limit induction function search trigonometry calculus

Let $0<x_{0}<1$, and $x_{n+1}=x_{n}(1-x_{n})$ for $n\ge1$. Prove that the limit $\lim_{n\to\infty}nx_{n}$ exist and is equal to 1.

Hi,
This is an ideea:

By induction you prove that $x_{n}>0$ and then by AGM it follows that $x_{n}<\frac{1}{4}$. thus the sequence is bounded.

$x_{n+1}-x_{n}=-x_{n}^{2}<0$ thus the sequence is monotonic from which it follows it converges.

We calculate your limit by an application of Cesaro Stolz lemma

$L=%Error. "limnx" is a bad command.
_{n}=\lim \frac{x_{n+1}-x_{n}}{1/(n+1)-1/n}$

$=\lim n(n+1)x_{n}^{2}=L^{2}$,
Thus $L=1$ or $L=0$.
It remains to show that $L=1$.

Please observe that we can show that $x_{n}<\frac{1}{n+1}$ for all $n\geq 1$.
We prove this by induction, definitely that for $n=1$ we have that $x_{1}<1/4<1/2$.

Now solve the equation and get that $\sqrt{1-4x_{n+1}}=1-2x_{n}$.
$\sqrt{1-4x_{2}}=1-2x_{1}>1-\frac{2}{4}=1/2$,
it follows that $x_{2}<\frac{3}{16}<\frac{1}{3}$.

assume that $x_{n}<1/(n+1)$ and prove that $x_{n+1}<1/(n+2)$
We get that $\sqrt{1-4x_{n+1}}=1-2x_{n}>\frac{n-1}{n+1}$,
It follows that
$x_{n+1}<\frac{n}{(n+1)^{2}}<\frac{1}{n+2}$ and we are done.
Now observe that the sequence $nx_{n}$ is an increasing sequence since

$\frac{(n+1)x_{n+1}}{nx_{n}}=\frac{(n+1)}{n}(1-x_{n})>1$ since
 $x_{n}<\frac{1}{n+1}$ which implies that $L$ cannot be 0.
Didi
I hope that there are no computational mistakes.

[quote]... thus the sequence is monotonic from which it follows it converges.[/quote]
There's a much shorter finish from this point. If the sequence converges, it can only converge to a fixed point of the function being iterated - that is, to a solution of $x=x(1-x).$ The only solution to this is $x=0.$

1) how do u use AMGM to show that $x_{n}<\frac{1}{4}$
2) what is the Cesaro Stolz lemma?

[quote="maokid7"]1) how do u use AMGM to show that $x_{n}<\frac{1}{4}$?[/quote]
We have that
\[x_{n+1}= x_{n}\left( 1-x_{n}\right) \leq \frac{\left( x_{n}+\left( 1-x_{n}\right) \right)^{2}}4 = \frac14 . \]
[quote="maokid7"]2) what is the Cesaro Stolz lemma?[/quote]
Use [url=http://www.mathlinks.ro/Forum/search.php]the search button[/url].
1. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=stolz+cesaro&t=4126
2. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=stolz+cesaro&t=57474

[color=darkblue][b]An easy extension.[/b] $\|\begin{array}{c}p\in N^{*}\ ,\ \{r_{1},r_{2},\ldots ,r_{p}\}\subset (0,1]\ ,\ 0<x_{0}<1\\\\ x_{n+1}=x_{n}(1-r_{1}x_{n})(1-r_{2}x_{n})\ldots(1-r_{p}x_{n})\ ,\ n\in N^{*}\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}x_{n}\searrow 0\\\\ nx_{n}\rightarrow \frac{1}{r_{1}+r_{2}+\ldots+r_{p}}\end{array}\ .$

[b]Particular case.[/b] $\|\begin{array}{c}p\in N^{*}\ ,\ 0<r\le 1\ ,\ 0<x_{0}<1\\\\ x_{n+1}=x_{n}(1-rx_{n})^{p}\ ,\ n\in N^{*}\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}x_{n}\searrow 0\\\\ nx_{n}\rightarrow \frac{1}{pr}\end{array}\ .$

[b]A similar exercise.[/b] $\|\begin{array}{c}0<a <1\ ,\ x_{1}>0\\\\ 1+x_{n+1}=(1+\frac{1}{a}\cdot x_{n})^{a}\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}x_{n}\rightarrow 0\\\\ nx_{n}\rightarrow\frac{2a}{1-a}\end{array}\ .$ [b]Particular case.[/b] $\|\begin{array}{c}p\in N\ ,\ p\ge 2\ ,\ x_{1}>0\\\\ 1+x_{n+1}=\sqrt [p]{1+p\cdot x_{n}}\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}x_{n}\searrow 0\\\\ nx_{n}\rightarrow \frac{2}{p-1}\end{array}\ .$[/color]

Nice generalizations Virgil,

Let's study the first particular case:

$0<a<1$. $x_{1}>0$ $1+x_{n+1}=(1+\frac{x_{n}}{a})^{a}$

By induction $x_{n}>0$. On the other hand by Bernoully we get that

$1+x_{n+1}<1+a\frac{x_{n}}{a}=1+x_{n}$ thus the sequence decreases.
It is bounded bellow by 0 above by $x_{1}$ and thus convergent.

Letting n converge to $\infty$ in the recurrence relation we get that the only possible value for the limit of the sequence is $0$.

By Cesaro Stoltz  we get that

I will use this ideea:

$(1+y)^{a}\approx 1+ay+\frac{a(a-1)}{2}y^{2}$ near $0$
$L=\lim (n+1)x_{n+1}=\lim-n^{2}\left[(1+x_{n}/a)^{a}-1-x_{n}\right]$
$=\lim n^{2}x_{n}\frac{1-a}{2a}=L^{2}\frac{1-a}{2a}$
Thus, $L=0$ or $L=\frac{2a}{1-a}$.
For showing that $L=\frac{2a}{1-a}$ the technique is the same as posted in a previous post.
Didi

[color=darkblue][b]Theorem.[/b] Let $r$ be a positive number and let $f: (0,b)\rightarrow (0,1)$ be a continue function so that $\lim_{x\to 0}f(x)=1$ and $\lim_{x\to 0}\frac{1-f(x)}{x^{r}}=L>0\ .$
Construct the reccurent sequence $x_{0}\in (0,b)\ ,\ x_{n+1}=x_{n}\cdot f(x_{n})\ ,\ n\in N\ .$ Then ${x_{n}\searrow}0$ and $nx^{r}\rightarrow\frac{1}{rL}\ .$

[b]Exercise.[/b] $\|\begin{array}{c}0<x_{0}<\frac{\pi}{2}\\\\ 2x_{n}x_{n+1}=\sin x_{n}\sin 2x_{n}\ ,\ n\in N\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}x_{n}\searrow 0\\\\ x_{n}\sqrt n\rightarrow\sqrt{\frac{3}{5}}\end{array}\ .$[/color]

[color=darkblue]Thank you, Perfect_Radio ! I forgot that I posted it. I published in G.M.B. long ago an article with this theorem and its aplications.[/color]

[quote="Virgil Nicula"][color=magenta][b]Theorem.[/b] Let $r$ be a positive number and let $f: (0,b)\rightarrow (0,1)$ be a continuous function so that $\lim_{x\to 0}f(x)=1$ and $\lim_{x\to 0}\frac{1-f(x)}{x^{r}}=L>0\ .$
Construct the reccurent sequence $ x_{0}\in \left( 0, b \right)$, $ x_{n+1}= x_{n}\cdot f \left( x_{n}\right), \forall n \in \mathbb N$. Then, $x_{n}\searrow 0$ and $n x_{n}^{r}\to \frac1{rL}$.[/color][/quote]

(i) $\left\{ x_{n}\right\}$ is strictly decreasing and positive, so it has a limit $\ell$.

If we suppose that $\ell > 0$, then by taking the limit of the recurrence we get that $\ell = \ell f \left( \ell \right)$, i.e. $f \left( \ell \right) = 1$, which is impossible. Hence, $\ell = 0$.

(ii) Since $\left\{ \frac1{x_{n}^{r}}\right\}_{n \geq 1}$ increases to $\infty$, by Cesaro Stolz we have that
\begin{eqnarray*}\lim_{n \to \infty}n x_{n}^{r}&=& \lim_{n \to \infty}\frac{n}{\frac1{x_{n}^{r}}}\\ \ &=& \lim_{n \to \infty}\frac{(n+1)-n}{\frac1{x_{n+1}^{r}}-\frac1{x_{n}^{r}}}\\ \ &=& \lim_{n \to \infty}\frac{x_{n}^{r}x_{n+1}^{r}}{x_{n}^{r}-x_{n+1}^{r}}\\ \ &=& \lim_{n \to \infty}\frac{x_{n}^{2r}f^{r}\left( x_{n}\right)}{x_{n}^{r}\left( 1-f^{r}\left( x_{n}\right) \right)}\\ \ &=& \lim_{n \to \infty}\frac{x_{n}^{r}f^{r}\left( x_{n}\right)}{1-f^{r}\left( x_{n}\right)}\\ \ &=& \lim_{n \to \infty}\left( \frac{x_{n}^{r}}{1-f \left( x_{n}\right)}\cdot \frac{f^{r}\left( x_{n}\right)}{1+f \left( x_{n}\right)+\ldots+f^{r-1}\left( x_{n}\right)}\right) \\ \ &=& \lim_{x \to 0}\left( \frac{x^{r}}{1-f \left( x \right)}\cdot \frac{f^{r}\left( x \right)}{1+f \left( x \right)+\ldots+f^{r-1}\left( x \right)}\right) \\ \ &=& \frac1{L}\cdot \frac1{r}.\end{eqnarray*}


linear algebra
matrix
trigonometry
geometry
circumcircle
linear algebra unsolved

Let ABC be an acute triangle. Show that
\[ \left(\begin{array}{ccc}cos\frac {\pi}{3} & \minus{} \frac {cosA}{2} & \minus{} \frac {cosC}{2} \\
\minus{} \frac {cosA}{2} & cos\frac {\pi}{3} & \minus{} \frac {cosB}{2} \\
\minus{} \frac {cosC}{2} & \minus{} \frac {cosB}{2} & cos\frac {\pi}{3}\end{array}\right)\]
is positive semidefinite. If ABC is not equilateral, then the matrix is positive definite.

This is nice, because it provides a nice and easy proof of http://www.mathlinks.ro/Forum/viewtopic.php?t=69200 . But the problem must be corrected: The matrix is [b]never positive definite[/b]. Its determinant is always $ 0$ (because $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$, where $ x \equal{} \cos A$, $ y \equal{} \cos B$, $ z \equal{} \cos C$ for a triangle $ ABC$). Its eigenvalues are $ 0$, $ \frac12\left(3 \minus{} \frac {OH}{R}\right)$ and $ \frac12\left(3 \plus{} \frac {OH}{R}\right)$, where $ R$ is the circumradius, $ O$ the circumenter and $ H$ the orthocenter of triangle $ ABC$, unless I am mistaken in computations. Maybe there is a geometric proof :)

Ah, and of course there is no reason in requiring triangle $ ABC$ to be acute.

  darij


geometry
incenter
analytic geometry
vector
ratio
angle bisector

Let $ A(a,b),B(c,d),C(e,f)$ triangle on Cartesian plane.
Let the external bisector of $ B,C$ meets internal bisector $ A$ at $ O_1$, defined $ O_2,O_3$ analogously
Can you define the coordinate of $ I$ (incenter) and $ O_1,O_2,O_3$ (excenter)?

Sorry, if I put this on Theorems and Formulas, because I'm don't konw where I must put this

You can use the formula: $ BC.\overrightarrow{IA} \plus{} CA.\overrightarrow{IB} \plus{} AB.\overrightarrow{IC} \equal{} \overrightarrow{0}$

What do you mean? Can you tell the exact value?
Sorry, I don't understand vector :blush:

[hide="hint"]Incentre I divides the internal angle bisector AD of A in the ratio $ \frac {b\plus{}c}{a}$, where a,b,c represent the side lenghts of BC,CA and AB respectively.

Excentre opposite to A divides the internal angle bisector AD of A externally in the same ratio.

Anyways, there is a standard formula that you can just commit to memory and use it whenever you want.
$ I_x\equiv \frac {ax_1\plus{}bx_2\plus{}cx_3}{a\plus{}b\plus{}c}$

$ I_y\equiv \frac {ay_1\plus{}by_2\plus{}cy_3}{a\plus{}b\plus{}c}$

$ E_A(x)\equiv \frac {bx_2\plus{}cx_3\minus{}ax_1}{b\plus{}c\minus{}a}$

$ E_A(y)\equiv \frac {by_2\plus{}cy_3\minus{}ay_1}{b\plus{}c\minus{}a}$

where $ A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), I(I_x,I_y)$ and $ E_A(x)$ represents the x-coordinate of the excentre opposite to A.
 [/hide]

What is $ a,b,c$ in $ \frac{ax_{1}\plus{}bx_{2}\plus{}cx_{3}}{a\plus{}b\plus{}c}$?

[quote="bambang"]What is $ a,b,c$ in $ \frac {ax_{1} \plus{} bx_{2} \plus{} cx_{3}}{a \plus{} b \plus{} c}$?[/quote]

a=BC, b=CA, c=AB


One-third of a 30-student class is absent today. One-half of
those were also absent yesterday. What percent of the class has
been absent for two straight days? Express your answer to the
nearest whole number.

Ignoring the number of students in the class, we can find that one-sixth of the students were absent both days. This rounds to 17%.


vector
induction
geometry unsolved
geometry

a) Prove that however we choose seven vectors in 3D, there are 2 which form an acute angle.

b) Are there similar results for higher dimensions?

A rather simple induction should show that there are at most $2n$ vectors in $\mathbb R^n$ with non-pozitive pairwise scalar products.

We may assume WLOG that one of the vectors is $e_1=(1,0,\ldots,0)$. If two of the other vectors are $(x_1,\ldots,x_n),(y_1,\ldots,y_n)$, and we assume them not to be collinear with $e_1$ (i.e. they're not pozitive multiples of $(-1,0,\ldots,0)$), then we have $x_1,y_1\le 0$, and this means that the projections $(0,x_2,\ldots,x_n),(0,y_2,\ldots,y_n)$ must also have non-pozitive scalar product.

