is defined recursively by
$ w(n,0):=1 $ and for $ m\ge 1 $
\[
w(n,m)=\sum\limits_{k=0}^{m-1}w(n,m-k-1)H_{n-1}^{(k+1)}\; . \]

Perhaps of interest are following :
        [b] R E F E R E N C E S[/b]
[1] M. Abramowitz, and I.A. Stegun,[b]  Handbook of Mathematical Functions with Formulas,
 Graphs and Mathematics Tables[/b], National Bureau of Standards, Applied Series 55, U.S. Government Printing Oce, Washington, D.C. (1964).
[2] V. Adamchik , [b]On Stirling numbers and Euler sums[/b] , Journal of Computational and Applied Mathematics 79 (1997) 119-130.
[3] D. Borwein, J. Borwein and R. Girgensohn,[i] Explicit evaluation of Euler sums[/i],
        Proc. Edinburgh Math. Sot. 38 (1995) 277-294.
[4] J.M. Borwein, D.M. Bradley and D.J. Broadhurst, [i]Evaluations of $k$-fold Euler/Zagier sums:
   a compendium of results for arbitrary $ k$ [/i], manuscript, 1996.
[5] P.L. Butzer and M. Hauss, [i]Stirling numbers of first and second kind[/i],
     in: Approximation, Interpolation and Summability, Israel Math Conf Proc.,  Vol. 4 (Weizmann Press, Israel, 1991) 89-108.
[6] P.L. Butzer and M. Hauss, [i]Riemann zeta function[/i], Appl. Math. Lett. 5 (1992) 83-88.
[7] R.Chelluri , L.B. Richmond and N.M. Temme, [i] Asymptotic Estimates for Generalized Stirling  Numbers[\i],  Centrum voor Wiskunde en Informatica , Report MAS-R9923 , Amsterdam , 1999 ,ISSN 1386-3703.
[8] L. Comtet,[i] Advanced Combinatorics[/i], (Reichel, Dordrecht, 1974).
[9] R.E. Crandall,[i]Fast evaluation of multiple zeta sums[/i], Exp. Math., to appear.
[10]  R.E. Crandall and J.P. Buhler,[i] On the evaluation of Euler sums[/i], Exp. Math. 3 (1994) 275-285.
[11] P. Erd\"os,[i]  On a Conjecture of Hammersley[/i],  J. Math. Soc., 28, 232-236, (1953).
[12] P.  Flajolet, and H. Prodinger, [i] On Stirling Numbers for Complex Arguments
and Hankel Contours[/i], SIAM J. of Discrete Math. 12, 155-159, (1999).
[13]  F. Flajolet and B. Salvy, [i]Euler sums and contour integral representations[/i],      Exp. Math., to appear.
[14] D. Gardy, [i]Some results on the asymptotic behaviour of coeffcients of large powers of     functions[/i], Discrete Math., 139, 189{217, (1995).
[15] R.L. Graham, D.E. Knuth and 0. Patashnik,[i]Concrete Mathematics[/i],          (Addison-Wesley, Reading, MA, 1989).
[16] E.R. Hansen, [b]A Table of Series and Products[/b],(Prentice-Hall, Englewood Cliffs, NJ, 1975).
[17] K.S. Kolbig,[i] Nielsens generalized polylogarithms[/i], SIAM J. Math. Anal. 17 (1986) 1232-1256.
[18] Y. Luke , [b]The Special Functions and their Approximations, Vol. 1,[/b] Academic Press , 1969.
[19] C. Markett, [i]Triple sums and the Riemann Zeta function[/i],   J. Number Theory 48 (1994) 113-132.
[20] L.Moser, and M. Wyman,[i] Asymptotic development of the Stirling numbers of the   first kind[\i], J. London Math. Soc. 33, 133-146, (1958).
[21] L.Moser, and M.Wyman,[i] Stirling numbers of the second kind[/i], Duke Math. J. 25, 29-43,   (1958).
[22] N. Nielsen ,[b]Trait\'e  \'el\'ementaires des Nombres de Bernoulli[/b],Gauthier-Villars,   Paris,1923
[23] A.M. Odlyzko,[i] Asymptotic methods in enumeration[/i], in Handbook of  Combinatorics, Vol. 2,   R. Graham, M. Gr\"otschel and L. Lovasz, Eds., Elsevier  (1995).
[24] F.W.J. Olver,[i] Asymptotics and Special Functions[/i], A. K. Peters (1997).
[25] A.P. Prudnikov, Yu.A. Brychkov and 0. Marichev, [b] Integrals and Series, Vol. 3: More Special Functions.[/b]   (Gordon and Breach, New York, 1990).
[26] B. Richmond and D. Merlini,[i] Stirling numbers for complex arguments[/i], SIAM J. of Discrete  Math., 10, 73-82, (1997).
[27] N.M. Temme, [b] Asymptotic estimates of Stirling numbers[/b],Studies in Applied Math. 89, 233-243, (1993).
[28] D. Zagier,[i] Values of zeta functions and their applications[/i], preprint, Max-Planck, 1994.

I found very usefull the works: [1],[2],[16],[18],[22],[25],[27].
 By the way , [25] is a very good book, but very expensive.
 In 1992, in Germany, the total cost of all three volumes was
 approx. 1500 DM (=approx.= 750 Euro !).
Also [16] is very interesting but not easy to find it !

As an  exercise, try to prove/or disprove/ the equality including
Stirling numbers of first kind $\; ||s(n,k)|| $ :
\[
\sum\limits_{k=0}^{n}\arctan{k}=\arctan{
\frac{-s(n,n-1)+s(n,n-3)-s(n,n-5)+\ldots}
{s(n,n)-s(n,n-2)+s(n,n-4)-\ldots}} \; .
\]


limit
function
calculus
derivative
trigonometry
real analysis
real analysis unsolved

Let $ f: [0,\infty) \minus{} > R$ be $ f(x) \equal{} e^{\sqrt {x}} \plus{} e^{ \minus{} \sqrt {x}}$.
Find $ \lim_{n \to \infty} \lim_{x \to 0, x > 0} f^{(n)}(x)$.

I'm sure I'm missing something, but...

you're looking for $ \lim_{n\to\infty, x\to 0^{\plus{}}} f^{(n)}(x)$.

But your function is $ f(x) \equal{} 2\cosh x$. It's derivative is $ 2\sinh x$, and the derivative of $ 2\sinh x$ is $ 2 \cosh x$. So you get into the cycle, for odd and even $ n$. As these functions are defined and continuous in $ x \equal{} 0$, we may take the value of $ f^{(n)}(0)$ as the limit. But it's sometimes 1, and sometimes 0. (just like in "ordinary" sines and cosines). Where's my mistake?

Sry, I forgot to put the $ \sqrt{}$ :oops: I've edited the post


What is the maximum number of points with integer coordinates which can lie on a circle with center at
$ (\sqrt{2},\sqrt{3})$

it's not hard to show that if $ a, b, c, d, e, f$ are integers such that $ a \plus{} b\sqrt {2} \plus{} c\sqrt {3} \equal{} d \plus{} e\sqrt {2} \plus{} f\sqrt {3}$, then $ a \equal{} d, b \equal{} e, c \equal{} f$.

the maximum number of lattice points on a circle centered at $ (\sqrt {2},\sqrt {3})$ is one. if $ (x,y), (u,v)$ are two lattice points on the circle, then by the above fact, $ (x \minus{} \sqrt {2})^2 \plus{} (y \minus{} \sqrt {3})^2 \equal{} (u \minus{} \sqrt {2})^2 \plus{} (v \minus{} \sqrt {3})^2\Rightarrow x^2 \plus{} y^2 \minus{} 2\sqrt {2}x \minus{} 2y\sqrt {3} \equal{} u^2 \plus{} v^2 \minus{} 2u\sqrt {2} \minus{} 2v\sqrt {3}\Rightarrow x \equal{} u, y \equal{} v$

extension: replace $ (\sqrt{2}, \sqrt{3})$ with $ (\sqrt{m}, \sqrt{n})$


inequalities
Cauchy Inequality
inequalities proposed

[quote="Ji Chen"]If $x,y,z$ be nonnegative real numbers such that $x+y+z=1$, then

$y\sqrt{2x+y}+z\sqrt{2y+z}+x\sqrt{2z+x}\leq1$,

with equality if $x=1,y=z=0$.[/quote]
Jensen.

