inequalities
function
inequalities theorems

How is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]Karamata[/url] weighted Inequality?

thanks!

I'm copying it from a paper. I have not worked on this yet, so I can't comment.

[quote="A. Hrabrov"]The Karamata inequality can be generalized. The generalization (L. Fuchs, [i]A new proof of an inequality of Hardy-Littlewood-Polya[/i], Mat. Tidsskr. B. 1947. p. 53-54) reminds of the [weighted] Jensen inequality.

[b]Fuchs inequality.[/b] Let f(x) be a convex function and $p_1$, $p_2$, ..., $p_n$ be positive numbers. Further, let $x_1$, $x_2$, ..., $x_n$, $y_1$, $y_2$, ..., $y_n$ be 2n real numbers such that $x_1\geq x_2\geq ...\geq x_n$ and $y_1\geq y_2\geq ...\geq y_n$ and such that

  $\sum_{i=1}^{k} p_ix_i \geq \sum_{i=1}^{k} p_iy_i$

for every natural k with $1\leq k\leq n-1$ and

  $\sum_{i=1}^{n} p_ix_i = \sum_{i=1}^{n} p_iy_i$.

Then,

  $p_1 f\left(x_1\right) + p_2 f\left(x_2\right) + ... + p_n f\left(x_n\right) \geq p_1 f\left(y_1\right) + p_2 f\left(y_2\right) + ... + p_n f\left(y_n\right)$.[/quote]

  Darij

Darij wrote where the proof can be found: L. Fuchs, [i]A new proof of an inequality of Hardy-Littlewood-Polya[/i], Mat. Tidsskr. B. 1947. p. 53-54

Sorry to bring back old topic, i had a question:
$\sum_{i=1}^{k}p_{i}x_{i}\geq \sum_{i=1}^{k}p_{i}y_{i}$

for every natural k with $1\leq k\leq n-1$ and

$\sum_{i=1}^{n}p_{i}x_{i}= \sum_{i=1}^{n}p_{i}y_{i}$.
This just means that $\{ p_{i}x_{i}\}$ maximizes $\{ p_{i}y_{i}\}$ right?
And this is proven by repeated application of karamata and continuity?

That's from [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18701[/url] :
[quote="darij grinberg"]I'm copying it from a paper. I have not worked on this yet, so I can't comment.

[quote="A. Hrabrov"]The Karamata inequality can be generalized. The generalization (L. Fuchs, [i]A new proof of an inequality of Hardy-Littlewood-Polya[/i], Mat. Tidsskr. B. 1947. p. 53-54) reminds of the [weighted] Jensen inequality.

[b]Fuchs inequality.[/b] Let f(x) be a convex function and $ p_1$, $ p_2$, ..., $ p_n$ be positive numbers. Further, let $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ be 2n real numbers such that $ x_1\geq x_2\geq ...\geq x_n$ and $ y_1\geq y_2\geq ...\geq y_n$ and such that

  $ \sum_{i \equal{} 1}^{k} p_ix_i \geq \sum_{i \equal{} 1}^{k} p_iy_i$

=>for every natural k with $ 1\leq k\leq n \minus{} 1$ and
  $ \sum_{i \equal{} 1}^{n} p_ix_i \equal{} \sum_{i \equal{} 1}^{n} p_iy_i$.

Then,

  $ p_1 f\left(x_1\right) \plus{} p_2 f\left(x_2\right) \plus{} ... \plus{} p_n f\left(x_n\right) \geq p_1 f\left(y_1\right) \plus{} p_2 f\left(y_2\right) \plus{} ... \plus{} p_n f\left(y_n\right)$.[/quote]

  Darij[/quote]

why do $ p_1$, $ p_2$, ..., $ p_n$ have to be positive ?
I tried to prove weighted karamta yesterday by myself (not exactly by myself . I used this idea [url]http://www.imocompendium.com/tekstkut/ineq_im.pdf[/url] page 11-12 ) and couldn't find reason why $ p_i$ have to be positive , i hadn't found Fuchs' book in my city , so i don't know the original proof , but i think that one is true too :-? :

Let $ f(x)$ be a convex function and $ p_1$, $ p_2$, ..., $ p_n$ be real(!?) numbers. Further, let $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ be 2n real numbers such that $ x_1\geq x_2\geq ...\geq x_n$ and $ y_1\geq y_2\geq ...\geq y_n$

$ \sum_{i \equal{} 1}^{k} p_ix_i \geq \sum_{i \equal{} 1}^{k} p_iy_i$

for every natural k with $ 1\leq k\leq n \minus{} 1$ and

  $ \sum_{i \equal{} 1}^{n} p_ix_i \equal{} \sum_{i \equal{} 1}^{n} p_iy_i$.
If $ f(x)$ is increasing , it's enough $ \sum_{i \equal{} 1}^{n} p_ix_i \geq \sum_{i \equal{} 1}^{n} p_iy_i$.
Denoting $ c_i\equal{}\frac{f(y_{i})\minus{}f(x_{i})}{y_{i}\minus{}x_i}$ which is non-increasing and $ A_i \equal{} p_1x_1\plus{}...\plus{}p_ix_i$ and $ B_i \equal{} p_1y_1\plus{}...\plus{}p_iy_i$

$ \equal{}> \sum^{n}_{i\equal{}1}p_{i}f(x_i) \minus{}\sum^{n}_{i\equal{}1}p_{i}f(y_i) \equal{} \sum^{n}_{i\equal{}1}p_{i}c_i(x_i\minus{}y_i)
\equal{}$
$ \equal{} \sum^{n}_{i\equal{}1}(c_i\minus{}c_{i\plus{}1})(p_1x_1\plus{}p_2x_2\plus{}...\plus{}p_ix_i) \minus{} \sum^{n}_{i\equal{}1}(c_i\minus{}c_{i\plus{}1})(p_1y_1\plus{}p_2y_2\plus{}...\plus{}p_iy_i) \equal{}$
$ \equal{} \sum^{n\minus{}1}_{i\equal{}1}(c_i\minus{}c_{i\plus{}1})(A_i\minus{}B_i) \plus{} c_n(A_n\minus{}B_n)$

1. $ c_i\geq c_{i\plus{}1}$ , $ A_i\geq B_i$ for $ 1\geq k\geq n \minus{} 1$ and
2. $ A_n\equal{}B_n$ or if $ f(x)$ is convex - $ c_n\geq 0$ and $ A_n\geq B_n$
so $ \sum^{n}_{i\equal{}1}p_{i}f(x_i) \minus{}\sum^{n}_{i\equal{}1}p_{i}f(y_i) \geq 0$

Finally my question is : Why do $ p_i$ have to be positive ?

Well, on http://www.imocompendium.com/tekstkut/ineq_im.pdf , page 11, proof of Karamata inequality, the author uses $ c_i$ as $ c_i\equal{}\frac{f(y_i)\minus{}f(x_i)}{y_i\minus{}x_i}$
How do you prove that $ c_i$ is non-increasing? I understand the whole proof except this small part...

Thanks

Let a>b and f(x) is a convex function .
$ \frac {f(a)\minus{}f(c)}{a\minus{}c}>\frac {f(b)\minus{}f(c)}{b\minus{}c}$ ??
There are 3 cases but they can be proven in a similar way .

So let just a>b>c :
$ (b\minus{}c)f(a)\minus{}(b\minus{}c)f(b)>(a\minus{}b)f(b)\minus{}(a\minus{}b)f(c)$ ??
$ (b\minus{}c)f(a)\plus{}(a\minus{}b)f(c)>(a\minus{}c)f(b)$ ??
$ \frac {b\minus{}c}{a\minus{}c} f(a)\plus{}\frac {a\minus{}b}{a\minus{}c} f(c)> f(b)\equal{}f(\frac {b\minus{}c}{a\minus{}c} a\plus{}\frac {a\minus{}b}{a\minus{}c} c)$ ??
$ \frac {b\minus{}c}{a\minus{}c}\plus{}\frac {a\minus{}b}{a\minus{}c}\equal{}1$
but this is the definition of the convex function :P


analytic geometry
geometry
vector
algebra unsolved
algebra

Five points of plane  A1, A2, A3,  A4, A5 lie on a plane and their coordinates are integer numbers. To prove that among all triangles with tops in these points is three or more triangles, the planes (squares) of which are expressed integers.
 :|

[quote="Vova-LFML"] the planes (squares) of which are expressed integers.
 :|[/quote]

 :?:
Do you mean area of the triangles?

In that case, it is easy : there are four types of integer points, namely (o,o),(o,e),(e,o),(e,e), where e means even and o means odd.
By the pigeon-hole principle, among the five points, there are two which must belong to the same type, say $ A_1$ and $ A_2$.
Now, for $ i\equal{}3,4,5$ the area of the triangle $ A_1A_2A_i$ is $ | \frac {det(A_1A_2,A_1A_i)}2 |$ which is clearly an integer since all the coordinates are integers and the coordinates of vector $ A_1A_2$ are both even.

Pierre.

Sorry, I have already solved it , so I dont need any more help.Thank everybody :)


Hi,

Can anyone tell me what package I need to install LaTeX on a linux webserver in order to use the latexrender package, like here on the forum? I'd need a minimal install for that, most packs I find are over 150 mb, but I cannot imagine really all the files are needed for the simple rendering like here... and my host wants to install it only if I can bring it to a smaller size.

Thanks!

[quote="Peter VDD"]but I cannot imagine really all the files are needed for the simple rendering like here... and my host wants to install it only if I can bring it to a smaller size.[/quote]LatexRender does compile a real tex document so you are are likely to need most of the files used in a LaTeX system, including the fonts, though you won't need many of the packages. Even a small MiKTeX distribution for Windows, which only includes the basic packages, needs 100MB.

In the Linux case, most distros include an automated installation (eg RPMs) giving you the large system, so a smaller system may need manual installation which could require specialist expertise.

It may be worthwhile asking in the [url=http://groups.google.co.uk/group/comp.text.tex?hl=en][b]comp.text.tex[/b][/url] newsgroup if the experts there know of a small installation. You will need to tell them which flavour of Linux the server uses.

Ok, thank you Steve. :)
I contacted the server owner and surprisingly he doesn't care about the space it takes.

Thing now is that I'd need to install it myself from distance. Can I do this through SSH + root access, or would it need local intervention?

I found a [url=http://www-mobile.ecs.soton.ac.uk/bjc97r/tips/latex-install.html]useful document[/url] on it, and I wondered if anyone more familiar with this could tell me if the info displayed there is trustworthy and correct? This is my first SSH/linux/distance install/... experience, so I'd rather ask than badly ruining something... it just seems TOO simple like they propose it there.

I haven't tried manually installing LaTeX since all the OS's I've used come with an automated installer, which is the easiest way to install it. See if you can find such an automated installer for your system.

A couple of points about the instructions you refer to:
1. it is for an upgrade of LaTeX not a new install (where for example are the binaries?)
The install.txt file included in the download at ftp://tug.ctan.org/pub/tex-archive/macros/latex/base/install.txt makes it clear that you also need to install a TeX system
2. it doesn't mention the need for refreshing the filename database, which is how LaTeX knows where to find the files. This is done using the texhash command.

So go for an installer where someone has done the work for you :). Ask your host about it - servers I've used with RedHat and SUSE have control panels which give access to RPMs which makes things easy. The [url=http://rpmfind.net/linux/rpm2html/search.php?query=tetex&submit=Search+...&system=&arch=]RPM Repository[/url] also has loads of such packages. [url=http://www.tug.org/tetex/]teTeX[/url] (unlike the download in the document you mention) is a complete TeX distribution.

Don't forget that for LatexRender you must have ImageMagick and Ghostscript installed as well. They are well-known enough to be listed as software that can be automatically installed for most distros.


rrusczyk wrote:I am thinking of a number.  I tell you that it is divisble by 9 and by 12.

a) Must it be divisible by 3?

[hide]Yes, because it is divisible by 9, which itself is divisible by 3. For that matter, it is divisible by 12, which itself is divisible by 3.[/hide]

b) Must it be divisible by 9 + 12 = 21?

[hide]I'll let someone else do this one.[/hide]

c) Must it be divisible by (9)(12) = 108?

[hide]See my answer for d. for a clue to the answer to this one.[/hide]

d) What is the largest number that must divide my number?

[hide]36, which is the product of the prime factors of 9 and 12.[/hide]

It's good to see some easy problems here from the founder of the site. I'll try to redo my challenge problem series as the Easy Problem Series when I get over the cold I have this weekend.

rrusczyk wrote:I am thinking of a number.  I tell you that it is divisble by 9 and by 12.

a) Must it be divisible by 3?

b) Must it be divisible by 9 + 12 = 21?

c) Must it be divisible by (9)(12) = 108?

d) What is the largest number that must divide my number?

a.[hide]Yes, becuase 9 and 12 are divisble by 3.  If x is divisble by y, and w is divisible by x, then w is divible by y.[/hide]

b.[hide]No.  108 is not divisble by 21, and 108 is divisble by 12 and 9.  If x is divisble by y and z, then x isn't necessarily divisble by (y+z)[/hide]

c.[hide]No.  72 is divible by 9 and 12, and it isn't divisble by 108.  If x is divisble by y and z, then x isn't necessarily divisble by (y*z)[/hide]

d.[hide]36.  9's prime factorization is 3^2, and 12's prime factorization is 3*4.  You take the number with the biggest exponent for every number from both sides (in this case, 3^2 and 4) and multiply them all.  #^2*4 = 36.[/hide]

[hide]a) Must it be divisible by 3?

yes, call the number x. x = 9a where a is an integer. x = 3(3a), x/3 = 3a, 3a is an integer so is it always divisible by 3.

b) Must it be divisible by 9 + 12 = 21?

no. counterexample, 108 = 21*5 + 3, and 108 = 9*12

c) Must it be divisible by (9)(12) = 108?

the prime factorisation os 9 and 12 is 3^2 and 3*2^2, os if a number x is divisible by 3^2 and 3*2^2, it must itself have in its prime facotrisation all of these, so it is divisible by (3^2)(2^2). but it is not nessecarily divisible by the product, since there are numbers which have the nessecary prime facotrs to be divisible by 9 and 12 but no those that are counted twice when multiplying the prime facotrisations.

eg. 36 = (2^2)(3^2), so it has the nessecary prime facotrs to be divisible, however if we multiply the prime factors: (3^2)(3*2^2) = (3^3)(2^2) and so an extra three is there that is not nessecary.

