superior algebra
superior algebra unsolved

Why the set of all $ n\times n$ matrices with rank = r is not an algebraic set while all $ (r \plus{} 1)\times (r \plus{} 1)$ monors are also vanished?

Thks in Advance.

[quote="HilbertThm90"]Can you explain the question in a little more detail?  :maybe: Thanks.[/quote]

Seconded.

I think he means algebraic as a subset of $ \mathbb{A}^n(k)$.

The condition that the rank is $ \le r$ is that a certain set of determinants vanish, which is an algebraic condition, so the set of matrices with rank $ \le r$ is algebraic. But the only way to single out the matrices with rank exactly $ r$ is to take out those with rank $ \le r\minus{}1$. Intuitively, the complement of an algebraic set shouldn't be algebraic, and I think some argument about Zariski open/closed sets should work.


linear algebra
matrix
algebra
polynomial
quadratics
vector
superior algebra

Let $ GL(2,q)$ be matrix group on the  finite field $ GF(q)$.  Its order is $ q(q\minus{}1)(q^2\minus{}1)$. There exists an element of order $ q^2\minus{}1$, I wonder that what is the  type of these elements of order $ q^2\minus{}1$.

What do you mean by "the type of these elements" ? What "type"?

Do you want to know how to proof the existence of such matrices? One (non-constructive) way is to consider the field $ \mathbb{F}_{q^2}$ and its multiplicative group. This group is cyclic so there is an element $ a\in\mathbb{F}_{q^2}$ with order $ q^2\minus{}1$. The multiplication map $ \mathbb{F}_{q^2}\to\mathbb{F}_{q^2}, x\mapsto ax$  is $ \mathbb{F}_q$-linear and has order $ q^2\minus{}1$ so we get an automorphism of the two-dimensional $ \mathbb{F}_q$-vectorspace with order $ q^2\minus{}1$. Choose a basis and you get the matrix we are looking for.

Thanks.

I want to know the matrix representation of elements of order $ q^2\minus{}1$. I'll try to do myself.

If you look closely at what Gockel said, you can find such a represenation.  First you need to find the minimal polynomial $ x^2 \plus{} ax \plus{} b$ for a primitive root in $ \mathbb{F}_{q^2}$ (there is probably no closed form for this) and then form its companion matrix: $ \left(\begin{array}{cc} 0 & \minus{} b \\
1 & \minus{} a \end{array} \right)$. (all appropriate matrices will be conjugate to such a matrix)  You can find this polynomial more explicitly by factoring the cyclotomic polynomial $ \Phi_{q^2 \minus{} 1}$ into quadratic factors in $ \mathbb{F}_q$.

For example, take $ \mathbb{F}_3$.  We have $ \Phi_8 \equal{} x^4 \plus{} 1\equal{}(x^2 \plus{}x \minus{}1)(x^2\minus{}x\minus{}1)$, giving us conjugacy class representatives of $ \left(\begin{array}{cc} 0 & 1 \\ 1 & \minus{}1\end{array}\right)$ and $ \left(\begin{array}{cc} 0 & 1 \\ 1 & 1\end{array}\right)$, both of which have order 8.

EDIT: Corrected significant typo, added an example.

Thanks,  LydianRain.

$ GF(q^n)$ can be regarded as a $ n$-dimmentional vectors space on $ GF(q)$. Let $ x: \equal{}(x_1,x_2,\cdots, x_n)\in GF(q^n)$, where $ x_i\in GF(q)$.  I wonder that what is the criterion of the product of $ (x_1,\cdots,x_n)(y_1,\cdots, y_n)$.


Let $ABC$ is an equaliteral triangle.$X$ any point in it.
Prove that $S_{X}\leq \frac{2}{3}S_{ABC}$
For $S_{X}$ see this
 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58188]www.mathlinks.ro/Forum/viewtopic.php?t=58188[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=388279#p388279]www.mathlinks.ro/Forum/viewtopic.php?p=388279#p388279[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65594]www.mathlinks.ro/Forum/viewtopic.php?t=65594[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65597]www.mathlinks.ro/Forum/viewtopic.php?t=65597[/url]
Nice colection,isn't it? :D  :D

[quote="Tiks"]Let $ABC$ is an equaliteral triangle.$X$ any point in it.
Prove that $S_X\leq \frac{2}{3}S_{ABC}$
For $S_X$ see this
 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58188]www.mathlinks.ro/Forum/viewtopic.php?t=58188[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=388279#p388279]www.mathlinks.ro/Forum/viewtopic.php?p=388279#p388279[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65594]www.mathlinks.ro/Forum/viewtopic.php?t=65594[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65597]www.mathlinks.ro/Forum/viewtopic.php?t=65597[/url]
Nice colection,isn't it? :D  :D[/quote]
Why have you forgote  this nice problem?
It is still unsolved! :P

Hey,this problem is still unsolved!
Try it,it is really a good problem. :wink:

Hmmm, I still see that this problem remains unsolved so I thought I would post a reminder. :)


Let $ f : R^{\plus{}}\to R^{\plus{}}$ be an increasing function. For each $ u\in R^{\plus{}}$ , we denote $ g(u) \equal{} \inf \{ f(t)\plus{}u/t | t>0\}$. Prove that
(a) If $ x\le g(xy)$ , then $ x\le 2f(2y)$
(b) If $ x\le f(y)$ , then $ x\le g(xy)$.

In analysis the infimum or greatest lower bound of a subset $ S$ of real numbers is denoted by $ \inf(S)$ and is defined to be the biggest real number that is smaller than or equal to every number in $ S$.

[quote="mustafa-yafes"]Let $ f : R^{ \plus{} }\to R^{ \plus{} }$ be an increasing function. For each $ u\in R^{ \plus{} }$ , we denote $ g(u) \equal{} \inf \{ f(t) \plus{} u/t | t > 0\}$. Prove that
(a) If $ x\le g(xy)$ , then $ x\le 2f(2y)$
(b) If $ x\le f(y)$ , then $ x\le g(xy)$.

In analysis the infimum or greatest lower bound of a subset $ S$ of real numbers is denoted by $ \inf(S)$ and is defined to be the biggest real number that is smaller than or equal to every number in $ S$.[/quote]

Posted some hours ago : see http://www.mathlinks.ro/Forum/viewtopic.php?t=316736

And give the contest reference, plase, to help searching.

you can find a nice solution from halil94's post  :idea:

[quote="pco"]
Posted some hours ago : see http://www.mathlinks.ro/Forum/viewtopic.php?t=316736

[/quote]

I wasn't knowing this is turkish question and posted before
sorry..


group theory
superior algebra
superior algebra unsolved

I came across this in a book and I am not sure how to prove it :(
Let $ C_2 \equal{} \{1,\sigma\}$ the cyclic group of order 2.
Prove that the group ring $ \mathbb{C}C_2$ is isomorphic, as a $ \mathbb{C} \minus{} algebra$ to $ \mathbb{C}\times\mathbb{C}$. The $ \mathbb{C} \minus{} algebra$ structure is given by $ z \mapsto (z,z)$. (What does that mean?)
(Hint: Consider $ \{\frac12 \plus{} \frac12 \sigma, \frac12 \minus{} \frac12 \sigma\}$).

Should I adhere to the isomorphism theorems? Or try to find a bijective homomorphism? I dont really know :(.
Thanks!

I assume "$ z\mapsto (z,z)$" is telling you the $ \mathbb C$-algebra structure that is being put on the ring $ \mathbb C \times \mathbb C$, in other words the specified copy of $ \mathbb C$ that sits inside it.

Consider the map $ f: \mathbb C[C_2]\to \mathbb C \times \mathbb C$ given by $ 1\mapsto (1,1)$, $ \sigma \mapsto (1,\minus{}1)$ (or, if you want to use your hint, $ \frac12 \plus{} \frac12\sigma \mapsto (1,0),\frac12\minus{}\frac12\sigma\mapsto(0,1)$). This is your isomorphism.

Maschke's theorem :P
(Theorem 3.1 [url=http://math.mit.edu/~etingof/replect.pdf]here[/url]).

  darij

How do you verify that it is an isomorphism? Say, the first function. I mean if $ 1\mapsto (1,1)$ then $ a\mapsto (a,a)$ but what does $ z \equal{} a\plus{}bi\mapsto$ mean? Is it $ (z,z)$? If so then what does $ f(z\plus{}w\sigma) \mapsto$ mean? :(

If you follow the suggestion of caffeineboy, then
$ f(a\plus{}bi) \equal{} a f(1) \plus{} bf(i) \equal{} (a,a) \plus{} (b,\minus{}b) \equal{} (a\plus{}b, a\minus{}b)$.

Are you considering the $ a,b$ to be complex? Because $ f(\sigma) \equal{} (\minus{}1,1)$ not $ f(i)$. Am I mixing something up? :(

Okay here's a full solution.
Let $ f: \mathbb C[C_2]\to \mathbb C \times \mathbb C$ be defined by $ 1\mapsto (1,1)$, $ \sigma \mapsto (1, \minus{} 1)$. In other words we define $ f$ by $ f(1)\equal{}(1,1)$ and $ f(\sigma)\equal{}(1,\minus{}1)$, and extend by the conditions that we need to be true to make $ f$ an algebra homomorphism.
So $ f(a\plus{}b\sigma)\equal{}f(a)\plus{}f(b\sigma)\equal{}af(1)\plus{}bf(\sigma)$ because we defined $ f$ that way, so $ f(a\plus{}b\sigma)\equal{}(a\plus{}b,a\minus{}b)$. (Here $ a,b\in \mathbb C$). It is also true that $ f(a\plus{}b\sigma)f(c\plus{}d\sigma)\equal{}f((a\plus{}b\sigma)(c\plus{}d\sigma))$, you can check.
Now we can prove that $ f$ is an isiomorphism:
Surjective: For $ (x,y)\in \mathbb C^2$ take $ \frac{x\plus{}y}2 \plus{} \frac{x\minus{}y}2\sigma$
Injective: If $ a\plus{}b\equal{}c\plus{}d$ and $ a\minus{}b\equal{}c\minus{}d$, then we find that $ a\equal{}c$ and $ b\equal{}d$

Ah so $ a,b$ were indeed complex. That was throwing me off since im used to $ a,b$ denoting the scalars of the complex number. Thanks :)


geometry
geometric transformation
induction
perimeter
combinatorics solved
combinatorics

Each square of a $ \left(2^{n}-1\right)\times\left(2^{n}-1\right)$ board contains either the entry $ +1$ or the entry $ -1$. Such an arrangement of entries is called [i]successful[/i] if each entry is the product of its neighbours (i. e. entries in squares sharing a side with it). Find the number of successful arrangements.

I think this was solved before (probably by kappa); you might want to try looking for it.

I couldn't solve it so far, but I think the following is true:

If we place a queen (from chess) in the center square, then all the squares she attacks are 1's.

This works for n=2 (3x3 square) and n=3 (7x7 square). There are really few such arrangements, by the way: for 3x3 there is only 1 (all 1's), and for 7x7 there are 2.

I don't have a complete proof for the lemma above, but I would bet my money on it :D. That might not be enough to make it true, though...

I think that there is only one arrangement for n>=2. I will prove this by induction. For n=2 it's easy to prove by hand that the only valid arrangement is the one with only ones.

Now, consider an 2^(n+1)-1 by 2^(n+1)-1 board. Define a "2neighbour" of a cell to be the image of a translation of 2 cell lengths in either one of the four directions. It's easy to prove, by taking cases after the number of 2neighbours of a square, that the number in a cell is the product of the numbers in its 2neigbhours.

We consider the 2^n - 1 by 2^n-1 formed like this : all cells (i,j) with i and j both even. By the property above, this clearly satisfies the property and by the induction step, it's all 1.

Now consider the 2^n by 2^n board formed like this : all cells (i,j) with i and j both odd. Should we prove that this is as well all ones, we are done.  Now using the fact that on the big board we have only 1 on the positions with (i,j) both even, we can easily prove that each number in a cell with (i,j) both odd is the product of its vertical 2neighbours AND the product of its horizontal 2neighbours. Now, this means that each number in the 2^n x 2^n board, except those on the perimeter, are 1 (because each cell in the big board is the product of all 2neighbours, that is the product of the vertical 2neighbours squared, or equivalently 1).  We are still working in the 2^n x 2^n board. Let x be the value in the cell (1,1) and y be the value in the cell (1,2). Then in (2,1) we also have y and so x = y*y*1 , so x=1.  Now it's easy to see that we have, for example, on line 1, the following pattern :

1 y y 1 y y 1 ...

Take a 1 below an y, which has all other 3 neighbours 1. (not 2neighbours!). Since that 1 is the product of its neighbours, we have that y=1. Therefore, y=1 and the first line is filled with 1. We do the same for the first column, last line and last column. Therefore the 2^n x 2^n board is also filled with 1.  If we color the 2^(n+1)-1 by 2^(n+1)-1 board as a chess table with 2 colors, (1,1) being white, we have that all white squares are 1. Therefore since a black square's value is the product of only white square values, we infer that the whole table is filled with 1.

Thus, only one solution for n>=2 and two solutions for n=1.

Looks like you're right. My second arrangement for the 7x7 board was wrong and that's what misled me :(.

does anyone know the solution of this problem using linear algebra?


[hide="answer"]Call the least integer n, the next n+1, the last n+2. Thus, the sum of their squares is 3n^2+6n+5=365. Solving the quadratic for n, we get either 10 or -12, so the answer is 10.[/hide]

you could put it in a quadratic or just kill it with brute force which is actually really easy. you know it has to be in the hundreds and it has to be pretty low. it can't be in the 200s because that will be too high. so try 11-12-13 or 10-11-12.

Or for relatively large values, divide the number by 3, take the square root of it, whatever number it is closest to, is the middle number, and then go from there.

[quote="vishalarul"][hide]Set the variables to be $x-1$, $x$, $x+1$ to get this equation:
$3x^{2}+2 = 365$

This method is a lot easier. [/hide][/quote]

actually in this case, it is much easier to guess in your head. since all the value must be in the 100-200 range, its 10-11-12 or 11-12-13. but 169 is 69 which is more than 65, so it must be 10-11-12. or 10.  i did that in my head in like 4 or 5 seconds.  the other way would have taken me like 20.

but with bigger numbers, quadratic is the way to go.

