[quote="pohoatza"]$\sum{\frac{2a^{2}+bc}{b+c}}\geq \sum{\frac{a^{2}+2bc}{b+c}}\geq \frac{(\sum{a})^{4}}{3\sum{ab(a+b)}}\geq \frac{9(a^{2}+b^{2}+c^{2})}{2(a+b+c)}$[/quote]
It's wrong for $a=b$ and $c\rightarrow0^{+}.$ :wink:

Indeed Ji Chen, you are right,
$4\sum x^{5}-\sum x^{4}(y+z)-3\sum x^{3}(y^{2}+z^{2})+4\sum xy^{2}z^{2}$
$\neq4\sum x^{3}(x-y)(x-z)+3\sum x^{4}(y+z)-3\sum x^{3}(y^{2}+z^{2})$. :P

I am waiting somebody post a solution without expanding.  :(

[quote="Vasc"]If $a,b,c$ are non-negative numbers, no two of which are zero, then

$\sum \frac{2a^{2}+bc}{b+c}\geq \frac{9(a^{2}+b^{2}+c^{2})}{2(a+b+c)}$.[/quote]
The inequality is equivalent to:

$\sum_{cyc}\left({2a^{3}+abc \over b+c}+2a^{2}+bc\right) \geq{9 \over 2}(a^{2}+b^{2}+c^{2}).$

Recalling the known inequality:

$\sum_{cyc}{a^{3}+abc \over b+c}\geq a^{2}+b^{2}+c^{2}.$

It suffices to prove that:

$\sum_{cyc}{a^{3}\over b+c}+bc \geq{3 \over 2}(a^{2}+b^{2}+c^{2})$
$\Leftrightarrow \sum_{cyc}{a^{2}+b^{2}-c^{2}\over (a+c)(b+c)}(a-b)^{2}\geq 0$
$\Leftrightarrow \sum_{cyc}S_{c}(a-b)^{2}\geq 0.$

Assuming $a \geq b \geq c$, hence:

$S_{c}\ge 0,$
$S_{b}\geq 0,$
$S_{b}+S_{a}={c^{2}(a+b+2c)+(a+b)(a-b)^{2}\over (a+b)(b+c)(c+a)}\ge 0.$

So that, the original inequality is proved.

Yes, Lovasz. This is also my solution. :lol:

[quote="Vasc"]If $ a,b,c$ are non-negative numbers, no two of which are zero, then

$ \sum \frac {2a^{2} \plus{} bc}{b \plus{} c}\geq \frac {9(a^{2} \plus{} b^{2} \plus{} c^{2})}{2(a \plus{} b \plus{} c)}$.[/quote]
$ \sum \frac {2a^{2} \plus{} bc}{b \plus{} c}\minus{} \frac {9(a^{2} \plus{} b^{2} \plus{} c^{2})}{2(a \plus{} b \plus{} c)}\equal{} \sum\frac{(b \minus{} c)^{2}[2(b\plus{}c\minus{}a)^2\plus{}bc]}{2(a \plus{} b)(a \plus{} c)(a\plus{}b\plus{}c)}\geq 0.$

[quote="Lovasz"]It suffices to prove that:

$ \sum_{cyc}{a^{3}\over b \plus{} c} \plus{} bc \geq{3 \over 2}(a^{2} \plus{} b^{2} \plus{} c^{2})$[/quote]
$ \sum_{cyc}{a^{3}\over b \plus{} c} \plus{} bc \minus{}{3 \over 2}(a^{2} \plus{} b^{2} \plus{} c^{2})\equal{} \sum_{cyc}{(b \minus{} c)^{2}(b\plus{}c\minus{}a)^2\over 2(a \plus{} b)(a \plus{} c)}\geq 0.$

[quote="Ji Chen"] $ \sum \frac {2a^{2} \plus{} bc}{b \plus{} c} \minus{} \frac {9(a^{2} \plus{} b^{2} \plus{} c^{2})}{2(a \plus{} b \plus{} c)} \equal{} \sum\frac {(b \minus{} c)^{2}[2(b \plus{} c \minus{} a)^2 \plus{} bc]}{2(a \plus{} b)(a \plus{} c)(a \plus{} b \plus{} c)}\geq 0.$

$ \sum_{cyc}{a^{3}\over b \plus{} c} \plus{} bc \minus{} {3 \over 2}(a^{2} \plus{} b^{2} \plus{} c^{2}) \equal{} \sum_{cyc}{(b \minus{} c)^{2}(b \plus{} c \minus{} a)^2\over 2(a \plus{} b)(a \plus{} c)}\geq 0.$[/quote]
Very nice identities, Ji Chen! :lol:

[quote="Ji Chen"][quote="Vasc"]If $ a,b,c$ are non-negative numbers, no two of which are zero, then

$ \sum \frac {2a^{2} \plus{} bc}{b \plus{} c}\geq \frac {9(a^{2} \plus{} b^{2} \plus{} c^{2})}{2(a \plus{} b \plus{} c)}$.[/quote]
$ \sum \frac {2a^{2} \plus{} bc}{b \plus{} c} \minus{} \frac {9(a^{2} \plus{} b^{2} \plus{} c^{2})}{2(a \plus{} b \plus{} c)} \equal{} \sum\frac {(b \minus{} c)^{2}[2(b \plus{} c \minus{} a)^2 \plus{} bc]}{2(a \plus{} b)(a \plus{} c)(a \plus{} b \plus{} c)}\geq 0.$

[/quote]

WHY?

[quote="Vasc"]If $ a,b,c$ are non-negative numbers, no two of which are zero, then

$ \sum \frac {2a^{2} + bc}{b + c}\geq \frac {9(a^{2} + b^{2} + c^{2})}{2(a + b + c)}$.[/quote]

$ LHS - RHS$
$ = \frac{4\sum{x^5}-\sum{yz(y^3+z^3)}-\sum{3y^2z^2(y+z)}+xyz(\sum{x^2}+3\sum{yz})}{2\sum{x}\prod{(y+z)}} \ge 0$

Lemma1 $ \sum{x^5}-\sum{yz(y^3+z^3)}+xyz\sum{yz} \ge 0$ (1)

$ LHS$
$ = \sum{x(x^2-yz)(x-y)(x-z)}$

Let $ x \ge y \ge z$,like Schur's inequality

$ LHS$
$ \ge (x-y)^2(x+y)(x(x-z)+y(y-z)) \ge 0$.

Lemma2 $ \sum{yz(y^3+z^3) \ge \sum{y^2z^2(y+z)}}$ (2)

Hence $ LHS - RHS = \sum{yz(y+z)(y-z)^2} \ge 0$.

So We have

Lemma3 $ \sum{x^5}-\sum{y^2z^2(y+z)}+xyz\sum{yz} \ge 0$ (3)

Lemma4 $ \sum{x^5}-\sum{yz(y^3+z^3)}+xyz\sum{x^2} \ge 0$ (4)

For (3)*3+(4), we obtain the answer. #

[quote=Vasc]If $a,b,c$ are non-negative numbers, no two of which are zero, then

$\sum \frac{2a^{2}+bc}{b+c}\geq \frac{9(a^{2}+b^{2}+c^{2})}{2(a+b+c)}$.[/quote]


I remember that problems like this one was posted on the forum:

compute the lim(x-->0) (1-cos(x)*cos(2x)*..*cos(nx))/x^2  for any n natural.

now.. there is a generalization for these kind of problems:

prove that:
If lim(x-->0) (1-f(x))/x^p=L, L finite, then we have that:
lim(x-->0) (1-f(x)f(2x)*..*f(nx))/x^p=L* (sum(k=1,n) k^(p+1)).

cheers! :D :D

Think again about that p+1.

We have lim{x->00}f(x)=1 and
The limit is lim (1-f(x)+f(x)-f(x)(f2x)+...-f(x)f(2x)...f(nx))/x^p.
Compute this and get what you said but with p instead of p+1.

yes.. u r right .. the problem is like so:

If lim(x-->0) (1-f(x))/x^p=L, L finite, then we have that:
lim(x-->0) (1-f(x)f(2x)*..*f(nx))/x^p=L* (sum(k=1,n) k^p).

sorry!

From the hypothesis we get

f(x)=1-Lx^p+x^p.e(x) where lim_(x->0)e(x)=0

1-f(x)f(2x)...f(nx)=
1-(1-Lx^p+x^p.e(x))(1-L2^px^p+2^px^p.e(x))...(1-Ln^px^p+n^px^p.e(x)) =
Lx^p \sum_(k=1 to n ) k^p + term with x^(greater than p)+...

divide by x^p we get

L \sum_(k=1 to n ) k^p


function
analytic geometry
logarithms
graphing lines
slope
geometry
perpendicular bisector

Some points are chosen on the graph of lnx function on coordinate system. Here we have several questions:
1) Is it possible to draw a isosceles triangle?
2) Is it possible to find a piece of regular n-gon?
3) Is it possible to draw a triangle whose biggest angle is 120?

(Note: Above, lnx denotes the logarithm function base e?(also known as natural logarithm.)

[b](1)[/b] Obviously, the graph is continuous and is not a straight line. So choose any point of the graph as a centre and draw a circle, then the point with the two intersections form an isoscele triangle.

[b](2)[/b] Draw the graph and then it will be very clear.
Assume that there is such a n-gon.Then we have the perpendicular bisector of any sides of the n-gon intesect the graph.

But Slope of any of these side is of the form $ \frac{x\minus{}y}{In x\minus{}In y}$ is positive,$ \Rightarrow$ the perpendicular is of negative slope,  $ \Rightarrow$ the intesection of the perpendicular and the graph
, say $ P(z,In z)$, satisfies $ \begin{cases}z>x \\ In z<In x \end{cases}$ or $ \begin{cases}z<x \\In z>In x \end{cases}$, impossible. Done.

[b](3)[/b] It is possible. Indeed it's possible not only for $ 120^{\circ}$, but also for any angle $ x^{\circ}$ saisfying $ 90^{\circ}<x^{\circ}<180^{\circ}$


Prove the among $16$ consecutive integers it is always possible to find one which is relatively prime to all the rest.

Let $ A$ be the arbitrary set of $ 16$ consecutive integers. It is clear that if $ p$ is a common prime factor of two elements of $ A$ then $ p \in \{2, 3, 5, 7, 11, 13 \}$. Thus, it is enough to show that we can find an element of $ A$ which is not divisible by any number from this set.
Consider the following residues mod $ 30$: $ 1, 7, 11, 13, 17, 19, 23, 29$. They are coprime with $ 30$ and note that no matter how we choose the $ 16$ consecutive residues there are always $ 4$ of them belonging to this set. The difference between two of this numbers is never divisible by $ 7$ and $ 13$ and the only differences divisible by $ 11$ are $ 23 \minus{} 1 \equal{} 22 \equal{} 29 \minus{} 7$ but this numbers are too far of each other to be in the same set of $ 16$ consecutive integers. Hence among those $ 4$ numbers there is at most one number divisible by $ 7$, at most one number divisible by $ 11$ and at most one number divisible by $ 13$, so we can choose a number which is not divisible by any of them and since this number is also coprime to $ 30$ we are done.

I have a question about this solution.  What if the first, second or third number in $ A$ is congruent to 17, 23 or 29 mod 30, and is also divisible by 7?  Then the 14th, 15th or 16th is also, but is relatively prime to 30, as is asserted is impossible by the previous post.

Here is a proposed fix: If the 1st, 2nd, 3rd number is 17 or 29 mod 30 and divisible by 7, then there are 3 other residues in $ A$ relatively prime to 30.  At most one is divisible by 11, and at most one is divisible by 13, so there is one not divisible by any prime less than 16 as sought.

Now if the 1st, 2nd, 3rd is 23 mod 30 and divisible by 7, then 29 and 1 are also in $ A$.  These are both relatively prime to every number in the set even if one of them is divisible by 11 or 13, because the nearest multiples of 11 or 13 are outside of $ A$, as 11 from 1 mod 30 is 20 mod 30 which is too far to be inside the bounds.

I'm sorry if I'm not clear, but there was a gaping hole in TomciO's proof I just had to fix.


Prove that the product of $ k$ consecutive integers must be divisible by $ k!$.

$ _nC_k \in Z$   $ \forall n \ge k$ Q.E.D.

How do you prove $ n(n\minus{}1)\cdots (n\minus{}k\plus{}1)$ is divisible by $ k!$ if $ n<k$.

Because the divisibility of $ n(n\minus{}1)(n\minus{}2)...(n\minus{}k\plus{}1)$ is implied by the divisibility of $ (n\plus{}mk!)(n\minus{}1\plus{}mk!)...(n\minus{}k\plus{}1\plus{}mk!)$ for any integer m. Taking m large enough, we have that the cases when $ n< k$ are implied by the cases when $ n \ge k$.

There's a less interesting reason than that:  it's just equal to zero.

rofler's solution is worth examining.  What it means for $ {n \choose k}$ to be an integer is that, among the possible ways to pick an [i]ordered[/i] list of $ k$ elements out of a set of $ n$ elements, we can "ignore the order" and every element behaves the same way when we do this - that is, for every [i]unordered[/i] list of $ k$ elements there are exactly $ k!$ ways to turn it into an ordered list.  In other words, the action of $ S_k$ is free.  This isn't always the case, so it's good to appreciate what about this problem allows a free action.