I square and use AM-GM once.
It would have been a nice problem for a JBMO 2007, don`t you think?

Yes, arqady and andyciup.

[quote="spider_boy"]If your aim is to win an olympiad medal, you can solve the IMO problems just using simple (standard) inequalities (AM-GM, Holder, Jensen, $\ldots$) but applied very ingeniously.

I haven't said that IMO problems are easy. Of course they require very deep analysis, to solve them you need to train for years.[/quote]
Using the Cauchy inequality, we have

$(x+y+z)^{3}$

$=(y+z+x)[y(2x+y)+z(2y+z)+x(2z+x)]$

$\geq(y\sqrt{2x+y}+z\sqrt{2y+z}+x\sqrt{2z+x})^{2}$.


The one that I can think of right now is when I walked into the girl's bathroom in 4th grade and actually went to potty. and then.. yeah..

another one is when I walked into my class and sat on a grape. and when i stood up... you know what happened then

or... or...  :lol:

lol, is this from the post Timecat8 made on my blog? :)

wow, I hope no one was in there at that time... :roll:


Δεν ξερω αν η ασκηση εχει ξαναμπει αλλα ελπιζω να σας αρεσει.

    Εστω τριγωνο  $ ABC$  εγγεγραμμενο σε κυκλο και εστω $ AD$  το υψος του.Εστω τυχαια ευθεια παραλληλη προς την $ BC$  που οριζει στην περιγεγραμμενη περιφερεια του τριγωνου τα σημεια   P,S.

    Δειξτε οτι οι ευθειες  Simson που αντιστοιχουν στα P,S  τεμνονται πανω στο υψος $ AD$.

Είναι γνωστό ότι τα συμμετρικά σημείου Ρ  του κύκλου ΑΒΓ ως προς τις πλευρές του τριγώνου ΑΒΓ βρίσκονται πάνω σε ευθεία στην οποία ανήκει και το Η (ευθεία Steiner). H Simpson του Ρ είναι // προς την Steiner του Ρ  ( ως ευθεία που ενώνει τα μεσα πλευρών τριγώνου ΡΕΖ , όπου Ε,Ζ τα συμμετρικά του Ρ ως προς ΑΒ,ΑΓ )
Αν Κ η προβολή του Ρ στην ΒΓ και Λ πάνω στην ΕΖ ώστε Ρ,Κ,Λ συνευθειακά τότε Κ είναι πάνω στην ΒΓ,  ΚΛ=ΚΡ και αν το υψος ΑΔ τεμνει την  Simpson του Ρ στο Μ το ΗΜΚΛ # οπότε ΗΜ=ΡΚ
Επειδή ομως ΡΣ//ΒΓ τότε ΡΚ=ΣΤ όπου Τ η προβολή του Σ στην ΒΓ και ΣΤ=ΗΜ άρα και η  Simpson του Σ θα περάσει από το Μ σημείο του ύψους ΑΔ


1.)
2+2=4

some say it equals 5. those who think that 2+2=5 are WRONG

2.)
DHMO=water

that is not just an opinion, it is a fact. sorry, but nobody shall ban dhmo

3.)
Unlike what AIME15 thought, i have no intention of sending a virus via a ".exe" file. do not be afraid if i happen to attach a ".exe" file

If anyone objects to these contreversial facts, feel free to state your reason here. I posted this after getting tired of seeing these facts still contreversial.

This does not belong here.
Where? I don't know.
I'm not familiar with the forums.
I thinks it goes somewhere in "mathematics"
FF is for FUN and frivolous discussions

fine. moderators, please move this to the appropriate math forum


quadratics
number theory unsolved
number theory

Is this ture?
P is a prime number so that 3 is a divisior of p-1.and suppose k an integer.prove that there are exactly 2 integers like x so that :p is a divisior of x^2+ax+a^2.
Thank you. :?:

Well, there are infinetely many such integers...
I think you consider them $\mod p$.
Then it's true (for $p \nmid a$):
Reformulate it as $\left( 2x+a \right)^2 \equiv -3a^2 \mod p$. This can be solved iff $-3$ is a square $\mod p$, which using quadratic reciprocity reduces to $p \equiv 1\mod 3$. It's also easy that there are two solutions.


geometry
geometric transformation
reflection
\/closed

Daylight Savings Time begins today.  To change your timezone setting to reflect DST, click Profile, then Preferences, then Internationalisation [sic], then set Summertime to yes.

[quote="Bicameral"]Just random thinking:  I anticipate parents not noticing and sending students to school at 1 hour after the start of school.[/quote]

Yeah, that would be bad.

What about the start of standard time again? Sending students to school one hour early? Bad.

[quote="Bicameral"]Just random thinking:  I anticipate parents not noticing and sending students to school at 1 hour after the start of school.[/quote]

Today is Sunday...

[quote="davidyko"]*cough* Chinese school *cough*
There always used to be at least one person who failed to notice. Every year.[/quote]

Hah. That happened to us today. To three people... :wink:

lol there was one person at ours too.
It was last year, though, b/c our class "graduated"....
I miss Chinese school.

Eh, I missed Chinese school today cause of DST.

lol I was late to chinese class, I opened up my computer, and I noticed the time was 1 hour ahead and then I remembered...

A few years ago my teacher forgot about it. She turned up an hour late and in the mean time the entire class messed around XD.

This year though nobody forgot.

lol I was also late 1 hour to school today


Can anyone give a counter example to the following:
If $ D$ is a decomposition of $ X$ into homeomorphic sets, say all homeomorphic to $ Y$, then $ X$ is homeomorphic to $ D \times Y$.

The real line can be decomposed into intervals $ [n,n\plus{}1)$, all of which are homeomorphic. This would mean $ \mathbb{R} \cong \mathbb{Z} \times [0,1)$, but the latter
space is not connected, and $ \mathbb{R}$ is.


[quote="Naphthalin"]yeah i did it - i qualified for my 2nd (and last :( ) IMO for the german team. the others will be posted soon anywhere here.

Naphthalin[/quote]

German IMO Team for 49th IMO in Spain given in format: first name, last name (age, school grade), city, state

Jessica Fintzen (19 years, 13. grade), Quickborn, Schleswig-Holstein
Malte Lackmann (17 years, 12. grade), Bordesholm, Schleswig-Holstein
Florentin Münch (18 years, 12. grade), Jena, Thüringen
Lisa Sauermann, (15 years, 9. grade), Dresden, Sachsen
Georg Schröter, (18 years, 12. grade), Dresden, Sachsen
Philipp Weiß, (17 years, 12. grade), Hoyerswerda, Sachsen

For your interest German team information for the 39th IPhO given in format: first name, last name, school name/city

1. Martin Lüders, Max-Steenbeck-Gymnasium Cottbus
2. Georg Schröter, St.-Benno-Gymnasium Dresden
3. Chang Liu, Gymnasium Ehingen
4. Pavel Zorin, Gymnasium Hechingen
5. Jessica Fintzen, Elsensee-Gymnasium Quickborn

Though both competitions overlap this year two German students will compete in both events: Georg and Jessica.