OR, do as I said, to divide by 3 and take the square root of the number, and the number is the middle number of the sequence.

[hide]
just eyeballing this and i get the answer of 10 noting that the middle is around 120 (don't do this for larger problems)
[/hide]
-jorian

[hide]

The integers are 10,11,12 so then answer is 10
and I got this by guess and checke which in my opinion is easier in this case since its pretty obvious where the numbers ought to fall[/hide]

but that's not good if you didn't get it on your first try...
jorian

[quote="anirudh"][quote="ckck"]The sum of the squares of three consecutive positive integers is 365. What is the smallest of these integers?[/quote]

[hide]
$x^{2}+(x+1)^{2}+(x+2)^{2}$
$x^{2}+(x^{2}+2x+1)+(x^{2}+4x+4)$
$3x^{2}+6x+5=365$

Using the Quadratic Formula:

$n$ is either 10, or -12, but you cannot use the negative value, so the answer is $10$.

[color=blue]kyyuanmathcount: NO BLUE FONT!  that is mine, thanks.  Actually, no colored font anyways, it gets confusing.  [/color]
I thought that I shouldn't do color if its small. I've seen other people post in Blue all the time
[/hide][/quote]
Was this blue font in the mc forum?

[quote="anirudh"]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=717578#717578

That's the post[/quote]
Ah, but that's dark blue. You posted in light blue.

[quote="anirudh"]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=717578#717578

That's the post[/quote]

First of all, you need not copy everythings others do, you have already done your share of copying competition problems.

Secondly, we must have missed that.  The policy is no colored font unless there is a good reason (word painting and the such).

Third, I must go to class now.

I like vishalarul's solution the best, and that's the way I did it. I really like it because all the terms cross out and you're left with an easy equation. Although guessing or the such works, it might take more time if you don't get it right away. But a solution involving making an equaiton (such as the ones other people have come up with :) ) is bound to give you the answer right away. You can count on it all the time. :)


If :an=(1/n+1)+...+(1/2n)  Then: Lim (an)=Ln2  when n be great enough.

[quote="N.T.TUAN"]$\text{LHS}\geq n.\frac{1}{n+n}=\frac{1}{2}$.[/quote]

can you explain your solution pleass!!what does lhs mean? :)


I am now a third-year undergraduate (which is also the last year in Hong Kong education system), and am about to start my master's degree next year.

A problem that has been bothering me is about research interest. I am supposed to sort of know where my interests lie and then to find an advisor and start to try out some research or something like that.... but a big problem is that I have no idea where my interests lie. I have taken courses in different areas of mathematics and so far, most of them are fine for me, with no special (or maybe particularly strong) feeling towards any area.

Can anybody share their experience about how they found their research interest? Or maybe give me some advice?

Unfortunately this will probably be very unhelpful to you, but I found my interest by taking a special topics class.

I'd say if you don't know what you want to do, then decide based on other criteria: try to find a professor that you like or think would be a good adviser, and then do what that professor suggests.


ARML
Vieta
quadratics
analytic geometry
algebra
polynomial
sum of roots

1) Compute all values of $b$ such that one root of $x^{2}+bx+6=0$ is twice the other.

2) The roots of $x^{3}-7x^{2}-21x+d=0$ form a geometric sequence.  Determine the value of $d$.

3) What is the coefficient of $x^{8}$ in the expansion of $(x-1)(x+1)(x-2)(x+2)(x-3)(x+3)(x-4)(x+4)(x-5)(x+5)$?

4) If $q$ and $r$ are the roots of $x^{2}+2x+5=0$, find the quadratic equation whose roots are $q+1$ and $r+1$.

5) For some real value of $p$, two of the roots of $x^{3}+px^{2}+12x-9=0$ add to $4$.  Compute the third root.

6) Determine the sum of the $y$-coordinates of the four points of intersection of $y=x^{4}-2x^{2}-5x+3$ and $y=x^{2}+9x$.

1
If a and 2a are the 2 roots of the equation
x^2+bx+6=0, then
by vieta's theorem (a+2a)=3a=-b and 2a^2=6
which implies a=3^.5 and b=-3^1.5.

2
Let the solutions of the equations be a,ak,ak^2.
then by vieta's formula
a(K^2+K+1)=-7,(1);d=-(ak)^3,(2);ka^2(1+k+k^2)=-21(3)
Then from 1 and 3, ak=3=-7/(k^2+k+1)
Which implies d=-27 and k=-3and-(1/3)
k=-1/3 yields another value of d,-1/(27)^3
So, d=-27,-1/(27)^3

4
x^2+2x+5=0 implies that (q+r)=-2 and qr=5.so the roots of the wanted equation are (q+1) and (r+1).So the sum of this roots are 0 that is  [b]the coefficient of x divided by the coefficient of x^2 [/b]in the wanted equation is 0, and therefore the coefficient of x in the equation is 0.
(q+1)(r+1)=qr+q+r+1=5-2+1=4 .And therefore [b]the constant term of the equation divided by the coefficient of x^2 [/b] is 4.
So, the desired equation is [b]x^2+4=0[/b].

6
P(x)|=x^2+9x; Q(x) =x^4-2x^2-5x+3.
We need to find the sum of P(x) where x are the solution to the equation P(x) =Q(x). (1)
Let the solutions to (1) be a, b, c and d. We need to find
{a^2+b^2+c^2+d^2+9(a+b+c+d)}
By vieta’s formula (a+b+c+d) =0 (1) ;( ab+ac+ad+bc+bd+cd) =-3.
By squaring both sides of 1 and then subtracting from it 2(ab+ac+ad+bc+bd+cd) we get the value of (a^2+b^2+c^2+d^2) which is 6.
And then we get the desired value which is 6.

5
If the third root is t then by vieta's formula t=-(p+4)

umm just wonderin so i can learn something
whats vieta's theorem?

take polynomial x^7+ax^6+bx^5+cx^4+dx^3+ex^2+fx+g
sum of roots is -a
sum of pairwise products of roots is b
et cetera
sum of all products of six roots is f
product of roots is -g

I took degree 7 as a random example

wha....? ok i get sum of roots is -a xD whats a pairwise product? and so this works for all polynomials? hmmmm.....
btw just wonderin, how much u gotta post/play to get unmuted...


algebra
polynomial
logarithms
calculus
limit
induction
algebra proposed

[quote="Nawahd Werdna"]Is there some polynomial with complex coefficients such that $ \log_{k}x \equal{} P(x)$?[/quote]
take $ f(x)\equal{}ln(x)$ if $ p(x)\equal{}log_{k}(x)\equal{}\frac{ln(x)}{ln(k)}$
then $ f^{(deq(p)\plus{}1)}\equal{}ln(k).p^{(deg(p)\plus{}1)}\equal{}0$ (where $ g^{1}\equal{}g'$ and $ g^{(k\plus{}1)}\equal{}(g^{(k)})'$)
(impossible)

I am not very familiar with calculus, could you please further explain your solution???

It doesn't exist, because if we differentiate the polynom $ n\plus{}1$ times where $ n$ is the degree, we get 0. But $ log$ isn't 0 if we differentiate it $ n\plus{}1$ times.

I have a ideea, which I think also works for the simple case.
if $ B(0)\neq 0$, then $ \lim_{x\rightarrow 0}\frac{A(x)}{B(x)}$ is finite, but $ ln$ has limit $ \minus{}\infty$ in $ 0$.
If $ B(0) \equal{} 0$, we have $ ln(x)\cdot B(x) \equal{} A(x)$. We differentiate and get $ \frac{B(x)}{x}\plus{}B'(x)lnx \equal{} A'(x)$. So, $ B'(x)lnx \equal{} A'(x)\minus{}\frac{B(x)}{x}$. But now $ B'$ has a lower degree than $ B$ and the right one is still a polynomial, because $ B(0) \equal{} 0$. By induction, we will lower $ B$'s degree,until $ B(0)\neq 0$ (it must happen once), and we fall into the first case.
So, $ ln$ can't be written as $ \frac{A}{B}$.

[quote="Nawahd Werdna"]Would this proof still work if we replaced $ P(x)$ with $ \frac{A(x)}{B(x)}$??[/quote]
now we we have $ B(x)ln(x) \equal{} ln(k)A(x)$
now it's easly to see that if we differentiate the polynom $ (ln(k)A(x))$ $ deg(A)\plus{}1$ times we get $ 0$.
but if we differentiate the polynom $ (B(x)ln(x))$ $ deg(A)\plus{}1$ times we don't get $ 0$.

[quote="Nawahd Werdna"]
but if we differentiate the polynom $ (B(x)ln(x))$ $ deg(A)\plus{}1$ times we don't get $ 0$.[/quote]
Prove it:)

[quote="Frozen"]By induction, we will lower $ B$'s degree,until $ B(0)\neq 0$ (it must happen once), and we fall into the first case.
So, $ ln$ can't be written as $ \frac{A}{B}$.[/quote]

How do you use induction to lower B's degree? Also, how do you know that $ B(0)\neq 0$ must happen once?

EDIT: 100th Post!!!! :D  :D

[quote="Frozen"]So, $ B'(x)lnx \equal{} A'(x)\minus{}\frac{B(x)}{x}$. But now $ B'$ has a lower degree than $ B$ and the right one is still a polynomial, because $ B(0) \equal{} 0$. By induction, we will lower $ B$'s degree,until $ B(0)\neq 0$ (it must happen once), and we fall into the first case.[/quote]
Because $ B'(x)ln(x) \equal{} A'(x)\minus{}\frac{B(x)}{x}$, if we note $ B_{1}(x) \equal{} B'(x)$ and $ A_{1}(x) \equal{} A'(x)\minus{}\frac{B(x)}{x}$, we have $ B_{1}(x)lnx \equal{} A_{1}(x)$. Now, $ B_{1}$ has lower degree than $ B$, because we obtained it by differentiating $ B$. Again, if $ B_{1}(0) \equal{} 0$ the problem is finished, else, we can do all this thing again. We get $ B_{2}$ and $ A_{2}$. Why once we must have an $ k$ such that $ B_{k}(0)\neq 0$?.
Because $ B_{k}$ is obtained by differentiating $ B$ $ k$ times. And if $ B(X) \equal{} uX^{m}\plus{}...$, where $ m \equal{} deg(B)$, we have $ B_{m}(x) \equal{} u\cdot m!$. so $ B_{m}(0)\neq 0$.


1.conduct 1000 independent tests, p(A)=0.75 in each test. find the probability of the event |f-p|< 0.02.
2. the probability of the event A in each test is 0.5. find n ( the number of tests) satisfies P(|f-p|<0.02)=0.7698
here f is the frequency and p is the probability.

$ f\equal{}\frac{X_1 \plus{} \ldots \plus{} X_n}{n}$ isn't it, where $ P(X_i\equal{}1) \equal{} 1 \minus{} P(X_i \equal{} 0) \equal{} P(A)$. So you know $ E[X_i]$ and $ \sigma(X_i)$, and I think that you know that $ \sqrt{n} [\frac{\sum_i X_i}{n} \minus{} E[X_1] ]$ is almost $ \mathcal N (0, \sigma(X_1)^2])$. Can you conclude from that ?


Find all ordered pairs of positive integers $ a,b$ such that $ ab\minus{}1$ divides $ a^4\minus{}3a^2\plus{}1$. Prove no other solutions exist.

[quote="mystic_crusader88"]Find all ordered pairs of positive integers $ a,b$ such that $ ab \minus{} 1$ divides $ a^4 \minus{} 3a^2 \plus{} 1$. Prove no other solutions exist.[/quote]
Did you mean a^4-3a^2+1 is divisible by ab-1?

Yes. The way I said it and the way you phrased it both mean the same thing.

Since $ gcd(ab\minus{}1, a) \equal{} 1$, we can rewrite this condition as $ ab\minus{}1$ divides $ a^2 \plus{} b^2 \minus{} 3$.  I really don't know where I'm going there, but this form shows us that the relationship between $ a$ and $ b$ is symmetric.


I had an Alcumus score of 68.43 and then after I got a problem CORRECT, it went down to 68.27.... :mad:  :huh:

I'm not sure but it said that scores are not updated in real time.

Yeah; it doesn't change after every question.

I still don't get it, it went DOWN. It didn't say the same...

I think it also depends on whether you got it right the first time or the second time, and how it compares relative to others' results. I'm not sure though...

lol, it just happened again! (I got the problem correct on the first try)

Remember that the Alcumus Score is not computed in real time. You may have fallen sometime prior to the problem you just got correct and the overall score was just recently updated for the report or statistics page. Also, other people could be answering harder problems correctly, upping their score while lowering yours slightly. Overall, it's a very complicated computation and ups and downs should be expected.

remember that alcumus is more of a teaching and learning thing rather than a competitive thing

use it to get better at math, not to brag to your BFF Jill about your rank

[quote="jbatterson"]I am noticing the same thing... I can get a run of 10-15 correct in a row (some on the second try) and watch my Alcumus score drop half a point.  A run of 5-6 on the first try sometimes drops my Alcumus score.  I started the day with the goal of breaking 70,  I think I have missed 1 of the 25 or so problems I have done... my Alcumus score is lower than it was when I started.  Still fun practice and problems, but top 10 seems out of reach at this point, I suppose if I get a run of really hard problems correct I will be able to make up some ground.[/quote]

We fixed a bias in the Alcumus scoring system.  Top 10 not so out of reach now.

[quote="levans"]It's a super secret machine learning algorithm.[/quote]

Of course I do not know what they did. But for people who are interested in ranking wrt games I recommend to look at the [url=http://research.microsoft.com/osa/apg/default.aspx]Microsoft Research, Applied Games Group.[/url] For example you might be interested in Bayesian ranking then have a look [url=http://research.microsoft.com/~rherb/slides/Bayesian%20Ranking.ppt]here[/url] or [url=http://www.iro.umontreal.ca/~lisa/seminaires/07-11-2006.ppt]there.[/url] If you want to have more details check out [url=http://research.microsoft.com/~rherb/pub/group_all_.aspx]Ralf Herbrich's publication page.[/url]

I am kind of confused about the recalibration of Alcumus scores. I understand that it is not updated in real time, but does the formula change?