In general, LCM(a,b) is divisible by x if a and b are.

d) What is the largest number that must divide my number?

looking at the prime factorisations, (3*3), and (2*2*3). your number must contain these factors= (3*3)(2*2), or 36. [/hide]

Mikey:

1) Use spoiler so that others won't read your answers if they don't want to.

2) Think about your part D some more.

a) Must it be divisible by 3?

b) Must it be divisible by 9 + 12 = 21?

c) Must it be divisible by (9)(12) = 108?

d) What is the largest number that must divide my number?

[hide]

a) It must be divisible by 36, for that matter.

b) 36 is divisible by 9 and 12, but not 21.

c) 36 is divisible by 9 and 12, but not 108.

d) 36.

[/hide]


If $a,b,c,d>0$ then:

$ \frac{a}{b+c+d}+\frac{b}{c+d+a}+ \frac{c}{d+a+b} +\frac{d}{a+b+c} \geq \frac{4}{3}$. GENERALIZE IT!!

cheers! :D

I think it can be done by Jensen. just like ineq P2.12 and P2.13. and other similiar one's

For n positive numbers a1, a2, ..., an, we have

a1 / (a2 + a3 + ... + an) + a2 / (a3 + a4 + ... + a1) + ... + an / (a1 + a2 + ... + an-1) >= n / (n-1),

or equivalently

a1 / (p - a1) + a2 / (p - a2) + ... + an / (p - an) >= n / (n-1),

where p = a1 + a2 + ... + an.

The proof can be done by Jensen, just as you proposed, or by applying Chebyshev for the similarly sorted sequences (a1, a2, ..., an) and (1 / (p - a1), 1 / (p - a2), ..., 1 / (p - an)) and then using the AM-HM inequality for the numbers 1 / (p - a1), 1 / (p - a2), ..., 1 / (p - an).

  Darij

[quote="darij grinberg"][quote="siuhochung"]a1 / (p - a1) + a2 / (p - a2) + ... + an / (p - an) >= n / (n-1)

A stronger result?
Let m = a1 / (p - a1) + a2 / (p - a2) + ... + an / (p - an)
[...]
m>=4/n[/quote]

Well, this is a weaker result, since n / (n-1) >= 4 / n for every n.

In fact, my inequality ... >= n / (n-1) is sharp for every n > 1, so there is no stronger one (at least not if n / (n-1) is replaced by a constant).

[quote="siuhochung"]Back to the original one.
m[a1(p-a1)+a2(p-a2)+....+an(p-an)]>=p^2
m[p^2-a1^2-...-an^2]>=p^2
a1^2+...+an^2>=p^2/n by cauchy
mp^2(n-1)/n>=p^2[/quote]

I am not sure about this step: From a >= b and c >= d, it doesn't follow that a - c >= b - d. (But I may be mistaking, too.)

  Darij[/quote]

well... a-b>=c
b>=d
a-d>=a-b>=c?

This the way I did it :D:
Let  $a_1, a_2,..., a_n>0$ be real numbers. So the following inequality holds:
\[\frac{a_1}{\displaystyle\sum{a_i}-a_1}+{\cdots}+\frac{a_1}{\sum{a_i}-a_n}\geq\frac{n}{n-1}\]
So here is how I made it:
Sum $n$ at both sides, make $\sum{a_i}$ a common factor and we get:
\[\left(\sum{a_i}\right)\left(\frac1{\sum{a_i}-a_1}+\cdots\frac1{\sum{a_i}-a_n}\right)\geq\frac{n^2}{n-1}\]
So now we multiply both sides by $n-1$
\[\left((n-1)\sum{a_i}\right)\left(\frac1{\sum{a_i}-a_1}+\cdots\frac1{\sum{a_i}-a_n}\right)\geq n^2\]
And the last is true by trusty Cauchy :D

by mistake a posted the same thing twice, and now I dont know how to erase it (It does'nt appear the erase option when I edit it)

[quote="siuhochung"]well... a-b>=c
b>=d
a-d>=a-b>=c?[/quote]

Sorry! I had misinterpreted your argument.

  Darij


function
algebra
polynomial
real analysis
real analysis unsolved

Prove that a continuous real function which tends to 0 at infinity can be uniformly approximated with functions of the form $e^{-x} p(x)$ with $p$ real polynomial.

consider the algebra $A$ of function generated by the functions $P(x)e^{-n|x|}$, $n \geq 1$ and the constant functions, defined on the Alexandrof compactification $K$ of $\mathbb{R}$. One knows (Stone Weierstrass) that $A$ is dense in $(K,|.|_{\infty})$.

Hence, if $f$ is a continuous function that tends to $0$ at infinity, there exists some polynomials $P_1, P_2,..,P_n$ and integers $1\leq k_1<k_2<..<k_n$ such that $|f(x) - (P_1(x)e^{-k_1|x|}+P_2(x)e^{-k_2|x|}+..+P_n(x)e^{-k_n|x|})|_{\infty}<\epsilon$. So if one proves that for every integer $k \geq 2$ and every polynomial $P(x)$ there exists a polynomial $Q$ such that $|P(x)e^{-k|x|} - Q(x)e^{-|x|}|_{\infty}<\epsilon$, the conclusion will follows, which is easier.

Alekk, are you sure the existence of $Q$ is so simple? I really have some doubts.

http://www.mathlinks.ro/Forum/viewtopic.php?t=20804 will surely help  :)

Thanks, Alekk. Here is another one: prove that $f$ can be uniformly approximated by functions of the form $e^{-x^2}P(x)$. Does it remain true for functions of the form $e^{-x^{1/4}}P(x)$?  ;)


linear algebra
matrix
LaTeX
combinatorics

The integers $1,2, ... , n$ each appear $n$ times in an $nxn$ matrix. Show that there is a row or column containing at least $n^{1}/2$ distinct integers.

A familiar problem (I think) to many. I still can't solve it

Not sure what this means... $n^{1}/2$  :maybe:

I think $\sqrt{n}$; it's happened to me a couple of times: you can write an exponent without brakets, but it can only work if it consists only of one symbol; otherwise I'd say the program "drops" all but the first one.

That's pretty standard $\LaTeX$ ;)

What I find interesting is that he used brackets, but wrote {1}/2 instead of {1/2} (and \frac 12 without any brackets would have worked, too).

Actually, I think the forum automatically adds the brackets in if you don't include them.

As for the solution, let's count the number of times a number is repeated in each row and column - that is, if number $m$ appears $k$ times, that counts as $k-1$ "repeats".  Say number $m$ appears in $i$ rows and $j$ columns - then the sum of the number of repeats over all rows is $n-i$, and the sum of the number of repeats over all columns is $n-j$.  So the sum of the number of repeats over all rows ans columns is

$2n-i-j \leq 2n-2 \sqrt{ij}\leq 2n-2 \sqrt{n}$.

Summing over all numbers, you get $2n^{2}-2n^{3/2}$. So there is some row or column with no more than $n-\sqrt{n}$ repeats, which means that there are at least $\sqrt{n}$ different numbers in that row.


geometry
3D geometry
sphere
topology
advanced fields
advanced fields unsolved

I need help in this problem:

¿does the torus and the esphere (in $\mathbb{R}^3$ ) are homeomorphic?

please... in the most elementary way :oops:

$Tipe$

You'll need to find some way to involve homotopy or homology.

The simplest argument I can think of: the sphere is simply connected. That is, every closed curve is homotopic to a point. But it is possible to find a closed curve on the torus that is not homotopic to a point. This property - whether the space is simply connected or not - would be preserved by any homeomorphism.


Which problems did you guys like best and why?  Personally I liked #21 (chair line) and whichever one the 100 edged polyhedron was.  I hadn't seen recursion applied in such a way (though I had in other combinatorics problems) and liked the simplicity of that problem.  I liked the polyhedron one largely because it scared me away at first ("Uh oh, that looks like a nasty 3D Geo problem"), but like #21 it had a very nice, simple solution.

I like #17
it is very simple but looks very difficult yet yields a beautiful solution


function
induction
vector
algebra unsolved
algebra

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

I think that we need the condition $ f$ is continuous

[quote="zaya_yc"]$ \forall x,y\in R$

$ f(x \plus{} y)f(x \minus{} y) \equal{} (f(x)f(y))^{2}$ .

Find all $ f(x)$[/quote]

I think that this problem is rather hard without any supplementary condition (continuity for example).
Here is a partial solution :

Let $ P(x,y)$ be the assertion $ f(x \plus{} y)f(x \minus{} y) \equal{} (f(x)f(y))^2$

1) Let us first have a look on the case where $ \exists u$ such that $ f(u) \equal{} 0$
================================================
Let $ A \equal{} \{x$ such that $ f(x)\ne 0\}$
If $ A \equal{} \emptyset$, then $ f(x) \equal{} 0$ $ \forall x$ and we got a solution. We'll from now consider $ A\ne\emptyset$

$ P(x,y)$ shows that $ x,y\in A$ $ \implies$ $ x \plus{} y,x \minus{} y\in A$, so $ 0\in A$
For $ x\in A$, $ P(x,0)$ $ \implies$ $ f(0)^2 \equal{} 1$.
If $ f(x)$ is a solution, then $ \minus{} f(x)$ is a solution too. So wlog say from now $ f(0) \equal{} \plus{} 1$
$ P(\frac x2,\frac x2)$ $ \implies$ $ f(x) \equal{} f(\frac x2)^4$ and so $ f(x)\ge 0$ $ \forall x$

$ \forall x\in A$, $ P(x,x)$ $ \implies$ $ f(2x) \equal{} f(x)^4$
$ \forall x\in A$, $ P(2x,x)$ $ \implies$ $ f(3x) \equal{} f(x)^9$
And an immediate induction shows that $ \forall x\in A$, $ f(nx) \equal{} f(x)^{n^2}$ $ \forall n\in\mathbb Z$
From there we immediately get $ \forall x\in A$, $ f(px) \equal{} f(x)^{p^2}$ $ \forall p\in\mathbb Q$

So $ x\in A$ $ \implies$ $ px\in A$ $ \forall p\in\mathbb Q$
So $ A$ is a $ \mathbb Q$-vector space.

The reciprocal assertion may easily be established : if we have a solution $ f(x)$ we can find a new solution $ g(x)$ by taking $ g(x)$ as the restriction of $ f(x)$ on any $ \mathbb Q$-subvector space of $ \mathbb R$ and $ g(x) \equal{} 0$ anywhere else.

2) Let then $ A$ any $ \mathbb Q$-subvector space of $ \mathbb R$ over which $ f(x)\ne 0$ and let us have a look on $ f(x)$ over A.
====================================================================================
From the above elements, we can say :
$ f(x) > 0$ $ \forall x\in A$ (we considered the solutions where $ f(0) \equal{} \plus{} 1$)
$ f(px) \equal{} f(x)^{p^2}$ $ \forall p\in\mathbb Q$

Since $ f(x) > 0$ and $ f( \minus{} x) \equal{} f(x)$, we can set $ f(x) \equal{} e^{h(x^2)}$ for some $ h(x)$ and then :

$ h(x^2 \plus{} 2xy \plus{} y^2) \plus{} h(x^2 \minus{} 2xy \plus{} y^2) \equal{} 2h(x^2) \plus{} 2h(y^2)$

And we have at least all solutions of Cauchy's equation $ h(x \plus{} y) \equal{} h(x) \plus{} h(y)$

[u][i] And here, we should study if it exists some other solution .....[/i][/u]

3) Synthesis of solutions we found up to now :
==============================
The solutions we found up to now are :

3.1) : $ f(x) \equal{} 0$ $ \forall x$

3.2) : Let $ A$ any $ \mathbb Q$-subvector space of $ \mathbb R$ and $ h(x)$ any solution of Cauchy's equation $ h(x \plus{} y) \equal{} h(x) \plus{} h(y)$
$ f(x) \equal{} e^{h(x^2)}$ $ \forall x\in A$
$ f(x) \equal{} 0$ $ \forall x\notin A$
example : $ f(x)\equal{}1$ $ \forall x\in\mathbb Q$ and $ f(x)\equal{}0$ $ \forall x\notin \mathbb Q$

3.3) : Let $ A$ any $ \mathbb Q$-subvector space of $ \mathbb R$ and $ h(x)$ any solution of Cauchy's equation $ h(x \plus{} y) \equal{} h(x) \plus{} h(y)$
$ f(x) \equal{} \minus{} e^{h(x^2)}$ $ \forall x\in A$
$ f(x) \equal{} 0$ $ \forall x\notin A$

It would we rather easy to show that the only continuous solutions are $ f(x) \equal{} 0$ and $ f(x) \equal{} e^{ax^2}$ and $ f(x) \equal{} \minus{} e^{ax^2}$ but unfortunately, continuity is not a constraint.

Please, zaya_yc, could you kindly tell us where is this strange problem coming from ?


logarithms
function
calculus
derivative
limit
integration
complex analysis

What do think about these two results ?

$\sum_{k\geq 1} \ln k= 1/2\ln(2\pi)  $

$\sum_{k\geq 1}(-1)^{k}\ln k=1/2\ln(\pi/2)    $

http://mathworld.wolfram.com/LogarithmicSeries.html

I read on the site that they are divergent in the classical sense (which is pretty obvious), and have to be interpreted as zeta-regularized sums (whatever that means).

Yes, both series diverge. The first one makes no sense at all unless you bring in heavy complex analysis: define $\zeta(s)=\sum_{n=1}^\infty \frac1{n^s}$ for all $s$ with $\mathrm{Re}(s)>1$, and extend this function by analytic continuation to all of $C$ except 1. Then $\sum_n \frac{\ln n}{n^s}$ is the negative of the derivative of $\zeta$. Evaluate at $s=0$ to get the "sum".

