geometry
logarithms
analytic geometry
combinatorics unsolved
combinatorics

Lets try this one:

Given n points in the plane , what is the maximum number of times a given area determined by k for k <= n can occur?

According to R. Guy, 'Unsolved problems in Geometry', Springer, pb. F7, we have the following, which deals only with triangles :

Let $T(n)$ be the maximum number of triangles of a given area that can occur with vertices among a set of $n$ points in the plane. Erdös (what a surprise!) showed that :
\[c_1n^2 \ln (\ln (n)) < T(n) < c_2n^{\frac 5 2}\]
for some positive constants $c_1,c_2$.

The lower bound comes from the set of the points with integer coordinates $(r,s)$ such that $1 \leq r \leq a = [ \sqrt { \ln (n) } ]$ and $1 \leq s \leq \frac n a $ (and with the area $\frac {a!} 2$ ).

The minimum number of different areas of triangles that can occur with vertices among $n$ points in the plane, not all on a straight line is $mn)$ with :
\[0.32n-1 \leq m(n) \leq [ \frac {n-1} 2 ],\]
where the lower bound is due to Burton and Purdy, whereas the upper bound is due to Erdös, Purdy and Straus. They conjecture that the upper bound is the correct answer, at least for sufficientely large $n$.

I don't know if these results have been improved.

Pierre.

In the case where the points are supposed to be the vertices of a convex $n$-gon, you may also see :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=maximum+polygon+area&t=5891

Pierre.


In how many distinct arrangements of the letters of APPLY does
the letter A occupy the middle position?

A has 1 choice.
Then there are 4! permutations.

However we double-counted the two "P"'s. So divide by 2.
4!/2=12.


Let $n$ is a positive integer which is not a power of any prime number. Show that we can always find a permutation $(a_1, a_2, .... , a_n)$ of $(1, 2, .... , n)$ such that the following equality holds :
          $\sum a_i.cos(\frac{2\pi i}{n})=0$ , here $i$ runs from $1$ to $n$

Dear [b]treegoner[/b],
This source is not unknown,this problem is from Sweden 1983.I had a long solution.And you?

It's also from some ISL, it's been posted before by sam-n (but wasn't solved, I think), and the solution can be found on Kalva.

I don't think the solution is very long, is it ? ;) This problem is quite a tricky one as it is difficult to build the permutation.
    I have a question, is there any permutation when $n= p^k$ ?

You can check the Kalva solution for an answer to that (partial answer, at least :)).


calculus
integration
function
complex analysis
complex analysis unsolved

prove that a fn which is harmonic and bounded in the upper half plane , cont on the real axis can be represented as a poisson integral.

Subtract the Poisson integral from the given harmonic function: the difference (call it $ v(x,y)$) is harmonic, bounded, and vanishes on the boundary. You can use the Phragmen-Lindelof principle to conclude that $ v\equiv 0$. Or just apply the maximum principle to $ v(x,y)\minus{}\epsilon y$ with $ \epsilon>0$: it implies $ v(x,y)\le \epsilon y$.


calculus
integration
linear algebra
matrix
topology
real analysis
real analysis unsolved

The Itzykson-Zuber integral is well known: Given two $ N\times N$ diagonal matrixes $ A$ and $ B$ with diagonal elements $ \{a_i\}$ and $ \{b_i\}$, and an $ N\times N$ random unitary matrix $ U$ that is uniformly distributed over the Steifel Manifold (often referred to as isotropic or Haar matrix). Then we have
\[ \int_{U} \exp(tr (UAU^{\ast}B)) dU \propto \frac{ \det E(A,B)}{\det V(A) \det V(B)},\]
where $ V( )$ is the Vandermonde matrix and the $ ij$th element of $ E(A,B)$ equals $ \exp(a_ib_j)$.

Now relax the conditions to an $ M\times N$, where $ M>N$ matrix U such that $ U^{\ast}U=I_N$, an $ N\times N$ diagonal matrix $ A$ and an $ M\times M$ matrix $ B$.
Is there a simple formula for
\[ \int_{U} \exp(tr UAU^{\ast}B) dU.\]

Also, only the first $ N$ diagonal elements of $ B$ are non-zero

hmmm...should be possible by taking the limit somehow...?


A passanger walk for 3.5 hours, covering exactly 5 km in any period of one hour. Is it possible that his average speed during his walk exceeded 5 km/h?

[quote="towersfreak2006"]Average speed for any interval?

He can alternate "walking" 5 km (at a uniform rate) in 30 minutes and resting for 30 minutes. That gives him 10 km/hour the first 30 minutes.[/quote]

Yes, this question can very well be manipulated like that. But if you mean $without$ stopping and at a constant speed, then no

[quote="towersfreak2006"]Average speed for any interval?

He can alternate "walking" 5 km (at a uniform rate) in 30 minutes and resting for 30 minutes. That gives him 10 km/hour the first 30 minutes.[/quote]

Even with resting, I think.  His average pace here is 5 km/h not above it (even with the resting).

Well the question didn't specify what time interval to work with - i.e. average time in 1 hour, 30 minutes, any time. So depending how the question was meant to be interpreted, the answer could be yes or no. Could the poster of this topic please clarify? Thanks!

It's meant to be read as average speed over the whole walk.  It's pretty clearly worded, actually.

Here's a hint:

[hide]Let $v=5+\sin(2\pi t)$.  [/hide]


Let $ABC$ be a triangle, $AD$ the altitude from $A$ and $AM$ the median from $A$.
If $\angle{BAD}=\angle{MAC}$ then prove that $ABC$ is a right triangle.

 :)

The statement is wrong.  For example, take an equilateral triangle.  It satisfies, but it doesn't work.

What you mean is $\angle BAD = \angle DAM = \angle MAC$.

--

Solution:  let $\angle BAD = x$.

Put BM = y, AB = AM = c.

Sine law:

In ABM:
c/sin(90-x) = y / sin 2x

In AMC:
c / sin(90-2x) = y / sin x

So $\sin 2x / \sin (90 - x) = \sin x / \sin (90 - 2x)$

It implies $\sin 2x \cos 2x = \sin x \cos x$

or $\sin 4x = \sin 2x$.

We have $\sin(3x+x) = \sin(3x -x)$ so that $2\cos 3x \sin x = 0$.

Restrict $x$ to $[0, 60]$ (A < 180).  Then only x = 30 satisfies.  This imples A = 90 as desired.

No.  For example, consider  a triangle AB = 3, AC = 4, BC = 5.

Then BD = \sqrt{3^2 - (12/5)^2}, while BM = 5/2.

But these quantities are not equal, and thus $DM \ne 0$.

Whoops I was thinking in a isosceles right triangle.  :rotfl:

My solution in the problem which singular correct it.
$AMB$ triangle is isosceles.
Additionally $MD=DB=\frac{1}{2} MC$
From this at the right angled $ACD$ angle $C=30$ and angle $CAD=60$ so angle $A=90$ angle $B=60$ angle $C=30$

It kinda looks like the TST problem of Hong Kong we discussed several days ago.


function
analytic geometry
vector
trigonometry
Euler
geometry
complex numbers

Could some1 help me understand the relationship between the 2?

Additionally, what type of questions could i form from the union of these 2 concepts?

A way to look at something in polar coordinates is as a vector in component form, $( \cos \theta \ , \ \sin \theta )$.
Getting the imaginary form $r \cos \theta+r i \sin \theta$, where $r$ and $\theta$ are from the polar coordinate $(r \ , \ \theta)$.
From Euler then that equals $r e^{i \theta}$.
To place that into the $a+b i$ form then just multiply the trigonometric functions and $r$ together.

Possible questions, are converting from the two coordinate systems or performing addition, mutiplication, or taking roots of a few of them.

Those questions you suggested, haven't they been done before?  i need somthing original....

could some1 help me think of an unique investigation in this area T_T

Ummm. Nearly everything in math has been done before. Unless you're working on an unsolved problem, you're pretty much guaranteed to be using the ideas of some famous mathematicians.

What exactly do you mean by "original"? Is this a school project or something?

Well, polar form of complex numbers is very helpful for easily deriving trigonometric identities that are much more tedious to derive with synthetic geometry. You could consider trying to derive various trig identities on your own.

i know that nearly everything has been done.   When i mean original, i mean like applying a different twist to it.

For example the "nim " game is players take 1 turn per time.  A twist would be to give them 2 turns per time.  ( this is just a ex)

An complex # takes form of z= x+iy where x is Re and iy is Im.

This can be written in the form (cos theta + i sin theta) = z = 1 as polar coordinates.

my problem is how would you write z = 4+5i in polar coordinates.

You're missing an important part.

In polar, you have an "angle" (argument) and a radius. So $(r, \theta)$ would imply a distance of $r$ from the "origin" (the value 0) and an angle of $\theta$ against the real axis. Thus, the number given by $(r, \theta)$ is equivalent to $r \cos \theta+ir \sin \theta=r(\cos \theta+i \sin \theta)$

Now, if you have $4+5i$, you can use the distance formula to get $r$: $r=\sqrt{4^{2}+5^{2}}=\sqrt{41}$

Since $\tan \theta$ is gonna equal $\frac{\text{opposite}}{\text{base}}=\frac{\text{imaginary part}}{\text{real part}}=\frac{5}{4}$, you can say $\theta = \tan^{-1}\frac{5}{4}$.

$(r, \theta)= \left( \sqrt{41},\tan^{-1}\frac{5}{4}\right)$

Ah thank you zuton.

Slightly related question,

Euler's formula, cos theta + i sin theta = e^ ( i theta).  What are the applications of this in polar functions?

I am trying to devolop a research focus/problem centered around the ocncept of polar functions and imaginery #s and Euler's formula seems to be a link between the two.  Any suggestions?

thanks

Sorry, but you're going to have a lot of trouble developing original research based on Euler's formula.  The underlying mathematics have been very well studied, and to do any original research in the area would likely require a graduate degree in complex analysis.

It doesn't have to be original... I just don't want to be regurgitating a textbook.

There's not much to say about Euler's formula that isn't already in a textbook; it won't be a high-school textbook, but it will still be a textbook.  If regurgitating a college textbook in disguise is okay...

Otherwise, game theory itself might be a better place to make some "twists."

Since I seem to be @ a deadend, what is an aspect of polar functions that i could concentrate on that would result in a [b]good[/b] 1000 word + project...


a,b,c are positive a+b+c=1
prove that
2/(ab+bc+ca)>=(1+a)/(1-a)+(1+b)/(1-b)+(1+c)/(1-c)
stronger than 2003JAPANESE

Clear denominators.  It suffices to show that

$\sum ab\geq 2\sum \frac {abc}{a+b}$

which is equivalent to

$\sum \frac {1}{a}\geq 2\sum \frac {1}{a+b}$

and since

$\frac {1}{a}+\frac {1}{b}\geq \frac {4}{a+b}$

by Jensen's inequality

the result follows.

i have an idea, and i hope it works :(
put $ a = \frac 1x , b \frac 1y , c = \frac 1z $
the inequality is equivalent to
$ \sum \frac {x+1}{x-1} \leq \frac {2xyz}{x+y+z} $ with $xy+yz+xz = xyz $
since $ \frac {x+1}{x-1} = 1 + \frac {2}{x-1} $
we have $ \sum \frac {x+1}{x-1} = 3 + 2 \sum \frac {1}{x-1} $
hence $ \sum \frac {1}{x-1} = \frac {3+xyz-2(x+y+z)}{x+y+z-1} $
put $ u = x+y+z $ and $ n = xyz $
the inequality is equivalent to
$ 3 + 2(\frac {3+n-2u}{u-1}) \leq \frac {2n}{u}$
which is equivalent to
$\frac {3+n-2u}{u-1} \leq \frac nu - \frac 32 $
which is equivalent to
$ 3+n-2u \leq n - \frac nu - \frac 32 u + \frac 32 $
which is equivalent to
$ 3 - 2u^2 \leq -n - 3/2u^2 + 3/2u $ (there is a mistake here, look down)
which is equivalent to
$ n + 3 \leq 1/2u^2 + 3/2u$
applying Mac Laurin inequality we get
$ (\frac u3)^2 \geq \frac n3 $ then $ u^2 \geq 3n $
hence it remains to prove that
$ u \geq 2 $ which is obvious since $ x \geq 1 , y \geq 1, z \geq 1 $
and it's done ( i wish  :P )

:mad:  for an other time , i made computation mistake!!
but when i correct it it leads me to a contradiction !!
ok, we have $ 3 - 2u \leq - \frac nu -3/2u + 3/2$ is equivalent to $ 3u - 2u^2 \leq - n -3/2u^2 + 3/2u $
after applying mac laurin and simplification we get  $ 3/2 \geq 3 $ !!!
can any body help me :?


let ABCD be a paralellogram,(A<90),(C) is a circle with the diameter AC.(C) intersects BC,CD in points E,F.

the tangent to (C) in point A,intersects BD in point P.prove that:E,F,P are colinear.

i think we can solve it with seva's theorem easily. :D
(you are so nice ashegh with your nice problem) :D

yes,yes,...

all the things that u do with ceva and menelaus ,i do with pole and polar... :(

im not satisfied with it.

