linear algebra
matrix
algebra
function
domain
polynomial
linear algebra unsolved

If $ A\in M_{n \times m}{(\mathbb{R})}$  and $ B\in M_{m \times n}{(\mathbb{R})}$. Note m & n are distinct natural numbers.

How could we know the determinant of the Matrix $ AB$ without calculating?

Also, How could know that for any matrix is invertible or not without calculating the determinant and eigenvalues of it ?

i.e. fastest way to know if it has a inverse { surely for square matrices }.

 :maybe:  :roll:

That's "determinant". Also, "calculating" is the appropriate form in both places.

$ AB$ is an $ n\times n$ matrix. If $ n>m$, it's never invertible. If $ m<n$, I don't know any shortcuts.

[quote="jmerry"]That's "determinant". Also, "calculating" is the appropriate form in both places.

$ AB$ is an $ n\times n$ matrix. If $ n > m$, it's never invertible. If $ m < n$, I don't know any shortcuts.[/quote]

Thank you jmerry I corrected  :blush:

You said if $ n>m$ is never invertible. could you tell me why?

And Is there anybody knows shortcuts about another case.? :)

[quote="Soarer"][quote="Hidden Scofield"][quote="jmerry"]The rank of a product is no greater than the smallest rank of any term.[/quote]

that's very nice theorem.

do you have proof of it ? Sorry if I make noise for you dude :blush:[/quote]

For two matrices A,B where multiplication can be performed, show that rank (AB) \leq rank(A) and rank(AB) \leq rank(B).[/quote]

Hey Soarer , that's what jmerry wrote ;

i.e we need to show : $ rank(AB) < min\{ rank(A) , rank(B) \}$. if $ n > m$.

Also, need shortcuts for $ n < m$.

[quote="Hidden Scofield"][quote="jmerry"]The rank of a product is no greater than the smallest rank of any term.[/quote]

that's very nice theorem.

do you have proof of it ? Sorry if I make noise for you dude :blush:[/quote]
It is clear that the image of a linear map can't have a higher dimension than its domain, so we have $ \rm{rank}(F\circ G)\leq \min\{\rm{rank}(F),\rm{rank}(G)\}$.  Now think of $ A$ and $ B$ as linear maps.

That's great Eigenray.

But What about if $ n < m$. :huh:

Do you have any shortcuts.?

Is there a relation between the eigenvalues of $ A,B, and \ AB$ ? Surely all previous matrices square.

I'm pretty sure there is..

But what is it?

Thx

[quote="Hidden Scofield"]Is there a relation between the eigenvalues of $ A,B, and \ AB$ ?[/quote]

What exactly do you mean when you talk about the eigenvalues of a non-square matrix?

In that case there's no general relationship between the eigenvalues, but of course $ det AB\equal{}(det A)(det B)$, and the determinant is the product of the eigenvalues.

thank you , I know that..

back to original question.

You are looking for something that doesn't exist. Non-square matrices don't have eigenvalues, and there are just too many things going on with the product of two non-square matrices.

Jmerry I Understand...

I edited my post before (I forgot sm information only ). I know there is no eigenvalues for Non-Square Matrices.

Thank you guys about your help .

I'm gonna try to find shortcuts about it.

 :D

I think what you want is the [url=http://en.wikipedia.org/wiki/Cauchy-Binet_formula]Cauchy-Binet formula[/url] or its generalization [url=http://en.wikipedia.org/wiki/Minor_%28linear_algebra%29]here[/url].

If you want the characteristic polynomial of $ AB$ for $ A\in \mathcal{M}_{n \times m}{(R)},B\in \mathcal{M}_{m \times n}{(R)}$
over any commutative ring $ R$, you can use the formula
$ \det (XI_n\minus{}AB)\equal{}\sum_{J\subseteq\{1,\ldots,n\}\atop {K\subseteq\{1,\ldots,m\}\atop |J|\equal{}|K|}}(\minus{}1)^{|J|}(\det A_{J,K})(\det B_{K,J})X^{n\minus{}|J|}$

where $ A_{J,K}$ is the $ J\times K$-minor of $ A$ for all $ J\subseteq\{1,\ldots,n\},K\subseteq\{1,\ldots,m\}$. (and $ B_{K,J}$ analog)
(Does that help?  :P )

[quote="olorin"]I think what you want is the [url=http://en.wikipedia.org/wiki/Cauchy-Binet_formula]Cauchy-Binet formula[/url] or its generalization [url=http://en.wikipedia.org/wiki/Minor_%28linear_algebra%29]here[/url].

If you want the characteristic polynomial of $ AB$ for $ A\in \mathcal{M}_{n \times m}{(R)},B\in \mathcal{M}_{m \times n}{(R)}$
over any commutative ring $ R$, you can use the formula
$ \det (XI_n \minus{} AB) \equal{} \sum_{J\subseteq\{1,\ldots,n\}\atop {K\subseteq\{1,\ldots,m\}\atop |J| \equal{} |K|}}( \minus{} 1)^{|J|}(\det A_{J,K})(\det B_{K,J})X^{n \minus{} |J|}$

where $ A_{J,K}$ is the $ J\times K$-minor of $ A$ for all $ J\subseteq\{1,\ldots,n\},K\subseteq\{1,\ldots,m\}$. (and $ B_{K,J}$ analog)
(Does that help?  :P )[/quote]

It's nice information , but somewhat complex.

does not matter.

Nevertheless , waiting to nice solution.

Thx :D  :wink:


linear algebra
matrix
induction
number theory open
number theory

de pus pe mathlinks
si pe justlinux.com
subiect

problem on matrix problem
-------------------------

A guy who went by the name of Heineken had a weird
habbit when he was drunk.He usually threw all the
bears off his table,took out his knife and started
to write on the table a matrix.
He wrote on the upper left edge 0,after that he put
on the next right cell the smallest number that wasn't
on the row and column and so forth in this manner

(upper left)-> -> -> ->
-> -> -> -> -> -> -> ->
 .
 .
||
\/
.
.
-> -> -> -> (R,C)

When he gets to (R,C) wich is the cell on the row R
and column C he stops because he is too drunk and
cannot write any more.

For R,C given,find the number lying at cell (R,C)

restrictions:

       10^9<=R,C<=2*10^9

time :      0.1 sec .

-------------------------

So now comes a formula wich one friend of mine told
me that should work but wich i do not understand,this
beeing the reason i wrote this subject.

(R-1)^(C-1) wich is actually (R-1)XOR(C-1).

If there is another more logical solution to this problem
i'd be glad to know it,because the only explanation i can find
for this formula is connected to the case where R=2 and C=2:
here it goes:

for this case we have the matrix:

   C 1 2
 R|----------
 1|  0 1
 2|  1 0

so,
    C-1  0 1
R-1-----------
  0|     0 1
  1|     1 0

this resembles if i'm right with the
binary table for XOR

OR has the table

  0  1
 -----
0|0  1
 |
1|1  1

but XOR has table

  0  1
 -----
0|0  1
 |
1|1  0

wich is exactly what i have for R=2 and C=2

Further i cannot proove the corectness of the
formula.I hope that someone will help please!

Also i have some code done by myself wich is
ofcourse verry naive but i think works well
for small R and C

for (I=2;I<=L;I++)
for (J=2;J<=C;J++)
	{
	INT mark[max(L,C)+2];
	memset(mark,sizeof(int)*(max(L,C)+2),0)
	for(k=1;k<=I-1;k++)
			mark[M[k][J]]=1;
	for(k=1;l<=J-1;k++)
			mark[M[I][k]]=1;
	k=0;
	while(mark[k++]==1);
	M[I][J]=k;
	}

this builds the matrix cell by cell untill it reaches
what is needed.
Also i found some nice simetries wich appear
for a particular case when R=8 and C=8

where the matrix is like this:

01234567
10325476
23016745
32107654
45670123
54741032
67452301
76583210

About the graphs i made using this matrix in excel
it seems to me that the numbers switch in an ordered
manner across the columns and rows,i can't figgure yet
out why this is in a simetrical.
These graphs i made with excel using the matrix above.

[quote="spx2"]A guy who went by the name of Heineken had a weird
habbit when he was drunk.He usually threw all the
bears off his table,took out his knife and started
to write on the table a matrix.
He wrote on the upper left edge 0,after that he put
on the next right cell the smallest number that wasn't
on the row and column and so forth in this manner

(upper left)-> -> -> ->
-> -> -> -> -> -> -> ->
 .
 .
||
\/
.
.
-> -> -> -> (R,C)

When he gets to (R,C) wich is the cell on the row R
and column C he stops because he is too drunk and
cannot write any more.

For R,C given,find the number lying at cell (R,C)

restrictions:

       10^9<=R,C<=2*10^9

time :      0.1 sec .

-------------------------

So now comes a formula wich one friend of mine told
me that should work but wich i do not understand,this
beeing the reason i wrote this subject.

(R-1)^(C-1) wich is actually (R-1)XOR(C-1).[/quote]
The formula is easy to prove by induction on $(R,C)$ (sorted lexicographically), employing the following lemma

[b]Lemma[/b]. For nonnegative integers $A,B,C$, if $C < (A\oplus B)$ then either $(C\oplus A)<B$ or $(C\oplus B)<A$.
In other words, $C = A'\oplus B$ or $C=A\oplus B'$ for either $A'<A$ or $B'<B$, where $A'=(C\oplus B)$ and $B'=(C\oplus A)$.
(Here $\oplus$ denotes XOR operation.)

Proof of Lemma. Let $i$ be the highest bit position at which the binary representations of $C$ and $A\oplus B$ differ. Since $C < (A\oplus B)$, the i-th bit in $C$ is zero while the $i$-th bit in $A\oplus B$ is 1. Then it is 1 in one of $A$ or $B$, and 0 in the other. Without loss of generality assume that it is 1 in $A$ and $0$ in $B$. Then we claim that $(C\oplus B)<A$. Indeed, all bits higher than $i$ are the same in both $C\oplus B$ and $A$, while the $i$-th bit is 0 in $C\oplus B$ and 1 in $A$, implying that  $(C\oplus B)<A$. QED

Now, how the induction works. Suppose that Heineken have filled up every cell $(R',C')$ lexicographically preceding $(R,C)$ with the value $(R'-1)\oplus (C'-1)$. We need to show that the smallest number not used $R$-th row and $C$-th column is $(R-1)\oplus (C-1)$. Indeed, for any smaller number $X<((R-1)\oplus (C-1))$, we have by Lemma that $X=((R'-1)\oplus (C-1))$ or $X=((R-1)\oplus (C'-1))$ for some $R'<R$ or $C'<C$, meaning by induction that $X$ has been already used in the $C$-th column or $R$-th row. Therefore, Heineken writes in the cell $(R,C)$ the number $(R-1)\oplus (C-1)$.


inequalities
trigonometry
inequalities unsolved

If you love inequalities , try this :
  Prove that for every acute triangle ABC , we have :
     (cos A)^3 + (cos B)^3 >= 2 (cos ((A+B)/2))^3 [/tex]

Obviously looks like a candidate for Jensen's inequality.

if we have [tex]f(x) = \cos^3(x)[/tex] then [tex]f''(x) = -\frac{3}{4}(\cos(x)+3\cos(3x))[/tex], and it is easy to see that we have [tex]f''(x) > 0[/tex].

The inequality can be written like:
[tex]\frac{\cos^3(A)+\cos^3(B)}{2} \geq \cos^3(\frac{A+B}{2})[/tex], which obviously holds by Jensen.

The result of f''(x) is right but f''(x) is not >0 with every acute triangle ABC .
   Let's do it again

hxtung , if this problem is easy , let's prove it :P

hxtung , this problem is not easy . If you can solve it , you are really good at maths  :P

I have heared it long time ago.
It was very easy problem.

Omid Hatami , can you post the solution ?

We prove a stronger statement:
\[\frac{\cos^2(A)+\cos^2(B)}{2} \geq \cos^2(\frac{A+B}{2})\]
This is stronger,because:
\[\frac{\cos^3(A)+\cos^3(B)}{2} \geq {\frac{\cos^2(A)+\cos^2(B)}{2}}^{3/2} \geq \cos^3(\frac{A+B}{2})\]
Now if $A,B \geq \frac{\pi}{4}$ inequality is obvious with Yen-sen inequality.
Now suppose $A \geq \frac{\pi}{4} \geq B$(Because $A+B \geq \frac {\pi}{2}$ we can suppose this)
Now suppose $A=\frac {\pi}{4}+x$ and $B=\frac {\pi}{4}-y$ that ${\pi}/4 x,y \geq 0$
We see $x \geq y$
With some computatation we see inequality is equivalent to:
$\frac{\sin(2x)-\sin(2y)}{2}\leq \sin(x-y)$
which is obvious because $\frac{\sin(2x)-\sin(2y)}{2}=\sin(x-y)\cos(x+y)$
So you see this problem is very easy and simple!!!! :D  :D  :D
____________________________________
No pain, no gain

:D  :D  I think you should be more carefully .  May be you are wrong :
     [ (cos A)^3 +(cos B)^3 ]  2 <= { [ (cos A)^2 + (cos B)^2 ] /2 }^3/2

Try it again , hxtung , Hatami!!! :)  :)

Of course you are making me upset.
Because of NEWTON INEQUALITY we have :
$\displaystyle (\frac {x^3+y^3}{2})^{\frac 13} \geq (\frac{x^2+y^2}{2})^{\frac 12}$
____________________
No,pain,no gain


There is no quick way!

-interesting_move

P.S. Muhahahahahahhahahahahahahahhahahahahahhaha!

You put 25! in a calculator and divide it by 10^6.

[quote="JBL"]That would be good, if only it didn't have 26 digits.  I would try writing out the product 1*2*...*24*25, cancelling out the numbers corresponding to 10^6, and then pairing things off to make a final digit of 1 -- for example, 3 and 7, 13 and 17, etc. -- and just go on like that.[/quote]

I tried your suggestion and it works nicely.  By the way, how do you know that 25! has 26 digits (without using a calculator)?