You can also prove this fact by computing the greatest power of $ p$ that divides the numerator and denominator; if you do this correctly, you can prove [url=http://planetmath.org/encyclopedia/KummersTheorem.html]Kummer's Theorem[/url], a much stronger result.


algorithm
quadratics
Vieta
number theory proposed
number theory

[quote="hinhhoc273"]Prove that there are many solution in Z+ of equation: (x+1)/y +(y+1)/x=4.[/quote]
[hide="Solution"]
Notice that a solution $ (1,1)$ exists to this.  Now, we will show that if we can find a solution $ (x,y)$ with $ x\le y$ then we can repeat an algorithm to find a solution $ (a,b)$ so that $ x \plus{} y < a \plus{} b$.  Notice that we would be done after proving that such an algorithm exists since we can repeat the algorithm infintitely many times to create infinitely many solutions.  Now, upon expansion, the given statement is equivalent to $ x^2 \plus{} x \plus{} y^2 \plus{} y \minus{} 4xy \equal{} 0$, so $ x^2 \minus{} x(4y \minus{} 1) \plus{} (y^2 \plus{} y) \equal{} 0$.  This is a quadratic in $ x$, so it has two solutions for $ x$.  The second solution, by Vieta's Theorem, is \[ 4y \minus{} 1 \minus{} x\ge 4x \minus{} x \minus{} 1 \equal{} 3x \minus{} 1\ge 2x > x > 0\]  Thus, the solution $ (4y \minus{} 1 \minus{} x, y)$ works to the equation.  Notice that the equation is symmetric in $ x$ and $ y$, so the solution $ (y, 4y \minus{} 1 \minus{} x)$ works.  Yet, $ y \plus{} 4y \minus{} 1 \minus{} x > y \plus{} x$, so if $ a \equal{} y$ and $ b \equal{} 4y \minus{} 1 \minus{} x$, we have found the desired solution $ (a,b)$.  [/hide]


A simple pendulum 1 m long has a bob of 10kg.
a)How much work will be required to move the pendulum from its vertical position to a horizontal position(it's easy)
b)If the pendulum swings from a horizontal position, what will be the velocity and kinetic energy of the bob at the instant it passes through the lowest position of its path?

a) The gravitational field is a conservative vector field so the line integrals are independent of the path chosen.  Therefore, work is simply the product of force and distance.  The force is simply that to oppose gravity ($ F=mg$) and the distance is 1 m so $ W=98.0\mbox{ N\cdot\text{m}}$
b) It will travel a total distance of 1 m.  I'm assuming the pendulum starts from rest so $ v_{i}=0$.  Then
$ v_{f}^{2}=0^{2}+2g$       (I substituted in the values)
and so $ v_{f}=\sqrt{2g}$
Then $ KE=\frac{1}{2}mv^{2}=10g=98.0\text{ J}$
which is what we hoped it would be.


is there anyone familiar with good software for capturing and burning web streams onto cds. Supposedly quick time pro is not very good, whats good for windows media player....is there a software program that is good for all of the players such as real player, itunes and windows media player that has gotten good reviews? this would be good for recording science lectures off the net, so it relates to this subject and others on this site

I hope these two links will help you in a way :):
http://all-streaming-media.com/streaming-media-faq/faq-record-stream-windows-media-player.htm
http://all-streaming-media.com/streaming-media-faq/faq-record-stream-real-player-realone.htm


linear algebra
matrix
superior algebra
superior algebra solved

1) Let A\in M_4(C) prove that A can be written as a sum at most of 4 matrices of rank 1
2)Prove that I_4 cannot be written as a sum of less than four matrices
of rank 1
Manuela Prajea, Ion Savu, Judet 2003
---------------------------------------------------------------------------
1) rank(A)=r, A is equivalent to
diag(1,0,0,0)=E_1 or
diag(1,1,0,0)= E_1+E_2 or
diag(1,1,1,0)=E_1+E_2+E_3 or
diag(1,1,1,1)=E_1+E_2+E_3+E_4
where E(i,j) is the matrix with 1 at the i-row, j-column when i=j,
I wrote E(i,i)=E_i
there exist P,Q invertible in M_4(C) such that
A=PE_1Q or
=PE_1Q +PE_2Q or
=PE_1Q +PE_2Q +PE_3Q  or
=PE_1Q +PE_2Q +PE_3Q +PE_4Q since P,Q are invertible
rank(PE_iQ)=rank(E_i)=1

2) We study each case
If I_4 = C with rankC=1 contradiction since rank(I_4)=4

If I_4=C+D with rank(C)=rank(D)=1
I_4-C=D it is known that
|rank(I_4)-rank(C)| \leq |rank(I_4-C)=rank(D)| so 3 \leq 1 contradiction

If I_4= C+D+F with rank(C)=rank(D)=rank(F)=1
|rank(I_4)-rank(C)| \leq  |rank(I_4-C)=rank(D+F)| \leq rank(D)+rank(F)
4-1 \leq 1+1 contradiction

For (a): If A=/=0_4 we can write it as sum of matrices which are all 0 except for a certain line (which is the same as in A). If A=0_4 we can write  A=[1111]                [1111]
     [0000]    +   (-1)*[0000]
     [0000]                [0000]
     [0000]                [0000]


function
integration
real analysis
real analysis unsolved

The function $ f: [\minus{}1,1] \rightarrow \mathbb{R}$ is even and integrable.

$ a)$ Prove that $ I\equal{}\displaystyle\int^1_{\minus{}1} f(x) \arccos xdx\equal{}\frac {\pi}{2} \displaystyle\int^1_{\minus{}1} f(x) dx$

$ b)$ Compute $ J\equal{}\displaystyle\int^1_{\minus{}1} \frac {\arccos x}{1\plus{}x^2} dx$

Just take into account that $ \arccos(\minus{}x) \equal{} \pi \minus{} \arccos x$, then make the change of variable $ x \rightarrow \minus{}x$ and you're done.
At b) make $ f(x)\equal{}\frac{1}{1\plus{}x^2}$ and use what you've learnt at a).  :wink:

Divide [-1，1] into n subinteral denoted by [x(i),x(i+1)]，i=1,2,...,n.记M(i)=sup{f(x),x belong to [x(i),x(i+1)] }and m(i)=inf{f(x),x belong to [x(i),x(i+1)]}.
\displaystyle\int^1_{-1} f(x) \arccos xdx<=\sum_{i=1}^\{i=n} {M(i)\over n \int_i\over n^i+1\over n arccos x\,dx}=\frac {\pi}{2}\sum_{i=1}^\{i=n}{M(i)\over n}
to be samely we have \displaystyle\int^1_{-1} f(x) \arccos xdx>=\frac {\pi}{2}\sum_{i=1}^\{i=n}{m(i)\over n}
since f(x) is itegrable ,let n tends to infinite .wo draw that \displaystyle\int^1_{-1} f(x) \arccos xdx=\frac {\pi}{2} \displaystyle\int^1_{-1} f(x) dx


inequalities
geometric inequality
inequalities proposed

[color=purple]That is a long time I don't post in this forum. Here is a nice geometric inequality I hope you will like it
  $ \frac{r}{R}+\frac{(a^{2}-b^{2})^{2}+(b^{2}-c^{2})^{2}+c^{2}-a^{2})^{2}}{16S^{2}}\le \frac{1}{2}$
 :D  :D  :P [/color]

[color=purple] What do you think about this problem gemath ? :D [/color]

Nothing for me to think about it!  :D it is only easy algebraic problem by Ravi subtitude however I will help you to rewrite it  :D  and others will kill it for you, and note that it has a stronger version
\[ \frac{r}{R}+\frac{(a-b)^{2}c^{2}+(b-c)^{2}a^{2}+(c-a)^{2}b^{2}}{16S^{2}}\ge \frac{1}{2}\]
and algebraic version is
\[ \frac{(x+y)(y+z)(z+x)}{4xyz}+\frac{(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}+(x^{2}-y^{2})^{2}}{16(x+y+z)xyz}\ge \frac{1}{2}\]
which is trivial with mathlinkers in inequalities box  :)


function
LaTeX
analytic geometry
trigonometry
calculus
calculus computations

The function $ f: \mathbb{R}^{2} \minus{} \{0\}\rightarrow\mathbb{R}^{2}$ is given by $ f(x,y) \equal{} \left(\frac {x}{x^{2} \plus{} y^{2}},\frac { \minus{} y}{x^{2} \plus{} y^{2}}\right)$.

Show that $ f$ is bijective. Find the inverse of $ f$. What are the images of the circles in their central position.

$ \text{\LaTeX}$ tip: the parentheses you want are \left( and \right). Don't use \( and \) - that does things you don't want to happen. (I edited for you.) And to indicate what is in set $ A$ but not set $ B$ use A \setiminus B, not A - B. (I didn't edit that.)

Now as for the problem: the bijection is from $ \mathbb{R}^2\setminus\{(0,0)\}$ to $ \mathbb{R}^2\setminus\{(0,0)\}.$ The function is actually its own inverse. This is one instance in which the algebra is much easier to see if you use complex arithmetic and think of $ f$ as mapping $ \mathbb{C}\setminus\{0\}$ to $ \mathbb{C}\setminus\{0\}.$ Thought of that way, it has a very nice algebraic form - can you see it?

One side effect: even the images of non-central circles are nice.

The denominator hints at the use of polar coordinates, so that $ r\equal{}x^2\plus{}y^2$

[quote="tmrfea"]The function $ f: \mathbb{R}^{2} - \{0\}\rightarrow\mathbb{R}^{2}$ is given by $ f(x,y) = \left(\frac {x}{x^{2} + y^{2}},\frac { - y}{x^{2} + y^{2}}\right)$.

Show that $ f$ is bijective. Find the inverse of $ f$. What are the images of the circles in their central position.[/quote]
Set ${{ \alpha = \frac {x^2}{(x^2 + y^2)^2}} + \frac {y^2}{(x^2 + y^2)^2}} = \frac {1}{x^2 + y^2}$. Then:
$ f(f(x,y)) = f \left(\frac {x}{x^{2} + y^{2}},\frac { - y}{x^{2} + y^{2}}\right) = \left( \frac {\frac {x}{x^2 + y^2}}{\alpha}, \frac {\frac {y}{x^2 + y^2}}{\alpha}\right ) = (x,y)$. So $ f$ is it's own inverse.
If we consider $ z = x + yi$.
Then $ g(z) = \frac{x}{x^2+y^2} + i \cdot \frac{-y}{x^2+y^2} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{1}{x+iy} = \frac{1}{z}$ ;)

Yes, that complex form makes it pretty obvious why this function is its own inverse. And (like all linear fractional transformations) it maps [b]all[/b] circles or lines (not just circles centered at the origin) into circles or lines.

So we would use the same approach, just that $ z$ would be in polar coordinates for the following task:

$ f: \mathbb{R}^{2} \rightarrow\mathbb{R}^{2}$ and $ f(x,y) = (e^{x}\cos{y},e^{x} \sin{y})$

Let $ z$ be $ z=|x|(\cos{y}+i\sin{y})$

Then
\begin{eqnarray*}
f(x,y)&=&e^{x}\cos{y}+ie^{x}\sin{y}=\\
 &=&e^{x}(\cos{y}+i\sin{y})=\\
 &=&
\{
\begin{array}{cc}
 e^{|z|}\frac{z}{|z|} & \mathrm{if} \quad x \geq 0\\
e^{-|z|}\frac{z}{|z|} & \mathrm{if} \quad x < 0

\end{array}
\end{eqnarray*}

That one is very simply $ z\mapsto e^z.$ It is periodic of period $ 2\pi i.$ (Assuming $ z\equal{}x\plus{}iy.$)

I see $ \left( z = x + iy \rightarrow f(x,y) = e^{x}(\cos{y} + i\sin{y}) = e^{x}e^{iy} = e^{x + iy} = e^{z}\right)$.

Another interesting task for the of complex arithmetic would be:

Simplify the following sums:
\begin{eqnarray*} A_{n} = 1 + r\cos{\alpha} + r^{2}\cos{2\alpha} + \ldots + r^{n - 1}\cos{(n - 1)\alpha} & = & \sum_{k = 0}^{n - 1}r^{k}\cos{k\alpha} \\
B_{n} = r\sin{\alpha} + r^{2}\sin{2\alpha} + \ldots + r^{n - 1}\sin{(n - 1)\alpha} & = & \sum_{k = 0}^{n - 1}r^{k}\sin{k\alpha} \end{eqnarray*}


I thought about starting a problem of the day here, so I make up a problem and people answer it.

Problem. Four students each earn one dollar and fifty cents a day. Each apple at the local grocery store costs two dollars. At the end of a standard three-hundred sixty-five day year, they all go and spend their savings on apples. Once each person can buy no more, they combine their funds and buy more apples, and if they can't share them equally, they cut them.
(a) How many apples do they need to cut?
(b) How many apples total do they buy?
(c) Do (a) and (b) again, with the exception that they now go through a leap year in savings(366 days).

[hide]Ok, four students at $ \$$1.50 a day. Multiply that so, $ \$$1.50 times 365 equals $ \$$547.50. But they earn that each, so $ \$$547.50 times 4 equals $ \$$2,190. Divide that by the cost of apples ($ \$$2) which equals 1,095 apples. Divide that by 4, and that equals 273.75 apples. However, they won't get a fair share, so you have to cut up an apple into four slices. The total amount of apples is now 272 apples, and 7 apple slices. I'm kind of stuck at this point...Doing the math was just fun. Can someone also explain how to do the rest of it? (if I got the above information correct)[/hide]

[hide]Each student earns $ \$$547.50 in the year, which is 273 whole apples. Pooling the remainder of their money, there is a total of $ \$$6, or 3 apples. To divide these up, I think they would have to cut all 3 of them...maybe not.  Is the answer to part a [b]3[/b]?
(b)So, in total they bought (273*4)+3 apples = 1095 apples.
(c) don't really want to do it right now
[/hide]

@mathcountsfreak: Your answers are correct.

