Final result is up now.

Check your  country score here:
http://erdos.fciencias.unam.mx/results.htm

Congarates to everybody.....:D :D

the second team is USA, 213
third one: Russia 212

totally there are 17 perfect scores this year.

17 perfect scores means that the problems are much easier than last year's.
China has 5 gold and 1 silver, including 4 perfect scores.
Total 235.
Rank 1st.

and Iran is 4th with 2 gold (one full-mark) and 4 silvers and a total of 201! EXCELLENT! Perhaps 1st next year!

Vietnam did very poorly this year. I think they are much stronger than Sinapore but they score less than Singapore. Strange.

Congratulation CHINA . but VIETNAM is too bad this year ,  :(  , so disappointed

[quote="nmtruong1986"][quote="beo_chay_so"][quote="iandrei2"]I counted 17 scores of 42  :roll:[/quote]
I counted only 16.  May be you counted total score of Sweden.  ;)[/quote]
Are you Quoc Khanh???[/quote]
Hix, khong phai dau a, em con chua bao gio tham thi doi tuyen toan ca. Em la dan chuyen tin  :P

Can you please put your sentence into English?

[quote="shobber"]Hongkong got one gold this year. Is he a Mathlinker?[/quote]
yes, he's yptsoi.

The Philippines improved but still in the bottom part of the list. You guys are so good. Congratulations...

Kyrgyzstan have got 2 bronze medals, but its better than last year!
Congrats China

[quote="amir2"]Congradulation to Saudi Arabi .
It so hard that they get 3 all togheter.[/quote]
That's not nice.

[quote="amir2"]Congradulation to Saudi Arabi .
It so hard that they get 3 all togheter.[/quote]Well at least they participated in the competition.

PS Please stop making such ugly comments in the future  :|  :ninja:

[quote="amir2"]Congradulation to Saudi Arabi .
It so hard that they get 3 all togheter.[/quote]

but maybe their english is better than yours (two errors in one sentence) :o  :o

Before making such comments,  look first to yourself Mr Amir (prince :P  :P )

[quote="blahblahblah"]Am I the only one who didn't read that statement as offensive?  I think it's admirable to put in an effort like that, especially when you know that you don't really have a chance of doing well.[/quote]

Ditto.  I read it the same way.

Olympiad is not a competition for countries but for individuals .It is for this reason that we dont have an official country ranking.
Olympaid and others international competitions would help to keep the peace between the lands.
I apolozise for what amir has said(close friends).
Our team in its first attemps in imo , didnt get lots of points too,but it is not something to be embarrassed of .
Thanks to the the imo instructors and countries instructors that gived us such a chance to have a peacefull competition.
May all be in peace.

lomos_lupin, actually where are you from? Originally I thought you are from Iran, but now it seems that you are from Saudi Arabia.

I am from iran .
My home is iran, be sure my friend.

PS."It seems" , what an interesting word it is , isnt it?
All of the great results and proofs  followed by this simple word.[b]It seems[/b]


inequalities
Functional Analysis
real analysis
real analysis unsolved

Hi, I have a problem:
Let E be a Hilbert space, $ E_j$ be a subspace of E, $ P_j: E\mapsto E_j$ be an orthogonal projection and $ 0\leq c_j\leq 1$ with $ j\equal{}1,\cdots, n$ such that
\[ \sum_{j\equal{}1}^n c_j P_j\equal{}Id_E\]
Is the inequality following true or false?
\[ \parallel{}\sum_{j\equal{}1}^n c_j x_j\parallel{}^2\leq \sum_{j\equal{}1}^n c_j\parallel{}x_j\parallel{}^2\]
for all $ x_j\in E_j$.

A good inequality! It took me 3 hours to check it.

Now the solution. Denote $ T_j \equal{} c_j P_j$. Then $ T_j \ge 0$, $ \sum T_j \equal{} I$. To solve the problem it is sufficient to prove that for every collection of $ x_1, \ldots, x_n \in E$ the following inequality holds true:
\[ \eqno{(*)} \,\,\,\,\,\,\,\,\,\,\,
\|\sum_{k\equal{}1}^n T_k x_k\|^2 \le \sum_{k\equal{}1}^n \langle T_k x_k, x_k\rangle.\]

In order to prove this let us introduce an auxilliary linear space $ H \equal{} E^n$. For elements $ \bar x, \bar y \in H$, $ \bar x \equal{} (x_1, \ldots, x_n )$, $ \bar y \equal{} (y_1, \ldots, y_n )$ denote
\[ [\bar x,\bar y] \equal{} \sum_{k\equal{}1}^n \langle T_k x_k, y_k\rangle.\]
The above expressione defines a non-negative Hermitian bilinear form on $ H$, so the Cauchy-Schwartz inequality
\[ |[\bar x, \bar y]|^2 \le [\bar x, \bar x] [\bar y,\bar y]\]
is applicable. So
\[ |\sum_{k\equal{}1}^n \langle T_k x_k, y_k\rangle|^2 \le \sum_{k\equal{}1}^n \langle T_k x_k, x_k\rangle \sum_{k\equal{}1}^n \langle T_k y_k, y_k\rangle.\]
Let us consider the particular case when all the $ y_k$ are equal to one the same norm-one element $ y \in E$. Then we obtain
\[ |\langle \sum_{k\equal{}1}^n T_k x_k, y \rangle|^2 \le \sum_{k\equal{}1}^n \langle T_k x_k, x_k\rangle.\]
This implies (*) because $ \sup_{y \in E: \|y\|\equal{}1} |\langle \sum_{k\equal{}1}^n T_k x_k, y \rangle| \equal{} \|\sum_{k\equal{}1}^n T_k x_k\|$.

Thank you very much, your proof is very intelligent.


1.) How many pairs of vertical angles are created when 15 lines intersect at the same point?

[hide="Answer"]210[/hide]

2.) The [i]hailstone number problem [/i]is one of the best known unsolved problems.  Start with an integer [i]n[/i] and replace it by n/2 if [i]n[/i] is even and by 3n + 1 if [i]n[/i] is odd. Continue this process.  For example, if [i]n[/i] = 11, we get the sequence 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, ..., and end with the cycle 1, 4, 2.  It is not known whether there exist any positive values n for which the sequence does not ultimately lead to the cycle 1, 4, 2.  Among the numbers 11 to 20, which generates the longest pre-1 sequence?  The sequence generated by 11, for example has 14 terms before the first 1 is encountered.

(Is there a quick solution to this problem?) [hide="Answer"]18 & 19[/hide]

3.) The value of the expression 1 / 2 / 3 / 5 / 7 / 11 / 13 can be altered by including parentheses.  If we are allowed to place as many parentheses as we want, how many distinct values can be obtained for this expression?

[hide="Answer"]32[/hide]

[hide]1. 15*14/2 but because 2 pairs of vert. angles are created at an intersection the answer is 15*14/2*2=15*14=210.

3. 2^6-(5C1+5C+5C3+5C4+5C5)=2^6-2^5=32.[/hide]

[hide="1"]use combinatorics: vertical angles need 2 lines each, and they make 2 pairs of verticals each, so (15 ncr 2)*2 is the answer, which is 15*14=210[/hide]

unimpossible -

Why did you do 2^6 - 2^5 = 32 ?


email
calculus
linear algebra
number theory

im going to be a college freshman this coming fall. and have taken calc BC already. i am wondering what kind of math classes you guys would suggest i take.
(i have been accepted as a mathematics major btw)

Speaking as a math department undergraduate advisor ...

How about finding out who does advising for the Math Department of the college you will be attending and contacting that person (most advisors do a lot of communicating by email) and asking for advice?

I'd recommend any of the following if they're offered and you haven't taken them:
AP Statistics
Linear Algebra
Calculus D
That's all I know from what's offered at my school.

What is Calc D?

The standard sequence for you to take now would be linear algebra, multivariable calc, and differential equations.  Normally, that's a one-year sequence.

You could also take introductory courses in number theory or (if you're a bit more daring) abstract algebra/group theory.  But I'm not the one to ask about other options.  Talking to a math department advisor is a good idea.


geometry
inequalities
inequalities proposed

[quote="jarod"]Let $a,b \in R$ such that :
 $a^2+b^2+6=4(a+b)$ and
 $c^2+d^2+64=12(c+d)$
 Find the maximum and minimum value of $S =(a-c)^2+(b-d)^2$[/quote]

$(a-2)^2+(b-2)^2=2$ and $(c-6)^2+(d-6)^2=8$. So $a,b$ belong to a circle with center $(2,2)$ and radius $\sqrt2$, whereas $c,d$ belong to a circle with center $(6,6)$ and radius $2 \sqrt2$.
$S$ denotes the square of the distance between two random points of these circles. It's easy to see that these circles intersect at two points so the minimum of $S$ is $0$. The maximum is $2(6+2\sqrt2-2+\sqrt2)^2=2(4+3\sqrt2)^2$ (we only need to find the intersections of the above circles with the line $x=y$)

[quote]  so the minimum of S is 0[/quote]
Sorry, I still don't understand it. Because:
$S =(a-c)^2+(b-d)^2$
If $S=0$ then $a=c$ and $b=d$ so put them in the system:
$a^2+b^2+6=4(a+b)$ and $a^2+b^2+64=12(a+b)$
According to my caculation, there is no real root at all. Why is that?
I think your sols in very nice and it seemed correctly to me. But.. what's happen? :D

At first sight I don't think the circles intersect. The distance between their centres is $4 \sqrt{2}$ which is "too large" (as the radii are $\sqrt{2}$ and $2 \sqrt{2}$).

The circles don't intersect, but using your idea about the line $y=x$, I got $S_{\text{min}}=2$ and $S_{\text{max}} = 98$, since the first circle intersects the line $y=x$ at $(1,1)$ and $(3,3)$ and the second circle intersects the line $y=x$ at $(4,4)$ and $(8,8)$.

[quote="Arne"]Yeah, that should work. :)

However, can you *prove* that these points give the minimal and maximal distances? :D[/quote]

I think it's because the line joining the centers of the circles is $y=x$, so farthest (from the origin) point of the first circle along that line is $(3,3)$ and the closest (to the origin) point of the second circle along the line is $(4,4)$ and a similar argument works for the maximal distance. I know this isn't really a proof, but if I come up with one, I'll post it.

[quote="Arne"]
However, can you *prove* that these points give the minimal and maximal distances? :D[/quote]

first of all, sorry for my original miscalculation.
Secondly, it's easy to prove by geometry arguments that these points give the minimal and maximum.
Name the circles $(K,R)$ and $(L,R_1)$. Take $A$ arbitrary point on $K$ and $B$ on $L$. Name the points that we believe that give the minimal $C,D$. We want to show that $AB\geq CD$. But from $ABLK$ we have $AB+BL+KA>KL$ or $AB+R_1+R>R+R_1+CD$ which proves the "minimality" of $CD$. "Maximality" can be proven in a similar way using initially $2$ times the triangle ineq and then the inequality $AC+CD+DB>AB$


Find the no. of unordered pairs{A,B} of subsets of an n-element set X which satisfy the condition:
(a)$A\neq B$  , (b)$A\cup B=X$

It should be $2^n-1$. (A=X, B=any subset of X other than X itself.)

Sorry, the answer is different. I actualy don't know the answer but I know only this much that the answer contains $3^n$. :oops:

So, this is a question in an examination and your answer is correct if you give  a good reason for it.HiHi..

Oh, I'm really very sorry. There was  a slight mistake in the question (actually $LA_TEX$) So see for the question again :|

[quote="K.W.M.A.N"]my guess:

$2^{n-1}$ if n is odd
$2^{n-1} - \frac{n!}{2(n/2)!(n/2)!}$ is n is even

it does not contains $3^n$ however... :|[/quote]

It is wrong then. Well you can tell your approach may be then we find error in it and who knows if we reach to the solution through yours. :roll:

it is wrong indeed... I assumed $A \cap B = \emptyset$, it will give like for {1,2,3,4,5} as possibilities
{{1}{2,3,4,5}} , {{2},{1,3,4,5}} etc = 5  possibilities
{1,2}{2,3,4}, {{1,3}{2,4,5}} etc = 10 possibilities
doing this more general will provide the answer I gave...
i think a correct solution may be found by noticing that these are all 'minimal' solutions from which all others can be obtained by adding elements to one of the sets, but i'm afraid it will be difficult to calculate the number of doublecounts, triplecounts, ...
Maybe there's some other, more obvious solution

Let the elements of the set $X$ be $x_i$ for $i=1,2,...,n$

Then there are 4 possibilities:
(1) $x_i$ belongs to both A and B
(2) $x_i$ belongs to A not B
(3) $x_i$ belongs to B not A
(4) $x_i$ neither belongs to A nor B

But $A \cup B=X$, thus the possibilities are (1),(2),(3) for each $x_i$
Thus total number of ways of arranging the elements is $(3^n-1)/2$, because one case is when all elements are in both A and B, which is not to be considered.
Also, since we need unordered pairs, thus we divide by 2.

Thus my answer is $\frac{3^n-1}{2}$.
Please tell if it is correct or not. :?:

Hmm... not so hard?
For each number we have 4 options: in A or not and in B or not, and 3 out of 4 lead to the number being in the union. So that's $3^n$ fulfilling that property. Now we need to filter out where A=B. But that's only possible if A and B both contain all numbers, so there are $3^n-1$ ordered couples (A,B), and thus $\frac{3^n-1}2$ unordered couples.