I cheated and used a calculator :).  Without a calc, it's pretty messy -- you need to approximate the logarithm function (since the number of digits of a number n is 1 + [Log(n)], where [x] is the largest integer not bigger than n).  (If you know logs but haven't seem the theorem I just stated, try to prove it for yourself!)  In order to do it without a calc, you'd just have to take approximations as good as you can of the product.  Or, if you know some really fancy stuff, there's a good approximation of the factorial function that is relatively easy to take the logarithm of and gets you within 1 digit of what you want.  25 actually is pretty nice for that particular approximation because $\frac{25}{e} \approx 10$ (go figure!).

[quote="jen7"]What is a quick way to figure out the last digit of (25!/10^6)?[/quote]

Since 25! has 6 5s in its prime factorisation and 23 2s, we can remove 10^6 by taking 6 2s out and 6 5s out.

this is the product of all numbers from 1 to 25, excluding multiples of 5 and divided by 2 6 times

this is equal to (1*2*3*4)(1*2*3*4*6*7*8*9)^2 / 2^6  (because you go through ten twice and upto 25)
=(1*2*3*4)(1*3*6*7*8*9)^2
=(2*3*4)(3*6*7*8*9)^2

== (4)(1*6*9)^2 (mod 10)
== 4*4*4 (mod 10)
== 4 (mod 10)

SO I think it's 4... but im not certain

In general I guess this could be done (if it works) with any number, say "what is the last nonzero digit of the number 2004! ?, by grouping the digits:

2004! with no multiples of 5 left in = (1*2*3*4)(1*2*3*4*6*7*8*9)^200
then
= (4)(2^200)(3^200)(4^200)...(9^200) (mod 10)

which you can work out by finding patterns in 9^1, 9^2, 9^3 for example


Hi all! :)

Show that : $\forall a$ such that $\gcd(a,2) = 1$ and $\gcd(a,5) = 1$, then, there exists a multiple of $a$ that can be written only with ones.

Good luck! :D

It is well-known fact.
Let's take numbers:$1, 11, ..., 11...11$(last with $a+1$ ones). 2 of them are congruous modulo $a$(for example with $k$ and with $l$ ones($k<l$) ) by Dirichlet's principle.
Since $gcd(a,10)=1$ number $11...11$ with $l-k$ ones is divisible by $a$.

Nice proof.
I looked at $\frac{1}{a}$ as a decimal, and cut it off just before it starts to repeat, (we know it repeats because its prime to 2 and 5)

Then taking this decimal number as a new integer, multiplying it by $a$ will give 99999...99999    From here its not too hard. Just looking at factors and how it dosnt matter if 9 divides it etc...

[quote="ZetaX"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=67696[/quote]
 :oops: Sorry...

Anyway, thank you Pavel Kozlov, Seamusboyle for your solutions/ideas, and ZetaX (for the notification) ;)


[color=darkred]
find all [size=150] X [/size] more than zero [/color]
[img]http://photofile.ru/photo/fox610/3423991/large/73919977.jpg[/img]

We have: $ x^x \equal{} \frac {1}{\sqrt {2}}$
Let $ x \equal{} \frac {1}{2^m}$, we get:
$ x\cdot \log_{2}{x} \equal{} \minus{} \frac {1}{2}$, $ x\cdot m \equal{} \frac {1}{2}$, $ m \equal{} 2^{m \minus{} 1}$.
Last equation have only roots $ m_1 \equal{} 1$ and $ m_2 \equal{} 2$. Easy to prove that equations of this type has $ \leq 2$ roots. So the answer is $ x_1 \equal{} \frac {1}{2}$ and $ x_2 \equal{} \frac {1}{4}$.


[quote="Ulanbek_Kyzylorda KTL"]There are six positive consecutive integers.Prove that there exists a prime number that divides one of these integers.[/quote]

Huh ? Just choose any prime divisor of the first, for example ....

You should check your problem statement

[quote="dosums"]Well the first could be 1  :)[/quote]
Are you joking ? Then take any prime divisor of the sixth !

sorry :blush: i will post it correctly later

$ (\forall)a\in\mathbb{Z},(\exists)k\in\mathbb{Z}\mbox{ so that we } \\
 \{a;a\plus{}1;a\plus{}2;a\plus{}3;a\plus{}4;a\plus{}5\}\equal{}\{6k;6k\plus{}1;6k\plus{}2;6k\plus{}3;6k\plus{}4;6k\plus{}5\} \\
\mbox{and}\\
(6k)\vdots 2 \mbox{ and } (6k)\vdots 3;\\
(6k\plus{}2)\vdots 2;\\
(6k\plus{}3)\vdots 3;\\
(6k\plus{}4)\vdots 2.$


Hey, I was looking through the MathCount Warm-Up's and I found this weird question. (Well maybe it's not weird to you XP) It's on Warm-Up 10 and it goes like this:

Noel earns $25 each week mowing lawns. Each week he spends 2/5 of his earnings on himself and has the rest of his earnings placed directly into asavings account. Every third week he with draws $20 from his savings account to take his girlfriend to the movies. If he continues this same spending pattern, how many weeks will it take Noel to first exceed $125 in his savings acocount?

For some reason, the answer says 13. But any way I do it, it comes out 14!

I agree with your answer. Must have been a typo or something?

I believe this was discussed on the mathcounts webforum -- go look there and see what it says.  Check the coaches room, back more than a month maybe.


Prove or disprove:

Given any polynomial function with degree of at least 2, a local minimum or maximum occurs at x=0 if the coefficient of the linear term is 0.

If proven true, does the local minimum or maximum occur [i]if and only if[/i] the linear term has coefficient 0?

Let $ P(x) \equal{} a_nx^n \plus{} \cdots \plus{}a_1x \equal{} a_0$, with $ n \geq 2$, so $ P'(x) \equal{} na_nx^{n\minus{}1} \plus{} \cdots a_1$, with $ P'(0) \equal{} a_1$.
So, in fact, it's the exact opposite: a local minimum or maximum occurs at x=0 [i]only if[/i] the coefficient of the linear term is 0. An easy counterexample for the [i]if[/i] part is $ P(x) \equal{} x^n$ with odd $ n$.

The phrase "exactly the opposite" is probably not the best description -- [b]mavropnevma[/b] means to say that the order 1 term being 0 is necessary but not sufficient for there to be a local extremum at x=0.

The full condition is that [b]the lowest-order non-constant term must be of an even order[/b].

[hide="Heres a heuristic explanation that does not use calculus"]

We'll say there's a local extremum if, the function either increases both just to the left and to the right, or decreases both to the left and the right.

Suppose we represent a function like such:

$ f \equal{} \sum_{n \equal{} 0}^\infty a_n x^n$.

This has a local extremum if and only if $ g \equal{} f \minus{} a_0$ does (ie, a vertical shift so that the graph goes through the origin).

$ g \equal{} \sum_{n \equal{} 1}^\infty a_n x^n$

So, a local extremum at $ f \equal{} 0$ occurs iff $ g$ has the same sign just to the left and just to the right of $ x \equal{} 0$.

Let $ N$ be the index on the first non-zero $ a_n$, and let $ 2T$ be the greatest even integer not exceeding $ N$.

$ g \equal{} x^{2T} \sum_{n \equal{} 2T}^\infty a_n x^{n \minus{} 2T}$

Note that the $ x^{2T}$ factor doesn't change the sign, so $ g$ has the same sign on both sides of the origin iff $ h: \equal{} \frac {g}{x^{2T}}$ does.

[b]Case 1:[/b] If $ N$ is even, then $ N \equal{} 2T$, and:
$ h \equal{} \sum_{n \equal{} N}^\infty a_n x^{n \minus{} N}$

Note that the lowest-order term here is a constant, so around $ 0$, $ h \approx a_N$, and thus $ h$ has the same sign around both sides of $ x \equal{} 0$, and consequently $ f$ has an extremum.

[b]Case 2:[/b] If $ N$ is odd, then $ N \equal{} 2T \plus{} 1$, and:
$ h \equal{} \sum_{n \equal{} N}^\infty a_n x^{n \minus{} N \plus{} 1}$

The lowest order term of $ h$ is $ a_N x$, so just around $ x \equal{} 0$, $ h \approx a_N x$, and thus $ h$ changes signs at $ x \equal{} 0$ and there is not a local extremum.[/hide]


I know there are Canadian Competition and they are very good.. Does anyone know where I can view those sample problems?

Also, is there any competition like AMC in Australia? Just curious

There is. I remember TripleM talking about it, and one of the more difficult problems was posted in a forum. Sadly, I dont remember much about it right now.

[quote="Silverfalcon"]Also, is there any competition like AMC in Australia?[/quote]

Google "Australian Mathematics Trust" and you'll find what is basically Australia's equivalent of the AMC organization.

cemc.uwaterloo.ca

past contests
there r past contests

The website for the AMC of the east is here: [url]http://www.amt.edu.au/[/url]

I think the Australian one is slightly more difficult than the American one but I'm not sure. I have a collection of the questions.

camel.math.ca is where you can find Canadian contests

In Australia, for IMO selection it's AMO by invitation, then top 25 go to APMO, and then top 12 plus 12 juniors (up to Year 11 or 10, I guess) go to a camp in the April holidays.

Otherwise there are various competitions- AMC which most people enter in high school, AIMO for up to year 10's, and AMOC Senior Contest, the latter two invitational.


function
trigonometry
complex numbers
complex analysis

I've seen complex numbers represented as $\cos \theta+i\sin \theta$, but is it possible to take say, $\sin(i+1)$?

I'm wondering this because of a few homework problems that gave "no solution".  I think I solved them using the Taylor series for sine, but is that valid?

I've been wondering about this too. You beat me to it. :)

My calculator gives an "ERR:DATA TYPE" when I try to do $\sin(i+1)$. But that isn't a very satisfactory answer.

Wikipedia: http://en.wikipedia.org/wiki/Sine#Relationship_to_exponential_function_and_complex_numbers

Yes, there's a unique analytic continuation, i.e. you can have complex numbers as the argument.

But that gets into complex analysis, and frankly, I don't know much about it either.

The analytic continuation is very simple.  A short demonstration:

Given $e^{ix}= \cos x+i \sin x$, note that

$e^{-ix}= \cos x-i \sin x$

And therefore that

$\cos x = \frac{ e^{ix}+e^{-ix}}{2}$
$\sin x = \frac{ e^{ix}-e^{-ix}}{2i}$

Using the example of evaluating $\sin (i+1)$, we see that

\begin{eqnarray*}\sin (i+1) &=& \frac{ e^{-1+i}-e^{1-i}}{2i}\\ &=& \frac{i}{2}\left( e^{1-i}-e^{-1+i}\right) \\ &=& \frac{i}{2}\left( e ( \cos 1-i \sin 1 )-\frac{\cos 1+i \sin 1}{e}\right) \\ &=& \left( \frac{e \sin 1}{2}+\frac{\sin 1}{2e}\right)+\left( \frac{e \cos 1}{2}-\frac{\cos 1}{2e}\right) i \\ \end{eqnarray*}

Cool.   :)   That's the same answer I got with the Taylor series.


inequalities
function
inequalities proposed

[quote="arqady"]$ a\geq0$, $ b\geq0$, $ c\geq0$, $ d\geq0$.

then:

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \minus{} 4abcd\geq \sqrt {2}(a^2b^2 \plus{} b^2c^2 \plus{} c^2d^2 \plus{} d^2a^2 \minus{} 4abcd)$[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-49-m-rozenberg.html

[quote="arqady"]Thank you can_hang2007 for your interest! But I can not read your file.  :([/quote]
I am sorry dear Arqady. I uploaded my files on http://skydrive.live.com where I have an account there. A little note is, when I upload file type .RAR, it will be renamed as .RAR_. So, when you download it, please rename it as .RAR and then you can open it normally.

I am very sorry for this inconvenience.

[quote="FelixD"]I can't open the file either...
Why don't you post the proofs dircetly on you hp as before? :)[/quote]
Thanks for your suggestion, FelixD. I like to write my solution in file now. It will look nicer than posting directly on the blog. :lol:

And I have changed to use blogspot instead of wordpress now, because it has more functions than wordpress.

[quote="arqady"]Thank you can_hang2007 for your interest! But I can not read your file.  :([/quote]
Dear arqady,you can download here :)

Thank you, vuthanhtu_hd!
Nice proof, can_hang2007!


function
trigonometry
calculus
derivative
integration
advanced fields
advanced fields theorems

I was wondering if someone could give me some more insight on the Trace theorem.  The version I am referring to is the following:

U-bounded with $ \partial U$ - $ C^1$
Then there is a bounded linear operator:  $ T: W^{1,p}(U) \rightarrow L^p(\partial U)$
with:

(i)  $ Tu\equal{}u$ on $ \partial U$ if $ u \in W^{1,p}(U) \cap C(\overline{U})$
(ii)  $ \|Tu\|_{L^p(\partial U)} \leq C \| u \|_{W^{1,p}(U)}$  for all $ u \in W^{1,p}(U)$ with $ C\equal{}C(p,U)$.

My questions are the following:

1.  Why do we need $ W^{1,p}$?  Why doesn't this work for functions $ L^p(U)$?  The only references I have  fail to address this.
2.  The case $ p\equal{}\infty$ is not addressed.  Does that case not make sense, or what is the problem with it?
3.  I had a conversation earlier today where it came up that this theorem works for Lipschitz boundary.  Is it still for $ W^{1,p}$ functions?  What is a good reference for this proof?

Thanks in advance for clearing these issues up!

1. Let $ U$ be the unit ball $ \{x\colon |x|<1\}$. The function $ u(x)\equal{}\log(1\minus{}|x|)$ is in $ L^p$ for any $ p<\infty$, but what happens on the boundary? For $ p\equal{}\infty$ consider $ v(x)\equal{}\sin u(x)$ -- this function is bounded in $ U$ but how would you extend it to the boundary?

2. I don't see a problem with $ W^{1,\infty}\to L^{\infty}$. If a function belongs to $ W^{1,\infty}$, then it has a Lipschitz representative which naturally extends to a Lipschitz (hence bounded) function on the boundary.