This means that we may assume that all the vectors in our set which are different from $(1,0,\ldots,0)$ and $(-1,0,\ldots,0)$ are situated in the hyperplane determined by the equation $x_1=0$. By the induction hypothesis, there are at most $2(n-1)$ such vectors, and adding the extra two we get at most $2n$.


[quote="conejita"]Aki pongo un problem, k a mi parecer ta dificil.
Sea ABC un triangulo no isosceles con H su ortocentro. Sea D el pie de la altura desde A y E el pie de la altura desde B. Sea K la interseccion de la linea DE con AB. Si M es el punto medio de AB, demuestra que CM es perpendicular a KH.

Espero soluciones!!!  :)[/quote]
Me parece que es  conocido que si se tiene ABCD: un cuadrilatero inscrito en una circunf. si M es la intersec. de AB con DC. N es la intersec. de BC con AD y P es la intersec. de AC con BD. El centro de la circunf. es el ortocentro de MNP. Si no es tan conocido me dicen, y pongo su demostracion. Con eso sales pues AEDB es inscriptible en una circunferencia de centro M. Por lo anterior: M es el ortocentro de KHC. Entonces KH es perpend. a MC. Mejor pongo su demostracion de una vez porque nunca escuche esa propiedad como teorema:
sean R y S las intersec. de MN con BC y AD respectivamente. Esto si es conocido: P, A, S y D forman una cuaterna armonica. Luego la polar de P pasa por S. Ademas las rectas MP, MA,MS y MD forman un haz  armonico, entonces P,B,R y C tambien forman una cuaterna armomica. Luego la polar de P pasa por R.
Es decir la polar de P es la recta RS=recta MN. Entonces MN es perpend a PO(donde O es el centro de la circunf.) Analog. NP es perpend. a MO. entonces O es el ortocentro de MNP.

Teorema(H. Brocard): En todo cuadrilátero cíclico ABCD , su circuncentro es ortocentro de su triángulo autopolar.
Nota: El triángulo autopolar es el triangulo formado por los puntos de corte de las diagonales (AC, BD) y los dos pares de lados opuestos (AB,CD), (BC,AD).
Se puede ver una demostración en pdf:
http://www.mathlinks.ro/Forum/viewtopic.php?t=15133
En español:
http://www.aloj.us.es/rbarroso/trianguloscabri/sol/sol168sal.htm

Ola chen!!
Oye eres resultado es my bueno, peo yo no soy muy weno n geometria proyectiva, te agradeceria que si puedes, pongas una solucion un pokito mas euclidiana, si es que la conoces.
O si no poner una solucion del problema origimnal diferente (kreo k una ayuda seria usar la circunferencia de los nueve puntos del triangulo ABC).
Espero cualquiera de las 2 cosas-!!!

Ah y grax x la otra solucion!!! :P

Esta es mi solución:
Tracemos una perpendicular a KH por C que intersecte a AB en M’, como M es punto medio de AB entonces lo que queremos demostrar es equivalente a M=M’

Sea F el punto de intersección de CH con AB como H es ortocentro entonces CF perpendicular a AB.
Denotemos a R como la intersección de M’C y KH.
Entonces <HRM’=<HFM’=90º por lo que cuadrilátero FHRM’ es cíclico,
Como <HRC=<HEC=<HDC=90º entonces pentágono  EHRDC es cíclico. Entonces HR es el eje radical de las circunferencias circunscritas a esos polígonos. Como K esta en el eje radical, entonces KF • KM’ =KE • KD, y como EFMD es cíclico (circunferencia de los nueve puntos) entonces KF • KM=KE • KD entonces KF • KM’=KF • KM entonces KM’=KM por lo tanto M=M’ lo cual queríamos demostrar.
jeje :lol:

:wink: SEA F EL PIE DE LA ALTURA DE C y T EL PUNTO DE CORTE DE CM CON KH.
sea: KB = c, BF=a, FM=b, MA=a+b, CF=h, HD=d.
NOTAMOS QUE K,B,F,A SON CUATERNA ARMONICA POR TEOREMA YA QUE K ES LA INTERSECCION DE ED CON AB, Y CF,AD,EB CONCURREN.
=>c(2b+a)=a(2(a+b)+c)->2bc=2a(a+b)->bc=a(a+b), ab+bc=a(a+2b).
b(a+c)=a(a+2b)....(1).
POR LA SEMEJANSA ENTRE LOS TRIANGULOS, CBF Y AHF , TENEMOS QUE:
a/n=h/a+2b->hn=a(a+2b), pero de (1). hn=b(a+c)...(2)
SEA <HKF=X, <FCM=Y, DE 2 TENEMOS QUE: Tang(x) =Tang(y), por lo tanto X=Y, => <CTK=<CFK=90º.  :D

[quote="conejita"]
 yo no soy muy weno n geometria proyectiva, te agradeceria que si puedes, pongas una solucion un pokito mas euclidiana, si es que la conoces.[/quote]
una pregunta eso es geo proyectiva o  polos y polares??, :huh:


$ a$ is real number such that $ a \ge 1$. Find the Min value of: $ y \equal{} \sqrt {a \plus{} \cos x}  \plus{} \sqrt {a \plus{} {\mathop{\rm s}\nolimits} {\rm{inx}}}$

PS: I hate to solve because the solution is too long!!! :(

$ y' = \frac {c}{2 \sqrt {a + s}} - \frac {s}{2 \sqrt {a + c}} = 0$

$ \iff c \sqrt {a + c} = s \sqrt {a + s}$

$ \implies c^3 + ac^2 = s^3 + as^2$

$ \iff (c - s)(1 + cs + (c + s)a) = 0$

In the first case, we get solutions of the form $ \left( N + \frac {1}{4} \right) \pi$.

For the second case, we can write $ s$ in terms of $ c$ and solve. It is ugly to compute. The second case only applies for $ x$ that's slightly above $ 1$ though; after that, it becomes unsolvable in the reals.

When the second case is solvable in the reals it produces the minima.

When it is not, then we can easily show that $ x = \left( 2 N + \frac {1}{4} \right) \pi$ produces maxima and $ x = \left( 2 N - \frac {3}{4} \right) \pi$ produces minima.


On the blogs page it says that to create a blog, you click "Blog options" but there was no such button.  Can somebody help me???

Go to your profile and you should see this. See the blog options menu?

Ooooooh.  That would probably make sense.  People can lock/delete this now.


geometry
perimeter
algebra
function
domain
trigonometry
real analysis

I need to find a formula that calculates the the largest equal distance between points of a perimeter, while not letting the bisection of a line between two congruent points become larger than 2 feet from the edge of the perimeter. The initial problem I know how to do.  Is there a formula that does this?  :surrender:

Perimeter of what? If you are talking about an arbitrary (say, convex) domain, then, alas, there is no simple formula and the problem is, most likely, not solvable analytically. If you are talking about something simple like a triangle or a square or a circle, then the problem is fairly easy. Or, maybe, I completely misunderstood what you said: some expressions you used like "congruent points" remain a mistery to me. My interpretation of the question was that you take a fixed curve, consider all pairs of points on the curve such that the distance from the middle of the interval joining them to the curve does not exceed $2$, and ask what the maximal distance between such a pair of points may be.

The object is a circle. As an example: a circle with a diameter of 32 feet is broken into 6 equal parts to keep the distance between points below 20 feet.  Take point A and draw a straight line to point B.  Bisecting AB and measuring the distance from the bisection to the edge of the perimeter.  That distance must be equal to or less than 2 feet.  The formula I am trying to find is:  If I plug in different diameters for the circle.  To come up with the [b]Least Number[/b] of (equal distance) points, that keep AB's center from being greater than 2 feet away from the perimeter edge.  I hope I answered this correctly for you?  I wish I had more mathmatical terms for you but I don't. Sorry

Now it is clear :) As to mathematical terms, if you are unsure about their exact meaning, the fewer you use, the better :P. Now back to the question. Let $r>2$ be the radius of the circle. Let $2\alpha$ be the angle between the radii ending at the points $A$ and $B$. Then the distance from the middle point of the interval $AB$ to the circumference is $r(1-\cos\alpha)$. To make it less than or equal to $2$, we must take $\alpha\le \arccos\frac{r-2}{r}$. Thus, the desired number of points is just the least integer exceeding or equal to $\frac \pi{\arccos \frac{r-2}{r}}$. Hope, this helps :).


[quote="byrdgyrl"]Help explain this to me please.  How come when a nine is to an even power the last digit is 1 and when nine is to an odd power the last digit is a 9?  I would also welcome other tricks and an explination of how they work.[/quote]

Well... I guess it's just 9*1*9*1*9*1 because it goes... 9^1=9 9^2=8[b]1[/b]  then you have the 9*1 which would result in another 9 and then 9*9=8[b]1[/b]

It's not a very good explanation but i think it should help you.

okay, 9^1 = 9.... and when you multiply the last digit (which is 9) by 9, you get 81, and when you get the last digit of 81, it's 1, and when you multiply that by 9, you get  9 ... and it starts all over again

okay, after I read my explanation, it sounds kind of confusing but I don't know how to explain it the way I want to explain it, but if you don't understand, then just totaly ignore it

[quote="Treething"]Because the math gods stated it should be so.[/quote]

Thanks everyone!


find all function f: R----->R such that for all x in R.

f(f(f(x)))-3f(f(x))+6f(x)= 4x+3

Mods, please move this to Algebra section...


[quote="lovemathvn"]Let ABCD be a quadrilateral where $ \widehat{ABC} \equal{} \widehat{ADC} \equal{} 90^0$ and $ \widehat{BCD} < 90^0$.Choose a point E on the opposite ray of AC such that DA is the angle-besector of BDE.Let M be the chosen arbitrarily between D and E.choose another point N on the opposite ray of BE such that $ \widehat{NCB} \equal{} \widehat{MCD}$.Prove that MC  is the angle-besector of DMN

 my solution for this problem uses cyclic quadrilateral.Now could you find some another solutions for me?thanks:D[/quote]
this topic hasn't had any responds yet:(

Let $P \equiv AD \cap BC$ and $Q \equiv BA \cap DC.$ Then $A$ becomes the orthocenter of the acute $\triangle CPQ$ and $\triangle BED$ is its orthic triangle. Thus, $EC$ bisects $\angle BED$ and $C$ becomes the E-excenter of $\triangle EBD.$ If $\angle MCD=\angle NCB,$ it follows that $\angle MCN=\angle DCB$ $\Longrightarrow$ $\angle DCB=\angle MCN=\angle90^{\circ}-\frac{_1}{^2}\angle BED,$ which implies that $C$ is common E-excenter of $\triangle BED$ and $\triangle NEM$ $\Longrightarrow$ $MC$ bisects $\angle DMN.$


Arne wrote:Solve in integers :

mn + 1 = m + 2mn + 2m + n.

(n+1)(m+1):^2:=2(m:^2:n+1). hmmm...that doesn't go anywhere. [hide]Let' try...n=(m:^2:+2m-1)/(m:^2:-2m-1)=1+4m/(m:^2:-2m-1). If n is an integer, then (m:^2:-2m-1)|4m. So, either m=0, or the magnitude of 4m is greater than the magnitude of m:^2:-2m-1. Ok, i'll find where m:^2:-2m-1 is positive. m:^2:-2m-1:ge:0 => (m-1):^2::ge:2 => m:le:1-:sqrt:2 or m:ge:1+:sqrt:2. As m is an integer, this is equivalent to saying m:le:-1 or m:ge:3. Now, if m:ge:3, 4m will be positive. So, saying that the magnitude of 4m is greater than the magnitude of m:^2:-2m-1 is equivalent to saying that m:^2:-2m-1:le:4m, as both quantities are positive. This is equivalent to  (m-3):^2::le:10 => 3-:sqrt:10:le:m:le::sqrt:10+3. As m is an integer, this is equivalent to saying 0:le:m:le:6. Thus, for m:ge:3, the only possibilities are in the range 3:le:m:le:6. Now, we do a similar analysis for the case m:le:-1 to find the only possibilities there are in the range -2:le:m:le:-1. Thus, all possibilites for m must lie in the range -2:le:m:le:6. We try each value to find the solutions. [/hide]


Euler
algebra
polynomial
modular arithmetic
search
function
number theory unsolved

let P(x) be poly. with integer coeff. s.t. P(2^i) is perfect m th powerfor all natural i.(m natural and fixed)show P is itself m th power. :!:

:D ,igot.
  std. pro. is if result  is true for all naturals  then we are done.for reference barbeau.
  so,it suffices to prove set of prime divisors of nos. of p(2^i) is infinite,whichcan be shown by eulers extension of fermat.
 willgive details,if want.

Yes, I really see easily how to prove that there are infinite distinct primes dividing the numbers $p(2^{i})$, but how does this help solve the problem?

I hope you will supply the details of how to prove that $p$ is the mth power of a polynomial from this infinity of primes.

Thanks in advance.

do youhave barbeau bookon polynomials?
there is given proof of result,if truefor alll naturals.
will post complete sol. if you dont have.please ask. :D

Yes, I have Barbeau's Polynomials... a problem book.

In what page of this book is solved the problem?

Nonetheless it would be nice when someone posts it here ;)

Edit (deleted some nonsense):

Yes, ociretsih, you are right. Sorry. Unfortunately, I do not have the time to try to fix it now.

Grobber, I take a look at the post where you prove that given any polynomial $g(x)$ irreducible in $\mathbb Z[x]$, there are infinite primes $p$ and values $n$ such that $p\|g(n)$.
But this DOES NOT APPLY TO OUR CASE!
Imagine that $p^{2}|g(2^{a})$, and that e is the order of 2 modulo p, that is, e is the smallest natural number such that $2^{e}\equiv 1\pmod p$. What if $p^{2}|2^{e}-1$? Then $p^{2}$ divides $g(2^{a+e}), g(2^{a+2e}), g(2^{a+3e}),\dots$ and you cannot proof there is a single value such that $p\|g(2^{x})$.