[u]Lemma.[/u]

$\triangle\ ABC,X\in BC,m(\angle XAB)=x,m(\angle XAC)=y\Longrightarrow$ $\frac{XB}{XC}=\frac{AB\sin x}{AC\sin y}.$

[u]The my solution for the proposed problem.[/u]

I note $m(\angle ACD)=x,\ m(\angle ACB)=y$ and I will apply the Menelaus's

theorem to the transversal $\overline {PDB}$ in the triangle $BCD$. Therefore,

$\frac{PD}{PB}\cdot \frac{EB}{EC}\cdot \frac{FC}{FD}=\frac{AD\cos y}{AB\cos x}\cdot \frac{AB\cos (x+y)}{AC\cos y}\cdot \frac{AC\cos x}{AD\cos (x+y)}=1\Longrightarrow P\in EF.$

thanks for all of ure posts...

really nice solutions ;)


inequalities
real analysis
real analysis unsolved

Let k(n) be the smallest m such that H(m)>=n where H(m) is the harmonic sum(1+1/2+...1/N).Find the limit of k(n+1)/k(n) when n-->infinity.

[I've alrady written a large part of the message and I don't want to modify it now: I took $k(n)$ to be the largest $m$ s.t. $H_m\le n$; I'm sure you don't mind :)]

We write $1+\frac 12+\ldots+\frac 1{k(n)}=\ln(k(n))+c+\epsilon_n$ and $1+\frac 12+\ldots+\frac 1{k(n)+1}=\ln(k(n))+c+\epsilon'_n$, with $\epsilon_n,\epsilon'_n\to 0$. We can now use the inequalities $[k(n+1)+1]e^ce^{\epsilon'_{n+1}}>e^{n+1}\ge ek(n)e^ce^{\epsilon_n}$, so $\frac{k_{n+1}+1}{k_n}>ex_n$ with $x_n\to 1$. We can obtain a similar upper bound, and together these inequalities show that $\frac{k_{n+1}}{k_n}\to e$.

Voila une jolie question: prove that $ k_{n+1}-ek_n$ is bounded.


find the vertex of $y = 3x^2 + x - 2$

hint $y = ax^2 + bx + c$
$y = a (x - h)^2 + k$

[hide]x-coordinate of vertex: -b/2a = -1/6.

Substitute into the original equation: y coordinate is thus $-2 \frac{1}{12}$. [/hide]

[hide] I did it in the same way as 236factorial, and I got the same result too[/hide]


How do you find the equation of the tangent line to $ f(x)\equal{}x^3$ at the point $ x\equal{}2$?

The equation of the tangent line of $ y\equal{}f(x)$ at $ x\equal{}a$ is given by $ \boxed{y\equal{}f'(a)(x\minus{}a)\plus{}f(a)}$.

$ f'(x)\equal{}3x^2\implies f'(2)\equal{}3\cdot 4\equal{}12$.
$ y\equal{}12x$.

We must translate the line to the right until it is directly underneath (or above) the point of tangency, $ x\equal{}2$, so we now have:

$ y\equal{}12(x\minus{}2)$.

We must translate up until it's coinciding with $ f(2)\equal{}8$.

$ y\equal{}12(x\minus{}2)\plus{}8\implies \boxed{y\equal{}12x\minus{}16}$.

By the way, divide $ x^3$ by $ (x\minus{}2)^2$ to find the quotient and remainder.

Ya... $ 3x^2$... but where from there? I seem to be having issues with finding tangent lines.


Which of the numbers $2^{2006!}$ and $2^{2006}!$ is greater ?

Nice problem. First put  $2^{2006}=x$.
Then we have to compare the numbers   $A=x^{2006!}$  and  $B=x!$

Or  $A=\underbrace{x\cdot x.....\cdot x}_{2005!\ times}$

and  $B=\underbrace{1\cdot 2....\cdot x}_{2^{2006}\ factors }$

As $2005!>2^{2006}$   and every time   $x>a$  where  $a=1,....,x-1$  we have that

$A>B$

I hope you understand my solution because my expression is not as good as it should be.sorry.


Compute $E=\sqrt{\binom{n}{1}\cdot\binom{n}{2}\cdot\ ... \cdot\binom{n}{n}}$ :?

uhh wat form does it have to be?

cuz if we note that $\binom{n}{n} = 1$ and that $\binom{n}{1} = \binom{n}{n-1}$, then we can reduce our thing to:

$E = \sqrt{\left(\binom{n}{1}\cdot\binom{n}{2}\ldots\binom{n}{n/2}\right)^2} = \binom{n}{1}\binom{n}{2}\ldots \binom{n}{n/2}$

but yea i ono if u want it even more simplified

True, but what if n is odd?
It may be better just to express it in pi notation with factorials.

oh... this isnt much of a problem then tho if the thing doesnt simplify nicely

cuz then u just have some ugly stuff over some more ugly stuff

wait actually for case of $n$ that is odd then we get a perfect square for the expression... doesnt seem like it follows a pattern or anything though

$E=\sqrt{\binom{n}{1}\cdot\binom{n}{2}\cdot\ ... \cdot\binom{n}{n}}=$
$\sqrt{\binom{n}{1}\cdot\binom{n}{2}\cdot\ ... \cdot\binom{n}{n-1}}=$
$\sqrt{\frac{(n!)^{n-1}}{(1!)^2 (2!)^2 (3!)^2 ... ((n-1)!)^2}}$
$=\frac{1}{1!2!3!...(n-1)!} \sqrt{(n!)^{n-1}}$
$=\frac{(n!)^{\frac{n-1}{2}}}{1!2!3!...(n-1)!}$

That's what I'm getting... donno how to latex pi notation

So yeah, you would only get a perfect square when n is odd, or else the exponent ${\frac{n-1}{2}}$ is not integral

Yeah! That's a good form for me. Anyway, I wrote wrong the problem. The order for radical is $n$ not $2$ :blush:


A square has vertices (1,1),(1,-1),(-1,-1),and (-1,1).Find the probability that a randomly selected point inside the square will have the sum of its coordinates greater than $\frac{1}{2}$. Express answer as a common fraction in lowest term.

[quote="Treething"][hide="answer"]
9/32
[/hide][/quote]

yea that's correct
do you have a solution?

[hide]
Since the sum of the coordinates $> \frac{1}{2}$,
$x+y>\frac{1}{2}$
$y>-x+\frac{1}{2}$
If you draw the figure, you region you want is a isosceles right triangle with sides $1 \frac{1}{2}$ (area=$\frac{9}{8}$.

The area of the square is $2\cdot 2=4$.

So is the probability $\frac{\frac{9}{8}}{4}=\boxed{\frac{9}{32}}$?
[/hide]


LaTeX
function
calculus
derivative
real analysis
real analysis unsolved

OK:
let $f(x)$ and $g(x)$ be functions of $x$ defined for $x\geq 0$.
Let $f(x)=-dh/dx$ for the function $h=h(x)$.
Now look at the equation $f^{2}(x)+Ch(x)=g^{2}(x)$, for some constant $C$. Eliminate, if possible, dependancy on $h(x)$ in this equation, so that it expresses a relation between $f$ and $g$ only, the $h$ not an unknown.

I think it's just calculate implicit derivative:
$\frac{d}{dx}\left( f^{2}+Ch \right) = \frac{d}{dx}\left( g^{2}\right)$
$2f\frac{df}{dx}+C\frac{dh}{dx}= 2g\frac{dg}{dx}$
$2f\frac{df}{dx}-Cf = 2g\frac{dg}{dx}$
$f\left(f\frac{df}{dx}-C\right) = 2g\frac{dg}{dx}$

The last line should be $f(2 \frac{df}{dx}-C)=2g \frac{dg}{dx}$


Find all functions $f: \Re^+\rightarrow\Re^+$ satisfying for all $x,y>0$

$x^2(f(x)+f(y))=(x+y)f(f(x)y)+1$

hi,
here is my solution :
$x= 0 \Rightarrow  yf(f(0)y) +1 = 0$

1)$f(0) = 0$
$\Rightarrow 0+1 != 0$[size=200][color=red]which is impossible[/size][/color]

2)lets assume now that $f(0) = \alpha$and$f(0)y = x$
$\Rightarrow  y f(x) = -1$
$\Rightarrow  f(x) = -\frac{1}{y}(1)$
$\Rightarrow  \alpha y = x \Rightarrow  \frac{1}{y} = \frac{\alpha}{x}(2)$
$(1),(2) \Rightarrow  f(x) = \frac{\alpha}{x}$and$\alpha != 0$
 :D .

But $x=0$ is not allowed!!!
$x>0$ & $y>0$

For $x=y$ we have:
$xf(x) = f(f(x)x) +1$ and:
$f(x) = a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_{0}$
Degree of $f(x)$ ezual $n$. Degree of $xf(x)$ equal $n+1$ and degree of $f(f(x)x) +1$ equal $n^2+n$. So...:
$n^2 + n = n+1$ => $n=1$ So:
$f(x) = ax+b$ and we have equal: $xf(x) = f(f(x)x) +1$. So we get function:
$f(x) = x-1$
but if we come to equal:
$x^2(f(x)+f(y)) = (x+y)f(f(x)y) +1$ this function is false. So, what's wrong?

We can still use the idea: If $x=y$
$\Rightarrow xf(x) = f( xf(x) )+1$
Denote: $\Rightarrow g(x) = xf(x), g(x) = f(g(x))+1$
Since we also have (by the definition of $g(x)$) $f(g(x)) = \frac{g(g(x))}{g(x)}$
$\Rightarrow g(x) = \frac{g(g(x))}{g(x)} + 1$
$\Rightarrow g(x)^2 = g(g(x)) + g(x)$
$\Rightarrow g(g(x)) = g(x)(g(x)-1)$
$\Rightarrow g(k) = k(k-1)$ where $k = g(x)$
For the rest, I dunno...

For $x=y$ we have $2x^2f(x)=2xf(xf(x))+1$

Very nice problem.
All what i have found its such function dont exist if $f(R^+)=R^+$
If that then just take $x$ such that $f(x)=1$
Can someone give a hint for not surjection function ie if $f(R^+)<>R^+$ :?

I have probably made a mistake while rewriting it here from a paper.. so I suppose there is a mistake.. :maybe:

I think maybe it is $x^{2}(f(x)+f(y))=(x+y)f(f(x)y)$
And this one is not to solve.
Let $x=y$
then $f(f(x)*x)=f(x)*x$
Let $f(x)*x=g(x)$
then Let $y=1,x=g(t)$
then we get that $g(t)^{2}+(f(1)-1)g(t)-1=0$
Notice that $g(t)>0$
then $g(t)=Const$
So $f(x)=\frac{g(x)}{x}=\frac{C}{x}$
And we easily get that $C=1$


Hey. Well, I am trying, but there are 2 problems: 1.) I know barely anything about fractals. 2.) I don't know VB.

Is there a certain part of the VB program that differs greatly from the gw-basic one? If so, tell me. Effort is a constant and as the number of lines of code increases, so does effort per unit.

Also, don't forget to CHECK your units. I know VB does some strange things with all of the different types (twips and all of that.)

i found the mistake. in gw basic black cricles don't apppear because dos background is black so i needed to make the some of the cirlce's colors (the ones that failed to build the fractal) invisble.

Hmm....I'm not sure why this is in Intemediate.  I'll move it to the proper forum.

your code didn't work
When i ran the program it said syntax error in "730: If A1 geq 0.1 Then Goto..."

[quote="IPERFERPI"]I am trying to program this Flag Chaser program using VB---i have a chaser, 3 flags, 2 speed, and a hunter in the form, now I want to make a control array of two hunters, but i cant aviod to make them go the same direction, can anyone help me? :?[/quote]

What exactly is a flag chaser?  Sorry for my ignorance if it is a well-known programming concept.

If you have actually tried the program using Visual Basic, you'd know what it is, it's just a common game like snake that every one plays ;)


topology
real analysis
real analysis unsolved

Let $ A,B$ nonempty bounded subsets of a metric space $ M$. Prove that:

$ diam(A\cup B)\leq diam(A)\plus{}diam(B)\plus{}d(A,B)$

Let $ \epsilon>0$ be given.
There are $ a\in A$ and $ b \in B$ so that $ d(a,b) \le d(A,B) \plus{} \epsilon$.
Let $ a_1\in A$ and $ b_1 \in B$, then
$ d(a_1,b_1) \leq d(a_1,a) \plus{} d(a,b) \plus{} d(b,b_1) \leq \mbox{ diam }(A) \plus{} d(A,B) \plus{} \epsilon \plus{} \mbox{ diam }(B)$
$ d(a_1,b_1) \leq \mbox{ diam }(A) \plus{} d(A,B) \plus{} \epsilon \plus{} \mbox{ diam }(B)$  for every $ \epsilon$ and trivially also true when $ a_1$ and $ b_1$ are in the same set $ A$ or $ B$.
And so $ diam(A\cup B)\leq diam(A)\plus{}diam(B)\plus{}d(A,B)$


Given that 1$ \%$ of 1$ \%$ of a number is 5, what is the number?


The harmonic mean, $m$, of $a$ and $b$ is given by $\frac{2}{\frac{1}{a} + \frac{1}{b}}$.
What is the harmonic mean of 3 and 6?

[quote="hwenterprise"]How did you do that?![/quote]
[hide]
$\frac{2}{\frac13+\frac16}=\frac{12}{2+1}=4$[/hide]

shobber, why did you do the problem the way you did...how did you interpret it?

I believe shobber multiplied both the numerator and denominator of that fraction by 6.

I have a question relating to harmonic mean:

Say, you drive to work at 60mph, and drive home from work at 40mph, would you find the average speed for the entire trip using the harmonic mean or the arithmetic mean?

I believe its the harmonic mean, but I'm jut amking sure.