I have noticed that a few times my score has gone down despite no wrong answers since the last score change. For example, my score changed from 90.02 to 90.22, I got seven questions right, and none wrong, and then it changed to 89.94.

Can anybody help explain this? Thanks.

So if Bob had a percent of 100%, and if he goes on a trip for a year without touching Alcumus, his percent will go down?

[quote="levans"]Other people could be answering harder problems correctly, upping their score while lowering yours slightly. Overall, it's a very complicated computation and ups and downs in your score should be expected.[/quote]

That would make sense if there is some sort of bayesian algorithm that is calculating weights for each problem.  Assume, for instance, that your score is not recomputed on the basis of solving new problems but on the basis of the past problems that you have solved.  Then, as more and more people encounter those same problems (and/or rate them), the system has a more accurate idea of the "difficulty" of those problems.  I assume more "difficult" (those with larger variance in correctness and/or answers given) problems will be given higher weighting.  Therefore, unless you have gotten each problem correctly (on the first try), then as other people encounter and solve problems, your rating could fluctuate.  Similarly, I would imagine if you do have all the problems solved correctly, but you've only seen a small number of them, the calculated variance for your "real" rating will be higher and that may be reflected in the actual rating.

Let's hope this does not give people any ideas on gaming the rating system by rating the questions they get right as really difficult and the questions they get wrong as really easy.  (Though such a problem could be easily solved by taking the user-ratings out of the equation for computing "difficulty".)

Given the cleverness of the AoPS people, I imagine they also have some sort of hidden "carelessness" statistic that is based on answering "difficult" questions correctly and "easy" questions incorrectly.

But one thing that doesn't seem right is that even if somebody finishes all of the problems, then they're Alcumus Score will not be the same as their percentage correct, and might not even be close (even though the purpose of the Alcumus Score is said to be an approximation of the expected percentage of problems that one will get correct!).

Yeah.

If you answer all of the problems, and get some wrong, and get some of the ones that you got wrong again, and get them right that time, it'll count as you never got it wrong.

That really made sense.


calculus
integration
calculus computations

Hey all,

I could use some help evaluating this integral. I would appreciate if you evaluated it as an indefinite intelgral first, just so I can see how that process works.

$ \int_0^1 x^3\sqrt {1 \plus{} 4x^2}dx$

EDIT:

solved:

several integration by parts later:

$ \int x^3\sqrt {1 \plus{} 4x^2}dx \equal{} \frac{1}{12}x^2(1\plus{}4x^2)^{3/2}\minus{}\frac{1}{120}(1\plus{}4x^2)^{5/2}$

Integration by parts is unnecessary.  With the substitution

$ u \equal{} 1\plus{}4x^2$,

$ du \equal{} 8x \, dx$,

we have

$ \int x^3 \sqrt{1\plus{}4x^2} \, dx \equal{} \frac{1}{32} \int (4x^2)(8x) \sqrt{1\plus{}4x^2} \, dx \equal{} \frac{1}{32} \int (u\minus{}1)\sqrt{u} \, du$.

Distributing, we obtain

$ \frac{1}{32} \int u^{3/2} \minus{} u^{1/2} \, du \equal{} \frac{1}{32} \left(\frac{2}{5} u^{5/2} \minus{} \frac{2}{3} u^{3/2}\right) \plus{} C\equal{} \frac{(1\plus{}4x^2)^{3/2}(6x^2\minus{}1)}{120} \plus{} C$.

Cool, that way is certainly easier. Thanks.

hello, we get
$ \int_{0}^{1}x^3\sqrt{1\plus{}4x^2}\,dx\equal{}\frac{5}{24}\sqrt{5}\plus{}\frac{1}{120}$
Sonnhard.


Suppose there are $ n$ seats on an airplane and $ n$ passengers.  Each passenger has a ticket that lists a specific seat he is supposed to sit in.  The first passenger boards the plane and selects a seat randomly.  Every subsequent passenger boards the plane and sits in his own seat if its available, but randomly sits in any available seat if his seat is taken.  What is the probability that the last passenger sits in his own seat?

Source: Question posed during SIMUW 2006.

[hide]Consider the seat of the last person to board the plane. It will be either empty or taken. Since everything that happens before this moment can produce both results without one tending to be more likely than the other, the probability is $ \tfrac{1}{2}$.[/hide]

[hide="hint?"]
Assume the last passenger does NOT sit in his own seat.  That means one of the other seats is empty.
That seat has to belong to a passenger other than the last one.  So this passenger must have not sat in his own seat.  He had to have been the 1st passenger.

[/hide]

No, the probablilities probably are not the same.  You have fallen into the principle of indifference trap mwahahaha

[hide="very wrong looking"]
Combinatorics
How can we make the first n-1 passengers sit in n-1 seats?
$ \frac{(n\minus{}1)!}{n(n\minus{}1)!}\equal{}\frac{1}{n}$?

[/hide]

[quote="CircleSquared"]Assume the last passenger does NOT sit in his own seat. That means one of the other seats is empty. That seat has to belong to a passenger other than the last one. So this passenger must have not sat in his own seat. [b]He had to have been the 1st passenger.[/b][/quote]
The first person could have sat in somebody's seat, and that somebody then randomly sat in the last person's seat. The solution I provided is not complete, but it is definitely the beginning of a solution that I heard.

Oopy, forgot about that.  Oddly enough, I can now change the statement to
[hide]
Either the last person is sitting in his own seat or he is sitting in the 1st person's seat.
[/hide]

[hide]The probability is [i]1/2[/i] for cases when [i]n ≥ 2[/i], [i]1[/i] for [i]n = 1[/i].  You can prove the former by using induction.  Without loss of generality, assume that passengers board the plane in order of their seat number (ie. the first person is supposed to sit in seat [i]1[/i] etc).

For the case when [i]n = 2[/i], simply observe that the first person to enter the plane has an equal likelihood of sitting in their own or the second person's seat.

For our induction hypothesis, assume that for some [i]n[/i], the probability is [i]1/2[/i] for [i]m = 2, ..., n[/i].  Then for [i]n + 1[/i], we have that there is a [i]1/(n +1)[/i] chance of the first person sitting in a given seat.  Let [i]k[/i] be the seat in which the first person sits; if [i]k = 1[/i], then all passengers can sit in their own seats; if [i]k = n + 1[/i], the [i]n + 1[/i]-th passenger can't sit in their own seat.  For [i]2 ≤ k ≤ n[/i], we know that the passengers after the first but before the [i]k[/i]-th (should there be any) can sit in their own seats; we simply have to consider passengers [i]k[/i], [i]k +1[/i], ..., [i]n[/i] and [i]n + 1[/i].  If we now consider seat [i]1[/i] to be the correct seat for the [i]k[/i]-th passenger (since it isn't desired by later passengers), this reduces to the previous problem for [i](n + 1 - k)[/i] passengers and seats, and so our induction hypothesis tells us this has probability [i]1/2[/i].

Putting this altogether, we find that the probability for the case for [i]n + 1[/i] is given by

[i]1/(n +1) + 0/(n +1) + ∑[size=59]i[/size](1/2)/(n + 1)[/i], (*)

where the final sum is taken over [i]i = 2[/i] to [i]i = n[/i], and so expression given by (*) is equal to

[i]1/(n +1)  + (n-1) / 2(n + 1) = 2 / 2(n + 1) + (n-1) / 2(n + 1) = [2 + n - 1]/2(n + 1)

= (n + 1) / 2(n + 1) = 1/2.[/i]

Hence, by induction, the result holds for all [i]n ≥ 2[/i].[/hide]


Five men are shipwrecked on an island with a lone monkey as the native inhabitant. The first day the five men spend all thier time gathering coconuts. During the night one man woke up and decided to take his share of the coconuts. He divided them into five piles and there as one left over which he gave to the monkey. He hid his share and went to sleep. Soon a second man woke and did the smae thing. After dividing the coconuts into five piles, one coconut was left over and he gave it to the monkey. He hid his pile and went to sleep. The third, fourth, and fifth man followed exactly the same procedure. The next morning, after they all woke up, they diided the remaining coconuts evenly. This time, there were no coconuts left over. What is the least amount of coconts they could have collected on the first day?

This took me forever. Good luck.

First of all, notice what the units digit must be. It must be [hide]either 1 or 6, because if the monkey takes 1, the rest is divisible by 1[/hide]
Then, notice how each person takes 1/5 of some remainder. Hint: [hide]you will want to manipulate powers of 5...[/hide]
Set the number of coconuts that the people start with as "x," and work down from that. You'll want a calculator.

I think that the answer is either [hide]15621 or 15626. I'm leaning a lot towards the former.[/hide]

I disagree. I think its
[hide]3121[/hide]
I used basically the same process you did but I think there was a least common denominator somewhere in there.


I mean some of those which are held by the University of Waterloo

the problems and solutions are posted on the website

Can we?????????????????

Sure, but I'd discuss them in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=320]Canada forum[/url]. You can discuss them here, if you want, though.


Three point charges $ Q_1$, $ Q_2$, and $ Q_3$, lie in three vertices of a square with side lenght 1 m.

(a) Are there points or surfaces in the space in which the potential is zero (excluding infinity)? Which?

(b) There are two points in which the electric field is zero. Estimate, with explanation, their location. If we don't assume that the potential is zero at infinity, would those points be the same?

(c) Make a rough drawing of the field lines.

For part (a) it is enough to give a rough description of the possible points/surfaces where the potential might be zero.

Part [b]c[/b] is easy. Field lines originating from positive charges terminate in negative charge. And the $ \plus{}5mC$ dominates over the $ \plus{}2mC$ charge.

Even so you can make a rough drawing. What about the other questions?

b) Yes they would be the same. E field doesnt depend on the actual value of potential at a point.

(a)The pt must be close to $ Q_2$  most  probably towards $ P_3$

(b)conjurer's answer is right(E depends on rate of change of potential...)

and 1 pt where E looks zero  lies on $ P_1 P_2$
other pt.. :maybe:

pt where E is zero shud rather be in tringle $ P_1 P_2 Q_3$

Those answers still look very vague...  :(


[quote="toan hoc"]Suppose that $ a;b;c$ are three lengths of sides of triangle.Prove that:
$ \frac{a}{\sqrt{a^{2}\plus{}3bc}}\plus{}\frac{b}{\sqrt{b^{2}\plus{}3ac}}\plus{}\frac{c}{\sqrt{c^{2}\plus{}3ab}}\geq\frac{3}{2}$[/quote]
Let $ a\plus{}b\plus{}c \equal{} 1.$
Hence, by Jensen we obtain: $ \sum_{cyc}\frac{a}{\sqrt{a^{2}\plus{}3bc}}\geq\frac{1}{\sqrt{a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}9abc}}.$
Id est, it remains to prove that $ 4(a\plus{}b\plus{}c)^{3}\geq9(a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}9abc).$
Let $ a \equal{} y\plus{}z,$ $ b \equal{} x\plus{}z$ and $ c \equal{} x\plus{}y.$ Then
$ 4(a\plus{}b\plus{}c)^{3}\geq9(a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}9abc)\Leftrightarrow$
$ \Leftrightarrow32(x\plus{}y\plus{}z)^{3}\geq9\sum_{cyc}\left((x\plus{}y)^{3}\plus{}3(x\plus{}y)(x\plus{}z)(y\plus{}z)\right)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(7x^{3}\minus{}6x^{2}\minus{}6x^{2}z\plus{}5xyz)\geq0\Leftrightarrow$
$ \Leftrightarrow6\sum_{cyc}(x^{3}\minus{}x^{2}y\minus{}x^{2}z\plus{}xyz)\plus{}\sum_{cyc}(x^{3}\minus{}xyz)\geq0,$
which true by Schur and Muirhead.

For those who are interested, this is the same problem that nmtvn posted and which (at the time) was from an active contest of some sort; see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=914160#914160


Sa se demonstreze ca intr-un corp finit,orice element se scrie ca suma de doua patrate. ;)

Daca are un numar par de elemente, $2^k$, toate elementele sunt patrate, pentru ca satisfac $x^{2^k}-x=0$.

Pe de alta parte, daca are un numar impar de elemente, $p^k$, cu $p$ prim impar, atunci $\frac{p^k+1}2$ elemente sunt patrate: solutiile ecuatiei $x^{\frac{p^k+1}2}-x=0$. Fie $Q$ multimea patratelor. Daca luam $a$ arbitrar in corp, atunci multimile $Q$ si $a-Q$ au suma cardinalelor mai mult de $p^k$, deci se intersecteaza, si de aici e evident.

Problema iese si din urmatoarea lema:
Daca $G$ este un grup finit, iar $A,B\subseteq G$ verifica $|A|+|B|>|G|$, atunci $G=AB$.

Cum o folosim? Fie $K$ corpul tau finit, $(K,+)$ grupul sau aditiv, iar $A=B=K^2$. Verifici ca are loc inegalitatea(de exemplu pentru ca $x^2=0\Longrightarrow x=0$) si tragi concluzia.

PS Spatii dupa semnele de punctuatie, Cezar!


The ratio of boys to girls in Jamal's class is 3:2. If four more girls join the class, there will be the same number of boys and girls. What is the number of boys in the class?

Hi;

[hide] There are 12 boys and 8 girls[/hide]

[hide]

Call the number of boys $ 3x$ and the number of girls $ 2x$.

Thus $ 3x\equal{}2x\plus{}4$ and $ x\equal{}4$.

Thus there are $ \boxed{12\ boys}$ and $ \boxed{8\ girls}$[/hide]


The sum of the squares of two positive integers is 73. What is
the sum of the two positive integers?

We have:

$ x^2 \plus{} y^2 \equal{} 73$

and we wish to find $ x \plus{} y$.

One way is to use brute force. The squares under 73 are:

$ 1, 4, 9, 16, 25, 36, 49, 64$

We note that $ 64 \plus{} 9 \equal{} 73$, so our answer is $ 8 \plus{} 3 \equal{} \boxed{11}$.


inequalities trigonometry function integration algebra polynomial calculus

So in my textbook, it uses Bessel's inequality and Fourier coefficients to state
$ \frac{\pi^{2}}{6}\geq\sum_{n\equal{}1}^{k}\frac{1}{n^{2}}$
which is of course $ \zeta(2)$.  I was wondering if there was a way to expand Bessel's inequality for fourth powers so that we could come up with an expression of the form
$ \frac{\pi^{4}}{90}\geq\sum_{n\equal{}1}^{k}\frac{1}{n^{4}}$
which is $ \zeta(4)$?  Any ideas?

Yes, of course, but you'll need a different function.

For instance: Let $ f(x)\equal{}\frac{\pi\minus{}x}{2}$ for $ x\in(0,2\pi).$ We can write

$ f(x)\equal{}\sum_{k\in(\mathbb{Z}\setminus\{0\})}\frac{e^{ikx}}{2ik}\equal{} \sum_{k\equal{}1}^{\infty}\frac{\sin kx}{k}$

Bessel's inequality and/or Parseval's identity applied to this function will tell you what you need to know about $ \zeta(2).$

Now find a function $ g$ satisfying the following conditions:

$ g'(x)\equal{}f(x),$ $ g$ is periodic of period $ 2\pi,$ and $ \int_0^{2\pi}g(x)\,dx\equal{}0,$

[This will take a little computation, but it's all polynomial calculus.]

It should be directly evident what the Fourier coefficients for $ g$ would be - just integrate $ f$ term by term. Try applying Bessel and/or Parseval to this function. (You'll need to find the $ L^2$ norm of $ g,$ which is a little more polynomial calculus.) The result will tell you what you need to know about $ \zeta(4).$

Integrate $ g$ (and adjust the constant of integration to get mean value zero) an you'll get $ \zeta(6).$ This process can be continued for as far as you have patience for the computation.

I don't have any experience with analysis yet and I don't know what an $ L^{2}$ norm is so could you perhaps get me started for $ \zeta(4)$?  Then I could try to use that to get to $ \zeta(6)$ and so on.  Thanks.

All I meant by "$ L^2$ norm" was the norm in the inner product space. That is,

$ \|g\|^2\equal{}\langle g,g\rangle\equal{}\frac1{2\pi}\int_0^{2\pi}|g(x)|^2\,dx.$

In this case, that involves computing a polynomial integral.


ve van de nay tot nhat la nen vao mathnfriend.net

làm thế này cũng tốt chứ mắc gì đâu,THTT nên được phổ biến ở càng nhiều diễn đàn càng tốt chứ bạn

Thiết nghĩ không nên so sánh diễn dàn này với diễn đàn kia làm gì dieuhuynh à,diễn dàn nào cũng có mặt mạnh và yếu.Mạnh của ML là thành viên đông và có người Việt mà không ở Việt Nam,trong tương lai thì chưa biết,nhưng hiện giờ ML phổ biến hơn MNF đấy,dù gì cũng rất hoan nghênh bạn đã góp ý,tôi cũng là một thành viên của MNF mà  :) để bổ sung cho khiếm khuyết trong bài post của dieuhuynh tôi đưa link của MNF ra cho các bạn tham khảo link của MNF [url=http://mathnfriend.net/]MNF[/url] còn đây là
ĐỀ RA KÌ NÀY SỐ 346
[img]http://www.nxbgd.com.vn/toanhoctuoitre/tapchi/Images/20064189238_D1.JPG[/img]

Here is the English version
[img]http://www.nxbgd.com.vn/toanhoctuoitre/tapchi/Images/200641894048_D2.JPG[/img]
[img]http://www.nxbgd.com.vn/toanhoctuoitre/tapchi/Images/2006419162444_D3.JPG[/img]

đưa link luôn cho đỡ mất công load hình
http://www.nxbgd.com.vn/toanhoctuoitre/?p=7&id=6&ReportID=116&ph=11

em thấy tại sao mà trên nxbgd chỉ có mấy số thôi,ra số mới lại mất đi 1 số!!! :maybe: ,chán thật!!!em mún đọc mấy số cũ ơi là cũ thì fải làm sao đây anh???

web THTT chỉ mới xuất hiện vài tháng,nên chỉ bao gồm THTT những số gần đây thôi.
 Nếu cần những số đời xưa ơi là xưa thì phải hỏi mấy thầy thôi em à,em có thể hỏi bác Ngô Trần Ái thử xem

[quote="heongo"]Dạ??em đâu có biết nick của bác Ngô Trần Ái đâu ạ!!!tại em mới vào mà!!!! :blush:[/quote]
khong,"xin xo" truc tiep chu,nghiem tuc ma,khong nen thong qua nick (ma linkgreencold cung dau co biet nick cua bac Ngô Trần Ái dau)


The sum 498+499+500+501=1998 is one way of expressing 1998 as a sum of consecutive positive integers. Find all ways of expressing 1998 as a sum of two or more consecutive integers. Prove your list is complete.

[hide="Hint"] The sum of $2k+1$ consecutive numbers with middle term $a$ is $(2k+1)a$.  And the sum of $2k$ consecutive numbers with middle terms $a, a+1$ is $k(2a+1)$. [/hide]


“Isn’t that a rather large number?” I wondered when I looked at the number.
“May be, but you said your pup is trained, and I want  to see him  multiply   this number  by 29”
 I guess he didn’t really believe my [url=http://www.artofproblemsolving.com/Forum/images/avatars/45961890041e4a6671adbd.jpg]dog[/url]can do math.
The pup, lazy as ever, meanwhile did what he always does, (he complains that it is difficult to write/type with paws):  he puts a digit in front of the number and the same digit at the end.
(In other words, if the original number, in decimal notation, looked like:
$(a_1a_2a_3\dots a_n)$  and asked to be multipleid by 29,
 now looked like $(b a_1a_2a_3\dots a_nb)$

 (where  $b$ is a decimal digit. - needless to say so are $a_1,a_2$ etc)

As usual, the pup was  right. And we came to know that the number was smallest such number  which had that property.

What was that number?
For other adventures of the pup see:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81602

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=469324#p469324

As usual please hide your answers.

[hide] Wasn't this on the AIME? :? I think the answer is $\boxed{25}$, which becomes $725$ when multiplied by 29. :D[/hide]

It has to have the digit on both ends. It would work if 25*29 = 1251 or something like that. Not only in front of the number.

[quote="lotrgreengrapes7926"]Can you use a calculator?? :?
[hide] Call the number being multiplied by 29: $x$; $b$ is the number added to the front and the back.
If $x$ is a 1-digit number, we can set the equation $29x=101b+10x \Longrightarrow 19x=101b$. This means b must be 19 to satisfy the equation while b and x are both integers; although, in our problem, b cannot be 19.
If $x$ is a 2-digit number, we can set the equation $29x=1001b+10x \Longrightarrow 19x=1001b$. We can now see the pattern for the equations $19x=10...01b$. To find $19x$, we need to find the smallest number for $10...01$ such that $10...01$ is divisible by $19$. Using a calculator, we can see that the first such number is $1000000001$. $\frac{1000000001}{19}=52631579$. So the answer is $\boxed{52631579}$.

OK, even I'm confused after explaining that.  :roll: [/hide][/quote]

Well, I  expected it to be done without the use of a calculator, but  as you do know the answer,  you can work backwards and see if you could  get that without a calculator..
[hide="hint"]Fermat's  little theorem [/hide]
Good job in anycase.

[hide]Let the number be $x$ and length $n$ . Then

$29x=10x+(10^{n+1}+1)b$ $\Longleftrightarrow x=\frac{b(10^{n+1}+1)}{19}$

But we check that $10^9+1\equiv 0\mod 19$ , so the smallest such $x$ such be when $n=8$ and $b=1$

$\therefore x=\frac{10^9+1}{19}=52631579$   :)[/hide]


geometry
circumcircle
power of a point
radical axis

Three circles of equal radii intersect at $ A$, $ A_1$, $ B$, $ B_1$, $ C$, $ C_2$, as shown. Prove that $ AA_1$, $ BB_1$, $ CC_1$ are concurrent. [asy]defaultpen(fontsize(8));
pair A, A1, B, B1, C, C1;path CA=circle((0,0),10), CB=circle((13,0),10), CC=circle((2,12),10);
draw(CA);draw(CB);draw(CC);
C=intersectionpoints(CA,CB)[0];C1=intersectionpoints(CA,CB)[1];
A=intersectionpoints(CC,CB)[0];A1=intersectionpoints(CC,CB)[1];
B=intersectionpoints(CA,CC)[0];B1=intersectionpoints(CA,CC)[1];
label("$C$",C,(0,1));label("$C_1$",C1,(0,-1));
label("$A$",A,(-1,-1));label("$A_1$",A1,(1,1));
label("$B$",B,(1,0));label("$B_1$",B1,(-1,0));[/asy]

one liner:

[hide]
It is the circumcenter of the triangle formed by the three centers of the circles[/hide]

And this is a specific case of:
[hide]
the fact that the radical lines formed between three circles are concurrent[/hide]

They are concurrent at the 'radical center' of those three triangles.

The locus of points with equal power (power is the distance from the center squared minus the radius of the circle squared) to two circles is a line. If a point has equal power to C_1,C_2, and C_2,C_3, then it has equal power with C_3,C_1, so it lies on that radical axis too


Sarah and Betty both run the 400 meter dash in track. Sarah's average time is 71 seconds with a standard deviation of 6 seconds. Betty's times have a mean of 70 seconds with a standard deviation of 8 seconds. The distribution of times for both runners are approximately normal.

ok so basically i got
$S(71,6)$
$B(70,8)$

the question is

"If they were to race, what is the probability that Betty wins?"

$P(B<S)$ seems to be the question.

i did find the probability for each person that they will run under 65 seconds but i don't know how to find the probability that Betty wins.

is this too hard?
if someone has any ideas please do not hesistate to post

I don't know a thing about standard deviation, etc.

That said, I would reccomend drawing a rectangle with one side using length to represent the time of one girl, the other side for the other girl. So for betty it would go from 62 to 78 seconds and be 16 units long. For Sarah it would be 65 to 77 and be 12 units long. then you draw a line so that on one side of it, Betty beats sarah, on the line is a tie and on the other side Sarah beats Betty. find the area of the appropriate region out of the entire region. (the line has no area so technically there is no probability of a tie...)

Is the answer 21/64? If it is i'll post the solution (though I may have made an arithmetic mistake).

Let $S(71,6)-B(70,8) = A(1, \sqrt{6^{2}+8^{2}}) = A(1,10)$, which is another normal distribution. Then, when $A$ is positive, Betty wins.


probability
Mafia
geometry
geometric transformation
reflection

wow in one day 7 players Thats pretty good :D
So you can choose your door now if you want to and teams, I will put it like

1.Team:king! and cygao,1,100
2.etc. but remember, since you share points, someone can lose 135 points for the team(which isn't good)
 Addition*: In the later levels I might let some teams of 3 or 4

Remember:

|Door 1|   |Door 2|
what's behind them?
an epic scene?
It all depends on what you pick :lol:

Players:(name)(health kits)(points left)
1.bluecarneal,1,100
2.King!,1,100
3.tennis123,1,100
4.spaceguy524,1,100
5.ernie,1,100
6.jjfun1,1,100
7.cygao15,1,100
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
14 spots to go ><
Game will start on sunday if the 20 spots aren't filled

[quote="kianex"]
Example of a team
1.Team:king! and cygao,1,100
2.etc. but remember, since you share points, someone can lose 135 points for the team(which isn't good)
 Addition*: In the later levels I might let some teams of 3 or 4
[hide=" Rules"]This game will be about doors. You will start out with 100 points(like hp), the event happens and if you get like a dragon you might lose like 25 points, if you face a thief, you might be lucky and escape, but no healing. If you end up with a bad event behind the door(the harsher the event the more points you lose) hope you have an alliance. After you choose your door, if you get it right, there will be a fountain to heal 5 points,(cannot go over) Every two levels the # of doors will increase and you will have to choose one. There will be hints to help you cancel out which door is wrong. You have one life and one health kit. If you get it wrong, I will post a story on what's behind the door. The health kit revives one person with 25 points(including yourself). If you win(like the whole game), you can choose the next # of doors to start with again. After every 2 levels to make it easier on you, you will get one health kit back. Ok so there are 20 slots pm me when all the slots are filled, I might consider adding more.
*NOTE: you can make teams of TWO only,but if you do, you may share points, but you can have only one health kit
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
And you can either pm me(not recommended) or post here
Right now there are two doors after every round I will say who's alive and how many health kits you have left

|Door one| |Door two| What's Behind them? Life or Death For the first two rounds you wont die, but you will get a strike. Three strikes for the first two rounds and your out until the real one starts. I might ask a person to help if I cant help make clues[/hide]
Remember:
[hide="Doors"]
|Door 1|   |Door 2|
what's behind them?
an epic scene?
It all depends on what you pick :lol: [/hide]

[hide="Players"]Players:(name)(health kits)(points left)
1.bluecarneal,1,100
2.King!,1,100
3.tennis123,1,100
4.spaceguy524,1,100
5.ernie,1,100
6.jjfun1,1,100
7.cygao15,1,100
8.EggyLv.999,1,100
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.[/hide]
Game will start on sunday if the 20 spots aren't filled[/quote]

hm....

this seems kinda confusing but ima try it.

join

I believe this game has been mod-abandoned.. Let's just let it die.

@ jjfun:no
@ spceguy: I went to say the same as jampa(I was in a hurry)

heh sorry I can't always go online, so I forget about this eventually but yeah I'll start trying to remember again.
Probably I will post after the class.

How about from now on, every week a new round will start so you guys don't expect too much out of me.

[hide=" DOOR 1"] Spaceguy524 and zheng venture into Door 1 which is known to be batty. *step, step, step, flutter* they hear as they venture on. They quickly zip around to find that nothing is there. So they keep going and they hear *step, step, step, SCREECH*. They instantly spin around, but they can't stop, they keep on turning, and they realize they are flying. While they are still spinning they crash into each other.(-2 points) They keep on going, eyes not working, but they can still find their way around, they suddenly are embathed in a bright white light. Zheng screeches and spaceguy screamsand they stare at each other. They both scream," WHY DO YOU LOOK LIKE A BAT", except in a weird scratchy voice. The people who saw bats must've seen the victims of this curse.
"How do we go back to being humans?" zheng questions.
"How would I know, I'm in the same situtation as you are," spaceguy replies.
Instantly, they are blinded again(-2 more points), by white light(except spaceguy's is black),but this time for real, they hear a slight thump as they both fall to the ground unconscious. The next thing they know, they're back in the cave, except still bats. they both fly out and when they get out of within 50 feet of the cave's entrance, they feel heavier and they are falling. They fall out of the sky both half-bat half-human, and they hit a soft ground with a slight thump.
"Wow where are we?" they both ask.
And they instantly recognize it, they both are in the lobby of the cave center, lying on the heavily carpeted floor.[/hide]

Points:
zheng,2,83(-2 for bumping into spaceguy and for being blinded)
Spaceguy524,2,96(same thing, except hid blinding was a black light)

have to go will do tomorrow, but heed what I said, the next round's stories(scenarios) will not be made till next week.

[quote="kianex"]
Instantly, they are blinded again(-2 more points), by white light(except spaceguy's is black),
[/quote]

How are you blinded by black light?

Should we pick doors now?
If so, [b]Door 2.[/b]

INDEED, PLZ LOCK IT.
I HATE WHEN PEOPLE REVIVE THREADS BTW, IT WASTES MY TIME.
WELL I GUESS BEING IN THE GAME SECTION IS WASTING TIME :D  :D

[size=84][color=red]Seriously, please stop with the all caps. We're definitely aware of what's happening. [/color][/size]


modular arithmetic
number theory
least common multiple
superior algebra
superior algebra unsolved

Assume we have a nonempty set M with an associative binary operation satisfying the following two laws for all elements $x,y$, with fixed integer exponents m, n >= 1:
1m. $x x^{m}= x$
2n. $x y^{n}x = y x^{n}y$
For some combinations $(m,n)$ we can conclude that the operation is commutative, i.e. $x y = y x$ for all $x,y \in M$.
The question is: Which are these combinations?

LiMa

The cases $m = n = k \ge 2$ have been answered affirmative in this thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=117632

For $n > m$, applying law 1m to law 2n results in law 2n' with $1 \leq n' \leq m$ and $n' \equiv n \pmod m$. So it is enough to check the cases where $n \leq m$.

For $n=1$, one can show $x^{2}= x$ and so one is reduced to the case $m=1$, which can be proven.

For $3 \le m$ not a power of 2, $n = 2$ the conclusion does not hold. The nonabelian groups of order $p^{3}$ with exponent $p$, for $p$ an odd prime, fulfill the law $x y^{2}x = y x^{2}y$. If $p$ is an odd prime divisor of $m$, then $x^{p}= e$ implies $x x^{m}= x$, and so we have a counterexample.
For $4|m$, $n=2$ one can take the dieder group with 8 elements, since $x^{2}$ is in the center for all $x$.

So all the cases $m \ge 3$, $n=2$ are negative.

LiMa

Until now, we have:
YES:
(m,n) = (k,k), k >= 2
(m,n) = (k,1), k >= 1
NO:
(m,n) = (k,2), k >= 3

For any $m \ge 1$, $n \ge 3$, from laws (1,m) and (2,n) follows $x = x x^{n-2}$, i.e. law (1,n-2). From that we get (2,2). With $d : = \mathrm{lcm}(m, n-2)$ we get $x = x x^{d}$, i.e. (1,d).

From (1,d) and (2,2), with d as above, we can get (1,n-2), (2,n) and (1,m).

Therefore, for any $n \ge 3$, the two laws (1,m) and (2,n) are equivalent to the two laws (1,d) and (2,2).

So, my conclusion is:
If n=1, or if n=2 and m in {1, 2}, then commutativity is enforced.
If n=2 and m > 2, then there are noncommutative examples.
If n>2, we calculate $d : = \mathrm{lcm}(m, n-2)$; commutativity is enforced iff d in {1, 2}.

LiMa


Consider a triangle $ABC$ in the plane.Let $P,Q$ and $R$ be the midpoints of the sides $BC,CA$ and $AB$ respectively.Let $P_{1}$ be the midpoint of $CP$.Let ${AP_{1}}$ and $CR$ meet at $D$ and $PQ$ and $R$ meet at $E$. then compute the ratio of the area of the triangle $ABC$ to the area of the quadrilateral ${EDPP_{1}}$

Here is my solution. It's not a hard problem, unless I missed something and made a mistake.

I denote area by $[...]$

It's clear that $QE=EP$. Thus, we can say that $[P_{1}PE]=\frac{1}{16}[CBA]$.
Call $[CP_{1}D]=\alpha$ and $[P_{1}DE]=\beta$.

Now, $\alpha+\beta = \frac{[ABC]}{16}$.
Also, $\alpha+[CDA] = \alpha+16 \beta = \frac{[ABC]}{4}$.

This gives us $\beta = \frac{\frac{1}{5}}{16}[ABC]$, so
\[[P_{1}DEP]=[P_{1}DE]+[P_{1}EP] =\frac{3[ABC]}{40}\]

[quote="Jan"]Thus, we can say that $[P_{1}PE]=[CBA]$.[/quote]

 :huh:   :huh:

Of course I mean $\frac{1}{16}$ I'll edit it, sorry.

I didn't understand how you got ${CDA}$ = $16$$\beta$

That's because $EP_{1}\parallel AC$, and because $\left|EP_{1}\right| = \frac{1}{4}\left|AC\right|$.

This uses of course the fact we already proved, i.e. $\Delta P_{1}PE$ and $\Delta CBA$ are similar, with factor $4$.

$\triangle P_{1}DE\sim\triangle ADC$, and since $P_{1}E={1\over 2}CQ={1\over 4}CA$, the conclusion follows.

EDIT: Jan got here first.

Apparently you have a different proof from the one I posted ashwinkrishnan. I'm interested in it, could you please show it?

Let us consider $A$ as the origin and assign position vectors to all points with reference to this origin, then the position vectors of different points become as follows:-
$C$ = $c$ , $B$=$b$
$Q$ = $\frac{c}{2}$ $R$ = $\frac{b}{2}$
$P$ = $\frac{b+c}{2}$
$P_{1}$ = $\frac{3c+b}{4}$
we now find out the position vectorss of $D$ and $E$ in terms of $b$ and$c$.let us write the equations of the lines $CR$,$PE$ and $AP$
$CR$ $r = c+x({\frac{b}{2}}-c)$
$PE$ = $PQ$  $r ={\frac{c-yb}{2}}$
$AP$ - $r = z{\frac{3c+b}{4}}$
                               [b] for some arbitrary constants $x,y,z$[/b]Solving the first two equations gives the position vector of the point $E$ and the first and the last give the position vector of $D$.
we get the values as
$E$ = ${\frac{2c+b}{4}}$
and $D$ =${\frac{b+3c}{5}}$
Hence are of the quadrilateral = 1/2$DP*{EP_{1}}$ and on substiuting the values of $DE$ and $EP_{1}$ we get the area of the quadrilateral as $3/20$ $b*c$ .But area of the triangle $ABC$ is $1/2$ $b*c$
[u][b]Hence the required ratio is [/b][/u]$3/{40}$
[b][u]NB[/u][/b] here $b*c$ denotes the cross product of the vectors $b$ and $c$


Find all real numbers a such that

[Sqrt(n) + Sqrt(n+a)] = [Sqrt(4n+1)]

holds for all positive integers n.

First step, I think:

[ :sqrt:(4n + 1)] = [2 :sqrt:n] unless  n = m2 + m for positive integral m, in which case [ :sqrt:(4n + 1)] = 2m + 1.  Then we need to proceed with two cases.

The second case there (n = m2 + m) gives us some bound like 3 > a > k, where k is some number just a little bigger than 1/2 -- in fact, it approaches 1/2 from above, so we get a minimum value there when m = 1 (n = 2).  (I haven't proven that yet.)  Actually, the 3 comes from plugging in n = 1.  The k comes from plugging in values of m.

For the other case, let :sqrt:n = Z + f, Z integral, f non-negative fractional.  Consider the case where f < 1/2.  Then [2 :sqrt:n] = 2Z.  Thus, plugging into the given expression, we get bounds on a, in particular that a < 2I - 2f - 4If + 1, which only gets small if f gets to 1/2 fast and I gets to infinity slow.  The only lower bound that this case sets on a is going to be zero, which we could have figured out anyway by setting n = 1 in the original equation.  Hmm ... I'm not so impressed by this.

I believe the solution set is the interval [hide]9 - 6:rt2: :le: a :le: 2[/hide].  For now, I'll prove necessity, that is, every solution lies in that interval.  Either someone else can prove sufficiency, or another day I'll try to prove it.

Proof of Necessity:  Suppose that a satisfies the condition of the problem.  To prove the lower bound on a, [hide]set n = 2.  The original equation becomes [:rt2: + :sqrt:(2 + a)] = 3.  That implies :rt2: + :sqrt:(2 + a) :ge: 3.  Solving for a gives a :ge: 9 - 6:rt2:.[/hide]

To prove the upper bound on a, [hide]let k be an integer at least 2.  Set n = k2 - 1.  The original equation becomes [:sqrt:(k2 - 1) + :sqrt:(k2 + a - 1)] = 2k - 1.  That implies :sqrt:(k2 - 1) + :sqrt:(k2 + a - 1) < 2k.  Now let k approach infinity.  By studying the first two terms of the binomial expansion of each square root on the left, we must have a - 1 :le: 1, or a :le: 2.  Thus we have proved the upper bound.[/hide]


If a,b and c are the roots of the equation x^3-x^2-x-1=0

show that

(a^1982-b^1982)/(a-b)+(b^1982-c^1982)/(b-c)+(c^1982-a^1982)/(c-a)

is an integer

[hide="Hint"]
Write all of the expressions as one fraction. Use vieta equations after expansion to simplify the result. If you want to put in the effort, you can even find the integer using Newton's sums because this is a symmetric sum.[/hide]
Let me know if this helps.

[hide="solution"]
First, we have to be aware of the factorization

$a^{n}-b^{n}= (a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$.

Thus, each group of terms can be simplified. After simplifying, we get:

$\sum_{i=1981,j=0}^{1981}a^{i}b^{j}+\sum_{i=1981,j=0}^{1981}b^{i}c^{j}+\sum_{i=1981,j=0}^{1981}c^{i}a^{j}$

Now, we note that these terms can be grouped as follows:

$\sum_{i=1,j=1980}^{1980}(a^{i}+b^{i}+c^{i})(a^{j}+b^{j}+c^{j})+2(a^{1981}+b^{1981}+c^{1981})-1979(a^{1981}+b^{1981}+c^{1981})$

So all that is left is to prove that $a^{n}+b^{n}+c^{n}$ where $a,b,c$ are the roots of $x^{3}-x^{2}-x-1 =0$ is an integer for every $n$.
We observe that $a+b+c = 1$ and $a^{2}+b^{2}+c^{2}= 3$.  Since $a,b,c$ are the roots of the equation, we have: $(a^{3}+b^{3}+c^{3})-(a^{2}+b^{2}+c^{2})-(a+b+c)-1=0$ so $a^{3}+b^{3}+c^{3}$ is obviously an integer. Now, we multiply the original equation by $x$ to get $x^{4}-x^{3}-x^{2}-x=0$ Using the same strategy, we see that $a^{4}+b^{4}+c^{4}$ is an integer. Repeating this process, we immediately see that $a^{n}+b^{n}+c^{n}$ is an integer for all positive integers $n$ so we are done.
[/hide]

[quote="Phelpedo"]How do you use Newton's sums for powers higher than the degree of the equation?[/quote]

[hide="Like this"] $x^{3}-x^{2}-x-1 = 0 \implies x^{3+k}-x^{2+k}-x^{1+k}-x^{k}= 0$

Summing symmetrically with $a, b, c$ we have

$s_{3+k}-s_{2+k}-s_{1+k}-s_{k}= 0$

The generalization should be clear. [/hide]


trigonometry
geometry
geometric transformation
reflection

Here's a better diagram.  (It may look a bit weird in the preview, but just zoom in and it'll be fine.)

why did you draw it again i don't understand it doesn't need to be taken pains i think it's understandable.:)

if there is nobody to solve this i will send solution.It is not  hard question.

[quote="Altheman"]extend $BC$ to intersect $AE$ at $F$, it is evident that $AE=EF$, $BF=AF$, and $CE=AD=AF$. Now I don't see how to finish exactly...i want to show that $BC=CD$...but i am stuck now...[without trig]...i have to sleep...maybe more later...[/quote]

How do you know that $CE = AD$?

[hide]
Erm, I got $\alpha=140$

Extend $BC$ and $AE$ and call the intersection point $F$. We realize that $AFB = 80$. Now, draw a perpendicular line from vertex $A$ down to $BF$, and call this point $I$. If we reflect $F$ about $AI$, we get that $F'AD = 60$. However, since $F'A = FA$, $\triangle F'AD$ is equilateral.

Since $\triangle F'AD$ is equilateral, $AD = DF'=DC$

$\angle ADF' =60 \rightarrow \angle F'DG = 50 \rightarrow \angle DF'G = 40 \rightarrow \alpha = 140$
[/hide]

EDIT: fixed.

it is true zellex.thank you...

[hide][img]http://img389.imageshack.us/img389/3189/sadie2.png[/img][/hide]


floor function
logarithms
function
algebra proposed
algebra

Let $x_1=1$, $x_2$, $x_3$, $\ldots$ be a sequence of real numbers such that for all $n\geq 1$ we have \[ x_{n+1} = x_n + \frac 1{2x_n} . \] Prove that \[ \lfloor 25 x_{625} \rfloor = 625 . \]

With this sequence we can find out the general of sequence and use this to prove result.
$ v_n\equal{}2x_n$
And consider the function:
$ f(x)\equal{}\frac{x\plus{}\sqrt{2}}{x\minus{}\sqrt{2}}$

We wish to show that $25 \le x_{625} < 25 + \frac{1}{25}$.

Consider the sequence $(y_n)_{n \ge 1}$, where $y_n = x_n^2$. Our recurrence relation then becomes \[ y_{n+1} = y_n + 1 + \frac{1}{4y_n}.\] Since $y_1 = 1$, it follows that \[ y_n = n + \sum_{i = 1}^{n-1} \frac{1}{4y_i}. \] We see immediately that $y_n \ge n$ for all $n$. Thus $625 \le y_{625}$, so $25 \le x_{625}$.

Now we need to show that $x_{625} < 25 + \frac{1}{25}$. This is equivalent to $y_{625} < 625 + 2 + \frac{1}{625}$. In fact, we will show that $y_{625} \le 625 + 2$; the desired conclusion follows.

Note that $y_{625} = 625 + \sum_{i=1}^{624} \frac{1}{4y_i}$. Since $y_i \ge i$, we see that $\frac{1}{4y_i} \le \frac{1}{4i}$, so \[ y_{625} \le 625 + \sum_{i=1}^{624} \frac{1}{4i}. \] It suffices to show that $\sum_{i=1}^{624} \frac{1}{4i} \le 2$, or equivalently \[ \sum_{i=1}^{624} \frac{1}{i} \le 8. \]

Let $H_n$ denote $\sum_{i=1}^n \frac{1}{i}$, so we want to show that $H_{624} \le 8$. We have the following Lemma:

[b]Lemma.[/b] $\log (n+1) < H_n \le 1 + \log n$
[i]Proof.[/i] Note that $H_n = \sum_{i=1}^n \frac{1}{i} > \int_1^{n+1} \frac{1}{t}\, dt = \log (n+1)$ and $H_n = 1 + \sum_{i=2}^n \frac{1}{i} \le 1 + \int_1^n \frac{1}{t} \, dt = 1 + \log n$. $\blacksquare$

Now using Lemma, we see that $H_{624} \le 1 + \log 624$, so it suffices to show that $\log 624 \le 7$. Thus it suffices to show that $\log 625 \le 7$. Since $\log 625 = 4 \log 5$, we wish to show that $\log 5 \le \frac{7}{4}$, or equivalently $e^{7/4} \ge 5$.

Now we use the Taylor expansion of $e^x$; namely, \[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots \] This means that \[ e^{7/4} > 1 + (\frac{7}{4}) + \frac{(7/4)^2}{2} + \frac{(7/4)^3}{6} + \frac{(7/4)^4}{24}. \] We easily evaluate the RHS as \[1 + (1 + \frac{3}{4}) + (1 + \frac{17}{32}) + \frac{343}{384} + \frac{2401}{6144}\] \[ = 3 + (\frac{3}{4} + \frac{2401}{6144}) + (\frac{17}{32} + \frac{343}{384})\] \[ > 3 + 1 + 1 = 5,\] as desired. $\blacksquare$


calculus
integration
trigonometry
calculus computations

Here are some nice integrals you might want to play with.

1) $\int_{0}^{\pi/2}\cos\left(\pi\sin^{2}x\right)dx=0$

2) $\int_{0}^{\pi/2}\cos\left(2\pi\sin^{2}x\right)dx=-\int_{0}^{\pi/2}\cos\left(\pi\sin x\right)dx$.

3) $\int_{0}^{\pi/2}\cos\left(\frac{2\pi}{3}\sin^{2}x\right)dx=\frac{1}{2}\int_{0}^{\pi/2}\cos\left(\frac{\pi}{3}\sin x\right)dx$.

Probably for many of you these integral equalities are not complicated to solve, however I just wanted to let you know about some of the integrals Ramanujan was playing with.

yes, that is the technique.

So, it seems that not all of Ramanujan's formulas are so hard to prove, the other too integrals can be handled by a similar technique.


algebra
polynomial
geometry
geometric transformation
rotation
Galois Theory
number theory unsolved

I have stuck with this problem

$ x_1 \equal{} 0$, $ x_2 \equal{} 2$ and $ x_3 \equal{} 3$

Prove that $ p|x_p$ if $ p$ is prime number, where $ x_n \equal{} x_{n \minus{} 2} \plus{} x_{n \minus{} 3}$

I also want your here what do you think about this aproach (which is, I think, totaly wrong beacuse roots are not integer):
It is easy to get $ x_n \equal{} a^n \plus{} b^n \plus{} c^n$ where $ a,b,c$ are (diffrent) roots of $ t^3 \minus{} t \minus{} 1 \equal{} 0$.

Then by Fermat we have:

 $ a^p\equiv a\; (p)$
 $ b^p\equiv b\; (p)$
 $ c^p\equiv c\; (p)$

Then $ x_p \equiv a \plus{} b \plus{} c \equiv 0\; (p)$  :rotfl:  :D  :blush:  :oops:

That approach can be salvaged, but you have to work over the splitting field of $ t^3 \minus{} t \minus{} 1$ over $ \mathbb{F}_p$ and the approach will fail if $ t^3 \minus{} t \minus{} 1$ has multiple roots over $ \mathbb{F}_p$.  What will be true in the good case is that $ x \mapsto x^p$ is an element of the Galois group, so $ \{ a^p, b^p, c^p \}$ will be a permutation of $ \{ a, b, c \}$.

But you should look for an elementary proof  :P

[quote="Number1"]I have stuck with this problem

$ x_1 \equal{} 0$, $ x_1 \equal{} 2$ and $ x_2 \equal{} 3$

Prove that $ p|x_p$ if $ p$ is prime number, where $ x_n \equal{} x_{n \minus{} 2} \plus{} x_{n \minus{} 3}$

I also want your here what do you think about this aproach (which is, I think, totaly wrong beacuse roots are not integer):
It is easy to get $ x_n \equal{} a^n \plus{} b^n \plus{} c^n$ where $ a,b,c$ are (diffrent) roots of $ t^3 \minus{} t \minus{} 1 \equal{} 0$.

Then by Fermat we have:

 $ a^p\equiv a\; (p)$
 $ b^p\equiv b\; (p)$
 $ c^p\equiv c\; (p)$

Then $ x_p \equiv a \plus{} b \plus{} c \equiv 0\; (p)$  :rotfl:  :D  :blush:  :oops:[/quote]

Since $ x_1$ cannot equal $ 0$ and $ 2$ at the same time i'm guessing that you meant to put $ x_0 \equal{} 0$, $ x_1 \equal{} 2$ and $ x_2 \equal{} 3$.

Also I'm having problems making sense of the problem, take for example :$ x_3 \equal{} x_1 \plus{} x_0 \equal{} 2 \plus{} 0\equal{}2$ and $ 3$ is a prime number but it doesn't seem to divide $ 2$, this immediately proves it's wrong.

Did I misunderstand the problem?

Number1's solution only makes sense if the initial conditions are $ x_0 \equal{} 3, x_1 \equal{} 0, x_2 \equal{} 2$.  Then we get $ x_3 \equal{} 3, x_4 \equal{} 2, x_5 \equal{} 5$, and so forth.

Anyway, I'll bite.  The discriminant of $ t^3 \minus{} t \minus{} 1$ is $ 4 \minus{} 27 \equal{} \minus{} 23$, so for every prime $ p \neq 23$ the polynomial has three distinct roots in $ \mathbb{F}_{p^3}$ and the first approach carries through.  When $ p \equal{} 23$, we can factor out the root $ t \equal{} 3$ and get the factorization

$ (t \minus{} 3)(t^2 \plus{} 3t \plus{} 8) \equal{} (t \minus{} 3)(t \minus{} 10)^2$

so we can verify that $ x_n \equiv 3^n \plus{} 10^n \plus{} 10^n \bmod 23$ and again the first approach carries through.

Another idea would be the combinatorial approach:  $ x_n$ is the number of ways to tile a circular ring of size $ n$ with tiles of size $ 2$ and $ 3$, so when $ n$ is prime every orbit under rotation has the same size (which is $ n$), hence the total number of tilings is divisible by $ n$.


Hypothesis says that $ A,B$ are quasi-commutative ; then $ A,B$ are simultaneously triangularizable (Mac Coy 1934). Therefore $ AB\minus{}BA$ is nilpotent. Now the proofs of 1),...,4) are easy.


[quote="mdk"]prove that n^3+n is even for all n[/quote]

If we are tricky...

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

It is also

$6\binom{n+1}{3}+2n$.

That was a pretty pointless post...but it is O K.

Another way which is longer.

If $n$ is odd. Then $n^{3}$ is odd. An odd number plus an odd number is even.

If $n$ is even. Then $n^{3}$ is even. An even number plus an even number is even.

yeah, I just factored and noticed that it was even either way.

A similar but more difficult problem:

For how many positive integers $n$ is $n^{3}-8n^{2}+20n-13$ prime?

Well consider: n=1 -> 0 n=2 -> 3 n=3 -> 2 ? n=4 -> 3 ? n=5 -> 12
We prove it is either div. by three or two. Indeed, mod 2 its n^3-1 which is div by two for n odd
Mod 3 it is (if 3|n) -1
else it is 3n hence div by three
thus the only case is even div by three, i.e. n=0 mod 6 which is div by 5
plug in: 216n^3-288n^2+120n-13 mod 5 n^3+2n^2+2 which I believe can be factored...
mod 7 n^2-n^3+n+1 which is nice since 3|7-1

[quote]Well consider: n=1 -> 0 n=2 -> 3 n=3 -> 2 ? n=4 -> 3 ? n=5 -> 12
We prove it is either div. by three or two. Indeed, mod 2 its n^3-1 which is div by two for n odd
Mod 3 it is (if 3|n) -1
else it is 3n hence div by three
thus the only case is even div by three, i.e. n=0 mod 6 which is div by 5 [/quote]
So n must be divisible by 6, otherwise the polynomial would be divisible by 2 or 3. But I don't quite get your last step. Why find "216n^3-288n^2+120n-13 mod 5 n^3+2n^2+2"?

I didn't finish the last step... basically I noticed that the only trouble is the 6k=n case, taking k=1 I find it is 35 in this case so we must either prove 5 divisibility in this case of 7 divisibility.


http://onemorelevel.com/game/the_package

How long does it take you?

you don't need a calc to get it that fast. If black is odd and red is even, cut using that knowledge. If you know the divisibility rules for 3 and 5, you could use that.

There was a puzzle on the wire part?

I just randomly decided to cut the black wire...

HAHAHAHAH! I knew the numeric keypad thingy was a backwards magic square! (which is still a magic square)!!!! :D

basically yeah. But the trick is [color=white]you don't need to go 9999, 9998, 9997, etc.[/color]

it took me like 13 mins
i know thats a long time.  i got stuck on the kypad and switches.

[quote="MustItAlwaysBeMe"]That was pretty awesome.  It was much easier than I thought.  And I did the last part in like .4 seconds :) (I had a calculator in front of me)[/quote]
i didnt need to. i just did the units digits in my head.


Let ${a_{1}, a_{2}, ... , a_{n}}$ be $n$ odd positive integers such that their pairwise mutual differences are all distinct.
Show that $\sum_{i=1}^{n}a_{i}\ge \frac{n(n^{2}+3)}{4}$

The claim as stated is false.  For $n=2$ let the positive integers be $a_{1}=a_{2}=1,$ and $1+1 \geq \frac{2(2^{2}+3)}{4}=3.5$ is false.  So we modify the claim and insist that the odd integers be distinct.

We prove the claim by induction.  If the odd positive integers have distinct pairwise differences, then the sum$\sum_{i=1}^{n}a_{i}$ will be least when the integers are the first $n$ terms from the sequence $1,3,7,13,21,31,43,\dots.$  Here the consecutive differences are $2,4,6,8,10,12,\dots.$  The $n$th term of the sequence is

\[a_{n}=1+(2+4+\cdots+2(n-1))=1+2(1+2+\cdots+(n-1))=1+(n-1)(n-2)=n^{2}-3n+3.\]

For $n=1$ through $n=6$ it is easy to verify the claim:

\[1\geq\frac{1(1^{2}+3}{4}=1\]
\[1+3=4\geq\frac{2(2^{2}+3)}{4}=3.5\]
\[1+3+7=11\geq 9\]

and so forth up to

\[1+3+7+13+21+31=76\geq \frac{6(6^{2}+3)}{4}=58.5\]

Let $S_{n}=\sum_{i=1}^{n}a_{n},$ and suppose that $S_{n}\geq \frac{n(n^{2}+3)}{4}$ for some positive integer $n \geq 7.$  Then we have

\[S_{n+1}\geq S_{n}+(n+1)^{2}-3(n+1)+3\geq \frac{n(n^{2}+3)}{4}+(n+1)^{2}-3(n+1)+3\]
\[=\frac{n^{3}+3n+4(n+1)^{2}-12(n+1)+12}{4}\]
\[=\frac{n^{3}+4n^{2}-n+4}{4}\]
\[=\frac{n^{3}+3n^{2}+6n+4+(n^{2}-7n)}{4}\]
\[\geq\frac{n^{3}+3n^{2}+6n+4}{4}\]
\[=\frac{(n+1)((n+1)^{2}+3)}{4}.\]

This completes the induction.

[quote="dabug"]The claim as stated is false.  For $n=2$ let the positive integers be $a_{1}=a_{2}=1,$ and $1+1 \geq \frac{2(2^{2}+3)}{4}=3.5$ is false.  So we modify the claim and insist that the odd integers be distinct.
[/quote]

You are right. There is a small problem for the case of $n=2$. (Although one could argue that for $\{1, 1\}$, $a_{2}-a_{1}= a_{1}-a_{2}$,  so it doesn't satisfy the conditions...)
However for $n\ge3$ we can never have $a_{i}= a_{j}$ because then for some $a_{k}$ with $i\neq j \neq k$ we would have $a_{k}-a_{i}= a_{k}-a_{j}$ which contradicts the problem statement. So distinct integers are implied in this case.


darkquantum wrote:Show that if n is a positive integer,

GCF(2^2^n-2^2^(n-1)+1 , 2^2^n+2^2^(n-1) +1)=1

[hide] Let a=2^2^(n-1). Then your expression simplifies to: GCF(a^2-a+1, a^2+a+1). Well, if two numbers are relatively prime, then any two of their factors must be relatively prime (for example, since 18 and 35 are relatively prime, any of their factors, such as 9 and 7, must also be relatively prime). Okay, now a^3-1 and a^3+1 are relatively prime, since they are two odd integers differing by two. So, thus, any two of their factors, such as a^2-a+1 and a^2+a+1, must also be relatively prime. [/hide]

MysticTerminator is correct yet again.

Yet again, a different solution:

[hide]We know that (a,b), given b>a, is equal to (b-a,a).

Now, I won't spoil the fun, so the hint is to use this principle on this problem, and the solution is quite nice. [/hide]


inequalities
number theory unsolved
number theory

Prove that inequality $ f(n)\leq g(1)\plus{}g(2)\plus{}...\plus{}g(n)$

where $ f(n)$ is the sum of all positive divisors of $ n$, and $ g(n)$ is the number of positive divisors of $ n$.

[quote="nguyentrang"]Prove that inequality $ f(n)\leq g(1) \plus{} g(2) \plus{} ... \plus{} g(n)$

where $ f(n)$ is the sum of all positive divisors of $ n$, and $ g(n)$ is the number of positive divisors of $ n$.[/quote]

Nice problem!

In the sum $ g(1) \plus{} g(2) \plus{} ... \plus{} g(n)$ , every number $ k\le n$ is counted as much as he has multiples $ k\cdot l \le n$. Therefore, $ k$ is counted $ \Big[\frac n k\Big]$ times. So $ g(1) \plus{} g(2) \plus{} ... \plus{} g(n) \equal{} \Big[\frac n 1\Big] \plus{} \Big[\frac n 2\Big] \plus{} ... \plus{} \Big[\frac n n\Big]$. Every divisor of $ n$, call it $ d$, is counted in this sum as $ \Big[\frac n {n/d}\Big]$. So this sum is greater or equal the sum of all divisors of $ n$.


What's the minimum number of square that is needed to get a complete $n$x$n$ grid?

This has definitely been posted before here. I believe it's a question from an Italian olympiad (possibly 2000 or 2002), that this question has also been given in the french correspondence problems for IMO (2004?) and that the answer is $2(n - 1)$.


integration
limit
real analysis
real analysis unsolved

$ f: [0,1] \rightarrow R$ continous and $ F: [0,1]\rightarrow R$, $ F(x)\equal{}\int_{0}^{1}min(x,t)f(t)dt$
a)Prove that $ F^{\prime\prime}(x)\plus{}f(x)\equal{}0$
b) find $ \lim_{x\to1}\frac{F(x)}{x\minus{}1}$

you just need to write it as :
$ F(x)\equal{}\int_{0}^{1} \min(x,t) f(t)dt \equal{} \int_{0}^{x}tf(t) dt \plus{}x\int_{x}^{1}f(t)dt$
Now is easy to answer on your question  :)

a) $ F(x)\equal{}\int_{0}^{x} tf(t) dt \plus{} x\int_{x}^{1} f(t)dt$.
    So $ F'(x)\equal{}xf(x)\minus{}xf(x)\plus{}\int_{x}^{1}f(t)dt \Rightarrow F''(x)\equal{}\minus{}f(x)$

b) by using l'Hospital if $ \int_{0}^{1}tf(t)dt \equal{}0$ we get $ \lim_{x \rightarrow 1} \frac{F(x)}{x\minus{}1}\equal{}\lim_{x \rightarrow 1} F'(x) \equal{} \lim_{x \rightarrow 1} \int_{x}^{1}f(t)dt \equal{}0$; if $ \int_{0}^{1} tf(t) dt >0$  then the limit is  $ \minus{}\infty$ or if  $ \int_{0}^{1} tf(t)dt<0$ then limit is  $ \plus{}\infty$.


function
real analysis
real analysis unsolved

Let $ f: \mathbb{R} \to \mathbb{R}$, such that $ f(x)\equal{}\frac{x\plus{}2} {x^2\plus{}2x\plus{}2}$

$ a)$ Show that $ f$ is bounded

$ b)$ Find the minimum and the maximum of $ f$

hello, we have $ \minus{}\frac{1}{2}\frac{\sqrt{2}}{2\plus{}\sqrt{2}} \le \frac{x\plus{}2}{x^2\plus{}2x\plus{}2}\le \frac{1}{2}\frac{\sqrt{2}}{2\minus{}\sqrt{2}}$ for all $ x \in \mathbb{R}$.
Sonnhard.


There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and
1 orange peg to be placed on a triangular peg board. In how many
ways can the pegs be placed so that no (horizontal) row or
(vertical) column contains two pegs of the same color?  (Any two pegs of the same color are indistinguishable.)

[asy]draw((0,0)--(5.5,0)--(0,5.5)--cycle);
for (int i=0; i<5; ++i) {
dot((0.5,i+0.5));
}
for (int i=0; i<4; ++i) {
dot((1.5,i+0.5));
}
for (int i=0; i<3; ++i) {
dot((2.5,i+0.5));
}
for (int i=0; i<2; ++i) {
dot((3.5, i+0.5));
}
dot((4.5,0.5));[/asy]

There's only one way.

[code]Y
R Y
G R Y
B G R Y
O B G R Y[/code]

That post is considered spam. Don't post if you don't have anything to to post/say. Agreeing and saying ya uh-huh this dude is right is spam thx.


What is the best constant $ k_n $ such that for any real nmbers $ a_1,a_2,...,a_n $ we have $ (a_1-a_2)^2+(a_2-a_3)^2+...+(a_n-a_1)^2\geq k_n\sum_{i<j} (a_i-a_j)^2 $ ?

I found in my collection the following similar problem: What is the greatest constant  $q_n$ such that for any real numbers $a_i$ , the inequality holds
  $a_1^2+...+a_n^2+\frac{q_n}{n}(a_1+.....+a_n)^2\geq (1+q_n)(a_1a_2+...+a_na_1)$ ?

I found the particular results:
   $q_n=\sqrt{5}$ for $n=5$  ;
   $q_n=1$ for $n=6$ ;
  $q_n=\sqrt{2}-1$ for $n=8$ .

The problem can be reformulated in the nicer form: What is the best constant  $p_n$  such that for any real numbers $a_i$  with $a_1+...+a_n=0$, the inequality holds
	$\frac{a_1a_2+...+a_na_1}{a_1^2+...+a_n^2}\leq p_n$ ?

Actually, the last inequality is known as Fan inequality (1955):
If $a_1, ... , a_n$ are real numbers such that $a_1+...+a_n=0$, then
     $\frac{a_1a_2+a_2a_3+...+a_na_1}{a_1^2+a_2^2+...+a_n^2}\leq cos\frac{2pi}{n}$.

This is similar to Harazi proof

[quote="harazi"]I used the following result of Ozeki:
  $ max\frac{x_1\cdot x_2+x_2\cdot x_3+...+x_{n-1}\cdot x_n}{ x_1^2+x_2^2+...+x_n^2} =cos\frac{pi}{n} $. Old members probably remember a problem posted by S_Alex which reduces immediately to this one. But then I didnt know this and I struggled a lot to solve that problem.[/quote]

To Vasc:
I am your fans.I wish see you here more.That is all.
Tex.


A rectangular tank of mass $ m_o$ and charge Q over it is placed   over a smooth horizontal floor. A horizontal electric field E exists in the region. Rain  drops are falling vertically in the tank at the constant rate n drops per second. Mass of each drop is m. Find the time taken by the tank to reach half the maximum speed.

[hide="Ans"] $ \frac{m_o}{n m}$ [/hide]

The Equation of motion is given by
$ EQ\equal{}vmn\plus{}(m_0\plus{}mnt)\frac{dv}{dt}$
which, on solving gives,
$ \frac{EQ}{EQ\minus{}vmn}\equal{}\frac{m_0\plus{}mnt}{m_0}$

At terminal velocity, $ \frac{dv}{dt}\equal{}0$ and hence, $ v_T\equal{}\frac{EQ}{mn}$

Thus substituting $ v\equal{}\frac{EQ}{2mn}$, we have$ 2m_0\equal{}m_0\plus{}mnt$, that is

$ t\equal{}\frac{m_0}{mn}$

$ F \equal{} QE \equal{} {{d(Mv)}\over {dt}} \equal{} {{dM}\over {dt}}v \plus{} M{{dv}\over {dt}}\;\;\;\;\;\;$        where $ M\equal{}m_0 \plus{}mnt$    as     $ {{dM}\over {dt}}\equal{}nm .$
So
  \[ {{dv}\over{QE\minus{}nmv}} \equal{} {{dt}\over{m_0\plus{}nmt}}.\]

Integration leads to

\[ v \equal{}{{QE}\over{nm}}( 1 \minus{} {1 \over{1\plus{}n{m \over m_0} t}})\]

which for t->$ \infty$ gives the  maximum ( or limit ) speed of $ {{QE}\over{nm}}.$

If v is half of it,

 $ \;\;\;\;\;\;   1 \plus{} {{nm} \over m_0} t  \equal{} 2 \;\;\;\;\;$   and

$ t \equal{} {m_0 \over {nm}}$


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $A,B \in \mathcal M_3 \left( \mathbb R \right)$ s.t. $\det A = \det B = \det \left( A+B \right) = \det \left( A-B \right)$.

Compute $\det \left( xA+yB \right)$, where $x,y$ are some numbers in $\mathbb R$.

Is it $det(xA+yB)=0$? :)  If not I think I misscalculated something. :?

Hello,it's so late again i hope that what i'm gonna say would be logic,at first we have to prove that $A$ isn't invertible which is sort of easy to prove:
let's consider a base where the matrix $C$ similar to $A$ is triangular and let's consider $\lambda_1$,$\lambda_2$ & $\lambda_3$ the diagonal elements of $C$ .
Thus $\lambda_1$ * $\lambda_2$ *$\lambda_3$=($\lambda_1$+1)*($\lambda_2$+1)*($\lambda_3$+1)
                                                                       =($\lambda_1$-1)*($\lambda_2$-1)*($\lambda_3$-1)
we study the case where the three elements are equal and when two are equal we have direct contradictions.
so if they're distinct then:let's suppose that $\lambda_1$ =($\lambda_2$+1)
                                                        then $\lambda_1$ =($\lambda_3$-1)  ==> $\lambda_2$=$\lambda_3$-2  (1)

so                                                            $\lambda_2$ =($\lambda_1$-1)
                                                        then $\lambda_2$ =($\lambda_3$+1)   (2)
Thus we have a contradiction and this way $\det(A)$=0.
the conditions of the exercice lead to a multivariable polynomial  expression of $\det(xA+yB)$=$\alpha$*x*y*(x+y)*(x-y)    $\alpha$ is a real number .

My solution is :
assuming $x\neq 0$ then   $\det(xA+yB)=x^3\det(A+\frac{y}{x}B)$. Let $P(x)=\det(A-xB)=\det(A)x^3+ a_1x^2+a_2x+\det(A)$
From $P(1)=P(-1)=P(0)$ we get that $a_1=0,a_2=-\det(A)$ so $P(x)=\det(A)(x^3-x^2+1)$. So $\det(xA+yB)=x^3\det(A)[(\frac{y}{x})^3-(\frac{y}{x})^2+1]$. How can I prove that $\det(A)=0$? :blush:

HAHAHA i was sure that i was going to make a stupid comment as i was so sleepy well the  matrix $A$ whereof i proved that $\det(A)$=0 isn't the matrix A of the exercice it's $AB^-1$ if we consider that $A$ is invertible because in that case :
$\det(AB^-1)$=$\det(I)$=$\det(AB^-1+I)$=$\det(AB^-1-I)$
then we consider AB^-1 as a complex matrix and according to the proof i've already given we deduce that $\det : (AB^-1)$=0 Thus A & B aren't invertible :).
Regarding the solution of Spix now i guess that you 've proved that way that :
$\det(xA+yB)=\det(A) (y^3-y^2+x^3)=\det(A) (x^3-x^2+y^3)$ which leads  to $\det(xA+yB)=0$ .
Myproblem is that i still don't c what is wrong in the polynomial expression i've given could someone help plz :) ?

Well, since nobody bothered to give a complete solution, I'll do that for you.

 [i] Solution.[/i]

   We have  \[ det(xA+yB)=f(x, y)=a_{1}x^{3}+a_{2}x^{2}y+a_{3}xy^{2}+a_{4}y^{3}. \]
    We also have \[ f(0, 1)=0=detB=0\Leftrightarrow a_{4}=0, \]
                        \[ f(1, 0)=det A=0\Leftrightarrow a_{1}=0, \]
                        \[ f(1, 1)=det(A+B)=0\Leftrightarrow a_{2}+a_{3}=0\Leftrightarrow, \]
                        \[ f(1, -1)=det(A-B)=0\Leftrightarrow a_{2}-a_{3}=0. \]

     So we have $a_{2}=a_{3}=0$. The conclusion follows. :)

Hey Spix i guess i already proved this wasn't my proof clear enough??

Your proof is clear. Only that Cezar`s "complete solution" is not so complete after all. Maybe he proved that $\det(A)=0$ through another way....

[quote="MoHaMeD"]Hey Spix i guess i already proved this [i]wasn't my proof clear enough??[/i][/quote]

Well, actually, no  :(  The three eigenvalues are $\lambda_{1,2,3} = -1,0,1$, and I can't derive a contradiction from this.

Can you/someone else/Cezar Lupu help me?

I considered that $A$ is invertible ($B$ is invertible too) and i supposed that $\lambda_1,\lambda_2$and $\lambda_3$ are the diagonal elements of the matrix $C$ which similar to $AB^-1$ in an other base:The mistake that i have made is that i considered::${\lambda_1,\lambda_2,\lambda_3}={\lambda_1-1,\lambda_2-1,\lambda_3-1}={\lambda_1+1,\lambda_2+1,\lambda_3+1}$  which is obviously false in fact we have $\lambda_1$ * $\lambda_2$ *$\lambda_3$=($\lambda_1$+1)*($\lambda_2$+1)*($\lambda_3$+1)=($\lambda_1$-1)*($\lambda_2$-1)*($\lambda_3$-1)=1. let's consider P the unitarian polynomial having ${\lambda_1,\lambda_2,\lambda_3}$ as roots we find that $P=X^3-X+1$ this polynomial has only one real root which is less than $-1/\sqrt 2$ thus $\lambda_1$ * $\lambda_2$ *$\lambda_3$=$\lambda_1$ * $|$$\lambda_2|^2$ =1  so $\lambda_1>0$ which is false then our hypothesis was wrong :)

[quote="perfect_radio"]But I think that $\lambda_1 \cdot \lambda_2 \cdot \lambda_3$ is equal to $-1$, not to $1$.[/quote]
Umm but  i guess that $\lambda_1 \cdot \lambda_2 \cdot \lambda_3$=$det(AB^{-1})=det(I)$

[quote="perfect_radio"]The characteristic polynomial of a matrix is $\det \left( A - x I_n \right)$, which in this case is $x^3-x+1$. BUT, the free term of $\det \left( A - x I_n \right)$ if $(-1)^n \lambda_1 \lambda_2 \ldots \lambda_n$, and this, in our case, means that $- \lambda_1 \lambda_2 \lambda_3 = \det A$ ! (I wrote $A$ instead of $AB^{-1}$ due to laziness)[/quote]
i think  it's $\lambda_1 \lambda_2 \ldots \lambda_n$ check the case of n=3 with $A=2I$ :)

Well, I was wrong... we always have $\prod \lambda_i =$ free term; but there's still a problem: the characteristic polynomial of $AB^{-1}$ is actually $-x^3+x-1$ !!! Am I still wrong?

I think the characteristic polynomial should be $\det(xI - A)$ in general -- this convention is chosen so that the leading coefficient is always 1.

you're right in fact it seems that it was like i was forcing the result to be a contradiction  :blush:  notice that thanks to the $C=AB^{-1}$ that we have found for any invertible matrix $B$ we can find a suitable $A=CB$ thus $det(xA+yB)=det(xC+yI)det(B)$ i think that by using the density of invertible matrixes we can even generalise this for $B$ which isn't invertible .
moreover $det(xC+yI)=(y+x\lambda_1)(y+x\lambda_2)(y+x\lambda_3)=y^3-yx^2+x^3$
what do you think??

By the way do u know that because of your message i didn't go to a school party ;) anyway thank you so much because thanks to you i am back to the forum again :)


Ύστερα από επικοινία με τον Βορνίκου , κάναμε μια μικρή αναβάθμιση.
 Αυτό οφείλεται στη δική σας μεγάλη συμμετοχή και διάθεση για μάθηση και δημιουργία!

 Από δω  και πέρα να στέλνουμε  όσο μπορούμε τα μηνύματα στο σωστό χώρο.

 [color=red][b]Καλή δύναμη και ακόμα μεγαλύτερη επιτυχία στο Ελληνικό φόρουμ ![/b][/color]

 Μπάμπης

[color=red]Πολύ καλή και χρήσιμη πρωτοβουλία κ . Στεργίου :) [/color]

Και επειδή πάντα η αρχή είναι δύσκολη προτείνω (και για πρακτικούς λόγους) να μετακινήσετε (αν είναι δυνατόν αυτό) το sticky με τα προβλήματα φοιτητών στον αντίστοιχο χώρο :wink:

Συγχαρητήρια!!! :thumbup:  :thumbup:  :thumbup:
Η δημιουργία υποφόρουμ ήταν ότι καλυτερο αυτή τη στιγμή.

[color=red][b]Ο Σιλουανός  Moderator ![/b][/color]

Ευχόμαστε  στο Σιλουανό ΚΑΛΗ ΣΥΝΕΧΕΙΑ και με το μεράκι του - ως φοιτητής σε λίγο - να αναβάσει το  φορουμ που τόσο αγάπησε  ακόμα ψηλότερα ! Ας του χαρίσει επιπλέον η Θεά των Μαθηματικών ένα διεθνές ολυμπιακό μετάλλιο . Τόσες θυσίες έχει έχει στο βωμό της  !!!

 Μπάμπης

[quote="xp"]Μπάμπη το  "μικρή αναβάθμιση" κάντο πιο λιανά...
Sil συγχαρητηρια για το Moderator[/quote]

 Χρήστο,
 φτιάξαμε τα υποφόρουμ! Αυτό θεωρώ ... '' μικρή αναβάθμιση'' - από τεχνικής απόψεως(ας είμαστε και λίγο ταπεινοί!).Διότι από πλευράς συμμετοχής και δράσης είμαστε το καλύτερο στον κόσμο νομίζω.
 Μεγάλη αναβάθμιση θα είναι  μόνο αυτή  , στην οποία θα έχουμε καθημερινά  εκατοντάδες επισκέπτες και μαθητές με ολυμπιακό προσανατολισμό !Για αυτό το φόρουμ αγωνιζόμαστε .

 Μπάμπης

Συγχαρητήρια και στους 2 moderators
 :lol: Ρ.Μ

Συγχαρητήρια για αυτή την σημαντική αναβαθμιση στο forum αλλά βεβαίως και στο φίλο μου τον Σιλουανό . Εύχομαι φέτος να είναι η χρονιά των μεταλλίων όπως σου αξίζει Σιλουανέ .


Find out all the values of n such that there exists a choice of the signs for the following equality to be true:
1(+-)2(+/-)3...(+/-)n=0

possible as long as there's an even amount of odd numbers.
so as long as n is 0 or 3, mod 4.

I recognise some Romanian olympiad problem here :D (I think from a TST)

lucky romenians, getting that easy questions ;)

:P

But the original Romanian question was [b]much[/b] more difficult than this one :D


floor function
calculus
integration
ratio
number theory unsolved
number theory

Let $r\geq1$ be areal number that holds with the property that for each pair of positive integer numbers $m$ and $n$, with $n$ a multiple of $m$, it is true that $\lfloor{nr}\rfloor$ is multiple of $\lfloor{mr}\rfloor$. Show that $r$ has to be an integer number.

[b]Note: [/b][i]If $x$ is a real number, $\lfloor{x}\rfloor$ is the greatest integer lower than or equal to $x$}.[/i]

My solution suffices for $r\ge 0$.  Denote $\{x\}$ to be the fractional part of $x$, in other words, $x=\lfloor x\rfloor+\{x\}$.

First, I hope its obvious that if $r\in\mathbb{Z},$ then the theorem holds.

We now proceed by contradiction.

If $r$ is not an integer, then $r$ can be written in the form $z+\epsilon$ where $z\in\mathbb{Z}$ and  $0<\epsilon<1$.  Now, using the bounds on $\epsilon$, we have that there exists an integral $m\ge 3$ such that

$2<m\epsilon<3$.

Then $\lfloor mr\rfloor=\lfloor (z+\epsilon)m\rfloor=zm+\lfloor m\epsilon\rfloor=mz+2\ge 2$.

Now, lets consider the number $\{m\epsilon\}$. Since $2<m\epsilon<3,$ we know that $0<\{m\epsilon\}<1$.

Thus, there exists an integral $k\ge 2$ such that $1\le k\{m\epsilon\}<2$.  Let $n=mk$. Then:

$\displaystyle\lfloor nr\rfloor=\lfloor (km)(z+\epsilon)\rfloor=kmz+\lfloor km\epsilon\rfloor=kmz+\lfloor k\lfloor m\epsilon\rfloor + k\{m\epsilon\}\rfloor=k(mz+\lfloor m\epsilon\rfloor) +1$

But then $\displaystyle\frac{k(mz+\lfloor m\epsilon\rfloor)+1}{mz+\lfloor m\epsilon\rfloor}=k+\frac{1}{mz+\lfloor m\epsilon\rfloor}$

and since the denominator is greater than or equal to $2,$ the ratio cannot be an integer.  Therefore $\epsilon=0$ and $r\in\mathbb{Z}$.

Let $ m,k$ be positive integers and $ r\ge 0$ with the mentioned property. So there exists a integer $ l$ such that $ \lfloor mkr\rfloor\equal{}l\lfloor mr\rfloor$

Let $ p\equal{}\lfloor mr\rfloor$ and $ q\equal{}mr\minus{}p$, then $ kp\plus{}\lfloor kq\rfloor \equal{}lp\Rightarrow p(l\minus{}k)\equal{}\lfloor kq\rfloor$. If $ p\equal{}0$ we must have $ q\equal{}0$ so $ r$ is an integer.

Suppose $ p>0$. Since $ 0\le q<1$ we have $ 0\le \frac{\lfloor kq\rfloor}{p}\le\frac{k\minus{}1}{p}$.

This must hold for all $ m$ and $ k$ so let's take $ k\equal{}p$, it becomes $ 0\le \frac{\lfloor kq\rfloor}{p}\equal{}l\minus{}k\le\frac{p\minus{}1}{p}<1$ then $ q\equal{}0$, again $ r$ is an integer.
This completes the proof.

:)

Let $k$ be an integer and $r$ with the desired property. Pick $m=1$ and write $\lfloor kr \rfloor = \lfloor k \lfloor r \rfloor + k \langle r \rangle \rfloor $, where $\langle r \rangle$ is the decimal part of $r$, notice that since $k \lfloor r \rfloor$ is and integer, we can take it out from the floor function. Rearranging the terms we get:
\[ \lfloor kr \rfloor - \lfloor k \langle r \rangle \rfloor = k \lfloor r \rfloor \].
If $\lfloor k \langle r \rangle \rfloor=1,$ then $\lfloor kr \rfloor$ is not a multiple of $\lfloor r \rfloor $. But we can always find $k$ that satisfies this if $\langle r \rangle >0$, just by taking the least number such that $\langle r \rangle>1$ we instantly see that $2>k \langle r \rangle>1$ and so, no matter what $r$ we take, if it has a non-zero decimal there exists a $k$ for which the divisibility condition doesn't hold, thus $r$ is an integer.

Let $r\ge 1$ be a real number that holds with the property that for each pair of positive integer numbers $m$ and $n$, with $n$ a multiple of $m$, it is true that $\lfloor nr \rfloor$ is multiple of $\lfloor mr \rfloor$. Show that $r$ has to be an integer number.

Note: If $x$ is a real number, $\lfloor x \rfloor$ is the greatest integer lower than or equal to $x$.


Sorry if this does not belong here. I am not sure where to put it.

When I try to do an emoticon in my posts, they show up as  :| instead of the yellow smiley face. How can I do the yellow smiley face emoticons?

Go to the [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=prefs&sub=posting]User CP[/url] to permantly turn your usage of smilies on.  :)

Thank you.

I tried that last night. but I think I forgot to hit submit......

does it work? it should be like this... :D  :lol:


inequalities calculus AMC USA(J)MO USAMO

Prove that for $ x, y \in \mathbb{R}^\plus{}$, (Is that the right way to express "x and y such that both x and y are positive real numbers"?),

$ \frac{x}{y} \plus{} \frac{y}{x} \ge 2$

Yes, that notation works.

[hide="Solution"]This is a one-step, straightforward, cut-and-dried, duck-soupy application of AM-GM.

$ \frac{x}{y}\plus{}\frac{y}{x}\geq 2\sqrt{\frac{xy}{yx}}\equal{}2$.[/hide]

We begin with the trivial inequality and go from there. Note that we don't have to worry about division and multiplication, as it's given that $ x$ and $ y$ are positive.
\begin{align*} (x - y)^2 & \ge 0 \\
x^2 - 2xy + y^2 & \ge 0 \\
x^2 + y^2 & \ge 2xy \\
\frac {x}{y} + \frac {y}{x} & \ge 2 \end{align*}
As for your question about notation, I have absolutely no clue. :roll:

EDIT: Beaten, but didn't use the "cut-and-dried" AM-GM solution. :P

[hide="Comment"]
Is there a way to do it without applying AM-GM? Maybe it won't be faster, but this seems like a simple enough fact that there has to be more than one explanation, and maybe some that will be more intuitive/graphical.

EDIT> Beaten by another post; maybe there are more.

P.S. Just a general, completely unrelated question, calculus is allowed on the USAMO, right?
[/hide]

Yes, calculus is allowed, but never necessary and only rarely useful.

AM-GM in two variables is equivalent to the trivial inequality, so there's no reason to make such a fuss.  You can also use Rearrangement, for example, or AM-HM.  Generally, you'll find AM-GM to be by far the simplest "explanation" for most inequalities compared to some other methods.

As for "graphical" explanations, what does the graph of $ y \equal{} x \plus{} \frac {1}{x}$ look like?  Alternately, what line passing through the point $ (1, 1)$ minimizes the taxicab distance between the $ x$-intercept and the $ y$-intercept?

(I prefer the method in the 3rd frame for more than
one reason, including the method I'm going to show next.)

I'm showing this for more variety/creativity, not for its relative
lack of economy of typing.

<><><> I'm going to split this into two cases:

Case I
---------

Let y = x

x/y + y/x becomes

x/x + x/x =

1 + 1 =

2 >= 2

===================================================
Case II
----------

Without loss of generality,
let x > y . . . . (1)

Then x - y > 0 . . . . (2)

From (1), we have

y < x.

Flipping each side we get
1/y  > 1/x  . . . . (3)

Multiply (3) by the positive quantity (x - y)**
on each side:

(x - y)/y > (x - y)/x

Separate:

x/y - y/y > x/x - y/x

x/y - 1 > 1 - y/x

x/y + y/x > 1 + 1

x/y + y/x > 2

**  It is not allowed to multiply each side by
zero, so (x - y) is kept to being strictly greater
than zero here, and x and y being equal to each
other is handled separately in Case I.

[quote="yanglunj"]$ x, y \in \mathbb{R}^ \plus{}$, (Is that the right way to express "x and y such that both x and y are positive real numbers"?)[/quote]  There may be ambiguity about whether $ {\bf R}^ \plus{}$ contains zero or not -- $ {\bf R}_{ > 0}$ is unambiguous.

Also, in the future, please give your posts more meaningful titles.