If $ a$, $ b$ and $ c$ are positive reals prove the following inequality:

$ \frac{\sqrt{a\plus{}b} \plus{}\sqrt{b\plus{}c} \plus{} \sqrt{c\plus{}a}}{\sqrt{3(ab\plus{}bc\plus{}ca)}}\leq 2 \max \{\frac{1}{\sqrt{a\plus{}b}} ,  \frac{1}{\sqrt{b\plus{}c}}, \frac {1}{\sqrt{c\plus{}a}}\}$

Suppose \frac{1}{\sqrt{a+b}} is max. Then a and b are less or equal than c. Multiply  the both sides by \sqrt{3(ab+bc+ac)} and  \frac{1}{\sqrt{a+b}} .You will get :
(a + b) +(\sqrt{ab+bc+ac+bb})+(\sqrt{ab+bc+ac+aa}) on the left side. Now you just  need to use AQ inequality for 3 numbers in small parentheses and the fact : aa<=ac and bb<=bc.

[quote]Suppose $ \frac {1}{\sqrt {a \plus{} b}}$ is max. Then $ a$ and $ b$ are less or equal than $ c$. Multiply the both sides by $ \sqrt {3(ab \plus{} bc \plus{} ac)}$ and $ \frac {1}{\sqrt {a \plus{} b}}$ .You will get :
$ (a \plus{} b) \plus{} (\sqrt {ab \plus{} bc \plus{} ac \plus{} b^{2}}) \plus{} (\sqrt {ab \plus{} bc \plus{} ac \plus{} a^{2}})$ on the left side. Now you just need to use AQ inequality for 3 numbers in small parentheses and the fact : $ a^{2}\leq ac$ and $ b^{2}\leq bc$.[/quote]

I have made just a small typo to this very nice and simple solution.

Thank you [b]zeta[/b]

Also try to prove or disprove stronger inequality:

If $ a$, $ b$ and $ c$ are positive reals then:

$ \frac {\sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a}}{\sqrt {3(ab \plus{} bc \plus{} ca)}}\leq \frac{2}{3} (\frac {1}{\sqrt {a \plus{} b}} \plus{} \frac {1}{\sqrt {b \plus{} c}}\plus{} \frac {1}{\sqrt {c \plus{} a}})$

[quote="delegat"]Also try to prove or disprove stronger inequality:

If $ a$, $ b$ and $ c$ are positive reals then:

$ \frac {\sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a}}{\sqrt {3(ab \plus{} bc \plus{} ca)}}\leq \frac {2}{3} (\frac {1}{\sqrt {a \plus{} b}} \plus{} \frac {1}{\sqrt {b \plus{} c}} \plus{} \frac {1}{\sqrt {c \plus{} a}})$[/quote]
I think: $ \frac {\sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a}}{\sqrt {3(ab \plus{} bc \plus{} ca)}}\ge \frac {2}{3} (\frac {1}{\sqrt {a \plus{} b}} \plus{} \frac {1}{\sqrt {b \plus{} c}} \plus{} \frac {1}{\sqrt {c \plus{} a}})$
We have:
$ \frac {2}{3}\sqrt {3(ab \plus{} bc \plus{} ca)}(\frac {1}{\sqrt {a \plus{} b}} \plus{} \frac {1}{\sqrt {b \plus{} c}} \plus{} \frac {1}{\sqrt {c \plus{} a}}) \equal{} \frac {2}{\sqrt {3}}(\sum \sqrt {\frac {ab \plus{} bc \plus{} ca}{a \plus{} b}})$
$ \equal{} \frac {2}{\sqrt {3}}(\sum \sqrt {\frac {ab}{a \plus{} b} \plus{} c})$
By Cauchy-Schwarz inequality:
$ (\sum \sqrt {\frac {ab}{a \plus{} b} \plus{} c})^2 \le 3(a \plus{} b \plus{} c \plus{} \sum \frac {ab}{a \plus{} b})$
We will prove that:
$ 4(a \plus{} b \plus{} c \plus{} \sum \frac {ab}{a \plus{} b}) \le (\sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a})^2$
$ \Leftrightarrow \sum \sqrt {(a \plus{} b)(a \plus{} c)} \ge a \plus{} b \plus{} c \plus{} \sum \frac {2ab}{a \plus{} b}$
By Cauchy-Schwarz inequality, we have:
$ \sum \sqrt {(a \plus{} b)(a \plus{} c)} \ge \sum (a \plus{} \sqrt {bc})$
We need to prove:
$ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \ge \sum \frac {2ab}{a \plus{} b}$
Which is obivious true by AM-GM: $ a \plus{} b \ge 2ab$


The game of Technopoly ends this way:
At the end of every round after (and including) the 10th, a 20-sided die is rolled. If the number on the die is lower than the round number, The game ends.

How many rounds is the game expected to last?

Hint
[hide]
Geometric distribution  I think,...
[/hide]

The probability that it will end on the tenth round is $.45$.
The probability that it will end on or before the eleventh round is $.45+.55\times .5=.725$.  Since that is the first probability that is over $.5$, the expected value is $11$.

Expected value is where you take the mean; I don't think you just take the value that is greatest. For example, the expected number of heads when flipping 3 coins is 1.5

The expected value is the value that is obtained with $50\%$ probability.  That value will be somewhere between $10$ and $11$.

[edit]To find the exact expected value, you would need to take the weighted average of the probabilities of all the terms between $10$ and $21$, but I'm too lazy to do this right now.


An acute triangle $ABC$ has circumcircle with center $O$. A point $P$ is taken on the shorter arc $AB$. The perpendicular from $P$ to $BO$ intersects $AB$ at $S$ and $BC$ at $T$. Likewise, the perpendicular from $P$ to $AO$ intersects $AB$ at $Q$ and $AC$ at $R$.

(a) Show that triangle $PQS$ is isosceles

(b) Show that $PQ^{2}=(QR)(ST)$

(a) trivial angle chase

(b) We can easiley prove that $CRQB$ and $ASTC$ are cyclic. Hence it follows that $QR.ST = AQ.\frac{BC}{AC}.BS.\frac{AC}{BC}$
So all we have to prove is that $\Delta PQA$ and $\Delta BSP$ are similar.
This is easy, because $\angle APQ = \angle PBA$ after an easy angle chase.


Prove that  for all  positive integers  [tex] n\geq3 [/tex]  and for all  positive  real  numbers  [tex] x_1 , x_2 , ...,x_n [/tex]  at  least one of  the  following inequalities

[tex]\sum\limits_{i=1}^{n}\frac{x_i}{x_{i+1}+x_{i+2}}\geq\frac{n}{2} [/tex]  ,

[tex]\sum\limits_{i=1}^n\frac{x_i}{x_{i-1}+x_{i-2}}\geq\frac{n}{2}[/tex]

holds  .  (Where  [tex]x_{n+1}=x_1 , x_{n+2}=x_2 , x_0=x_n , x_{-1}=x_{n-1} [/tex]

Thank  you  :)


Putnam
modular arithmetic
induction
limit
AMC
USA(J)MO
USAMO

okay so i've managed to pull this off too :D

first of all the statement of the problem
=============================
Let $n$ be a fixed positive integer. How many ways are there to write $n$ as a sum off positive integers, $n = a_{1}+a_{2}+\cdots+a_{k},$ with $k$ an arbitrary positive integer and $a_{1}\le a_{2}\le \cdots a_{k}\le a_{1}+1?$ For example, with $n=4$ there are four ways: $4, 2+2, 1+1+2,1+1+1+1.$
=============================

denote by $f(n)$ the required number. we shall prove that $f(n)=n.$ obviously $n=ka_{1}+r,$ where $0 \leq r<k,$ which means that $a_{1}=[n/k]$ where $[x]$ denotes the integer part of the real $x.$

but one can see that $n=[n/k]+[(n+1)/k]+[(n+2)/k]+\cdots+[(n+k-1)/k].$ just take $n=qk+r$and make the computations.

that means that $f(n) \geq n$ (as $k$ passes from 1 to $n$). let us now prove that for each $k$ there is exactly one $a_{1}$ (and namely $[n/k]$) with the property that $n = ka_{1}+r.$ suppose we also have a $b_{1},b_{2},...,b_{k}$ so that $n=kb_{1}+r'.$ Then $r\equiv r' \pmod{k},$ but being residues we get that $r = r'$ and thus $ka_{1}=kb_{1}\Rightarrow  a_{1}= b_{1},$ so in fact $f(n) \leq n$ which gets us to $f(n)=n.$ :D very nice problem.

This problem can also be done by induction on n.

There is a bijective mapping between the solutions of n and n+1 (excluding one point).

Here is another solution:

\begin{eqnarray*}f(n)&=&[x^{n}]\sum_{k=1}^{\infty}\frac{x^{k}}{1-x^{k}}\frac{1}{1-x^{k+1}}\\ &=&[x^{n}]\frac{1}{(1-x)^{2}}\sum_{k=1}^{\infty}\frac{x^{k}}{(1+x+\cdots+x^{k-1})(1+x+\cdots+x^{k})}\\ \ &=&[x^{n}]\frac{1}{(1-x)^{2}}\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left(\frac{1}{1+x+\cdots+x^{k-1}}-\frac{1}{1+x+\cdots+x^{k}}\right)\\ \ &=&[x^{n}]\frac{1}{(1-x)^{2}}\lim_{n\rightarrow\infty}\left(1-\frac{1-x}{1-x^{n+1}}\right)\\ \ &=&[x^{n}]\frac{x}{(1-x)^{2}}\\ \ &=&n \end{eqnarray*}

One advantage to this solution is its lack of words, since sometimes people don't articulate things well and lose points.  Plus its nice to not having to worry about the combinatorics and number theorey, all of that is done automatically.

[quote="Negative_3"]One advantage to this solution is its lack of words, since sometimes people don't articulate things well and lose points.[/quote]
Would that solution receive full credit?  All answers are supposed to be justified thoroughly, I thought.  Even though your method works, shouldn't there be some explanation?  If the work is sufficient itself, could bad wording in an explanation really hurt?

Surely this soultion world receive full credit as is.  As for the last question, I would think sometimes.

And I noticed you are also from Memphis.  Is this some kind of major coencidence?  I am a student at the University of Memphis.

I did a background check on you.  You are definiately from Memphis TN (not Memphis Eygpt), and I have your name narrowed down to two possibilities.  However, it is likely that you are the guy moving back to New Orleans this Fall.

And, I saw that you were working on this problem at one point:

\begin{eqnarray*}\prod_{n=2}^{\infty}\frac{n^{3}-1}{n^{3}+1}\end{eqnarray*}

Less than a week ago, my roommate randomly selected this problem out of a Putnam book for us to do.

And you say you like starting math clubs.  But why do you not come to the U of M's math club?  I know that you know about it.  Plus you would probably like to meet Cecil Rousseau.  He is the chair of the USAMO, and since you took the USAMO maybe you want to talk about it.

BTW - Here are some problems given to the math club to solve by Cecil Rousseau and others:

[b]1)[/b]  Evaluate

\begin{eqnarray*}\sum_{k=1}^{\infty}(-1)^{k}\frac{\ln k}{k}\end{eqnarray*}

[b]2)[/b]  Evaluate

\begin{eqnarray*}\sum_{k=0}^{n}(-1)^{k}\frac{\binom{2n}{2k}}{\binom{n}{k}}\end{eqnarray*}

[b]3)[/b]  Evaluate

\begin{eqnarray*}\sum_{k=0}^{2n}\binom{k}{n}\frac{1}{2^{k}}\end{eqnarray*}

[b]4)[/b]  Suppose $\Pi$ is a partition of $[n]$ into $k$ blocks and $\Pi'$ is an unrelated partition of $[n]$ into $k+m$ blocks.  Prove at least $m+1$ elements of $n$ are in a smaller block in $\Pi'$ than in $\Pi$.

[b]5)[/b]  Fill the squares of an $n\times n$ grid with the integers $\{1,2,\ldots,n^{2}\}$. Prove there exists two adjacent squares that differ by at least $n/2$. Also prove there exists two adjacent squares that differ by at least $n$ (a stronger result).

BTW - Adjacent means two squares share an edge (not just a corner).

[b]6)[/b]  There are $n$ vertices on a circle, $p$ of which are colored blue. Two blue vertices are adjacent if fewer than $k$ vertices lie between them. Prove that making the blue vertices consecutive maximizes the number of pairs of adjacent blue vertices.

Can you explain how you obtained the "crucial" formula $f(n) = [x^{n}] \sum_{k=1}^{\infty}\frac{x^{k}}{1-x^{k}}\frac{1}{1-x^{k+1}}$? It seems pretty strange to me (a person that hasn't much experience with generating functions) that you directly obtained terms like $\frac{x^{k}}{1-x^{k}}$.

Sure.

\begin{eqnarray*}\frac{x^{k}}{1-x^{k}}\frac{1}{1-x^{k+1}}=(x^{k}+x^{k+k}+\cdots)(1+x^{k+1}+x^{(k+1)+(k+1)}+\cdots) \end{eqnarray*}

If we expand this product, we would get terms that look like $x^{k+k+k+(k+1)+(k+1)}$.  If $k=2$ this would be $x^{2+2+2+3+3}$ which contributes 1 to $[x^{12}]$.  And $2+2+2+3+3$ is a valid representation of $n=12$.

Heopfully it is clear that everything gets counted, and there is no double counting.

[quote="Negative_3"]I did a background check on you.[/quote]
wow, okayyyy....
[quote="Negative_3"] You are definiately from Memphis TN (not Memphis Eygpt), and I have your name narrowed down to two possibilities.  However, it is likely that you are the guy moving back to New Orleans this Fall.[/quote]
Don't quit your day job, but keep at it! [hide] Close, but that's Tong.[/hide]

[quote="Negative_3"]And you say you like starting math clubs.  But why do you not come to the U of M's math club?  I know that you know about it.  Plus you would probably like to meet Cecil Rousseau.  He is the chair of the USAMO, and since you took the USAMO maybe you want to talk about it.[/quote]
Maybe I'll make usamo next year, but sadly I only managed to get to the AIME this yr.  Just tell me when the math club meetings are.  I feel kinda awkward replying to this on a public, we can pm from now on, so...

Here's the way I wrote up the answer for my group back then:

There are always exactly $n$ ways. Since each $a_{i}$ is either $a_{1}$ or $a_{1}+1,$  we have the inequalities $ka_{1}\le n < k(a_{1}+1),$ or $a_{1}\le \frac{n}{k}< a_{1}+1.$ Hence, $a_{1}= \lfloor(n/k)\rfloor.$ For each positive integer $k \le n,$ let $n = ak+r,$ where $a$ is a positive integer and $0 \le r < k.$ This is simply the division algorithm, and $a \ge 1$ since $k \le n.$ We then form the sum of $(k-r)$ copies of $a$ (this is at least one, since $r < k$) and $r$ copies of $a+1.$ The sum of this string of numbers is $a(k-r)+(a+1)r = ak+r = n.$ This string of numbers is unique for each $k,$ and since we can do this for each $k,$ there are exactly $n$ possibilities.