[quote="Dcadu9lvr"]Would you find the average speed for the entire trip using the harmonic mean or the arithmetic mean?[/quote]
The harmonic mean does the trick.  Here's another way to write the formula for harmonic mean that helps me remember the formula.

\begin{eqnarray*}
\frac{1}{h}+\frac{1}{h}&=&\frac{1}{a}+\frac{1}{b}\\
\\
\frac{1}{h}+\frac{1}{h}&=&\frac{1}{40}+\frac{1}{60}\\
\\
\frac{2}{h}&=&\frac{60+40}{40\times 60}\\
\frac{1}{h}&=&\frac{1}{2}\times\frac{100}{2400}\\
\frac{1}{h}&=&\frac{1}{48}\\
\\
h&=&48
\end{eqnarray*}


everyone help!!!!
people from alabama are coming back... maybe not but they soon will be.
so even if you are not an alabamian help join the cause of making this forum a  undead and better place to
live *ahem* i mean post

Dont tell people what to do gene, ask them politely.

(If you didnt catch on to that, it was supposed to be ironic that I commanded you not to command people...)

Sigh...

But weird, in my opinion, I don't think AL forum activity matters that much. What matters is that we all try our best and have fun! High five!  :thumbup: (And thumbs up!)

UHHHHHHH.....
im goign to do this FTW! style
/me is getting scarred by dynamo's enthusasim


dar che soorat rabeteye zir dorost ast?
agar $x \to a$ aangah $f(x) \to f(a)$

be nazar mirese age tabe $f(x)$ peyvaste bashe in hokm doroste.


It was a piece of cake for us!! Tomorrow we will upload some photos too!

Incidentally, did you happen to take our teddy too?

I d rather say the moose is [b]snatched[/b] instead of stolen...

i think our technique was amazing! We were quite unprepared but still very organised!! :wink:
Our pin deserves a place on the Canamoose!!! :lol:

Okay, but you guys used brute force.  That's no fun, because then they know you have the moose, and they come after you, rather than just being confused.  Remember, every good problem has an elegant, creative solution.  You could say stealing the moose is IMO #7.

[quote="probability1.01"]We had them really confused for a while on the last day. Alas, elegant approaches do not always work...[/quote]

Yes, but we'd just hit a hitch in our solution.  We did manage to pull it back together and take the moose without force, right?  Without physical force, at least... :D

I am very happy that THE tradition continues...
next year try to steal the German tiger... :roll:


Coloring
number theory
greatest common divisor
relatively prime
combinatorics
IMO
IMO 1985

Let $n$ and $k$ be relatively prime positive integers with $k<n$. Each number in the set $M=\{1,2,3,\ldots,n-1\}$ is colored either blue or white. For each $i$ in $M$, both $i$ and $n-i$ have the same color. For each $i\ne k$ in $M$ both $i$ and $|i-k|$ have the same color. Prove that all numbers in $M$ must have the same color.

We will start from some element $a$ and show that we can equate it with any other element of $M$.

First, we note that if $a \equiv b \mod{k}$, $a$ and $b$ have the same color, so it is only necessary to show that all residues modulo $k$ are the same color.

Also, $a \mod{k}$ and $n-a \mod{k}$ have the same color, as given. (1)

By the second condition, we can also infer that $a \mod{k}$ and $-a \mod{k}$ (2) are the same color because we just take an element $x \equiv a \mod{k}$ such that $x < k$ and we have $x \equiv a \mod{k}$ and $|x-k| = k-x \equiv -x \equiv -a \mod{k}$ the same color.

So we generate residues (all $\mod{k}$) as follows.

If $mn+a$ is blue for some integer $m$, we have $-mn-a$ is blue by (2) and $n-(-mn-a) = (m+1)n+a$ is blue by (1). Hence by induction all $mn+a$ are blue. But noting that $(n,k) = 1$, $mn+a$ clearly takes on all residues modulo $k$ and the proof is complete.

If $a$ is blue and $a>k$ then we see that $a-k$ is also blue. If $a<k$ then $a$ $k-a$ and $n+a-k$ have this same color. We can write that for all $a \not = k$ $a$ and $a-k$ (mod n) have that same color.
All the numbers we will consider $(mod n)$. Let's assume that $-k (mod n)$ is blue. By induction we have that $-lk (mod n)$ is also blue for $l=1...,n-1$. If $-l_{1}k \equiv-l_{2}k (mod n)$ we have that $l_{1}=l_{2}$ and for only l=n-1 $-lk=k (mod n)$. So all the numbers $-lk (mod n)$ are diffrent and there are n-1 of them.

We know that $ ka\bmod n$, $ k$ and $ n$ relatively prime and $ a$ ranging from $ 0$ to $ n \minus{} 1$ constitutes a complete residue system mod $ n$.
Let $ k(n \minus{} 1) \equiv q \bmod n$, $ 1 \leq q \leq n \minus{} 1$ (since q is nonzero).

Let $ q$ be blue. Consider $ q \minus{} ka \bmod n$. For different $ a$, we have different residues $ \mod n$. Also, by the definition of $ q$, $ q \minus{} ka$ is nonzero $ \mod n$ for $ 0 \leq a \leq n \minus{} 2$. Therefore we may produce $ n \minus{} 1$ distinct, nonzero residues $ \mod n$, all the same color, completing the proof.

[quote="limac"]I think I am missing something as this seems too easy.

Suppose that 1 is Blue. Then note that for all $k \in \{ 3,\dots,n-1 \},\text{gcd}(1,k) = 1.$ So, $1$ and $|1-3| = 2$ have same color. In this way we find that all the numbers upto n-2 have the same color as 1. Then by the first condition, 1 and n-1 have the same color, and hence all the elements have the same color.[/quote]

$k$ was one fixed number.

Two words: careful Bezout

Because of the second rule, if $a \equiv b$ (mod $k$) and both are in $M$, then they have the same color ($ b > a \Rightarrow b > k$). Using this then with $a < k$ we see that if $a \equiv -b$ (mod $k$) and both are in $M$, then they have the same color (since $k<n$). Also, since $i$ and $n-i$ have the same color, we see that if $a+b \equiv n$ (mod $k$), and both are in $M$, then they have the same color. Since $k<n$, all congruences modulo $k$ occur in $M$. The congruence of $i$ will have the same color as that of $-i$ which will have the same color as $n+i$. Because $(n,k)=1$ we see that the congruence of $b=a+nk$ (where $b$ is anyone) will have the same color as the congruence of $a$. So $a$ and $b$ will always have the same color if both are in $M$. Done.

Extend the coloring to all integers in a way such that 0 is colored the same way as $k$, and otherwise the coloring of $i+n$ is identical to the coloring of $i$.

The rest of the proof is a careful argument that the coloring has period $k$; let the set of all numbers $i$ such that $i$ and $i-k$ have the same coloring be $S$. Obviously by the given condition, any numbers congruent to $k, k+1, k+2, \cdots, n-1$ mod $n$ are in $S$.

Now, suppose $i$ is congruent to 0 through $k-1$ modulo $n$. Define $a \sim b$ between two positive integers $a, b$ to show that $a, b$ have the same coloring. Then, $$i-k \sim n+i-k = n-(k-i)\sim k-i \sim i$$ by the second condition, so these numbers are in $S$ as well.

This means that the coloring is periodic with periods $n$ and $k$. But $\gcd(n, k) = 1$, so this implies all numbers have the same color, as desired.

To remove the absolute value, extend the coloring by adjoing [i]ad infinitum[/i] the block in brace to the left
$$ \cdots [\bullet, n-1, n-2, \dots, 1, \bullet, 1, \dots, 1, \dots, n-1][\bullet, n-1, n-2, \dots, 1, \bullet, 1, \dots, 1, \dots, n-1]\cdots $$
Then, it is given that the elements in positions $i-k$ and $i-n$ are painted the same color as the element in position $i$, as long as none of these is in a position with a $\bullet$.

Since $(n, k)=1$, it is now clear that starting from 1, we may reach every other residue $\pmod n$ by subtracting $k$ and $n$: we just need to be careful and subtract $n$ whenever subtracting $k$ would land us on a $\bullet$.


function
combinatorics unsolved
combinatorics

1.    how many distinct terms are there in the expansion of a(a + b)(a + b + c)(a + b + c + d)(a + b + c + d + e) ?
2.    Is there a way to generalize this problem? [/code]

The product will have the form
$ \sum a^{p_a} b^{p_b} c^{p_c} d^{p_d} e^{p_e}$
with the restrictions
[i] $ p_a\leq 5 ,p_b\leq 4,\dots, p_e\leq 1$
[ii] $ p_a\plus{}\dots\plus{}p_e\equal{}5$

For small integer like $ n\equal{}5$ we can find number of solutions by inspection.

The genearlisation asks for the non negative solutions of the equation,
$ x_1\plus{}x_2\plus{}\dots\plus{}x_n\equal{}n$
with the restrictions $ x_i\leq i \quad \forall i$

As of now I donot know how to work this general case  :)

[quote="JBL"]There are more restrictions than that (and this is very good, because your rendering of the problem gives us the answer of sequence [url=http://www.research.att.com/~njas/sequences/A000707]A000707[/url] in the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url], not particularly nice to work with): we must actually have $ p_e \plus{} p_d \leq 2$ (not just $ p_d \leq 2$), $ p_e \plus{} p_d \plus{} p_c \leq 3$ (not just $ p_c \leq 3$), etc.  After I've pointed this out, it should be not-too-hard to see that this gives rise to the Catalan numbers.  To biject with (say) paths from $ (0, 0)$ to $ (n, n)$ with steps of the form $ (1, 0)$ or $ (0, 1)$ that never pass above the diagonal $ x \equal{} y$, send each solution to (the properly restricted version of) srikanth's equation to the unique such path that has horizontal steps at $ (0, 0)$, $ (1, x_1)$, $ (2, x_1 \plus{} x_2), \ldots, (n \minus{} 1, x_1 \plus{} \ldots \plus{} x_{n \minus{} 1})$.[/quote]

that sounds better. thank you. :roll:

Just to wrap something up: we can generalize srikanth's problem to ask for the number of solutions of $ x_1 \plus{} \ldots \plus{} x_k \equal{} n$ with $ 0 \leq x_i \leq i$ for each $ i$.  This is given by [url=http://www.research.att.com/~njas/sequences/A008302]A008302[/url], the coefficients in the product $ \prod_{i \equal{} 1}^n \sum_{j \equal{} 0}^i x^j$.  This is also equivalent to the enumeration of permutations by number of inversions.

Thank you Joel.
Does the sequence A008302 have a closed form ? (the OEIS page gives a gf but no closed form). For example, I could find a closed form a [url=http://www.mathlinks.ro/viewtopic.php?t=184271]related[/url] problem with similar restrictions.

[quote="srikanth"]The product will have the form
$ \sum a^{p_a} b^{p_b} c^{p_c} d^{p_d} e^{p_e}$
with the restrictions
[i] $ p_a\leq 5 ,p_b\leq 4,\dots, p_e\leq 1$
[ii] $ p_a \plus{} \dots \plus{} p_e \equal{} 5$

For small integer like $ n \equal{} 5$ we can find number of solutions by inspection.

The genearlisation asks for the non negative solutions of the equation,
$ x_1 \plus{} x_2 \plus{} \dots \plus{} x_n \equal{} n$
with the restrictions $ x_i\leq i \quad \forall i$

As of now I donot know how to work this general case  :)[/quote]
[i] $ p_a\leq 5 ,p_b\leq 4,\dots, p_e\leq 1$
[ii] $ p_a \plus{} \dots \plus{} p_e \equal{} 5$ is not enough

[quote="srikanth"]Does the sequence A008302 have a closed form ?[/quote]
I think generating functions and recursions are the best you're going to get.  (The (equivalent) question of enumerating permutations by inversions is sufficiently well-studied that if there were a simple formula, it would almost certainly be in the OEIS.)


For all positive integer $ n\geq 2$, prove that product of all prime numbers less or equal than $ n$ is smaller than $ 4^{n}$

[hide]
The result is clearly true for $ n\equal{}2,3$. Induct by proving for $ 2n,2n\plus{}1$ given true for $ n$.

If $ M$ is equal to either $ 2n$ or $ 2n\plus{}1$, then the product of primes $ p$ such that $ n<p\leq M$ divides $ \binom{2n\plus{}1}{n}$. Now we have
\[ \binom{2n\plus{}1}{n}\leq \sum_{k\equal{}0}^n \binom{2n\plus{}1}{n}\equal{}2^{2n}\equal{}4^n\]
So the product of primes $ p$ such that $ n<p\leq M$ is at most $ 4^n$, and the product of primes less than or equal to $ n$ is less than $ 4^n$, so the product of primes less than or equal to $ M$ is less than $ 4^{2n}\leq 4^M$.
[/hide]


Write the number \[ \frac 1{\sqrt{2}-\sqrt[3]{2}}  \] as the sum of terms of the form $2^q,$ where $q$ is rational. (For example, $2^1 + 2^{-1/3} + 2^{8/5}$ is a sum of this form.) Prove that your sum equals $1/(\sqrt{2}-\sqrt[3]{2}).$

I multiplied top and bottom by $\frac{2^{1/3}}{(2^{1/6}+1)(2^{2/3}+2^(1/3)+1)}$ and it came out nicely.

[quote="pkerichang"]uhh...can i ask why you post the problem here???they can go to the official website and get them...[/quote]

It's for the [url=http://www.artofproblemsolving.com/Forum/resources.php]contest resources.[/url]


advanced fields
advanced fields unsolved

Does anyone know how to classify primes of the form $ 2p \plus{} 1$? where $ p$ itself is a prime.