Edit: apparently varun found that out before.

so it did have an obvious solution, and one that easily can be generalised...  for s sets $A_1,A_2,\ldots,A_s$with union X and not all equal I got
\[ \frac{(2^s-1)^n-1}{n!} \]
correct?
what would it be if all sets had to be pairwise different, meaning $A_i \neq A_j$ if $0<i,j\leq s$ and $i\neq j$ ?

[quote="K.W.M.A.N"]correct?[/quote]isn't it $s!$ in the denominator?

Please, donot ask me any solution because I don't know the correct solution.

Well I saw Peter VDD got the same answer as i got. So I'm  :D


integration
trigonometry
calculus
calculus computations

[quote="pureblack2003"]calculate $\int_{0}^\frac{\pi}{16}\frac{dx}{\cos^{4}x}$[/quote]
In generally let $\int\frac{dx}{\cos^{n}x}=I_{n}$
 then
$I_{n}=-\frac{\sin{x}}{(n-1)\cos^{n-1}x}+\frac{n-2}{n-1}I_{n-2}$ Remember!!!

The integral is very easy!  I am calculating $\tan \frac{\pi}{16}$. :lol:

Answer: $\tan \frac{\pi}{16}+\frac{1}{3}\tan^{3}\frac{\pi}{16}$.

$\tan \frac{\pi}{16}=\sqrt{4+2\sqrt{2}}-1-\sqrt{2}$.

$\int \frac{1}{\cos^{4}x}\,dx = \int \sec^{4}x\,dx = \int \sec^{2}x(1+\tan^{2}x) = \int \sec^{2}x\,dx+\int \tan^{2}x\sec^{2}x\,dx$
$= \tan x+\frac{\tan^{3}x}{3}+C$,

and so your definite integral equals $\tan\left(\frac{\pi}{16}\right)+\frac{1}{3}\tan^{3}\left(\frac{\pi}{16}\right)$

$\tan \frac{\pi}{8}=\tan 2\cdot \frac{\pi}{16}=\frac{2\tan \frac{\pi}{16}}{1-\tan^{2}\frac{\pi}{16}}$

$\Longleftrightarrow \sqrt{2}-1=\frac{2x}{1-x^{2}}$, where $x=\tan \frac{\pi}{16}$.

$\Longleftrightarrow 2x=(\sqrt{2}-1)(1-x^{2})$
$\Longleftrightarrow 2(\sqrt{2}+1)x=1-x^{2}$
$\Longleftrightarrow x^{2}+2(\sqrt{2}+1)x-1=0$, since $x=\tan \frac{\pi}{16}>0$, yielding $\tan \frac{\pi}{16}=-(1+\sqrt{2})+\sqrt{4+2\sqrt{2}}$.

thanks . all i wanted to see was how to calculate $\tan \frac{\pi}{16}$


Find all the reals $ x\geq 998$ for which the expression $ \frac{\left[\frac{x}{1998}\right]}{\left[\frac{x}{998}\right]}$ is maximal.

Yes, that's right answer, but not correct on 100%. My solution maybe a little simpler:
$ x\equal{}1998k\plus{}m$. Where if $ k\geq 1$ then $ 0\leq m\leq 1997$ and if $ k\equal{}0$ then $ 998\leq m\leq 1997$.

$ \frac{k}{\left[\frac{1998k\plus{}m}{998}\right]}\to max$

$ \frac{1998k\plus{}m}{998}\equal{}2k\plus{}\frac{k}{499}\plus{}\frac{m}{998}$
$ \frac{k}{2k\plus{}\left[\frac{k}{499}\plus{}\frac{m}{998}\right]}\to max$

Now easy to see that maximum will be when $ \left[\frac{k}{499}\plus{}\frac{m}{998}\right]\equal{}0$ or $ 2\leq 2k\plus{}m\leq 997$ (I delete $ k\equal{}0$ because in that case $ m\geq 998$).

We get $ x_k\equal{}1998k\plus{}m$ where $ 1\leq k\leq 498$ and $ 0\leq m\leq 997\minus{}2k$ for current fixed $ k$. :)


$a,b$ and $c$ are the side lengths of a triangle, prove that :

$\frac{3a^{2}-c^{2}}{c^{2}+ab}+\frac{3b^{2}-a^{2}}{a^{2}+cb}+\frac{3c^{2}-b^{2}}{b^{2}+ac}\geq 3$

I have a nice simple solution for it without expanding !

[quote="Vasc"]The following stronger inequality holds:

$\frac{a^{2}-b^{2}}{a^{2}+bc}+\frac{b^{2}-c^{2}}{b^{2}+ca}+\frac{c^{2}-a^{2}}{c^{2}+ab}\le 0$.[/quote]
The inequality is equivalent to
\[\sum\frac{b^{2}+bc}{a^{2}+bc}\ge 3 \]
By Cauchy Schwarz inequality, we have
\[\sum\frac{b^{2}+bc}{a^{2}+bc}\ge \frac{(a^{2}+b^{2}+c^{2}+ab+bc+ca)^{2}}{\sum (b^{2}+bc)(a^{2}+bc)}\]
Hence, it suffices to show that
\[(a^{2}+b^{2}+c^{2}+ab+bc+ca)^{2}\ge 3\sum (a^{2}+bc)(b^{2}+bc) \]
or equivalently
\[\sum a^{4}+2\sum ab^{3}+\sum a^{2}bc \ge 3\sum a^{2}b^{2}+\sum a^{3}b \]
Using Ravi substitution, we can easily prove this inequality. :)

[quote="toanhocmuonmau"] Hence, it suffices to show that
\[\sum a^{4}+2\sum ab^{3}+\sum a^{2}bc \ge 3\sum a^{2}b^{2}+\sum a^{3}b \]
Using Ravi substitution, we can easily prove this inequality. :)[/quote]
This inequality follows also adding Schur's Inequality of fourth degree
\[\sum a^{4}+abc\sum a \ge \sum ab^{3}+\sum a^{3}b \]
to the known inequality
\[3\sum ab^{3}\ge 3\sum a^{2}b^{2}. \]
Moreover, there is a direct solution without using Cauchy-Schwarz, Schur's or Ravi substitution. :)

[quote="Vasc"]
to the known inequality
\[3\sum ab^{3}\ge 3\sum a^{2}b^{2}. \]
[/quote]

but this ain't true. i think you've mistaken something. :(

[quote="silouan"][quote="Vasc"]The following stronger inequality holds:
$\frac{a^{2}-b^{2}}{a^{2}+bc}+\frac{b^{2}-c^{2}}{b^{2}+ca}+\frac{c^{2}-a^{2}}{c^{2}+ab}\le 0$.[/quote]

The triples $a^{2},\ b^{2},\ c^{2}$ and $\frac{1}{a^{2}+bc},\ \frac{1}{b^{2}+ca},\ \frac{1}{c^{2}+ab}$

are opposite arranged . So by rearrengement inequality we have
$\frac{a^{2}}{a^{2}+bc}+\frac{b^{2}}{b^{2}+ca}+\frac{c^{2}}{c^{2}+ab}\leq \frac{b^{2}}{a^{2}+bc}+\frac{c^{2}}{b^{2}+ca}+\frac{a^{2}}{c^{2}+ab}$

so QED[/quote]

nice work silouan! its a shock that noone found this surprisingly easy proof!

[quote="nayel"][quote="Vasc"]
to the known inequality
\[3\sum ab^{3}\ge 3\sum a^{2}b^{2}. \]
[/quote]

but this ain't true. i think you've mistaken something. :([/quote]

This is a very known inequality for $a,b,c$ side-lengths.

[quote="Vasc"] This is a very known inequality for $a,b,c$ side-lengths.[/quote]

Yeah , someone can prove it with Ravi subtitution .
BTW , Vasc did you like my solution and what's your solution ?

Your solution is nice, Silouan.
But a more elementary solution is to write $c^{2}-a^{2}=(c^{2}-b^{2})+(b^{2}-a^{2})$.


Hello!
Any KVPY fellows here?
I know Rushil is one because he was in my batch.
Anyone else?

Hi! I didnt recognise you -- whats your full name?

I am Ritam.
KVPY is Kishore Vaigyanik Protsahan Yojana.

I s'pose this is only for 2004-06 awardees??
 i got KVPY last year i.e. 2005-07.
 seems they've reduced the scholarship amt. or something like that.
 makes me go [size=150][color=red]GAH! [/color][/size]

Does anybody has the question papers of KVPY from year-1999 to year-2007 ? I really need it

[quote=Prabh2005][quote=nathantareep][quote=Rushil]Mind telling us your name then! , basurp?!!

Hi Ritam!![/quote]

I am preparing for kvpy too.[/quote]

saar haw so pro telling that to people who gave kvpy before u were born
:salut:[/quote]
Just to keep it going... I'm a KVPY fellow too.
14-15 years after the guy who started this thread, tho. lol.


Let  $A$   be  a  given point  outside  a  given  circle. Determine  points  $B, C, D$ on the  circle  such that the quadrilateral $ABCD$ is  convex  and has  the  maximum area .

Will these problems appear in resources project,too?

This would be a great idea!! :lol:

Let $BD = b$ and $C$ between $B$ and $D$.
So the area is $A = b(h_{1} + h_{2})$.
The maximum occurs when $C$ is the second intersection of $AO$ with the circle and $BD$ is the diameter $\bot$ to $AC$.


geometry
perpendicular bisector
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14907[/img]

 :(

we have $ \frac{a\plus{}b\plus{}c}{2}\equal{}r_a \cot \frac{A}{2}$, so we have $ a\plus{}b\plus{}c$, and we have $ a$, and we have $ p\minus{}a$. Let a circle with radius p-a and center A that meet the side line AB on D and AC on E. The perpendicolar line from D to AB meet the perpendicolar line from E to AC on I. The circle with radius p and center A meet AB on D' and AC on E'. The perpendicolar line from D' to AB meet the perpendicolar line from E' to AC on P. The circle with radius ID and center I is the incirlce, the circle with radius PD' and center P is the A-excircle. Then 2 internal tangent with incircle and A-excircle are the 2 possibility for the side line BC.

contruct segment b+c  say MN and angle 180 - A on M
take distance 2M(a) and from N mark it on the ray making angle 180-A with MN
consider any one point(K) of the 2 marks won't matter
 draw the perpendicular bisector of MK let it intersect MN at L
mak distance LN on the other side of L on MN
LNK is the required triangle
justification
appolonius theorem and midpoint theorem

"contruct segment b+c say MN and angle 180 - A on M
take distance 2M(a) and from N mark it on the ray making angle 180-A with MN
consider any one point(K) of the 2 marks won't matter "

2ra=2M(a) ?
2ra is upper than 180-A arc capable of b+c segment

 :(

sorry i posted this in wrong place
actually it was for your other problem in which the length of median from a the length b+c and angle A were given
once again sorrry

"Let a circle with radius p-a and center A that meet the side line AB on D and AC on E. [b]The perpendicolar line from D to AB [/b]meet the perpendicolar line from E to AC on I"   ??

 :(

[quote="jrrbc"]"Let a circle with radius p-a and center A that meet the side line AB on D and AC on E. [b]The perpendicolar line from D to AB [/b]meet the perpendicolar line from E to AC on I"   ??[/quote]

what is the problem?

D and E are the tangency point of the incircle with AB and AC.

the [b]a[/b] is correct but the [b]I[/b] is incorrect

p=a+b+c ?

something wrong with the p-a

 :(


geometry
parallelogram
analytic geometry
vector
geometry unsolved

Opposite vertices define only a set of parallelograms with minimum area = 0 and maximum = |a-c|.|b-d|...

:!: Immanuel
i cannot understand what you want to say
this question's a lot lot more messy than you would be thinking

Seams that the position of two opposite vertices doesn't define the parallelogram's area: the height is given, but what about the base?

:roll: there 's a very messy construction .
the question you are pondering over can be solved if we take one of the vertices as the  origin.

[quote="nishantkumar"]:rotfl: if the opposite vertices of a paralleogram have coordinates (a,b) and (c,d)  ,

find the area of the parallegram in terms of a,b,c,d.[/quote]

this seems like a spam post, given that it occasionally shows up at the AoPS website, with these given, and one other vertex on the origin... $ |ad\minus{}bc|$ is the answer on your mind, I assume :P

That would come from the vector product modulus, of the vectors position of the given points. That would be OK if the problem states that the origin is one of the vertices. But that[b] wasn't in the original data[/b]...


Thanks to those of you who have emailed or posted to ask how we're doing out here near San Diego.  Our main office is pretty far from the fires (at least 10 miles), though the air quality is starting to deteriorate.  None of the AoPS staff has had to evacuate their homes, nor have any of our homes been threatened yet.  We're not entirely out of the woods, but so far, we're OK.  Monday and Tuesday classes were postponed, but for now, the classes are on for tonight.  Book shipments were delayed for a couple days, but we're catching up now.

I'm glad that AoPS and employees are OK! :)

I'm looking at NOAA visible weather satellite images. The smoke plumes are still visible, but they aren't the dramatic features they were a day or two ago. I guess one thing you had on your side is that the burn area of the 2003 Cedar fire lies between Alpine and the bulk of the Witch fire, which is the worst one. And the Harris fire to your south started out burning away from you and across the US-Mexico border. With what winds there are (thankfully, down, but the air is still dry) now coming from the southeast, we're seeing smoke from San Diego County and Orange County drifting over our heads in Long Beach.

All of my Irvine students in the SoCal ARML team: are you guys OK? Did they cancel school (at one point, it seems the Santiago fire wasn't too far from Northwood HS.)

Glad to here that AoPS is okay.  :)
AoPS ftw!

[quote="Valentin Vornicu"]The real danger in the office is the flu that's been roaming around for weeks now  :mad:[/quote]

It was rrusczyk, wasnt it. :o

I am very happy you guys are safe.  I was worrying when I saw the news on ESPN.  I'm sure there's hundreds of kids in every portion of this world that are delighted that you guys are unharmed, and that you can continue to teach us the world of math.  (You too, Mr. Lehoczky, but you're at Wall Street!)

:huh:                                         .

I didn't even know that you guys moved around....

But I am glad that you guys are safe..
Smart people should feel safe..... :lol:

Glad to hear that everyone from AoPS in San Diego is safe!

[quote="Z-coli"] (You too, Mr. Lehoczky, but you're at Wall Street!)[/quote]

Mr. Lehoczky still works with AoPS?

Wow...I totally live under a rock...the first time that I heard about the fires was a few hours ago when I logged onto AoPS...

I would probably explode if these forums were gone.   :D

I am so happy to hear that all of you are safe!

Glad to Hear AoPS and all who work there are ok!  :)

I am relieved to heat that AoPS is safe! :)

[quote="Valentin Vornicu"]The real danger in the office is the flu that's been roaming around for weeks now  :mad:[/quote]

Don'tcha hate it when that happens?

Glad to hear that everything is all right.

[quote="Altheman"]Wow...I totally live under a rock...the first time that I heard about the fires was a few hours ago when I logged onto AoPS...[/quote]

same...

[quote="Temperal"][quote="Z-coli"]
(You too, Mr. Lehoczky, but [b]you're at Wall Street![/b])[/quote]
Mr. Lehoczky still works with AoPS?[/quote]

You should read more carefully.

[quote="Z-coli"]I am very happy you guys are safe.  I was worrying when I saw the news on ESPN.  I'm sure there's hundreds of kids in every portion of this world that are delighted that you guys are unharmed, and that  [b]you can continue to teach us the world of math.  (You too, Mr. Lehoczky, [/b]but you're at Wall Street!)[/quote]

Due to this, I was assuming that that he was including Mr. Lehoczky in the "you can continue to teach us the world of math" group, though I knew he was at Wall Street.

AoPS is in San Diego? Is that the main office or something else

[quote="vv123"]Glad you guys are ok! I would have been pretty mad if anything happened! :rotfl:[/quote]

Yeah! No forum. don't know what migh'ta happened... :rotfl:

[quote="TWY"]AoPS is in San Diego? Is that the main office[/quote]Yes.

Out of curiousity and impatience, when is the new expected timetable for WOOT grading? Next week?

One of my first reactions to the fire in California was wondering about the safety of the AoPS admins and their families and I'm glad to hear that everyone is okay.  :)

I'm really glad that everybody is fine, and aops is ok.

Levator:(for a minute there, i missed the right half of "fine" and thought it said "fire")

I'm glad to hear that the AoPS staff hasn't been hit by the wildfires at all.

Remember to breathe  :P

[color=red]D3m0n Shad0w[/color]


When using the D'Alembert method for calculating the solution to the wave equation

$ \frac{\partial^2 u(x,t)}{\partial t^2}\equal{}c^2\frac{\partial^2 u(x,t)}{\partial x^2}$

we do the substitution:

$ \xi \equal{} x \plus{} c t$
$ \eta \equal{} x \minus{} c t$

and so, we get:

$ \frac{\partial^2 u(\xi,\eta)}{\partial \xi \partial \eta}\equal{}0$

This is obtained by calculating:

$ \frac{\partial u}{\partial x}\equal{}\frac{\partial u}{\partial \xi}\plus{}\frac{\partial u}{\partial \eta}$
$ \frac{\partial u}{\partial t}\equal{}c(\frac{\partial u}{\partial \xi}\minus{}\frac{\partial u}{\partial \eta})$
$ \frac{\partial^2 u}{\partial x^2}\equal{}\frac{\partial^2 u}{\partial \xi^2}\plus{}2\frac{\partial u}{\partial \eta}\frac{\partial u}{\partial \xi}\plus{}\frac{\partial^2 u}{\partial \eta^2}$
$ \frac{\partial^2 u}{\partial x^2}\equal{}c^2(\frac{\partial^2 u}{\partial \xi^2}\minus{}2\frac{\partial u}{\partial \eta}\frac{\partial u}{\partial \xi}\plus{}\frac{\partial^2 u}{\partial \eta^2})$

Can anybody explain to me how are the four equations obtained? I know it uses the chain rule, but since the book I'm using is in English (and this isn't my native language) I do not know to what should I reference in the books I have.

Thanks in advance!

[url=http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node12.html]This page[/url] gives a bit more details. Or [url=http://mathworld.wolfram.com/dAlembertsSolution.html]this one[/url].


function
quadratics
modular arithmetic
number theory open
number theory

What is known about $ F(x)\equal{}\sum_{i\equal{}0}^{p\minus{}1} \left( \frac{i}{p} \right) x^i$ and $ G(x)\equal{}\sum_{i\equal{}0}^{\infty} \left( \frac{i}{p} \right) x^i\equal{}\frac{F(x)}{1\minus{}x^p}$?
I know the proof of quadratic reciprocity that uses values of $ F$ at roots of unity of order $ p$. But are $ F$ and $ G$ have some other interesting properties? Were they studied?
(for example, a recent problem here showed that $ F'(1)\equal{}0$ when $ p \equiv 1 \pmod 4$, $ p$ a prime, or: $ (x\minus{}1)^2|F(x)$)

Sorry if this question is not very focused  :blush:

Since $ \left( \frac{i}{p} \right)$ is a multiplicative function it really makes more sense to study its Dirichlet $ L$-series

$ \sum_{i\equal{}1}^{\infty} \frac{ \left( \frac{i}{p} \right)}{i^s}$

especially since the Legendre symbol is a [url=http://en.wikipedia.org/wiki/Dirichlet_character]Dirichlet character[/url].  Studying the zeroes of this function is a key step in the proof of Dirichlet's theorem on arithmetic progressions, where the Legendre symbol is the only nontrivial real-valued character and usually treated separately from the complex case.  One recovers $ G(x)$ by taking the (inverse?) [url=http://en.wikipedia.org/wiki/Mellin_transform]Mellin transform[/url].


Harvard college MIT Princeton AMC USA(J)MO USAMO

I have read some of the discussions below and one thing cought my mind...is there any opportunity to go to Harvard, Mit or Princeton or other if one doesn't achieve anything spectacular in the IMO, but just in the national olimpiad ? Cause it seems strange a bit that usually only IMO sweapers get invited to these universities.
Last year I achieved 7 place in the Polish National Olimpiad in Maths, is there any possibility for people in my situation to get a place in the best scientific universities ? I suppose that for people in US it's not a problem to get a scolarship and attend any university he wants (in Poland all olimpiad members with some succes get free entrance) with these kinds of titles, but despite the fact that Mit,Harvard and others invite talented students from the world- is there any chance for people like us ?

In my opinion, everybody has an opportunity to get into the colleges you have mention.
Yes, if you have won some IMO medal, it is definitely helpful, but will not guarantee a spot.
On the other hand, other things that show your characteristics - essays, recommendations, taking challenging courses, ... might get you into those colleges.

You probably are better than me in math.
but the question is, do you love math more than I do?
if you can say yes in less than a second,
I'm sure most of the colleges will want you.

It will be hard, but I hope your passion will surpass the difficulty.
Good luck.
I have a Polish mentor who I admire a lot. :)

love for maths is (for me) a bad word..."passion" fits better :).Obviously, if I wasn't dedicated enough to math I wouldn't participate in national contests. As a matter of fact, some of my friends-contestans perceive scientific contest as sport..not pure math (I agree with them on this spot).
By the way, are you studying in US ?

ps.And who's that math mentor from Poland ;)? (may I ask)

Each of those Universities admit thousands of people each year. Not all of them could have been Olympiad winners!

Of course, a lot of math majors at Princeton did not attend the IMO, and probably about 1/3 of the American math majors did not make the USAMO when they were in high school. Nevertheless, some of the students that I was the most impressed with and who were the most successful, both in and after college, were non-Olympiad participants.

So national math contests are just one form of assessment. They do not gaurantee or prevent admission to anyone. Things like summer research opportunities and USAMTS may possibly be even more impressive, because they introduce students to the process of research and show initiative on the part of the student to participate.

[quote="Rafal"]By the way, are you studying in US ?[/quote]
Yes, I'm from Korea and I have lived in US almost four years.
I went to a public school in New York, and I'll be attending MIT starting this fall.

[quote="Rafal"]ps.And who's that math mentor from Poland ;)? (may I ask)[/quote]
[hide="I hope he doesnt mind"]Jan Siwanowicz, putnam fellow :) [/hide]

Well, i never made USAMO (though i did take the AIME 3 times and was school winner for the AMC for 3 years)
i also participated in USAMTS and went to summer programs (HCSSiM and CCNY summer scholars)
i will be going to MIT in the fall :)

thanks all for reply

ps. Jan Siwanowicz you say...i found him on the top scorers list  [url=http://www.om.edu.pl/45.html]www.om.edu.pl/45.html[/url]...8th place in Poland and Bronze at IMO..nice ... i suppose he's the the only Pole who participated in the national olympiade being abroad :). My friend must no him cause he's on the same list, was he your tutor at high-school ?

new york city math team coach (I was in the team in 2003, 2004, 2005)
olympiad problem solving class teacher at CCNY summer program (2003)

but he's more than that.  :)

New York city math team...sounds nice :), was Aaron Pixton in your team ?, i suppose the USAMO is harder than our contest, being a finalist is probably equal to qualifying for the Polish squad for IMO ( maybe I'm a bit exaggerating )

as far as I know, aaron pixton is not from the city, but the upper part of new york state.

Hmmm...I could have sworn that Aaron was from somewhere out in Long Island...

And Jan is definitely someone that one should look up to as a mentor.  Jan was a member some time ago (I can't remember), and he has coached for as long as I knew of the NYC Math Team (which is about 5 years).  Maybe he wants to share some of his experiences?

[quote="Midnight"]somewhere out in Long Island...[/quote]
I thought he was on upstate new york team at ARML with 314159265358979323846?

He's from upstate New York.  Specifically the Binghamton area, due north-west from the city.  It is along the southern border of New York State, just north of Pennsylvania.  In any event, a good 150 miles or more away.

Hmmm. I do not know what to say. It's all Gee Hoon's work.

I feel real shy now.

Private message me if anyone wants to talk. :)


calculus
Vieta
function
symmetry
algebra
polynomial
system of equations

On a calculus problem, we were asked (through Newton's Method) to find the real zero of

\[ x^3 + 21x + 7.88\]

However, I was wondering if it was possible to apply Vieta's Formulas to this, such that

\begin{eqnarray*}
r_1 + r_2 + r_3 &=& 0 \\
r_1r_2 + r_1r_3 + r_2r_3 &=& 21 \\
r_1r_2r_3 &=& 7.88
\end{eqnarray*}

I have no idea how I would go about solving this system of equations; the appropriate section in AoPS volume 2 simply only referenced Vieta's in passing, so that the equation may be factored. The function I am given, however, cannot be factored as easily.

(Note: Descartes' Rule of Signs tells us there is 2 imaginary roots and 1 real, negative root. Newton's Method approximates the zero at $ -0.373$.)

If your book asked for Newton's Method, you should probably use it, for practice if nothing else. (And in that case, this should be moved to the calculus forum.)

And yes, you could use Vieta's, but you will notice that, in general, for equations of higher degrees than 2, Vieta's would be less effective, except for a few special cases.

You should use Newton's Method so you are familiar with how to approximate these roots, and how calculators do it.

Oh; I was just curious if there was a potential algebraic method of determining zeros - I did end up using Newton's Method for the problem, however.

There is a cubic formula, and there is a sense in which looking at the symmetry of Vieta's formulas helps you figure out what it is, but let's make one thing clear: the easiest way to solve the system of equations that Vieta's formulas gives you is to use the cubic formula on the corresponding cubic polynomial.

And if you insist on using the cubic and quartic formula's (and they're already too complicated), you'll have to know Newton's Method to solve for anything higher without Vieta's.

That's not really true; there are other numerical methods for solving polynomials.  I'm thinking in particular of the eigenvalue method.

Strictly speaking, yes, but I was trying to limit this to Newton's and Vieta's. I will not deny that there are other methods, however.

In a nutshell: solving the Viete system of equations ALWAYS brings you back to the original polynomial from which they originated. It is a vicious circle  :(


$\sum_{cyc}\ln\left(\frac{a+b+c}{3}\right)\leq\ 4\ln\left(\frac{a+b+c+d}{4}\right)$
   E bine daca aplic Jensen pentru functia concava $\ f(x)=\ln x$?


We have a square of side length 1. We inscribe two smaller squares inside the first square such that they do not overlap. The two squares have side lengths a, b respectively.

Prove that a+b <= 1

I think I can do it messily, reducing to cases.  Anyone see a quick way to do it?

This is my first post, sorry if this is in the wrong place.  Original source: [url]http://www.math.ucsd.edu/~fan/ron/papers/75_06_squares.pdf[/url]

Welcome to the Forum!
This problem is folklore, and to my understanding attributed to Erdos. It happens to have been posted before in
http://www.mathlinks.ro/viewtopic.php?search_id=828185957&t=194102
Enjoy! :)

Thanks for the reply, I think there might be an easier way than the way you linked to though..

We can seperate the two squares by a line.  If the line is parallel to the any sides of the square, clearly the original conditions are satisfied.  Otherwise we orient the square so that the gradient of the line is negative, [url=http://imgur.com/4KgpJ.jpg]Figure 1[/url]

Then we extend this line to meet the one or more of the sides externally [url=http://imgur.com/D0B7h.jpg]Figure 2[/url] .  So now we have two right angled triangles.  The biggest square we can fit in a right angled triangle is one with a vertex at the right angle.  So the biggest squares we can fit in the two right angled triangles are squares with vertices at opposite corners. [url=http://imgur.com/fwfeD.jpg]Figure3[/url], and now you can easily see that a+b <=1.  Is this OK?


geometry
rectangle
quadratics
analytic geometry
absolute value
calculus
calculus computations

1)Let $x$ be irrational and $r$ rational nonzero number.
Show that $rx$ and $r+x$ are irrational.

2) Show that $(z+\frac{1}{z})^{2}=12$ lacks rational solutions.

3) Let $B$ be a set of real numbers with a lower bound and let
\[-B=\{-x,x\in B\}\]
Show that $\mbox{inf}(B)+\mbox{sup}(-B)=0$

4) Show that every rectangle contains a point with rational coordinates.

1 is by contradiction.

2 follows from one. take square root of both sides. consider your absolute value signs. multiply through by z. quadratic formula. use 1.

3 follows by routine check. you can think of B as a set of numbers on one side of the number line (to the right of zero, say). and of -B as flipping that set of numbers over to the other side of the number line so that -(inf B) remains closest to zero.

or maybe someone would like to solve 3 more explicitly.

4 you can have the lower left corner of the rectangle at the origin and then look at the point x=p(1,1) for 0<p<1. infinitely many points x must be in your rectangle. this can be extended to where the lower left cornder is located at a point with rational coordinates (a,b) by considering x = (a,b) + p(1,1). of course you can easily pick p = 0 but I am saying this because it may give you an idea of how to handle the case where x is a point with an irrational component. Also because it's geometrically intuitive.

Otherwise you can always do the old there exists a rational point in any interval to find your point with rational coordinates (and this is probably what is expected  :o ). But I surmise that you probably don't know how to prove that.

Someone will come along and show you how to prove it if you can't extend the previous method. :)


A rectangular aquarium 26 inches long, 12 inches wide, and
10 inches tall is filled with water to a level of $ 8\frac12$ inches. When a rock is placed in the aquarium, the water level rises $ \frac38$ inch. What is the number of cubic inches in the volume of the rock?

the volume of the rock is equal to the amount of water displaced

so, if we find the volume of the water displaced, we find the volume of the rock

the aquarium is $ 26$ in. long and $ 12$ in. wide
the water level goes up $ \frac{3}{8}$ of an in.

so, the volume of water displaced is $ 26\times12\times\frac{3}{8} \equal{} 117$

thus, the volume of the rock is $ 117 \text{in}^2$


prove that there exists many infinitely n such that
Is it $ n \plus{} 1|3^n \minus{} 2^n$
 :o  :o  :o

Equivalently, $ \left( \frac{3}{2} \right)^n \equiv 1 \bmod (n\plus{}1)$, which occurs whenever $ n \plus{} 1$ is prime (and not equal to $ 2$ or $ 3$).

I donnot think it is true
I think it is a carmichael number +1!!

[quote="kihe_freety5"]I donnot think it is true
[/quote]
It's true. Use Fermat's Little Theorem for $ n\plus{}1\equal{}p$ - prime, $ p \neq 2, 3$.

The Carmichael numbers are among the [i]composite[/i] solutions (except $ 561$, which is divisible by $ 3$).

it is too basic
and more $ n \plus{} 1|p^n \minus{} q^n$ where $ p,q\in \mathbb N$
and Fermat Little theorem works here


Consider a distribution of charge in the form of a cylinder, like a long charged pipe.  Prove that the field inside the pipe is zero.  Prove that the field outside is the same as if the charge were all on the axis.  Is either statement true for a pipe of square cross-section on which the charge is distributed with uniform surface density?

I know it's distributed along the surface, and I know the field inside is 0.  I'm just not really sure how to rigorously prove it.  I'm thinking that if I could prove it for an infinitesimal circle, then I could integrate 0 from $ \minus{}\infty$ to $ \infty$ and re-obtain 0.  Any help please?  Especially for the second part?

It a consequence of gauss' law... :wink:

put a soup can around the axis of the cylinder.
if the soup can is smaller than teh cylinder, there is no charge inside => no field going inside or out.
if is larger, then by symmetry it looks as if it's all on teh axis.

it doesnt' work on square pipe because its not perfectly symmetrical.
the soup can could intersect the square

I like your logic.  The only problem is it isn't a mathematical proof.


My number of posts just dropped by about 30. I don't know why, but I am guessing some thread got deleted. Is this possible? It just changed today. I don't really care about my # of posts anyway, I just want to see why it changed.

I think its maybe cause of the recent happening, The games forum had A TON of posts moved to the Fun Factory.

That happened to me, too.

Why were they moved? They were perfectly fine and nearly spamless...

Yes, Why were they moved (like my game)?

they weren't games, they were discussions...

They were moved for not having any intellectual content ;)
And definitely, they weren't discussions.

The number game was abundant with intellectual content. I still don't get it.

You mean Dojo's game? Well only [b][i][u]SOME[/u][/i][/b] people were doing intellectual posts.

Come on people, the game was silly.

1000=1001-1
1002 is divisible by 2 (example, I know it hasn't gotten to 1000 yet). What's the point?

Dojo made this game to try to list (with the help of the other users) some NICE properties of numbers, not like my example above, which many people followed.

ZetaX is totally right about this, as soon as such posts started, it wasn't a game, it was a ''collecting posts topic''

And it's not like the game lost any "content" by being moved.

[quote="jmerry"]Yes, it was silly from the start- and I say that as someone who posted in the thread.[/quote]This is becoming silly too :)


function
geometry
geometric transformation
vector
inequalities
Support
invariant

two questions

(1) let $ (X, \Sigma, \mu )$ be a measure space and $ p,q \in [1, \infty ], \frac{1}{p} \plus{}\frac{1}{q} \equal{}1$
and $ \{ f_n \} \cup \{f \} \in L^p(X, \Sigma, \mu )$, and $ \{ g_n \} \cup \{g \} \in L^q(X, \Sigma, \mu )$

given $ \|f_n \minus{} f \|_p \to 0$ and $ \|g_n \minus{} g \|_q \to 0$ show $ \|f_ng_n \minus{} fg \|_1 \to 0$

having issues finding a way to bound the mixed term...

(2) let $ f \in L^p(R, M_L, \mu_L )$ for $ p \in [1, \infty )$ for each real h, define
$ f_h(x)\equal{}f(x\plus{}h)$, show $ \|f \minus{} f_h \|_p \to 0$ as $ h \to 0$

showing they are in the Lp space is pretty trivial, but i cant seem to bang out a proof of the norm convergence.

(1) is a fairly standard piece of analysis. Would it help you see things if you wrote $ f_n\equal{}f\plus{}(f_n\minus{}f)$ and $ g_n\equal{}g\plus{}(g_n\minus{}g)$ ?

For (2) you need a "nice" dense subspace of $ L^p(\mathbb{R}).$ There are several workable choices for that nice dense subspace. You could use the set of all integrable step functions, or you could use the set of compactly supported continuous functions - both are dense subspaces of $ L^p$ as long as $ p<\infty.$ First prove what you want for the "nice" functions, then make direct use of the density.

Side note: $ L^{\infty}$ does not have "nice" dense subspaces, and this result (continuity of translation) is false in $ L^{\infty},$ as you can directly show with a simple example.

(1) seemed easy, but i kept running into an issue. here is what i tried:

$ \| f_ng_n \minus{} fg \|_1 \equal{} \| f_ng_n \minus{} fg_n \plus{} fg_n \minus{} fg \|_1 \le \| (f_n \minus{} f)g_n \|_1 \plus{} \| f(g_n \minus{} g) \|_1$

$ \| (f_n \minus{} f)g_n \|_1 \le \| f_n \minus{} f \|_p \| g_n \|_q$

$ \| f(g_n \minus{} g) \|_1 \le \| f\|_p \| g_n \minus{} g \|_q$

the second one we can control by convergence since $ \| f\|_p$ is fixed for each n,

but the first one has a changing value of $ \| g_n \|_q$ so i dont know what to do =(

edit: oh i see you mean do both at once!

$ \|f_ng_n \minus{} fg \|_1\le \| f(g \minus{} g_n)\|_1 \plus{} \|g(f \minus{} f_n)\|_1 \plus{} \|(f \minus{} f_n)(g \minus{} g_n)\|_1$

thank you much km

[quote="kenn4000"](1) seemed easy, but i kept running into an issue. here is what i tried:
but the first one has a changing value of $ \| g_n \|_q$ so i dont know what to do =(
[/quote]

If $ g_n\to g$ in $ L^q$, then for large $ n$, $ \|g_n\|_{L^q(X)}$ is very close to $ \|g\|_{L^q(X)}$, so for a uniform bound it suffices to take $ \max\{\|g\|_{L^q(X)}\plus{}\varepsilon,g_1,\ldots,g_N\}$ where $ N$ is chosen so that $ \|g_m\minus{}g\|_{L^q(X)}<\varepsilon$ for $ m>N$.

Generic version of the question:

$ X,Y,$ and $ Z$ are normed vector spaces. $ L: (X\times Y)\to Z$ is a bilinear mapping that satisfies the inequality $ \|L(x,y)\|_Z\le C\|x\|_X\|y\|_Y$ for some fixed constant $ C$ and all $ x\in X,y\in Y.$ Prove that if $ x_n\to x$ in $ X$ and $ y_n\to y$ in $ Y,$ then $ L(x_n,y_n)\to L(x,y)$ in $ Z.$

There's probably a way to be even more general than that, as well as some questions about possible partial converse statements.

Of course, the main work involved is probably what went into proving $ \|L(x,y)\|_Z\le C\|x\|_X\|y\|_Y$ in the first place. In the case of the original problem, that's Hölder's inequality.

Its a standard fact that $ {C_c}^{\infty}$ which is the space of all continuous functions with compact support is dense in $ L^p , p\geq1$.

So given $ \epsilon > 0$ there exists a continuous function $ g$ with compact support such that $ \parallel f \minus{} g \parallel_p < \epsilon$, and since the $ L^p$ norm is translation invariant we also have that for any $ h > 0$,  $ \parallel f_h \minus{} g_h \parallel_p < \epsilon$

We also have that $ {\parallel g_h \minus{} g \parallel_p}^p \equal{} \int_{ \minus{} A}^A {\mid g(x \plus{} h) \minus{} g(x) \mid}^p dx$ which can be made less than $ \epsilon$ for small enough $ h$ by the uniform continuity of $ g$ on $ [ \minus{} A,A]$

Now putting all this together, we get that $ \parallel f \minus{} f_h \parallel_p \leq \parallel f \minus{} g \parallel_p \plus{} \parallel g \minus{} g_h \parallel_p \plus{} \parallel g_h \minus{} f_h \parallel_p < 3\epsilon$ for sufficiently small $ h$.

[quote="msaid"]and since the $ L^p$ norm is translation invariant we also have that for any $ h > 0$,  $ \parallel f_h \minus{} g_h \parallel_p < \epsilon$[/quote]
Can you show this?


Let $ x_{1},x_{2},\dots ,x_{n}$ be positive real numbers.
Prove that $ \sum_{i\equal{}1}^{n}(S\minus{}x_{i})^{x_{i}}\geq n\minus{}1$,where $ S\equal{}\sum_{i\equal{}1}^{n}x_{i}$.

[hide="hint"]There are two cases:
First Case:$ max\{x_{1},x_{2}\dots ,x_{n}\}\geq 1$.
Second Case:$ max\{x_{1},x_{2}\dots ,x_{n}\}\leq 1$.
Solution for the first case is obvious.But for second case it isn't easy,but it's very nice :wink:
[/hide]

It's a very known inequality, Erken.
Your inequality follows from the following reasoning:
if $ x > 0$ and $ 0 < y < 1$ then $ x^{y}>\frac{x}{x\plus{}y}.$ :wink:


Let $a,b,c$ and $d$ be odd integers such that $0<a<b<c<d$ and $ad=bc$. Prove that if $a+d=2^{k}$ and $b+c=2^{m}$ for some integers $k$ and $m$, then $a=1$.

[quote="nicetry007"]We can write $ d \equal{} 2^{k} \minus{} a$ and $ c \equal{} 2^{m} \minus{} b$.

$ ad \equal{} bc \Rightarrow a(2^{k} \minus{} a) \equal{} b(2^{m} \minus{} b) \Rightarrow b^{2} \minus{} a^{2} \equal{} 2^{m}b \minus{} 2^{k}a \Rightarrow (b \plus{} a)\;(b \minus{} a) \equal{} 2^{m}b \minus{} 2^{k}a$ ----------(*)

Claim: $ m \leq k$.
$ b\; < \; d,\; c\; < \; d\; \Rightarrow\; 2^{m}\; \equal{} \; b \plus{} c\; < \; 2d\; < \;2^{k \plus{} 1}\; \Rightarrow\; m\; < \; k \plus{} 1$.

Using $ m \leq k$ in (*), we get

$ (b \plus{} a)\;(b \minus{} a) \equal{} 2^{m}\;(b \minus{} 2^{k \minus{} m}a) \;\Rightarrow\; 2^{m}| (b \plus{} a)\;(b \minus{} a)$

Both $ b \plus{} a$ and $ b \minus{} a$ are even. $ gcd(b \plus{} a,\;b \minus{} a) \equal{} gcd(b \plus{} a \minus{} (b \minus{} a),b \minus{} a) \equal{} gcd(2a, b \minus{} a) \equal{} 2\cdot gcd(a, (b \minus{} a)/2)$.

The highest power of $ 2$ that divides both $ b \plus{} a$ and $ b \minus{} a$ is $ 2$ as $ a$ is an odd integer.  In other words, one of $ \{b \plus{} a, \;b \minus{} a\}$ is divisible by $ 2$ and the other is divisible by $ 2^{m \minus{} 1}$.

Claim: $ b < 2^{m \minus{} 1}$
$ 2b < b \plus{} c \equal{} 2^{m}\Rightarrow b \; < \; 2^{m \minus{} 1}$.

Hence, $ b \minus{} a \; < \;2^{m \minus{} 1}$ and $ 2^{m \minus{} 1}\;|\;(b \plus{} a)$.

But $ b \plus{} a \; < \; b \plus{} c \equal{} 2^{m}\Rightarrow b \plus{} a \equal{} 2^{m \minus{} 1}\Rightarrow b \plus{} c \plus{} a \equal{} 2^{m \minus{} 1} \plus{} c \Rightarrow 2^{m} \plus{} a \equal{} 2^{m \minus{} 1} \plus{} c \Rightarrow c \minus{} a \equal{} 2^{m \minus{} 1}$

$ ad \equal{} bc \Rightarrow ad \equal{} (2^{m \minus{} 1} \minus{} a)c \Rightarrow a(d \plus{} c) \equal{} 2^{m \minus{} 1}c$.

Suppose $ a > 1$. Since $ a$ is an odd integer $ a | c$. Let $ c \equal{} pa$.

Now $ c \minus{} a \equal{} 2^{m \minus{} 1}\Rightarrow a(p \minus{} 1) \equal{} 2^{m \minus{} 1}\Rightarrow a | 2^{m \minus{} 1}$. But $ a$ is an odd integer greater than $ 1$. Hence, our assumption that $ a \; > \;1$ is wrong.

$ a \equal{} 1, \;b \equal{} 2^{m \minus{} 1} \minus{} 1,\; c \equal{} 2^{m \minus{} 1} \plus{} 1$ and $ d \equal{} 2^{2(m \minus{} 1)} \minus{} 1$.[/quote]

mod: checked and found correct 20071102,0:51:04


How many isosceles triangles are there whose vertices are the vertices of a regular nonagon?

Answer

Let's consider a regular nonagon, with vertices 1 to 9 arranged in a clock-like pattern, with 9 at the top.

Let's consider all possible isosceles triangles with an apex at 9.

We have 9-1-8, 9-2-7, 9-3-6 and 9-4-5

The triangles with vertices 9-1-8, 9-2-7 and 9-4-5 have rotational symmetry of order 1, hence each of the triangles formed when the apex is moved to each point in turn will be different giving 27 different triangles.

However the triangle with vertices 9-3-6 is also equilateral, and has rotational symmetry of order three. Moving the apex by three points will give the same triangle. Hence there are only three distinct possible triangles of this type.

Hence there are 30 possible isosceles triangles whose vertices are the vertices of a regular nonagon.


Ba arze sobh bekheir be hameye doostan (chon man sobh in post ro ferestadam :D )
Agar $ a,b,c,d > 0$ bashand sabet konid
\[ \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{d^2} \plus{} \frac {d^3}{a^2}\ge\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{d} \plus{} \frac {d^2}{a}
\]
Ba hamon shart sabet konid(va inke $ n$ tabiyi ast)
\[ \frac {a^n}{b^{n \minus{} 1}} \plus{} \frac {b^n}{c^{n \minus{} 1}} \plus{} \frac {c^n}{d^{n \minus{} 1}} \plus{} \frac {d^n}{a^{n \minus{} 1}}\ge\frac {a^{n \minus{} 1}}{b^{n \minus{} 2}} \plus{} \frac {b^{n \minus{} 1}}{c^{n \minus{} 2}} \plus{} \frac {c^{n \minus{} 1}}{d^{n \minus{} 2}} \plus{} \frac {d^{n \minus{} 1}}{a^{n \minus{} 2}}
\]

Baraye Ghesmate avvalesh ba miangine $ 65$ tayi darim :
$ 46\frac {a^3}{b^2} \plus{} 9\frac {b^3}{c^2} \plus{} 6\frac {c^3}{d^2} \plus{} 4\frac {d^3}{a^2} \geq 65\frac {a^2}{b}$
hala age inaro dori benevisin jam konin hokm bedast miad.

Baraye hallate kollisham kafie ke $ 4$ adade $ x_{1} \equal{} (n \minus{} 1)(4n^2 \minus{} 5n \plus{} 2)$ , $ x_{2} \equal{} n^2$ , $ x_{3} \equal{} n(n \minus{} 1)$
va $ x_{4} \equal{} (n \minus{} 1)^2$ va alave bar oon farz konid $ A\equal{}\frac{a^n}{b^{n\minus{}1}}$ , $ B\equal{}\frac{b^n}{c^{n\minus{}1}}$
$ C\equal{}\frac{c^n}{d^{n\minus{}1}}$ , $ D\equal{}\frac{d^n}{a^{n\minus{}1}}$ oon moghe in miangino benevisid :
$ x_{1}A\plus{}x_{2}B\plus{}x_{3}C\plus{}x_{4}D\geq (x_{1}\plus{}x_{2}\plus{}x_{3}\plus{}x_{4})\frac{a^{n\minus{}1}}{b^{n\minus{}2}}$
ke age inaro dori jam konin bazam hokm bedast miad .
baraye peyda kardane in adadam kafie ye dastgahe moadelat hal konin ke kare asoonie.

Rahe halle khobi bod Amma aya in ravesh baraye inham javab mide  :?:
Agar $ a_{1},a_{2},...,a_{m}>0$ Sabet conid ($ m,n$ tabiyi va $ m\plus{}1\equal{}1$ farz shavad)
\[ \sum_{cyc}\frac{a_{1}^n}{a_{2}^{n\minus{}1}}\ge\sum_{cyc}\frac{a_{1}^{n\minus{}1}}{a_{2}^{n\minus{}2}}\]

dar morede soale dovvomam ba miangin mishe kar kard faghat oon zarayebe $ x_{1}$ , $ x_{2}$ , $ x_{3}$ va $ x_{4}$ ro bayad be zarayebe $ x_{2} \equal{} n^{n \minus{} 2}$ , $ x_{3} \equal{} n^{n \minus{} 3}(n \minus{} 1)$ , $ \cdots x_{i} \equal{} n^{n \minus{} i}(n \minus{} 1)^{i \minus{} 2}$
baraye har $ 2 \leq i \leq m$ .
va alave bar oon $ x_{1} \equal{} (n \minus{} 1)(x_{2} \plus{} x_{3} \plus{} \cdots \plus{} x_{m \minus{} 1} \plus{} 2x_{m})$ ro dar nazar begirin va miangin ro intori benevisin :
$ \sum x_{i}\frac {a_{i}^n}{{a_{i \plus{} 1}}^{n \minus{} 1}} \geq (\sum x_{i}) (\frac {a_{1}^{n \minus{} 1}}{a_{2}^{n \minus{} 2}})$
bad inaro bazam dori benevisino jam konid hokm bedast miad


I just got an email yesterday from the USAMTS saying that I won a $ \$$200 partial scholarship for 2009 Canada/USA Mathcamp (since I was one of the top 30 females in this year's USAMTS scoring). Since my parents kinda think it's still too expensive even with just a $ \$$200 dollar discount, they are hesitating to let me go. If I couldn't attend Mathcamp this year, can that scholarship money be converted to cash money, or can it be cumulative (meaning if I get additional scholarship money for Mathcamp then I can use it to pay for and apply for future years if it's worth the money)? Or do I just have to forfeit that scholarship to no use if I don't attend Mathcamp this year? :(

I'm almost certain that you can not exchange it for cash, because, like really all scholarships, the point is to convince people to attend. I don't know if it is cumulative. Mathcamp does seem to provide decent financial assistance though, http://www.mathcamp.org/fin.php

[quote="pianoforte"]I think she did, but there aren't that many girls to begin with...[/quote]
That just annoys me because I got a silver and I'm not even close to Top 30; I need like a 94.

Yeah it does (and so does MPfG), but this is not the place for it.

Mathcamp issues its scholarships separately to boys and girls, to try to encourage more girls to attend.

All students who were awarded a Mathcamp scholarship were notified privately via email.

The scholarship is not convertible into cash.  However, as the email explained, it is in addition to any need-based financial aid that you might qualify for (that is, it won't reduce your need-based aid at all, so you'll end up paying 200 dollars less than you would have regardless).

The scholarship is also contingent on your being accepted to Mathcamp in the first place; not all scholarship winners will necessary be accepted.

Finally, don't discuss other people's scores in public without their permission; that's really rude.

So it's like you'd still get $ \$$200 as if it was [b]won money[/b] even if you don't apply to Mathcamp this summer?

EDIT: DPatrick, what did you mean by getting financial aid if we received that scholarship but not going to the camp?

[quote="DPatrick"]
The scholarship is [b]not[/b] convertible into cash. [/quote]

Is it cumulative, though? (If so I would have 400 now and likely will become 600 next year should I apply next year).

The email that was sent out about the scholarships also said to contact Mathcamp with your questions.  Mathcamp awarded the scholarships, not the USAMTS.  I cannot answer for Mathcamp's policies.

The short answer to all these questions is that it is a $ \$200$ scholarship to 2009 Mathcamp.  That's all.  If you don't attend 2009 Mathcamp, then you get no benefit.  If you need more details, please contact Mathcamp as instructed in the email.

[quote="DPatrick"]Finally, don't discuss other people's scores in public without their permission; that's really rude.[/quote]
My apologies


inequalities
LaTeX
algebra
polynomial
function
inequalities proposed

This is maybe a hard inequaltity but i have a very nice solution for it.How about you,let's try:
Let a,b,c are three sides of a triangle.Prove that:
[tex] (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+3\frac{(a-b)(b-c)(c-a)}{abc} \geq 9+A [/tex]
where
[tex] A=\frac{3abc}{4[(ab)^2 +(bc)^2 +(ca)^2] }.[\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}] [/tex]. :D

Hmz...

looks a little artificial and long to me...

Thanks about your idea.But sure,you don't prove it if you find boring,but if you are successful, I swear that you will be very happy.The way to do is so fantasy,I think so.
And about the length of the inequality,you can have some condition to make it nicer.For example:
 \[ 1/a^2+1/b^2+1/c^2=1 \] .Hope that everybody have fun.  :)

hei Anh Cuong , i have used a lot of times to find your problem
you have posted too much
and i have tried , as you say
but i think your inequality will be stronger when 3/4==>3/2
is it right?

i will give you all them in monday
it is too long to type there and " i don't know to write in Latex"

This one is maybe a hard problem,but it is too boring,ugly and tastes disgusting.
So throw it away.However I have had a solution of Master. :(
(But I don't mind if someone else can find it too :D )

Dear Anh Cuong, I don't find this inequality ugly but instead I think that it is very interesting.

I have tried but I cannot solve it.

Can you, please, post your solution or Master's solution?

I am very curious to see it.

Thank you very much.

Dear manilo,the inequality is equal to:
[tex] (a+c-b-\frac{3}{4b})(a-b)^2+(a+b-c-\frac{3}{4c})(b-c)^2+(b+c-a-\frac{3}{4a})(c-a)^2 \geq 0 [/tex]
where [tex] \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1 [/tex]
Now,we must do an important step.Maybe you will find it if you try... :)

Dear Anh Cuong, I am in trouble with proving your relation.

