function
topology
real analysis
real analysis theorems

given the function:

$ f: R\rightarrow R$ ,PROVE:  f is continuous over R(=real Nos) iff $ f(\overline A)\subseteq\overline {f(A)}$ ,WHERE A is a subset of the real Nos ,by using the ε-δ definitions

First assume $ f$ is continuous, and we have $ \bat{A}=A\cup\{\text{accumulation points of }A\}$.  If $ \textbf{p}$ is an accumulation point of $ A$ then there is a sequence $ x_{k}$ in $ A$ which converges to $ \textbf{p}$.  By continuity, $ f(x_{k})\rightarrow f(\textbf{p})$ so $ f(\textbf{p})\subset\overline{f(A)}$ since $ f(x_{k})$ are all points in $ f(A)$.  This proves the first direction.

Now assume $ f(\bar{A})\subset\overline{f(A)}$.  Let $ (x_{k})\rightarrow\textbf{p}$ be a sequence in $ A$.  If $ f(x_{k})\not\rightarrow f(\textbf{p})$, then there exists some $ r>0$ and a subsequence $ x_{k_{j}}$ such that $ \left\|f(x_{k_{j}})-f(\textbf{p})\right\|\geq r$.  Let $ B_{J}=\{x_{k_{j}}\mid j\geq J\}$.  We have $ \textbf{p}\in\overline B_{J}$ for any $ J$ so $ f(\textbf{p})\in\overline f(B_{J})$.  Since $ f(x_{k_{j}})\neq f(\textbf{p})$ and $ f(\textbf{p})\in\overline{\{f(x_{k_{j}})\mid j\geq J\}}$ for any $ J$, there exists a further subsequence $ x_{k_{j_{i}}}$ such that $ f(x_{k_{j_{i}}})\rightarrow f(\textbf{p})$.  This contradicts the fact that the first subsequence was chosen so that $ \left\|f(x_{k_{j}})-f(\textbf{p})\right\|\geq r$.  Hence we must have that $ f(x_{k})\rightarrow f(\textbf{p})$, which is the definition of continuity.

Here's an alternative (shorter) proof for the second direction: let $ C\subset\mathbb{R}$ and be a closed set.  Let $ A=f^{-1}(C)$ so $ f(\bar{A})\subset\overline{f(A)}=\overline{f(f^{-1}(C))}=\bar{C}=C$.  Therefore $ \bar{A}\subset f^{-1}(C)=A$, so $ A$ is closed.  Since the inverse image of a closed set is closed, $ f$ is continuous.

[quote="JRav"]First assume $ f$ is continuous, and we have $ \bat{A} = A\cup\{\text{accumulation points of }A\}$.  If $ \textbf{p}$ is an accumulation point of $ A$ then there is a sequence $ x_{k}$ in $ A$ which converges to $ \textbf{p}$.  By continuity, $ f(x_{k})\rightarrow f(\textbf{p})$ so $ f(\textbf{p})\subset\overline{f(A)}$ since $ f(x_{k})$ are all points in $ f(A)$.  This proves the first direction.

.[/quote]

I am sorry i said in my post to prove the above by using ε-δ definitions ,meaning not to use sequences and open or closed sets.

Now in your proof you write that :  $ f(\textbf{p})\subset\overline{f(A)}$.

How can you have f(p) which is the image of the accumulation point p being a subset of the set, $ \overline{f(A)}$


calculus
integration
trigonometry
logarithms
calculus computations

Let $I$ be the value of your integral. Using the substitution $x=\frac{1}{t}$ you get that:

$2I=\frac{\pi}{2}\int_{0}^{\infty}\frac{\ln(1+t^{2})}{1+t^{2}}dt-\pi\int_{0}^{\infty}\frac{\ln t}{1+t^{2}}dt+2\int_{0}^{\infty}\frac{\arctan t\ln t}{1+t^{2}}dt$.

The second integral is $0$ and the other two integrals can be calculated via the substitution $t=\tan u$.

[quote="didilica"]$\int_{0}^{\pi/2}x\ln \tan x dx=-\frac{\pi}{2}\int_{0}^{\pi/2}\ln \sin x dx+2\int_{0}^{\pi/2}x \ln \sin x dx$.

For calculating these integrals use the following Fourier expansion:

$-\ln 2\sin \frac{x}{2}=\sum_{n=1}^{\infty}\frac{\cos nx}{n}$, $0<x<2\pi$.[/quote]

i get $\int_{0}^{\pi/2}\;x\;\cdot\;\ln{(\tan{x})}\; dx\;=\;\frac{7}{8}\;\zeta{(3)}\;\;\;\;\;$

hmmmm.... wonder if that's correct!  :|


$ e^{i(i\pi)} \equal{} r(cos(i\pi) \plus{} isin(i\pi))$
$ r \equal{} \ ?$
$ cosi\pi \equal{} \ ?$
$ sini\pi \equal{} \ ?$

If you know the exponential definitions of sine and cosine, you should have no trouble.  Recall that
$ \cos{z}\equal{}\frac{e^{iz}\plus{}e^{\minus{}iz}}{2}$ and $ \sin{z}\equal{}\frac{e^{iz}\minus{}e^{\minus{}iz}}{2i}$
Now we have that $ z\equal{}\pi i$, so plugging in yields
$ \cos{i\pi}\equal{}\frac{e^{\pi}\plus{}e^{\minus{}\pi}}{2}$ and $ \sin{i\pi}\equal{}\frac{e^{\pi}\minus{}e^{\minus{}\pi}}{2i}$


with median point of AB i have de circumcenter
with locus from this median point to Lg i have the line that intersects circumcircle at C vertex

 :lol:


linear algebra
matrix
Functional Analysis

Let  [tex]A[/tex] - linear operator in Hilbert space, [tex]A^2=A[/tex] and [tex]AA^*=A^*A[/tex]  ( [tex]A^*[/tex] is adjoint operator).
Prove that [tex]A=A^*[/tex]

Very interesting result !

Just to make sure : it's true also in infinite dimension ?

[quote]Just to make sure : it's true also in infinite dimension ?[/quote]
Yes it is. In fact it is the most interesting part of the problem
BTW, we consider complex Hilbert space

[quote]Is it sufficient to prove that the eigenvalues are real?[/quote]
This fact alone is not sufficient for operator to be self-adjoint even in finite dimension space. Example: take [tex]A=\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}[/tex], then eigenvalue is 1 - real, yet adjoint operator is   [tex]A^*=\begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}[/tex].
However, in n-dimension space we can prove that there are n (linearly independent) eigenvectors, and n real eigenvalues. This would be sufficient in finite dimension case. But I don't think it is possible to use this approach in general case. In infinite dimensions eigenvectors and eigenvalues may have many unusual properties..
Just an example. Consider Hilbert space [tex]\ell_2[/tex], which is space of complex sequences [tex]\ell_2=\{z=\{z_1,...,z_n,...\}|\sum_{i=1}^{\infty}|z_i|^2<\infty\}[/tex] with [tex](x,y)=\sqrt{\sum_{i=1}^{\infty}x_i y_i^*}[/tex]. Consider operator [tex]A: Ax=\{0,x_1,...,x_{n-1},...\}[/tex]. Then
1)[tex]A^*x=\{x_2,x_3,...,x_{n+1},...\}[/tex]
2)[tex]A[/tex] has no eigenvalues
3)Eigenvalues of [tex]A^*[/tex] are [tex]\lambda : |\lambda|<1[/tex], so set of eigenvalues is uncountable, although there is countable basis in [tex]\ell_2[/tex]

I think I got it at last ....

Let x in E and y = Ax - A*x, <,> is the scalar product

Since A and A* commutes, AA*y = A*Ay = 0 and then <Ay,Ay> = <A*Ay,y> = 0 and <A*y,A*y> = 0. This implies Ay=A*y=0

Ay + A*y = Ax - AA*x + A*Ax - A*x = Ax - A*x = Ax - A*x = y = 0

So we have proven Ax=A*x for all x or A=A*.

Nice proof (you wanted to say AA*y=A*Ay=0)

My approach to this problem was slightly different, although  it leads to similar computations

[hide]For each operator A there are self-adjoint operators B and C s.t. A=B+iC (take B=(A+A*)/2, C=(A-A*)/2i). Now, from A*A=AA* we obtain BC=CB, so (BC)*=C*B*=CB=BC. From A^2=A and A*^2=A* we obtain 2 equations on B,C. From these equations we can obtain C=0   [/hide]

[quote="JBL"]Can you show how you get from the given information that A* :^2: = A*?[/quote]
We start from (AB)*=B*A* which follows directly from definition of adjoint operator: <x,(AB)*y>=<ABx,y>=<Bx,A*y>=<x,B*A*y>
Now, (A*)^2=A*A*=(AA)*=(A^2)*=A*


The set
$A \cup B \cup C - (A - (B \cup C)) - (B - (A \cup C)) - (C - (A \cup B)) - \\ ((A \cap B) - C) - ((A \cap C) - B) - ((B \cap C) - A)$
is:

a)$A - B - C$

b)$B - A - C$

c)$C - A - B$

d)$A \cap B \cap C$

e)$A \cup B \cup C$

yes, (d) indeed it is the union of A,B and C of wich you subtract first all elements belonging to A only, then to B only, then to C only and then all elements beloning to A&B but not to C, then ... so that only elements in A, B and C remain

[hide][quote="zanttrang"](d) is correct - it follows directly from a simple re-arrangement of inclusion-exclusion.[/quote]

Could you explain how do you do this simple re-arrangement of inclusion-exclusion?
thanks[/hide]

or you could just draw the venn diagrams...

I draw Venn diagram to find the answer but there is an ever quicker way :P since there is a choice list

Equation is symetric relative to A,B and C so the answer must also be and can only be $A \cap B \cap C$ or $A \cup B \cup C$

Of course result must be smaller then $A \cup B \cup C$ so it is $A \cap B \cap C$   :D


i think we have the stronger one:

$1+\frac{3}{a+b+c}\geq  \frac{6}{ab+ac+bc}$

to prove this one we use the notations : $a=\frac{1}{x}...$ and we obtain that our ineq is equivalent with : $(1-\frac{3}{x+y+z})^{2}\geq  0$

[quote="pohoatza"]i think we have the stronger one:

$1+\frac{3}{a+b+c}\geq \frac{6}{ab+ac+bc}$
[/quote]
$1+\frac{8}{a+b+c}\geq \frac{11}{ab+ac+bc}\Leftrightarrow$
$\Leftrightarrow(a+b+c-3)(ab+ac+bc)\geq11(a+b+c-ab-ac-bc),$
$1+\frac{3}{a+b+c}\geq \frac{6}{ab+ac+bc}\Leftrightarrow$
$\Leftrightarrow(a+b+c-3)(ab+ac+bc)\geq6(a+b+c-ab-ac-bc).$
If $a+b+c-ab-ac-bc\leq0$ then they are trivial.
But if $a+b+c-ab-ac-bc\geq0$ then
$(a+b+c-3)(ab+ac+bc)\geq11(a+b+c-ab-ac-bc)\geq6(a+b+c-ab-ac-bc).$
Thus, the mine is stronger. :wink:
[quote="pohoatza"]
to prove this one we use the notations : $a=\frac{1}{x}...$ and we obtain that our ineq is equivalent with : $(1-\frac{3}{x+y+z})^{2}\geq 0$[/quote]
I think, you are wrong.
Your inequality follows from $(1-\frac{3}{x+y+z})^{2}\geq 0,$ but it's not equivalent to your inequality. :wink:

[quote="arqady"][quote="Sunjee"][quote="arqady"]Let $a,$ $b$ and $c$ are positive numbers such that  $abc=1$. Prove that
$1+\frac{8}{a+b+c}\ge \frac{11}{ab+bc+ca}$[/quote]
I have very simple proof, what's your proof, arqady?[/quote]
By $uvw$ it's very easy.[/quote]
To prove inequality using some theorem's like uvw, abc,... it is not easy. As a Canhang say's we have to improve our skills. Your inequality is equivalent to: (a->\frac{1}{a^3},...)
\[\frac{a^3+b^3+c^3-3abc}{a^3+b^3+c^3}+8\sum\frac{bc(a^2-bc)^2}{(\sum a^3)(\sum a^3b^3)}\geq 0\]

[quote="arqady"]Let $a,$ $b$ and $c$ are positive numbers such that  $abc=1$. Prove that
$1+\frac{8}{a+b+c}\ge \frac{11}{ab+bc+ca}$[/quote]
In fact, this can be proved easily by mixing variables. I will post here the mixing variables step.

Write the inequality as
$(a+b+c)(ab+bc+ca)+8(ab+bc+ca) \ge 11(a+b+c),$
$(a+b+c)(ab+bc+ca)-3(a+b+c) \ge 8(a+b+c-ab-bc-ca).$
Since $abc=1,$ it follows that
$a+b+c-ab-bc-ca=abc-(ab+bc+ca)+a+b+c-1=(a-1)(b-1)(c-1),$
and so the last inequality is equivalent to
$(a+b+c)(ab+bc+ca-3) \ge 8(a-1)(b-1)(c-1).$
Without loss of generality, assume that $c=\max\{a,\; b,\;c\}.$ Since $abc=1,$ we get $c \ge 1.$ Now, setting $t=\sqrt{ab},$ then it is easy to verify that
$ab+bc+ca -3 \ge t^2+2tc-3 \ge 0,$
$a+b+c \ge 2t+c,$
so
$(a+b+c)(ab+bc+ca-3) \ge (2t+c)(t^2+2tc-3) .$
On the other hand,
$(a-1)(b-1)=ab-(a+b)+1 \le ab -2\sqrt{ab}+1=(t-1)^2,$
and since $c-1 \ge 0,$ we get
$(a-1)(b-1)(c-1) \le (t-1)^2(c-1).$
Finally, we come up with the inequality
$(2t+c)(t^2+2tc-3) \ge (t-1)^2(c-1).$
By replacing $c=\frac{1}{t^2},$ it is not hard to check the last inequality.