Are there infinitely many such primes?

Such primes are called Sophie Germain primes. There are maybe infinitely many Sophie Germain primes, but it is still an open problem.
You can find some interesting facts about these numbers on wiki. Google it


vector
linear algebra
matrix
geometry solved
geometry

Given two vectors $\vec{a}$ and $\vec{b}$ , find an arbitrary operation on them in the terms of dot products, cross products , additions etc ... , that yields \[ a_x b_x \vec{i} + a_y b_y \vec{j} + a_z b_z \vec{k} \]. However, the same 'operation' ( the one derived above) is applicable irrespective of the choice of axes!

Yes, but that is NOT independent of the choice of axes!

What I'm looking for is something like ... say $\vec{a} \times \vec{b} + \vec{a} \cdot \vec {b} \vec{i} + ....$ etc.... i.e.. a combination of crosses, dots, sums , differences etc with no reference to $\vec{i}$ etc!

Comeon guys, I need your help immediately. Please dont limit yourself to cross and dot products --- dels etc. will do just fine!! Please help!! Can it be proved that there is no elementary formula as I want????

Let's close an old topic:

[quote="Rushil"]Given two vectors $\vec{a}$ and $\vec{b}$ , find an arbitrary operation on them in the terms of dot products, cross products , additions etc ... , that yields
\[a_{x}b_{x}\vec{i}+a_{y}b_{y}\vec{j}+a_{z}b_{z}\vec{k}\]
. However, the same 'operation' ( the one derived above) is applicable irrespective of the choice of axes![/quote]

Do I correctly understand that you define a binary operation $f: \mathbb{R}^{3}\times\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ by setting

$f\left( \left(\begin{array}[c]{c}a_{x}\\ a_{y}\\ a_{z}\end{array}\right) ,\left(\begin{array}[c]{c}b_{x}\\ b_{y}\\ b_{z}\end{array}\right) \right) =\left(\begin{array}[c]{c}a_{x}b_{x}\\ a_{y}b_{y}\\ a_{z}b_{z}\end{array}\right)$

and you want a basis-independent formula for this operation?

Actually, "basis-independent" can mean different things here, but neither of them will hold, since your operation $f$ is one of the most basis-dependent operations one can imagine:

Consider the vectors $\overrightarrow{a}=\left(\begin{array}[c]{c}1\\ 0\\ 0\end{array}\right)$ and $\overrightarrow{b}=\left(\begin{array}[c]{c}0\\ 1\\ 0\end{array}\right)$ and the orthogonal matrix $U=\left(\begin{array}[c]{ccc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ \frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ 0 & 0 & 1 \end{array}\right)$. Orthogonal matrices define congruent transformations, so the vectors $U\overrightarrow{a}$ and $U\overrightarrow{b}$ are just our two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ written in a different orthonormal basis of $\mathbb{R}^{3}$. If the operation $f$ would have any kind of basis-independent formula, then we would have $\left\vert f\left(U\overrightarrow{a},U\overrightarrow{b}\right) \right\vert =\left\vert f\left( \overrightarrow{a},\overrightarrow{b}\right) \right\vert$. But this is not the case: $\left\vert f\left( \overrightarrow{a},\overrightarrow{b}\right) \right\vert =0$ and $\left\vert f\left( U\overrightarrow{a},U\overrightarrow{b}\right) \right\vert =\frac{\sqrt{2}}{2}$. So there cannot be any basis-independent formula for your operation $f$.


[quote="pda"]A train breaks down an hour after starting its journey The engineer takes half an hour to complete repairs (while at rest), but the train can continue at only half its usual speed and arrives at its destination two hours late. Had the delay occured 100 km further on the journey, the train would ony have been an hour late. What is the total length of the journey?[/quote]
[hide="There must me a more elegant way than mine"]Let $ s$ be the speed in $ \textrm{kph}$, and $ x$ be the length in kilometers. Then we can set up the equations:

$ 1+\frac{1}{2}+\frac{x-s}{\frac{s}{2}}= \frac{x}{s}+2$

$ 1+\frac{1}{2}+\frac{100}{s}+\frac{x-s-100}{\frac{s}{2}}= \frac{x}{s}+1$

Now, we can subtract both equations:
$ -\frac{100}{s}+\frac{200}{\frac{s}{2}}= 1 \implies s = 100 \textrm{ kph}$

Now, we may substitute into the first equation:
$ \frac{2(x-100)}{100}-\frac{x}{100}= \frac{1}{2}\implies x = \boxed{250\textrm{ kilometers}}$[/hide]

I *think* there is a faster way to solve for $ s$...
[hide]
Because the train still arrived late on both situations, we know that both the actual and ideal situation will still experience the full speed, repair, and half speed phases of the journey.

Therefore the only difference between the actual and ideal situation is the delay occurring 100km farther, where the ideal situation would take $ \frac{100}{s}$ hours and the actual situation would take $ \frac{100}{\frac{s}{2}}$ hours to travel this 100 km difference. The difference in time between the ideal and actual situation is 1 hour, so:

$ \frac{100}{\frac{s}{2}}-\frac{100}{s}= 1 hr.$
$ s = 100 \frac{km}{hr}$

This way, you can avoid the system of equations to solve for $ s$, and only need
$ 1+\frac{1}{2}+\frac{x-s}{\frac{s}{2}}= \frac{x}{s}+2$
to plug in $ s$ and find that $ x = 250 km$
----
Upon previewing my post, I still dont get how this solution seems longer than vishalarul's. Its [i]suppose[/i] to be faster, I guess I made my solution too wordy and verbose...
[/hide]


Joe sold his dryer for $\$60$ and made a profit of 25% over the estimated value. He also sold his washer for $\$60$ but had a loss of 20% of its estimated value. What is the difference indollars between the total estimated value for both items and the total sale price for both items?

[hide]
For the washer, $\$60$ is a $25\%$ profit.
Thus $1.25x=60$
$x=48$

For the dryer, $\$60$ was a $20\%$ loss.
Thus $.8x=60$
$x=75$

He made $\$120$ and the estimated value was $48+75=123$

A 3 dollar difference[/hide]


Find all $ n$ integers such that $ n^3 \plus{} 23$ is a perfect square.

Hi ! I just give some hint bellow  :) ...

[hide="Hint"]Assume $ n^3 \plus{} 23 \equal{} m^2$

First prove that $ n \equiv 1 (mod4)$

Then wright the equation in this form : $ n^3 \plus{} 3^3 \equal{} m^2 \plus{} 2^2$ , And use this fact that if $ p$ is a prime number and $ p \equiv 3 (mod4)$ , and $ p|a^2 \plus{} b^2$ then $ p|a$ , $ p|b$[/hide]

By the way I don`t think this problem due to algebra. :wink:

[quote="soumik"]What becomes obvious, is that $ y$ can't be even.

So let y be odd....

$ y^3 \plus{} 23 \equal{} x^2$

$ y^3 \plus{} 8 \equal{} x^2 \minus{} 15 \implies (y \plus{} 2)(y^2 \minus{} 2y \plus{} 4) \equal{} x^2 \minus{} 15$

Now we know that if a number be of the form $ 4k \plus{} 3$, it must have a prime factor of the said form....

If the above equation is to hold true, then we have $ (y \minus{} 1)^2 \plus{} 3 \equal{} 4m \plus{} 3 \equal{} (y^2 \minus{} 2y \plus{} 4)$

Which further gives $ (x^2 \minus{} 15)$ must have a factor (prime factor) of the form $ 4p \plus{} 3$....

Happy?

Because if y is odd, x is even...So that gives $ (x^2 \minus{} 15)\equiv 1(mod4)$....thus contradicting existance of solutions.[/quote]

$ {8^2 \minus{} 15}\equiv 1(mod4)$ and $ 7 \equiv 3(mod4)$

but $ 7 |8^2 \minus{} 15$  :wink:


linear algebra
matrix
algorithm
Gauss
ARML

Suppose M is a square matrix, and L is a left inverse of M. How do we show that it is also a right inverse of M?

Let $ R$ be the right inverse.
\[ R \equal{} IR \equal{} (LM)R \equal{} L(MR) \equal{} LI \equal{} L.\]

That can't be a complete proof, because it uses nothing specific to matrices. There are rings in which elements can have a one-sided inverse but not a two-sided inverse; the standard example uses shift operators on an infinite-dimensional space.

In order to use MellowMelon's idea, we also need an existence result: if $ M$ is a $ n\times n$ matrix over some field $ F$ with a left inverse $ L$, it also has a right inverse $ R$.

For something very specific to matrices, I tend to think in algorithmic terms. A matrix is invertible iff it can be completely factored into a product of elementary matrices. (And there's an algorithm for doing exactly that: Gaussian elimination.) Once you have that the whole left inverse/right inverse thing pretty much takes care of itself.

Thanks to all who responded.

Kent, I'm not seeing how "the whole thing takes care of itself." Could you explain this more in-depth?

I've just recalled another argument that goes like this: if this is indeed the case, then det(LM) = det(L) det(M) = 1, so det(M) is nonzero, so M must be invertible, so it must have some inverse N. If so, then ML = ML(MN) = M(LM)N = MN = I, so L is a left inverse as well, and L = I, as desired. Is this valid? Or am I implicitly using this fact when invoking the multiplicity of determinants, or the fact that det(A) = 0 iff A isn't invertible?

(I'm just starting to learn linear algebra, so I apologize if some of my questions/responses seem unintelligible.)

We have a square matrix $ A$ and we apply row operations to it according to the Gaussian elimination algorithm (or possibly Gauss-Jordan, as we go ahead and clear out the upper triangle as well as the lower triangle). One of two possible things happen:

Either we eventually arrive at a matrix with at least one row of zeros, in which case we know for sure that $ A$ is not invertible, or we arrive at the identity matrix.

In the latter case, we have that $ E_kE_{k\minus{}1}\cdots E_2E_1A\equal{}I,$ where each $ E_j$ is an elementary matrix - that is, a matrix which performs an elementary row operation. Elementary matrices are invertible, since for each of the three types of elementary matrix, we can explicitly describe the inverse.

If $ E_kE_{k\minus{}1}\cdots E_2E_1A\equal{}I,$ then $ A\equal{}E_1^{\minus{}1}E_2^{\minus{}1}\cdots E_k^{\minus{}1}.$

Now suppose $ BA\equal{}I.$ The $ BE_1^{\minus{}1}E_2^{\minus{}1}\cdots E_k^{\minus{}1}\equal{}I,$ which we can then solve to get $ B\equal{}E_kE_{k\minus{}1}\cdots E_2E_1.$ It follows that $ BA\equal{}E_1^{\minus{}1}E_2^{\minus{}1}\cdots E_k^{\minus{}1}E_kE_{k\minus{}1}\cdots E_2E_1\equal{}I.$

[quote="Zhero"]if this is indeed the case, then det(LM) = det(L) det(M) = 1, so det(M) is nonzero, so M must be invertible, so it must have some inverse N. [/quote]
The fact that you said "inverse" instead of either "left inverse" or "right inverse" shows that you're using a result that is stronger than the result you want to prove.  The point is that at this point you don't know or need any of the properties of the determinant.  One way to flesh this argument out is to construct the adjunct matrix, which lets you define both a left and a right inverse, but Kent's solution is much more elementary.

There's a [url=http://www.axler.net/DwD.html]certain school of thought[/url] which says that linear algebra is better done without determinants, and while I don't know how I feel about it in general it certainly applies here.  The construction of the determinant is somehow very "high-level" - exterior algebra and a whole mess of difficult concepts - while most of linear algebra is relatively "low-level," especially the sort of arguments that Kent likes, and those arguments are very illuminating here.  For example, it's obvious why Kent's argument doesn't generalize to the infinite-dimensional setting (where the result is false):  Gaussian elimination doesn't terminate there.  (You could argue that it's "obvious" that determinants don't exist in infinite dimensions, but this is a [b]much[/b] more subtle point.)

[quote="Kent Merryfield"]
Either we eventually arrive at a matrix with at least one row of zeros, in which case we know for sure that $ A$ is not invertible, or we arrive at the identity matrix.[/quote]
I'm a bit iffy with this part. That a matrix does not have an inverse is not equivalent to saying that a matrix does not have a left inverse.

If we were instead making an argument for right inverses, the result would be obvious. We get that $ A \equal{} E_k E_{k \minus{} 1} \ldots E_1 B$, where $ B$ has a row of zeroes in it. Clearly, $ BX$ has a row of zeroes in it for any matrix $ X$, and therefore cannot be the identity. Therefore, $ A$ cannot have a right inverse; otherwise, $ I \equal{} AR \equal{} (E_k E_{k \minus{} 1} \ldots E_1) (BR) \neq I$.

With left inverses, I think we'll have to use another form of elementary matrices that operate on the  columns instead of rows, and then revert to using the elementary matrices that operate on columns, and proceed as Kent did.

With right inverses, I think we'll have to do take the row operation matrices, and then finish it off with column operation matrices, and proceed in a manner analogous to Kent's.

Thanks to everyone who responded. t0rajir0u, I never thought of determinants that way. I guess Algebra II has poisoned my mind, where the only things we learned were matrix addition, multiplication, and determinants.

[quote="Zhero"]That a matrix does not have an inverse is not equivalent to saying that a matrix does not have a left inverse. [/quote]
A row of zeroes implies that the matrix can't have full rank, so it can't have either a left or a right inverse.