Can anyone do (c)?

C is the same as A and B only with 366 instead of 365.
[hide="Solution"]
Each person earns $ \$1.50*366$, or $ \$549$. This is 274 whole apples each. They end up with 4 more dollars, so that is 2 more apples. There is a total of $ 274*4\plus{}2$ apples bought, so they must $ \boxed {\text{cut } 2 \text{ apples and buy } 1,098 \text{ apples}}$ [/hide]


non asian gets -3; asian gets +2

starting off....

non-asian

-3

-16

@BOGTRO: Add 2 to the number in the previous post if you are Asian, and subtract 3 from the number if you are not Asian.
@#H34N1: So, Indian counts as non-Asian for the purposes of this game?

-12

The message is too small. Please make the message longer before submitting.


Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?

$\text{(A) }29 \qquad \text{(B) }42 \qquad \text{(C) }45 \qquad \text{(D) }47 \qquad \text{(E) }50$

Ohh! I remember this from last year!! I put "47" but I never found out from other people if it was right. Can someone explain how to do this because I think I got this wrong!!

Solution:

[hide]
Converting their speeds to "revolutions/minute" we see that they have the same revolutional speed of about .796 rev/min, which means that they take 1.256 minutes per lap.

This means that they will cross when they come back to the start line and at the point directly across the start line.

30/1.256 x 2 = 47.76...

We knock off the decimals because that just means that they were [i]about[/i] to cross again.

Answer is 47, D[/hide]


Solve the equation:

$ (x\minus{}2xy\minus{}y^2)y' \plus{} y^2 \equal{} 0$.

I have the solution in the book, but I'm stuck at the following thing:

I'll give the start also.

$ y\equal{}0$ is a solution. For $ y \neq 0$ we have $ y' \neq 0$ we will look at the equation having as unknown $ x(y)$. Hence the equation becomes: $ y^2 x' \plus{} x(1\minus{}2y)\minus{}y^2 \equal{} 0$.

Here I'm stuck. Why does the equation become $ y^2 x' \plus{} x(1\minus{}2y)\minus{}y^2 \equal{} 0$ if we consider it into $ x(y)$? I can't see how they come to this. Can someone do it step by step or explain a little bit in detail?

Please help. Thank you very much!  :blush:

It's a little algebra plus the fact that $ y' \equal{} \frac{1}{x'}$.  (This is just the chain rule applied to $ \frac{dy}{dx}$ and $ \frac{dx}{dy}$.)


I had a question in the "hard" section questions of my textbook. Here's a variation of the problem:

The probability of Bob's team winning when Bob is tired is $\frac{12}{25}$. The probability of Bob's team winning when Bob has lots of energy is $\frac{14}{25}$. The probability of Bob's team winning when Bob isn't there is $\frac{13}{25}$.

On any day, the probability of Bob being tired is $\frac{1}{2}$ (guess his teachers torture him). The probability of Bob having lots of energy is $\frac{1}{3}$. The probability of Bob not coming is $\frac{1}{6}$.

(1) On January 26, what is the probability that Bob's team won?
(2) If Bob's team won, what is the probability that he:
(a) was tired
(b) had lots of energy
(c) didn't come

[hide="Weird thing to notice about problem (if you are bored)"]If Bob doesn't come, the probability of them winning is greater than $50\%$![/hide]

Here is my solution, but I am not sure if it is right:
[hide]
1. The probability that Bob is tired is 1/2 and that the team will win when he is tired is 12/25.  This means that the chance that his team wins while he is tired is (1/2)(12/25)=6/25.  You can do a similar thing for the rest and add everything up.  So, for Bob having a lot of energy is (1/3)(14/25)=14/75.  Also, for not showing up is (1/6)(13/25)=13/150.  This means that when you add everything up, you get $\frac{13}{150}+\frac{14}{75}+\frac{6}{25}=\frac{36+28+13}{150}=\frac{77}{150}.$
2's.  Take 150 games played.  The expected number of wins is 77 (since the chance of winning is 77/150).  Out of these, the expected number of games that they would win with Bob tired is 36 (since 6/25=36/150).  Similarly we can calculate that they probably won 28 games while Bob was energetic and 13 games when he was not there.  Since the team won 77 games, and 36 of which were when he was tired, the probability that he is tired when they won would be $\frac{36}{77}$, when they he was energetic and they would win is $\frac{28}{77}=\frac{4}{11}$, the probability that he wasn't there when they would won is $\frac{13}{77}$. [/hide]

that's a cool way to solve #2. i think the answers are correct.


geometry
geometric transformation
geometry solved

I would politely ask grobber (or anyone else), to please prove this "easy-to-prove lemma":

Given a center T and pairs of points (U,U'), (V,V') s.t. there is a similarity of center T which takes U to U' and V to V', then if UU' $\cap$ VV' = P, the circles (UVP),(U'V'P) intersect again at T.

It is just an angle chase, man. The angle between $UU'$ and $VV'$ is equal to the angle between $TU$ and $TV$, because $TVV'$ is obtained from $TUU'$ through a spiral similarity centered at $T$. From here you get that $T,U,V,P$ are concyclic. Same for $T,U',V',P$.


Prove that if all $ x_i$ are positve, but less than 1, and $ \displaystyle\sum_{i \equal{} 1}^{n}x_i \equal{} \frac {1}{y}$, then $ \displaystyle\prod_{i \equal{} 1}^{n}\frac {1 \minus{} x_i}{1 \plus{} x_i}\ge \frac {y\minus{}1}{1 \plus{} y}$.

EDIT: Sorry, Jiang, you are correct.  My post has been edited.

By induction on $ n$,
the inequality is trivial in case $ n\equal{}1$.
Assume $ \prod^{n}_{i\equal{}1}\frac{1\minus{}x_i}{1\plus{}x_i}\geq\frac{y\minus{}1}{y\plus{}1}$
First, $ \sum^{n\plus{}1}_{i\equal{}1}x_i\equal{}\sum^{n}_{n\equal{}1}x_i\plus{}x_{n\plus{}1}\equal{}\frac{1}{y}\plus{}x_{n\plus{}1}\equal{}\frac{1}{y/(1\plus{}x_{n\plus{}1}y)}\equal{}\frac{1}{y'}$.
Then, $ \prod^{n\plus{}1}_{i\equal{}1}\frac{1\minus{}x_i}{1\plus{}x_i}\equal{}\prod^{n}_{n\equal{}1}\frac{1\minus{}x_i}{1\plus{}x_i}\times\frac{1\minus{}x_{n\plus{}1}}{1\plus{}x_{n\plus{}1}}\geq\frac{y\minus{}1}{y\plus{}1}\times\frac{1\minus{}x_{n\plus{}1}}{1\plus{}x_{n\plus{}1}}\equal{}\frac{y\minus{}1\minus{}x_{n\plus{}1}y\plus{}x_{n\plus{}1}}{y\plus{}1\plus{}x_{n\plus{}1}y\plus{}x_{n\plus{}1}}$
Note that $ y\minus{}1\minus{}x_{n\plus{}1}y<y\plus{}1\plus{}x_{n\plus{}1}y$, so $ \frac{y\minus{}1\minus{}x_{n\plus{}1}y\plus{}x_{n\plus{}1}}{y\plus{}1\plus{}x_{n\plus{}1}y\plus{}x_{n\plus{}1}}\geq\frac{y\minus{}1\minus{}x_{n\plus{}1}y}{y\plus{}1\plus{}x_{n\plus{}1}y}\equal{}\frac{y'\minus{}1}{y'\plus{}1}$


limit
real analysis
real analysis solved

Suppose that $\lim u_n=l \in \mathbb{R}$.
Prove that : $\lim_{n\rightarrow+\infty} \frac{1}{n}\sum_{k=1}^{n}(-1)^ku_k=0$.

for example use cesaro for $u_{2n}$ and $u_{2n+1}$ and sum.


geometry
MATHCOUNTS
algebra
polynomial
trigonometry
quadratics
group theory

Anyone plan on going to the JHMT on Nov 8th?  :)

I'll be there.

I've been so busy with college applications that I haven't done much math lately . . . I'd better practice or I might not win.  :o

You might have a better chance than I do since I haven't studied math for two years.

Seriously: I looked at last year's problems and I was like :help:

[quote="7h3.D3m0n.117"] I haven't studied math for two years.[/quote]

 :huh:

Actually, JHMT problems are often pretty accessible even if you haven't studied math recently - a lot of them depend on one trick and are otherwise fairly easy.

Oh shoot - we have a Tree Planting event that I DEFINATELY have to go. But I already signed up for the JHMT.  :(

I'm wondering if we can go to JHMT, sign in, drive back to the tree plant, sign in, plant a seed or two, and drive back and hopefully not completely miss a few rounds.

Sigh...

I doubt they'd want you to do that, and I highly doubt your team would want you missing the team rounds.  Are you sure you can't skip the tree planting?

Skipping the tree planting is like getting on the bad side of a bunch of seniors, plus putting my $ \$15$ membership dues to waste.

Skipping the JHMT is like getting on the bad side of my [potential] calc teacher, pissing off my teammates, and putting their $ \$5$ tournament entry fee to waste.

I'm not sure what I can do. I don't have to spend the entire day at tree planting, but I have to do an hour at least. I'm just trying to see if I can compromise and what's the best time to tree-plant.

Thanks to everyone involved with running JHMT for another great tournament!  I'm glad that other strong teams showed up this year - while winning every year is not terrible, it is nice to have some of the other top students from the region to compete against.  Congratulations to Matt Superdock for winning!

A few points: as I'm sure everyone has realized, the Intellectual Whirlwind simply does not work with that many teams.  I understand that this year you didn't realize there would be so many, but please remove this round for next year.  Go back to the nongraded fun rounds that were present previously, or if you want to be hardcore, try a Guts Round.

I have mixed feelings about the Synergy Round.  It is nice to have a different kind of round from what is normally seen, but I don't really find that kind of emphasis on puzzle-solving in a math tournament appropriate.  The puzzles were interesting, though.

The Algebra Round definitely had some very nice problems on it.  #10 was great.  I'm interested to see a non-brute force method to #9, though.

On the other hand, I found the Combinatorics problems a bit uninspiring, although #10 is probably cool. I just didn't see how to do it.

Overall, I enjoyed it.  I'm sad that this is my last year to participate.  Thanks for the hard work you put into this every year.

Synergy was okay. I'm thinking about writing a brute-force program for the Number Play one. Did anyone find a easy way to do Numbers in Networks?

The Algebra Round seemed rather easy. I'm still miffed that I didn't get #10, probably because I combined everything into one big fraction, and I'm still mad that I missed one solution for #9 and missed getting a perfect score.

Geometry and Combinatorics was fun, although I neglected the last name for #7. There was an argument about #8 with the ambiguous wording with a fellow teammate; it worked out for both of us.

Intellectual Whirlwind was fun to watch, but it did take too long. I did like some of the problems like the Snakes on a Plane one, the Jackie Chan one, and the NOOB one.

I'm disappointed I didn't place in the top 5 for Algebra, even though I'm certain I should have. In any case, it was very fun.

Yeah, I have no idea why you didn't place on Algebra. I'm sure you tied with me.  I was getting pretty worried that they didn't call us to do a tiebreaker - I thought that maybe 83 was out of the top 5.

I also forgot the last name for Combo #7.  Would have been in the big tiebreaker for 2nd if I hadn't.

I'm glad they gave us both interpretations for Combo #8.  That really was ambiguous.

And yes, I'm still mad at Numbers in Networks.  That puzzle is EVIL.

I actually skipped tree planting to be at the JHMT.

The Exploration Round with the financial mathematics was nasty. My team gave up halfway through and started a tic-tac-toe tournament.

Synergy round was pretty fun. Although the problem with the number grid where we had to fill in with numbers 1-9 to sum up to the outside numbers was gay

Intellectual Whirlwind was pretty fun - it reminded me of the old Countdown system of MATHCOUNTS, but yes, it took way too long. In fact, I didn't even stay for the whole round (but I did get to be my team speaker) because it took so long and my parents were already waiting.

Algebra: this section was evil. I spend the majority of my time on two problems and I got a couple of the easier ones wrong because I didn't know one of the trick polynomial factorizations.

As for combinatorics, I didn't know how to simplify the problem with the "naming the baby." I simply left my answer as a summation (I mean, who has the time to multiply $ 25 \times 24 \plus{} 24 \times 23 \dots$?) and hope my grader had a sense of humor.

For naming the baby, you can just do $ \binom{25}{2}$.  Once you have chosen two distinct letters, there is only one way to order them.

Hm, the Exploration Round was actually okay if you did things right, except for the last problem, which was horrendous.

By the way, I figured out a way to do Algebra #9 without brute force!
[hide]
We have

$ \sin x\plus{}\cos x\plus{}\sin 2x\plus{}\cos 2x\equal{}0$.

So

$ \sin x\plus{}\cos x\plus{}2\sin x\cos x\plus{}1\minus{}2\sin^2 x\equal{}0$.

