logarithms
LaTeX
inequalities
function
algebra
polynomial
floor function

assume that k is a positive integer and A ={p1,p2,...,pn} is a subset of prime number . prove that the number of the solution of x-y=k, which the prime divisors of them are in A ,is finite?
(I found it realy hard after 4 hours struggling)

As I remember it is only conjecture. Where did you find this problem?

I found it in a famous book in iran as a hard problem
i asked it several times but no one could answer so i post it here.
are you , Myth, sure that it's an open question?

I do not know about Myth references. Clearly, this problem is an open problem if we do not consider a finite set A of primes.
But, maybe the fact that A is supposed to be finite leads to a solvable problem....

Pierre.

I think exact text of conjecture is "equation ax-by=c has finite number solution in positive integers, where a, b and c are coprime integers". It was published in some book, and I read it 10 years ago. Yesterday I tried to find the source but I failed. Maybe I am wrong (?).

Well, to know if each natural number can be represented as the difference of two primes is an open problem.

Pierre.

Well, it looks to me like the conjectures and open questions you guys have mentioned aren't really the same thing as the problem posted by sam-n, but now I'm afraid to even give it a try..

is this hard .but I think I can prove this .this is problem not very hard with every body  :D

I found the foillowing problem to be connected to this problem (for the case n=2):

Let $k\g>0$ be a fixed real and $r>0,r\in R-Q$. Show that $2^n\{nr\}<k$ only for finitely many $n$.

I don't know if it's true, but it seems so. In our problem we can put $r=\log_pq$, where p and q are our primes and we get something very similar (not exactly the same though). Don't mind the way I reached it (it's not that hard anyway). I'm just curious about this new fact.

Opinions? It looks impossible to tackle (to me):D

[quote="grobber"]Let $k\g>0$ be a fixed real and $r>0,r\in R-Q$. Show that $2^n{nr}\lk$ only for finitely many n.[/quote]
Just fixing grobber's $\text{\LaTeX}$
...Let $k>0$ be a fixed real and $r>0,r\in R-Q$. Show that $2^n{nr}<k$ only for finitely many n...

Terribly sorry! There was another mistake: I meant $\{nr\}$ instead of just $nr$. I've edited it.

[quote="grobber"]Let $k\g>0$ be a fixed real and $r>0,r\in R-Q$. Show that $2^n\{nr\}\lk$ only for finitely many n.[/quote]
Sorry that I'm being picky but, grobber, your statement is not clear. This is your code: 2^n\{nr\}\lk
So, which one of the followings do you mean?

$1)$ The [u]expression[/u] $2^n\{nr\}\lk$ produces finitely many values for finitely many $n$.

$2)$ The [u]inequality[/u] $2^n\{nr\}<k$ is satisfied for finitely many $n$.

My opinion: I think you mean #2 because $k$ would be [b]obsoletely[/b] defined (in other words, $k$ would be useless). Again please clarify it for our readers (and me  ;) ) Thanx

PS: What do you mean by the operation $\{\cdot\}$? Is $\{A\}$ the fractional part or decimal part of $A$? i.e. Is it $\{A\}=A-[A]$, where $[A]$ is the greatest integer function?

$\{x\}$ is, indeed, the fractional part of x.

The statement is that the inequality is satisfied for finitely many n, so it's 2) in your line of questioning :)

I hope it's clear now and I hope someone has an idea..

Actually, the problem samn posted is not at all open and it follows from a very hard theorem in transcendental number theory. I will state the theorem ( I know almost nothing in this field, so I will write it as I found it):
  Suppose $ a_i, i=1,r $ are non-zero algebraic numbers with degrees at most d and heights at most A. Also suppose $ b_i, i=0,r $ are algebraic numbers with degrees at most d and heights at most B, where $ B>1$. Let us suppose that $ K=b_0 +b_1\cdot log a_1+...+b_r\cdot log a_r $ is non-zero. Then we find $ C=C(r,d)>0 $ and $ w=w(r)\geq 1 $ such that
 $ |K|\geq B^{-c\cdot (logA)^w} $. We say here that an algebraic number a  has degree d and height A if it is root of an irreducible polynomial with integer coefficients, of degree d, the common divisor of its coefficients is 1 and the maximum among all absolute values of the coefficients is A.
  Now, this theorem easily implies the following result:

  Let A a finite set of prime numbers and let $ 1=n_1<n_2< $ the set of numbers which have all their prime factors in A. Then there exists a constant $ c=c(P) $ such that for all j we have $ n_{j+1}-n_j>\frac{n_j}{(log n_j)^c} $ and from here the problem samn posted.
  I still wonder what elementary solution could that problem have.

I'm just the one who wrote it. No contribution. By the way, what have you done at IMC? It's a gold medal, isn't it?

nobody know yet, at this point the scores are like this

me - 123 points, 36th
claudiu - 119 points, 42nd
kappa - 109 points, 65th (or something like that)
ana caraiani - 72 points (!!)

some of the people have complaints about the gradings, and we shall see later how the final score table will look.

Anyway, if it matters for something, I found the problems quite tough. Especially the 5 and 6 problems from each day. They are really monstruous.

could you,Harazi give us more reference about your Lemma where can find it an its proof,btw it shouldn't be such a hard that we cant solve it with elementary soln because the book that mentioned it is elementary book.!!!

Tomorrow I will post the refference. I have to leave now. Bye! Ps: if there is an elementary solution, then find it. The problem seems impossible for me.


modular arithmetic
Euler
induction
algebra
polynomial
function
number theory

Show that
\[ \sum_{d|n,d>0}\mu(d) x^{n/d}\equiv 0\pmod{n}\]
for any positive integer $ n$ and any integer $ x$.
For example,
\[ x^{9}-x^{3}\equiv 0\pmod{9}\]
\[ x^{6}-x^{3}-x^{2}+x\equiv 0\pmod{6}\]
for any $ x$.

Nice result!

First we prove it for prime powers. It suffices to show that, for any prime p and integer x:
$ x^{p^{k}}-x^{p^{k-1}}\equiv 0 \pmod{p^{k}}$
$ x^{p^{k-1}}\left( x^{(p-1)p^{k-1}}-1 \right) \equiv 0 \pmod{p^{k}}$
$ x^{p^{k-1}}\left( x^{\phi(p^{k})}-1 \right) \equiv 0 \pmod{p^{k}}$
If x is coprime to p, then $ \left( x^{\phi(p^{k})}-1 \right) \equiv 0 \pmod{p^{k}}$ by Euler-Fermat's theorem. If $ p \mid x$, then $ p^{k}\mid x^{p^{k-1}}$ because $ p^{k-1}\ge k$ ($ p \ge 2$ and then trivial by induction)

Now for the general n. If p is a prime that divides n, and $ n = p^{k}m$, with m coprime to p, it suffices to show that:
$ p^{k}\mid \sum_{d \mid n}\mu(d) x^{\frac{n}{d}}$
$ p^{k}\mid \sum_{d \mid p^{k}m}\mu(d) x^{\frac{n}{d}}$
$ p^{k}\mid \sum_{d \mid m}\sum_{e \mid p^{k}}\mu(d) \mu(e)x^{\frac n{de}}$
$ p^{k}\mid \sum_{d \mid m}\mu(d) \sum_{e \mid p^{k}}\mu(e) x^{\frac n{de}}$
$ p^{k}\mid \sum_{d \mid m}\mu(d) \sum_{e \mid p^{k}}\mu(e) \left(x^{\frac{m}{d}}\right)^{\frac{p^{k}}e}$.
But if $ P(x) = \sum_{e \mid p^{k}}\mu(e) x^{\frac{p^{k}}e}$, we know that $ p^{k}\mid P(x)$ for every x, and we can write the previous formula as:
$ p^{k}\mid \sum_{d \mid m}\mu(d) P(x^{\frac{m}{d}})$

And this is of course true.  :)

Nice work edriv.
If $ x$ and $ n$ are both positive, one can show using Mobius inversion that
\[ \sum_{d|n,d>0}\mu(d) x^{n/d}\]
is the number of mappings $ f$ of $ Z/nZ$ into a set of $ x$ elements such that for any nonzero $ y\in Z/nZ$ there exists $ x\in Z/nZ$ such that $ f(x)\ne f(x+y)$. This means, a little roughly, that $ f$ has no period except multiples of $ n$. The mappings with that property fall into equivalence classes each of which has $ n$ elements; $ f$ and $ g$ are in the same equivalence class if $ f(x+y)=g(x)$ for all $ x$ and some $ y$.

(edit:)
We can go from the case of positive $ x$ to the general case of all $ x$ using this lemma:
If $ k$ is a polynomial in one variable over $ Z$, of degree $ a$, and $ k(x)\equiv 0\pmod{n}$ for $ a+1$ consecutive values of $ x$, then $ k(x)\equiv 0\pmod{n}$ for all $ x$.
A proof is easy by induction on $ a$; the polynomial $ k(x+1)-k(x)$ has smaller degree than $ k$ and is zero mod $ n$ at $ a$ consecutive points.

There's an easier way to go from the case of $ x>0$ to the general case: Add a multiple of $ n$ to $ x$. :oops:

 :D

Let $ f(A,B,...)$ be a polynomial in any number of variables over $ Z$, and let $ n$ be a positive integer. Define a polynomial $ F$ by

\[ F(A,B,...) \equal{} \sum_{d|n} \mu(d) f(A^d, B^d, ...)^{n/d}\]

Then all the coefficients of $ F$ are divisible by $ n$. Fermat's little theorem is the special case in which $ n$ is a prime and $ f$ is a constant!  :showoff:

I discuss this result at length [url=http://qchu.wordpress.com/2009/08/23/newtons-sums-necklace-congruences-and-zeta-functions/]here[/url] and [url=http://qchu.wordpress.com/2009/11/04/newtons-sums-necklace-congruences-and-zeta-functions-ii/]here[/url].

Thanks Qiaochu, but in #5 I'm saying every /coefficient/ of $ F$ is multiple of $ n$, not just the values of $ F$. But the stronger statement is not much deeper. In the special case
\[ f(A,B,...)\equal{}A\plus{}B\plus{}...\]
the coefficients of $ F$ have an interpretation in terms of aperiodic functions $ Z/nZ \to \{A,B,...\}$, as you know, and the general case can then be reached by replacing each monomial in $ f$ with a sum of new indeterminates.

By the way, there is no harm in allowing the number of variables to be infinite.

Yes, this is all clear from the combinatorial interpretation.


How many ways can the word $ FATHER$ be arranged such that no vowel is in an even position? (e.g. FATHER would not be allowed because the letter A is occupying position 2)

Okay, so we have a restriction. We deal with that first.

In FATHER, the vowels are A and E. We place them first.

_ X _ X _ X

So there are 3 spots for the two vowels, so $ P(3, 2) \equal{} 6$ ways to place the vowels.

Then we move on to the consonants. We have four spots left over for four consonants; that's just $ 4!\equal{}24$.

So the number of ways to place the vowels AND consonants is just the product of the above two numbers, or $ 6 \cdot 24 \equal{} \boxed{144}$.


I accidentally cracked it.  :?

I was doing a project that used pattern recognition, and it could successfully predict the windows random number generator with 50% accuracy (out of three choices). It had not been specifically designed for that purpose.

Just warning you.

well, is it possible to make a code to generate random number?

It has of cource got a pattern...

The "rand()" function for C/C++ (and probably for other languages too) is only "pseudorandom".

A simple way to see that, is to place rand() in a loop and print out the numbers... They will repeat after a while. ;)

True random functions is nearly impossible to create... Usually programs like encryption programs measure mouse movement, time, CPU usage and other things like that to get a "random" number. But even those number-generators isn't "completely random"... :D

well, thanks a lot. i did not knew those methods of generating so called "random number".

yeah, the built-in random number generator is pseudorandom, and pretty bad at that.  also, it differs from compiler to compiler, which is why you have to make use of the constant RAND_MAX

of course it's possible.  With easy javascript too.  Math.random for a number from 0-1 like .563585149722415... and multiplied by 10 gives you from 0-1 take the ceiling and you get integers from 1-10

There is no such thing as real "computer-generated random" numbers.  They are not random.  They usually have a seed of the time.  A professor told me that some people use radiation to generate random numbers.  These numbers are a good bit more random and much less predictable, but it's pretty dangerous and probably hard to use :lol:

[quote="xtrememath8"]I realized this a while ago. I program with Visual Basic. The "random" numbers were the same every time I ran the program. I was able to make a random number when a button was pushed, however, by creating a timer object that increased a value by one every fraction of a second. When the user pressed the button, the timer stopped, and the value was relatively random. Thus, the user generated the random number, not the computer.[/quote]

I haven't used VB in a long time (ditched that many years ago and switched to C/C++),  but don't you have to seed the random number first?  Can't you re-seed the random number generator when the button is clicked?

No clu WHAT that means. I'm not really a hardcore, gungho programmer.  :wallbash_red:  :weightlift: I just know that you press the button, and the value stored at that time is relatively random.

The only way to make a truly random number would be to look at a radioactive atom* and if it split (or something  :what?:) would do something. Either that of have a computer role millions of dice and have them return the answers every second.

*anyone with physics, go ahead and correct this.

Not sure about this.

I think there actually are quantum random-number generators available! It did sound weird to me, but you can google some out:

[url=http://www.idquantique.com/products/quantis.htm]www.idquantique.com/products/quantis.htm[/url]

It can be done (as IntrepidMath said) with decaying atoms, measuring electron spin, checking if a single photon passed through a half-reflecting mirror or similar experiments, which are believed to be truly physically random...


superior algebra
superior algebra unsolved

Let $ F$ be a set of elements with two well-defined binary operations, $ \oplus$ and $ \otimes$ on its elements. Now, suppose that $ (F,\oplus)$ is a group with identity element $ 0$ and $ (F\backslash 0,\otimes)$ is a group.  In addition, for any $ x$, $ y$, $ z\in F$, $ x\otimes (y\oplus z)\equal{}(x\otimes y) \oplus (x\otimes z)$ and  $ (x\oplus y)\otimes z\equal{}(x\otimes z) \oplus (y\otimes z)$.

Show that $ \oplus$ is commutative.

Hint: Expand $ (1 \oplus x) \otimes (1 \oplus y)$ in two ways.


geometry
3D geometry
number theory proposed
number theory

hallo,
prove that ther doesnt exist a perfect cube of the form $2a(a^{2}+3b^{2})$ where $a, b \in N$.  :)

[quote="PTynan89"]What, just try $b = a$, for any $a \in \mathbb{N}$.

Then you have $2a(4a^{2}) = 8a^{3}= (2a)^{3}$,
which is obviously a perfect cube.[/quote]

hallo,
o sorry i forgot to say $a \ne b$ sorry  :)

If a=bmod2 then there are two possitive integers x,y with a=x-y and b=x+y so the equation is 2a(a^2+3b^2)=z^3 => 2(x-y)(4x^2+4y^2+4xy)=z^3 =>(2x)^3-(2y)^3=z^3  and this equation hasn't  possitive integers solutions. :maybe:

i have the same solution as Αρχιμήδης 6. do you have another one ?

unfortunately I don't have  a solution  :(

It's better to set a=(x+y)/2 and b=(x-y)/2 so the equation has been solved. :wink:


logarithms
inequalities
inequalities unsolved

$ a\geq1$ , $ b\geq1$ prove:

$ alna\plus{}e^{b\minus{}1}\geq ab$

[$ lna\equal{}log_{e}a$]

Consider :
$ f(b) \equal{} e^{b \minus{} 1} \minus{} ab \plus{} a\ln{a}$
$ f'(b) \equal{} e^{b \minus{} 1} \minus{} a$
$ f''(b) \equal{} e^{b \minus{} 1} > 0$
So $ f(b)\geq f(\ln{a} \plus{} 1) \equal{} 0$
So the inequality claim.
The equalities hold when $ b \equal{} \ln{a} \plus{} 1$

[hide="A bit another solution"] Your problem equivalent with:
$ \frac{e^{b\minus{}1}}{a}\geq b\minus{}lna\Leftrightarrow e^{b\minus{}lna\minus{}1}\geq b\minus{}lna$
It is OK because $ e^x\geq x\plus{}1$[/hide]


geometry
parallelogram
geometry unsolved

The point $ K$ is selected inside parallelogram $ ABCD$ such that the midpoint of $ AD$ is equidistant form $ K$ and $ C$ and such that the midpoint of $ CD$ is equidistant from $ K and$A. Let $ N$ be midpoint of $ BK$. Prove that angle $ NAK =$ angle $ NCK$

Very nice problem.
Solution:
Let $ P$ and $ Q$ be the midpoints of $ AD$ and $ BC$ respectively.Let $ d$ be the line through $ D$ and $ d\|AC$.
Denote $ A_{1}=BA\cap d$ and $ C_{1}=BC\cap d$.Since $ d\|AC$ we have that $ ACC_{1}D$ and $ ACDA_{1}$ are parallelograms.
Since $ Q$ is midpoint of $ CD$ then $ A,Q,C_{1}$ lies on one line.The same way we prove that $ C,P,A_{1}$ lies on one line too.
So we have that $ CP=PA_{1}=PK$ and $ QC_{1}=QA=QK$ $ \Rightarrow$ $ \angle C_{1}KA=\angle CKA_{1}=90$ $ \Rightarrow$ $ \angle AKA_{1}=\angle C_{1}KC$
Now since $ N$ is midpoint of $ BK$ and and $ A$ is midpoint of $ BA_{1}$ we have that $ \angle AKA_{1}=\angle NAK$.The same way $ \angle CKC_{1}=\angle NCK$.
So we have $ \angle NAK=\angle NCK$.


How do you prove that the polar equations r = 2 cos (x) - 1 and r = 2 cos (x) + 1 are the same curve?

I tried plugging in points, but that doesn't work. So I dont know how to start this problem...any help will be greatly appreciated.

Thanks

Multiply through $ r$ and we get

$ r^2 \equal{} 2rcos(\theta) \plus{} r$ and $ r^2 \equal{} 2rcos(\theta) \minus{} r$

Note that in the cartesian plane $ x^2\plus{}y^2 \equal{} r^2 \Longrightarrow r\equal{}\pm \sqrt{x^2\plus{}y^2}$

$ \Rightarrow r\equal{}\minus{}r$

but how does that prove that the 2 original functions are equivalent?
 r = -r just shows that it is an even "function", right?

It suffices to show that the following two parametric polar curves are equivalent:
$ (2 \cos t \minus{} 1, \; t), \; t \in \mathbb{R}$
$ (2 \cos s \plus{} 1, \; s), \; s \in \mathbb{R}$

[hide="Hint"]What happens when $ s \equal{} t \plus{} \pi$?[/hide]


geometry
congruent triangles
combinatorics unsolved
combinatorics

Hello you all, I am new here. I wish to open with a question...Given a triangle, where each one of it's sides equals exactly 1. Also given 5 dots. Prove that the maximum length between every two dots is 0.5. All dots should be located inside the triangle or on one of its sides.
Clue:
use the pigen hole principle!

Awaiting answers...

I think that you mean that there are two points whose max distance is $\frac{1}{2}$.  This problem doesn't belong in this forum either, it should probably be moved to the intermediate one.

Divide the triangle into four congruent triangles with side length $\frac{1}{2}$.  One of these triangles has two points.  The result follows.

Thanks a lot zgorkster.
And yes it should be in the intermidiate section - I did not realize that the site was divided into difficulty levels as well.
Anyway, thanks again... nice solution! :lol:


How would I figure out how many multiplications it would take to evaluate this polynomial:

[url=http://img488.imageshack.us/img488/9946/untitled4nh.jpg]Polynomial[/url]

when they are preformed as indicated.  Also how would I write the number of multiplications as a function of n?

This is a for a Calculus project come up that I have to study for.. we have not yet gone over it (6 chapters away).

If u use synthetic division, there won't be very many multiplications:

i.e. $2+3x+4x^2+5x^3+6x^4=2+x(3+x(4+x(5+x(6))))$ which has only 4 multiplications. In general, the number of multiplications for degree $n$ will be $n$ this way, except for leading coefficient 1 which has $n-1$ multiplications. The question is, is there a faster way...?


I want a proof without induction for the following statement:
$ 5^n \equiv 5 \pmod{10} \quad \forall n \in \mathbb{N}^*$

I think it's easy, but it's now beyond my skills...  :blush:
Thanks in advance!

[hide="inductionish"]
We are trying to prove that the last digit of any positive power of 5 is 5.
Base: n=1
$ 5^n\equal{}5$ true
Inductive Step: If $ 5^n$ ends in 5, then $ 5_(n\plus{}1)$ ends in 5.
This is easy to see because if the last digit of $ 5^n$ is 5 and we multiply that by 5, it's still 5.
[/hide]

[hide="without induction"]10 is even, and 5 is odd. We prove that a power of an odd number cannot be an even number:

$ (2x+1)^n=2^n x^n+n*2^{n-1} x^{n-1}+\cdots+(-1)^{n-1}n*2x+(-1)^n$

We can group some of the terms:

$ 2(2^{n-1}x^n +n*2^{n-2} x^{n-1}+\cdots+(-1)^{n-1}*x) \pm 1$

That is obviously odd. Therefore, the power of any odd number is never even.

Now a multiple of 5 must have a units digit of 0 or 5. If the units digit is 0, then the multiple is even, but $ 5^n$ can't be even, and therefore must have a units digit of 5.

$ \mathtt{QED}$[/hide]


parameterization
trigonometry
calculus
calculus computations

Hint: Use the variation of parameters to find the solution.
You will first have to solve the equation $ \frac{d^2x}{dt^2} \plus{} \omega^{2}x \equal{} 0$.
The complimentary solution should be something like $ x_{c}(t) \equal{} c_{1} \cos \omega t \plus{} c_{2}\sin \omega t$.


An 80-pound box of apples will have 10$ \%$ of the apples spoil. At the same rate, how many pounds of spoiled apples will there be for every 2160 pounds of unspoiled apples?

The answer is actually 240.

At least it is on FTW.

Yes, because dragon96 you only calculated the pounds of apples.  There will be $2400 - 2160 = \boxed{240}$ unspoiled apples.

[hide=Solution]$2160=80\cdot 27$, and in an $80$ pound box, there will be $8$ pounds of spoiled apples, so there are $8\cdot27=\boxed{216}$ pounds of spoiled apples.[/hide]


linear algebra
matrix
vector
linear algebra unsolved

Is it true that every symmetric matrix is diagonalizable? If yes, how can it be proved? Also can we say that a symmetric matrix has only real eigenvalues? Also i guess determinant of a symmetric matrix is the product of its eigenvalues, and trace of that is the sum of its eigenvalues. Is it true? Why? Lastly, does the transpose of any square matrix have some special properties related to diagonalization?

Yes a $real$ symetric matrix is diagonalisable in $M_n(R)$ ,it might be proved by recursion , and its eigenvalues are real ,regarding the trace and the determinant the answer comes from the fact that matrix has got the same trace and det of the diagonal matrix as they're similar.
There is nothing special related to the transpose ,but in general $^{t}A$ is similar to $A$

Thanks for your interest. Actually i proved that trace of a symmetric matrix is the sum of and determinant of that is the product of its eigenvalues. It wasn't that hard, but i couldn't prove that every real symmetric matrix is diagonalizable although i did it for 2x2 and 3x3 matrices. Anyway, here is another question which i don't know where to start : If the sum of entries in each row of a matrix is c, prove that c is an eigenvalue of this matrix.

For the second question: consider the vector $v=[1\,1\,\dots\,1]^T.$ Note that $Av$ produces a vector, each element of which is the sum of a row of $A.$

There are a number of different ways of proving that a real symmetric matrix is diagonalizable.

Here's one argument. Suppose $A$ is real and symmetric. $A$ has at least one complex eigenvalue and eigenctor.

Suppose that $Av=\lambda v.$ Use the $A^*$ to mean the conjugate traspose of a complex matrix $A.$

$v^*(\lambda v)=\lambda v^*v=\lambda\|v\|^2$

But $v^*(\lambda v)=v^*Av$. That's a $1\times 1$ matrix; it's conjugate transppose is its conjugate.

Hence $\overline{\lambda}\|v\|^2=\overline{v^*(\lambda v)}=\overline{v^*Av}$

$=(v^*Av)^*=v^*A^*v=v^*Av=\lambda\|v\|^2$ since $A^*=A.$

Thus, $\lambda$ is real. We can then choose $v$ to be a real vector. Also choose $v$ so that $\|v\|^2=1,$ which we can easily do.

Find an orthonormal basis of $\mathbb{R}^n$ that has $v$ as its first element. We can do that because we can always extend an independent set to a basis, and we can make that basis orthonormal by the Gram-Schmidt process. Let $P$ be the matrix whose columns are this orthonormal basis. $P$ is an orthogonal matrix - that is, $P^TP=I.$

Compute $P^TAP.$ The first column of $AP$ is $Av,$ which equals $\lambda v.$ From this we get that we can write $P^TAP$ in block-structured form as

$P^TAP=\begin{pmatrix}\lambda &w^T\\0 & A'\end{pmatrix}.$

But $(P^TAP)^T=P^TA^TP=P^TAP$ is symmetric. From that we see that $w^T=0$ and that $A'$ is itself symmetric.

Now work the same proof on the $(n-1)\times(n-1)$ matrix $A'.$ (This is, at heart, a proof by induction). We eventually work down to the existence of $Q$ such that $Q^TQ=I$ and $Q^TAQ=\text{diag}(\lambda_1,\lambda_2,\dots,\lambda_n).$ And the first argument shows that all of those eigenvalues are real.

What a nice and simple solution for the second question! Thanks Mr. Merryfield! But i think the proof of the fact that every real symmetric matrix is diagonalizable is anadvanced one. That is, (at least for now) it's beyond the scope of my linear algebra knowledge. But i have been studying hard, and i really like algebra. I will frequently try to post interesting questions or proofs here.


I really don't get this problem. the teach explained it, but i never really got it, and it remained as the unsolved mystery, soo somebody explain it please!

            x(2y+z)-(zw-yw)x/y

Umm... Could you explain what the problem is about. Cause if you just give us that we don't know if we're suppose to simplify or to solve for something.  :?

could you make the equation a little bit clearer by adding parentheses where required because it if har to tell if you want the whole equation as a fraction.  I assume not but in math its best to be certain

umm...can someone delete this thing please?? i posted this like a long time ago....


MATHCOUNTS
number theory
least common multiple

*phhhhhht* :) That was obvious :)
Maybe calling them base x would help. That's how I first learned the concept.

You don't need to 0(mod 2) condition. It's already clear from the other two.

MathFiend wrote:[hide]10...10/6=R4...10/4=R2...10/2=R0[/hide]

Mmmkay

and you're right, Simon, but it makes the whole thing easier. (well, minisculy, because it's really really obvious from the first two)

I managed to figure out the solution to this one, but how would one find the number n that satisfied

n = a (mod b) = c (mod d) = e (mod f)?

look...you don't need mods, you just need common sense. Think about what they are saying.

Hint:[hide]if I say something has remainder 4 when divided by 5, that means it is 1 less than a multiple of 5. Do the same thing here. Does that help?[/hide]

Or you could do [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=773]this[/url] method.
but modular arithmatic is much easier.

MysticTerminator:

Ok, so I can construct a bunxh of phrases saying n = 5a + 1, 7b + 3, 10c + 2, etc... where a, b, c are constants. But now what?

Tare:

Ok, I see...but that problem is slightly different...

no, fyty, i did not say to write them in that form. I said, write it in the form, [hide]n is 2 less than a multiple of 4, etc.[/hide]. The solution will be easier this way.

Ok...so say we do it your way. What then?

sigh, i changed around the question without thinking. its cause im perdy sure i posted the original question before. they were prime number thingies 5,7,3 were the mod things. i think tare linked it, but im too lazy

Trial and error? or it could be finding a pattern, which also takes a lot of time...

and I meant generically, like you could solve this problem MysticTerminator's way, but I really think modular arithmatic comes in handy, not only for straightout problems like these, but in other problems too.

Tare...hold that thought and we will discuss it later after this problem is solved.

This is a MATHCOUNTS question, and MATHCOUNTS questions will usually be of this form. For this type of question, my method is much easier.

Fyty, well, don't you notice a pattern? spoiler:[hide]the statements will say, n is 2 less than a multiple of 6 and n is 2 less than a multiple of 4. Well now, that means n is 2 less than something that is both a multiple of 6 and a multiple of 4. The smallest such number will be the LCM of 6 and 4, or 12. n will be 2 less than this, or 10. do you follow me? if you do, try this problem:

when n is divided by 14, there is a remainder of 11. when n is divided by 7, there is a remainder of 4. and when n is divided by 6, there is a remainder of 3. what is the smallest such n that satisfies these conditions?[/hide]

Yes, but you can't solve my more general problem to find the number n that satisfies

n = a (mod b) = c (mod d) = e (mod f)

That is, as in your latest example, 14 - 11 = 7 - 4 = 6 - 3 = 3. If the numbers were not so nice, it wouldn't have worked out.

fyty, what you ask is a very interesting question that has been studied in number theory extensively and culminates in the Chinese Remainder Theorem. google/mathworld it. or wait for an explanation from me tomorrow.

the reason i posted an "easy" problem is because this thread is preparing you for mathcounts. mathcounts would not ask you for the chinese remainder theorem; it would ask you for comparatively easy problems such as the one i posted. this is why i am teaching you the method for this sort of problem.

Ok, I will look up the Chinese Remainder Theorem...sometime (throws it into to-do text file of 35 lines)...  8-)

[hide]Its ten. Just use common sense[/hide][/hide]

people.....this topic was over a year old

*googles the chinese remainder theorum*
*reads what mathworlds has to say*
*eyes pop in confusion*
*tries another site*
*still doesn't get it*
*sighs, then decides that he'll ask his math teacher sometime later*
*But for now, he'll just stick to mods*

When you see a problem like that just find LCM and subtract 2...

This topic is old; it is large; was large before reviving it; had good solutions; the problem is a simple modulus problem, of which there are many newer ones; and nobody has contributed anything new since revival. All of these are not good for our image.

Please be careful about reviving topics this old. And this one, please let fade into oblivion. If you need more problems, post some...!

Don't take this personally, I used to do this myself before a moderator set me straight.

Billy

Syntax Error wrote:What is the least positive integer that meets the following conditions?

a)when divided by 6, the remainder is 4
b)when divided by 4, the remainder is 2
c)when divided by 2, the remainder is 0

Hint:[hide]use mods[/hide]

Answer10

-_O

To make this harder...see if you can do this one.

Jane has a lot of cookies.  When she has them in bags of 13 each she has 10 left over.  When she bags them in boxes of 50 each, she has 47 left over.  When she puts them in bags of 7 each she has 4 left over. Find, with fully written proof, the least number of cookies she can have.

[hide="My Answer"]Find the LCM and subtract 3
13x50x7=4550
4550-3=4547[/hide]

38 lol i didn't have to use mods i did a logical guess


Evaluate:

Integral from 0 to infinity x/(e^x-1) dx.

If you have seen this problem before, please do not post the solution. Thanks.

Uh, if you substitute u=e^x then it becomes the integral from 0 to infinity of ln u/(u-1)(u) du, which is the integral of ln u/(u-1) - ln u/u, and the latter term is easy by parts; as for the former, since ln u is an infinite polynomial in (u-1) (though I have forgotten what it looks like) it should not be that hard. But I think this solution is pretty lame.

[color=cyan]This looks like a complex analysis problem . . .[/color]

For the substitution, I mean to say the integral from 1 to infinity, duh.

You don't have to use complex analysis to solve this problem. You also don't need to substitute u=e^x.

I have a solution to this one that's not particularly elegant.

[hide]

First do an integration by parts letting u = x and dv = dx/(e^x - 1). Doing a substitution for e^x - 1, you can get that v = ln (1-e^-x).

So integral from 0 to infinity x/(e^x-1) dx

= lim (x -> inf) [x ln (1-e^-x)] - lim (x -> 0) [x ln (1-e^-x)] - integral (0 to inf) [ln (1-e^-x) dx]

= 0 - 0 (using L'Hopital) - integral (0 to inf) [ln (1-e^-x)]

Then making the substitution y = e^-x this last integral transforms to, integral (0 to 1) [ln (1-y)/y dy]. And the integrand of this can be expanded by taylor series to give you:

- integral (0 to 1) [ln (1-y)/y dy] = sum (n=1 to inf) [(-1)^(n+1)/n^2] = (pi^2)/12 (using the fact that sum 1/n^2 = (pi^2)/6).

[/hide]

I would do this way :

1/(e^x-1) =  :Sigma: e^(-nx)

int xe^(-nx) = 1/n:^2: (by parts)

inverting int and :Sigma: gives :Sigma:1/n:^2: = :pi::^2:/6

[quote="ComplexZeta"]Hm I think you went wrong somewhere because it should be :pi:^2/6. If anyone is interested in my solution, let me know. I sort of went the other way, so I don't think it's very easy to come up with the solution that I used given the integral.[/quote]

Oops. My taylor series had some erroneous minus signs in it. It should be:
- integral (0 to 1) [ln (1-y)/y dy] = sum (n=1 to inf) [1/n^2] = (pi^2)/6.

i/3's method is much better though.

The fact that the inverse of e^x-1 is e^-x + e^-2x + ..., where did you learn that? Is it supposed to be taught in a first year calculus course? Did you discover it yourself whilst solving the problem?

[quote="i/3"]Oups, sorry I forgot the spoiler  :([/quote]

instead of posting a message saying you forgot the spoiler... why don't you edit your first one and ADD THE SPOILER TAG?

[quote="Spoon"]why ... ADD THE SPOILER TAG?[/quote]

Probably because I'm [color=blue][b]far less [/b][/color]clever than you are  :P !


limit
induction
inequalities
number theory open
number theory

$s_n$ is sum of the first n prime .Prove that $n^2=<s_n\leq (n+1)^2$

This statement is true when $n \leq 9$ and false when $n > 9$. The last part can easily be proven by induction on the inequality $S_{n-1} > n^2$ (n $\geq$ 10), which combined with the fact that $p_n \geq 2n+1$ for all $n >4$ implies that

$S_n = S_{n-1} + p_n > n^2 + 2n + 1 = (n + 1)^2$.


If $ n \in \mathbb{N}$ let $ r(n)$ denote the sum of the remainders when $ n$ is divided by $ 1,2,3,4,....n$ respectively. Prove that $ r(k)\equal{}r(k\minus{}1)$ for infinitely many positive integer $ k$

[quote="rajelaz"]If $ n \in \mathbb{N}$ let $ r(n)$ denote the sum of the remainders when $ n$ is divided by $ 1,2,3,4,....n$ respectively. Prove that $ r(k) \equal{} r(k \minus{} 1)$ for infinitely many positive integer $ k$[/quote]

I' will show that $ r(2^n \minus{} 1) \equal{} r(2^n)$

Let $ r$ be The remainder of $ 2^n$ when we divide $ 2^n$ by $ k$ and $ k < 2^n$

if $ r \equal{} 0$ Then we have $ 2^n \minus{} 1 \equal{} k(p \minus{} 1) \plus{} k \minus{} 1$ and since $ k \minus{} 1 < k$ the remainder of $ 2^n \minus{} 1$ is $ k \minus{} 1 \equal{} 2^c \minus{} 1$.But the number of $ k$ for which the remainder is equal to $ 0$ is the number of divisor  of $ 2^n$  excluding $ 2^n$($ i.e$ because we are working on the numbers which are least that $ 2^n \minus{} 1$) which is $ n$

if $ r \ge 1$ Then we have $ 2^n \minus{} 1 \equal{} pk \plus{} r \minus{} 1$ hence the remainder of $ 2^n \minus{} 1$ is $ r \minus{} 1$ when we divide $ 2^n \minus{} 1$ by $ k$ .On the other hand the number of $ k$ for which the remainder that we get is greater than is $ 1$ is $ 2^n \minus{} (n \plus{} 1)$

Hence we have :

The sum of the remainders that satisfy the case $ r \ge 1$ is :

$ \sum_{r_k\ge1}(r_k \minus{} 1) \equal{} \left(\sum_{r_k\ge1}r_k\right) \minus{} (2^n \minus{} n \minus{} 1)$

The sum of the remainders which satisfy the case $ r \equal{} 0$ is :

$ \sum_{k \equal{} 1}^{k \equal{} n \minus{} 1}(2^k \minus{} 1) \equal{} 2^n \minus{} n \minus{} 1$

and Thus we have $ r(2^n \minus{} 1) \equal{} \left(\sum_{r_k\ge1}(r_k \minus{} 1)\right)$$ \plus{} \sum_{k \equal{} 1}^{k \equal{} n \minus{} 1}(2^k \minus{} 1)$ $ \equal{} \left(\sum_{r_k\ge1}(r_k \minus{} 1)\right) \plus{} \sum_{k \equal{} 1}^{k \equal{} n \minus{} 1}(2^k \minus{} 1) \plus{} \sum_{r_k \equal{} 0}r_k \equal{} r(2^n) \minus{} (2^n \minus{} n \minus{} 1) \plus{} (2^n \minus{} n) \minus{} 1 \equal{} r(2^n)$

and Finaly there exists an infinitly numbers that satisfy our condition  :)

Ps: I found  this solution when I was in Class correct me if there is any mistake  :D

As in the previous post, we prove that $r(2^k) = r(2^k - 1)$.Note that each remainder is essentially one less for $2^k - 1$ than $2^k.$ The exceptions are the integers that divide $2^k,$ in which case if $2^i, i<k$ divides $2^k,$ then the remainder for $2^k - 1$ with $2^i$ is $2^i - 1.$ $2^k$ has $k+1$ factors, but we don't use $1$ and $2^k$ as for $1$ the remainder is always zero, and $2^k$ is too big for $2^k - 1$. So we essentially use only $k+1 -2 = k-1$ of the factors. There are $2^k - (k-1) - 1 - 1$ remaining factors left over, and for each of them, the value at $2^k - 1$ is one less than $2^k,$ (note that they are all positive remainders for $2^k$.)

Therefore, we get the sum $r(2^k - 1) = \displaystyle \sum_{i=1}^{k-1} (2^i - 1) - (2^k - (k-1) - 1 - 1) + r(2^k).$ We can easily see that $\displaystyle \sum_{i=1}^{k-1} (2^i - 1) = (2^k - (k-1) - 1 - 1),$ and thus we have the desired result.


calculus
integration
trigonometry
algebra
polynomial
derivative
calculus computations

Ok here's the solution I've come up with which works, but look at the part that says "Identity states that". I'd like to solve the problem but without making use of that integration identity, any suggestions?
[img]http://grimpirate.t35.com/work3.gif[/img]

Substitute $t=\sqrt{2}\cdot u+3$. Look at the integral in that form.

I would never have used trig substitution here- square roots of linear polynomials are best handled by algebraic substitutions.

Ahh, I see thanks for the insight jmerry, using a dummy variable I can mess with the fraction to get what I need in the numerator. Very nice. Would you care to expand on what you mean by algebraic substitutions in this case? I'm not quite sure I'm following you. I'll post the new solution here after this post.

We can start by substituting $u=\sqrt{1+2x},x=\frac{u^{2}-1}{2}$. This transforms the integral into the integral of a rational function.

Here's the change according to what you had shown me previously, assuming I can just substitute a dummy variable like that, though I don't feel so confident about it. However, it does yield the correct result, that may just be a coincidence.
[img]http://grimpirate.t35.com/work4.gif[/img]

You're a genius jmerry that's all I have to say about that. I had originally started the problem that way but didn't see it going anywhere when I tried to take the derivative as  du/dx, rather it's more apparent when you take the derivative as dx/du. This is a much cleaner approach to the problem.
[img]http://grimpirate.t35.com/work5.gif[/img][/img]


Show that for all integer $ k > 1$ , there are infinitely many natural numbers $ n$ such that $ k2^{2^{n}}\plus{}1$ is composite.

Sorry. I made a mistake but now it is ok.

[quote="Rust"]It is open problem.[/quote]

look at that.

[quote="kathi;)"]

look at that.[/quote]

What is it ?? :(

[quote="kathi;)"]its an open problem -> atm, there doesn't exist a solution (and even "good" mathematicans dont have one)[/quote]

I believe Rust was referring to the problems before it was edited.  Though I could be wrong...

Anyways, it suffices to show that for any $ k$, there is an odd prime $ p$ such that there are infinitely many $ n$ with $ 2^{2^{n}}\equiv\minus{}\frac{1}{k}\mod{p}$.  Well then it suffices to show there is some $ a$ which is a power of $ 2\mod{p}$ such that for some $ r$, $ a^{2^{r}}\equiv a\equiv\minus{}\frac{1}{k}$.  We then have $ a^{2^{r}\minus{}1}\equiv 1$..

So if there is a prime $ p|k\plus{}1$ such that for some $ n$, $ p|2^{2^{n}}\minus{}1$, then it's easy (those are the "easy" k, easily classified)...otherwise, I don't know...

It is easy to prove for odd k. Let $ P\equal{}k2^{2^{m}}\plus{}1$ is prime (else all members of sequence are composite) and let $ T_{P}$ is period 2 mod $ P$, then $ P|2^{2^{n}}\plus{}1$ if
(1) $ 2^{n}\equiv 2^{m}\mod T_{P}$.
If k is odd $ ord_{2}(T_{P})\le 2^{m}$, therefore (1) had infinetely many solutions $ n\equal{}m\plus{}ar,a\in N$, were $ r$ is period 2 mod $ T_{P}$.


Find a formula for the determinant of the n-by-n matrix:
\[
\left|\begin{array}{cccccc}
0 & 1 & 0 && 0 & 0 \\
1 & 0 & 1 & \cdots & 0 & 0 \\
0 & 1 & 0 && 0 & 0 \\
& \vdots & & \ddots & \vdots & \\
0 & 0 & 0 && 0 & 1 \\
0 & 0 & 0 & \cdots & 1 & 0
\end{array}\right|
\]
This matrix has two diagonals of 1's, directly above and directly below the main diagonal, and 0's everywhere else.  For example, if n=5, the matrix is
\[
\left|\begin{array}{ccccc}
0&1&0&0&0 \\
1&0&1&0&0 \\
0&1&0&1&0 \\
0&0&1&0&1 \\
0&0&0&1&0
\end{array}\right|
\]

Call the given determinant $D_n$ expand it from the first row, you will
get $D_n$ in term of $D_{n-1}$ and $D_{n-2}$.
Something like $D_n=aD_{n-1}+bD_{n-2}$  for $n\geq 2$

[quote="Moubinool"]Call the given determinant $D_n$ expand it from the first row, you will
get $D_n$ in term of $D_{n-1}$, $D_{n-2}$...[/quote]
I don't believe that this is accurate (unless I'm misinterpreting what you mean):

Take the 3 x 3 case:
\[\left|\begin{array}{ccc} 0&1&0\\1&0&1\\0&1&0 \end{array}\right|\]

If you expand using the 1st row, you get
\[-\left|\begin{array}{cc} 1&1\\0&0 \end{array}\right| = 0\]

(And no, the answer is not that it's 0 for all n; it's not 0 for n=2 for example)

Hmm... this is my guess. Expand by minors on the first row and get
\[
-\left|\begin{array}{ccccccc}
1&1&&&&&\\
&&1&\cdots&&&\\
&1&&&&&\\
&\vdots&&\ddots&&\vdots&\\
&&&&&1&\\
&&&\cdots&1&&1\\
&&&&&1&\\
\end{array}\right|_{(n-1)\times(n-1)}
\]

Expand by minors on the first column and get
\[
-\left|\begin{array}{ccccccc}
&1&&&&&\\
1&&1&\cdots&&&\\
&1&&&&&\\
&\vdots&&\ddots&&\vdots&\\
&&&&&1&\\
&&&\cdots&1&&1\\
&&&&&1&\\
\end{array}\right|_{(n-2)\times(n-2)}
\]


integration trigonometry limit algebra polynomial function calculus

[hide="rough solution"]
Let $y = y(s) : = \int_{0}^{\infty}\frac{\sin x}{x}e^{-sx}\, dx$. Then

$y' =-\frac{1}{1+s^{2}}$

so $y(s) = c-\arctan(s)$.

to determine $c$, evaluate $\lim_{s\to\infty}y(s)$. then we have $c = \frac{\pi}{2}$ and

$\int_{0}^{\infty}\frac{\sin x}{x}\, dx = y(0) = \frac{\pi}{2}$

of course this solution is indeed correct, but logic is imperfect. interchanging two limits may fail to give the same value, so you have to verify it when needed.[/hide]

This problem has been discussed in this forum many times before.

As for the logic of sos440's solution:

For any $s>s_{0}>0,$ we have $e^{-s_{0}x}$ or some polynomial times $e^{-s_{0}x}$ available for use as a dominating function in the Dominated Convergence Theroem.

That (the DCT) justifies differentiating under the integral sign and hence the calculation of $y'(s)$ for $0<s<\infty.$ Another use of the DCT shows that $\lim_{s\to\infty}y(s)=0,$ and thus establishes the formula

$y(s)=\frac{\pi}2-\arctan(s)$ for $s>0.$

So what part of this argument requires the most delicate analysis? The part that sos440 didn't acknowledge when he said "$y(0)$." None our our arguments directly defined $y(0)$ or showed anything about continuity. We need the following theorem, which I'll name for the more familiar theorem it is closely analogous to:

[b]Abel's Theorem for Integrals[/b]

Suppose $f(x)$ is absolutely integrable on $(0,b)$ for every $0<b<\infty$ and that

$\lim_{b\to\infty}\int_{0}^{b}f(x)\,dx=L$ exists.

For $s>0,$ define $F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx.$

Then $\lim_{s\to0^{+}}F(s)=L.$

[We may also need an additional growth hypothesis - something like $\int_{0}^{b}|f(x)|\,dx\le Cb^{k}$ for some $C$ and $k.$]


Let $ n\in\mathbb{N}$ and $ n\geq2$, Prove that:

$ \sum^{\infty}_{k=1}\frac{1}{k\sqrt[n]{k+1}}>n$


inequalities
trigonometry
inequalities proposed

Let $a, b, c > 1$ be such that $a+b+c=abc$. Prove the inequality

$\left( a^{2} -1 \right) \left( b^{2} -1 \right) \left( c^{2} -1 \right) \leq 8$.

The classica trigonometric substitution works ($a = \tan A$, ...)


Princeton
college
summer program
Mathcamp
articles
search

[color=violet]

Which SAT I review book do you guys like the most Barrons, Kaplan, or Princeton? I found Barrons to be the most comprehensive, but tedious. Princeton is somewhat incomplete. Kaplan seems to be somewhere in between. Any suggestions?

Also, is memorizing the 3000 words in the Barron's Book necessary to guarantee a 750+ score?[/color]

Careful, you don't wanna provoke the wrath of The Evil One (a.k.a. Simon's Alter Ego, ComplexZeta).  He feels quite strongly that one should not study for the SAT or take it twice.  But I'll let him express his views on the matter himself, as he inevitably finds this post.

[quote="mathnovice"][color=violet]
Also, is memorizing the 3000 words in the Barron's Book necessary to guarantee a 750+ score?[/color][/quote]

Nothing is guaranteed.  You can get yourself high on altoids, or forget your altoids, or you can just make a couple of bonehead mistakes.  But 3,000 words is pretty hefty.  I have Kaplan's, and it has about 2500, which seems clunky but is fairly close.

Oh, and since we're on the subject of SAT Prep, there is a free SAT preparation website - yes, ABSOLUTELY FREE, no annoying ads, free account, etc. - called Number2.com.  Visit it at [url]http://www.number2.com[/url]

My brother always bugs me about the site's name, but don't pay attention. It's quite helpful.  And it's on Simons Ban List.  :D

Just kiddin' with you, Simon...don't start headpatting me or anything...

[Edit] Oh, I'm verry sorry, mathnovice.  I am guilty of the worst oversight of mathematics: assuming things that aren't necessarily true.  *bangs head on table*.  I also posted twice.  Silly me.

lol. i want this "my life in terms of pi" book. i think that the princeton review is a book that kinda speaks to you. but since ive only read some of the verbal part in that book, i really wouldnt know.  at least get 10 Real Sats. that would help. even though i didnt use it. but hey, i was doing other things.

Hmm... http://www.number2.com seems to be a good website.

(By the way I am not a girl- I accidentally chose "violet" instead of "dark blue" lol     :cry: )

Anyway, thanks!

[quote="Fierytycoon"][quote]Um...hehe. Maybe I should fool Fierytycoon into thinking I'm a girl by posting in violet. That would be funny.[/quote]

Nope, not a very fool-proof method. A person's writing style can easily give him/her away.  8-)[/quote]
You also need to change your avatar into violet :P (a little insider joke)

[quote="ComplexZata"]Why am I evil? *patpat* mathfanatic.[/quote]
What's with people and head-patting :?:
a.) Does it hurt?
b.) Why not pat someone on their back?
c.) I though head-patting was rippledance's territory, or is it common thing among AoPSers (I think it sounds cool...AoPS[b]er[/b]s:))

[quote](I think it sounds cool...AoPSers[/quote]

Aye, yes, a compensation title when I can't yet become a MOPper ;)

[quote="Tare"][quote="Fierytycoon"][quote="ComplexZata"]Why am I evil? *patpat* mathfanatic.[/quote]
What's with people and head-patting :?:
a.) Does it hurt?
b.) Why not pat someone on their back?
c.) I though head-patting was rippledance's territory, or is it common thing among AoPSers (I think it sounds cool...AoPS[b]er[/b]s:))[/quote][/quote]

It's something that's common at rippledance's school. It was originally something that Jimmy Jia did to Z9R4C3 (I guess most of you don't know who those people are), but then others started stealing it. (Ok rippledance will probably tell me I'm completely wrong, but this is what I think might be true.) Then mathfanatic stole it from rippledance, and since rippledance and Z9R4C3 often *patpat* me, I shall pass it along to others.

z9r4c3 and Jimmy were head-patting each other throughout the whole of mathcamp and then z9r4c3 started headpatting me b/c I'm shorter than she is and I started headpatting some people and the headpatting syndrome spread. *shrug* Jimmy will tell you that z9r4c3 and I definitely deserve headpatting. Then z9r4c3 and I decided to headpat ComplexZeta for fun. And now he's continuing it. Yay Simon *patpat*.
Headpatting was originally used b/c of the connotations of superiority etc etc...now we're just carrying on the tradition.
Tsk...it's a little too late to pretend to be a girl, mathfanatic.

And of course they did not *invent* the headpatting, they merely reintroduced it to the mathematics world (such as rippledance).  It's generally considered demeaning.

On the topic of SAT prep:  No, you shouldn't memorize the words.  You should [i]already know[/i] nearly all of them.  If not, you should read more - memorizing words doesn't really tell you what sort of context to use them in and so forth (better than nothing, I'm sure).  The only SAT book that I recommend is Ten Real SATs, because it's good to be able to test yourself on an actual SAT to see generally how you'll do.  And then if you really want to study you'll know what and how much.

It's certainly not necessary to memorize *all* of the words.  Some of those words are downright weird, and even people with 1600s often don't know them.

[quote="rippledance"]Tsk...it's a little too late to pretend to be a girl, mathfanatic.[/quote]

Well, too late for me, definitely, and why would I want to?  But it's not too late for some newbie like, for instance, ElectricSnowman.  Just an idea...
 :wink:

As long as we're on the subject of head-patting, I have a somewhat relevant story.  My friend and this girl kept bonking each other on the head with ski poles while we were skiing, so we had this joke that this was a form of flirting.  I have to wonder if headpatting is similar...

[quote]The only SAT book that I recommend is Ten Real SATs, because it's good to be able to test yourself on an actual SAT to see generally how you'll do. And then if you really want to study you'll know what and how much. [/quote]

I have that book.  Simon chewed me out for buying it.   :D

[quote="mathfanatic"] I have that book.  Simon chewed me out for buying it.   :D[/quote]

Unsurprisingly. But I agree that that's about the only book I would buy if I did.

Oh and btw, hello, shadowedreamer. =) Post! [/quote]

[quote="mathfanatic"][quote="rippledance"]Tsk...it's a little too late to pretend to be a girl, mathfanatic.[/quote]

Well, too late for me, definitely, and why would I want to?  But it's not too late for some newbie like, for instance, ElectricSnowman.  Just an idea...
 :wink: [/quote]
I can, I think. Although I naturally act like a hermaphodite :)

[quote="mathfanatic"]As long as we're on the subject of head-patting, I have a somewhat relevant story.  My friend and this girl kept bonking each other on the head with ski poles while we were skiing, so we had this joke that this was a form of flirting.  I have to wonder if headpatting is similar...[/quote]
 :shock: Ouch...

So you do head-patting to short people...I'm kinda short...

back to the topic: does anyone like Princeton? i tried the Physics SAT II one and was kinda disappointed.

(this is my first post here, sorry if i somehow manage to mess it up)

I don't really practice much for the SAT or ACT, just the vocabulary... i use Arco's vocabulary quiz book to study that.

Wow, applepuppy, do I know you?  :wink:

It's impossible to mess up worse than I have.  *Thinks back to dem AoPS Story posts...*

I have taken tests from Princeton, Barrons, and some others and I have found that if you want to trick yourself, buy Barrons.  It nearly ALWAYS underestimates your score.  I went into the test thinking I would get a extremely low score.  However, I ended up getting just a low score.

Sometimes, tricking yourself is good.

KAPLAN!!!

ALL MY SAT/ACT BOOKS ARE KAPLAN.. .well maybe except this 10 Real SAT's (Barron)

chinaboy, no offense or anything...its juss that i was never a fan of kaplan

[quote="MithsApprentice"]I've always hated Barron's and Kaplan for some reason...and i think princeton review doesnt give problems that are similiar to the difficulty lev on the real SAT's for some reason...thats probly juss me though[/quote]

I didn't like them much either - they seemed too out of touch with what the test was really like. I read an article in the Chicago Tribune about an online test company that wrote their own SAT and uses it as a diagnostic, but I can't remember what exactly it was.. Prep something.. the guy who wrote the test wrote a book too, but I don't remember that either. I'll try to google search it and post it if I figure it out..

Kaplan/Barrons/PR/REA/Arco/5steps really depends on which test you take!

Like I know for AP CompSci, Litvin's "Be Prepared" is way better than Barrons or any other APCS prep book.

But on the other hand, for AP Stats, Barrons is clearly superior to every other review.

[quote="aznphatso"]Kaplan/Barrons/PR/REA/Arco/5steps really depends on which test you take! [/quote]
I heard cliffnotes is used almost like a textbook (since its far better) for AP bio, but it is not true for other subject from cliffnotes.

[quote="pdub87"]
I didn't like them much either - they seemed too out of touch with what the test was really like. I read an article in the Chicago Tribune about an online test company that wrote their own SAT and uses it as a diagnostic, but I can't remember what exactly it was.. Prep something.. the guy who wrote the test wrote a book too, but I don't remember that either. I'll try to google search it and post it if I figure it out..[/quote]

The company I was thinking of is PrepMe.com for the online stuff. The book writer is Joe Jewell, The Underground Guide to the SAT.. I love google haha what would we do without it?


anyone know how to prove the Seydewitz theorem : If a triangle is inscribed in a conic show any line conjugate to one side meets the other sides in conjugate points.

Let $ABC$ be the triangle, and let $P$ be the pole of $BC$. Also, let $X\in AB$ be a variable point, and $Y=AC\cap XP$.

The transformation $X\mapsto Y$ is projective, and so is the transformation sending $X$ to the point where the polar of $X$ cuts $AC$. All you need to prove is that they coincide, and in order to do this, you can check that this holds for three positions of $X$, properly chosen so that's it's easy to achieve your goal :). Take $X=A,X=B$, and $X=PC\cap AB$. The statement of the theorem is easy to check in these cases, and since the two projectivities from $AB$ to $AC$ coincide in three points, they must coincide everywhere.

What is meant by "conjugate" in the statement of the problem?

I was puzzled too at first :). I believe it means this: two elements are said to be conjugate iff one of the elements contains the dual of the other element (and vice-versa). So, for example, two lines are conjugate iff one of them contains the pole of the other one.

Thanks.   I arrived at the same solution after your explanation, but I think that for one of the vertices it is necessary to verify the following statement (not difficult):

if the circumscribed triangle $ A'B'C' $ corresponding to $ ABC $ is drawn, then $ AA', BB', CC' $ are concurrent.

For the other two vertices there is nothing to check.


Show that for every integer $n \geq 0$ and $m \geq 1$, $\sum_{k=1}^{m}{\frac{1}{n+k}}>\sum_{k=1}^{m+1}{\frac{1}{n+m+k}}$.

This seems to be wrong. :huh:
For example take $m=1$, $n \ge 1$.

You're quite right, this doesn't work. I tried to extrapolate a result from another context, but it seems I cannot do it like this ....


function
limit
calculus
calculus computations

Find $ lim_{n\to \infty}a_n$ if $ a_1 \equal{} 200$,$ a_{n \plus{} 1} \equal{} \frac {2}{2^{a_n}}$
Help me ,please

[hide="hint"]For any function $ f: \mathbb{R}^{mn}\to \mathbb{R}^n$ and any sequence $ \{ x_k\}_{k\in \mathbb{N}}\subset \mathbb{R}^n$ satisfying $ x_{k} \equal{} f\left( x_{k \minus{} 1},x_{k \minus{} 2},...,x_{k \minus{} m}\right)$ for $ k > m$, if the limit $ \lim_{k\to \infty}x_k \equal{} x$ exists, it must satisfy $ x \equal{} f(x,...,x)$.

So in your example it suffices to show that a limit exists, as from the above it follows that if it exists it must be $ 1$.[/hide]

To show the limit exists, suffices to show it is bounded


search
number theory proposed
number theory

let $ a_1,...,a_p$ are  natural and different numbers  that p is prime.prove that :
$ \exists i,j: \frac{a_i}{gcd(a_i,a_j)} \geq p$

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1653692415&t=303646[/url] for a very similar topic.

Assume that $ gcd(a_{1}, a_{2}, a_{3}, ... , a_{n})\equal{}1$. Suppose there were an $ a_{i}$ divisible by $ p$. Then there is an $ a_{j}$ not divisible by $ p$. Then $ \frac{a_{i}}{gcd(a_{i},a_{j})}\equal{}pk \geq p$

Now assume that none of the $ a_{i}$'s are divisible by $ p$. Then there exist i,j s.t. $ a_{i}\equal{}a_{j}(modp)$ with $ a_{i}>a_{j}$ . Now, $ a_{i}\equal{}a_{j}\plus{}pt$ for some positive integer $ t$ . We observe $ \frac{a_{i}}{gcd(a_{i},a_{j})}\geq \frac{a_{i}}{t} > p$


Question:

Find the Sum upto n terms:

[img]http://www.goiit.com/equations/2008/10/4/a73234a2-5e13-4031-a902-2531db58a385.png[/img]

Will rate each answer

It's pretty tough when the general term goes unspecified ;)

He wants you to find a closed form for $ \sum_{k\equal{}1}^n \prod_{j\equal{}1}^{k} \frac{4j\minus{}3}{4j\minus{}1}$

[hide]Let $ a_n \equal{} \frac {1 \cdot 5 \cdot ... \cdot (4n \minus{} 3)}{3 \cdot 7 \cdot ... \cdot (4n \minus{} 1)}$. Then $ \frac {a_{k \plus{} 1}}{a_k} \equal{} \frac {4k \plus{} 1}{4k \plus{} 3}$, or $ (4k \plus{} 3) a_{k \plus{} 1} \equal{} (4k \plus{} 1) a_k$. Summing both members between 1 and n, we have

$ \sum_{k \equal{} 1}^n (4k \plus{} 3) a_{k \plus{} 1} \equal{} \sum_{k \equal{} 1}^n (4k \plus{} 1) a_k$

$ \sum_{k \equal{} 1}^n 4k a_{k \plus{} 1} \plus{} 3 \sum_{k \equal{} 1}^n a_{k \plus{} 1} \equal{} \sum_{k \equal{} 1}^n4ka_k \plus{} \sum_{k \equal{} 1}^n a_k$

$ 4 \sum_{k\equal{}1}^n [(k\plus{}1)a_{k\plus{}1} \minus{} ka_k] \minus{} 4 \sum_{k\equal{}1}^na_{k\plus{}1} \plus{} 3 \sum_{k \equal{} 1}^n a_{k \plus{} 1} \equal{} A_n$

$ 4[(n\plus{}1)a_{n\plus{}1} \minus{} a_1] \minus{} \sum_{k\equal{}1}^na_{k\plus{}1} \equal{} A_n$

$ 4na_{n\plus{}1} \plus{} 4a_{n\plus{}1} \minus{} 4a_1 \minus{} (A_n \minus{} a_1 \plus{} a_{n\plus{}1}) \equal{} A_n$

$ 2A_n \equal{} 4na_{n\plus{}1} \plus{} 3a_{n\plus{}1} \minus{}3a_1$

$ A_n \equal{} 2na_{n\plus{}1} \plus{} \frac32 a_{n\plus{}1} \minus{} \frac12$,

where $ A_n \equal{} \sum_{k\equal{}1}^na_k$.[/hide]

We can write $ a_{n} \equal{} \prod_{j\equal{}1}^{k} \frac{4j\minus{}3}{4j\minus{}1} \equal{} \frac {\Gamma \left( \frac {3}{4} \right) \Gamma \left( k \plus{} \frac {1}{4} \right)}{\Gamma \left( \frac {1}{4} \right) \Gamma \left( k \plus{} \frac {3}{4} \right)}$ using the [url=http://en.wikipedia.org/wiki/Gamma_function]Gamma function[/url].

Can we write the general form without the Gamma function, somehow?

I believe my previous post answers that question.

Oh, I meant, can we write $ a_k$ closed form without Gamma?
(without using $ \Pi$, $ \Sigma$, $ \ldots$, or $ \Gamma$)


trigonometry
inequalities
inequalities proposed

prove that for any  $ \Delta ABC$,the inequlity
\[ \sin^2{A}(\tan{\frac{B}{2}}\plus{}\tan{\frac{C}{2}})^2\plus{}\sin^2{B}(\tan{\frac{C}{2}}\plus{}\tan{\frac{A}{2}})^2\plus{}\sin^2{A}(\tan{\frac{B}{2}}\plus{}\tan{\frac{C}{2}})^2>2\]
holds.

[quote="Tourish"]prove that for any  $ \Delta ABC$,the inequlity
\[ \sin^2{A}(\tan{\frac {B}{2}} \plus{} \tan{\frac {C}{2}})^2 \plus{} \sin^2{B}(\tan{\frac {C}{2}} \plus{} \tan{\frac {A}{2}})^2 \plus{} \sin^2{A}(\tan{\frac {B}{2}} \plus{} \tan{\frac {C}{2}})^2 > 2\]
holds.[/quote]

do you want to write that\[ \sin^2{A}(\tan{\frac {B}{2}} \plus{} \tan{\frac {C}{2}})^2 \plus{} \sin^2{B}(\tan{\frac {C}{2}} \plus{} \tan{\frac {A}{2}})^2 \plus{} \sin^2{C}(\tan{\frac {A}{2}} \plus{} \tan{\frac {B}{2}})^2 > 2 ?\]?

if is, let $ a,b,c$ be the sides of $ \Delta ABC$, and setting $ a \equal{} y \plus{} z,b \equal{} z \plus{} x,c \equal{} x \plus{} y,x,y,z > 0$, that we easy to prove that
\[ \sin^2{A}(\tan{\frac {B}{2}} \plus{} \tan{\frac {C}{2}})^2 \equal{} \frac {{4x^2 \left( {y \plus{} z} \right)^2 }}{{\left( {x \plus{} y} \right)^2 \left( {x \plus{} z} \right)^2 }},\]so inequality equivalent to
\[ \sum {\frac {{4x^2 \left( {y \plus{} z} \right)^2 }}{{\left( {x \plus{} y} \right)^2 \left( {x \plus{} z} \right)^2 }}} > 2\]
\[ \Longleftrightarrow \sum {2x^2 \left( {y \plus{} z} \right)^4 } > \left( {x \plus{} y} \right)^2 \left( {y \plus{} z} \right)^2 \left( {z \plus{} x} \right)^2\]
\[ \Longleftrightarrow (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \plus{} 32x^2y^2z^2 > 0,\]obvious true.

similarly, we can prove that
\[ \sin^2{A}(\tan{\frac {B}{2}} \plus{} \tan{\frac {C}{2}})^2 \plus{} \sin^2{B}(\tan{\frac {C}{2}} \plus{} \tan{\frac {A}{2}})^2 \plus{} \sin^2{C}(\tan{\frac {A}{2}} \plus{} \tan{\frac {B}{2}})^2 < 4.\]

Sorry for my typo....But you are really great,exactly,my solution is the same.


geometry
similar triangles
geometry unsolved

Let a, b, c be the sides of a reference triangle. Show that the sides of the excentral triangle

a' = acsc(1/2 A)
b' = bcsc(1/2 B)
c' = ccsc(1/2 C)

Remember the reference triangle is the orthic triangle of the excentric triangle.  At this point, take similar triangles and it's trivial.  (Or you could just angle chase from the start and use the same method, but I find this easier)


geometry
incenter
trigonometry
analytic geometry
geometry proposed

Let a triangle ABC. H is the orthocenter, I is the incenter of triangle ABC. Known that $ \cos B \plus{} \cos C \equal{} 2 \cos A$, prove that IH is perp to the A-median.

Using some of the hunderts of trigonometrical formulas of triangles, I think it is possible reformulate the condition on the cosinuses in terms of the sides, in the following way:

(s-b)(s-c)/a(s-a) = cos 45º.

Maybe it is possible now establish the relation between the angle formed by the lines IH and the median AM.

I have an ugly solution using trilinear coordinates (that is, numbers which are proportional to the directed distances from a point to the sides of the triangle). I am searching some other more nice, although if I do not find out, I will post later...

You can see at [url]http://www.mathlinks.ro/viewtopic.php?p=1179706[/url]

Here is a pdf file with my solution with trilinear coordinates.


Let ABCD be a convex quadrilateral such that $ DAC\equal{}DBC$. M and N be the midpoints of AB and CD, respectively. $ O\equal{}AC\cap BD$, and P and Q be the projections of O on AD and BC, respectively. Prove that $ MN\bot PQ$
 :)

[quote="Rose_joker"]Let ABCD be a convex quadrilateral such that $ DAC \equal{} DBC$. M and N be the midpoints of AB and CD, respectively. $ O \equal{} AC\cap BD$, and P and Q be the projections of O on AD and BC, respectively. Prove that $ MN\bot PQ$
 :)[/quote]
See all the links [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=213068]here[/url].

Kostas Vittas.


number theory
relatively prime
prime numbers

Fin all possible pairs of relatively prime numbers such that the square of one of the integers when increased by 125 is divisible by the other. :)  (within the first thousand) :P

Από τον άλλο ακέραιο, ή από το άλλο τετράγωνο;

Λοιπόν εκ των προτέρων sorry αλλά δεν πάει άλλο . ΕΛΕΟΣ !!!! :ninja:  STOP SPAMING (ούτε o Cesar Lupu  :P )

Σιλουανέ, αν και δεν είναι Forum γλώσσας θα σου πω το εξής:

Find all possible pairs of relatively prime numbers such that the square of one of the integers when increased by 125 is divisible by the other.

Στο other μπορεί να θεωρείται προσδιορισμός και η λέξη square ως εννοούμενη και παραλλειπόμενη για να μη γίνει επανάληψη και οι λέξεις integer ή number για τον ίδιο ακριβώς λόγο.

Δηλαδή καταλαβαίνω:
1. Βρείτε όλα τα δυνατά ζευγάρια πρώτων μεταξύ τους αριθμών τέτοια ώστε το τετράγωνο του ενός από τους ακεραίους όταν αυξηθεί κατά 125 να διαιρείται με [u]ΤΟ[/u] άλλο (τετράγωνο).
ή
2. Βρείτε όλα τα δυνατά ζευγάρια πρώτων μεταξύ τους αριθμών τέτοια ώστε το τετράγωνο του ενός από τους ακεραίους όταν αυξηθεί κατά 125 να διαιρείται με [u]ΤΟΝ[/u] άλλο (αριθμό).

Φίλε maniopas, χωρίς να θέλω να παρεξηγηθώ ο Σιλουανός [i]προφανώς δεν αναφέρεται μόνο στο συγκεκριμένο[/i]...Σε αυτό πιθανόν έχεις δίκιο ότι θα χρειαζόταν μια διευκρίνιση. Ομως θα ήταν πολύ καλύτερο αν προσπαθούσες να βρεις λύση για μια απ' τις 2 περιπτώσεις (ή και για κάθε μια ξεχωριστά) και να βάλεις τη λύση και να πεις:"αν εννοείς αυτό, η λύση είναι..." κ.τ.λ.
Απ' την άλλη παρατηρώ ότι επανειλημμένα γράφεις posts τα οποία δεν βοηθούν στη ευρυθμία του "ελληνικά". Για παράδειγμα, ρώτησες στο πρόσφατο πρόβλημα θεωρίας αριθμών ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180685&sid=be1ef5981d7d67c13bca5541fdd06973[/url]) αν εννοεί για άπειρο πλήθος όρων λόγω των αποσιωπητικών που υπήρχαν στο τέλος. Αν όμως το κοιτούσες λίγο πιο προσεκτικά θα ήταν τελείως παράλογο να σε ρωτάει για απειρους όρου γιατί θα ήταν σαν να σε ρωτούσε να βρεις ότι κάθε $ p$ διαιρεί το άπειρο (αφού θα ήταν άπειροι οι όροι).
Ευχομαι να έγινα ξεκάθαρος και επαναλαμβάνω ότι τα λεώ αυτά με όλη την καλή πρόθεση, απλά θα ήταν καλύτερο να ελλατωθούν (όσο γίνεται) τα posts που είναι απλά σχόλια και δεν γράφουν κάποια λύση ή προτείνουν κάποιο πρόβλημα.

Ναι, συγνώμη αν γίνομαι σπαστικός, αλλά αυτό που ήθελα να πω (το είχα διατυπώσει εκεί λάθος) ήταν αν έχουμε τυχαίου πλήθους όρους ή μόνο τους δύο που αναφερόταν. (Δεν κάνω συχνά τέτοιες παρατηρήσεις πάντως, αλλά σήμερα ήταν εξαίρεση.) Απλά, μία αλλαγή στο θέμα μπορεί να μου φάει μέχρι και 2 ώρες δουλειά (δεν αφήνω ποτέ άσκηση παρά μόνο αν τη λύσω - ε, καλά ίσως για να φάω και να πάω για ύπνο) στην περεταίρω διερεύνηση του προβλήματος (αν ας πούμε βελτιώνεται η τιμή ή ισχυροποιείται η ανισότητα) και γι' αυτό ζητάω επίμονα ακριβείς εκφωνήσεις. (π.χ. σε αυτή την άσκηση σπάω το κεφάλι μου να δω αν γίνεται γενίκευση με n αριμούς και πώς θα ήταν η μορφή της τότε.)
Βέβαια, τουλάχιστον έτσι πιστεύω, στα μαθηματικά πρέπει να προωθείται ο διάλογος και όχι μόνο ξερές ερωταποκρίσεις. (Αν δε πεις και το αστείο πώς θα σου αρέσει το αντικείμενο;)
Πάντως, αφού ενοχλεί δε θα ξαναρωτήσω παρά μόνο αν είναι απολύτως αναγκαίο.

Πάντως επανεχρχόμενοι στην άσκηση αυτού του post:

είναι $ l^2y^2 \equal{} x^2 \plus{} 125$ και $ x \equal{} ky \plus{} m$
άρα $ x \equal{} k\frac {\sqrt {x^2 \plus{} 125}}{l} \plus{} m < \equal{} > l(x \minus{} m) \equal{} k\sqrt {x^2 \plus{} 125} < \equal{} >$
$ \frac {l}{k}(x \minus{} m) \equal{} \sqrt {x^2 \plus{} 125} < \equal{} > (\frac {l}{k})^2(x^2 \minus{} 2xm \plus{} m^2) \equal{} x^2 \plus{} 125$
σε προσπάθεια να συνεχίσω καταλαβαίνετε γιατί εκνευρίστηκα (αν δεν ήταν αυτό θα πελάγωνα τζάμπα στις πράξεις)

είναι $ ly \equal{} x^2 \plus{} 125$ και $ x \equal{} ky \plus{} m$
άρα $ x \equal{} \frac {x^2 \plus{} 125}{l} \plus{} m < \equal{} > x^2 \minus{} lx \plus{} lm \plus{} 125 \equal{} 0$
εδώ ζητάμε να έχει ακέραιες άρα η διακρίνουσα τέλειο τετράγωνο αριθμού που να αφήνει υπόλοιπο με το 2 όσο και ο l (είμαι πτώμα στην κούραση και θα συμπληρώσω το πρωί)
Ίσως να μη λύνεται έτσι, αλλά από αυτό το σημείο και μετά η διαδικασία έχει απλουστευτεί, αφού έχουμε 3 εξισώσεις, 4 αγνώστους και το δεδομένο των ακεραίων.

Φιλε maniopas δεν εξετασες την πιθανη περιπτωση:$ x^2\plus{}125\equal{}y^2z$ με $ (x,y)\equal{}1$.Πιστευω οτι αυτη θελει ο Γιωργος να λυσουμε.

Είναι η πρώτη μου περίπτωση (αντί για $ z$ έχω βάλει $ l^2$ γιατί η εκφώνηση είναι για θετικούς αριθμούς).

Ορίστε η πλήρης λύση της (η προηγούμενη προσπάθειά μου ήταν βλακεία):

$ (x,y) \equal{} 1 (I)$
$ x^2 \plus{} 125 \equal{} y^2l^2 < \equal{} > (ly)^2 \minus{} x^2 \equal{} 125 < \equal{} > (ly \minus{} x)(ly \plus{} x) \equal{} 125< \equal{} >$(έστω $ ly \equal{} z$)

1. $ (z \minus{} x)(z \plus{} x) \equal{} 5(25) < \equal{} >$(επειδή $ z \minus{} x <\equal{} z \plus{}x$) $ z \minus{} x \equal{} 5, z \plus{} x \equal{} 25 < \equal{} >$
$ z \equal{} 15, x \equal{} 10 < \equal{} > y \equal{} {1, 3, 5, 15}, x \equal{} 10$, που από την (I) έχει τις δεκτές λύσεις
$ y \equal{} {1, 3}, x \equal{} 10$
ή
2. $ (z \minus{} x)(z \plus{} y) \equal{} 1(125) < \equal{} >$(επειδή $ z \minus{} x <\equal{} z \plus{}x$) $ z \minus{} x \equal{} 1, z \plus{} x \equal{} 125 < \equal{} >$
$ z \equal{} 63, x \equal{} 62 < \equal{} > y \equal{} {3, 7}, x \equal{} {62}$, που είναι σύμφωνες με την (I)

Βρείτε όλα τα δυνατά ζευγάρια πρώτων μεταξύ τους αριθμών τέτοια ώστε το τετράγωνο του ενός από τους ακεραίους όταν αυξηθεί κατά 125 να διαιρείται με ΤΟ άλλο (τετράγωνο).
Και παλι δεν σε καταλαβαινω...πρεπει $ y^2/x^2\plus{}125$.Πως απεδειξες οτι $ z\equal{}l^2$

Λάθος έκανα (έλυνα μια πορόμοια άσκηση που ήταν για θετικούς αριθμούς και μου έμεινε το ότι είναι θετικοί). Τώρα θέλουμε και τα αντίθετα αυτών που έδωσα. Δηλαδή z(αυτό που χρησιμοποιείς στη δική σου εκφώνηση) = $ \pm l^2$.

Μα δν λεε να ειναι ισοι αλλα να τον διαιρει αρα $ y^2/x^2 \plus{} 125 \equal{} > zy^2 \equal{} x^2 \plus{} 125$ Δεν σημαινει οτι ο $ z$ ειναι τετραγωνο...Για να καταλαβεις αν $ a^2/b$ τοτε δεν σημαινει οτι το πηλικο $ b/a^2$ οτι ειναι τετραγωνο....


There are three boxes of marbles. Box 1 has  30 marbles, box 2 has 37 marbles,  and box 3 has 46 marbles. Two people play the following game:

Each in turn picks either 1, 2, 4, or 6 marbles in one of three boxes.  The objective of the game is to force the opponent to pick up the last marble.

What is the winning startegy?

Go second. If your opponent picks any even number out of a box, you pick any even number. If your opponent picks 1, you pick 1. Reducing this way, you would eventually force your opponent to pick the last marble.

That doesn't guarantee a win: after your turn, there might be 7 marbles in one box and none in the other, in which case your opponent might take six and win.

mathblitz, there's a "Delete" button for your post (next to the edit button) when it's the most recent post on a thread.

Oh whoops I meant the first time take 1 if he/she takes even and take even if he/she takes odd. On next turns, go with that strategy.


[hide] Since 2x=5+3y, we have that y=(2x-5)/3.  Since 3x+4y=-18, we have that y=(-3x-18)/4.  Since both of these equations are equal to y at the intersection, we have that (2x-5)/3=(-3x-18)/4.  Solve for x to get that x=-2  Plug this in for one of our equations to get that y=-3.  Thus the intersection point is $(-2,-3)$.  [/hide]

[quote="xxxxx"]At what point do these lines intersect: 2x=5+3y and 3x+4y=-18?[/quote]
[hide]Just use systems. The first one is $2x-3y=5$ and the second one is $3x+4y=-18$. So multiplying the first one equation by 3 and the second by -2 we now have $6x-9y=15$ and $-6x-8y=36$. So adding these two up we get $-17y=51$. So y=-3. Solving for x equals -2. So the point is $(-2,-3)$.[/hide]

[quote="7h3.D3m0n.117"][quote="Quevvy"][quote="#H34N1"][hide]

Graph it.

By looking at grpah, you find the answer.
[/hide][/quote]
But if the intersection is not at an integer pair, then how would you find it? you would use systems or substitution[/quote]

you can use both but id prefer substitution that way it is easier and you can get a fraction as an answer (besides i think systems is more work)

the EASIEST way though is to graph it on your TI-83+ calculator and just hit 2nd Trace Intercept and it'll give u an answer (in decimal form though  :( )[/quote]

i don't know of ANY middle school curiculum math problem that would require the use of a TI-83.....
in middle school math clas you should use either substitution, elimination, or graphing

@Quevvy: I was trying to give an alternative solution :wink:

[quote="davidlizeng"][/quote][quote="7h3.D3m0n.117"][quote="Quevvy"][quote="#H34N1"][hide]

Graph it.

By looking at grpah, you find the answer.
[/hide][/quote]
But if the intersection is not at an integer pair, then how would you find it? you would use systems or substitution[/quote]

you can use both but id prefer substitution that way it is easier and you can get a fraction as an answer (besides i think systems is more work)

the EASIEST way though is to graph it on your TI-83+ calculator and just hit 2nd Trace Intercept and it'll give u an answer (in decimal form though  :( )[/quote][quote="davidlizeng"]

i don't know of ANY middle school curiculum math problem that would require the use of a TI-83.....
in middle school math clas you should use either substitution, elimination, or graphing[/quote]

yeah they teach you the three main ways and then they usually don't show u u can graph...however being the calculator geek i am i found out how to do it easily (As for some reason our materials list required a TI-83)


I'm about to submit my project to Siemens Westinghouse competition, and time is running low and I still need to write up my paper. I already have some of it on MS Word, but I noticed that the judging criteria says:

Presentation: Is the method of presentation consistent with the nature of the work and with scientific practice in the discipline involved?

Although LaTex is standard among mathematics papers, could I use MS Word? Or, do you think they will deduct from "presentation?" Could I type it up on Mathematica and print it like that?

The reason I don't want to use LaTex is that I don't have enough time to learn it and do it well.

Hmm... well I'd say it depends. If you're really good at doing things in Word you may attempt it, but generally Word isn't an option at all. I don't know about mathematica, but PDF's with $\LaTeX$ are generally the best method on earth :)

I used the equation editor that comes along with Word Perfect which is essentially the same thing as the Word equation editor.  Didn't seem to cause me any troubles.  Just try and get it done and looking as nice as possible.  You should look your paper over carefully after printing it and before submitting it, of course.

I vote for TeX. Takes 30 mins to learn, and saves hundreds of hours of typing compaired with doing the same thing in word.

[quote="mathnovice"]Presentation: Is the method of presentation consistent with the nature of the work and with scientific practice in the discipline involved?
[/quote]
[i]Disclaimer: I am [u]not[/u] a Westinghouse expert![/i]

I suspect the above text refers more to the style of the paper (for example, using "we" in a math paper, versus the passive-voice style typically used in, say, biology papers; stating your facts as theorems, lemmas, etc.; properly citing sources) than the typesetting.

Having said that, LaTeX is the gold standard for mathematics typesetting.  Your paper will look much, much nicer and more professional if it is typeset in LaTeX than if it is typeset in MS Word.

Although I must respectfully disagree with my colleague Valentin in stating that you can learn LaTeX "in 30 minutes" (I think it will take a few hours of playing around with it to get the hang of it), it is definitely worth the investment of time, and I do agree with Valentin that if you are planning to continue to write mathematics, it will save you hundreds of hours in the long run.

Good luck!

if you have mathtype (Advanced version of equation editor in word assuming thats what u used, although i am not sure if this is also plausible in equation editor)
you can set the translator to latex and then copy/paste.  it will do the job for you.  better yet, just export your word document to pdf


I was trying to prove this case in my head all day yesterday, but I was pretty stumped.
I am looking for a short, simple proof of this in which we don't prove AM-GM for all n, just 3 (similar to n=2, we use the Trivial Ineguality, which can't be applied for all n.).
Any ideas :?:

[quote="seamusoboyle"]$\frac{a^3+b^3+c^3}{3} \geq abc$[/quote]

Wouldn't AM-GM here be $\frac{a + b + c}{3} \geq \sqrt[3]{abc}$?

Another fun way is to  use  $\frac{x +y} 2 \ge \sqrt{xy}$

$\frac{a + b} 2 \ge \sqrt{ab}$  (when (x,y) are (a,b)

$\frac{c + d } 2 \ge \sqrt{cd}$

so $\frac{a+b+c+d} 4 \ge \frac{\sqrt{ab} + \sqrt{cd} } 2$   and now using sqrt (ab) and  sqrt(cd) as our x and y .. we get
 $\ge \sqrt [4] {abcd }$

So we have am-gm for 4 numbers, all we have to do is ..

to  select $d$ such that $d = \frac{a+b+c} 3$ in above ...

You get $(1/4)(4/3)(a+b+c) = (abc)^{1/4} (a+b+c)^{1/4)}$  which nicely reduces too ..

and you get $\frac{a+b+c} 3 \ge \sqrt [3] {abc}$

Neatly bundled!

(You can use same way to do for other numbers too...) :)

[quote="seamusoboyle"]$\frac{a^3+b^3+c^3}{3} \geq abc \leftrightarrow a^3+b^3+c^3-3abc \geq 0 \leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\geq 0$

Can you getit from there?[/quote]

to dot the i's and cross the t's the above is  equal to $(a+b+c)\frac{((a-b)^2+(b-c)^2+(c-d)^2)} 2$

The second factor (all squares ..) can not be negative , and $a,b,c$ are non negative  etc .....


Μια  ασκηση Α' Λυκείου μου εδωδσε την ιδέα για την κατασκευη της..

Αν $ a,b,c>0$ και $ abc\equal{}1$ να βρεθεί το

$ (\frac{a\plus{}(ac\plus{}c\plus{}1)}{ab\plus{}a\plus{}1}\plus{}\frac{b\plus{}2(ab\plus{}a\plus{}1)}{bc\plus{}b\plus{}1}\plus{}\frac{c\plus{}3(cb\plus{}b\plus{}1)}{ac\plus{}c\plus{}1})_{min}$

Θετω $ a \equal{} \frac {x}{y}, b \equal{} \frac{y}{z}, c \equal{} \frac {z}{x}$
Μετα απο πραξεις η αρχικη σχεση καταληγει στην εξης:

$ \frac {2xyz(xy \plus{} yz \plus{} zx) \plus{} xyz(x^2 \plus{} y^2 \plus{} z^2) \plus{} (x^2y^3 \plus{} y^2z^3 \plus{} z^2x^3)}{xyz(xy \plus{} yz \plus{} zx)}$

Αν σπασουμε το κλασμα στα 3 εχουμε:
$ 2 \plus{} \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {x^2y^3 \plus{} y^2z^3 \plus{} z^2x^3}{x^2y^2z \plus{} xy^2z^2 \plus{} x^2yz^2}$

Το $ x^2 \plus{} y^2 \plus{} z^2$ ειναι μεγαλυτερο ή ισο του $ xy \plus{} yz \plus{} zx$ και η ισοτητα ισχυει για $ x \equal{} y \equal{} z$.
Το $ x^2y^3 \plus{} y^2z^3 \plus{} z^2x^3$ ,απο muirhead, ειναι μεγαλυτερο ή ισο του $ x^2y^2z \plus{} xy^2z^2 \plus{} x^2yz^2$ και η ισοτητα ισχυει επισης για $ x \equal{} y \equal{} z$.

Αρα για $ x \equal{} y \equal{} z$ η αρχικη παρασταση εχει ελαχιστο 4.

hint
[hide]ειναι $ ac \plus{} c \plus{} 1 \equal{} c(ab \plus{} a \plus{} 1) \equal{} ac(bc \plus{} b \plus{} 1)$[/hide]
Σακη δες οτι για ολα 1 βγαινει το αποτελεσμα 7 και οχι 4. Ισως κατα λαθος να το ελυσες χωρις του συντελεστες 2,3 στο δευτερο και στο τριτο κλασμα  :wink:

Sakis, o Athinaios έχει δίκιο... με το μετ/μό $ a \equal{} x/y$ κλπ. η παράσταση παίρνει την πολύ κομψή μορφή:
$ 1 \plus{} \frac {z}{x} \plus{} \frac {2x}{y} \plus{} \frac {3y}{z} \equal{} 1 \plus{} c \plus{} 2a \plus{} 3b$
οπότε... :D

Cheerio,

Durandal 1707

Edit: A, προς αποφυγή παρεξηγήσεων, δεν εννοώ ότι το ελάχιστο είναι 7... π.χ. για $ c\equal{}2, b\equal{}1/2, a\equal{}1$ παίρνουμε $ 6.5$.

Πραγματι, λογω στραβομαρας, παρελειψα το 2 και το 3 :blush: .

[quote="Durandal"] η παράσταση παίρνει την πολύ κομψή μορφή:
$ 1 \plus{} \frac {z}{x} \plus{} \frac {2x}{y} \plus{} \frac {3y}{z} \equal{} 1 \plus{} c \plus{} 2a \plus{} 3b$
οπότε... :D

[/quote]

Nαι ακριβως εκει ηθελα να οδηγησω και εγω με την υποδειξη. μετα το min ευκολα με ΑΜ-ΓΜ για τους 3 τελευταιους ορους.  :)

[hide]BTW, σωστά βγάζω $ 1 \plus{} 3\sqrt [3]{6}$?[/hide]

Cheerio,

Durandal 1707

οχι μια χαρα αυτο ειναι. :)

[quote="Athinaios"]οχι μια χαρα αυτο ειναι. :)[/quote]

Ναι, απλώς το έβγαζα με το μάτι και όταν πρωτοείδα την άσκηση για κάποιο λόγο μου είχε κολλήσει ότι το αποτέλεσμα θα προέκυπτε ρητό... Meeeh :P

Cheerio,

Durandal 1707

Σάκη προσοχή! Η muirhead ισχύει σε συμμετρικές παραστάσεις. Δεν μπορείς να την επικαλεστείς εδώ!


Let abc be a 3-digits  positive number base 10.Find  max(abc-(a^3+b^3+c^3))=?

can it be 396
i thought that abc must be 620

[hide]The three digit number is written as
$100a + 10b + c$

We want to find the maximum of
$100a - a^3 + 10b - b^3 + c - c^3 = a(100-a^2) + b(10-b^2) + c(1-c^2)$

Now, we can find the maximum of each of the terms, for $0\le a, b, c \le 9$.

$\text{max}(a(100-a^2)) = 384 (a = 6)$
$\text{max}(b(10-b^2)) = 12 (b = 2)$
$\text{max}(c(1-c^2)) = 0 (c = 0, c = 1)$

So, the maximum is obtained for 620 or 621, and the maximum is 396.[/hide]


integration
calculus
function
calculus computations

Find the volume of the solid region R bounded above by the paraboloid

[tex]z=1-x^2-y^2[/tex] and below by the plane [tex]z=1-y[/tex]

The solution to this problem is:

[tex]V=\int_{0}^{1} \int_{-\sqrt{y-y^2}}^{\sqrt{y-y^2}} (1-x^2-y^2)dxdy -\int_{0}^{1} \int_{-\sqrt{y-y^2}}^{\sqrt{y-y^2}}(1-y)dxdy[/tex]

I thought that i understood how to solve these types of problems but I don't understand this solution. This is the procedure that I would use to solve this problem, please correct my mistakes.

1. set z=o in the paraboloid. this reduces the problem from 3-d to 2-d
2.  graph the curve of the paraboloid in 2-space (it's a circle)
3. set z=0 in the plane. this reduces the plane to 2-space ( it's a line)
4 what i have now is something like a half circle or hemisphere
5. find the limits of integration in the x and y directions

Here is where my solution fails. If y runs from the  lower limit of 1 to the upper limit of [tex]\sqrt{1-x^2}[/tex] (top of paraboloid), how do i find the limits of x?

As another note I thought that I could use one double integral to solve this. What type of volume problem requires two double integrals as opposed to one double integral?

some help from anyone would be appreciated.

The region in the $x$-$y$ plane over which you integrate is not given by the intersection of either surface with the plane $z=0$; rather it is given by the intersection of the two surfaces with each other.

$1-x^2-y^2=1-y$

$x^2+y^2-y=0$

$x^2+(y-\frac12)^2=\frac14.$

This is recognizably a circle, centered at $(0,\frac12)$ of radius $\frac12.$

The function we are integrating over the interior of that circle is the difference between the two heights. We might as well complete the square in that, too:

$(1-x^2-y^2)-(1-y)=\frac14-(x^2+(y-\frac12)^2).$

So, Volume $=V=\int\int_{x^2+(y-\frac12)^2\le\frac14}\frac14- (x^2+(y-\frac12)^2)\,dx\,dy$

Now, make the simple change of variables $v=y-\frac12,\,dv=dy.$

$V=\int\int_{x^2+v^2\le\frac14}\frac14-(x^2+v^2)\,dx\,dv$

Now switch to polar coordinates. $x^2+v^2=r^2$ and $dx\,dv=r\,dr\,d\theta.$

$V=\int_0^{\frac12}\int_0^{2\pi}(\frac14-r^2)r\,d\theta\,dr$

$V=2\pi\int_0^{\frac12}(\frac14-r^2)r\,dr=-\frac{\pi}2(\frac14- r^2)^2|_0^{\frac12}=\frac{\pi}2(\frac14)^2=\frac{\pi}{32}.$

One could make some generalizations about the nature of the intersection of a paraboloid of revolution with a plane.

Thanks for the reply I really appreciate it.


calculus
integration
conics
geometry
3D geometry
sphere

There is a dielectric spherical shell radius R,mass $m_{1}$,chargew $q_{1}$ with two small holes along the diameter.A small mass $m_{2}$ with charge $q_{2}$ is coming from infinity along that diameter.Find the time that the mass $m_{2}$ passes through the the shell.

is the dielectrical shell fixed, or can it move also?
and should we count with gravity too?

If it was fixed I wouldn't give the mass of the shell :wink:.There is no gravity.

This question cannot be resolved by direct integration i think because the integral will involve infinity. Try to see it as a degenerate case of orbital motion (a straight line instead of a nice conic curve perhaps), and set up a nice ODE by N2L and considering the relative motion with CM as the origin(Easy). Then you can get the answer.

PS: Actually I have resolve a similar problem, but the attraction force in gravitational force instead of electrostatic force... but computation-wise should be similar since you can just treat a charge sphere as a charge point mass....

Here is what I got:
$T=2R \sqrt{\frac{2R\pi \varepsilon_{0}}{q_{1}q_{2}}\frac{m_{1}m_{2}}{m_{1}+m_{2}}}$
Is that ok?

Did you forget one term in the last answer? :blush:  that ok?[/quote]

Is it
\[t = \frac{2R}{v_{0}\sqrt{1-\frac{2kq_{1}q_{2}(m_{1}+m_{2})}{m_{1}m_{2}Rv_{0}^{2}}}}? \]

$v_{0}$ is the initial velocity of the body and it should be given...

YEp right I for got to post the part with the initial velocity, but I counted with it, :mad:  anyway it was just one conservation of energy and one conservation of momento.


Tommy wants to paint each (outer) wall of his four-sided house with one of five colors such that no two adjacent walls have the same color. How many ways can he do this? (The different walls are distinguishable.)

[quote="nutz_for2.718281828"][hide="solution"]There are 5 choices for the first wall, 4 choices for the next, 4 for the third and three or four for the fourth. We need casework here. First,  there's the case where the third wall is the same color of the first wall. Then there are 5*4*4*4=320 ways to paint the house. The second case is when the third wall ISN'T the same color as the first wall. Then there are 5*4*4*3=240 ways to paint the house. We add each case to get 560 ways to paint the house.[/hide][/quote]
Sorry, that's not correct.  :(
Your idea of using cases was good, but you need to check your numbers again and recount both of the cases.  :wink:

[hide="solution"] We need casework here. First,  there's the case where the third wall is the same color of the first wall. Then there are 5*1*4*4=80 ways to paint the house. The second case is when the third wall ISN'T the same color as the first wall. Then there are 5*4*3*3=180 ways to paint the house. We add each case to get 260 ways to paint the house.[/hide]

[quote="bpms"][quote="nebula42"][quote="BFG"][hide="solution"] We need casework here. First,  there's the case where the third wall is the same color of the first wall. Then there are 5*4*1*4=80 ways to paint the house. The second case is when the third wall ISN'T the same color as the first wall. Then there are 5*4*3*3=180 ways to paint the house. We add each case to get 260 ways to paint the house.[/hide][/quote]
Yes, that's correct.  :)
By the way, in case anyone's interested, I believe there was a problem like this in the 2006 National MathCounts Competition.[/quote]
Yes but there were only four kinds of wood in the nationals problem (unless I don't remeber correctly...).[/quote]
Yes, that's why I said there was a problem [i]like[/i] this one.  :wink:
By the way, try to solve that version. (How many ways are there to paint the house if there are only four colors available?)

[quote="bpms"][hide]Casework. Two paint colors are used...
6 color choices/2 possible arrangments=
6*2=12.
3 colors
4 color choices/12 arrangements
48
4 colors
1 color choice/ 4 arrangements
4.
12+48+4=64 is probably wrong...[/hide][/quote]
Sorry, that's incorrect.  :(
Hint - try doing these two cases:
[hide]1. Call one side the first one. Consider the case when the side opposite from it is the same color.
2. Consider the case when the side opposite from it is the same color.[/hide]

[quote="SplashD"][hide]so $(4)(3)(1)(3)+4(2)(3)(2)=\boxed{60}$?[/hide][/quote]
Almost. You have the correct expression for the total, but you made a calculation mistake somewhere that got you the wrong answer.  :wink:

[quote="lotrgreengrapes7926"]Generalization: Tommy wants to paint each (outer) wall of his four-sided house with one of $x$ colors such that no two adjacent walls have the same color. How many ways can he do this? (The different walls are distinguishable.)[/quote]
[hide="Hint"]Try looking at what is similar in the solutions for the four color problem and the five color problem, and apply this for $x$. The problem actually isn't much different from the ones for four and five.[/hide]


analytic geometry
conics
ellipse
hyperbola
trigonometry

I need the answers to these problems, I'm not sure if mine are correct  :blush: .

1. find (1-i)^6 express result in rectangular form.
2. find the distance between P(10,7) and Q(-4,8)
3. Find the equation of the circle with center (1,2) and radius 3
4. what are the cordinates of the foci of the ellipse with equation (3x^2)+(4y^2)-6x+16y+7=0?
5.write this (y^2)-6y-4x+17=0 in standard form
6. what conic section is this...(3x^2)-(2y^2)-5x+8y=0
7.Find the equation of the hyperbola with foci (1,-5) and (1,1) and whose tranverse axis is 4 units long.

[hide="Problem 1"]Convert into polar coordinates: $ \sqrt{2} (\cos \minus{}45 \plus{} i \sin \minus{}45)$. When we take it to the sixth power, we then obtain $ 2^3 (\cos \minus{}270 \plus{} i \sin 270) \equal{} 8i$ via DeMoivre's Theorem.[/hide]

[hide="Problem 2"]Distance Formula: $ \sqrt{14^2 \plus{} 1^2} \equal{} \sqrt{197}$[/hide]

[hide="Problem 3"]A circle with center $ (h, k)$ and radius $ r$ has formula $ (x\minus{}h)^2 \plus{} (y\minus{}k)^2 \equal{} r^2$. Therefore, the answer is $ (x\minus{}1)^2 \plus{} (y\minus{}2)^2 \equal{} 9$.[/hide]

[hide="Problem 5"]A simple factorization yields $ (y\minus{}3)^2 \equal{} 4x \minus{} 8$[/hide]

[hide="Problem 6"]It is a hyperbola because the two squared terms have differing signs.[/hide]

[hide="Problem 4"]
We can solve this by completing the square. We start off with
$ 3x^2\plus{}4y^2\minus{}6x\plus{}16y\plus{}7\equal{}0$
$ 3x^2\minus{}6x\plus{}3\plus{}4y^2\plus{}16y\plus{}16\equal{}12$
$ 3(x\minus{}1)^2\plus{}4(y\plus{}2)^2\equal{}12$
$ \frac{(x\minus{}1)^2}{4}\plus{}\frac{(y\plus{}2)^2}{3}\equal{}1$
Therefore, the center of the ellipse is $ (1,\minus{}2)$, $ a\equal{}2$ and $ b\equal{}\sqrt{3}$. For ellipses, the foci are located $ c$ away from the center along the major axis, where $ a^2\equal{}b^2\plus{}c^2$. Hence, $ c\equal{}\sqrt{7}$. Therefore, the foci are located at $ (1\pm\sqrt{7}, \minus{}2)$.
[/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299920]Problem 4[/url] was solved already. By the way, foci are of the length $ \sqrt{a^2\minus{}b^2}$, where $ a$ and $ b$ are the lengths of the semimajor and semiminor axes, respectively.


function
probability
Support
abstract algebra
probability and stats

Let $ p$ be a probability distribution for a symmetric random walk on $ \mathbb{Z}$ with finite range.

If $ f: \mathbb{Z}\to\mathbb{R}$ is a function satisfying $ f(0)\equal{}0$, $ f(x)\equal{}o(|x|)$, and $ \mathcal{L}f(y) \triangleq \minus{}f(y)\plus{}\sum_{x\in\mathbb{Z}} p(x)f(x\plus{}y) \equal{} 0$ for $ y\not\equal{}0$ (e.g., $ f$ is harmonic and does not grow too fast at infinity), then $ f\equal{}0$.

This is more or less trivial in the case where $ p$ is the probability distribution for simple random walk on $ \mathbb{Z}$, but the situation is less clear when $ p$ assigns positive probabilities to several points.  In fact, I really have no idea on how to approach the general case.

Do you need to make some sort of assumption on $ p$ for this to work?  For example, if $ p$ only takes on even values, then $ f$ would only have to be zero on the even integers, and could be nonzero and constant on the odd integers.

A natural assumption to make would be that $ p$ is such that every integer is reached with positive probability at some point.  I'm not sure, but it might be possible to show that this is equivalent to showing that your walk will almost surely reach every $ a$ at some point.  You could then try and argue as follows:

Let $ a$ be fixed, and consider the stopping rule where you stop once you reach $ a$.  If the random walk statement is true, then the expectation of the stopped $ f$ is equal to $ f(a)$.  However, the expectation of $ f$ at any finite time is $ 0$ by the harmonic assumption.  Now use your assumptions on $ f$ to show that the limit of the expectation at time $ t$ is equal to the expectation of the stopped $ f$.

I haven't actually worked through this (and it's been a long time since I did anything related to this), so there might be a snag somewhere.  However, it feels like the above should work.

[quote="kevinatcausa"]Do you need to make some sort of assumption on $ p$ for this to work?  For example, if $ p$ only takes on even values, then $ f$ would only have to be zero on the even integers, and could be nonzero and constant on the odd integers.[/quote]

the support of $ p$ is assumed to contain some sort of generating set for $ \mathbb{Z}$.

[quote="kevinatcausa"]Let $ a$ be fixed, and consider the stopping rule where you stop once you reach $ a$.  If the random walk statement is true, then the expectation of the stopped $ f$ is equal to $ f(a)$.  However, the expectation of $ f$ at any finite time is $ 0$ by the harmonic assumption.  Now use your assumptions on $ f$ to show that the limit of the expectation at time $ t$ is equal to the expectation of the stopped $ f$.[/quote]

this seems like a good idea, but we need to take into account that, for example, $ f(x)\equal{}x$ is harmonic, so we need the growth condition, and in two dimensions multiples of the potential kernel will work.


Let a;b;c>0
We have
$ \frac{abc}{a^3\plus{}b^3\plus{}c^3}\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{ac\plus{}ab\plus{}bc}\ge \frac{4}{3}$

Assume $ p\equal{}1$
<=>$ (3q\minus{}1)(10q\minus{}9r\minus{}3) \ge 0$
By schur we have :$ 9r \ge 4q\minus{}1$
$ (3q\minus{}1)(10q\minus{}9r\minus{}3) \ge 2 (3q\minus{}1)^2 \ge 0$
Done :)

<=>
\[ \sum (a\minus{}b)^2 (\frac{1}{ab\plus{}bc\plus{}ca}\minus{}\frac{a\plus{}b\plus{}c}{3(a^3 \plus{}b^3 \plus{}c^3)})\geq 0\]
wich is true because \[ 3(a^3 \plus{}b^3 \plus{}c^3)\geq (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\]


calculus
integration
algorithm
calculus computations

is it possible to integrate [size=150][b] e^cos 2x[/b] [/size]with respect to x  :?:

According to the online Mathematica integrator, no.
[url=http://integrals.wolfram.com/index.jsp?expr=e^cos(2x)&random=false]Integrator[/url]

hello, see here
http://en.wikipedia.org/wiki/Risch_algorithm
Sonnhard.


geometry
3D geometry
tetrahedron
search
Putnam

This is a problem that I do not know the solution to

Suppose that every point in the plane is colored either red, white, or blue.  Show that there exists two points in the plane which are exactly 1 inch apart and have the same color.

I think this problem is well-known. It is a part of unsolved problem: what is the minimal number of colors for which this statement (there are 2 points withs the same color and distance of 1) is false.

[quote="grobber"]Isn't it rather easy, though? Just take the four vertices of a regular tetrahedron of side [tex]1[/tex]. There must be two of them which have the same color, and each pair is one inch apart, so the conclusion follows.[/quote]
But they are not "on plane"

Yes, cotan.  This problem is well-known (and has a name).  There is a simple device (which also has a name) for proving gauss202's statement.

[If nobody comes up with the solution in a couple of days, I shall reveal the name of the famous problem and post a picture of the famous device.]

Yes, I realize this is a famous problem. I saw the solution once when I was in high school. But I have no idea what it is now.

fuzzylogic - I can find a counterexample to your method. Color the plane with stripes of width 3/4 and alternating colors.
However, it is possible to prove that there exist a equilateral triangle with length n [i]or[/i] 2n

This configuration works nicely! (all segments shown have length 1)

Yep. That's the solution that I had seen. Had you seen this before also or did you come up with it on your own?

I had seen it years ago (in 7th grade) so I knew it had to do with some clever configuration, but I had to rediscover it just now.

Sorry, gauss.  I pm'ed you, but I guess you haven't receive it yet.

The problem is called [hide]"The Chromatic Number of the Plane"[/hide].  The device posted by zabelman is called a [hide]"Moser Spindle"[/hide].  A web-search will reveal a lot more information.

I think it has been proven that it takes at least 4 colors and that it takes no more than 7 colors.  The exact value has remained an open question for decades.

After seeing the device posted by [b]zabelman[/b] above, I recalled that I saw this problem long long time ago. The solution I knew was as follows which is just a restate of "the device" in "words".

So let's assume it's not true, then the vertices of any equilateral triangle of side length 1 inch have different colors.

By fliping the equilateral triangle along one side, we can prove that any two points of distance  :rt3: have the same color.

So any point on a circle of radius  :rt3: has the same color as the center.

Taking any chord of length 1 on the circle, the end points of the chord are exactly 1 inch apart and have the same color.

A contradiction!

a related problem:

Prove or disprove whether it is possible to color each point in the Euclidean plane with nine colors so that no two points with the same color are a unit distance apart.

edit: well i see RCV already posted about this, but its still a nice problem to try (its a putnam problem with a slick solution)

Terribly sorry: I thought the points were in space :(.

Take a look [url=http://www.mathpuzzle.com/chrompln.html]here[/url] for a coloring of the plane with 7 colors s.t. no two points are a unit distance apart (so it's better than nine :)). Anyway, the construction with nine colors could be as follows: cover the plane with squares of side slightly less than [tex]3\frac {\sqrt 2}2[/tex] and divide each such square in nine smaller squares in which you put the nine colors (in the same configuration for all the large squares). This might be another stupid mistake I make, but right now it seems to me that it works (although it's not really "slick" :)).

Actually, I think the squares need to be exactly of side length 3/ :rt2:, while containing only some of their edges -- for example, if we have square ABCD, let it contain edges AB and BC, including vertices A and B but excluding vertex C.  Anyhow, that's the idea.