[quote="darij grinberg"][quote="Omid Hatami"]I think he goes to university right now.[/quote]

I think you misunderstood Orl, as he explicitely said Gri[b]m[/b]berg. And to answer his question: there is no Darij Grimberg anymore. Sadly, with the departure of people like NoMane, ZeroX or our beloved Qwerewolf from the IMO training, an old tradition has been broken.

  darij[/quote]
 :lol:

that isn't fair - you didn't give me the chance to post the REAL german team - there had been big surprises again ;)

dg

for the people who are interested - you find the evaluations for the german TSTs at [url]http://wmls.no-ip.org/aimo[/url]. there was a final exam between thomas weiß and philipp weiß to decide the 6th team member, philipp gained 8 points, tom 1.
the TST's consist of several ISL problems which schouldn't be posted too soon here...

Naphthalin

[quote="Naphthalin"]
the TST's consist of several ISL problems which schouldn't be posted too soon here...

Naphthalin[/quote]
OK,but i guess not all the problems was from ISL :) Please post the others. :wink:
And also some problems from ISL was already officially published.So maybe we can post all the problems from Germany TST?

i don't think so. first, we don't know which problems are from ISL, and if the problem is already here, there is no need to post it. wait for the official release of the ISL please...

Naphthalin

[quote="kdano"]Only six problems have been officially published from this ISL:
Last year's IMO-problems....[/quote]

kdano, can you please post the Hungarian TST?
I have heard that you had one extremely difficult problem that was only solved by a single student.

[quote="kdano"][quote="Erken"]And also some problems from ISL was already officially published.[/quote]

Only six problems have been officially published from this ISL:
Last year's IMO-problems....[/quote]
I can't agree with you.I know that there are at least three problems,other than IMO problems,that were officially published,and two of them here at Mathlinks,one of them at Mathlinks contest and another one(i forgot the link).Also one or two problems from ISL was published at one book.And this book sold at BMO.

[quote="primoz2"]Could you please post this years German TST?[/quote]

Naphthalin, could you please post the German TST from last year?
The resources section contains the TSTs from 2004, 2005, and 2006,
but the problems from 2007 are missing:

http://www.mathlinks.ro/Forum/resources.php?c=64&cid=54

[quote="Naphthalin"]the TST's consist of several ISL problems which schouldn't be posted too soon here...[/quote]

I think it is okay to post these problems now.  :)

German results: http://www.imo-official.org/team_r.aspx?code=GER&year=2008

Lisa Sauermann, 35 points, gold medal
Georg Schröter, 28 points, silver medal
Malte Lackmann, 24 points, silver medal
Jessica Fintzen, 19 points, bronze medal
Florentin Münch, 18 points, bronze medal
Philipp Weiß, 15 points, bronze medal

I'm looking for the problems of the "german tst 2007" for a few days , but cannot find them in the web...
So does anybody know, where I can find them, or would anybody be so kind and post them here?

Thanks, Felix

[quote="Naphthalin"]give me some time and i will do it^^

Naphthalin[/quote]

This guy is too lazy to post them...

So, Naphthalin, please post the problems. :wink:


geometry
3D geometry
tetrahedron
trigonometry
parallelogram
trig identities
Law of Cosines

Let $ABCD$ be a tetrahedron with $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, and $CD=13$, as shown in the figure. Let $d$ be the distance between the midpoints of edges $AB$ and $CD$. Find $d^{2}$.

[asy]
pair C=origin, D=(4,11), A=(8,-5), B=(16,0);
draw(A--B--C--D--B^^D--A--C);
draw(midpoint(A--B)--midpoint(C--D), dashed);
label("27", B--D, NE);
label("41", A--B, SE);
label("7", A--C, SW);
label("$d$", midpoint(A--B)--midpoint(C--D), NE);
label("18", (7,8), SW);
label("13", (3,9), SW);
pair point=(7,0);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));[/asy]

[quote="4everwise"]Sweet, check out that image.  :jump:[/quote]

lol... I have to admit, it is pretty awesome :D

[hide="solution"]let x be the median from A to BC, and y be the median from B to CD, using the median formulas,
$2(7^2+18^2)=4x^2+13^2$
$2(27^2+36^2)=4y^2+13^2$
$2(x^2+y^2)=41^2+4d^2=7^2+18^2+27^2+36^2-13^2$
$d^2=137$, anybody have a good way to calculate that besides multiplying it out[/hide]

[quote="eryaman"][quote="4everwise"]Let $d$ be the distance between the midpoints of edges $AC$ and $CD$. Find $d^2$.[/quote]Wait...do you mean $AB$?[/quote]Of course.  :)

I expanded a bit on Altheman's solution, since mine was pretty much the same.

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=644[/img]

Let's find a formula for calculating median lengths given the sides. In the diagram above, let $BC=a$, $AC=b$, $AB=c$, and $AM=d$. The law of cosines yields two equations: \[ \begin{array}{l} \frac 14a^2+d^2-ad\cos \theta =c^2 \\ \frac 14a^2+d^2+ad\cos \theta =c^2\\ \end{array} .  \]Adding and simplifying, $4d^2=2b^2+2c^2-a^2$.

Now, let $x$ be the median from $A$ to $CD$, and let $y$ be the median from $B$ to $CD$. Using our newly derived median formulas, $2(7^2+18^2)-13^2=4x^2$ and $2(27^2+36^2)-13^2=4y^2$.

Applying our formula once more, $2(x^2+y^2)=41^2+4d^2=7^2+18^2+27^2+36^2-13^2$and $d^2=\boxed{137}$. [/hide]

Edit: Woah, this solution looks nice. It's all spaced out nicely and stuff. :)

[quote="eryaman"]Hmm...isn't the "median formula" basically Stewart's theorem?[/quote]Yep, except medians are more convenient. :)

[quote="Zellex"]Uh, what's the median formula?[/quote]Refer to my solution. :)

A better way to derive the median formula:

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=649[/img]

Consider $\triangle ABC$ with median $AE$. Extend $AE$ to $D$ so that $AE=ED$. It is easy to prove that $ABCD$ is a parallelogram, and it is also well-known that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. Thus $AD^2+BC^2=2AB^2+2AC^2 \Longleftrightarrow AD^2=2AB^2+AC^2-BC^2$. Hence, $AE^2=\frac{2AB^2+AC^2-BC^2}{4}$.


calculus
integration
complex analysis
complex analysis unsolved

Hi all

I am trying to solve a certain contour integration problem and I have reduced to showing the following:

Let $ N$ grow arbitrarily large, and let $ g(z) \equal{} z^{ \minus{} k} e^{ \minus{} 2\pi i z}$ with $ k \geq 2$. I want to show that for any fixed, strictly negative $ y$, the integral
\[ \int_{N \plus{} iy}^{ \minus{} N \plus{} iy} g(z) dz\]
goes to zero as $ N$ goes to $ \infty$ (Where z = x+iy and we are integrating along the straight line). Any help would be appreciated.