Let $ a\equal{}\sin x and b\equal{}\cos x$ for simplicity.  So

$ 2a^2\minus{}(2b\plus{}1)a\minus{}(b\plus{}1)\equal{}0$.

Use the quadratic formula to find $ a$:

$ a\equal{}\frac{2b\plus{}1\pm \sqrt{4b^2\plus{}4b\plus{}1\plus{}4(2)(b\plus{}1)}}{4}\equal{}\frac{2b\plus{}1\pm \sqrt{4b^2\plus{}12b\plus{}9}}{4}$

Thus

$ a\equal{}\frac{2b\plus{}1\pm (2b\plus{}3)}{4}\Rightarrow a\equal{}b\plus{}1$ or $ a\equal{}\minus{}\frac{1}{2}$.

The first gives $ x\equal{}\frac{\pi}{2}$ and $ x\equal{}\pi$.  The second gives $ x\equal{}\frac{7\pi}{6}$ and $ x\equal{}\frac{11\pi}{6}$.
[/hide]

I spent about 5 minutes doing the Synergy puzzle with inserting numbers to sum to the outside numbers.

Note that the second column is 2, and that the top row is 24. The two middle rows are then 3 and 4, so solve those rows, then solve the first column, then solve the last row/column.

I liked our last statement for the last problem in Exploration. I wonder how much credit we got for it.

Intellectual Whirlwind would have easily been faster had they done it in the individual rooms after Double Challenge or separated into smaller subgroups so multiple teams could have done it simultaneously; it would have taken an hour maximum at a guess and we could have gotten out half an hour earlier.

They haven't been released yet by the staff at JHU. Unless someone has uploaded them online, there currently aren't any copies available online.

[quote="buzzer11"]where can i find the JHMT problems?[/quote]
You can contact one of the staffs of the JHMT and let them know you want these problems. They probably will send all of them to you(I did get it on that way). :)
By the way, can anybody let me know what is the top 3 scores of the Algebra part in this year? I think these 10 problems are easier than they were last year, especially #9 and #10. I already miss one problem(#6) since I'm still sucked at counting and probability. I don't have enough time to work on other parts though. To be honest, the unlimited Round's problems look really bored. :(

I was third on Algebra with an 83.  First place was a perfect 100.  I don't remember second place, although I know it was strictly between 83 and 100, there weren't any tiebreakers.

If I remember correctly, 2nd place was an 85.

[quote="7h3.D3m0n.117"][quote="E^(pi*i)=-1"]For naming the baby, you can just do $ \binom{25}{2}$.  Once you have chosen two distinct letters, there is only one way to order them. [/quote]

I think you're overcounting - I thought the first two letters had to be in alphabetical order and the last one was constant?[/quote]

The first two letters did have to be in alphabetical order.

Each combination can only be ordered alphabetically in one way, so since there are $ \binom{25}{2}$ combinations (25 letters since Z cannot be used), that is also the answer.

whoa why didn't Matt Superdock come last year, when i was there.

who were the top teams?

Hi, I'm the problem director so if you have any particular questions, contact me and I can send you the questions (I have the solutions but I may not be allowed to release them yet, but if you want me to solve a problem, I will).

This year, I happended to think that the algebra round was particularly easy.  We originally had another batch of questions that were apparently too hard to give.  With regards to the financial economics round, as another poster pointed out, it was brutal.  The round was pretty difficult, but a lot of people lost stupid points.

First place team was Howard Area Homeschoolers Fighting 42's ( :D )

Second place was Eleanor Roosevelt's top team.

After that I don't remember.

Aww the Jr. Einstein's didn't place?  :P

@JRav: which official were you? I saw like three of them on Saturday.

I actually didn't spend too much time in the exam rooms since I spent a lot of time grading papers.  For the brief moments when I was in the Glass Pav, I was the one with the blue backpack, holding the Complex Analysis book.  Hopefully Daniel graduates this year so I can get a higher position. :D

Personally, when I saw the tests, I thought they were pretty easy and I'm not surprised to see such high scores.  For next year, do you think we should increase the difficulty a little bit, or do you think it was hard enough?  I have to admit, the Explorations Unlimited round might have been a bit difficult for high school students.  I graded the first page, which should have been the easiest.  Unfortunately, the highest score on my page was 14/16, with less than five 12's, a few more 11's, and a majority in the 4-8 range.

Feedback please.

The individual rounds could have been harder.  It seemed like there was a maximum of one question per topic that required serious thinking.

I didn't think Exploration Unlimited was too hard.  Most of the questions were straightforward.  I do think that the topic wasn't a great choice - I think that the scores would have been higher had it simply been more interesting.

About the Jr. Einsteins: Heh, weren't they awesome?  I loved how everyone suddenly got quiet as they started ripping through the questions.

[quote="E6(pi*i)=-1"]I didn't think Exploration Unlimited was too hard. Most of the questions were straightforward. I do think that the topic wasn't a great choice - I think that the scores would have been higher had it simply been more interesting. [/quote]
When I read through them, I didn't think they were that hard (but I'm in college).  However, had you seen some of the scores, you certainly would be thinking differently.  With the exception of maybe 3 teams, most of the scores were pretty bad.  I'll try to make the round more interesting next year.

The Exploration topic for next year should be much more interesting. I think that part of the reason why my team didn't work the last half of that round was due to the complications and difficulty of the subject.

Personally, for the Individual rounds, I thought they were hard enough, but don't listen to me since I haven't done competitional math in 2 years (disregarding JHMT last week). I did not, however, like the Algebra round, since we didn't get scratch paper! I had to write in small margins (not a fan of erasing) and this screwed some of my calculations.

(And this part is just me whining: I'm left handed and writing on those desk-chairs drove me nuts.)

There was no scrap paper?  That's mismanagement so I have no idea why that happened.  I'll try to find out for you.  There should have been a few left-handed desks in the room but if not, we apologize.

I have some ideas for next year's Explorations Unlimited round.  Whether or not they are implemented remains to be seen.  It's hard to make that round because we need to take a difficult subject and make it understandable for high school students, so we're somewhat limited in what we can do.

You had to specifically request scrap paper for the Algebra round in our room. They handed it out before Double Challenge.

I saw a few left-handed desks in that room. I almost sat down at one.

I have a question regarding #10. I wrote down an answer of [hide]$ \displaystyle\frac {\pi^2 \minus{} 9}{3}$[/hide] and I'm under the impression that it was marked wrong because I did not place in the top 5, yet I had the same correct answers as a fellow teammate who placed 3rd. Would such an answer be marked wrong?

Hey guys!  This is Matt Superdock if you didn't know, haha  :lol:

I had a really good time at the competition.  It seemed to drag on a little long, but it was extremely well-run compared to some of the local competitions in the Philadelphia area.  The synergy round is a really good idea IMO, although I was disappointed that a lot of the questions seemed to rely almost entirely on guessing and getting lucky.  The networks/numbers problem and the triangles problem were both really annoying (sorry), but I really liked the trap the diamond problem.  I also liked the idea of the second round (whatever it was called), but I didn't especially like the topic.  Still, it's cool to have some unique rounds that other competitions don't have.

Overall, I liked the problems from the individual rounds.  A few of them were really interesting.  I thought I had solved all the problems coming out of the test, but I ended up missing the naming problem because I didn't read the problem well enough and also the last combinatorics problem, even though I still don't know what I did wrong.  I'm looking forward to seeing a solution to that one.  :)

I was really amazed at the amount of applause during the award ceremony.  That was really cool.  :)

I can't come back next year (I'm a senior), but Radnor probably will!

BTW, better way to do alg #9: Write $ \sin x\plus{}\sin 2x$ as $ \sin (3x/2\minus{}x/2)\plus{}\sin (3x/2\plus{}x/2)$, and the same for cosine.  It comes out nicely.

I did recall seeing a few left-handed desks during the individual rounds, but during team rounds, where the rooms were smaller, had a less likely chance of left-handed desks. (But this is a trivial matter).

As for scrap paper, I didn't realize we had to request it for the Algebra round. Sorry!

[quote="7h3.D3m0n.117"]I didn't realize we had to request it for the Algebra round.
[/quote]

They didn't tell us that actually. It was a good thing I had scrap paper from Exploration Unlimited/Synergy. I saw people asking for scrap paper though. I know some people were complaining in our room because they wrote all their work in tiny font on their question paper.

I think scrap paper should have been provided so don't worry about that.  I was grading papers at the time so I'm not sure who was in charge of your room.

Who were the winners?  I wasn't able to stay for the results.

Here's what I remember of the individual results:

Algebra:

1. Matt Superdock (Radnor)
2. Holden Lee (Eleanor Roosevelt)
3. Alex Kandell (Howard Area Homeschoolers)
4. ???
5. ???

Geometry/Combinatorics:

1. Holden Lee (Eleanor Roosevelt)
2. Matt Superdock (Radnor)
3. Dennis Won (Langley)
4. Jonathan Horton (Fairfax)
5. Jacob Pritt (Frederick/Walkersville)

Overall:

1. Matt Superdock (Radnor)
2. Holden Lee (Eleanor Roosevelt)
3. Dennis Won (Langley)
4. Alex Kandell (Howard Area Homeschoolers)
5. ???

About scrap paper: bring your own!  It works quite well.  :wink:

About Exploration Unlimited: I don't think it should be too much easier . . . otherwise, it won't be challenging for top teams.  The winning scores were pretty high as it was.

No, he's always lived in Frederick.  But they started their own team this year.

[quote="JRav"][quote="darkdieuguerre"]I have a question regarding #10.[/quote]
Your answer is certainly correct.  Whether or not you were given credit for it, I don't know.  One of the coaches contacted us about this, so maybe it was yours.  I believe you would have placed, but I'm not sure.  Sorry.[/quote]
JRav, thank you for answering.  It's much appreciated.  That coach would be me.  It did look like he had the correct answer.  If it turns out that it changes his score, will that change the standings?  I do realize it's too late to do a tie-breaker, but perhaps both students could be given the award.

I thought you all did an excellent job on the tournament.  The Johns Hopkins contest has always been one of my favorites.  And congratulations on how much it's grown!  JHU is making a name for itself in high school math tournaments.  :-)

Catherine Asaro
Coach, Howard Area Homeschoolers

Alex, next year I think we'll have to retire the name The Fightin' 42's.  You and MathDancer were really the core of the team, and when you've both graduated, it wouldn't seem right to keep using the name.

And the Jr. Einsteins were indeed delightful.  Gabby has also qualified for Part II of the UMD Contest.  (!)  It turns out she's not quite the youngest ever, though.  Richard McCutchen was a year younger, in sixth grade, when he first qualified.

[quote="chess64"]Jacob moved?  or what[/quote]

His Dad is working on building a math club for the students out in Frederick, not just homeschoolers, but in public and private school, too.  Jacob still does a little with us, but mostly he's working with his Dad now.

It's too bad they had to leave from the JHU tournament.  Jacob (Mark) qualified for one of the tie-breakers, but he wasn't able to do it because they had to go, I think.

And hey!  You guys need to start a UMBC contest (like you haven't heard that before :-D).

Best,
Catherine

[quote="E^(pi*i)=-1"]About Exploration Unlimited: I don't think it should be too much easier . . . otherwise, it won't be challenging for top teams. The winning scores were pretty high as it was.[/quote]
Well not necessarily easier, but certainly we'll try to make it something you all enjoy.  I have some ideas that I think are really interesting (or at least they interested me when I was in high school, and still do), but they may be a little challenging.


geometry
3D geometry
tetrahedron
circumcircle
geometry unsolved

Find out whether it is possible to determine the volume of a tetrahedron knowing the areas of its faces and its circumradius.

What a stupid problem - it is NOT possible!

Consider a rectangular parallelepiped with sides $a,b,c$. Consider the tetrahedron you get by choosing four of its vertices that are pairwisely no neighbors.

It is clear that all its areas are the same. One may calculate that they are something like $a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}$ (upto squaring and maybe a constant). the circumradius is something like $a^{2}+b^{2}+c^{2}$. But it is somehow clear that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$ and $a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}=(ab+bc+ca)^{2}-2abc(a+b+c)$ do not determine $abc$ ($a+b+c$, $ab+bc+ca$, $abc$ are independent).


Basically, you just create a random cipher and the next person deciphers it. I'm wondering whether this is too complicated...
Example: EYDSNPMACTEGRESE
E Y D S
N P M A
C T E G
R E S E

ENCRYPTEDMESSAGE

Has anyone ever read The Adventures of the Dancing Men by Sir Arthur Conan Doyle?

That has a great cipher. Be warned, the Dover Thrift edition has a mistake in the code.

Hmm, you can make some pretty complicated ciphers and make it practically possible to figure out. Maybe there should be some restrictions on what you can do for the cipher.

lol the adventures of a dancing man had only a monoalphabetic substitution cipher


Check out this link
http://www.libfind.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2005-aa/05answersaime1.html
it was from a 2 year old forum post here. What are people's opinions about the AMC makeover?

Maybe not opinions, just comments..

edit:
http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2001-8a/01answers8.html
is the page from 2001.

I think it was actually nicer in 2001 than 05 but nicest now.

Yeah, same.
In order of preference:
1. 2007
2. 2001
3. 2005


function
trigonometry
limit
algebra
functional equation
algebra unsolved

Find all functions $f: R\rightarrow R$ such that:
$f(x).f(y)-f(x+y)=sinx.siny$
I know the solution that $f(x)=cosx$ but i don't find a completely solution. Please help me?

[Edited]

Even with the change of variable $f(x)=\cos (g(x))$ (which is already a restriction), it's not that easy.

But at least it made me notice there are at least the two solutions:
$f(x)=\cos (x)$ and $f(x)=\cos (-x)$.

In fact, if $f(x)$ is a solution, $f(-x)$ must be a solution too.

i got the whole solution
[hide]first put $y=0$
we get $f(x)f(0)-f(x) = 0$
which shows $f(0) = 1$
now it the main equation we put y = h
we get $f(x)f(h)-f(x+h) = sinx.sinh$
now we apply the limit h->0
hence we can say that $f(h) = 1$ $sinh = h$
hence we can write equation as $f(x)-f(x+h) = sinx.h$
we rearrange the equation as $\frac{f(x+h)-f(x)}{h}=-sinx$
now L.H.S is $f`(x)$. now we got the differential equation
$f`(x) =-sinx$. Integrating both the sides we get $f(x) = cos(x)+c$ now we use that $f(0) = 1$ we get $f(x) = cos(x)$[/hide]

Nice!
But you assume that $f$ is continuous at $0$,
$\lim_{h\to 0}f(h)=f(0)=1$.
Maybe it can be proved.

lordWings has proved that $f(-h) = f(h)$ hence we can say it is continious at x =0 .

Not every even function is continuous at $0$!

By the way, I didn't mean $f(x)=f(-x)$.
I only meant to say that if $f(x)$ is a solution, then the other function $g(x)=f(-x)$ is a solution too (proved substituting $x$ by $-x'$ and $y$ by $-y'$ in the functional equation).

I'm not sure if it's easy to prove $f(x)=f(-x)$.

It would be useful, because, taking $y=-x$,
$f(x)f(-x)-1=-\sin^{2}(x)$,
$f(x)f(-x)=\cos^{2}(x)$.

I'm very sory about my mistake. This problem have continuous condition. Thank you for your solution

Thank you. In this case, the solution is complete.
ashwinrk_jain proved that the only solution that is continuous at $x=0$ is $f(x)=\cos x$.

[quote="ashwinrk_jain"]
...
we get $f(x)f(h)-f(x+h) = sinx.sinh$
now we apply the limit h->0
hence we can say that $f(h) = 1$ $sinh = h$
hence we can write equation as $f(x)-f(x+h) = sinx.h$
we rearrange the equation as $\frac{f(x+h)-f(x)}{h}=-sinx$
...
[/quote]

I think this is almost a complete solution, but still not quite complete,
because it assumes differentiability of $f$ in $0$:

If the above argument is written down carefully, it reads as follows:

$\forall h \neq 0: \frac{1}{h}[f(x)f(h)-f(x+h)] = \sin x \frac{\sin h}{h}$
$\Rightarrow \lim_{h \to 0}\frac{1}{h}[f(x)f(h)-f(x+h)] = \sin x \lim_{h \to 0}\frac{\sin h}{h}= \sin x$
$\Rightarrow \sin x = \lim_{h \to 0}\Big[ \frac{1}{h}[f(x)-f(x+h)]+\frac{1}{h}[f(x)f(h)-f(x)] \Big]$
 $= \lim_{h \to 0}\Big[ \frac{f(x)-f(x+h)}{h}+f(x) \frac{f(h)-1}{h}\Big]$

In order to continue we now have to distribute the limit to both summands.
This however is possibly only if at least one of the limits exists.

The existence of the first limit means that $f$ is differentiable in $x$,
the existence of the second limit means that $f$ is differentiable in $0$.
So mere continuity of $f$ does not help us here to continue the solution,
and we either have to prove that the  given equation implies the differentiability
of $f$ in $0$, or we have to find a completely different solution...  :wink:

[quote="lordWings"][quote="solyaris"] ...
$\sin x= \lim_{h \to 0}\Big[ \frac{f(x)-f(x+h)}{h}+f(x) \frac{f(h)-1}{h}\Big]$[/quote]
But [b]if[/b] there's continuity at $x=0$,
$\lim_{h \to 0}f(h)=f(0)=1$ and then you have
$-\sin x= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$,
so the function is differentiable because the limit exists.[/quote]

I agree that $\lim_{h \to 0}f(h)=1$ by continuity,
but this does not imply $\lim_{h \to 0}\frac{f(h)-1}{h}=0$.
The numerator and the denominator both go to zero,
so the limit may be anything, and may even not exist.

You're right.  :blush:
Even if we have differentiability at $x=0$,
$\sin x= \lim_{h \to 0}\Big[ \frac{f(x)-f(x+h)}{h}+f(x) \frac{f(h)-1}{h}\Big]$,
$\Rightarrow\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=-\sin x+f(x)f'(0)$,
the function is differentiable for all $x\neq 0$ too because the limit exists, but then we have
$f'(x)=-\sin x+f(x)f'(0)$.

So, we still have some work to do.

I think we had better tkinking of continuity of the function. And the function can solve easyly by change this one by another function so reduce to Cauchy function equation

Here's another thought.
The only odd function that is a solution from the functional equation is $f(x)=\cos x$.
Notice that the proof doesn't involve continuity.
[hide="Proof"](a) $y=0$ $\to$ $f(x)(f(0)-1)=0$ $\to$ $f(0)=1$,
because $f(x)=0$ is not a solution to the functional equation.

(b) $y=-x$ $\to$ $f(x)f(-x)-f(0)=-\sin^{2}x$ $\to$ $f(x)f(-x)=\cos^{2}x$,
because $f(0)=1$ from (a).

(c) $y=x$ $\to$ $f^{2}(x)-f(2x)=\sin^{2}x$.

(d) = (b) - (c) $\to$ $f^{2}(x)-f(2x)-f(x)f(-x)=\cos^{2}x-\sin^{2}x$
$\to$ $f(x)(f(-x)-f(x))+f(2x)=\cos (2x)$.

So, if $f$ is odd, that is $f(x)=f(-x)$,
from (b) we already had $|f(x)|=|\cos x|$,
from (d) we have $f(2x)=\cos (2x)$.
[/hide]Can we prove that $f$ must be an odd function?

[quote="lordWings"]But at least it made me notice there are at least the two solutions:
$f(x)=\cos (x)$ and $f(x)=\cos (-x)$.too.[/quote]

Are these functions different?

No, of course not. :P
This is what I meant:
I was looking for $g(x)$ such that $f(x)=\cos (g(x))$ was a solution to the functional equation.
(It's a restriction because it assumes $|f(x)|\le 1$.)
And there are at least two different solutions for $g(x)$:
$g(x)=x$ and $g(x)=-x$.
(I got them with another restriction: assuming $g(x)$ is a solution to Cauchy's functional equation.)
Of course, both $g(x)$ lead to the same $f(x)$.

[Corrected]
Hello. I think I found short and nice solution without differentiability (I hope such word exists  :lol: ). OK the key idea is to show that exist $x$ such that $f(x)=0$.
First step is to show that $f(0)=1$. It can be achieved as you did before.
Next putting $y=-x$ we get $f(x)\cdot f(-x)-f(0)=\sin x\cdot\sin(-x)$ or $f(x)\cdot f(-x)=\cos^{2}x$. Then pluging $x=\pi/2$ we get $f(\pi/2)\cdot f(-\pi/2)=0$. So indeed $f(\pi/2)=0$ or $f(-\pi/2)=0$. Let us say $f(\pi/2)=0$. Assuming $y=\pi/2$ into main equation we get $f(x+\pi/2)=\sin x$ and after argument shifting $x \to x-\pi/2$ $f(x)=\cos x$ is achieved. Dealing with $f(-\pi/2)=0$ returns the same results.

[Thanks to hien for spotting out the mistyping]

Yes, very beatiful. I has checked carfully but didn'd find any error. Note that there is a mistake of typing.
Then pluging $x=\frac{\pi}{2}$ we get
 $f(\frac{\pi}{2}).f(\frac{-\pi}{2})=0$.


algebra
polynomial
trigonometry
algebra solved

Prove that the polynomial $x^n + 4$ can be written as the product of two polynomials (each with degree less than $n$) with integer coefficients if and only if $n$ is a multiple of 4.

One implication is trivial: $x^{4k}+4=(x^{2k}-2x^k+2)(x^{2k}+2x^k+2)$.

Assume now that $x^n+4=f(x)g(x)$. All roots of $x^n+4$ (and thus of $f,g$) have modulus $\sqrt [n]4$. This means that the free terms of $f,g$ must either be both $-2$ or $2$ (none of them can be $\pm 1$).

Assume $f,g$ have degrees $m,p$ respectively. Then $\sqrt[n]4^m=\sqrt[n]4^p=2\Rightarrow 2^{\frac {2m}n}=2^{\frac {2p}n}=2\Rightarrow \frac {2m}n=\frac {2p}n=1$. This means that $n$ is even and $m=p=\frac n2$. Now assume $m=p$ is odd. The roots of $x^n+4$ can be partitioned in two sets with $m=p=\frac n2$ elements each s.t. their products are equal (both $\pm 2$). This can easily be shown to be impossible considering that the roots are $\sqrt[n]4(\cos\frac {(2k+1)\pi}n+i\sin\frac{(2k+1)\pi}n),\ k\in\overline{0,n-1}$ and assuming $n=2(2t+1)$. Even more can be done, we canshow that the product of any $\frac n2$ roots (if \frac n2$ is odd) isn't real.


[quote="tdl"]Let $ a,b,c > 0$. Prove that:
$ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})^2\geq\frac32(\frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c})$[/quote]

$ \Leftrightarrow \sum_{cyc} \frac{a^2}{b^2} \plus{}2\sum_{cyc} \frac{a}{c} \ge \frac32\left((\frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c}\right)$

$ \Leftrightarrow 2\sum_{cyc} \frac{a^2}{b^2} \plus{}\sum_{cyc} \frac{a}{c} \ge 3\sum_{cyc} \frac{a}{b}$

By AM - GM inequality:

$ \frac{a^2}{b^2} \plus{}\frac{a^2}{b^2} \plus{}\frac{b}{a} \ge 3\frac{a}{b}$

$ \frac{b^2}{c^2} \plus{}\frac{b^2}{c^2} \plus{}\frac{c}{b} \ge 3\frac{b}{c}$

$ \frac{c^2}{a^2} \plus{}\frac{c^2}{a^2} \plus{}\frac{a}{c} \ge 3\frac{c}{a}$

Adding up then we are done!


Let $ x,y,z$ natural numbers so that $ xyz \equal{} 78$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 206.$ What is $ x\plus{}y\plus{}z$?

A. 18
B. 20
C. 30
D. 42
E. None of these

[hide]I tried doing manipulation, but it didn't work.  So I tried factoring 78.

$ xyz\equal{}78\equal{}(3)(2)(13)$

This limits the possibilities significantly.  We see that one of the three numbers must be 13 because if it were 26, 39, or 78, the sum of squares would be too big.  Thus $ z\equal{}13$ and $ x^2\plus{}y^2 \equal{} 206\minus{}13^2 \equal{} 37$.  Just looking at the numbers, it becomes evident that one of $ x,y$ is 6 and the other is 1.

$ x\plus{}y\plus{}z\equal{}13\plus{}6\plus{}1\equal{}20\equal{}>B$[/hide]

[hide="Why Manipulation Is a Bad Idea"]

The problem tells us that $ x,y,z$ are all natural numbers. So, let's think simple and break 78 down. Very easily, we see that 2*3*13 = 78. So, simple trial-and-error or two tells us that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1 \plus{} 36 \plus{} 169$, not necessarily in that order but it doesn't matter. So, $ x\plus{}y\plus{}z \equal{} 1\plus{}6\plus{}13 \equal{} 20$. B [/hide]


search
number theory unsolved
number theory

Can anybody please guide me to a set of good problems on the Chinese Remainder Theorem?

what do you mean with "guide"? give you problems? links? a summary?
try to search for this keyword here...

Naphthalin


Find $x,y>0$ such that
$\frac{1}{2}(x+y)=\sqrt{3y}-\frac{1}{\sqrt{x}}$

 :)

write as $(\sqrt{x}+1/x)^2+(\sqrt{y}-\sqrt{3})^2=3+1/x^2$, see that $(1,3)$ is a solution and then argue that it's the only possible value for $x$ based on how the function increases/decreases, like $(\sqrt{x}+1/x)^2>3+1/x^2$ for positive $x$ except for $x=1$.

Yes,that is!! :)
it can be solved either using your method or bringing in mind inequalities......

$\frac{2}{\sqrt{x}}+x=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}+x\geq 3$

while $2\sqrt{3y}-y\leq3$..... :)


geometry
3D geometry
tetrahedron
ratio
AMC
AIME

Create a formula for the volume of a tetrahedron. :D
this should be a lot of fun...

if this question isn't challenging enough:

say there is a tetrahedron. another tetrahedron can be formed by connecting the centers of each of the faces of the tetrahedron. The ratio of the volumes of the larger tetrahedron to the smaller tetrahedron can be expressed as m/n, in which m and n are relatively prime. Find m+n

[quote="8parks11"]i know
$\frac{\sqrt2}{12}*s^{3}$

what do you mean create?
make a new one?[/quote]
He never said regular.