My bad, I thought you were someone else.


Arthur's scores on his first five tests were 86, 80, 85, 75 and 84. What score must he get on the next test so that his average will be exactly 85?

He will need to get 100
If you add all the numbers together,
86, 80, 85, 75 and 84, you get 410
multiply 85 by 6 and you get 510
510-410=100

So basically, it's like this:
The arithmetic mean of the six numbers is $85$

Let $x$= the score on his next test

$\frac{86+80+85+75+84+x}{6}=85$

$x=\boxed{100}$

Or instead of adding them all together we could choose a different method. Let's count up the scores by saying how much each is larger than 85. We get 1, -5, 0, -10, -1. This sum is -15. Thus the last number is 15 larger than 85. Add 15 to 85 to get $\boxed{100}.$


analytic geometry
calculus
calculus computations

Find the minimum and maximum of $ f(x,y,z)\equal{}xyz$ with $ 6x\plus{}3y\plus{}2z\equal{}0$ and $ x^2\plus{}\frac{y^2}{4}\plus{}\frac{z^2}{9}\equal{}1$.

$ 6x \equal{} a,\ 3y \equal{} b,\ 2z \equal{} c\Longrightarrow a \plus{} b \plus{} c \equal{} 0,\ a^2 \plus{} b^2 \plus{} c^2 \equal{} 36,\ abc \equal{} 36xyz$
$ \Longrightarrow ab \plus{} bc \plus{} ca \equal{} \minus{} 18$. Thus $ a,\ b,\ c$ are the roots of the cubic equation  Since $ a,\ b,\ c$ are real numbers, $ \minus{} 12\sqrt {6}\leq abc\leq 12\sqrt {6}$.
$ \therefore \minus{} \frac {\sqrt {6}}{3}\leq xyz\leq \frac {\sqrt {6}}{3}$, yielding $ f_{max} \equal{} \frac {\sqrt {6}}{3}$ when $ \{a,\ b,\ c\} \equal{} \{ \minus{} \sqrt {6}, \minus{} \sqrt {6},\ 2\sqrt {6}\},\ f_{min} \equal{} \minus{} \frac {\sqrt {6}}{3}$ when $ \{a,\ b,\ c\} \equal{} \{ \minus{} 2\sqrt {6},\ \sqrt {6},\ \sqrt {6}\}$.

Comment: I should have put $ x \equal{} X,\ y \equal{} 2Y,\ z \equal{}3Z$. :o

Thank you, kunny.

Using your comment,
[quote="kunny"]Comment: I should have put x = X,\ y = 2Y,\ z =3Z. [/quote]
it is also simple to solve it with Lagrange multipliers (before the obtained system was uglier).

$ f(x,y,z) \equal{} xyz \equal{} f(X,Y,Z) \equal{} 6XYZ$  (1)
$ 6x \plus{} 3y \plus{} 2z \equal{} 0 \Rightarrow 6X \plus{} 6Y \plus{} 6Z \equal{} 0 \Rightarrow X \plus{} Y \plus{} Z \equal{} 0$  (2)
$ x^{2} \plus{} \frac {y^{2}}{4} \plus{} \frac {z^{2}}{9} \equal{} 1 \Rightarrow X^2 \plus{} Y^2 \plus{} Z^2 \equal{} 1$  (3)

$ F(X,Y,Z,\lambda_1,\lambda_2) \equal{} f(X,Y,Z) \minus{} \lambda_1(X \plus{} Y \plus{} Z) \minus{} \lambda_2(X^2 \plus{} Y^2 \plus{} Z^2 \minus{} 1)$

$ \frac {\partial F}{\partial X} \equal{} 6YZ \minus{} \lambda_1 \minus{} 2X\lambda_2 \equal{} 0$
$ \frac {\partial F}{\partial Y} \equal{} 6XZ \minus{} \lambda_1 \minus{} 2Y\lambda_2 \equal{} 0$
$ \frac {\partial F}{\partial Z} \equal{} 6XY \minus{} \lambda_1 \minus{} 2Z\lambda_2 \equal{} 0$

$ Y\frac {\partial F}{\partial X} \minus{} X\frac {\partial F}{\partial Y}: 6Y^2Z \minus{} Y\lambda_1 \minus{} 6X^2Z \plus{} X\lambda_1 \equal{} 0$  (4)
$ Z\frac {\partial F}{\partial X} \minus{} X\frac {\partial F}{\partial Z}: 6YZ^2 \minus{} Z\lambda_1 \minus{} 6X^2Y \plus{} X\lambda_1 \equal{} 0$  (5)

Now there are some options to consider.

1. If any two variables are same, then we have a unique solution (regarding conection of $ X$, $ Y$ and $ Z$; using (2)) of (1), giving $ T_1( \minus{} 2\frac {1}{\sqrt {6}},\frac {1}{\sqrt {6}},\frac {1}{\sqrt {6}} ), f(T_1) \equal{} \minus{} \frac {\sqrt {6}}{{3}}$ and $ T_2(2\frac {1}{\sqrt {6}}, \minus{} \frac {1}{\sqrt {6}}, \minus{} \frac {1}{\sqrt {6}} ), f(T_2) \equal{} \frac {\sqrt {6}}{{3}}$.

2. Any pair of variables $ X, Y, Z$ is different. Therefore from (4) we get $ X \equal{} 0$ or $ Y \equal{} Z$ (contradiction). For $ X \equal{} 0$ we get two other solutions $ T_3(0,\frac {1}{\sqrt {2}},\frac {1}{\sqrt {2}} ), f(T_3) \equal{} 0$ and $ T_4 (0, \minus{} \frac {1}{\sqrt {2}}, \minus{} \frac {1}{\sqrt {2}}), f(T_4) \equal{} 0.$

Therefore we get global extremes in $ T_1$ and $ T_2$. (and all permutations of coordinates)

I have one question here: How do we know whether there is an extreme or a cradle point in $ T_3$ and $ T_4$?


$ \sum_{i < j}^{n}\frac {a_{i}a_{j}}{a_{i} \plus{} a_{j}}\leq\bigl(\sum_{i < j}^{n}a_{i}a_{j}\bigr)\cdot\frac {n}{2(a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n})}$

Just hints please, not the complete solution.

Thanks you.

I think I found something but I need some help :

Let be $ f(a_1,a_2,...,a_n) \equal{} \sum_{i < j}^{n}\frac {a_{i}a_{j}}{a_{i} \plus{} a_{j}} \minus{} \bigl(\sum_{i < j}^{n}a_{i}a_{j}\bigr)\cdot\frac {n}{2(a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n})}$

We see that for every real $ k$ we have : $ f(ka'_1,ka'_2,...,ka'_n) \equal{} kf(a'_1,a'_2,...,a'_n)$. Then, the inequality in homogen and we can set : $ \sum_{i \equal{} 1}^{n}{a_i} \equal{} 1$

The inequality rewrites like :

$ \sum_{i < j}^{n}\frac {a_{i}a_{j}}{a_{i} \plus{} a_{j}}\leq\bigl(\sum_{i < j}^{n}a_{i}a_{j}\bigr)\cdot\frac {n}{2}$

From AM-Gm we have : $ \sum_{i < j}^{n}\frac {a_{i}a_{j}}{a_{i} \plus{} a_{j}}\leq \frac {\sum_{i < j}^{n}{\sqrt {a_ia_j}}}{2}$

From Cauchy we have : $ \frac {\sum_{i < j}^{n}{\sqrt {a_ia_j}}}{2}\leq\frac {\sqrt {n\bigl(\sum_{i < j}^{n}a_{i}a_{j}\bigr)}}{2}$

So we need to show that : $ \frac {\sqrt {n\bigl(\sum_{i < j}^{n}a_{i}a_{j}\bigr)}}{2}\leq\bigl(\sum_{i < j}^{n}a_{i}a_{j}\bigr)\cdot\frac {n}{2}$

But, how ?

Thanks.

[quote="zweig"]So we need to show that : $ \frac {\sqrt {n\bigl(\sum_{i < j}^{n}a_{i}a_{j}\bigr)}}{2}\leq\bigl(\sum_{i < j}^{n}a_{i}a_{j}\bigr)\cdot\frac {n}{2}$[/quote]

This is wrong, you have overshot here. Dehomogenization is rather misleading in problems like this.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=155691 anyway.

  darij


For those who do not know, National Assessment & Testing administers national math contests by mail.  This can be very appealing for schools that do not have many local face-to-face contests throughout the year, that do not have much money to pay registration fees, or that do not have a coach with time to attend competitions on the weekend.

Our contests are written to be accessible to all high school students, having both easy problems that inexperienced students can build confidence on as well as difficult problems that will challenge the most experienced competitor.  Because of this composition, our contests are also excellent for community college teams as well as advanced middle school teams.

2005-2006 Competition Schedule

Thursday, September 29th, 2005:  2005 Fall Startup Event

     This is a fun, fast-paced competition in which individuals have 30 minutes to solve 100 problems.  The questions cover a wide variety of topics, and are all quick and easy for students familiar with the subject.  Analogous to a musician practicing scales, this contest can help students and teams identify areas they want to focus on.

"Excellent time at the start of the year.  Excellent opportunity to interest new ninth-graders in math team - it's the competition that draws them in." - Jane Fenn, Corning-Painted Post West High School

Thursday, November 3rd, 2005:  2005 Team Scramble

     In this exciting collaborative experience, many students work together to answer 100 questions in thirty minutes.  Team leaders who understand the strengths of various team members will be able to divide the work to efficiently answer the most questions, while recruiting many students allows for more double-checking of the team's work.

"Kids went crazy when we gave them the solution key after the test." - Jim Ferguson, Dana Hills High School

Thursday, December 8th, 2005:  2005 Ciphering Time Trials

     This individual competition consists of ten rounds in which students have three minutes to solve three problems.  Students must decide which problems they will be able to answer quickly, and whether to attempt less familiar problems or double-check questions with which they are more comfortable.

"Kids loved the fast pace!" - Patrick Montague, Science Academy of South Texas

Thursday, February 9th, 2006:  2006 Four-by-Four Competition

     This is a team contest with ten rounds in which teams of four students have three minutes to solve four problems.  Students learn to work together and quickly distribute problems to team members with strengths in the corresponding subjects.

"It's fun to have a creative format like this." - Sue Forster, Bismarck High School

Wednesday, March 29th, 2006 - April 5th, 2006:  2006 Collaborative Problem-Solving Contest

     This competition consists of fifteen intricate multi-part questions that your entire school (including staff) can collaborate on for an entire week.  Students get experience using real-world problem-solving skills such as brainstorming, collaboration, research, and application of technology.

"We had a great time with this contest - we can't wait until next year!  With better planning, we hope to make this an even bigger event at our school.  As it was, I had to chase kids home tonight." - Tim Corica, The Peddie School

what about the 3-round individual tests??? (Log1 or somethign like that)

RedDevil,

     We may be administering a Mu Alpha Theta contest of some sort next year, but the format and schedule will be determined this summer if so.  We'll keep this forum updated as things progress.

hey this sounds interesting could someone please tell me where you can register. or just email me at nickjames8806@gmail.com. o and if anyone want gmail just pm me or something.

Go to their website: http://www.natassessment.com/

There is a link to the registration form.

MaoKid,

     Your coach can also register by sending me an e-mail at clymer@natassessment.com, letting us know what contest(s) they're interested in, as well as the

School Name:
School Address:
School City/State/Zip:
Coach's Name:
Coach's E-mail:
Coach's Phone:


topology
function
calculus
derivative
analytic geometry
graphing lines
slope

Let $ P\equal{}\{(x,y,z): z^3\equal{}x^4 \plus{} y^5\}$. Is this set a manifold?

What about a point $ (0,0,0)$? It seems that $ P$ is a manifold as it is a graph of a function $ (x,y)\rightarrow {(x^4 \plus{} y^5)}^{1/3}$ which is smooth ($ C^1$) since it has smooth partial derivatives. Am I right? Could you please help?

Is it a manifold? Of course it has a manifold structure, and that structure is homeomorphic to $ \mathbb{R}^2$.

Then again, that isn't really the question you wanted to ask. You really wanted to ask whether this can be a smoothly embedded submanifold of $ \mathbb{R}^3$. For that- it's obviously smooth away from the origin, with one variable locally analytic in the others. At the origin, projecting into the xy-plane shows it as $ C^1$, but we don't get any more derivatives. By the theory I know ($ C^{\infty}$), the answer would be no.

Thank you very much for the answer!

[quote="jmerry"]...that isn't really the question you wanted to ask. You really wanted to ask whether this can be a smoothly embedded submanifold of $ \mathbb{R}^3$.[/quote]

Well, actually I am not sure of terminology, but I think you're right. I wrongly used the word "smooth" -- what I mean by this is $ C^1$. According to [url]http://en.wikipedia.org/wiki/Differentiable_manifold[/url] i should ask: do $ P$ is a differentiable manifold? (to be precise -- $ C^1$-manifold).

If I understand you correctly, $ P$ is $ C^1$-manifold because of the reason I gave in the previous post. I think $ P$ is not $ C^2$-manifold - but I do understand precisely why. $ P$ is a graph of a function which is not $ C^2$ (see previous post), but why we are sure that we can't find another $ C^2$-function $ f$, such that $ P$ is a graph of $ f$?  :maybe:

Thanks for the advice!  :)

[quote="solenoid"]Let $ P \equal{} \{(x,y,z): z^3 \equal{} x^4 \plus{} y^5\}$. Is this set a manifold?

What about a point $ (0,0,0)$? It seems that $ P$ is a manifold as it is a graph of a function $ (x,y)\rightarrow {(x^4 \plus{} y^5)}^{1/3}$ which is smooth ($ C^1$) since it has smooth partial derivatives. Am I right? Could you please help?[/quote]

That function is not $ C^1$ in any neighborhood of $ (0, 0)$, as you can check by looking at points on the curve $ x^4 \plus{} y^5 \equal{} 0$. (In fact the partial derivatives don't exist anywhere on that curve except at $ (0, 0)$.)

It's not that simple- vertical slopes don't break things here, since we can just take a chart in a different direction. This is a $ C^{\infty}$ smooth embedded manifold if we remove the origin.

On the other hand, the graph chart clearly isn't good enough to cover the origin.


solve for x
$|2x|-|x+1|=1$

can anyone help me on this problem? this is what I have so far... [hide]
$2x=1+|x+1|$
$-2x=1+|x+1|$

$2x=1+|x+1|$
$2x=1+x+1$
$2x=1-x-1$  For absolute values you're suppose to take the negative and the positive value, but isn't it suppose to be $2x=1+1-x$.  because -3+1=-2=1-3.
[/hide]
Thanks :P

[hide="hint"]
If $2x$ is positive, then $x+1$ can't be negative.

If $2x$ is negative, then $x+1$ can be positive (or zero) only if $x\geq{-1}$ and $x+1$ can be negative only if $x<-1$
[/hide]

[quote="myc"]solve for x
$|2x|-|x+1|=1$
[/quote]

[hide="answer"]
break it up into three cases:

$0 < x$:
$2x-(x+1) = 1$
$\boxed{x = 2}$

$-1 < x < 0$:
$-2x-(x+1) = 1$
$-3x = 2$
$\boxed{x =-\frac23}$

$x <-1$
$-2x+(x+1) = 1$
$x = 0$, however, this does not apply because this equation only applies if x is less than -1.

Now we have to test the values x = -1 and x = 0:
$|0|-|1| = 1$
$1 = 1$ which means x = 0 works.

$|-2|+|-1+1| = 1$
$2+0 = 1$
$2 = 1$ which is false, meaning x = -1 doesn't work.

So we have the final answer as $\boxed{x = (-\frac23, 0, 2)}$.
[/hide]

for $x=0$ we have $|0|-|0+1|=1 \Rightarrow-1=1$, which is false.


function ratio calculus derivative trigonometry

Hi everybody!

I've got some problems with this exercise of quantum mechanics:

Consider the eigenvalue $ E_0(V_0)$ of the fondamental state of a unidimensional potential well:
$ V(x)\equal{} \plus{}\infty$ if $ x<0$ and $ x>2L$
$ V(x)\equal{} 0$ if $ 0<x<L$
$ V(x)\equal{} V_0$ if $ L<x<2L$

a) Studying the function $ E_0(V_0)$ for $ V_0>0$ demonstrate that exists a value $ V_0$ such that $ E_0(V_0)\equal{}V_0$

b)Describe qualitatively the wave function of the fondamental state for this specific value $ V_0$

c)Estimate numerically the ratio $ \frac{V_0}{E_0(0)}$

I can't find $ E_0(V_0)$...

Just write up the time independent Schrödinger equation for the different regions, set up the boundary conditions and solve it.
Where the potentional is infinite, the wavefunction is zero, hence the boundary conditions expressing the continuity of the wave function, is that it should be zero at the infinite boundary i.e. at $ 0$ and $ 2L$.

Then write up the Schrödinger for the $ (0,L)$ and seperately the $ (L,2L)$ regions and then fit the two wave functions with the boundary conditions at $ L$ i.e. the wavefunction itself should be continuous and the first derivative should be continuous.

Ok, writing the Schrödinger equations:

From 0 to L: $ \minus{}\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}\equal{}E\psi$ and i obtain $ \psi\equal{}A\sin k_1x\plus{}B\cos k_1x$ where $ {k_1}^2\equal{}\frac{2mE}{\hbar^2}$

From L to 2L:$ \minus{}\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}\equal{}(E\minus{}V_0)\psi$ but in this case what kind of solution should i have? I don't know if $ E\minus{}V_0$ is positive or negative...


You have a stack of coins facing heads up, numbered $1$ to $2006$, $\#1$ on the bottom and $\#2006$ on the top. A $\text{k-flip}$ $(1\le k\le n)$ is a procedure by which you take the top $n$ coins in the pile, and flip them over as a whole.[hide="For example"]A $\text{3-flip}$ applied to a stack of $4$ coins would result as follows: #1 remains heads up, but #4 is on top of that and is tails up, #3 is on top of that and is tails up, and #2 is on the very top and is tails up.[/hide]How many $\text{k-flips}$ are needed to rearrange these coins so that their order does not change, but they are all tails up?[hide="hint"]Hoani Tweor Pzlzue[/hide]
You can generalize if you want..

[quote="SplashD"]is your hint in code?[/quote]
I think it is  :D

He probably meant[hide="Hint"][url=http://en.wikipedia.org/wiki/Hanoi_towers]Hanoi Tower Puzzle[/url][/hide]

[hide="this is what I got"]
Let $n$ be the number of coins you are using.

If $n=2$, then doing a 1-flip,2-flip,1-flip,2-flip is the fastest way to do the flip.  It's small enough that you can check this by hand.

If it takes $x$ flips to complete the problem for $n=k$, call the sequence of flips to solve the puzzle the "flip sequence".

Now consider the case for $n=k+1$.  Perform the flip sequence for $n=k$ on the pile of coins, then do a $k+1$ flip, 1-flip, $k+1$ flip.  This will give you the flip sequence for $n=k+1$.  So the flip sequence for the pile of $k+1$ coins is no more than 3 flips longer than the flip sequence for $k$ coins.

So, in general, if there are $n$ coins, the flip sequence will be $3n-2$ flips long.  So for $n=2006$ coins, it needs 6016 flips. I realize this isn't a rigerous solution to the problem, but I think it's correct.
[/hide]

:clap2: i think its correct too  :D
but actually i don't how to prove that its the shortest way..
does anyone know?

[quote="minsoens"]If you're interested:
[url=http://www.mrc-cbu.cam.ac.uk/~mattd/Cmabrigde/]Cambridge Research[/url]
[quote]Aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it deosn't mttaer in waht oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist and lsat ltteer be at the rghit pclae. The rset can be a toatl mses and you can sitll raed it wouthit porbelm. Tihs is bcuseae the huamn mnid deos not raed ervey lteter by istlef, but the wrod as a wlohe.[/quote] :cool:[/quote]

That's a lie.

Try flipping every letter except the first and the last and see if you can understand a sentence of words like that.

Waht mneosnis setatd is [color=white]ctelpmoe bihsllut and ulretty[/color] fslae, if you wnere't arawe taht I was snilbmarcg tsehe wdros tihs way, yuo'd never gseus waht the legnor wdros wree.

Anyways I believe that the answer is

[hide]
$2^{2006}$
[/hide]

I'm pretty sure that Calculuslover is right.  That was the same type of thing I thought of, although I did it in a slightly different order after the first couple of coins.  Either way, it ends up being [hide="answer"]$3n-2$, so for 2006, $6018-2=\boxed{6016}$[/hide]


geometry
incenter
circumcircle
geometry proposed

Given an isosceles triangle $ABC$ with base $AB$, cirumcenter $O$, incenter $S$ and $\angle C<60^\circ$.
The circumcircle of $AOS$ intersects $AC$ at $D$. Prove that $SD\parallel BC$ and $AS\perp OD$.

Actually, this is the additional problem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101976]post #4 of http://www.mathlinks.ro/Forum/viewtopic.php?t=28336[/url]. You are right, it's all about angle chasing...

  darij


Four points $ A, B, C, D$ are chosen on a circle in such a way that arcs $ AB, BC,$ and $ CD$ are of the same length and the $ arc DA$ is longer than these three. Line $ AD$ and the line tangent to the circle at $ B$ intersect at $ E$. Let $ F$ be the other endpoint of the diameter starting at $ C$ of the circle. Prove that triangle $ DEF$ is equilateral.

I don't think that DEF is equilateral, it's only $ \overline{EF} \equal{} \overline{FD}$ (that is simple to prove)


In their base $ 10$ representation, the integer $ a$ consists of a sequence of $ 1985$ eights and the integer $ b$ consists of a sequence of $ 1985$ fives.  What is the sum of the digits of the base 10 representation of $ 9ab$?

$ \textbf{(A)}\ 15880 \qquad \textbf{(B)}\ 17856 \qquad \textbf{(C)}\ 17865 \qquad \textbf{(D)}\ 17874 \qquad \textbf{(E)}\ 19851$

Sum of the digits should be dividiable by $ 5$ and $ 9$ so the answer is $ 17865$ or $ \boxed C$

[hide="If one does not see that"]
$ 9ab \equal{} 9\cdot\underbrace{88\ldots88}_{1985}\cdot\underbrace{55\ldots55}_{1985} \equal{} \underbrace{99\ldots99}_{1985}\cdot\underbrace{44\ldots44}_{1985}0 \equal{} 10((10^{1985} \minus{} 1)\cdot\underbrace{44\ldots44}_{1985})$
$ \equal{} 10(\underbrace{44\ldots44}_{1985}\underbrace{00\ldots00}_{1985} \minus{} \underbrace{44\ldots44}_{1985}) \equal{} \underbrace{44\ldots44}_{1984}3\underbrace{55\ldots55}_{1984}60$
$ \implies4\cdot1984 \plus{} 3 \plus{} 5\cdot1984 \plus{} 6 \plus{} 0 \equal{} 9\cdot1984 \plus{} 9 \equal{} 9\cdot1985 \equal{} \boxed{\textbf{(C)}\ 17865}$. This generalizes nicely as well.
[/hide]


Find all pairs of integer numbers $ (p, q)$ such that  the roots of the equation: \[ (px \minus{} q)^2 \plus{} (qx \minus{} p)^2 \equal{} x
\]are also integer numbers.

We have discriminant$ \equal{}(4pq\plus{}1)^2\minus{}(2p^2\plus{}2q^2)^2 \ge 0$
Case 1:$ (4pq\plus{}1) \le \minus{}(2p^2\plus{}2q^2)$, $ 2(p\plus{}q)^2\plus{}1 \le 0$, which is impossible.

Case 2:$ (4pq\plus{}1) \ge (2p^2\plus{}2q^2)$, $ 1 \ge 2(p\minus{}q)^2$, so $ p\equal{}q$
So the equation becomes $ 2p^2(x\minus{}1)^2\equal{}x$
It's easy to check that $ x\equal{}2,p\equal{}1$ are the only solution.

[quote="Moonmathpi496"]Find all pairs of integer numbers $ (p, q)$ such that  the roots of the equation:
\[ (px \minus{} q)^2 \plus{} (qx \minus{} p)^2 \equal{} x
\]
are also integer numbers.[/quote]
@stephencheng: You are completely wrong. For $ p\equal{}q\equal{}1$ the equation is $ 2x^2 \minus{} 5x \plus{} 2 \equal{} (x\minus{}2)(2x\minus{}1)$ but $ \frac{1}{2}$ is not an integer.
$ (px\minus{}q)^2 \plus{} (qx\minus{}p)^2 \equal{} x \iff f(x) \equal{} (p^2\plus{}q^2)x^2 \minus{} (4pq\plus{}1)x \plus{} (p^2\plus{}q^2) \equal{} 0$. If $ p\equal{}q\equal{}0$ then we have $ x\equal{}0$. Assume otherwise. Then $ x$ is a solution iff $ \frac{1}{x}$ is. So we have $ x \equal{} \pm 1$ (But $ (px\minus{}q)^2\plus{}(qx\minus{}p)^2 \ge 0 \Rightarrow x \ge 0$) So $ x\equal{}1$. Which gives $ 2(p\minus{}q)^2 \equal{} 1$, which has no solution since $ p,q$ is integers. So $ (p,q)\equal{}(0,0)$

[quote="Mathias_DK"][quote="Moonmathpi496"]Find all pairs of integer numbers $ (p, q)$ such that  the roots of the equation:
\[ (px \minus{} q)^2 \plus{} (qx \minus{} p)^2 \equal{} x
\]
are also integer numbers.[/quote]
@stephencheng: You are completely wrong. For $ p \equal{} q \equal{} 1$ the equation is $ 2x^2 \minus{} 5x \plus{} 2 \equal{} (x \minus{} 2)(2x \minus{} 1)$ but $ \frac {1}{2}$ is not an integer.
$ (px \minus{} q)^2 \plus{} (qx \minus{} p)^2 \equal{} x \iff f(x) \equal{} (p^2 \plus{} q^2)x^2 \minus{} (4pq \plus{} 1)x \plus{} (p^2 \plus{} q^2) \equal{} 0$. If $ p \equal{} q \equal{} 0$ then we have $ x \equal{} 0$. Assume otherwise. Then $ x$ is a solution iff $ \frac {1}{x}$ is. So we have $ x \equal{} \pm 1$ (But $ (px \minus{} q)^2 \plus{} (qx \minus{} p)^2 \ge 0 \Rightarrow x \ge 0$) So $ x \equal{} 1$. Which gives $ 2(p \minus{} q)^2 \equal{} 1$, which has no solution since $ p,q$ is integers. So $ (p,q) \equal{} (0,0)$[/quote]

Sorry, I haven't checked carefully the case when $ p \equal{} 1$ and omit the case when $ p \equal{} 0$. :ninja:


calculus integration LaTeX trigonometry function algebra domain

hi, i am trying to prepare for a complex analysis exam

and one application was finding integral of x^(a-1)*sin(x) for x from zero to infinity

this can be done by integrating exp(z*i)*z^(a-1) over the contour,with 0<a<1

[r,0] to [R,0]
[R,0] to [0,R] (quarter of a circle)
[0,R] to [0,r]
and [0,r]  to [r,0] quarter of a circle

then you find that the value is gamma (a)*sin(Pi*a/2)

the next problem is finding the integral of sin(x^n)/x^n for x from zero to infinity
i'm stuck with that
i find that it is 1/n* the previous problem with a=1/n-1 , which is unfortunately lower than zero

however i should find as solution 1/(n-1)*gamma(1/n)*sin(Pi/2*(1-n)/n) which i do find by brutally substituting a=1/n-1 in the previous solution

any ideas anyone?

I will help you this way, just because I'm practicing LaTeX :D :

And one application was finding integral of $x^{a-1}\sin(x)$ for $x$ from $0$ to $\infty$

this can be done by integrating $e^{zi}z^{a-1}$ over the contour,with $0<a<1$

$[r,0]$ to $[R,0]$
$[R,0]$ to $[0,R]$ (quarter of a circle)
$[0,R]$ to $[0,r]$
and $[0,r]$  to $[r,0]$ quarter of a circle

then you find that the value is $\Gamma (a)\sin \left( \frac{\pi a}{2}\right)$

the next problem is finding the integral of $\frac{\sin (x^n)}{x^n}$ for $x$ from $0$ to $\infty$
i'm stuck with that
i find that it is $1/n$ the previous problem with $a=1/n-1$ , which is unfortunately lower than zero

however i should find as solution $\left(\frac{1}{n}-1\right)\Gamma \left(\frac{1}{n}\right) \sin\left(\frac{\pi} 2\cdot\frac{1-n}{n}\right)$ which i do find by brutally substituting $a=\frac{1}{n}-1$ in the previous solution

any ideas anyone?

Use \sin{ ... } to make the trig functions appear nice. Also use \left( .... \right) to make the parathesis be as tall as the numbers in them. Use \cdot to make a central dot, like 2 \cdot 3 would generate: $2 \cdot 3$.