That looks like a Fourier Transform.  Until someone comes up with a better way, we could possibly start out this way:  Consider the integral with respect to x and I won't change it's magnitude if I go from $ - \infty$ to $ \infty$.  Then for fixed y, I get:
\begin{align*}\int_{ - \infty}^{\infty}\frac {e^{ - 2\pi i(x + iy)}}{(x + iy)^k}dx & = e^{2\pi y}\int_{ - \infty}^{\infty}\frac {1}{(x + iy)^k} e^{ - 2\pi i x}dx \\
& = e^{2\pi y}\int_{ - \infty}^{\infty}\frac {1}{(x + iy)^k}e^{ - 2\pi i x z}dx\Biggr|_{z = 1} \\
& = e^{2\pi y}\, \mathfrak{F}\left\{\frac {1}{(x + iy)^k}\right\}\Biggr|_{z = 1} \end{align*}
Now, at this point, I'll appeal to Mathematica for starters until I could show the results below directly or until we can come up with a better way.  But Mathematica returns for positive integer k, an expression multiplied by the factor:
\[ z (\text{Sign}[z] ( - 1 + \text{Sign}[\text{Abs}[\text{Re}[y]]]) - 2 \text{HeavisideTheta}[2 \pi z \text{Sign}[\text{Re}[y]]] \text{Sign}[\text{Re}[y]])\]
And that expression is zero for z=1 and negative y.


function
complex analysis
topology
complex analysis theorems

we were trying to prove that meromorphic functions fin the extended complex plane are the rational functions. the prof just sketched a proof.

he said that we could consider $g(z)=f(\frac{1}{z})$ and  $g$ has either a pole at 0 or a removable singularity there and anyways it has to be holomorphic in a deleted neighbourhood of 0. he said thats why f can have only finitely many poles in the plane. I didnt get this part ( why finite?) . and then he said near each pole we could write f as a sum of the principal part and a holomrphic part, and consider $F= f-f_{1}-f_{2}-...-f_{k}$ where the $f_{k}'s$ are the principal parts. and then we can show that this is entire and finish off from there. can someone explain why we should have finitely poles. i dint understand. thanks.

common people, this is already my 2nd post . i am sure there are millions of ppl who know this. just give me a hint or something  :|

Hint: First, we know the extended plane is a compact space, therefore infinite sets have accumulation points. Now suppose that in the extended plane f(z) has infinitely many poles. Let p be the accumulation point for the set of poles. Consider the function 1/f(z) in a certain open set of p.

Here's another way to do it, which uses a more naive sense of compactness (but the same idea). Consider f(z) on the set |z|<=1 f(1/z) on the set |1/z|<=1.  These are both compact sets and determine f on the extended complex plane.  You can use the notion of f being expressable as a Laurent series to prove that there's always an epsilon ball around any pole (which shows that there are a finite number of poles on a compact subset of the plane) and you're done since f and f(1/z) are both meromorphic (in fact, the more interesting conclusion is that the sum of the residues on the complex plane has to equal the residue at the completion of the complex plane).

[quote="gka"]Consider f(z) on the set |z|<=1 f(1/z) on the set |1/z|<=1. .[/quote]

Do you mean:

Consider f(z) on the set |z|<=1/f(1/z) on the set |1/z|<=1?


On the sides $AB,BC,CD,DA$ of quadrilateral $ABCD$ are chosen points $K,L,M,N$ such that $KL\parallel MN$ and $KM$ is perpendicular to $NL$. Prove that $KM$ intersects $NL$ on $BD$.

I think something is wrong with the problem. In my sketch it doesn't seem true.

I think $ ABCD$ should be a rectangle  :wink:


algebra
system of equations
algebra proposed

hello, let $ \sqrt{3x}\equal{}a$ and $ \sqrt{7y}\equal{}b$, then we have the equations
(1)$ 1\plus{}\frac{1}{\frac{a^2}{3}\plus{}\frac{b^2}{7}}\equal{}2$
(2)$ 1\minus{}\frac{1}{\frac{a^2}{3}\plus{}\frac{b^2}{7}}\equal{}4\sqrt{2}$.
Adding both equations we have $ 1\equal{}\frac{1}{a}\plus{}\frac{2\sqrt{2}}{b}$. From here we have $ a\equal{}\minus{}\frac{b}{2\sqrt{2}\minus{}b}$.
By substituting this in the equation (2) and factoring we get
$ \minus{}\frac{(\minus{}3b^2\plus{}12b\sqrt{2}\minus{}28)(\minus{}b^2\plus{}4b\sqrt{2}\plus{}6)}{(\minus{}31\minus{}3b^2\plus{}12b\sqrt{2})b}\equal{}0$. It follows $ b\equal{}\sqrt{2}(2\plus{}\sqrt{7})$. By backsubstituting we  compute $ x\equal{}\frac{11}{21}\plus{}\frac{4}{21}\sqrt{7}$ and $ y\equal{}\frac{22}{7}\plus{}\frac{8}{7}\sqrt{7}$ .This is the solution of our system.
Sonnhard.

$ x \plus{} y \equal{} \frac 1m$
$ x \minus{} y \equal{} \frac 1n$

So we have
$ (\frac 1m \plus{} \frac 1n)(1 \plus{} m)^2 \equal{} \frac 43$
$ (\frac 1m \minus{} \frac 1n)(1 \minus{} m)^2 \equal{} \frac {32}{7}$

$ \frac 2m \equal{} \frac {4}{3(1 \plus{} m)^2} \plus{} \frac {32}{7(1 \minus{} m)^2}$

expanding gives
$ 21m^4 \minus{} 62m^3 \minus{} 110m^2 \minus{} 62m \plus{} 21 \equal{} 0$

$ m \plus{} \frac 1m \equal{} t$ gives

$ t^2 \minus{} 31t \minus{} 57 \equal{} 0$

[quote="mr.danh"]Solve the system of equations:
$ \sqrt {3x}(1 \plus{} \frac {1}{x \plus{} y}) \equal{} 2$
$ \sqrt {7y}(1 \minus{} \frac {1}{x \plus{} y}) \equal{} 4\sqrt {2}$[/quote]
let $ a\equal{}\sqrt x,b\equal{}\sqrt y$I found
$ x\equal{}\frac{1}{\sqrt 3}\plus{}\frac{2}{\sqrt 21},y\equal{}2\frac{\sqrt 2}{\sqrt 7}\plus{}2$
and 3 pairs $ x,y$which instead $ \plus{}$ ,we let $ \minus{}$
for example $ x\equal{}\frac{1}{\sqrt 3}\minus{}\frac{2}{\sqrt 21},y\equal{}2\frac{\sqrt 2}{\sqrt 7}\minus{}2$...

Since I saw no one provided a complete solution to this problem I decided to post my solution:
[hide="Solution"]
After squaring $(1)$ and $(2)$ and rearranging we have:
$1+\frac{1}{(x+y)^2}=\frac{4}{3x}-\frac{2}{x+y}(3)$
$1+\frac{1}{(x+y)^2}=\frac{32}{7y}+\frac{2}{x+y}(4)$
Now by equating we have:
$\frac{4}{3x}-\frac{32}{7y}=\frac{4}{x+y}$
$\Longleftrightarrow 28y^2-96x^2-152xy=0$
$\Longleftrightarrow (6x-y)(4x+7y)=0$
[hide="Case 1"]
$6x=y$
Plugging this into $(1)$,squaring and after simplifying  we get:
$49y^2-308y+36=0$.
From this equation we get $2$ solutions but only $y_1=\frac{2}{7}(11+4\sqrt{7})$ satisfies the given system of equations.
So we have the first solution:
$y_1=\frac{2}{7}(11+4 \sqrt{7}),x_1=\frac{1}{21}(11+4 \sqrt{7})$
$====================$End of case 1$======================$
[/hide]
[hide="Case 2"]
$y=\frac{-4}{7}x$
Plugging this into $(2)$,squaring and after simplifying we get:
$9x^2+30x+49=0$
From here we get 2 complex solutions which indeed satisfies the given system of equations.
$x_2=\frac{1}{3}(-5-2\sqrt{6}i)$,$y_2=\frac{-4}{21}(-5-2\sqrt{6}i)$
$x_3=\frac{1}{3}(-5+2\sqrt{6}i)$,$y_3=\frac{-4}{21}(-5+2\sqrt{6}i)$
$===================$End of case 2$========================$
[/hide]
Hence we have:
$x_1=\frac{1}{21}(11+4 \sqrt{7})$,$y_1=\frac{2}{7}(11+4 \sqrt{7})$
$x_2=\frac{1}{3}(-5-2\sqrt{6}i)$,$y_2=\frac{-4}{21}(-5-2\sqrt{6}i)$
$x_3=\frac{1}{3}(-5+2\sqrt{6}i)$,$y_3=\frac{-4}{21}(-5+2\sqrt{6}i)$