For the second, we again have a derivative involving the zeta function at zero, but this time the sum is (probably) also $\lim_{s\to0^+}\sum_{n=1}^\infty \frac{(-1)^n\ln n}{n^s}$, a limit of conditionally convergent sums.

In both cases, I wouldn't use anything so strong as an $=$ sign.

Although I've never seen the term "zeta-regularized sum" anywhere else, it must mean that we introduce a new variable by putting $n^s$ in the denominator, and then extend the definition to the region in question by analytic continuation.

The thing that bothers me is since they both diverge, shouldn't there be a different notation to denote the use of "zeta-regulation", since it can't converge and diverge at the same time.

[quote="Jimmy"]shouldn't there be a different notation to denote the use of "zeta-regulation", since it can't converge and diverge at the same time.[/quote]
Probably there should, but mathematicians are lazy people and try to keep notation to bare minimum (except, maybe, in low level textbooks). For instance, if you look closely, you'll find that the same $\int$ sign is actually used for several different things. Of course, sometimes it gets very confusing, like in this case. One has to remember that there are several "standard" methods of summation of divergent series and a few theorems saying that two different methods agree when both are applicable, though, to be honest, I do not remember whether and how $\zeta$-regularization and other "analytic continuation" methods fit into this general scheme. Anyway, I am inclined to agree with you on the account that one should put a couple of letters in front of the $\sum$ sign to specify which summation method was used but I, probably, also will be one of the first to forget to write those letters when actually working with summation of divergent series  :).

It's kind of like saying that $1+2+\ldots+\infty=-\frac {1}{12}$, which is what you get from plugging in to the functional equation for the zeta function.

A nice formula I like (which might have a few details wrong, since I've forgotten some of the derivation):
$\sum_{i=1}^n i^k=\frac{n^{k+1}}{k+1}\left(1+\sum_{j=1}^k \binom{k+1}{j}\frac{\zeta(-j)}{n^j}\right)$
for $k,n$ positive integers.

It doesn't have much to do with anything else here, but it's pretty neat.

I wonder if string formal works in this problem. :)
only they are very similar.
But I think I don't get meaning of this one. :(


someone tells me how to do 4bii and iii plz

one sec i will post the problems in 0.5 hr; i will create a different topic so the problems are in first post.

[quote="1234567890"]Since it's Friday, I guess we can start, right?[/quote]

I bet you qualified for CMO from Michigan...since you got practically everything except for 4b, and stupid error on 3c.

I didn't qualify since I didn't get any of 4, and I made a stupid error on 1c...I am hoping my score will be above 66...

To mathgeniuse^ln(x): From what you're saying it looks like you got somewhere from 66-69 (probably closer to 69) i think that will be enough for you to make the cutoff

Thank You So Much...

If you are right...you will be so SMART...not like you're not already

[quote="mathgeniuse^ln(x)"]
If you are right...you will be so SMART...not like you're not already[/quote]

Thx, i bet you are smarter, ya i believe the cutoff will be somewhere from 66-68.

[quote="Pineappleperson"]wtf...
why are michigan people on canada forum..???
oh well, didn't take COMC this year.[/quote]
They wrote COMC too, I think.


abstract algebra
algebra
polynomial
calculus
integration
function
domain

Let $ I \subset k[x_1,...,x_n]$ then $ \sqrt {I} \equal{} \bigcap_{I \subset \mathfrak{m}} \mathfrak{m}$.

and you had to use the fact that $ IA_f \equal{} A_f$ (localization)

because $ \sqrt {I}$ is the intersection of all prime ideals over $ I$, it is enough to show the claim for prime ideals $ \mathfrak{p}$, i.e. we have to show that $ J : \equal{} Jac(k[x_1,...,x_n]/\mathfrak{p}) \equal{} 0$.

the case $ \mathfrak{p} \equal{} 0$ is easy: if $ f \in Jac(k[x_1,...,x_n])$, $ f$ is contained in the kernel of $ x_i \mapsto a_i$ for all $ a_i \in \overline{K}$. fixing $ a_1,...,a_{n \minus{} 1}$, $ f(a_1,...,a_{n \minus{} 1},x_n) \in \overline{K}[x_n]$ has infinitely many zeroes, so it must be zero. continuing this argument yields $ f \equal{} f(x_1,...,x_n) \equal{} 0$.

to reduce the general case from this one, use noether normalization: we have a finite monomorphism $ B \to k[x_1,...,x_n]/\mathfrak{p}$, where $ B$ is a polynomial ring in finitely many variables over $ k$. in particular, $ Jac(B) \equal{} 0$. by "lying over", $ Spec(k[x_1,...,x_n]/\mathfrak{p}) \to Spec(B)$ restricts to a surjective map $ Spec_{max}(k[x_1,...,x_n]/\mathfrak{p}) \to Spec_{max}(B)$. thus we have $ B \cap J \equal{} Jac(B) \equal{} 0$. now assume there is some $ 0 \neq f \in J$ and choose an integral equation of $ f$ over $ B$ with minimal degree. then, using that $ k[x_1,...,x_n]/\mathfrak{p}$ is an integral domain, the constant term is $ \neq 0$. but on the other hand it's a multiple of $ f$, so that it lies in $ B \cap J \equal{} 0$, a  contradiction. thus we get $ J \equal{} 0$.

Thanks. I later figured out that

[quote]
and you had to use the fact that $ IA_f\equal{}A_f$ (localization)
[/quote]

is just a restatement of the Hilbert Nulstellensatz, i.e.

If $ I \subset k[X]$ and $ f$ vanishes at every algebraic zero of $ I$, then $ IA_f\equal{}A_f$ implies that some power of $ f$ is in $ I$.

what $ f$ should be? the equation $ I A_f \equal{} A_f$ holds when $ f \notin I$.


function
real analysis
real analysis unsolved

hi

is this true that arccotx=arcsinx/arccosx?

i have a function arccot(n^2+n+1) and i want to find a formula to write it as.  but i only know formulas for cot.

how do i do it? :idea:

[quote="misan"]$ \cot^{ \minus{} 1}(1 \plus{} n \plus{} n^2) \equal{} \cot^{ \minus{} 1}(n) \minus{} \cot^{ \minus{} 1}(1 \plus{} n)$ :([/quote]

hey how did you do it?

See [url=http://mathworld.wolfram.com/InverseCotangent.html]inverse cotangent[/url] and [url=http://mathworld.wolfram.com/Machin-LikeFormulas.html]Machin-like formulae[/url].

The proof of misan's identity (which is the key step in evaluating the sum $ \sum_{k \equal{} 1}^{n} \cot^{ \minus{} 1}(k^2 \plus{} k \plus{} 1)$, a well-known problem that motivates this question) is straightforward:  take the arccotangent of both sides and use the cotangent addition formula.  Its motivation involves the above links.


trigonometry
limit
algebra unsolved
algebra

Without some extra conditions on the sequence, this problem is wrong. (Is that the full original statement?)
As stated, the sequence could converge to $n\pi$ for any odd integer $n$, or it could go to infinity, or it could oscillate between positive and negative values, failing to converge to anything.

Assuming that $0\le u_n \le \frac{2^{n-1}}{\sqrt{2}}$ for $n>3$ so that $1-\frac{u_{n+1}^2}{2^{2n-1}} \ge 0$, we can prove convergence and find the limit. This assumption can be used to uniquely determine $u_{n+1}$ from $u_n$.

Let $\sin \theta_n = \frac{u_n}{2^{n-1}}$; then $\sin^2 \theta_n + (1-2\sin^2 \theta_{n+1})^2 = 1$, so $\cos^2 \theta_n = \cos^2 2\theta_{n+1}$. Obviously, we want $\theta_{n+1} = \frac12 \theta_n$ to make this work. The assumption above is equivalent to $0 \le \sin \theta_n \le \frac{\sqrt{2}}{2}$.
Let $\theta_1=\pi$, $\theta_2=\frac{\pi}{2}$, $\theta_3 = \frac{\pi}{4}$; then if $\theta_n = \frac{\pi}{2^{n-1}}$ we satisfy the assumption above, so this sequence generates the one we care about.
We have $u_n=2^{n-1} \sin (\frac{\pi}{2^{n-1}})$. Since $\lim_{x \to 0} \frac{\sin x}{x} = 1$, $\lim_{n \to \infty} u_n = \pi$.


i was jus wonderin how u wud solve this problem

An 80mL solution of sparkling cide in grape juice contains 25% sparkling cider. How much grape juice needs to be added to make this a 15% sparkling cider solution?

thx

[hide]In the original 80 mL solution with 25% cider, there was 80*.25=20 mL of cider. In the solution with added grape juice, we want 20 to be 15% of the solution. 20=.15(x), and x=133 1/3. By subtracting 80 from this answer, you get the number of mL of added grape juice: 133 1/3-80=160/3, or [b]53 1/3 mL [/b]of added grape juice.[/hide]

oooo kewl thx yo! gud 3xplanation by the way

i hate mixtures!!!!  When i was taking algebra i wanted to burn the chapter about mixtures!  Nice soultion too!

lol y didnt u burn it!!! now i ll hve to do tat chapter


probability LaTeX counting distinguishability

[quote="vikas"]5 people in a choose numbers between 1-10 independently.  What is the probability that exactly 3 of the people will have choosen the exact same number.[/quote]
[hide="Someone please check me on my answer"]There are $ 10^{5}$ total ways.

There are $ \binom{5}{3}= 10 \textrm{ triplets}$.

Pick one of these triplets. There are $ 10$ ways to choose the numbers inside the triplet, and $ 9^{2}= 81$ ways to choose the numbers outside the triplet.

Thus, the number of desired outcomes is:
$ 10 \cdot 10 \cdot 9^{2}$

The total is $ 10^{5}$; hence, our probability is:
$ \frac{10^{2}\cdot 9^{2}}{10^{5}}= \boxed{\frac{81}{1000}}$[/hide]

EDIT: Fixed $ \text{\LaTeX}$ error.

[quote="buzzer11"][quote="vishalarul"]There are $ \binom{5}{3}= 10 \textrm{ triplets}$.
[/quote]

Are you sure? Since the triplets are undistinguishable, I don't think you need to use combinations at all.

Shouldn't it be just 10?[/quote]

The $ \binom{5}{3}= 10$ represnets triplets of people.  They are distinguishable.

Lets say you get the case 1,1,1,3,4.  There are five people A,B,C,D,E.  There rae 5C3 ways to choose 3 pepople out of that group to get the three number 1s


search
linear algebra
linear algebra unsolved

$ A,B \in M_{3}(C)$ s.t $ (AB)^2\equal{}AB^2A$.Prove that $ (AB\minus{}BA)^3\equal{}0_{3}$

Using your asumption, you get easily:
$ (AB\minus{}BA)^2\equal{} (BA)^2\minus{}(AB)^2$, and $ (AB\minus{}BA)^3\equal{}(AB)^3\minus{}(BA)^3$,
So $ \mbox{tr}((AB\minus{}BA)^k)\equal{}0$, for $ k\equal{}1,2,3$, and $ AB\minus{}BA$ is nilpotent.

[quote="JC_math"]
So $ \mbox{tr}((AB \minus{} BA)^k) \equal{} 0$, for $ k \equal{} 1,2,3$, and $ AB \minus{} BA$ is nilpotent.[/quote]
why the problem is finished?

If $ A \in M_n(F)$, $ trA \equal{} ... \equal{} tr A^n \equal{} 0$, then $ A$ is nilpotent, thus $ A^n \equal{} 0$. This can be proved by using Jordan form.

[quote="Soarer"]If $ A \in M_n(F)$, $ trA \equal{} ... \equal{} tr A^n \equal{} 0$, then $ A$ is nilpotent, thus $ A^n \equal{} 0$.form.[/quote]

In order to prove this you may consider $ \lambda_{k}$ with $ k\equal{}\overline{1,n}$ the eigenvalues of $ A$ and use the fact that
if $ \lambda_{k}$ is eigenvalue of A then $ \lambda_{k}^p$ is eigenvalue of $ A^p$ and $ \sum_{k\equal{}1}^n \lambda_{k}\equal{}Tr(A)$

[quote="Svejk"][quote="Soarer"]If $ A \in M_n(F)$, $ trA \equal{} ... \equal{} tr A^n \equal{} 0$, then $ A$ is nilpotent, thus $ A^n \equal{} 0$.form.[/quote]

In order to prove this you may consider $ \lambda_{k}$ with $ k \equal{} \overline{1,n}$ the eigenvalues of $ A$ and use the fact that
if $ \lambda_{k}$ is eigenvalue of A then $ \lambda_{k}^p$ is eigenvalue of $ A^p$ and $ \sum_{k \equal{} 1}^n \lambda_{k} \equal{} Tr(A)$[/quote]
You could also run a search on this forum, it has been posted multiple times.


How easy is for someone to lie on a college application? I mean, if you think about it, the only concrete things colleges can check are GPA, AP scores, SAT/ACT scores and big, regional level competition results/awards. Other than those, can they really cross-check anything else? Couldn't you just make other things up? I mean, you obviously won't outright say that you won some kind of grand national award, but smaller things could pass. For instance, you could say that you've won state in piano or another musical instrument ever year (and you haven't). Or that you won the President's Volunteer Award for so many hours (but you really volunteered for less). Or you could raise your AMC scores by a good number of points (obviously you wouldn't say you got a perfect score if you really didn't) You probably can't modify your extracurricular activities in school because of teacher's signatures (like saying you're the president or vice-president of something) I'm not asking this because I'm planning on padding my app with false information, but because I want to know that my competition isn't doing it.

That's what recommendations and your school information are for.  There are a lot of things you can check by contacting a few teachers and your school.  (Or your parents ;) )

Obviously people are padding up their resumes, and they're not getting fished out and penalized for it. But admissions officers know that this happens.

The best thing you can do is keep your entire application coherent. Not only will it help show that you are probably not making stuff up, it will also make your application objectively superior to an application that has a padded resume.

Essays and short answers are potentially great places to include short phrases that allude to particular elements of your resume and put them into perspective. Of course, this must be artfully done -- you can't risk breaking flow, coherence, or concision within your essay.