3. Yes, it's still for $ W^{1,p}$. I think that the book of Evans and Gariepy may have this result, but I'm not sure.

Thanks Mlok for the $ L^p$ counterexample.  But what about if $ u \in L^p(U)$ and $ u \in C(\overline{U})$?  Is that enough to have a trace, or do you still need a condition on the derivative as given by the Sobolev spaces?

Btw, thanks for the reference.  The amazon preview looked pretty good.  I'll look for it at the library.

If $ u\in C(\overline{U})$, then the trace exists. Duh.

[quote] If $ u \in C(\overline{U})$, then the trace exists. Duh. [/quote]

Well, if that is the case, then why does the Theorem bother with the specific case:

$ u \in W^{1,p}(U) \cap C(\overline{U})$ ?

You could just drop the Sobolev part and conclude the result.  What am I failing to understand?

Some functions $ u$ extend to the boundary simply by continuity. We don't need a trace operator for them. The purpose of the trace operator is to extend $ u$ to $ \overline{U}$ in a meaningful way, even when there is no continuous extension. What does "in a meaningful way" mean? First of all, if $ u$ continuously extends to $ \overline{U}$, then the trace must equal to that extension. Second, the trace of a sum must be the sum of traces (lineariry). Finally, we must be able to say something about the trace, i.e. estimate its integral norm via the norm of $ u$.

To recap in symbols:

$ u \in W^{1,p} \cap C(\overline{U})$  :   trace exists by theorem

$ u \in L^p \cap C(U)$                        :   trace does not exist by your counterexample(s)

$ u \in L^p \cap C(\overline{U})$         :   trace trivially exists by $ C(\overline{U})$ part

I guess it's the third claim that I do not get ...

Okay, it was a hard day here. I'll shut up.


Fill in the two blanks in the expression $0.5\ \underline{\ \ }\ 0.2\ \underline{\ \ }\ 5$ with two different operation symbols from the list: $\times, +, \div$.  What is the greatest possible value of this expression?  Express your answer as a decimal to the nearest tenth.

[hide="solution"]
Divide by the smallest decimal, $.2$, and multiply by $5$, for the greatest value, $12.5$.
[/hide]

Does anybody else want to try this problem?  :|


Prove that, for a triangle $ ABC$,

$ (a) \tan(\frac{A}{2})\cdot \tan(\frac{B}{2})+\tan(\frac{B}{2})\cdot \tan({\frac{C}{2})+ \tan(\frac{C}{2})\cdot\tan(\frac{A}{2})=1}$;

$ (b) \tan(\frac{A}{2})\cdot\tan(\frac{B}{2})\cdot\tan(\frac{C}{2}) \le \frac{1}{8}$.

[hide="#1"]For convenience, let $ x = A/2$, $ y = B/2$, and $ z = C/2$. We have that $ \tan (x+y+z) = \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - (\tan x \tan y + \tan y \tan z + \tan z \tan x)}$. However, $ x+y+z = 90$, so $ \tan (x+y+z)$ is undefined, so $ 1 - (\tan x \tan y + \tan y \tan z + \tan z \tan x) = 0$, so $ \tan(\frac{A}{2})\cdot\tan(\frac{B}{2})+\tan(\frac{B}{2})\cdot\tan({\frac{C}{2})+\tan(\frac{C}{2})\cdot\tan(\frac{A}{2})=1}$, as desired.
[/hide]

Is #2 correct? If $ A = B = C = 60^\circ$, then we get $ \frac{1}{3\sqrt{3}} > \frac{1}{8}$.

[quote="AndrewTom"]Prove that, for a triangle $ ABC$,

$ (b) \tan(\frac {A}{2})\cdot\tan(\frac {B}{2})\cdot\tan(\frac {C}{2}) \le \frac {1}{8}$.[/quote]

I think it should be $ \frac{\sqrt{3}}{9}$ instead of $ \frac{1}{8}$ since I've posted that problem before. Here's the link http://www.mathlinks.ro/viewtopic.php?t=309472

Thanks Obel1x, I've just realised my mistake.


algebra
polynomial
geometry
geometric transformation
algebra proposed

I found this problem on kalva.demon.co.uk. It’ s Russia 1996. But there's obviously some mistake here, since polynomials $x^n$ give trivial solution. I don't think it can be so easy. Does anyone know the right formulation?

"Do there exist two finite sets such that we can find polynomials of arbitrarily large degree with all coefficients in the first set and all roots real and in the second set?"

Kalva's translation of the problem is dead wrong. I translate the problem from Kvant ([url=http://kvant.mccme.ru/1996/05/p23.htm]problem M1564 by E. Malinnikova[/url], problem 7 for the 11 class of [url=http://kvant.mccme.ru/1996/05/p53.htm]Allrussian 1996[/url]):

Does there exist a finite set M of nonzero real numbers such that for any natural number n, we can find a polynomial of degree $\geq$ n whose coefficients are elements of M, and whose roots are all real and also all belong to M ?

  Darij

oh, and you should inform John Scholes too :D


Does anybody know a formula for a coin problem?

A coin problem like this:
You have nickels, dimes, and quarters. How many ways can you total  $\$$ 1.00?
The answer is 29, but I need a formula to do this problem quicker.

I think somebody in this forum already mentioned this. I THINK.

If there is a formula, can somebody PLEASE explain it.

Using an example will help ALOT.

I don't think there is a formula.
There however, is a good way to divide the casework for this problem.
Divide the cases into the amount of quarters there are: 4,3,2,1,0
Case 1:Four quarters.
There is only one arrangement
QQQQ

Case 2: 3 quarter

We have used 75 cents, so we have 25 left.
You can have 0,1,2 dimes, and the rest nickels
3 more cases.

Case 3: 2 quarters

We have used 50 cents, so we have 50 more cents.
We can have 0,1,2,3,4,5 dimes and the rest nickels
6 more cases

Case 4: 1 quarter
We have used 25 cents so we have 75 more
0,1,2,3,4,5,6,7 dimes and the rest nickels
8 more cases.

Case 5: 0 quarters
We have 100 cents left
0,1,2,3,4,5,6,7,8,9,10 dimes and the rest nickels
11 more cases

Sum up the cases
1+3+6+8+11=29

I make charts when i did these problems, but using casework seems pretty good too.

I think if there was a formula for this kind of problem, it would be really similar to the one for the number of 0s at the end of $n!$. Try to find one based on the quarters/dimes thingy in my solution

I know who mentioned the formula for the coin problem.
It was "press alt f4"
He put it in Latex and said it could be simplified
Its in the "What we really need to know" topic
on page 9
Just go to search
and type in "dime"
make sure you narrow the forum to the Mathcounts forum.

geez that formula is hideous.
Now nearly useful enough for memorization.
Tonight I will try to come up with one for quarters,nickels,dimes

Heres the formula for dimes/nickels/quarters...
$\sum_{k=0}^{\lfloor\frac{n}{25}\rfloor+1}\lfloor\frac{n-25k}{10}\rfloor+1$

where $n$ is the number of cents
I think thats right

O yeah.
$n$ is the amount of cents

$k$ is the dummy variable

whats the use of "k" in the expression. Wouldn't 25k always equal 0

Basically you are starting with $k=0$ and evaluating it.
Then $k=1$ all of the way up to $\lfloor\frac{n}{25}\rfloor+1$

Can you give it in non-sigma notation? Thats usually easier for people that haven't learned summation to read.

I know sigma notation, but I don't understand why "k" is in the expression

How about just use casework and be done with the problem in like a minute rather than trying the remember an ultra-long fomula and then taking like 5 minutes trying to apply it.

Exactly. Don't actually use the formula. Its there for INFORMATIONAL PURPOSES :icecream:

biffanddoc, I especially like how you just wrote the possible amount of dimes and didn't bother to see how many nickels. I'll have to remember to do that next time.

[quote="biffanddoc"]i feel special *feels special*

That is because when you determine the amount of dimes / quarters, the amount left is always a multiply of 5 and thus you can make it out of nickels[/quote]

How would your casework change had the problem included pennies?

[quote="biffanddoc"]*has no idea*[/quote]

I'm going to assume that wasn't sarcasm. If so,

How would you go about solving if it had included pennies?

Well I think then you'd just be substituting even more.
Now, each amount of nickels can be replaced by pennies.
So this time you WOULD have to write out how many nickels there were, and just remind that with each nickel taken away 5 pennies can take the spot.

does the formula even work?

EDIT: when using your BIFFANDC's (sp?) formula, I got 28.5

No, almost all binomials are factorable...


calculus
derivative
real analysis
real analysis unsolved

[quote="fedja"][hide="Hint"]$\frac 1{x^2+1}=\frac i2\left(\frac 1{x+i}-\frac 1{x-i}\right)$[/hide][/quote]
It's Ok ! But which are the solutions ?

[quote="flip2004"]
It's Ok ! But which are the solutions ?[/quote]
[hide="Solution"]We are to solve $\left(\frac{x-i}{x+i}\right)^{n+1}=1$. Let $\zeta=e^{\frac{2\pi ik}{n+1}}$ ($k=1,2,\dots,n$) be any root of unity of degree $n+1$ different from $1$. Then $i\frac{1+\zeta}{1-\zeta}$ is a solution and there are no other roots. From here it is easy to see that the answer is $\cot \frac {\pi k} {n+1}$ ($k=1,\dots,n$)[/hide]

,,Fedja" solution =
We are to solve $\left(\frac{x-i}{x+i}\right)^{n+1}=1$. Let $\zeta=e^{\frac{2\pi ik}{n+1}}$ ($k=1,2,\dots,n$) be any root of unity of degree $n+1$ different from $1$. Then $i\frac{1+\zeta}{1-\zeta}$ is a solution and there are no other roots. From here it is easy to see that the answer is $\cot \frac {\pi k} {n+1}$ ($k=1,\dots,n$)

Indeed, nice solution !/Alex/proposer.


[b]EX1:[/b]
Let $ a, b, c \in R^3$ such as :
$ (\forall x \in R) : |ax^2\plus{}bx\plus{}c|\le 1$
1)Show that $ |c|\le 1$
2)Show that $ |a\plus{}c| \le 1$
3)Show that $ a^2\plus{}b^2\plus{}c^2 \le 5$
[b]EX2:[/b]
Let $ |x| \le \frac{1}{2}$ and $ |y| \le 1$
Prove that : $ |4x^2y\minus{}y\minus{}x| \le \frac{17}{16}$

I don't understand your first exercise. Your condition $ |ax^2\plus{}bx\plus{}c| \leq 1 \; \forall x \in \mathbb{R}$ immediately implies that $ a\equal{}b\equal{}0$ because polynomials of positive degree are, by nature, unbounded.


With x is positive real number, find the min value of $ P \equal{} x \plus{} \sqrt {x^2  \plus{} \frac{1}{x}}$

hello, let $ P(x)=x+\sqrt{x^2+\frac{1}{x}}$ you have to solve the equation $ P^'(x)=0$ with
$ P^'(x)=1+\frac{2x-\frac{1}{x^2}}{2\sqrt{x^2+\frac{1}{x}}}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, let $ P(x) = x + \sqrt {x^2 + \frac {1}{x}}$ you have to solve the equation $ P^'(x) = 0$ with
$ P^'(x) = 1 + \frac {2x - \frac {1}{x^2}}{2\sqrt {x^2 + \frac {1}{x}}}$.
Sonnhard.[/quote]
But equation $ P^'(x)=0$ result a 4th equation, that's difficult! :(

hello, but $ P^'(x)=0$ is equivalent with $ (2x-1)(4x^2+2x+1)=0$.
I think this will help you.
Sonnhard.

Right, in this case, it simplifies very nicely. If you isolate the radical and square, you quickly end up with $ x^3 \equal{} 1/8$.

Regardless, in general, setting the derivative equal to zero doesn't mean that you have to compute the derivative. In many cases, implicit differentiation can be cleaner and/or less scary:

[i]Start as Nguyen Ngoc Linh did:[/i]
$ (P \minus{} x)^2 \equal{} x^2 \plus{} \frac {1}{x} \implies xP^2 \minus{} 2x^2P \equal{} 1$

[i]Differentiate wrt x:[/i]
$ 2xPP' \plus{} P^2 \minus{} 4xP \minus{} 2x^2 P' \equal{} 0$

[i]Apply the conditions P'=0 and P>0:[/i]
$ P^2 \minus{} 4xP \equal{} 0 \implies P \equal{} 4x$

[i]Substitute that into the original equation.[/i]

Another idea to try is Lagrange multipliers; in other words, let $ y^2 \equal{} x^2 \plus{} \frac{1}{x}$ and minimize $ P \equal{} x \plus{} y$.  Of course, this is getting a little beyond Intermediate Topics.

[quote="Dr Sonnhard Graubner"]hello, but $ P^'(x) = 0$ is equivalent with $ (2x - 1)(4x^2 + 2x + 1) = 0$.
I think this will help you.
Sonnhard.[/quote]
Thanks very much, Dr Sonnhard Graubner ^^!


inequalities
limit
function
algebra unsolved
algebra

Does there exist $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(0)>0$ and $ f(x\plus{}y) \geq f(x)\plus{}yf(f(x))$ for all $ x,y \in \mathbb{R}$?

Anyone? Any ideas? :maybe:  Anything helps, please!

[quote="BanishedTraitor"]Does there exist $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(0) > 0$ and $ f(x \plus{} y) \geq f(x) \plus{} yf(f(x))$ for all $ x,y \in \mathbb{R}$?[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} y)\geq f(x) \plus{} yf(f(x))$

Let $ u > 0$.

$ P(x,u)$ $ \implies$ $ \frac {f(x \plus{} u) \minus{} f(x)}{u}\geq f(f(x)$
$ P(x \plus{} u, \minus{} u)$ $ \implies$ $ f(f(x \plus{} u)) \geq\frac {f(x \plus{} u) \minus{} f(x)}{u}$

And so the new assertion $ Q(x,u)$ : $ f(f(x \plus{} u)) \geq\frac {f(x \plus{} u) \minus{} f(x)}{u}\geq f(f(x))$ $ \forall x$, $ \forall u > 0$ and so $ f(f(x))$ is a non decreasing function.