[quote="Zhero"]With left inverses, I think we'll have to use another form of elementary matrices that operate on the  columns instead of rows, and then revert to using the elementary matrices that operate on columns, and proceed as Kent did. [/quote]
Why?  Kent's argument works with either row or column operations.

[quote]A row of zeroes implies that the matrix can't have full rank, so it can't have either a left or a right inverse. [/quote]
I think I also meant to say "not immediately equivalent," rather than "not equivalent," sorry. Also, my linear algebra book was confiscated by my math coach who wanted me to practice for ARML, and I JUST got to the part with ranks and left/right inverses. I'll think about what you said, though.

[quote]Why? Kent's argument works with either row or column operations.[/quote]
My bad. I was very unclear in my previous post. I was having trouble understanding why a row of zeroes automatically meant that a matrix had neither a left nor a right inverse. I "proved" it, and concluded my proof that matrices with left inverses must also have a right inverse by telling the reader to proceed as Kent did.

In any case, I just realized my previous post didn't make much sense at all. I don't know what was going through my head before. Apparently, to show that there exists a left inverse if and only if there exists a right inverse, it suffices to show that left/right-invertible matrices admit a factorization into elementary matrices; from there, we simply proceed as Kent did. To show this, we just use Gauss-Jordan row/column elimination (row for right inverses, column for left inverses); from there, it we see that if there is a row/column of zeroes, the matrix cannot be right/left invertible (as we would get [something with a row of zeroes][right inverse] = I, a clear contradiction.)

I hope this proof actually makes sense. :/


When is a^x - b^x a prime number ?
And a particular case:
How to prove that the number 3^277 - 2^277 is a prime number?

It surely is necessary that $ a\equal{}b\plus{}1$ and that $ x$ is prime. But both is fulfilled here.
With some effort, one could prossibly find some test similar to that for Mersenne numbers. But as this number is in reach of computer prime testing (and even the Mersenne test would require a computer here), it's probably simpler to use some standard test.

[quote="ZetaX"]It surely is necessary that $ a \equal{} b \plus{} 1$ and that $ x$ is prime. But both is fulfilled here.
With some effort, one could prossibly find some test similar to that for Mersenne numbers. But as this number is in reach of computer prime testing (and even the Mersenne test would require a computer here), it's probably simpler to use some standard test.[/quote]

Assuming your first bit, about $ a \equal{} b \plus{} 1, x \in \mathbb{P}$ then it is true, take $ b \equal{} 2$, $ x \equal{} 277, 227 \in \mathbb{P}$

Having said that, I don't understand where you got the first bit from, could you elaborate?

Thanks

ZetaX stipulated that those conditions are [i]necessary[/i], not sufficient.  (For example:  $ 2^{11} \minus{} 1^{11} \equal{} 23 \cdot 89$.)  The reasons:

If $ a \minus{} b > 1$ then $ a \minus{} b | a^x \minus{} b^x$.

If $ x$ is composite (in particular, has a nontrivial factor $ k$) then $ a^k \minus{} b^k | a^x \minus{} b^x$.

ZetaX,
one could prossibly find some test similar to that for Mersenne prime test. How?
How can this help ? Please elaborate.

I believe ZetaX is talking about the [url=http://mathworld.wolfram.com/Lucas-LehmerTest.html]Lucas-Lehmer Test[/url], although I don't immediately see how it could be generalized.


$\text{Find x if }2^{16^{x}}={16}^{2^{x}}$

The very top exponents are both x (obvious, but...it was hard to see on my monitor).

Your help is very appreciated!

slightly different
[hide]

from [b]rnwang's[/b] solution... if you dont know logs then you can just do this..

$\implies 2^{3x}=4$
$\implies 2^{3x}=2^{2}$
$\implies 3x=2\implies x=\boxed{\frac{2}{3}}$[/hide]

[quote="Kevin K."]In which order do the exponents go?

For the left side, is it:

2 to the 16th power, then that to the xth power?

Or is it 16 to the xth power, then 2 to the result of the previous?[/quote]

By convention, $a^{b^{c}}= a^{ (b^{c}) }$ since the alternate interpretation just gives $(a^{b})^{c}= a^{bc}$.


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Sorry folks for not adding an option for the Moon landing program being completely genuine (which I highly disagree with).

Here is the new poll.

Argument for the topic of discussion (Poll):

I believe there a lot of strong reasons to challenge the claim that NASA ever accomplished human landing on moon.

Please make an honest assessment of the filming and pictures of Moon and pictures of Earth as seen from moon, shown by NASA and draw your own unbiased conclusions as to whether we really made it to the moon.

The Van Allen Radiation belt that surrounds the Earth is so powerful and intense that it makes it highly unlikely to let any object pass through it without getting fried. The Moon is still 250,000 miles further away from the Van Allen Radioation belt. The Apollo program claims to have made the Journey from earth to moon in 10 days. In other words the space ship travelled at approx. 2600 miles/Hr for four days and stayed there orbitting the Moon for 2 days and made a return journey back to earth again at 2600 miles/hr for four days.

Imagine How much fuel is needed to accomplish this! How much precaution and calculation and skill is needed to actually land on moon while the spaceship is orbitting the moon.

Imagine how hot it is going to be when you are 250,000 miles closer to sun!

There are lot of unfeasibilities accompanying the whole moon landing program considering the technilogy that we pocessed in the 1960s.

However there are several motives suggested for the U.S. government to fake the moon landings - some of the recurrent elements are:

Distraction - The U.S. government benefited from a popular distraction to take attention away from the Vietnam war. Lunar activities did abruptly stop, with planned missions cancelled, around the same time that the US ceased its involvement in the Vietnam War.
Cold War Prestige - The U.S. government considered it vital that the U.S. win the space race with the USSR. Going to the Moon, if it was possible, would have been risky and expensive. It would have been much easier to fake the landing, thereby ensuring success.
Money - NASA raised approximately 30 billion dollars pretending to go to the moon. This could have been used to pay off a large number of people, providing significant motivation for complicity. In variations of this theory, the space industry is characterized as a political economy, much like the military industrial complex, creating fertile ground for its own survival.
Risk - The available technology at the time was such that there was a good chance that the landing might fail if genuinely attempted.

Sorry, but you have no idea...

[quote="KRIS17"]The Van Allen Radiation belt that surrounds the Earth is so powerful and intense that it makes it highly unlikely to let any object pass through it without getting fried.[/quote]
Sometimes, satellites get "fried", but still this is a rare event (as long as there are no intense solar flares). Claiming that "everything" gets fried is based on no reason (a couple of other human-made things have passed through, too!). Additionally, there were humans aboard that could repair something.

[quote="KRIS17"]The Moon is still 250,000 miles further away from the Van Allen Radioation belt. The Apollo program claims to have made the Journey from earth to moon in 10 days. In other words the space ship travelled at approx. 2600 miles/Hr for four days and stayed there orbitting the Moon for 2 days and made a return journey back to earth again at 2600 miles/hr for four days.

Imagine How much fuel is needed to accomplish this![/quote]
Not exactly the same, but imagine it's a car:
In fact, if you would use the fuel usage of a normal car, it would take only a couple of gallons of fuel. I randomly guess one needs $ \frac {1}{10}$ litres to reach 100mph with a car (maybe less, maybe a bit more, but surely not some hundreds of liters). So to reach their speed, one would need about $ 26^2$ that much, so about $ 70$ liters. Doesn't sound so much ;)

[quote="KRIS17"]How much precaution and calculation and skill is needed to actually land on moon while the spaceship is orbitting the moon.[/quote]
No argument given why this should be impossible (remember that a pack of probes landed on other planetes/asteroids without even having a human there that steers them).

[quote="KRIS17"]Imagine how hot it is going to be when you are 250,000 miles closer to sun! [/quote]
1. Who says it's closer to the sun and not wider away¿
2. Yeah, a $ \frac {1}{600}$th of earth's distance to the sun will be really very relevant... ;)

[quote="KRIS17"]There are lot of unfeasibilities accompanying the whole moon landing program considering the technilogy that we pocessed in the 1960s. [/quote]
No argument given where technology didn't suffice.

[quote="KRIS17"]However there are several motives suggested for the U.S. government to fake the moon landings - some of the recurrent elements are:

Distraction - The U.S. government benefited from a popular distraction to take attention away from the Vietnam war. Lunar activities did abruptly stop, with planned missions cancelled, around the same time that the US ceased its involvement in the Vietnam War.
Cold War Prestige - The U.S. government considered it vital that the U.S. win the space race with the USSR. Going to the Moon, if it was possible, would have been risky and expensive. It would have been much easier to fake the landing, thereby ensuring success.
Money - NASA raised approximately 30 billion dollars pretending to go to the moon. This could have been used to pay off a large number of people, providing significant motivation for complicity. In variations of this theory, the space industry is characterized as a political economy, much like the military industrial complex, creating fertile ground for its own survival.
Risk - The available technology at the time was such that there was a good chance that the landing might fail if genuinely attempted.[/quote]
Maybe, but still doesn't say it's fake.

You didn't give any reasonable argument, but instead showed that you didn't even think about most of your arguments(otherwise, yould know that e.g. the distance argument is ridiculous).

[quote="KRIS17"]Imagine how hot it is going to be when you are 250,000 miles closer to sun! [/quote]
Seriously?  Your argument is that I could do a calculation that might show ... something about how hot it is on the moon?  I have an alternate suggestion: you provide a calculation estimating exactly how much hotter it is 250,000 miles (or slightly more than 0.25%*) closer to the sun.  Then, we can discuss what this has to do with anything other than your idiocy.

* I think ZetaX neglected the miles-kilometers conversion.  To put this distance in further perspective: it's less than 1/10th of the variation in the Earth's orbit.  I.e., rather than going "all the way to the moon" in order to get this much closer to or further from the sun, it suffices to wait a month or two.

I have gathered the following information from internet, so please keep in mind that these are not "My Observations".

And you are free to choose any option. This is a poll. You may place any facts/images or demonstrations to either support the claim or refute it as you wish. Let us not turn this into a one sided debate.

9 SPACE ODDITIES:

1.  Apollo 14 astronaut Allen Shepard played golf on the Moon. In front of a worldwide TV audience, Mission Control teased him about slicing the ball to the right. Yet a slice is caused by uneven air flow over the ball. The Moon has no atmosphere and no air.

2.  A camera panned upwards to catch Apollo 16's Lunar Landerlifting off the Moon.  Who did the filming?

3.  One NASA picture from Apollo 11 is looking up at Neil Armstrong about to take his giant step for mankind. The photographer must have been lying on the planet surface. If Armstrong was the first man on the Moon, then who took the shot?

4.  The pressure inside a space suit was greater than inside a football. The astronauts should have been puffed out like the Michelin Man, but were seen freely bending their joints.

5.  The Moon landings took place during the Cold War. Why didn't America make a signal on the moon that could be seen from earth? The PR would have been phenomenal and it could have been easily done with magnesium flares.

6.  Text from pictures in the article said that only two men walked on the Moon during the Apollo 12 mission. Yet the astronaut reflected in the visor has no camera. Who took the shot?

7.  The flags shadow goes behind the rock so doesn't match the dark line in the foreground, which looks like a line cord. So the shadow to the lower right of the spaceman must be the flag. Where is his shadow? And why is the flag fluttering if there is no air or wind on the moon?

8.  How can the flag be brightly lit when its side is to the light? And where, in all of these shots, are the stars?

9.  The Lander weighed 17 tons yet the astronauts feet seem to have made a bigger dent in the dust. The powerful booster rocket at the base of the Lunar Lander was fired to slow descent to the moons service. Yet it has left no traces of blasting on the dust underneath. It should have created a small crater, yet the booster looks like it's never been fired.

There is no reason to believe what some crank sas on the internet. Maybe being a bit more crital on such informations would be helpful (You know, there are sites that say we are made by flying spaghetti monsters, right¿)

1.: Have you ever heard of the word "joke"¿¿¿

2.: Automated camera.

3.: same as 2.

4.: Based on no evidence, no reason, and especially: no facts.

5.: Any suggestion how to do it¿ By the way, I would hate that ;)

6.: We already had a discussion about cameras... additionally, show me the foto, maybe it's just not visible very well.

7.: Picture, please. But I guess it's an optical effect, e.g. from perspective view.

8.: They had lights with them (and without seeing the foto, there is no reason to assume that sun's light didn't suffice). For the stars: ever tried to see stars at daylight¿

9.: If there is about 1cm of dust, you get footsteps but no craters. And please show fotos ;)

All these answers are based on my common sense, not on any further research; and still, they sound much more reliable to me than your science fuction.
And if you think the debate is one-sided, then it's just because I just refuted all your arguments. You misunderstand the difference between "unfair debate" and "losing a debate" ;)

I've always wondered why the American flag was waving when there is a microscopic amount of air on the moon (but there's probably an explanation for it, I still believe that the landing was genuine).

Here's a picture
[img]http://www.evworld.com/images/apollo11_salute.jpg[/img]

I'm not sure, but wouldn't you be able to put the flag like that, then it would stay up?

There is a small metal rod on the top, holding it in place (I looked that one up). They thought of that problem, but found a simple solution. The waving is just a bit of movement made by the astronauts beforehand.

Actually the reason why the flag is moving is because it was prestarched.

The flag isn't actually moving, but it was set up slightly crumpled and that gives the illusion of movement.

As for the first man on the moon picture, here's how it goes.
A is the first guy on the moon
B is the second guy on the moon

A gets out (is first)
B gets out (is second)
B takes a picture of A.

Now we have a photo of the first man on the moon....

Agreed.  I think they did get to the moon, but some other stuff was exageertated.

OMG i spell really good.

Here a few some images of the astronauts on the moon taken from "the Project Apollo Image gallery".

In the pic from my previous post the moon appears to have a curvilinear shape at the eadge. How far is the lunar module from the moon's surface to capture the curvature of the moon's edge. This pic was taken just before the lunar module landed on the moon.

Look at another image of moon prior to approaching the moon's surface...

Seriously? OK, going through your pictures in order:

1. The only two shadows you know enough about to draw a conclusion from are that of the man with the pole, and the one on the left, and they look fine.

2. The moon doesn't have an atmosphere to diffuse the light.

3. What? Obviously the thing is going to be far from the moon before it gets close to the moon... the whole point of those pictures was to capture the curvilinear shape...

I believe the argument he's trying to make with picture #3 is trying to make is that, at the time the picture was taken, the module was too close to the moon's surface to capture the curvature of the moon.  Without knowing the exact time of the picture, and the exact distance the module was from the moon's surface at that time, it is impossible to determine whether or not this argument is valid.  The phrase "just before landing on the moon" is not enough to determine this.

Actually, I have seen a Mythbusters show on this. It challenged some well known points.

1. The flag would not flutter with no wind.
Scientists already know the moon is a vaccum. They went to a vaccum chamber, stuck in the flag, and video taped the movement. They seemed to almost completely be the same.

2. The footprint would not be possible without water.
Again, the moon is a vaccum. Using things that are close to the moon's surface, the imprinted a 'moon boot' at average stomping weight. It made a clear footprint.

3. The jumping on the moon was just a video slowed down.
When they did that, it looked completely unnatural, and when they used one of those planes that simulated moon's gravity, it looked more like the video.

4. The two rocks had a shadow pointing in different directions. This could only happen with two set lights.
When they made a set with only one light, they managed to tilt the rock's angles to make two different shadow directions.

5. The man climbing the ladder was too bright.
They again used one light to simulate the same thing.

Of course, that doesn't mean anything. They could all have been riding in a plane with all those set things on board. I say it was just a spoof.

OK, I am not KIDDING!

Please take a look at this:

http://www.news.com.au/story/0,23599,24070088-13762,00.html

Then you will know what is the point I am trying to make here.

And I'mnot kidding either if I deny that kind of stories for being conspiracy theories.
There are many much more reasonable explanaitions for this (being faked, being just a try to get famous, ...).

Why can't these kind of story tellers not just leave the internet alone... There are already enough cranks and trolls here...

From what I understand, the previous post seems to claim that we cannot trust everything we see or hear on the internet.

However, if you must care to know the fact, here it an excerpt from the interview with former Astronaut who walked on the moon for the longest number of hours, Dr Edgar Mitchell.

[i]FORMER NASA astronaut and moon-walker Dr Edgar Mitchell - a veteran of the Apollo 14 mission - has stunningly claimed aliens exist.
And he says extra-terrestrials have visited Earth on several occasions - but the alien contact has been repeatedly covered up by governments for six decades.
Dr Mitchell, 77, said during a radio interview that sources at the space agency who had had contact with aliens described the beings as 'little people who look strange to us.'
He said supposedly real-life ET's were similar to the traditional image of a small frame, large eyes and head.
He claimed our technology is "not nearly as sophisticated" as theirs and "had they been hostile", he warned "we would have been gone by now".[/i]

What is the point of this?

Let us consider two cases:

Case 1: The Astronaut Edgar Mitchell's claim that government covered up the Alien contact is FALSE:

It shows that people who claim to have walked on moon, who were nurtured by NASA are the people who can also believe in such childish concepts as existence of aliens.

Case 2: The Astronaut Edgar Mitchell's claim that government covered up the Alien contact is TRUE:

It makes one wonder, if government can cover up such an important event of Aliens contacting Earth for 60 years, how hard it is for the government to cover up faked Moon landings? how hard it is for the government to keep the mouths of NASA and all the people involved with moon landing experiment shut?

You don't have to look far to find evidence regarding that point. Check out the [url=http://en.wikipedia.org/wiki/Edgar_Mitchell]Wikipedia on Edgar Mitchell[/url]. He's been into the paranormal since childhood. A few things:

1. Believing that aliens exist is not childish.

2. A personal belief in the paranormal does not make a person less qualified to be an astronaut. It's not like he's going to suddenly steer the mission off course and go alien-hunting.


[quote="Xevarion"]In the following, assume $a, b, c > 0$.

Find the largest $k$ so that
$\sum_{sym}(a^{4}+ka^{2}b^{2}) \ge (k+1)\sum_{sym}a^{3}b$

I think it might be $k=3$ but I'm not sure.

More generally, what is the largest $k$ so that
$\sum_{sym}(a^{n}+ka^{n-2}b^{2}) \ge (k+1)\sum_{sym}a^{n-1}b$ ?
I could only prove $k=2$ for $n=5$, but I don't see yet why larger $n$ should mean smaller $k$.[/quote]

Applying $b=c=1$, we obtain $a^{4}-(k+1)a^{3}+2ka^{2}-(k+1)a+1 \geq 0$, or $(a-1)^{2}(a^{2}+(1-k)a+1) \geq 0$. So if $k$ is bigger than 3, as $a^{2}+(1-k)a+1=0$ has solutions, there is an interval in which the expression becomes negative. So the maximum of $k$ is 3.

In case of $k=3$, the inequality is equivalent to $(a-b)^{4}+(b-c)^{4}+(c-a)^{4}\geq 0$ :)

For the second one, I found that $k \leq \frac{n-1}{n-3}$. This can be obtained by the similar way to the above idea. Applying $b=c=1$, we obtain $a^{n}-(k+1)a^{n-1}+ka^{n-2}+ka^{2}-(k+1)a+1 \geq 0$, or $(a-1)^{2}(a^{n-2}+(1-k)a^{n-3}+\cdots+(1-k)a+1)\geq0$. So $(a^{n-2}+(1-k)a^{n-3}+\cdots+(1-k)a+1) \geq 0$ for any positive $a$.

Applying $b=c=1$, we get $k \leq \frac{n-1}{n-3}$. Otherwise, this is equivalent to $\frac{a^{n-1}-1}{a-1}\geq (k-1)\frac{a^{n-2}-a}{a-1}$. So $k-1 \leq \frac{a^{n-1}-1}{a^{n-2}-a}$. Note that the numerator of the derivate of $\frac{a^{n-1}-1}{a^{n-2}-a}$ is $a^{2n-4}-(n-2)a^{n-1}+(n-2)a^{n-3}-1$
$=(a-1)^{3}(a^{2n-7}+3a^{2n-6}+\cdots+\frac{(n-2)(n-3)}{2}a^{n-3}+\frac{(n-2)(n-3)}{2}a^{n-4}+\cdots+3a+1$.

This means that $\frac{a^{n-1}-1}{a^{n-2}-a}$ is increasing iff $a>1$. So the minimum is at $a=1$, and the limit is $\frac{n-1}{n-3}$. This tells us that the maximum $k$ is the value.

In case of $k = \frac{n-1}{n-3}$, the inequality is equivalent to $\sum_{cyc}(a-b)^{2}((n-3)a^{2}+2ab+(n-3)b^{2}\geq 0$. As $(n-3)a^{2}+2ab+(b-3)b^{2}\geq 0$, this is true.


I always hear people say that align should be used over eqnarray, but I cannot for the life of me figure out where to put the ampersands so that there isn't a ridiculous amount of space between the expressions.
Could someone please shed some light? What am I doing wrong?

\begin{align*}
&\phantom{\Rightarrow} &3(1 - ve^{-\mu}) &= 2(1 - v^2e^{-\mu})
\\&\Rightarrow &3 - 3ve^{-\mu} &= 2 - 2v^2e^{-\mu}
\end{align*}

\begin{align*}
\phantom{\Rightarrow} &3(1 - ve^{-\mu}) &= 2(1 - v^2e^{-\mu})
\\\Rightarrow &3 - 3ve^{-\mu} &= 2 - 2v^2e^{-\mu}
\end{align*}

\begin{align*}
3(1 - ve^{-\mu}) &= 2(1 - v^2e^{-\mu})
\\\Rightarrow 3 - 3ve^{-\mu} &= 2 - 2v^2e^{-\mu}
\end{align*}

Is this what you are looking for?
\begin{align*} 3(1 - ve^{ - \mu}) & = 2(1 - v^2e^{ - \mu}) \\
\Rightarrow \quad 3 - 3ve^{ - \mu} & = 2 - 2v^2e^{ - \mu} \end{align*}
See [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=voss-mathmode]Math mode[/url] for more about align and similar environments.


[quote="kunny"]For a constant number $ p>1$, let $ p\leq a\leq p^{2},\ p\leq b\leq p^{2},\ p\leq c\leq p^{2},\ p\leq d\leq p^{2}$ be real numbers such that  $ (a+1-p)(b+1-p)(c+1-p)(d+1-p)=1$.[/quote]

I have a better one:

Let $ a$, $ b$, $ c$ be nonnegative reals satisfying $ a\leq 0$, $ b\leq 0$, $ c\leq 0$. Assume further that $ \sqrt{a^{2}+b^{2}+c^{2}}=0$. Prove that

$ \left(\frac{a}{b}-\frac{a}{b}\right)^{2}+\left(\frac{b}{c}-\frac{b}{c}\right)^{2}+\left(\frac{c}{a}-\frac{c}{a}\right)^{2}\geq 0$.

Determine the cases of equality.

  darij

Sorry I can't understand your problem.

If possible, this time so trivial is my problem?

kunny

[quote="Peter"]Darij points out that there are no such $ a,b,c,d$ other than $ a=b=c=d=p$... and you don't even need the $ \le p^{2}$ for that.[/quote]

Yes, and his inequality even does not become an equality, unless my computations are wrong.

  darij

what reasoning? If $ a\ge p$ then $ a+1-p\ge1$...  :maybe:


If $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$, then

            $ a^2\plus{}b^2\plus{}c^2\geq 2\sqrt{3abc}\minus{}6abc\minus{}\frac{1}{9}$

[quote="Ligouras"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$, then

            $ a^2 \plus{} b^2 \plus{} c^2\geq 2\sqrt {3abc} \minus{} 6abc \minus{} \frac {1}{9}$[/quote]

It's obvious because $ 6abc\minus{}2\sqrt{3abc}\plus{}\frac{4}{9}\ge 0$, for $ abc\le \frac{1}{3\sqrt{3}}$. :P

[quote="Pain rinnegan"][quote="Ligouras"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$, then

            $ a^2 \plus{} b^2 \plus{} c^2\geq 2\sqrt {3abc} \minus{} 6abc \minus{} \frac {1}{9}$[/quote]

It's obvious because $ 6abc \minus{} 2\sqrt {3abc} \plus{} \frac {4}{9}\ge 0$, for $ abc\le \frac {1}{3\sqrt {3}}$. :P[/quote]

 :wink:


LaTeX
number theory proposed
number theory

[color=blue][size=150]PERU TST IMO - 2006[/size]
Saturday, may 20.[/color]

[b]Question 01[/b]

Find all $(x,y,z)$ positive integers, such that:

$\sqrt{\frac{2006}{x+y}} + \sqrt{\frac{2006}{y+z}} + \sqrt{\frac{2006}{z+x}},$

is an integer.

---
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88509]Spanish version[/url]

$\text{\LaTeX}{}$ed by carlosbr

Let a,b,c positive rational numbers. It is easy to show, that  $\sqt a +\sqrt b +\sqrt c$ rational  if and only if a,b,c perfect square of rational numbers. Therefore $x+y=2006r_1^2,y+z=2006r_2^2,x+z=2006r_3^2$. Because x,y,z is integer and 2006 free square rational numbers $r_1,r_2,r_3$ are natural. It give $x=1003(r_1^2+r_3^2-r_2^2),y=1003(r_1^2+r_2^3-r_3^2),z=1003(r_2^2+r_3^2-r_1^2),$ satisfied condition: $\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$ is an integer.
Let $x\le y\le z$. We have $r_1\le r_3 \le r_2, r_1^2+r_3^2>r_2^2$. There are 3 solutions $1.r_1=r_2=r_3=1, 2. r_1=1,r_3=r_2=2, 3. r_1=r_2=r_3=3$ or $1. x=y=z=1003, 2.x=1003=y,z=7021, 3.x=y=z=9027.$

Sorry for reviving, but I think Rust is missing 1 set of ordered triples in his solution.

Note that $ \frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{4}\equal{}1$ also works, which generates $ r_1\equal{}2$, and $ r_2\equal{}r_3\equal{}4$, giving $ x\equal{}y\equal{}2*2006$ and $ z\equal{}12*2006$ if you let $ x \le y \le z$, but really you can also switch around the integers in the ordered triples.


function
real analysis
real analysis unsolved

(a) Is there a non-decreasing function $ f : \mathbb{R} \rightarrow \mathbb{R}$ which no point on it that $ f$ is continuous at?

(b) The same question but include a condition that $ f$ has a lower-bound?

Thanks, I would appreciate any helps.