In terms of your resume, choose a structure that highlights your strengths rather than a structure that makes it look like it has more entries. Feel free to include short, concise descriptions or clarifications where you think they will give the reader a better idea of the important things you have done. For example, if you did a science fair project and won a regional award, a state award, and two less significant awards, include the regional and state awards in your awards section, but instead of mentioning the less significant awards, include a short one or two line description of what your science project was.

What's to stop people from lying?

Answer:  Nothing, except a genuine desire to be decent human being, and a healthy fear of getting caught.

But as you point out, if you try to lie about something big and important, you run a big risk of getting caught.  If you lie about little things, you probably won't get caught, but it's almost certainly not going to help you much.  What's the point of lying about stuff no one cares about anyway?  In other words, the risk is low, but the reward is even lower, giving a very poor risk-to-reward ratio.

Interesting little fact:  Lying on your application is essentially the only thing that can get you officially "expelled" from Harvard.  Even if you get caught plagiarizing your term paper, cheating on your chem final, knocking over a liquor store, raping a fellow student, etc., the worst punishment you would receive is a "forced withdrawal."  Under such circumstances, you can always ask to come back, and you can continue to truthfully say that you are still a Harvard student.  But only if you get admitted under false pretenses would you be completely purged from the system, as if you never existed.

You might also be surprised at how much can be checked. AMC scores, in particular, are verifiable; any school that cares enough to ask for them probably knows how to look them up.

More likely, it is people doing things like misrepresenting their responsibilities in a position*, saying they are an active participant in some group that they are only on the mailing list for, downplaying the roles of others in a group accomplishment, inflating community service hours, and things like that. Essentially, they make many reasonably-perceivable tweaks to their overall image that add up.

* - [i]Not so often for a job, but perhaps for an internship, a leadership position in a club/team, a volunteer position, a personal responsibility at home, etc.[/i]

There was a kid in my school who lied about how many houses his parents owned on his applications.  The schools found out, and in addition to not giving him any scholarships he wasn't even admitted to any of the schools.  So there are some things that the schools will find out, no matter how well you try to hide them...  Just be truthful.


limit
function
calculus
derivative
logarithms
integration
calculus computations

How to prove that $ \lim_{x\to\infty}\left(1\plus{}\frac{1}x\right)^{x}\equal{}e$?

Thanks.

What are you using as your definition for $ e$?  I've sometimes seen that $ e$ is defined to be that limit...

[quote="Zakary"] ... I've sometimes seen that $ e$ is defined to be that limit...[/quote]In the my opinion, the definition of the number $ e$ is $ \boxed{\ \lim_{n\to\infty}\left(1\plus{}\frac{1}{n}\right)^{n}\equal{} e\ }$.

Afterwards, using the [b]Heine[/b]'s definition, we'll prove that $ \lim_{x\to\infty}\left(1\plus{}\frac{1}{x}\right)^{x}\equal{} e$, i.e. for any sequence $ x_{n}$, $ n\in N^{*}$

such that $ \lim_{n\to\infty}x_{n}\equal{}\infty$ we have $ \lim_{n\to\infty}f(x_{n})\equal{}e$ (show easily with the function "integar part" : $ [x]\le x<[x]\plus{}1$).

[quote="Virgil Nicula"][ (show easily with the function "integar part" : $ [x]\le x < [x]\plus{}1$).[/quote]
 :)  and this you have already solve your problem #4 from "10 limits with ... problems (without derivatives, please !)."
therefore $ \lim_{x\minus{}\plus{}oo}\frac{[x]}{x}\equal{}1$

How do you prove that there is a limit, and it doesn't increase indefinetely with greater n?

[quote][size=150][b][u]Definition of the number[/size] [size=200]e[/size].[/u][/b]

[size=117][color=darkred]$ 1\blacktriangleright$ [b]For any $ n\in N^{*}$ denote $ e_{n}\equal{}\left(1\plus{}\frac{1}{n}\right)^{n}$ and $ E_{n}\equal{}\left(1\plus{}\frac{1}{n}\right)^{n\plus{}1}$. Observe easily that $ \boxed{e_{n}<\ e_{n}\cdot\left(1\plus{}\frac{1}{n}\right) \equal{} E_{n}\ }\ \ (1)\ .$

$ 2\blacktriangleright$ Apply G.M.$ \le$A.M. to $ x_{k}: \equal{}\left\{\begin{array}{ccc}1\plus{}\frac{1}{n}&\mathrm{\ if\ }& k\in\overline{1,n}\\ \\ 1 &\mathrm{\ if\ }& k \equal{} n\plus{}1\end{array}\right\|$ $ \implies$ $ \frac{n\left(1\plus{}\frac{1}{n}\right)\plus{}1}{n\plus{}1}\ge\sqrt [n\plus{}1]{\left(1\plus{}\frac{1}{n}\right)^{n}\cdot 1}$ $ \implies$

$ \left(1\plus{}\frac{1}{n\plus{}1}\right)^{n\plus{}1}\ge\left(1\plus{}\frac{1}{n}\right)^{n}$ $ \implies$ $ \boxed{\ e_{n}\le e_{n\plus{}1}\ }\ \ (2)\ .$

$ 3\blacktriangleright$ Apply G.M.$ \le$A.M. to $ y_{k}: \equal{}\left\{\begin{array}{ccc}1\minus{}\frac{1}{n\plus{}1}&\mathrm{\ if\ }& k\in\overline{1,n\plus{}1}\\ \\ 1 &\mathrm{\ if\ }& k \equal{} n\plus{}2\end{array}\right\|$ $ \implies$ $ \frac{(n\plus{}1)\left(1\minus{}\frac{1}{n\plus{}1}\right)\plus{}1}{n\plus{}2}\ge\sqrt [n\plus{}2]{\left(1\minus{}\frac{1}{n\plus{}1}\right)^{n\plus{}1}\cdot 1}$ $ \implies$

$ \left(\frac{n\plus{}1}{n\plus{}2}\right)^{n\plus{}2}\ge\left(\frac{n}{n\plus{}1}\right)^{n\plus{}1}$ $ \implies$ $ \left(1\plus{}\frac{1}{n\plus{}1}\right)^{n\plus{}2}\le\left(1\plus{}\frac{1}{n}\right)^{n\plus{}1}$ $ \implies$ $ \boxed{\ E_{n\plus{}1}\le E_{n}\ }\ \ (3)\ .$

*******************************************************************************************

In conclusion, $ 2 \equal{} e_{1}\le e_{n}< e_{n\plus{}1}<\ldots < E_{n\plus{}1}< E_{n}< E_{1}\equal{}4$ for any $ n\in N^{*}$ and $ E_{5}<3$ $ \implies$

$ (e_{n})$ and $ (E_{n})$ are monotonously and bounded $ \implies$ exists $ \lim_{n\to\infty}e_{n}\equal{} e$ and $ \lim_{n\to\infty}E_{n}\equal{} E$. Thus, $ (1)$, $ (2)$, $ (3)$ $ \implies$

$ E \equal{}\boxed{\ e\in (2,3)\ }$ and $ \boxed{\ \left(1\plus{}\frac{1}{n}\right)^{n}< e <\left(1\plus{}\frac{1}{n}\right)^{n\plus{}1}\ }$, i.e. $ \boxed{\ \frac{1}{n\plus{}1}<\ln (n\plus{}1)\minus{}\ln n <\frac{1}{n}\ }$ for any $ n\in N^{*}$.[/b][/color][/size][/quote]

You could also apply the mean-value theorem to $ \log(1\plus{}x)$ on the interval $ \left[0,\frac{a}{n}\right]$ with $ |n|>a$. This will actually give a limit for $ \exp(a)$.

[quote="EBz89"]You could also apply the mean-value theorem to $ \log(1\plus{}x)$ on the interval $ \left[0,\frac{a}{n}\right]$ with $ |n| > a$. This will actually give a limit for $ \exp(a)$.[/quote]
Often you prove first that $ \log x$ and $ e^{x}$ are continuous and differentiable as $ x > 0$ by using the limits $ \lim_{n\to\infty}\left (1\plus{}\frac{1}{n}\right )^{n}\equal{} e$ and $ \lim_{n\to\infty}\left (1\plus{}\frac{x}{n}\right )^{n}\equal{} e^{x}$. How do you prove that $ \ln$ is differentiable without using those limits anywhere? Okay, it is just some series manipulation if you define $ e^{x}$ and $ \ln x$ by using series.

The way I did it is by just applying the binomial expansion on all of the terms
$ \sum_{j \equal{} 1}^{n}\binom{n}{j}1^{n\minus{}j}n^{\minus{}j}$ for integer n

as n increases without bound, this becomes

$ \sum_{k \equal{} 1}^{\infty}\frac{1}{k!}$

If you define e as the value at 1 of a function that equals it's own dirivative and passes through 0,1, then you just take the taylor series about x=0, plug in x=1 and you get the same infinite sum.

[color=darkblue]Denote $ f(x) \equal{}\left(1\plus{}\frac{1}{x}\right)^{x}$, $ x > 0$. [u]Let $ \underline{(x_{n})}$ be a sequence for which $ \underline{x_{n}> 0}$ and $ \underline{x_{n}\rightarrow\infty}$[/u]. From $ [x_{n}]\le x_{n}< 1\plus{}[x_{n}]$ obtain :

$ \frac{1}{1\plus{}\frac{1}{1\plus{}[x_{n}]}}\cdot u_{n}\equal{}$ $ \left(1\plus{}\frac{1}{1\plus{}[x_{n}]}\right)^{[x_{n}]}\le$ $ \left(1\plus{}\frac{1}{x_{n}}\right)^{x_{n}}\le\left(1\plus{}\frac{1}{[x_{n}]}\right)^{1\plus{}[x_{n}]}\equal{} v_{n}$ $ \implies$ $ \frac{1}{1\plus{}\frac{1}{1\plus{}[x_{n}]}}\cdot u_{n}\le f(x_{n})\le v_{n}$,

where $ u_{n}\equal{}\left(1\plus{}\frac{1}{1\plus{}[x_{n}]}\right)^{1\plus{}[x_{n}]}$, $ v_{n}\equal{}\left(1\plus{}\frac{1}{[x_{n}]}\right)^{1\plus{}[x_{n}]}$ are subsequencies of $ e_{n}$, $ E_{n}$ respectively $ \implies$ $ u_{n}\rightarrow e$, $ v_{n}\rightarrow e$ $ \implies$ $ \underline{f(x_{n})\rightarrow e}$.

Prove similarly (with the "integer part") that for any $ x > 0$, $ \boxed{\ \frac{1}{1\plus{}x}<\ln\left(1\plus{}\frac{1}{x}\right) <\frac{1}{x}\ }$.[/color]

@Puuhikki; I would either define $ \exp$ via the infinite series and then define $ \ln$ to be it's inverse or define $ \ln(x)$ to be the integral between $ 0$ and $ x$ of $ \frac{1}{t}$ with respect to $ t$ and define $ \exp$ to be it's inverse - the two definitions imply the differentiability of $ \exp$ and $ \ln$ respectively and via a well-known theorem $ \ln$ and $ \exp$ respectively. Since the only facts that the idea I posted above uses are  that $ \ln$ and $ \exp$ are inverses at that the derivative of $ \ln$ is $ \frac{1}{t}$ this would avoid circularity. The theorem I mention is that the inverse of a differentiable function is differentable (given that it is well defined).

If you define $ \ln x\equal{}\int_{1}^{x}\frac{1}{t}\ dt$ and $ e$ to be the positive real number such that $ \ln e\equal{}1$:
[hide]
This is equivalent to proving $ \lim_{x\rightarrow\infty}x\ln\left( 1\plus{}\frac{1}{x}\right) \equal{}1$. However, for positive $ x$, we have
\[ t\in\left[ 1,1\plus{}\frac{1}{x}\right]\Rightarrow\frac{x}{x\plus{}1}\leq\frac{1}{t}\leq 1\]
\[ \frac{1}{x\plus{}1}\leq\int_{1}^{1\plus{}\frac{1}{x}}\frac{1}{t}\ dt\leq\frac{1}{x}\]
\[ \frac{x}{x\plus{}1}\leq x\ln\left( 1\plus{}\frac{1}{x}\right)\leq 1\]
from which the limit follows.
[/hide]

[color=darkblue]Denote $ f(x)\equal{}\ln x$, $ x>0$. Thus,

$ \lim_{x\to\infty}\left(1\plus{}\frac{1}{x}\right)^{x}\equal{}e\Longleftrightarrow$ $ \lim_{x\to\infty}x\ln\left(1\plus{}\frac{1}{x}\right)\equal{}1\Longleftrightarrow$ $ \lim_{x\searrow 0}\frac{\ln(1\plus{}x)}{x}\equal{}1$ $ \Longleftrightarrow$ $ \boxed{\ \lim_{x\to 1}\frac{\ln x}{x\minus{}1}\equal{}1\ }$ $ \Longleftrightarrow$ $ f'(1)\equal{}1$.

Otherwise, from the upper "[b]clenching[/b]"  (see the my last reply) $ \frac{1}{x\plus{}1}<\ln\left (1\plus{}\frac{1}{x}\right)<\frac{1}{x}$

obtain for $ x>0$ that $ \frac{x}{x\plus{}1}<x\ln\left(1\plus{}\frac{1}{x}\right)<1$ $ \implies$ $ \lim_{x\to\infty}x\ln\left(1\plus{}\frac{1}{x}\right)\equal{}1$.[/color]


What is the sum of the two values of $ x$ for which $ (x\plus{}3)^2 \equal{} 121$?


May be use Am-Gm
Because $a+b+c=1$ so $abc \le \frac{1}{27}$
We have $(1+a)(1+b)(1+c)\ge 64\sqrt[4]{\frac{1}{3^{9}}abc}\ge ...\ge RHS$


Euler
function
number theory unsolved
number theory

Suppose $a \in Z$ is an integer not divisible by p . Prove that

$a^ \phi^({}^{p{}^r}^) \equiv 1 mod p^r$.

Any ideas on how to approach this ??