Let's suppose $ f(f(x))\leq 0$ $ \forall x$, then $ Q(x,u)$ shows that $ \frac {f(x \plus{} u) \minus{} f(x)}{u}\leq 0$ $ \forall u > 0$ and so $ f(x)$ is a non increasing function. So $ f(f(x))\leq 0$ $ \implies$ $ f(f(x)) < f(0)$ $ \implies$ $ f(x) > 0$ $ \forall x$, in contradiction with $ f(f(x))\leq 0$ $ \forall x$.

So, $ \exists a$ such that $ f(f(a)) > 0$ and so (since $ f(f(x))$ is non decreasing), $ f(f(x)) > 0$ $ \forall x\geq a$

Then $ P(a,x)$ $ \implies$ $ f(x \plus{} a)\geq xf(f(a)) \plus{} f(a)$ and so $ \lim_{x\to \plus{} \infty}f(x) \equal{} \plus{} \infty$ and so $ \lim_{x\to \plus{} \infty}f(f(x)) \equal{} \plus{} \infty$.

So, $ \exists b$ such that $ f(f(b)) > 2$ and then $ P(b,x)$ $ \implies$ $ f(x \plus{} b)\geq xf(f(b)) \plus{} f(b)$ and so $ f(x \plus{} b) > 2x \plus{} f(b)$ $ \forall x > 0$

So, $ \exists c$ such that $ f(f(x)) > 0$ and $ f(x) > x \plus{} 1 > 0$ $ \forall x > c$

Then $ \forall x > c$ $ P(x,f(x) \minus{} x)$ $ \implies$ $ f(f(x))\geq f(x) \plus{} (f(x) \minus{} x)f(f(x))$ and so $ 0\geq f(x) \plus{} (f(x) \minus{} x \minus{} 1)f(f(x))$.

But this is impossible, since, $ \forall x > c$ :  $ f(f(x)) > 0$, $ f(x) > 0$ and $ f(x) > x \plus{} 1$

So, no such function exists.


inequalities
trigonometry
vector
geometry
invariant
trig identities
Law of Cosines

I've seen the following proof of the Cauchy-Schwarz inequality many times:

Letting $ \vec{a} \equal{} (a_1, a_2, \ldots , a_n)$ and $ \vec{b} \equal{} (b_1, b_2, \ldots , b_n)$, we have
\[ \parallel{}a\parallel{} \parallel{}b\parallel{} \equal{} \sqrt {(a_1^2 \plus{} a_2^2 \plus{} \ldots \plus{} a_n^2) (b_1^2 \plus{} b_2^2 \plus{} \ldots \plus{} b_n^2)}
\]
and
\[ a \cdot b \equal{} a_1b_1 \plus{} a_2b_2 \ldots \plus{} a_nb_n
\]
Since $ a \cdot b \equal{} \parallel{}a\parallel{} \parallel{}b\parallel{} \cos{\theta}$,
\[ \sqrt {(a_1^2 \plus{} a_2^2 \plus{} \ldots \plus{} a_n^2) (b_1^2 \plus{} b_2^2 \plus{} \ldots \plus{} b_n^2)} \geq a_1b_1 \plus{} a_2b_2 \ldots \plus{} a_nb_n
\]
I've also heard that this proof is bogus. Can someone explain?

[quote="darij grinberg"]Where do you know $ a \cdot b \equal{} \parallel{}a\parallel{} \parallel{}b\parallel{} \cos{\theta}$ from?

  darij[/quote]

err...I don't know if you are actually asking (probably not), but I'll say anyway,

Consider the triangle formed by the two vectors a, b and use law of cosines and the n-dimensional distance formula.

[quote="Altheman"]Consider the triangle formed by the two vectors a, b and use law of cosines and the n-dimensional distance formula.[/quote]

What is an angle between three points? How do you know that it exists? How do you define it otherwise than through the formula $ a \cdot b = \parallel{}a\parallel{} \parallel{}b\parallel{} \cos{\theta}$, which only makes sense if you can prove that $ |a\cdot b|\leq \parallel{}a\parallel{} \parallel{}b\parallel{}$ ?

The proof is bogus because it takes the existence of geometric notions in $ \mathbb{R}^n$ - for instance, angles, along with their properties (like the one that angles are real and thus have cosines in the interval $ [ - 1,1]$) - for granted. But their existence has to be proven like everything else in mathematics, and the proof is not particularly difficult (it is done in most Linear Algebra I courses) but it uses the very same Cauchy-Schwarz inequality that you "prove" using these notions! So your proof of Cauchy-Schwarz suffers from circular reasoning.
Similarly you could "prove" the inequality $ \sqrt {a^2 + b^2} + \sqrt {c^2 + d^2}\geq\sqrt {(a + c)^2 + (b + d)^2}$ by interpreting it as the triangle inequality in the plane - but actually you have to prove the triangle inequality in the plane first, and this is more or less done by proving the inequality $ \sqrt {a^2 + b^2} + \sqrt {c^2 + d^2}\geq\sqrt {(a + c)^2 + (b + d)^2}$ (actually, in a simpler form, using vectors - but it is still deduced algebraically from Cauchy-Schwarz). Algebra preceeds geometry, at least in $ \mathbb{R}^n$ for $ n\geq 4$ (and $ n\geq 3$ actually, though some people don't realize it).

In brief: You cannot use geometry to prove Cauchy-Schwarz, because you need Cauchy-Schwarz to make sure that geometry (in the form you want it) exists at all.

  darij

[quote="Phelpedo"]I was thinking something along those lines, but the usual counterargument I see is "Just take the plane containing the two vectors in question."[/quote]

This should be done more concretely - by proving that you can indeed reduce the situation to a plane (with $ a\cdot b$, $ \left|\left|a\right|\right|$ and $ \left|\left|b\right|\right|$ remaining invariant), and then you still have to prove Cauchy-Schwarz on a plane ($ \sqrt{x^2\plus{}y^2}\cdot\sqrt{a^2\plus{}b^2}\geq xa\plus{}yb$), unless you take plane Euclidean geometry for granted.

  Darij


:o

What is the mass in grams of 0.250 moles of Mg

and I need the steps for this too.PLZZZZZZZZZZZZZZZZZZ
and Which of the fallowing nitrogen-contain compounds would supply the most nitrogen per gram of fertilizer???

1) Potassium azide, KN3, Molecular Weight= 81.1g/m
2)There is not enough info to answer this problem.
3) Potasium nitrite, KN2, MW=85.1g/m
4)iron Azide, Fe(N3)2, MW=140g/m
5) Sodium Azide, NaN3, MW= 65g/m

a) Since the atomic mass of $Mg$ is $ 24 \frac{gm}{mol} $, the  weight of $ 0.25 $ moles is $ 0.25 \times 24 $ gm = $6 $gm

b)I'm not giving the answer . But the method is to find the weight of Nitogen in the compund. Do this by multiplying $14$ (atomic mass of $N$ ) by the no. of atoms in that compound. Divide this my the total weight of the compound and multiply by 100. Findwhich is the largest out of given options!
Hope this helps!! Which class are you in?


i don't do music, but i like to listen.

my favourite composer is edward grieg.

but i like c. debussy too. his prelude to the afternoon of a faun [i don't know the name of it in french  :( ] was quite revolutionary at that time, and it still hasn't lost it's beauty.

I enjoy the music by the Romantic composers most, Chopin is my most favourite composer:lol:. I particular enjoy his nocturnes, they are really sweet! :P

Chopin's Nocturnes are a joy to hear, as are some Beethoven Sonatas.

 :lol:

Beethoven's Sonatas are pretty good as well, I particularly enjoy the Pathetique Sonata, I think I should learn it sometime. :P

[quote="now a ranger"]I learned it last year and was state winner and would have played it at some university recital if I had not gone to summer camp.

I loved the beginning, slow, and the end of the 1st page, where you play down the keys on the piano before beginning the allegro part was really fun for me. I listened to a tape for a while, too.[/quote]

Oh! You are a state winner, How impressive! :o  Did you perform the Pathetique Sonata in the competition? Are you going to study (or studying) music in University?

I performed it in the competition. I was really nervous, but I didn't mess up. Really relieved afterwards. I was a 7th grader in 2006, so yeah... I think I won in Division 2 which is 7th grade to 10th grade.

Haha I have no idea what I'll do in the future

[quote="kate25"]This is mostly geared toward Baroque, Classical, and Romantic periods.[/quote]

I would say it's mostly geared towards piano music.

I voted for someone else, but I like Debussy a lot.

I thought Debussy's L'isle Joyeuse is almost perfectly written.  I played it for a competition.

I also like Prokofiev's Sonata #3, you can look both of these up on YouTube.

Uhm...none? Rachmaninov died in 1943, and he is the latest on the list. No one from our generation? Or at least semi-close to it. Look up Cage, Ligeti, Corigliano, Crumb, Adler, etc... The list goes on. They will change your lives forever, and you won't think the listed people make a good list for favorite composers anymore.

Now I'm not saying they weren't influential, or weren't important. I'm saying that as favorite composers you should have some people that are actually writing music that speaks to our generation.

I would vote for Brahms or Ravel, but of these, I think I like Beethoven the most.

I also echo Hilbert's sentiments in your exclusion of twentieth century composers;  I'm a huge fan of Ligeti and Corigliano...  I'm also surprised more people on this forum aren't interested in minimalism, given its structure and often mathematical basis.  Look up Glass, Reich, and John Adams sometime :).

Bach is LAME.

Ok well I think Bach was an amazing composer, but for me I'll definitely pass. Just don't find it interesting enough. I'll have to admit that I like the chromatic fantasy and fugue though.

My favorite composer is Rachmaninoff with Chopin following closely behind. For modern music it is Crumb (I believe that modern music is a very different kind of music than what has come before so I tend to separate them).

By the way I play piano rather seriously and am a composition major (ok really a double major in applied math and music composition).

I voted for Bach, but Eric Whitacre is my favorite modern composer: I've played a few of his pieces, and they are among my favorite songs, both to play and to listen to.  Additionally, his choral pieces are beyond amazing.

Hard to pick one, would probably go with Brahms, Beethoven, Chopin, Copland, Gershwin, Rachmaninoff (I know, I have trouble deciding)

I can't help but ask where Shostakovich is.

[size=200]Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak! Dvorak![/size]

Dojo, if you mean where Shostakovich is from, he is Russian.I played a piece by him at my first STate competition when I was nine. I had so much fun with it.

I mean where is he on the poll? (Along with Dvorak)

The poll dint really work for me but I :love: Beethoven sonatas

My favorites are the romantic/impressionistic composers, especially Rachmaninoff, Chopin, and Debussy.

My favorite composer is Chopin. (And the 200th anniversary of his birthday just passed approximately four months ago.)

I really enjoy listening to Mozart's pieces, his music is pretty exciting to me.  :thumbup:

I like Bach, but I appreciate Romantic composers more.
The ornaments and improvisations in Baroque are nice, though...

[quote=brianapa]IKR
There is no Dvorak or Mendelssohn OR TCHAIKOVSKY!!!
Rip[/quote]

or schumann


[quote="thomaskurian89"]Let $ n$ be a positive integer with all digits equal to $ 5$ such that $ n$ is divisible by $ 2003$. Find the last six digits of $ \frac {n}{2003}$.[/quote]
We know that $ n \equal{} \frac{5}{9} \cdot (10^m \minus{} 1)$ for some $ m$. But it's easy to see that $ m \ge 6$. So:
$ \frac{n}{2003} \equiv \frac{\minus{}5}{9 \cdot 2003} \equiv \minus{}5 \cdot 9^{\minus{}1} \cdot 2003^{\minus{}1} \bmod 10^6$

First we will find $ 9^{\minus{}1} \bmod 10^6$:
$ \frac{1}{9} \equiv \frac{\minus{}10^6\plus{}1}{9} \equal{} \frac{\minus{}999999}{9} \equiv \minus{}111111 \bmod 10^6$

Know to $ 2003^{\minus{}1} \bmod 10^6$:
$ \frac{1}{2003} \equiv \frac{k \cdot 10^6\plus{}1}{2003} \bmod 10^6$. And we want to find k such that $ 2003 \mid k10^6\plus{}1$. So $ k \equiv \minus{}(10^6)^{\minus{}1} \equiv \minus{}(503)^{\minus{}1} \bmod 2003$. $ k \equiv \minus{}\frac{1}{503} \equiv \minus{}\frac{1\plus{}2003j}{503} \bmod 2003$ So we want to find $ j$ such that $ j \equiv \minus{}(2003)^{\minus{}1} \equiv 9^{\minus{}1} \equiv \frac{1}{9} \equiv \frac{504}{9} \equiv 56 \bmod 503$.
Setting $ j \equal{} 56$ we get:
$ k \equal{} \minus{}\frac{1\plus{}2003j}{503} \equal{} \minus{}\frac{1\plus{}2003 \cdot 56}{503} \equal{} \minus{}223$.
And then $ \frac{1}{2003} \equiv \frac{1\minus{}223 \cdot 10^6}{2003} \equiv \minus{}111333$

So $ \frac{n}{2003} \equiv \minus{}5 \cdot 9^{\minus{}1} \cdot 2003^{\minus{}1} \equiv \minus{}5 \cdot 111111 \cdot 111333 \equiv \minus{}604815 \equiv 395185 \bmod 10^6$.

So the last six digits is: 395185

A good solution would probably look like:

Since $ n$ consists of only $ 5$ digits we can write $ n \equal{} \frac{5}{9} (10^m\minus{}1), m \in \mathbb{N}$. Since $ 2003 \nmid 5,55,555,5555,55555,555555$ we conclude $ m \ge 6$.
Thus $ \frac{n}{2003} \equiv \frac{\frac{5}{9} (10^m\minus{}1)}{2003} \equiv \frac{5 \cdot 10^6\minus{}5}{9 \cdot 2003} \bmod 10^6$.
First we note that $ 9 \cdot (\minus{}111111) \equiv 1 \bmod 10^6$ and $ 2003 \cdot (\minus{}111333) \equiv 1 \bmod 10^6$.

Thus $ \frac{n}{2003} \equiv \frac{5 \cdot 10^6\minus{}5}{9 \cdot 2003} \cdot 9 \cdot (\minus{}111111) \cdot 2003 \cdot (\minus{}111333) \equal{} (5 \cdot 10^6 \minus{} 5) \cdot (\minus{}111111) \cdot  (\minus{}111333) \equiv \minus{}5 \cdot (\minus{}111111) \cdot  (\minus{}111333) \equiv \minus{}604815 \equiv 395185 \bmod 10^6$.
From here it's easy to see that the last six digits (base 10) of $ \frac{n}{2003}$ is $ \boxed{395185}$


For $ n \geq 2$, we note $ p_{n}$ the number of permutations $ \sigma$ of the set {$ 1, . . . , n$} such that there is a single
integer $ i \in$ {$ 1, 2, . . . , n \minus{} 1$} such that $ \sigma$(i) > $ \sigma$(i + 1). Find $ p_{n}$.

 :)

If $ \sigma(n) \equal{} n$ we count $ p_{n\minus{}1}$ such permutations. If $ \sigma(n) \neq n$, there exists $ 1\leq i \leq n\minus{}1$ such that $ \sigma(i) \equal{} n$, hence $ \sigma(i) > \sigma(i\plus{}1)$, and this has to be uniquely such, therefore all the other elements can be chosen in $ \genfrac {(} {)} {0pt} {} {n\minus{}1} {i\minus{}1}$ ways: $ i\minus{}1$ increasing elements to the left, and $ n\minus{}i$ increasing elements to the right.

Hence $ \displaystyle p_n \equal{} p_{n\minus{}1} \plus{} \sum_{i\equal{}1}^{n\minus{}1} \genfrac {(} {)} {0pt} {} {n\minus{}1} {i\minus{}1} \equal{} p_{n\minus{}1} \plus{} 2^{n\minus{}1} \minus{} 1$. Since $ p_2 \equal{} 1$, this yields $ \displaystyle p_n \equal{} p_2 \plus{} \sum_{k\equal{}2}^{n\minus{}1} (2^{k} \minus{} 1) \equal{} \boxed{2^n \minus{} n \minus{} 1}$.

I have a question that I may have misunderstood this problem.
[b]1.How to understand the number of permutations $ \sigma$ of the set ?
How to cauculate?
2.$ \sigma(i) > \sigma(i \plus{} 1)$
What is $ \sigma(i)$[/b] Here?
Can it be expressed?

:roll:

$ \sigma$ is a bijection from $ [n] \equal{} \{1, \ldots, n\}$ to itself.

[b]Thanks to all.
I have got it with your help.
By the way.
[quote="Agr_94_Math"][color=darkred]Bulgarian Math Olympiad problem[/color].[/quote]
[color=red]Can you refer which year or post the links?[/color]
 :) [/b]


analytic geometry
geometry
vector
conics
ellipse
power of a point
radical axis

Given any non-trapezoidal quadrilateral ABCD (no two opposite sides are parallel). Extend each of AB, BC, CD, and DA in both directions. Let E and F be the points of intersection of these rays.  Then the midpoints of AC, BD, and EF are collinear.

[i]Alternative formulation:[/i] Let ABCD be a quadrilateral such that two pairs of opposing sides are not parallel.
E is the intersection of AB and CD, F is the intersection of AD and BC. Let L, M and N be the midpoints of the segments AC, BD and EF, respectively. Prove that L, M and N lie on one straight line.

I have already found a way to prove this with the Cartesian method of assigning coordinate variables and then matching the line equation of the 3 midpoints.

But, I am wondering: is there a way of doing this proof under Euclidean methods (ie. angles, congruence theorems, etc.)?

The link below provides with a diagram, and two ways of proving this theorem. I don't understand either of them.
 :blush:

http://www.cut-the-knot.org/Curriculum/Geometry/Quadri.shtml#explanation

Thank you for your help!

[b]Moderator edit:[/b] Picture included.

There is a nice proof using radical axis.

[b]Lemma:[/b]
We have a $\triangle{ABC}$. Then orthocentre of  $\triangle{ABC}$ has the same power of a point with respect to all circles, which diamaters are all cevians from points $A,B,C$.

Let's come back to our problem
Consider four triangles : $\triangle{FAD},\triangle{EDC},\triangle{FBC},\triangle{EAB}$ and three cevians: $FE,AC,BD$ and three circles, which diametres are these cevians. Now notice, that each cevian is a cevian of each traingle and these tree cevians are cevians of each trangles, hence
orthocentres of abovementioned trangles lie on a line and this line a radical axis of three aforementioned circles.
If three circles has radical line, then their centres lies on a line, but their centres are midpoints of $EF, AC, BD$. The end.

[quote="Ottoman"]Given any non-trapezoidal quadrilateral ABCD (no two opposite sides are parallel). Extend each of AB, BC, CD, and DA in both directions. Let E and F be the points of intersection of these rays.  Then the midpoints of AC, BD, and EF are collinear.

I have already found a way to do this with the Cartesian method of assigning coordinate variables and then matching the line equation of the 3 midpoints.

But, I am wondering: is there a way of doing this proof under Euclidean methods (ie. angles, congruence theorems, etc.)?

The link below provides with a diagram, and two ways of proving this theorem. I don't understand either of them.
 :blush:

http://www.cut-the-knot.org/Curriculum/Geometry/Quadri.shtml#explanation

Thank you for your help![/quote]

I had the same problem, but I got the solution, but I have it in spanish...

[quote="dondigo"]There is a nice proof using radical axis.[/quote]

I appreciate your help; it's a shame that we are not at the stage of learning radical axis yet. So far, we have only used angular theorems. If you have a solution which uses more basic geometry concepts, I would love to hear them.

[quote="hucht"]I had the same problem, but I got the solution, but I have it in spanish...[/quote]

That's fine, since I only need to look at the general idea rather than needing to copy. Worst case scenario, I have a program for translation. :)

All the proofs using radical axis or power of a point can be translated into proofs with similar triangles. I've proven that theorem long ago but I can't really remember it. I'm ashamed to say that now when I look at it I have no clue how to solve it. Anyway, I think there are 3 or 4 different proofs.

[quote="M@re"][color=red]
$L$ and $M$ are the midpoints of the diagonals $AC$ and $BD$ of a quadrilateral $ABCD$. The lines $AB$, $CD$ intersects at $E$, and the lines $AD$, $BC$ intersects at $F$. Let $N$ be the midpoint of $EF$. Then points $L$, $M$, $N$ are collinear. (Line $(L,M,N)$ is called Newton line)
[/color]
Solution: Let $P$,$Q$,$R$ be the midpoints of segments $AE$, $AD$, $DE$ respectively. Then $L$, $M$, $N$ are on the lines $PQ$, $QR$, $RP$ respectively. Apply the Menelaus theorem to $\triangle EAD$ and points $B,C,F$, that is the same like Menelaus theorem on triangle $PQR$ and points $L,M,N$  and there you go.

bye[/quote]

but how would you show, if for example you were using triangle ABC, and points D, E, F lie on lines BC, AC, AB, then points D, E, F are colinear when

   AF/FB x BD/DC x CE/EA = -1

which is what menelaus theorem is basically, i guess what i am asking is how would you proceed to show that points D, E and F are collinear after you prove the above statement.

Because, please correct me if i am wrong, but isnt the menelaus theorem assuming that points D,E,F are already collinear, in a way.

Please tell me how to proceed,

Or is there a much simpler way to prove the question below

Given any non-trapezoidal quadrilateral ABCD (no two opposite sides are parallel). Extend each of AB, BC, CD, and DA in both directions. Let E and F be the points of intersection of these rays. Then the midpoints of AC, BD, and EF are collinear.

First of all, I don't know how old are you, and I suppose that I am
not old enough (and smart enough) to give advice of this type but...
it's very important that you know some other techniques for solving
geometry problem except elementary ways, so eventually you will have
to learn some other stuff.

Second: Menelaus theorem is "IFF" statement, in other words it's
says that points $D,E,F$ on lines $AB,BC,AC$ are collinear IFF
$\frac{AD}{DB}\cdot \frac{BE}{EC}\cdot \frac{CF}{FA}=1$ (it's not
$-1$ when you don't use vector which will be the case most of the
time). Proof of this theorem is strictly elementary (one direction
is to construct perpendicular lines from $A,B,C$ to line $DEF$, and
other is to suppose that intersection point of line $DE$ with $AC$
is $F'$ and to prove that from $\frac{AD}{DB}\cdot \frac{BE}{EC}\cdot \frac{CF'}{F'A}=1$ (which you have already
proven) and $\frac{AD}{DB}\cdot \frac{BE}{EC}\cdot \frac{CF}{FA}=1$
you obtain $F'=F$). So Menelaus theorem is also an elementary
theorem.

bye

Thank you for clarifying this for me,

its been a while since i have done discrete math, i am just trying to solve a similar problem for my son. i guess i a have lost my touch.

thanks again

and by the way,
[quote]L and M are the midpoints of the diagonals AC and BD of a quadrilateral ABCD. The lines AB, CD intersects at E, and the lines AD, BC intersects at F. Let N be the midpoint of EF. Then points L, M, N are collinear. (Line (L,M,N) is called Newton line)

Solution: Let P,Q,R be the midpoints of segments AE, AD, DE respectively. Then L, M, N are on the lines PQ, QR, RP respectively. Apply the Menelaus theorem to \triangle EAD and points B,C,F, that is the same like Menelaus theorem on triangle PQR and points L,M,N and there you go. [/quote]

how can i prove (justify) that the points L, M and N, end up being points on the lines PQ, QR, RP. Would you just say that the reason is "By Construction", or would there he a different method.
Just curious as to how you would do it.

thanks again

Another simpler proof:
We known, if $K,L$ midpoints of $BD,EF$ by an old theorem: $[KAL]=[KLC]=\frac{[BFDE]}{4}$. Then the altitudes $AP, CQ$ from $A,C$ to $KL$ line are congruent, further $KL$ cut $AC$ at $M$, therefore: the right triangles $AMP$ and $CQM$ are congruent, so: $AM=MC$, hence $M$ is midpoint of $AC$. QED.

If you have triangle $ABC$ and midpoints $M,N$ of sides $AB,AC$ then line $MN$ intersects every segment $AD$ where $D\in BC$ in its midpoint. You prove that by Talles theorem.

You can find 30 Euclidean proofs of this theorem in the book intitled "Méthodes et techniques en géométrie, A propos de la droite de Newton" by Jean-Louis Ayme, Editions Ellipses, Paris (2003).
He did tell me that my proof is in his book.

[hide="hint"]Use lemma follow

Let $A$, $B$, $C$, $D$ be four points in the plane. Denote $O$ is common point of $AC$ and $BD$. $H$ is orthorcenter of triangle $OAB$, $K$  is orthorcenter of triangle $OCD$. $X$ is midpoint of $AD$, $Y$ is midpoint of $BC$. Then $HK\perp XY$.[/hide]

[quote="Problem"]Let ABCD a convex quadrilateral such two pairs of opposing sides are not parallel.
E is an intersection of AB and CD, F is an intersection of AD and BC. Let L, M and N
the average points of the segments AC, BD and EF, respectively. Prove that L, M and N are in one same straight line.[/quote]

I got this solution yesterday, but i don't know how can i put a picture here.. sorry !

Let a parallel on $CD$ for $N$, determining $K$ on $DF$ and $H$ on $CF$. As $N$ it is average point of $EF$, then in the triangle $DEF$, $KM$ it is average base of $OF$ and therefore $K$ is average point of $DF$. With this, in triangle $CDF$, $KH$ is average base of $CD$ and thus $H$ is average of $CF$. In triangle $ACF$, $H$ and $L$ are the respective average points of $CF$ and $AC$. Thus, $HL$ is average base of $AF$ and then $HL$ is parallel of $AF$. Either $G$ the intersection of $HL$ with $CD$. In triangle $CDF$ as $H$ it is average point of $CF$ and $HG$ is parallel of $FD$ (since $Hl$ is parallel the $AF$) then $HG$ is average base of $DF$ and thus $G$ is average point of $CD$. In the same way, in triangle $BDF$, $K$ and $M$ are the respective average points of $DF$ and $BD$. Thus $KM$ is average base of $BF$ and then $KM$ is parallel of $BF$. Either $G'$ the intersection of $KM$ with $CD$. In triangle $CDF$, as $K$ is average point of $DF$ and $KG'$ is parallel $FC$ (since $KM$ is parallel the $BF$) then $KG'$ is average base of $CF$ and thus $G' \equiv G$ is average point of $CD$. By Menelaus in the triangle $CDF$ with the straight line $ABE$:
$1=\frac{AD.BF.EC}{AF.BC.ED}=\frac{2GL.2KM.2HN}{2HL.2GM.2KN}=\frac{GL.KM.HN}{HL.GM.KN}$

Then by Menelaus on triangle $GKH$ with the straight line $ABE$ we have that $L, M, N$ are collinear.

[sorry about my english]

[quote="santosguzella"]

Then by Menelaus on triangle $ GKH$ with the straight line $ ABE$ we have that $ L, M, N$ are collinear.

[/quote]

Better "triangle $ GKH$, reciprocal of Menelaus for points $ l$,$ M$ and $ N$ we have that these points are collunear" :wink:

I've made your figure ! and ...you have (re) discovered the solution from [url]http://www.cut-the-knot.org/Curriculum/Geometry/Quadri.shtml[/url]  It's the Proof#2 !

P.S. sorry for revival but ...NEWTON IS VERY IMPORTANT ! :oops:


superior algebra
superior algebra unsolved

$ H$ is a finite group.

$ G \equal{} H*H$. For $ z$ in $ H$ and $ (x,y)$ in $ G$, the following is defined:

$ (x,y)*z \equal{} x*z*y^{ \minus{} 1}$

Need to show that $ H$ is a $ G$-set under this action.

Let $ g_1 \equal{} (x,y)$ and $ g_2 \equal{} (m,n)$.

Then $ g_1* ((g_2)*z) \equal{} (x,y)* ( (m,n)* z) \equal{} (x,y)* (m*z*n^{ \minus{} 1}) \equal{} x*m*z*n^{ \minus{} 1}*y^{ \minus{} 1}$

$ (x*m)*z*(n*y)^{ \minus{} 1} \equal{} ((x*m),(n*y))*z$.

But $ ((x*m),(n*y))*z$ $ \not \equal{}$ $ ((x,y),(m,n))*z$

Did I miss anything in the above?

Need to show that $ (x*m)*z*(n*y)^{ \minus{} 1} \equal{} ((x,y),(m,n))*z$

Thank you in advance!

[quote="Soarer"][quote="Orbit"]
$ (x*m)*z*(n*y)^{ \minus{} 1} \equal{} ((x*m),(n*y))*z$.

But $ ((x*m),(n*y))*z$ $ \not \equal{}$ $ ((x,y),(m,n))*z$
[/quote]

So unless I misunderstand you, you seem to make a mistake here by thinking that $ (xm,ny)*z \equal{} g_1*(g_2*z)$ (because of the inverse thing I mentioned in the previous post)[/quote]

Yes, I think that there is a mistake in my calculations above. The method seems to be correct. I am just not sure how to get around it to show that $ g_1*(g_2*z) \equal{} (g_1*g_2)*z$.

Following your calculation,
$ g_1 * (g_2 *z) \equal{} x*m*z*n^{\minus{}1}*y^{\minus{}1} \equal{} (xm)*z*(yn)^{\minus{}1}$ (This is probably where you made your mistake)
$ (g_1g_2)*z \equal{} (xm,yn)*z \equal{} xm*z*(yn)^{\minus{}1}$.
So they are equal.

[quote="Soarer"]Following your calculation,
$ g_1 * (g_2 *z) \equal{} x*m*z*n^{ \minus{} 1}*y^{ \minus{} 1} \equal{} (xm)*z*(yn)^{ \minus{} 1}$ (This is probably where you made your mistake)
$ (g_1g_2)*z \equal{} (xm,yn)*z \equal{} xm*z*(yn)^{ \minus{} 1}$.
So they are equal.[/quote]

Thank you!!!


i know this girl, whos birthday is on thursday.
she is now on her first collage year, but in high school in sweden she won 3 bronze medals, and shes the only one that has ever been to IMO from sweden in their first year i high school.
now i wanted to give her a really hard problem (as hard as the most difficult IMO problems) on thursday that she havent seen before, but i didnt know where to post so i posted in this forum.
i doubt problems from russia are any good couse her father and brother are former russian IMO medal winners and she might have seen it then.
anyone know a problem that isnt so well known and that might be good?

Hello Peter,

I guess you are talking about Valentina. Send my regards on her birthday. You can, among other places on the forum, find difficult and interesting problems under the "Contest" button on the top of the page.

/Jacob

[quote="Pascual2005"]You shouldnt give her a problem ask her out and invite her to eat something in avery nice place....I believe it will have a better result :)[/quote]

i know but the thing is that she has a lot of study to do and she and i only got the relationship as she as a leader of a math group (math circle we call it here, a sort of group training problem solving) which contains of two people, me and one more. and im 14, she is about 20 i think so it would be kind of ridicolous.

[quote="siuhochung"]had she done all imo past paper?
If not, why don't you give her a imo q3 /q6? Supposingly a bronze medalist cannot do them. (Except James Bond  :lol: )[/quote]

i think she has seen almost all IMO problems, but thanks anyway.

[quote="Valentin Vornicu"]You can give her some ML Contest problems from the previous editions. Try Round 3 from the first contest. Any problem should give her a very long head-ache :P :D[/quote]

what is ML??

[quote="pezeta90"][quote="Valentin Vornicu"]You can give her some ML Contest problems from the previous editions. Try Round 3 from the first contest. Any problem should give her a very long head-ache :P :D[/quote]

what is ML??[/quote]ML= MathLinks EveryOne (the story is complicated, basically I am talking about this site :) ).
MathLinks Contest - http://www.mathlinks.ro/Forum/contest.php

Try this problem posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33734]here[/url]. It seems to be very easy, but in fact none has found a clear solution up to now  :lol:

I remember Valentina from IMO 2004 :) , but I think that she probably doesn't know who I am because I didn't get the chance to talk to her.

Do you happen to know what her favourite topics are?  It would probably be easier to suggests problems from a particular category.  Initially, I would have suggested that you look for some problems from the Olympiad Problems section, but from your later posts I guess you probably want some help with judging their difficulty.

Of course, I guess any problem which receives a lot of praise from some posters such as Pierre, Myth, harazi, Namdung, Vasc etc. would probably make an excellent choice :P , but it's also not that easy to find those...

[quote="pezeta90"]her favorite topic is number theory[/quote]You can give her the 2nd problem in [url=http://www.mathlinks.ro/Forum/contest/3rd%20Edition/3rd%20edition%20-%20round%206.pdf]this file[/url].


Wow, this is like really really really old and may have been posted a million times by now. You're dividing by zero somewhere.  :wink:

[quote="Chocoboom"]Whats wrong with this proof?

$a=b$
$a^{2}=ab$
$a^{2}-b^{2}=ab-b^{2}$
$(a+b)(a-b)=b(a-b)$
$a+b=b$
$a+a=a$
$2a=a$
$2=1$[/quote]
You go from $(a+b)(a-b)=b(a-b)$ to $a+b=b$, which isn't valid because $a=b$ and $a-b=0$.

The error is much clearer when you put in numbers.  Say that $a = 2$.  Then

$2^{2}= 2 \cdot 2$
$2^{2}-2^{2}= 2 \cdot 2-2^{2}$
$(2+2)(2-2) = 2(2-2)$
$2+2 = 2$

Uh-oh, we did something bad there.


function
ceiling function
induction
pigeonhole principle
graph theory
algebra unsolved
algebra

Let $n$ be a positive integer and let $S$ be a set of $2^n+1$ elements. Let $f$ be a function from the set of two-element subsets of $S$ to $\{0, \dots, 2^{n-1}-1\}$. Assume that for any elements $x, y, z$ of $S$, one of $f(\{x,y\}), f(\{y,z\}), f(\{z, x\})$ is equal to the sum of the other two. Show that there exist $a, b, c$ in $S$ such that $f(\{a,b\}), f(\{b,c\}), f(\{c,a\})$ are all equal to 0.

[b]Lemma[/b] Given a complete graph on $k\ge 3$ vertices whose edges are colored in black and white s.t. every triangle has an even number of black edges (either $0$ or $2$), we can find a complete subgraph on $\left\lceil\frac k2\right\rceil$ vertices all of whose edges are white.

[b]Proof[/b]

It’s easy to see that the partial graph generated by all the vertices and the black edges is a complete bipartite graph. If $A,B$ is its bipartition, the subgraphs induced on $A,B$ are both complete graphs with white edges only, and at least one of $|A|,|B|$ is $\ge\left\lceil\frac k2\right\rceil$, so we’re done.

We can now prove the assertion by induction on $n$. Suppose it's true for $n-1$. Construct a graph $G$ on a set $S$ with $k=2^n+1$ vertices, and color the edge $(a,b)$ white if $f(\{a,b\})$ is even and black if it's odd. By applying the lemma, we conclude that $S$ has a subset $\bar S$ with $2^{n-1}+1$ elements s.t. all the pairs of elements in $\bar S$ are mapped onto even numbers. Now divide these numbers by $2$ to get a map from a set with $2^{n-1}+1$ elements to $\{0,1,\ldots,2^{n-2}-1\}$ which still satisfies the condition that for any $x,y,z$, one of $f(\{x,y\}), f(\{y,z\}), f(\{z, x\})$ is the sum of the other two, and we can apply the induction hypothesis.

Also in the condition

$2^{n} +1$ can be replaced by $2n +1$ and
$2^{n-1} -1$ with $n -1$

:D.

This conculsion is based on another solution- not by induction but directly !

We'll prove the result for the problem using $2n+1$ and $n-1$.

We consider a graph with $2n+1$ vertices and label each one with a number; in any triangle, the sum of two labels equals the third. We want to prove that there is a triangle with all labels zero. If there are more than $n$ labels that are zero, then some vertex has at least edges labeled zero, which gives us a triangle with all labels zero. If there are at most $n$ labels that are zero, and no vertex has two edges that are labeled zero, then there is a complete subgraph of at least $n+1$ vertices that contains no zero edges (to see this, for each edge labeled zero, take one of the endpoints and throw away the other, and then take all remaining vertices).

It remains to derive a contradiction from the existence of this graph of $n+1$ vertices. Indeed, consider the edge AB with the largest label $k \le n-1$. For each vertex C, the labels on CA and CB add up to $k$. By the pigeonhole principle, there exist vertices X and Y that are not B such that AX and AY have the same label, which we'll call $a$. Then BX and BY also have the same label $b$, and since the label on XY is nonzero, it must be both $2a$ and $2b$. It follows then that the label on XY is $k$, and $a = b = \frac{k}{2}$.

Now consider some fifth vertex Z. Out of the labels on ZA, ZB, ZX, and ZY, suppose WLOG that ZA and ZX are the biggest. They are both more than $\frac{k}{2}$, so their difference must equal $\frac{k}{2}$, since that's the label on AX. But then one of the labels ZA and ZX must be at least $k$, which means one of them equals $k$, WLOG the label on ZA. But then the label on ZB must equal $0$, since the label on $AB$ is also $k$. This is our desired contradiction, so the result is proven.

[quote="imortal"]Also in the condition

$ 2^{n} \plus{} 1$ can be replaced by $ 2n \plus{} 1$ and
$ 2^{n \minus{} 1} \minus{} 1$ with $ n \minus{} 1$

:D.

This conculsion is based on another solution- not by induction but directly ![/quote]
What about the induction ?  :wink:  How we can prove it by induction ?

solved with graph theory but didnt include graph theory in sol

[hide=Solution]
Assume the contrary. Let $m\leq 2^{n-1}-1$ be the maximum element in the range of $f$. Observe that there exists no $\{a,b,c\}\subseteq S$ such that two of $f(\{a,b\})$, $f(\{b,c\})$, $f(\{c,a\})$ are equal to $0$, as the third one would also have to equal $0$. Additionally, there exists no $\{a,b,c,d\}\subseteq S$ such that $f(\{b,a\})=f(\{c,a\})=f(\{d,a\})>\frac{m}{2}$, as we would have $f(\{b,c\})=f(\{c,d\})=f(\{d,b\})=0$. Take $x$ and $y$ such that $f(\{x,y\})=m$. All $z\in S\setminus\{x,y\}$ satisfies $f(\{z,x\})+f(\{z,y\})=m$. There are two cases.

[b]Case 1: [/b]$m$[b] is odd.[/b] Observe that for all $z\in S\setminus\{x,y\}$, one of $f(\{z,x\})$, $f(\{z,y\})$ is at least $\frac{m+1}{2}$. Thus for all $k>0$, there are at most $4$ $z\in S\setminus\{x,y\}$ such that one of $f(\{z,x\})$, $f(\{z,y\})$ is equal to $k$ (otherwise WLOG there exists $\{z_1,z_2,z_3\}\subseteq S$ such that $f(\{z_1,x\})=f(\{z_2,x\})=f(\{z_3,x\})=\max\left(k,m-k\right)\geq\frac{m}{2}$, a contradiction). Additionally, there are at most $2$ $z\in S\setminus\{x,y\}$ such that one of $f(\{z,x\})$, $f(\{z,y\})$ is equal to $0$ (otherwise WLOG there exists $\{z_1,z_2\}\subseteq S$ such that $f(\{z_1,x\})=f(\{z_2,x\})=0$, a contradiction). Since $|S\setminus\{x,y\}|=2^n-1$, it follows that
\[
2^n-1\leq 2+4\left(\frac{m-1}{2}\right)\leq 2^n-2,
\]
a contradiction.

[b]Case 2: [/b]$m$[b] is even.[/b] Observe that for all $z\in S\setminus\{x,y\}$, one of $f(\{z,x\})$, $f(\{z,y\})$ is at least $\frac{m}{2}$. Thus for all $k>0,k\neq\frac{m}{2}$, there are at most $4$ $z\in S\setminus\{x,y\}$ such that one of $f(\{z,x\})$, $f(\{z,y\})$ is equal to $k$ (otherwise WLOG there exists $\{z_1,z_2,z_3\}\subseteq S$ such that $f(\{z_1,x\})=f(\{z_2,x\})=f(\{z_3,x\})=\max\left(k,m-k\right)\geq\frac{m}{2}$, a contradiction). Additionally, there are at most $2$ $z\in S\setminus\{x,y\}$ such that one of $f(\{z,x\})$, $f(\{z,y\})$ is equal to $0$ (otherwise WLOG there exists $\{z_1,z_2\}\subseteq S$ such that $f(\{z_1,x\})=f(\{z_2,x\})=0$, a contradiction).

We also claim that there exists at most $4$ $z\in S\setminus\{x,y\}$ satisfying $f(\{z,x\})=f(\{z,y\})=\frac{m}{2}$. Indeed, if there exists $\{z_1,z_2,z_3,z_4,z_5\}\subseteq S\setminus\{x,y\}$, for all $i\neq j$, $f(\{z_i,z_j\})$ is equal to $0$ or $m$. Clearly there must exist WLOG $z_i,z_j$ satisfying $f(\{z_i,z_j\})=0$. WLOG this is $z_1,z_2$. It follows that $\forall i\in [3,5]:f(\{z_i,z_1\})=f(\{z_i,z_2\})$. Assume they are all $m$ (otherwise we are done). it follows that $f(\{z_3,z_4\})=f(\{z_4,z_5\})=f(\{z_5,z_3\})=0$, a contradiction.

Since $|S\setminus\{x,y\}|=2^n-1$, it follows that
\[
2^n-1\leq 2+4\left(\frac{m}{2}-1\right)+4\leq 2^n-2,
\]
a contradiction.
[/hide]


invariant
vector
function
group theory
abstract algebra
superior algebra
superior algebra unsolved

Consider simple example of Lie group: $R$.
$f_{1}(x)=x+1$, and $f_{2}(x)=x+2$ are two smooth field on $R$, (actually, invariant field).
Then by a general thm, $[f_{1}, f_{2}]$ should still be a invariant field (actually should be zero field).