A monotone function is differentiable almost everywhere. It is continuous except at countably many points.

for a: you can take the f in an interval [a,b]  f is bounded in [a,b] (becasue is non deacreasing in R) so f can be discountinious in countably points for b: you can have dirichlet function where it is 1 for rationals and -1 for irrationals you can prove it is continuous on no points(by taking two sequences conevrgin in a     where one is rationg sequence and the other irrational)


A waiter served $ 650.00 worth of dinners in a busy restaurant. She received tips in the amount of$75.00. How much less did the waiter receive in tips than she would have if she receiver her expected rate of 15% of the cost of the dinners?


calculus
integration
function
calculus computations

I encountered this while solving someother problem. My friend and I broke our head over this but sadly no avail, so any suggestions are welcome. :P

$ \frac {dy}{dx} \equal{} \frac {1}{x^2 \plus{} y^2}$

Solve for y as a function of x.

Sorry I posted this here by mistake. Can somebody move it..

and i suppose it is the probem i gave...well where do u want to get it moved :huh:

Change the variable $ y(x) \rightsquigarrow x(y) \rightarrow y'\equal{}\frac{dy}{dx} \rightarrow x'\equal{}\frac{dx}{dy}\equal{}x^2\plus{}y^2$. This is a Ricatti type equation.

csabafarkas-- after this how will ya proceed


[quote="Andrej_pp"]Prove that the equation
$x(x+1)=4y(y+1)$
has infinitely many solutions in rational numbers.[/quote]

$\forall a\in\mathbb{Q}$, take :

$x=\frac{-a^{2}-2a+3}{4a}$
$y=\frac{a^{2}-4a+3}{8a}$

And then you have $x(x+1)=\frac{(-a^{2}-2a+3)(-a^{2}+2a+3)}{16a^{2}}=\frac{a^{4}-10a^{2}+9}{16a^{2}}$
and $4y(y+1)=\frac{(a^{2}-4a+3)(a^{2}+4a+3)}{16a^{2}}=\frac{a^{4}-10a^{2}+9}{16a^{2}}$

So $x(x+1)=4y(y+1)$ $\forall a\in\mathbb{Q}$

And we have infinitely many solutions in rational numbers.


probability
function
probability and stats

a classical approach to this kind of problem is to introduce the generating function $ f(z)\equal{}E[z^{X_n}]$ where $ X_n$ is the number of offspring at the $ n$-th generation (keyword: branching process)

[quote="neworder"]We start with a population of $ N$ bacteria. In each step every bacterium reproduces - a single bacterium has $ k$ offspring with probability $ p_k$. A bacterium dies after reproducing. Prove that the population never dies out with probability $ > 0$ iff $ \sum_{k} kp_{k} > 1$ or $ p_1 \equal{} 1$.

There are many approaches possible, I'm interested in your ideas.[/quote]

I don't think this is true. If we look at the number of bacteria after one step, then we have

P(X(1) > 0 ) = [Every bacteria reproduces] = 1 - [No bacteria reproduce]= 1-(P(0))^N

As long as P(0)>0, there will be a non-zero chance of extinction.

You misunderstand the question. It is about [b]positivity[/b] of the probability that the colony survives forever, not about that probability being equal to $ 1$. You are right, it is strictly less than $ 1$ if $ P(0)>0$, but this alone doesn't prevent it from being positive.


function
calculus
derivative
inequalities
search
inequalities proposed

Let x,y,z be non-negative number such that x+y+z=3 and xy+yz+zx=1.
Find the maximum of
$ P\equal{}x^4\plus{}y^4\plus{}z^4$

$ P\equal{}x^4\plus{}y^4\plus{}z^4\equal{}(x^2\plus{}y^2\plus{}z^2)^2\minus{}2(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)$

$ \equal{}\{(x\plus{}y\plus{}z)^2\minus{}2(xy\plus{}yz\plus{}zx)\}^2\minus{}2\{(xy\plus{}yz\plus{}zx)^2\minus{}2xyz(x\plus{}y\plus{}z)\}$

$ \equal{}47\plus{}12xyz$.

[quote="kunny"]$ P \equal{} x^4 \plus{} y^4 \plus{} z^4 \equal{} (x^2 \plus{} y^2 \plus{} z^2)^2 \minus{} 2(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)$

$ \equal{} \{(x \plus{} y \plus{} z)^2 \minus{} 2(xy \plus{} yz \plus{} zx)\}^2 \minus{} 2\{(xy \plus{} yz \plus{} zx)^2 \minus{} 2xyz(x \plus{} y \plus{} z)\}$

$ \equal{} 47 \plus{} 12xyz$.[/quote]
Can you continue?

[quote="kunny"]I'm stuck here, AM-GM can't be used.[/quote]
Why???
$ x\plus{}y\plus{}z \geq 3\sqrt[3]{xyz} \Leftrightarrow xyz \leq 1$
So $ 47\plus{}12xyz \leq 47 \plus{}12$
So we found the max dear friend :)

We generalize the problem to all real numbers.

Essentially, $ m\equal{}xyz$ has to make the equation $ t^3\minus{}3t\plus{}t\plus{}m$ have 3 roots. If the function has 3 distinct roots, then we can raise m, since by looking at the graph of the cubic there must be a small part under the curve that when raised, will eventually become tangent to the x-axis.

Therefore, the max will be attained when the function has (at least) a double root. WLOG, let $ x\equal{}y$. We have $ 2x\plus{}z\equal{}3$ and $ x^2\plus{}2xz\equal{}1$.

$ z\equal{}3\minus{}2x$ so $ x^2\plus{}2x(3\minus{}2x)\equal{}1$

$ \minus{}3x^2\plus{}6x\minus{}1\equal{}0$
$ 3x^2\minus{}6x\plus{}1\equal{}0$
$ 3(x\minus{}1)^2\equal{}2$

$ x\equal{}1\pm\sqrt{2/3}$.

Taking $ x\equal{}1\plus{}sqrt(2/3)$ however yields a negative $ xyz$ so the max must be at $ x\equal{}1\minus{}\sqrt{2/3}$. Then $ z$ is easily found to be $ 3\minus{}2x\equal{}1\plus{}\sqrt{2/3}$

[quote="rofler"]We generalize the problem to all real numbers.

Essentially, $ m \equal{} xyz$ has to make the equation $ t^3 \minus{} 3t \plus{} t \plus{} m$ have 3 roots.[/quote]

My solution seems to be same as yours, but that should be $ t^3 \minus{} 3t \plus{} t \minus{} m\equal{}0$ and the original problem is more harder than the generalization.

@8826, Is there solution without using differentiation?

[quote="ElChapin"]Look what values should $ x$, $ y$ and $ z$ have in the equality case Dimitris X. kunny was right :wink:

:)[/quote]
[quote="kunny"]Is the answer $ \frac {105 \plus{} 16\sqrt {6}}{3}$ for $ \{x,\ y,\ z\} \equal{} \left\{\frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \plus{} 2\sqrt {6}}{3}\right\}$?[/quote]
[quote="Dimitris X"][quote="kunny"]I'm stuck here, AM-GM can't be used.[/quote]
Why???
$ x \plus{} y \plus{} z \geq 3\sqrt [3]{xyz} \Leftrightarrow xyz \leq 1$
So $ 47 \plus{} 12xyz \leq 47 \plus{} 12$
So we found the max dear friend :)[/quote]
Don't you think it to be strange?
Dimitris_X continued kunny's solution using AGM this way:
$ 3 \equal{} x \plus{} y \plus{} z\ge3\sqrt [3]{xyz}\implies xyz\le 1$
So naturally equality holds iff $ x \equal{} y \equal{} z \equal{} 1$.
But that does NOT satisfy the condition $ xy \plus{} yz \plus{} zx \equal{} 1$;
so I think the expression has NO maximum, as equality does not hold.
  :huh:
That's funny! :(  Please help me, kunny..........

[hide="edit/Paradoxical thought"]
Wait, if we work the other way , $ xy \plus{} yz \plus{} zx \equal{} 1\ge 3\sqrt [3]{x^2y^2z^2}\implies xyz\le\frac {1}{3\sqrt {3}}$   :huh:  which is stronger than $ xyz\le 1$.
If we use this then $ P\le 47 \plus{} \frac {4}{\sqrt {3}}$.  But, again, the equality case($ x \equal{} y \equal{} z \equal{} \frac {1}{\sqrt {3}}$) DOES NOT satisfy the condition $ x \plus{} y \plus{} z \equal{} 3$. Then again, working this way we see that $ P$ has NO Maximum. Thus it must be the correct logic.

Then what about rofler's solution ? I found no flaw in it. We see that an interesting paradox arises .
[b]My conclusion[/b]
[i]1)Either the question is wrong, or $ P$ DOESN'T HAVE ANY MAX.
2) $ P<47\plus{}\frac{4}{\sqrt{3}}<59$ if the question is correct.[/i]

Can anyone please tell me why this sort of inconvenience arises? :blush:
[/hide]

[quote="8826"]Let x,y,z be non-negative number such that x+y+z=3 and xy+yz+zx=1.
Find the maximum of
$ P = x^4 + y^4 + z^4$[/quote]
The following generalization holds.

Let $ x_1,x_2,...,x_n$ be nonnegative real numbers such that $ x_1+x_2+...+x_n=constant$ and $ x_1^2+x_2^2+...+x_n^2=constant$.
For $ k>2$, the sum $ S=x_1^k+x_2^k+...+x_n^k$ is maximal when ${ x_1=...=x_{n-1}\le x_n}$.

[quote="kunny"]Is the answer $ \frac {105 \plus{} 16\sqrt {6}}{3}$ for $ \{x,\ y,\ z\} \equal{} \left\{\frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \plus{} 2\sqrt {6}}{3}\right\}$?[/quote]

Regrettably my result is wrong, because $ \{x,\ y,\ z\} \equal{} \left\{\frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \plus{} 2\sqrt {6}}{3}\right\}$ doen't satisfy $ xy\plus{}yz\plus{}zx\equal{}1$.

But I feel like to kill the problem. I will post the solution.

[quote="kunny"][quote="kunny"]Is the answer $ \frac {105 \plus{} 16\sqrt {6}}{3}$ for $ \{x,\ y,\ z\} \equal{} \left\{\frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \plus{} 2\sqrt {6}}{3}\right\}$?[/quote]

Regrettably my result is wrong, because $ \{x,\ y,\ z\} \equal{} \left\{\frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \plus{} 2\sqrt {6}}{3}\right\}$ doen't satisfy $ xy \plus{} yz \plus{} zx \equal{} 1$.

But I feel like to kill the problem. I will post the solution.[/quote]
You 're right, kunny.$ \{x,\ y,\ z\} \equal{} \left\{\frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \minus{} \sqrt {6}}{3},\ \frac {3 \plus{} 2\sqrt {6}}{3}\right\}$ satisfy $ xy \plus{} yz \plus{} zx \equal{} 1$. :P

My answer was correct. :lol:

@8826 What's the source of the problem?

We can just use the standard trick.. Let $ 3u \equal{} x\plus{}y\plus{}z, 3v^2 \equal{} xy\plus{}yz\plus{}zx, w^3 \equal{} xyz$. Then remember the wellknown ineq:
$ w^3 \le 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt{(u^2\minus{}v^2)^3}$ (Follows from $ (x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2 \ge 0$)
So we have $ xyz \le \frac{4\sqrt{6}\minus{}9}{9}$(Since $ u\equal{}1, v^2 \equal{} \frac{1}{3}$)

[quote="kunny"]How can we guarantee $ x\geq 0,\ y\geq 0,\ z\geq 0?$[/quote]
What do you mean? The inequality $ xyz \le \frac {4\sqrt {6} \minus{} 9}{9}$, is proved for all $ x,y,z$. Not just $ x,y,z \ge 0$.
Finding equality is trivial..(Two of them must be equal) Set $ x \equal{} y \equal{} a$, then $ z \equal{} 3 \minus{} 2a$, so we have $ xy \plus{} yz \plus{} zx \equal{} 1 \iff 3a^2 \minus{} 6a \plus{} 1 \equal{} 0$ which gives $ a \equal{} \frac {3 \minus{} \sqrt {6}}{3}$, so we have equality with $ (x,y,z) \equal{} \left ( \frac {3 \minus{} \sqrt {6}}{3}, \frac {3 \minus{} \sqrt {6}}{3}, \frac {3 \plus{} 2\sqrt {6}}{3}, \right )$. (But you already wrote this, so I figured that it didn't matter if i wrote it again :))

[quote="Mathias_DK"][quote="kunny"]How can we guarantee $ x\geq 0,\ y\geq 0,\ z\geq 0?$[/quote]
What do you mean? The inequality $ xyz \le \frac {4\sqrt {6} - 9}{9}$, is proved for all $ x,y,z$. Not just $ x,y,z \ge 0$.[/quote]

I can understand it.

[quote="Mathias_DK"][quote="kunny"]How can we guarantee $ x\geq 0,\ y\geq 0,\ z\geq 0?$[/quote]

Finding equality is trivial..(Two of them must be equal)[/quote]

I can't understand it, because I'm not an expert of Inequality Solver. :wink:

That is to say, I can't understand what Vasc said as well.

[quote="Vasc"][quote="8826"]Let x,y,z be non-negative number such that x+y+z=3 and xy+yz+zx=1.
Find the maximum of
$ P = x^4 + y^4 + z^4$[/quote]
The following generalization holds.

Let $ x_1,x_2,...,x_n$ be nonnegative real numbers such that $ x_1 + x_2 + ... + x_n = constant$ and $ x_1^2 + x_2^2 + ... + x_n^2 = constant$.
For $ k > 2$, the sum $ S = x_1^k + x_2^k + ... + x_n^k$ is maximal when ${ x_1 = ... = x_{n - 1}\le x_n}$.[/quote]

In IMO, can we use freely the fact that has been argued in MathLinks without proof?