Let \[p(x) = x^{n}+a_{n-1}x^{n-1}+...+a_{0}\] be complex polynom. Prove that all his zeroes are inside the circle \[|x| = 1+max|a_{i}|\]

But it is just the triangular inequality! If $M$ is $1+\max(...)$ then if $|z|>1$ you have $|z|^{n}\leq (M-1)(1+|z|+...+|z|^{n-1})<(M-1)\frac{|z|^{n}}{|z|-1}$ and you are done.


Suppose that $ n\geq 2$ be integer and $ a_1,\ a_2,\ a_3,\ a_4$ satisfy the following condition:

i) $ \ n$ and $ a_i\ (i \equal{} 1,\ 2,\ 3,\ 4)$ are relatively prime.

ii) $ \ (ka_1)_n \plus{} (ka_2)_n \plus{} (ka_3)_n \plus{} (ka_4)_n \equal{} 2n$ for $ k \equal{} 1,\ 2,\ \cdots ,\ n \minus{} 1$.

Note that $ (a)_n$ expresses the divisor when $ a$ is divided by $ n$.

Prove that $ (a_1)_n,\ (a_2)_n,\ (a_3)_n,\ (a_4)_n$ can be divided into two pair with sum $ n$.

Sorry, I made a typo. I correct the problem as follows.

Suppose that $ n\geq 2$ be integer and $ a_1,\ a_2,\ a_3,\ a_4$ satisfy the following condition:

i) $ \ n$ and $ a_i\ (i \equal{} 1,\ 2,\ 3,\ 4)$ are relatively prime.

ii) $ \ (ka_1)_n \plus{} (ka_2)_n \plus{} (ka_3)_n \plus{} (ka_4)_n \equal{} 2n$ for $ k \equal{} 1,\ 2,\ \cdots ,\ n \minus{} 1$.

Note that $ (a)_n$ expresses the [color=red]remainder[/color] when $ a$ is divided by $ n$.

Prove that $ (a_1)_n,\ (a_2)_n,\ (a_3)_n,\ (a_4)_n$ can be divided into two pair with sum $ n$.

hey, nobody answer this. I tried the problem and the only thing I have to prove for solve the problem is:
If we have 2 residues $ a$ and $ b$ such that:
$ (ka)_n\ge k  \Longleftrightarrow (kb)_n\ge k$ for all $ 1\le k \le n\minus{}1$
Then $ a\plus{}b\equal{}n\plus{}1$ or $ a\equal{}b$.
Please [color=blue][i]kunny[/i][/color], if you have a solution, post it.

This problem is quite similar to (and more general than) this problem
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=340038&sid=502505c7edf25f7b8c81bb871cdd3609#p340038[/url]
It's very hard...


Este foro en específico es para Discusiones Generales. Si se trata de un problema de nivel de secundaria, tratelo en el foro [url=http://www.mathlinks.ro/Forum/index.php?f=288]Matemáticas de Preparatoria[/url], si se trata de un problema perteneciente a una competencia (o de nivel semejante a las mismas), tratelo en el foro [url=http://www.mathlinks.ro/Forum/index.php?f=287]Olimpiadas de Matemática[/url]. Básicamente cualquier otro tema deberia ser tratado en este foro.

Modere el tono de la conversación. Al igual que el resto de los foros en este sitio, este foro es visitado por miembros de potencialmente todas las edades, incluyendo miembros de edades inferiores a los 13 años, por lo tanto algunos temas deben ser tratados con mucho tacto, y otros no deberían ser tratados del todo. En caso de aparecer un mensaje inapropiado, [b]será editado o borrado a discreción del moderador[/b] o de darse el caso, alguno de los administradores.

[b]Trate también de moderar el lenguaje[/b]. Palabras que suelen considerarse ofensivas o inapropiadas en una conversación semiformal, lo serán también aquí. Del mismo modo, trate de ser [b]tolerante[/b]. Recuerde que nuestro idioma es hablado por una población bastante numerosa y en una región bastante extensa, por lo tanto, algunos modismos son distintos, algunas palabras pueden ser consideradas ofensivas en una región y significar algo apropiado y distinto en otra. De todas maneras, en caso de aparecer un mensaje inapropiado, [b]será editado o borrado a discreción del moderador[/b].

Cualquier consulta, puede enviarme un mensaje personal.

Hola,

A raíz de algunos post recientes, uno de ellos en los foros en Español, se ha tomado una decisión, avalada por los administradores:

Se considerará spam cualquier post en el cual se este procurando publicitar y vender un producto nuevo a varias personas, y en dicho caso, el mensaje será borrado sin ningún previo aviso.

En el caso de una venta única, de un articulo no nuevo, relacionado con la actividad del foro (ejemplo: Algún libro usado de AoPS), se recomienda el uso de portales directamente dedicados a la venta de este tipo de productos (ejemplo: [url=http://www.ebay.com/]eBay[/url]), y en el caso de ser publicado en el foro, el thread será cerrado o borrado a discreción del moderador.

Hola Comunidad Latina:

Como ya sabran soy el nuevo moderador

Me alegra esta posicion y espero poder contribuir mas a la comunidad

Continuan las mismas reglas: En Español y MUCHA MATEMATICA!

Espero cuelguen sus problemas aca:

[b]En Discusiones Generales [/b]se discuten temas del folclor Matematico: Noticias, Olimpiadas, Eventos, Informacion, Fechas, Curiosidades, esperamos sean de Matematicas, no?

[b]En Olimpiadas de Matematicas[/b] se ponen problemas de olimpiadas y se resuelven, favor poner la fuente del problema, el area y una descripcion para posiblemente organizarlos en un futuro. (y evitar el span!)

[b]En Matematica de Preparatoria[/b] se ponen temas relacionados con las matematicas a nivel escolar, a modo de introduccion o nivel mas basico que el de olimpiadas.

Otras actividades que tenemos son las de [b]EL reto de la semana[/b], simplemetne se trata de enviar soluciones a los problemas que semanalmente estamos proponiendo a mi mail (pascual2k1@hotmail.com) o al de Manuel u otro de los miembros del grupo creador.

Tambien pueden colgar articulos en español y aceptamos cualquier tipo de sugerencia para hacer de esta comunidad algo bien grande.[/b]

Pascual, gracias por administrar el foro.. esperamos aprender y disfrutar mucho de la buena matematica latina....

[b]Saludos a todos...[/b]

 Bueno, como ya lo habran notado...
[b]Valentin V.[/b] nos ha designado a Pascual y a mi como moderadores de la comunidad ...

Ya Pascual lo ha dicho ...
esperamos de parte de todos los miembros de la comunidad su participacion y colaboracion
para mejorar este forum ...

[b]Opinion..[/b]
La formacion de estas comunidades se dio como manera de evitar y/o reducir la barrera de los idiomas
que aun mucha gente tiene y fomentar la colaboracion entre sus miembros ... eso es lo que esperamos
tambien de esta comunidad, como podran notar, [b]ya somos la 2da Comunidad[/b] (numero de posts) en
el Area Europa y estamos creciendo rapidamente ... esto amerita mayor organizacion y planificacion de
 nuestra parte ..

[b]Finalidad...[/b]
Estoy convencido que este forum generara un desarrollo en el nivel academico de la comunidad iberoamericana...
y todo esto en pos de mejorar la participacion de las delegaciones latinoamericanas en las proximas olimpiadas (IMO, APMO, etc..)

[b]Pedido..[/b]
- Hago un pedido a todo miembro de esta comunidad a compartir la informacion con la que cuentan y asi difundirlo
aqui entre sus miembros, dentro de poco se habilitara una seccion donde podremos cargar articulos que escribamos, articulos traducidos, ebooks encontrados, etc ..
- Pido difundir este forum entre los circulos de estudio en cada pais a los que tengamos acceso, pues mientras seamos más, MathLinks tendra mas consideraciones y apoyo a esta comunidad..

[b]Finalmente..[/b]
Estamos abiertos a sugerencias y nuevas propuestas en pos de mejorar esta comunidad, favoreciendo directamente a sus miembros y como no ... buscando que esta comunidad aporte al forum MathLinks en su conjunto, de inciativas de mejora..

Carlos Bravo :)
Lima -PERU

Just for the sake of it, it's MathLinks (one word, together) not "Math Links" which means a different thing. MathLinks = MathInterConnects, not a collection of links ;) :D

Creo que ya no hay actividad en el foro... parece que el moderador es Luis Gonzalez.  Sr. Gonzalez, como cree que podemos generar mas actividad en el foro?  Me falta practica de usar la computadora en espanol.


limit
real analysis
real analysis unsolved

Let $ x_0,x_1\in\mathbb{R}$ and
\[ x_{n}\equal{}\frac{(n\minus{}1)g}{1\plus{}(n\minus{}1)g}x_{n\minus{}1}\plus{}\frac{1}{1\plus{}(n\minus{}1)g}x_{n\minus{}2}\]
for some constant $ g>0.$
Find $ \lim_{n\to\infty}{x_n}.$

Let $ p_n\equal{}\frac{1}{1\plus{}(n\minus{}1)g}$, we have $ x_n\minus{}x_{n\minus{}1}\equal{}(\minus{}1)^{n\minus{}1}p_2p_3\cdots p_n(x_1\minus{}x_0)$.


limit
calculus
integration
function
complex analysis
complex analysis unsolved

I wish to document a related problem in here as a separate item and hope that's alright:

Suppose $ f(z)$ has an infinite set of isolated poles which accumulate to the non-isolated singularity $ z_0$ all of which are bounded by the contour $ \Gamma$, that is:

$ \{z_i\}$ is a sequence of poles of $ f(z)$such that:

$ \mathop\lim\limits_{n\to\infty} z_i\to z_0$

When is the following true:

$ \mathop\oint\limits_{|z|=\Gamma} f(z)dz=2\pi i\sum_{n=1}^{\infty}\mathop\text{Res}\limits_{z=z_n} f(z)$

Just seems to me this will always be true.  Can someone give a counter-example in which it is not?

Thanks.

Take any $ f$ for which it is true and add $ \frac 1{z\minus{}z_0}$ to it. Then it'll become false.

Ok thanks.  I'll work through it so that I can understand why.

Hello Fedja, I think I got it.  I wonder if that construct is the only way?  My justification of why I think that might be the case is:

Suppose $ f(z)$ has a non-isolated singularity at $ z_0$ with the accumulation of isolated poles as described above, then (perhaps) the sum above is valid unless $ f(z)$ can be expressed as:

$ f(z)\equal{}g(z)\plus{}h(z)$

with $ g(z)$ having the non-isolated singularity at $ z_0$ and $ h(z)$ having a pole at $ z_0$ with a non-zero residue.  Take for example:

$ f(z)\equal{}\frac{1}{Sin[1/z]}\plus{}1/z$

The $ Sin[1/z]$ terms prevents $ f(z)$ from having a pole at the origin. However:

$ \oint f(z)dz\equal{}\oint \frac{1}{Sin[1/z]}dz\plus{}\oint 1/z dz$ and thus the integral includes the residue of $ 1/z$.  From this perspective, $ h(z)$ could have poles distributed elsewhere throughout $ \Gamma$ and the sum I think would still apply.  It's only when $ h(z)$ includes a pole at the non-isolated singularity does the idea break down.  Or no?

Not trying to prove myself right or nothing, just trying to better explore a concept I find interesting.

That is, indeed, essentially the only way just because if the formula is not true for $ f$ and the series of residues converges, then you can subtract $ \frac c{z\minus{}z_0}$ with appropriately chosen $ c\in\mathbb C$ from $ f$ and the formula will become true. On the other hand, it may happen that you encounter a function $ f$ for which the series of residues diverges and then you are in trouble from the very beginning. Try to construct such a function yourself :).


Find the last three digits of $2007^{2006^{2005^{\cdots^{1}}}}$.

I'm not sure how useful most of that exponentiation is.

[hide]We take it $\pmod{8}$ and $\pmod{125}$.
Since $2007 \equiv-1 \pmod{8}$, and $2006^{\cdots}\equiv 0 \pmod{2}$, $2007^{\cdots}\equiv 1\pmod{8}$.

Next $2007^{\cdots}\equiv 7^{\cdots}\pmod{125}$.
Then, by Euler's totient theorem,
$7^{2006^{\cdots}}\equiv 7^{6^{\cdots}}\pmod{125}$
(Since $\phi{125}= 100$, and $2006 \equiv 6 \pmod{100}$.)
Apply again to get $7^{6^{2005^{\cdots}}}\equiv 7^{6^{5^{\cdots}}}\pmod{125}$... and one last time - this time using Carmichael's theorem to cut the work -

$7^{6^{5^{2004^{\cdots}}}}\equiv 7^{6^{5^{0}}}\equiv 7^{6}\equiv 24 \pmod{125}$

So using CRT, we find the last three digits to be $649$. [/hide]

[quote="white_horse_king88"]
$7^{2006^{\cdots}}\equiv 7^{6^{\cdots}}\pmod{125}$
(Since $\phi{125}= 100$, and $2006 \equiv 6 \pmod{100}$.)
Apply again to get $7^{6^{2005^{\cdots}}}\equiv 7^{6^{5^{\cdots}}}\pmod{125}$[/quote]

I don't believe this is true, as you can't apply that trick to powers, as they don't necessarily repeat in cycles of the totient function...

Oops... sorry.  Actually, both of us were wrong, you being to the lesser extent:

[hide]
We have:

$x = 2006^{2005^{\cdots}}$

$2007^{x}\equiv 7^{x}\pmod{125}$

Now this is where it should be changed.

Looking at it $\pmod{100}$, we need to use CRT since 6 and 100 aren't coprime.

So we need to find:

1. $2006^{2005^{\cdots}}\bmod{25}$

and

2. $2006^{2005^{\cdots}}\bmod{4}$

1.
$2006^{2005^{\cdots}}\equiv 6^{2005^{\cdots}}\pmod{25}$
Using Carmichael,
$2005^{2004^{\cdots}}\equiv 5^{2004^{\cdots}}\equiv 5 \pmod{20}$, since $\lambda(25) = 20$.
So $6^{5}\equiv 1 \pmod{25}$

2.
$2006^{2005^{\cdots}}\equiv 2^{2005^{\cdots}}\equiv 0 \pmod{4}$

Using CRT on $x \equiv 1 \pmod{25}$ and $x \equiv 0 \pmod{4}$, we find $x \equiv 76 \pmod{100}$.

Now we plug this back in: $2007^{x}\equiv 7^{76}\equiv 101 \pmod{125}$.

Now we use CRT on $2007^{\cdots}\equiv 101 \pmod{125}$ and $2007^{\cdots}\equiv 1 \pmod{8}$, and we see that:

$2007^{\cdots}\equiv 601 \pmod{1000}$

[/hide]

Oh, wow. I don't know why I thought $(6,100) = 1$.  :blush:


geometry
geometric transformation
rotation
search
combinatorics proposed
combinatorics

Some $ m$ squares on the chessboard are marked. If among four squares at the intersection of some two rows and two columns three squares are marked then it is allowed to mark the fourth square. Find the smallest $ m$ for which it is possible to mark all squares after several such operations.

I guess $ m\equal{}15$. Consider the marking where all the squares on the leftmost column and the topmost row are marked. I don't know how to prove.

A different question:If we consider that board to be same after rotation, how many different markings are there it is possible to mark all squares after several such operations (of course with minimum number of markings) ?

here I assume the initial marking is already given and can't be changed by us.
if it is an $ \ n^2 $ checkboard then$ \  m =  (n-1)^{2}+1 $
see $ \  m >(n-1)^{2} $
because set all the row black apart from the last one.then we can't mark the whole board.
now to see why m=(n-1)^{2}+1
work consider any white point.call it a. as there is atmost (n-2) unmarked left there is atleast one marked in that column call it b. now conside all other (n-1) column and search for c,d such that c, d are in same rows corr to a,b.then atleast
one such c,d is there s.t. both c,d are marked. so consider the square abcd.so now we can mark a.
as a is arbitrary e are done
so
$ \  m =  (n-1)^{2}+1 $ is minimum no to guarranty the colouring of all the squares.

Suppose that we mark all cell.if it has a rows and b column,we prove m min=a+b-1.we will clean many cell to max such that it can recovery to mark all. Hence the first row we can clean max:b-1 cell.we prove that min cell of 3 rows consecutive=b+1.with 2 conditions we use fist condition for a-3 rows then use second conditions.( because don’t have 4 cell with some shape and around white


function
number theory unsolved
number theory

Find the smallest integer of the form $|f(m,n)|$ with

$(a) f(m,n)=36^m-5^n.$    $(b) f(m,n)=12^m-5^n.$
[color=indigo][Please what does that mean : $f(m,n)$ is the problem asking for the minimum value of the function?][/color]

Davron

I think it is for $m, n > 0$, otherwise, see Rust's post.
In this case, the answers are :
(a) $(m, n) = (1, 2) \leadsto |f(m, n)| = 11$
and
(b) $(m, n) = (1, 1) \leadsto |f(m, n)| = 7$.

Indeed, the problem asks you to find the minimal value $|f(m, n)|$ can be. So, now that you know what you're asked to do, try to solve it. You have the answers if you want to check. If you cannot solve it, just ask, and I'll help you. (But maybe someone else will post a proof soon ;).)


If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{6}{7}$, where a, b, and c are positive integers, what is the smallest value of a+b+c?

[hide]Let one of them be $\frac12$.  That leaves $\frac5{14}$.

After guess-and-check, I get 1/3 and 1/42.

So, I get $47$.

[/hide]

[hide]well i do guess and check, get 42, 3, and 2
therefore, 47[/hide]

[quote="redsmartie"][hide][i]Supposing a<=b<=c,[/i]
1/a + 1/a + 1/a >= 6/7
 :arrow: 3/a >= 6/7; a <= 7/2, a = 1,2,3
1/a < 6/7, a > 7/6 > 1
[b]a = 2,3[/b]

[b]a = 2[/b]
1/b + 1/c = 5/14
1/b + 1/b >= 5/14
[b]b <= 28/5 :arrow: b = 3,4,5[/b]

a=2, b=3, c=42
a=2, b=4, c=28/3
a=2, b=5, c=20/11

[b]a = 3[/b]
1/b + 1/c = 11/21
1/b + 1/b >= 11/21
[b] b <= 42/11 :arrow: b=1,2,3[/b]
But b connot equal 1 or 2 because it's has to be greater or equal to than a.
It also connot equal 3.

In the end, a=2, b=3, c=42 :arrow: so a+b+c=47. [/hide][/quote]

Here is your answer written in $\LaTeX$ so it is easier to read (I hope I didn't make any mistakes  :D ):

[hide][i]Supposing $a\leq b\leq c$,[/i]
$\frac{1}{a} + \frac{1}{a} + \frac{1}{a}\geq \frac67$
 :arrow: $\frac{3}{a} \geq \frac67$; $a \leq \frac72, a = 1,2,3$
$\frac{1}{a} < \frac67$, $a > \frac76$ $> 1$
[b]a = 2,3[/b]

[b]a = 2[/b]
$\frac1{b} + \frac1{c} = \frac5{14}$
$\frac1{b} + \frac1{b} \geq \frac5{14}$
[b]$4b \leq \frac{28}{5}$ :arrow: b = 3,4,5[/b]

a=2, b=3, c=42
a=2, b=4, $c=\frac{28}{3}$
a=2, b=5, $c=\frac{20}{11}$

[b]a = 3[/b]
$\frac1{b} + \frac1{c} = \frac{11}{21}$
$\frac1{b} + \frac1{b} \geq \frac{11}{21}$
[b] $b \leq \frac{42}{11}$ :arrow: b=1,2,3[/b]
But b connot equal 1 or 2 because it's has to be greater or equal to than a.
It also connot equal 3.

In the end, a=2, b=3, c=42 :arrow: so a+b+c=47. [/hide]

[quote="moogra"][hide]47 I used guess and check (sad huh)[/hide][/quote]

It's actually not that sad, because this problem actually requires some luck. A systematic way of doing this problem would take a while and is probably too advanced for a chapter level problem.

[hide]
I think that guess and check is the easiest way to solve this problem:
If you have $\frac{1}{2}$, you have $\frac{5}{14}$ left. After guess and check, $\frac{1}{3}$ and $\frac{1}{42}$
$\boxed{47}$[/hide]


Find a formula for the n[i]th[/i] term of the sequence

$ 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...$

Where the integer $ m$ appears precisely $ m$ times.

[hide]
when n is a triangle number (1, 3, 6, 10, 15...), this problem becomes trivial

just reverse solve

$ n = \frac{a(a+1)}{2}$ for a

$ 2n = a^{2}+a$
$ a^{2}+a-2n = 0$

$ a = \frac{-1+\sqrt{1+8n}}{2}$

now since each triangle number term is the final term with that given number, using the ceiling function will fill in all the in betweeners.

For example. solving for the 5th term will give you something between 2 and 3, which the ceiling function will resolve to 3

thus:
$ a_{n}= \left\lceil{\frac{-1+\sqrt{1+8n}}{2}}\right\rceil$
[/hide]


Let's say i have two forms: Form1 and Form2. On Form1 i have
a label1 and on Form2 i have a button.
Now i want ,when i press the button from Form2 to write in label1
from Form1 the text "HELLO".

How do i make this??? Please give complet answer and source code.
Thank you!

You need some way to access the controls of the first form from the second form, right? One way to do this is to make the forms "global variables" or more precisely, static members of your main class:

[code]class myapp {
	public static form1 f1;
	public static form2 f2;
	static void Main () {
		f1 = new form1 ();
		f2 = new form2 ();
		f1.Show ();
		f2.Show ();
		Application.Run (f1);
	}
}[/code]

form1 and form2 are the classnames of Form1 and Form2. Now you can freely access the members of the two forms from anywhere in your application (provided that the members are public):

[code]class form1 : Form {
	public Label label1;
	...
}

class form2 : Form {
	void buttonclick (object sender, EventArgs e) {
		myapp.f1.label1.Text = "HELLO";
	}
	...
}
[/code]

I attach a complete sourcecode (I had to change the extension to .tex because of the restrictions....). Feel free to ask any questions you have about my program.

Well, i couldn't compile it because i'm not familliar with classes. Look my forms are:

[code]
/* form1.cs */
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Text;
using System.Windows.Forms;

namespace forms
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {
            Form2 f2 = new Form2();
            f2.Visible = true;
        }
    }
}
[/code]

and

[code]
/* form2.cs */
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Text;
using System.Windows.Forms;

namespace forms
{
    public partial class Form2 : Form
    {
        public Form2()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {

        }
    }
}
[/code]

Now : i have a label and a button in form1  and a button in form2.
Buton from form1 is for opening form2 and button from form2 is for put label1 = "HELLO".
Now please help me to do this in my forms. I attached u my entire project. So please modify it to see the changes.
Thank you!


hi guys
need ur help
if u guys know about site having some exceptionally good c++ program.
Plz don't post ordinary program.
I have to give it for my externals in just one week.
Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz  tell me some really nice programs.

not a single suggestion.
I TEACHER will hang me if I will not submit it next Monday
.

Any type of application based C++ program used in the field of technology , mathematics ,medical etc.

u can post any c++ program used to make thing quite easy.
Like I have seen one program on how to update  Criminal Datas?
Believe me
It was the worst program I have seen in my life .
I m finding a program related to mathematics something dealing with graphs,functions ,calculas or even geometry but not does one dealing with how u can draw a 2-d graph or find the area with given sides blah..blah

plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz post it if u have something.


On a certain math exam, $ 10 \%$ of the students got 70 points, $ 25 \%$ got 80 points, $ 20 \%$ got 85 points, $ 15 \%$ got 90 points, and the rest got 95 points. What is the difference between the mean and the median score on this exam?

$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 1\qquad
\textbf{(C)}\ 2\qquad
\textbf{(D)}\ 4\qquad
\textbf{(E)}\ 5$

[hide="Answer"]Simplest case, there are $20$ students. $2$ scored $70$, $5$ scored $80$, $4$ scored $85$, $3$ scored $90$, and $6$ scored $95$. Multiplying those out, the average score is $\frac{2(70)+5(80)+4(85)+3(90)+6(95)}{20}=86$, while the median is $85$, so the answer is $86-85=1\Rightarrow \boxed{B}$.[/hide]


The graph of $ f(x) \equal{} x^4 \plus{} 4x^3 \minus{} 16x^2 \plus{} 6x \minus{} 5$ has a common tangent at $ x \equal{} p$ and $ x \equal{} q$. Compute the product $ pq$.

How do you know that the common tangent is $ 46x \minus{} 105$?


Three girls A, B and C and nine boys are to be lined up in a row. In how many ways can this be done if B must be between A and C and A,B must be separated by exactly 4 boys?

[hide="Boys are Indistinguishable"]
Case 1
A_ _ _ _B_ _ _ _ _
5 possible spots for C
$ 5 \cdot 9=45$

Case 2
_A_ _ _ _B_ _ _ _
4 Possible spots for C
$ 4 \cdot 9 = 36$

Case 3
_ _A_ _ _ _B_ _ _
$ 3 \cdot 9= 27$

Case 4
_ _ _A_ _ _ _B_ _
$ 2 \cdot 9 = 18$

Case 5
_ _ _ _A_ _ _ _B_
$ 9$

Add them $ \boxed{9 (5+4+3+2+1) = 135}$.
[/hide]
[hide="Boys are Distinguishable"]
Case 1
A_ _ _ _B_ _ _ _ _
5 possible spots for C
$ 5 \cdot 9!$

Case 2
_A_ _ _ _B_ _ _ _
4 Possible spots for C
$ 4 \cdot 9!$

Case 3
_ _A_ _ _ _B_ _ _
$ 3 \cdot 9!$

Case 4
_ _ _A_ _ _ _B_ _
$ 2 \cdot 9!$

Case 5
_ _ _ _A_ _ _ _B_
$ 9!$

Add them $ \boxed{15 \cdot 9!=5443200}$.
[/hide]


function
topology
real analysis
real analysis theorems

A function is f : X -> Y is continuous iff f(Closure(A)) is contained in Closure(f(A)) for every subset of A of X. How can we prove this? Then
f : X -> Y is cont. iff f-inv(Int(A)) is contained in Int(f-inv(A)), right?

$(\Rightarrow)$ Assume $f$ to be continuous. Take $\overline{x}\in \overline{A}$, and $V$ an open neighborhood of $f(\overline{x})$. We want to show that $V \cap f(A) \neq \emptyset$. By continuity, there is a neighborhood $U$ of $\overline{x}$ such that $f(U) \subseteq V$. But we also have $U \cap A \neq \emptyset$, whence $f(U \cap A) \cap V \neq \emptyset$, and $f(A) \cap V \neq \emptyset$.