[quote="Sunjee"] As a Canhang say's we have to improve our skills.[/quote]
I agree with this statement.
[quote="Sunjee"] Your inequality is equivalent to: (a->\frac{1}{a^3},...)
\[\frac{a^3+b^3+c^3-3abc}{a^3+b^3+c^3}+8\sum\frac{bc(a^2-bc)^2}{(\sum a^3)(\sum a^3b^3)}\geq 0\][/quote]
Are you sure?


complex analysis
function
algebra
domain
complex analysis unsolved

Hello!

This statement has some resemblance with the Maximum Principle, but the second condition with the exponent got me stuck:

Let $ \minus{} \infty < a < b < \infty$ and $ G: \equal{} \{z\in{\mathbb{C}}: \, a < Re(z) < b\}$

Let $ f$ be continuous on $ \bar{G}$ and holomorphic on $ G$, so that
$ |f(z)|\le{}K$ for all $ z\in\partial{G}$ AND $ |f(z)|e^{ \minus{} |z|}\le{}C$ for all $ z\in{}G$

Show that then holds: $ |f(z)|\le{}K$ for all $ z\in{}G$

What does the seocnd condition imply?
Does anybody have an idea how to prove this thing?

This is a part of the general Phragmen-Lindelof principle: if the function grows not too fast in an unbounded domain and is bounded on the boundary, it is bounded in the entire domain. The exact admissible rate of growth heavily depends on the domain and in the case of the strip, it is really double exponential. The trick in the proof is always the same: multiply your function by a very fast decaying function in the domain that is close to $ 1$ on a big set and apply the classical maximum principle to a big bounded portion of the domain. In the setting of this particular problem, multiplying by $ e^{\delta z^2}$ with small $ \delta>0$ does the trick. Note that some growth restriction is necessary: $ e^{e^{iz}}$ is bounded on $ \text{Re\,}z\equal{}\pm\pi$ but not in the strip $ \minus{}\pi\le\text{Re\,}z\le\pi$.

fedja you must be a professor , so impressed


trigonometry
ratio
geometry
geometric transformation
reflection
symmetry

Can anyone help me understanding the De Moivre's Theorem?

I don't understand how ${1+i = \sqrt {2} (\cos {\frac {\pi}{4}} + i \sin \frac {\pi}{4})}$

O_o...

Ah. I feel so stupid about not realizing it.

It's been since I did a trig. How do you translate $\cos \theta$ in a fraction form? Like how do you know $\cos \frac {\pi}{4} = \frac {\sqrt 2}{2}$ Because I don't understand how you do that.. Do you memorize it? I know the conversion of radians to the degrees but..

Thanks.

EDITE: Understand now.

if you can't memorize them, think about them visually... a pi/4 angle would be the same as 45 degrees... and cos :theta:  would be adjacent over hypotenuse... and if you know your 45-45-90 triangles, you should know the ratios to be  :sqrt:2/2

for future reference, check this out:
http://www.math.com/tables/graphs/unitcircle.gif

Most frequently used angles that I have encountered are those of the form $\frac{k\pi}{d}$ for $d=1,2,3,4,6$. So a useful-to-remember-angles list of trigonometric values are

$\sin 0=\frac{\sqrt{0}}{2}=\cos\frac{\pi}{2}$

$\sin\frac{\pi}{6}=\frac{\sqrt{1}}{2}=\cos\frac{\pi}{3}$

$\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}=\cos\frac{\pi}{4}$

$\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}$

$\sin\frac{\pi}{2}=\frac{\sqrt{4}}{2}=\cos 0$

Subsequently, other values for $\sin\frac{k\pi}{d}$ can be found by using trigonometric identities. Try to exploit symmetries around the circle like mirror reflections. For example, $\cos\theta=\cos(-\theta)$ and that is a reflection on the $x$-axis (cosine is an even function). $\sin \theta$ satisfies reflection on the $y$-axis (sine is an odd function). Try investigating symmetry for $\tan$ !!!
Study harder Silverfalcon, you can do it.


There are $ n$ football teams in a round-robin competition where every 2 teams meet once. The winner of each match receives 3 points while the loser receives 0 points. In the case of a draw, both teams receive 1 point each. Let $ k$ be as follows: $ 2 \leq k \leq n \minus{} 1$.  At least how many points must a certain team get in the competition so as to ensure that there are at most $ k \minus{} 1$ teams whose scores are not less than that particular team's score?

Please give me some guidance.
Thank you.

The question is in fact asking about what the maximum score of the kth team could be.

And the answer should be [3n-2/3(k+1)]+1
Just consider the first k teams as a group and count the total score they could get. The construction is simple.


Some may already know this, but  middle school students in Ohio/Cincinnati Area may be interested in:

March 11, 2006 is Ohio - Mathcounts  State competetion (Columbus, Ohio)

T Divison competition for Cincinnati  is on Feb 25, 2006.

There are two other events which might be of interest:

[b]Feb 9, 2006 - Mini Mathcounts Competition at St Margaret of York . [/b]
[b]March 5, 2006 -  University of Cincinnati Math Contest.[/b]
Also: Check out our Cincinnati website address
 [url]http://www.saseasschool.org/mathcounts.htm[/url]
Check out my interactive Rubick's cube on the site. and photos etc...

Please see below  cut-and-paste of the e-mail I have received.
 [quote]Subject: Mini Mathcounts Competition (Very Relaxed and Informal)

 Hope this finds the right group of people.  (Grades 6-8-- or younger if they want)

We would like to host another mini-mathcounts competition this year at St. Margaret of York.  We want to do it after school l 4:00 -6:00 on Thursday Feb 9th.  It serves as a bit of a warmup for the big Mathcounts Competition Which I believe is on the 18th.  Basically I just take an old Mathcounts competition and cut the whole thing in half.  We will have trophies and pizza.  Cost is  $\$15$  dollars  per school and you can bring as many teams as you want (assuming we don't we fill up, which we didn't last year) We hope to have teams of 4 to make it work like the real mathcounts, but if you only have a group of 3 we can score them on an average. Just e-mail me with your coaches e-mail and phone # and let me know if you think you can come. You can just bring the money with you when you come.
That's Thursday Feb. 9th St. Margaret of York
9495 Columbia Road Loveland, OH 45140
4: 00-6: 00pm but we probably won't actually start until around 4: 15 since I know it is tough to get up here.
Pete Tarnowski Cell (503-3930)
If you know of others who might be interested please forward this to them. [/quote]  [quote] Subject: 21st Annual UC Math Contest - Saturday March 4th - register now! Dear Colleagues, Math Team Competition - 3 to 5 students work together in teams to solve problems! Great for all your mathcounts students. University of Cincinnati and Dr. James Osterburg are again sponsoring the 21st Annual Academic Competition in Mathematics at the University of Cincinnati on Saturday, March 4, 2006. Please email me your registration. The contest has become an excellent way to continue your Mathcounts team and training by allowing more students to participate in this TEAM event. There are 3 Levels: Level 1 Calculus (Please forward this to your high school math teachers) Level 2 Algebra 2 and Geometry Level 3 Pre-Algebra and Algebra 1 I take teams of 3 to 5 students to participate in Level 3 and they love it! Below is the registration form. You may EMAIL me your registration and then send a check later. The cost is $\$10$  per team.  Below are three items:  Registration Form, Instructions, and Math Competition Schedule for the Day
 [/quote]

Any one taking part in Chapter comp. at Cincinnati  coming Saturday  (Feb 25) ?

Columbus – Cincinnati area schools swept the state honors at the Ohio State MATHCOUNTS competition held yesterday at Columbus State College. Seven of the top ten students were from the Cincinnati area, including the entire top 4 who will advance to nationals. Sycamore’s Mathletes earned the highest score in the competition, which consisted of two individual exams and a team problem-solving test.

Three students from Sycamore JHS, Ashvin Sravatsa, Emily Hodun, and Chris King, and Dennis Tseng from Mason MS have earned an expense paid trip to Virginia to represent Ohio at the National MATHCOUNTS competition on May 11-14.  Dan Mirus, the coach of Sycamore will coach the Ohio team at the National competition.

Results for the top five schools from the Cincinnati competition that competed in the Ohio were:

                        Sycamore Junior High School 1st place

Mason Middle School 2nd place

Greene Intermediate School, 7th place

Indian Hill Middle School 13th place

St. Ursula Villa, 21st place

Cincinnati Country Day, 22nd place

Cincinnati area mathletes took the top five spots and 14 of the top 33 spots at the Ohio State competition.  This strong showing underscores that the Cincinnati area continues to establish its regional and national reputation as a problem-solving powerhouse.

Students advancing to the National MATHCOUNTS competition will vie to for top honors, scholarships, NASA camp programs, and computers. In addition, all national mathletes receive a substantial gift from the MATHCOUNTS foundation industry sponsors.

Congrats!

I'll be at the Cincinnati chapter MathCounts...I was there last year too.


induction
combinatorics unsolved
combinatorics

We have on the plane finity set of red and blue lines. Such that none two of this lines are parallel. By every point of intersection two blue lines pass red line, and by every point intersection of two red lines pass blue line. Prove that all lines have common point.

Isn't it just induction on the number of lines?   :huh:


integration
trigonometry
calculus
symmetry
algebra
polynomial
geometric series

Using residues, show that if $ a>|b|>0$, then
$ \int_{0}^{2\pi}\frac{\sin^{2}{\theta}}{a\plus{}b\cos{\theta}}\ d\theta\equal{}\frac{2\pi}{b^{2}}\left(a\minus{}\sqrt{a^{2}\minus{}b^{2}}\right)$

Is there a "trick" to doing this or must I actually brute force this and make stuff go away?

I know you said you wanted a solution using residues, but here's a non complex approach if you are interested.

We expand, in a geometric series,  $ \frac {1}{a \plus{} b\cos{\theta}}$, justified by our conditions on a and b, in our integral to find that
\[ I \equal{} \int_{0}^{2\pi}\frac {\sin^{2}{\theta}}{a \plus{} b\cos{\theta}}\ d\theta \equal{} \frac {1}{a} \int_{0}^{2\pi}{\sin^{2}{\theta} \sum_{k \equal{} 0}^{\infty} \left(\frac {b}{a}\right)^k ( \minus{} 1)^k \left(cos{\theta}\right)^k d\theta}
\]
We next exchange the integral and sum, using the usual uniform convergence argument and see that we must consider integrals of the form
\[ J_k \equal{} \int_{0}^{2\pi}{sin^2{\theta}cos^k{\theta}d\theta}
\]
Symmetry tells us that $ J_k \equal{} 0$ for k odd.

Noting the fact that this integral is also symmetric about $ \frac {\pi}{2}$ means
\[ J_{2m} \equal{} 2*2\int_{0}^{\frac {\pi}{2}}{sin^2{\theta}cos^{2m}{\theta}d\theta} \equal{} 2\frac {\Gamma(1/2)\Gamma(\frac {2m \plus{} 1}{2})}{\Gamma(2 \plus{} m)}
\]
We can simplify this and note
\[ J_{2m} \equal{} 2\frac {\sqrt {\pi}}{2}\frac {(m \minus{} \frac {1}{2})!}{(m \plus{} 1)!}
\]
Recall
\[ \frac {1}{(1 \minus{} x)^{1/2}} \equal{} \sum_{m \equal{} 0}^{\infty} \binom{m \minus{} \frac {1}{2}}{m} x^{m}
\]
so if we consider
\[ \int_{0}^{x}{\frac {1}{(1 \minus{} x')^{1/2}}dx'} \equal{} \int_0^x{\sum_{m \equal{} 0}^{\infty} \binom{m \minus{} \frac {1}{2}}{m} x'^{m}dx'}
\]
we find that
\[ 2\left(\frac {1 \minus{} \sqrt {1 \minus{} x}}{x}\right) \equal{} \sum_{m \equal{} 0}^{\infty} \binom{m \minus{} \frac {1}{2}}{m}\frac { x^{m}}{m \plus{} 1} \equal{} \frac {1}{\pi} \sum_{m \equal{} 0}^{\infty} J_{2m}x^m
\]
From above we see that
\[ I \equal{} \frac {1}{a} \sum_{m \equal{} 0}^{\infty} \left(\frac {b}{a}\right)^{2m}J_{2m}
\]
So letting $ x \equal{} \left(\frac {b}{a}\right)^2$ We have
\[ I \equal{} \frac{2\pi}{a} \frac {1 \minus{} \sqrt {1 \minus{} \left(\frac {b}{a}\right)^2}}{\left(\frac {b}{a}\right)^2}
\]
as desired.

Well thanks anyway.  I did indeed evaluate this using residues, which were not particularly difficult to compute.  The difficulty lies in actually expanding the polynomials that result and re-package them in the neat form desired.

npizzo very interesting solution  :) .

more general :
suppose we need to find integral :
$ I: \equal{} \int_{0}^{2\pi}R(\cos x , \sin x ) dx$ by substitution $ e^{ix} \equal{} z$ we have
\[ \int_{|z| \equal{} 1} R(\frac {z \plus{} z^{ \minus{} 1}}{2},\frac {z \minus{} z^{ \minus{} 1}}{2i})\frac {dz}{iz}
\]
and after by residue ...

First, I don't really need this problem anymore since I already solved it.  Second, the problem wasn't that I don't know how to do residue calculations, the problem was that it got really ugly so I was hoping that there was possibly some algebraic shortcut.


MIT
college
geometry
geometric transformation
reflection

Hey all,

I just applied EA to MIT. Out of curiosity, just how important is the personal statement because the most important qualities that MIT looks for are character and personality.

I assume that the personal statement is the one big thing that can truly reflect a student's character and personality.

Any thoughts on this issue are welcomed.

Here's what I've never understood about questions like this: are you going to write a [i]worse[/i] personal statement if I tell you it's less important than you think?  Write the best personal statement that you can, and the admissions office will apply the semi-inscrutable filters and analyses and random subjective judgements that they apply, and then hopefully you'll get in.  Which is to say, good luck with your application :).

i'm pretty sure personal statements and things like that dont matter

If they didn't matter then why are they being asked for? ;)

Personal statements matter enough to knock you out of some programs if you write a lousy one. A famous saying about application college essays, which would apply to this kind of personal statement, is "They can heal the sick, but they can't raise the dead." The rest of your application has to be strong, of course, but there are always applicants who are right on the borderline based on everything else in their application, and you might as well, as JBL said, do the best you can on all elements of your application.

[quote="Elemennop"]Your middle name doesn't really matter but they still ask for that, do'nt thye?[/quote]

This is a great analogy :P.

On a more relevant note, I think personal statements can be important just because they're on the application. I don't feel like they do a great job in general of reflecting character and personality at least nowadays because so much effort is put into making them "good." But yeah they're not something to blow off (although maybe Elemennop did and he still ended up at the same place as me? :wink:).

Or read what Matt McGann and Ben Jones time and time again say - the essay is the only way a college admissions officer is going to get to know [i]you[/i]. It makes a lot of sense to write one well and straight from the heart.


Mary has 50 dollars. She goes to the mall and buys lipstick and then she buys shampoo, which is half the price of the lipstick. She then spends half of what she has left on a purse, leaving her with 15 dollars.
How much did the shampoo cost?

good job, answering most of my questions. :)

[hide]x=cost of lipstick

$x+\frac12x+\frac14x+15=50$

$\frac74x=35$

$x=20$

price of shampoo is $10[/hide]


The sequence 3, 15, 24, 48, ... is those multiples of 3 which are one less than a square. Find the remainder when the 1994th term is divided by 1000.

I think the answer is [color=#FF0000]80[/color], [color=#FF0000]but I'm not sure. Here is what did: each member equals to  n^2-1, which factors into (n+1)(n-1). So n must be all positive integers except multiples of 3.
To find 1994th n   set up equation n-[n/3]=1994. n should be 997*3=2991. the last 3 digit of 2991^2-1 will be the answer.[color]

Kalva says the answer is 63. Does anyone else have an idea on how to approach this?

Sorry about that, it had to be wrong. 997*3 can not be a possible value of n since it's a multiple of n. Also n can't be 1. So the equation should be n-[n/3]-1=1994. So n=2992, the answer should be 63


Excuse me could someone explain to me the way in which is chosen the German MEMO(Middle European Mathematical Olympiad)team
If somebody knows something about team or questions that were used for choosing the team.

Regards

hello, this is what i have found
http://www.oemo.at/en/info/memo.php
Sonnhard.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=366767&start=0]here[/url] for MEMO 2010.

I described that [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2002405#p2002405]here[/url] before.


During the rest hour, one of five students (A,B,C,D and E) dropped a glass of water. The following are the responses of the children when the teacher questioned them:

A: It was B or C who dropped it.
B: Neither E nor I did it.
C: Both A and B are lying.
D: Only one of A or B is telling the truth.
E: D is not speaking the truth.

The class teacher knows that three of them NEVER lie while the other two ALWAYS lie. Who dropped the glass?

[hide]is it... C who dropped it? A,B,and E are telling the truth while C and D are lying.[/hide]
[color=darkred][size=75]Please use hide tags.
-nebula42[/size][/color]

[hide]start with c (lying or truth)
If hes telling the truth, than D also has to be true, but it isn't. So C lies.
If C lies, than A and B are both telling the truth so it is either B or C. D is not speaking the truth so ABE are the truthful people. Since B didn't do it, it has to be c[/hide]

To BrendanKearney: The logic that C lies => A and B are both telling the truth is faulty.

[hide]
Just go by cases.
But first, we notice that at least one of C and D must be lying, as the two can't both be telling the truth.
I. C is telling the truth
i. D is lying
Both A and B must be lying, which is not possible since only 2 are lying.
II. C is lying--at least one of A and B must be telling the truth
i. D is lying
Both A and B must be telling the truth, and E must be telling the truth also. So, either B or C dropped it, but B says that B didn't do it. So the only possible one is C.
ii. D is telling the truth--one of A,B are lying and one is not lying--E must be telling the truth. Contradiction.
So it must be $ \boxed{C}$.
[/hide]

[hide="Took me a while"]
C and D are either both lying (A and B both tell the truth) or one is lying and one is telling the truth.

We can eliminate C telling the truth and D lying, because that would cause 3 lying people.

We can eliminate C lying and D telling the truth, because that would cause C, one of A or B, and E to lie, so we would have 3 liars again.

So, C and D are both lying, which means that A and B are both telling the truth. That means E tells the truth, which he/she does.

Since A and B are both telling the truth, it's B or C that did it, but B did not do it, so $ \boxed{C}$ did it.

[/hide]

is there a general trick to logic problems, or is it mainly just common sense? :huh:

i'll probably post some more when i get the time...

1. Look at the most restrictive things (in this case, the fact that it's impossible for both C and D to be true at the same time).
2. Organize your casework.

These usually can solve logic problems.

found another logic problem:
"Arrange the digits 1-9, without repeating, in such a way that
All the digits between 1 and 2 add up to 6.
All the digits between 2 and 3 add up to 14.
All the digits between 3 and 4 add up to 38.
All the digits between 4 and 5 add up to 9.
Find the smallest value of this 9-digit number."

I used trial/error for 15 min to get the right answer... :|

$ \boxed{371628594}$

Let's start with the first clue.

There is only one way that this could happen, and it would be if the sequence is $ 162$.

Then, we can tell that $ 3$ and $ 4$ are the numbers at the end, because $ 1 \plus{} 2 \plus{} ... \plus{} 8 \plus{} 9 \minus{} 3 \minus{} 4 \equal{} 38$

$ 3$_$ 162$_$ 4$

It becomes clear that the only sum between $ 4$ and $ 5$ is $ 9$, so let's add that.

$ 3$_$ 162$_$ 594$

Only $ 2$ digits left!

Since the sum between the $ 2$ and $ 3$ is $ 14$, we can now finish.

$ 371628594$

Lol I used guess and check too and I got it in about 15 secs...

did u use 162 instead of 261 cuz u want to minimize? cuz i included the 261 case cuz later, when u add more numbers in, it still might become smaller, but idk... :|

there are only 2 solutions:
that number forwards and that number backwards
obviously ernie's answer is right b/c the last digit is greater than the first 1

--new problem--
a right triangle has sides $ 2\sqrt2,6\sqrt2$ and an integer for the third side
what is the area of the right triangle?

Uhh, this isn't the mathcounts marathon steven...

We have two cases: where $ 6\sqrt{2}$ is the hypotenuse or when the unkonwn integer side is the hypotenuse.  Assuming 6sqrt2 is the hypotenuse:

$ \sqrt{72\minus{}8}\equal{}8$.

Thus, the unknown side is length 8.  So the area is $ 24\sqrt{2}$.


A mass $m$ is converted into energy, which is used to accelerate another mass $m$. Find its velocity if it began at rest.

A mass $m$ must be propelled to a portion $x$ ($0\le x<1$) of the speed of light. The energy consumed is equivalent to a portion $y$ ($y\ge0$) of mass $m$. Find $y$ in terms of $x$.

To solve this one would assemble the needed information:

e=Mc^2  (Total energy transferred to the the second mass m)

Assume that there is some way to have 100% efficiency since nothing else is given in the problem.  It may be closer to a theoritical 1% if we have to start with atomic energy and end up with a significant velocity relative to c.

Use the special relativity equations to see how much the rest mass m, increases at the velocity reached with this amount of energy, e.  Since e also equals Mv^2, absent relativistic effects the velocity would be c.

The rest is merely looking up the details of the equations and solving for the velocity.

Hope this helps you reach the answer you want.


Do I [i]have[/i] to do a sport to get into a place like Harvard?
Would it be so bad if I didn't?

People with no sports interest at all can and do get into Harvard. But Harvard is one of the colleges that originated intercollegiate sports in the United States, so sports are very much part of Harvard's culture. Most Harvard alumni I know played a sport before college, and kept up that sport at least at the intramural level in college. DARE TO BE YOURSELF is the main advice, for any college, but if you don't mind sports and have time to take one up, it can't hurt.

I did not do any sports in high school (well I swam a little bit, but not too seriously, not on the school team really). I don't play an instrument either. And I got into Princeton.

Now I play intramural frisbee, basketball, and soccer, and I love them all. (Well not basketball that much, but still, playing sports is really fun!)

What grade are you in?  If you're still young (gradewise) I would suggest at least giving a sport a try for one season.  Unless you've tried it before how do you know you won't enjoy it?  Being biased, I would suggest giving cross country and/or track and field a try :D

Does doing a sport means you're good at it? Or that you just play on a team or something?

Just joining a team. I don't know...sports just don't seem appealing to me at all. I really wish I didn't feel that way, but I do.

Try something unusual, if you can, like ultimate frisbee or rugby or whatever. Go according to your preferences -- there are plenty of good sports that are non-contact if you don't like rough games. And you can pick according to how much running or strength or whatever is necessary too. There are so many sports out there! Have fun! :D

Try tennis, it's a lifetime sports, and it's a sport you can spend a lot of money on. pretty awesome :P

Yeah,I have trained BOX last four years :P ,unfortunately I have stoped doing that because I just have no enough time :( .
So what is my question,I want to know your opinion,will it help me if I include that information in my applications? :maybe:

[quote="Tiks"]So what is my question,I want to know your opinion,will it help me if I include that information in my applications? :maybe:[/quote]
Probably it will help. You want to describe everything you do that could possibly be interesting, or that could show your positive qualities. ;)

[quote="Tiks"]Do you think I need to presant any documant to prove the fact that I was training BOX? :ninja:[/quote]
No, surely not. They will believe you if you just list it as one of your activities or interests. If you ever went to any interesting boxing-related event or contest or something, then you could mention this.

just say ur in jv tennis, soccer, track blah blah nobody knows :D :lol:

^ Yeah don't listen to him. You get into that mindset and eventually your application is giving admissions officers a vibe that's not you.

Just be yourself; be up front with the colleges and don't try to cut corners to get ahead. If you take tennis lessons, you can list that in the activities section along with scuba diving and skiing (as long as they're not 1 time per year) and admissions would love to see this.

Yeah, be yourself on your app.  If you have to put someone else's stats on the app then you probably aren't fit for that school and quite honestly I hope you fail out.

And don't take a sport just to put it on your college app.

If anything, do as others said and try a sport to see if you like it. If you've already done this, and you just happen to dislike sports, then don't force yourself to do so.

I personally immensely enjoy soccer, but I am not particularly good at it, and I highly doubt that it will have any impact upon whether a college accepts me. Colleges aren't looking for someone that does everything. If sports aren't your thing, they aren't your thing. To be honest, I highly doubt they will have any significant impact unless you are like in a high division on a select team.

the point is to make ur background strong and stand out.  If you did something say it, if u never ever did it then don't.  :D You can lie of course but that's moral issue, I'll let u judge.  If you did play tennis for example like an hour / week, don't put "I play tennis 1 hr/week", say like "I used to play tennis and have an interest in it" or something like that (This is not a lie, you did play). Don't exaggerate too much like "I used to teach michael jordan how to play basketball", then they might do a check by calling him up to clarify if this is true  :roll:

[quote="aznphatso"][quote="dkushnir"]Does once a week tennis lesson / pleasure downhill skiing / pleasure scuba diving count?
Or should it be high school team in tennis and downhill skiing etc.?[/quote]

On my Stanford app, I put "Recreational sports" under extracurriculars and had "JV volleyball (10), leisure basketball, volleyball, swimming (9-12)" 5 hrs/week because that's just how it is for me. They didn't seem to mind.[/quote]

Same, except mine were even less conventional: Tennis, Frisbee, DDR :P. I put them on the same line so as to not waste my extracurricular space.

[quote="aahmadi"]Did both of you (paladin and ryan) apply for stanford [b]this[/b] year?
and out of curiousity, are you the same ryan ko who won mathcounts?[/quote]

One of the sad things about being me is that that Ryan Ko (Mathcounts champ, the one that shows up on all the google searches w/ the picture meeting the President) is already Stanford '10 -- so no, I'm another Ryan Ko, potentially Stanford '11. :blush:  D'oh!  :oops:

And yes, Paladin and Ryan (aznphatso) did apply to stanford this year, we both got in early action.

[quote="aahmadi"]Did both of you (paladin and ryan) apply for stanford [b]this[/b] year?[/quote]

Yeah, we both did.

EDIT: Darn, too slow.

[quote="paladin8"]
EDIT: Darn, too slow.[/quote]

Not by much ;)


Find all the functions $f: R^+\to R^+$ so that for every $x,y\in\mathbb{R^+}$,

$(x+y)f(f(x)y)=x^2f(f(x)+f(y))$

[quote="ehsan2004"]It has been discussed before[/quote]
Can you please give me the link?

This is the 6th problem of the 2004 Czech and Slovak Republics Math Olympiad.

And there are no such functions.

Suppose $ f(a)\equal{}f(b)$ for $ a\ne b$.
Compare $ P(a,y)$ and $ P(b,y)$, we get for any $ y>0$,
$ \frac{a^2}{a\plus{}y}\equal{}\frac{b^2}{b\plus{}y}$. This cannot be true.
So $ f(x)$ is injective.

Let $ x$ be the positive root to $ x^2\minus{}x\minus{}1\equal{}0$, plug in $ y\equal{}x^2\minus{}x\equal{}1$, we have
$ f(f(x))\equal{}f(f(x)\plus{}f(1))\Rightarrow f(x)\equal{}f(x)\plus{}f(1)\Rightarrow f(1)\equal{}0$. Contradiction.


what happens to ketone obtaied on  reductive ozonolysis of alkene when treated with $ \ce{H2O2}$

That would be an oxidative ozonolysis. Regarding your question: what happens to a "normal" ketone when treated with hydrogen peroxide?

[hide="Answer"]Nothing happens: the ketone remains unchanged. Only an aldehyde is oxidized to the carboxilic acid.[/hide]

thanks thats what i wanted to know.What are "abnormal" ketones.also by oxidative i meant ozonlysis without zinc

[quote="Pardesi"]What are "abnormal" ketones[/quote]

In this context, an abnormal ketone would be one with a functional group that reacts with H2O2, such as an alpha,beta-unsaturated ketone.

now that's rare because an ozonolysis product still having a double bond.
what would happen to the double bond then?My guess water addition

[quote="Pardesi"]now that's rare because an ozonolysis product still having a double bond.
what would happen to the double bond then?[/quote]

I don't undertstand what you mean by that. Explain yourself.

i meant that if there were a double bond in a product obtained by ozonolysis then why didn't the double bond break during the ozonolysis itself


[quote="espark52"]in the sequence 1,6,15,28,45, and so on, the nth tertm of the sequence is givin by a polynomial, $an^{2}+bn$. wat's the ordered pair (a,b)?[/quote]

The sequence can be expressed as the quadratic $2n^{2}-n$.

$\boxed{(2,-1)}$.

...and I don't think this (or any of your other problems) is Intermediate level.

[quote="espark52"]in the sequence 1,6,15,28,45, and so on, the nth tertm of the sequence is givin by a polynomial, $an^{2}+bn$. wat's the ordered pair (a,b)?[/quote]
[hide="A Different Solution"]$a\cdot1^{2}+b\cdot1=1;\,a\cdot2^{2}+b\cdot2=6$

$a+b=1;\,2a+b=3$

Subtract the first equation from the second to get $a=2.$ Then, substitute $a=2$ into the first equation to find $b=-1.$

$\boxed{(2,-1)}$[/hide]


analytic geometry
combinatorics unsolved
combinatorics

A super-rook sits on a $ 2009 \times 2009$ chessboard. A move of this rook consists of moving $ a$ squares in one direction and $ b$ squares in the other $ (a, b > 0)$. Thus, from square $ (p, q)$ the rook can move to any of the eight squares $ (p\pm a, q\pm b)$ and $ (p\pm b, q\pm a)$.

Prove that the rook can only return to its starting position after an even number of moves.

for the $ x$ coordinate to stay the same, the number of $ (x,y)\rightarrow (x\plus{}a,y\pm b)$ moves must be the same as the number of $ (x,y)\rightarrow (x\minus{}a,y\pm b)$ moves...

[quote="pleurestique"]for the $ x$ coordinate to stay the same, the number of $ (x,y)\rightarrow (x \plus{} a,y\pm b)$ moves must be the same as the number of $ (x,y)\rightarrow (x \minus{} a,y\pm b)$ moves...[/quote]
I think you could not conclude this since we still have moves $ (x \pm b,y\pm a)$. Correct me if I'm wrong!

you're right... the solution goes like this then... after $ k$ moves, the coordinates are of the form $ ((r\minus{}s)a\plus{}(t\minus{}u)b,(x\minus{}y)a\plus{}(z\minus{}w)b)$, where

(1) $ r\plus{}s\equal{}z\plus{}w$
(2) $ t\plus{}u\equal{}x\plus{}y$
(3) $ r\plus{}s\plus{}t\plus{}u\equal{}x\plus{}y\plus{}z\plus{}w\equal{}k$

if the rook is back to its original square, we also need $ (r\minus{}s)a\equal{}(u\minus{}t)b$ and $ (x\minus{}y)a\equal{}(w\minus{}z)b$. this implies

(4) $ (r\minus{}s)(w\minus{}z)\equal{}(u\minus{}t)(x\minus{}y)$.

(1) and (2) imply that $ r\minus{}s, w\minus{}z$ have the same parity and $ u\minus{}t, x\minus{}y$ have the same parity. because of (4), $ r\minus{}s, w\minus{}z, u\minus{}t, x\minus{}y$ in fact all have the same parity, which implies that $ r\plus{}s, w\plus{}z, u\plus{}t, x\plus{}y$ all have the same parity as well. so $ k\equal{}(r\plus{}s)\plus{}(t\plus{}u)$ must be even.


I need some quick help and explanation for some AP Biology questions.

Which statement is NOT true about a hypothesis?
a. experiments or observations are undertaken to test a hypothesis
b. a hypothesis cannot be proven true.
c. a hypothesis cannot be proven false.
d. data that support a hypothesis do not actually prove it.

Which statement is NOT true about experimental design?
a. All conditions are held the same except for the condition being tested by the experiment
b. it's best to use genetically identical mice to reduce the uncontrolled variables
c. the condition being tested in an experiment is the controlled variable.
d. the dependent variable is observed the end of the experiment when statistical comparisons are made between groups.

Which of the following terms best describes the collection of scientific data through observation in the field, such as observing the behavior of birds?
a. a scientific model
b. an experiment
c. descriptive research
d. a scientific theory or principle
e. experimental results

thanks very much for the help

The first one is C: you can prove a hypothesis false if even one experiment yields contradictory results. There is a famous quote by Einstein that goes "No amount of experimentation can ever prove me right; a single experiment can prove me wrong."

The second one is C: the controlled variable (also referred to as the independent variable) is the thing you change to see what effect it has on some other variable, called the dependent variable. The dependent variable is the one being observed.

The third one is B.


calculus
integration
geometry
3D geometry
analytic geometry
calculus computations

Hello, I have a problem here. Actually its at task where i dont understand one thing. And that is how to find the boundries for the triple integral.

Its like this: $D$ is a area that both lies inside the circular cone $z=\sqrt{x^2+y^2}$  and on the inside of  the sphare $x^2+y^2+z^2=1$. Find $\int\int_D\int z dV$.

Now i am not so good with cylindrical coordinates, but I think they can be used here. Anyway, what I have come up with is that $r \leq z\leq \sqrt{1-r^2}$ and $0 \leq\theta\leq 2\pi$, but then I have to find the boundries for $r$ and I have not been able to find out in the last three hours.

So I would appriciate if anyone could explain to me how I find the boundries for $r$.

Tnx in advance!

Here's a pdf that answers the exact question you have (look at problem #27):
http://www.math.ucdavis.edu/~kouba/Math21CHWDIRECTORY/HW13.pdf

If for some reason you can't access it, here are the boundaries for D:
\[ D: \left\{ \begin{array}{cc} 0 < r < \frac{1}{\sqrt{2}} \\ 0 < \theta < 2\pi \\ r < z < \sqrt{1-r^2} \end{array}\right.  \]

Oh! This was nice. Now i got it.. Thanks swapnillium.  :)

Actually I wonder if you know where thoes tasks are from? I would really appriciate it if you knew where I could get the original tasks for the solutions in the .pdf file. Cuz I really need the training.. Its propobly from a book, but it is worth a try asking :)

Yeah, sure thing. That pdf was from my old MAT 21C class at UC Davis (that's why I knew where to look :D ). That's the class where I learned all about multivariate calculus! The professor who teaches it (Dr.Kouba) is really nice and has a lot of practice problems and stuff for students. So... here is the home page for that professor's classes:
http://www.math.ucdavis.edu/~kouba/
Just go to this URL and check out the MAT21C website or the MAT21C homework solutions. Unfortunately, like you said, the problems are from a specific book. But there are a lots of other things there that you can check out and improve you mutivariate calculus skills.  :)

Also, here is a website that has a lot a math applets. The ones you might be interested in would be under "Three-Dimensional Graphing."  That has links to applets that can graph in rectangular/cylindrical/spherical coordinates!! A superb java applet(s) that helps you visulize those beautiful 3D graphs (you just need to have java installed on your pc if you don't have it already).
http://cs.jsu.edu/mcis/faculty/leathrum/Mathlets/

-good luck learning!  ;)

Thanks man! This is some good stuff. This will help me alot I think. Especially the applets, cuz i need to learn how to draw the figures before i do the tasks. As it is now, I almost allways manage to draw stuff upsidedown if the equations are to hard  :blush: ''

Thanks again...


what do you think it will be like? Predictions on your placement? Team placement? Test difficulty?

honestly, i hope to get around 50-150 so ill click on both of them :P  my expectations r sooooooooooo high! for out team...[hide]AL FTW!!! jk but its a possibility....
realisticly, i think that we'll place in the middle, 20-40  because we're beast like that [/hide]

my place: I think the test will be crazy hard (i like those) and i think our team will get top 20 (hopefully!)

What was your chapter score (total)?

:noo:  :noo:  :noo:  nats......= me be emo.....

Lol what did we get as a team? And who did the best?

All of us failed!

Al got 43rd :(  :blush:  :(

who do you think got first???? (Sid) Who do you think got last????? (me)


probability
expected value
probability and stats

Imagine an infinite bee hive built by hexagons.
[img]http://bee.ryex.net/hive.png[/img]

A bee is laying eggs in empty hexagons and climbing to a new one. She continues to lay eggs and climb into new cells, as long as there is empty cells surrounding her. After a while she will probably walk in a circle and block herself with her own eggs.

[img]http://bee.ryex.net/paths.png[/img]
This picture shows different paths that the bee can follow. The first path with 7 eggs is the absolute minimum of eggs that the bee must lay down before she is trapped. The next two paths is the result of random simulation.

What I want to find out is how many eggs she lays out before she is trapped. What is the expected value for number of eggs? Using random simulations I have made a graph showing the distribution of probability for different number of eggs.
[img]http://bee.ryex.net/graph_100_000_000.png[/img]

From this distibution I have calculated the expected value to be approximately 78 eggs. This value is based on
random simulations, and I want to find the exact number. Maybe it is possible to find an expression for the probability for a given number of eggs.

I hope somebody can helo me on this one.
For more information visit [url]http://bee.ryex.net/[/url]

Not sure if this helps, but that looks a lot like the graph of a  [url=http://en.wikipedia.org/wiki/Gamma_distribution]Gamma Distribution.[/url]


Question: There are 20 people on the basketball team. How many ways are there to pick the starting lineup of 8 players if Joe must be in the starting lineup?

we have 1*19*18*17*16*15*14*13= 25395520 ways.

P.S. I know this is spamming, but tank u for posting an ez question.

If positions do not matter, then it is simply $ \binom{19}{7}$ (19 choose 7) or $ \boxed{50388}$.

If they do, then your answer is correct. :)


function
trigonometry
integration
limit
calculus
logarithms
real analysis

Hi,

as is well known, if
$ f: [ \minus{} \pi,\pi] \longrightarrow \mathbb{C}$
is twice continuously differentiable, then
the for the Fourier partial sums,

$ S_n(f,t) \rightarrow f(t)$

as $ n\rightarrow \infty$, uniformly on $ [ \minus{} \pi,\pi]$.

Does anyone know a good, simple example
of a function $ f: [ \minus{} \pi,\pi] \rightarrow \mathbb{C}$
which is once continuously differentiable, but $ \emph{not}$
twice, and for which this type of convergence does not hold?
What exactly can go wrong if it only once continuously differentiable,
but not twice??

Thanks,

.,.

One quick comment: $ f\in C^1$ does imply that the partial sums converge pointwise everywhere to the function. Note that the Fourier coefficients would be $ o(n^{\minus{}1}).$ Since for a continuous function the Cesaro means or Abel means of the Fourier series (pick either one) converge uniformly to $ f,$ then Tauber's Theorem implies that $ S_nf(x)\to f(x)$ everywhere.

Note also for building examples: if you try an example in which $ f''$ is bounded and has only jump discontinuities (more broadly, if $ f'$ is of bounded variation), then that would imply that the Fourier coefficients are $ O(n^{\minus{}2}),$ which is more than enough to ensure uniform convergence.

... haven't answered your question yet, of course.

If $ f\in C^1,$ then $ S_nf\to f$ uniformly. The answer to your question is that nothing can go wrong if $ f\in C^1\setminus C^2,$

Define $ D_n(x) \equal{} \sum_{k \equal{} \minus{} n}^ne^{ikx} \equal{} \frac {\sin\left(\left(k \plus{} \frac12\right)x\right)}{\sin\left(\frac12\,x\right)}.$

One thing to note about $ D_n$ is that $ \frac1{2\pi}\int_{ \minus{} \pi}^{\pi}D_n(x)\,dx \equal{} 1$ for all $ n.$

Define $ h_n(x) \equal{} \int_0^xD_n(t)\,dt.$ We have these properties:

$ h_n(\pi) \equal{} \pi$ and $ h_n( \minus{} \pi) \equal{} \minus{} \pi,$ and $ \lim_{n\to\infty}h_n(x) \equal{} \pi\,\text{sgn}(x)$ pointwise on $ [ \minus{} \pi,\pi],$ where $ \text{sgn}(x)$ is $ 1$ for $ x > 0,$ $ \minus{} 1$ for $ x < 0$ and $ 0$ for $ x \equal{} 0.$

Furthermore, we have that there exists a constant $ M$ such that $ |h_n(x)|\le M$ for all $ n$ and all $ |x|\le\pi$ and for each $ \delta > 0$ the convergence of $ h_n$ to the signum function us uniform on the set $ [ \minus{} \pi,\pi]\setminus ( \minus{} \delta,\delta).$

We can write $ S_nf(x) \equal{} \frac1{2\pi}\int_{ \minus{} \pi}^{\pi}D_n(t)f(x \minus{} t)\,dt.$

Subtract off $ f(x)$ and use that property about the integral of $ D_n$ to get

$ S_nf(x) \minus{} f(x) \equal{} \frac1{2\pi}\int_{ \minus{} \pi}^{\pi}D_n(t)[f(x \minus{} t) \minus{} f(x)]\,dt.$

Now split this into $ [ \minus{} \pi,0]$ and $ [0,\pi]$ and separately integrate each half by parts. We use that $ f$ is assumed to be $ C^1.$ Here's the right half:

$ \frac1{2\pi}\int_0^{\pi}D_n(t)[f(x \minus{} t) \minus{} f(x)]\,dt \equal{}$

$ \left.\frac1{2\pi}(h_n(t) \minus{} \pi)[f(x \minus{} t) \minus{} f(x)]\right|_0^\pi \plus{} \frac1{2\pi}\int_0^{\pi}(h_n(t) \minus{} \pi)f'(x \minus{} t)\,dt$

Note that the boundary terms are zero at both ends. The integration by parts on the other half proceeds similarly. We arrive at:

$ S_nf(x) \minus{} f(x) \equal{} \frac1{2\pi}\int_{ \minus{} \pi}^{\pi}[h_n(t) \minus{} \pi\,\text{sgn}(t)]f'(x \minus{} t)\,dt$

$ |S_nf(x) \minus{} f(x)|\le \frac1{2\pi}\int_{ \minus{} \pi}^{\pi}|h_n(t) \minus{} \pi\,\text{sgn}(t)|\, |f'(x \minus{} t)|\,dt$

$ \le \|f'\|_{\infty}\cdot\frac1{2\pi}\int_{ \minus{} \pi}^{\pi}|h_n(t) \minus{} \pi\,\text{sgn}(t)|\,dt$

Note that this estimate is now uniform - it has no dependence on $ x.$ All we need to do is to show that the limit of the right hand side of this estimate goes to zero, which we can do by dividing the integral into two pieces, one for $ |t|\le\delta$ and one for $ |t|\ge\delta$ and applying separate estimates to each.

Thank you very much for this answer! You are
right, another way to prove it is to first prove,
as you already stated, that $ f\in C^1(\mathbb{T})$
implies $ \widehat{f}(n) \equal{} o(1/n)$, then observe
that this implies that
$ \sum_{n\equal{}\minus{}\infty}^\infty |\widehat{f}(n)|$
converges, and then appeal to the well-know
convergence theorem, which is e.g. proved
in Körner's book "Fourier Analysis" at the beginning
of chapter 9. This theorem ensures uniform convergence.

.,.

No, $ o(\frac1n)$ is not enough for absolute convergence. The simple argument doesn't work.

[quote="fedja"]But if you notice that $ n\widehat f(n)\in \ell^2$ and use Cauchy-Schwarz, it will. ;)[/quote]
D'oh! I was looking at exactly that, and now I can't remember why I talked myself out of it.

Thanks for your corrections!

And what is an example of sequence
$ (a_n)_{n\in \mathbb{N}}\subseteq \mathbb{R}_{\geq 0}$
such that
$ \lim_{n\rightarrow \infty} n\cdot a_n \equal{} 0$
but
$ \sum_{n\equal{}0}^{\infty} a_n$
diverges to $ \infty$?

@fedja: How exactly do you want to employ
Cauchy-Schwartz? And what is your argument
for $ n \widehat{f}(n)\in \ell^2$? Do you mean
the simple argument that if
$ \widehat{f}(n) \equal{} n^{\minus{}1\minus{}\epsilon}$, then $ n \equal{} o(\frac{1}{n})$
because $ \lim_{n\rightarrow \infty} n\cdot n^{\minus{}1\minus{}\epsilon} \equal{} n^{\minus{}\epsilon}\longrightarrow 0$
for $ n\rightarrow \infty$, and then $ \biggl(\widehat{f}(n)\biggr)^2 \equal{} n^{\minus{}2\minus{}2\epsilon}$,
so $ \sum_{n\equal{}1}^\infty \biggl(\widehat{f}(n)\biggr)^2$ converges?

$ a_n\equal{}\frac{1}{n\log n}$ (starting appropriately at $ n\equal{}2$.)

$ in\widehat f(n)$ are the Fourier coefficients of $ f'\in C\subset L^2$. As to Cauchy-Schwarz, just write
\[ \sum_n |\widehat f(n)|\le \left(\sum_n n^2 |\widehat f(n)|^2\right)^{1/2}
\left(\sum_n \frac 1{n^2}\right)^{1/2}\,,\]
where the sum is taken over all $ n\ne 0$.


$F_{n}=F_{n-1}+F_{n-2}$ is the $nth$ term in the  $Fibonacci$ serie.

Lucas numbers are define as follow :
$L_{n}=F_{n+1}+F_{n-1}$

Prove : $F_{2n}=F_{n}*L_{n}$

I try to use induction but not results yet...any hint???...

Isn't this a little too difficult for the getting started forum?

was this in a textbook? because i think I saw it somewhere...

 :maybe:

Don't you first have to prove that $L_{n}=L_{n+1}+L_{n+2}$ by induction?

You can prove $F_{m+n}=F_{n-1}F_{m}+F_{n}F_{m+1}$ by induction on $m$ or $n$ and then set $m=n$.

Well, you could probably do it combinatorically.  You may not know this, but Fibbonacci numbers and the number of ways to tile a board of length $n$ with squares and rectangles ($f_{n}$) are closely related.  In fact, $F_{n}= f_{n-1}$.  On the other hand, $L_{n}$ counts the number of ways to tile a circular board with $n$ spaces with squares and rectangles, so it becomes $f_{2n-1}=f_{n-1}*L_{n}$.  If you could find a bijection a board of length $2n-1$, and a board of length $n-1$ combined with a loop of length $n$, that would be a proof.

So... look at the first $n-1$ spaces, if it ends with a square or rectangle (as opposed to half a rectangle, with the other half on space $n$), turn the first $n-1$ spaces into a board of length $n-1$ (that's the $f_{n-1}$), and the remaining $n$ spaces into a loop of length $n$ (that's the $L_{n}$) which is "in-phase".  That means whatever object occuppies the last space doesn't cross over to the first space.  Basically, just wrap the $n$ spaces around into a circle.  On the other hand, if a rectangle occuppies spaces $n-1$ and $n$, turn the first $n$ spaces into an "out of phase" $L_{n}$, which means the rectangle occuppies spaces the first and last space.  Wrap everything around and then shift it over one.  Turn the last $n-1$ spaces into the $f_{n-1}$.

I think that does it.  Every combination of an $n-1$ board and $n$ loop can be made into exactly one  $n-1$ board, and vice-versa.

It's not induction, but after reading the book "Proofs that Count" (where I learned this proof method), I've seen that Fibbonacci and Lucas number problems are so much easier to do with combinatorics.


Is Analytic geometry worth investing my little time in? By analyitc, I mean parametric stuff, vectors...

I've heard that it's good to study it a bit, but I personally haven't seen many problems that requires it in the AMCs.

I meant for the AIME too. In my opinion, the kinds(not difficulty)  of problems on the AIME and AMCs really aren't that different.

Analytic geometry has tradeoffs with regular geometry; it occasionally trivializes problems, and it is more universal, but it's usually far uglier. If you're going to learn more of it, the most important thing to know is when to use it.


If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, then the ratio of the smaller base to the larger base is:
$ \textbf{(A)}\ \frac{1}{2} \qquad\textbf{(B)}\ \frac{2}{3} \qquad\textbf{(C)}\ \frac{3}{4} \qquad\textbf{(D)}\ \frac{3}{5} \qquad\textbf{(E)}\ \frac{2}{5}$

Let the base and diagonals be r, the smaller base and height be h. Let the length of the legs be x. Since all isoceles trapezoids are cyclic, by Ptolemy, $ rh\plus{}x^2\equal{}r^2$, so $ x\equal{}\sqrt{r^2\minus{}h^2}$. Drop both altitudes from the endpoints of the smaller base, we get a right triangle with sides $ \sqrt{r^2\minus{}h^2}$, h, and $ \frac{r\minus{}h}{2}$. Pythag simplifies to $ 5h^2\minus{}3r^2\plus{}2rh\equal{}0$, which factors as $ (5h\minus{}3r)(h\plus{}r)$, so the answer is $ \frac{3}{5}$.

Here's a solution without using Ptolemy:

Let $ s$ be the smaller base and $ b$ be the bigger base.
$ b \minus{} (b \minus{} s)/2 \equal{} (b \plus{} s)/2 \equal{} x$
$ x^2 \equal{} (b \plus{} s)^2/4 \equal{} b^2 \minus{} s^2$ - using $ b$ as hypotenuse and $ s$ as leg

Letting $ b \equal{} 1$ we get $ 5s^2 \plus{} 2s \minus{} 3 \equal{} 0 \implies s \equal{} 3/5$


How many axes of symmetry does a rectangle have given that it is not a square?

2. The horizontal and vertical lines through the center.


algorithm
number theory
number theory unsolved

$ x,y$ are positive integer;
$ x^{3}\plus{}45\equal{}y^{2}$

First all of, if $ x$ is even then $ y^2 \equiv 5 \bmod{8}$ which is not possible
If $ x \equiv 1 \bmod{4}$ then $ y^2 \equiv 2 \bmod{4}$, again not possible

Then we have $ x \equiv 3 \bmod{4}$

Now we transform our originial problems into:

$ x^3\plus{}21^3\equal{}y^2\plus{}96^2$
$ (x\plus{}21)(x^2\minus{}21x\plus{}21^2)\equal{}y^2\plus{}96^2$
Now $ x^2\minus{}21x\plus{}21^2\equiv 3 \bmod{4}$ so this expression has a prime factor of the form $ p\equal{}4k\plus{}3$ (we have to check that the number
$ x^2\minus{}21x\plus{}21^2$ is positive for $ x>0$, and for $ x<0$ we have that $ x>\minus{}4$ so we check this numbers and we found no solutions)
Now it is well known that if $ p\equal{}4k\plus{}3$ divides a sum of integer squares, then $ p$ divides both squares (I can provide a proof if you want)
So $ p$ has to divide $ 96$ so $ p\equal{}3$
Then $ x^2\minus{}21x\plus{}21^2\equiv 0 \bmod{3}$ so $ x\equal{}3l$
We have that $ 21^3\plus{}x^3\equiv y^2\plus{}96^2\equiv 0 \bmod{27}$
Then $ y\equal{}3a$ and we have   $ a^2\plus{}32^2\equiv a^2\plus{}1\equiv 0 \bmod{3}$ which is clearly not true, so we have found that there are not any integer solutions.

[quote="lambruscokid"]First all of, if $ x$ is even then $ y^2 \equiv 5 \bmod{8}$ which is not possible
If $ x \equiv 1 \bmod{4}$ then $ y^2 \equiv 2 \bmod{4}$, again not possible

Then we have $ x \equiv 3 \bmod{4}$

Now we transform our originial problems into:

$ x^3 \plus{} 21^3 \equal{} y^2 \plus{} 96^2$
$ (x \plus{} 21)(x^2 \minus{} 21x \plus{} 21^2) \equal{} y^2 \plus{} 96^2$
Now $ x^2 \minus{} 21x \plus{} 21^2\equiv 3 \bmod{4}$ so this expression has a prime factor of the form $ p \equal{} 4k \plus{} 3$ (we have to check that the number
$ x^2 \minus{} 21x \plus{} 21^2$ is positive for $ x > 0$, and for $ x < 0$ we have that $ x > \minus{} 4$ so we check this numbers and we found no solutions)
Now it is well known that if $ p \equal{} 4k \plus{} 3$ divides a sum of integer squares, then $ p$ divides both squares (I can provide a proof if you want)
So $ p$ has to divide $ 96$ so $ p \equal{} 3$
Then $ x^2 \minus{} 21x \plus{} 21^2\equiv 0 \bmod{3}$ so $ x \equal{} 3l$
We have that $ 21^3 \plus{} x^3\equiv y^2 \plus{} 96^2\equiv 0 \bmod{27}$
Then $ y \equal{} 3a$ and we have   $ a^2 \plus{} 32^2\equiv a^2 \plus{} 1\equiv 0 \bmod{3}$ which is clearly not true, so we have found that there are not any integer solutions.[/quote]

 Is there any special algorithm or techniques to find some relationship such that 45+96^2 = 21^3 ?
 It looks impossible to find such equation only by some feeling  :o

In fact, it seems magical, but when you have done lots of problems similar to this, you find out that is perfectly reasonable. This trick  has been used in several problems, in which the main idea is to provide a factorization and then use the lemma about sum os squares.(it is my favorite lemma in number theory ) So when I first saw this problem I thought that this "trick"  a good idea, and luckily I found that particular solution which allowed me to solve the problem

I found a particular solution of t^3-45=z^2, I did not have any guarantee that it would work, but I insist on the point that having known that trick, it was worth trying

I hope you like the solution, and if you wish I can tell you other problems that can be done in a similar way


limit
integration
calculus
real analysis

If $f(x)=e^{x}$, then

$L=\lim_{n\to\infty}\sum_{k=1}^{n}{1\over n}f\left({k\over n}\right)$, which is Riemman sum for $f(x)$ on $[0,1]$ divided into $n$ equal subintervals. Hence

$L=\int_{0}^{1}e^{x}\mathrm{d}x=e^{x}\bigg|_{0}^{1}=e-1$

I got the Riemman sum and I know how to convert it to an integral, but I don't understand how to get the interval $[a,b]$ to equal $[0,1]$ and the $n$ equal subintervals (did you get it from the $\frac{1-0}{n}$?)

[quote="g1sk8terz"]Find $\lim_{n\rightarrow\infty}\frac{e^{\frac{1}{n}}+e^{\frac{2}{n}}+e^{\frac{3}{n}}+...+e^{\frac{n}{n}}}{n}$

using Riemann Sums.[/quote]
$\lim_{n\rightarrow\infty}\frac{e^{\frac{1}{n}}+e^{\frac{2}{n}}+e^{\frac{3}{n}}+...+e^{\frac{n}{n}}}{n}= \lim_{n\to\infty}\sum_{i=1}^{n}(e^{\frac{i}{n}})(\frac1n)$

The $\frac1n$ tells us that the interval has length of 1 and we know that $f(x)=e^{x}$ so the right hand endpoints are $0+\frac{i}{n}$, so the left part of the interval is 0. Therefore, the limit is $\int_{0}^{1}e^{x}dx$ and you can proceed as in the other post.

Was that clear?

On an interval $[a,b]$ take $n-1$ inner points $x_{1},\dots,x_{n-1}$ and denote $x_{0}\equiv a, x_{n}\equiv b$. Then Riemann sum for $f(x)$ on $[a,b]$ is defined as

$R=\sum_{k=1}^{n}f(x_{k})\Delta x_{k}$ or $R=\sum_{k=1}^{n}f(x_{k-1})\Delta x_{k}$

where $\Delta x_{k}=x_{k}-x_{k-1}$. If the limit of that sum exists for any sequence of inner point sets which satisfy $\max\Delta x_{k}\to 0$, then we call it Riemann integral of $f(x)$ on $[a,b]$.

Obviously, the simplest case is when we split the interval evenly. Then $x_{k}=a+{k\over n}(b-a)$ and $\Delta x_{k}={1\over n}$, hence the result.

NOTE: From ${k\over n}=a+{k\over n}(b-a)$, which must hold FOR ALL integer $k$ in $[0,n]$, you very easily get $a=0, b=1$.


Billy Bob is going to see the fireworks tonight! There are 64 seats. Only 8 are good seats. Since 63 other people are going to come, He will need to get there early. He happens to take the last seat that is a good one. If he is not on the edge of the row he is in, how many possible seatings are there, if everyone is considered different?

[hide="really confusing problem"]There are 6 places he can sit, assuming that the 8 good seats are just the first row, and 63! for all the rest.
$6 \times 63!$ is some big number[/hide]

[quote="samath"]Maybe the first 7 people took good seats, in which case its $6\times 7!\times 56!$[/quote]

But we do not know who are the first 7 people so they can be interchanged with the other 56 people.

!!! I don't like the wording of this problem.

For all we know there is one row and the eight central ones are considered "good".


Each month in the country of Kyzur, residents wait until the first Thursday, then have festivals on every subsequent Tuesday until the end of the month.  If a given month has 31 days, what is the probability that there are four festival days during that month?

[hide="Just an Idea"]
Think that they are actually asking you that, if a moth has 31 days, what is the probability to have five Tuesdays, and that is easy to compute. If you want to know, see the next hide
[/hide]

[hide="The Next Hide"]\[\frac37\][/hide]

[hide]
Consider the following:
Day of Week of 1st, Days Remaining after first Thursday, Remaining days after end of the week
Sunday, 26, 24
Monday, 27, 25
Tuesday, 28, 23
Wednesday, 29, 27
Thursday, 30, 28
Friday, 24, 22
Saturday, 25, 23

$31 \equiv 3 \pmod {7}$. So the month ends on the day of the week two days after the day of the week it began on.

For months that begin on any day from Sunday to Thursday, we only need to ensure that there are three full weeks (from Sunday to Saturday) remaining, because we know that the month will end on Tuesday or later. This means that the partial week will still give us an opportunity to hold a festival. Examining the remaining days at the end of the week, we see that a month starting on any day, Sunday-Thursday will allow for 4 fesitvals to be held.

Months that begin on Fridays or Saturdays will end on Sunday or Monday, respectively. So for them to have 4 festival days, they must have three full weeks after the week of the first Thursday, and in addition, there must be two extra days for months beginning on Friday and three for months beginning on Sunday. This sums to a total of 23 and 24 days after the completion of the first week containing a Thursday. This isn't satisfied.

Thus the probability is 5/7?[/hide]

[size=59]I edited my LaTeX.[/size]

[hide]There are seven combinations of the three days after 28.  Only 3 have Thursday in them, so the answer is 4/7[/hide]

[hide]A Thursday and four Tuesdays count for 27 days, leaving four days. However, the month can start on a Thursday, so there are five days out of the week that the month can start on, so the answer should be 5/7.[/hide]

[quote="djimenez"]Sorry, in my last post I missread, I said Tuesday instead of Thursday.

[hide="Read this Mel"]
I think there is an error in your reasoning (actually I didn't understand pretty well what your aregument was :blush: ) but if we assume that there is an equal probability for each day to start the month, then, in order to have the five Thursdays, the first one and the other four for the festivals, then the month needs to start in Tuesday, Wednesday or Thursdays, otherwise there will be only four. Then, from seven posible first days of the month, there are just three that work, the probability has to be $\frac37$.
[/hide][/quote]

[hide="djimenez"]
Sorry about the bad explanation. The truth is that I approached the problem by looking at the calendar in front of me, put one finger on the start day, one on the last, and just examined every day of the 7 days to see which ones would allow for 4 festival dates to happen. Obviously, that didn't come out too well in words. [/hide]

[hide="Festival date confusion?"]
Maybe I'm misreading the question, but I thought that it said that the people would wait for the first Thursday to pass before holding any festival. Then, after that, each and all festivals would be held on Tuesdays. So all you need are 4 Tuesdays to occur in that month after the first Thursday. [/hide]

[hide]I agree with what Tarquin said... since one only needs 27 days to have 4 Tuesdays in a month, the first thursday can fall on the 1st, 2nd, 3rd, 4th or the 5th. So, I think the answer is 5/7
[/hide]


What's your mathcounts goal this year?

For example, My goal is:

Top 5 in chapter, Top 6 in state, and if I advance to nats, top 150 in nats.

My goal???

School: 1st
Chapter: 1st
States: 1st or 2nd or 3rd or 4th
Nationals: Top 57 (so my name is on the list of top 57)

Those are my goals.  Not like that would ever happen to me.  :D

School: 1st

Chapter: 1st/2nd

State: top 4 (Lol, robinhe and I cant both get 1st!)

National: top 40

school- 1st

chapter- top 5

state- a 30 or higher, actually get to countdown

nationals-

basically I just want to do better than last year. And go to nationals.

Last year:
school probably 4th, possibly higher, no idea
chapter 3rd
state 29th or 21st don't know actually but suspect 29th.

first at nationals =ooo

realistic goal: countdown at nationals

Goal:
1st school
1st state (written countdown team)
1st national (written countdown team masters)

Actually I just realistically hope to make countdown.

My Goal:
Make nationals!(or at least do better than last year)

and....
[hide]not to have a nervous breakdown.jkjk :D [/hide]


A four-digit number uses each of the digits 1,2,3 and 4 exactly once. What is the probability that the number is a multiple of 4? Express your answer as a common fraction.

Note: Again, please give the most efficient way of doing this problem, any tricks, etc.

[hide]The divisibility by 4 rules is if the last two digits are divisible by 4, then the whole numbers is divisibly by four (because it all multiples of 100 are divisible of four)

there are 4*3 ways of selecting the last two digits

if the tens digit is 1 then the number that is divisibly by 4 is 12 (then if the tens digit is 2 and so on)

the numbers are

12, 24 32, 42

four numbers give 4/12 or [b]1/3[/b]

or if you think about it, another way is to know that there is a multiple of 4 in any pair of consecutive evens.  From there you can figure out (because of the 2 and 4) that there will be one number that is a multiple of 4 that starts with any of the four numbers...[/hide]

[hide]
There are 24 possible numbers (4! since order counts)
The last two digits must be 12, 24, or 32 for it to be divisible by 4.
There are two numbers that end in 12, two that end in 24, two that end in 32.
That gives you $\frac{6}{24}$ or $\boxed {\frac{1}{4}}$[/hide]

[hide]Well here is how I thought to do it at first, even if it isn't the most efficient way.

When finding if a number is divisible by 4, it only matters if the number the first (right) 2 digits are divisible by 4.

Thus we do the permutation of '4 take 2'.     $\frac{4!}{2}=$ $\frac{24}{2}=$ $\112$

We can elminate 6 to be divisble by 4, because they are odd.  Both the 2 and 4 with 3 and 1 at the end.  And then 1 and 3, and 3 and 1.

The remaining ones can be sorted into 'pairs'.  1 and 2 which is divisible by 4 and then 1 and 4 which is not.  3 and 2 which is divisible by 4, and 3 and 4 which is not.  Lastly 2 and 4 which is divisible by 4, and 4 and 2 which are not.

Since we had half of half of them that were divisible by 4, we do  $\displaystyle{\frac{1}{2}}$  $\displaystyle{\cdot}$ $\displaystyle{\frac{1}{2}=}$ $\displaystyle{\frac{1}{4}}$    :lol:[/hide]

to[i][b][u] ehehheehee[/u][/b][/i]:    42 isn't a multiple of 4!  40 and 44 are though... ;)

[hide]There are 4!=24 total four-digit numbers. If the number is divisible by 4, then it must end in
12
24
32
and for each, there are 2 ways to arrange the other two numbers. 2x3=6.
$\frac{6}{24}=\boxed{\frac{1}{4}}$[/hide]


trigonometry
LaTeX
geometry
trig identities
Law of Cosines
Law of Sines
Pythagorean Theorem

:D Call this square is ABCD.From Pytagore we have:
$\ {AB^2}+{BC^2}= {AC^2}$
$\ {3^2}+{3^2}={AC^2}$
$\sqrt{18}={AC}$

 :rotfl: Math is my life

or... you could multiply by $\sqrt{2}$ (don't ask any questions) to get $3\sqrt{2}$

Since the sides of the square are equal, we get an iscoceles right triangle or a 45-45-90. These always have sides with the proportion $x: x: x\sqrt2$. $x=3$, so the diagonal is $3\sqrt2$.
You can prove they always have that proportion with the Pythagorean Theorem.

[quote="Thoragoros"]I am not quite sure how to do this, the side of a square is 3, what then is the diagonal of the square? Need to know how to do this. :oops:[/quote]

As other posters have already said, Pythagoras is the best way:
$x=\sqrt{3^2+3^2}=\sqrt{18}$
$x=3\sqrt{2}$

[hide]I did it the same way as other people and I got

$3\sqrt{2}$[/hide]

[hide]
if the side of a square is x then the diagonal is x root 2 so the answer is $3\sqrt{2}$[/hide]

In general, if the side of a square is $s$, then the diagonal is $s\sqrt2$

[quote="Treething"]
(don't ask any questions)
[/quote]

Do ask questions... :mad: :D

Proof:
You have a right triangle with side $s$ and diagonal $d$, then we have
$s^2+s^2=d^2$
$2s^2=d^2$ (collecting like terms)
$s\sqrt2=d$ (taking square roots)

Although this proof is very easy and simple, please avoid using proofs here in this kind of sections, it'll be a bit advanced for some users....

using pythogorean, you get 3 squared + squared which is 18. take the square root of that to get 3 square root of 2

$3 \sqrt{2}$.  The diagonal is always the side length $\times$ the square root of 2.

[quote="funcia"]Since you have all these nice people willing to help you...I think you should reading The Number Devil you might feel that I am putting you down...but I think I can remember it having something on this and much, much more.  :idea: Oh, you may have read it already, or I might be thinking of the wrong book.[/quote]
I read that in 5th grade, and it really improved my math skills.

3 sqrt 3 ...if you don`t know pitagora..use the area formulas...if S = area of the square we have...l^2 = d^2 / 2
...

[quote="mathnerd314"]You don't need to download LaTeX to use it on the forum.[/quote]

If you want to put x=2 in latex put before and after it dollar signs, to have $x=2$

[quote="Padmavathi"]we have this formula in our textbook-diagonal of a square= 1 sideX root2
soo..3root 2 which gets 4.242
Sorryfor not using Latex, i haven't downloaded it yet[/quote]
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php

some basic latex

nobody has explained what pythogean is, this is what it is:

$a^{2}+b^{2}=c^{2}$ where $a$ and $b$ are sides of the triangle and $c$ is the hypotenuse
it is really helpful

for this problem, you should divide the squrare in half diagonally, getting a triangle. use pythogean and you will get the answer: $\sqrt{18}$

[quote="laughinghead505"]It has to be a right triangle true. :D[/quote]

"The side of a square is 3". This mean the triangle has an angle of $90^\circ$.

yea

i noticed my mistake

btw, does the theorem work on non-right triangles?

[quote="akalra1"]btw, does the theorem work on non-right triangles?[/quote]
No, you can only say $a^{2}+b^{2}<c^{2}$ when $\angle C < 90^\circ$ and $a^{2}+b^{2}>c^{2}$ when $\angle C > 90^\circ$.

[quote="lotrgreengrapes7926"]No, you can only say $a^{2}+b^{2}<c^{2}$ when $\angle C < 90^\circ$ and $a^{2}+b^{2}>c^{2}$ when $\angle C > 90^\circ$.[/quote]

You "can" come up with a general formula for all triangles, but it would involve trigonometry.

[quote="akalra1"]btw, does the theorem work on non-right triangles?[/quote]

Think about it for yourself; take a right triangle, and stretch the point that has the right angle out.  You've increased the length of both of the legs without increasing the length of the hypotenuse, so the Pythagorean theorem obviously fails.

[quote="mathnerd314"][quote="bpms"]If you use trig,[/quote]

The point is that you have to use mathematics beyond the Classroom Math level.[/quote]

...exactly

i am going into seventh grade, our school doesnt teach trig yet

[quote="bpms"][quote="akalra1"]never thought of that that way

is there a formula to get all the sides of a triangle?[/quote]
What do you know about the  triangle? If you use trig, then the law of cosines will be enough with two sides and an angle opposite to one of the sides. Law of sines is good enough if you know two angles and one side opposite to one of the angles.[/quote]

what are those?  can you pm me or post them on getting started or something?

[hide]pyth thrm = 4^2+4^2=4 root 2[/hide]

diagonal=side lenght $\times \sqrt{2}$  :)


modular arithmetic
algebra unsolved
algebra

Prove that a positive integer can be expressed in the form $3x^2+y^2$ iff it can also be expressed in form $u^2+uv+v^2$, where $x,y,u,v$ are all positive integers.

i think the constraint should be $x,y,u,v$ are [b]non negative[/b] integers. otherwise the following case is not accounted for

$3 = 3x^2 + y^2 \implies x = 1, y = 0, u = 1, v = 1$

We have that
$3x^2 + y^2 = (x-y)^2 + (x-y)(x+y) + (x+y)^2 = (y-x)^2 + (y-x)2x + (2x)^2$,
so if $n$ is of the form $3x^2 + y^2$, we can express it as $u^2 + uv + v^2$ since one of the pairs $(x-y,x+y)$, $(y-x,2x)$ works for $(u,v)$.

Assume then that $n$ is of the form $u^2 + uv + v^2$.
If $u \equiv v \pmod{2}$, we may choose $x = (u+v)/2$, $y = |u-v|/2$ so that $n = 3x^2 + y^2$. Else we may assume that $u$ is even (and $v$ odd). Choose $x = u/2$ and $y = v + u/2$. Thus $n = 3x^2 + y^2$.


group theory
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ \sigma\equal{}[1 2 3 \cdots n]$ in $ S_n$. Show that the centralizer of $ \sigma$ is the cyclic group generated by $ \sigma$.

I think I need to show that $ \sigma^k$ are cycles of length $ n$ for certain $ k$ (I'm guessing $ k$ relatively prime to $ n$), since disjoint cycles are commutative. What is a nice way to do this, or is there a better way to answer this problem?

The centralizer of $ \sigma$ is the set of elements $ \phi$ such that $ \phi^{\minus{}1}\sigma\phi\equal{}\sigma$.

There's a simple way to represent the action of a permutation by conjugation using the cycle decomposition: permute the elements in your listing of the cycle.
Two lists are the same cycle if and only if they're the same elements in the same order, except that we may start in a different place: $ (123)$ and $ (231)$ are the same cycle, but $ (123)$ and $ (132)$ aren't.

That means that the only elements in the centralizer act by moving the elements forward uniformly in the cycle- and those are the powers of $ \sigma$.

or direct calculation:

Let $ \tau\in C_{S_n}(\sigma)$, then working modulo $ n$ with representatives $ 1,\ldots,n$ we observe:

$ \tau(i)\equiv\tau\sigma (i\minus{}1)\equiv\sigma\tau (i\minus{}1)\equiv\tau(i\minus{}1)\plus{}1\bmod{n}$ which implies $ \tau(i)\equiv\tau(1)\plus{}i\minus{}1\bmod{n}$, i.e. $ \tau\equal{}\sigma^{\tau(1)\minus{}1}\in <\sigma >$


induction
number theory proposed
number theory

The sequence $ u_n$, $ n\equal{}0,1,2,...$ is defined by $ u_0\equal{}0, u_1\equal{}1$ and for each $ n \ge 1$, $ u_{n\plus{}1}$ is the smallest positive integer greater than $ u_n$ such that $ \{ u_0,u_1,...,u_{n\plus{}1} \}$ contains no three elements in arithmetic progression. Find $ u_{100}$.

[quote="moldovan"]The sequence $ u_n$, $ n \equal{} 0,1,2,...$ is defined by $ u_0 \equal{} 0, u_1 \equal{} 1$ and for each $ n \ge 1$, $ u_{n \plus{} 1}$ is the smallest positive integer greater than $ u_n$ such that $ \{ u_0,u_1,...,u_{n \plus{} 1} \}$ contains no three elements in arithmetic progression. Find $ u_{100}$.[/quote]

We can prove by induction the following statement:

"For $ n \ge 1$, let $ n\equal{}\overline{a_ma_{m\minus{}1}...a_2a_1}_2$, then $ u_n\equal{}\overline{a_ma_{m\minus{}1}...a_2a_1}_3$"

So $ 100\equal{}2^6\plus{}2^5\plus{}2^2 \Rightarrow u_{100}\equal{}3^6\plus{}3^5\plus{}3^2\equal{}981$

A famous related open question is this one:
http://en.wikipedia.org/wiki/Erd%C5%91s_conjecture_on_arithmetic_progressions

In the simplest case, we consider whether every set of natural numbers with a divergent reciprocal sum contains infinitely many 3-term arithmetic progressions.

To go about that simpler question, we can ask, if we were trying to construct a set with as many positive integer elements as possible below a given bound N, while still avoiding 3-term arithmetic progressions of elements, just how much better than this "greedy" sequence could we do?

I know for a fact that you can get the number of elements below N to be arbitrarily close to $ (2N)^{\ln(2) \over \ln(3)}$ for infinitely many specific values of N. The proof of that is short and pretty. Can you find it? :P


generally, would it be alright to use a calculator to find a maximum of a function on a proof problem, or do we have to prove that that is the maximum algebraically?

would you need to "demonstrate that the tool completed the step correctly"?

[quote="math154"]:P

Read [url=http://www.usamts.org/Rules/U_RulesComplete.php]Rule 9.[/url][/quote]


Hi all, I have almost completed my under-graduation; I had attempted the GRE, and I managed to secure a pretty good score, I didn't attempt TOEFL yet. Though I managed to secure 1400 in the GRE, I am contemplating of re-taking the GRE to bolster my Quants score (I feel it's pretty low) . I have some queries, they are:

[b] Query 1) [/b]

I actually planned to pursue an M.S. in Quantitative/Computational Finance, but at present, with the financial crisis snowballing, and markets plummeting, I am constantly reviewing my decisions. My baccalaureate degree is in computer science.My G.P.A. isn't great, it is somewhere between 3.0 and 3.1.

At present, Is pursuing an M.S. in Quantitative/Computational Finance feasible? Does it land me in a lucrative job at present? (I have asked a similar query before, but I didn't receive a pertinent reply).

[b] Query 2) [/b]

Since my baccalaureate degree is in computer science, and with my GPA/GRE score/Co-curricular activities I may land up in a good/above average school, but my query is:

After I finish my M.S. in computer science, can I get a lucrative/good programming/software job in the U.S.? (I mean in 2012, when I may finish my M.S.)

[b] Query 3) [/b]

While applying for an M.S. (in USA), what fields/courses ( for example-Computer Science, Mathematical/Quantitative/Computational Finance)  should I pursue in order to be guaranteed that I will land up in a lucrative job? ( By lucrative I mean , a job that will result in a comfortable life, it doesn't connote that I maybe earning millions, hundreds of thousands of dollars after I finish my M.S.)

I ask people in this forum to reply, or provide their valuable suggestions/advice to these queries of mine. ( Please provide a pertinent reply).

Thanks!

[b] P.S. [/b]:  I am in an imbroglio (at present) and I am confused on what kind of career I would like to pursue (I like Computer Science as well as finance), so I am constantly seeking replies/advice to my queries.

I surmise there are plenty of experienced, very intelligent people in here. Why doesn't anyone reply to my queries? Do they appear fatuous?

An M.S. in Comp Sci could open some additional doors especially if you were tempted to go into research.

That said, you are probably best off expanding your career options and giving yourself a nice vertical niche  by choosing to master in Finance.

Financial organizations tend to be on the bleeding edge of systems development. It is a nice synergy.

Okay, Are you suggesting that I pursue an M.S. in Mathematical/Quantitative Finance? But with the present financial crisis, I think there doesn't exist any "quant" jobs.

[b] @ Everyone [/b] :Can anyone give more suggestions?

See here: [url]http://terrytao.wordpress.com/career-advice/[/url]
Is general but helps... :oops:

My hunch would be (and others may want to verify this) that background in computer science is more applicable to finance than background in finance is to computer science. So, unless you are [i]particularly[/i] interested in going into finance, it may be better to stick with computer science. And you don't seem that particularly interested in either path, since your goal seems to be to get a "lucrative" job.


function
real analysis
real analysis solved

$g:R\rightarrow R$ continous $E=\{(x,y)\in R^2; g(y)=g(x)+x\}$
Prove that $E$ is not bounded in   $R^2$

Take $h:\mathbb R^2\to R,\ h((x,y))=g(x)-g(y)+x$. This is a continuous map, and assume it's never $0$ for $|x|+|y|\ge k$ (i.e. assume $E$ is bounded). Take $x$ large enough and $y$ close enough to $x$ in order to have $|x|+|y|\ge k$ and $|g(x)-g(y)|<|x|=x$. This means that $h$ takes only positive values on $S=\{(x,y)\in\mathbb R^2||x|+|y|\ge k\}$ because this set is connected and $h$ is continuous, so the intermediate value theorem holds. In the same manner we can show that $h$ takes only negative values on $S$, and we have a contradiction.


To minimize the product, we place the smallest numbers in the tens digit, like so:

6 __

7 __

Now we pair up the smaller with the larger, and the two middle numbers, so that we have 69 and 78. Therefore, the product is 5382.

Strangely the answer appears to be 5372 for some reason. I believe this might be a problem.

I used a calculator and did $68*79$ and got $\boxed{5372}$.
Is there a solid approach to this problem?
I just picked a random set.

Um yes, that is correct.

As isabella2296 said, the two integers should be $6$_ and $7$_.  Now, all we do is test $68 \cdot 79$ and $69 \cdot 78$.  We see that $68 \cdot 79 = \boxed{5372} < 69 \cdot 78 = 5382$.

I guess you want to maximize the difference of the two numbers after finding what the tens digits are.


Prove that for every $n$ there is natural $a$  such that $a^2+a+1$  has $n$ prime divider.

If you meant 'at least' $n$ prime divisors, the desired result is a consequence of
http://www.mathlinks.ro/Forum/viewtopic.php?t=49211

Let $p_1, \cdots, p_n$ be $n$ pairwise distinct primes of the form $3k+1$ (which is possible from Dirichlet's theorem about primes in arithmetical progressions).
Now, for each $i$, let $a_i$ be such that $a_i^2 + a_i + 1 = 0 \mod [p_i]$ (see the link)
Now, use the chinese remainder theorem to find $x$ such that $x = a_i \mod [p_i]$ for each $i$. Thus $x^2+x+1 = a_i^2+a_i+1 = 0 \mod [p_i]$ for each $i$ which ensures that $x^2+x+1$ has at least $n$ distinct prime divisors.

Pierre.

Well, in the "at least" version, it also trivially follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=20203 for $f\left(n\right)=n^2+n+1$. The "exactly n prime divisors" variant is very plausible, but I guess it will be very hard to prove or disprove it...

  darij


number theory
least common multiple
number theory proposed

Find all a for the sequence $ \{n^{n}\}_{n\geq 1 }$ is periodic mod a from the first term .More than find the smallest T such that $ (n\plus{}T)^{n\plus{}T}\equiv n^n (\mod a ),\forall n\in N$
Note : problem is true .
$ \{n^n\} (\mod a )$ is periodic from $ n\geq n_0$

If $ a\equal{}a_1*a_2*...*a_k$ , were $ (a_i,a_j)\equal{}1$, then the ring $ Z/aZ\equal{}\sum_i (Z/a_iZ)$.
Therefore sequense $ f(n)$ is periodic mod a, if and only if $ f(n)$ periodic $ \mod a_i \ \forall i$ and period $ T_a\equal{}lcm(T_{a_i})$.
Therefore sufficiently consider $ a\equal{}p^n$, were p is prime. Obviosly in these case $ p|T$ and $ T_{p^k}|T_{p^n}$ if k<n.
1. $ a\equal{}p$ we get $ (n\plus{}kp)^{n\plus{}kp}\mod p \equal{}n^{n}n^{kp}\mod p$, therefore $ p\minus{}1|k$. Obviosly $ T\equal{}p(p\minus{}1)$ is period mod p.
2. $ a\equal{}p^n,n>1$ we get $ (n\plus{}kp)^{n\plus{}kp}\mod p^n\equal{}n^{n\plus{}kp}\plus{}n^{n\minus{}1\plus{}kp}(n\plus{}kp)p\plus{}C_{n\plus{}kp}^2n^{n\minus{}2\plus{}kp}p^2\plus{}...$
It give, that $ T_{p^n}\equal{}p^{n\minus{}1}(p\minus{}1)$ or $ T_{p^n}\equal{}lcm(p,\phi (p^n)), n\ge 1$.
In general case \[ T_a\equal{}lcm_{p|a}((p\minus{}1)p^{min(1, \ ord_p(a)\minus{}1)}).\]

You just prove the sequence is period but for which a it period from the first term .
 :wink:

I calculate $ T_a$ for all a, and prove, that $ (n\plus{}T_a)^{n\plus{}T_a}\equal{}n^n\mod a \ \forall n\ge 1.$

But for example a=16 then the sequence mod 16 is not periodic from the first term  :wink:

I am sorry, I forget $ p \equal{} 2$. In these case for $ a \equal{} 2^n$ we get $ T_a \equal{} 2, n \equal{} 1, \ T_a \equal{} 4, n \equal{} 2,3, T_a \equal{} \ T_a \equal{} 2^{n \minus{} 2},n > 3$.
First member from with sequence is periodic is $ 2k \plus{} 1$, suth that $ 2^n\not |(2k)^{2k}, \ 2^n| (2m)^{2m} \ \forall m>k$.
For $ a\equal{}p^n$ p is odd prime first member from with sequence is periodic is $ pk\plus{}1$, suth that $ p^n\not |(pk)^{pk}, \ p^n|(pm)^{pm} \ \forall m>k$.
Therefore sequense is periodic from first member if and only if $ ord_2(a)\le 3$ and $ ord_p(a)\le p$ for odd prime divisors.

When I did this problem I got that the solution was all $ a$ such that $ a$ didn't have any "bad factors" of the form $ b^c$, $ 2\leq b < c$. I don't know if that's the same thing you're saying, but in a different way. And then I calculated the period as $ lcm(a,\phi(a))$. I could like, try to explain my logic and stuff but I want to know if that's the same answer as you're all writing cause i can't tell.
(slightly edited)

All a must be a number satisfy :
$ v_{p}(a)\leq p,\forall p\in P$


Don't give me brute force solutions... :D

1. a, b, c, d are positive reals s.t. ab+bc+cd+da+ac+bd=6
Prove a^2+b^2+c^2+d^2+2abcd>=6

2. a, b, c are positive reals
Prove 9(a^2+bc)(b^2+ca)(c^2+ab)<=8(a^3+b^3+c^3)^2

I have a stupid question.. everybody talks about brute force solutions!!
I don't understand what they are! Can someone explain it to me?  :?

bye bye

I've a nice soln :D
And you? Try to solv :D

I know you asid no 'brute force', but for the second one it works just fine, the calculations are easy. You get to 18*a^2*b^2*c^2+9abc(a^3+b^3+c^3)<= 8(a^6+b^6+c^6)+ 7(a^3*b^3+b^3*c^3+c^3*a^3). We have 6(a^3+b^3+c^3)>= 18a^2*b^2*c^2; 2(a^3b^3+a^3c^3+b^3c^3)+a^6+b^6+c^6>= 3abc(a^3+b^3+c^3); a^6+b^6+c^6>=abc(a^3+b^3+c^3), all by AM-GM, and we just combine them. See ya!:D

I remember my teammate on the Singapore team gave a totally brute-force solution for the 1 Yen Qn. 1 :)

That was embarrassing.  I couldn't really bear to look at the completed work.

i think i have a proof of the 1st inequality using (guess wot?) muirhead,though i'm not sure if it's correct .i'll post it later.but i doubt if it is not a brute force solution :D

I also prooved 1. But at the moment a have no time to write solution.
My proof do not use Murhead. Only your favorite Shura's inequality.

For 1.
First step: heterogeneous inequality --> homogeneous inequality
LHS
a2+b2+c2+d2+2abcd=(ab+bc+ca+ad+bd+cd)(a2+b2+c2+d2)/6+2abcd
RHS
6=(ab+bc+ca+ad+bd+cd)2/6

Thus new inequlity is
(ab+bc+ca+ad+bd+cd)(a2+b2+c2+d2)+12abcd \geq (ab+bc+ca+ad+bd+cd)2
or
\sum a3b(12 items) + \sum (abc2)(12 items) + 12abcd \geq \sum a2b2(6 items) + 2\sum (abc2)(12 items) + 6abcd
or
\sum a3b(12 items) + 6abcd \geq \sum a2b2(6 items) + \sum (abc2)(12 items).

Now we apply small old trick
\sum a3b(12 items)=1/2\sum a3b(12 items)+1/2\sum a3b(12 items) \geq 1/2\sum a3 b(12 items)+\sum a2b2(6 items)
(since a3b + b3a \geq 2a2b2).

So we must prove
(1) \sum a3b(12 items) + 12abcd \geq 2\sum (abc2)(12 items)

Finally (1) is a sum
(Shura's inequality for a,b,c)*d + (Shura's inequality for a,b,d)*c + ...


modular arithmetic
trigonometry
LaTeX
calculus
derivative
number theory
prime factorization

[22:41] Samson: i think you have to make a thread in AoPS though
[22:41] Mith: you can make thread this time
[22:42] Samson: negative

Alrighty folks, Samson's too shy to make a thread...

[21:57] Samson: Teh Pwning Susan: but daniel said u were shy!
[21:57] Samson: ...
[21:57] Mith: HAHAHAHAHAHA
[21:59] Samson: i am not shy.

So I did. Round 1 solutions and discussion here? :D

(23:42:14) Nathan: 7 was Dimitar
(23:42:31) Nathan: 1, 5, 8, and 9 were Joel
(23:42:33) Nathan: the rest were Cody
(23:42:51) Nathan: well Joel had the idea for 8 and Cody made the actual question
(23:43:00) Samson: let's see...
(23:43:04) Samson: my favorite question this round
(23:43:07) Samson: has to be...
(23:43:33) Samson: actually
(23:43:38) Samson: can you make a thread on AoPS

Too shy?  Making topics like these is a great way to get your post count up.  :D

[hide="Number 6"]
Notice that if n is not equivalent to 0 mod 9 then $a_n=0$.  So all we have to do is consider that cases where $n \equiv 0\mod 9$.

So notice that $a_9=8$, $a_{19}=8$, etc. $a_{89}=8$.  We can prove that if $n\equiv 9\pmod 9$, but is not congruent to 99 mod 100 then $a_n=8$.

Notice $a_{99}=17$, $a_{199}=17$, etc.  We can similaly prove that if $n\equiv 99\pmod {100}$ but is not congruent to 999 mod 1000 then $a_n=17$.

Notice $a_{999}=26$, $a_{1999}=26$, etc.  Like the above cases, this can be proved that if $n\equiv 999\pmod {1000}$ then $a_n=26$.

Since $8+9=17$, and $17+9=26$, whenever another $9$ digit is "added" to the end then $a_n$ is increased by 9.

So there are 200 total numbers ending in 9, so $200\cdot 8=1600$.
There are 20 more of these numbers ending in 99, so $20\cdot 9=180$.
There are 2 more of these numbers ending in 999, so $2\cdot 9=18$.

$1600+180+18=\boxed{798}$
[/hide]

[hide="Number 7"]
By the half angle formula $\sin^2 \frac{x}{2}=\frac{1-\cos x}{2}$, so the value we are actually looking for is $\frac{1-\cos A}{2}+\frac{1-\cos B}{2}+\frac{1-\cos C}{2}$.  (I'm too lazy to look up how to do alpha, beta, and gamma with latex.)

Using the law of cosines we find:
$400=361+441-2(19)(21)\cos A$, which gives $\cos A=\frac{67}{133}$.
$361=441+400-2(21)(20)\cos C$, which gives $\cos C=\frac{4}{7}$.
$441=361+400-2(19)(20)\cos B$, which gives $\cos B=\frac{8}{19}$.

So plugging in these numbers we have $\frac{1-\frac{67}{133}}{2}+\frac{1-\frac{4}{7}}{2}+\frac{1-\frac{8}{19}}{2}$.  Simplifying gives $\frac{100}{133}$, and $100+133=\boxed{233}$.[/hide]

[hide="Number 8"]
Notice that $104!=103!\cdot 104$, and $104=2^{3}\times 13$.

So if the prime factorization of $103!$ is $2^{a}13^{b}p_3^cp_4^d...$, then the prime factorization of $104!$ is $2^{a+3}13^{b+1}p_3^cp_4^d...$.

So $103!$ has $(a+1)(b+1)(c+1)(d+1)...$ factors, and $104!$ has $(a+4)(b+2)(c+1)(d+1)...$ factors.

Now we must figure out the values of $a$ and $b$.  So $a=51+25+12+6+3+1=98$, and $b=7$.

Therefore our answer is $\frac{(99)(8)(c+1)(d+1)...}{(102)(9)(c+1)(d+1)...}=\frac{(99)(8)}{(102)(9)}$.  Simplifying gives $\frac{44}{51}$, so the answer is $44+51=\boxed{095}$.[/hide]

[quote="joml88"]Too lazy to look up alpha beta and gamma :?:  Why, they are just \alpha, \beta, and \gamma :P

$\alpha \ \beta \ \gamma$[/quote]Wow I feel really stupid.  :blush:

Edit: oh and someone needs to put a solution for number 9.  I didn't get how to do it.

[quote="eryaman"]Edit: oh and someone needs to put a solution for number 9.  I didn't get how to do it.[/quote]

Nathan said he'd do it tomorrow. I would but I've got a sneaky feeling I'll something newb up the division again.

im not sure how to use latex on here, so here's what i did using plain text:

take derivative of f(x), you get f'=4x^3-c
so the critical point is (c/4)^(1/3)

put it on a number line, you find that this critical point is a minimum

so do f((c/4)^(1/3))>0, solve for c.

you should get 60*(15)^(1/2)

the final answer would be 60+15=75

[quote="chesswonk"]
so do f((c/4)^(1/3))>0, solve for c.
[/quote]This is where I got stuck at when doing the problem, the algebra got [i]way[/i] too messy.

[quote="Xevarion"]You can also just solve for $c$ and then take the derivative. Also, to use LaTeX, just put the code between dollar signs like you usually would. Cody had some awesome way to do this without calculus; I'll post it if I can remember it.[/quote]

yeah, I did it the same way with calculus and I was wondering if there was a more elementary way to solve it.

[quote="Xevarion"]All right, here it is: Cody's pwnage non-calculus solution.

We want to minimize $x^3 + \frac{675}{x}$. Let's write it as $x^3 + \frac{225}{x} + \frac{225}{x} + \frac{225}{x}$. That expression is four times the arithmetic mean of its four terms, so it must be greater than or equal to four times the geometric mean of the four terms. Thus, the expression is $\ge 4\cdot\sqrt{\sqrt{225^3}} = 4\cdot15\sqrt{15}$. The LHS is minimized when we have equality, so $c = 60\sqrt{15}$.

(sorry, I was too lazy to look up how to do a fourth root with LaTeX)

This is why Cody is more awesome than the rest of us.[/quote]

Ugh, that's how I did it.  Apparently this was my week for making stupid mistakes though.

Btw, \sqrt[4]{x} gives $\sqrt[4]{x}.$

isnt AM-GM only holds for positive real number ? $x$ isnt always positive real though ... plus the question also want $x$ to be in imaginary  :ninja:

AM-GM works for negative real $x$ provided that there are an even number of terms and we take the absolute value of both sides. Also, we're looking for the border between where $x$ would be real and where complex, so that aspect of it is okay (i.e., we're approaching from the real side).

I must say, that's pretty clever.   :lol:


In acute triangle $ABC$,the point$D$is the foot of height$BD$.There is a point$E$on side$AB$ satisfying$\angle AEC=45^{o}$and$BD=2CE$.

Prove that the two statements below $DE//BC$and$CE=AC+AD$ are equavilent.