Thanks

$gcd(a,p)=1\Rightarrow gcd(a,p^r)=1$
so euler theorem can be applied as ZetaX mentions :)


linear algebra
matrix
group theory
abstract algebra
vector
superior algebra
superior algebra solved

a) Let $G$ be the group of all $2 \times 2$ matrices $\left(\begin{array}{rc} a & b \\ c & d \end{array}\right)$ where $a$,$b$,$c$ and $d$ are integer modulo $p$, $p$ a prime number, such that $ad-bc \not= 0$. $G$ forms a group relative to matrix multiplication. What is the order of $G$?
b) Let $H$ be subgroup of the $G$ of part (a) defined by \[ H= \left\{ \left(\begin{array}{rc} a & b \\ c & d \end{array}\right) \in G \, | ad-bc=1 \right\}  \]
what is the order of $H$?

The set of all matrices $n \times n$ with non-zero determinants and where elements of matrices are $\mod p$ is a group called $GL_n(Z_p)$ (the group you mentioned in (a)).

$|GL_n(Z_p)|=(p^n-p^{n-1})(p^n-p^{n-2})...(p^n-1)$.

The order of this group is such because in the first step you choose one from
$p^n-1$ possible rows in the matrix. We can choose the next row on $p^n-p$ possibilities (this time less because of linear independence) and the same with the third row and etc...

Subset of $GL_n(Z_p)$ including matrices with determinant equaling $1$ is called
$SL_n(Z_p)$ (your subgroup $H$). Because we can divide $GL_n(Z_p)$ into $p-1$ equal subsets
(according to their determinants ($p-1$ possible)  (why?) ; only $SL_n(Z_p)$ is a subgroup from those subsets) thus $(p-1)|SL_n(Z_p)|=|GL_n(Z_p)|$ and we have the answer.

Sorry maybe i'm dumb. I don't undertsand that. Its looks like you generalize this for $n \times n$ matrix. I still don't get the course and study this by myself.

I am self-studying this book too!  :D
Anyway, what you need to do is to translate the notion of invertible. One equivalent understanding is rank A = n (let's say A is a n*n matrix) Now we try to choose the columns, one by one.
For the first column, there are $p^n$ choices, but the zero vector should be excluded, so there are $p^n-1$ choices.
for the second column, there are $p^n$ choices, but it cannot lie in the span of the first column, so there are $p^n - p$ choices.
Using the same reasoning we arrive at the order of general linear group of order n.

For $SL_n (Z_p)$, all you need to do is to show that the (p-1) subsets of $GL_n(Z_p)$ (each of different det), is one-to-one, thus all of them having same number of elements. In particular, $o(SL_n (Z_p)) = \frac{o(GL_n(Z_p))}{p-1}$

Thanks Soarer, now I can understand. This is one of the problems that i can't solve in that book ( section 2).
It will become easier if we have learned Linear Algebra. :)


MATHCOUNTS
articles
geometry
analytic geometry
Gauss
AMC
AIME

I observed xpmath using it in MATHCOUNTS forum so I'll assume it's appropriate for Classroom Math.
I don't understand the AoPS Wiki article on this and couldn't find it with search.

So...

What are mass points and how do you use them? And in what types of problem?
In "Geometry Problem" in "MATHCOUNTS" xpmath uses mass points and this seems much easier than my solution. Also, the AoPS Wiki states it to be a shortcut.

I want this to become one of my problem-solving techniques.  :)

I too have been wondering the same thing...

[url=http://www.google.com]Google[/url] is your [url=http://www.lmgtfy.com/?q=mass+points]friend.[/url]

I don't understand any of those; all those websites barely scratch the surface of the concept (then of course I can't use it), other's don't talk about it at all, and yet more use language much too advanced for me to understand and try to use it with tethrahedrons (simple people I'm just starting!) and it all makes no sense.

That's why I posted on this forum.  :)

[url=http://www.artofproblemsolving.com/Wiki/index.php/Mass_Point_Geometry]This[/url] is a good introduction, written in simple language. Additionally, the first external link seems okay to me.

[quote="batteredbutnotdefeated"]that was great! I fully understood the concept![/quote]
Is that sarcasm?

not sarcasm, just overly expressed joy...

Not the wiki article, the google thingy. The mouse moves by itself and typing just randomly appears without me doing anything. What more could you ask for?

oh one more thing, how come my signature does not show up?
can someone please help me out on my signature problem?

[url=http://mathcircle.berkeley.edu/archivedocs/2007_2008/lectures/0708lecturesps/MassPointsBMC07.ps]This[/url] was a good handout; I remember the lecture was pretty interesting.

[quote="PowerOfPi"]I still don't really get it :?: . I am dumb. :huuh:[/quote]

Ditto here.

[quote="isabella2296"][quote="jonathanchou711"]Um. I don't see what was so exciting about that wiki article. It didn't really make sense to me.[/quote]

What doesn't make sense about it?[/quote]
It's a stub and thus very incomplete. It also doesn't explain the purpose of Mass Points. Examples with full explanations could be added to flush out the article. I encourage those who understand Mass Points to expand on that wiki article [url=http://www.artofproblemsolving.com/Wiki/index.php/Mass_Point_Geometry]here[/url].

Yes, but I also suggested

[quote="isabella2296"] Additionally, the first external link seems okay to me.[/quote]

which is the same article AIME15 pointed out. I actually learned from that one myself, and it's very good.

The article being referred to OK, but I think it's meant to go along with a lecture. So some concepts and calculations are not fully explained, and the answers to the examples are not given. I haven't found any really good explanations for Mass Points on the web.

[u]A beginning explanation for Mass Points:[/u]
Mass Point Geometry is a geometry where points (in the plane or space) are assigned a positive real number which we call the Mass of the Point. We denote points/masses by $ (P, m)$, where P is the name or coordinates for the point and m is the numerical Mass assigned to that point.

There is an "addition" rule defined for the points/masses in this geometry given by $ (P,m) \plus{} (Q,n) \equal{} (R, m \plus{} n)$, where the point $ R$ is a point on the line joining $ P$ and $ Q$ at a location such that $ PR: RQ \equal{} n: m$. This location is uniquely determined by m and n. [Read this again and draw a picture. It is the key to understanding Mass Point Geometry. If you know anything about physics, $ R$ is essentially located at the "balancing point" of segment $ PQ$]

So for example, if we have the points $ P \equal{} (1,2)$ with mass 2 and $ Q \equal{} (4,8)$ with mass 1, what would be the location and mass of $ (P,2) \plus{} (Q,1)$? [hide="See here for the answer:"] $ R$ would be located at $ (2,4)$ because this is a point such that segment $ PR$ is half segment $ RQ$, and the point $ R$ would have a mass of 3. So $ ((1,2),2) \plus{} ((4,8),1) \equal{} ((2,4),3)$[/hide]

Another example, if we have the points $ P \equal{} (6,3)$ and $ Q \equal{} (1,18)$, what masses would YOU assign to $ P$ and $ Q$ so that $ R$ (which comes from the sum of $ P$ and $ Q$) will have the location $ (4,9)$? [hide="See here for the answer:"] Since $ PR: RQ \equal{} 2: 3$, you could assign the mass 3 to $ P$, and 2 to $ Q$ (or any other positive numbers that have this same ratio).[/hide]

Now that we've defined an "addition" for points, what do you think would be the way to do "subtraction" of points??
What would $ ((8,2),5) \minus{} ((2, \minus{} 1),3)$ be? [hide="Answer:"] $ ((12,4),2)$[/hide]

Notice that given the masses of two points in the plane, the addition/subtraction rule only determines the masses of certain points in the plane. It does not determine the masses of many other points in the plane, and in fact those points' masses can be assigned in any way you like.

Now the question is, what can you do with this addition?? ...to be continued

Mass points are really just a fast, easy, clean way to incorporate Ceva's theorem and [url=http://planetmath.org/encyclopedia/ProofOfVanAubelsTheorem.html]Van Aubel's theorem[/url] together.

Basically, if you have two points with masses (call them aA and bB, where a, b denote the masses and A, B denote the points), their mass point sum is the point C with mass a + b such that, when you have a lever with masses a and b, the location of the fulcrum would be at C. Formally, AC/CB = b/a.

I'm not an expert at mass points, but if I recall correctly, we may assign arbitrary masses to the vertices of a triangle. If we have a triangle ABC, we'll call these mass points aA, bB, and cC. If we are, say, already given that some point E lies on AB and some point F lies on BC, we may pick our masses appropriately such that the point part of aA + bB is E and the point part of bB + cC is F. For instance, if we choose E and F to be the midpoints, masses of 1 on A, B, and C would result in E and F being the sum of the mass points.

What of the last point, cC + aA? If we draw cevians AF and CE, and draw a cevian through B going through the point of intersection, and let G be the intersection of that cevian and CA (sorry for the wordiness), we can see that G = cC + aA (this is a simple consequence of Ceva's theorem.)

One nice property of mass points is the associative property: (aA + bB) + cC = aA + (bB + cC). I'm not going to prove it here, but it should be noted that that sum is the intersection of the cevians AF, BG, and CE.

Some examples will probably clear this up. Take medians, for instance. In triangle ABC, if we have midpoints M of AB, N of BC, and P of CA, and if we assign masses of one to each vertex, the sum of the masses will be M, N, and P (with M, N, and P each having a mass of 2.) The centroid, G, lies on AN, and A has mass 1 and N has mass 2. Therefore, G is situated two-thrirds along the line AN, with AG / GN = 2.

Another example. Say we have a triangle ABC, and M is a trisection point of AB closer to A, and N is a trisection point of BC closer to C. If we let A have mass 2, B have mass 1, and C have mass 2. From here, we can see that 3M = 2A + 1B, and 3N = 2C + 1B. From here, we can deduce that if the cevian through B and the intersection of cevians AN and CM intsercts AC at P, 4P = 2C + 2A, that is, P is the midpoint of A and C.

In addition, if we let O be the intersection of the cevians, since O lies on a line segment AN such that A has mass 2, and N has mass 3, we get that AO/ON = 3/2. Similarly, CO/OM = 3/2, and BO/OP = 4/1 = 4.

Diagrams would probably be helpful, but I don't know how to make them. :/

Ah, Gauss beat me to it. -_-

So who discovered mass point, and did the person provide a proof?
Anways how would we actually PROVE mass point works? (i saw [url=http://www.artofproblemsolving.com/Wiki/index.php/1992_AIME_Problems/Problem_14]an aime problem[/url] where we use the ratio of side length is equal to the ratio of the masses but i don't really get WHY...  :(  :huh:


function
algebra
functional equation
algebra unsolved

Function $f$:$R \to R$ satisfies $f(x^3+y^3)=(x+y)((f(x))^2-f(x)f(y)+(f(y))^2)$, where $x$, $y$ are arbitrary natural numbers. Prove: For any real number $x$, $f(1996x)=1996f(x)$.

Setting $x = y = 0$ we get $f(0) = 0$ and setting $x = 1, y = 0$ we get $f(1) = f(1)^2 \implies f(1) = 0, \pm 1$.

Multiply the RHS by $f(x) + f(y)$ and we get $( f(x) + f(y) ) f(x^3 + y^3) = (x+y)( f(x)^3 + f(y)^3 )$.  Set $y = 0$ and we have $f(x) f(x^3) = x f(x)^3$, so either $f(x) = 0$ for naturals $x$ or $f(x^3) = x f(x^2)$ for naturals.

I do not understand how to find the solution for reals, as it is entirely possible to define the function, for example, $f(x) = x, x \in \mathbb{N}$ and $f(x) \neq x, x \not\in \mathbb{N}$ and then the condition $f(1996x) = 1996 f(x)$ would no longer be met but the given condition would?

Edit:  From the condition, we can find that $f'(1996x) = f'(x)$ for all real $x$, which implies $f'(x) = c$ for some constant, so $f(x) = cx + b$ for some constants (plugging back into the original equation quickly gives $b = 0$ and plugging into the givens gives $c = 0, 1$) - however, how can we determine from the given functional equation, which holds only for naturals, the value of $f(x)$ for reals?  As before, I am stumped.


A circle is inscribed in a quadrilateral $ ABCD$ . Let $ I$ be the centre of the circle. Suppose that:
$ (AI\plus{}DI)^2  \plus{}  (BI\plus{}CI)^2 \equal{}(AB\plus{}CD)^2$.

Prove that $ ABCD$ is an isoscles trapezoid.


algebra
polynomial
symmetry
geometric sequence

$ a,b,c \in R\minus{}\{0\}$ ;If $ (ab\plus{}ac\plus{}bc)^3\equal{}abc(a\plus{}b\plus{}c)^3$ then show $ \{a,b,c\}$ has a geometric sequence order.

[hide="Hint"]Suppose the condition were re-written as $ P(a,b) \equal{} 0$. $ \{a, b, c \}$ is in geometric order iff. one of $ a^2 \minus{} bc$ and permutations is a factor of $ P$. What does symmetry tell you?[/hide]

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=323597[/url]


calculus
integration
inequalities
trigonometry
calculus computations

Prove the following inequality for $x\geq 0.$ \[\int_{0}^{x}(t-t^{2})\sin^{2004}t\ dt <\frac{1}{2006}\]

[hide]
This is a very weak bound. First, it is clear that the integral attains its maximum when $x=1$. Now upper bound $\sin x$ by $\sin 1$. We get that the integral is at most \[\int_{0}^{1}(t-t^{2}) (\sin 1)^{2004}dt = \frac{1}{6}(\sin 1)^{2004}< 10^{-150}.\] [/hide]


Cac ban cho toi hoi:Lam the nao de co chu ki o duoi moi bai viet cua minh,tieng Anh cua to kem lam.

[quote="nttu"]NTTuan co the vao profile , tim signature , roi ky ten vao do  ;)[/quote]

Toi da thu roi nhung khong duoc :(

Hay la toi chon che do khong hien ra nhi?

Xin loi duoc roi thi phai ,cam on nhe!

[quote="N.T.TUAN"][quote="vnmathboy"]co phai nttuan o tpHCMko nhi???[/quote]

Nham roi ban a :D[/quote]

Gia Lai?