I'll try to explain my idea very clearly :)

Starting from the begining: We have set $ 3u \equal{} x\plus{}y\plus{}z, 3v^2 \equal{} xy\plus{}yz\plus{}zx, w^3 \equal{} xyz$.
It is obvious that $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \ge 0$.
What is not so obvious is $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \equal{} 27(\minus{}(w^3\minus{}(3uv^2\minus{}2u^3))^2\plus{}4(u^2\minus{}v^2)^3)$. But it is true, and I know this identity by heart :)
Hence $ 4(u^2\minus{}v^2)^3 \ge (w^3\minus{}(3uv^2\minus{}2u^3))^2$. I want to show $ w^3 \le 3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}$. If $ w^3 \le 3uv^2\minus{}2u^3$ it is obvious, so assume $ w^3 > 3uv^2\minus{}2u^3$.
Then we have (by taking squareroots):
$ 2\sqrt{(u^2\minus{}v^2)^3} \ge |w^3\minus{}(3uv^2\minus{}2u^3)| \equal{} w^3\minus{}3uv^2\plus{}2u^3 \iff$
$ w^3 \le 3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}$.
This does, as before yield the inequality $ xyz \le \frac{4\sqrt{6}\minus{}9}{9} \forall x,y,z \in \mathbb{R}$. :)

Of course, that is not enough to prove that it is the maximum. But we can "guess" the number $ x,y,z$ such that the equality holds relatively easy. :)
$ w^3 \le 3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}$ is a consequence of $ (x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2 \ge 0$, so if $ w^3 \le ..$ attains equality, then $ (x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2 \ge 0$ must also attain equality! There's equality e.g. when $ x\equal{}y$. So we put $ x\equal{}y\equal{}a$, and then $ z \equal{} 3\minus{}2a$, and we proceed as in the post before.
Then we get the triple $ (x,y,z) \equal{} \left ( \frac {3 \minus{} \sqrt {6}}{3}, \frac {3 \minus{} \sqrt {6}}{3}, \frac {3 \plus{} 2\sqrt {6}}{3}, \right )$. And we put it in and check that $ xyz \equal{} \frac{4\sqrt{6}\minus{}9}{9}$ and that $ x,y,z > 0$. Then $ \frac{4\sqrt{6}\minus{}9}{9}$ certainly is a maximum, and we are done :)

For Vasc's generalisation: I don't think that you can use such things at IMO unless you have a more clear reference :) The safest is probably to memorize the proof :P

i think all we have to do is the following:
we have that $ x^2\plus{}y^2\plus{}z^2\equal{}7$, so $ x^4\plus{}y^4\plus{}z^4 \plus{}2(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)\equal{}49$
but we also have that $ 2(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)\equal{}2\minus{}12xyz$ so
$ x^4\plus{}y^4\plus{}z^4\equal{}47\plus{}12xyz$ but $ xyz\le1$(by AM-GM) so
 max $ x^4\plus{}y^4\plus{}z^4\equal{}59$  :lol:

[quote="v235711"]$ xyz\le1$(by AM-GM) [/quote]

This implies that equality happens when $ x \equal{} y \equal{} z$ but, this is impossible if $ x \plus{} y \plus{} z \equal{} 3$ and $ xy \plus{} yz \plus{} zx \equal{} 1$. Then $ 59$ is not the minimun.   Read the posts at the begining of the thread. :wink:

[quote="Mathias_DK"]I'll try to explain my idea very clearly :)

Starting from the begining: We have set $ 3u \equal{} x \plus{} y \plus{} z, 3v^2 \equal{} xy \plus{} yz \plus{} zx, w^3 \equal{} xyz$.
It is obvious that $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \ge 0$.
What is not so obvious is $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} 27( \minus{} (w^3 \minus{} (3uv^2 \minus{} 2u^3))^2 \plus{} 4(u^2 \minus{} v^2)^3)$. But it is true, and I know this identity by heart :)
Hence $ 4(u^2 \minus{} v^2)^3 \ge (w^3 \minus{} (3uv^2 \minus{} 2u^3))^2$. I want to show $ w^3 \le 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$. If $ w^3 \le 3uv^2 \minus{} 2u^3$ it is obvious, so assume $ w^3 > 3uv^2 \minus{} 2u^3$.
Then we have (by taking squareroots):
$ 2\sqrt {(u^2 \minus{} v^2)^3} \ge |w^3 \minus{} (3uv^2 \minus{} 2u^3)| \equal{} w^3 \minus{} 3uv^2 \plus{} 2u^3 \iff$
$ w^3 \le 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$.
This does, as before yield the inequality $ xyz \le \frac {4\sqrt {6} \minus{} 9}{9} \forall x,y,z \in \mathbb{R}$. :)

Of course, that is not enough to prove that it is the maximum. But we can "guess" the number $ x,y,z$ such that the equality holds relatively easy. :)
$ w^3 \le 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$ is a consequence of $ (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$, so if $ w^3 \le ..$ attains equality, then $ (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$ must also attain equality! There's equality e.g. when $ x \equal{} y$. So we put $ x \equal{} y \equal{} a$, and then $ z \equal{} 3 \minus{} 2a$, and we proceed as in the post before.
Then we get the triple $ (x,y,z) \equal{} \left ( \frac {3 \minus{} \sqrt {6}}{3}, \frac {3 \minus{} \sqrt {6}}{3}, \frac {3 \plus{} 2\sqrt {6}}{3}, \right )$. And we put it in and check that $ xyz \equal{} \frac {4\sqrt {6} \minus{} 9}{9}$ and that $ x,y,z > 0$. Then $ \frac {4\sqrt {6} \minus{} 9}{9}$ certainly is a maximum, and we are done :)

For Vasc's generalisation: I don't think that you can use such things at IMO unless you have a more clear reference :) The safest is probably to memorize the proof :P[/quote]

Thank you so much!

kunny

[quote="Mathias_DK"]I'll try to explain my idea very clearly :)

Starting from the begining: We have set $ 3u \equal{} x \plus{} y \plus{} z, 3v^2 \equal{} xy \plus{} yz \plus{} zx, w^3 \equal{} xyz$.
It is obvious that $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \ge 0$.
What is not so obvious is $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} 27( \minus{} (w^3 \minus{} (3uv^2 \minus{} 2u^3))^2 \plus{} 4(u^2 \minus{} v^2)^3)$. But it is true, and I know this identity by heart :)
Hence $ 4(u^2 \minus{} v^2)^3 \ge (w^3 \minus{} (3uv^2 \minus{} 2u^3))^2$. I want to show $ w^3 \le 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$. If $ w^3 \le 3uv^2 \minus{} 2u^3$ it is obvious, so assume $ w^3 > 3uv^2 \minus{} 2u^3$.
Then we have (by taking squareroots):
$ 2\sqrt {(u^2 \minus{} v^2)^3} \ge |w^3 \minus{} (3uv^2 \minus{} 2u^3)| \equal{} w^3 \minus{} 3uv^2 \plus{} 2u^3 \iff$
$ w^3 \le 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$.
This does, as before yield the inequality $ xyz \le \frac {4\sqrt {6} \minus{} 9}{9} \forall x,y,z \in \mathbb{R}$. :)

Of course, that is not enough to prove that it is the maximum. But we can "guess" the number $ x,y,z$ such that the equality holds relatively easy. :)
$ w^3 \le 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$ is a consequence of $ (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$, so if $ w^3 \le ..$ attains equality, then $ (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$ must also attain equality! There's equality e.g. when $ x \equal{} y$. So we put $ x \equal{} y \equal{} a$, and then $ z \equal{} 3 \minus{} 2a$, and we proceed as in the post before.
Then we get the triple $ (x,y,z) \equal{} \left ( \frac {3 \minus{} \sqrt {6}}{3}, \frac {3 \minus{} \sqrt {6}}{3}, \frac {3 \plus{} 2\sqrt {6}}{3}, \right )$. And we put it in and check that $ xyz \equal{} \frac {4\sqrt {6} \minus{} 9}{9}$ and that $ x,y,z > 0$. Then $ \frac {4\sqrt {6} \minus{} 9}{9}$ certainly is a maximum, and we are done :)

For Vasc's generalisation: I don't think that you can use such things at IMO unless you have a more clear reference :) The safest is probably to memorize the proof :P[/quote]
Sorry for my spam but i'm really curious to learn more about this powerfull method...Does anybody have notes for that???Please pm me :)

:D  As Mathias_DK answered to that question few days ago: search for aqady's posts.

:)

[quote="tuandokim"][quote="Mathias_DK"]I'll try to explain my idea very clearly :)

Starting from the begining: We have set $ 3u \equal{} x \plus{} y \plus{} z, 3v^2 \equal{} xy \plus{} yz \plus{} zx, w^3 \equal{} xyz$.
[/quote]
with $ x,y,z\in R$
I don't think we can assume:
$ 3v^2 \equal{} xy \plus{} yz \plus{} zx$
sorry if it's wrong :)[/quote]
We can. Just let $ v \equal{} \sqrt {\frac {xy \plus{} yz \plus{} zx}{3}}$. Since $ x,y,z > 0$ we get $ v^2 > 0$. When some of $ x,y,z$ is negative it is possible that $ v \notin \mathbb{R}$, but we always have $ v^2 \in \mathbb{R}$.


Does anyone know the theory behind solving equations (or approximating the solutions) of this form ($x$ is the variable):

$\sum a_{i}e^{-xb_{i}}= 1$,

where $a_{i}$ and $b_{i}$ are reals satisfying $a_{i}\ge 0$, $\sum a_{i}= 1$, and $\sum a_{i}b_{i}\ge 0$. The equation has exactly two solutions (one of which is zero) except in the case in which the last condition is equality. Any sort of reference would be great. I know it has something to do with exponential families but I can't find anything that really deals with the solutions.

Consider the function $f(x)=\sum_{i=1}^{n}a_{i}e^{-xb_{i}}$. Then $f'(0)<0$ and $f''(x)=f(x)=\sum_{i=1}^{n}a_{i}b_{i}^{2}e^{-xb_{i}}>0$ that when $x\rightarrow-\infty$ converges to $+\infty$. Obviously we don't have any positive solutions, and if there are 3 solutions for $f=1$ then $f'$ will be zero at 2 points and $f''=0$ has a solution, Contradiction. So $f=1$ has at most 2 solutions and because there exist $\epsilon>0$ that $f'(-\epsilon)<1$ and because $\lim_{x\rightarrow-\infty}f(x)=+\infty$, thus $f(x)=1$ has a negative solution. We are done.

Actually, the solution is positive (since $f'$ is negative at zero and $f''$ is always positive). But what I'm interested in is the actual solution and comparing the values of solutions; for instance whether $b_{i}$ or $b_{i}'$ will produce a smaller solution for some $a_{i}$. I've gotten some estimates using Taylor series, but they are not conclusive.


abstract algebra
superior algebra
superior algebra solved

It is well known that if $R$ is Noetherian and $M$ is a finitely generated $R$-module then $M$ is Noetherian.

Anybody knows a counterexample when $R$ is not Noetherian?


안녕!
I am from the United States of America. i tried asking this question to a couple guys in the SoCal ARML forum but they're a bit too advanced for me(freshman). I was hoping you'd understand and answer it without using any freaky theorems or postulates no one's ever heard of. just use simple polynomial factoring.

Prove:
if $n\in\&\&Z$
then prove $n^{7}-n$ is divisible by $7$.

If it works for $n = 0, 1, 2, 3, 4, 5, 6$ it works for all integers. [hide="Because..."] $(n+7)^{7}-(n+7) \equiv n^{7}-n \bmod 7$ [/hide]

For all prime $p$ and all integer $a$ , the femats little theorem implies that $p|{a^p} - a$


function
complex analysis
complex analysis unsolved

Let $f$ be entire complex function such that $\Re{f(z)}\geq 0$ for all complex $z$. Prove that $f$ is constant.

$e^{-f(z)}=e^{-Re(z)}e^{-iIm(z)}$, whence $|e^{-f(z)}| \leq 1$, and $e^{-f}$ is entire. Now the result follows by Liouville's theorem.


Let $ a,b,$ and $ c$ be real numbers such that $ a\minus{}7b\plus{}8c\equal{}4$ and $ 8a\plus{}4b\minus{}c\equal{}7$. What is $ a^{2}\minus{}b^{2}\plus{}c^{2}$?

[hide="hint"]
Solving for a and b, we get $  a\equal{}\frac{13\minus{}5c}{12},\ b\equal{}\frac{13c\minus{}5}{12}$.
Then $ a^{2}\minus{}b^{2}\equal{}1\minus{}c^{2}\Rightarrow a^{2}\minus{}b^{2}\plus{}c^{2}\equal{}1$
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[quote="erdogankerem123"]Let $ a,b,$ and $ c$ be real numbers such that $ a\minus{}7b\plus{}8c \equal{} 4$ and $ 8a\plus{}4b\minus{}c \equal{} 7$. What is $ a^{2}\minus{}b^{2}\plus{}c^{2}$?[/quote]

I think this has been posted before but I cant find it

[hide="solution"]Rearrange the equations and square them to get:

$ a^{2}\plus{}64c^{2}\minus{}16ac\equal{}16\plus{}49b^{2}\plus{}56b$

$ 64a^{2}\plus{}c^{2}\plus{}16ac\equal{}49\plus{}16b^{2}\minus{}56b$

Add them and you get $ a^{2}\minus{}b^{2}\plus{}c^{2}\equal{}\boxed{1}$.[/hide]