$(\Leftarrow)$ If there is a point $x$ at which $f$ is not continuous, then there is an open neighborhood $V$ of $f(x)$ such that for every open neighborhood $U$ of $x$, $f(U) \cap (Y-V) \neq \emptyset$. Assign a point $y_{i}$ in this intersection for every such $U_{i}$, and write $y_{i}=f(x_{i})$, $x_{i}\in U_{i}$. Clearly enough $f(\overline{\{x_{i}\}})$ contains $f(x)$ but $f(x)$ does not belong to $\overline{f(\{x_{i}\})}\subseteq Y-V$. Contradiction.

Thanks for the solution. It's simple and pretty good.

Well, one could actually wonder whether this problem can be solved without using the axiom of choice. I remember having this problem for my general topology exam a few years ago, but I don't remember solving it this way ... there must be something else without the axiom of choice.


inequalities
induction
inequalities unsolved

Define $d_i=a_{i+1}-a_{i}.$ Then $|d_{k+1}-d_{k}| \le 1$ for $k=0,1,...,n-1$ and $d_0+d_1+...+d_n=0.$ It suffices to prove that
$$|d_0+d_1+...+d_k| \le \frac{(k+1)(n-k)}{2}$$
for all $k$. For the sake of contradiction, assume
$$d_0+d_1+...+d_k > \frac{(k+1)(n-k)}{2}$$
for some $k$. It easy to see that
$$(d_k+k)+(d_k+k-1)+...+(d_k+1)+d_k \ge d_0+d_1+...+d_k > \frac{(k+1)(n-k)}{2} \Leftrightarrow 2d_k+2k > n. $$
Hence,
$$0=d_0+d_1+...+d_n \ge \frac{(k+1)(n-k)}{2} + (d_k-1)+(d_k-2)+...+(d_k-n+k) = \frac{(n-k)(2d_k+2k-n)}{2} > 0,$$
which is a contraction. In like manner, we can show that
$$d_0+d_1+...+d_k < -\frac{(k+1)(n-k)}{2}$$
leads to a contradiction. Therefore, we are done.

[quote=quangminh1173]
$$(d_k+k)+(d_k+k-1)+...+(d_k+1)+d_k \ge d_0+d_1+...+d_k > \frac{(k+1)(n-k)}{2} \Leftrightarrow 2d_k+2k > n. $$
[/quote]
In 6th line you used inequality $d_0\le d_k+k$. I wonder how do you know it's true since $d_0$ is included only in equation $d_0+d_1+...+d_n=0$, apart from inequalities $|d_{k+1}-d_{k}| \le 1$.

[quote=WolfusA]
In 6th line you used inequality $d_0\le d_k+k$. I wonder how do you know it's true since $d_0$ is included only in equation $d_0+d_1+...+d_n=0$, apart from inequalities $|d_{k+1}-d_{k}| \le 1$.[/quote]
edited


function
trigonometry
irrational number
real analysis
real analysis unsolved

I am learning signals and systems from oppenheim book.
In one page has a table about signals $ e^{(j\omega _0 t)}$ and $ e^{(j\omega _0 n)}$ for continuous and discrete time respectively
and is says that for continious time the exponential is periodic for any choice of $ \omega _0$ and for discrete time that is periodic ONLY if $ \omega _0 \equal{} m \frac {2\pi}{N}$
with $ m$ and $ N$ integers and $ N > 0$

i believe that
in continuous time we have also $ \omega _0 \equal{} k \frac {2\pi}{T_0}$ is this right??

where is the difference?
i can't understand and i am really confused
Thanks!

In discrete time, the exponential is periodic with periodicity $ P$ iff

\[ \exp\left(j\omega_0n\right)\equal{}\exp\left(j\omega_0(n\plus{}P)\right)\]
which implies that
\[ \exp\left(j\omega_0P\right)\equal{}1\]

which in turn implies that $ \omega_0P$ must be a multiple of $ 2\pi$.
But in discrete time, unlike in continuous time, $ P$ is restricted to be an integer. Hence, $ \omega$ must be a rational number $ m/N$ times $ 2\pi$.
If $ \omega_0$ is an irrational number times $ 2\pi$, there is no way that $ P\omega_0$ can ever be a multiple of $ 2\pi$.

[quote="DK"]
 Hence, $ \omega$ must be a rational number $ m/N$ times $ 2\pi$.
[/quote]

Did you mean  "Hence, $ \omega$ must be a rational number $ m/P$ times $ 2\pi$" ?

Generally i think that
we define $ \omega \equal{} \frac {2\pi}{T}$ and here $ \frac {2\pi}{P}$

$ \exp\left(j\omega_{0}P\right) \equal{} 1$ means tha we want $ \cos\left(\omega_{0}P\right) \equal{} 1$
so we have $ cos(\omega P) \equal{} cos(\frac {2\pi}{P} P) \equal{} cos(2\pi) \equal{} 1$
why $ \omega \equal{} \frac {2\pi}{P} P$ isn't allowed?


calculus
integration
trigonometry
function
calculus computations

Find

$\int{\frac{x^{2}}{\sqrt{2-x^{2}}}}\,dx$.

Use the [b]result[/b], rather than the integrand, to show that $-\sqrt{2}\le{x}\le\sqrt{2}$ for the integral to be defined.

Make the substitution $x^{2}= 2\sin^{2}u \ du$, then we have

${\int \frac{x^{2}}{\sqrt{2-x^{2}}}\ dx = \int\frac{2\sin^{2}u \sqrt{2}|\cos u|}{\sqrt{2-2\sin^{2}u}}\ du = 2\int\frac{\sin^{2}u \sqrt{2}|\cos u|}{\sqrt{2}|\cos u|}}\ du = 2\int\sin^{2}u \ du$$= u-\frac{1}{2}\sin 2u$

$=\sin^{-1}\frac{x}{\sqrt{2}}-\frac{x\sqrt{2-x^{2}}}{2}$

From this, we see that $\sqrt{2-x^{2}}$ and $\sin^{-1}\frac{x}{\sqrt{2}}$ is defined for $x\in[-\sqrt{2},\sqrt{2}]$.

I don't see how the result in any way implies anything about where the integrand is defined. You must realize that finding an integrand is nonsense when it isn't defined...

[quote="coffeym"]Perfect.

As an aside on notation, would you not tend to write $\frac{1}{\sqrt{2}}$ in rationalised form ($\frac{\sqrt{2}}{2}$) if you did this by hand?  I ask because I've heard a few people swear by the rule "never leave a denominator irrational".  :lol:[/quote]

Actually, it was just the opposite; when I did it by hand I wrote $\frac{1}{\sqrt{2}}$, then when I checked the answer on my calculator it was $\frac{\sqrt{2}}{2}$. Personally, I don't really have a preference about rationalizing the denominator.

[quote="coffeym"]Use the [b]result[/b], rather than the integrand, to show that $-\sqrt{2}\le{x}\le\sqrt{2}$ for the integral to be defined.[/quote]

There could be many ways to write the result, and some of those may have meaningful extensions to values of $x$ outside the stated interval.  When that happens it is often a feature, not a bug;  for example, when showing that some function unexpectedly has an analytic continuation.

[quote="coffeym"]Perfect.

As an aside on notation, would you not tend to write $\frac{1}{\sqrt{2}}$ in rationalised form ($\frac{\sqrt{2}}{2}$) if you did this by hand?  I ask because I've heard a few people swear by the rule "never leave a denominator irrational".  :lol:[/quote]

They only cared about that when I was in the 8th grade. So I sometimes do it by habit.


vector
Ring Theory
superior algebra
superior algebra theorems

$K$ a field, $A$ a commutative $K$-algebra of finite dimension. Prove that there is only finitely many maximal ideal.

It's always exciting when I get to use tough theorems, such as the [url=http://planetmath.org/encyclopedia/WedderburnArtinTheorem.html]Artin-Wedderburn Theorem[/url], in this case :).

Let $\mathcal J$ be the Jacobson radical of $A$. Now, $\tilde A=A/\mathcal J$ is a [url=http://planetmath.org/?op=getobj&from=objects&name=SemiprimitiveRing]semiprimitive[/url] ring, which is also a finite-dimensional algebra over $K$. Moreover, the maximal ideals of $\tilde A$ are in one-to-one correspondence with those of $A$ via the canonical projection $A\to\tilde A$. It thus sufficces to prove that $\tilde A$ has finitely many maximal ideals.

Notice that $\tilde A$ is also [url=http://planetmath.org/encyclopedia/Artinian.html]Artinian[/url]. This is because ideals of $\tilde A$ are subspaces of the vector space $\tilde A$ over $\mathbb K$, and there are no infinite descending chains of subspaces of a finite-dimensional vector space.

Since $\tilde A$ is a semiprimitive and Artinian ring, we can apply the above-mentioned theorem to conclude that it must be a direct sum of rings of matrices over division rings that contain $K$. We also know that $\tilde A$ is commutative, so it's in fact a fintie product of extension fields of $K$. Now it all comes down to proving that a finite direct product of fields , $\prod_{i=1}^nF_i$, has finitely many maximal ideals. If an ideal of such a ring contains an element $(x_i)_{i=1}^n$ with $x_j\ne 0$, then it contains all such elements, so the ideals of our ring are the products of several of the fields $F_i$, which means that the maximal ideals are precisely $M_j=\prod_{i\ne j}F_i$.

I hope it's Ok.


A master is chasing after his dog. For every three steps the dog takes, the master
takes two. However one of the master’s steps is equal to two of the dog’s in length. The
dog runs 10 steps before the master begins chasing. When the master finally catches up
with the dog, how many steps has the dog taken?

The man catches up one step on the dog for every 3 steps the dog takes. Thus the man catches up with the dog after 20 of his steps. In this time, the dog has traveled $ \boxed{40}$ steps.


abstract algebra
algebra unsolved
algebra

From the line segment $[-1; 1]$ $k\geq 2$ different points have been choosen. For each point $i$ define $P_i$ as the product of distances to the other $k-1$ points. Define $S= \sum \frac{1}{P_i}$. Prove that $S \geq 2^{k-2}$.

Actually, I have a solution to this problem but I didn't solved it by myself so I'm posting it here...

Given n points $x_1,x_2,...,x_n$
Consider the Trebusep'polynomial T:degT$=n-1$
The coefficient of $x^{n-1}$is $2^{n-2}$
By using the Lagrange's theorem:
$T_{n-1}(x)=\sum_{i=1}^{n}\frac{\prod_{j\neq i}(x-x_j)}{\prod(x_i-x_j)}.T_{n-1}(x_{i})$
The coefficient of $x^{n-1}$ is $\sum_{i=1}^{n}\frac{T_{n-1}(x)}{\prod(x_i-x_j)}$
$\Rightarrow 2^{n-2}=|\sum\frac{T_{n-1}(x)}{\prod(x_i-x_j)}|=\sum\frac{|T_{n-1}(x)|}{\prod|x_i-x_j|}$
$\Rightarrow 2^{n-2}\leq \sum\frac{1}{\prod|x_i-x_j|}=S_n$(q.e.d)

Yes, it's also a official solution :). Good work.


Find the greatest $ n$ for which there are points $ P_1$, $ P_2$, ... , $ P_n$ in the plane such that each triangle whose verticies are among $ P_1$, $ P_2$, ... , $ P_n$ has a side less than $ 1$ and a side greater than $ 1$.

I felt like this was not a good problem, but I'm probably wrong, as I have no right to be contradicting all these super-smart people.  :oops:

There is no limit to the value $ n$. For any $ n$, no matter how large, we may arrange points $ P_1$ to $ P_{n \minus{} 1}$ so that they are collinear and the greatest distance between them is less than 1 (i.e., they are all on, but not on the endpoints of, a segment of unit length). Then let $ P_n$ be far far away, at a distance greater than $ 1$ from any of the points.

Thus any triangle formed must include $ P_n$ and two of the points from $ P_1$ to $ P_{n \minus{} 1}$, because if it did not include $ P_n$, the "triangle" would degenerate into a line. Now we placed $ P_n$ so that any line to it from another point is greater than $ 1$, and we also placed the other points so that any line between them is less than $ 1$.

So there is no greatest possible $ n$.

By triangles, I think they also included degenerate triangles (eg, any three points).

[hide="Then"]It's a well-known graph theory problem; consider a graph with $ \ge 6$ points. We pick an arbitrary point $ P$, then of the remaining five points at least three belong to either $ S_1 \equal{} \{\mathrm{points\ less\ than\ 1\ from\ P}\}, S_2 \equal{} \{\mathrm{points\ greater\ than\ 1\ from\ P}\}$. Without loss of generality, let the first set have size $ |S_1| \ge 3$; if any two points in $ S_1$ are less than $ 1$ apart those two points and $ P$ form a triangle with all sides less than $ 1$, contradiction. But otherwise at least three points in $ S_1$ are greater than $ 1$ apart, and the triangle with those three points generate a contradiction. Then just show $ n \equal{} 5$ works.[/hide]

I really should stop posting here ..  :maybe:

Edit: Wait, why not? Edit: Just read sticky ..

@ leoxnlin: When you are doing a problem, you assume that it isn't trivial...Your assumption that degenerate triangles are not triangles trivializes the problem. I suppose that it was not worded the best, but generally, you should attempt to solve a non-trivial problem if you see something on a contest or whatever.

@azjps: what is greater or less than 1? Why can you say without loss of generality that it is the greater set...?

Does this question ask for the $ n$ for which such an arrangement is [b]possible[/b]?

The largest $ n$ such an arrangement of points exists...