[quote="Spring2005"][quote="N.T.TUAN"]Cung khong phai :D[/quote]Anh Tuan hoc Le Quy Don o Nha Trang, Khanh Hoa dung ko?[/quote]

Cung khong phai :D

Cac ban dung doan nua, mat thoi gian lam.

[quote="Lovasz"]Mấy ông tiện thể cho tôi hỏi cái này.
Tại sao mỗi lần gửi bài tôi lại thấy có cái biểu tượng "edit" và "X"(delete) ở dưới góc phải. Cái này không cần cho lắm vì có thể vào "quote" để chỉnh sửa nội dung mà. Làm cách nào để xóa nó đi vậy?[/quote]

Khong co cach nao dau,de vay con sua nua chua,cach ma ong noi khong sua duoc daU :D

[quote="Lovasz"]Mà kể cũng kì lạ thật. Ai không có, chỉ mình tôi có :D[/quote]

Ai chang co,khi dang nhap roi thi chi duoc phep sua bai cua minh thoi :D

[color=blue]Đúng vậy. Mình chỉ được quyền sửa bài viết của mình thôi chứ, sao có thể sửa bài của người khác (trừ admin)[/color]


without worrying abt tthe yield outline a method to synthesise 2 methyl pyridine.
u have available to u aceatldehyde as the only starting material. and any inorganic reagent

[hide]source: who but me:P  [/hide]

Looking at number of carbons of the product, its obvious this synthesis will involve amonia, acid treatment, and some heat in order to make dehydration and aromatization easier.

The actual is simple: just heat a mixture of acetaldehyde and ammonia and 2-picoline is obtained directly.


Source: USC Math Competition 2002

Two intersecting circles each have radius $6$ and the distance between the two centers is $6(3^{1/2}$, find the area of the region that lies inside both circles.

EDIT: Sorry for calling it the remainder problem, I accidently put the wrong name :oops_sign:

I wish I could draw a diagram here cuz it would be much easier to explain...

Let $P$ and $Q$ be the centers of two circles.
Let $A$ and $B$ be the intersecting points of two circles, and let $C$ be the intersection of $PQ$ and $AB$.

Since $PA=6$ and $PC=CQ=3\sqrt{3}$, $\angle APC=30^\circ$, which means $\angle APB=60^\circ$ and $\triangle PAB$ is an equilateral triangle.

The area we want to find is $2(sector PAB-\triangle PAB)$.
The area of the sector is $\frac{1}{6}(6^2\pi)=6\pi$
The area of the triangle is $\frac{6^2\sqrt{3}}{4}=9\sqrt{3}$

Therefore, the area of the region is $2(6\pi-9\sqrt{3})=12\pi-18\sqrt{3}$

[quote="frt"]I wish I could draw a diagram here cuz it would be much easier to explain...

Let $P$ and $Q$ be the centers of two circles.
Let $A$ and $B$ be the intersecting points of two circles, and let $C$ be the intersection of $PQ$ and $AB$.

Since $PA=6$ and $PC=CQ=3\sqrt{3}$, $\angle APC=30^\circ$, which means $\angle APB=60^\circ$ and $\triangle PAB$ is an equilateral triangle.

The area we want to find is $2(sector PAB-\triangle PAB)$.
The area of the sector is $\frac{1}{6}(6^2\pai)=6\pai$
The area of the triangle is $\frac{6^2\sqrt{3}}{4}=9\sqrt{3}$

Therefore, the area of the region is $2(6\pai-9\sqrt{3})=12\pai-18\sqrt{3}$[/quote]
frt, did you check what numerical value is that area? it is a negative number, please check your solution

edit: i think i found what is the problem with your solution, \pai is not $\pi$ \pi  :!:  :!:  :!:


real analysis
integration
function
real analysis unsolved

For wich Lebesgue-measurable subset $ E$ of the real line does a positive constant $ c$ exist for which $ \displaystyle{\sup_{\minus{}\infty < t < \infty}\Big{|}\int_{E}e^{itx}f(x)dx\Big{|} \leq c \cdot \sup_{n\equal{}0,\pm 1,...}\Big{|}\int_{E}e^{inx}f(x)dx\Big{|}}$, for all integrable funtcion $ f$ on $ E$?

Remark: Tomorow starting the Schweitzer Comp.

It is not [i]very hard[/i], but requires reasonably good knowledge of duality and some idea of what functions are representable by absolutely convergent Fourier series. Here is the [hide="answer"]
Take finitely many closed subintervals $ I_k\subset \mathbb R$ that are disjoint $ \mod 2\pi$ (in particular, no interval cannot be of length $ 2\pi$ or more). Take any measurable subset of their union and add an arbitrary set of measure $ 0$. The resulting set will satisfy the condition and every set satisfying the condition can be obtained this way.
[/hide]
I do not want to spoil the fun too soon, so I'm not posting the solution now (even if you know the answer, finding the proof remains a non-trivial task) ;)


Frank takes a $\$400,000$ loan from the bank to purchase a house. Each month, the bank charges $6\%$ interest on the money Frank still owes the bank. Also, each month Frank gives a payment of $x$ dollars to the bank ($x$ is constant for all the months).

Compute the value of $x$ if Frank pays off the loan completely after $30$ years.

$6\%$ [i]monthly[/i] interest is awfully high.  These questions will usaully define interest annually and make you divide it into monthly payments.  Anyway,
[hide] we have interest payments of $400,000(.06)$ for the first month and $400,000(.06^{2})$ for the second month.  We see the pattern forming.  We have the geometric series (interest payments only) $400,000(.06+.06^{2}+.06^{3}+\cdots+.06^{n})$.  In this case $n$ is finite.  $12$ monthly payments for $30$ years gives $n = 360$.  The sum of a finite geometric series is given by $a\left(\frac{1-r^{n}}{1-r}\right)$. In this case $.06^{360}$ approaches zero.  So we have $400,000\left(\frac{1}{1-.06}\right) \approx 425,531.91$ in total interest paid.  So the total payment on the mortgage is $400,000+425,531.91 = 825,531.91$.  $12$ monthly payments of $x$ over $30$ years gives $360x = 825,531.91 \implies x \approx \boxed{2,293.14}$.[/hide]

I don't believe that answer is correct. The bank only charges interest each month on the principle still owed, not on the original amount borrowed.

I also believe that the problem probably meant that there was 6% annual interest (ARP) [i]compounded[/i] monthly. You will see why in a minute.

[hide]So assuming that to be correct, consider this:

The total future value of the money borrowed is $400,000\left(1+\frac{.06}{12}\right)^{360}$. Each time you make a payment of x dollors, you reduce the future value of the money borrowed by x dollars along all of the future interest associated with that x dollars.

So $400,000\left(1+\frac{.06}{12}\right)^{360}-x\left(1+\frac{.06}{12}\right)^{359}-x\left(1+\frac{.06}{12}\right)^{358}-\ldots-x = 0$

$400,000*(1.005)^{360}= x\left[\frac{(1.005)^{360}-1}{1.005-1}\right]$ by sum of a geometric series.

$x = \frac{400,000*1.005^{360}*.005}{1.005^{360}-1}= 2,398.20$ dollars per month.

If it really was 6% interest charged each month, you would have $x = 24,000$ dollars per month.[/hide]

Also, the correct spelling is Mortgage.

Sorry for the ambiguity. The interpretation I had in mind was the one gauss202 used.

Oh, and  :blush: about the spelling of "mortgage."

thanks...nice post
[url=http://www.interestonly-mortgage.com][color=#F8F8F8][u]interest only mortgage[/u][/url][/color]


Suppose that chord a intersects a circle at two pts to create an arc which subtends an angle of 30 degrees. Similarly, chord b intersects a circle at two pts to create an arc which subtends an angle of 90 degrees. Given that c=a+b, what angle would c "subtend."


vector
analytic geometry
linear algebra
matrix

On page 102 of the Hoffman and Kunze linear algebra, the authors derive the following result using properties of linear functionals and dual bases:

If W is a k-dimensional subspace of an n-dimensional vector space V, then W is the intersection of n-k *hyperplanes in V.
*where a hyperplane is an n-1 dimensional subspace of V

However they never again refer to this result in the section and I'm quite confused about what the relevance or significance of this is. After deriving this, they go on to explain that the corrolary implies that "if we select some ordered basis for the space, each k-dimensional subspace can be described by specifying n-k homogeneous linear conditions on the coordinates relative to that basis." However the discussion itself has nothing to do with the result that W is the intersection of n-k hyperplanes in V; it could very well be carried out without even mentioning this corrolary.

So. Can anyone explain the relevance of this? My own initial thoughts were that if A is an m x n matrix and rref(A) has k non-zero rows then the dimension of the 'null-space' of A is n-k. But I'm a little fuzzy on the details and whether this is at all related. The row space of A is annihilated by n-k independent linear functionals on R(n) and I suppose I'm asking whether that is equivalent to saying the row space of A is the intersection of n-k hyperplanes. Besides, the authors don't even mention this. Does it mean that W is the intersection of [i]exactly [/i]n-k hyperplanes in V (I'm assuming they are distinct)? Are these hyperplanes unique?

This is not a very deep and exciting result: it only means that a $k$-dimensional subspace can be described by $n-k$ linear equations. Think of lines in three-dimensional space ($k=1$, $n=3$). The symmetric equations of a line is a system of $2=3-1$ equations. The geometric meaning of this is that a line is an intersection of two (2-dimensional) planes. Of course, such planes are not unique.

It seemed that it wasn't very interesting so I was wondering why they bothered to mention it if they weren't even going to apply it anywhere. Am I wrong then about the connection between the theorem and matrices?


inequalities
calculus
inequalities proposed

For $ a,b,c$ positive reals
$ \sum\frac{1}{a^{2}\plus{}bc}\leq\frac{(a\plus{}b\plus{}c)^{2}}{3(ab\plus{}bc\plus{}ca)}(\sum\frac{1}{a^{2}\plus{}b^{2}})$

It is a known inequality of [i]Pham Huu Duc[/i]. You can find an approach [url=http://reflections.awesomemath.org/2007_2/3variable.pdf]here[/url] (page 6). However, I'm also interested in a non-computational solution, without calculus (if there is one).


Να υπολογισθεί το άθροισμα

Από 0 μέχρι το 2006 της παράστασης, γραμμή κλάσματος, αριθμητής 2ν + 3, παρονομαστής ν^4 + 6ν^3 +13ν^2 +12ν +7

[quote="kostas"]μπραβο Ραμανατζαν. η λυση σου ειναι υποδειγματικη. :)[/quote]

 :)

Καλή η λυση σου Ραμανατζαν. :)

Μπράβο Ramanujan!

Προβλήματα αυτού του τύπου κατασκευάζονται με την εξής συνταγή: Θεωρούμε μια παράσταση, π.χ. ρητή, παίρνουμε την προς τα μπρος διαφορά της (forward difference) με βήμα (step) 1 ή 2 ή 3 ή… , αναδιοργανώνουμε το αποτέλεσμα για να καλύψουμε τα ίχνη μας…και κοπανάμε ένα Σ μπροστά με κατάλληλα όρια άθροισης! Εκθετοποιείς αν θέλεις και κάνεις Π το Σ (αντίθετα, βέβαια, το Π γίνεται Σ με λογαριθμοποίηση). Άλλος τρόπος υπολογισμού ή εκτίμησης αθροισμάτων είναι ο περίφημος τύπος (ασυμπτωτικός) Euler- Maclaurin. Έχει πέσει στην αντίληψή μου ότι οι αναλογιστές (actuaries) έχουν μεγαλύτερη από τους μαθηματικούς ευχέρεια με τέτοιου είδους αθροίσματα επειδή τα φαινόμενα ζωής και θανάτου υπολογίζονται σε διακριτό χρόνο (ημέρες, μήνες, έτη).

Lance

[quote="DuLac"]
 Άλλος τρόπος υπολογισμού ή εκτίμησης αθροισμάτων είναι ο περίφημος τύπος (ασυμπτωτικός) Euler- Maclaurin. [/quote]

Θα ήθελα μια περιγραφή  :)

[quote="DuLac"]Έχει πέσει στην αντίληψή μου ότι οι αναλογιστές (actuaries) έχουν μεγαλύτερη από τους μαθηματικούς ευχέρεια με τέτοιου είδους αθροίσματα επειδή τα φαινόμενα ζωής και θανάτου υπολογίζονται σε διακριτό χρόνο (ημέρες, μήνες, έτη). [/quote]

Μην βάζεις και στοίχημα :D . Τί ακριβώς είναι οι αναλογιστές???  :?

[quote="DuLac"]
 Άλλος τρόπος υπολογισμού ή εκτίμησης αθροισμάτων είναι ο περίφημος τύπος (ασυμπτωτικός) Euler- Maclaurin. [/quote]

Εννοείτε αυτόν  $R_n=\frac{f^{(n)}(\xi)(x-\xi)^{n-1}(x-a)}{(n-1)!}$    ?

Αν ισχύει  $lim_{n\rightarrow \infty}R_n=0$  και a=0 η σειρά καλείται Mclaurin .
Aλλά δεν είναι το μόνο πράγμα στα μαθηματικά που λέγεται  Mclaurin  ;)
οπότε δεν ξλερω αν είναι αυτό που αναφέρεται παραπάνω ......

Εννοώ

βλ. http://wolfram.mathworld.com/Euler-MaclaurinIntegrationFormulas.html

More on sums later!


Να βρείτε τη μέγιστη τιμή της παράστασης $ x^2y\plus{}y^2z\plus{}z^2x$ όπου $ x,y,z$ πραγματικοί με $ x\plus{}y\plus{}z\equal{}0$ και $ x^2\plus{}y^2\plus{}z^2\equal{}6$.

Γιωργο μπορεις να δωσεις κανενα hint? Γιατι εχει μεινει αρκετο καιρο αλυτη αυτη η ασκηση. Εχω βρει διαφορες μεγιστες τιμες χρησημοποιοντας σχεδον ολες τις κλασικες ανισοτητες και ολα τα τεχνασματα, αλλα το προβλημα ειναι οτι σε ολες αυτες τις μεγιστες τιμες ετσι οπως ειναι οι συνθηκες η ισοτητα δεν ισχυει ποτε!

Δεν εχω δοκιμασει βεβαια πολλαπλασιαστες Lagrange αλλα αν ηταν θεμα BritishMO θα πρεπει να υπαρχει σιγουρα μια καθαρα αλγεβρικη λυση

Nick ένα hint
[hide]Το μέγιστο της παράστασης είναι 3 και επιτυγχάνεται για παράδειγμα για $ x \equal{} y \equal{} \minus{} 1$ και $ z \equal{} 2$[/hide]
Υπόδειξη και για τη λύση
[hide]Θεώρησε ότι $ abc \equal{} k$ και εξέφρασε την παράσταση συναρτήση του $ k$ . Μετά έχεισ συνάρτηση μιας μεταβλητής και ξεμπλέκεις εύκολα .  :wink: [/hide]
Αν αυτά δεν βοήθησαν πείτε μου να βάλω αναλυτικά τη λύση

Thanx Σιλουανε, ειχα προσπαθισει σημερα το πρωι να μετατρεψω την παρασταση σε παρασταση μιας μεταβλητης κανοντας αντικαταστασεις απο τις συνθηκες, αλλα αυτο που προεκειψε ηταν μια τεραστια παρασταση με ριζες μεσα κλπ στην οποια ηταν αδυνατον να βρεθει η μεγιστη τιμη (η αν υπηρχε περιπτωση να βρεθει η μεγιστη τιμη, ηθελε πολλες παραγωγισεις, πραξεις κλπ)

Αυριο θα προσπαθησω μηπως καταφερω να εκφρασω την παρασταση συναρτηση του $ k \equal{} xyz$ οπως λες γιατι τωρα ειναι αργα. Προφανος ομως στη λυση σου εχεις θεωρισει οτι $ xyz \neq 0$, διοτι για $ xyz \equal{} 0$ λογο των 2 συνθηκων η παρασταση παιρνει τις τιμες $ \minus{} 3\sqrt {3}$ και $ 3\sqrt {3} > 3$

Btw μια ερωτηση απο απλη περιεργια. Σε ποια σχολη περασες? Μαθηματικων Αθηνας?


logarithms
inequalities
analytic geometry
graphing lines
slope
LaTeX
probability

Okay.  Well this competition is posted for people who are interested in mathematics and enjoy learning from these problems.  This is how the competition works.  First, I post a series of 10 problems and call them Rond #1.  You then Pm me the answers and solutions.

[b]PM ME THE SOLUTIONS & ANSWERS[/b] When you send me the solutions, you will receive the points for the correct answers you get (including solutions).  If you get a problem wrong, I will PM you back and you may try sending only those again to me before the deadline.  Then it continues on until you either stop PMing me or the deadline is reached. The points will be revealed after every round.

[b] The deadline for round #1 is met.[/b]

[hide="Round #1"]
1. Prove: -0=0.
2. What is the 3765th number in the sequence 10, 19, 28, 37,...
3. Prove: If $A,B,C,D$ are sets such that $A\subseteq C$ and $B\subseteq D$, then $A\times B\subseteq C\times D$.
4. Expand: $(1+x)^6$.
5. The sum of the digits of a 2 digit number is 8.  If the first and last digits of the number are interchanged, the number is increased by 36.  Find the number.
6. We have 2 cables and a router that has 4 ports.  How many different arrangements of these 2 cables are possible?
7. Solve for x, y, and z:
\[ 6x-4y-7z=17\\ 9x-7y-16z=29\\ 10x-5y-3z=23  \]
8. The risistance R of two resistors in parallel is given by $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$.  R is found to be $283\Omega$, but if $R_1$ is doubled and $R_2$ is halved, then $R$ is found to be $291\Omega$.  Find $R_1$ and $R_2$.  Note: $\Omega$ is just a symbol known as omega/ohms that is the unit used for resistence.
9. What is the value of $i^{17}$?
10. If
i) $f(1)=1$
ii) $f(2x)=4f(x)+6$
iii) $f(x+4)=4f(x)+4f(1)+8f(2)$
Find $f(6)$.
[/hide]

[b] The deadline for round #2 is October 13th, 2005[/b]
[hide="Round #2"]
1. Solve for x: $\left|\frac{6-5x}{3+x}\right|\le\frac 12$.
2. Solve $2x^2-5x-3=0$.
3. Rewrite without logorithms: $\log (a^2-b^2)+2\log a=\log (a-b)+3\log b$.
4. If a person has $\$85,000$ in a retirement fund, how much can be withdrawn each year so that the fund will be exhausted in 20 years if the amound remaining in the fund is earning 6.5% per year?
5. Find the sum of infinitely many terms of the GP $9, 3, 1, \frac 13$.
6. Find the eighth term of $(3a+b^5)^11$
7. Suppose that the heights of 2000 students have a mean of $69.3 in.$ with a standard deviation of $3.2 in.$ Assuming the height to have a normal distribution, predict the number of students (a) shorter than $65 in.$; (b) taller than $70in.$; (c) between $65in.$ and $70in.$.
8. If $a$ and $b$ are any numbers, prove that $|ab|=|a|\cdot|b|$.
9. ([i]The Triangle Inequality[/i]) If $a$ and $b$ are any numbers, prove that $|a+b|\le |a|+|b|$.
10. Prove that if neither line $l_1$ nor line $l_2$ are vertical, then $l_1$ and $l_2$ are perpendicular iff the product of their slopes is $-1$.
[/hide]

[b][color=red]By the way, please do not discuss problems until the deadline is met. You may now discuss round #1![/color][/b]

i just signed up to post in this thread, couldnt justify signing up before, now i have something to post.

question 6

[hide]

4! / (4 - 2)!

= 24/2 = 12

[/hide]

edit : i think this round is a good at a good level, it could be a little harder, but not too much.

question 9

[hide]

to find i^17 you must multiply squareroot of -1 17 times. giving you i again.

therefore the awnser to i^17 is i

[/hide]

:\ sorry if this looks terrible i will edit it later, im new to latex

question 4

[hide]

$x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1$

i knew pascals triangle would come in handy one day  ;)

[/hide]

[color=green][size=75]$\LaTeX$ed by...you know who :D[/size][/color]

im assuming in question 7 that the pronumerals are constants?

im very intrested in seeing how one would solve this.

[quote="chess64"][quote="boxedexe"][b]NOTE: YOU ARE ONLY ALLOWED TO POST ONE SOLUTION IN EVERY POST YOU MAKE.[/b][/quote]

Why? :?[/quote]

The probability of people getting points will decrease if one person posts all answers at once.

can't people keep double, triple, quadruple posting? That's even worse,

[quote="DanielKirchner"]question 2
[hide]
S(a) = 10+9(a-1)
S(3765) = 10+9*3764=33877[/hide][/quote]
Incorrect

[quote="chess64"]Just make people PM you...[/quote]

Well, if people PM me, people would not be able to.... lol, okay, that'll work better.

PEOPLE!  PM ME TEH ANSWERS.

[quote="chess64"][quote="boxedexe"]7. Solve for x, y, and z:
\[ 6x-4y-7z=17\\ 9x-7y-16z=29\\ 10x-5y-3z=23  \][/quote]

[hide]$\left[ \begin{tabular}{c} x\\y\\z \end{tabular} \right] = \left[ \begin{tabular}{ccc} 6&-4&-7\\ 9&-7&-16\\ 10&-5&-3\\ \end{tabular} \right] \cdot \left[ \begin{tabular}{c} 17\\29\\23 \end{tabular} \right] ^{-1}=\left[ \begin{tabular}{c} 3\\2\\-1 \end{tabular} \right]$[/hide][/quote]

You may also use linear systems.

wouldnt it make more sense to post it in the thread? the aim is to show others how you got to the awnser, not to win

[quote="236factorial"][quote="chess64"]Matrices are always the fastest...if you have access to a calculator :D[/quote]

Your first answer was wrong... then you edited it. You calculator was wrong  :o[/quote]

That's because I accidentally put a $5$ instead of a $6$ when I was posting it :blush:

just if anyone wants to see how i solved question 2

[hide]

take a = 10
take d(difference) = 9
take n = 3765

plug into this formula i made up x = a + ( n - 1 ) x d

awnser been 33886

[/hide]

[quote="complex"]plug into this formula i made up $x = a + (n-1)d$[/quote]

Sorry to disappoint you, but this is a well-known formula. Good job on deriving it, though! :)

for number one...

 :D

we have x=x where x is any number

if we add the same number on both sides, the numbers are equal

x+(0)=x+(-0)

subtract x on both sides

we get 0=-0

 :D  :D

sorry if this is cheesy...i counlt find anything else :D

I'll have to say #1 is a poorly written problem.

There exists no "positive" or "negative" in zero. Zero is just a zero. If you have to be technical, note that:

$0 = -0$ can be written as $0 \cdot 1 = 0 \cdot -1$ and the definition of zero states that anything multiplied to 0 is 0.

I think we shouldn't discuss the problem until the deadline is met but still, this is a poorly written problem and I think it can use replacement.

[color=red][size=150]RESULTS FOR ROUND ONE[/size][/color]

Here are the results for round one, however, some people did not get high marks due to not answering enough questions. Besides, thsi round was little messed up. :P

danne89...8/10
complex...4/10
DanielKirchner...1/10
Chess64...1/10

Btw, second roudn will add up to thsi round, so if danne89 participates next time and he gets 10/10, his score will be 18/20.

When will the next round be posted? I'm very nervous about the next round.  :D

The next round will be up in about a few hours.  If not, tomorrow. :)

can someone explain to me how the matrices work? i dont understand the way chess64 did it, (6 x -4 x -7)(1/17) != 3

Ummm...Can we still join if we didn't participate in the first round?

Sure, I'll consider your mark as n/10 instead of n/20. :)

Roughly half the problems in this contest are too hard for getting started (typically the last half of each set).  I suggest breaking the contest in half and putting half of it in intermediate.  We'll move the thing entirely if it continues being too hard.

i agree, the last half of each set is a bit too hard, its good as a challange, but only the last two question should be at that level.


logarithms
inequalities
inequalities proposed

Let $ a,b,c>0$ proove that holds $ a^a b^b c^c>= (abc)^{\frac{a+b+c}{3}}$ $ e^{a+b+c-3\sqrt[3]{abc}}$

[quote="giannis"]Let $ a,b,c > 0$ proove that holds $ a^a b^b c^c > = (abc)^{\frac {a + b + c}{3}}$ $ e^{a + b + c - 3\sqrt [3]{abc}}$[/quote]
My ugly proof:
If $ b = c$ then we assume $ a = kb.$
Then $ a^a b^b c^c \geq (abc)^{\frac {a + b + c}{3}}e^{a + b + c - 3\sqrt [3]{abc}}\Leftrightarrow k^{\frac {2k - 2}{3}}\geq e^{k + 2 - 3\sqrt [3]k},$ which easy to check out.
Let $ a > b > c$ and $ a = x^3,$ $ b = y^3$ and $ c = z^3.$
Then $ a^a b^b c^c \geq (abc)^{\frac {a + b + c}{3}}e^{a + b + c - 3\sqrt [3]{abc}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3 - y^3)(\ln x - \ln y)\geq x^3 + y^3 + z^3 - 3xyz\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x - y)^2\left(\frac {x^2 + xy + y^2}{\theta_z} - \frac {x + y + z}{2}\right)\geq0,$
for some $ \theta_x\in(z,y),$ $ \theta_y\in(z,x)$ and $ \theta_z\in(y,x).$ We obtain:
 $ S_z = \frac {x^2 + xy + y^2}{\theta_z} - \frac {x + y + z}{2} > \frac {x^2 + xy + y^2}{x} - \frac {x + y + z}{2} > 0,$
$ S_y = \frac {x^2 + xz + z^2}{\theta_y} - \frac {x + y + z}{2} > \frac {x^2 + xz + z^2}{x} - \frac {x + y + z}{2} > 0$
and $ x^2S_y + y^2S_x > x(x^2 + xz + z^2) + y(y^2 + yz + z^2) - \frac {(x^2 + y^2)(x + y + z)}{2} > 0.$
Thus, $ \sum_{cyc}(x-y)^2S_z>(x-z)^2S_y+(y-z)^2S_x=(y-z)^2\left(\frac{(x-z)^2}{(y-z)^2}S_y+S_x\right)>$
$ >(y-z)^2\left(\frac{x^2}{y^2}S_y+S_x\right)>0.$
Id est, our inequality is proved.


function
calculus
integration
quadratics
inequalities
complex analysis
calculus computations

Just out of interest, are there elementary (just single-variable calculus, no Fourier series or more advanced techniques) methods to prove
$ \zeta(4)\equal{}\sum_{n\equal{}0}^{\infty}\frac{1}{n^{4}}\equal{}\frac{\pi^{4}}{90}$?

More generally, is there a closed form for $ \zeta(2k)$ where $ k$ is a positive integer, and an elementary way to prove this?

Question 2: yes, there is, but I haven't seen an elementary proof: http://www.mathlinks.ro/viewtopic.php?t=180928

Define what you mean by "elementary proof."  Single-variable calculus only helps you apply summations in closed form if you have information about some corresponding power series (if you are disallowing Fourier series, which are really power series on the unit circle anyway), and the function $ \sum_{n \equal{} 0}^{\infty} \frac {x^n}{n^4}$ is provably not "elementary"; it can be expressed as repeated integrals that are not expressible in terms of elementary functions.

Edit:  I suppose you are looking for a proof along [url=http://en.wikipedia.org/wiki/Basel_problem#The_proof]these lines[/url].  You could probably use the same bounding trick by using Newton sums to compute the sum of fourth powers of cotangents and cosecants, but it gets pretty ugly.  I don't like proofs like this; computational tricks that "pop out" the answer magically obscure what's really going on.  In particular, the use of de Moivre's formula can be considered a "hidden" use of Fourier-analytic concepts.