So, grobber, if you haven't this idea in your head, but another one, please post it.
And a message to "who": please post the solution.

sorryfordelay,infactforall n thiswas some previous romanian pro.,
let, f is poly. with int. coeff. and f(n)=Kn^p ,p fixed>=2.
suppose f^1/p=h (f1^a1*f2^a2*...*fk^ak)^1/p where 1<=a1,a2,...,ak<=p-1,fi,fjmutually coprime(1,=i,j,+k) and fi irreducible.
let Uijfi+Vijfj=dij(1)(i not equal j)allso (fi,fi#)=1 in Q(x), let wifi+zifi3=c(2)i(1,=i<=k) here dij,ciare non zero integers.
let q be prime s.t.
1.q >max(mod ci,mod dij) i,j.
2.q is prime divisor og f1(r) for some r a natural no.
note 2. followsfrom obvious fact that set of prime divisors of f1(1),f1(2),... is infinite,
we have q does not divide any of fj(r) where j=2,3,..,k.
from (1)
so,q^p divides (f1(r))^a1,q^2 dividesf1(r).
now using f1(r+q)-f1(r) cong. to q*f1#(r)  (mod q^2)(3)
we see q divides f1(r+q) and as before
q^2 dividesf1(r+q).
again from (3)we get q divides f1#(r),which contradicts (2)
q,.e.d.
inour case proofis same except to prove thing which i mentioned earlier. :)

You're trying to do the same kind of proof that grobber proposed before, but it does not work for the same reasons.
Take a look at my post above entitled "It's not that easy, grobber" that explains why this proof DOES NOT WORK.
If you don't want to bother reading the post above here I detail where your attempted proof fails: your congruence

q*f1#(r) (mod q^2)

is replaced in our case by

(2^(a+ke)-2^a)*f1#(2^a) (mod q^2)

where e is the least natural such that 2^e-1 is a multiple of q, but WHAT IF 2^e-1 IS A MULTIPLE OF q^2???

Hi, guys. Some very long time ago I posted the problem with $p(2^{n})$ perfect square and after quite a long time two beautiful solutions were found: by Peter Schoelze and Iura. Iura generalized the problem to this point. I suggest you search on the forum that discussion and see whether one can adapt the proof (I am sure one can do that).

I finally found the solutions to the square case from Peter Sholze and iura as an off-topic discussion [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15102]here[/url]. Now comes the proof for the mth power case. We start with the following

LEMMA. If $P(x)\in\mathbb Z[x]$ has all non-zero complex roots multiple a multiple of m times (that is, $P(x)=ax^{r}Q(x)^{m}$ in $\mathbb C[x]$) and $P(2^{i})$ is a perfect mth power for all natural i, then $P(x)=R(x)^{m}$ in $\mathbb Z[x]$.
PROOF. The irreducible factors of $P(x)$ in $\mathbb Z[x]$ have simple and distinct roots, so each one should appear a multiple of m times as factor, unless it is the factor $x$, and so $P(x)=ax^{r}S(x)^{m}$ with $a\in\mathbb Z$, $S(x)\in\mathbb Z[x]$. This implies that $ax^{r}=P(x)/S(x)^{m}$ is also a perfect mth power for $x=2^{i}$, and from this follows easily that $a$ is a perfect mth power and $m\mid r$.

Now, using the lemma above, it is sufficient if we proof that every non-zero complex root of $P(x)$ has a multiplicity multiple of m.

FIRST CASE. Suppose $P(x)$ has degree multiple of m and its leading coefficient is a perfect mth power.
We have the Taylor series around $\infty$: $P(x)^{1/m}=\sum_{i=\deg P(x)/m}^{-\infty}a_{i}x^{i}\in\mathbb Q[[x,x^{-1}]]$. Now putting in this series the values $x=2^{i}$ it is straightforward to prove that the sum of the negative power terms shoud vanish for every $i$ big enough, what implies that those negative power terms are null, so that $P(x)^{1/m}\in\mathbb Q[x]$.

SECOND CASE. Suppose $m\not|\deg P(x)$ and/or the leading coefficient of $P(x)$ isn't a perfect mth power.
For this case we have two proofs, following Peter Sholze or iura.

iura-LIKE PROOF. The polynomial $P(x)P(2^{m}x)P(2^{2m}x)\cdots P(2^{m(m-1)}x)$ is in the first case, so it's a perfect mth power of another polynomial and so, its every complex roots have multiplicity multiple of m. Take $\alpha\in\mathbb C$ a complex root of $P(x)$ whose multiplicity isn't a multiple of m, if any, with $\|\alpha\|$ greatest. Then $\alpha$ should be a root of some $P(2^{mk}x)$ of multiplicity not a multiple of m, but now $2^{mk}\alpha$ is a root of $P(x)$ of multiplicity not a multiple of m, so that $\|2^{mk}\alpha\|\leq\|\alpha\|$ implying that $\alpha=0$.

Peter Sholze-LIKE PROOF. The functions $P(x)^{1/m}$, $P(2^{m}x)^{1/m}$, $P(2^{2m}x)^{1/m}$, $P(2^{3m}x)^{1/m}$,... have Taylor series around $\infty$ belonging to $a_{\deg P(x)}^{1/m}x^{\deg P(x)/m}\mathbb Q[[x^{-1}]]$, where ${a_{\deg P(x)}\in\mathbb Z}$ is the leading coefficient of $P(x)$. Now it is evident that exists a non-trivial linear combination over $\mathbb Z$ $\sum_{j=0}^{j=l}c_{j}P(2^{jm}x)^{1/m}=o(1)$ where we can suppose wlog that $c_{0}\neq0$, $c_{l}\neq0$, $l>0$ (just eliminate the positive powers of $x$ in the Taylor series by an integer linear combination). But evaluating at $x=2^{i}$, for $i$ big enough, we see the left hand side of this equality, being integer, should be zero, and from this we have the identity $\sum_{j=0}^{j=l}c_{j}P(2^{jm}x)^{1/m}=0$. Take $\alpha\in\mathbb C$ a complex root of $P(x)$ whose multiplicity isn't a multiple of m, if any, with $\|\alpha\|$ greatest. This is equivalent to $\alpha$ being a brach point of $P(x)^{1/m}$, that is, if we take a small loop around $\alpha$ in the complex plane and we do analytic prolongation of $P(x)^{1/m}$ we obtain different values corresponding to other complex mth roots, but then at least one of the $P(2^{jm}x)^{1/m}$, let's say $P(2^{mk}x)^{1/m}$, must take a different value because we have a zero sum. So $\alpha$ should be a root of some $P(2^{mk}x)$ of multiplicity not a multiple of m, but now $2^{mk}\alpha$ is a root of $P(x)$ of multiplicity not a multiple of m, so that $\|2^{mk}\alpha\|\leq\|\alpha\|$ implying that $\alpha=0$.


In how many different ways can 12 dimes be divided into three
piles with an odd number of dimes in each pile?

You can't divide an even number integrally into an odd number of groups odd numbers added together, since two odd numbers multiply to make an odd number. The situation is therefore impossible, making the answer $ \boxed{0}$.


How can I tex the following function:

$ f(x,y)\equal{}(x^2\plus{}y^2)\sin{(\frac{1}{\sqrt{x^2\plus{}y^2}})}$ when $ (x,y)\neq(0,0)$
$ f(x,y)\equal{}0$ when $ (x,y)\equal{}(0,0)$.

I'd like to be able to tex it using the big $ \{$ symbol, as in $ f(x,y)\equal{}\{\dots$ with upper and lower entries, but I don't know how.

Help please? Thanks.

Use the cases enviroment:
\[ f(x,y)\equal{}
\begin{cases}
(x^{2}\plus{}y^{2})\sin{\left(\cfrac{1}{\sqrt{x^{2}\plus{}y^{2}}}\right)} &\text{ when } (x,y)\neq(0,0)\\
0 &\text{ when } (x,y)\equal{}(0,0)
\end{cases}
\]
Note: To get the larger fraction in a document you should use \dfrac rather than \cfrac which I had to use above for this forum.

Strangely, my TeXnicCenter doesn't recognize several of the control sequences you used in that coding. Is there a particular packet I need for it to work?

You need \usepackage{amsmath} in the preamble, but then you should have that in all your documents.


induction
trigonometry
algebra unsolved
algebra

[quote="suedenation"]Can anyone prove the following using induction?

[color=green]Suppose that x is a real number that is not an integer multiple of "pi". Prove using induction that

cos(x)cos(2x)...cos(2^(n-1)x) = [sin(2^n)x]/(2^n)(sinx)

for all integers n>= 1. [/color]

Thanks very much!!!  :lol:[/quote]
The correct formula should be $\cos x \cos 2x \cdots \cos 2^{n-1}x=\frac{\sin 2^nx}{2^n\sin x}$

I think this is what you need only.
$\frac{\sin 2^k x}{2^k\sin x}\cdot\cos 2^k x=\frac{\frac12 \sin 2(2^k x)}{2^k\sin x}=\frac{\sin (2^{k+1}x)}{2^{k+1}\sin x}$

In fact, your formula can also be proved without applying induction. Consider
$\sin x \cos x \cos 2x \cdots \cos 2^{n-1}x$
$=\frac12 \sin 2x \cos 2x \cdots \cos 2^{n-1}x$
$=\frac14 \sin 4x \cos 4x \cdots \cos 2^{n-1}x$
$\cdots$
$=\frac{1}{2^n}\sin 2^nx$
Dividing both sides by $\sin x$ gives the result. ;)


hwo know how to download the china tst 2008 Quizzes and solution?

http://www.mathlinks.ro/Forum/resources.php?c=37&cid=47&year=2008
Maybe this is useful

Daniel

Daniel I think he is not referring to the China IMO Team Selection Tests but rather to the quizzes which we only can offer for the [url=http://www.mathlinks.ro/resources.php?c=37&cid=47&year=2006]year 2006[/url] unfortunately. If some Chinese members want to contribute you are more than welcome.  :)

Can Fang-jh post solution China TST 2008 and China TST 2008 quizzes?the solution can be made in Chinese.

Fang-jh:I hope you 'll come and join this topic:http://www.mathlinks.ro/viewtopic.php?t=247236&sid=4af50436ab29194e496b801d25dd37d5;http://www.mathlinks.ro/viewtopic.php?t=101694

[quote="Fang-jh"]All China IMO 2008 team all quizzes in [url]http://www.mathlinks.ro/viewtopic.php?t=198095[/url]#4, but  nobody put them in resources section![/quote]

Sorry I did not notice. I just added them. Just send me a pm next time! Do you have the quizzes for any other year?

[quote="math10"]Can orl post problem China tst 2007 quizzes ?[/quote]

No, I cannot. This is because I do not have them.  :wink: We kindly need to wait for Fang-jh. But I can add them to the resources section then.

[quote="math10"]Can Fang-jh post solution China tst 2008, 2007 and quizzes in this topic?[/quote]

No, the problems have been posted to their own threads where the discussion should take place. You can find the links from the resources section.

Hello orl,

There are quizzes from China TST 2003 and 2005:
http://www.mathlinks.ro/viewtopic.php?p=561072#561072
http://www.mathlinks.ro/viewtopic.php?p=287839#287839

And TST Quiz 2004: http://www.mathlinks.ro/Forum/viewtopic.php?t=5283

Why don't you add them to the resources section?  :)

By the way, the CMO problems were translated by a group of Chinese guys a long time ago. Can we add them to the resources section now?

[quote="orl"]Have the problems from the thread for China TST 2004 quizzes been posted posted to the forum yet? [url=http://www.mathlinks.ro/viewtopic.php?t=154052]Here[/url] we discuss mock problems. Is it that mock problems are just used for training but not for selecting the team which is based on weighting of the CMO, TSTs and Qiuzzes?![/quote]
The mock problems are used for training. I'll post the TST 2002 Quizzes in a few days when I have time.


I have a photo in my PC, but it is a file .bmp, but here it says: The avatar filetype must be .jpg, .gif or .png  .  How can I solve this?

Another: Can I put a photo in my signature? Thanks.

1. copy paste or something to the picture to save it in a different format. or upload it to a website or something.

2. Yes you can use the image buttons. [code][img]http://www.artofproblemsolving.com[/img][/code]

2. And if I have it on my own PC? Because I can put it if it is on Internet

upload it as an attachment in the test forum

Done. Here's the link. http://www.mathlinks.ro/Forum/viewtopic.php?p=592234#592234

Now?? I want to put it in my signature

Let see if it works..EDIT:no it doesn't! is that the right code?

Dunno why it didn't work. I think the code is wrong.

The link to the picture is seems not to be seen as picture by the system.
Could also be that the admins don't want pictures to be stored on MathLinks...
At least, I suggest to store the picture somewhere else on the web and link there.

That photo you posted is way too big to be in an avatar.

[quote]Only one image can be displayed at a time, its width can be no greater than 80 pixels, the height no greater than 100 pixels, and the file size no more than 21 KB.[/quote]The photo you posted is 96x150 pixels.

Use imageshack to make the picture avatar size, and upload it to the internet as well.  This is a free service, with no signup required.

http://www.imageshack.us

Just upload the file, and resize the image to avatar using the appropriate button.

mysmartmouth: Thank you so much!!!! Now, after trying without success for 2 days, I could post this photo... thanks again!!!

And thanks to everyone that helped me.

The photo is nice, isn't it?? (I'm not him)

[quote="MichaelFaraday"]How do I change the Pixel Count in an image[/quote]
I recommend Adobe Photoshop 7.0.

baah. too complicated for such a simple thing. Picasa will do it also

ok, I used MS Photo editor, thanks for the advice


inequalities
function
inequalities proposed

Welcome back Ligouras, I have corrected the problem in a different way, hope you like it

$ \prod(a{}^{1}{}^{0}-2a^{7}+5a^{5}+a^{4}-7a^{2}+7a+4)\geq(a+b+c)^{6}$

there is a nice proof for this inequality as follows

$ \prod(a{}^{1}{}^{0}-2a^{7}+5a^{5}+a^{4}-7a^{2}+11) \ge (a^3+2)^2$

After expanding all the expressions above, we get the following function which is positive for all $ a >0$

$ (a-1)^2(a^7+2a^6+3a^5+2a^4+3a^2+7a+7) \ge 0$

Then, Holder does help to find out the solution for this inequality

$ \prod (a^3+1+1) \ge (a+b+c)^3$

We are done!

[quote="great math"]Welcome back Ligouras, I have corrected the problem in a different way, hope you like it

$ \prod(a{}^{1}{}^{0} - 2a^{7} + 5a^{5} + a^{4} - 7a^{2} + 7a + 4)\geq(a + b + c)^{6}$

there is a nice proof for this inequality as follows

$ \prod(a{}^{1}{}^{0} - 2a^{7} + 5a^{5} + a^{4} - 7a^{2} + 11) \ge (a^3 + 2)^2$

After expanding all the expressions above, we get the following function which is positive for all $ a > 0$

$ (a - 1)^2(a^7 + 2a^6 + 3a^5 + 2a^4 + 3a^2 + 7a + 7) \ge 0$

Then, Holder does help to find out the solution for this inequality

$ \prod (a^3 + 1 + 1) \ge (a + b + c)^3$

We are done![/quote]

Dear Great-math, has not corrected the problem.
Thanks for the change and for the solution.
I propose you the following problem:

$ \prod(a{}^{1}{}^{0} - 2a^{7} + 7a^{5} + a^{4} - 7a^{2} + 12)\geq(a + b + c)^{6}$

Thanks for your inequality. My idea is to use the same idea as above

$ a{}^{1}{}^{0}-2a^{7}+7a^{5}+a^{4}-7a^{2}+12 > (a^3+2)^2$

It is proven by the following estimation

$ (a-1)^2(a^8+2a^7+3a^6+a^5-3a^4+2a^3+8a^2+10a+5) + a^7+ a^6 + 3 >0$

[quote="great math"]Thanks for your inequality. My idea is to use the same idea as above

$ a{}^{1}{}^{0} - 2a^{7} + 7a^{5} + a^{4} - 7a^{2} + 12 > (a^3 + 2)^2$

It is proven by the following estimation

$ (a - 1)^2(a^8 + 2a^7 + 3a^6 + a^5 - 3a^4 + 2a^3 + 8a^2 + 10a + 5) + a^7 + a^6 + 3 > 0$[/quote]

They agree. The problem is also resolvable with a decomposition. Aspect :wink:


It's not possible...$ \sqrt {p^4 \plus{} 8p^3 \plus{} 16p^2}\ge0$, so $ \sqrt {p^4 \plus{} 8p^3 \plus{} 16p^2} \equal{} \minus{} 4$ is impossible if $ \sqrt{p^4\plus{}8p^3\plus{}16p^2}\in\mathbb R$.

But $ \minus{} 4\in \mathbb{R}$ :wink:

While it's true that in most conventions it's preferred that $ \sqrt {x^2} \equal{} |x|$, it's not always the case. If I was to say $ \sqrt {4} \equal{} 2$, I can also say it equals $ \minus{} 2$ since $ 2^2 \equal{} ( \minus{} 2)^2 \equal{} 4$.

But to avoid being ambiguous, we generally let $ \sqrt {m}$ be positive.

[hide="Answer"]$ \sqrt {p^2(p \plus{} 4)^2} \equal{}p^2 \plus{} 4p \equal{} \minus{} 4 \implies (p \plus{} 2)^2 \equal{} 0 \implies \boxed{p \equal{} \minus{} 2}$

$ \sqrt {q^2(q \minus{} 6)^2} \equal{} q^2 \minus{} 6q \equal{} \minus{} 9 \implies (q \minus{} 3)^2 \equal{} 0 \implies \boxed{q \equal{} 3}$

$ \sqrt {r^2(r \minus{} 12)^2} \equal{} r^2 \minus{} 12r \equal{} \minus{} 36 \implies (r \minus{} 6)^2 \equal{} 0 \implies \boxed{r \equal{} 6}$

So substituting the values, we get $ \frac {pq}{rq \plus{} 2p} \equal{} \frac { \minus{} 6}{18 \minus{} 4} \equal{} \boxed{ \minus{} \frac {3}{7}}$

Again, this may still be ambiguous. Does $ \sqrt {p^2(p \plus{} 4)^2} \equal{} \minus{} (p^2 \plus{} 4p)$ as well? In this case, I'm assuming they're keeping the standard convention with the given and similar examples with evaluating $ \sqrt {\text{something}}$.[/hide]

[b]Patterns_34[/b],I agree to your point. Anyway... Your answer Is what I'm expecting and it's right!

[quote="Patterns_34"]But $ \minus{} 4\in \mathbb{R}$ :wink: [/quote]

My point was that square roots are nonnegative in the reals, if you looked at the whole thing...


probability
number theory
relatively prime

Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[hide="hint"]Find the probability that there is a 2x2 red square using Inclusion-Exclusion Principle.  Then subtract that from 1 to find the probability there is a 2x2 red square.[/hide]

Very very unsure about this answer, but here goes anyways.
[hide="Ugly counting and junk"]The number of ways to obtain a 2x2 square (with no other 2x2 squares) is 2 colors for middle x 4 corners to choose from = 8 different ones. (Top-left red, top-left blue ... bottom-left blue.) Now for the other squares that aren't included in this 2x2, which is a total of 5 squares. These squares can be any color so long as another 2x2 is not formed (or else the number of combinations will be overcounted). There are 3 combinations for each pair of middle-corner squares that are right next to the 2x2 already formed, so a total of 9 combinations there. The last corner may not be the same as the center if the two squares next to it are the same as the center, which is 1 of the 9 combinations. For the other 8, it can be either color, so 8x2+1=17 combinations for the other squares. So now we have 8x17 combinations of working 2x2 squares. Lastly, count the overlaps. Not sure if it's right, but I counted 12 overlaps, not too sure how to explain other than count the cases of multiple squares and multiply them by the possibilities for the blank squares. So 8x17+12=146 total combinations for 2x2 squares, and of course $2^9$ total combinations, and $\frac{146}{512}$ reduces to $\frac{73}{256}$ so $m+n = 73+256 = 329$?
[/hide]
Once again, very unsure.

Hello? Does anyone have an answer for this? I'm curious about how you'd do it nicely...

[hide]So if you have (sorry for the bad picture)
[code]
_______
|  |   |   |
---1---2---
|   |   |  |
---3---4---
|   |   |  |
------------
[/code]
As your 3x3 grid:

First consider a 2x2 square centered at location "1".  Then the other 5 spaces have 5 options, so the probability of there being a 2x2 square centered at location "1" is $\frac{2^5}{2^9}$.  Since there are four locations, we have $4\left(\frac{2^5}{2^9}\right)$.

However, we have to subtract the probability of there being 2 2x2 squares. If there are squares in both places 1/2, 1/3, 3/4, 2/4, those have probability of $\frac{2^3}{2^9}$ each, so $4\left(\frac{2^3}{2^9}\right)$.  Then squares in places 2/3 and 1/4 have probability $\frac{2^2}{2^9}$ each.

Now we add the probability of 3 2x2 squares.  There are 4 such options: 1/2/3, 1/2/4, 1/3/4, 2/3/4, each with probability $\frac{2}{2^9}$.

Now we subtract the probability of there being 4 2x2 squares.  The probility of this is $\frac{1}{2^9}$.

Therefore it is $4\left(\frac{2^5}{2^9}\right)-\left(4\left(\frac{2^3}{2^9}\right)+2\left(\frac{2^2}{2^9}\right)\right)+4\left(\frac{2}{2^9}\right)-\frac{1}{2^9}$

This evaluates to $\frac{95}{417}$, and $1-\frac{95}{512}=\boxed{\frac{417}{512}}$.[/hide]

wow, that's a nice way to approach that problem... :D


calculus
integration
trigonometry
function
geometry
geometric transformation
reflection

\[ \int {\frac {{dx}}{{\sqrt [4]{{1 \minus{} x^4 }}}}}
\]

\[ \int {\sqrt [3]{{x^3 \minus{} 6x}}} dx
\]

\[ \int\limits_{ \minus{} 1}^1 {\frac {{dx}}{{\sqrt {1 \plus{} x^2 } \plus{} \sqrt {1 \minus{} x^2 } }}}
\]

\[ \int {\frac {{\sin x \minus{} \cos x}}{{x \plus{} \sin 2x}}} dx
\]

\[ \int\limits_0^\pi {\frac {{a \minus{} b\cos x}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx
\]
Some problem I can't solve :blush:

for the first one,

$ \int^1_0 \frac {dx}{\sqrt[4]{1\minus{}x^4}} \equal{} \int^1_0 \frac {dx}{x}$

making use of the property that

$ \int^1_0 \ f(1\minus{}x)dx \equal{} \int^1_0 \ f(x)dx$

but I don't know the general indefinite integral :(

hello, the integral can't expressed in elementary functions, the primitive function contains the hypergeometric function. Maple 12 gives the following answer
$ \int\frac{1}{\sqrt[4]{1-x^4}}\,dx=x{\2F1(1/4,1/4;\,5/4;\,{x}^{4})}$
Sonnhard.

$ I = \int\limits_{ - 1}^1 {\frac {{dx}}{{\sqrt {1 + x^2 } + \sqrt {1 - x^2 } }}} = \frac {1}{2}\left[ \int\limits_{ - 1}^1 {\frac {\sqrt {1 + x^2 }}{x^2}dx + \int\limits_{ - 1}^1 {\frac {\sqrt {1 - x^2 }}{x^2}dx\right]}}$ (That was just by rationalizing the bottom portion).

Setting $ x = \tan \theta$, we get:
$ J = \int\limits_{ - 1}^1 {\frac {\sqrt {1 + x^2 }}{x^2}dx = \int\limits_{ - \frac {\pi}{4}}^{\frac {\pi}{4}} {\frac {1}{\cos \theta \sin^2 \theta}}d \theta = \int\limits_{ - \frac {\pi}{4}}^{\frac {\pi}{4}} {\frac {\cos^2 \theta + \sin^2 \theta}{\cos \theta \sin^2 \theta}}d \theta}$

$ = \int\limits_{ - \frac {\pi}{4}}^{\frac {\pi}{4}} {\frac {1}{\cos \theta}d \theta + \int\limits_{ - \frac {\pi}{4}}^{\frac {\pi}{4}} {\frac {\cos \theta}{\sin^2 \theta}}d \theta = 4 \tanh^{ - 1}( \tan (\frac {\pi}{8})) - 2 \sqrt {2}}$

Setting $ x = \cos \theta$, we get:
$ K = \int\limits_{ - 1}^1 {\frac {\sqrt {1 - x^2 }}{x^2}dx = - \pi}$.

Hence, $ I = \frac {1}{2}[J + K] = 2 \tanh^{ - 1}( \tan (\frac {\pi}{8})) - \sqrt {2} - \frac {\pi}{2}$

Note: Attila the Hun's post is incorrect. There's an algebra mistake in it.

If you make the substitution (reflection) that he suggests, then

$ \int_0^1\frac {dx}{\sqrt [4]{1 \minus{} x^4}} \equal{} \int_0^1\frac {dx}{\sqrt [4]{1 \minus{} (1 \minus{} x)^4}} \equal{} \int_0^1\frac {dx}{\sqrt [4]{4x \minus{} 6x^2 \plus{} 4x^3 \minus{} x^4}}$

Which is certainly not the same thing as $ \int_0^1\frac {dx}{x}$.

In fact, $ \int_0^1\frac {dx}{x} \equal{} \infty$ while $ \int_0^1\frac {dx}{\sqrt [4]{1 \minus{} x^4}}$ converges to a finite number.

If you make the substitution $ u \equal{} x^4$ then you get $ \frac14\int_0^1u^{ \minus{} 3/4}(1 \minus{} u)^{ \minus{} 1/4}\,du,$

which is a beta integral and can be written as $ \frac14\Gamma\left(\frac14\right)\Gamma\left(\frac34\right).$

Whether that constitutes a "closed form" depend on your attitude towards such expressions as $ \Gamma\left(\frac14\right).$

[quote="Kent Merryfield"]
Whether that constitutes a "closed form" depend on your attitude towards such expressions as $ \Gamma\left(\frac14\right).$[/quote]
And on whether you remember what $ \Gamma(x)\Gamma(1\minus{}x)$ is in terms of more elementary functions ;)

Ah yes - I didn't have my eyes open there. So we actually have

$ \int_0^1\frac{dx}{\sqrt[4]{1\minus{}x^4}}\equal{}\frac{\pi\sqrt{2}}{4}.$

That is to say, $ \Gamma(x) \Gamma(1 - x) = \frac {\pi}{\sin(\pi x)}$.
Which, together with what Kent Merryfield did before gives:

${ \int_0^1\frac {dx}{\sqrt [4]{1 - x^4}} = \frac14\Gamma\left(\frac14\right)\Gamma\left(\frac34\right) = \frac14 \frac {\pi}{\sin(\frac14\pi)} = \frac {\pi }{4 \cdot\frac12 \sqrt2} = \frac {\pi}{2 \sqrt2}} = \frac {\pi\sqrt2}{4}$, as he said.

Setting $ t = \tan \frac {x}{2}$, then we get $ dx = {\frac {2}{1 + t^2}} dt, \cos x = \frac {1 - t^2}{1 + t^2}$ so that we get the following:
${ \int _0^\pi {\frac {{a - b\cos x}}{{a^2 + b^2 - 2ab\cos x}}} dx = \int_0^\infty{\frac {{a - b(\frac {1 - t^2}{1 + t^2})}}{{a^2 + b^2 - 2ab (\frac {1 - t^2}{1 + t^2})}}} \frac {2}{1 + t^2}} dt$

We may need partial fraction decomposition as the next step.


MATHCOUNTS
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I am making a solutions document for a MathCounts team member using TexnicCenter (latex) and I need help with some of the problems!!  Please post nice solutions that I can quote, copy, and paste.  Thanks!  Also, I an numbering these problems with their actual 1994 numbers...

6) The numbers from 1 through 8 are seperated into two sets, A and B.  The numbers in A are multiplied together to get A.  The numbers in B are multiplied together to get b.  The larger of the two numbers a and b is written down.  What is the smallest possible number that can be written down using this procedure?

8) Given the equation $\frac{1}{m}+\frac{1}{n}=\frac{2}{15}$, where m and n are positive integers, find the value of $m+n$ such that m is greater than n and m is not a multiple of n. [hide="Note"]My solution had a bunch of messy equations.  Easier way?[/hide]

10) The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?  [hide="Note"]Once again, my solution had a bunch of messy equations.  Easier way?[/hide]

11) Suppose a number x satisfies $x^2-4x-1=0$.  How much greater is x than its reciprocal? [hide="Note"]I solved by completing the square, finding the positive root, taking its reciprocal, and subtracting.  I got the right answer; is there an easier way?[/hide]

21) Solve for N: $\frac{7!8!9!}{7!+8!+9!}=\frac{8!N!}{9}$. [hide="Note"]I solved this a while back, but I can't solve it now.  Mental block...[/hide]

[quote="chess64"]Since $n\nmid m$...[/quote]

What does that symbol mean?

$a|b$ means $a$ divides $b$.

Likewise $a\nmid b$ means $a$ does not divide $b$.

[hide="Number 21"]Turn every term on left side all into multiples or $7!$ like this:

$\frac{(7!)^3\times 8\times 72}{7!(1+8+72)}$

Cancel the terms to make the denominator 9 and you will end up with this (I hope):

$\frac{(7!)^2\times 8\times 8}{9}$

Then divide the numerator by $8!$ and you will get this:

$\frac{7!\times 8}{9}$

which is simply $\frac{8!}{9}$ so $N=\boxed{8}$[/hide]

Thanks!  Anyone want to try numbers six or ten?


Given two intersecting lines, and a point $P$.
Construct a circle throug $P$, tangent to the two lines.

Hmm...

Construct any circle tangent to the two lines (its center must lie on the angle bisector of the angle containing $P$) and then map it homothetically along the bisector to include $P$? (If $P$ is on any line, then one solution per angle i.e. two solutions total; If $P$ is not on any line, then two solutions within the same angle.)

Can't you can just invert wrt a circle centered at P, and then it's easy?

Here's the diagram:

Take an arbitrary point $O$ on the angle bisector, drop the perpendicular to one side and construct the tangent circle. Line $AP$ intersects that circle in points $X$ and $Y$. Draw $PO_{1}\parallel YO$ and $PO_{2}\parallel XO$ where $O_{1}$ and $O_{2}$ lie on the bisector. Circles $(O_{1},O_{1}P)$ and $(O_{2},O_{2}P)$ are the solutions to the problem.

[quote="Jan"]Exactly what I had in mind, Farenhajt.

K81o7, please describe your construction, it sounds interesting to know how to draw all those inverses.[/quote]

Construct any circle centered at P.  To invert one of the lines, you can invert three points on the line and then draw the circle through the three inverses, or you can use a more specific construction.  So you now want to construct the four tangent lines (in most cases, there are four) of the two inverses of the lines.  So you construct the centers of similitude, construct the tangent lines, and invert back....

Inversion is an extremely useful tool in constructions considering that it's conformal...


What is the units digit of 367^654?

[hide=" Hint"]

7^4 = 1 (mod 10).

[/hide]

I could guess that this should be moved to the middle school contests, or HSB.

[hide="Solution"]
The units digit of $ 367^{654}$ is also the units digit of $ 7^{654}.$ The units digit of powers of $ 7$ cycle in cycles of 4, so \[ 7^{654}\equiv7^{4\cdot163\plus{}2}\equiv7^2\equiv49\equiv\boxed{9}\pmod{10}.\][/hide]


algebra
polynomial
superior algebra
superior algebra solved

Let $k$ be a field, $f(X)$-irreducuble polynomial in $k[x]$ and $K$-finite normal extension over $k$. Prove that if $g,h$-irreducuble factors of $f(X)$ in $K[X]$, then there exists an automorphism $\sigma$ of $K$ over $k$ such that $g=h^{\sigma}$.
Give a counterexample when $K$ isn't normal over $k$.

An easy couterexample: Let $k=\mathbb Q$, $K=\mathbb Q(\sqrt[3]2)$ and $f(x)=x^3-2$. $f$ is irreducible over $k$, and it factors as $(x-\sqrt[3]2)\cdot(x^2+\sqrt[3]2x+\sqrt[3]4)$ over $K$. Two irreducible polynomials of different degree apear, so...

Take any two factors $g,h$ of $f$. When $x,y$ are any roots of $g,h$ over $K$, then there is a $k$-automorphism of the closure of $K$ that sends $x$ to $y$ (since $x,y$ are both roots of $f$ over $k$). This automorphism sends $K$ to $K$ since it is normal, so it sends the minimal polynomial of $x$ to the minimal polynomial of $y$, thus the desired property.


LaTeX
trigonometry
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calculus
calculus computations

im sorry first because i speak a little english and second because i dont know use the latex,howevwe i hope that you help me with thats problems.

1) evaluate the limit when x-0 de  (sin(a+2x)-2sin(x+a)+sina)/x2  without use L'Hopital

2) show that for each positive real numbers x,y    e^(x+y)/xy>=e^2
3) show that if f(x)=a1sinx+a2sin2x+....+ansin(nx) and |f(x)| <=|senx| then   |a1+2a2+....+nan| <=1

1) $ \sin(a\plus{}2x)\minus{}2\sin(x\plus{}a)\plus{}\sin a\equal{}2\sin(x\plus{}a)\cos x\minus{}2\sin(x\plus{}a)\equal{}$

$ \equal{}2\sin(x\plus{}a)(\cos x\minus{}1)\equal{}\minus{}4\sin(x\plus{}a)\sin^2\frac{x}{2}$

Then $ \lim_{x\to 0}\left(\minus{}4\sin(x\plus{}a)\sin^2\frac{x}{2}\right)\equal{}\minus{}\sin a\lim_{x\to 0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\equal{}\minus{}\sin a$

2) Let $ f(x)\equal{}\frac{e^x}{x}, \ x>0$

$ f'(x)\equal{}\frac{e^x(x\minus{}1)}{x^2}\Rightarrow x\equal{}1$ is a minimum point.

Then $ f(x)\geq f(1), \ \forall x>0\Rightarrow\frac{e^x}{x}\geq e$

$ \frac{e^y}{y}\geq e$

Multiply the inequalities:$ \frac{e^{x\plus{}y}}{xy}\geq e^2$

Show that for all positive real numbers $ x$, $ y$:
$ \frac {e^{x \plus{} y}}{xy}\ge e^2$
[hide="My Answer"]$ \frac {e^x}{x}\frac {e^y}{y}\ge e^2$
This is equivalent to proving that $ \frac {e^x}{x}\ge e \forall x \in \mathbb R^ \plus{}$

Let $ f(x) \equal{} \frac {e^x}{x}$
$ f'(x) \equal{} \frac {(x \minus{} 1)e^x}{x^2}$
$ f'(x) \equal{} (x \minus{} 1)\frac {e^x}{x^2}$
The function is strictly decreasing on $ (0, 1)$ and strictly increasing on $ (1,\infty )$.
Thus, the global minimum of the function is at $ x \equal{} 1$. (Since the function is only exists for $ x > 0$)
And from $ f(1) \equal{} e$, the conclusion follows[/hide]

Show that if $ f(x) \equal{} a_1\sin x \plus{} a_2\sin 2x \plus{} a_3\sin 3x... \plus{} a_n\sin nx$ and $ |f(x)| \le |\sin x|$ then $ |a_1 \plus{} 2a_2 \plus{} 3a_3... \plus{} na_n| \le 1$

(Did the first question without realizing it was done already, !\$%?& multiple tabs distracting me :P)


Select 2 cards from a deck of 52.
Find P(at least one is not a heart [u]or[/u] at least one is a face card)

By De Morgan's Law.

P(at least one is not a heart [u]or[/u] at least one is a face card)=1-P(both of two are heart [u]and[/u]  both of two aren't  face card).

[hide="Solution"]$ P(\text{at least one is not a heart } or \text{ at least one is a face card})$

$ \equal{} 1 \minus{} P(\text{both are hearts } and \text{ neither are face cards})$

$ \equal{} 1 \minus{} P(\text{both are hearts})P(\text{neither are face cards})$

$ \equal{} 1 \minus{} \left(\frac {1}{4}\right) \left(\frac {1}{4}\right) \left(\frac {10}{13}\right) \left(\frac {10}{13}\right)$

$ \equal{} \boxed{\frac {651}{676}}$[/hide]

Edit: Whoops, kunny beat me to it.

I got a different answer:

[hide="Solution"]
Like the previous posts said, we want to find $ 1 \minus{} P$, and $ 1 \minus{} P$ is the probability that both cards are hearts and neither card is face card.

The probability that the first card is a heart and not a face card is $ \frac{10}{52}$, since there are ten cards in a deck that fit these requirements. The probability that the second card is a heart and not a face card is $ \frac{9}{51}$, so $ 1 \minus{} P \equal{} \frac{10}{52} \cdot \frac{9}{51} \equal{} \frac{15}{442}$. Therefore, $ P \equal{} \boxed{\frac{427}{442}}$.
[/hide]

I think mathwise is correct.  It seems like the problem wants no replacement

Yes, miyomiyo has the right idea, but forgot to take into account the conditional probability on the second card. Mathwise's answer is correct.


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I've just made a very interesting and seemingly quite useful discovery. I hope I've not reinvented the wheel (again). I'm not exactly sure what field of math. it falls into.

Anyway, it's a simple method to develop a formula for the length of overlap between whatever number (the larger the number, the larger the formula, obviously) of segments on the abscissa (of course, you can consider these segments to be the projections of segments aligned to whatever line in the plane and use the slope to correct the result for the projected segments). It's actually quite simple. I'm not even sure if it deserves the attention I give it, but seeing as how I've found nothing on the subject either on the web or in my math books...

My problem now is how to register this work as to obtain some sort of recognition for it. Not money, obviously.

I want to make it public. But I would very much like to ensure I collect the credit for what is essentially my work. How does one go about achieving this? Should I write some article on the matter along with explanations and proof? But to whom do I submit it to? Do I need to write to some official mathematics journal? How can leisure time math researchers, like me, claim credit for their discoveries? I am no more involved in academia than being a student but this project of mine has nothing to do with the faculty curriculum. pls. help.

What the heck!? You actually did it!?

You deserve more credit than I gave...

No. I haven't solved the problem of area of overlap between whatever 2 triangles, if that's what you mean. Meanwhile I tried something else. Which I have managed to solve.

I have found the formula for the length of overlap between whatever 2 segments (obviously on the same line, with a reference point on that line) and a method to theoretically finding the formulas for the length of overlap between 3, 4, 5, etc. segments.

I'm sure I'll eventually solve the triangle overlap area problem too. I only shifted to the segments problem in lack of any significant progress on that one for some time now and to perhaps gain insight and new ideas on how to best approach it.

Err... if you don't mind my asking, what's so hard about finding the overlap between two segments?

[quote="maersklandro"]Nothing really. You can program a function to do it. But this is a mathematical expression that does the same thing.[/quote]
Then why is it worth publishing?

just because you can code a function to do this doesn't mean it a math. solution shouldn't be researched, does it?

If you think you have discovered a result worthy of publication, you should write it up in article form and submit it to a journal.  There are journals designed to accept papers from casual (that is, non-specialist) mathers.  I know of one in particular, the name of which temporarily escapes me -- I'll find it by Monday, hopefully.  In general, though, "solving a math problem" is not the sort of thing that you can get legal protection/mandated recognition for.

well no. but i find most journals go to lengths to retain the copyright to the various works they publish. of course, math discoveries are not inventions to be readily patented. that would be ridiculous.

By the way, since you still seem to be interested in your triangle problem, let me give you a function that replaces part of the casework:

I don't know if you consider "min" and "max" and "sign" to be "mathematical" in the sense you are looking for (I presume not -- a solution to the question you say you've solved using min and max is very, very simple), but we can create equivalent functions starting from continuous functions.  Min and max arise from a clever use of the function $\frac 12 (x + |x|)$, while the sign function on positive and negative numbers arises from $sign(x)=\frac{x}{|x|}$ (although of course this is problematic for x = 0).  The case I can give you is "one triangle is entirely inside of another."  This is clearly important since in every other case, we have edges of the two triangles intersecting, which in theory we can detect with another function similar to the one I'm about to construct.

What we really need is the function, "a point is entirely within a given triangle."  We can rework this condition: a point is inside a given triangle when, for each pair of vertices of the triangle, it is on the same side of the line they determine as the third vertex.  (This is the same as saying that the inside of a triangle is the only part of the plane to the left of all sides of the triangle, travelling counter-clockwise.)  Say we have points $P=(a, b)$ and $Q=(c, d)$.  The following function gives 1 if $(x, y)$ and $(w, z)$ are on the same side of the line $PQ$ and 0 if they are on opposite sides:

$\frac 12 |sign(x_1y_2 - y_1x_2 + w(y_2-y_1) - z(x_2 - x_1)) +$ $sign(x_1y_2 - y_1x_2 + x(y_2-y_1) - y(x_2 - x_1))|$

(I leave this to you to prove.)  Now, if you have a triangle with known vertices and another point, you get a "detector function" for "point is inside the triangle" by multiplying 3 copies of this expression together.  (In each case, $(x_1, y_1)$ and $(x_2, y_2)$ will be two vertices of the triangle, one of $(w, z)$ and $(x, y)$ will be your point, the other will be the third vertex.)  Now, if you have two triangles, you get the detector function for "one triangle inside the other" by multiplying three of [i]those[/i] expressions together, one for each vertex of the triangle you're hoping will be inside.

The upshot of this is as follows:  if this resulting function is $F(T_1, T_2)$, giving value 1 if $T_1$ is entirely inside $T_2$ and 0 otherwise, [i]your[/i] final function has the form:
$F(T_1, T_2)\cdot Area(T_1) + F(T_2, T_1)\cdot Area(T_2) + (1 - F(T_1, T_2))(1-F(T_2,T_1))\cdot G(T_1,T_2)$
where $G$ is a yet-to-be-determined function that gives the area of intersection when they don't overlap.  (I'm also not proving this, but it's true and you should be able to without too much trouble.)

(I still don't think this is a sensible project of yours, by the way, but your post made me realize that this case at least I could build in to a function.  Now all you need to find is $G$, so you're that much closer.)

I don't really understand what it is that you have discovered.  However, if you discover a physical property or a math foumula that is novel and useful, it would be patentable.  I know that there are at least four patents using the bimodal particle size distribution to allow continued flow when the particles are a bit sticky.

That's the basis behind the Morton Salt Company's slogan, when it rains, it pours.  (They used sand in salt, just enought to stop the caking when the humidity gets high.  You may notice rice put into restruant's salt cellers for the same purpose (but this breaks up the slightly sticky particles by impact.

If someone really solved the traveling salesman problem they could become very rich.  Airlines, military and many others would appreciate improvments from "good" solutions to "best" solution, even if it were less than a 10% increase in efficiency.

How to get the patent protection is a difficult question.  Ususally the patent one obtains is as good as the search and legal work done to prosecute it.  (One can always hope that one's best idea is the second or third one, after one has developed the contacts and money necessary to obtain good patent protection and knowledge which conuntries to file and the critical aspects of non-disclosure and deadlines for international filing.

Remember, many "trusty" folks turn out to be overrated when real money is at stake.  (I remember a king of England saying, in Richard III I believe, "My kingdom for a friend.")  If it isn't worth the friendship, don't test it.  Many have destroyed friends and family by mixing them with money (family businesses or windfall of money from gambling or lawsuit recoveries).

Take care and enjoy your work, if someone else had already done it, it doesn't take anything from your work.  I would suggest publication, so others can build on it if it truly is unique, unlike Newton who didn't publish for decades.

Take care and congradulations.

Well. I've discovered a formula for the common length of 2 segments on the abcissa.

Then I found the way to extend it to however many segments I wish.

I'm sure it works. However, I've still to test it thoroughly. I haven't the nerve. But I have tested 2 random 3 segment overlap cases as well as all the possible 2 segment overlap cases and it worked, so I am quite sure it's correct. Also, neither segment tip order nor segment order matters to the result.

And I don't think I'll bother to try and find a proof. I'll just leave it at theorem level. the next step is actually writing the general form for it, so people who have never seen it before can apply it straight to the specific number of segments they need.

The next project still is a formula for the overlap area of whatever 2 triangles.

I thought about mailing everything relevant about this formula to myself, so I can have some sort of assurance against plagiarism. Is that any good, legally speaking? 10q a lot.


Find the distance between $ P_1(3, \minus{}185^{\circ})$  and $ P_2(\minus{}4, \minus{}94^{\circ})$ on the polar plane.
I don't know how to do it. please help.

Hello use the following lemma

The distance $ d$ between two points $ (r_1,\theta_1)$ and  $ (r_2,\theta_2)$ in the polar plane is:

$ d \equal{} \sqrt {r_1^2 \plus{} r_2^2 \minus{} 2r_1r_2cos(\theta_1 \minus{} \theta_1)}$

This comes from the cosine law  :)

[quote]Hello use the following lemma

The distance  between two points $ (r_{1},\theta_{1})$ and $ (r_{2},\theta_{2})$  in the polar plane is:

 $ d \equal{}\sqrt{r_{1}^{2}\plus{}r_{2}^{2}\minus{}2r_{1}r_{2}cos(\theta_{1}\minus{}\theta_{1})}$

This comes from the cosine law  [/quote]

Just to make sure, did you mean $ d \equal{}\sqrt{r_{1}^{2}\plus{}r_{2}^{2}\minus{}2r_{1}r_{2}cos(\theta_{1}\minus{}\theta_{2})}$

And thank you for the help.

[quote="Kenshin24"]Hello use the following lemma

The distance $ d$ between two points $ (r_1,\theta_1)$ and  $ (r_2,\theta_2)$ in the polar plane is:

$ d \equal{} \sqrt {r_1^2 \plus{} r_2^2 \minus{} 2r_1r_2cos(\theta_1 \minus{} \theta_2)}$[/quote]

Oh yes , I did mean that  :blush:


limit
trigonometry
function
calculus
calculus computations

hi,
compute this limit : $\lim{x\to 2}{\frac{Arcsin(\frac{2\sqrt[2]{x-1}}{x})-\frac{Pi}{2}}{x-2}}$

[quote="toetoe"]thank you so much benorin for your answer .i have got the same result using $x=\frac{1}{cos^{2}(a)}$.now ,i think that every hard limit can computed using the hospital's rule . but is there any condition before using it,please?[/quote]

You can only apply [url=http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule]l'Hôpital's rule[/url] if the limit is an indeterminant form of the type $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, but some [url=http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms]simple manipulations[/url] easily admit the additional indeterminant forms of type $1^{\pm\infty},0^{0},\infty^{0},\infty-\infty,$ and $0\cdot{\pm\infty}$.

Also,

[quote="toetoe"]an other thing ,i guess you have computed one case that is when $2>x$ .if $x>2$then f'(2) is $\frac{1}{2}$.am i right?[/quote]

Are you investigating the differentiability of the function $f(x)=\left\{\begin{array}{cc}\sin^{-1}\left(\frac{2\sqrt{x-1}}{x}\right),&\mbox{ if }x\neq 2\\ \frac{\pi}{2}, & \mbox{ if }x=2 \end{array}\right.$ at the point $x=2$?

[quote="toetoe"]no ,i am just studying the derivability of the function at x=2.if i  understand the limit equalize $\frac{-1}{2}$ at $2^{+}$ and $\frac{1}{2}$at $2^{-1}$,please ,am i right because i might have an examen on a similar function ?[/quote]

The limit computed above is two-sided, and so the result holds for both sides (i.e. they are both -1/2)

\[\lim_{x\to 2}\frac{\sin^{-1}\left(\frac{2\sqrt{x-1}}{x}\right)-\frac{\pi}{2}}{x-2}=-\frac{1}{2}\]

thanks for your answer but i think that $\sqrt{(x-2)^{2}}$depends on the bigger number between 2 and x .so every side has its own limit.

yes ,and that means that :

$x>2 \Rightarrow |x-2| = x-2 \Rightarrow \lim_{x\to 2^{+}}\frac{f(x)-\frac{Pi}{2}}{x-2}= \frac{-1}{2}$

$x<2 \Rightarrow |x-2| =-(x-2) \Rightarrow \lim_{x\to 2^{-}}\frac{f(x)-\frac{Pi}{2}}{x-2}= \frac{1}{2}$

am i right ? please ,help me tomorrow i am having the exam  :blush: .


[quote="*210*"]hi!
$ f$ and $ g$ two fonctions defined by $ f(x) \equal{} ax^{2} \plus{} bx \plus{} c$ and $ g(x) \equal{} cx^{2} \plus{} bx \plus{} a$ such as $ | f( \minus{} 1) |\leq 1$ and $ | f(0) |\leq 1$ and $ | f(1) |\leq 1$
1) prouve that $ (\forall x\in [ \minus{} 1,1]) : | f(x) | \leq \frac {5}{4}$   and $ : (\forall x\in [ \minus{} 1,1]) : | g(x) | \leq 2$[/quote]

maybe i am misinterpreting the question

consider $ \left(\frac{10}{9},\frac{10}{9},\minus{}\frac{11}{9}\right)$

I don't know whether or not you're misinterpreting, but $ 11/9, 10/9 > 1$, so that triple of values fails $ |f(0)| \leq 1$.

[quote="facis"][hide="Hint"]$ g(x) \equal{} f(1/x)$[/hide][/quote]
Correct me if I'm wrong, but I think the above assertion is true only when $ x$ is a particular root of $ g$ such that the two roots $ x$ and $ 1/x$ for their respective function is a reciprocal of each other (since the coefficients are reversed for the two functions).

[quote="Lazarus"]maybe i am misinterpreting the question

consider $ \left(\frac {10}{9},\frac {10}{9}, \minus{} \frac {11}{9}\right)$[/quote]
What does this triplet mean?

oops... the relationship I meant to point out is $ g(x) \equal{} x^2 f(1/x)$

I think Lazarus means $ a\equal{}10/9$, $ b\equal{}10/9$, $ c\equal{}\minus{}11/9$, but these do not satisfy the conditions.

Now, my stab at it:
$ f(\minus{}1) \equal{} a\minus{}b\plus{}c$ so $ \minus{}1 \le a\minus{}b\plus{}c \le 1$
$ f(0) \equal{} c$ so $ \minus{}1 \le c \le 1$
$ f(1) \equal{} a\plus{}b\plus{}c$ so $ \minus{}1 \le a\plus{}b\plus{}c \le 1$

Subtracting the first inequality from the third and dividing by two, we have $ \minus{}1 \le b \le 1$.  Adding the first and third inequalities and then dividing by two, $ \minus{}1 \le a\plus{}c \le 1$; thus $ \minus{}2 \le a \le 2$.

now I'm stuck.

While I was writing my last post, I noticed that the conditions I developed were necessary but not sufficient for $ a$, $ b$, and $ c$ to meet the given restrictions.  So here is another try.

$ |f(\minus{}1)| \le 1 \implies \minus{}1 \le a\minus{}b\plus{}c \le 1 \implies b\minus{}1 \le a\plus{}c \le b\plus{}1$
But also $ |f(1)| \le 1 \implies \minus{}1 \le a\plus{}b\plus{}c \le 1 \implies \minus{}1\minus{}b \le a\plus{}c \le 1\minus{}b$.  Thus we have $ max(b\minus{}1 , \minus{}b\minus{}1) \le a\plus{}c \le min(1\minus{}b, b\plus{}1)$.  If $ b \ge 0$, we have $ b\minus{}1 \le a\plus{}c \le 1\minus{}b$, and if $ b \le 0$, $ \minus{}b\minus{}1 \le a\plus{}c \le b\plus{}1$.

Still not getting anywhere...


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Show that $\sum_{k=1}^n d_1(k)-d_3(k)=\sum_{j=0}^{\infty}(-1)^jE(\frac{n}{2j+1})$,where $d_i(n)$ is the number of dividers of n which are congruent to $i(modulo 4)$.

Let's work by induction:
Clearly the LHS converges for all $n$ and fulfills the property for $n=1$.
If $n-1$ grows by one to $n$, $E(\frac{n}{2j+1})$ grows by one iff $2j+1 | n$.
So after subtracting the induction hypothesis, we only have to prove: $d_{1}(n)-d_{3}(n) = \sum_{2j+1|n}(-1)^{j}$
But this can be done by induction over the multiplicativity of all prime factors of $n$ and looking at their residues $\mod 4$.
Additionally, both sums give $0$ iff not all prime factors $p \equiv 3 \mod 4$ of $n$ occure an even number of times, and give $\sum_{\substack{ p|n \\ p \in \mathbb{P}\\p\equiv 1\mod 4}}(v_{p}(n)+1)$ otherwise.

Note that $4(d_{1}(n)-d_{3}(n))$ gives the number of representations of $n$ as sum of two squares, so ($4$ times the LHS) $+1$ describes the number of lattice points inside the circle with radius $\sqrt n$. Dividing the equation through $n$ gives by this (since $\frac{E(\frac{n}{2j+1})}{n}\to \frac{1}{2j+1}$) the Leibniz series for $\frac{\pi}{4}$ for the RHS, so both sides describe more or less the area of a circle.


inequalities
quadratics
LaTeX
algebra
algebra solved

Find all  $n\in \mathbb{N}$, for which there is a set of positive numbers $x_{1}, x_{2},\ldots, x_{n}$ that satisfies the system

\[\begin{cases}x_{1}+x_{2}\ldots+x_{n}=9\\ \frac{1}{x_{1}}+\ldots+\frac{1}{x_{n}}=1\end{cases}\]

Using Cauchy-Schwarz inequality or twice AM/Gm, it is easy to prove that, for all positive integer $n$ and real numbers $x_1,...,x_n$, we have :
$(x_1+...x_n)( \frac {1} {x_1} + ... + \frac {1} {x_n}) \geq n^2$.
It follows that if $n$ is a solution then $n \leq 3$.

Conversely, it is easy to verify that $n=1$ is not a solution, and that $n=3$ is a solution (take x_1=x_2=x_3=3).
For $n=2$ the system reduces to
$x_1+x_2=9$ and $x_1+x_2=9$ which leads to a quadratic equation $X^2-9X + 9=0$ which has two positive real solutions, which are $\frac {9+ \sqrt{55}} {2}$ and  $ \frac {9- \sqrt {55}} {2}$, thus $n=2$ is a solution too.

Pierre.

Pierre, this is nothing but you write both $x_1+x_2=9$ instead of $x_1x_2=9$. indeed, we have $\displaystyle \frac{1}{x_1}+\frac{1}{x_2}=1 \iff \frac{x_1+x_2}{x_1x_2}=1 \iff x_1x_2=9$
l wonder why l write those stupid calculations, everyone know that !

I try to improve my Latex style everyday....The road seems to be still long.

Pierre.

finally, this is not a problem for the "advanced section".


Anyone from New Jersey interested in joining a math team of kids who's schools don't have teams or have pathetic teams? We would probably compete for iTest, PuMaC, HMMT, ARML, etc. Please post or PM me if you are interested. Let's go New Jersey.

maybe; but there doesn't seem to be a lot of ppl responding to this thread.

I'm interested, but I'm not really on the same level as you guys, like, I'll prolly make AIME this year and get like 4-6ish, but I really want more contest exposure, so I'd put in the effort and try to learn... that sounded kinda gay.

I made into state last year but not national in MC.


Of the 2400 jellybeans in a jar, 1650 are purple. Lucia then
removes 150 purple jellybeans from the jar. What fraction of
the jellybeans left in the jar are purple? Express your answer
as a common fraction.

this is basically a fraction:

1650-150=1500
2400-150=2250

so the fraction is 1500/2250, which can be reduced to 2/3.


The functions A(t) and B(t) are solutions to this system of differential equations:

x' = y
y' = x
x(0) = 1
y(0) = 0

Find and classify all equilibrium points of this system.

I don't see any difficulty in this problem IF you know all your definitions and still remember some bits of linear algebra. This forum is not for routine homeworks.


What is $ \left(\sqrt2\plus{}\sqrt8\plus{}\sqrt{18}\plus{}\sqrt{32}\right)^2$?

it simplifies to
$ \left(1\sqrt{2}\plus{}2\sqrt{2}\plus{}3\sqrt{2}\plus{}4\sqrt{2}\right)^2$
this is easily turned into $ 10^2\cdot 2 \equal{}200$

200.
i miss ftw :(


geometry
advanced fields
advanced fields unsolved

Consider five distinct points in $ \mathbb{A}^2(\mathbb{R})$ that fail to impose independent conditions on $ P_{2,3}$.
Show that these points are collinear.

I'd like to understand for 'impose independent conditions'
Please give me some hint.

You might want to consult the definition you have given [url=http://www.mathlinks.ro/viewtopic.php?t=229847]here[/url]. Then try to show that the evaluation map given in the proof given there fails to be surjective.


1. Can someone show me a combinatorial proof of the hockey stick identity? $ \binom{r}{r}\plus{}\binom{r\plus{}1}{r}\plus{}\binom{r\plus{}2}{r}\plus{} \cdots \plus{} \binom{f}{r} \equal{} \binom{f\plus{}1}{r\plus{}1}$. I've done it for a specific case say when $ f \equal{} 5$ and $ r \equal{} 2$. But I don't know how to generalise it.

2. How many different ordered triples $ (a,b,c)$ of non-negative integers are there such that $ a\plus{}b\plus{}c\equal{}50$?

How would you do this question using partitions?

Many thanks!

For 1: The right hand side counts how many ways there are to mark $ r \plus{} 1$ squares on a strip of length $ f \plus{} 1$.  Suppose I told you additionally that the last marked square MUST be at location $ k$.

First of all, what are the possible values for $ k$?  Secondly, for each of those values, how many ways are there to mark the squares so the last marked square is in fact at $ k$?


AMC
AIME
function
number theory
relatively prime

I just took the 2000 Aime, and this was one of the questions:

Find [S/10], where S is the sum of all numbers m/n, where m and n are relatively prime positive divisors of 1000.

(The following concealed part has what their answer is and why I find it fishy. So if you want to do it first, then dont reveal this stuff.)

[hide]
They basically say that its a generating function in the form of (2^-3 + 2^-2 + ... + 2^3)*(5^-3 + ... + 5^3), which I agree with. But then they dont subtract out 1/1. I would think that 1 is not relatively prime to itself?
[/hide]

Haha, sorry about wasting this space. Gcd(1,1) = 1. Duh!


Please do not share A/B/C homework problems or practice materials with anyone who might end up on the team 4 years from now. We tend to cycle our material every 4 years and release of the material would make our jobs of setting up practices much more difficult.

In particular, all the posters from AoPS Illinois forum who are not yet in high school (and not on the team this year) should NOT be involved in discussion of the HW or practice materials.

I am not sure, but I suspect the D-team has a similar policy.  For all you D-teamers, please clarify this with Coach Karafiol.  I'll send him a note.

thanks,

coach fritz

is it okay if we post problems we write for the homework assignments? (sorry I should've asked before posting...)

not a problem.  I usually need to edit them into something solvable anyway  :P .

good luck on saturday.

coach fritz


See the Classification section [url=http://en.wikipedia.org/wiki/Projection_(linear_algebra)]here[/url].


[quote="brian_gold"]For any natural n
Prove that $ (1 \plus{} \frac {1}{3})(1 \plus{} \frac {1}{8})(1 \plus{} \frac {1}{15})*...*(1 \plus{} \frac {1}{n^2 \plus{} 2n}) < 2$[/quote]
$ \left(1 \plus{} \frac {1}{3}\right)\left(1 \plus{} \frac {1}{8}\right)\left(1 \plus{} \frac {1}{15}\right)\cdot...\cdot\left(1 \plus{} \frac {1}{n^2 \plus{} 2n}\right)\equal{}$
$ \equal{}\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{n^2}{(n\minus{}1)(n\plus{}1)}\cdot\frac{(n\plus{}1)^2}{n(n\plus{}2)}\equal{}\frac{2(n\plus{}1)}{n\plus{}2}<2.$

Arqady! Thank you for your solution. And who is Michael Rozenberg?

Nice solution, arqady! It is the Telescoping product method, I think.
 :)
[quote="brian_gold"]Arqady! Thank you for your solution. And who is [b]Michael Rozenberg[/b]?[/quote]
 :rotfl:  :rotfl:


[quote="hpe"][quote="color"]Solve
$u_{x}-u_{t}-u=0$ and $u(x,0)=e^{-x^{2}}$

$x\in \mathbb{R}$ and $t>0$[/quote]What is the pde that is solved by $v(x,t) = e^{t}u(x,t)$? Can you solve this with the method of characteristics?[/quote]

No. Pleas explain more. How about Laplace transforms?


linear algebra
matrix
linear algebra unsolved

[quote="viking"]$A=\left(\begin{array}{ccccc}A_1&B_{12}&B_{13}&...&B_{1k}\\0&A_2&B_{23}&...&B_{2k}\\...&...&...&...&...\\0&0&0&...&A_k\end{array}\right)$ ,where $A_j$ in $Mr_j(K)$;$r_1+r_2+..+r_k=n$
Prove that : $detA=detA_1.detA_2...detA_k$[/quote]
Note the generalization. Use elementary row operations and observe that the matrix remains upper block triangular throughout.


Four different coins are used in Australia with values of 5, 10, 20 and 50 cents. The Australian coins in Sampson's pocket have a total value of $ \$3.95$, and he has at least one of each coin. What is the smallest number of coins that he could have in his pocket?

well, we use the 50 as much as possible, so that gives us 3.50, with 7 coins. each must be used once in 45 cents, so we use 1 20, 2 10s, and 1 5. 7+1+2+1=11


Are there any Scientific Units (or just units) named after women? This question was brought up in a video lecture (MIT OCW) about kinetics/radioactivity (with a nice poem 'Days of Our Half-Lives'). So yeah, I was just curious. :)

I know but she (the lecturer) said that the $ \text{curie}$ was named after Pierre.

http://www.sizes.com/units/curie.htm

[quote] The curie was named for the then recently deceased Pierre Curie (1859–1906), French physicist and husband of Marie[/quote]

That was a line from the article.


function
real analysis
real analysis unsolved

Let $ f: I\minus{}>R$ be a function with primitives and with $ f(\frac{x\plus{}y}{2})<\equal{}\frac{f(x)\plus{}f(y)}{2}$ for any $ x,y$ in $ I$. Prove that $ f$ is continuous.

Ok..these are easy...but how do you conclude from $ g$ increasing that $ f$ is continuous? :maybe:


modular arithmetic
combinatorics proposed
combinatorics

Consider the collection of all three-element subsets drawn from the set $ \{1,2,3,4,\dots,299,300\}$.
Determine the number of those subsets for which the sum of the elements is a multiple of 3.

$ \pmod{3}$ such a three element subset is $ (0,0,0),(1,1,1),(2,2,2)$ or $ (0,1,2)$
We have $ 100$ elements $ i\pmod{3}$ for $ i\equal{}0,1,2$
so the desired result is
\[ 3\binom{100}{3}\plus{}100^3\equal{}1485100\]

If $(x_1,x_2,x_3)$ is a triplet such that $3 \mid x_1+x_2+x_3$ then there are four possibilities:
 All $x_i \equiv 0$ (mod 3)
All $x_i \equiv 1$ (mod 3)
 All $x_i \equiv 2$ (mod 3)
$x_1 \equiv 0$ (mod 3) $x_2 \equiv 1$ (mod 3) $x_3 \equiv 2$ (mod 3)
1st, 2nd and 3rd are similar in counting. There are $300$ numbers in total, $100$ of each category. So we need to choose $3$ numbers from $100$ in each case which is simply.    $3 \cdot {100\choose 3}$. Now there 100 options for each $x_i$ and we need to choose 3, so it becomes ${100\choose 1}$.Hence, our answer becomes
\[3 \cdot {100 \choose 3} + {100 \choose 1}^3 \]


Define a sequence <a(n)> of integers by a(1) = 2, a(2) = 7, and

-1/2 < a(n+1) - {a(n)^2/a(n-1)} <= 1/2

for n >= 2.

Prove that a(n) is odd for all n >= 2.

PS: Sorry for not being able to type maths...but this looks clear enough.

Situation, when $ a_{n\plus{}1}\equal{}\left[\frac{a_{n}^{2}}{a_{n\minus{}1}}\plus{}\frac{1}{2}\right]$ give $ a_{n}\equal{} cx_{1}^{n}\plus{}dx_{2}^{n}, |x_{2}| < x_{1}$, because $ a_{n}$ are integer, must be $ a_{n\plus{}1}\equal{} aa_{n}\plus{}ba_{n\minus{}1}$ with integer a,b. I found $ a_{3},a_{4}$, then $ a \equal{} 3,b \equal{} 2$ and chek, that
$ a_{n\plus{}1}\equal{} 3a_{n}\plus{}2a_{n\minus{}1}$ give same sequence..


Welcome to Song of the Week! I have seen this done in many other forums, and this will probably work well here.

[b]How It Works[/b]: Starting today, Wednesday, people will post in thread a filled-out template:

[code]
[b]Song Title[/b]:
[b]Song Artist[/b]:
[b]YouTube Link[/b]:
[b](Optional) Comments[/b]:
[/code]

[i]Note that you should not post the YouTube video itself in the thread, just the link.[/i] This will reduce clutter.

After 5 days (so until the end of Sunday), people can keep posting their ideas. The next two days, Monday and Tuesday, will be reserved for voting. If you decide to vote, please listen to any song you are unfamiliar with to make this as fair as possible. To vote, fill out the following template in the thread:

[code]
[b]I Vote For[/b]: [Song Title] by [Song Artist], posted by [User] on post [Post Number].
[/code]

I will tally up the votes on Wednesday as soon as possible, post the winning video in the thread and the user who suggested it, and then the new SotW will begin.

Anyone can participate at any time!

[b]Song Title[/b]: Vanilla Twilight
[b]Song Artist[/b]: Owl City
[b]YouTube Link[/b]: http://www.youtube.com/watch?v=aENY16Mjw6k
[b](Optional) Comments[/b]: A nice, not-so-well-known song. :) I actually discovered this from an SotW on some other forum.

[b]Song Title[/b]: Music using ONLY sounds from Windows XP and 98!
[b]Song Artist[/b]: SomethingUnreal (Youtube user)
[b]YouTube Link[/b]: http://www.youtube.com/watch?v=dsU3B0W3TMs
[b](Optional) Comments[/b]: Who doesn't like Windows?
 :D

Just a comment, anyone who posts Weird Al should automatically be disqualified.

[b]Song Title[/b]: Glass from Jasper
[b]Song Artist[/b]: Roedelius
[b]YouTube Link[/b]: http://www.youtube.com/watch?v=XJTA9EVKXdk
[b](Optional) Comments[/b]: Hmmm... Betcha haven't heard a music that flows like this before!

i post song lyrics on my facebook every day, so this is right in my league... today's posting:
Song Title: Rise Above This
Song Artist: Seether
YouTube Link: http://www.youtube.com/watch?v=NO2nqcN3EGg
(Optional) Comments: don't commit suicide :P
i heard this song on a cruise ship :P but nowadays i listen to [url=http://www.pandora.com]Pandora Radio[/url] to find new music. My station is a mix of alternative rock, hard rock, and heavy metal with some bands removed and others added

[b]Song Title[/b]: On the Bright Side
[b]Song Artist[/b]: Nevershoutnever!
[b]YouTube Link[/b]: http://www.youtube.com/watch?v=aF5tisuEMRs
[b](Optional) Comments[/b]: It's so inspirational, for all of those short, tall... big and small, rich or poor, humble or spoiled... this songs for you  :)

P.S. Songs by Owl City are beautiful poetry, look at the lyrics I tell you... LOOK! Hehe... :)

[b]Song Title:[/b]  Bring on the Rain
[b]Song Artist: [/b] Jo Dee Messina (with Tim McGraw)
[b]YouTube Link:[/b]  [url]http://www.youtube.com/watch?v=xWYRfsjBNQk[/url]
[b](Optional) Comments:[/b]  An emotionally uplifting song that helps you get through those times when it seems like everything's going wrong.

Hopefully, some people can really understand this one.

Song Title: Superb
Song Artist: Fahrenheit
You Tube link: http://www.youtube.com/watch?v=31hHqBa2qxE&feature=channel_page

(optional) Comments: This is a Taiwaneese band that i love to listen too. Uh YES I am from Taiwan.  :)

PM me if you like it or if you are from Taiwan too. Oh and if this isn't appropriate for some reason.

[b]Song Title[/b]: My life be like ohh ahh
[b]Song Artist:[/b] Gritz
[b]You Tube link[/b]: http://www.youtube.com/watch?v=t-yCg-0-baE

Song Title: Viva La Gloria
Song Artist: Green Day
YouTube Link: http://www.youtube.com/watch?v=DG_jYaIBXFw
Comments: My favorite song off the latest album.

[b]Song Title[/b]: All Star
[b]Song Artist[/b]: Smash Mouth
[b]YouTube Link[/b]: http://www.youtube.com/watch?v=WKMyj4N45x0
[b](Optional) Comments[/b]: I love this song!