For example you wrote \[ \displaystyle \frac{1}{n-1}\gamma \frac{1}{n} \sin(\Pi /2.\frac{1-n}{n}) \]
but it would be better to look like
\[ \displaystyle \frac{1}{n-1}\gamma \frac{1}{n} \sin {\left(\frac \pi 2 \cdot \frac{1-n}{n}\right)} \]

I just did a moderator's edit of dickchimney's edit of what fredbel6 said. I hadn't realized that Valentin was going to post what he did. Some of the changes are upper case Gamma's and lower case  pi's. (Trying to read what fredbel6 should have written instead of what he actually wrote.)

Maybe now one of us will start looking at the mathematics of what fredbel6 is asking.

:( sorry guys

i'm not so good with mathematical software
thanks for making my question look better

greetz

fred

You can't substitute brutally like this , because this affects all the convergence, and moreover, out of the domain of $Euler$ function

1. Let's assume that $ 0 < \alpha < 1 $
 The two quantities $\int_{0}^{\infty}{\cos{x}\cdot x^{\alpha-1}}$ and $\int_{0}^{\infty}{\sin{x}\cdot x^{\alpha-2}}$ have some relation.
 So you just have to compute the former!

2. There is still a special case which need to be considered separately : $\alpha = 1$
  Prove that $\int_{0}^{\infty}{\frac{\sin{x}}{x}} = \frac{\pi}{2}$

:D good idea!

the relation can be found with partial integration
and when finding the first integral, with the sinus, you get that of cos(x)*x^(a-1) for free with it!

so now it seems so easy!

btw, the n=1 case is weird as the solution doesn't allow n=1 but that's no problem because that is a known integral in or course

thx!


conics
parabola
function
complex analysis
complex analysis unsolved

hi
find a conformal  map   $ \phi : P \rightarrow D $ ,
$ P = \{ z = x+ iy : y^2 > 4(x+1) \}  \quad ; \quad D = \{ |z| <1 \} $

wl

1) w=z^(1/2).
If z=x+iy, w=u+iv you get x=u^2-v^2, y=2uv. So the line v=1 become y^2=4(x+1). The part of the complex plane outside the parabola is mapped into the half-plane Im(w)>1, so the function w= z^(1/2)-i map  y^2>4(x+1) into the upper half-plane.

2)Apply the well-known trasformation that map the upper half-plane to the unit disk (for example, w=(z-i)/(z+i))


this year is a special year. romania will win IMO2005...
we're waiting the subjects  :wow:

This year is an ordinary year. China will win.

If in the second day the subjects are also so common (3-rd problem an inequality which practically solves itself???), then it is absolutely clear who will win the olympiad.  :(

I just do not understand, why not all mathlinkers managed to solve the problem 3 on IMO, which is easier then every preliminary lemma for some of a million 3 variables inequality discussed here at the inequality forum (most of them I even do not know how to try).

Fedor do you think almost all of the members of the USA team got 5-7 points in this inequality?

At first, I am not an expert in USA team at all. It has too little Chineses this year, I do not know what may we expect from them.

At second, was it your guess, Manuel, or did all they already claim that they solved it?

i haven't speak to any  of them so it was just a guess..

i know one of the chineese.. sherry.. i know her and i know she is really good in geometry and in geometric inequalities.. but dunno how she did in this one..

what do you suppose why are the problems so easy this year?
wil the next day be harder?

does anyone know someone from croatia or can find out how they did?
I'd appreciate if someone could contact CRO2.

bok matija

kaj znaci concurrent?
kaj znaci positive i negative terms?
 aj mi sam to napisi , al nemoj rijesit zadatak plizzzzzzzzz

concurrent znaci konkurente: sijeku se u jednoj točki
postive and negative terms: pozitivni i negativni clanovi niza

I hope the US did well on problem 3, since one of its members actually proposed a problem using similar techniques (Cauchying the denominator to make it a common denominator) for an olympiad written by the Winners at MOP for the rest of the MOPpers, and we all saw and admired the solution. (It certainly made #3 quite easy for me)


function
logarithms
calculus
calculus computations

Compute

$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\plus{}1}\sin(n)}{n}$

I have my own solution but I would like to see if there are others. I omit my solution until this is answered so that people who wish to try it won't get any unwanted hints.

[hide="First hint."]$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\plus{}1}z^n}{n}\equal{}\log(1\plus{}z)$[/hide]

[hide="Second hint."]Let $ z\equal{}re^{i\theta},$ and eventually $ \theta\equal{}1$ and $ r\to 1.$[/hide]

[hide="Unrelated to the first two hints, starting over."]Try finding the Fourier series of the function $ f(x)\equal{}x$ for $ \minus{}\pi<x<\pi,$ extended to be periodic of period $ 2\pi.$[/hide]

Dear Mr. Merryfield, thank you very much for your reply. For your first hint I assume you mean the following because it works

$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \plus{} 1}\sin(n)}{n} \equal{} \text{Im}\bigg[\sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \plus{} 1}e^{in}}{n}\bigg] \equal{} \text{Im}\left[\ln\left(1 \plus{} e^{i}\right)\right] \equal{} \frac {1}{2}$

As for your second hint, I am not quite sure as to what you mean.

As for your third hint, this is how I solved the problem prior to posting except a slightly different function and a different interval.

Thanks again,
Mathstud


When the repeating decimal $ 0.363636\ldots$ is written in simplest fractional form, the sum of the numerator and denominator is:

$ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 114 \qquad \textbf{(D)}\ 135 \qquad \textbf{(E)}\ 150$


group theory
abstract algebra
geometry
geometric transformation
topology
advanced fields
advanced fields unsolved

Let $X$ be a compact topological group and $G$ a closed subgroup acting on $X$ by left translation, show that $X/G$ is Hausdorff.

First a

[i]Lemma[/i]

Let $X$ be a compact Hausdorff topological space, and $\sim$ an equivalence relation on $X$ with the property that the set $S_\sim=\{(x,y)\in X\times X\ |\ x\sim y\}$ is closed in $X\times X$. Then the quotient space $X/\sim$ is Hausdorff.

To prove the result we want, notice that in our case, with $X$ being the group and $\ x\sim y\iff xy^{-1}\in G$, we have $S_\sim=\{(x,y)\in G\times G\ |\ xy^{-1}\in G\}$. Since $G$ is closed in $X,\ S_\sim$ is obviously closed in $X\times X$, so the conclusion that $X/G=X/\sim$ is Hausdorff follows from the lemma.

[i]Proof of the lemma[/i]

It's very convenient to use the language of [url=http://planetmath.org/encyclopedia/Net.html]nets[/url]. $\pi: X\to X/\sim$ will be the canonical projection map, and for any $x\in X$, we’ll denote $\pi(x)$ by $\bar x$.

First let’s notice that in the given hypotheses (we can drop the condition of $X$ being compact, and keep only the condition that $S_\sim$ be closed), $X/\sim$ is a $T_1$ space. Indeed, every equivalence class under $\sim$ is closed, so given $\bar x\ne\bar y\in X/\sim$, the subset of $X$ consisting of all equivalence classes different from that of $y$ is an open set partitioned into full equivalence classes, so its image through $\pi$ is an open set around $\bar x$ which does not contain $\bar y$.

Suppose now that $(\bar x,\bar y)\in X/\sim\times X/\sim$ lies in the closure of the diagonal $\Delta=\{(\bar z,\bar z)\ |\ \bar z\in X/\sim\}$ of $X/\sim$. This means that given any neighborhood $\bar U\times\bar V$ of $(\bar x,\bar y)$, there is a point $(\bar z,\bar z)\in\bar U\times\bar V$. After applying $(\pi\times\pi)^{-1}$ to the whole thing, this means that there is a point $(z^1,z^2)\in S_\sim$ is any neighborhood of the form $(\pi\times\pi)^{-1}(\bar U\times\bar V)$ of $(x,y)$. The system of neighborhoods of the type $\bar U \times\bar V$ (which is a local basis at $(\bar x,\bar y)$) forms a net when ordered by reverse-inclusion, so we can form the corresponding net consisting of elements of the form $(z^1,z^2)$. Since this is a net in a compact subset of $X\times X$ (namely $S_\sim$), it has a subnet convergent to some $(w^1,w^2)\in S_\sim$. After applying $\pi\times\pi$ to this again, we conclude that a subnet of the net of neighborhoods of the type $\bar U\times\bar V$ of $(\bar x,\bar y)$ converges to $(\pi(w^1),\pi(w^2))$, so, since $X/\sim\times X/\sim$ is $T_1$ and $\pi(w^1)=\pi(w^2)$, we get $(\bar x,\bar y)=(\pi(w^1),\pi(w^2))\Rightarrow\bar x=\bar y$, i.e. $(\bar x,\bar y)\in \Delta$.

This shows that $\Delta$ is closed, which is the necessary and sufficient condition for $X/\sim$ to be Hausdorff.

the condition $X$ compact is not necessary. let $G$ a closed subgroup of $X$.
the topology on $X/G$ is the minimal top such that $p: X\to X/G$ is continuous.
let $U=X-G$ open subset of $X$. Then $p^{-1}(p(U))=U=$ open, showing that $V=p(U)$ is open in $X/G$.
let $x,y\in X$ such that $\bar x=p(x)$ and $\bar y=p(y)$ are distinct in $X/G$.
in other words $y^{-1}x\notin H\Leftrightarrow y^{-1}x\in U$.
Then $xV$ is an open neighborhood of $\bar y$, $yV$ is an open neighborhood of $\bar x$, and $xV\cap yV=\emptyset$.


Two circles C and C' have centers O and O' and meet at M and N. The common tangent closer to M touches C at A and C' at B. The line through B perpendicular to AM meets the line OO' at D. BO'B' is a diameter of C'. Show that M, D and B' are collinear.

Let $MN\cap{AB}=\{X\}$, then $\angle{AXM}=\angle{OO'B}$ and $\angle{MAX}=\angle{DBO'}$. Therefore
$\triangle{MAX}\sim\triangle{DBO'}\Longleftrightarrow\triangle{MAB}\sim\triangle{DBB'}$. From here it can be concluded that $\angle{AMB}=\angle{BDB'}$ and the conclusion follows.


probability
combinatorics unsolved
combinatorics

Alphonse and Beryl, as usual, are playing one of their amusing games.  :D
The game is the following:

Alphonse throws a die, and Beryl throws it twice, and Alphonse now throws it twice, and so on. (so the order would be A, B, B, A, A, B, B, A, etc.)
The winner is the person who gets the number 1 for the first time.

Find the probability that Alphonse will win.

I guess this is really easy.. I solved it just now..  :blush:

Let $p$ be the desired probablity.
Then $p = \frac 1 6 + \frac 1 6 (\frac 5 6 )^3 + (\frac 5 6 )^4 p$
since at his third throw, knowing that the game has not been won before, the first player return back to the start of the game.
Thus $p = \frac {31} {61}.$

Pierre.


YAY SEPTEMBER!!! ONLY JUST OVER TEN MONTHS TILL IMO!!!!

SCHOOL IN FOUR DAYS! :)

I like September.

[quote="nutz_for2.718281828"]IT's SEPTEMBER!!!  I'm gonna go crazy!!! *goes crazy with delight*

less than 4 months until I get my macintosh!!![/quote]

Why get a mac? They're overpriced and mac OS is uncompatible with way too much stuff.  :huh:

On the other hand, it's schooltime! NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

uhh... I usually go back to school in July, but my dad made it August. What's so terrible about school? :huh:

P.S. My Windows computer is do &*^(ed crappy that I can't play ANY disc!!! THAT'S the matter with buying a mac. Also, My screen is also %#^@ed crappy, that all of the math problems I do I misread.

Within four months, Nutz will be in the upper right corner :
[img]http://www.ctrlaltdel-online.com/comics/20031226.jpg[/img]

hahaha...nutz_fore you sure you don't have spy ware?*eyes suspiciously*

YAY IT'S SEPT! I'm already thinking of Trick or Treating(only the dressing up part)

I'll dress up like pi.

[quote="funcia"]
I'll dress up like pi.[/quote]
Don't get me the wrong way but again I can't resist  :wink:(especially with that signature) :
[img]http://users.telenet.be/frederic_bel6/Fun/refusepi.jpg[/img]

that is so funny i actually freak out my peers with my recitations of pi...

fredbel my dog says he can't resist that man either

[quote="nutz_for2.718281828"]P.S. My Windows computer is do &*^(ed crappy that I can't play ANY disc!!![/quote]
Wow, I'm having the same problem. My computer will read disks, but very slowly. I assume it's a faulty drive, as Spybot and Mcafee didn't catch anything.

[url]http://www.albinoblacksheep.com/flash/apple.php[/url]

check that out! not very funny though ...

[quote="funcia"]I want to put one but I think it's too many KBs[/quote]

host it to imageshack or photobucket or putfile or something like that and link it or [img] or something like that, you don't need to host it onto aops itself, do the aops people a favor

[quote=1=2]IT's SEPTEMBER!!!  I'm gonna go crazy!!! *goes crazy with delight*

less than 4 months until I get my macintosh!!![/quote]

/bump lol

[quote=artofproblemsolvinguser1]bro this was 15 years ago... These guys all probably have jobs rn.[/quote]

OP still likes math tho
look at his recent posts


inequalities
limit
function
logarithms
calculus
geometry
geometric transformation

The number $\frac{2}{1}+\frac{2^{2}}{2}+\frac{2^{3}}{3}+...+\frac{2^{n}}{n}$ is represented as a non-reducible fraction $\frac{p_n}{q_n}$ ("non-reducible" means that the numbers $p_n$ and $q_n$ are coprime).

[b]a)[/b] Prove that $p_n$ is an even number.
[b]b)[/b] Prove that if $n > 3$, then $p_n$ is divisible by 8.
[b]c)*[/b] Prove that for every natural k, one can find a natural number n such that the numbers $p_n$, $p_{n+1}$, $p_{n+2}$, ... are all divisible by $2^k$.

  Darij

VERY nice problem, darij! anyway i can't give a solution for c) you will really understand, sorry!

a) and b) are easy: all rationals are divisible by 2, which gives a), and for b), just look at the first 4 numbers: their sum is divisible by 8, and every other rational of that sum is divisible by 8 as well.

so, let's come to c). i will give a very short introduction to p-adic number theory.
introduce a new norm in the rationals: $|x|_p:=p^{-m}$ where $p^m||x$ is the exact power of $p$ in $x$. furthermore, define $|0|_p=0$(giving rise to the fact that 0 is infinitely often divisible by $p$). obviously, $|xy|_p=|x|_p|y|_p$.
define the distance $d(x,y)=|x-y|_p$. let's show it is a distance. it is obvious that $d(x,y)\geq 0$ with equality iff $x=y$ as well as $d(x,y)=d(y,x)$. one great thing about p-adic number theory is that the triangle inequality $d(x,z)\leq d(x,y)+d(y,z)$ may be replaced by the sharper inequality $d(x,z)\leq \max \left(d(x,y),d(y,z)\right)$. this is equivalent to $|x+y|_p\leq \max \left(|x|_p,|y|_p\right)$ after trivial substitutions. now, if $x$ is at most as often divisible by $p$ as $y$, then $x+y$ will be as well at least as often divisible by $p$ as $x$. that's what the inequality says.
so, we get a new distance in the rationals! in fact, one can show, that trivial generalizations of the above distances and the standard distance give all distances in the rationals.
now, let's complete the rationals with respect to new norm $|\cdot |_p$. that means, let's look at the set of all cauchy sequences modulo all sequences converging to zero. cauchy sequences are sequences $(a_n)$ such that $\forall_{\epsilon>0} \exists_{N} \forall_{m,n\geq N} |a_m-a_n|_p<\epsilon$. sequences converging to zero are those sequences $(a_n)$ such that $\forall_{\epsilon>0} \exists_{N} \forall_{n\geq N} |a_n|_p<\epsilon$. we call that completion $\mathbb{Q}_p$.
what do we want to show in c)? in fact we want to show that our sequence $a_n=\sum_{i=1}^{n} \frac{2^i}{i}$ converges to zero with respect to the 2-adic distance. this sounds rather weird since we always imagine our sequence in the rationals with the normal distance. in fact, it is trivial to show that the sequence $p^n$ converges to zero in $\mathbb{Q}_p$, whereas $\frac{1}{p^n}$ diverges! in the following, you always have to keep in mind that a number is small iff it is often divisible by $p$.
the next thing we need, is that a p-adic series converges iff it's n-th term goes to 0:

let $(a_n)$ be a sequence, $s_n=\sum_{i=1}^{n} a_n$. then $s_n$ converges iff $a_n$ goes to zero.
to converge means here to be a member of $\mathbb{Q}_p$ which means that our sequence is a cauchy sequence.
so, let's assume $\lim_{n\to \infty} a_n=0$. then $|s_n-s_m|_p=|\sum_{i=m+1}^{n} a_n|_p\leq \max_{m+1\leq i\leq n} |a_i|_p$. since $a_i$ goes to zero, there exists $N$ such that for all $m\geq N$ $|a_m|_p<\epsilon$. we may take that $N$ in order to have $|s_n-s_m|_p<\epsilon$ for all $m,n\geq N$.
if $s_n$ is convergent, then $|a_n|_p=|s_n-s_{n-1}|_p<\epsilon$ for sufficiently large $n$ giving that $a_n$ goes to zero.

introduce the function $\log_p(1+x)=\sum_{i=1}^{\infty} (-1)^{i-1} \frac{x^i}{i}$. it converges iff $|x|_p<1$ by the above theorem.
the next thing to prove is that
$\log_p(xy)=\log_p(x)+\log_p(y)$ if $|x-1|_p,|y-1|_p<1$(this implies $|xy-1|_p<1$).
the above form is intuitive, but harder to prove, so let's prove $\log_p(1+x+y+xy)=\log_p(1+x)+\log_p(1+y)$ if $|x|_p<1$, $|y|_p<1$. it is clear that $|x+y+xy|_p\leq \max \{|x|_p,|y|_p,|xy|_p\}<1$. everything is convergent, so we just need to check that the power series are identical(this is well-known since we know the identity holds in real numbers, but let's have the fun of proving it).
the coefficient of $x^i$ is $\frac{(-1)^{i-1}}{i}$ since $x^i$ appears only in $(-1)^{i-1}\frac{(x+y+xy)^i}{i}$. this is sure equal to the right. the same for $y^i$.
the coefficient of $x^iy^j$ for $i,j\neq 0$ is 0. assume $i\geq j$. $x^iy^j$ appears in $(-1)^{k-1}\frac{(x+y+xy)^k}{k}$ where $k=i,...,i+j$ with quantity $\frac{(-1)^{k-1}}{k}\frac{k!}{(k-j)!(k-i)!(i+j-k)!}$. it is left to show
$\sum_{k=i}^{i+j} (-1)^{k-1}\frac{(k-1)!}{(k-j)!(k-i)!(i+j-k)!}=0$.
now, after that much analysis, let's try some combinatorics. hmm...i don't see it! someone finds a nice proof for that one?
[LOL: mathematica just told me $\sum_{k=i}^{i+j} (-1)^{k-1}\frac{(k-1)!}{(k-j)!(k-i)!(i+j-k)!}=(-1)^{i-1}\frac{\sin(j\pi)}{ij\pi}$ :D]

the rest is clear: $\log_2(-1)+\log_2(-1)=\log_2(1)=0$, thus $\log_2(-1)=0$ giving exactly the desired result.

Peter

Sorry Peter, can you explain a little bit the last part, I understood the definitions but i dont know how do you use it and i dont get the logarithmic thing....

Also if it is not a problem, i would like to know more about p-adic theory....

Thanks.

I hope Darij will translate original solution for this problem ;)
I have only small hint for it, and I can't understand how it works.

hmm...

first, i have to say that after introducing $\mathbb{Q}_p$, i had to do some more things first, like extending the norm to all elements in $\mathbb{Q}_p$ (this can be done by taking the limit of the norms...actually, either that limit is 0 or it is constant from a certain point on), and then showing that any cauchy sequence of elements of $\mathbb{Q}_p$ is indeed convergent. however, proving this is straightforward. you might look at elements of $\mathbb{Q}_p$ in the form $\sum_{i=n}^{\infty} a_ip^i$, for some $n$ and $a_i=0,...,p-1$. this is the way i think of them.
yeah, then for the problem, it is good to introduce the analogous of the logarithm(the definition as a power series is exactly the same). then one proves the functional equation and can use that equation for proving that the series of the problem converges to 0 2-adically.
i had much problems when first confronted to p-adic number theory since it is quite unintuitive first. but that gets better. however, i can't give a full introduction to p-adic number theory, try to look for a good book. everything i new is out of the book "Introduction to Cyclotomic Fields" by Washington which needs it to do some other things(the theory of p-adic L-functions etc.). that book is really hard stuff, "introduction" is maybe the wrong word for it.

Peter

I downloaded from http://kvant.mccme.ru/ all "Kvant" volumes of 1970—2002. So I have them on 9 CD-Rs now and I made a little text file so that I can easily find all problems and solutions. M434 solution is here:
http://kvant.mccme.ru/1977/12/gif/77_12-36.gif
http://kvant.mccme.ru/1977/12/gif/77_12-37.gif
Sorry, I have no time to translate them, because I'm downloading Athens 2004 results and images right now :D

I think, it will be better if you willn't hurry next time.  ;)
Anyway, solution is very nice (though I had not doubts about it :) ).

hmm...as far as i understand it(i speak no word russian...ok, i now how to pronounce "what" and "yes" in russian, but i have no idea how to write) it's more or less the same thing, though more or less "fallen from heaven". after taking a look again at my solution i saw that one can do it elementary in a quite similar manner to what they did. there's no real need to introduce p-adic numbers, one can just do an argument by how often numbers are divisible by 2 in our case. anyway, i think my solution is more enlightening if one understands the solution :D

Peter

Peter, looks like you have understood the idea of the proposed solution quite well... Actually, I would love to translate it, but I have not too much time now. Yes, it is "fallen from heaven"; in fact, the last sentence of the solution reads: "The origin of this solution is explained in the paper <<2-adic numbers>>".  ;)  Alas, I have no time to look up this paper now.

Thanks for your solution, too, but I think I will try to learn some more fundamental number theory first :-) Actually, I don't even understand the beginning: What is the "exact power" of a prime number in a [b]rational[/b] (not integer) number?

[i]Remark.[/i] The statement of problem [b]b)[/b] in Kvant 3/1977 erroneously claimed it to be true for $n\geq 3$ instead of $n>3$. This is wrong since [b]b)[/b] is not true for $n=3$. I have corrected this flaw when translating the problem.

  Darij

Since any rational can be represented as $\prod p_i^{a_i}$ where $p_i$ are the primes and $a_i$ are [b]integers[/b] (most of which are $0$, of course), the exact power of $p_i$ is $p_i^{a_i}$.

Thanks, Grobber, now I understand...

[quote="grobber"]Since any rational can be represented as $\prod p_i^{a_i}$ where $p_i$ are the primes and $a_i$ are [b]integers[/b] (most of which are $0$, of course),[/quote]

Aah... that is nice! And, of course, this representation is unique. Why did I never hear anything about this representation?

  Darij

I was requested to translate the [url=http://kvant.mccme.ru/1977/12/p36.htm]proposed solution (from Kvant 12/1977, p. 36-37)[/url], so here I do. The solution uses a bit of elementary calculus but no preknowledge from number theory. Here is a translation:

[b]M434.[/b]

The number $\frac21+\frac{2^2}{2}+\frac{2^3}{3}+...+\frac{2^n}{n}$ is represented as a non-reducible fraction $\frac{p_n}{q_n}$ ("non-reducible" means that the numbers $p_n$ and $q_n$ are coprime).

[b]a)[/b] Prove that $p_n$ is an even number.

[b]b)[/b] Prove that if n > 3, then $p_n$ is divisible by 8.

[b]c)*[/b] Prove that for every natural k, one can find a natural number n such that the numbers $p_n$, $p_{n+1}$, $p_{n+2}$, ... are all divisible by $2^k$.

[i]Solution.[/i] [b]a), b)[/b] If an irreducible fraction $\frac{p}{q}$ is the sum of some irreducible fractions $\frac{p_1}{q_1}$, $\frac{p_2}{q_2}$, ..., $\frac{p_n}{q_n}$ whose numerators $p_1$, $p_2$, ..., $p_n$ are all divisible by $2^k$, then p is also divisible by $2^k$. For the solution of problem [b]a)[/b], it remains to note that the numerators of the fractions $\frac21$, $\frac{2^2}{2}$, $\frac{2^3}{3}$, ... remain even after reduction, since any natural number m satisfies the inequality $2^{m-1}\geq m$. For the solution of problem [b]b)[/b], it would be enough to have the inequality $2^{m-3}\geq m$. This inequality, however, holds only beginning with m = 6. Hence, we must show that the numerators of the reduced forms of the sums $\frac{2}{1}+\frac{2^{2}}{2}+\frac{2^{3}}{3}+\frac{2^{4}}{4}$ and $\frac{2}{1}+\frac{2^{2}}{2}+\frac{2^{3}}{3}+\frac{2^{4}}{4}+\frac{2^{5}}{5}$ are divisible by 8. These sums equal $\frac{32}{3}$ and $\frac{256}{15}$, respectively, so the proof is complete.

[b]c)[/b] Consider the polynomial $L_{n}\left( x\right) =\frac{x}{1}+\frac{x^{2}}{2}+...+\frac{x^{n}}{n}$. We will show that the polynomial f(x) defined by the equation $f\left( x\right) =L_{n}\left( 2x-x^{2}\right) -2L_{n}\left( x\right) $ can be written in the form

$\textbf{(1)}\;\;\;\;\;\;\;\;\;\; f\left( x\right) =x^{n+1}\cdot \left( \frac{a_{0}}{n+1}+\frac{a_{1}}{n+2}x+...+\frac{a_{n-1}}{2n}x^{n-1}\right) $,

where $a_0$, $a_1$, ..., $a_{n-1}$ are integers.

For this end, it is enough to show that the derivative f'(x) has the form

$f^{\prime }\left( x\right) =x^{n}\left( a_{0}+a_{1}x+...+a_{n-1}x^{n-1}\right) $

(then, noting that f(0) = 0, we obtain [b](1)[/b]).

Note that

$L_{n}^{\prime }\left( x\right) =1+x+x^{2}+...+x^{n-1}=\frac{1-x^{n}}{1-x}$.

Thus,

$f^{\prime }\left( x\right) =L_{n}^{\prime }\left( 2x-x^{2}\right) \cdot \left( 2-2x\right) -2L_{n}^{\prime }\left( x\right) $
$=\frac{1-\left( 2x-x^{2}\right) ^{n}}{1-2x+x^{2}}\cdot \left( 2-2x\right) -\frac{2\left( 1-x^{n}\right) }{1-x}$
$=2\cdot \frac{x^{n}-x^{n}\left( 2-x\right) ^{n}}{1-x}=x^{n}\cdot \frac{2\left( 1-\left( 2-x\right) ^{n}\right) }{1-x}$

and

$1-\left( 2-x\right) ^{n}=\left( 1-\left( 2-x\right) \right) \left( 1+\left( 2-x\right) +\left( 2-x\right) ^{2}+...+\left( 2-x\right) ^{n-1}\right) $
$=-\left( 1-x\right) \left( 1+\left( 2-x\right) +\left( 2-x\right) ^{2}+...+\left( 2-x\right) ^{n-1}\right) $.

This proves that the polynomial f'(x) can be written in the form

$f^{\prime }\left( x\right) =x^{n}\left( a_{0}+a_{1}x+...+a_{n-1}x^{n-1}\right) $.

Hence, as said above, it follows that the polynomial f(x) can be written in the form [b](1)[/b].

[Explicitely,

$a_k=\left( -\frac{1}{2}\right) ^{k-1}\cdot \sum_{u=0}^{n-1}\left( \binom{u}{k}\cdot 2^{u}\right) $;

this formula makes it possible to prove [b](1)[/b] without calculus by induction, but believe me, the inductive proof is a horror. Darij]

Since $f\left( 2\right) =L_{n}\left( 0\right) -2L_{n}\left( 2\right) =-2L_{n}\left( 2\right) $, we can set x = 2 in the equality [b](1)[/b] and get

$L_{n}\left( 2\right) =-2^{n}\cdot \left( \frac{a_{0}}{n+1}+\frac{2a_{1}}{n+2}+...+\frac{2^{n-1}a_{n-1}}{2n}\right) $
$=-\left( \frac{2^{n}a_{0}}{n+1}+\frac{2^{n+1}a_{1}}{n+2}+...+\frac{2^{2n-1}a_{n-1}}{2n}\right) $.

From the inequality $n<2^{\frac{n+1}{2}}$, which holds for every natural number n, we have $2n<2\cdot 2^{\frac{n+1}{2}}=2^{\frac{n+3}{2}}$. Thus, all denominators on the right hand side are smaller than $2^{\frac{n+3}{2}}$. Hence, none of these denominators is divisible by $2^{\left[ \frac{n+3}{2}\right] +1}$; consequently, every of the numerators on the right hand side, even after being reduced, contains the 2 at least in the power $2^{n-\left( \left[ \frac{n+3}{2}\right] +1\right) }$ (of course, this only makes sense for $2^{n-\left( \left[ \frac{n+3}{2}\right] +1\right) } \geq 1$, i. e. for $n\geq 4$). Thus, the number $L_{n}\left( 2\right) =\frac{2}{1}+\frac{2^{2}}{2}+...+\frac{2^{n}}{n}$ itself must be a fraction whose numerator, in reduced form, is divisible by $2^{n-\left( \left[ \frac{n+3}{2}\right] +1\right) }$.

To complete the solution of [b]c)[/b], it remains to note that the exponent $n-\left( \left[ \frac{n+3}{2}\right] +1\right) =n-1-\left[ \frac{n+3}{2}\right] \geq n-1-\frac{n+3}{2}=\frac{n-5}{2}$ increases unboundedly with increasing n.

The paper "2-adic numbers" explains the origin of this solution.

[i]D. Faddeev[/i]

  Darij

i'm pretty sure that somewhere on the site is a solution to the $\sum$ {ugly fraction} $=0$ (which appears at the end of Peter's solution). But I didn't anything with the Search button. can you help me?

If you are talking about this one:

[quote="Peter Scholze"]$\sum_{k=i}^{i+j} (-1)^{k-1}\frac{(k-1)!}{(k-j)!(k-i)!(i+j-k)!}=0$.[/quote]

then see http://www.mathlinks.ro/Forum/viewtopic.php?t=16482 .

  darij

Just an easier prove for $\ln(x)+\ln(y)=\ln(xy)$:
It's true for $x=1$.
Now the functions $\ln(x)+\ln(y)$ and $\ln(xy)$ as functions of $x$ have the same derivation everywhere where they are defined.
Since they are the same at some point, they are identical everywhere.

Could you give me examples of numbers that are in $\mathbb{Q}_p$, but not in $\mathbb{Q}$? I'm almost understanding Peter's post...

There are lot's of examples:
First, note that every $p$-adic number can be written as (infinite) sum $\sum_{k=m}^{\infty} a_k p^k$ with $m \in \mathbb{Z}$ and $a_k \in \{0,1,2,...,p-1\}$ (this is just nicer to work with, especially to write them down, since you can then write it as a 'infinite' number to the base $p$).

Now as in decimal fractions, exactly those are rational whose sequence $(a_k)$ gets periodic, so as good as all are not in $\mathbb{Q}$ (try to prove it when you want, it's not too hard).
Some examples are then the transcendental numbers $\sum_{k=0}^{\infty} p^{k!}$ and $\sum_{k=0}^{\infty} p^{k^2}$.
Another kind of example is given by the following:
When $p$ is odd, then exactly those $n \in \mathbb{Z}, p \nmid n$ are squares in $\mathbb{Q}_p$ (respectivly in $\mathbb{Z}_p$ too, the set of $p$-adic integers), that are squares $\mod p$.
E.g., since $-1$ is a square $\mod 5$, it's also a square in $\mathbb{Q}_5$, the square of $3+3\cdot 5+2\cdot 5^2+3\cdot 5^3+1 \cdot 5^4+2\cdot 5^6+1\cdot 5^7+4\cdot 5^8+1\cdot 5^9 + ...$ (such roots can be found using Hensels Lemma) and it's negative.


This is kind of a random question, but can't AoPS just somehow program LaTex into the website, so like you don't have to download it or anything? Is it possible?

Latex is incorporated into AOPS. You don't have to install anything to use Latex  :rotfl:

See:
$ LaTeX$
$ \lim_{x \to 0}\frac{\sin x}{x} \equal{} 1$

Just enclose the code within Dollar Signs.
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php

AoPS does have an online "$ \text{\LaTeX}$ Program".
http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php .

The only reason to download $ \text{\LaTeX}$ is if you want to make full documents and use the typesetting features. If all you want is the pretty math, the TeXer suffices.

oh.... sry i only looked at the "Download LaTex" sign...

thanks for letting me know


The whole numbers are written consecutively in the pattern
below. In which row (A, B, C, D or E) will the number 500 be
written?
\begin{tabular}{llllllll} Row A & 5 & 6 & 7 & & 17 & 18 & 19 \\
Row B & 4 & & 8 & & 16 & & \cdot \\
Row C & 3 & & 9 & & 15 & & \cdot \\
Row D & 2 & & 10 & & 14 & & \cdot \\
Row E & 1 & & 11 & 12 & 13 & & \\
\end{tabular}

The $ \text{\LaTeX}$ dollar signs were missing.

Solution: The pattern repeats every 12 numbers.

500 mod 12 is 8.

The 8th term is in row $ \boxed{B}$, and so is the 500th.


Hi, I am trying to prove (a+b+c)² ≥ ab + ac+ bc.
a,b,c are real numbers. They can be negative.
I could prove it in two cases:

[b]Case 1:[/b] ab+ac+bc ≥ 0:

(a+b+c)² - (ab+ac+bc)
= a² + b² + c² + 2(ab+ac+bc) - (ab+ac+bc)
= a² + b² + c² + (ab+ac+bc) ≥ 0

Therefore it is true.

[b]Case 2:[/b] ab+ac+bc < 0:

(a+b+c)² - (ab+ac+bc)
= (a+b+c)² + ( -(ab+ac+bc) ) > 0

Therefore it is also true.

My question is: is there a way to prove it in one go without using two cases?

- Thanks

Yes you can kill it easily by one :wink:
We have to prove that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \geq 0$,which is true for all $ a,b,c \in R$(just multiply with $ 2$ and see for squares)

Where a,b and c are of the real number system and a>b>c:

(a-b)>=0 therefore a^2-2ab+b^2>=0, a^2+b^2>=2ab

Similarly for (a-c)^2: a^2+c^2>=2ac
              &
                (b-c)^2: b^2+c^2>=2bc

By adding each of these:

(a^2+b^2)+(a^2+c^2)+(b^2+c^2)>=2ab+2ac+2bc

2(a^2+b^2+c^2)>=2(ab+ac+bc)

therefore a^2+b^2+c^2>= ab+bc+ac

Thats one method.... i think this method is pretty cool actually


funny exercice:
find $ \lim_{x \to \infty} \exp( \minus{} n \lambda) \sum_{k \equal{} 0}^{[nx]} \frac {(n \lambda)^{k}}{k!}$ for $ x \geq 0$, where $ \lambda>0$ is fixed.

Isn't it Central Limit Theorem for Poisson?  :)


calculus
integration
logarithms
calculus computations

Let  A(n) = (log n)/n
Then, is there an easy way to estimate the infinite sum:
A1 + A2 + ....

Perhaps i'm just rusty (been reviewing my calculus as i'm about to start a post-graduate course that requires it. Haven't touched this material in years...)
Some directions I'm considered:
- Tried replacing to equivalent notation:   A(n)= m / 2^m
- Tried bounding it with an integral, but wasn't sure how to calculate Integral[(log n)/n].

Any help or hints would be greatly appreciated.
Thanks :)

Thanks so much.
Of course it should have been obvious... sorry for being a headful of rust...

thanx again.

Assuming your question is the straightforward "does it converge or diverge?", focus first on size, not on the mechanics of any tests. Since $\ln n$ is (eventually) greater than 1, we have the direct (one-sided) comparison

$\sum\frac{\ln n}n>\sum\frac1n$

And the harmonic series diverges. Since this is larger than the harmonic series, it diverges.

The only reason I can see for bringing the integral test into this would be if you asked a different and more detailed question, such as the rate of growth of the partial sums $\sum_{k=2}^n\frac{\ln k}k.$

yeah like kent said, since the harmonic series is divergent and $\sum \frac{\ln{x}}{x} > \sum\frac 1x$ then this one is divergent also.


trigonometry
search
geometry proposed
geometry

If the angles of a triangle verify the relation :
\[
\displaystyle \frac {\sin^2 \hat {A}+\sin^2 \hat {B}+\sin^2 \hat {C}} {\cos^2 \hat {A}+\cos^2 \hat {B}+\cos^2 \hat {C}}=2
\]
then the triangle is right triangle.

Before Orl posts an angry remark about old ISL problems and the "search" option  :) , I give the link to the thread where this one was solved: http://www.mathlinks.ro/Forum/viewtopic.php?t=21148 .

  darij


inequalities
trigonometry
inequalities proposed

If $ a, b, c$ are real numbers with $ abc \neq 0$ and  $ x, y, z$ are real numbers, no two of them are zero, then

$ \frac{(a^2c^2 \plus{} b^2x^2)(a^2b^2 \plus{} c^2y^2)(c^2b^2 \plus{} a^2z^2)}{a^2b^2c^2(b^2x \plus{} c^2y \plus{} a^2z)^2} \ge \frac{3}{4}$.

My very elementary solution, not hard :)
I don’t like de-homogenizing, but here it really simplifies the thing...

Let $ X: \equal{}\frac{bx}{ac}$ etc. Then the inequality becomes $ \frac{(1\plus{}X^2)(1\plus{}Y^2)(1\plus{}Z^2)}{(X\plus{}Y\plus{}Z)^2}\ge\frac34$. (*)

This suggests to put $ X\equal{}: \tan\alpha$ etc. (but no triangle here!) Then
$ \Leftrightarrow \frac{1 }{ \prod\cos^2\alpha (\sum\tan\alpha)^2}\ge\frac34$
$ \Leftrightarrow \prod\cos\alpha \sum\tan\alpha \le\frac2{\sqrt{3}}$
$ \Leftrightarrow \sin(\alpha\plus{}\beta\plus{}\gamma)\plus{} \sin\alpha\sin \beta\sin \gamma \le\frac2{\sqrt{3}}$.
If not $ \alpha\equal{}\beta\equal{}\gamma$, the LHS of this can be increased. So it suffices to prove (*) for $ X\equal{}Y\equal{}Z$.

But by AM-GM, $ \frac{(1\plus{}X^2)^3}{9X^2}\equal{} \frac{(\frac12\plus{}\frac12\plus{}X^2)^3}{9X^2}\ge \frac{27\frac{X^2}4}{9X^2}\equal{} \frac34$.
Equality occurs iff $ b^2x\equal{}c^2y\equal{}a^2z\equal{} \frac{abc}{\sqrt{2}}$. [Why didn't you follow the convention and rather combine $ a$ with $ x$ etc?]

Note that any three of $ a,b,c,x,y,z$ can be chosen as arbitrary positive numbers and will yield a (unique) equality 6-tuple.. :D


Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that \[ f(x^3)\minus{}f(y^3)\equal{}(x^2\plus{}xy\plus{}y^2)(f(x)\minus{}f(y))\forall x,y\in\mathbb{R}.\]

The solution space are all affine-linear functions:

Setting $ y\equal{}0$ yields $ f(x^3)\equal{}f(0)\plus{}x^2(f(x)\minus{}f(0))$ and resubstitution into the original equation gives after equivalent transformation: $ (y^2\minus{}x^2)f(0)\equal{}(xy\plus{}y^2)f(x)\minus{}(x^2\plus{}xy)f(y)$. Evaluation of the special cases $ x\equal{}\pm 1$ yields the identities $ f(y)\equal{}\pm y[f(\pm 1)\minus{}f(0)]\plus{}f(0)$ (with $ y\not\equal{}\mp 1$ in the respective cases). For any $ y\not\in\{0,\pm 1\}$ we get by subtraction of the two identities: $ f(1)\plus{}f(\minus{}1)\equal{}2f(0)$ from which we see that $ f(y)\equal{}y[f(1)\minus{}f(0)]\plus{}f(0)$ actually holds for all $ y$. Conversely, any affine-linear function fulfills the required property, since $ (x\minus{}y)(x^2\plus{}xy\plus{}y^2)\equal{}x^3\minus{}y^3$ holds.


Find all functions $f:\mathbb{R}\to\mathbb{R}$ for which
\[ x(f(x+1)-f(x)) = f(x), \]
for all $x\in\mathbb{R}$ and
\[ | f(x) - f(y) | \leq |x-y| , \]
for all $x,y\in\mathbb{R}$.

[i]Mihai Piticari[/i]

We immediately obtain $\frac{f(x+n)}{x+n}=\frac{f(x)}{x}$, $0\not\in \{x,x+1,...,x+n\}$. So if $\frac{f(y)}{y}\neq \frac{f(x)}{x}$ then we easily obtain contradiction with the inequality.

[quote="Myth"] So if $\frac{f(y)}{y}\neq \frac{f(x)}{x}$ then we easily obtain contradiction with the inequality.[/quote]

How? I can't see.

[quote="bodom"]how do we obtain contradiction? :blush:[/quote]

Well, that's what I was asking...

after working a bit indeed we have a contradiction but it is not that obvious i guess :? .if u want i will post my solution :)

here is an idea:
First of all, from the second condition, we find that f is a particular case of Lipschitz function wich is continuous.
We can easily proove f(n)=nf(1) for every integer n.
The problem is that i cannot extend this to any rational number.
If f(x)=xf(1) for every rational x we have cauchy and it's easy.

i do not think a 9th grader would know what a continuos function is

and i do not see how do u get that kf(x)=xf(k) for every reals k,x

ok.here we go(excuse me but i do not know yet how to use latex :blush: ).just like myth said we obtain that(*)  f(x+n)/(x+n)=f(x)/x for every real number x and every integer n.then it is clear that f(n)=na for every integer n,where f(1)=a and |a|<=1.f(0)=0 and then |f(x)|<=|x|.we want to prove that f(x)=ax for every real x.let us assume that there is a number x from R-Z with f(x)=bx,|b|<=1.
1) b<a
writing the ineq we get:
|na-bx|<=|n-x|.where x is the number from above and n is an integer,let's say a positive one.from (*) we know that f(x+k)/(x+k)=f(x)/x=b.let us choose k and n so that n=[x]+1.
we now get that (1-a)/(1-b)>=x/n.we see that t=(1-a)/(1-b)<1.if t is very small. let us choose a big n and x and we have a contradicttion.if t-->1 then we choose n--->infinit( :blush: ) and again we have contradiction because for n big enough x/n>t.
2)b>a.it's about the same thing.i'm to lazy to complete :D .i hope it is correct.

[quote="bodom"]what do u mean?isn't that clear enough?oh and i managed to see the official one.it's longer then mine but more beautiful[/quote]

Well, maybe it's clear. I didn't understand one part, but now I do. I'll complete reading it later.

Could you post the official solution?

no i can not.i only said that i sow the solution(in a book that was not mine but i only remembered that it was longeeer and somehow more beautiful).sorry.


Cho x,y,z>0 thõa x+y+z=1. CMR:
$ \frac{1\plus{}x}{y\plus{}z}\plus{}\frac{1\plus{}y}{z\plus{}x}\plus{}\frac{1\plus{}z}{x\plus{}y}$     :leq   2$ {(\frac{x}{z}\plus{}\frac{z}{y}\plus{}\frac{y}{x})}$
      Các bạn xen luôn cả đề nha.Nhớ là phải cho ý kiến đấy nhé
1/i/GPT: $ sinx\plus{}\sqrt[2]{1\minus{}sinx}\equal{}cosx\plus{}\sqrt[2]{1\minus{}cosx}$
   ii/CMR hệ sau có đúng 3 nghiệm:$ x^3\plus{}1\equal{}3y;y^3\plus{}1\equal{}3x$

2/Nhận dạng tam giác
cosA+2cosB+3cosC+2sin$ \frac{A}{2}$+4sin$ \frac{B}{2}$+6sin$ \frac{C}{2}$=9

3/  CMR tổ hợp chập(n+1)của(2n+2)   chia hết cho (n+2)

4/Cho hình chóp S.ABC có cạnh bên SA vuông góc với mp(ABC).Gọi  $ B_1;C_1$theo thứ tự là hình chiếu của A trên SB;SC.Giả sử BC cắt $ B_1C_1$tại I.Gọi AD là đường kính đường tròn ngoại tiêp tam giac  ABC  CMR:
  i/SD vuông góc với AI
  ii/ :exists điểm K cách đều 6 điểm  A,B,C,D,$ B_1,C_1$
5/như trên

Posts in the high school forum should be made in English.  Post in Vietnamese should be made [url=http://www.artofproblemsolving.com/Forum/index.php?f=256]here[/url].


inequalities
function
inequalities proposed

Suppose that $a$ is a real postive number bigger than one and $n$ is a natural number. Prove that:

$a^{n+1}+(2n+1)(1-a)\geq\frac{1}{a^{n}}$

That is not true , when $n=1$

$a^2+3(1-a)\geq \frac{1}{a}$ is not always true . Try to pick very small $a$ then RHS will be infinity will LHS is finite .

sorry. my mistake. $a$ is bigger or equal than one.

Here's my solution:

According to the AM-GM inequality we have:

$1+a+a^2+...+a^{2n}\geq(2n+1)a^n$

therefore we have

$\frac{a^{2n+1}-1}{a-1}\geq(2n+1)a^n$

by multiplying by $\frac{a-1}{a^n}$ we will get the desired inequality.


number theory
algebra
Combinatorial Number Theory
Sequence
IMO Shortlist

A nonempty set $ A$ of real numbers is called a $ B_3$-set if the conditions $ a_1, a_2, a_3, a_4, a_5, a_6 \in A$ and $ a_1 \plus{} a_2 \plus{} a_3 \equal{} a_4 \plus{} a_5 \plus{} a_6$ imply that the sequences $ (a_1, a_2, a_3)$ and $ (a_4, a_5, a_6)$ are identical up to a permutation. Let $A = \{a_0 = 0 < a_1 < a_2 < \cdots \}$, $B = \{b_0 = 0 < b_1 < b_2 < \cdots \}$ be infinite sequences of real numbers with $ D(A) \equal{} D(B),$ where, for a set $ X$ of real numbers, $ D(X)$ denotes the difference set $ \{|x\minus{}y|\mid x, y \in X \}.$ Prove that if $ A$ is a $ B_3$-set, then $ A \equal{} B.$

[hide]Because $D(B) = D(A)$, we can find functions $f$ and $g$ such that $b(m) - b(i) = a(f(m,i)) - a(g(m,i))$ for all $m, i$ with $0 \leq i \leq m-1$ and $f(m,i) > g(m,i)$. We shall show that $a(f(m,i)) - b(m)$ is constant.

Let $m_1 ,i_1, m_2, i_2$ be any nonnegative integers with $m_1 \geq i_1 + 1$ and $m_2 \geq i_2 + 1$. For any $k > i_1$, we have that $(b(k) - b(i_1)) - (b(m_1) - b(i_1)) = a(p) - a(q)$ for some integers $p > q$. Let $s_k$ be a sequence of positive integers such that $b(k) - b(i_1) = a(s_k) - a(u)$ for some $u < s_k$. We also have $b(m_1) - b(i_1) = a(f(m_1,i)) - a(g(m_1,i))$.
Substituting these, we get that

$a(s_k) + a(g(m_1,i)) + a(q) = a(p) + a(f(m_1,i_1)) + a(u).$

If we take $k$ to be sufficiently large, we can force $s_k$ to grow sufficiently large as well (or else $b(k) - b(i_1)$ will be unbounded, while $a(s_k) - a(u)$ will be bounded, since $u < s_k$.) Hence, for sufficiently large $k$, $s_k > f(m_1, i_1)$. $s_k > u$, so since $A$ is $B_3$, we must have $s_k = p$. It follows that $f(m_1, i_1) = q$ and $u = g(m_1, i)$. Hence, we have $b(k) = a(s_k) + (b(m_1) - a(f(m_1, i_1))$ for all sufficiently large $k$. In a similar way, we can find a sequence $t_1, t_2, \ldots$, such that for all sufficiently large $k$, $b(k) = a(t_k) + (b(m_2) - a(f(m_2, i_2)))$. Subtracting, we see that $a(t_k) - a(s_k) = a(t_{k+1}) - a(s_{k+1})$ for all sufficiently large $k$.

Rearrange that to get $a(t_k) + a(s_{k+1}) + a(0) = a(0) + a(s_k) + a(s_{k+1})$. $s_k \neq s_{k+1}, 0$, so $s_k = t_k$. Hence, $0 = a(t_k) - a(s_k) = (b(m_1) - a(f(m_1, i_1))) - (b(m_2) - a(f(m_2, i_2)))$, so our claim is proven.

For $x > 0$, let $h(x) = f(x,0)$. Since $a(f(m,i)) - b(m)$ is constant and $a$ is increasing, $f(m,i)$ is independent of $i$, so $f(m,i) = h(m)$ for all $m \geq i \geq 0$. Because $a$ and $b$ are increasing functions, $h$ must be increasing as well in order for $a(h(m)) - b(m)$ to remain constant.

For any $n > 0$, we have $a(n) - a(0) = b(x) - b(y)$ for some integers $x,y$. We also have $b(x) - b(y) = a(h(x)) - a(g(x,y))$, so $a(g(x,y)) + a(n) = a(h(x)) + a(0)$. $n \neq 0$, so $h(x) = n$, so $h$ is surjective over the positive integers. Since $h$ is increasing and surjective over the positive integers, we must have $h(n) = n$ for all positive integers $n$, so $a(n) - b(n)$ is constant when $n \geq 1$.

It follows easily that $a(m) - a(n) = b(m) - b(n)$ whenever $m > n \geq 0$. For any $n \geq 0$, we have $a(n) - a(0) = b(x) - b(y) = a(x) - a(y)$ for some $x > y \geq 0$. $a(x) + a(0) + a(0) = a(n) + a(y) + a(0)$. $y \neq x$, so $y=0$ and $x = n$, whence $a(n) = b(n)$ for all $n$, as desired. [/hide]

[color=#00f]CLAIM 1.[/color] If $a_i+a_j=a_k+a_l$ then $\{i,j\}=\{k,l\}$
[color=#00f]Proof. [/color]
Otherwise take $p\neq i,j,k,l$ then $a_p+a_i+a_j=a_p+a_k+a_l$, but $\{p,i,j\}\neq \{p,k,l\}$, contradiction. $\blacksquare$
Hence each pair of integer $(i,j)$ with $i>j\geq 0$ unique determine an element in $D(A)$.
$\noindent\rule{15cm}{0.2pt}$
Now for some $i>j>0$ let
\begin{align*}
b_j&=a_{x_1}-a_{x_2}\\
b_i-b_j&=a_{x_3}-a_{x_4}\\
b_i&=a_{x_5}-a_{x_6}\\
\end{align*}
Then $a_{x_1}+a_{x_3}+a_{x_6}=a_{x_2}+a_{x_4}+a_{x_5}$
Now $x_1\neq x_2$. If $x_1=x_4$ then $a_{x_3}+a_{x_6}=a_{x_2}+a_{x_5}$. From CLAIM 1 we have $\{x_3,x_6\}=\{x_2,x_5\}$. Since $x_5\neq x_6$, we have $x_5=x_3$ and $x_2=x_6$. Hence
\begin{align*}
b_j&=a_x-a_y\\
b_i-b_j&=a_z-a_x \qquad(1)\\
b_j&=a_z-a_y\\
\end{align*}
for some $z>x>y$.
Similarly if $x_1=x_5$ we have
\begin{align*}
b_j&=a_x-a_y\\
b_i-b_j&=a_y-a_z \qquad (2)\\
b_i&=a_x-a_z\\
\end{align*}
for some $x>y>z$.
$\noindent\rule{15cm}{0.2pt}$
Now if $(i,j)$ satisfies $(1)$ we will call it good and bad if it satisfies $(2)$.
[color=#00f]CLAIM 2. [/color]$(i,1)$ is good for all $i>1$.
[color=#00f]Proof.[/color]
Let $b_i=a_x-a_y$ for some $x,y$.
Firstly if there exists $j,k$ such that $(j,1)$ is good while $(k,1)$ is bad, then $b_j=a_p-a_y$ for some $p$ while $b_k=a_x-a_q$ for some $q$. Now from $(1),(2)$ that $q\neq y$ and $p\neq x$, hence $(j,k)$ is neither good or bad, contradiction.
Now if all $i>1$ is bad then $b_j=a_x-a_z<a_x$, hence $B$ is bounded, contradiction. $\blacksquare$
$\noindent\rule{15cm}{0.4pt}$
Now for all $i.1$ let $b_i=a_{f(i)}-a_y$. Now we can compute all the elements in $D(B)$, indeed we have
$$b_i-b_j=a_{f(i)}-a_{f(j)}$$
where $f(1)=x$ and $f(0)=y$. Since $f$ is strictly increasing, and that it is surjective, we must have $f(i)=i$, hence $a_i=b_i$ as desired.


[quote="ductrung"]The following is stronger \[\sum_{cycl}(a^{2}-ab+b^{2})(b^{2}-bc+c^{2}) \ge \frac{1}{3}(a^{2}+b^{2}+c^{2})^{2}\] Also, it is true for all reals $a,b$ and $c$.[/quote]
Good job, [b]ductrung[/b]. For positive $a,b,c$, the reverse inequality of yours is as following
\[\sum_{cycl}(a^{2}-ab+b^{2})(b^{2}-bc+c^{2}) \le (a^{2}+b^{2}+c^{2})^{2}\]

I think $ 2(a^2-ab+b^2)  \ge a^2+b^2 $ help here.

[quote=ductrung]The following is stronger \[\sum_{cycl}(a^{2}-ab+b^{2})(b^{2}-bc+c^{2}) \ge \frac{1}{3}(a^{2}+b^{2}+c^{2})^{2}\] Also, it is true for all reals $a,b$ and $c$.[/quote]

[img]http://s3.sinaimg.cn/middle/006ptkjAzy76shwmPSi22&690[/img]


superior algebra
superior algebra theorems

Hi all,

I know its looks like dumb question,

In the definition of ring, the distributive law is defined to be $a(b+c)=ab+ac$.

Will it lead to a contradiction if we define $(a+b)(a+c)=a+bc$?

my junior ask me, but i don't know how to explain that :huh:

please help

thx.

Unless your ring is zero (or maybe not unitary), this will never happen since $0=(1-1)(1+0)=1$.


algebra
polynomial
linear algebra
matrix
search
linear algebra solved

let $A$ and $B$ be 2 n*n matrices with real entries. Suppose that there exist $n+1$ pairwise distinct reals $c_1,c_2,..,c_{n+1}$ such that
$A+c_iB$ is nilpotent for every $1 \leq i \leq n+1$. Prove that $A$ and $B$ are nilpotent.

For each $1\le k\le n$, $Tr((A+xB)^k)$ is a polynomial of degree $\le n$ in $x$. Since all of these polynomials have at least $n+1$ roots, it means that they are all null, so, on the one hand, $Tr (A^k)=0$, and, on the other hand, $Tr(B^k)=0$ (this is obtained by dividing $A+xB$ by $x$ for $x\ne 0$ and making $x\to\infty$) for all $1\le k\le n$. It's well-known that this implies the nilpotence of $A$ and $B$.

A matrix A is nilpotent if there is k such that $A^k=0$. Alekk, the problem appeared in oral ENS last year, but I think it appeared some years ago in IMC. The idea with the trace is magnificient.

another solution :
write $det(xId-(A+yB))=x^n+a_{n-1}(y)x^{n-1}+..+a_{1}(y)x^{1}+a_{0}(y)Id$
where $a_{k}(y)$ is a polynomial of degree at most $n$. Since the caracteristic
polynomial of a nilpotent matrix is $x^n$; we can conclude that each $a_{k}(y)$
is the nul poynomial. Conclusion follows.. (in fact, it is almost the same solution  :D  )

let $A$ and $B$ be 2 n*n matrices with real entries. Suppose that there exist reals $c_1,c_2,..,c_{n}$ such that
$A+c_iB$ is nilpotent for every $1 \leq i \leq n$. Are $A$ and $B$ nilpotent? I can't find counter examples ..

[quote="alekk"]let $A$ and $B$ be 2 n*n matrices with real entries. Suppose that there exist reals $c_1,c_2,..,c_{n+1}$ such that
$A+c_iB$ is nilpotent for every $1 \leq i \leq n+1$. Prove that $A$ and $B$ are nilpotent.[/quote]

Is the problem stated correctly?  If $A+cB$ is nilpotent for 2 different values $c_1 \neq c_2$ then $A,B$ are nilpotent.  $c_3 ... c_{n+1}$  are not necessary.

[quote="grobber"]Why is that? Shouldn't they commute in order for use to be able to say that? Or maybe there's something deeper that I'm missing? :?[/quote]

Sorry, yes, I was thinking of the lemma for commutative rings.

For the noncommutative problem, it is not necessary to use traces; an nxn matrix $M$ is nilpotent if and only if $M^n = 0$.    Each entry of  $(A+tB)^n$ is a polynomial of degree $n$ vanishing at all $c_i$, so identically 0.   For the second problem in the thread one needs a solution of

$(A+tB)^n = (t-c_1)...(t-c_n)C$  where $C$ is a constant matrix.

[quote="Makaveli"]How do we know that those polynomials have at least n+1 roots? I guess  c1,c2,3..are roots (by Cayley-Hamilton, when we look at the characteristic polynomial of the matrix) but how come we are considering (A+xB)^k (why is the power exactly k) I am confused[/quote]

A property of nilpotent matrix is that if $ A$ is a n by n matrix, then A is nilpotent iff $ Tr (A^k) \equal{} 0$ for $ k \equal{} 1,2,...,n$.


function
probability
analytic geometry
probability and stats

Let  $ B_n$ be the set of all bijective functions from {1,2,......n}   to   (1,2,.......n} such that for each element f

(a)  f(k) $ \leq$ k+1             (k=1,2,.........n)

(b)  f(k) $ \neq$ k                (k=2..........n)

Find the probability that f(1) $ \neq1$  for an f chosen randomly from $ B_n$

Let $ (a_1, \ldots, a_n)$ denote $ i \mapsto a_i$ for $ i \equal{} 1, \ldots, n$.

First let $ f_k$ denote the set of all $ f$ satisfying (a) and (b), as well as $ f(1) \equal{} 1$, with $ n \equal{} k$. We consider the base cases $ n \equal{} 3$ and $ n \equal{} 4$, noting that $ f_3 \equal{} \{(1, 3, 2)\}$ and $ f_4 \equal{} \{(1, 3, 4, 2)\}$. Subsequently, $ f_n$ for $ n > 4$ can be found inductively as follows: given $ (a_1, \ldots, n \minus{} 1, a_{n \minus{} 1}) \in f_{n \minus{} 1}$, we have $ (a_1, \ldots, n \minus{} 1, n, a_{n \minus{} 1}) \in f_n$, and given $ (a_1, \ldots, a_{n \minus{} 2}) \in f_{n \minus{} 2}$, we have $ (a_1, \ldots, a_{n \minus{} 2}, n, n \minus{} 1) \in f_n$. Moreover $ f_n$ contains no other functions since all valid possible permutations of the first $ n \minus{} 2$ coordinates are determined by those found in $ f_{n \minus{} 1}$ and $ f_{n \minus{} 2}$ and the $ (n \minus{} 1)$ th coordinate in $ f_n$ must be $ n$. Thus $ |f_k| \equal{} |f_{k \minus{} 1}| \plus{} |f_{k \minus{} 2}|$, with $ |f_3| \equal{} |f_4| \equal{} 1$.

Similarly let $ g_k$ denote the set of all $ f$ satisfying (a) and (b), as well as $ f(1) \ne 1$, with $ n \equal{} k$. The base cases are now $ n \equal{} 2$ and $ n \equal{} 3$, with $ g_2 \equal{} \{(2, 1)\}$ and $ g_3 \equal{} \{(2, 3, 1)\}$, and again $ |g_k| \equal{} |g_{k \minus{} 1}| \plus{} |g_{k \minus{} 2}|$ with $ |g_2| \equal{} |g_3| \equal{} 1$. Thus notice that $ |g_{n \minus{} 1}| \equal{} |f_n|$.

With $ B_1 \equal{} \emptyset$, we thus have for $ n > 1$ that the desired $ p_n \equal{} \frac {|g_n|}{|f_n| \plus{} |g_n|} \equal{} \frac {|g_n|}{|g_{n \plus{} 1}|} \equal{} \frac {F_{n \minus{} 1}}{F_n}$, where $ \{F_n\}_{n \equal{} 1}^{\infty} \equal{} \{1, 1, 2, 3, \ldots\}$ is Fibonacci.


I'll rename them a, b, c, d. What we have to prove is this:

a) a^3+b^3+c^3+d^3>=1/a+1/b+1/c+1/d=bcd+acd+abd+abc. By adding ineqs of the type a^3+b^3+c^3>=3abc this is true;

b)a^3+b^3+c^3+d^3>=a+b+c+d. By adding ineqs of the type 1/2*a^3+1/6*b^3+1/6*c^3+1/6*d^3>=a (we use AM-GM and abcd=1) this is also true.

From a) and b) we get the conclusion.


For what value of $ n$ does $ \frac{n}{12} \plus{} \frac{n}{6} \equal{} \frac{n}{n}$?

Multiplying everything by $ 12$, we get $ n\plus{}2n\equal{}12$, and $ 3n\equal{}12$, which means $ n\equal{}\boxed{4}$.


I was just wondering...does anyone's geometry class require two-column proofs, e.g. statements and reasons? My geometry class did, and I was always skeptical about their value, especially since they, or at least the teacher, required statement of things like the reflexive property that I thought were obvious. Additionally, I have never seen any two-column proofs to a geometry problem outside of that class.

Is this a common geometry teaching method?

Two-column proofs are almost always used in classrooms, even though they are not as popular elsewhere.

Later on, two-column proofs will not be necessary, and you won't have to show every single step, but I consider showing every step in a proof important as a fundamental skill while learning.

Like I said before in another thread, two column proofs are a great way of organizing your thoughts before you right out an actual proof. I find that writing a paragraph proof from scratch can usually end up messy and disorganized, but if I start with a two column, my ideas are presented where they need to be.

What made me laugh was my Geometry teacher calling two columns "real or formal proofs" while paragraph form was deemed "informal."

Two-column proofs can be useful for beginning geometers but I think the move should be made to paragraph proofs later on.

But that's made up for (manyfold) by the lack of flexiblity in terms of space usage, inability to comment or place emphasis on specific sections, and inability of using lemmas.

Also, there are symbols you can use, such as $ \implies$ and $ \therefore$ if you don't want to use too many logic words :)

Two column proof help organize thoughts for some but I prefer paragraph

My geometry teacher would dock off points if it wasn't 2 column proof...


Could someone help me proceed with this inequality. thanks
Find the smallest positive integer n such that:
(n)^1/2+(n-1)^1/2 ≥100

It's $\subseteq \mathbb{R}$... so your best bet is to algebraically solve the equation.

$x^{\frac{1}{2}}+(x-1)^{\frac{1}{2}} \ge 100$
$x \ge \frac{100020001}{40000} = 2500.5$
$\lceil 2500.5 \rceil=\boxed{2501}$

I don't have time to show all my work right now, but if you need help solving an equation involving fractional exponents just ask.


limit
function
real analysis
real analysis unsolved

If $ S_n,_m$ is a double sequence of nonnegative reals for $ n\ge 1$ and $ m\ge 1$, under what condition(s), $ lim_{n\to \infty} lim_{m\to \infty} S_n,_m \equal{} lim_{m\to \infty} lim_{n\to \infty} S_n,_m ?$

Thanks in advance.

(i)$ \lim_{m \to \infty} S_{n,m} \equal{}r_n$ exists for every $ n\in N$

(ii)$ \lim_{n \to \infty} S_{n,m} \equal{}t_m$ exists for every $ m\in N$

(iii)$ \lim_{n\to \infty} r_n \equal{} \lim_{m \to \infty} t_m$

=)

If you're searching for an applicable test for interchanging the order of limits, Lebesgue's dominated convergence theorem will give you a sufficient condition. If

1) $ \tilde{S}_n \equal{} \lim_{m\to\infty} S_{m,n}$ exists for each $ n$,
2) there exists a bounded monotone increasing sequence $ b_n$ such that $ |S_{m,n} \minus{} S_{m,n\minus{}1}| \leq b_n \minus{} b_{n\minus{}1}$ for all $ m$, $ n$,
(that is, the increment of $ b_n$ dominates the change of $ S_{m,n}$ in a uniform mannar)

then both $ \lim_{m\to\infty} \lim_{n\to\infty} S_{m,n}$ and $ \lim_{n\to\infty} \lim_{m\to\infty} S_{m,n}$ exists and coincides.

Of course, we can interchange the role of $ m$ and $ n$.

The best-known sufficient condition for being able to say that $ \lim_{x\to x_0}\lim_{n\to\infty}f_n(x)\equal{}\lim_{n\to\infty}\lim_{x\to x_0}f_n(x)$ is that of the uniform convergence of the sequence of functions $ f_n.$

Can you use that to create an analogous condition for double sequences?

I was thinking along these lines only but I could not get an analogous condition for double sequences.

Try this: Suppose $ a^{(n)}$ is a sequence of sequences that converge pointwise to a sequence $ b.$ That is, for each $ k,$ $ \lim_{n\to\infty}a_k^{(n)}\equal{}b_k.$

We say that $ a^{(n)}$ converges uniformly to $ b$ if and only if $ \forall\,\epsilon>0\,\exists\,N$ such that if $ n>N,$ then $ |a_k{(n)}\minus{}b_k|<\epsilon$ for every $ k.$

Now, try proving this theorem:

Suppose each $ a^{(n)}$ is a convergent sequence and $ a^{(n)}$ converges uniformly to $ b.$ Then $ \lim_{k\to\infty} b_k$ exists and
\[ \lim_{k\to\infty} b_k\equal{}\lim_{n\to\infty}\left(\lim_{k\to\infty}a_k^{(n)}\right).\]


There are following three numbers:
a) a nine digit number
b) a single digit number
c) again a nine digit number.

$ Question 1$
Place them one after another in the same sequence to get a nineteen digit number .... and lo .... it is a palindrmic number.

$ Question 2$
Add the three numbers and lo .... the resultant number is also a palindrome.

What are the three numbers ?

Let the digits be (a, b, c, d, e, f, g, h, i) , ( j ) , (i, h, g, f, e, d, c, b, a)

We have $ a \neq 0$ and $ i \neq 0$

We also have $ 10^{8}(a+i)+10^{7}(b+h)+10^{6}(c+g)+10^{5}(d+f)+10^{4}(e+e)+10^{3}(f+d)+10^{2}(g+c)+10(h+b)+(i+a+j)$ is a palindrome.

Which means that J must be zero, because $ a+i = a+i+j$

Now we have the following:
(e + e) < 10
meaning $ e < 5$

I'm stuck from here.


Show that there are no rational numbers $ a,b,c,d$ satisfying
$ (a\plus{}b\sqrt 2)^{100}\plus{}(c\plus{}d\sqrt 2)^{100}\equal{}7\plus{}5\sqrt 2$

From condition we must have
$ (a\minus{}b\sqrt{2})^{100}\plus{}(c\minus{}\sqrt{2}d)^{100}\equal{}7\minus{}5\sqrt{2}<0$ (contradition !)

[quote="TTsphn"]From condition we must have
$ (a \minus{} b\sqrt {2})^{100} \plus{} (c \minus{} \sqrt {2}d)^{100} \equal{} 7 \minus{} 5\sqrt {2} < 0$ (contradition !)[/quote]
How so? I didn't get it  :(

Since the brackets are raised to 100-th power, for all rational $ a,b,c,d$ neither of them can be negative (because 100 is even), hence the left side of your equality is positive for all such numbers. And since the right side is negative ($ 7 < 5\sqrt {2}$), it is not possible for such numbers to exist. Or in simpler words, the left side is always positive, and the right side is "always" negative - contradiction.

I think he's asking how the second equality was determined.  TTsphn conjugated both sides in $ \mathbb{Q}[ \sqrt{2} ]$.  Equivalently, expanding both sides and equating rational and irrational parts, we simply flip the sign of the irrational part of both sides.

[quote="t0rajir0u"]I think he's asking how the second equality was determined.  TTsphn conjugated both sides in $ \mathbb{Q}[ \sqrt {2} ]$.  Equivalently, expanding both sides and equating rational and irrational parts, we simply flip the sign of the irrational part of both sides.[/quote]

Yes that's what I asked, thanks :lol:
I thought of rationalizing and was getting stuck all the time.
Conjugating never occurred to me, damn  :mad:


i didn't :o  :o

oh snap man

i've only seen like [size=150]1[/size] steak 'n shake in my life

[quote="MysticTerminator"]I saw a "steak n shake" yesterday[/quote]
whoa is it because you were traveling? I think it might only exist in the midwest... at least haven't seen any elsewhere

um... what is a steak n shake?

sorry if this is a stupid question :D

...I see one everyday while driving to school.  :maybe:

They're around here:
[img]http://www.steaknshake.com/pics/newmap.gif[/img]

Arnav I'm going to kick your butt for spamming. -.-


Solve for $ |xy|:\log{x^{2}}\plus{}2\log{y}\equal{}10$

[hide]$ \log x^{2}\plus{}2\log y\equal{}2(\log x\plus{}\log y)\equal{}2\log(xy)\equal{}10\Rightarrow\log(xy)\equal{}5$, so $ |xy|\equal{}10^{5}\equal{}100000$.[/hide]

[quote="Xaero"][hide]$ \log x^{2}\plus{}2\log y \equal{} 2(\log x\plus{}\log y) \equal{} 2\log(xy) \equal{} 10\Rightarrow\log(xy) \equal{} 5$, so $ |xy| \equal{} 10^{5}\equal{} 100000$.[/hide][/quote]
Not exacly, I mean $ \log x^{2}\neq 2\log x$ (for $ x\in\mathbb{R}\backslash\{0\}$), It's $ \log x^{2}\equal{} 2\log |x|$ (for all $ x$ except $ x \equal{} 0$)

[quote="Xaero"]Is the logarithm of a negative number even defined? I thought $ \log x$ can only have a definition when $ x\in\mathbb{R}^{\plus{}}$. Do we even need to consider $ x\in\mathbb{R}^{\minus{}}\cup\{0\}$? I would automatically assume that $ x$ is positive...[/quote]

Xaero, you just have to slightly adapt your simple and good demo :

$ y>0$ and $ x\neq 0$. So : $ \log x^{2}\plus{}2\log y \equal{} 2(\log |x|\plus{}\log y) \equal{} 2\log(|x|y) \equal{} 10\Rightarrow\log(|x|y) \equal{} 5$, so $ |x|y\equal{}|xy| \equal{} 10^{5}\equal{} 100000$.


from a variable point in the plane of triangle $ABC$.draw three lines to its vertices.

if the sum of these three segments is equal to $8$,which of these sentenses is true?

$1)P>16$

$2)8<P<16$

$3)P>17$

$4)16<P<24$

$P=AB+BC+CA$ ;)

This question is just triangle inequality.

\[ AB +BC + AC < 2AP+2BP+2CP = 16 \]

So $(2)$ is the only possible answer


Could you please tell me where I can find the problems and solutions of the Asian Physics Olympiad 2002?
Unfortunately, the site http://www.cite.nie.edu.sg/apho2002/ isn't working...

try  here [url]http://www.lookforbook.com[/url]

why are sending these useless. are yo ucongrating all other who have 2 stars??????????????????????????????????

why do you discuss other topics in this post. this is physics post. we must discuss about physics.. understand ashwinrk_jain :diablo:

you start congratulating not i am!!!!!!!!!!!!!!!!!!!!!!!!
 :w00tb:  :w00tb:  :w00tb:

well you should reply politly in that case :mad:

[code]	why do you discuss other topics in this post. this is physics post. we must discuss about physics.. understand ashwinrk_jain Diablo
[/code]
Did you forget this.  :!:  :!:  :!:

no but you should not post for your rank.  :mad:  :mad:  :mad:   do not post more replys

I want to tell you that only that you do not post replies like this for just increasing the posts.

no only you post to get higher rank. stop posting immorally

hahaha what the hell is happening here.Both you babies stop fighting over this and stop being jealous of each other's ranks.Let the mods ake care of this.