vector
analytic geometry
graphing lines
slope

In Section 12.3 of Aops vol. 2, there's a section about vectors and the dot product.

It states that the dot product of $a$ and $b=(AO)(OB)\frac{OX}{AO}=(OB)(OX)$ (diagram included).

Then it says that
[quote]Thus the dot product of $a$ and $b$ is the lengh of $b$ times the length of the projection of $OA$ onto $OB$ (segment $OX$ above).[/quote]

I don't understand what that means.  Clearly $OA$ is longer than $OX$...

[quote="t0rajir0u"][quote="brady00n"]I don't understand what that means.  Clearly $OA$ is longer than $OX$...[/quote]

$OX$ is the projection of $OA$ onto $OB$ (imagine a light shining perpendicular to $OB$; then $OX$ is the shadow of $OA$).

Then the dot product of $a$ and $b$ is the length of $b$ times the length of $OX$.[/quote]

Oh, ok I was mistaken in definition of projection.

Thanks!  I have some more questions

1. How is property 3 or dot products relevant?

[quote="t0rajir0u"]What do you mean?  What is property 3?[/quote]

He means how is dot product ("or property 3") relevant.

As in, how is it useful?  Well, here's one way.

Given a point $P$ on a line, and given its normal vector $\vec{v}$ (the vector perpendicular to the slope vector), we can think of the line as being the set of all points $X$ such that $\vec{PX}$ is perpendicular to $\vec{v}$.  In other words, their dot product is zero.

So let $\vec{v}= \left< a, b \right>$.  Let our point $P = (x_{0}, y_{0})$.  Then our general line equation is given by

$\vec{v}\cdot \left( (x, y)-P \right) = 0$

Or, in other words,

$\left< a, b \right> \cdot \left< x-x_{0}, y-y_{0}\right> = 0$

Or, even better,

$a(x-x_{0})+b(y-y_{0}) = 0$

Now, this might not look so impressive, but we can extend this to finding equations for [b]planes[/b].  (In three dimensions, a plane can also be defined as the set of points with vectors from a given point perpendicular to a given normal vector.)  In three dimensions, an arbitrary plane is given by

$a(x-x_{0})+b(y-y_{0})+c(z-z_{0}) = 0$

There are [b]many other uses[/b] of the dot product, but this is a nice one, I think.

Cool stuff!

One more follow-up...

In Example 12-7, doesn't following the procedure of the example 12-6 by saying two points are on the line wrong?  In 12-7, the two points are clearly not on the identical line.  If not, how can you show that the normal is $(A B)$?

You'll need to supply a few more details.  I'm not sure exactly what you mean.


I can throw a football x yards.  Dan Marino can throw a football 2x+5 yards.  Our combined total yards is 110 yards.  How far can Dan Marino throw a football?

I believe you misread the question.  It asks how far can Dan Marino throw a football?  Not me


Here's a problem that I don't know where to start...

Can the set of positive integers be partitioned into Fibonacci sets? This means finding infinitely many pair-wise disjoint Fibonacci sets whose union contains every positive integer.

Yes they can. That was ARML #10 on the power round 2003.

Hmm right, but what was the solution? I lost that packet... (pretty pathetic I know, but still...)

Here's a way to get started. If you want to cover all of the positive integers, how do you get the numbers 1 and 2 in there. After that, start trying to make sets that cover the smallest integers not already covered and look for a pattern in the first *two* memebers of each of those sets. Once you think you found it, try proving it by induction.

This was problem number 10 on this years ARML Power Question.


function
number theory unsolved
number theory

find all functions $f:N^{*}\rightarrow N^{*}$ satisfy two condition:
                 a)$f(1)=1$
                  b)$f(f(n))*f(n+2)+1=f(f(n+1))*f(n+1)$   for all $n \in N^{*}$

The lack of a closing bracket on the RHS of (b) makes it difficult to work out what you mean. Also, what does $*$ stand for?

it means $product$ .And i had the solution with this problem last night.i used $IMO 1977$ and I have the solution


conics
LaTeX
projective geometry
geometry unsolved
geometry

Let P and P' be two points in the plane of a triangle ABC, and let A'B'C' and A"B"C" be the cevian triangles of P and P', respectively. Denote

X = B'C' /\ B"C"; Y = C'A' /\ C"A"; Z = A'B' /\ A"B";
X' = B'C" /\ B"C'; Y' = C'A" /\ C"A'; Z' = A'B" /\ A"B'.

Hereby, the sign /\ means "point of intersection" (of two lines).

Prove that

[b](1)[/b] The lines A'X, B'Y, C'Z meet at a point Q;
[b](2)[/b] The lines A"X, B"Y, C"Z meet at a point Q';
[b](3)[/b] The points X', Y', Z' lie on the line PP';
[b](4)[/b] The lines AX, BY, CZ concur;
[b](5)[/b] The points A, X', Y, Z are collinear;
[b](6)[/b] If X" = YZ /\ BC; Y" = ZX /\ CA; Z" = XY /\ AB, then the points X", Y", Z" lie on the line QQ'.
[In other words, the axis of perspective of the triangles ABC and XYZ is the line QQ'.]

[b]The crux here is to prove (5).[/b] All the rest is easy once [b](5)[/b] is proven.

Actually, [b](5)[/b] is very easy using Pascal's theorem for conics. [b]My question is to establish (5) elementarily[/b] (this means: without conics or projective / affine transformations).

Any ideas?

Many thanks (already for having read this all)!

  Darij

[quote="KangHyungJune"]Well, you can prove (5) by using only Menelaus.[/quote]

Could you please show me this proof?

  Darij

sorry, i am late :)
 it's school exam season ,so i couldn't post this.

actually, you need to use pascal's theory but it's simple. :)
well, i'll show that Y,A,Z  are collinear.

Let Y_0,Z_0 be the intersection of CX and A''C'', BX and A''B'', respectively.

we want to show Y_0=Y, Z_0=Z
then since C'', X, B" are collinear, you can apply Pascal's theorem  .
you get A, Y, Z are collinear (line YAZ and BA"C).

so all we have to do is to show Y_0=Y. (you can show Z_0=Z in same way.)

it's sufficient to show Y_0, C',A' are concurrent.
let D,E be the intersection of B'C' and BC, B"C" and BC, respectively.
(i'll assume two lines are not parellel. if parallel, it becomes more simple)
from triangle BA"C" and Y_0, C', A' , using the inverse version of Menelaus,

it's sufficient to show BC'/C'C" * C"Y_0/Y_0A" * A"A'/A'B = -1    .....(*)
 (sorry, i don't know how to use latex :(  )
from triangle EC"A" and line Y_0XC, using Menelaus,
 A"Y_0/Y_0C" * C"X/XE * EC/CA" = -1  (1)
from triangle C"BE  and line DC'X, using Menelaus,
C"C'/C'B * BD/DE * EX/XC" = -1 (2)

then by (1) and (2)
(*) becomes EC/CA" * BD/DE  * A"A'/A'B = -1    (**)
you can show this by simple caculations :)

i.e, let AC'/C'B = x/y, BA'/A'C=y/z, CB'/B'A=z/x,
AC"/C"B=a/b, BA"/A"C =b/c, CB"/B"A=c/a

then by some menelaus,
 EC = c/c-b *BC    (assume the length has directions)
BD = y/y-z *BC

CA" = -c/b+c *BC
A'B = -y/y+z * BC

A"A' = (cy-bz)/(y+z)(b+c) *BC
A'B = (-cy+bz)/(y-z)(c-b) *BC

by multiplying all these things, you get (**)

it was a long proof :)

Of course, I didn't mean killing it by calculations...

Anyway, the problem is already solved: Part [b](5)[/b] follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24392]the Pascal Permutation Theorem[/url].

  darij

actually, not pascal, it's pappus in my proof.
my mistake.

Dear Darij Grinberg:
Except (3), all the rest have been proven elememtarily, could you show me how to prove (3)?
thank you a lot!

Actually, [b](3)[/b] is the easiest of all: The points C, B', B" are collinear, and the points B, C', C" are collinear. Thus, by the Pappos-Pascal theorem, the points B'C" /\ C'B", B"B /\ C"C and CC' /\ BB' are collinear. But B'C" /\ C'B" = X', B"B /\ C"C = P' and CC' /\ BB' = P. Hence, X' lies on PP'. Similarly, Y and Z lie on PP'.

[Note to self: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=168693 when writing up for book.]

  Darij

Dear all,
a proof of this first five propositions is presented on my web site

http://perso.orange.fr/jl.ayme/

Look: contenu, volume 2 (2008) "Les deux points de Schroeter"
Jean-Louis


function
integration
calculus
logarithms
derivative
real analysis
real analysis solved

Does there exists continious non-constant function $f:[0,1]\rightarrow\mathbb{R} $
such that $f(x)=\frac{1}{x}\int_{0}^{x}f(t)dt$ for all x from (0,1]?

No,it doesn't.

We will argue about the following problem.

Let $k (\neq0)$ be a constant number.Find the continuous function $f(x)$ at $ [0,1]$ which satisfies integral equation as follows.

\[\frac{1}{k}f(x)=\frac{1}{x}\int^x_0 f(t)dt\]

The Given equation rewrites $\int^x_0 f(t)dt=\frac{1}{k}xf(x)$. Since LHS is differntiable,so does RHS.
Thus in using fundamental theorem of calculus ,differentiating both sides of the equation,we have $f(x)=\frac{1}{k}\left[f(x)+xf'(x)\right]$,
as $(0,1]$, rewriting $\frac{f'(x)}{f(x)}=\frac{k-1}{x}$ and integrating both sides of the equation,we obtain $\ln f(x)=(k-1)\ln x+Const.$

Putting $x=1$ into this equation,yielding $Const.$ is $\ln f(1)$ .Thus we obtain $f(x)=f(1)x^{k-1}\ (0,1]$.

Now for the problem which posted by [b]eugene[/b],we put $k=1$ into this result,

yielding $f(x)=f(1)$ i.e.$ f(x)$ is constant function.Consequently the function which satisfies the condition in the problem doesn't exist.

Thank you.

kunny

This could be made a little simpler. By the Fundamental Theorem of Calculus, $f(x)$ is differentiable for $x>0.$ Take eugene's original equation, first multiply by $x$ and then differentiate. The result is:

$xf'(x)+f(x)=f(x)$ which implies that $f'(x)=0$ for $x>0.$

This shows that $f$ is constant. Furthermore, an arbitrary constant solves this equation.

ok, thanks kunny,Kent...my solution is similar to Kent-it is evident that there is one way to figure out with this problem-differentiation


linear algebra
matrix
linear algebra unsolved

Let $A_1,A_2,....,A_m$ be matrix such that $A_iA_j=A_jA_i=0$ with all i,j .
Prove that $rank(A_1)+....+rank(A_n)\leq n$

That statement can't be right. What if the $A_i$'s are all just constant multiples of the identity matrix.

What is the correct statement of the problem?

The condition $A_iA_j=A_jA_i=0$ simply says that if $V$ i the $n$-dimensional subspace on which $A_i$ act as endomorphisms, then the images of $A_i$ are mutually orthogonal. This means that the sum of spaces $A_i(V)$ is a direct sum, so the dimension of this sum is equal to the sum of the dimensions. The sum is, of course, a subspace of $V$, so it has dimension $\le n$, so we're done.

Could you explain more clearly? I really don't understand.

$\forall j\neq 1$ $ImA_j \subset KerA_1$
so $\oplus_{j \neq 1}ImA_j \subset KerA_1$
so $\sum_{j\neq 1} rank(A_j) \leq dim(KerA_1)=n-rank(ImA_1)$
so $\sum_i rank(ImA_i) \leq n$


A class has 5 students.

It wants to create two teams using all the students, with each team having atleast one person.  How many different pairs of teams can be created?

yea thats right..

I thought someone was going to do 5C1+5C2+5C3+5C4...

[hide]
Case 1
One Team has 1 person and the other has 4, there is only $ 5$ possible arrangments.

Case 2
One Team has 2 people which is $ \binom{5}{2}=10$ arrangments.

Teams of $ 3$ and $ 4$ have been counted already.
Then $ \boxed{5+10 = 15}$ possible arrangments.
[/hide]


Given $ a, b, c \geq\ 0$ satisfy $ a\plus{}b\plus{}c\equal{}3.$
Prove that: $ (\frac{1}{2a^2\plus{}1})^{\frac{2}{3}} \plus{} (\frac{1}{2b^2\plus{}1})^{\frac{2}{3}} \plus{} (\frac{1}{2c^2\plus{}1})^{\frac{2}{3}} \geq\ \sqrt[3]{3}$

@can_hang: Bai` nay`em dinh. tang. anh nhung anh lai lam` mat' bai` cua? em tang Loc roi`.
The nen bai` nay` em tang. Loc tiep', lan` nay` anh ko dc lam` dau nhe'. Em se~ tang anh bai khac  :)

PS: I wait for a nice solution, not use Jensen ineq (think about Am-Gm).
And the stronger still trues:
$ \frac{1}{\sqrt[]{2a^2\plus{}1}} \plus{} \frac{1}{\sqrt[]{2b^2\plus{}1}} \plus{} \frac{1}{\sqrt[]{2c^2\plus{}1}} \geq\ \sqrt[]{3}$
Sorry for spam  :oops:

[quote="nguoivn"]PS: I wait for a nice solution, not use Jensen ineq (think about Am-Gm).
And the stronger still trues:
$ \frac {1}{\sqrt []{2a^2 \plus{} 1}} \plus{} \frac {1}{\sqrt []{2b^2 \plus{} 1}} \plus{} \frac {1}{\sqrt []{2c^2 \plus{} 1}} \geq\ \sqrt []{3}$
Sorry for spam  :oops:[/quote]
Note that with $ x \in \left[ {0,3} \right]$ we have
\[ \frac{1}{{\sqrt {2x^2  \plus{} 1} }} \ge \frac{5}{{3\sqrt 3 }} \minus{} \frac{{2x}}{{3\sqrt 3 }}
\]
Thus,
\[ \sum {\frac{1}{{\sqrt {2a^2  \plus{} 1} }} \ge \sum {\left( {\frac{5}{{3\sqrt 3 }} \minus{} \frac{{2a}}{{3\sqrt 3 }}} \right)} }  \equal{} \sqrt 3
\]

[quote="Honey_S"]
\[ \frac {1}{{\sqrt {2x^2 \plus{} 1} }} \ge \frac {5}{{3\sqrt 3 }} \minus{} \frac {{2x}}{{3\sqrt 3 }}
\]
[/quote]

How did you get it? I mean the idea, not the computations :wink;

Yes, nice proof, Honey_S.
But I think your method can't solve the following problem:
Given $ a, b, c \geq\ 0$ satisfy $ a\plus{}b\plus{}c\equal{}3.$
Prove that:
$ (\frac{1}{a^2\plus{}1})^{\frac{5}{6}} \plus{} (\frac{1}{b^2\plus{}1})^{\frac{5}{6}} \plus{} (\frac{1}{c^2\plus{}1})^{\frac{5}{6}} \geq\ \frac{3}{\sqrt[6]{32}}$

Because the following stronger isn't trues:
$ \frac{1}{\sqrt[]{a^2\plus{}1}} \plus{} \frac{1}{\sqrt[]{b^2\plus{}1}} \plus{} \frac{1}{\sqrt[]{c^2\plus{}1}} \geq\ \frac{3}{\sqrt[]{2}}$

PS: Both of them can be solved by Am-Gm, my friend  :)

[quote="nguoivn"]Yes, nice proof, Honey_S.
But I think your method can't solve the following problem:
Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3.$
Prove that:
$ (\frac {1}{a^2 \plus{} 1})^{\frac {5}{6}} \plus{} (\frac {1}{b^2 \plus{} 1})^{\frac {5}{6}} \plus{} (\frac {1}{c^2 \plus{} 1})^{\frac {5}{6}} \geq\ \frac {3}{\sqrt [6]{32}}$

Because the following stronger isn't trues:
$ \frac {1}{\sqrt []{a^2 \plus{} 1}} \plus{} \frac {1}{\sqrt []{b^2 \plus{} 1}} \plus{} \frac {1}{\sqrt []{c^2 \plus{} 1}} \geq\ \frac {3}{\sqrt []{2}}$

PS: Both of them can be solved by Am-Gm, my friend  :)[/quote]
Actually, this inequality is weaker than this inequality:
\[ \frac{1}{5a^2\plus{}7}\plus{}\frac{1}{5b^2\plus{}7} \plus{}\frac{1}{5c^2\plus{}7} \ge \frac{1}{4}.\]
;) :)


integration
limit
function
real analysis
real analysis unsolved

Let $ f_n(x) \equal{} \frac{n}{2}1_{[\minus{}1/n,1/n]}$.  Prove that for $ g \in L(\mathbb{R})$,

$ \int \mid {\int f_n(y\minus{}x)g(x) dx \minus{} g(y)} \mid dy \to 0$ as $ n \to \infty$

First hint:

$ \int f_n(y\minus{}x)g(x)\,dx\equal{}\int f_n(x)g(y\minus{}x)\,dx$

$ \int f_n(x)\,dx\equal{}1$ so $ g(y)\equal{}\int f_n(x)g(y)\,dy$

$ \int f_n(y\minus{}x)g(x)\,dx\minus{}g(y)\equal{}\int f_n(x)(g(y\minus{}x)\minus{}g(y))\,dx$

Second hint. Let $ \gamma_g(x)\equal{}\int|g(y\minus{}x)\minus{}g(y)|\,dy$ (Note: $ dy$ is precisely what I mean.)

Prove that

(a) $ \gamma_g(x)\le 2\|g\|_1$ for all $ y.$

(b) $ \lim_{x\to 0}\gamma_g(x)\equal{}0.$

Hint for the proof of (b): first show it is true for some special category of "nice" functions $ g,$ then use that fact that those "nice" functions are dense in $ L^1.$ There are several possibilities for "nice" that all work.


(1/2)sin(2x)[tan^2(x)–cos(2x)+2cos^2(x)]

Please help simplify this as much as possible.

[hide="Hint."]Identities used:
$ \sin(2x)\equal{}2\sin(x)\cos(x)$
$ \cos(2x)\equal{}\cos^{2}(x)\minus{}\sin^{2}(x)$
$ \sin^{2}(x)\plus{}\cos^{2}(x)\equal{}1$
$ \tan^{2}(x)\plus{}1\equal{}\sec^{2}(x)$[/hide]

[hide="Solution."]Well...$ \left(\frac{1}{2}\right)\sin(2x)[\tan^{2}(x)\minus{}\cos(2x)\plus{}2\cos^{2}(x)]\Leftrightarrow$
$ \sin(x)\cos(x)[\tan^{2}(x)\minus{}\cos^{2}(x)\plus{}\sin^{2}(x)\plus{}2\cos^{2}(x)]\Leftrightarrow$

$ \sin(x)\cos(x)[\tan^{2}(x)\plus{}1]\Leftrightarrow$

$ \sin(x)\cos(x)\sec^{2}(x)\Leftrightarrow$

$ \tan(x)$[/hide]

[quote="ysk1"]$ \frac{1}{2}\sin2x[\tan^{2}x$– $ \cos2x\plus{}2\cos^{2}x]$

Please help simplify this as much as possible.[/quote]

use latex looks nicer
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php

The problem with that link is it fails to tell you the most crucial part of usign LATEX, the dollar signs.

To use latex, just put your math stuff in a pair of dollar signs and you're good.


How i can show that $ X\equal{} \left\{(x,y) \in \mathbb{R}^2 : y>\frac{1}{x}>0\right\}$ is open in $ \mathbb{R}^2$?
I've seen a proof for the case $ X'\equal{}\{(x,y) \in \mathbb{R}^2 : y>0\}$ where for $ P\equal{}(a,b)\in \mathbb{R}^2$ just take the the open ball centered at $ a$ with radius $ b$. But for the first one i arrived at a messy expression for the radius... i used a triangle involving $ a,b$ and i'm trying to find a clever way to do it.

Consider the complement, and show that it is closed. Do you know how to do this?

Let $ (a,b)\in X$, set $ \{d_2((a,b),(x,\frac{1}{x})): x>0\}$ has minimal element. Why?

CaptainCanada: Ok, then i need to show that every limit point of $ X$ is in $ X$. but how to do this?

Ognjen09: I've thinked at that. why i can guarantee that always i have a minimum distance... i know it exists, but no clue on what argument i need to use to prove that. I've tried to imagine when this fails... wondering continuity


inequalities trigonometry algebra function domain

Here's my problem:

$ -1 \leq \sin(x) \leq 1$

$ \frac{-1}{\sin(x)}\leq \frac{\sin(x)}{\sin(x)}\leq \frac{1}{\sin(x)}$

$ -\csc(x) \leq 1 \leq \csc(x)$

$ \underline{\csc(x) \geq-1}\; \bigcup \; \csc(x) \geq 1$

the underlined part is false, as the inequality sign should be facing the other way. What am I doing wrong?

These are corollarys of definitions of $ \sin x, \csc x$.

I'm not sure what you meant by that. I know the answer is supposed to be:

$ \csc(x) \in \; ]-\infty,-1[ \;\;\bigcup \;\; ]1,\infty[$ which is equivalent to saying $ -1 \geq \csc(x) \geq 1$ but for some reason I can't prove it using the method I showed in my first post, which usually works for proving the domain of a trigonometric function. I'm just curious as to where I'm screwing up because if it's something algebraic, I can avoid similar situations in the future.

[quote="Prime factorization"]What am I doing wrong?[/quote]

For instance, you have divided an inequality by $ \sin x$, but $ \sin x$ can be negative.

  Darij

Sorry , I mean that ''the underlined part is false, as the inequality sign should be facing the other way. What am I doing wrong?'' is your signature  :oops: . Ok! Read what darij grinberg said!

[quote="Prime factorization"]Ok thanks, that explains it. It seems the only valid method is to use the multiplicative inverse property of inequalities. Thus:

$ -1 \leq \sin(x) \leq 1$

$ \frac{1}{-1}\geq \frac{1}{\sin(x)}\geq \frac{1}{1}$

$ -1 \geq \csc(x) \geq 1$

which is correct and leads to the right result  :lol:[/quote]

This method is not generally valid either, as you see by the result...

What is valid is:
$ a\geq b \wedge c\geq 0\Longrightarrow ac\geq bc$;
$ a\geq b \wedge c\leq 0\Longrightarrow ac\leq bc$.

  Darij


AMC
USA(J)MO
USAMO
number theory solved
number theory

Hello!

This problem is from the USAMO-99.

I've tried to solve it for over a year.

Let p>2 be a prime and let a,b,c,d are integers not divisible by p, such that
{ra/p}+{rb/p}+{rc/p}+{rd/p} = 2 for any integer r not divisible by p.

Prove that at least 2 of the pairwise sums of a,b,c,d are divisible by p.
({x} is the fractional part of x)

hey
this problem pestered me for that long as well.and i chekd the solution on [url=http://www.kalva.demon.co.uk/usa/usoln/usol993.html]at site HERE[/url] - it gives a solution which dint satisfy me somehow.i think there must be sum official solution,which might be slicker and more enlightening
regards

i have a solution to this problem stashed somewhere here in my room, gimme some time and i'll find it! :)

Hmm... Just going through the archives and really wanted to have a clean solution for this one. Actually I have not been able to come up with any solution but the one vineet is talking about really does not settle me. Thanks.

... a four real old problem to haunt us ...
Sorry to bump this, but seeing as Valentin became active again I was wondering if he had a solution or anybody else?

[quote="oleg"]Hello!

This problem is from the USAMO-99.

I've tried to solve it for over a year.

Let $p>2$ be a prime and let $a,b,c,d$ be integers not divisible by $p$, such that
\[\left\{\frac{ra}{p}\right\}+\left\{\frac{rb}{p}\right\}+\left\{\frac{rc}{p}\right\}+\left\{\frac{rd}{p}\right\}= 2 \]
for any integer $r$ not divisible by $p$.

Prove that at least $2$ of the pairwise sums of $a,b,c,d$ are divisible by $p$.
($\{x\}$ is the fractional part of $x$)[/quote]

Over [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340038#p340038]here[/url] they only have the Kalva solution also but a solution by Zuming is referenced...[/url]


If $ \sin^6 x \plus{} \cos^6 x \equal{} \frac{2}{3},$ find all possible values of $ \sin(2x)$.

An interesting general principle that can be used to solve problems of this type.

[b]Lemma:[/b]  Prove that any symmetric polynomial in $ \sin x, \cos x$ can be written as a polynomial in $ u \equal{} \sin x \plus{} \cos x$.

[hide="Hint"] First prove that a symmetric polynomial in $ a, b$ can be written as a polynomial in $ u \equal{} a \plus{} b, v \equal{} ab$.  Is there a relationship between $ u, v$ in this case? [/hide]


inequalities
vector
function
geometry
trigonometry
triangle inequality
geometric inequality

[quote="Lovasz"]hmm... I don't have time now. Here's some notes:
-Squaring.
[b]-Huygens's inequality.[/b]
-Homogenize.[/quote]
What's Huygens's inequality ? Can you tell me about it ?

Thank you very much . :)

[quote="Lovasz"][quote="Tiks"]
Lovasz have you solved this problem?
And what is the Huygens's ineq?I don't know it too.[/quote]
Yes, I solved it.

Huygens's inequality looks like Minkowski triangle inequality.[/quote]
$Lovasz$ please,post your solution,I really want to look at it.
I have spent much time to solve this problem in that way,but I can't finished it :(.

[quote="Tiks"]$x,y,z>0$ and $x+y+z=1$
Prove that
\[ \sqrt{x^2+xyz}+\sqrt{y^2+xyz}+\sqrt{z^2+xyz}\geq \sqrt{x^2+y^2+z^2+xy+xz+yz+2\sqrt{3xyz}}  \][/quote]
$\sqrt{x^2+xyz}+\sqrt{y^2+xyz}+\sqrt{z^2+xyz} \geq \sqrt{x^2+y^2+z^2+xy+xz+yz+2\sqrt{3xyz}}$
After squaring, we have:
$3xyz+2\left(\sqrt{x^2+xyz}\sqrt{y^2+xyz}+\sqrt{y^2+xyz}\sqrt{z^2+xyz}+\sqrt{z^2+xyz}\sqrt{x^2+xyz} \right) \ge xy+yz+zx+2\sqrt{3xyz}\ \ (*)$
You know that: $\sqrt{a+x}\sqrt{b+y} \ge \sqrt{ab}+\sqrt{xy}$, then:
$3xyz+2\left(\sqrt{x^2+xyz}\sqrt{y^2+xyz}+\sqrt{y^2+xyz}\sqrt{z^2+xyz}+\sqrt{z^2+xyz}\sqrt{x^2+xyz} \right) \ge 9xyz+2(xy+yz+zx)$
Now, we must prove:
$9xyz+xy+yz+zx \ge 2\sqrt{3xyz}$
Homogenize works.

[quote="Ramanujan"][quote="Lovasz"]
Now, we must prove:
$9xyz+xy+yz+zx \ge 2\sqrt{3xyz}$
Homogenize works.[/quote]
We can also use triangle inequality ( $\|a\| +\|b\|+\|c\| \geq \|a+b+c\|$) for vectors $(x,\sqrt{xyz}) \, ,(y,\sqrt{xyz})\, ,(x,\sqrt{xyz})$ to obtain this.[/quote]
Show your reply in more details. I'm not understand why you can get $9xyz+xy+yz+zx \ge 2\sqrt{3xyz}$ from $\|a\| +\|b\|+\|c\| \geq \|a+b+c\|$.

This  $\|. \|$ is the length of a vector in ${\mathbb{R}}^2$ with the usual inner product $<\, , \,>$

defined for arbitrary vectors $u=(x,y) ,\, v=(a,b)$ as $<u,v> \, = xa+yb$.

For $u=(a,b)$ it is $\|u \| = \sqrt{<u,u>} =\sqrt{a^2 +b^2}$.

Triangle inequality holds, that is $\| u+ v \| \leq \| u \|+\| v \|$.

 :)

Oh, $|\vec{a}|$ is more familiar than $\|a\|$, with me.

[quote="Tiks"][quote="Lovasz"]hmm... I don't have time now. Here's some notes:
-Squaring.
-Huygens's inequality.
-Homogenize.[/quote]
Lovasz have you solved this problem?
And what is the Huygens' ineq?,I don't know it too.[/quote]
   Here is Huygen's Inequality

[quote="Tiks"][quote="Lovasz"]
Now, we must prove:
$9xyz+xy+yz+zx \ge 2\sqrt{3xyz}$
Homogenize works.[/quote]
I am afride $Lovasz$,but you are wronge,this ineq. isn't true for all $x,y,z>0$ such  that$x+y+z=1$ :(  :) .
Try to take $x=y=\frac{1}{4}$ and $z=\frac{1}{2}$ ;).[/quote]
So guys,any other ideas? :P

[quote="silouan"]The well-known  $a^2+b^2+c^2+2\sqrt{3abc}\leq 1$  could be usefull ?[/quote]
I don't think so,I am sure the proposed(by me)problem is stronger! ;)

[quote="Albanian Eagle"]jensen of course ;)[/quote]
How with Jensen?
Don't forget that $f(x)=\sqrt{x}$,isn't convex function! ;)
So please,show me for which function you want to apply Jensen's ineq?

[quote="Albanian Eagle"]$\sqrt{x}$ is not but $\frac{1}{\sqrt{x}}$ is convex ;)[/quote]
$Albanian$ $Eagle$,if you have solved this,problem please post it with all detals.
Thank you.

[quote="Tiks"]$x,y,z>0$ and $x+y+z=1$
Prove that
\[ \sqrt{x^2+xyz}+\sqrt{y^2+xyz}+\sqrt{z^2+xyz}\geq \sqrt{x^2+y^2+z^2+xy+xz+yz+2\sqrt{3xyz}}  \][/quote]

Is Jensen's necessary?  The proof looks doable by Minkowski's,

$\sum_{cyc} \sqrt{x^2 + xyz} \ge \sqrt{ (x + y + z)^2 + (3 \sqrt{xyz})^2 }$

Then we need to prove

$xy + yz + zx + 9xyz \ge 2 \sqrt{3xyz}$,

Which holds true at $x = y = z$ so I'm inclined to believe it holds in general :P  Any ideas?

Edit:  Oh, this is the same as the one Lovasz posted... I didn't notice :blush:

[quote="t0rajir0u"][quote="Tiks"]$x,y,z>0$ and $x+y+z=1$
Prove that
\[ \sqrt{x^2+xyz}+\sqrt{y^2+xyz}+\sqrt{z^2+xyz}\geq \sqrt{x^2+y^2+z^2+xy+xz+yz+2\sqrt{3xyz}}  \][/quote]
$xy + yz + zx + 9xyz \ge 2 \sqrt{3xyz}$,

Which holds true at $x = y = z$ so I'm inclined to believe it holds in general :P  Any ideas?

Edit:  Oh, this is the same as the one Lovasz posted... I didn't notice :blush:[/quote]

But this one is false as Tiks said.

Albanian Eagle we are waiting for your solution :| .

[quote="fuzzylogic"]Tiks, It doesn't take the best inequality solvers here to find a solution.

Basic idea is translate it into a triangle inequality, it's equiv. to

(*) $(AI+BI+CI)^2 \ge \frac{1}{2}(a^2+b^2+c^2) + 2\sqrt{3}S$, where $I$ is the incenter.

I don't have time to write a detailed solution now (It's about 7pm PDT, I am hungary now). So be patient. :-)

Actually,  1/4 of credit will be given to  the person who proves (*) is equiv. to the original inequality. Another 1/4 of credit will be given to the person who proves (*). And I will claim 1/2 of credit for writing this post.

Is it fair enough?

 :D  :D  :D[/quote]
Ok,you got this problem :) ,if you want to know a story of this problem,here it,the geo. ineq. wiche you talking about,was proposet in Armenian TST 2006(look here[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=85355]Geo. ineq.[/url]),and trying to solve it I get to this ineq with $x,y,z$ and after spending 3 hours on it :wallbash_red:  I finally unterstoud that I am not able to solve it  :stretcher:  ,but luckily,when I had only 20 minuts untill finish of the olimpiad :noo: ,I found synthetical proof for geo ineq   :heli:  :10:  .

[quote="fuzzylogic"]OK, I am back.

Instead of using geometry, here is an algebraic proof which basically comes from the insight of the geometric inequality posted above.  I will leave the geometric inequality to the AoPSers (Mathlinkers) who are interested in it.

[quote]
After squaring, it suffices to prove:

$2\sqrt{x^2+xyz}\sqrt{y^2+xyz}+2\sqrt{y^2+xyz}\sqrt{z^2+xyz}+2\sqrt{z^2+xyz}\sqrt{x^2+xyz}$

$\ge (xy+yz+zx - 3xyz) +2\sqrt{3xyz}$
[/quote]

We will prove that:

(*) $2\sqrt{x^2+xyz}\sqrt{y^2+xyz} \ge (x+y) (xy + \sqrt{3xyz}) = xy - xyz + (x+y)\sqrt{3xyz}$

$LHS = 2 \sqrt{x(x+y)(x+z)}\sqrt{y(y+x)(y+z)} = 2(x+y) \sqrt{xy(y+z)(x+z)}$

$= 2(x+y) \sqrt{(xy)^2 + xyz}\ge (x+y) (xy + \sqrt{3xyz})$

The last inequality follows from C-S: $\sqrt{a^2+b^2} \ge \frac{1}{2}a + \frac{\sqrt{3}}{2}b$.

OK, I think this should be enough, the rest should be pretty obvious.[/quote]
Congratulations very nice solution ;) ,but anyway I cant understand how you have found it :?
P.S. You one of the Bests in Ineq. Solvers ;) .

[quote="fuzzylogic"][quote="Tiks"]Congratulations very nice solution ;) ,but anyway I cant understand how you have found it :?
[/quote]

Thank you!

I am not sure how you solved the triangle inequality mentioned above:

$(AI+BI+CI)^2 \ge \frac{1}{2}(a^2+b^2+c^2) + 2\sqrt{3}S$.

Here is how I proved it:

First use the law of cosines: $a^2 = BI^2+CI^2 - 2 (BI) (CI)\cos(\angle{BIC})$, etc.

Then I divided the triangle area into three piece: $S_{IBC} = \frac{1}{2}(BI) (CI)\sin(\angle{BIC})$, etc

If I put these pieces together, it reduced to prove

$2 (BI) (CI) \ge \sqrt{3}(BI) (CI)\sin(\angle{BIC}) - (BI) (CI)\cos(\angle{BIC})$, etc

That's obvious.

My algebraic proof was constructed from this idea. Hope you understand now. :-)[/quote]
Now I understand how you found algebraic proof :) .
I proved this ineq with constucting equilateral triangles on the sides of $ABC$  ;) .

[color=darkblue][b]Remark.[/b] $x^2+xyz=x(1-y)(1-z)=x(x+y)(x+z)\ !$
[b][u]Yes, Fuzzylogic ![/u][/b] The proposed inequality is equivalently with the inequality
$\sum \sqrt{\frac{x(x+y)(x+z)}{x+y+z}}\ge \sqrt{\sum x^2+\sum xy+2\sqrt{3xyz(x+y+z)}}\ .$
Using the substitutions $x=p-a$, $y=p-b$, $z=p-c$ we obtain
a geometrical inequality $\left(\sum\sqrt{\frac{bc(p-a)}{p}}\right)^2\ge 2pr\sqrt 3+p^2-r(4R+r)\ .$
But $AI^2=\frac{bc(p-a)}{p}$ a.s.o.$,\ \frac 12\cdot \left(a^2+b^2+c^2\right)=p^2-r(4R+r)$
and $pr=S\ .$ Therefore, the previous inequality becomes
$\boxed {\ \left(\sum AI\right)^2\ge \frac 12 \left(a^2+b^2+c^2\right)+2S\sqrt 3\ }\ \ \ (*)$
The algebraical form of the inequality $(*)$ [u]is a its ugly "paint"[/u], [b]Tiks[/b] ![/color][hide="Some classical and theoretical knowledge."][color=red] [u]The Fermat's point[/u] $F$ is the inside point for which $m(\widehat {BFC})=$ $m(\widehat {CFA})=$ $m(\widehat {AFB})=120^{\circ}\ .$ This point has the property $\min_{X\in \pi}\sum XA=\sum FA\ ,$ i.e. for any point $X$ from the plane $\pi$ of the triangle $ABC$ there is the inequality $\sum XA\ge \sum FA\ .$The construction of the remarkable Fermat's point is easily. Consider the outside equilateral triangles $BCA_1$, $CAB_1$, $ABC_1$. Prove easily that $F\in AA_1\cap BB_1\cap CC_1\ ,$  $AA_1=BB_1=CC_1=\boxed {\ \sum FA=\sqrt{\frac 12\left(a^2+b^2+c^2\right)+2S\sqrt 3}\ }\ .$ Indeed, from the [u]generalized Pitagora's theorem[/u] applied in the triangle $AA_1C$ obtain $AA_1^2=a^2+b^2-2ab\cdot \cos (C+60^{\circ})=$ $a^2+b^2-ab\cdot \cos C+ab\sqrt 3\cdot \sin C=$ $\frac 12\left(a^2+b^2+c^2\right)+2S\sqrt 3\ .$[/color][/hide]

I added in the previous reply the definition of the Fermat's point for a triangle, its construction and some properties.

[quote="Albanian Eagle"]I used jensen for $\sqrt{(x+y)(y+z)(z+x)}(\sum \frac{x}{\sqrt{x(y+z)}})$ ;)
[/quote]
I think, so  turns out the false inequality. ;)

[quote="Albanian Eagle"]I used jensen for $\sqrt{(x+y)(y+z)(z+x)}(\sum \frac{x}{\sqrt{x(y+z)}})$ ;)
but the solution above is the best.
Quite shorter than i was expecting for this ineq.
Didn't see a geometric aspect behind it :)[/quote]
Albanian Eagle can you give us more detals of your solution?
I am not sure about it too.

after using jensen it reduces to:
$(x+y)(y+z)(z+x)(x+y+z)^2\geq (x^2+y^2+z^2+xy+yz+zx+2\sqrt{3xyz(x+y+z)})(\sum x^2 y) \leftrightarrow (\sum x^2 y)(\sum xy-2\sqrt{3xyz(x+y+z)})+2xyz(x+y+z)^2\geq 0\leftrightarrow (\sum (x^3 y^2 +2x^3 y z +3x^2 y^2 z))^2\geq 12(\sum x^2 y)^2 xyz(x+y+z)$

[quote="Albanian Eagle"]after using jensen it reduces to:
$(x+y)(y+z)(z+x)(x+y+z)^2\geq (x^2+y^2+z^2+xy+yz+zx+2\sqrt{3xyz(x+y+z)})(\sum x^2 y) \leftrightarrow (\sum x^2 y)(%Error. "sumxy" is a bad command.
-2sqrt{3xyz(x+y+z)})+2xyz(x+y+z)^2\geq 0\leftrightarrow (\sum x^3 y^2 +3x^3 y z +5x^2 y^2 z)^2\geq (\sum x^2 y)^2 3xyz(x+y+z)$
i think I have a proof for this but now
I'm not sure either :blush:
I will post it after checking it.[/quote]
We will wait :) .

I edited a small typing mistake above!
the last inequality is true by muirhead!(after expansion of course :D )

[quote="Tiks"]$x,y,z>0$ and $x+y+z=1$
Prove that
\[\sqrt{x^{2}+xyz}+\sqrt{y^{2}+xyz}+\sqrt{z^{2}+xyz}\geq \sqrt{x^{2}+y^{2}+z^{2}+xy+xz+yz+2\sqrt{3xyz}}\]
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145158


[hide="Solution"]
Notice that the three terms that add up to zero cannot be negative, as they are square. So, the only three positive numbers that add up to zero are zero, zero, and zero (of course.) So, the x, y, and z that make these terms zero are $x=3, y=4, z=5$. Thus, $x+y+z=12$[/hide]

Correct me if I'm wrong.

[quote="ashegh"]we have :

$(X-3)^2>0$

[/quote]

Do you mean $(X-3)^2\ge0$? Same with your other two.

[hide]
Since they are squared, they will be positive.
x-3=0, so x=3
y=4 and z=5
3+4+5=12, so, the answer is D.[/hide]


let$X={{1,2,3,....,n}}$ .
  find the minimal of value $m$ such that :
. $\abs{A} =m$and$\forall a,b \in A$, we have$a \vdots b$  ($A\in X$)

[quote="tranthanhnam"]let$X={1,2,3,....,n}$ .
  find the minimal$m$ such that :
. $/A/ =m$and$\forall a,b \in A$, we have$a \vdots b$[/quote]

I no understand problem of you.

I guess it means that $b \left| a \right.$...

[quote="tranthanhnam"]let$X={1,2,3,....,n}$ .
  find the minimal$m$ such that :
. $/A/ =m$and$\forall a,b \in A$, we have$a \vdots b$[/quote]
$X$ not in problem of you!

I think the correct problem is the same this one.
Given $n$ is positive integer.
Let $X=\{1;2;...;n\}.$
Find the largest value of m such that there exists a set $A \subset X$ which satisfies following conditions:
i) $|A|=m$
ii) $\forall a;b\in A, a\le b$ then $a|b$ .

if I am correct, I guess $m=[\log_2{n}]$


inequalities
combinatorial geometry
Functional Analysis
rearrangement inequality
inequalities unsolved

Suppose that the sequence $(s_1, \ldots, s_n) \succ (t_1, \ldots t_n)$. That is, both sequences are non-increasing, and $s_1 + \cdots s_m \geq t_1 + \cdots t_m$ for each m with equality if $m=n$. Then show that for each permutation $\sigma$ of ${1, 2, \ldots n}$, we can assign a positive real $k_\sigma$, such that $\sum_\sigma k_\sigma = 1$ and

\[
\sum_\sigma k_\sigma (s_{\sigma(1)}, \ldots, s_{\sigma(n)}) = (t_1, \ldots, t_n)
\]

This Lemma is very hard to prove and it is used in Kiran Kedlaya's notes about inequalities : "A less than B"  like main fact of Muirhead theorem. :)

Why hard? We have to prove that a point $ T = (t_1,t_2,\dots,t_n)$ lies in a convex hull of points $ S_{\sigma} = (s_{\sigma(1)},s_{\sigma(2)},\dots,s_{\sigma(n)})$. For proving this assume a contrary. Then by separation (Hahn-Banach) theorem there exists a linear functional $ f(X = (x_1,\,x_2,\dots,x_n)) = a_1x_1 + a_2x_2 + \dots + a_nx_n + a$ such that $ f(T) > 0 > f(S_{\sigma})$ for any permutation $ \sigma$. Let $ c_1\ge c_2\ge \dots \ge c_n$ be a rearrangement of $ a_1,a_2,\dots,a_n$. Then $ \max_{\sigma} f(S_{\sigma}) = c_1s_1 + c_2s_2 + \dots + c_ns_n + a$ (by trans-inequality, or rearrangement inequality, or how is it called else), also we have $ c_1t_1 + c_2t_2 + \dots + c_nt_n + a\ge f(T)$ by the same inequality. So, to get a contradiction we only need to prove that $ c_1s_1 + c_2s_2 + \dots + c_ns_n\ge c_1t_1 + c_2t_2 + \dots + c_nt_n$. It follows from identity (Abel transform, or summation by parts): $ c_1s_1 + c_2s_2 + \dots + c_ns_n = c_n(s_1 + s_2 + \dots + s_n) + (c_2 - c_1)(s_1 + s_2 + \dots + s_{n - 1}) + \dots (c_1 - c_2)s_1$ and the same for $ t$-s.

[quote="BG Yoda"]what means "lies in a convex hull of points" ?[/quote]

We say that the point $ A$ lies in the convex hull of the points $ B_1$, $ B_2$, $ \dots$, $ B_n$, if there exist non-negative reals $ \alpha_1$, $ \alpha_2$, $ \dots$, $ \alpha_n$ such that $ \sum \alpha_i\equal{}1$ and $ A\equal{}\sum \alpha_iB_i$.


trigonometry
inequalities unsolved
inequalities

Let triangle $ ABC$ and $ x\in(o;\frac {\pi}{2})$
such that $ \sin^{3} x \equal{} sin(A \minus{} x)sin(B \minus{} x)sin(C \minus{} x)$.Find  max $ x$

 :lol:

hello, do you meant the equation $ \sin^3(x)\equal{}\sin(A\minus{}x)\sin(B\minus{}x)\sin(C\minus{}x)$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you meant the equation $ \sin^3(x) \equal{} \sin(A \minus{} x)\sin(B \minus{} x)\sin(C \minus{} x)$?
Sonnhard.[/quote]
Yes,thanks you!


Find n prime iff n+2, n*n+4,7n+2,9n+2 are also prime.

[quote="i_like_pie"][hide]$n=3$[/hide][/quote]

[hide="What about"]$n=5$?[/hide]

now , i will prove that $n=3$ and $n=5$ are the only solutions:

if $n>5$ , there are $4$ cases ($n$ prime):

$1.$ $n=5k+1 \Rightarrow n^{2}+4\equiv 0(\mod{5})$
$2.$ $n=5k+2 \Rightarrow 9n+2\equiv 0(\mod{5})$
$3.$ $n=5k+3 \Rightarrow n+2\equiv 0(\mod{5})$
$4.$ $n=5k+4 \Rightarrow 7n+2\equiv 0(\mod{5})$


$a,b,c \in \mathbb{C}^*$ such that:
$i)|a|=|b|=|c|
\newline ii)a+b+c=a^4+b^4+c^4
\newline iii)a^2+b^2+c^2=a^8+b^8+c^8 \newline$
Prove that a,b,c are the affixes of an equilateral triangle.

And what is solution? I don't see it  :blush:


Consider two particles arbitrarily placed on a plane. Particle A moves linearly in a constant direction with constant speed M. Particle B moves at the same speed M, but at any given moment, moves towards particle A. Given that B doesn't lie on the path of A, prove that particle A and B never meet.

No calculus.

Nice Problem. Solution
[hide]
Draw a coordinate plane with $x$ and $y$ axis and locate particle $A$ on the origin. It's straight path is on the $x$ axis.]
$B$ is chasing it. Let the instantaneous angle the velocity of $B$ makes with $x$ axis be $u$

Case 1:
[img]http://onlinecauchy.com/1.GIF[/img]

$B$ is on the y axis or on the left of the y axis.  In either case the horizontal velocity of $B$ is $M*\cos(u)$, while the horizontal velocity of $A$ is is $M$. Therefore, $B$ will never catch up.

Case 2:
[img]http://onlinecauchy.com/2.GIF[/img]

Let the coordinates of $B$ be $(j,k)$, this is the case where $k\ge j$
 by the time $A$ gets to $(j,0)$ If $B$ was traveling down in a straight line to $(j,0)$ it wouldn't make it in time $A$ gets there, but this time its a curve, so definitely not make it.

Case 3:
[img]http://onlinecauchy.com/3.GIF[/img]
This the case where $j>k$. As $B$ comes closer to $A$ case 2 happens again (too lazy to explain why but think about it)

Case 2 ultimately becomes case 1 if u reposition $A$ at the center of a new origin.
[/hide]


Let A = the sum of the positive factors of 2009 and B = the product of the positive factors of 2009. If $ Ax^2\plus{}Bx\plus{}C\equal{}0$ has one solution, find C, if A, B, and C are relatively and A is positive.

(please include the formula for the sum and product)

The positive factors of $ 2009$ are $ 1$, $ 7$, $ 41$, $ 49$, $ 287$, and $ 2009$. The sum of these is $ 1\plus{}7\plus{}41\plus{}49\plus{}287\plus{}2009\equal{}2394$. The product is $ (1)(2009)(7)(287)(41)(49)\equal{}(2009)(2009)(2009)\equal{}2009^3$. In order for the equation to have one solution, we must have $ B^2\minus{}4AC\equal{}0$. Solve this for $ C$.


Asymptote
function
calculus
calculus computations

Find the horizontal asymptotes for each function below.
I have worked on these two questions but to no avail.
I have to calculate the limit of + and - infinity but don't know how to do this successfully.

Question 1:

f(x) = $4e^{\frac2x}$

Question 2:

f(x) = e^(-3x) + 4e^(-x)

Do you mean $\frac{e^2}x$ or $e^{\frac2x}$ ?

Som parentheses would make things clearer.


Functional Analysis
real analysis
real analysis theorems

Find an example to prove that $ A(X)$ is not closed subspace of $ X$ although $ A: X \longrightarrow X$ is a continuous linear operator.

Hmmm... Do you really expect we'll answer all the functional analysis homework exercises you are posting? I should say that the course you are taking (or the book you are reading) is quite good but it'll be useful for you [b]only[/b] if you spend some time working on the problems yourself. Here is a hint though. Take any Hilbert space $ H$ and choose an orthonormal basis $ e_k$. Now define $ A$ by $ Ae_k\equal{}a_ke_k$. If $ \sup_k|a_k|<\plus{}\infty$, the operator is continuous. If ..., then the image is dense in $ H$ but not equal to $ H$ (I leave finding the condition ... to you :))


counting
distinguishability
geometry
geometric transformation
rotation

How many ways can you make a necklace with 8 red beads and 32 blue beads (the beads are indistinguishable) such that between any two red beads there are at least two blue beads?

[quote="SplashD"]is there a clasp on this necklace?[/quote]
The question didn't say anything about a clasp, so I would assume that there isn't one.  :wink:

[quote="nebula42"][quote="Phelpedo"]Put the 8 red beads down with 2 blues in between all of them. Now we have to put 16 more blue beads in 8 slots, which can be done in $\binom{23}{16}$ ways. Then we need to divide by something if rotation is allowed.[/quote]
That's what I first thought of, but it doesn't work because accounting for rotation is [i]way[/i] too complicated, unfortunately.  :wink:[/quote]

you would get a non-integer if you divided by 8 which gets rid of the rotation

so [img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/e/b/6/eb6c09d0a6efed1c28ccd790452452f76c7ad607.gif[/img] is wrong? what is the answer then?


function
calculus
derivative
limit
real analysis
advanced fields
advanced fields unsolved

I think so.  I assume you mean pointwise convergence.

Let $ f_{k}(x) =-x$ for $ x\in[0,1/2-1/k]$ and $ f_{k}(x)=2-x$ for $ x\in[1/2-1/(2k),1]$, and $ f_{k}$ is whatever on $ (1/2-1/k,1/2-1/(2k))$.

Yeah, pointwise convergence is useless.

[b]yenlee[/b], in your example the limit $ f$ is not a continuous function.

Oops!  I completely missed that condition when I read the problem.  Definitely harder with that condition :)

Suppose that there is a sequence of open sets $ U_{n}\subset [0,1]$ with finite overlap of closures $ \overline{U_{n}}$ and such that the connected components of $ [0,1]\setminus \overline{U_{n}}$ have diameter at most $ 1/n$. If such a sequence exists, then let $ f_{n}(x)=2a-x$ on each connected component $ (a,b)$ of $ [0,1]\setminus \overline{U_{n}}$ and define $ f_{n}$ within $ \overline{U_{n}}$ to make it $ C^{1}$. We have $ f_{n}(x)\to x$ uniformly, but $ f_{n}'(x)\to-1$ pointwise.

But I don't know if such a sequence $ (U_{n})$ actually exists.  :(

Such sequence $ (U_{n})$ does not exist. The condition "the diameter at most $ 1/n$" makes the limsup of $ (U_{n})$ a dense subset of [0,1]. (Baire category theorem)

Can we turn the argument around then? Let $ U_{n}=\{x\colon f_{n}'(x)>0\}$. The sets have finite overlap, even with closures. On the other hand, if $ f_{n}(x)\to f(x)=x$ pointwise, then for each $ \epsilon>0$ there is $ N$ such that any interval of length $ \epsilon$ meets $ U_{n}$ for all $ n>N$.

Not sure how to generalize this to the case of general $ f$ (which might not have any intervals of monotonicity).

[quote="fedja"]Let $ f_{k}\in C^{1}[0,1]$. Assume that $ f\in C[0,1]$ and, for every $ t\in [0,1]$, $ \lim_{k\to\infty}f_{k}(t)=f(t)$. Suppose also that $ \lim_{k\to\infty}f_{k}'(t)$ exists and is negative at every point $ t\in [0,1]$. Is it possible that $ f(1)>f(0)$?[/quote]

sorry I dont have latex...but if you take Fn(x)=-1/(x+1) + 1/(n+1)  , Fn belongs to C1[0,1] , Fn--->F , lim F'n exist and its negative , and F(1)>F(0) I think you have ur(possitive) answer  :)

[quote="mlok"]If you think you found a simple solution of a problem posted by [b]fedja[/b], look for a mistake.  :) $ F_{n}'(x)=\frac{1}{(x+1)^{2}}>0$.[/quote]

Heh srry ...its 05:00 am here  :oops:   :blush:

It is enough to find an example such that $ \lim_{k\to\infty}f_{k}'(t)$ is identically zero while $ f(1)>f(0)$. Gut feeling tells us that $ f$ should be the Cantor function.  :)

The Cantor function is defined as a limit of a sequence of functions: Start with $ f_{0}(x) = x$. For each $ n$ inductively, let $ f_{n+1}(x) = 0.5 f_{n}(3x)$ when $ 0 \le x \le 1/3$, $ 0.5$ when $ 1/3 \le x \le 2/3$ and $ 0.5+0.5 f_{n}(3(x-2/3))$ when $ 2/3 \le x \le 1$. The limit of $ (f_{n})$ is the Cantor function. (from wikipedia definition).

Is this sequence what we are looking for? No.  $ \lim_{k\to\infty}f_{k}'(t)$ is infinite or undefined at each point in the Cantor set. Need to repair $ (f_{n})$.

Let $ f_{0}$ be a smooth function such that it is 0 on [0,1/3] and 1 on [2/3,1]. Start with this new $ f_{0}$ instead of $ x$. Define $ f_{n+1}$ from $ f_{n}$ in the same way. The limit of $ f_{n}$ is the same Cantor function. Sets $ U_{n}=\{x\colon f_{n}'(x)>0\}$ are mutually disjoint. Therefore $ \lim_{k\to\infty}f_{k}'(t)$ is identically zero.

:coolspeak:  And the convergence of $ f_{k}$ is even uniform.


1)  Suppose A is a $ m x m$ matrix and has SVD  $ A \equal{} USV*$.  Find an eigenvalue decomposition of the 2m x 2m hermitian matrix

$ \begin{matrix}
0 &&A*\\
A && 0
\end{matrix}$

$ \begin{pmatrix}U^*&0\\ 0&V^*\end{pmatrix} \begin{pmatrix}0&A^*\\ A&0\end{pmatrix} \begin{pmatrix}U&0\\ 0&V\end{pmatrix}\equal{} \begin{pmatrix}0&S^*\\ S&0\end{pmatrix}$

so the original matrix is unitarily similar to $ \begin{pmatrix}0&S^*\\ S&0\end{pmatrix}.$

This in turn is similar via a permutation to the diagonal block matrix whose $ 2\times 2$ diagonal blocks are $ \begin{bmatrix}0&\overline{\sigma_i}\\ \sigma_i&0\end{bmatrix}$ where the $ \sigma_i$ are the singular values.

The eigenvalues of that $ 2\times 2$ block are $ \pm|\sigma_i|.$ Hence the eigenvalues of our original matrix are $ \{|\sigma_1|,|\sigma_2|,\dots,|\sigma_m|,\minus{}|\sigma_1|,\minus{}|\sigma_2|,\dots,\minus{}|\sigma_m|\}.$

Note that the original matrix is self-adjoint (Hermitian) and hence its eigenvalues must be real; and also that it has trace zero so its eigenvalues must sum to zero.


find x,y,z is a positive number so that
$ \sum x^{2}+\sum x^{3}= \sum x^{4}+xyz$

Sorry , x,y,z are positive integers or just positive?
If x,y,z are positive integers, the problem is quite simple . If x,y,z just are positive real numbers then U post the prob. to the wrong topics!  :D

oh.thank you .sorry but can you  prove it ?


integration
real analysis
real analysis unsolved

find all $ t,k\in \mathbb{R}$ that:
$ \int_0^{\infty} \frac {x^k}{(1 \plus{} x)^t} dx$  be converge.

[quote="bookworm_vn"]$ t \minus{} k > 1$[/quote]
please send your proof.

It's actually $ k>\minus{}1,t\minus{}k>1.$


[quote="tetrahedr0n"]You don't honestly expect anyone to sincerely answer this question, right? And it seems pretty clear, right off the bat, that you are very much against drug use (the "are you on" instead of a more reasonable "do you ever use" and the capitalization of "drugs.")[/quote]

I dont care about people's replies but rather more about the poll results. And yes I am against drugs.

[quote="eryaman"]I assumed you meant illegal drugs.  I'm sure most people here have used drugs of some sort before.  :lol:[/quote]

Ofcourse i meant illegal

I hate to be picky, but I would like some definitions:

Define 'on' Is that addicted, or been intoxicated a few times?

You meant illegal drugs, but do you mean alcohol and cigarettes? THey are legal, just not for minors.

Ok, that was not the point, but if you want an honest answer: I have sampled illegal drugs (for USA), but they were not illegal where I had them, hence the fuzzyness of the question.

So? Some of my mom's colleagues went to the Netherlands and got some Heroin. Obviously that's illegal, but not in the Netherlands. That's where it gets fuzzy.
Anyways, never have never will.
I'm not that stupid - though my brain has recently gone through a dramatic degenerative process whereby I start to forget simply things randomly, like Heroin's Formula... I meant Heron's Formula.

Really? I thought heroin was illegal in the Netherlands as well. I believe only marijuana is legal (in certain small quantities, and to be used only in certain places, like the famous hash bars.)

Maybe it was marijuana, then. I'm not sure. But, my point is same.

Look, you can say illegal, illegal, illegal all you want, but there really is no distinct 'illegal' line that drugs cannot cross.

And you did not really answer my question.

I would rephrase the poll question: Have you ever taken mind-altering substances when it was illegal for you to do so at that time and place?

Then I would have answered: NO

[quote="white_horse_king88"]So? Some of my mom's colleagues went to the Netherlands and got some Heroin. Obviously that's illegal, but not in the Netherlands.[/quote]You are [b]wrong[/b]. Heroin is [b]illegal[/b] in Netherlands, as well as in all EU. All "adictive" drugs (ie drugs that are not medicaments) are illegal ("weed", "cocaine", "heroine",...) in Netherlands. However it seems that the policemen there don't care too much when it comes to marijuana, but [b]it's still illegal[/b] and we strongly discourage such behavior.

I personally think that you don't need to test everything to be sure it's a very bad thing for you, and for other people near you, so [b]NO DRUGS![/b].

drugs fry brain

need brain for math

Actually, no, I've never done anything like that, never smoked, never drank. (although I almost drank vodka instead of my glass of water at a party when I was 4-picked up the wrong glass)

I think I had smoked a cigarette when I was a small child(due to curiosity)

You are wrong Valentin, marijuana (again, in small amounts) can be legally consumed in the Netherlands. It's not just a case of police indifference. The same is true of prostitution - it exists in only a few places, it is heavily regulated, but it is legal.

no drugs, just small amounts of alcohol as a teen. but then i decided that i didnt like it, so i still dont drink. ah well.

[quote="tetrahedr0n"]You are wrong Valentin, marijuana (again, in small amounts) can be legally consumed in the Netherlands.[/quote]Read [url=http://europa.eu.int/comm/health/ph_projects/2000/drug/fp_drug_2000_frep_18_en.pdf]here[/url] (warning: large .pdf file) at page 23 about the drug legislation in Netherlands.

If you're lazy I quote

[quote]Regulations on drugs are laid down in the Opium Act The Act draws a distinction
between hard drugs, (e.g. heroin, cocaine and XTC) which pose an unacceptable
hazard to health, and soft drugs (e.g. hashish and marihuana), which constitute a far
less serious hazard. The possession of drugs is an offence. However, the possession
of a small quantity of soft drugs for personal use is a minor offence.
Importing and exporting drugs are the most serious offences under the provisions of
the Opium Act, although manufacturing, selling and attempting to import drugs are
also offences. As is the case in other countries, the cultivation of hemp is prohibited,
except for certain agricultural purposes (e.g. to form windbreaks, and for the
production of rope). New legislation is currently being drafted to raise the maximum
penalty for commercial hemp production from two to four years imprisonment.
On the principle that everything should be done to stop drug users from entering the
criminal underworld where they would be out of the reach of the institutions
responsible for prevention and care, the use of drugs is not an offence.[/quote]

and

[quote]The sale of small quantities of soft drugs in coffee shops (which are not allowed to
sell alcohol) is therefore technically an offence, but prosecution proceedings are only
instituted if the operator or owner of the shop does not meet the following criteria:
• no more than five grams per person may be sold in any one transaction;
• no hard drugs may be sold;
• drugs may not be advertised;
• the coffee shop must not cause any nuisance;
• no drugs may be sold to minors (under the age of 18), nor may minors be
admitted to the premises;
• The mayor may order a coffee shop to be closed.[/quote]

Thus the conclusion is that [b]it is illegal[/b] but it's not prosecuted as I said before.

I've taken drugs around twenty or thirty times. Tylenol, Robitussin, prescription drugs :P ...

No. I've never taken illegal drugs, and I don't plan to. I've taken small sips of wine at New Year's Day celebrations, but that's about it. Scented markers are perfectly legal to sniff, and they make me happy enough.

have you ever rode a rollercoaster while you are high? or if you take acid, there will be a little green guy on your right shoulder and a little red guy on your left shoulder and they will start fighering each other. and getting so drunk that you wake up in the front lawn so totally stellar. and there is a shipment of needles coming in today. this will be a great weekend.

[quote="GoBraves"][quote="mcalderbank"]I hate to be picky, but I would like some definitions:
You meant illegal drugs, but do you mean alcohol and cigarettes? .[/quote]
Like i said....ILLEGAL DRUGS[/quote]

If you are under 18 and in the US (as I believe you are) both, alcohol and cigarettes are illegal for you.

[quote="Valentin Vornicu"]Thus the conclusion is that [b]it is illegal[/b] but it's not prosecuted as I said before.[/quote]

I am reading the document, and it is somehow ambiguous. The prosecution that they describe define more a [i]restringed substance[/i] than an [i]illegal[/i] one, at least marihuana and the soft drugs.

Bleh - Valentin, tetrahedron already corrected me. And still, unless laws are enforced, they are meaningless. I think that marijuana, like tetrahedron said, is legal in limited amounts. I've had dozens of people who've visited the Netherlands tell me that. Plus - here: [hide="Legal"]http://www.worldwideseeds.com/news.php[/hide].

If you want to get it into people's heads that taking illegal, mind-altering drugs is a bad idea, get them to read [i]Fear and Loathing in Las Vegas[/i] by Hunter S. Thompson. It portrays the drug culture in a way.... well, let's just say, what that stuff does to you isn't pretty. Take it from the man that knows: Thompson.

[quote="Valentin Vornicu"]That's an official document of the EU. You can look up the Act itself and check. It is illegal, however it's not prosecuted for small amounts. Explanation: they had their jails full, and there was no place to put people in. Think about the fact that the Netherlands is one of the countries (if not "the" country) with the highest density of urban population in Europe and possibly in the world. In such a small space there are too many people, not even counting the countless visitators, and obviously not too many spots in jail ...[/quote]
Well, I honestly don't think that the overpupulation on the jails is the actual problem. I think that the Netherlands law point of view is more like: [i]Is it really a crime to get high?[/i] I mean, automutilation is not a crime, but very unhealthy, in a similar way, I would say that to do any drug is not a crime, but very unhealthy. Also, it is totally obvius that the human and economical cost of the total prohibition is quite high, then, probably they think that is cheaper from many points of view to have a controlated [i]legal[/i] market and a small black market than a total prohibition and a big black market.

On the other hand, what I say is that [i]illegal[/i] means forbiden by law, and the document itself says that it is allowed (even when they talk about minor offence, but that sounds to me too ambiguous to have a totally reincident legal offence that will never be prosecuted) the sell of small quantities of drugs, and that sound to me more like a hard restriction than a prohibition.

Basically what the document says is that if the cops really want to do it, they can bust you for doing it no matter how much you have on yourself. [b]It is[/b] illegal (required by the EU laws!), but it's not prosecuted.

And jails are a problem in all countries, especially in those in which the density of the urban population is very very high.

Besides [i]required[/i] prescription drugs (yes, you can get legally addicted to prescription drugs and they can thus become harmful), I have had a couple sips of wine - mind you, just sips, and only on a New Year's a couple years ago and on one Passover - and I suffer from secondhand smoke. :(

[quote="white_horse_king88"]So? Some of my mom's colleagues went to the Netherlands and got some Heroin. Obviously that's illegal, but not in the Netherlands. That's where it gets fuzzy.
Anyways, never have never will.[/quote]

There are indeed some very awful and inconsequent laws in the Netherlands. Some coffy shops in our country are allowed to sell softdrugs, but are forbidden to buy drugs. So coffee shops has to use illegal ways to obtain the drugs. There are sometimes discussions in our country about legalizing the cultivation and production of drugs. Supporters of the legalization claim that legalization decreases the crime rate, for drug addicts don't have to steal money. Personally I strongly disagree both legalizing selling and buying drugs.

[quote="Valentin Vornicu"]I don't get this. If it's illegal to buy drugs, how can they sell them?!!  :?  :?[/quote]
I also don't get it, but it is really true. Coffee shops are allowed to sell drugs, but the both the production and the import of drugs are forbidden. So there is no way for them to get the drugs. I hope the law will soon be changed...

I don't know, I think some of us might be on drugs.  I mean, I hear it all the time.  "You frequent a [i]math[/i] forum?  Are you on drugs?"

So there you go. :)

On a related note, here is something very funny:

It is legal to sell opium poppy seeds (think about: poppy seed bagels), but it is illegal to put those seeds (Papaver somniferum poppy seeds) in the ground and then grow them, because you could theoretically score the poppy and create opiates, like opium (duh), morphine, and heroin. :wacko:  :wacko:  :wacko:

In fact here the government tells you how to make opium! : [url=http://www.usdoj.gov:80/dea/pubs/intel/20026/20026.html#cooking]Cooking Opium[/url]

Also, seed house(s) carry it: [url=http://www.parkseed.com/webapp/wcs/stores/servlet/StoreCatalogDisplay?langId=-1&storeId=10101&catalogId=10101&RequestType=PageNumRequest&PageNumber=3&SearchText=seed%20poppy&SearchUnion=Y&mainPage=textsearchresults]Poppy Seeds at Parks Seed House[/url] (Look for the papaver somniferum variety)

And....why do I know all this????? Because of a certain class: How many ways can you get high off of things already in your house...............formerly known as Health.

This was ridiculously easy (a 5 minute google search).

On the legal addiction: Remember Rush Limbaugh's debacle with the painkillers?

I've smelled marijuana...it's gross...
you just feel like vomiting...

I don't wanna get addicted to it as I did to smokes...
It's hard to give up...

I made it though! :lol:

hahaha
Now I look like a good girl--can spend more time doing physics and math

[quote="Connie"]I've smelled marijuana...it's gross...
you just feel like vomiting...[/quote]
It is interesting, I have smelled too, and I prefer that smell than the smell of regular cigarettes, but that is my personal opinion.

[quote="Connie"]I don't wanna get addicted to it as I did to smokes...
It's hard to give up...[/quote]
Well, accordingly to medical studies, if by smokes you refer to regular tobacco, well, cannabinoids, as THC, are not really phisically addictive, the addiction is mainly psicological. Nevertheless, tobacco has nicotine, that is phisically very addictive, actually, I remember that in 1999 I read in a magazine (it is called "Conozca Más", spanish) that some people had discovered that it was 5 times more addictive than Heroin, but I don't have any idea of how they can messure that!

[quote="Connie"]Now I look like a good girl--can spend more time doing physics and math[/quote]
That sound very wise!

You can actually get high by scraping the black stuff off the peels of bananas and drying them. You need about a dozen bananas to get a decent amount, though. Perfectly legal as well.

A note of warning- a few years ago, I went to a concert, and was able to get up to the front rows with one of my friends. I thought I smelled something a little funky but I didn't make anything of it. However, I was VERY LUCKY, because my friend knew what it was- marijuana- and we got the heck out of there. If you plan on being anywhere where there MAY be drugs, know what to look for. If the place gets busted, they aren't going to ask questions- you'll be there too. Also, trace amounts of marijuana- enough to get a positive urine test- can be found just by being around the stuff. So be careful.


Suppose

$ a\not = b\not =c$

$  \{ \begin{array}{c}{1}
a^2x+(b+c)y+z=a^3 \
b^2x+(c+a)y+z=b^3 \
c^2x+(a+b)y+z=c^3
\end{array}$

 Prove that $ x = a+b+c$.

 :oops:  :D  :roll:

Subtracting first and second equation:
$ (a^2\minus{}b^2)x \minus{} (a\minus{}b)y \equal{} a^3\minus{}b^3$ and since $ a$ does not equal $ b$, we can conclude

$ (a\plus{}b)x \minus{}y \equal{} a^2 \plus{} ab \plus{} b^2$. Similarly,
$ (b\plus{}c)x \minus{} y \equal{} b^2 \plus{} bc \plus{} c^2$
$ (a\plus{}c)x \minus{}y \equal{} a^2 \plus{} ac \plus{} c^2$.

Subtracting second equation from first,
$ (a\minus{}c)x \equal{} a^2 \plus{} ab \minus{} bc \minus{} c^2$. Substitute $ x \equal{} a \plus{} b \plus{} c$ and see it works, and do the same thing with the other two pairs of equations.


I found this problem in the Canadian Grade 8 Gauss competition. I like it a lot :). There's quite a clever solution once you see how to count the blocks right.

A large block, which has dimensions $n$ by $11$ by $10$, is made up of a number of unit cubes and one $2$ by $1$ by $1$ block. There are exactly $2362$ positions in which the $2$ by $1$ by $1$ block can be placed. What is the value of $n$?

The answer is
[hide]Are you sure you want to see it?
[hide]the number which is a  perfect cube, and if you add 1 to it, it is a perfect square.[/hide].
..Hint ..
[hide]The  odd-cube can be any direction (three possible) .. and for  any one particular direction the choices would be (a-1) * b * c  where a,b.c are dimension...
[/hide]
.[/hide]


inequalities
triangle inequality
calculus
calculus computations

Let $ B$ be the set of all bounded sequences $ \textbf{x} \equal{} (x_1,x_2,\ldots)$, and define $ d(\textbf{x},\textbf{y}) \equal{} \sup\{|x_j \minus{} y_j|: j \equal{} 1,2,\ldots\}$.

How do you prove the triangle inequality to show that $ d$ is a metric for B?

Apply the triangle inequality to each component and take supremums.


Explain why $ c$ and $ d$ are the $ x$ and $ y$ intercepts, respectively, of the line $ \frac{x}{c}\plus{}\frac{y}{d}\equal{}1$.

Explanation 1:Let $ y \equal{} 0\Longrightarrow x \equal{} c$, $ x \equal{} 0\Longrightarrow y \equal{} d$

Explanation 2: The equation of the line is $ y \equal{} \minus{} \frac {d}{c}x \plus{} d.$

Expalnation 3: You can perform a transformation to the line $ x \plus{} y \equal{} 1.$

@amathlearningstudent:
This is your post in $ \text{\LaTeX}$.
[quote="amathlearningstudent"]$ x=0 \implies y=d$
$ y=0 \implies x=c$[/quote]
It looks like this:
[code]$x=0 \implies y=d$
$y=0 \implies x=c$[/code]
Anyway, kunny posted the exact same thing. You should try to avoid posting unless you're contributing somehow.

-----

I'm interested in Explanation 3. If we take $ x+y=1$, which has the $ x$ intercept at $ x=1$, and $ y$ intercept at $ y=1$, and apply some transformation (translation/rotation, I guess), we can change the $ x$ and $ y$ intercepts, of course. How would we show that this ends up giving $ \frac{x}{c}+\frac{y}{d}=1$?

Dividing $ x$ by something stretches the graph in the $ x$ direction, and dividing $ y$ by something stretches in the $ y$ direction (consider the standard equation for an ellipse, which is really a stretched circle), so I guess that justifies this somehow.

Clearly, the s and y intercepts of $ x\plus{}y\equal{}1$ are $ 1$ and $ 1$. Now, if we stretch this graph horizontally by a factor of $ a$, the $ x$ intercept becomes $ a$. Similarly, the $ y$ intercept is multiplied by $ b$ to become $ b$ when we stretch the graph horizontally by $ b$.


invariant
function
search
algebra
polynomial
complex analysis
complex analysis unsolved

Searching for something unrelated I stumbled upon this:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5327
it's an old high school olympiad combinatorics problem, but I think there's an elegant complex analytic proof to be had that wasn't originally considered, though briefly mentioned in a related thread.

the original thread is one of my all time favorite AoPS threads and i'd be interested in seeing your solution.

actually, with Markov Chain theory, it is quite straightforward:
consider the symmetric random walk on $ \mathbb{Z}^2$: it is recurrent so every $ 2$ non zero invariant measures are scalar multiple of one another.
End of the story  :)
(hence boundedness can be relaxed to lower boundedness)
Rq: it is weird that this probabilistic argument works in $ \mathbb{Z}^2$ but not in $ \mathbb{Z}^3$. However the result is still true in $ \mathbb{Z}^n$

The random walk proof appears in the original topic, on the second page (posted by Ravi B; it's message #23). Also, Peter Scholze posted a proof (which I never read) for the most general statement in this context: there are no non-constant positive harmonic functions on $ \mathbb Z^n$. Orl posted a proof by Donald J. Newman using local convexity, the Krein-Milman Theorem, and such. There is another solution entirely analogous to what fedja did in [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1021187269&t=28827]the continuous version of the problem[/url], using periodic distributions, i.e. distributions on the torus, and Fourier transforms to switch between the torus and $ \mathbb Z^n$ (but in order to apply this method directly you do need boundedness or at least polynomial growth; I've been meaning to post this proof in the original topic, but apparently I'm too lazy to do that :)). I'm sure there are many more proofs out there. However, I'd also like to see the complex analytic one that JoeBlow had in mind :?.


algorithm
Support
search
geometry
function
trigonometry
Pythagorean Theorem

what about math? math doesnt exist but in our thoughts, it is the base of EVERYTHING and ANYTHING existing, but is more an idea than something real. but if most of real things are based in some mathemathical ideas: optimization, maximization, algorithms,.., etc. how can it be only and idea if it is what real world supports in?

Well math is a creation by man to explain the world.  Since basically everything in the real world follows some sort of mathematics, I guess we did a good job creating and developing it.

[quote="JRav"]Well math is a creation by man to explain the world.[/quote]

Is it?  I'm sure that once upon a time, theoretical math was developed and used to explain physical phenomena, but is that true any more?  Concepts which were once abstract and detached from the world have become applicable in many places.  However, that might just be due to the progress of computers...

Math WAS developed to explain phenomena.  We happened to do a very good job developing and applying it such that it DOES effectively describe these things.

The way I look at it is that Math is a method that was developed to interpret discoveries in everything that humans find.

Math wasnt "created" by man. For example, before Pythagoras discovered the Pythagorean Theorem, it's not as if it wasn't always true. Like if someone proves the Goldbach conjecture in the future, that doesn't mean that it isn't true right now.

Random Analogy: For example, when people discovered that lobsters were edible, lobsters didn't suddenly become edible. Lobsters were already edible before they discovered they were.

[quote="Nerd_of_the_Ages"]Math wasnt "created" by man. For example, before Pythagoras discovered the Pythagorean Theorem, it's not as if it wasn't always true. [/quote]

This is a far deeper philosophical idea than you make it seem.  The statement of the Pythagorean theorem is a statement about objects called "right triangles" and a property of a part of those objects called "length."  Were these concepts created by man?  Do right triangles "exist"?

The question asked in this thread is a deep one that is the subject of the field of [url=http://www.google.com/search?hl=en&q=mathematical+philosophy&btnG=Google+Search]mathematical philosophy.[/url]  Don't expect to find a simple answer.  Mathematicians have been grappling with this concept since Plato.  :)

[quote="Nerd_of_the_Ages"]Math wasnt "created" by man. For example, before Pythagoras discovered the Pythagorean Theorem, it's not as if it wasn't always true. Like if someone proves the Goldbach conjecture in the future, that doesn't mean that it isn't true right now.[/quote]

Well any mathematical statement follows from some set of assumptions.  So something can't simply be "true": it has to be based off of something (as t0r said).  So because we defined objects and made assumptions which seemed to fit with what we observed, we found statements which seemed to fit with our world.  But, for example, Euclidean geometry is not the only kind of geometry: the nature of the system depends on the initial axioms.

Some people here have been discussing what is commonly referred to as the "unreasonable effectiveness of mathematics" in describing physical phenomena. Anyone wanting to find out more may be interested in googling the term.

math should also been have made for practical purposes as well, ie for easier calculating stuff

Why should calculations be a product of mathematics¿ Calculations were needed in daily life already a long time ago. From this and other things, mathematics evolved.

[quote="jimhu"][quote="JRav"]Calculations are a result of math.  Math was not invented to "simplify calculations" since calculations are themselves products of math.[/quote]

I see, so what you're saying is that math was never created, since people have been doing simple calculations since who knows how many thousands of yrs ago before they even thought of the specialized concept of
math"[/quote]

I consider "calculation" an aspect of mathematics.  Mathematics is first and foremost a [b]tool[/b] to help us understand and organize [b]patterns.[/b]  It derives its usefulness from the fact that many aspects of the macroscopic world fall into patterns that we can understand.

[quote="t0rajir0u"] Mathematics is first and foremost a [b]tool[/b] to help us understand and organize [b]patterns.[/b]  [/quote]
I would agree if not for one caveat: the assumption that the world follows "patterns" is just a cunning restatement of the assumption that the world follows "mathematical laws". Just replace the name "mathematics" with "patternology" and you are back to square one, i.e., to the questions a) why there exist any patterns in the nature and b) why those patterns are (at least approximately) understandable to humans. I'm far from insisting that the mathematics is really primary and the Universe as we know it is merely a perceptable by 5 senses manifestation of it but when you think that you are essentially nothing but an electrochemical machine and your five senses just allow you to measure certain numeric quantities (intensity of heat, wavelength of light, etc.), the ground becomes a bit shaky. Indeed, as far as I can tell, there is no experiment that would prove beyond doubt that our peception of the world and reactions to it are any different from those of a sufficiently cleverly written computer program that sees nothing but numerical values generated by some simple mathematical laws and tries to fit its observations into several standard preset templates. Of course, you may say that the numbers just allow us to describe "physical reality" but if [b]everything[/b] that is physically observable can be given a numerical value and the laws of addition do not change whether we count eggs or stars or electrons, isn't it simpler and more logical to assume that "five" in "five eggs" is more "real" than the "eggs" themselves because, for all we know, the eggs might not even be there if the reproductive system in birds and reptiles were just slightly different, but you would have to destroy the entire Universe to eliminate the notion of "five" (or, if you prefer, "three" or "four") from it?
Again, let me say that I do not insist on this point of view: everyone has to decide for himself what he wants to believe. I just wanted to say that the question in the original post has about as many chances to be answered in a satisfactory manner as the one about the existence of God :)

[quote="fedja"][quote="t0rajir0u"] Mathematics is first and foremost a [b]tool[/b] to help us understand and organize [b]patterns.[/b]  [/quote]
I would agree if not for one caveat: the assumption that the world follows "patterns" is just a cunning restatement of the assumption that the world follows "mathematical laws". [/quote]

I would actually disagree.  To say that the world follows mathematical laws is to imply a degree of precision that is not always corroborated by actual data.  When I say that the world follows patterns, I merely mean that there is order that can be extracted from natural phenomena.  The progression of the seasons, for example, is not a neat, precisely mathematical periodic function, but the seasons [b]do[/b] repeat.

Math is neither real nor imaginary, it's complex.

Ah, but the repetition of seasons is an almost precise periodic function governed by the change of the angle at which the sun rays hit the ground which is obscured by various other factors that just follow completely different patterns that do not need to have the same period or be periodic at all but which still are comprehensible to a certain extent and follow precise mathematical laws. Your objection can be exaggerated to the claim that the function $ \sin x\plus{}\frac 1{x^2\plus{}1}$ is not described by mathematical laws because it "sort of repeats" but is not an exact periodic function. A mathematical law governing a phenomenon is not obliged to be a single clean formula; to look for that would be utterly naiive. My question is whether there is anything in the nature at all that escapes mathematical laws (be it simple laws of elementary logic or stochastic differential equations or whatever else) and is governed by some completely different principles that require a different way of thinking to comprehend (not because we currently just don't know what exactly we should "measure" and "compute" or because "measuring" and "computing" it requires resources far exceeding those within our reach, etc., but because there is nothing to "measure" and "compute" in principle)? You seem to mean that there is some underlying "reality" and some "patterns" in it partially describable by "mathematics" while I suspect that there is no other reality than the patterns themselves and those patterns are mathematics (I do not mean symbolic or verbal representation of mathematics here, of course, those are the product of human civilization; what I mean is the essence of it like the content of some theorem that doesn't change if you say it in English, or in Russian, or write it in some symbols on the board, etc.)

[quote="fedja"]My question is whether there is anything in the nature at all that escapes mathematical laws (be it simple laws of elementary logic or stochastic differential equations or whatever else) and is governed by some completely different principles that require a different way of thinking to comprehend [/quote]

Ultimately, I think this would depend on your definition of mathematics.  Because all mathematics, to my mind, ultimately derives from some way in which we have evolved to understand the real world (the existence of discrete objects, for example, driving our notion of cardinality), I do not think there is any aspect of the real world that cannot be incorporated into what we call "mathematics."


inequalities
function
limit
algebra unsolved
algebra

Find all functions $f: \mathbb R \to \mathbb R$ such that
\[ f(x) \ge x \plus{} 1\] and \[ f(x\plus{}y) \ge f(x)f(y)\] for all reals $x$ and $y$.

[quote="moriarti"]Find all functions $ f: R \minus{} > R$ such that

$ f(x) \ge x \plus{} 1$ and $ f(x \plus{} y) \ge f(x)f(y)$ for any x, y.[/quote]

$ f(\frac x2\plus{}\frac x2)\ge f(\frac x2)^2\ge 0$ and so $ f(x)\ge 0$

If $ f(u)\equal{}0$, then $ f(\minus{}\frac 12)f(u\plus{}\frac 12)\le f(u)\equal{}0$ and so $ f(\minus{}\frac 12)f(u\plus{}\frac 12)\equal{}0$. But $ f(\minus{}\frac 12)\ge \minus{}\frac 12\plus{}1>0$ and so $ f(u\plus{}\frac 12)\equal{}0$
So $ f(u\plus{}\frac n2)\equal{}0$ and this is impossible since $ f(x)\ge x\plus{}1>0$ $ \forall x>\minus{}1$

So $ f(x)>0$ $ \forall x$

$ f(0\plus{}0)\ge f(0)^2$ and so $ f(0)\le 1$ but $ f(0)\ge 0\plus{}1$ and so $ f(0)\equal{}1$
$ \forall y>0$, $ f(y)\ge y\plus{}1>1$ and so $ f(x\plus{}y)\ge f(x)f(y)>f(x)$ (since $ f(x)>0$ and $ f(y)>1$ and so $ f(x)$ is increasing.

Let $ x>0$ : $ f(x\plus{}y)\ge f(x)f(y)\ge f(y)(x\plus{}1)$ and so $ \frac{f(x\plus{}y)\minus{}f(y)}x\ge f(y)$
Let $ x>0$ : $ f(y)\ge f(\minus{}x)f(x\plus{}y)\ge f(x\plus{}y)(\minus{}x\plus{}1)$ and so $ \frac{f(x\plus{}y)\minus{}f(y)}{x}\le f(x\plus{}y)$

So $ f(x\plus{}y)\ge \frac{f(x\plus{}y)\minus{}f(y)}x\ge f(y)$

Setting $ x\to 0$ in this inequality, and remembering that $ f(x)$ is increasing, we get that $ \lim_{x\to 0}f(x\plus{}y)\equal{}f(y)$ and so $ f(x)$ is continuous.

More, since $ f(x)$ is continous, we see that $ \lim_{x\to 0}\frac{f(x\plus{}y)\minus{}f(y)}x\equal{}f(y)$ and so $ f(x)$ is $ C_1$ and $ f'(x)\equal{}f(x)$
And, since $ f(0)\equal{}1$, we get $ f(x)\equal{}e^x$

And it is easy to check that this mandatory condition is enough and that this indeed is a solution.

Hence the unique solution : $ \boxed{f(x)\equal{}e^x}$ $ \forall x$


[quote="alex2008"]For positive real numbers $ a,b$ and $ c$, prove that
\[ \frac {\sqrt {b \plus{} c}}{a} \plus{} \frac {\sqrt {c \plus{} a}}{b} \plus{} \frac {\sqrt {a \plus{} b}}{c} \ \ge\ 3\sqrt {2}\sqrt {\frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}}
\]
[/quote]
By AM-GM inequality,we have:
$ LHS \ge 3 \sqrt[6]{\frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{a^2b^2c^2}}$
So,we only need to prove that: $ (a\plus{}b)(b\plus{}c)(c\plus{}a)(ab\plus{}bc\plus{}ca)^3 \ge 8a^2b^2c^2(a\plus{}b\plus{}c)^3$
We have:
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge \frac{8}{9}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$
$ (ab\plus{}bc\plus{}ca)^4 \ge 9a^2b^2c^2(a\plus{}b\plus{}c)^2$
then,we have Q.E.D :)

[quote="quykhtn-qa1"][quote="alex2008"]For positive real numbers $ a,b$ and $ c$, prove that
\[ \frac {\sqrt {b \plus{} c}}{a} \plus{} \frac {\sqrt {c \plus{} a}}{b} \plus{} \frac {\sqrt {a \plus{} b}}{c} \ \ge\ 3\sqrt {2}\sqrt {\frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}}
\]
[/quote]
By AM-GM inequality,we have:
$ LHS \ge 3 \sqrt [6]{\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a^2b^2c^2}}$
So,we only need to prove that: $ (a \plus{} b)(b \plus{} c)(c \plus{} a)(ab \plus{} bc \plus{} ca)^3 \ge 8a^2b^2c^2(a \plus{} b \plus{} c)^3$
We have:
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge \frac {8 : lol: }{9}(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$
$ (ab \plus{} bc \plus{} ca)^4 \ge 9a^2b^2c^2(a \plus{} b \plus{} c)^2$
then,we have Q.E.D :)[/quote]

Your solution is very nice;quykhtn-qa1  :lol:


When two consecutive whole numbers are randomly selected, what is the probability that one of them is a multiple of 4? Express your answer as a common fraction.

The mod residue combinations can be:

0, 1
1, 2
2, 3
3, 0

So the probability is simply $ \boxed{\frac{1}{2}}$.


logarithms
limit
calculus
calculus computations

[quote="Kent Merryfield"]If $f(x)=\ln x,$ then $f'(0)=1.$

That means that $\lim_{h\to 0}\frac{\ln(1+h)}h=1.$[/quote]

Thanks a lot for the answer   :)

I have some general questions about equivalences:
a) what's the practical use of them?
b) how does one proceed to find an equivalence (we're supposed to use something called "asymptotic development")?
c) could you, please, illustrate with the help of this:
${(\frac{1}{2}+\frac{1}{n})^{n}}$ ?


linear algebra
matrix
probability and stats

Hi!

I am working with a matrix whose entries each converge to a number. It is difficult for me to say why they converge, because the convergence of each entry of the matrix depends on [b]every [/b]entry of the matrix-the convergence involves a lot of variables. I don't know where to start with proving the convergence. Can someone tell me what to start reading so I can find out what I need to learn, or do you need more information to answer my question? I will be happy to post the details!

Los

More specificity would help- I have no idea what sort of problem you're talking about given what you just wrote.


real analysis
calculus
derivative
limit
function
algebra
domain

I wasn't really sure where to post this so I put it here. I want to study real and complex analysis on my own and was wondering if anyone could recommend some good books (English only). Thanks

Rudin's analysis book?
Serge Lang's complex analysis? (I wonder if it is difficult for a beginner)

Does Rudin's Real and Complex Analysis book cover both real and complex analysis satisfactorily?

You're not close to being ready for adult Rudin (its essentially a graduate level text).  Try baby (Principles of Mathematical Analysis) Rudin first.

If anything, I might even go for a more basic analysis book if you have absolutely no background.  Steven Lay's book is not too bad [url=http://www.amazon.com/exec/obidos/tg/detail/-/0131481010/ref=pd_sbs_b_3/103-3605750-2021466?%5Fencoding=UTF8&v=glance]link to amazon[/url]

although I'm definitely biased towards Rudin.  But it requires a lot more mathematical ability, and if you can't solve the problems, there's not much point in buying it.

It also contains functional and harmonic analysis, even a little operator theory.
I didn't read the whole book so I do not know if it introduces some basic concepts, such as residues.
Maybe it would be a bit difficult for self-studying.
For complex analysis, you may take Ahlfor's.

[quote="blahblahblah"]You're not close to being ready for adult Rudin (its essentially a graduate level text).  Try baby (Principles of Mathematical Analysis) Rudin first.

If anything, I might even go for a more basic analysis book if you have absolutely no background.  Steven Lay's book is not too bad [url=http://www.amazon.com/exec/obidos/tg/detail/-/0131481010/ref=pd_sbs_b_3/103-3605750-2021466?%5Fencoding=UTF8&v=glance]link to amazon here[/url]

although I'm definitely biased towards Rudin.  But it requires a lot more mathematical ability, and if you can't solve the problems, there's not much point in buying it.[/quote]

Can you give me a problem (not from analysis) that requires the same level of insight as a problem from the graduate level Rudin, so I have some idea of what it's like?  Or when you say mathematical ability do you mean experience?  I read an excerpt and it was easy to understand, but I'm sure it gets much harder.  Another question, what prerequisites are necessary, I've taken Multivariate calc and Differential Equations, is this enough?

Thanks.

Jimmy,

Unfortunately, having multivariable calculus and differential equations under your belt probably won't help you much in analysis.  As for judging your level, I'll give you a few relatively basic problems from baby Rudin and you can see if you can understand/solve them (they're on differentiation)

Easy (almost trivial):

Suppose that $f'(x)>0$ on $(a,b)$.  Prove that $f$ is strictly increasing on $(a,b)$ and let $g$ be its inverse function.  Prove that $g$ is differentiable, and that $g'(f(x))=1/f'(x)$

Suppose $g$ is real on $\mathbb{R}$ with bounded derivative.  Fix $\epsilon>0$ and define $f(x)=x+\epsilon g(x)$.  Prove that $f$ is injective (one-to-one) if $\epsilon$ is small enough.

If $C_i, 1\leq i\leq n$ are real, and $\sum \frac {C_i}{i}=0$, prove that $\sum C_ix^i$ has a real root between $0$ and $1$.

Harder:

Suppose that $f$ is twice differentiable on $[a,b]$ and $\epsilon>0$.  Prove that there exists $\delta>0$ such that

$\left\vert \frac {f(t)-f(x)}{t-x}-f'(x)\right\vert<\epsilon$ whenever $0<\vert t-x\vert<\delta$, and $x,t\in [a,b]$

Suppose that $a$ is real, $f$ is twice differentiable on $(a,\infty)$, and $M_0,M_1,M_2$ are the least upper bounds of $\vert f\vert, \vert f'\vert$, and $\vert f''\vert$ on the same interval.

Prove that $M^2_1\leq 4M_0M_2$

Hope this gives you some idea.

[quote="blahblahblah"]I would guess (just looking at tables of contents) that adult Rudin does a decent job at least of covering complex analysis.  I like Ahlfors for complex analysis, but a) it's really pricey, and b) it's also a graduate level text.  If he were buying Rudin (either one) + Ahlfors in hardcover new, it would be around 270 USD.[/quote]

Wow, that is expensive, lucky for me my birthday is in a few weeks :lol:

The first two problems are pretty simple, (I can't read the third), but I think I will start with "baby" Rudin, so I fully understand everything.  Thanks for your help.

I fixed the tex (had to run off for dinner earlier).  Why don't you post proofs for the first few and I can tell you if they're right (or 'right enough')

For the first one, since $f(x)$ is increasing and only increasing on the interval then its inverse must be "smooth" and thus differentiable. (I know this isn't formal justification)
By definition:
$g(f(x))=x$
$f'(x)g'(f(x))=1$  (Chain Rule)
$g'(f(x))=\frac{1}{f'(x)}$

For the second one, I think it's saying $\lim_{\epsilon \to 0} f(x) = x$

I'll look at the others later.

Thanks again.

But why is $f$ increasing?

[hide]
for any $b>a$, $f(b)-f(a)=(b-a)f'(x)>0$ for $x\in (a,b)$ by the mean value theorem
[/hide]

Next we need to show that $f^{-1}$ is continuous, but since you don't know a formal definition of continuity, we'll omit this part and assume continuity of the inverse.  But you can't just differentiate both sides - we don't know that $g$ is differentiable, so

[hide]
$\lim_{t\to x} \frac {g(f(t))-g(f(x))}{f(t)-f(x)}=\lim_{t\to x} \frac {t-x}{f(t)-f(x)}=\frac {1}{f'(x)}$
[/hide]

For the second one, if $f$ is a differentiable function which has $f(a)=f(b)$ for $a\not=b$, then we need $f'$ to vanish somewhere on $(a,b)$ (this is Rolle's Theorem)

So with $\vert g'\vert<M$, taking $\epsilon>0$ small enough that $M\epsilon<1$, we get

[hide]$\vert f'(x)\vert =\vert 1+\epsilon g'(x)\vert\geq 1-\vert \epsilon g'(x)\vert>1-M\epsilon>0$ so that the derivative cannot vanish[/hide]

As you can see, analysis is pretty subtle.

Yes, you're right analysis is very subtle.
But, by justifying each step, you know that you fully understand what you're saying.

If I remember from calc, the formal definition of continuity at a point $c$ is that the limit of the function as $x \to c$ exists and equals $f(c)$.  I aslo remember Rolle's Theorem from calc, but at the time just dismissed it as common sense.

Well thanks for everything.

The formal definition of continuity at $x$ (in $\mathbb{R}$) which I'm referring to is that for any $\epsilon>0$, there exists $\delta>0$ such that $\vert x-y\vert<\delta$ implies $\vert f(x)-f(y)\vert<\epsilon$

Your definition of continuity is fine, Jimmy- as long as you know the formal definition of limit. There are some other definitions that come up as well, especially when a function is continuous on all of its domain; once you prove that the definitions are equivalent, you can use whatever's most convenient.

Oh- the wide pages are coming from the text in those links to the book.
A shorter alternate text for the links would be useful.

Yes, that was my fault.  How do you put the little tags in URLs to shorten them?

Here's one way to do it:

[url=http://whatever]link text[/url]

I need to brush up on that, I only vaguely remember the epsilon delta arguments.

Hey guys whats wrong with Apostol.Its my personal favourite after Rudin(baby).I feel its a great book...


geometry
analytic geometry
graphing lines
slope

I finally found a good Geometry problem so here goes:

Point B is on (-4, 0). Point A is on y=(1/2)x:^2: and AB is perpendicular to the x axis. Point P is on y=(1/2)x:^2: also. If the area of the triangle ABP is 24 miles:^2:, then what is the equation for the line AP?

[hide]ok, so A is on the line x=-4 so it is the point (-4,8). THE area of ABP is AB*hieght*1/2=24  48/AB=hieght  hieght=6   P is on the line x=2. so it is the point (2,2). plug in slope, we get equation for AP is 2/6(x+4)=y or 1/3x+4/3=y[/hide]

Mmm, you're correct till the very last part...and it's point A you need to figure out since the problem already gives you the coordinates of point B.

Tare, check over the wording of the problem -- it looks like you translated it quickly or something.  Where did miles come from?

Well originally it said "centimeters" but the unit doesn't matter anyways since you're trying to figure out an equation.

Well it does sorta matter if you're not told what scale the axes and stuff are to, doesn't it?


quadratics
algebra
polynomial
product of roots

I found this problem quite an interesting application of a certain theorym:

Solve the equations:
$x-y-xy=11$
$x^2{y}-y^2{x}+30=0$

Enjoy :)  :D  :lol: !!!

[hide="hint"]let $a=x-y$ and $b=xy$.[/hide]

Alright, here is my solution:

[hide]Solution consists of 4 ordered pairs: $(3, -2) , (2, -3) , (5, -1) , (1, -5)$

Method:
As FMako said, substitute a for x-y and b for xy, so $a-b=11$ and $ab+30 = 0$
$a(-11+a) + 30 = 0$
$a^2 - 11a + 30 = 0$
$a = 5, 6 = x-y$
$5b = -30$ so $b = -6 = xy$
$6b = -30$ so $b = -5 = xy$
$(5+y)y = -6 = y^2 + 5y + 6 = (y+2)(y+3)$, so $y = -2, -3$, substituting gives $x = 3, 2$
$(6+y)y = -5 = y^2 + 6y + 5 = (y+1)(y+5)$, so $y = -1, -5$, substituting gives $x = 5, 1$
Thus you obtain the ordered pairs $(3, -2) , (2, -3) , (5, -1) , (1, -5)$ as the solution
[/hide]

btw, what theorem were you eluding at?

So this is what I did.
[hide]
Incidentally, I chose $a$ and $b$ for $x-y$ and $xy$ as well.

So there's the equations $a-b=11$ and $ab=-30$
Setting $a=11+b$, I substitute into the second equation, getting
$b^2+11b=30$, and factoring, obtain that $b=-5$ and$-6$

From this, we see then that when $b=-5, a=6$, and when $b=-6, a=5$

So return to the definitions of $a$ and $b$, and we have
for the pair $(a, b)=(6, -5)$, we have
$x-y=6$ and  $xy=-5$.  So substituting that $y=-5/x$, we plug in to the former equation and obtain a quadratic, with $(5, -1)$ and $(1, -5)$ as answers.
Repeat that for the other values of $(a, b)$ and we obtain the ordered pairs $(2, -3)$ and $(3, -2)$
[/hide]
edit- oh man, this was my first post with LaTex.  so slow. darn.

By the way, what was the theorem you were speaking of? I'm interested.

I solved it the same way as scorpius119  :)


For USAMO preparation, I'm thinking about working on either Mathematical Olympiad Challenges or Problem-Solving Strategies. Because I haven't had much time to prepare, I am looking towards making Red MOP. I know PSS is a much higher level book, but it would take a long time to finish it, and the other one is relatively shorter. Which book is better?

It seems you want a short-run book, and for that, mathematical olympiad challenges fits the bill.(ACoPS does nicely as well) PSS takes a good 200+ hours (maybe?) of working at to truly reap its benefits.


calculus
integration
probability
geometry
trigonometry
calculus computations

f(x,y)= 3(y^2)/ 2    if 0<x<2  , 0<y<1   and 0 else.

Finding the the Probability that (X+Y>2)

My question which I think is more a calculus question than a probability one which is why it's here:

I drew a nice little picture defining the area on the plane I'm concerned with. It's bounded by the line x+y=2, y=1 and x=2.

So the next thing I try to do is to perform the double integral first with respect to x and then with respect to y

My limits for integration with respect to x are 2 and 2-x and with respect to y are 1 and 0. The answer I got was 3/8.

The solution page in my book is integrating from 2 to 1-x which confuses me and gets the answer 7/8.

Am I doing something wrong?

[quote="Mally Thaznian"]My limits for integration with respect to x are 2 and 2-x [/quote]
That can't be right. $ x$ cannot range from $ 2$ to $ 2\minus{}x.$ A variable cannot have limits of integration which include itself.

If a double integral over a non-rectangular region, the outer integral should have constant limits of integration, while the inner integral has limits of integration which depend on the outer variable - but not the inner variable.

My first hint is to draw the picture of the region you're integrating over. I think you've already done that. It's a triangle. The right hand edge is given by $ x\equal{}2,$ the upper edge by $ y\equal{}1,$ and the lower left edge by an equation which we will write as whichever of $ y\equal{}2\minus{}x$ or $ x\equal{}2\minus{}y$ suits our purposes.

Now, remembering what I said about inner an outer variables, we have two ways of writing our integral:

$ P(X\plus{}Y>2)\equal{}\int_1^2\int_{2\minus{}x}^1\frac32y^2\,dy\,dx\equal{}\int_0^1\int_{2\minus{}y}^2\frac32y^2\,dx\,dy$

Try both of those - they should give you the same answer either way.

Here's a question from a calculus test I gave last spring. It seems pertinent:

1. Some of these expressions are correctly written, meaningful expressions that you could evaluate (assuming you were told what $ f, g,$ [i]etc.[/i] were). The others are meaningless; incorrect as written. Circle the ones that are [i]correct[/i].

(a) $ \int_0^xf(x)\,dx$

(b) $ \int_0^xf(t)\,dt$

(c) $ \int_0^{\pi}\int_0^{r\cos\theta}u(r,\theta)\,dr\,d\theta$

(d) $ \int_0^1\int_y^1\int_{x\plus{}y}^2g(x,y,z)\,dz\,dx\,dy$

(e) $ \int_0^{10}\int_{\minus{}y}^y\int_{\minus{}\sqrt{z^2\minus{}x^2}}^{\sqrt{z^2\minus{}x^2}}w(x,y,z)\,dy\,dx\,dz$


Every integer is painted in black or white. Prove that for at least one of the colors, it is possible to represent every even integer in form of $ a\plus{}b\minus{}c\minus{}d$, where $ a,b,c,d$ have this color.

A nice problem :gleam: . Could you tell us what the source is? Do you require $ a,b,c,d$ to be distinct? :ddr:

[quote="Akashnil"] Do you require $ a,b,c,d$ to be distinct? [/quote]

No.

It's from a Ukranian contest.  :)

Let the set containing the White integers be $ W$,the set containing the Black integers be $ B$.

The converse statement is:

a)More Mathematical:$ \exists$ $ B$ and $ W$ and $ \exists$ integers $ m$ and $ n$ such that $ \not \exists$ $ a,b,c,d\in W$ and $ \not \exists$ $ e,f,g,h\in B$ such that $ a\plus{}b\minus{}c\minus{}d\equal{}2m$ or $ e\plus{}f\minus{}g\minus{}h\equal{}2n$.

b)Less mathematical: There exists $ B$, $ W$, $ m$, $ n$ such that $ 2m$ can't be expressed as $ a\plus{}b\minus{}c\minus{}d$ and $ 2n$ can't be expressed as $ e\plus{}f\minus{}g\minus{}h$ where $ a,b,c,d\in W$ and $ e,f,g,h\in B$.

Let us assume it is true.

Let $ t\in W$
$ (t\plus{}2m) \not \in W$
[If $ (t\plus{}2m)\in W$, let $ a\equal{}t\plus{}2m,b\equal{}c\equal{}d\equal{}t$... we get contradiction.]
$ (t\plus{}2m)\in B$
$ (t\plus{}2m\plus{}n) \not \in B$
[If $ (t\plus{}2m\plus{}n)\in B$, let $ e\equal{}f\equal{}(t\plus{}2m\plus{}n),g\equal{}h\equal{}(t\plus{}2m)$... we get contradiction.]
$ (t\plus{}2m\plus{}n) \in W$
$ (t\plus{}m\plus{}n) \not \in W$
[If $ (t\plus{}m\plus{}n)\in W$,let $ a\equal{}b\equal{}(t\plus{}2m\plus{}n),c\equal{}d\equal{}(t\plus{}m\plus{}n)$... we get contradiction.]
$ (t\plus{}m\plus{}n)\in B$
$ (t\plus{}m)\in W$
[Similarly]
$ t\in B$

Contradiction. Thus our assumption was wrong and we are done.

Please tell me if it is correct or not, if you like my solution or not. :)

It looks good to me, with the following (minor) complaints:
1) That's not the converse, it's the negation.
2) The statement [quote]$ \exists$ $ B$ and $ W$ and $ \exists$ integers $ m$ and $ n$ such that $ \not \exists$ $ a,b,c,d\in W$ and $ \not \exists$ $ e,f,g,h\in B$ such that $ a \plus{} b \minus{} c \minus{} d \equal{} 2m$ or $ e \plus{} f \minus{} g \minus{} h \equal{} 2n$.[/quote] is not in any meaningful sense "more mathematical" than [quote]There exists $ B$, $ W$, $ m$, $ n$ such that $ 2m$ can't be expressed as $ a \plus{} b \minus{} c \minus{} d$ and $ 2n$ can't be expressed as $ e \plus{} f \minus{} g \minus{} h$ where $ a,b,c,d\in W$ and $ e,f,g,h\in B$.[/quote]  Use of the symbol $ \exists$ does not make a sentence more rigorous (nor, in general, does it increase ease of reading or clarity).

Thanks. :)  I really need that kind of advice because I hope to get through INMO next year and I will have to write in English. This year, I thought I had cracked all the 6 problems but it turned out that I wasn't in the list. :(  I really suck. May be my solutions weren't rigorous enough.


Calculate
\[ \mathop {\lim }\limits_{\begin{subarray}{l}
  x,y > 0 \\
  x \plus{} y \to  \plus{} \infty
\end{subarray} } \left( {\left( {x^2  \plus{} y^3 } \right) \minus{} \left( {x \plus{} y} \right)} \right)
\]
and
\[ \mathop {\lim }\limits_{\begin{subarray}{l}
  x,y > 0 \\
  x \plus{} y \to  \plus{} \infty
\end{subarray} } \frac{{x^2  \plus{} y^3 }}
{{x \plus{} y}}.
\]

it seems like the limit must be infinity...

if the limits exist at all, then they should exist for all (x+y) that approach infinity, including the case where you fix x or y...

so we should have

$ \lim_{x \plus{} y \to \plus{} \infty} \left( (x^2 \plus{} y^3 ) \minus{} (x \plus{} y) \right) \equal{} \lim_{y \to \infty} \left( c(c\minus{}1) \plus{}y(y^2\minus{}1) \right)\equal{}$
$ \lim_{x \to \infty} \left( d(d^2\minus{}1) \plus{}x(x\minus{}1) \right) \equal{} \infty$

?