Song title:Aerials
Song artist:system of a down
youtube link: http://www.youtube.com/watch?v=UJ6FbEmDDcU
comments: they were so good but they went on hiatus

[b]Song Title[/b]: Mr. Boombastic
[b]Song Artist[/b]: Shaggy
[b]YouTube Link[/b]:[url]http://www.youtube.com/watch?v=FwLCnh_EnEQ[/url]

[b]Title[/b]:You belong with me
[b]Artist[/b]: Taylor Swift
[b]YouTube Link[/b]:[url]http://www.youtube.com/watch?v=Fw9QMSl9Xic[/url]
[b]Comments[/b]: I know this is a girl song but it rox.

i just heard kanye west's voice saying "yo taylor, i'm really happy for you, and imma let you finish, but beyonce had one of the best videos of all time!"
Title:Somebody told me
Artist: The killers
YouTube Link:http://www.youtube.com/watch?v=5gjlNn0mG8Y
Comments: the killers.

hmmm maybe i'll decide to post a song

Title: Prettiest Friend
Artist: Jason Mraz
YouTube: [url]http://www.youtube.com/watch?v=rIZmDHPqGl4[/url]
Comments: Its an awesome song :]

[b]Title:[/b] Evolution
[b]Artist:[/b] KoRn
[b]YouTube Link:[/b] http://www.youtube.com/watch?v=L99UMoLvkrI
[b]Comments:[/b] Yeah.

Title: Ghost Opera
Artist: Kamelot
YouTube Link: [url]http://www.youtube.com/watch?v=S9hGHnXPEi4[/url]
Comments: Seriously one of the most beautiful symphonic metal songs I've ever heard. (Ignore the video haha).

can we start voting now?

if so, vote somebody told me.

[b]Title[/b]:Love Story
[b]Artist[/b]: Taylor Swift
[b]YouTube Link[/b]:[url]http://www.youtube.com/watch?v=5wRkoGKQ8qQ[/url]
[b]Comments[/b]: None necessary :P

I'm sorry, what's with the Swift domination?

[b]Vote: Evolution[/b].  Self voting all the way!

Then I shall do the same. Though I admit Evolution is a great song.

[b]Vote: Ghost Opera[/b]

why haven't we decided the song of the week yet??

Because as of now we have a tie for first.

Votes:
You belong with me (0)
Somebody told me (0)
Prettiest Friend (0)
Evolution (1) - ernie
Ghost Opera (1)- cyberspace
Love Story (0)

(the vote for somebody told me didn't count because another song was posted afterwards. If you still want to, vote it again)

Keep voting people!

I guess I'll be the tiebreaker and vote for [b]Ghost Opera[/b].

kk, I'm guessing that that parentheses was pointed to me so, vote [b]somebody told me[/b]


I'm writing a latex document, using "\documentclass[12pt]{report}" so the class where you have \chapter{} and \section{}. But the thing is: When starting a new chapter about the top 1/3 of the page is lost. (just blank space)

Is there any parameter I can set to reduce that space? The one of \section is ok, but the one from \chapter is really way too much...

There's a short discussion of this problem in the comp.text.tex newsgroup at http://tinyurl.com/5cgcb


Incorrect
Problem:
Automobile license plates for a state consist of four letters followed by a dash and two single digits. How many different license plate combinations are possible if exactly one letter is repeated exactly once, but digits cannot be repeated?

Solution:
There are 26 ways to choose the repeated letter, \binom{25}{2} ways to choose the other two letters, \binom{4}{2} ways to choose in which two of the four positions to put the repeated letters, 2 ways to choose how to arrange the remaining two letters, 10 ways to choose the first digit and 9 ways to choose the second digit for a total of $ (26)\binom{25}{2}\binom{4}{2}(2)(10)(9)\equal{}\boxed{8,\!424,\!000}$ combinations.

Your response: 8424000

This problem provided to Alcumus courtesy of MATHCOUNTS

YES! I have this problem too! Every time it says i got it wrong, and my rating falls. I've sent an error report already. I'm waiting until this problem is solved before continuing with Alcumus, as when I get it wrong, it reappears a few questions later.

Admins: Please atleast take this problem off for now. Thanks!

It seems we had the wrong answer input for this problem. It should work now. Sorry, guys!

[quote="pinkmuskrat"]I think there is a similar problem with:

What is the value of $ 91^2$?

Could you please check that? Thanks. =)[/quote]It's a bit hard to find the question just given the text. The best way to report these problems is through the bug report system. At the bottom of every solution page, you will see a "Click here to report an error with problem or solution" link. By clicking this link, we will be able to address these issues more quickly, and plus, you will gain karma :)

[quote="pinkmuskrat"]well its mainly because i dont want my alcumus rating do go down like a bunch by giving up to get to the solution page  :P

since the problems very easy, i answered once and i have a second try, which is waiting on my screen[/quote]

When we fix a broken problem, we generally wipe the trials for that problem clean, so your Alcumus rating won't be hurt by missing a broken problem once we have fixed it.


calculus
integration
geometry
calculus computations

Expand and simplify $ (\sqrt{x\minus{}1} \plus{} 1)^{2}$.

(i) Evaluate

$ \int_{5}^{10}\, \frac{\sqrt{x\plus{}2\sqrt{x\minus{}1}} \plus{} \sqrt{x\minus{}2\sqrt{x\minus{}1}}}{\sqrt{x\minus{}1}}\,dx$.

(ii) Find the total area between the curve

$ y\equal{}\frac{\sqrt{x\minus{}2\sqrt{x\minus{}1}}}{\sqrt{x\minus{}1}}$

and the $ x$-axis between the points $ x\equal{}\frac{5}{4}$ and $ x\equal{}10$.

(iii) Evaluate

$ \int_{\frac{5}{4}}^{10}\, \frac{\sqrt {x\plus{}2\sqrt{x\minus{}1}} \plus{} \sqrt{x\minus{}2\sqrt{x\plus{}1} \plus{} 2}}{\sqrt{x^{2}\minus{}1}}\,dx$

[color=red] Moderator edit: it never makes sense to quote the post directly preceding yours in full![/color]

$ \begin{array}{l} \int\limits_{\frac {5}{4}}^{10} {\frac {{\sqrt {x \plus{} 2\sqrt {x \minus{} 1} } \plus{} \sqrt {x \minus{} 2\sqrt {x \plus{} 1} \plus{} 2} }}{{\sqrt {{x^2} \minus{} 1} }}} dx \equal{} \int\limits_{\frac {5}{4}}^{10} {\frac {{\sqrt {x \minus{} 1} \plus{} 1 \plus{} \sqrt {x \plus{} 1} \minus{} 1}}{{\sqrt {{x^2} \minus{} 1} }}} dx \equal{} \\
\equal{} \int\limits_{\frac {5}{4}}^{10} {\frac {{\sqrt {x \minus{} 1} \plus{} \sqrt {x \plus{} 1} }}{{\sqrt {x \minus{} 1} \cdot \sqrt {x \plus{} 1} }}} dx \equal{} \int\limits_{\frac {5}{4}}^{10} {\left( {\frac {{\sqrt {x \minus{} 1} }}{{\sqrt {x \minus{} 1} \cdot \sqrt {x \plus{} 1} }} \plus{} \frac {{\sqrt {x \plus{} 1} }}{{\sqrt {x \minus{} 1} \cdot \sqrt {x \plus{} 1} }}} \right)} dx \equal{} \\
\equal{} \int\limits_{\frac {5}{4}}^{10} {\left( {\frac {1}{{\sqrt {x \plus{} 1} }} \plus{} \frac {1}{{\sqrt {x \minus{} 1} }}} \right)} dx \equal{} 2\left[ {\sqrt {x \plus{} 1} \plus{} \sqrt {x \minus{} 1} } \right]_{\frac {5}{4}}^{10} \equal{} 2\sqrt {11} \plus{} 2 \\
\end{array}$


search
advanced fields
advanced fields unsolved

Let $ f: X\to Y$ be a continuous map of topological spaces. Let $ \mathcal{F}$ be a sheaf on $ X$ and $ \mathcal{G}$ a sheaf on $ Y$.

Show that there exist canonical morphisms of sheaves $ \mathcal{G}\to f_*f^{\minus{}1}\mathcal{G}$, $ f^{\minus{}1}f_*\mathcal{F}\to \mathcal{F}$. and the first is surjective if $ f$ is a closed immersion, isomorphism if $ f$ is open immersion.

How these implies that there exists a canonical bijection
\[ \text{Mor}_Y(\mathcal{G},f_*\mathcal{F})\simeq\text{Mor}_Y(f^{\minus{}1}\mathcal{G},\mathcal{F})?\]

you first have to prove the adjunction $ \text{Mor}_Y(\mathcal{G},f_*\mathcal{F})\simeq\text{Mor}_X(f^{ \minus{} 1}\mathcal{G},\mathcal{F})$, which is rather trivial (use the universal properties of garbification and direct limit to evaluate the right side). this adjunction yields the two morphisms (unit and counit):

$ \mathcal{G} \to  f_* f^{\minus{}1} \mathcal{G} , f^{\minus{}1} f_* \mathcal{F} \to \mathcal{F}$. the [i]second[/i] is an isomorphism for open immersions, just look at the definitions for that. and for the closed immersion you may consult http://www.mathlinks.ro/Forum/viewtopic.php?p=1011572&search_id=586740084#1011572

What is garbification and its universal property? I tried to search it in Google and in Liu.

I guess that by the garbification of a presheaf, he means the sheaf associated to the presheaf.

  darij

It looks like garbification=sheafification. I didn't realize that Garbe=sheaf in German.

:rotfl: sorry my fault.

I also used this term here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=181423

 :blush:


LaTeX
probability
combinatorics unsolved
combinatorics

sum(k=0 -> 2n-1 : nCr(i-1, n-1) * 2 ^(1 - i) ).

Combinatorial argument or whatever cleverness is welcome.

Latex use and a clear question are welcome too  :mad:

Pierre.

Ok. After several tries, I finally got it. Here it is:

$\sum_{i=0}^{2n-1} {i-1 \choose n-1} * 2^{1- i}$

Sum it.

the answer is $1$.
 one way to see this(quasi-combinatorial) is to interpret this as a probability after rewriting the given expression as $\sum_{k=0}^n \frac{\binom{n+k-1}{n-1}}{2^{n+k-1}}$.

Consider the grid $[0, n-1]\times [0,n]$ and consider paths that start from the origin and go at each step, with probability $\frac{1}{2}$ upward or rightward, and independently at each step, and stops when it hits the line $y=n$. Clearly any path stops within $2n-1$ steps. Alternately the probability that a path stops after being at the point $(k,n-1)$ is given by $\binom{n+k-1}{k}\cdot (\frac{1}{2})^{n-1}\cdot (\frac{1}{2})^k$ and that is the general term in the given expression.


What proof technique do you need to be good at to get into a prestigious math club?

Induction.  :P

[quote="lingomaniac88"]Woo-hoo, my 100th post!

Here, we post math jokes and humorous things that relate to mathematics (like something you may find online).

I'll post one that's pretty funny:
[b]Proof: The less you know, the more you make.[/b]
Let $K$=Knowledge, $P$=Power, $T$=Time, $M$=Money, and $W$=Work.
Postulate: $K=P$
Postulate: $T=M$
As every engineer knows: $P=\frac{W}{T}$
Substituting in: $K=\frac{W}{M}$
Solving for money: $M=\frac{W}{K}$
So as $K$ approaches 0, $M$ approaches $\infty$, regardless of $W$.
The rest, well, you can figure it out.[/quote]

What if $W$ was equal to 0?

there's a bunch of cool jokes in this topic: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44429[/url]

[hide=this quote][quote=lingomaniac88]Well, assuming that $W\ne0$.

[b]Dictionary of Definitions of Terms Commonly Used in Mathematics Lectures[/b]
The following is a guide to terms which are commonly used but rarely defined. In the search for proper definitions for these terms we found no authoritative, nor even recognized, source. Thus, we followed the advice of mathematicians handed down from time immortal: "Wing It."

CLEARLY:
I don't want to write down all the "in-between" steps.
TRIVIAL:
If I have to show you how to do this, you're in the wrong class.
OBVIOUSLY:
I hope you weren't sleeping when we discussed this earlier, because I refuse to repeat it.
RECALL:
I shouldn't have to tell you this, but for those of you who erase your memory tapes after every test...
WLOG (Without Loss Of Generality):
I'm not about to do all the possible cases, so I'll do one and let you figure out the rest.
IT CAN EASILY BE SHOWN:
Even you, in your finite wisdom, should be able to prove this without me holding your hand.
CHECK or CHECK FOR YOURSELF:
This is the boring part of the proof, so you can do it on your own time.
SKETCH OF A PROOF:
I couldn't verify all the details, so I'll break it down into the parts I couldn't prove.
HINT:
The hardest of several possible ways to do a proof.
BRUTE FORCE (AND IGNORANCE):
Four special cases, three counting arguments, two long inductions, and a partridge in a pair tree.
SOFT PROOF:
One third less filling (of the page) than your regular proof, but it requires two extra years of course work just to understand the terms.
ELEGANT PROOF:
Requires no previous knowledge of the subject matter and is less than ten lines long.
SIMILARLY:
At least one line of the proof of this case is the same as before.
CANONICAL FORM:
4 out of 5 mathematicians surveyed recommended this as the final form for their students who choose to finish.
TFAE (The Following Are Equivalent):
If I say this it means that, and if I say that it means the other thing, and if I say the other thing...
BY A PREVIOUS THEOREM:
I don't remember how it goes (come to think of it I'm not really sure we did this at all), but if I stated it right (or at all), then the rest of this follows.
TWO LINE PROOF:
I'll leave out everything but the conclusion, you can't question 'em if you can't see 'em.
BRIEFLY:
I'm running out of time, so I'll just write and talk faster.
LET'S TALK THROUGH IT:
I don't want to write it on the board lest I make a mistake.
PROCEED FORMALLY:
Manipulate symbols by the rules without any hint of their true meaning (popular in pure math courses).
QUANTIFY:
I can't find anything wrong with your proof except that it won't work if $x$ is a moon of Jupiter (Popular in applied math courses).
PROOF OMITTED:
Trust me, It's true.

I found that one right off Google.[/quote]

[/hide]

This was so funny! :rotfl: