Εστω $ x,y$ ακεραιοι.Να λυσετε την εξισωση
$ (x\plus{}2)^4\minus{}x^4\equal{}y^3$
Εχω μια λυση αλλα πιστευω οτι θα υπαρχει μια αρκετα απλουστερη.....

[quote="Dimitris X"]Εστω $ x,y$ ακεραιοι.Να λυσετε την εξισωση
$ (x \plus{} 2)^4 \minus{} x^4 \equal{} y^3$3
Εχω μια λυση αλλα πιστευω οτι θα υπαρχει μια αρκετα απλουστερη.....[/quote]

Με λιγες πραξεις το αριστερο μελος γραφεται: $ 2^3(x^2 \plus{} x \plus{} 2)(x \plus{} 1) \equal{} y^3$, αρα $ (x^2 \plus{} x \plus{} 2)(x \plus{} 1) \equal{} z^3$ για καποιο ακεραιο $ z$, οι αριθμοι στις παρενθεσεις ειναι σχετικα πρωτοι αφου ο μεγιστος κοινος διαιρετης τους διαιρει τον $ (x\plus{}1)^2$ και τον $ x^2 \plus{} 2x \plus{} 2 \equal{} (x\plus{}1)^2 \plus{} 1$ αρα και τη διαφορα τους που ειναι ιση με $ 1$, αρα για καποιους ακεραιους $ a,b$ εχουμε: $ x^2 \plus{} x \plus{} 2 \equal{} (x\plus{}1)^2 \plus{} 1 \equal{} a^3$, και $ x \plus{} 1 \equal{} b^3 \Rightarrow (x \plus{} 1)^2 \equal{} c^3, c\equal{}b^2\geq 0$ οποτε $ 1 \equal{} (a \minus{} c)(a^2 \plus{} ac \plus{} c^2)$ απο οπου ευκολα εχουμε: $ c\equal{}0$ και $ a\equal{}1$, αρα $ x\equal{}\minus{}1$, και $ (x,y) \equal{} (\minus{}1, 0)$.


what?? then you are saying that 2ab*Sqrt(6) is an integer, which I don't think it is.... :?

[quote="ttd"]Find the smallest integer $N$ satisfying: for every integer $n \geq N$ there exist
positive integers $a,b$ satisfying: $n=[a\sqrt{2}+b\sqrt{3}]$.where [x] denotes the
integer part of x

[color=red]Edited by Megus: I've changed the wording to make it more readable[/color][/quote]


Determine all functions f(x) which are defined for x <> 0 and satisfy tue condition f(x) - 3f(1/x) = 3^x.

I feel sorry to post here but I am really a beginner in everything accept of plane euclidean geometry :-(

But this one was really simple: Write down the equation for x, then the one for 1/x, multiply the second one by 3 and add it to the first one, and you get

  -8 f(x) = 3^x + 3 * 3^(1/x),

what yields

   f(x) = - (3^x + 3 * 3^(1/x)) / 8.

Now it is simple to check that the equation is fulfilled.

At an IMO training exam this year, we had a similar problem:

A function f is given by the implicite condition f(x) + f(1 - 1/x) = 1 + x for x element of R \ {0; 1}.
Find an explicite formula for f.

  Darij


Apologies if this has already been asked, but I didn't find such a topic upon searching this forum:

I would like to teach myself probability theory and am interested in well-regarded books in this field, and would be greatly interested in any recommendations y'all might provide.

Possible ways to be more specific if there are a ton of books out there or something and you need a better idea of what I'm looking for - I have a strong background in analysis, e.g. measure theory, so I would most likely be interested in something at least at that level. (I would suppose most books would make use of measure theoretic concepts, but just to be specific.) Of course, I would like a text that is fairly mathematically rigorous, but I'll probably need to study this subject just as much for other fields (economics, statistical physics ?) as for my own mathematical edification.

Thank you in advance.

One of the classic textbooks on probability is William Fellers [url=http://www.amazon.com/Introduction-Probability-Theory-Its-Applications/dp/0471257095/ref=sr_1_1?ie=UTF8&s=books&qid=1206242361&sr=1-1]An Introduction to Probability Theory and Its Applications[/url].

For a more axiomatic treatment try [url=http://www.amazon.com/Course-Probability-Theory-Revised/dp/0121741516/ref=pd_bxgy_b_text_b]A Course in Probability Theory[/url] by Kai Lai Chung or [url=http://www.amazon.com/Theory-Probability-Random-Processes-Universitext/dp/3540254846/ref=sr_1_1?ie=UTF8&s=books&qid=1206242791&sr=1-1]Theory of Probability and Random Processes[/url] by Koralov and Yakov Sinai.

as a graduate student in probability I would recommend Rick Durrett's [url=http://www.amazon.com/Probability-Theory-Examples-Richard-Durrett/dp/0534243185]Probability: Theory and Examples[/url], it's the clearest measure-theoretic-probability text I've read (and i've gone through most of it now so i feel pretty comfortable saying that :) ).  it seems to be one of the most popular texts for graduate courses in probability these days, and has lots of (fairly interesting) exercises.

i would probably stay away from Feller and Breiman - both texts used to be very popular but seem very dated in my experience.

if you really like the analytic side of things you might glance through [url=http://www.amazon.com/Probability-Theory-Analytic-Daniel-Stroock/dp/0521663490/ref=pd_sim_b_img_23]Stroock[/url], but it seems brutally hard for a first go at the subject in my opinion.

i think billingsley and chung are somewhat common texts as well, but i don't have any experience with them.

In a third semester intro course (at the second undergraduate level, there are four) to probability we used grimmet and stirzaker Probability and random processes. I thought it was readable and pretty good, but I haven't read anything else in the subject.

it seems like a nice text, but there isn't much measure theory.  incidentally i am taking [url=http://www.pims.math.ca/science/2008/08ssprob/courses.html]a course[/url] from grimmett this summer.

on the plus (or minus?) side it has a solution manual.


Sa se calculeze :
$I_n =\int\limits_{\Gamma} { (y^{2}-z^{2})dx+(z^{2}-x^{2})dy+(x^{2}-y^{2})dz}$ unde $\Gamma$ reprezinta intersectia planului$x+y+z=\frac{3a}2$ cu cubul $0 \lneq x,y,z \lneq a$
direct si cu formula lui Stokes.

Dau rezolvarea pentru intersectia planului $x+y+z=\frac{3}2$ cu cubul $\ [0,1]^{3}$, cealalta facandu-se in mod analog. Conform formulii Stokes, $I=\int\limits_{\Gamma} { (y^{2}-z^{2})dx+(z^{2}-x^{2})dy+(x^{2}-y^{2})dz}$ = $\int\limits_{\Sigma} { (-2y-2x)dx}$^${dy + (-2y-2z)dy}$^${dz + (-2z-2x)dz}$^${dx}$, unde $\Gamma$ este bordul suprafetei $\Sigma$ (din desen se poate observa ca intersectia cubului cu planul respectiv este un hexagon regulat, asadar $\Gamma$ este hexagonul, iar $\Sigma$ - hexagonul "plin").
Folosim acum parametrizarea:
 $X=x , Y=y , Z=\frac{3}2-x-y$
Rezulta $I = -2\int\int\limits_{D} { [(x+y) + (y+\frac{3}2-x-y) + (x+\frac{3}2-x-y)]dxdy } = -3A_{D}$, unde D este domeniul pe care oscileaza (x,y), iar $A_{D}$ este aria proiectiei hexagonului pe planul (XOY). In final, obtinem $I=-\frac{9}4$.

Stiu ca'i intrebare de n00b dar ... am cautat mai prin multe locuri insa la stockes nu am gasit cum sa calculez dx ^ dy , dy ^ dz si respectiv dz ^ dx. La intregala tripla stiu ca se calculeaza cu iacobieni insa acolo determinantii sunt D(X,Y)/D(u,v) . Aici cum se procedeaza ?!?! caci aceia's parametri sferici?!?!
Merci in avans


[quote="jensen"]kasi hast ke raje be khavase adade haghighi chizi bedoone?[/quote]
yani chi? chejoor khavasi ro migi?? ;)

like this :
if $x,y,z$ are the real numbers then we have:
$x+(y+z)=y+(x+z)$ and ....
something that we can't proove them and they are our osoole.

pas donbale "axiom"hayi hasti ke dar jabr stefade mishand? Axioms be chizhayi migan ke kheli vaze hastand, manande $1+1=2$ ya $1=1$.

Masoud Zargar

fekr nakonam "axiom" ha oonghadr ha ham ke to fekr mikoni vaseh bashan masalan inke \[ xy=yx \] aslan vaseh nist ama chon ma azash estefade mikonim fekr mikonim vaseh ast.anywaY kesi in "axziom" ha ra midoone? :D

man ina ro midoonam va fekr mikonam ke dige baraye Q,R chizi tarif nashe.mesle inke baraye C ham hamina hast ba chan ta asle mozooe dige.

1.number+number=number
2.a+(b+c)=(a+b)+c
3.a+b=b+a
4.(There is something like zero): a+0=a
5.$\forall x \exists -x$ : x+(-x)=0
6.number*number=number
7.a(b.c)=(a.b)c
8.ab=ba
9.(There is something like one): 1.a=a
10.$\forall a \exists a^{-1}$ : a.a^-1=1
11.a(b+c)=ab+ac

thanks alot  mahmood to be man kheili komak kardi,vaghean mamnoon. :D

ketabe analiz e apostol ro bekhoon.moshkelet hal mishe. ;)  :D

merci kheili mamnoon az lotfe hamatoon. :D


A simple pendulum with length $l$ hang on a car.
The car accelerate horizontally with acceleration $a$
Can the bob do SHM?
If yes,What will the period be?

It's the same situation as if the pendulum was in a grav field with grav acceleration $\sqrt{g^2+a^2}$, so the period will be

$T=2\pi\sqrt{\frac{L}{\sqrt{g^2+a^2}}}$,

I think.


[quote="joml88"]Uh, where's Opera?  I know there are some devoted Opera users on this site who'll get a little ticked :D  I use mozilla firefox though.[/quote]
Opera? Is that the one Myth recommended in his signature?

Yes.  There are some other people who  really like it as well.

[quote="MysticTerminator"]what who voted for IE ... it's like half the votes right now[/quote]IE's all I've ever used... Why does everyone seem to dislike it so much?  :?

I use firefox. It simply features most settings. ^^ (about 600 times as much as IE, I guess)

Yea, I use firefox, unless there's a website that won't work the right way on it.

i use IE because it comes with windows and i'm too lazy to go seeking other web browsers. it satisfies my needs pretty well, i just sometimes feel like it should have tabs... otherwise that's about it!

firefox here

all you IE users don't know what you're missing out on.

Seems most math people like firefox, it is the better browser.

I'm too used to IE to use firefox a lot, but i AM using it right now.

Never used netscape much.

IE bugs up certain things, but i like the layout, its comfy.

I use Firefox, but sometimes you need IE to show some pages correctly.

Definitely Firefox here. It's by far the best browser in terms of security and with all of the extensions (get IE view if it only renders correctly in IE) it makes everything faster. I currently have 14 AoPS tabs open with a few Amazon and Googles. I have to say, tabular is the way to be. Opera is ok, but it has no plugins.

firefox all the way!
although I've had problems lately... firefox won't load most flash animations...

I use firefox and internet explorer, depending on what I feel like.


I reicently did a math contest of the university of McMaster. I have four questions that I get stuck on.... I'd appreciate a soultion or some help or hints on how to solve these:

1. Whats the next number in the sequence?
$7, 33, 29, 59, 51, 86, 73,_{_}_$

-- I use finite differences but it gets me nowhere...

2. Let P be a point inside a square S so that the distances from P to the four vertices (corners), in order, are 7, 35, 49, $x$. What is $x$?

-- No clue on where to begin

3. A father in his will left all his money to his childern in the following manner:

1000 dollars to the first born and $\frac{1}{10}$ of what then remains, then
2000 dollars to the seccond born and $\frac{1}{10}$ of what then remains, then
3000 dollars ............

When he was done, each child had the same ammount. How many childern were there?

-- by guessing randomly i got to 402.. which gets me to a few cents off but cant get any closser...

4 Two people stand back to back next to the rails in a small railway station. As the head of the express train (which does not stop at the station) reaches them, they start to walk in opposite directions parallel to the rails. As the tail of the train reaches each of them, they stop, having walked 30m and 40m respectively. If they both walked with idenctical, constant spreeds and the trains speed was constant, how long was the train?

-- I got my answer as 240m this was done by calculating the speed of the train and then the speed and time that the persons walked at. Im not sure its right...

Any hits/help would be appreciated.

[quote="Jeremic"]I reicently did a math contest of the university of McMaster. I have four questions that I get stuck on.... I'd appreciate a soultion or some help or hints on how to solve these:

1. Whats the next number in the sequence?
$7, 33, 29, 59, 51, 86, 73,_{_{}}_$

-- I use finite differences but it gets me nowhere...

2. Let P be a point inside a square S so that the distances from P to the four vertices (corners), in order, are 7, 35, 49, $x$. What is $x$?

-- No clue on where to begin

3. A father in his will left all his money to his childern in the following manner:

1000 dollars to the first born and $\frac{1}{10}$ of what then remains, then
2000 dollars to the seccond born and $\frac{1}{10}$ of what then remains, then
3000 dollars ............

When he was done, each child had the same ammount. How many childern were there?

-- by guessing randomly i got to 402.. which gets me to a few cents off but cant get any closser...

4 Two people stand back to back next to the rails in a small railway station. As the head of the express train (which does not stop at the station) reaches them, they start to walk in opposite directions parallel to the rails. As the tail of the train reaches each of them, they stop, having walked 30m and 40m respectively. If they both walked with idenctical, constant spreeds and the trains speed was constant, how long was the train?

-- I got my answer as 240m this was done by calculating the speed of the train and then the speed and time that the persons walked at. Im not sure its right...

Any hits/help would be appreciated.[/quote]
for number 2, draw a diagram and draw the horizontal and verrtical lines going though P.   The let the side be S and lable other distances

[hide="For #3"]If there were $x$ dollars, you have the equation for the first two kids:

$1000+{x-1000\over 10}=2000+\frac{x-\left(1000+{x-1000\over 10}\right)-2000}{10}$

Solve it, find out how many dollars each child received and divide to get the number of children.[/hide]

Your answer for #4 is correct.

[quote="Jeremic"]What do you mean.. like a cross trough point P? I tried that, but im not sure what you're getting at since there are no angles in trangles created.

[url=http://imageshack.us][img]http://img182.imageshack.us/img182/1699/squarewf0.jpg[/img][/url][/quote]
When you do that you make a bunch of right triangles, and right triangles mean pythagorean theorem.

[quote="Jeremic"]1. Whats the next number in the sequence?
$7, 33, 29, 59, 51, 86, 73,_{_{}}_$
[/quote]
try splitting it up into two different sequences: $7,29,51$ and $33,59,86$. Now, use finite differences.

[quote="13375P34K43V312"]Well we can't really say that, because we aren't given the order of the lengths. We can't find a unique answer, but we can find a finite number of possible answers. (I believe 3)[/quote]
[quote="Jeremic"]in order, are 7, 35, 49, [/quote]

Thanks for the help guys!!! Btw, thats an aswome thoerem!

[quote="bpms"][quote="jli"][quote="fireemblem13"][hide]how about use the british flag theorem(name?).
$7^{2}+49^{2}=35^{2}+x^{2}$[/hide][/quote]

there seriously is a theorem called that? :rotfl:

pretty useful[/quote]
Yes, it is a real theorem. It can be derived easily by using pythagorean.[/quote]

oh yeah...i see how it works now.

funny signature bpms

[quote="Altheman"]remember that the british flag theorem if for rectangles, the problem uses the word: square, hence, you must choose the ordering of the distances that produces a square[/quote]
All squares are rectangles.

[quote="jli"][/quote][quote="bpms"][quote="jli"][quote="fireemblem13"][hide]how about use the british flag theorem(name?).
$7^{2}+49^{2}=35^{2}+x^{2}$[/hide][/quote]

there seriously is a theorem called that? :rotfl:

pretty useful[/quote]
Yes, it is a real theorem. It can be derived easily by using pythagorean.[/quote][quote="jli"]

oh yeah...i see how it works now.

funny signature bpms[/quote]
What is so funny about my signature?


In $\triangle{ABC}$, $\angle{C}=90^o$. The incircle of triangle ABC touches $AB$ at $D$.
Prove $AD \cdot BD = S_{ABC}$

Have you ever seen this problem before?

Using standard notation, $AD\times BD = (s-b)(s-a)$ and $S_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}$, so it is sufficient to prove that $(s-b)(s-a)=s(s-c)$.  Now, $4(s-b)(s-a)=(c+a-b)(c+b-a)=c^2-(b-a)^2$, and $4s(s-c)=(a+b+c)(a+b-c)=(a+b)^2-c^2$.  This is a right triangle, so $a^2+b^2=c^2$, and hence $4(s-b)(s-a)=4ab=4s(s-c)$ as required.

Cute, and new to me.

Here is my solution.

AB = x+y, BC = y+z, AC = x+z.

$(x+y)^2 = (x+z)^2 + (y+z)^2$

$xy = z^2 + yz + xz$

$2xy = z^2 + yz + xz + xy$

$xy = \frac{1}{2} (x+z) (y+z)$

$AD \cdot BD = S$

Still, I want to know if anyone has any other proofs? Also, has anyone seen this problem before?


number theory
Diophantine equation
number theory proposed

[quote=marko avila]this diophantine equation has its level of difficulty because  it needs a lot of very ugly and enormous computations to solve it!! and at the end (if you solve it) it has a one line solution :D  here it is:

prove that the equation:

  $x^3+y^3=z^4-t^4$

has infinitely many solutions  with $mcd(x,y,z,t)=1$[/quote]

I really don't understand some bit of this problem. like if you take $y = -y$ then the problem become like this $x^3 - z^4 = y^3- t^4 $. can't we simply take $x=-y  and z=t$


Find the number of roots of
\[ \sin \pi x \equal{} x^2  \minus{} x \plus{} \frac{5}
{4}
\]

[hide]$ x^2\minus{}x\plus{}{5\over 4}\leqslant 1\iff x^2\minus{}x\plus{}{1\over 4}\leqslant 0\iff \left(x\minus{}{1\over 2}\right)^2\leqslant 0\iff x\equal{}{1\over 2}$

Checking holds, so it's the only solution.[/hide]


I made one in another topic but nobody participated. Let's try and see if GFF has cooler people  :lol:

By now you all should know about Sudoku and the rules but for the benefit of those who haven't I'll explain them.
You can't repeat numbers in a square (which I'll leave a space in between) or in a row or column. I'll start off.

??? ?41 ??3
2?? 3?? 14?
??? ??? 59?

??? 6?? 4??
?8? 794 ?5?
??4 ??3 ???

?67 ??? ???
?25 ??7 ??6
9?? 81? ???

Please solve then post a new one.


function
calculus
integration
real analysis
real analysis solved

Find $f:I\to(0,\infty)$ continuous and $I\subset \mathbb{R}$, if
$f(0)=1$ and $\frac1f\in\int f(x)dx$

What exactly do you mean by "$\in$"? I read that as "is an element of", which doesn't make sense here.

The differential equation $\left(\frac1f\right)^{\prime}=f$ is equivalent to $f^{\prime}(x)=-(f(x))^3,$ which can be routinely solved to get $f(x)=\pm\frac1{\sqrt{a+2x}}$, where $a$ is a yet-to-be-determined constant.

Given $f(0)=1$, we have $a=1$ and take the positive sign; therefore $f(x)=\frac1{\sqrt{1+2x}}$, which is defined for all real $x>-\frac12$.


induction
complex analysis
complex analysis unsolved

Hi

I have a mathproblem I can't solve.....

[color=magenta]
Explain to a skeptical person why the fact that a statement is true for the number n = 0 toghether with the fact that it is proven that if this is true for the number n = p, then it is also true for n = p+1 and it means that the statement is true for n = 100.
[/color]

I don't have a clue how to solve this and I need to learn this before my big exam....

I would be grateful if anyone could explain this for me in a way that I can understand it..

Thanks

This is the [url=http://en.wikipedia.org/wiki/Mathematical_induction]principal of mathematical induction[/url].  It states that if a statement is true for some base case, then you can assume it is true for $ n$ and use that to prove it is true for $ n\plus{}1$, where $ n$ is a positive integer.  Think of it like dominoes: If you can make the first one fall into the second, the second into the third, the third into the fourth, and so on, you can knock down all of the dominoes.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Find the finite groups $G$ with the property that the cardinalities of the conjugacy classes of $G$ are precisely the divisors of $|G|$ (smaller than $|G|$), each taken precisely once.

Well, I have an idea: |G| has to be a perfect number, because it is the sum of all smaller divisors of itself (since G is the disjoint union of its conjugacy classes).
|G| must be even, because there I say so  :wink:  :wink:
So $|G|=2^{n-1}\cdot(2^{n}-1)$ for some n s.t. $2^{n}-1$ is prime. But I have no idea how to to prove that or how to find the groups.

ok, we got it..

let me denote the conjugacy class of $g$ with $C_{g}$, and it's centralizer with $Z_{g}$ (for every group we're using), and let me call $p$ the smallest prime dividing the order of $G$.

observe that there exist $g$ such that $|C_{g}| = p$, so $G$ has a subgroup of index $p$, $H$, which is normal. since $G/H$ is abelian, $C_{\pi(g)}= \pi(C_{g})$ ($\pi$ is the projection of $G$ onto $G/H$), so in particular there exist $k\in G$ such that $C_{k}= kH$. $|Z_{k}| = p$, and so $Z_{k}= (k)$. but $Z_{k}\subset Z_{k^{2}}$, and if $p\neq 2$, $Z_{k}= Z_{k^{2}}$, but then $|C_{k}| = |C_{k^{2}}|$, and that's an undesired thing :P .

so $p=2$, and $|G|$ is an even perfect number: $|G| = 2^{q-1}(2^{q}-1)$. let $r: =2^{q}-1$. let $x$ be an element of order $r$, which generates a $r$-sylow $N$, which is normal (for a matter of cardinality, by sylow theorem). there's a $2$-sylow $T$, and $TN = G$, so $G = N \rtimes_\varphi T$, where $\varphi: T \to \text{Aut }N \cong \mathbb{Z}/(r-1)\mathbb{Z}$. but $T$ is a 2-group, and the only elements of order a power of two in $\text{Aut }N$ are identity and the map $\phi_{g}: h \mapsto h^{-1}$.
but identity doesn't work, since $|C_{(h,y)}| = |C_{(h',y)}|$ for any $h,h' \in N, y\in T$.
now, let's consider $\phi$. if $r>3$, there exist $h' \neq h,h^{-1},e \in N$, and it's easy to check that (= i'm almost totally sure that) $|C_{(h,y)}| = |C_{(h',y)}|$ for any $y\in T$.
so $r=3$, and it's easy to check (= i've checked it and i'm sure about it) that $S_{3}$ is working.

a special thank to alex_zeta and jack_daniel's for solution of this problem...

[size=75][[color=red]EDIT[/color]: some bugs fixed, something made more clear... i've just changed most of the proof :p ][/size]


$ P(x) \equal{} a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}ax\plus{}a_{0}$
Let $ P(x)$ be a 5th degree polynomial,each coefficient can only take one value from $ 1$ to $ 30$

what values can the roots of this polynomial take (between what numbers)? i mean is it possible to say the polynomial's roots can only take values between $ z$ and $ t$.(because the coefficients are known).

All [i]real, rational[/i] answers are in the the range of $ \minus{}30$ to $ \plus{}30$. I am not entirely sure if that restriction applies to the imaginary plane of radius 30 about the origin, nor am I sure if that applies to irrational roots (although I think it does). This can be proven by Rational Zeros Test in which all positive and negative factors (not just prime factors) of $ a_{0}$ (called "$ p$") and the same is taken of $ a_{n}$ (called "$ q$") are taken $ p/q$ for all possible positive and negative answers that are both real and rational.


We have positive integer $n$ and the set $M$ which has $n^{2}+1$ positive integers.This set has the following property:For each $n+1$ different numbers chosen from $M$ there exist pair $a_{k},a_{l}$ such that $a_{k}$ divides $a_{l}$.Prove,that in set $M$ there exist different numbers $b_{1},b_{2},\ldots,b_{n+1}$such that $b_{i}$ divides $b_{i+1}$ for $i=1,2,\ldots ,n+1$.

Hmm... I don't think this is Polish MO 1999.

http://www.mathlinks.ro/Forum/viewtopic.php?p=198301#p198301


If $x=-y \mod k-1$ and $x=-2y \mod k$, what, if anything, can you conclude about x, y, and k?

use $\equiv$ when dealing with mods..

Im not too hot on this topic but heres my shot
[hide]
from $x\equiv-y \bmod{k-1}$ we can conclude that:
$k-1|x-(-y) \Rightarrow k-1|x+y$

And from $x\equiv-2y \bmod{k}$:
$k|x+2y$

from here I have a hunch theres something I can figure out, but like I said I suck at number theory stuff.. for now :gleam:
[/hide]

[hide="continuing off of Glen"]
since $k$ and $k-1$ are consecutive, one of either $x+y$ or $x+2y$ must be even, since an even number can divide only even numbers

Looking at cases:
x must be even
or
x and y must both be odd

so what cannot happen is:
$y$ is odd while $x$ is even

not sure what else we can get
[/hide]

I'm pretty sure this works to find $y$:
[hide]$\text{Let A}=$
$x \equiv-y \mod (k-1)$
$x \equiv-y-1 \mod k$

$\text{Let B=}$
$x \equiv-2y \mod k$

$-2y=-y-1$
$-y=-1$
$\boxed{y=1}$

EDIT: Also $x \equiv-2 \mod k$ so $x$ is $2$ less than $k$.  That is, $x=k-2$.

If you look at the step I did in A, I'm pretty sure it works.  For example:
$3 \equiv 1 \mod 2$
$3 \equiv (1-1) \mod (2+1)$
$3 \equiv 0 \mod 3$[/hide]

$x \equiv-2 \mod k$ does not mean that $x=k-2$.  What it means is that $x-(-2)$ is some multiple of $k$, say $nk$.  So that $x+2 = nk$.  The method that you used does not hold.
 Suppose $a \equiv b \mod k$. Then $a-b = nk$ or $a = nk+b$ for some integer $n$.
But $a \equiv b-b \mod (k+b) \implies a-0 = (k+b)n$. These two equations contradict each other:
$n(k+b) \neq nk+b$.
Example:
$17 \equiv 1 \mod 8 \implies 17-1 = 8n$. Which is true.
But $17 \equiv 1-1 \mod (8+1) \rightarrow 17 \neq 9n$.

Oh okay, I see.  I'm pretty new on this whole number theory / mod thing but i figured it may work.


function
LaTeX
integration
trigonometry
calculus
calculus computations

Hey guys, this might be an easy problem for you but i am not sure where to start.  Please help.

Let k(t), t>=0, be a continuous function with k>=0. Prove that the curve

a(t) = [(integral(cos(integral k(s) ds,0,u)du,0,t),(integral(sin(integral k(s) ds,0,u)du,0,t)]

has at the moment t, the curvature k(t) and satisfies  the initial condition
a(t)=(0,0) and a'(t)=(1,0)

Please help, i am completely lost.  [/code]

We use latex here (the tags are dollarsigns, so use $\$\ldots\$$ to typeset latex).

[quote="deftone"]Hey guys, this might be an easy problem for you but i am not sure where to start.  Please help.

Let $k(t), t\geq 0$, be a continuous function with $k\geq 0$. Prove that the curve
\[\vec{a}(t) = \left< \int_{0}^{t}\cos\left( \int_{0}^{u}k(s) ds\right) du, \int_{0}^{t}\sin\left( \int_{0}^{u}k(s) ds\right)\right> \]
has at the moment $t$, the curvature $k(t)$ and satisfies  the initial condition
$\vec{a}(t)=<0,0>$ and $\vec{a}^\prime (t)=<1,0>$

Please help, i am completely lost.[/quote]

Thank you fot that, i wasnt sure what the hell to do or whether it was legible.

[quote="deftone"]Hey guys, this might be an easy problem for you but i am not sure where to start.  Please help.

Let $k(t), t\geq 0$, be a continuous function with $k\geq 0$. Prove that the curve
\[\vec{a}(t) = \left< \int_{0}^{t}\cos\left( \int_{0}^{u}k(s) ds\right) du, \int_{0}^{t}\sin\left( \int_{0}^{u}k(s) ds\right)\right> \]
has at the moment $t$, the curvature $k(t)$ and satisfies  the initial condition  $\vec{a}(0)=<0,0>$ and $\vec{a}^\prime (0)=<1,0>$
Please help, i am completely lost.[/quote]
Denote $\lambda(u)=\int_{0}^{u}k(s)\; ds$ and assume that the (plane) curve has the (vectorial) equation
\[\vec{a}(t) = \left< \int_{0}^{t}\cos{\lambda(u)}\; du , \int_{0}^{t}\sin{\lambda(u)}\; du \right>: =\left< X(t) ,Y(t)\right>\; . \]
Then the curvature at moment $t$ is  ${{\tau(t)= \frac{\parallel \vec{a}^{\; \prime}(t)\times \vec{a}^{\; \prime\prime}(t)\parallel }{{\parallel \vec{a}^{\; \prime}(t)}\parallel^{3}}}}=\frac{|X^{\prime}(t)Y^{\prime\prime}(t)-Y^{\prime}(t)X^{\prime\prime}(t)| }{\left( X^{\prime}^{2}(t)+Y^{\prime}^{2}(t)\right)^{\frac{3}{2}}}=...$

[size=75]
[b]Remark:[/b] Because many engineers used curvatures, in your spirit,in many books there  is considered that $\tau(t): = \frac{X^{\prime}(t)Y^{\prime\prime}(t)-Y^{\prime}(t)X^{\prime\prime}(t) }{\left( X^{\prime}^{2}(t)+Y^{\prime}^{2}(t)\right)^{\frac{3}{2}}}\; .$

A possible reason is that  engineers consider that if  $\vec{r}=a\cdot\vec{z}$, then $\parallel \vec{r}\parallel =a\cdot \parallel \vec{z}\parallel$ instead $\parallel \vec{r}\parallel =|a|\cdot \parallel \vec{z}\parallel$ .
[/size]


Is there a method for  finding the determinant of this 4x4 matrix?

\[ \left(
\begin{array}{cccc}
1 & \minus{}2 & 0 & 0\\
1 & \minus{}1 & 0 & 0 \\
0 & 0 & 5 & \minus{}3\\
0 & 0 & 3 & \minus{}1\\
\end{array}
\right)\]

Yeah; in the case where a matrix is of the form
\[ \begin{bmatrix} A & 0 \\
0 & B \end{bmatrix}
\]
, (where $ A, B$ are square matrices and the two $ 0$'s are zero-matrices of the right dimension to make the overall matrix square... you know what that means) then the determinant is simply $ \det{A}\det{B}$.

If you'd like a proof of this, let me know :)


http://web.mit.edu/hmmt/www/datafiles/solutions/2006.shtml

In the solution to problem 2, I don't see how $ 2AB\equal{}BC$.

This is geometry, right? Anyway, use the Angle Bisector Theorem.

Since $ <ABP\equal{}45^{\circ}$ and $ <ABC\equal{}90^{\circ}$, we know that $ AP$ bisects $ <ABC$.  Then, just use the angle bisector theorem.


Hi everyone,

Could you please tell me if this is correct?

lim as t goes to infinity [e^-2t(t/-2-1/4)-.034]=

-.034

I think this is the correct answer, but I'm not sure why it would go to -.034. The first piece is going to infinity, so it's infinity -.034. Wouldn't that just be infinity?

Thank you very much

[quote="chocolatelover"]The first piece is going to infinity[/quote]

Are you sure?  :)  As you have written it (it is somewhat unclear), the first term is

$ e^{\minus{}2t} \left( \frac{t}{\minus{}2} \minus{} \frac{1}{4} \right)$

which actually goes to $ 0$ (the $ e^{\minus{}2t}$ factor goes to $ 0$ much faster than the linear factor goes to $ \infty$).

Learn [[LaTeX]] (at that link and [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690[/url]).  Until you learn latex, learn how to use parentheses so as to make what you write comprehensible to others.


A triangle has sides $ b=4,\ c=7$. Given that its area is $ 6\sqrt 5$, find all possible values for the length of $ a$


Do you think graphing is your favorite subject of algebra?

No way.

I don't mind graphing actually, though it's a little tedious. Once you know calculus it's not so bad.

Also, by the way, [hide="BBCode/emoticon abuse"]this is the Algebra Unsolved [b][i][u][size=200][color=red]Problems[/color][/size][/u][/i][/b] forum.   :o :o :o  :rotfl:  :rotfl:  :rotfl: [/hide]

But seriously, dude. Use the right forum.


geometry
3D geometry
sphere
vector
geometric transformation
homothety
ratio

Let $A_1,\dots,A_n$ be $n$ points in the space, and let $P_{ij}$ the plane normal to the line $A_iA_j$ and  passing by the barycenter of the other $n-2$ points .

Prove that the intersection of all the $P_{ij}$ planes is non empty if and only if the $n$ points are on a sphere .

You can generalize in a $m$-dimensional space :D

 :cool:   Do you know Monge  :cool:

Suppose that the $n$ points belong to a sphere $S$ with center $O$. Let $G$ be the isobarycenter of the $n$ points, thus as vectors : $OG = n\sum_{i=1}^n OA_i$.
Let $G_{i,j}$ be the isobarycenter of the $n-2$ points others than $A_i$ and $A_j$.
Let $P_{i,j}$ be the plane perpendicular to the line $A_iA_j$ through $G_{i,j}$.
Let $M_{i,j}$ be the midpoint of the segment $A_iA_j$.

Let $X$ be the point such that $OX = \frac n {n-1} OG$ (vectors)
We will prove that $X$ is a common point of all the planes $P_{i,j}$.

We have (as vectors) : $OG_{i,j} = \frac 1 {n-2} \sum_{k \notin \{i,j \}} OA_k$
Thus $XG_{i,j} = - \frac 1 {n-2} {i=1}^n OA_k + \frac 1 {n-2} \sum_{k \notin \{i,j \}} OA_k = - \frac 1 {n-2} (OA_i + OA_j) = - \frac 1 {n-2} OM_{i,j}$.

Since $A_i,A_j \in S$, it follows that the vector $OM_{i,j}$ is orthogonal to the vector $A_iA_j$. Thus, $X$ belongs to a line through $G_{i,j}$ and orthogonal to the line $A_iA_j$, so that $X \in P_{i,j}$, and we are done.

Note that this proof can be generalized to higher dimensions.

Anyone for the converse?

Pierre.

I will use the word [b]centroid[/b] instead of the words [b]barycenter[/b] and [b]isobarycenter[/b].

Then the problem from Diogene reads:

[i]Let $A_1$, $A_2$, ..., $A_n$ be n points in the space. For any two i and j with $1\leq i\leq n$, $1\leq j\leq n$ and $i \neq j$, let $P_{ij}$ be the plane perpendicular to the line $A_iA_j$ and passing through the centroid of all the points $A_1$, $A_2$, ..., $A_n$ except of the points $A_i$ and $A_j$.

Prove that all the planes $P_{ij}$ have a common point if and only if the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere.[/i]

Ah yes, this one is really easy. In fact, we begin with a preliminary.

For any two i and j, let $M_{ij}$ denote the midpoint of the segment $A_iA_j$. Furthermore, let $S_{ij}$ denote the plane through the point $M_{ij}$ perpendicular to the line $A_iA_j$, i. e. the perpendicular bisector plane of the segment $A_iA_j$. This plane is the locus of all points P satisfying $PA_i = PA_j$. Using this property, we can easily show:

[b]Lemma 1.[/b] All the planes $S_{ij}$ have a common point if and only if the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere.

[i]Proof.[/i] At first, we will prove one direction: If all the planes $S_{ij}$ have a common point, the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere.

In fact, if all the planes $S_{ij}$ have a common point, then we can denote this common point by O. Since the plane $S_{ij}$ is the locus of all points P satisfying $PA_i = PA_j$, and since our point O lies on all such planes, we have $OA_1 = OA_2 = ... = OA_n$. Therefore, the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere, namely on the sphere with center O and radius $OA_1 = OA_2 = ... = OA_n$.

Now, we will prove the converse direction: If the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere, then the planes $S_{ij}$ have a common point.

In fact, if the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere, then, denoting the center of this sphere by U, we have $UA_1 = UA_2 = ... = UA_n$. Since the plane $S_{ij}$ is the locus of all points P satisfying $PA_i = PA_j$, this entails that the point U lies on all planes $S_{ij}$. Hence, all the planes $S_{ij}$ have a common point, namely the point U.

This completes the proof of Lemma 1.

Now, for any two i and j, let $G_{ij}$ denote the barycenter of all of the points $A_1$, $A_2$, ..., $A_n$ except for the two points $A_i$ and $A_j$. Then, the plane $P_{ij}$ is the plane through the point $G_{ij}$ perpendicular to the line $A_iA_j$. We have to prove that all the planes $P_{ij}$ have a common point if and only if the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere.

Let G denote the centroid of all n points $A_1$, $A_2$, ..., $A_n$, and let h be the homothety with center G and factor $\displaystyle - \frac{n-2}{2}$.

We have the following constellation of points:

- The point $M_{ij}$ is the centroid of the two points $A_i$ and $A_j$.
- The point $G_{ij}$ is the centroid of all of the points $A_1$, $A_2$, ..., $A_n$ except for the two points $A_i$ and $A_j$.
- The point G is the centroid of all n points $A_1$, $A_2$, ..., $A_n$.

Hence, a well-known property of centroids yields that the point G lies on the line $M_{ij} G_{ij}$ and divides the segment $M_{ij} G_{ij}$ in the ratio $\displaystyle \frac{M_{ij}G}{GG_{ij}} = \frac{n-2}{2}$, with signed distances. Hence, $\displaystyle \frac{GM_{ij}}{GG_{ij}} = - \frac{M_{ij}G}{GG_{ij}} = - \frac{n-2}{2}$. But this entails that the point $M_{ij}$ is the image of the point $G_{ij}$ in the homothety with the center G and the factor $\displaystyle - \frac{n-2}{2}$, i. e. in the homothety h. Now, the image of the plane $P_{ij}$ in this homothety must be a plane which

[b]1.[/b] is parallel to the plane $P_{ij}$ (since a homothety maps every plane to a parallel plane), and thus perpendicular to the line $A_iA_j$ (since the plane $P_{ij}$ is perpendicular to the line $A_iA_j$);
[b]2.[/b] passes through the point $M_{ij}$ (since the plane $P_{ij}$ passes through the point $G_{ij}$, and this point $G_{ij}$ is mapped to the point $M_{ij}$ by the homothety h).

Hence, the image of the plane $P_{ij}$ in our homothety h must be the plane through the point $M_{ij}$ and perpendicular to the line $A_iA_j$. In other words, the image of the plane $P_{ij}$ in our homothety h is the plane $S_{ij}$. This must hold for all i and j, remember! Since homotheties preserve incidence, it follows that all the planes $P_{ij}$ have a common point if and only if all the planes $S_{ij}$ have a common point. But we know from Lemma 1 that all the planes $S_{ij}$ have a common point if and only if the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere. Hence, all the planes $P_{ij}$ have a common point if and only if the points $A_1$, $A_2$, ..., $A_n$ lie on one sphere.

This completes the solution.

[quote="Diogene"]:cool:   Do you know Monge  :cool:[/quote]

Yes, I do know Monge. Monge actually proved the analogue of this theorem in the plane:

[i]Let $A_1$, $A_2$, ..., $A_n$ be n points in the plane. For any two i and j with $1\leq i\leq n$, $1\leq j\leq n$ and $i \neq j$, let $l_{ij}$ be the line perpendicular to the line $A_iA_j$ and passing through the centroid of all the points $A_1$, $A_2$, ..., $A_n$ except of the points $A_i$ and $A_j$.

Prove that all the lines $l_{ij}$ have a common point if and only if the points $A_1$, $A_2$, ..., $A_n$ lie on one circle.[/i]

Actually, you can prove this theorem analogously to my above proof of the 3D theorem, but you can also obtain this theorem as a special case of the 3D theorem above.

  Darij

Ah really, I've heard about Cantor for the 2-D version.

Pierre.

Oh yes, sorry. It is called Cantor theorem.

  http://www.cut-the-knot.org/Curriculum/Geometry/CantorPoint.shtml

My error was probably that I confused [b]Cantor[/b] with [b]Carnot[/b] (similar words) and subsequently [b]Carnot[/b] with [b]Monge[/b] (both French geometers from the same era). You know, human brains work in a very strange way, and those of mathematicians yet more.

  Darij

Generalization for a vectorial space with a inner product and the induced metric :
{M} = the commun points of all the planes $P_{ij}$,  {S}= the centers of all the spheres passing through the n points, G =  the barycenter of  the n points .
:P   (2*S +(n-2)*M)/n = G     :P

:cool:  Monge theorem : For a non degenerating tetrahedron, all the planes passing through the middle of a side and normal to the opposite side have a common point  :cool:


Yesterday, my report showed all 20-something types of problems i had done, my accuracy, and the badges for each.  It showed the videos for all of them as well.

Now it only shows two types of problem and the videos for those two types, and it doesn't seem to remember what subjects i have mastered or passed.

Does it reset every day or something?

Background info:
Opera 9.61 Build 5252 Platform Mac OS X System 10.4.11 Java Java Runtime Environment installed

Let me know if it's still like that after you do a half-dozen more problems.  I suspect it thinks you need a little more work on a fairly early subject because you only were given 1-2 problems in it during the Diagnostic.  After you do a couple more, it should open up the other subjects.

No, this isn't it.  Something went a little awry on restart this morning with your account.  We'll get it sorted out (we still have all your trials, so you won't have to start from scratch!)

This is fixed.  You are now rightly shown as having passed many subjects.


calculus
integration
limit
calculus computations

How can i compute the fallowing limit :
$\lim_{n\to\infty} n^{5} \int_{n}^{n+2} \frac{x^2}{x^7+2} dx$


there are $ 100$ people in a row to buy a ticket
$ 50$ of them have $ 50$ $ paise$
and the other $ 50$ have a $ 1$ $ rupee$ each
the ticket costs $ 50$ $ paise$ how many ways can you arrange the people such that
each gets a ticket?
assume that the counter has no change at the beginning and people dont borrow or lend

Firstly, 100 paise= 1 rupee.

This is a interpretation of Catalan number. If you arrange the people in the queue (here for our convenience from left to right) , at any position we have to have at least as many 50 paise people as 1 rupee people. Now we can see the bijection between this and the Dyck's paths.

@Gaurav: भाय , one suggestion. If you are cut pasting the text from somewhere ensure you remove extra spaces. एह तो इन्टेरनाशनल फ़ॊरम है, आप्को रुपी और पैसे के conversion देना था.


function
real analysis
real analysis solved

Let $f : [a, b] \rightarrow [a, b]$ be a continuous function and let $p \in [a, b]$. Define $p_0 = p$ and $p_{n+1} = f(p_n)$ for $n = 0, 1, 2,...$. Suppose that the set $T_p = \{p_n : n = 0, 1, 2,...\}$ is closed, i.e., if $x \not\in T_p$ then $\exists \delta > 0$ such that for all $x' \in T_p$ we have $|x'-x|\ge\delta$.

Show that $T_p$ has finitely many elements.

This is a very cute problem :).

Since $T_p$ is a closed subset of $[0,1]$, it's actually compact, so $f^{(n)}(T_p)$ is also compact and thus closed for all $n$ ($f^{(n)}$ is the $n$'th iterate of $f$). If $T_p$ were infinite, it would have at least an accumulation point $p_m$. But then $p_m$ would also be an accumulation point of $f^{(m+1)}(T_p)=\{p_{m+1},p_{m+2},\ldots\}$ which does not contain $p_m$ (if $p_i=p_j$ for some $i\ne j$, it's clear that $T_p$ is finite), contradicting the fact that $f^{(m+1)}(T_p)$ is closed.

Suppose that $T_p$ is infinite. By the nested interval theorem, there must be at least one point $x$ such that $\forall \delta > 0$ there are infinity many points from $T_p$ in $(x-\delta, x+\delta)$. If $x \notin T_p$ we are done. If not WLOG suppose that $p_0 = x$. Consider a sequence $[a_i]_{i\in \mathbb{N}}$ of elements of $T_p$ converging to $x$. Let $S = \{f^{-1}(a_i): i \in \mathbb{N}\} \cap T_p$. Since $S$ is infinite it contains at least one limit point $y$. By continuity $f(y) = x$ so if $y \in T_p$ then $T_p$ is finite. Therefore $y$ is a limit point that is not in $T_p$ so $T_p$ is not closed.


modular arithmetic
number theory unsolved
number theory

The sum of two positive integers is 2310, prove that the product of these two integers can not divisible by 2310.

Indeed, $a+b=2310$ inmplies $ab=-a^2\pmod{2310}$, but 2310=2*3*5*7*11 has not equal prime factors, so $a$ must be divisible by 2310. Contradiction.


 I believe there is a glitch on alcumus. On ocasion, it recognizes that I? have answered a question incorrectly when I never even answered it, and enters that into its database.

Any suggestions why? Any help is appreciated.

What is making you believe this is the case?


function
real analysis
real analysis unsolved

Prove that if $f$ is twice differentiable function with $f(0)=0\/$, $f(1)=1\/$ and $f'(0)=f'(1)=0\/$, then $f''(x) \ge 4\/$ for some $x\in [0,1]\/$.

Either $f(\frac12)\ge\frac12$ or $f(\frac12)\le\frac12$. The problem is symmetric (replace $f(x)$ by $1-f(1-x)$), so WLOG we have the first.
By Taylor's theorem, $\frac12\le f(\frac12)= f(\frac12)-f(0)-\frac12f'(0)= \frac12f''(t)\cdot(\frac12-0)^{2}= \frac18f''(t)$ for some $t\le\frac12$. Consequently, $f''(t)\ge4$.

We can approach equality arbitrarily closely by approximating $g(x)=\begin{cases}2x^{2}&x\le\frac12\\-2x^{2}+4x-1&\frac12\le x\end{cases}$

[quote="jmerry"]
We can approach equality arbitrarily closely by approximating $g(x)=\begin{cases}2x^{2}&x\le\frac12\\-2x^{2}+4x-1&\frac12\le x\end{cases}$[/quote]

Is there any function where $f''=4$ (with conditions above)?

If you want the function to be twice differentiable everywhere, no. If $f$ satisfies the conditions and $|f'|\le 4$ everywhere, show (using the MVT) that $f-g$ is nonincreasing on $[0,\frac12]$ and nondecreasing on $[\frac12,1]$ That forces $f-g$ to be identically zero, and $g''$ does not exist at $\frac12$.


vector
geometry
calculus
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
Let,
$ {\vec{v}} \equal{} {\langle{{{v}_{x}},{{v}_{y}}}\rangle}$

Given,
$ {\vec{v}} \equal{} {{{\vec{v}}_{x}} \plus{} {{\vec{v}}_{y}}}$

Derive, without using geometry,
$ {\lVert{\vec{v}}\rVert} \equal{} {\sqrt {{{{v}_{x}}^{2}} \plus{} {{{v}_{y}}^{2}}}}$

[hide="Remark."]
I have always seen the derivation for the above done geometrically, but never any other way. So, I wanted to know another way to derive the above.
[/hide]
[b]2. Relevant equations.[/b]
Knowledge of Vectors and Calculus.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
I think one could derive,
$ {\lVert{\vec{v}}\rVert} \equal{} {\sqrt {{{{v}_{x}}^{2}} \plus{} {{{v}_{y}}^{2}}}}$

by using calculus or maybe as part of the process of solving a specific differential equation. I'm not sure how though, any ideas?
[/hide]
Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

This supposes that you have [i]some[/i] definition of magnitude. One definition is that $ \parallel{}\!\!v\!\!\parallel{}^2 \equal{} v \cdot v$, from which your equation is immediate. All the definitions end up being equivalent for this in one way or another, which one are you using? Your result is not a result unless you
already have meaning for the symbol $ \parallel{}\!\!v\!\!\parallel{}$.


Following Hatcher, a proof of homotopy invariance of induced homomorphisms in singular
homology relies on what looks like an ad-hoc and computational dissection of a prism
$ \Delta^n \times I$ (where $ I \equal{} \Delta^1$ is the unit interval), which is used to construct
a chain homotopy between the two induced chain maps. Furthermore, the definition of chain
homotopy looks a bit ad-hoc as given $ \partial H \pm H \partial \equal{} f \minus{} g$.

I'd like to make the homotopy invariance a bit more intuitive while still rigorous, by abstracting away the computational ugliness.

I had read the more intuitive looking definition of chain homotopy between two chain maps $ f,g : A \to B$ as a chain map

$ \zeta : A \otimes C_*(I) \to B$

where $ \zeta( \sigma \otimes e_0 ) \equal{} f(\sigma)$ and $ \zeta( \sigma \otimes e_1 ) \equal{}  g(\sigma)$, where $ e_i$ is the constant map from $ \Delta^0$ to $ i \in I$.

If $ X$ and $ Y$ are spaces, then a homotopy $ H$ provides an induced chain map

$ H_* : C_*(X \times I) \to C_*(Y)$,

so if there is a natural chain map $ \times : C_*(X) \otimes C_*(I) \to C_*(X \times I)$ then
we obtain a composite map

$ H_* \circ \times : C_*(X) \otimes C_*(I) \to C_*(Y)$,

which is hopefully a chain homotopy between $ f_*$ and $ g_*$. (For a ``nice enough'' chain map, this should be the case since $ \sigma \otimes e_i$ would include onto $ X \times \{i\}$.)

The question is: it is easy to construct such a chain map $ C_*(X) \times C_*(Y) \to C_*(X \times Y)$ in an intuitive way? Certainly a construction exists for cellular chains (and I believe for singular chains as well). But I'm unsatisfied until all the rigor is sufficiently motivated --- I'd hate to forget it all again in a couple of weeks. (This is my second time through and I'd love to not skim over the details this time).

By "nice" I mean that if $ *$ is the one - point space, then the induced map

$ C_*(X) \to C_*(X) \otimes C_*(*) \to C_*(X \times *) \to C_*(X)$

is the identity map.

Update: This comes out of the Eilenberg-Zilber theorem, which says that $ C(X) \otimes C(Y)$ and $ C(X \times Y)$ are actually chain homotopy equivalent. See "On Products of Complexes", by Eilenberg & Zilber


x,y,z are positive real numbers and x+y+z=sqrt{xyz}
Prove that xy+yz+zx≥9(x+y+z)

This inequality is quite trivial. :D.  Well, we know that $x+y+z=\sqrt{xyz}$ or $(x+y+z)^{2}=xyz.$
The inequality is equivalent with this one: \[ xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9(x+y+z),  \]
or \[ (x+y+z)^{2} (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9(x+y+z).  \]
Finally, our inequality is equivalent with this well-known one:

                                    \[ (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9.  \]

I think from here it is clear how to proceed. ;)

Another solution is
From the condition and $AM-GM$  we  have that  $xyz\geq 3^6$.  (1)

Now we want to prove that $xy+yz+zx\geq 9\sqrt{xyz}$.   From AM-GM again it

suffices to prove that $3\sqrt[3]{(xyz)^2}\geq 9\sqrt{xyz}$  which is equivalent to

$xyz\geq 3^6$ which is true from (1) and we are done


How many straight lines are in a saddle in 2D?

Equation = x^2 - y^2

2 or infinite?

Every plane of the form $x-y=a$ or $x+y=a$ intersects the surface $z=x^2-y^2$ in a line. No other plane does.


To get a 5 for a computer program, your solution needs to be rigorous - you can't just give the code and output.  A well-written computer solution would explain the exact algorithm your program uses and prove that it finds every solution.  (I don't think you need to get philosophical about it, though - you can assume that the computer evaluates arithmetic expressions correctly.)

Oh nuts, I didn't really explain my code very well (other than the comments inside the source).  I think it's pretty obvious what the program does though, is this okay? :\  I'd say more on this but I don't believe I'm supposed to discuss details of the problems at this time.  I hope I don't get a 1 for not rigorously explaining _how_ my program works...

[quote="diophantus"]What kind of scores are you getting.  I've always had the impression that computer solutions always get a 1.  If you're getting higher, perhaps I should begin doing computer solutions to save time.[/quote]

Computer-generated solutions earn 0 or 1 points automatically only if the problem requested no computer solutions. On a typical problem, a well-presented computer-generated solution is capable of earning 5 points.

The weakness of computer solutions is that they often solve the problem without mathematical insights. Such insights earn the commendations, so computer-generated solutions almost never earn commendations.

[quote="Myself"]So far, I have tried several times to solve problems with a computer program. However, I have never succeded to get a 5 on any of the solutions. The comments say that I need to present the program. How do I do this?[/quote]

The submission for such a solution should include:
1. A clear description of why you think your computer program will answer the question. This will briefly describe your algorithm, but more importantly, it will describe what your algorithm is supposed to calculate and how that relates to the problem.
2. The source code of your program. (Actually, if you wrote a good description of your algorithm, we graders won't read the source code. Include it for completeness and in case the output is in error.)
3. The output of your program.
4. Your answer to the question.

Part 1 is the key. Now that a significant number of USAMTS participants have access to good computational resources, we select problems that require thinking before the calculation can be set up, or thinking after the results are calculated. We judge the solution on the quality of that thought. Answering a problem with, "You asked for the smallest positive integer n that fit the requirements, so I checked all n starting at n=1, and n=987654321 was the first number that worked," is no fun.

Erin Schram
USAMTS grader

[quote="zanttrang"]One question I have about computer programs is whether Mathematica is a valid aid for problem-solving.  My friend has frequently used Mathematica as part of her solutions and gotten perfect scores, so I've assumed it was okay, but I'd still like the official word.  For example, on this last round (I won't get too specific right now, as that would be against the rules), for one problem I ended up with several nasty high-degree polynomials that I had to solve.  I just used Mathematica's "solve[..." command to do so, and put as part of my proof the Mathematica command I used to get the output (and then explained which of the answers it gave was the one that applied to my situation).  I assume this is okay, since polynomial equation-solving isn't a special algorithm I'd have to explain, and Mathematica is generally acknowledged to be a high-quality program.[/quote]

Mathematicia is a valid aid. Just credit Mathematica properly for the work it did, and there will be no objections. The only difficulty is that some of us graders are unfamiliar with Mathematica--I use Magma instead--so too brief a description of what Mathematica did could confuse us.

Mathematica can also be a crutch. You said you ended up with several nasty high-degree polynomials to solve. None of the problems were designed to require solving several nasty high-degree polynomials, so you must have stumbled across an especially inelegant solution. Because Mathematica let you finish the inelegant solution, you overlooked the possibility of finding an elegant solution.

Erin Schram
USAMTS grader

[quote="diophantus"]What kind of scores are you getting.  I've always had the impression that computer solutions always get a 1.  If you're getting higher, perhaps I should begin doing computer solutions to save time.[/quote]
I've used a total of 3 computer solutions, and all of them got a 5, but now I try to avoid using computer programs and look for an elegant solution instead. (One of mine had a run-time of 15 hours, so I know that I had a very inelegant solutionn) :D


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

I am trying to determine all groups G of order 63. I have a problem in case G has a normal non-cyclic subgroup of order 9, it seems that there are a lot of variants..

Actually, there aren't. $ GL_2(F_3)$ has order $ 8\cdot 6\equal{}48$, so elements of order 7 must act trivially on that subgroup (by conjugation), and the group is abelian.

[quote="jmerry"]Actually, there aren't. $ GL_2(F_3)$ has order $ 8\cdot 6 \equal{} 48$, so elements of order 7 must act trivially on that subgroup (by conjugation), and the group is abelian.[/quote]
I dont clearly understand your approach, it would be nice to see some more comment.
However, i also found this: suppose $ P$ is normal p-Sylow of order $ 9$, and let $ p, l$ be generators of P and $ y$ generator of Q. Then group multiplication is defined by $ t_1, t_2\in P$: $ y^{ \minus{} 1}p y \equal{} t_1, \ y^{ \minus{} 1}l y \equal{} t_2$. Consider group action of $ Q$ on $ P$ by conjugation. Then orbits of $ p$ and $ l$ can't intersect. If Q acts trivially we are done; suppose not, then wlog orbit of $ p$ consisits of 7 elements( $ |O(p)| \equal{} |(Q: Q_p)|$). Then orbit of $ l$ can be only $ l$, but then it is easy computation to check that orbit of $ p$ must be $ p$.

My approach? The automorphism group of $ (\mathbb{Z}/p\mathbb{Z})^n$ is naturally isomorphic to the general linear group $ GL_n(F_p)$, since we can represent the group as the additive structure of the vector space. The other elements of our big group act by conjugation on this group- we get a homomorphism from $ G$ to $ \text{Aut }H$ with kernel $ H$.

Alternately, you can save yourself the trouble and note right away that the group of order 7 is normal. That leads to four cases- two abelian, two nonabelian.

[quote="jmerry"]
Alternately, you can save yourself the trouble and note right away that the group of order 7 is normal. That leads to four cases- two abelian, two nonabelian.[/quote]
Well, i don't see any short argument for this, only what is written above..it does not follow from Sylow theorems directly: we only know that one of Sylow subgroups is normal.


linear algebra
matrix
geometry
rectangle
combinatorics proposed
combinatorics

Given a matrix $\{a_{ij}\}_{i,j=0}^{9}$, $a_{ij}=10i+j+1$. Andrei is going to cover its entries by $50$ rectangles $1\times 2$ (each such rectangle contains two adjacent entries) so that the sum of $50$ products in these rectangles is minimal possible. Help him.
[i]A. Badzyan[/i]

Combining these two observations you can't miss it (first one obvious, second one a little surprising):

${ a_{ij}\cdot a_{i(j+1)}+a_{(i+1)j}\cdot a_{(i+1)(j+1)}>a_{ij}\cdot a_{(i+1)j}+a_{i(j+1)}\cdot a_{(i+1)(j+1)}}$ and

${ a_{(i-1)j}\cdot a_{(i-1)(j+1)}+a_{ij}\cdot a_{(i+1)j}+a_{i(j+1)}\cdot a_{(i+1)(j+1)}}=$
$ =a_{(i-1)j}\cdot a_{ij}+a_{(i-1)(j+1)}\cdot a_{i(j+1)}+a_{(i+1)j}\cdot a_{(i+1)(j+1)}$

Hello.
I tried to solve this problem and I think I understood your hint, but still I cannot solve it. I tried to prove that in any configuration I can use such an alorhitm that uses only these two manipulations until I get a matrix in which all the rectangles are vertical. But I cannot find such an algorhitm. I also tried to discover some new manipulations and I did, but I still couldn´t universally come to an end.
Thanks for any reasonable replies.

I think lorincz answer is not Completely can someone say a prove for minimality

Here's a new approach for proving the minimality,  let change the problem in the following way, given a 10*10 table with numbers 1 to 100 labeling on the cells of the table as the problem says it above,we're gonna use 50 dominoes to tile the table so that if we add the product of numbers in each dominoe up,we get the minimal number.

first: our claim is that dominoes should tile vertically in each column,so that the very left column's dominoes tile (1,11),(21,31),...,(81,91) and saim tiling for other columns.

[b] Lemma:[/b]

we use UU as we want to show that the numbers which are hidden behind U&U belongs to One  dominoe, for example  in the following matrix :

a,b
c,d
e,f

where a&b,c&e,d&f belongs to same dominoe, we represent the matrix in the following way :

XX
YZ
YZ

plus, if we say that :

XX        X'Y'
YZ   =   X'Y'
YZ        Z'Z'

we try to show that the position of dominoes for  numbers in the following order in the table :
a,b
c,d
e,f
 in which a&b,c&e,d&f belong to same dominoe is not different with the position of dominoes as a&c,b&d,e&f belong to same dominoe.

this is obviouse by calculating as we can define a=a,b=a+1,c=a+10,e=a+20,d=1+11,f=a+21 . Lemma proved actually $/black triangledown$
:)

[b]Lemma [/b]:
there should be no
XX
YY

in the table as we are assuming the minimality,because we can change it  to

XY
XY

and get a smaller value, it can be easily proven as we can define that into

a,a+1
a+10,a+11 :)

now we are going to show 1&11 should belongs to a same dominoe,assume the contrary, so 1&2 should belong to a same,(from now we'll define u&v as u and v belong to same dominoe) so we'll get 11&21(by lemma), we should have 12&13 (if else,then by lemma we can make 1&11 and so contradiction),so 3&4,22&32(by lemma2),31&41 and ..., here at some point you can easily get contradiction by finding

OO
QQ

as we can get contradiction by lemma2, so we have 1&11, doing the same process for the dominoe whic contains number 2, we can get easily to contradiction and get 2&12, and doing this and this and so on, we can get that at the first two rows, we have 1&11,2&12,3&13,...,10&20.

now the remaining rows(which are even ) can be tiled as the first two rows tiled by dominoes in the vertical position by the same argument....

so it's Done!
=====
also we proved the problem for any table m*n which atleast one of m,n are even. :)  $/square$

The original statement for this problem is this:
The numbers from 1 to 100 are arranged in a 10×10 table: In the first row 1. through 10 from left to right,in the second row 11 through 20 from left to right, etc. Andrey wants to cut the table into1×2 rectangles, compute the product of numbers in each rectangle and sum these products up. How should he cut the table in order to get the least possible sum?
And this is my solution:
Clearly that: $ ab=\frac{a^2+b^2-(a-b)^2}{2} $ (1)
And in each rectangle, the difference between two numbers in it just can be 1 or 10 ( this is straight from the way Andrey arranges numbers in the board ). So from (1), we have that the sum min iff the product min, which means the,  difference between two numbers in an rectangles must be maximum. So the difference between two numbers in a rectangle equal to 10. So Andrey should cuts the rectangles vertically in each columns


I just thought of something and I want to hear your opinions on it...
I think Voldemort, the night that he "tried to kill Harry", actually made Harry the 7th horcrux!
Think about it, it's not impossible, and it would certainly add an interesting twist to the tale.

hmmmmmmmmmmmmm

also , didnt the snape killing dumbledore and the "his face showed revulsion" as he cried Avada kedavra  which bumped off dumby  , well didnt it seem fake ? Almost like J.k rowling purposely wants to show the whole guilt of snape , that he is truly a bad guy , But i think snape's gonna save harry in the 7 th book.

hey was that holocrux or horcrux :huh:  :huh: ( hey my memory's gettin weak , i read the 1/2  blood prince just 2 months ago :( )

Ya, it is horcrux, sorry :oops:  Also, I just realized that there are just 6 horcruxes, so yeah Harry is the 6th horcrux.

So you thinks he's gonna go look for the horcrux and then realize its him??
I don't agree.

And anyways I believe you must actually kill the [person in order to make a horcrux from their death. Right?  :?

Erm, why would he want to destroy his own horcrux? Theory debunked methinks.

No, Voldemort's horcrux is in Harry.  The theory makes sense (if magic and wizards make sense, that is).  In fact, it seems plausibly true.  Almost like you just ruined the last book!  In all seriousness, I think that theory has a lot going for it.  Although there is the point, that why would Voldemort want to kill Harry then?  And what about 'one can live only if the other dies' or whatever?  Instead, it would be 'either both live or both die'.

I actually somehow thought that Dumbledore might have been one of Voldemort's horcruxes, somehow.  Thus, he realized, and made Snape kill him.  There is a lot of evidence pointing towards Dumbledore's being a horcrux, but it seems very weird and unimaginable, so I don't know...

Well, it seems the only problem with my theory is, why would Voldemort try to kill Harry then? I think maybe at first Voldemort did not want to make Harry a horcrux, but something went wrong that night and he did. However, that was not his plan and he was not aware of this till later (or maybe he still doesn't know about it).

Er...as a former Harry Potter fan I feel the need to put two cents into this one.

As uh...magical as Rowling's world must be, there's some reality factors she uses. Like, for example, it would be completely ridiculous for Voldemort to want to kill Harry if he was a horcrux. Likewise, it would be completely ridiculous for Voldemort to not know that Harry is a horcrux. Also, what about the prophesy. mark him as an equal...neither can live while the other is alive...etc.? That clearly doesn't work out. Contradiction. And uh...QED and stuff.

Now, let's not anyone start saying Dumbledore is a horcrux. We all know that's wishful thinking: Voldemort shot at least three death spells at Dumbledore in the 5th book, Dumbledore happens to be quite a bit more powerful, and Dumbledore, I'm sure, doesn't really want to be a horcrux.

Now I do admit that I believe Rowling has already introduced what the last Horcrux is and that I have no idea what the last is. But clearly, we've got 3 down and a snake, a cup, and something of Gryffindor or Ravenclaw's left. Dumbledore has never made an incorrect conclusion and I highly doubt Rowlings would have him start making mistakes in book 6. Thus, we're left to guess what the 6th item is. Dumbledore also said that the sword was the last remaining possession of Gryffindor, so Ravenclaw is clearly the best bet. Recalling that the sword did destroy the diary in Chamber of Secrets, we know that's not a horcrux. As for Ravenclaw's object...I have no idea.

[quote="MithsApprentice"] Dumbledore has never made an incorrect conclusion and I highly doubt Rowlings would have him start making mistakes in book 6.[/quote]
That's not true... if I recall correctly there was something in the 5th book about him making mistakes, but I can't remember what it was.

[quote="MithsApprentice"] Thus, we're left to guess what the 6th item is. Dumbledore also said that the sword was the last remaining possession of Gryffindor, so Ravenclaw is clearly the best bet. Recalling that the sword did destroy the diary in Chamber of Secrets, we know that's not a horcrux. As for Ravenclaw's object...I have no idea.[/quote] Well, I still think Harry is the horcrux, but if he isn't, have you ever thought of the sorting hat? It said in it song (in book 2 I think) that it once belonged to Godric Gryffindor :|

Sorting hat could be a good guess.
And, I think you mean RAB?
People claim it's regalus black, but I have my doubts

Mith apprentice said that a snke was a horcrux which i think Dumbo said isn't true,instead he( Mith apprentice )forgot to mention the locket RAB took and i think he gave it to Voldo,regardless of the note and i think draco might die saving harry as well and the death eaters migght turn against voldo for some reason,i don't know which....

Any heir of gryffindor alive?And about Neville?....many questions. :roll:

what about Snape? I'm 80% sure that he is good, but... if he is, why would Dumbledore trust him
 ( Dumbledore must have a better reason than Snape feeling guilty about the death of James/Lily Potter)?

I really doubt that Harry is a horcrux. That would mean that in order to kill Voldemort he would have to die, and while J.K. is no stranger to killing off characters (Dumbledore! *sniff*) I think bumping off Harry himself would be a step too far. There would be hordes of traumatised 10 year olds needing counselling.

[quote]another thing . So who's REM[/quote]
You mean R.A.B.? Regulus Black (Sirius' brother) is the most popular theory.

REM is Rapid Eye Movement, as far as I know. It is also one of my favorite bands :D

I'm guessing certain translators guessed that RAB=Regulus Black and changed the initials to fit Regulus's name in the other translations. I saw that discussed on another message board. But apparently they have no actual information about who RAB is, so things will be a bit messed up if it isn't Regulus.

And I don't get how Harry could be a horcrux. You don't accidentally make a horcrux. You need complex spells, potions, et cetera. How did Voldemort manage to do that *accidentally* in the split second before the killing curse rebounded on him? It completely doesn't make sense.

so sorry , i meant RAB instead of the REM.

hmmm, Regulus black, huh ? Well where did he jump from? He wasnt anywhere in the story till now, was he??

and what is draco goin to do in the 7th one??

Well, can you think of anybody else in the story with those initials? Unless JK brings in a new character that was never even mentioned before... but I doubt she'll do that.

[quote] Dumbledore has never made an incorrect conclusion and I highly doubt Rowlings would have him start making mistakes in book 6.

That's not true... if I recall correctly there was something in the 5th book about him making mistakes, but I can't remember what it was.
[/quote]

I recall something either he said or someone said about him: that he is cleverer than most men, and therefore his mistakes tend to be more extravagant than most men.

We know that Regulus Black was killed by Voldemort or one of his followers (Book 5), this could explain why he was killed.

i think harry is the last horcrux
and i think voldemort comes back tom riddle style after harry "kills" him so harry needs to suicide before voldemort can become more "solid"

ever wondered if the scar was the mark of his horcuxosity?

I think it is quite possible that Harry's scar actually [i]is[/i] the horcrux :D

that's quite possible! especially since the last book is supposed to end with the word "scar." like...

there was no scar.

lol :D

there will be NO new characters in book seven.  ;)

thank god... i think there's too many as it is.
good thing she's killing them off... jk

[quote="la mafioso"]PLastic surgery is the way to go for harry

So they can shoot people out of buildings and blast fire out of their wands but they can't get rid of his scar?[/quote]

We have things called knives...

Harry HAS to be the last horcrux.

I mean, Dumbledore did say that "Voldemort left a part of him in you, Harry", to explain the scar hurting and how he was able to speak Parseltongue, right?

Isn't that the definition of a horcrux?  :?

well no.. a horcrux is a piece of ur soul, but it doesnt mean u necessarily change the object in any other way..

Though thinking about it, his scar seems like an interesting idea.

See, since no one knows the reasoning for voldemort's curse backfirign, we can assume that harry had some power already or.. that voldemort was trying to do something else. But then harry shouldnt remember green light, as he does, so i think that it was not a horcrux.

Anyway i dont think Harry or any body part/mark of harry is the horcrux. That would just complicate everything wayyyyy too much, and as a general thing, i dont think Rowling would kill off harry. Besides... that would destroy the prophecy, considering if harry killed voldemort but a fraction of voldemort survives - that would be harry living and voldemort surviving without end. And harry wouldnt be killing voldemort. And if any part of voldemort was in harry, then voldemort could not be KILLED BY HARRY. Yeah, and of course voldemort wouldnt try to kill him. Either way, i dont think theres any indication for scar or body of harry being a horcrux. And i think rumors of 'scar' being the last word are false, considering the book isnt finished...?

Hmm.. Didn't the prophecy say something like "If one of them die, then so does the other"?  If Harry was the horcrux, then that just might happen...

I guess that would make sense...

A book is always very interesting when it is unfinished......

I mean, when the 7th book comes out, the whole Harry Potter series won't be as exciting as now...

[quote="bleumoose"][quote="Melissa"]No, I think JK will finish the story off in the 7th book, because she does not want to write more sequels or prequels, she's got enough money already.[/quote]

its not the money. she's said she wants to write other books- which, i mean, that makes sense. obviously she's had potter on her mind for what? fifteen years or something now?

but yes, there is the money. k, i've got this fact and i'm not sure about it. i heard somewhere one of the following, and i cant remember which is true.
1. jk rowling is richer than the queen, and the richest woman in britain.
2. jk rowling has slightly less than the queen, making her the second richest woman in britain.
anybody want to verify?[/quote]

JK is the richest woman in Britain. Yup.  :)

[quote="slamchess"]my thoughts-

1.  ya harry is a horcrux, that is why he inherits some of voldys powers...

2. rab is a new unknown character which gives jk room to work with, or jk is throwing us off and those are the initials of a person's original name...

and where did rab appear again???[/quote]

number 1: that is absurd.
for number 2, that makes sense
it appeared at the end where harry finds note

for the last time, THERE WILL BE NO NEW [UNKNOWN] CHARACTERS IN BOOK SEVEN. yeeshus. in case you don't believe me, i read that in a j.k. interview.


calculus
integration
calculus computations

Suppose that $ P_1(x)\equal{}\frac{d}{dx}(x^2\minus{}1),\ P_2(x)\equal{}\frac{d^2}{dx^2}(x^2\minus{}1)^2,\ P_3(x)\equal{}\frac{d^3}{dx^3}(x^2\minus{}1)^3$.

Find all possible values for which $ \int_{\minus{}1}^1 P_k(x)P_l(x)\ dx\ (k\equal{}1,\ 2,\ 3,\ l\equal{}1,\ 2,\ 3)$ can be valued.

[quote="kunny"]Suppose that $ P_1(x) = \frac {d}{dx}(x^2 - 1),\ P_2(x) = \frac {d^2}{dx^2}(x^2 - 1)^2,\ P_3(x) = \frac {d^3}{dx^3}(x^2 - 1)^3$.

Find all possible values for which $ \int_{ - 1}^1 P_k(x)P_l(x)\ dx\ (k = 1,\ 2,\ 3,\ l = 1,\ 2,\ 3)$ can be valued.[/quote]

We have
\[ \displaystyle \begin{eqnarray*} P_k(x) & = & \frac {d^k}{dx^k}(x^2 - 1)^k = \frac {d^k}{dx^k}\left[ (x - 1)(x + 1)\right]^k \\
& = & \sum_{j = 0}^{k}\binom{k}{j} \frac {d^j}{dx^j}(x + 1)^k\cdot \frac {d^{k - j}}{dx^{k - j}}(x - 1)^k \\
& = & \sum_{j = 0}^{k}\binom{k}{j} \frac {k!}{(k - j)!}(x + 1)^{k - j}\cdot \frac {k!}{j!}(x - 1)^j \\
& = & k!\sum_{j = 0}^{k}\binom{k}{j}^2 (x + 1)^{k - j}(x - 1)^j \\
\end{eqnarray*}\]
so that
\[ \displaystyle \begin{eqnarray*} \int_{ - 1}^{1}P_k(x) P_{\ell }(x)dx & = & \int_{ - 1}^{1} k!\sum_{j = 0}^{k}\binom{k}{j}^2 (x + 1)^{k - j}(x - 1)^j \cdot {\ell }!\sum_{r = 0}^{{\ell }}\binom{{\ell }}{r}^2 (x + 1)^{{\ell } - r}(x - 1)^r dx \\
& = &  k! {\ell }!\sum_{j = 0}^{k}\sum_{r = 0}^{{\ell }}\binom{k}{j}^2 \binom{{\ell }}{r}^2 \int_{ - 1}^{1} (x + 1)^{k + \ell - j - r}(x - 1)^{j + r} dx \\
& = & k! {\ell }!\sum_{j = 0}^{k}\sum_{r = 0}^{{\ell }}\binom{k}{j}^2 \binom{{\ell }}{r}^2 \frac {(k + \ell - j - r)! (j + r)!}{(k + \ell )!} \\
\end{eqnarray*}\]
Denote the value of the integral by ${ M_{k\ell}}$ and the particular values are
\[{ M_{k\ell}} = \left(\begin{array}{ccc}3 & {\scriptstyle \frac {32}{3}} & 60 \\
{\scriptstyle \frac {32}{3}} & 36 & {\scriptstyle \frac {984}{5}} \\
60 & {\scriptstyle \frac {984}{5}} & {\scriptstyle \frac {5292}{5}} \end{array}\right)\]
I had Maple do the arithmetic, so if they're wrong it's my math above.


ΜΠΟΡΕΙ ΚΑΠΟΙΟΣ ΝΑ ΜΕ ΒΟΗΘΗΣΕΙ ΣΕ ΑΥΤΟ??

ΕΣΤΩ ΓΡΑΜΜΙΚΗ ΑΠΕΙΚΟΝΙΣΗ F:R^17->R^64, ME ΤΥΠΟ F(X)=AX ΟΠΟΥ Ο ΑΝΑΣΤΡΟΦΟΣ ΤΟΥ ΠΙΝΑΚΑ Α ΕΧΕΙ ΒΑΘΜΟ 11.ΠΟΙΟΣ ΕΙΝΑΙ Ο ΜΕΓΙΣΤΟΣ ΑΡΙΘΜΟΣ ΓΡΑΜΜΙΚΩΣ ΑΝΕΞΑΡΤΗΤΩΝ ΔΙΑΝΥΣΜΑΤΩΝ ΠΟΥ ΠΕΡΙΕΧΟΝΤΑΙ ΣΤΟΝ ΥΠΟΧΩΡΟ kerF.

ΕΥΧΑΡΙΣΤΩ[/b]

Ο πίνακας $A$  είναι ένας $64 \times 17$ πίνακας και αφού οι βαθμοί των  $A^{t}$ και $A$ είναι ίσοι $rank A = 11$ .

Τώρα από το θεώρημα $rank( A)+nullity( A) = 17 \Rightarrow nullity( A)= 6$ ή $dim (Ker (F)) =6$ .

Άρα ο μέγιστος αριθμός ανεξάρτητων διανυσμάτων είναι $6$ όσο και η διάσταση του $Ker (F)$ .

Ραμανατζαν is back.γεια χαρα φιλε.αληθεια rankA ειναι ο βαθμος του Α? και κατι αλλο χρησιμοποιεις οτι dimR^17=dimf(R^17)+dimKerf ?

ΑΠΟ ΠΟΙΟ ΘΕΩΡΗΜΑ ΕΧΟΥΜΕ ΟΤΙ RANK(A)+NULLITY(A)=17??

αν καταλαβα nullity=dim , rankA=βαθμΑ ιση ο τυπος που εγραψα στο πανω ποστ και αφου με προσσεγγιση ισομμορφισμου ισχυει dimf(R^17)=βαθμΑ και kerf=kerA , kai dimR^17=17 an αν κανω λαθος ραμανατζαν διορθωσε με :maybe:

[b]@drg1599[/b]

Αν λέγοντας βαθμό εννοείς το πλήθος των ανεξάρτητων στηλών ή γραμμών του πίνακα
τότε όλα προκύπτουν από το θεώρημα :[url]http://en.wikipedia.org/wiki/Rank-nullity_theorem[/url]

Μπορείς να βρείς αποδείξεις εδώ στην κατηγορία Άλγεβρα : [url]http://mathstuff.rootforge.org/gr/downloads.php[/url]

[b]@kostas[/b]

Γεια χαρά Κώστα επέστρεψα  :) . Δεν κάνεις λάθος. Νομίζω ότι [b]βαθμός = rank = τάξη [/b]

Για να μην μπερδεύουμε τα πράγματα: rank = βαθμίδα, order = τάξη, degree = βαθμός.

Lance

γεια σε ολους!μηπως μπορει να με βοηθησει κπ?
Αν η F:V->W γραμμικη απεικονιση,να δειξετε οτι ImF υποχωρος του W και KerF υποχωρος του V.
 ευχαριστω!


i agree with dragon. we should have like a ctc type thing that decides on the competitions we have and games and stuff. just like the tourney mania its kinda getting crowded here. if you don't know what ctc is go to the ftw other stuff.

I am now founding The Fun Factory Commitee, FFC for short. This is a group that will create games and polls, etc for the fun factory. THe only people allowed to join these games are FFC members. Joining FFC is easy! All you need to do is apply by pming me and agreeing to follow these rules:

1. DOn't join any non-FFC games
2. Don't make any games unless told to by a high-ranked member of FFC

Positions within FFC:

1 president

1 vice president

10 board members

normal members :D

Isn't this kinda rediculous. No ones going to think: hmm...I want to have a fun little noncompetitive game. Lets get it approved by some board. :/. Besides most games are in the GAMES forum.

I think it'll settle down here, just like FTW as the school progresses.
If it doesnt, I think it can start.

well things are like getting out of hand. we have so many games and stuff going around. and anywayhow come when people did tourney mania they didn't like that and started ctc. gahh that was worded wrong.w/e. you don't have to join. i'm just putting this up. and when i mean games i mean teh competitions we have on fun factory.

ok
this'll be quick

i think that the mania will calm down once school starts
people will spend more time doing schoolwork instead of using this forum
BTW, y do people play games on AoPS????
there are so many better websites out there

second thing
this is fun factory
it's supposed to be fun
have a commitee watching everything that's happening is not fun

third thing
if this mania continues (and it is getting annoying) this'll be a good idea
i'll join

however, wait until after school starts and see if this is still going on
after that we'll talk about this some more
BTW, i never check this forum anywayz so....

i agree it will slow down once school starts. same with ctc, but still just in case i want to have a commitee ready. so pm if you want to join or post here. you could be high-ranked ffc member!

Umm... I was kidding about CTC. But I think it will calm down when sept starts.Just make sure you don't make too many of the same type games like others. one or two is fine, but I think we have top staff, mod, math, and soda. I'll join if it doesn't calm down

okay i know it will probably calm down. but i'm jsut starting it jsut in case. getting the board members and stuff. so joined so far is

jjx1

vallon22

Dojo

Dragon

jsut to let you know i am running for president. if you would like to run for prez or vice prez pm me. also prez and vice prez nominates the board members.

At summer, there was a horrible flood at G&FF. Lots and lots of nonsense stuff replaced some good games. I will obv /in to this society...

BUT, please don't make it too restrictive. Make it as a control system, not as an Inquisition.

Yea, uh...
I don't support an FFC, but I do agree all the tug-of-wars are annoying.
However, my Russian Roulette game (stolen from another site) is totally awesome and involves some thought and not too much posting, so you should totally join that. :D
(Yay for advertising!)

[quote="jjx1"]these are the people joined so far.

jjx1

vallon22

Dojo

Dragon

Brut3?

remember if you want to run against me for prez or run for vice prez pm me. the vp and p nominate the 10 board members. and Brut3 i already joined your russian game. it sounds cool![/quote]

I joined :mad:

ok no Brut

jjx1

vallon22

Dojo

Dragon

pacman

everoyne i have heard all of the complaints you have made. i am ignoring them. i am going along with this plan. even if it doesn't work its for fun.SO CAN ALL OF YOU STOP LECTURING ME HOW THIS IS BADD AND ISN'T GONNA WORK. wait a second your post isn't as bad as everyone elses james. thank you for the advice and explanation. even though i'm still going along with this your post is nicer than others.

okay mousy has voted for me. if you don't beleive me ask her pacman and almath.
oh and i vote for myself
so prez results so far:

jjx1:3
pacman:0
almath:0

members:
guinea
vallon
dragon
dojo
pacman
almath
jjx1
mousy

people still to vote:vallon, dojo, pacman, almath, dragon, and new members

vp election updates:

guinea is the only person to sign up so far. if you would like to run for vp contact me

[quote="james4l"]Problem 1: How in the world will you control the forum: personally "top math subject" has slightly more relevance on AOPS than the "top soda" poll.

Problem 2: As opposed to then CTC the Fun Factory isn't as controllable. Also there are far more people who go on F&F. While it is possible to ban all non-FFC people from ever joining FFC games, the converse is rarely the case. Also, as opposed to FTW tournaments which happen every, say month, FF posts happen about twice a day. All I can say to your group: control with moderation.[/quote]

Problem 1: top math subject does have more to do with AoPS, but it takes the idea from another idea, which happens to be the 4th game about hurt and heal. To me, top soda and top math are exactly the same, they both come from the same place and both are redundant and unnecessary ideas.

Problem 2: I agree. Fun factory is uncontrolable because this is what people do for fun, and they feel like having fun. CTC is different because CTC controls the amount of games played. CTC was agreed by many people, and this isn't going quite well (in fact its split into thirds about disagree/not sure/agree).

i k this isn't going well. but if it doesn't work out this is just for fun i guess. its fun like voting for prez and stuff like that.

If you want a game of voting president then we can have a game of that.

okay the prez race for ffc is

jjx1:4
pacman:0
almath:0

members:
vallon
pacman
zoro
dojo
dragon
almath
guinea
jjx1
mousy

still needs to vote:
pacman
almath
dojo
dragon
vallon

vice prez election update:
guinea piggy and zoro have signed up to run for vice prez. if you would like to join please contact me

okay the prez race for ffc is. I'm gonna assume they vote for themselves.I have clinched teh president spot unless new members come on. pacman and almath if you would like to run for vice prez please post that.

jjx1:5
pacman:1
almath:1

members:
vallon
pacman
zoro
dojo
dragon
almath
guinea
jjx1
mousy
hurdler

still needs to vote:
dojo
dragon
vallon

vice prez election update:
guinea piggy and zoro have signed up to run for vice prez. if you would like to join please contact me

kiseki voted for me. if you don't beleive me ask her. .I have clinched teh president spot unless new members come on. pacman and almath if you would like to run for vice prez please post that.

jjx1:6
pacman:1
almath:1

members:
vallon
pacman
zoro
dojo
dragon
almath
guinea
jjx1
mousy
hurdler
kiseki

still needs to vote:
dojo
dragon
vallon

vice prez election update:
guinea piggy and zoro have signed up to run for vice prez. if you would like to join please contact me

okay I have won the prez race unless new members come on. pacman and almath, you can drop out now or keep on running if you would like. Also you can run for vp. I vote guineapiggy

vp race:

guinea piggy:1

zoro:0

Whoa.  No double (or triple) posting.

Edit your posts.

okay vice prez race:

guinea piggy:1

zoro:0

almath:0

pacman:0

[quote="jjx1"]I am now founding The Fun Factory Commitee, FFC for short. This is a group that will create games and polls, etc for the fun factory. [/quote]

First of all, you don't own Fun Factory - therefore, you have no right in forming a committee that decides what topics are posted. [i]Moderators[/i] decide which topics acceptable, not a bunch of self-elected posters.

Locked.


trigonometry
rotation
analytic geometry
function
geometry unsolved
geometry

Hi there... I am not good at maths... but I need a problem solved if possible.

I was wondering if anyone can help me.

Thanks in advance.

[b]What is the Problem?[/b]

If I know the locations of [b]2 points in space X,Y=A and X,Y=B[/b], as well as [b]know the ANGLES P and K and that they are equal to each other[/b].

[size=200]Q: How do I work out [b]the X,Y Value of H[/b][/size]

[img]http://www4.picturepush.com/photo/a/1292757/img/1292757.jpg[/img]

[b]What am I trying to do?[/b]

What I am really trying to do is work out the X,Y points along the arc between the two given points. So I can draw the arc. If you know a better way of doing this then please let me know.

[b]What I need[/b]

What I need is a way to calculate a formula using the [b]A and B x,y values[/b] and the [b]P and K Degree Values (witch are equal)[/b] to work out all the points on the arc that connect [b]A.B and H[/b]

Dose that even make sense?

Thanks in advance.

[quote="JRav"]Do you know the value of the segment $ \overline{AB}$?  If so, then you can take $ \triangle ABH$ (which is isoceles by your construction) and break it into two congruent right triangles.  Then, since you know the angles, you can just use some basic trig to find $ H$.[/quote]

Well, I do not know much about maths... I just haverun into this problem that i need a way to work out the x,y points of a arc like i said above.

So while you may be right I am not sure i understand those symbols....

Mabey if you wrote it again but in words?

Like say this is the values I have.... (looking at the diagram i posted on first page)

[b]A = x,y  where X = 100 and Y = 10
B = x,y where X=100 and Y = 200

Angle P = 20 Degrees
Angle K = 20 Degrees

What is the X,Y Vales of H?[/b]

[b]A = x,y  where X = 100 and Y = 10
B = x,y where X=100 and Y = 200

Angle P = 20 Degrees
Angle K = 20 Degrees

What is the X,Y Vales of H?[/b][/quote]

I am pretty sure this will work in teh case I have stated at the start....

Hy value = y of A + [(y of B - y of A)/2]
Hx value = x value (of A or B) +  {[(y of B - y of A)/2] * tanP}

10 + [(200 - 10 /2)  = 190 / 2 = 95] 105

So that is 10 + [200 - 10 / 2] = 105

and

100 + {[(200 - 10)/2] 8 tan20} = 134 (Just roudned tan result to .3 for simpel sake) = 134

So H x,y = 134,105

????

If so... how would that work.. if the values are like this? (eg.. say rotate that image i did so the triangle is on an angle instead of having the base line flat?

Say

[b]A = x,y  where X = 100 and Y = 10
B = x,y where X=50 and Y = 400

Angle P = 20 Degrees
Angle K = 20 Degrees

What is the X,Y Vales of H?[/b]

How would you work out H x,y then?

Am i even goign about this the right way? What I am tryign to do is get the points on teh CURVE so i can plot a curve that passes though the two poinys A and B?

I think i worked it out......

[size=150][b]Ok, working out the location of H from the angles P and K[/b][/size]

1) work out distance between A and B, let's call this D1:

  D1 = sqrt((Xa - Xb)^2 + (Ya - Yb)^2)

2) work out midpoint of AB line, let's call this M

  (Xm ,  Ym) = ((Xa + Xb) / 2 , (Ya + Yb) / 2)

3) work out gradient of line AB, let's call this g1

  g1 = (Ya - Yb) / (Xa - Xb)

4) work out gradient of line perpendicular to AB, let's call this g2

  g2 = -1 / g1    (note: this value won't exist if g1 =  0, need a special case, see below)

5) work out angle between line perpendicular to AB and horizontal, let's call this phi

  phi = atan(g2), if g2 exists (i.e. if g1 is not 0)
       = 90 degrees + Pi/2 radians, if g1 is 0

6) get distance between midpoint (M) and H, let's call this D2

  D2 = D1 * tan(K) / 2

7) get coordinates for C

  (Xh , Yh) = (Xm + D2 * cos(phi), Ym + D2 * sin(phi))

note, you'll get a semi circle when K and P are 45 degrees.

[size=150][b]and once you get the location of H..... . . . . ., find the X,Y Locations of the points along the curve of teh arc between A and B[/b][/size]

1) Work out the centrepoint of the arc (let's call this point C) - this is the point that is the same distance from points A,B and H. I haven't done the maths for the y coordinate yet, but for the x coordinate it is:

Xc = [(Ya - Yh)(Xb^2 + Yb^2) + (Yh - Yb)(Xa^2 + Ya^2) + (Yb - Ya)(Xh^2 + Yh^2)]
       -------------------------------------------------------------------------------
                      2*[Xb*Ya + Xa*Yh + Xh*Yb - Yb*Xa - Ya*Xh - Yh*Xb]

I think you can get the equation for the Y coordinate by swapping all the X's and Y's, but haven't checked that yet

2) Work out the radius of the arc, this can be done by calculating the distance between C and A (or C and H, or C and B).

r = sqrt((Xa - Xc)^2 + (Ya - Yc)^2)

3) Work out the the start and finish angles around the arc. The easiest way to do this is using a function called ATAN2 if you are using C or something similar.

Start angle = angle between horizontal and the line joining C and A = atan2(Ya - Yc, Xa - Xc)
Finish angle = angle between horizontal and the line joining C and B = atan2(Yb - Yc, Xb - Xc)

4) any angle between the start angle and finish angle will give you a point along the arc, so you can use code like the following pseudo code to generate these points:

  for each angle theta between the start angle and finish angle do the following
    calculate the x coordinate as Xp = Xc + r * cos(theta)
    calculate the y coordinate as Yp = Yc + r * sin(theta)
    output the point on the arc as (Xp,Yp)

Would somone be able to check this... or is there a way to do it that is simpler?


Let $(f_i)$ be a sequence of functions defined by: $f_1(x)=x, f_n(x) = \sqrt{f_{n-1}(x)}-\dfrac14$. ($n\in \mathbb{N}, n\ge2$)
(a) Prove that $f_n(x) \le f_{n-1}(x)$ for all x where both functions are defined.
(b) Find for each $n$ the points of $x$ inside the domain for which $f_n(x)=x$.

(a) put $f_{n-1}(x) = y$.  Then we have, by definition, to show $\sqrt{y} - 1/4 \le y \Leftarrow (y-1/4)^2 \ge 0$ which establishes it for $y \ge 1/4$, and $y \le 1/4$ is obvious (LHS negative.)

(b) $f_1(x) = x$.  From the first result, $f_1(x) \ge f_2(x) \ge \dots \ge f_n(x)$ so that we must have equality everywhere if $f_n(x) = x$.  So $\sqrt{x} - 1/4 = x$ is a quadratic in x with solution x= 1/4 only.


geometry
3D geometry
sphere
geometry proposed

On a circle of radius n, with n natural , are m chords such that for any point in the circle there is a chord such that the distance from that point to the chord is   <= 1. Prove that m >= n.

This idea is on the verge of becoming too old, unfortunately :).

Consider the circle as the projection of a sphere of radius $n$ on the plane, and in this sphere, consider some spherical segments which are projected onto bands in our circle, each band being contained between two parallel lines, each line a distance $1$ from one of our chords.

The hypothesis tells us that the segments cover the sphere (the area of the sphere), so the sum of their areas must be at least $4\pi n^2$. However, the lateral area of a segment with height $h$ is $2\pi nh$, and since all our segments have height $\le 2$, we get the desired conclusion.


Please , is this true or false ? Give an elementary answer.

This is a rewording of the following theorem:
[quote="[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18472][/url]"][b]Theorem 1.[/b] If ABC is a triangle, and P is a point in its plane, then the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA, respectively, concur at one point.[/quote]
applied to P = D.

A proof can be found e. g. in [url=http://www.cip.ifi.lmu.de/~grinberg/Isogonal.zip][i]Isogonal conjugation with respect to a triangle[/i][/url].

  Darij

Vielen Dank Darij !  I will read your paper carefully .

[u] Babis[/u]


Show that $ 2\sqrt {n \plus{} 1} > \frac {1}{\sqrt {1}} \plus{} \frac {1}{\sqrt {2}} \plus{} \cdots \plus{} \frac {1}{\sqrt {n \plus{} 1}} > \sqrt {n \plus{} 1}$


[color=blue] [size=150]$9^{th}$  Iberoamerican Olympiad[/size]
Fortaleza - Ceara, BRAZIL. [1994]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote] [b]Problema 2.i)[/b] Sea $ ABCD$ un cuadrilátero cíclico y suponga que existe una circunferencia centrada en el lado $ BC$ y tangente al resto de sus lados, probar que $ AB \equal{} AD \plus{} BC.$[/quote]
Sea $ O$ el centro de la circunferencia centrada en $ AB$ tangente a $ AD,DC,CB.$ Los rayos $ OC$ y $ OD$ son bisectrices de $ \angle BCD \equal{} \theta$ y $ \angle CDA \equal{} \varphi.$ Tomando un punto $ M$ en $ AB$ tal que $ AD \equal{} AM$ se forma el $ \triangle ADM$ isósceles del cual

$ \angle DMA \equal{} \frac {\pi \minus{} \angle DAB}{2} \equal{} \frac {\pi}{2} \minus{} \frac {\pi \minus{} \theta}{2} \equal{} \frac {\theta}{2}$

Por tanto, como $ \angle DCO \equal{} \frac {\theta}{2}$ $ \Longrightarrow$ El cuadrilátero $ DCOM$ es cíclico y por consiguiente $ \angle CMB \equal{} \angle ODC \equal{} \frac {\varphi}{2}.$ Por otro lado en el $ \triangle CMB$ se tiene:

$ \angle MCB \equal{} \pi \minus{} \frac {\varphi}{2} \minus{} (\pi \minus{} \varphi) \equal{} \frac {\varphi}{2} \Longrightarrow \ \angle CMB \equal{} \angle MCB \equal{} \frac {\varphi}{2}$

$ \triangle CMB$ es pues isósceles con base $ CM,$ así  $AB \equal{} AM \plus{} BM \equal{} AD \plus{} BC.$

El problema #4 puede ser generalizado de la siguiente manera:

[b]Problema:[/b] Construir el punto $ P$ interior al $ \triangle ABC$ de modo que los rayos $ PA,PB,PC$ corten los arcos $ BC,CA,AB$ de su circunferencia circunscrita en vértices de un triángulo acutángulo $ \triangle XYZ$ (triángulo circunceviano de P) que sea  semejante a otro dado.

[b]Solución:[/b] Denotemos $ \alpha' , \beta' , \gamma'$ los ángulos dados en $ X,Y,Z.$ Usando el Teorema generalizado de Simson, una inversión con centro en $ P$ y potencia arbitraria transformará los vértices de $ \triangle ABC$ en vértices de otro triángulo semejante al  triángulo pedal $ \triangle A'B'C'$ de $ P$ con respecto a $ \triangle ABC.$ Particularmente $ X,Y,Z$ son pues los inversos de $ A,B,C$ a través de la inversión centrada en $ P$ con potencia igual a la que tiene $ P$ respecto a el circumcírculo $ (O)$ de $ \triangle ABC.$ Así, el problema consiste en hallar el $ P$ cuyo triángulo pedal sea semejante al dado.

Las perpendiculares a $ A'B'$ y $ A'C'$ desde $ C$ y $ B$ concurren el en conjugado isogonal $ Q$ de $ P$ en $ \triangle ABC,$ entonces de los cuadriláteros cíclicos $ PA'BC'$ y $ PA'CB'$ tenemos que

$\angle QBA = \angle PBC'= \angle PA'C' \ , \ \angle QCA = \angle PCB' = \angle PA'B'$

$ \Longrightarrow \angle B'A'C' = \angle PBC = \angle PCB = \pi \minus{} \angle BQC$

Analogamente obtenemos $ \angle CQA \equal{} \pi \minus{} \angle C'B'A' \ , \angle AQB \equal{} \pi \minus{} \angle B'C'A'.$

Así, $ Q$ es construible ya que mira a los lados $ BC,CA,AB$ bajo ángulos suplementarios a los dados $ \alpha' , \beta' , \gamma'.$ Bastará pues con construir, por ejemplo, los arcos que miran a $ BC,CA$ bajo $ \pi \minus{} \alpha'$ y $ \pi \minus{} \beta',$ éstos se cortan en $ Q.$ Luego se sigue la construcción de $ P$ y su triángulo circunceviano buscado $ \triangle XYZ.$

[quote][b]Problema 5.[/b] Sean $n,r$ dos enteros positivos. Se quiere formar $r$ subconjuntos $A_1,A_2,...,A_r$ del conjunto $\{0,1,...,n-1\},$ de modo que todos los subconjuntos contengan exactamente $k$ elementos tal que, para todo entero $x$ con $0 \le x \le n-1$ exista $x_1 \in A_1,$ $x_2 \in A_2,..., x_r \in A_r$ (un elemento de cada conjunto) con $x=x_1+x_2+...+x_r.$ Además, hallar el valor mínimo de $k$ en función de $n$ y $r.$[/quote]
Para todo $x_i\in A_i,$ considere la función $f(x_1,x_2,\ldots,x_n)=x_1+x_2+\cdots+x_n.$ Se quiere encontrar el valor más pequeño de $k$ tal que la imagen de $f$ incluya a todos los números $0,1,2,\ldots,n-1.$

El tamaño óptimo ocurre pues cuando $f$ is injectiva, pero inclusive asumiendo ésto, se tiene que la imagen de $f$ es a lo más  $k^r.$ Entonces, si $k^r<n,$ existen $x\in [0,n-1]$ que no pueden ser calculados. Ahora, considérese $k=\lceil n^{\frac{1}{r}}\rceil.$  Si probamos que ésto es posible, entonces el problema se habrá resuelto. Entonces, consideremos los $r$ conjuntos

$A_1=\{0,1,\ldots,k-1\},$
$A_2=\{0,k,2k,\ldots,k(k-1)\},... $
$A_r=\{0,k^{r-1},2k^{r-1},\ldots,(k-1)k^{r-1}\}.$

Finalmente, si ninguno de los elementos de $A_r$ son mayores que $n-1,$ entonces simplemente remplazelos, por decir, con 1. Es facil probar que ésto es equivalente a escribir $x$ en base $k$ con $r$ dígitos y de hecho todo $x \in [0,n-1].$


counting derangement geometry MATHCOUNTS AoPSwiki factorial search

Does anybody know what derangements are? I don't understand them.

Someone JUST gave links for derangements in the marathon:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=104081&start=1840[/url]

What math are they located in? :?: I mean, are they under algebra, geometry, number theory, etc. And what math course do you usually learn them in? i.e. geometry, algerbra II, etc.

I'd say they would have to be considered Combinatorics.

I read the link but I don't really understand why.

There are 2 links to the Wolfram MathWorld site in the post: did you read those?

Derangements are basically the ways that you can organize some items so NONE of them are in the correct places.

[quote="CircleSquared"]There are 2 links to the Wolfram MathWorld site in the post: did you read those?[/quote]Mathworld pages are written at the level of a professional mathematician.  Although they are a good resource for advanced material, one can hardly expect Mathcounts-level students to be able to read and understand them.

Why not instead link to the [url=http://www.artofproblemsolving.com/Wiki/index.php/Derangement]derangement entry on the AoPSWiki[/url] and, if necessary, edit it to make it more accessible for students?

I meant I don't know why the formula works.

I've never seen the proof of the derangements theorem, but my guess is that it uses pretty advanced math.  If you want to find the "why it works", maybe try a higher level forum?

One thing I've never understood is why it's written as if it's the inverse of factorial.  :roll:

It's probably because factorial is often the number of ways to arrange some items
and derangements (aka subfactorial) is the number of ways to arrange the items so that they're all in the wrong places.

I heard the proof of the formula is intuitive.

Hmm....

Ok, I'll try to find it on the higher level forums and see.

EDIT: Here it is!!!  Well, kinda, at least.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1021143590&t=98943]A post with a good amount of information, and a problem involving derangements[/url]

Unfortunately I don't quite get the recurrence relation.

For example what if you have the n-1 = 5 derangement 23451, then there is only 1 way to insert the 6, not n-1.

Going backwards, if you have the n = 6 derangement 263145, when you take out the 6 and put it on the end, you get (23145)6, which is not a derangement.

What level math is this stuff?

Dont feel bad if you are having trouble with derangements.  Derangements are above the level of MATHCOUNTS.  Once in a while, there will be a problem that can be classified as a "derangements problem", but it will involve very small numbers.  Like "How many ways can you put 3 letters in 3 envelopes, such that everybody receives the wrong letter?"

Technically, the above is a "derangements" problem, but you can enumerate both possibilities quite easily.

If you want to learn something a little beyond MATHCOUNTS, the only thing I would suggest is learning that $!n\approx \frac{n!}{e}.$  Read this as "The number of derangements of n is approximately equal to the factorial of n, divided by the constant, e."  In fact, the number of derangements is the closest integer to the factorial of n, divided by the constant, e.

$!3 = \frac{3!}{e}= \frac{6}{2.718}= 2.21 \approx 2$

$!4 = \frac{4!}{e}= \frac{24}{2.718}= 8.83 \approx 9$

$!5 = \frac{5!}{e}= \frac{120}{2.718}= 44.1 \approx 44$

I doubt MATHCOUNTS would take you any higher.

BTW, a detailed derivation of this particular formula requires an understanding of infinite series, which you will learn in AP Calc BC, or similar.  [This is one reason why you need to look in a higher-level forum for a comprehensive discussion.]

derangements... ummm, there is an equivalent formula, but it definitely isn't as easy, and i have no idea how to get the other formula D(n)=(n-1)(D(n-1)+D(n-2)) from this one.  So, it is basically finding all possible arrangements, subtracting the ones with one in the right place, adding the ones with two in the right places, subtracting the ones with three in the right place, and so on.  We do this because, as we subtract the ones with one in the right place, we are also subtracting the ones with two, three, four, ... in the right place.  So we alternately add and subtract to even it out.