[/hide]

In the version I have $x$ and $y$ are given as positive real numbers. Here is a solution there:

Let $x = a^2$ and $y = b^2$. The idea is to add $\frac{1}{\sqrt3}$ times the first equation to $\frac{1}{\sqrt7} i$ times the second equation to get \[ (a + bi) + \frac{a-bi}{a^2+b^2} = \frac{2}{\sqrt3} + \frac{4\sqrt2}{\sqrt7} i. \] Letting $z = a+bi$ be a complex number, this equation rewrites as \[ z + \frac 1z = \frac{2}{\sqrt3} + \frac{4\sqrt2}{\sqrt7} i. \] We may now solve directly to get \[ z = \left( \frac{1}{\sqrt3} \pm \frac{2}{\sqrt{21}} \right) + \left( \frac{2\sqrt2}{\sqrt7} \pm \sqrt 2 \right) i. \] Thus, we have \[ x = \left( \frac{1}{\sqrt3}\pm\frac{2}{\sqrt{21}} \right)^2, \qquad y = \left( \frac{2\sqrt2}{\sqrt7} \pm \sqrt 2 \right)^2.  \] Plugging back in, we find only the $+$ solution works. Hence the final answer is only one pair: \begin{align*} x &= \left( \frac{1}{\sqrt3} + \frac{2}{\sqrt{21}} \right)^2, = \frac{11+4\sqrt7}{21} \\ y &= \left( \frac{2\sqrt2}{\sqrt7} + \sqrt 2 \right)^2 = \frac{22+8\sqrt7}{7}. \end{align*}

Technically, there are three solutions, $x=\left(\frac{11+4\sqrt{7}}{21},\frac{22+8\sqrt{7}}{7}\right)$ above, and $\left(\frac{-5\pm 2i\sqrt{6}}{3},\frac{20\mp 8i\sqrt{6}}{21}\right)$.

The latter can be verified by using the fact that the principal branch of the square root is $\sqrt{z}=\sqrt{\frac{|z|+\Re(z)}{2}}+i\operatorname{sgn}(\Im(z))\sqrt{\frac{|z|-\Re(z)}{2}}$.

Verification:

$\sqrt{3x}=\sqrt{-5\pm 2i\sqrt{6}}=1\pm i\sqrt{6}$, $\sqrt{7y}=\sqrt{\frac{20\mp 8i\sqrt{6}}{3}}=2\sqrt{2}\mp\frac{2\sqrt{3}}{3}i$, and $\frac{1}{x+y}=\frac{1}{\frac{-5\pm 2i\sqrt{6}}{3}+\frac{20\mp 8i\sqrt{6}}{21}}=\frac{7}{-5\pm 2i\sqrt{6}}=\frac{-5\mp 2i\sqrt{6}}{7}$.

$\sqrt{3x}\left(1+\frac{1}{x+y}\right)=\left(1\pm i\sqrt{6}\right)\left(\frac{2\mp 2i\sqrt{6}}{7}\right)=\frac{2}{7}|1\pm i\sqrt{6}|=\frac{2}{7}\cdot 7=2$
$\sqrt{7y}\left(1-\frac{1}{x+y}\right)=\left(2\sqrt{2}\mp\frac{2\sqrt{3}}{3}i\right)\left(\frac{12\pm 2i\sqrt{6}}{7}\right)=\frac{\sqrt{2}}{3}\cdot\frac{2}{7}|6\mp i\sqrt{6}|=\frac{\sqrt{2}}{3}\cdot\frac{2}{7}\cdot 42=4\sqrt{2}$


Let M be the intersection point of the diagonals AC
and BD of a convex quadrilateral ABCD. The bisector of angle ACD
hits ray BA at K. If MA ·MC +MA · CD = MB ·MD, prove that
BKC = CDB.

REALLY a GOOOOOOOOD one ! I enjoy it ! :D  Here is my proof: extend $ {AC}$ to $ {Z}$ such that :$ {CZ\equal{}CD}$; with $ {MA *MC \plus{}MA * CD \equal{} MB*MD}$ ; We easily get :$ {MA*(MC\plus{}CZ)\equal{}MB*MD}$ $ {MA*MZ\equal{}MB*MD}$ so,obviously :$ {A,B,Z,D}$ are concyclic; so $ {\angle CZD\equal{}\angle ABD}$;Notice that $ {\angle ACK\equal{}\angle DCK }$ and $ {CD\equal{}KC}$;so $ {CK //AD}$ we get $ {\angle KCD\equal{}\angle CDZ\equal{}\angle CZD\equal{}\angle ABD}$; so easy to get points $ {K,B,C,D}$ are concyclic .so $ {\angle BKC\equal{}\angle BDC}$ $ {Q.E.D.}$


MATHCOUNTS
AMC
AIME
AMC 10
USA(J)MO
USAMO
AMC 12

Ive been hearing a lot about this book called Art and Craft of Problem Solving, and I was wondering what exactly it was. What does it focus on: Mathcounts level questions or beginner proofs? How good at math do you have to be to benefit from (or even understand) this book?

From what i have seen on other discussions, etc, I think it benefits those who are beginning with Olympiads.

Okay, but what level would you need to be to benefit from the book? In other words, how hard are the concepts it teaches? Because I am just starting olympiad problems, but so could a 7 year old kid, and the kid wouldnt benefit at all from the book (unless he was david yang, but he wouldnt be 7, he would be like 5).

EDIT: What I mean by "what level do you need to be to benefit from the book," I mean like what is the approx. AIME score, amc10 score,.... you need to have to even understand the concepts of the book

It's for basic Olympiad training, so you could probably start absorbing basic material at approximately a score of 5 on the AIME, but a USAMO qualifier would probably get more out of the book.

It is a very hard book, and it is like the Art of Problem Solving volume 2, except it is harder and presents topics more indepth. I am not sure what AIME score/AMC 10 score you would need to have, but I think your AIME score should at least be a 4-5, and AMC 10 is around 145.

Im not trying to judge anyone, but it was pretty hard for me.

Good Luck!!
 :rotfl:

The AcOPS book is for Olympiad prep, way beyond Mathcounts. You should be able to score at least 5-6 consistently on AIME before doing that.

[quote="AwesomeToad"] You should be able to score at least 5-6 consistently on AIME before doing that.[/quote]

Hmm so I given that AIME score I guess I will have to wait to get this book until I am at least a sophmore....

......in college......

What stage are you in? Remember, ACoPS is for serious Olympiad Prep. This is way beyond Mathcounts, as I said before, and if you're posting here and doing Mathcounts, you really shouldn't worry now.:)

well I may have been exaggerating a bit. My AIMEII score was, ummm, not bad but not great (top 7th grade score in state, but thats not saying much [i]at all[/i]), so I probably need to wait a bit before I get this book

Really, what score was it? Of course, as you said, in some states, it's not saying much at all that you were first.

The AIME II was significantly harder than the AIME I. Don't worry too much about embarrasing your score since (hopefully) most people here don't know who you are :)

While obviously USAMO is harder than AIME, AIME is designed to be similar to Olympiad Problems, so, as I said. if you can score 5-6 on AIME, you probably are ready for this.

ACoPS is a good book, but reading it while you're mathcounts level is like reading AoPS volume 2 before AoPS volume 1. :wink:

When you can score quite high on the AMC10, the AMC12, and AIME, then you should go buy ACoPS. As others have mentioned, it is a difficult book if you are not at the right level, so you might have to wait...

Okay, enough with AcOPS.

Basically, for Mathcounts/AMC 10/12 use the Introduction Series of texts and Vol. 1, although some of the AMC 12 is in the Intermediate Series (Ex: Trig)

For AIME, make sure you mastered the Introductory material. Then study the Intermediate Series very thoroughly, it will help with Olympiad also. When you are done with Intermediate, do Vol. 2 and after that, do AcOPS (assuming you can get at least 5-6 without too much trouble. If you can't, it probably means you need to get the Intermediate Series more.

Do not do too much at a time of course it will exhaust you. And also, of course do some practice problems along the way.

Now, I have to say, this discussion is now way beyond Mathcounts, and probably doesn't belong here. But in the future this belongs to the AMC forum.

Good luck everyone.:)


An aircraft seam requires 25 rivets. The seam will have to be reworked if any of the rivets is defective. Suppose rivets are defective independently of one another, each with the same probability.

a) If 20% of all seams need reworking, what is the probability that a rivet is defective?

b) How small is the probabilty of a defective rivet be to ensure that only 10% of all seams need reworking?

I didn't even questions like these could be answered. It just doesn't make much sense.

a) The answer is given by $ (1\minus{}\sqrt[25]{1\minus{}0.2})$

b) The answer is given by $ (1\minus{}\sqrt[25]{1\minus{}0.1})$

[quote="aleph0"]a) The answer is given by $ (1 \minus{} \sqrt [25]{1 \minus{} 0.2})$

b) The answer is given by $ (1 \minus{} \sqrt [25]{1 \minus{} 0.1})$[/quote]

Not quite.

I don't think none of these answers are correct. The answer to part b is .00421

lets do it on a scale of 10.
part a )
2 seams are defective so, 2 or more rivets are defective. and total rivets =25*10=250.
hence, least probability of defective rivet is $ \frac{2}{250}\equal{}0.008$
and highest probability is 50/250 = 1/5 = 0.2
so, net probability is $ 0.008<\equal{} and <\equal{}0.2$

part b}
only 10% of seams need reworking is to be ensured , so only 1 rivet should be defective  as ,if there are more that need reworking then they could be in the same seam or different seam , so then we would not be sure that 10% only need reworking.
hence probability is 1/250 = 0.004 which is very close to your answer also.

aleph0 is definitely correct.

Let $ p$ be the probability that a rivet is defective.  Then $ 1 \minus{} p$ is the probability that the rivet is good, and thus $ (1 \minus{} p)^{25}$ is the probability that $ 25$ rivets are good.  Then the probability that at least one of these rivets is bad (the seem needs reworking) is just $ 1 \minus{} (1 \minus{} p)^{25}$.

a) $ 1 \minus{} (1 \minus{} p)^{25} \equal{} .2$
$ (1 \minus{} p)^{25} \equal{} .8$
$ 1 \minus{} p \equal{} \sqrt [25]{.8}$
$ p \equal{} 1 \minus{} \sqrt [25]{.8} \approx .008886$ or $ .8886\%$

b) $ 1 \minus{} (1 \minus{} p)^{25} \equal{} .1$
$ (1 \minus{} p)^{25} \equal{} .9$
$ 1 \minus{} p \equal{} \sqrt [25]{.9}$
$ p \equal{} 1 \minus{} \sqrt [25]{.9} \approx .004206$ or $ .4206\%$

when i thought about it, and worked through it, i actually got this one right.


calculus
integration
real analysis
function
limit
LaTeX
real analysis unsolved

Dear folks,

Here are my questions, thanks.

(1)if Fn converge to f in measure and  h is measurable and the measure is finite, how to prove h*Fn converge to h*f. I think if Fn converge to f in measure, it can not converge to f almost everywhere (since there is a couterexample-"wandering function"). Otherwise we can use Egoroff's. But if we prove it using the definition, how to prove that h is bounded (or maybe it is not bounded)?

(2) lim(n->inf.) Integral {from a to inf.  [n/(1+n^2*x^2)] dx }
want to use Lebesgue Dominant Convergent Theorem to get the limit inside the integral, but can we do that when a<0, since it is hard to find such an function g that dominate the sequence. Or, I need to use other convergence theorem?
 :(

On problem #2, you may not use $a<0.$ If $a\le0,$ the limit of the integral and the integral of the (pointwise) limit differ.

On thing you can do with this is to make a change of variables:

$\lim_{n\to\infty}\int_a^{\infty}\frac{n\,dx}{1+n^2x^2} = \lim_{n\to\infty}\int_{na}^{\infty}\frac{du}{1+u^2}.$

This limit is $0$ for $a>0$, $\frac{\pi}2$ for $a=0$ and $\pi$ for $a<0.$

Is there a reasonable dominating function? What we're looking for in a dominating function is an easily manipulated $g(x)$ that has the same essential behavior as $\sup_{n}f_n(x).$

In this particular case, I suspect that pointwise supremum of behaving like $\frac1x.$ Indeed, that is the case. By simple algebra:

$\frac{n}{1+n^2x^2}\le\frac1{2|x|}$ for all $n$ and $x.$ We use this estimate for $|x|\le 1.$

Behavior at $\infty$ is not really the problem; for $|x|>1$ it suffices to note that $\frac{n}{1+n^2x^2}\le\frac1{x^2}.$

So that's our dominating function:

$g(x)=\{\begin{array}{c}\,\frac1{2|x|} \text{ for }|x|\le1 \ \frac1{x^2}\text{ for }|x|>1\end{array}$

If $a>0$, $g$ is integrable on $[a,\infty)$ and the DCT applies.

If $a\le 0$, $g$ is not integrable on $[a,\infty)$ and the DCT does not apply. Since the conclusion of the DCT is also false, we cannot find a "better" dominating function.

As for question #1, I'm not sure of your notation. What do you mean by "*"? Pointwise multiplcation? Convolution?

Kent Merryfield, sorry about the confusion (coz I am a new meber and don't know that I can use Latex here). "*" just the multiplication of the two measureable functions.   When h is bounded it is trivial, but when it is not bounded, I can not find an appropriate sequence...


What is the greatest integer less than $ \minus{}\frac{15}4$?

hello, since $ 16>15$ we get $ \minus{}4<\minus{}\frac{15}{4}$.
Sonnhard.

[b]-15/4 = -3.75. On a number line, the greatest integer less than -3.75 is [color=red]-4[/color].[/b] :rotfl:


function
induction
limit
continued fraction
algebra unsolved
algebra

Find all functions $f: \mathbb R \to \mathbb R$ such that
(1) $f(x+1)=f(x)+1$ for all $x \in \mathbb R$, and
(2) For any non-zero real $x$,
\[\frac{f(x)}{x^2}=f\left(\frac 1x\right).\]

[quote="Lee Sang Hoon"]find all function f : R→R such that
(1)f(x+1)=f(x)+1
(2)f(x)/x^2=f(1/x)[/quote]

Very interesting ...

[hide="My solution"]

Let $ \sim$ the relation defined as : "$ x\sim y$ $ \iff$ either $ x$ and $ y$ both have a finite continued fraction, either both have an infinite one with the same trailing part from some point"

Obviously, the relation $ "\sim"$ is an equivalence relation. Let $ \mathcal C(x)$ be the equivalence class of $ x$.

Let $ k_n(x)\equal{}x\plus{}n$ ($ n\in\mathbb Z$) and let $ i(x)\equal{}\frac{1}{x}$
Let us define a "path" from $ x$ to $ y$ as any sequence $ p_x^y\equal{}k_a\circ k_b\circ i\circ k_c \circ i ...$ of compositions of $ k_n$ and $ i$ such that $ y\equal{}p_x^y(x)$.

Clearly, $ x\sim y$ $ \iff$ $ \exists$ a path $ p_x^y$

If we know a path from $ x$ to $ y$, then, using $ f(k_n(x))\equal{}k_n(f(x)$ and $ f(i(x))\equal{}\frac{1}{x^2}f(x)$, we can compute $ f(y)$ from $ f(x)$

Now are the three important properties :
P0 : If $ f(x)\equal{}x$ for some $ x$, then $ f(y)\equal{}y$ $ \forall y\sim x$
P1 : If the continued fraction of $ x$ is infinite periodic, then $ f(x)\equal{}x$
P2 : If the continued fraction of $ x$ is finite or infinite not periodic, all the paths from $ x\to y$ give the same relation between $ f(x)$ and $ f(y)$

1) P0 : If $ f(x)\equal{}x$ for some $ x$, then $ f(y)\equal{}y$ $ \forall y\sim x$
Take any path from $ x\to y$. Since $ f(k_n(x))\equal{}k_n(f(x))\equal{}k_n(x)$ and $ f(i(x))\equal{}\frac{1}{x^2}f(x)\equal{}\frac{1}{x}\equal{}i(x)$ it's easy to show with induction on elements of the path that $ f(y)\equal{}y$
So $ f(x)\equal{}x$ $ \forall x\in\mathcal C(x)$ and the two rules are always true over $ \mathcal C(x)$

2) P1 : If the continued fraction of $ x$ is infinite periodic, then $ f(x)\equal{}x$
Let $ y\equal{}[0;\overline{a_1,a_2,...,a_j}]$ (where $ a_1,...,a_j$ is the periodic part (starting anywhere).
Using as much as necessary the fact that $ f([b_0;b_1,b_2...])\equal{}b_0\plus{}\frac{1}{[b_1;b_2...]^2}f([b_1;b_2...])$, we have so $ f(y)\equal{}\alpha f(y)\plus{}\beta$ with $ \alpha < 1$ ($ \alpha$ and $ \beta$ depend on $ y$).
So $ f(y)$ has an unique value for $ y$ given. And, since $ f(y)\equal{}y$ is clearly a solution, this is the unique one.
So, using $ P0$, $ f(x)\equal{}x$.
Q.E.D.

3) If the continued fraction of $ x$ is finite or infinite not periodic, all the paths from $ x\to y$ give the same relation between $ f(x)$ and $ f(y)$
Let us call a path "canonical" if it contains neither subsequence $ i\circ i$ nor $ f_a\circ f_b$. Clearly there is a unique canonical path from $ x$ to $ y$.
Replacing $ k_n\circ k_m$ by $ k_{n\plus{}m}$ in a path gives the same relation between $ f(x)$ and $ f(y)$
Suppressing $ i\circ i$ in a path gives too the same relation between $ f(x)$ and $ f(y)$
Unicity of caninical path gives the final result.
Q.E.D.

And now the final result :
========================
Let $ c(x)$ a choice function associating to any $ x$ a representant (unique per class) of its equivalence class.
Let $ h(x)$ any function from $ c(\mathbb R)\to \mathbb R$

$ \forall x$ whose continued fraction is infinite periodic, $ f(x)\equal{}x$
For any other $ x\in c(\mathbb R)$, $ f(x)\equal{}h(x)$
For any other $ x$, take any path $ p_{c(x)}^x$ and compute $ f(x)$ from $ f(c(x))\equal{}h(c(x))$

And so infinitely many solutions.

Nota : If we required continuity, then :
We have $ f(n)\equal{}n\plus{}f(0)$ and $ f(\frac{1}{n})\equal{}\frac{n\plus{}f(0)}{n^2}$ and so $ \lim_{n\to \plus{}\infty}f(\frac{1}{n})\equal{}0$ and so (continuity) $ f(0)\equal{}0$
So (using $ P0$), $ f(x)\equal{}x$ $ \forall x\in\mathcal C(0)\equal{}\mathbb Q$
So, using continuity again and since $ \mathbb Q$ is dense in $ \mathbb R$, $ f(x)\equal{}x$ $ \forall x\in \mathbb R$

[/hide]

[quote="N.N.Trung"]Dear pco,
My knowledge in algebra is not good as yours. And I will be pleasant for a elementary solution ...  :lol:[/quote]

I did not find any elementary solution (in the case where $ f(x)$ is not continuous) and, looking at the result, I'm not sure we can find one.  :(

[quote="Lee Sang Hoon"]find all function f : R→R such that
(1)f(x+1)=f(x)+1
(2)f(x)/x^2=f(1/x)[/quote]

[hide]
if f(0)=a then a=f(0)=f(-1+1)=f(-1)+1=f(-2+1)+1=f(-2)+2=f(-3+1)+2=f(-3)+3=.........=f(n)-n
and since a=f(0)=f(1)-1=f(2)-2=f(3)-3=f(4)-4=.........=f(n)-n
thus a=f(n)-n or f(n)=n+a for all integers n
Thus f(x) is any function such that the integers follow f(x)=x+a; this function we will call Intish(x) so f(x)=Intish(x)+a

In (2)
f(x)-x^2*f(1/x)=0 so:
Intish(x)+a-x^2*Intish(1/x)-x^2*a=0  but if x is an integer then Intish[x]=x and more simplifying can be produced:
x+a-x^2*Intish(1/x)-x^2*a=0 or further simplified for Intish(1/x):
Intish(1/x)=a/x2-1/x-a so if we solve for as x->Infinity:
Intish(0)=-a

and finally:
a = f(0) {for integers} = Intish(0)+a = a-a = 0
a = 0

so at least for all integers x the function f(x) = x satisfy (1) + (2) {which can easily be varified for all reals by pluging f(x)=x into both (1) + (2)}

Now Intish(x) must be defined to see if any more functions exist that f(x)=x for all reals:
if x is a fraction then 1/x is a real number so by implementing (2):
f(x)=x^2*f(1/x)=x^2*(1/x)=x

but to prove that f(x)=x is the only function for real irrationals a irrational x such that 0<x<1 {or by (1) it has been reduced until it was less then one} then
f(x)=x^2f(1/x) but f(1/x)>1 so if 1/x=z+r where z>1 and 0<r<1 then
f(x)=x^2f(z+r)=x^2(f(r)+z)=x^2(f(r)+1/x-r)=x+x^2(f(r)-r)=x+x^2(r^2*f(1/r)-r); and if 1/r=a+b such that a>1 and 0<b<1 then
f(x)=x+x^2(r^2*f(1/r)-r)=x+x^2(r^2*f(a+b)-r)=x+x^2(r^2*(f(b)+a)-r)=x+x^2(r^2*(f(b)+1/r-b)-r)=x+x^2(r^2*(f(b)-b)+r-r)=x+x^2(r^2*(f(b)-b))
Notice that "x+x^2(f(r)-r)" was simplified to x+x^2(r^2*(f(b)-b)) and that if I picked any random two variables to equal 1/b where one was between 0 and one and the other greater then 1 then the simplification would then be x+x^2*r^2*b^2*(f(random variable between 0 and 1)- (random variable between 0 and 1)) then at infinity we would get the simplification x+x^2*r^2*b^2*(variable between 0 and 1)^2*(another variable between 0 and 1)^2*............
It is clear that the second term would approach 0 because of the continuous multiplication of a number between 0 and 1 squared.
thus the equations would become f(x)=x+0=x and thus the only function which satisfies the given two conditions is:

[b] f(x)=x [/b]

[/hide]

[quote="ScienceJar"] Notice that "x+x^2(f(r)-r)" was simplified to x+x^2(r^2*(f(b)-b)) and that if I picked any random two variables to equal 1/b where one was between 0 and one and the other greater then 1 then the simplification would then be x+x^2*r^2*b^2*(f(random variable between 0 and 1)- (random variable between 0 and 1)) then at infinity we would get the simplification x+x^2*r^2*b^2*(variable between 0 and 1)^2*(another variable between 0 and 1)^2*............
It is clear that the second term would approach 0 because of the continuous multiplication of a number between 0 and 1 squared.
thus ... [/quote]

This "thus ..." needs the function to be continuous.
So, you only proved that the only [u][b]continuous [/b][/u]solution is $ f(x)\equal{}x$

But other non continuous solutions exist (see my post)

I see your point it would have infinite solutions since continuity is not given.

That is very cool. Thanks :)


Let $P(x)$ be a nonzero polynomial of degree less than $2006$, $P(-1) \neq 0, P(0) \neq 0, P(1) \neq 0$. Let

$(\frac{d}{dx})^{2006}(\frac{P(x)}{x^{3}-x}) = \frac{f_{2006}(x)}{g_{2006}(x)})$
for polynomials $f_{2006}(x)$, $g_{2006}(x)$. Find the smallest possible degree of $f_{2006}(x)$

It means $2006$ applications of $f$ (I think).


Find all positive integer $a$, $b$ such that $2a-1$ and $2b+1$ are relative primes and $a+b|4ab+1$

Daniel

Let $p$ be $2a-1$, $q$ be $2b+1$, $s$ be $a+b=\frac{p+q}{2}$.
then we can rewrite our condition:
1) $(p,q)=1$;
2) $s|(p+1)(q-1)+1=pq+q-p$.
Since $q=2s-p$ then $s|p(2s-p)+2s-p-p$ :arrow: $s|p(p+2)$.
Since $p=2s-q$ then $s|q(2s-q)+q-2s+q$ :arrow: $s|q(q-2)$.
But $(p,\frac{p+q}{2})=\frac{(2p,p+q)}{2}=\frac{(2p,2q)}{2}=(p,q)=1$.
Hence $\frac{p+q}{2}|p+2$, i.e. $p+q|2p+4$. Analogously $p+q|2q-4$.
If $q=1$ then $b=0$ - contradiction.
If $q=2$ then $b=\frac{1}{2}$ - contradiction.
If $q\geq 3$ then fractions $\frac{2p+4}{p+q}$ and $\frac{2q-4}{p+q}$ are positive integer and their sum is equal to $2$. Hence $2p+4=2q-4=p+q$ :arrow: $q-p=4$ :arrow: $2b+1-2a+1=4$ :arrow: $b=a+1$.
And really all positive integer numbers which satisfy condition $b=a+1$ also satisfy condition of our problem because:
1) $(2a-1,2b+1)=(2a-1,2a+3)=(2a-1,4)=1$ because $2a-1$ is odd;
2) $a+b=2a+1|(2a+1)^{2}=4a(a+1)+1=4ab+1$.


limit
real analysis
real analysis unsolved

help me to simplify this:

$ ....... \plus{} 2^n/8 \plus{} 2^n/4 \plus{} 2^n/2 \plus{} 2^n$

what is the general form of this equation?? thanks

These are finite geometric series. The general form involves a difference of two exponentials.

[quote="EquinoX"]how about if I have this:

$ \sum\limits_{n \equal{} 0}^{n \minus{} 1} \left({\frac {1}{2}}\right) ^ n$

what is that equal to[/quote]

This is quite a tricky notation, since you used $ n$ as an iterator as well as the summation limit. Still:

$ \sum^{n\minus{}1}_{i\equal{}0}x^i\equal{}\frac{1\minus{}x^n}{1\minus{}x}$

[quote="milin"][quote="EquinoX"]how about if I have this:

$ \sum\limits_{n \equal{} 0}^{n \minus{} 1} \left({\frac {1}{2}}\right) ^ n$

what is that equal to[/quote]

This is quite a tricky notation, since you used $ n$ as an iterator as well as the summation limit. Still:

$ \sum^{n \minus{} 1}_{i \equal{} 0}x^i \equal{} \frac {1 \minus{} x^n}{1 \minus{} x}$[/quote]

Moreover, if $ |x| < 1$ (as it is in Equinox's question), then the sum converges to $ \frac {1}{1 \minus{} x}$ (derivation is simply computing the $ \lim_{n \rightarrow \infty}$)

Therefore, $ \sum_{i \equal{} 0}^{n\minus{}1} \left(\frac {1}{2}\right)^{i} \equal{} \frac {1}{1 \minus{} \frac {1}{2}} \equal{} \boxed{2}$

no, that's the infinite sum.  you should not expect a sum of $ n$ distinct terms to have a result independent of $ n$.


induction
number theory proposed
number theory

Let $(1 + \sqrt 3)^{2n+1} = A_n + B_n \sqrt 3$ where $A_n,B_n$ are positive integers.
Note that $A_0=B_0 = 1$ and that $A_n$ is the given sum.

It is easy to see that $A_{n+1} = 4A_n + 6B_n$ and $B_{n+1} = 2A_n + 4B_n$.
Therefore, an easy induction leads to $A_n = B_n = 0 \mod [2^n].$

Pierre.


OA & OB are perpendicular radii to circle O, whose radius is 2. the midpoint of OB, and ACD is a straight line segment. How long is the segment CD?


calculus
inequalities unsolved
inequalities

Find the natural number $\ n$ such that can exist $\ x,y,z>0$ whith the properties
$\ x^3+y^3+z^3=x+y+z$
$\ x^2+y^2+z^2=nxyz$.

It's obvious that no other positive set of numbers could have this property.

Let $f(x) : = x^3 - x$ for $x >0$. Observe that $f(1) = 0$, $f$ is negative for $0<x<1$ and
positive for $x>1$. In order for the first condition to hold one of $f(x),f(y),f(z)$ has to
be $\le 0$, say $f(z)$. So $0<z<1$ and we have that

$n xy > nxyz = x^2 + y^2 + z^2 > x^2 + y^2 \ge 2 xy$.

So $n > 2$. But every $n \ge 3$ can be attained: Let $y = z = \epsilon$, where $0 < \epsilon \le 1$.
Then $f(\epsilon) \le 0$ and we choose $x = x(\epsilon) \ge 1$ such that $f(x) = - 2f(\epsilon)$. By
calculus this can be done in a unique way. Furthermore $x(\epsilon)$ is continuous in $\epsilon$ and
we have $x(1) = 1$.  Now consider

$g(\ep) : = \frac{ x(\epsilon)^2 + 2 \epsilon^2}{x(\epsilon)\epsilon^2}$.

$g$ is continuous for all $0<\epsilon \le1$. We have $g(1) = 3$ and $g(\epsilon) \to \infty$
for $\epsilon \to 0$. Hence every natural number $n \ge 3$ is in the range of $g$.
So for every $n \ge 3$ there are $x,y,z >0$ satisfying the two conditions.