But my calculation shows:
$[f_{1}, f_{2}]=Df_{1}\bullet f_{2}-Df_{2}\bullet f_{1}=1 \times (x+2)-1 \times (x+1)=1$, this is
not zero, not even invariant. Where wrong? Thanks.

[quote="mlok"]As far as I understand invariant vector fields, $f_{1}$ and $f_{2}$ are not invariant.[/quote]

Tangent space of $R$ is still $R$, so a field is $f: R\to R$.

$R$ is an additive group, $f_{1}(t+x)=t+f_{1}(x)$ for any $t \in R$.

[quote="mlok"]More precisely, the tangent space $T_{x}$ at each point $x\in\mathbb R$ is isomorphic to $\mathbb R$. These are separate copies of $\mathbb R$ that together form a two-dimensional tangent bundle.

The field $f_{1}(x)=x+2$ consists of vectors of variable length: length $2$ at $x=0$, length $3$ at $x=1$, etc. In particular, $f_{1}(0)$ does not go into $f_{1}(1)$ under group translation.[/quote]

Thanks.  my mistake is I confused $R$'s action on $R$ and $T_{x}(R)$.
consider function $m: R \to R$ defined by$x \mapsto s \bullet x=s+x$ for fixed $s$,
the tangent linear mapping of this function is the action on $T_{x}(R)$, that is
$T_{x}(m): T_{x}(R) \to T_{s+x}(R)$ defined by $u \mapsto Dm \bullet u=1 \times u=u$.

So an example of invariant is $f_{1}(x)=2$, not$f_{1}(x)=x+2$.


a,b,c>0
prove\[ \frac{{a \plus{} \sqrt {ab}  \plus{} \sqrt[3]{{abc}}}}{3} \ge \sqrt {a.\frac{{a \plus{} b}}{2}\frac{{a \plus{} b \plus{} c}}{3}}
\]

[quote="nam_tran1237"]a,b,c>0
prove
\[ \frac {{a \plus{} \sqrt {ab} \plus{} \sqrt [3]{{abc}}}}{3} \ge \sqrt {a.\frac {{a \plus{} b}}{2}\frac {{a \plus{} b \plus{} c}}{3}}
\]
[/quote]
It can be $ \frac {{a \plus{} \sqrt {ab} \plus{} \sqrt [3]{{abc}}}}{3} \le \sqrt {a.\frac {{a \plus{} b}}{2}\frac {{a \plus{} b \plus{} c}}{3}}.$
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1820


[quote="astuteverbosity"]I think it's up to me to revive this forum
here, try this problem:

There are 3 rivers and after each river lies a grave. So there are 3 rivers and 3 graves. A man wants to leave the SAME amount of flowers at each grave, and be left with none at the end. What happens though is that each time he passes through one of the rivers the number of flowers he has doubles. So he has to start off with what number of flowers, taking into consideration that they double, so that he is left with no flowers whatsoever at the end?

Post your solution with answer an hide both
for example:[hide="answer"]answer[hide="solution"]solution[/hide][/hide][/quote] Do you mean the least possible amount of flowers he can start with?

[quote="astuteverbosity"]no, there is only one possible answer (i think)[/quote]

It works when you start with any multiple of 7 and use 8/7 of that number each time.


There are 2+6=8 faces
There are 6*2=12 vertices
There are 8+12-2=18 edges

Thus, the total is 8+12+18=38

Quicker way.

If the number of sides in a prism is $ s$, then the sum of faces, vertices, and edges will always be $ 6s \plus{} 2$.

The edges will always be $ 3s$, the vertices will always be $ 2s$, and the sides will always be $ s \plus{} 2$.

Try it if you don't understand.

Example: Find the sum of edges, vertices, and sides of a cube.

There are $ 12$ edges on a cube (you should know that xD, also $ 3s$).  There are $ 8$ vertices on a cube ($ 2s$).  There are $ 6$ faces on a cube ($ s \plus{} 2$).

The answer is $ 26$, which is also $ 6 (4) \plus{} 2$.

Also, one could note that E+2=F+V

Thus we are loking for 2E+2 which is 2(12)+2

[quote="vallon22"]mewto got that equation from the more famous way of writing: F+V-E=2

this equation says that the number of faces plus the number of verticies minues the number of edges always equals 2 for any 3-dimensional figure[/quote]
Isn't there one geometric figure that does not go by with this rule?

[quote="ernie"]The edges will always be $ 3s$, the vertices will always be $ 2s$, and the sides will always be $ s \plus{} 2$.[/quote]

Ernie is defining s as the number of sides on the n-gonal face of the prism.  I'll define n as this instead, because... it's an n-gon.

An n-gonal prism will always have n edges on the top and bottom faces, and n edges connecting the n vertices of these two faces.  So the total number of edges will be 3n.

An n-gonal prism will always have n vertices on the top and bottom faces, and no vertices elsewhere.  So the total number of vertices is 2n.

An n-gonal prism will always have n rectangular faces connecting the n edges of the top and bottom faces.  So the total number of faces is n+2.

Add them all up and get 6n+2.

[quote="vallon22"]
mewto got that equation from the more famous way of writing: F+V-E=2

this equation says that the number of faces plus the number of verticies minues the number of edges always equals 2 for any 3-dimensional figure
[/quote]

Actually, that equation (Euler's Formula) only works for simply connected polyhedra--3-dimensional shapes with straight edges, faces bounded by edges, and each edge connecting two vertices.

As some 3-dimensional figures such as cylinders, spheres, and non-simply connected polyhedra (more than 2 faces connected by each edge or some other discrepency) do not contain straight edges or faces bounded by edges, Euler's Formula cannot be used for them.


I'm a novice in C programming ,,,

I wanna create a C program which uses the ternary operator to print -1 or 1 if the value input to it is negative or positive.
Here's my program code:
[code]
#include<stdio.h>
#include<conio.h>
void main ()
  {
   int input;
   clrscr();
   printf("Input the value:");
   scanf("%d",input);
   input<=-1?"-1":"1";
   getch();
  }[/code]

But it's not working ,,, ! :(

Help ,,, ??

First of all there is an error in your scanf syntax
 It should be
scanf("%d",&input);
instead of input use &input( address of input variable)

In the ternary operator the line should be:

input<= -1 ? printf("-1") : printf("1");

so that it will print -1 for input values less than -1 and 1 for those greater
Giving just "-1" doesn't do anything.

Excluding ampersand is my mistake ,,, beside I didn't know about inserting printf there ,,, !
Thanks alot for your help ,,, !  :)

First, I know the input is an integer, but it's more correct to test whether input < 0 instead of input <= -1 (say you later change it to double).

Also, more readable (and efficient?) code:

[code]
printf(input < 0 ? "-1" : "1");
[/code]

Thanks Yongyi781 for that readable code ,,, !  :)


function
real analysis
real analysis unsolved

Find all the continuous functions f:R->R with the property
f(f(x))-3f(x)+2x=0 for each real x.

Solutions are f(x)=x for each x or f(x)=2x-c with c real, for every x.
Are there other solutions?

This type of functionals equations is treated in the course

Les Equations fonctionnelles

http://boumbo.salle-s.org/xavier/maths/stmalo/

choose

Les Equations Fonctionnelles

 5.3 Points Fixes page 77

I looked there Moubi, but didnt find anything to help in this situation.

I think those are the only ones. You can see that f is injective so it's monotonous. I took 2 cases. When it's increasing it's all nice and easy (f(x)=x). My teacher solved this for me last year but I forgot what he did
:(. I think he considered the sequence a_n=f(f(..(f(x)))..)) and he used that recurrence (that's obvious! :? ), but I can remember nothing more... I'm trying, though...

I dont think that when f is incresing results f(x)=x because also f(x)=2x-c
is a solution and its also increasing.
if we consider gn=f(f(..(f(x)))..)) it results (by adding some relations)
that f(x)-2x=(gn(x)-2^(n-1)f(x))/2^n-1 but from there i got nowhere.

Oh, sorry! I saw c-2x. It might be a different problem from the one I was speaking abt.

Consider F_n(x)=f(f...(f(x)...) n iterations.
We have as grobber said F_n+2=3F_n+1-2F_n
So F_n(x)=a(x)(2^n)+b(x).
And F_0=x and F_1=f(x), so
a(x)+b(x)=x and 2a(x)+b(x)=f(x), so a(x)=f(x)-x and b(x)=2x-f(x).
Try to continue from here.

Ive done this myself before, but i didnt obtain anything starting from here


In a building there are 6018 desks in 2007 rooms, and in every room there is at least one desk. Every room can be cleared dividing the desks in the oher rooms such that in every room is the same number of desks. Find out what methods can be used for dividing the desks initially.

Solution:
Let $ n$ be the maximum number of desks in a room.We have $ n\geq3$, because if $ n\leq 2$, then we have
$ 6018\leq 2007n\leq 4014$, false.

If $ n\geq 4$, then choose another room to be cleared.After the redistribution, in all "full"(not cleared) rooms we have at least
$ n$ desks, so we have $ 6018\geq 2006n\geq 8024$, false. So $ n\equal{}3$.

We observe that there are missing $ 3$ desks so that in all rooms there would be $ 3$ desks. We take two cases:

$ 1.$ If there is a room with $ 1$ desk, then the distribution is $ 1,2,3,3,3,...,3$(and permutations).Then this kind of distribution satisfies all the tasks, because we can do moves like this: if we clear the room with $ 1$ desk, then move this desk into the room with $ 2$ desks; if we clear the room with $ 2$ desks, then move these $ 2$ desks into the room with $ 1$ desk; if we clear any of the rooms with $ 3$ desks, then move $ 2$ of these desks into the room with $ 1$ desk and the other $ 1$ desk into the
room with $ 2$ desks.

$ 2.$ If there is no room with $ 1$ desk, then the distribution is $ 2,2,2,3,3,...,3$ (and permutations).Then this kind of distribution satisfies all the tasks, because we can do moves like this: if we clear any of the rooms with $ 2$ desks, then take the other $ 2$
rooms with $ 2$ desks and put in each of them $ 1$ desk; if we clear any of the rooms with $ 3$ desks, then take all the rooms with $ 2$ desks and put in each of them $ 1$ desk.

So the answers are:  $ 1,2,3,3,3,...,3$ and $ 2,2,2,3,3,...,3$ (and permutations).


geometry
geometric transformation
reflection
rectangle

Given a point $P$ inside a circle $\omega$, two perpendicular chords through $P$ divide $\omega$ into distinct regions $a,b,c,d$ clockwise such that $a$ contains the centre of $\omega$. Prove that $[a]+[c]\ge [b]+[d]$, where $[a]=\text{area of }a$.

[hide]Reflect one of the chords over the center of the circle.  Now look at the signs on the areas of the six regions formed.  Four of them cancel out, and the remaining two clearly have a positive sum.  In fact, that positive sum is just the rectangle formed by reflecting P over the diameter perpendicular to the first chord, reflecting P over the diameter perpendicular to the second chord, and reflecting P over the center of the circle![/hide]

nice solution. :D  that's exactly one of my two solutions. the trick here is to use symmetry. my second solution isn't very nice.


Let a,b,c$ \geq$0 .Prove that:
$ \frac{1\plus{}\sqrt{a}}{1\plus{}\sqrt{b}}\plus{}\frac{1\plus{}\sqrt{b}}{1\plus{}\sqrt{c}}\plus{}\frac{1\plus{}\sqrt{c}}{1\plus{}\sqrt{a}}\leq 3\plus{}a\plus{}b\plus{}c$
Let x,y,z$ \geq$0   such that  x+y+z=1 .Prove
$ \frac{1\plus{}x^2}{1\plus{}y^2}\plus{}\frac{1\plus{}y^2}{1\plus{}z^2}\plus{}\frac{1\plus{}z^2}{1\plus{}x^2} \leq \frac{7}{2}$

[quote="tranquoc"]Let a,b,c$ \geq$0 .Prove that:
$ \frac {1 \plus{} \sqrt {a}}{1 \plus{} \sqrt {b}} \plus{} \frac {1 \plus{} \sqrt {b}}{1 \plus{} \sqrt {c}} \plus{} \frac {1 \plus{} \sqrt {c}}{1 \plus{} \sqrt {a}}\leq 3 \plus{} a \plus{} b \plus{} c$
[/quote]
It's obviously true after expanding.
Try $ a\equal{}x^2,$ $ b\equal{}y^2$ and $ c\equal{}z^2,$ where $ x,$ $ y$ and $ z$ are non-negative. :wink:

[quote="arqady"][quote="tranquoc"]Let a,b,c$ \geq$0 .Prove that:
$ \frac {1 \plus{} \sqrt {a}}{1 \plus{} \sqrt {b}} \plus{} \frac {1 \plus{} \sqrt {b}}{1 \plus{} \sqrt {c}} \plus{} \frac {1 \plus{} \sqrt {c}}{1 \plus{} \sqrt {a}}\leq 3 \plus{} a \plus{} b \plus{} c$
[/quote]
It's obviously true after expanding.
Try $ a \equal{} x^2,$ $ b \equal{} y^2$ and $ c \equal{} z^2,$ where $ x,$ $ y$ and $ z$ are non-negative. :wink:[/quote]
It does not need expanding here. :)
Put $ a\equal{}x^2,b\equal{}y^2,c\equal{}z^2$ where $ x,y,z \ge 0$, then the inequality becomes
\[ x^2\plus{}y^2\plus{}z^2\plus{}3 \ge \frac{x\plus{}1}{y\plus{}1}\plus{}\frac{y\plus{}1}{z\plus{}1}\plus{}\frac{z\plus{}1}{x\plus{}1}\]
\[ \Leftrightarrow x^2\plus{}y^2\plus{}z^2\minus{}x\minus{}y\minus{}z \plus{}\left(x\plus{}1 \minus{}\frac{x\plus{}1}{y\plus{}1}\right) \plus{}\left( y\plus{}1\minus{}\frac{y\plus{}1}{z\plus{}1}\right) \plus{}\left( z\plus{}1\minus{}\frac{z\plus{}1}{x\plus{}1}\right) \ge 0\]
\[ \Leftrightarrow x^2\plus{}y^2\plus{}z^2\minus{}x\minus{}y\minus{}z\plus{}\frac{y(x\plus{}1)}{y\plus{}1}\plus{}\frac{z(y\plus{}1)}{z\plus{}1}\plus{}\frac{x(z\plus{}1)}{x\plus{}1} \ge 0\]
By Cauchy Schwarz Inequality, we have that
\[ \frac{y(x\plus{}1)}{y\plus{}1}\plus{}\frac{z(y\plus{}1)}{z\plus{}1}\plus{}\frac{x(z\plus{}1)}{x\plus{}1} \ge \frac{y}{y\plus{}1}\plus{}\frac{z}{z\plus{}1}\plus{}\frac{x}{x\plus{}1} \ge \frac{(x\plus{}y\plus{}z)^2}{x^2\plus{}y^2\plus{}z^2\plus{}x\plus{}y\plus{}z}\]
Hence
\[ x^2\plus{}y^2\plus{}z^2\minus{}x\minus{}y\minus{}z\plus{}\frac{y(x\plus{}1)}{y\plus{}1}\plus{}\frac{z(y\plus{}1)}{z\plus{}1}\plus{}\frac{x(z\plus{}1)}{x\plus{}1} \ge x^2\plus{}y^2\plus{}z^2\minus{}x\minus{}y\minus{}z \plus{}\frac{(x\plus{}y\plus{}z)^2}{x^2\plus{}y^2\plus{}z^2\plus{}x\plus{}y\plus{}z}\equal{}\frac{(x^2\plus{}y^2\plus{}z^2)^2}{x^2\plus{}y^2\plus{}z^2\plus{}x\plus{}y\plus{}z} \ge 0\]
:)


[quote="rachid"]a,b,and c are positive real numbers,prove that:
1/a(1+b) + 1/b(1+c) + 1/c(1+a) >= 3/1+abc.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161059

$ a,b,c\in\mathbb{R}\plus{}$ prove that $ \dfrac{1}{a(1\plus{}b)}\plus{}\dfrac{1}{b(1\plus{}c)}\plus{}\dfrac{1}{c(1\plus{}a)}\geq{\dfrac{3}{1\plus{}abc}}$


inequalities function integration geometry rectangle calculus induction

I'm just trying to understand analysis again ... :wallbash: ... and I come across this:

Minkowski's Inequality: Suppose $1 \le p < \infty$, $\Omega \subset \mathbb{R}^{n}$, $G \subset \mathbb{R}^{m}$, and $F$ is an appropriate function (the book doesn't say, but I guess it is on $\Omega \times G$ and $L^{1}$  on $\Omega$ for almost all points in $G$ and $L^{p}$ on $G$ for almost all points in $\Omega$... does that make sense?). Then
$\left[ \int_{G}\left( \int_\Omega |F(x, y)| dy \right)^{p}dx \right]^{\frac{1}{p}}\le \int_\Omega \left( \int_{G}|F(x, y)|^{p}dx \right)^{\frac{1}{p}}dy$
or in norm symbols $\|\ \|F(\cdot, *)\|_{L^{1}(\Omega)}\|_{L^{p}(G)}\le \|\ \|F(\cdot, *)\|_{L^{p}(G)}\|_{L^{1}(\Omega)}$ (I guess...)

Is this somehow equivalent to what Wikipedia thinks is "Minkowski's Inequality" $\|f+g\|_{p}\le \|f\|_{p}+\|g\|_{p}$ ? It looks like a totally distinct inequality to me. How do we prove it? (I am equally open to a complete proof or to hints so that I can prove it.)

it seems like it works by approximation: for any positive measurable $F$ (of course you can ask for $F$ to be positive), you can find a monotone sequence $F_{n}$ of simple functions that converge pointwise to $F$ a.e.. moreover, you can choose such $F_{n}$ to be a sum of characteristics of rectangles (which is, a sum $\sum a_{j}\chi_{A_{j}\times B_{j}}$, with $A_{j}, B_{j}$ in the right $\sigma$-algebras).
then inequality reduces to minkowski one.

i just hope my calculations weren't vain ;)

A way to keep the terminology straight: use "Minkowski integral inequality" for this one, and "Minkoski's inequality" for the simpler  one.

Minkowski's inequality says that the norm of the sum is less than or equal to the sum of the norms.

You stated that for two functions, but you can easily extend it by induction to any finite sum of functions, and take limits to get series of functions.

Now, just replace the word "sum" by the word "integral": "the norm of the integral is less than or equal to the integral of the norms." So you can see the relationship, but it is a generalization that requires its own proof.

As for proof: it helps greatly to linearize the problem.

Let $q$ be the dual exponent: $\frac1p+\frac1q=1.$

We have the following equivalent expression for the $L^{p}$ norm:

$\|f\|_{p}=\sup_{\|h\|_{q}=1}\int f\cdot h$

Pursue that idea and see how far it gets you.

(The exact nature of $\Omega$ and $G$ aren't important. They're measure spaces, and assume whatever measurability properties for $F$ that you need to quote Fubini's Theorem.)

--

Using the Minkowski integral inequality is one way to prove Young's inequality for convolution.

Something I've used with students:

Let me describe three bags of tools for a "hard" real analyst. These tools are small and sharp and of use in divide-and-conquer estimations. A "soft" analyst would have some different bags of tools, definitely including one full of consequences of the Baire category theorem. (The styles cannot exclude each other; the more tools from everyone's bags you master, the better off you are.)

Bag #1: The Cauchy-Schwarz inequality. Hölder's inequality. Minkowski's inequality.

Bag #2: You can change the order of summation of a double series (if nonnegative or absolutely convergent). The Dominated Convergence Theorem and the Monotone Convergence Theorem. Fubini's Theorem (and Tonelli's Theorem). If you see a double sum, or a double integral, or a sum and an integral, you are going to switch the order somewhere!

Bag #3: The fundamental theorem of calculus. Integration by parts. Green's theorem. The divergence theorem (and a vaguely integration-by-parts like theorem involving Laplacians usually called Green's identity). Stokes' theorem and its generalizations. (Bag #1 and Bag #2 are "positive": they estimate things by their absolute values - Bag #3 requires smoothness but doesn't need positivity. Don't be afraid to integrate the thing that oscillates.)

I'd probably put this topic - the Minkowski integral inequality - in Bag #1, although it has a whiff of Bag #2 about it.[/quote]


function
limit
topology
advanced fields
advanced fields unsolved

Assume that $X$ is a seperable metric space. Prove that if $f: X\longrightarrow\mathbb R$ is a function that $\lim_{x\rightarrow a}f(x)$ exists for each $a\in\mathbb R$. Prove that set of points in which $f$ is not continuous is countable.

Assume the contrary. Since $X$ is a separable metric space, it is also second countable. Write $(X_{n})$ a countable basis for its topology. Similarly, write $(Y_{n})$ any countable basis for the topology of $\mathbb{R}$. Let $(x_{i})$ be a non countable family of points of discontinuities of $f$, and let $(y_{i})$ be such that $y_{i}= lim_{x \rightarrow x_{i}}\ f(x)$. For every $i$, let $Y'(i)$ (resp. $X'(i)$) be any element of $(Y_{n})$ (resp. $(X_{n})$) such that $y_{i}\in Y'(i)$, $f(X'_{i}-\{x_{i}\}) \subseteq Y'(i)$ (by definition of limit) and $f(x_{i}) \notin Y'(i)$ (this is possible by the mere definition of a discontinuity, and the fact that $\mathbb{R}$ is Hausdorff). Moreover, let $Z_{i}$ be any element of $(Y_{n})$ such that $f(x_{i}) \in Z_{i}$ and $Y'_{i}\cap Z_{i}= \emptyset$. Now since the $X'(i)$ are in countable number, we must have one of them, say $X'(I)$, such that $x_{j}\in X'(I)$ for a non countable number of $j$. By the same token, you can extract a non countable subfamily $(x_{p})$ from such $(x_{j})$ such that $f(x_{p}) \in Z_{J}$ for some fixed $Z_{J}$. But now for $p \neq J$, $f(x_{p}) \notin Z_{J}$ by definition of $Z_{J}$. This is a contradiction.

[i]Solution[/i]:Consider the function $g(a)=f(a)-\lim_{x\rightarrow a}f(x)$. Easily for each $a\in X,\lim_{x\rightarrow a}g(x)=0$. Consider $n\in\mathbb N$. So for each $a\in X$ there exists an open set $U_{a}$ containig $a$ that $|g(U_{a}-\{a\})|<\frac1n$. Beacuse $X$ is second countable and Lindelöf, So $U_{a}$ contain an open countable subcovering of $X$. Therefore $g^{-1}([\frac1n,+\infty))$ is countable. The points that $f$ is not continuous exactly is $g^{-1}(0,\infty)=\bigcup_{i=1}^\infty g^{-1}([\frac1n,+\infty))$ which is countable union of countable sets and so it's countable.

[quote="Omid Hatami"][i]Solution[/i]:Consider the function $g(a)=f(a)-\lim_{x\rightarrow a}f(x)$. Easily for each $a\in X,\lim_{x\rightarrow a}g(x)=0$.[/quote]

Huh ... saying that is already equivalent to saying that $f$ is continuous on $X$, isn't it ?

No! \[f(x)=\left\{\begin{array}{cc}0&x\neq0\\ 1&x=0 \end{array}\right.\] Then $\lim_{x\rightarrow a}g(x)=0$ for each $a$, but $g(x)\neq0$. $g$ is not zero at point $0$ which $f$ is not continious. Also $\lim_{x\rightarrow a}g(x)=\lim_{x\rightarrow a}f(x)-\lim_{x\rightarrow y}\lim_{y\rightarrow a}f(y)=0$

You're right ... since you used the same variable $x$ in both definitions of your expressions, I got misleaded  :blush:

Basically, this is a slightly weaker version of Froda's theorem.


logarithms
calculus
integration
quadratics
real analysis
real analysis solved

Do there exist natural $k$ such that $\frac{1}{2ln2}+\frac{1}{3ln3}+...+\frac{1}{klnk}\ge 2003$

The series $\sum \frac{1}{n\ln n}$ disverge due to the integral convergence test.

Being quantitative about the integral test ...

$\sum_{j=1}^k\frac1{j\ln j} >\int_2^{k+1}\frac{dx}{x\ln x} = \ln\ln(k+1)-\ln\ln 2.$

If you want the sum to be greater than 2003, you may choose

$k>e^{e^{2003}\ln2}.$

An interesting question is how to compute this sum at high (!) accuracy. For example, what is "exact" value for $\sum_{k=2}^{10^{20}}\frac{1}{k\ln k}$ ;)

$\sum_{k=2}^n\frac1{k\ln k}=\int_{\frac32}^{n+\frac12}\frac{dx}{x\ln x}- \sum_{k=2}^n\int_{k-\frac12}^{k+\frac12}\left(\frac1{x\ln x}-\frac1{k\ln k}\right).$

The last sum can be estimated by Taylor polynomials.

$x\ln x = k\ln k + (1+\ln k)(x-k)+\frac1{2k}(x-k)^2+O(x-k)^3$

$\frac{1}{x\ln x}=\frac1{k\ln k} -\frac{1+\ln k}{k^2\ln^2k}(x-k) + \frac{2(1+\ln k)^2-\ln k}{2k^3\ln^3k}(x-k)^2+O(x-k)^3.$

The linear term integrates to zero, so we concentrate on the quadratic term. This gives an estimate for the error as

$\frac13\sum_{k=2}^n\frac{2+\ln k+2\ln^2k}{k^3\ln^3k}, $ which is less than $\frac13.$

Myth didn't define what he meant by "high" accuracy in, so we can interpret it any way we want to. If you want better than I've shown here, you're welcome to do the additional work that would require.

Using this, the estimate

$\sum_{k=2}^{10^{200}} \frac1{k\ln k}\approx \ln\ln(10^{200}+.5)-\ln\ln(1.5)\approx 7$

is accurate to within $\frac13.$

There's a cheap way to improve on what I wrote, and that is to add up the first few terms. So, (using the handy spreadsheet),

$\sum_{k=2}^{20}\frac1{k\ln k}\approx 1.90011983.$

Then,

$\sum_{k=2}^{10^{200}}\frac1{k\ln k}\approx 1.90011983 + \int_{20.5}^{10^{200}+.5}\frac{dx}{x\ln x} + \frac13\sum_{k=21}^n\frac{2+\ln k+2\ln^2k}{k^3\ln^3k}.$

This gives$ \sum_{k=2}^{10^{200}}\frac1{k\ln k}\approx 6.9271,$ accurate to within about $.0002.$

Yes, Myth, I know you said $10^{20}.$ I just thought $10^{200}$ would be more fun.

Yes, you used method which I meant for this problem ;) :)
It was a standart excercise in the course of asymptotic methods.

Of course, not! The question is to find partial sums at high accuracy.


Prove that:

$ a)$ $ \sqrt{16\minus{}5\sqrt{7}}\minus{}\sqrt{8\minus{}3\sqrt{7}}\equal{}\sqrt{2}$

$ b)$ $ \sqrt{3\minus{}2\sqrt{2}}\plus{}\sqrt{5\minus{}2\sqrt{6}}\plus{}\sqrt{7\minus{}2\sqrt{12}}\equal{}1$

[hide="Part A"]$ \sqrt {16 \minus{} 5\sqrt {7}} \equal{} \frac {32 \minus{} 2\sqrt {175}}{\sqrt {2}} \equal{} \frac {5 \minus{} \sqrt {7}}{\sqrt {2}}$
$ \sqrt {8 \minus{} 3\sqrt {7}} \equal{} \frac {16 \minus{} 2\sqrt {63}}{\sqrt {2}} \equal{} \frac {3 \minus{} \sqrt {7}}{\sqrt {2}}$
Thus $ \sqrt {16 \minus{} 5\sqrt {7}} \plus{} \sqrt {8 \minus{} 3\sqrt {7}} \equal{} \frac {5 \minus{} \sqrt {7}}{\sqrt {2}} \minus{} \frac {3 \minus{} \sqrt {7}}{\sqrt {2}} \equal{} \frac {2}{\sqrt {2}} \equal{} \sqrt {2}$[/hide]