Is it really a closed form when there's no closed form for the Bernoulli numbers? It's still an efficient form, and the Bernoulli numbers can be calculated recursively using the generating function. The recursion requires all previous values, so the work needed for the $ n$th Bernoulli number by this method is quadratic in $ n$.

If you want to speed up calculations of the Bernoulli numbers for large $ n$, use Stirling's formula and approximate values of the $ \zeta$ function to get close, and bound the denominator to find the number uniquely.

As for proving the formula with the Bernoulli numbers, I think I first saw a derivation in Ahlfors, based of course on complex analysis.

[quote="jmerry"]Is it really a closed form when there's no closed form for the Bernoulli numbers?[/quote]

You're right, it isn't a closed form, but it still is the most applicable form. If we're talking about real closed forms, I doubt that one would ever be found.

[quote="jmerry"]As for proving the formula with the Bernoulli numbers, I think I first saw a derivation in Ahlfors, based of course on complex analysis.[/quote]
This one? $ \zeta(n) \equal{} \frac{|B_{n}|2^{n \minus{} 1}\pi^{n}}{n!}$
Does this proof exist online?  I am familiar with most of the fundamentals of complex analysis so if this proof involved line integrals and residues, it wouldn't be much of a problem.

If one were to use Fourier series, is there a general way to evaluate $ \zeta(2k)$?  That is, is there some pattern to picking functions $ f(t)$ in the formula
$ \langle f,f_{n}\rangle\equal{}\frac{1}{2\pi}\int_{0}^{2\pi}f(t)e^{\minus{}int}\ dt$
such that we get $ \zeta(2k)$?  I know for $ \zeta(2)$, we can use $ f(t)\equal{}t$ but that's it.  I assume Bessel's inequality plays some role...


Source: Furman University Wylie Math Tournament Jr. Exam 1997 #15

A parabola passes through the points (-1,4), (1,0), and (2,7).  Find the sum of the x-coordinates associated with y = 64.

[hide]A standard forrmula for a parabola is Ax:^2:+Bx+C=y.

Substitute the 3 points on the parabola for x and y and you get

1.) A+4B+C

2.) A+B+C=0

3.) 4A+2B+C=7

1 can be expressed as A=4-4B-C and 2 can be expressed as A=-B-C. Substitution gives us 4-4B-C=-B-C:arrow:4=3B:arrow:B=4/3

Subtracting 3 from 1 gives us -3A+2B=-3. Substitute 4/3 for B gives us -3A+2*4/3=-3:arrow:-9A+8=-9:arrow:-9A=-17:arrow:A=17/9

Substituting 17/9 for A and 4/3 for B in 2 gives us 17/9+4/3+C=0:arrow:C=-29/9

Substituting 17/9 for A, 4/3 for B, and -29/9 for C in Ax:^2:+Bx+C=y gives us (17/9)x:^2:+(4/3)x-29/9=y:arrow:17x:^2:+12x-29=y.

Substituting 64 for y gives us 17x:^2:+12x-29=64:arrow:17x:^2:+12x-93=0:arrow:x=(-12:pm::sqrt:D)/34

(D stands for a constant)

The problem asks for (-12+:sqrt:D)/34+(-12-:sqrt:D)/34 which equals -24/34 which reduces to -12/17.[/hide]

I hope I didn't make an algebraic mistake...

Quote:1.) A+4B+C

[hide]There's something wrong with this.  This equation corresponds to the point (-1,4); hence, the equation should read A - B + C = 4.  Subtracting this from 2.) A+B+C=0  yields 2B = -4, so B = -2.

Plugging -2 for B into 3.) 4A+2B+C=7 gives 4A+C=11.

Plugging -2 for B into 2.) A+B+C=0 gives A+C=2.

Subtracting the first from the second, we find that 3A=9, or A = 3.

Subbing A and B into A + B + C = 0, we find that C = -1.

So the equation is y = 3x:^2: - 2x - 1

When y = 64,

3x:^2: - 2x - 1 = 64

3x:^2: - 2x - 65 = 0

The sum of the roots is -b/a = -(-2)/3 = 2/3.[/hide]

[quote="Tare"]I knew it! 'Tis too late...In some Oriental countries they wake up about now...[/quote]

Wait, I'm confused.  How does this have to do with the problem at hand?

Oh, I see...you're saying that it's way too late to be doing math...well, it's only 11:00 here...


Hi,
Do any aopsers go to boarding school?  In particular Exeter, Hotchkiss, Middlesex, or Andover.  If so, what do you think about them?

Thanks,
lyra

This thread belongs to the Round Table forum.  :wink:

my school has athletic competiions against some of those schools...

[quote="b-flat"]Ask a moderator or administrator  :D

Regular users cannot move topics (think of what the forum would be like if anyone could...)

Moved to Round Table.[/quote]
I know I'm repeating myself over and over but THAT would look like this :D
[hide][img]http://upload.wikimedia.org/wikipedia/commons/f/f0/Rwandan_refugee_camp_in_east_Zaire.jpg[/img][/hide]


Let $ ABC$ be a triangle, all of whose angles are greater than $ 45^{\circ}$ and smaller than $ 90^{\circ}$.
(a) Prove that one can fit three squares inside $ ABC$ in such a way that: (i) the three squares are equal (ii) the three squares have common vertex $ K$ inside the triangle (iii) any two squares have no common point but $ K$ (iv) each square has two opposite vertices onthe boundary of $ ABC$, while all the other points of the square are inside $ ABC$.
(b) Let $ P$ be the center of the square which has $ AB$ as a side and is outside $ ABC$. Let $ r_{C}$ be the line symmetric to $ CK$ with respect to the bisector of $ \angle BCA$. Prove that $ P$ lies on $ r_{C}$.

Wow, this is pretty interesting! Can anybody solve it?

[quote="discredit"][quote="¬[ƒ(Gabriel)³²¹º]¼"]if some one can read italian [url=http://olimpiadi.dm.unipi.it/downloader.php?id=107]here[/url] there are the officiale solution of all ITAMO08[/quote]

Unfortunately I can't...[/quote]
http://translate.google.com :P

Obviously it's not that easy, MathiasDK.

Somebody has a solution?

a)I will prove that $K'=[cos(A-45):cos(B-45):(cosC-45)]$ in trilinear coordinate works and only this point works.
called $B,A',A",C,B',...,C",$ the points in clockwise order that should form the squares +the vertices and $D,E,F$ the feet of the perpendicular from K and l the square. You get that $DKA'=A-45$ solving the systems that $DKF=180-B$ and cyclic using the fact that $A'K=A"K$, so you get that the trilinear coordinate are the one you want using the fact that l is costant.
b) take the two perpendiculars from $P$ to the side $AC,BC$ that intersects in $F,G$. We have that $PB=PC$ so $PG/PF=cos(GPB)/cos(FPA)$. $$GPB=90-GBP=90-(180-b-45)=b-45$$ similarly $FPA=A-45$. This all implies that the line CP and $CK$ are symmetric as ther trilinear are one the inverse of the other


Geometric Proofs

Question :

Draw a square ABCD. Draw the line segments connecting the midpoints on each side. Label the new quadrilateral EFGH.

Prove that quadrilateral EFGH is also a square.

I must use the statement and reason chart.  Help...

draw one of the diagonals of the larger square, and notice that you form similiar triangles

Can you atleast set it for me and get me started?

you have to look at it, [b]draw it out[/b], the line you drew and the side of the smaller square are parallel, so you have similiar triangles, then you can use the corresponding angles to show that the inner angles pf the inside quadrilateral are 45+45 = 90

see, you can quite easily prove it is a parallelogram by the mid point theorem , and as for the right angles ,just apply something alternate interior angles and stuff , i think it should  come then........

Yeah...eminem's solution is pretty easy,the most basic and nice one.

Symmetry, once you're done with this, try to prove that:

If we cut the sides in a fixed ratio, the resulting figure is a square.  Try to find the proportion of the areas.

I'll try to come up with a formal proof to the OP's question today.

[quote="fourierseries"]The proportion of the areas.[/quote]
A famous geometry theorem and a famous inequality.  :)


show that:

$\displaystyle\sum_{i=1}^{n}i^3 = ( \displaystyle\binom{n+1}{2})^2$

i dont really write any proofs, so someone please correct me if it's faulty..

okay we first show that it holds true when n=1, and it does, b/c $1^3 = 1 = {2\choose2}^2 $

now showing its true for n+1...

now k choose 2 = k(k+1)/2.  so for n+1, we have (n+1)(n+2)/2

sooooo.... since we have $1^3 + 2^3+....n^3 = (\frac{n(n+1)}{2})^2$

we're trying for n+1..... so.... we're trying to show $1^3 + 2^3 + ...(n+1)^3 = (\frac{(n+1)(n+2)}{2})^2$

but since $1^3 + 2^3+....n^3 = (\frac{n(n+1)}{2})^2$, we just need to show that

$(\frac{n(n+1)}{2})^2 + (n+1)^3 = (\frac{(n+1)(n+2)}{2})^2$

This can be done by the following:

$\frac{n^2(n+1)^2}{4} + (n+1)^2(n+1) =
(n+1)^2(\frac{n^2}{4} + n + 1) = (n+1)^2(\frac{n}{2} +1)^2 = $

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

yay. i hope that works

ah sweet. i think that's my first real proof. w00t for me.


Let $ a,b,c,x,y,z$ be positive real numbers such that $ ax \plus{} by \plus{} cz \equal{} xyz$. Prove that
\[ x \plus{} y \plus{} z > \sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a}\]

see here http://www.mathlinks.ro/viewtopic.php?t=298736

I just got my internet connection again so another return to forum.

my solution:
$ a \equal{} yz \alpha$
$ b \equal{} xz \beta$
$ c \equal{} xy \gamma$
$ \alpha \plus{} \beta \plus{} \gamma \equal{} 1$
$ \sum_{}^{} \sqrt {a \plus{} b} \equal{} \sum_{}^{} \sqrt {x(y\alpha \plus{} z\beta)} \le \sqrt {(x \plus{} y \plus{} z) \sum_{}^{} x(\alpha \plus{} \beta)} < x \plus{} y \plus{} z$ by Minkowski's inequality

this is my solution for this. please check.

the first try is to make cs ineq $ (\dfrac{b\plus{}c}{x}\plus{}\dfrac{c\plus{}a}{y}\plus{}\dfrac{a\plus{}b}{z})(x\plus{}y\plus{}z) \ge (\sqrt{b\plus{}c}\plus{}\sqrt{c\plus{}a}\plus{}\sqrt{a\plus{}b})^2$
now it suffices to prove that $ x\plus{}y\plus{}z>\dfrac{b\plus{}c}{x}\plus{}\dfrac{c\plus{}a}{y}\plus{}\dfrac{a\plus{}b}{z}$
but, after expanding $ xyz(x\plus{}y\plus{}z)>ax(y\plus{}z)\plus{}by(z\plus{}x)\plus{}cz(x\plus{}y)$
by $ ax\plus{}by\plus{}cz\equal{}xyz$, then the ineq is equivalent with $ (ax\plus{}by\plus{}cz)(x\plus{}y\plus{}z)>ax(y\plus{}z)\plus{}by(z\plus{}x)\plus{}cz(x\plus{}y)$ which is obvious by $ ax^2\plus{}by^2\plus{}cz^2>0$ after expanding


Let $n$ and $m$ positive integers. Find all the pairs $(a, b)$ such that :

$a$ and $b$ are positive integers
$ pgcd (a^m - b^m, a^n - b^n) = a ^ {pgcd(m,n)} - b ^ {pgcd(m,n)} $

It is widely known that for any integer [tex]a[/tex], the couple [tex](a,1)[/tex] is a solution but I don't know wheter it's the only one. (I'll post a soution if I can :P )

I'm quite sure (a,1) are the only solution's. I'll post a proof tomorow or Friday (I have to check it :) )

I  think it is $gcd(a,b)=1$.
Here my solution: If $gcd(a,b)$ #1,we have contraction easily.
If $gcd(a,b)=1$.We call $d=gcd(a^m-b^m,a^n-b^n) $then $a^{gcd(m,n)}-b^{gcd(m,n)}$ divide d.Put $c=gcd(m,n)$
Otherwise,there exists x,y such that$ mx-ny=c$ then we have $a^c-b^c$ divisible by d then we have done.

Yah !  :P I just realized (a,1) are [b]not[/b] the only solutions. Anyway I guess you're right.

Here's the sketch I had in mind (there was an error at the level of lemma 2 - being rather a supposition :))
I thinks it's the same idea as K09's

1. Let $\delta = ( m, n )$ then $m = \delta m'$, $n = \delta n'$ and we have
$( m', n' ) = 1$. Thus there does exist positive integers $u, v$ such that $1 = m' u - n' v$ and thus $\delta = mu - nv$.

2. Now let $d = ( a^m - b^m, a^n - b^n )$ then $d|a^m - b^m$ and $d|a^n -
b^n$. Thus from lemma 1, we have $d|a^{mu} - b^{mu}$ and $d|a^{nv} - b^{nv}$.
As a consequence $d|a^{mu} - a^{nv} + b^{nv} - b^{mu} = a^{nv} \cdot ( a^{mu - nv} - 1 ) + b^{nv} ( 1 - b^{mu - nv} ) = a^{nv} ( a^{\delta} - 1 ) + b^{nv} (1 - b^{\delta} )$.
Now note that $d|a^{nv} ( a^{\delta} - 1 ) + b^{nv} ( 1 -b^{\delta} ) \Rightarrow d|a^{\delta} - b^{\delta}$ if and only if $( a, d ) = ( b, d ) = 1$ and this happens (from lemma 2) if and only if $( a, b ) = 1$.

3. Finally (from lemma 3) $a^{\delta} - b^{\delta} |a^m - b^m$ and $a^{\delta}
- b^{\delta} |a^n - b^n$, thus $a^{\delta} - b^{\delta} | ( a^m - b^m, a^n - b^n ) = d$.

4. It follows from point 2. that if $(a,b) = 1$ then $d = a^{\delta} - b^{\delta}$.

(PS : I didn't prove any of those lemmas but I guess it mustn't be too difficult...)