We must prove $(1)=(2)$ where

$(1)$ $ (a+b+c)(\frac{1}{a} +\frac{1}{b} +\frac{1}{c}) +3\frac{(a-b)(b-c)(c-a)}{abc}  \geq 9 +\frac{3}{4abc}(\frac{(a-b)^2}{a} +\frac{(b-c)^2}{b} +\frac{(c-a)^2}{c}) $

$(2)$ $(a+b-c-\frac{3}{4c})(b-c)^2+(b+c-a-\frac{3}{4a})(c-a)^2 +(c+a-b-\frac{3}{4b})(a-b)^2 \geq0 $

We start with $(1)$ using also

$(3)$ $ \frac{1}{a^2} +\frac{1}{b^2} +\frac{1}{c^2} =1 $ which is equivalent to $(ab)^2+(bc)^2 +(ca)^2 = (abc)^2 $

now

$(4)$ $(a+b+c)(\frac{1}{a} +\frac{1}{b} +\frac{1}{c})= \frac{(3abc +a^2b+ab^2 +b^2c+bc^2+c^2a+ca^2)}{abc}  $

$(5)$ $(a-b)(b-c)(c-a) = a^2c-ac^2+bc^2-b^2c+b^2a-ba^2$

$(6)$  $(\frac{(a-b)^2}{a} +\frac{(b-c)^2}{b} +\frac{(c-a)^2}{c}) = \frac{a^3b+b^3c+c^3a-abc(a+b+c)}{abc} $

Using $(3),(4),(5),(6)$we obtain

$4abc(3abc+4(ab^2+bc^2+ca^2)-2(a^2b+b^2c+c^2a)) \geq 36((ab)^2+(bc)^2+(ca)^2) -abc(a+b+c) +a^3b+b^3c+c^3a $

Thus the final form of $(1)$ is

$(7)$ $16abc(ab^2+bc^2+ca^2)-8abc(a^2b+b^2c+c^2a) \geq 24(abc)^2 -abc(a+b+c) +a^3b+b^3c+c^3a $

Now we compute $(2)$

$ca(4ca+4bc-4b^2-3)(a^2+b^2-2ab) +ab(4ab+4ca-4c^2-3)(b^2+c^2-2bc) +bc(4bc+4ab-4a^2-3)(c^2+a^2-2ac) \geq 0$

Expanding this we obtain the final form of $(2)$

$(8)$ $3abc(a+b+c) +8abc(ab^2-a^2b+bc^2-b^2c+a^2c-c^2a) \geq 24(abc)^2 + 3(a^3c+c^3b+b^3a) $

I cannot make $(7)$(final form of $(1)$) equal to $(8)$(final form of$(2)$).

Can you please tell me where is my mistake or omission or the missing transformation?

I am also very interested with the original inequality and I would like also to know the reduced form of type

$  A(a,b,c)(a-b)^2 + B(a,b,c)(b-c)^2 + C(a,b,c)(c-a)^2 \geq 0 $

with $A,B,C$ cyclic rational polynomial functions of $a,b,c$

of your original inequality WITHOUT the condition

$ \frac{1}{a^2}  + \frac{1}{b^2} +\frac{1}{c^2} =1$

If possible another hint to solve $(2)$ because I have tried for all day and all night to solve it without success.

This is a VERY VERY HARD inequality and only Master said to have found a solution but he has not posted it.

Sorry for asking you for all these things but if you have not any time you can answer me another time.

Thank you very much, Anh Cuong.

if i said anything stupid, sorry Manlio
My solution come from Anh Cuong
So if you really want i will post my solution

Dear manilo,I really admire your effort.I had never thought that someone will like and waste a lot of time to do this ugly inequality before I posted it.
Thank you very much about your regard to my problem. :)
About your solution,it is not true from the start,because the RHS side must be:
[tex] \frac{3}{4abc}(\frac{(a-b)^2}{b} +\frac{(b-c)^2}{c} +\frac{(c-a)^2}{a}) [/tex]
Dear Map(Master's nick name) :D ,you can say that is your solution,because you have solve  almost it.I just gave you the last hint.However,you don't need to post it here,I will send a message to manilo about it.


The product of positive inttegers A and B is 6, and the sum of positive integers K and M is also 6. What is the positive difference between the probability that A=1 and B=6 and the probability that K=2 and M=4?

I believe the answer is $ 1/20$.  If $ (A,B)$ is the set of solutions for $ A$ and $ B$, then $ (A,B)\equal{}\{(1,6),(2,3),(3,2),(6,1)\}$.  For $ K$ and $ M$ we use similar notation: $ (K,M)\equal{}\{(1,5),(2,4),(3,3),(4,2),(5,1)\}$.  Then the probability of getting $ (A,B)\equal{}(1,6)$ is $ 1/4$ and $ 1/5$ for $ (K,M)\equal{}(2,4)$.

Hehe.  I would hope I can answer an SAT question correctly since I took it 2 years ago and am now in college. :D


geometry
circumcircle
trigonometry
projective geometry
geometry proposed

Let $H$ be the orthocentre of the triangle $ABC$. Denote the middlepoint $M$ of the side $[BC]$ and the intersections $E\in AC\cap BH$, $F\in AB\cap CH$, $L\in BC\cap EF$. Prove that $MH\perp AL$.

I have seen this problem in many competitions. This is a nice but difficult problem.I have the solution with powers - from romanian books. I hope that somebody will post it again.

 Babis

Here one general case:
http://www.mathlinks.ro/Forum/viewtopic.php?t=15668

H is the polar of AL with respect to M(BCEF)

[quote="Virgil Nicula"]Let $H$ be the orthocentre of the triangle $ABC$. Denote the middlepoint $M$ of the side $[BC]$ and the intersections $E\in AC\cap BH$, $F\in AB\cap CH$, $L\in BC\cap EF$. Prove that $MH\perp AT$.[/quote]
Hi $Virgil$ I thing you mean  $MH\perp AL$.
I solved more general and wery old problem,her it.
Let $ABCD$ is inscribed cuadrilateral,and $E\in{AB}\cap{CD}$,$M\in{AC}\cap{BD}$
$F\in{AD}\cap{BC}$,$O$ is center of circumcircle,then$OM\perp{EF}$
also we  can prove that $M$ is orthocentre of $EFO$.

Yessssssssssssssssssss Tiks, $\boxed {\ MH\perp AL\ }$[hide="For Mystic Terminator and Juan Carlos."]I wrote [b]Sursa : Classic [/b] and I wish a elementary solution and no ... an atomic bomb for to kill a butterfly ![/hide] [hide="The first elementary solution."] Denote the point $D\in BC\cap AH$, the power $p(X)$ of the point $X$

w.r.t. the circumcircle $w=C(O,R)$ and $x=m(\widehat {HLM})$, $y=m(\widehat {HAM})\ .$

Therefore, $MH\perp AL\Longleftrightarrow AM\perp HL\Longleftrightarrow \tan x=\tan y\Longleftrightarrow$

$\frac{HD}{LD}=\frac{MD}{AD}\Longleftrightarrow DH\cdot DA=DM\cdot DL\Longleftrightarrow p(D)=DM\cdot DL\Longleftrightarrow$

$DB\cdot DC=DM\cdot DL\Longleftrightarrow (B,C,L,D)$ is harmonic division, what is truly.[/hide] [hide="The second elementary solution."]Use the well-known relations : $DA\cdot DH=DB\cdot DC=DM\cdot DL$. Therefore,

$MH\perp LA\Longleftrightarrow LH\perp AM\Longleftrightarrow LA^2-LM^2=HA^2-HM^2\Longleftrightarrow$

$LD^2+(HA+HD)^2-LM^2=HA^2-(HD^2+DM^2)\Longleftrightarrow$

$LD^2+2\cdot HD\cdot AD=LM^2-DM^2\Longleftrightarrow$

$2\cdot DB\cdot DC=LM^2-DM^2-DL^2\Longleftrightarrow$

$2\cdot DB\cdot DC=DM(LM+LD)-DM^2\Longleftrightarrow$

$2\cdot DB\cdot DC=DM(LM+LD-DM)\Longleftrightarrow$

$DB\cdot DC=DM\cdot DL$, what is truly.
[/hide]


can anyone here tell me what are nedians (not medians)

The line which connects the vertix of a triangle with the niddle of the opposite side?

I was joking. I really don't know what it means. Maybe it is an error of writing.

Orbits of triangles obtained by interior division of sides

A New Look At Medians
Israel Koral
	Proof of a division ratio result.
	51, (1958)  123.


trigonometry
real analysis
real analysis unsolved


by using Bezout identity
for the first two fractions:
lets d divide the numerator and the denominator so d divide their difference which is 1 so d =1 and so is the gcd
For the third one:if d divide numerator and denominator so d divide n*(2n+1)-(2n^2+2n) which is equal to n :so d diide n and 2n+1(the numertor) so it divides their difference which is 1

[quote="fourierseries"]Show that for all natural numbers n, the fractions:

$\frac{n-1}{n},\frac{n}{2n+1},\frac{2n+1}{2n^2+2n}$

are irreducible.[/quote]

For the first, note that n-1 and n are relatively prime.  Thus the fraction is irreducable.
For the second, we note that if n/(2n+1) is reducable, then the the reciprocal is reducable.   However, the reciprocal is 2 + 1/n and we know that 1/n is irreducable.  So n/(2n+1) is irreducable.
Similarly, the reciprocal of the third fraction is n + n/(2n+1) and we know tha n/(2n+1) is irreducable.


topology
vector
geometry
geometric transformation
real analysis
real analysis unsolved

Let $ X$ be a vector space and $ \mathcal{B}$ the collection of all convex sets $ V$ containing 0 such that for each $ x \in X$ there is an $ \alpha > 0$ with $ \alpha x \in V$. Show that $ \mathcal{B}$ is a local base at 0 for a locally convex topology on $ X$ and this topology is stronger than any other locally convex topology on $ X$.

[In another problem, I have shown that the following conditions are sufficient for the translates of sets in a family $ \mathcal{N}$ of convex sets in a vector space $ X$ to form a base for a topology making $ X$ into a locally convex topological vector space.

(i) If $ N \in \mathcal{N}$, then each point of $ N$ is an internal point.
(ii) For $ N_{1},N_{2} \in \mathcal{N}$, there is an $ N_{3} \in \mathcal{N}$ such that $ N_{3} \subset N_{1} \cap N_{2}$.
(iii) If $ N \in \mathcal{N}$, then for each $ \alpha$ with $ 0 < |\alpha| < 1$ we have $ \alpha N \in \mathcal{N}$.

I was trying to use this earlier result for this question.]

Why specify that $ X$ is a topological vector space if you're not going to use the topology? Convexity is not a topological property for real vector spaces.

This means that the nature of the topology you get here depends only on cardinality. It's an exceedingly weird one, as well. If the span of $ x_i\minus{}x_j$ is infinite-dimensional, the sequence $ x_i$ diverges.

As for your question, the answer is almost immediately evident from the definitions. Locally convex topology? If it's a local base, it is. Strongest locally convex topology? It's the largest possible local base, so that's obvious.
Unfortunately, it's not a local base, because it doesn't exclude sets that should be closed. In $ \mathbb{R}$, $ \mathcal{B}$ includes $ [\minus{}1,1]$. That means that the translate $ [1,3]$ is also open, and thus the single point $ \{1\}$ must be. We get the discrete topology, and everything collapses.

A change that fixes this: $ \mathcal{B}$ is the collection of all convex sets $ V$ such that for each $ x\in X$, the intersection of $ V$ with $ \mathbb{R}x$ is $ Ix$ for some open interval $ I$ containing (or equivalently centered at) zero. This ensures that the intersection of any line with $ V$ is either empty or an open interval, and this property is preserved by intersecting translations of such sets.
The weird properties I mentioned still apply.

[quote] Why specify that $ X$ is a topological vector space [/quote]
It was a mistake. I have editted the original post.


There are 6 married couples at a party. At the start of the party, every person shakes hands once with every other person except his or her spouse. How many handshakes are there?

I was wondering what was wrong with my solution...
I thought since the first person shakes hands with ten people (excluding himself and spouse) the second person will shake hands with 10 people but since he already shook hands with the second person he would be shakind 9 different hands and so on until 0. So I thought it would be 10(11)/2 = 55 but the answer is 60. Please tell me what I did wrong...

My method is to notice that there are $ \dbinom62\equal{}15$ pairs of couples, and in each pair, there are a total of $ 4$ handshakes, and the answer is $ 15\cdot4\equal{}\boxed{60}$.

Alternatively, note that each person will shake hands with $ 12\minus{}2\equal{}10$ people, and each handshake is counted twice this way, so our answer is $ \frac{12\cdot10}2\equal{}\boxed{60}$.

I though it was 72 because if each person doesn't shake with it's spouse it should be 11*2+9*2+7*2+5*2+3*2+1*2

But this is the real answer[hide]All 12 people shake hands with 10 other people (everyone except themselves and their spouse). In multiplying $12 \times 10$, each handshake is counted twice, so we divide by two to get the answer of $\dfrac{12 \times 10}{2} = \boxed{60}$ handshakes.
[/hide]

[quote=Irshad]But then you must divide by two again to get thirty; the first time you eliminated the handshakes between each of the couples, and the second time you eliminate extra handshakes. Another solution: if you connect two dots, 1 line will be formed. If you connect three dots, 3 lines will be formed. If you connect four dots, 6 lines will be formed. If you connect five dots, 10 lines are formed. These numbers are called triangular numbers. Similarly if six people/dots connect, you get 15 lines, or handshakes. But then you must double it, since you counted the handshakes of only one person per couple. Both ways, you get thirty. The answer must be incorrect or have an error. :-D[/quote]

Um..this problem is 8 years old. Someone would have found an error.

I reported this as a bug. It's so confusing!!

Each person shakes 10 peoples hands, so 12 (the number of people) times 10 (the number of handshakes per person) = 120. However, we counted each handshake twice so we divide 120 by 2 = 60. Thus, 60 is the final answer. (Please note that it’s absolutely ridiculous to shake hands with yourself so the problem assumes that you can’t shake hands with yourself.) [size=50]Since you can’t shake hands with you or your spouse, there are only 10 people you can shake hands with.[/size]

Also, I agree that this problem may need an updated text.

It's really simple once you've figured out how many people one person shakes.

[quote=nickxiaopi]I don’t understand how to solve this problem  :help:[/quote]

What do you need help with?

[quote=Ljviolin11]I forgot that you can't shake hands with yourself :)[/quote]

 :rotfl:


Does there exist a non-constant function $ f: N\rightarrow N$ such that

$ f(x,y)\plus{}f(y,x)\equal{}f(x^2,y^2)\plus{}1$,

for all $ x,y\in N$?

[quote="t0rajir0u"]You mean $ f : \mathbb{N}^2 \to \mathbb{N}$.[/quote]

Ya, of course. Sorry for the mistake.  :)


trigonometry
calculus
integration
calculus computations

[quote="Farenhajt"]Determine $\prod_{n=3}^{\infty}\cos{\pi\over n}$[/quote]

Not much to do with this one, since $\prod_{n=3}^{\infty}\sec{\pi\over n}=K$ where
\[K\approx 8.7000366252081945032224098591130049711932979497428920921596672786834299641140251591185444140 \]
is the[url=http://mathworld.wolfram.com/PolygonCircumscribing.html]polygon circumscribing constant[/url] and hence the numerical value for this product

$\prod_{n=3}^{\infty}\cos{\pi\over n}=\frac1{K}\approx 0.11494204485329620070104015746959874283079533720086351684402339651896601282535305117794077$

I don't know what you expected someone to do with this, but if you had expected an exact value of $\frac1{12}$, know that there were at least two texts that incorrectly reported the value of $K$ as being 12.

P.S. You might be interested in this: the [url=http://mathworld.wolfram.com/InfiniteCosineProductIntegral.html]Infinite Cosine Product Integral[/url] which is this:
\[\int_{0}^{\infty}\cos (2x)\prod_{n=1}^{\infty}\cos{x\over n}\, dx \approx 0.39269908169 \]
-Ben


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

Why is a subgroup of a direct product of copies of $\mathbb{Z}/p\mathbb{Z}$ also such a direct product?

One way of seeing this is that such a group is a vector space over the finite field ${\mathbf F}_p$.

Can it be seen using the fundamental theorem of finitely generated abelian groups?

Yes

first of all, we do agree that p is a prime right?
because otherwise, you could take the subgroup in Z6*Z6 : (0,0) (2,2) (4,4) which contains only three elements and thus is not isomorphic to a direct product of copies of Z6

however : let G be a subgroup of H, a direct product of copies
well than the order of G will divide that of H, will thus be a power of p as well
also , G will be abelian and therefore be written  as a product of cyclic groups

the trick is now : in H, every element has order p or 1, therefore the same must hold for G, so if the decomposition of G would contain a Zp^2 or something, we would have an element of higher order
(example = in Zp*Zp^2 you could take (0,1) as an element with order p^2 )

i hope this helps you, if not ask, but the other poster suggested a good way to prove it as well


abstract algebra
Ring Theory
superior algebra
superior algebra solved

If A is a subring of B, and B is a subring
of C (assume commutative) such that B is a finitely generated
A-module, and C is a finitely generated B-module, then C is a
finitely generated A-module.

Let $b_i$ be generators for $B$ over $A$ and similar let $c_i$ be generators for $C$ over $B$.
Then the products $b_ic_j$ should generate $C$ over $A$.

Let the generators be defined as above.
Take any $c \in C$, then there are $\overline{b}_j \in B$ with $c=\sum \overline{b}_j c_j$.
For any of those $\overline{b}_j$ there are $\overline{a}_{i,j} \in A$ with $\overline{b}_j = \sum \overline{a}_{i,j} b_i$.
But then $c = \sum \overline{a}_{i,j} b_i c_j$ follows, thus the products $b_i c_j$ generate $C$ over $A$.
Since the number of $b_i$ and $c_j$ is finite, the number of these products is finite, too.


function
algebra
polynomial
parameterization
abstract algebra
domain
calculus

When we are decomposing some rational function with a linear factor to some power (i.e $ \frac{N(x)}{D(x)}$ where $ D(x)\equal{}(x\minus{}a)^k$) why do we consider the $ (x\minus{}a)^k$ term to be a repeated linear factor instead of a polynomial of degree $ k$? What are the underlying notions that cause us to treat these terms differently than irreducible polynomials? Any answers would be greatly appreciated. Thanks.

A good toy model for partial fraction decomposition is the integers (and while this is obvious in hindsight I am amazed that I have never seen this written down anywhere):  it's possible to write decompositions like $ \frac {5}{12} \equal{} \frac {2}{3} \minus{} \frac {1}{4}$, for example, and it's not hard to see why this works if you multiply by $ 12$.  Then one answer to your question is this:  partial fraction decomposition works like the Chinese Remainder Theorem, and $ (x \minus{} a)^k$ behaves like a prime power.

Another way to think about it is dimensionality.  If you restrict yourself to linear terms $ \frac {1}{x \minus{} a}$ you'll find that you simply don't have enough parameters to set in the partial fraction decomposition to cover all the possible coefficients of $ N$; the number of parameters you need grows like the total degree of $ D$, not its number of distinct roots.

A third way to think about it is Taylor series.  The series of $ \frac {1}{(x \minus{} a)^k}$ is geometric with a polynomial coefficient of degree $ k \minus{} 1$, so it can't be written as a linear combination of any terms with smaller $ k$.

Finally, there's the analytic view.  The term $ \frac {1}{(x \minus{} a)^k}$ describes a pole of order $ k$:  it goes to infinity faster than a pole with order less than $ k$ as $ x$ approaches $ a$, so to faithfully represent a rational function as a partial fraction decomposition you need to faithfully describe all of its poles.  (This is closely related to the first idea, but I'm not the one to explain why.)

Edit:  And because I think this question is genuinely interesting, you can also think about $ x$ as a linear operator on $ \mathbb{C}[x]/(D(x))$ (it acts by multiplication), and then partial fraction decomposition is closely related to the direct-sum decomposition of $ \mathbb{C}[x]/(D(x))$ into spaces of the form $ \mathbb{C}[x]/(x \minus{} a)^k$.  So in some sense this phenomenon is related to the [url=http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain]structure theorem[/url] (which is basically a huge generalization of CRT), the need for [url=http://en.wikipedia.org/wiki/Jordan_normal_form]Jordan blocks[/url] and the existence of nilpotent matrices; that's what happens when we take $ x$ to be a linear operator and $ D(x)$ to be its characteristic polynomial.

Edit #2:  So here's the question of the day, in lieu of the first thing I wrote: what is the relationship between [url=http://en.wikipedia.org/wiki/P-adic_number]$ p$-adic numbers[/url] and Taylor series expansion about $ x \equal{} a$?

Partial fractions is an intermediate step in some other problem, not an end in itself.

It would be perfectly valid to use \[ \frac{b_{k\minus{}1}x^{k\minus{}1}\plus{}b_{k\minus{}2}x^{k\minus{}2}\plus{}\cdots\plus{}b_1x\plus{}b_0}{(x\minus{}a)^k}\] in a partial fractions decomposition. Assuming you were starting with a proper rational function with that factor in the denominator, we should arrive at a system of linear equations that can be solved uniquely for the $ b_j.$

The problem with that comes with the end use. For many of the possible uses, most prominently computing an integral or finding an inverse Laplace transform, this isn't a particularly useful form, while\[ \frac{c_1}{x\minus{}a}\plus{}\frac{c_2}{(x\minus{}a)^2}\plus{}\cdots\plus{}\frac{c_k}{(x\minus{}a)^k}\]is a useful form.

Note that in either case, the $ k$th degree polynomial in the denominator must account for $ k$ constants to be solved for.

That's how I'd answer the O.P's original question. The reason for treating this case the way we usually do is utilitarian and driven by the end use of the technique.

Thanks for the answers.

tOrajirOu, I can make sense of most of the reasons you gave (I had some intuition that it would involve poles). But unfortunately I haven't studied Taylor Series in any great detail yet. However, from what I can comprehend, your explanation involving Taylor series is very interesting. I'm interested to hear an answer regarding your question.

Merryfield, I understand your point and agree that partial fraction decomposition is most useful when used as an intermediate for other problems. However, I don't quite follow your example. Maybe you can clarify.

I didn't give and example. [i]This[/i] would be an example:

$ \frac1{x(x\plus{}1)^2}\equal{}\frac Ax\plus{}\frac{Bx\plus{}C}{(x\plus{}1)^2}$

You can solve this for $ A,B,$ and $ C,$ right?

But in most cases, that's not the most useful thing you could do.

I seem to have slightly misinterpreted your original question.  What do you mean by considering $ (x \minus{} a)^k$ to be a polynomial of degree $ k$?

Merryfield, I tried your example and it does not give a correct decomposition. We must decompose a function of this form as follows:

$ \frac{N(x)}{(x\minus{}a)^k}\equal{}\sum_{i\equal{}1}^{k}\frac{A_i}{(x\minus{}a)^i}$

or for this specific case:

$ \frac{N(x)}{x(x\minus{}a)^k}\equal{}\frac{A_0}{x}\plus{}\displaystyle\sum_{i\equal{}1}^{k}\frac{A_i}{(x\minus{}a)^i}$

tOrajirOu, what I mean by "considering $ (x\minus{}a)^k$ to be a polynomial of degree $ k$" is:

Why can't we write the decomposition of $ \frac{1}{x(x\plus{}1)^2}$ (for example) as:

$ \frac{1}{x(x\plus{}1)^2}\equal{}\frac{A}{x}\plus{}\frac{Bx\plus{}C}{(x\plus{}1)^2}$

[quote="JZambrano201"]Merryfield, I tried your example and it does not give a correct decomposition. [/quote]
What do you mean, "not a correct decomposition"? Show me what you have.

[quote="JZambrano201"]Why can't we write the decomposition of $ \frac {1}{x(x \plus{} 1)^2}$ (for example) as:

$ \frac {1}{x(x \plus{} 1)^2} \equal{} \frac {A}{x} \plus{} \frac {Bx \plus{} C}{(x \plus{} 1)^2}$[/quote]
Oh.  Well, we can.  It's just not usually the useful thing to do, as Kent pointed out.

Merryfield, I looked over what I did and you're correct. my appologies.


Ummm...

2009 National Math Counts Count Down:

The difference of the squares of two positive integers is 2009. What is the greatest possible difference between the two integers?

Is there like a formula or something?

I guess you could just find perfect squares and add 2009......but I don't know if there's an easier way........

(x-y)(x+y) = 2009 = 7^2 x 41

Clearly, x-y < x+y, and you want to maximize x-y.
The maximum difference is [b]41[/b].

If you are wondering what the numbers themselves are, you see that:
x-y=41, and x+y=49.
Solving this, x=45 and y=4.

EDIT: Beaten. Also, how did you get the problem? From the video?

Meh just use the (x+y)(x-y)=x^2-y^2 then factorize 2009 and yea :D

Let the two integers be $ x$ and $ y$. Let's assume that $ x$ is greater than $ y$. Therefore we have;

$ x^2 \minus{} y^2 \equal{} 2009$

Considering this is all we know, we are going to have to consider all the different cases. And since $ 2009$ doesn't have many factors this shouldn't take too long. So why not?

Factoring the equation, we get;

$ (x \minus{} y)(x \plus{} y) \equal{} 2009$

Now we apply the casework by factoring $ 2009$ in as many ways as we can.

$ 2009 \equal{} 7 \times 7 \times 41$

$ 2009 \equal{} 7^2 \times 41^1$

We want $ x \minus{} y$ to be as large as possible, and once we find the largest value possible, we have our answer. We know that $ x \minus{} y$ is less than $ x \plus{} y$. So let's consider the possibilities. If $ x \minus{} y \equal{} 7$, then $ x \plus{} y \equal{} 287$. Surely we can think of a larger value of $ x \minus{} y$ then this, so lets try and find a larger value. Like the next smallest factor of $ 2009$, $ 41$. If $ x \minus{} y \equal{} 41$, then $ x \plus{} y \equal{} 49$, which is larger than $ 41$, so $ 41$ is a possibility. The next smallest factor of $ 2009$ is $ 49$. If $ x \minus{} y \equal{} 49$, then $ x \plus{} y \equal{} 41$. This is smaller than $ 49$, so it is not possible, because we know $ x$ and $ y$ are both positive integers, so $ x \minus{} y$ is $ 2y$ less than $ x \plus{} y$. So this means $ 41$ is the largest value.

[b]41[/b]


I am having trouble understanding how to write formulas (I should say create formulas) from a given sequence.

It does not matter what sort of sequence it is.  The confusion remains.

I have not found a textbook or easy math book that explains this process for the average Joe to grasp.

Here are the two questions:

(1) Write a recursive formula for the sequence

9, -18, 36, -72, ...

I understand that a_1 = first term given, which is 9 in this case.

I also found that -2 is multiplied by each term before to find the term that follows.

But how do I create the rest of the formula?

(2) Write a recursive formula for the sequence

3, 3(sqrt{3}), 9 9(sqrt{3})

Again, the first term is 3, which is written in the book as
a_1.

How do I create the needed formula?

I'll try to keep this simple. Your goal is to make a formula for $ a_n$ which will be "a function" of $ n$.

(1) the first one is $ a_1\equal{}9$. Second is $ a_2\equal{}\minus{}2\cdot 9$. Third is $ a_3\equal{}(\minus{}2)^2 \cdot 9$. Maybe if we change indexes to 0, 1 and 2, you'll see the pattern easier: $ a_0\equal{}9,\;a_1\equal{}\minus{}2\cdot 9,\;a_2\equal{}(\minus{}2)^2 \cdot 9...\;a_m\equal{}(\minus{}2)^m \cdot 9$. As we moved the indexes by one, we'll have to substract one from $ m$ to get our $ n$, i. e. $ a_n\equal{}(\minus{}2)^{n\minus{}1} \cdot 9$.

Second one is the variation on same theme...

These are actually geometric. $ t_n \equal{} ar^n$. A great way to solve this is $ \frac {t_{n \plus{} 1}}{t_n} \equal{} r$. Using that on the first one we find $ r \equal{} \minus{} 2\Rightarrow t_n \equal{} 9( \minus{} 2)^n$. And for the second, $ r \equal{} 3^{\frac {1}{2}}\Rightarrow t_n \equal{} 3(3^{\frac {n}{2}}) \equal{} 3^{\frac {n}{2} \plus{} 1}$.

The ratio $ \frac {t_{n \plus{} 1}}{t_n}$ is very useful. For geometric formulas, this ratio is constant. However, most other recursive formulas [i]converge[/i] to a ratio. The most common example is the fibonacci sequence whose term by term ratio converges to [url=http://en.wikipedia.org/wiki/Golden_ratio]the golden ratio[/url].
I think the best way for you to learn about recursive sequences is probably to do some problems from AMCs or the AOPS I&II. That way you'll see some variation.

[b]edit:[/b] post count = $ 7^3$

Err... I don't think either of the solutions presented are in the form of recursive sequences.
Recursive sequences define each subsequent term $ a_n$ as a function of the previous terms $ a_{n\minus{}1}, a_{n\minus{}2}, \dots$ , [b]not [/b]a function of $ n$ itself.

For example, the first sequence would just be defined as $ a_n \equal{} \minus{}2a_{n\minus{}1}$; $ a_0 \equal{} 9$.

You're right... I didn't pay enough attention to the question...

archimedes1 answered the question correctly.

However, I wanted to learn how to create these formulas without using textbook jargon.

Your answer is found in most textbooks but it does not make sense to people who are not a math major.

I don't believe many of us are math majors atm...

Anyway, that's the simplest way I can even think to explain this, so what is it you don't inderstand?

The basis of recursive formulas is, "In a sequence, how do I get the next term by manipulating previous ones?"  Therefore, simpler recursive formulas will say that $ a_n$ is some function of $ a_{n_1}$, and the first term, $ a_1$, is given.

For example, in the first question, the first term is 9.  Therefore $ a_1\equal{}9$.  To get the next term, you can multiply the previous term by $ \minus{}2$.  Therefore, for all integers $ n\geq 2$, $ a_n\equal{}\minus{}2a_{n\minus{}1}$.  And that's your recursive formula.

Delving further into the subject...
A recursive formula does not limit you to expressing $ a_n$ as a function of $ a_{n\minus{}1}$.  You can express any term $ a_n$ as a function of any combination of the previous terms, as archimedes1 mentioned before.  For example, the Fibonacci sequence is defined recursively as $ F_1\equal{}1$, $ F_2\equal{}1$, and for all integers $ n\geq 3$, $ F_n\equal{}F_{n\minus{}2}\plus{}F_{n\minus{}1}$.

[quote="perfect628"]I don't believe many of us are math majors atm...[/quote]

 :whistling:

Okay Sharkman...

[hide="Here goes"]Because you seem to be having trouble understanding all the notation and whatnot I'm not going to link you anywhere. I'll do the dirty work myself. To this end, I'm going to show you "The Babylonian" square root algorithm.

Consider some integer, $ n$ whose square root we wish to find. Denote it's square root as.. $ x$. So, $ \sqrt {n} \equal{} x$. After squaring both sides, $ n \equal{} x^2 \equal{} x\cdot x$. Now we divide both sides by $ x$ and get
\[ \frac {n}{x} \equal{} x
\]
This is the fundamental concept behind the algorithm. But what if $ n$ is not a perfect square? Well, I suppose we could always guess, right? For kicks let's assume $ n \equal{} 17 \equal{} 16 \plus{} 1 \equal{} 4^2 \plus{} 1$. A not so bad guess would be $ x_1 \equal{} 4$. Doing this we have
\[ \frac {17}{4} \equal{} 4.25\neq 4
\]
Now I don't want to get too technical, but it's important to know that $ x_1\le x\le \frac{n}{x_1}$. If the true value of $ x$ lies somewhere between $ x_1$ and $ \frac {n}{x_1}$, the the [b]average[/b] of $ x_1$ and $ \frac {n}{x_1}$ is closer to $ x$ than either. What we do now is let this average be $ x_2$. Formally stated,
\[ x_{k \plus{} 1} \equal{} \frac {x_k \plus{} \frac {n}{x_k}}{2}
\]
which in simplest form is
\[ \boxed{x_{k \plus{} 1} \equal{} \frac {x_k^2 \plus{} n}{2x_k}}
\]
Computing $ x_2$ for our example, we obtain $ x_2 \equal{} \frac {16 \plus{} 17}{8} \equal{} \frac {33}{8}$. Now we can compare this to $ x_1$ to show that it really IS much closer to $ 17$ when squared. $ x_1^2 \equal{} 16$ and $ x_2^2 \equal{} 1089/64 \equal{} 17.015625\neq 17$. Because $ x_2$ is not the true value of $ \sqrt {17}$ we can take another [i]recursion[/i] of the boxed formula to find that $ x_3 \equal{} 2177/528 \equal{} 4.123106061$ (I'm skipping compuations now because they are getting ugly fast) and that the square of this is $ \frac {4739329}{278784} \equal{} 17.00000359$.[/hide]
I hope that helps.


Study the convergence of sum(n>=1) (-1)^[sqrt(n)]/n, where [] is the integer part.
   Source: AMM (I don't know the year or issue).

You also have

 \sum (-1)^[sqrt(n)]/n^a

If a>1 the serie converge

If a \leq 0 the serie is divergent

If 0 < a  \leq 1/2 the serie is also divergent

If 1/2 < a \leq 1 the serie is convergent


[quote]Is atomic physics the study of lasers and how light is absorbed or emitted? thanks?[/quote]

Well, that's partially true... Most simplistically said, atomic physics is concerned with the atom as a whole: it investigates the properties of the electron hull and its interaction with the nucleus (note that there is a difference between nuclear and atomic physics, because nuclear physics deals with elementary particles, especially those within the nucleus). Absorbsion and emission of light is studied by atomic physics because it basically happens when electrons change their energy state in the atom, thus emitting or absorbing photons (light). While lasers are also closely related to this field, I think that they are a primary subject of quantum optics rather than atomic physics itself.


I offer to discuss here Olympic Games and its results.

:winner_first: Congratulations on getting Gold medal, Shizuka Arakawa.

Have you heard what happened to swiss Sara Fischer???? (That is really bad luck).


Prove that if $ a,b,c>0$, $ abc(a\plus{}b\plus{}c)\equal{}1$, $ bc\equal{}1$ then $ b\equal{}c\equal{}1$, $ a\equal{}\sqrt{2}\minus{}1$.

[quote="Spribo"]Prove that if $ a,b,c > 0$, $ abc(a \plus{} b \plus{} c) \equal{} 1$, $ bc \equal{} 1$ then $ b \equal{} c \equal{} 1$, $ a \equal{} \sqrt {2} \minus{} 1$.[/quote]

That's wrong : $ (a,b,c)\equal{}(\frac 13,\frac{4\minus{}\sqrt 7}3,\frac{4\plus{}\sqrt 7}3)$ is a counter example.

It is indeed, and I cannot understand why that book said that $ b\equal{}c\equal{}1$...


I would ask if someone know any e-book  :roll:  for introduction to Geometry and an introduction for Number Theory as well.

Thanks in advance  :lol:

I think he means the AoPs Intro books, not just intro books in general.
But to my knowledge, I believe that there are no e books available for any of the AoPs books.

[quote="isabella2296"][quote="enndb0x"]For free ebooks , use google and write[code]allinurl: +(rar|chm|zip|pdf|tgz) TheTitle of the book[/code]
In this way I found really good ebooks, in Number Theory and Geometry.I Hope this will work for you.
[i]Note: at "TheTitle of the book" you should write title of the book you need to find.[/i][/quote]

Are these e-books legal?  :maybe:[/quote]

hmm ...I am not sure but I really think these ebooks are ilegal , but some good books can be found by searching in this way and these books are hosted in websites of universities ,so there are legal books too.But I think these resources should be used for personal using (not coping )

and also try scribd.com, they might have a few. just a guess.


What would a fractional power mean and why??

$x^\frac{p}{q}=\sqrt[q]{x^p}$, so first find a number $n$ that is $x$ multiplied by itself $p$ times. Then your answer is a number that when multiplied to itself $q$ times gives $n$. (If you find that confusing, just ignore it!)

For example, $4^\frac{3}{2}=\sqrt[2]{4^3}=\sqrt[2]{64}=8$.

A fractional power can be like n^(x/y). Imagine taking the xth power of n, then taking the yth root of the result. (or the other way around works too.)

So 32 to the power of 3/5 would be 8.

Now for why:
Suppose you're given $x^{\frac {1}{n}}$.
Raise this number to the nth power and you get $(x^{\frac {1}{n}})^n=x^{\frac {1}{n} \cdot n}=x^1$.
And we know that to undo raising to a power, you take the root, so $\sqrt[n]{x}=\sqrt[n]{(x^{\frac {1}{n}})^n}=x^{\frac {1}{n}}$.

If the numerator is not 1, take it out, and proceed as before.
$x^{\frac{m}{n}}=(x^{\frac{1}{n}})^m=(x^m)^{\frac{1}{n}}$

[quote="Rushil"]What do iirational powers mean. What about complex powers????[/quote]

An irrational power is a limit of rational powers:

For example, $2^{\sqrt{2}} = \mbox{ the limit of the sequence } \{2^1,2^{1.4},2^{1.41},\cdots \}$.

For complex powers, $a^{x + yi} = a^x a^{yi} = a^x e^{yi \ln a} = a^x (\cos (y\ln a) + i\sin(y\ln a))$ using Euler's formula.


Prove that if "n" is a prime number, and $ n> a,b$, then the following expression is always divisible by "n".

$ a^{n\minus{}2} \plus{} a^{n\minus{}3}b \plus{}a^{n\minus{}4}b^{2}.......b^{n\minus{}2}$

Please explain it clear and step-by-step. Thanks! :lol:

[quote="ocha"]If $ a,b$ are positive integers then:
Let the expression be $ X$.
$ (a \minus{} b)X \equal{} a^{n \minus{} 1} \minus{} b^{n \minus{} 1}$

Since $ n$ is prime, by FLT: $ a^{n \minus{} 1} \equiv b^{n \minus{} 1} \equiv 1 \mod p \Leftrightarrow n | a^{n \minus{} 1} \minus{} b^{n \minus{} 1}$

However $ n \not | (a \minus{} b) \because n > a,b$ so $ n | X$[/quote]There is a major flaw from the OP and that is $ a\ne b$ is a missing condition as this solution exposes.


Sorry if this is spam, but I looked through the USAMTS site and couldn't find what score people need to get into AIME using USAMTS. If it changes, like qualification for AIME (like last year), then can somebody just give me an average score?

The last three years it was 68 for the first 3 rounds (at least I think so).

Just to back it up, yeah 68 is what it was for the past few years.

About what percentage of the contest-takers does it include, usually? Top x%?


At the moment, the thread called "Saddam Hussein sentenced to death..." is at the top of Round Table, thus shown in the full forum list as some shortcut-link. But if I click on it/tab it, it leads me to http://www.mathlinks.ro/Forum/viewtopic.php?p=668862#668862 .
Similar effects happen with most other such shortcuts.

Yes, I think they lead to the second latest post, though I could be wrong.

I've just noticed this on several of the college playground forums. When I click on the link in the full forum list, I'm sent to the second-newest thread in the same subforum as the thread allegedly linked to.

It definitely isn't fixed yet - it linked me to the age topic instead of Mafia game 21 in GFF.

Yeah clearly buggy. Let me see what is going on  :maybe: