quadratics
algebra
number theory proposed
number theory

Solve in positive integers:
x³-y³=2005(x²-y²)

See post #17 in
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x%5E3-y%5E3++2005%28x%5E2++y%5E2%29&t=35100[/url]

:? Well, I didn't see any solution. If you have a solution, please can you send it?

It's attached to Soarer's post (#17).

factor out $x-y$ when $x \neq y$, see it as quadratic equation that must have a square as discriminant; then do this again, then show that the new one can't be a square.


geometry
inequalities
analytic geometry
incenter
inradius
trigonometry
geometric inequality

Given are a triangle $ABC$ with a point $M$ in plane find min value :
$asin\frac{A}{2}MA^{3}+bsin\frac{B}{2}MB^{3}+csin\frac{C}{2}MC^{3}$
With cabri II, I can see it has min value when M is incircle centre but I can't give a proof.

Your Cabri conjecture is true. It wasn't that difficult after all. Here is a more general inequality:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle and P a point in its interior. Let $\left(x: y: z\right)$ be the homogeneous barycentric coordinates of the point P with respect to triangle ABC, such that x, y, z are all positive. Let M be any point in the plane. Then,

$\left(x+y+z\right)\left(\frac{x\cdot MA^{3}}{PA}+\frac{y\cdot MB^{3}}{PB}+\frac{z\cdot MC^{3}}{PC}\right)\geq yza^{2}+zxb^{2}+xyc^{2}$,

with equality if and only if M = P.[/color]

[i]Proof.[/i] According to the Fairly General Geometric Inequality ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=110333#110333]http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 post #4[/url]), we have

$\left(x+y+z\right)\left(x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\right) \geq yza^{2}+zxb^{2}+xyc^{2}$

with equality if and only if the point M has homogeneous barycentric coordinates $\left(x: y: z\right)$ with respect to triangle ABC. Since $\left(x: y: z\right)$ are the homogeneous barycentric coordinates of the point P with respect to triangle ABC, the point M has these barycentric coordinates if and only if M = P. So we obtain that

$\left(x+y+z\right)\left(x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\right) \geq yza^{2}+zxb^{2}+xyc^{2}$

with equality if and only if M = P. In other words,

$\left(x+y+z\right)\left(x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\right)$
$\geq \left(x+y+z\right)\left(x\cdot PA^{2}+y\cdot PB^{2}+z\cdot PC^{2}\right) = yza^{2}+zxb^{2}+xyc^{2}$.

Hence,

$x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\geq x\cdot PA^{2}+y\cdot PB^{2}+z\cdot PC^{2}$ and
$\left(x+y+z\right)\left(x\cdot PA^{2}+y\cdot PB^{2}+z\cdot PC^{2}\right) = yza^{2}+zxb^{2}+xyc^{2}$.

As a consequence of the latter identity, the inequality in question,

$\left(x+y+z\right)\left(\frac{x\cdot MA^{3}}{PA}+\frac{y\cdot MB^{3}}{PB}+\frac{z\cdot MC^{3}}{PC}\right)\geq yza^{2}+zxb^{2}+xyc^{2}$,

becomes

$\left(x+y+z\right)\left(\frac{x\cdot MA^{3}}{PA}+\frac{y\cdot MB^{3}}{PB}+\frac{z\cdot MC^{3}}{PC}\right)\geq \left(x+y+z\right)\left(x\cdot PA^{2}+y\cdot PB^{2}+z\cdot PC^{2}\right)$,

or, equivalently,

$\frac{x\cdot MA^{3}}{PA}+\frac{y\cdot MB^{3}}{PB}+\frac{z\cdot MC^{3}}{PC}\geq x\cdot PA^{2}+y\cdot PB^{2}+z\cdot PC^{2}$.

Now, adding $x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}$ to the inequality $x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\geq x\cdot PA^{2}+y\cdot PB^{2}+z\cdot PC^{2}$, we obtain

$2\left(x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\right)$
$\geq\left(x\cdot PA^{2}+y\cdot PB^{2}+z\cdot PC^{2}\right)+\left(x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\right)$,

what rewrites as

$2\left(x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\right)$
$\geq x\cdot\left(MA^{2}+PA^{2}\right)+y\cdot\left(MB^{2}+PB^{2}\right)+z\cdot\left(MC^{2}+PC^{2}\right)$
$\geq x\cdot 2\cdot MA\cdot PA+y\cdot 2\cdot MB\cdot PB+z\cdot 2\cdot MC\cdot PC$     (using the inequality $u^{2}+v^{2}\geq 2uv$ for any reals u and v)
$=2\left(x\cdot MA\cdot PA+y\cdot MB\cdot PB+z\cdot MC\cdot PC\right)$,

and thus $x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\geq x\cdot MA\cdot PA+y\cdot MB\cdot PB+z\cdot MC\cdot PC$. Hence,

$\left(\frac{x\cdot MA^{3}}{PA}+\frac{y\cdot MB^{3}}{PB}+\frac{z\cdot MC^{3}}{PC}\right)\left(x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\right)$
$\geq\left(\frac{x\cdot MA^{3}}{PA}+\frac{y\cdot MB^{3}}{PB}+\frac{z\cdot MC^{3}}{PC}\right)\left(x\cdot MA\cdot PA+y\cdot MB\cdot PB+z\cdot MC\cdot PC\right)$
$\geq\left(\sqrt{\frac{x\cdot MA^{3}}{PA}\cdot x\cdot MA\cdot PA}+\sqrt{\frac{y\cdot MB^{3}}{PB}\cdot y\cdot MB\cdot PB}+\sqrt{\frac{z\cdot MC^{3}}{PC}\cdot z\cdot MC\cdot PC}\right)^{2}$       (by the Cauchy-Schwarz inequality)
$=\left(x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}\right)^{2}$,

so that

$\frac{x\cdot MA^{3}}{PA}+\frac{y\cdot MB^{3}}{PB}+\frac{z\cdot MC^{3}}{PC}\geq x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}$
$\geq x\cdot PA^{2}+y\cdot PB^{2}+z\cdot PC^{2}$,

proving Theorem 1.

Now, apply Theorem 1 to the case when the point P is the incenter I of triangle ABC. Then, its homogeneous barycentric coordinates are $\left(x: y: z\right)=\left(a: b: c\right)$, so that we can set x = a, y = b, z = c. Then, according to Theorem 1, we have

$\left(a+b+c\right)\left(\frac{a\cdot MA^{3}}{IA}+\frac{b\cdot MB^{3}}{IB}+\frac{c\cdot MC^{3}}{IC}\right)\geq bca^{2}+cab^{2}+abc^{2}$,

with equality if and only if M = I. This simplifies to

$\left(a+b+c\right)\left(\frac{a\cdot MA^{3}}{IA}+\frac{b\cdot MB^{3}}{IB}+\frac{c\cdot MC^{3}}{IC}\right)\geq abc\left(a+b+c\right)$,

and upon division by a + b + c, this becomes

$\frac{a\cdot MA^{3}}{IA}+\frac{b\cdot MB^{3}}{IB}+\frac{c\cdot MC^{3}}{IC}\geq abc$.

Let r be the inradius of triangle ABC. If the incircle of triangle ABC touches its side CA at Y, then $IY\perp CA$ (since I is the center of this incircle), IY = r (since r is the radius of this incircle) and $\measuredangle IAY=\frac{A}{2}$ (since the point I, being the incenter of triangle ABC, lies on the angle bisector of its angle CAB). Thus, in the right-angled triangle AYI, we have $IA=IY/ \sin\measuredangle IAY$, what becomes $IA=r/ \sin\frac{A}{2}$. Similarly, $IB=r/ \sin\frac{B}{2}$ and $IC=r/ \sin\frac{C}{2}$. Hence, the inequality

$\frac{a\cdot MA^{3}}{IA}+\frac{b\cdot MB^{3}}{IB}+\frac{c\cdot MC^{3}}{IC}\geq abc$

becomes

$\frac{a\cdot MA^{3}}{r/ \sin\frac{A}{2}}+\frac{b\cdot MB^{3}}{r/ \sin\frac{B}{2}}+\frac{c\cdot MC^{3}}{r/ \sin\frac{C}{2}}\geq abc$.

Upon multiplication by r, this becomes

$a\sin\frac{A}{2}\cdot MA^{3}+b\sin\frac{B}{2}\cdot MB^{3}+c\sin\frac{C}{2}\cdot MC^{3}\geq abcr$.

As we know, equality holds if and only if M = I. Hence, we have proven the following result:

[color=blue][b]Theorem 2.[/b] If ABC is a triangle with incenter I and inradius r, and M is a point in the plane, then

$a\sin\frac{A}{2}\cdot MA^{3}+b\sin\frac{B}{2}\cdot MB^{3}+c\sin\frac{C}{2}\cdot MC^{3}\geq abcr$,

with equality holding if and only if M = I.[/color]

This clearly solves your problem.

  darij


Let $f(t)=(t-1)(t-9)(t-25)(t-49)\left(\frac{a}{t-1}+\frac{b}{t-9}+\frac{c}{t-25}+\frac{d}{t-49}-1\right)$

$f(t)$ is a polynomial with removable discontinuities. Let $f(t)=\lim_{x\rightarrow t}f(x)$, if necessary. Note that $f(t)$ is a fourth degree polynomial.

Since $f(t)=0$ for $t=4,16,36,64$. By the division algorithm, we can write $f(t)=q(t)*(t-4)(t-16)(t-36)(t-64)$. Since deg $f(t)$= deg $q(t)$+4, we have that: deg $q(t)$=0, so it is a constant. By only considering the leading term in the expansion of $f(t)$, it is clear that $c=-1$.

Consider the $t^{3}$ coefficient. We have that:

$a+b+c+d+(-1)(-1-9-25-49)=(-1)(-4-16-36-64)$
$a+b+c+d=3+7+11+15=\boxed{36}$

What’s the inspiration for the t-substituion?

It's really symmetric if we consider it as a function and find four roots; the degree seems like $4$, and $4$ roots are pretty useful.

I wish all AIME #15's were that easy right now ... and this was also the very last problem in Intermediate Algebra (20.86 lol)

[hide=@#10][url=aops.com/community/user/378892][b]Zorger74[/b][/url] · Nov 15, 2020, 9:35 PM [url=aops.com/community/p18925308](view)[/url][color=transparent]helo[/color]
Why isn't this in HSM? Anyway: [hide=Solution]Because I found this problem in a handout called polynomials, we are motivated to create one :P Since the only thing that changes in each equation is the first term in each denominator, we substitute a variable there. Multiplying through to clear denominators, and moving stuff to one side, we have that $(a-1)(a-9)(a-25)(a-49)-x^2(a-9)(a-25)(a-49)\dots=0$. Notice that this polynomial has roots $a=4,16,36,64$ and is monic. Thus we can write it as $(a-4)(a-16)(a-36)(a-64)$. We notice that the coefficient of $a^3$ contains $x^2+y^2+z^2+w^2$, so we expand that coefficient to find that $-a^3-9a^3-25a^3-49a^3-(x^2+y^2+z^2+w^2)a^3=-4a^3-16a^3-36a^3-64a^3$. Thus, the answer is $-1-9-25-49+4+16+36+64=\boxed{36}$.[/hide]

-----------
[color=#5b7083][aops]x[/aops] 10[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]
I was asking the same question.

Anyway, answering #7, the $w, x, y, z$ terms are sort of "bundled together" in such a way that the only thing that changes from equation to equation is the squared even powers. That allows us to essentially necessarily and sufficiently combine the information that is given directly in the problem into one polynomial in terms of $t$.

In fact, this isn't the first time I've seen such a polynomial interpolation-type SoE question, so I believe maybe in recent years this method became more well-known?

A final question: why did the testwriters use $x^2, y^2, z^2$ instead of $x, y, z$? There's no point in squaring the terms, if I could just go about and substitute $a=x^2$.


calculus
integration
conics
parabola
analytic geometry
geometry
calculus computations

Given the parabola $C: y=x^{2}.$ If the circle centered at $y$ axis with radius $1$ has common tangent lines with $C$ at distinct two points, then find the coordinate of the center of the circle $K$ and the area of the figure surrounded by $C$ and $K.$

[color=darkblue]If the point $w$ is the center of the circle $K$, then :
$\blacktriangleright 1.\ w\left(0,\frac{5}{4}\right)$ and the tangent points are $T_{k}\left((-1)^{k}\frac{\sqrt3}{2},\frac{3}{4}\right)$, where $k\in \overline{1,2}\ ;$
$\blacktriangleright 2.$ The found area is $2\int_{0}^{\frac{\sqrt 3}{2}}\left(\frac{5}{4}-\sqrt{1-x^{2}}-x^{2}\right)\ dx=\frac{3\sqrt 3}{4}-\frac{\pi}{3}\ .$
[b]Remark.[/b] For any $x\in (-1,1)$, $\left(\arcsin x+x\sqrt{1-x^{2}}\right)'=2\sqrt{1-x^{2}}\ .$

[b]Indication.[/b] The perpendicular to the tangent in the point $P\left(x,x^{2}\right)\in C$ is $Y-x^{2}=-\frac{1}{2x}(X-x)$ and the its intersection with the axis Oy is $w\left(0,x^{2}+\frac{1}{2}\right)$ which is the center of the circle $K$. Thus, $Pw=1$ $\Longleftrightarrow$ $(x-0)^{2}+\left[\left(x^{2}+\frac{1}{2}\right)-x^{2}\right]^{2}=1$ $\Longleftrightarrow$ $x\in \left\{\pm\frac{\sqrt 3}{2}\right\}$. Therefore, $T_{k}=\left((-1)^{k}\frac{\sqrt 3}{2},\frac{3}{4}\right)$, where $k\in \overline{1,2}$ and $w\left(0,\frac{5}{4}\right)$ a.s.o.[/color]

sorri ignore this post^^'''[since virgil corrected his post] :P

[quote="Virgil Nicula"][color=darkblue]If the point $w$ is the center of the circle $K$, then :
$\blacktriangleright 1.\ w\left(0,\frac{5}{4}\right)$ and the tangent points are $T_{k}\left((-1)^{k}\frac{\sqrt3}{2},\frac{3}{4}\right)$, where $k\in \overline{1,2}\ ;$
$\blacktriangleright 2.$ The found area is $2\int_{0}^{\frac{\sqrt 3}{2}}\left(\frac{5}{4}-\sqrt{1-x^{2}}-x^{2}\right)\ dx=\frac{3\sqrt 3}{4}-\frac{\pi}{3}\ .$
[b]Remark.[/b] For any $x\in (-1,1)$, $\left(\arcsin x+x\sqrt{1-x^{2}}\right)'=2\sqrt{1-x^{2}}\ .$ [/color][/quote]

That's right. :)

[color=darkblue][b][u]A nice remark.[/u][/b] For a fixed number $r_{0}\ge\frac{1}{2}$ and for any $n\in N$ define the number $r_{n}=r_{0}+n$. Then for any $n\in N$ the circle $K_{n}\left(w_{n},r_{n}\right)$, with the center $w_{n}\left(0,r^{2}_{n}+\frac{1}{4}\right)$ is tangent to the parabola $C$ with the equation $y=x^{2}$ and the circles $K_{n}$, $K_{n+1}$ are exterior tangent. The focus of the parabola $C$ is the point $F\left(0,\frac{1}{4}\right)$ and for $r_{0}=\frac{1}{2}$, the circle $K_{0}$ is tangent in the origin to the parabola $C$. If $r_{0}=1$ then the focus $F$ of the parabola $C$ belongs to $K_{0}$.[/color]


calculus
integration
derivative
limit
LaTeX
function
calculus computations

the derivative of e^(-1/x) is 1/x^2   e^(-1/x)
split the integral at zero

[quote]
Improper integral
Calculate the integral
$ \int\limits_{ \minus{} 1}^1 {\frac{{{e^{ \minus{} \frac{1}{x}}}}}{{{x^2}{{\left( {1 \minus{} {e^{ \minus{} \frac{1}{x}}}} \right)}^2}}}} dx$[/quote]

let $ I \equal{} \int\limits_{ \minus{} 1}^1 {\frac{{{e^{ \minus{} \frac{1}{x}}}}}{{{x^2}{{\left( {1 \minus{} {e^{ \minus{} \frac{1}{x}}}} \right)}^2}}}} dx    \equal{}  \minus{} \int_{\minus{}1}^{1}\frac{dx}{ x^{2} (1\minus{} e^{\frac{1}{x}} )}$

ALSO $ I \equal{} \int\limits_{ \minus{} 1}^1 {\frac{{{e^{  \frac{1}{x}}}}}{{{x^2}{{\left( {1 \minus{} {e^{  \frac{1}{x}}}} \right)}^2}}}} dx$

Adding        $ 2I \equal{} \minus{} \int_{\minus{}1}^{1}\frac{dx}{ x^{2} }$
therefore     $ 2I \equal{} 2$

hence

$ I \equal{} 1$

There are several different ways that that doesn't make any sense. To pick on one:

$ \minus{}\int_{\minus{}1}^1\frac{dx}{x^2}\equal{}\minus{}\infty,$ not $ 2.$

vijaymenon and akech, how can we change variable since u(x) isnt continious at [0,-1] and [0,1]  ?

i have not changed any variables rather i have jus applied a property of definite integrals ie

$ \int_{a}^{b}f(x)dx \equal{} \int_{a}^{b}f(a \plus{} b \minus{} x) dx$

AT KENT:
Sorry made a  blunder .

Does that mean the answer is $ \minus{} \infty$

Mhmhmhm...
Mine approach
\[ \begin{array}{l} \displaystyle \int\limits_{ \minus{} 1}^1 {\frac {{{e^{ \minus{} \frac {1}{x}}}}}{{{x^2}{{\left( {1 \minus{} {e^{ \minus{} \frac {1}{x}}}} \right)}^2}}}dx \equal{} \mathop {\lim }\limits_{a \to {0^ \minus{} }} } \int\limits_{ \minus{} 1}^a {\frac {{{e^{ \minus{} \frac {1}{x}}}}}{{{x^2}{{\left( {1 \minus{} {e^{ \minus{} \frac {1}{x}}}} \right)}^2}}}dx \plus{} } \mathop {\lim }\limits_{b \to {0^ \plus{} }} \int\limits_b^1 {\frac {{{e^{ \minus{} \frac {1}{x}}}}}{{{x^2}{{\left( {1 \minus{} {e^{ \minus{} \frac {1}{x}}}} \right)}^2}}}dx \equal{} }  \\
\equal{} \mathop {\lim }\limits_{a \to {0^ \minus{} }} \left( {\frac {1}{{1 \minus{} {e^{ \minus{} \frac {1}{a}}}}} \minus{} \frac {1}{{1 \minus{} e}}} \right) \plus{} \mathop {\lim }\limits_{b \to {0^ \plus{} }} \left( {\frac {1}{{1 \minus{} {e^{ \minus{} 1}}}} \minus{} \frac {1}{{1 \minus{} {e^{ \minus{} \frac {1}{b}}}}}} \right) \equal{} \\
\equal{} 0 \minus{} \frac {1}{{1 \minus{} e}} \plus{} \frac {1}{{1 \minus{} {e^{ \minus{} 1}}}} \minus{} 1 \equal{} \frac {2}{{e \minus{} 1}} \\
\end{array}\]
Is the above wrong?

I have no idea, but you could greatly improve the legibility of your mathematics using the align* package.  Also, are you putting in all this "\mathop" and "\limits" stuff, or is the forum doing it for you?  It's totally unnecessary: \lim_{a \to 0} gives $ \lim_{a \to 0}$ perfectly nicely, and you've made your integrals look worse than they otherwise would have.

I seem to remember [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1507884#1507884]having this conversation before[/url].

JBL
i dont use latex code. I am familiar to mathtype so with the translator i get the above type result.
Thank you for your "latex" code advices but i wont follow cause i found lot easier to use something that im already good at it.
Im interesting about the math part of this type.
Anyone can tell me if i am wrong and where?

Thank you in advance

Let $ {g_1}\left( x \right) \equal{} \frac{1}{{1 \minus{} {e^{ \minus{} \frac{1}{x}}}}},x \in \left[ { \minus{} 1,0} \right) \cup \left( {0,1} \right]$
 and $ {g_2}\left( x \right) \equal{} \left\{ {\begin{array}{*{20}{c}}
   {1 \plus{} \frac{1}{{1 \minus{} {e^{ \minus{} \frac{1}{x}}}}},x \in \left[ { \minus{} 1,0} \right)}  \\
   {1,x \equal{} 0}  \\
   {\frac{1}{{1 \minus{} {e^{ \minus{} \frac{1}{x}}}}},x \in \left( {0,1} \right]}  \\
\end{array}} \right.$

Try to answer to the following multiple choice question
An antiderivative of $ \frac{{{e^{ \minus{} \frac{1}{x}}}}}{{{x^2}{{\left( {1 \minus{} {e^{ \minus{} \frac{1}{x}}}} \right)}^2}}},x \in \left[ { \minus{} 1,0} \right) \cup \left( {0,1} \right]$
 is:
A: $ {g_1}\left( x \right)$
B: $ {g_2}\left( x \right)$
C: none of $ {g_2}\left( x \right)$,$ {g_1}\left( x \right)$ but function has other antiderivative $ {g_3}\left( x \right)$
D: we cant answer, since we searching antiderivative only at one interval and not at a union
Please [b]justify[/b] your answer


integration
trigonometry
calculus
calculus computations

Complete the square on the bottom and let $ x\plus{}2\equal{}\sin(\theta)$

hello, writing $ 4x\minus{}x^2$ in the form $ 4x\minus{}x^2\equal{}\minus{}(x\minus{}2)^2\plus{}4$ and let $ t\equal{}x\minus{}2$ so we have the integral $ \int \frac{(t\plus{}2)^2}{\sqrt{(4\minus{}t^2)^3}}\,dt$. Now you can substitute $ t\equal{}2\sin(z)$.
Sonnhard.

[hide]Considering the change of variables $ x \equal{} \frac{4}{u^2\plus{}1}$ we have

$ P \equal{} \int \frac{x^2}{(4x\minus{}x^2)^{3/2}}\,dx \equal{} \minus{}2\int \frac{1}{u^2(u^2\plus{}1)}\,du \equal{} 2 \int \minus{}\frac{1}{u^2}\,du \plus{} 2 \int\frac{1}{u^2\plus{}1}\,du$

$ \equal{} \frac2u \plus{} 2 \arctan u \equal{} 2 \sqrt{\frac{x}{4\minus{}x}} \plus{} 2 \arctan \sqrt{\frac{4\minus{}x}{x}} \plus{} C$.[/hide]

tried something....

[hide]$ Find \int \frac {x^{2}}{(4x\ \minus{}\ x^{2})^{\frac {3}{2}}}\ dx$

$ s \equal{} 4x \minus{} x^{2}$
$ x^2 \equal{} 4x \minus{} s$
$ ds \equal{} \minus{}2x \plus{} 4$
$ \minus{}\frac{1}{2}ds \minus{} 4 \equal{} x$

$ \int \frac{4x \minus{} s}{{s}^\frac{3}{2}} \equal{}\int \frac{4 * (\minus{}\frac{1}{2}ds \minus{} 4) \minus{} s}{{s}^\frac{3}{2}}$
$ \equal{}\minus{}2\int \frac{ds \minus{} 16 \minus{} s}{{s}^\frac{3}{2}}\equal{}\minus{}2 \int \frac{ds \minus{} 16 \minus{} 1}{s^\minus{}\frac{1}{2}}$
$ \equal{}\minus{}2 \int s^\frac{1}{2} \ ds \minus{} 15 \equal{}\minus{}3 s^\frac{3}{2} \minus{} 15s \plus{} c$

?????[/hide]

That is nonsense. In particular, it is not true in any sense that $ ds\equal{}\minus{}2x\plus{}4$; the two sides have different units.

Follow Mathstud28's approach and complete the square first, on the way to a trigonometric substitution.

$ \int \frac {x^{2}}{(4x\ \minus{} \ x^{2})^{\frac {3}{2}}}\ dx \equal{} \int \frac {x^{2}}{(\sqrt {2^2 \minus{} x^2})^3}\ dx$

$ x \equal{} 2 sin \theta$
$ dx \equal{} 2 cos\theta \ d\theta$

$ \int \frac {x^{2}}{(\sqrt {2^2 \minus{} x^2})^3}\ dx \equal{} \int \frac {(2sin\theta)^2 . 2 cos\theta \ d\theta}{(\sqrt{4 \minus{} 4 sin^2\theta})^3} \equal{} \int \frac {4sin^2\theta\ . \ 2cos\theta d\theta}{(4^\frac {1}{2}(1 \minus{} sin^2\theta)^\frac {1}{2})^3}$
$ \equal{} \int \frac {8\ sin^2\theta cos\theta d \theta}{8\ cos^3 \theta}$
$ \equal{} \int tan^2 \ \theta \ d\theta \equal{} \int (sec^2\ \theta \minus{} 1)\ d\theta$
$ \equal{} tan\theta \minus{} \theta \plus{} c$

$ sin\ \theta \equal{} \frac {x}{2} \equal{} \frac {opp.}{hyp.}$
$ adj. \equal{} \sqrt {2^2 \minus{} x^2}$

$ Therefore, tan\theta \minus{} \theta \plus{} c$
$ \equal{} \frac {x}{\sqrt {2^2 \minus{} x^2}} \minus{} sin^{ \minus{} 1} \frac {x}{2} \plus{} c$

[b]Wazika[/b], you forgot an $ x$ in the square root on the right hand side of your first equality.

[hide]wow thanks.....that's the correction my teacher gave us yesterday in class.... lol... I'll have to tell him about it before he gaves everyone our question wrong....we completed the square[/hide]


What does LaTeX stand for? If it stands for anything

I think the best way to answer this is to quote the first paragraph from [url=http://en.wikipedia.org/wiki/Knuth]Donald Knuth's[/url]The TeXbook:

[quote]English words like 'technology' stem from a Greek root beginning with the letters $\tau\epsilon\chi\dots$; and this same Greek word means art as well as technology. Hence the name $\text{\TeX}$ which is an uppercase form of $\tau\epsilon\chi$.[/quote]
In the 1980's [url=http://en.wikipedia.org/wiki/Leslie_Lamport]Leslie Lamport[/url] extended $\text{\TeX}$ using macros which made the system easier to use. His surname led to the extended system being called $\text{\LaTeX}$


Say a sentence or common saying without using the actual words. Also decode the sentence before yours. For example, "Goat ooze leap." means "Go to sleep."

I will start:

You night heads tastes of um Erica.

United States of America.

Sandy Egg Kazoo

San Diego Zoo?

Imp Eye Er... Stay Tub Ill Dee Ink

Empire State Building?

 Picks pan loafs

Since no one is getting it, here is a new one:

Tee eye ate in I'm tight Annie um.

The I eight in I'm tight Annie um??
huH??
Eye'm board!

I'm bored.

Isle goan ow.

I fail. Hard. :P

meewhee's is I'll go now. (Are you telling us what Picks pan loafs was? No idea.)
PowerOfPi's is TI-89 Titanium. You know, the graphing calculator?

Tie muff lie swan yo rehab info on.

Time flies when your having fun.

Hunt Ed how says.

Haunted houses

Mice who smelt "[i]sss[/i]"

Grin exam dam.

[hide="hint"]Dr. Seuss[/hide]

Green eggs and ham (nice one)

 mat in aha at

same hint

Cat in a hat.

A mayor ick an math um a ticks come pet ish on.

A mayor in mathematics come petition?
um..?

Eye are very co un fuse d.

I are very confused.

Close, you got mathematics right. Since I'm too lazy to make up a new one, that one's still the new problem.

A mayor ick an math um a ticks come pet ish on.

[hide="hint"]3 words[/hide]

American Mathematics Competition
(AKA AMC) :P

Eye wannabe con fuzz uld.

Just so you know..
the last word in my sentence is my OWN unique world :D
+

heh

I wanna be confuzzled.
lol

Math e mat icks some thymes is ard.

Mathematics sometimes is hard.

Trip aunt won for on fly hive.

[hide="hint"]it's a number[/hide]


this problem was marked difficult in this one book i have:

we have a 20x30 rectangle. is it possible to draw a straight line which intersects the interiors of 30 boxes?

now, i thought you could just draw the line along one of the sides of length 30, which makes sense, but seems too obvious to be the answer to the question. am i right, or am i missing something?

I am correctly assuming your 20 by 30 rectangle is comprised of unit squares, right?

[quote="JMI"]Does the problem mean just unit squares or does it  mean unit squares, and 2x2 squares, 3x3 squares,...?[/quote]

Well it really doesn't matter unless the problem meant 300 or something (in which case, I would guess "box" means rectangle).

I have Mathematical Circles. What section was it in? Also, aren't there answers in the back?

Well if you draw a line one vertex to the opposite vertex then the problem is equivalent to finding the number of boxes a diagonal would pass through in a 2 x 3 rectangle and then multiply by 10.  Doing this I get a line that passes through 40 interiors.  Is the problem asking for a line that only passes through 30 interiors?

It would make more sense for it to ask for 49 (the maximum).

yasha, its in Problems For the First Year, so there aren't any answers in the back. And no, the problem says what it says, that the line has to pass through the interiors of 30 squares.

Yes, I found it.

[quote="Mathematical Circles: Ch. 8 #115"]A 20 X 30 rectangle is cut from a sheet of graph paper. Is it possible to draw a straight line which intersects the interiors of 30 boxes of the rectangle?[/quote]

It makes no sense to me either. It's probably just a typo.

no, i don't think so, because if you just made a line through just one box, that could translate into making a line through many many boxes, definitely more than 30. either that or i misunderstood your question.

[quote="s0mp"]ah! maybe they are talking about boxes as in 1x2's, 1x1's 2x2's, 3x4's, etc., which would definitely make this a lot harder.[/quote]

Uhh... wouldn't that be a lot easier?

Oh, I know!  It means exactly 30 boxes.

yeah, well, um, yeah, actually, must have been a typo, since it looks like what i said earlier is pretty much impossible.

Could you make a line that passes through exactly 30 unit squares out of the 600 in the rectangle?

Yes, very easily. You just draw a line parallel to the side that is 30 half a unit away. It will go through exactly 30 squares.


An aquarium has a rectangular base that measures $ 100$ cm by $ 40$ cm and has a height of $ 50$ cm. It is filled with water to a height of $ 40$ cm. A brick with a rectangular base that measures $ 40$ cm by $ 20$ cm and a height of 	$ 10$ cm is placed in the aquarium. By how many centimeters does the water rise?

$ \textbf{(A)}\ 0.5 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 1.5 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 2.5$

[hide] The volume of the brick is 40*10*20=8000 cubic centimeters.  We also know that the area of the base of the aquarium is 400*10=4000 square centimeters.  Thus, the depth increases by \[\frac{8000}{4000}=2 cm\]  Thus, the answer is $D$. [/hide]

[quote="4everwise"]An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. It is filled with water to a height of $40$ cm. A brick with a rectangular base that measures $40$ cm by $20$ cm and a height of 	$10$ cm is placed in the aquarium. By how many centimeters does the water rise?

$\text{(A)}\ 0.5 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 1.5 \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 2.5$[/quote]
[hide]So the volume of the aquarium with the height of the water is going to be 160000cm. The volume of the rock is going to be 8000 cm. So the total volumes is going to be 168000 cm. We want the new height so divide by the area of the base or 4000 cm to get that the new height is 42cm. Since the original height was 40 it increased by 2 cm. So D.[/hide]

[quote="4everwise"]An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. It is filled with water to a height of $40$ cm. A brick with a rectangular base that measures $40$ cm by $20$ cm and a height of 	$10$ cm is placed in the aquarium. By how many centimeters does the water rise?

$\text{(A)}\ 0.5 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 1.5 \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 2.5$[/quote]

[hide="solution"]The volume of the brick is $8000$.  the areaof the base of the aquarium is $4000$.  $\frac{8000}{4000}=\boxed{2}$[/hide]

This is also on the 10A, it was the 5 question, we can see the difficulty lever

lame, you spammed in every thread made about amc12

Is it just me, or was this question, and the 12B's first question, the hardest first two questions ever?

The first question on the 12B took me a good three minutes to solve, whereas the next 9 took maybe 8 minutes total ;-)

Something seems odd there, to me.


function
real analysis
real analysis unsolved

Assign to each real $ x$ a [b]finite[/b] set $ F(x)$ (which doesn't contain $ x$ and is contained in $ \mathbb{R}$). Call a set $ S$ independant if $ x \in F(y)$ holds for no elements $ x,y$ of $ S$. Can such a set have continuum cardinality? Is this the correct place to post such problems?

Let $ \delta(x)=\min_{y\in F(x)}|x-y|$. This is a strictly positive function on $ \mathbb R$. Therefore, for some $ \epsilon>0$ the set $ \{x\in\mathbb R\colon \delta(x)>\epsilon\}$ contains a set of positive measure, say $ A$. Choose an open interval $ I$ of length $ \epsilon$ such that $ A\cap I$ has positive measure. Set $ S=A\cap I$.

Erm... Who said that the function is measurable? :? On the other hand, it seems like the solution can be repaired (just using cardinality instead of measure) if we could only show that a countable union of sets of cardinality less than continuum each is of cardinality less than continuum. It is certainly true if one assumes the continuum hypothesis but I do not immediately see why it is true unconditionally. :(

A solution: [hide]For each $ x$ there exists an interval with rational endpoints (which contains $ x$) that doesn't intersect $ F(x)$, lets call this $ I_{x}$. Clearly there are only countably many distinct such intervals however there are uncountably many $ x$ so uncountably many $ I_{x}$ must coincide. That is to say there is an uncountable set in which each $ I_{x}$ is the same for all elements of the set. Such a set has the desired property[/hide]

As to your statement about less than continuum sets I  really have no idea how to start it (or if it is true) without assuming CH. :(

I thought about Baire's Teorem but couldn't see how to use it. How have you done so? As to the thing about measure I assume you were asking Fedja since I have just left high school and really have a limited knowledge of such things.

[quote="mlok"]I was under impression that any Lebesgue nonmeasurable set contains a measurable subset of positive measure... is this true? If not, we can use Baire category instead.[/quote]
Well, the outer measure is positive. Does the outer measure have any type of inner regularity property?

[quote="mlok"]I was under impression that any Lebesgue nonmeasurable set contains a measurable subset of positive measure... is this true?
[/quote]
Of course, not! The standard construction of a non-measurable set (choosing one representative in each class of $ \mathbb R/\mathbb Q$ gives you a counterexample).
[quote]If not, we can use Baire category instead.[/quote]
I believe you'll run into the same problem: why is a set of second category of cardinality continuum?

@EBz89 Your "solution" is identical to my modification of mlok's one and has the same problem with it: we can easily conclude that one of the sets is uncountable but it doesn't automatically imply cardinality continuum you asked about.

@Kalle If I remember it right, the positivity of outer measure does not imply continuum cardinality without something like CH but here my memory is shaky :(.

Yes, the relation between nonmeasurable and positive-measure sets was the other way around. :blush:

EBz89 may have invented a new form of the continuum hypothesis...

Forget it, I was stupid... I got the problem from a book that did actually ask for continuum cardinality however I think it probably either assumed CH or just made a silly mistake. Sorry  :blush:

It is known that $ \kappa\mu = \max\{\kappa,\mu\}$ holds for infinite cardinal numbers. it requires Axiom of Choice, but not CH. i read it from wikipedia and sci.math. but haven't seen a proof.


In triangle $ ABC$  $ m(\angle{BAC}) \equal{} 90^\circ$
$ P \in [AC]$ , $ Q \in [BC]$ , $ R \in [AB]$ and $ APQR$ is a square.
$ N,K \in [BC]$ , $ M \in [AB]$ , $ L \in [AC]$ and $ KLMN$ is a square.

 given that $ |AP|\equal{}3 , |KL|\equal{}2\sqrt2$  find $ |AB|\plus{}|AC|$

From similar triangles, we have $ \frac {BR}{RQ} \equal{} \frac {QP}{PC}, or (AB \minus{} 3)(AC \minus{} 3) \equal{} 9$
Also, from similar triangles it is easy to see $ BN \equal{} 2\sqrt {2} \frac {AC}{AB}$ and $ KC \equal{} 2\sqrt {2} \frac {AB}{AC}$, thus we have from Pythag
$ AB^2 \plus{} AC^2 \equal{} 8(\frac {AB}{AC} \plus{} \frac {AC}{AB} \plus{} 1)^2$
Let $ AB\plus{}AC\equal{}x$ and $ AB \cdot AC \equal{}y$
We have
$ y\equal{}3x$
$ x^2\minus{}2y\equal{}8(\frac{x^2\minus{}y}{y})^2$
From this we get x=12.


inequalities logarithms limit search Euler function integration

Is there any formula that finds nth term of 1/1+1/2+1/3+1/4+...+1/n ?

If not, is there any proof that there is no formula for this?

Also..is there any wonderful explanation why this is infinite?

I just saw a proof that 1/1+1/2>1/2, 1/3+1/4>1/2, 1/5+1/6+1/7+1/8>1/2, and so on.

So 1/1+1/2+1/3+1/4+...+1/n>1/2+1/2+1/2+1/2+...

This doesn't come to my mind..

I'm sorry for asking too much questions..

[quote="car202"]
Also..is there any wonderful explanation why this is infinite?
[/quote]

Verify for yourself the following inequality, given without proof.

$\ln n < 1 + \frac{1}{2} + ... + \frac{1}{n} < \ln n + 1$

Clearly $\lim_{n \to \infty} \ln n = \infty$, and so the sum must diverge.

Alternately, the following formula (also given without proof):

$\ln (1 + x) = \sum_{k=1}^{\infty} \frac{-(-x)^k}{k}$

Gives the sum of the harmonic series as $- \ln 0$, undefined.

As for the partial sums:

http://mathworld.wolfram.com/HarmonicSeries.html

The question why there is or not a closed formula already occured.
Please search for it.
There where also given a lot of proofs for the divergency.

Yes Arne you are right
But i did not found where i made a mistake!
Normaly this link talk about euler cst,the limit to infinity of
$1+\frac{1}{2}+....+\frac{1}{n}-ln(n)$
Which till now,we don't know if it's algebric ,irrational or transcendent!

For a decreasing function $f(x)$ where $a_1$ is the first term in the sequence,
\[ \int^{n}_{1}f(x) dx \leq \sum^{n}_{i=1}f(n) \leq a_1 + \int^{n}_{1}f(x) dx \]
You can easily see that if you sketch an arbitrary decreasing function. Now,
\[ \int^{n}_{1}\frac{1}{x}dx \leq \sum^{n}_{i=1}\frac{1}{i} \leq 1 + \int^{n}_{1}\frac{1}{x}dx \]
Now as $n \rightarrow \infty$,
\[ \int^{n}_{1}\frac{1}{x}dx = \left[ln x\right]^{n}_{1} = ln n - ln 1 = ln n = \infty  \]
But,
\[ \int^{n}_{1}\frac{1}{x}dx \leq \sum^{n}_{i=1}\frac{1}{i} \]
Therefore, $\sum^{n}_{i=1}\frac{1}{i}$ is also infinite as $n \rightarrow \infty$.


Find all triples $ (x,y,z)$ of natural numbers satisfying $ 2xz\equal{}y^2$ and $ x\plus{}z\equal{}1987$.

[quote="moldovan"]Find all triples $ (x,y,z)$ of natural numbers satisfying $ 2xz \equal{} y^2$ and $ x \plus{} z \equal{} 1987$.[/quote]

[hide="brutal force.........."]

$ x \plus{} z \equal{} 1987$ is prime $ \Rightarrow x,z$ coprime.

So WLOG, from $ 2xz \equal{} y^2$ we let $ x \equal{} 2a^2, y \equal{} b^2$

So now we are going to find $ a,b$ so that $ 2a^2 \plus{} b^2 \equal{} 1987$

mod 4 and then mod 3 we have $ a \equal{} 6a_1 \plus{} 3$, $ b \equal{} 6b_1 \plus{} 1$, where $ a_1,b_1$ are integers.

$ a_1 \equal{} 0 \Rightarrow b^2 \equal{} 1969$, no solution.
$ a_1 \equal{} 1 \Rightarrow b^2 \equal{} 1825$, no solution.
$ a_1 \equal{} 2 \Rightarrow b^2 \equal{} 1537$, no solution.
$ a_1 \equal{} 3 \Rightarrow b^2 \equal{} 1105$, no solution.
$ a_1 \equal{} 4 \Rightarrow b^2 \equal{} 529 \Rightarrow b \equal{} 23$
$ a_1 \ge 5 \Rightarrow 1987 \equal{} 2a^2 \plus{} b^2 > 2a^2 \ge 2(33)^2 \equal{} 2178 > 1987$, contradiction.

So $ a_1 \equal{} 4, b \equal{} 23 \Rightarrow (x,z) \equal{} (1458,529),(529,1458)$
$ \Rightarrow (x,y,z) \equal{} (1458,1242,529),(529,1242,1458)$ as the only solutions.

[/hide]


inequalities
algebra
function
domain
analytic geometry

Is it possible that two operators have a nonzero commutator relation, but a zero uncertainty relation?

In particular I found that $ [\hat{x}, \hat{L_z}]\equal{}\hat{y}\hat{p_x}$. It makes sense that these shouldn't commute, yet it also makes sense that there shouldn't be an uncertainty in $ \Delta x\Delta L_z$ since why should knowing $ x$ position limit knowledge about $ z$ angular momentum.

I may just be finding the expectation $ <\hat{y}\hat{p_x}>$ incorrectly though.

From $ \vec L \equal{} \vec r \times \vec p,$ $ L_z \equal{} xp_y \minus{} yp_x \Longrightarrow$ $ \widehat L_z \equal{} \widehat x \widehat p_y \minus{} \widehat y \widehat p_x.$ Everything but $ \widehat x, \widehat p_x$ comutes $ \Longrightarrow$ $ [\widehat x, \widehat L_z] \equal{} \minus{} \widehat y [\widehat x \widehat p_x \minus{} \widehat p_x \widehat x] \equal{} \minus{} \text i \hbar \widehat y$ and similarly, $ [\widehat y, \widehat L_z] \equal{} \text i \hbar \widehat x.$ The expectation values of $ x, L_z$ of a system in some state $ |\psi\rangle$ are $ \overline x \equal{} \langle \psi | \widehat x | \psi \rangle,\ \overline L_z \equal{} \langle \psi | \widehat L_z | \psi \rangle.$ All operators are Hermitian, $ \widehat \xi \equal{} \widehat x \minus{} \overline x,\ \widehat \Lambda_z \equal{} \widehat L_z \minus{} \overline L_z$ are also Hermitian $ \Longrightarrow$ uncertainties of $ x, L_z$ are given by

$ (\Delta x)^2 \equal{} \langle \psi | \widehat \xi^2 | \psi \rangle \equal{} \langle \widehat \xi \psi | \widehat \xi \psi \rangle \equal{} \left| | \widehat \xi \psi \rangle\right|^2$
$ (\Delta L_z)^2 \equal{} \langle \psi | \widehat \Lambda_z^2 | \psi \rangle \equal{} \langle \widehat \Lambda_z \psi | \widehat \Lambda_z \psi \rangle \equal{} \left| | \widehat \Lambda_z \psi \rangle\right|^2$

[hide="By standard computation:"]
Schwartz inequality yields

$ (\Delta x)^2 (\Delta L_z)^2 \equal{} \left| | \widehat \xi \psi \rangle \right|^2 \left| | \widehat \Lambda_z \psi \rangle \right|^2 \ge \left| \langle \widehat \xi \psi | \widehat \Lambda_z \psi \rangle \right|^2 \equal{} \left| \langle \psi | \widehat \xi \widehat \Lambda_z | \psi \rangle \right|^2$

Since $ \widehat \xi, \widehat \Lambda_z$ are both Hermitian, their anticomutator $ \{\widehat \xi, \widehat \Lambda_z \} \equal{} \widehat \xi \widehat \Lambda_z \plus{} \widehat \Lambda_z \widehat \xi$ is also Hermitian. Their comutator is still $ [\widehat \xi, \widehat \Lambda_z] \equal{} [\widehat x, \widehat L_z] \equal{} \minus{} \text{i} \hbar \widehat y.$

$ \widehat \xi \widehat \Lambda_z \equal{} \frac {_1}{^2}\{\widehat \xi, \widehat \Lambda_z \} \plus{} \frac {_1}{^2} [\widehat \xi, \widehat \Lambda_z] \equal{} \frac {_1}{^2}\{\widehat \xi, \widehat \Lambda_z \} \minus{} \frac {_1}{^2} \text{i} \hbar \widehat y$

The anticomutator $ \{\widehat \xi, \widehat \Lambda_z \}$ and $ \widehat y$ are both Hermitian $ \Longrightarrow$ their expectation values are both real and the expectation value of $ \text{i} \widehat y$ is imaginary. Therefore,

$ (\Delta x)^2 (\Delta L_z)^2 \ge \left| \langle \psi | \frac {_1}{^2} \{\widehat \xi, \widehat \Lambda_z \} \minus{} \frac {_1}{^2}\text{i} \hbar \widehat y| \psi \rangle \right|^2 \equal{}$

$ \equal{} \frac {_1}{^4} \left| \langle \psi | \{\widehat \xi, \widehat \Lambda_z \}| \psi \rangle \right|^2 \plus{} \frac {_{\hbar^2}}{^4} \left| \langle \psi |\widehat y| \psi \rangle \right|^2 \ge \frac {_{\hbar^2}}{^4} \left| \langle \psi |\widehat y| \psi \rangle \right|^2$
[/hide]
$ \Delta x \Delta L_z \ge \frac {_{\hbar}}{^2} \left| \langle \psi |\widehat y| \psi \rangle \right| \equal{} \frac {_{\hbar}}{^2} \left|\overline y\right|$ and similarly, $ \Delta y \Delta L_z \ge \frac {_{\hbar}}{^2} \left| \langle \psi |\widehat x| \psi \rangle \right| \equal{} \frac {_{\hbar}}{^2} \left|\overline x\right|$

This computation of $ \Delta x \Delta L_z$ is valid only if the system state $ |\psi\rangle$ is in the domains of both $ [\widehat x, \widehat L_z]$ and $ \widehat y.$ Similarly for the computation of $ \Delta y \Delta L_z.$ (The inequality might not hold otherwise.) This uncertainty relation depends on the system state $ |\psi\rangle.$ Only if the comutator of two non-comuting operators is an imaginary constant times the identity operator, such as $ [\widehat x, \widehat p_x] \equal{} \text{i} \hbar,$ we get a fixed limit $ \Delta x \Delta p_x \ge \frac {\hbar}{2}$ independent of the system state.

After some physical realization of the z-axis, say by homogenious magnetic field or by particle beam direction, measurement of $ L_z$ projects any system state $ |\psi\rangle,$ including states with the expectation values $ \overline x \neq 0,\ \overline y \neq 0,$ to some eigenstate $ |\phi_m\rangle$ of $ \widehat L_z$ corresponding to the measured value $ m$, a real eigenvalue of this operator. Eigenstates of $ \widehat L_z$ are in the domains of $ [\widehat x,\widehat L_z],\ \widehat y.$ In this eigenstate $ |\phi_m \rangle$, $ \overline L_z \equal{} m,$ $ \Delta L_z \equal{} 0,$ and

$ \overline y_m \equal{} \langle \phi_m | \widehat y | \phi_m\rangle \equal{} \frac {_{\text i}}{^{\hbar}} \langle \phi_m | \widehat x \widehat L_z \minus{} \widehat L_z \widehat x | \phi_m\rangle \equal{}$
$ \equal{} \frac {_{\text i}}{^{\hbar}} \left(\langle \phi_m | \widehat x \widehat L_z | \phi_m \rangle \minus{} \langle \phi_m | \widehat L_z \widehat x | \phi_m\rangle \right) \equal{}$
$ \equal{} \frac {_{\text i}}{^{\hbar}} \left(m \langle \phi_m | \widehat x | \phi_m \rangle \minus{} m^*\langle \phi_m | \widehat x | \phi_m \rangle \right) \equal{} \frac {_{\text i}}{^{\hbar}}(m \minus{} m^*) \overline x_m \equal{} 0$

and similarly, $ \overline x_m \equal{} 0.$ Therefore, measurement of $ L_z$ upsets both $ \overline x$ and $ \overline y.$ The eigenstates of $ \widehat x$  are delta functions in the coordinate representation, so that a subsequent measurement of $ x$ on a system in the eigenstate $ |\phi_m\rangle$ of $ \widehat L_z$ can be only approximate. Since $ |\phi_m\rangle$ is not an eigenstate of $ \widehat x$ with zero eigenvalue, the result can be $ x \neq 0$, such that $ \Delta x \ll |x|$ for a sufficiently accurate position measurement. The measurement of $ x$ projects the system state out of the eigenstate $ |\phi_m\rangle$ of $ \widehat L_z,$ because $ \overline x_m \equal{} 0$ before and $ \overline x \neq 0$ after the measurement of $ x.$ The measurement of $ x$ therefore upsets the expectation value $ \overline L_z.$ But it is possible to prepare a bound stationary system with certain eigenvalue of $ \widehat L_z$ by measuring $ L_z.$ After the measurement, $ \Delta L_z \equal{} 0$ and $ \Delta x > 0,\ \Delta y > 0$ but finite (because the system is bound), so that $ \Delta x \Delta L_z \equal{} \Delta y \Delta L_z \equal{} 0.$


inequalities
number theory unsolved
number theory

Let $ a,b\in\mathbb{N}$ and $ d\equal{}\gcd(a,b)$ such that $ a\plus{}b|a^2\plus{}b^2$
Prove that $ ab\plus{}4\leq d^4$ and when the equality occurs

It doesn’t hold for $ a\equal{}b\equal{}1$.

$ d \equal{} \gcd(a,b)$
$ a \equal{} dm,b \equal{} dn$ where $ \gcd(m,n) \equal{} 1$
So $ d(m \plus{} n)|d^2(m^2 \plus{} n^2)$
We have $ \gcd(m \plus{} n,m^2 \plus{} n^2) \equal{} 1,2$ so we need to condider 2 case .
[u]Case 1[/u] $ \gcd(m \plus{} n,m^2 \plus{} n^2) \equal{} 1$ then
$ m \plus{} n|d$
The inequalities becomes :
$ d^2mn \plus{} 4\leq d^4$
$ \leftrightarrow d^2 (d^2 \minus{} mn )\geq 4$
It true because $ d > 1$ and $ d^2 \minus{} mn\geq (m \plus{} n)^2 \minus{} mn > 0$
[u]Case 2[/u] $ \gcd (m \plus{} n,m^2 \plus{} n^2) \equal{} 2$
So $ m \plus{} n|2d$
The inequalities equivalent :
$ d^2(d^2 \minus{} mn)\geq 4$
If $ m \equal{} n \equal{} 1$ the it is not true when we chose $ d \equal{} 1$
If one of m or n is great than 1.
Supose $ m > 1$ so $ d\geq 2$
But from $ m > 1$ so $ m,n$ is diferent .
So $ d^2 \minus{} mn \geq (\frac {m \plus{} n}{2})^2 \minus{} mn > 0$
This inequalities claim.


[quote="7h3.D3m0n.117"]what is the difference  between

$\pm$ and $\mp$?

my alg 2 book explains it kinda wierd...any comments?[/quote]

i think it only depends when u have 2 of either of those symbols in the same expression.

$\pm 2 \mp 3=-1$ or $1$

you can either take all the top ones or all the bottom ones.

$\pm a\pm b=a+b$ or $-a-b$
$\pm a\mp b=a-b$ or $-a+b$

bascially what max_tm said... the top corresponds to top and bottom corresponds to bottom

[quote="max_tm"]Yes, it only really matters when it's conditional, a good example is writing addition formulas like this:

$sin(a\pm b) = sinacosb \pm cosasinb$
$cos(a\pm b) = cosacosb \mp sinasinb$

The top rows correspond to the top rows and the bottoms to the bottoms. I know I'm not explaining this well but I hope it got the point across  :P.[/quote]

yeah the only time i saw the $\mp$ is in the triganometry section of my book and no where else

it's the progeny of supreme laziness and unwillingness to write out more than what is necessary

[quote="7h3.D3m0n.117"]what is the difference  between

$\pm$ and $\mp$?

my alg 2 book explains it kinda wierd...any comments?[/quote]

[hide]
In Some cases, you would use like this;

$1\pm 2$ means that you can have 2 cases: $1+2$, $1-2$.

If you do it the other way, then you do the minus first, like this:

Say $\cos x\mp sin x=tan x\pm \cot x$ means that;

$\cos x-\sin x=tan x+\cot x$ and $\cos x+\sin x=\tan x-\cot x$.

Thats all.
[/hide]


vector
linear algebra
matrix
linear algebra theorems

Can anyone give me a good reference about linear algebra over finite fields?

More specifically, is it still true that
 - $ HH^T$ and $ H^TH$ have (except for maybe zeroes) the same set of eigenvalues?
 - the multiplicity of $ 0$ as eigenvalue of $ HH^T$ (or $ H^TH$) is the dimension of the nullspace of the rows/columns of $ H$?

For the latter, do we take the normal eigenvalue (over characteristic $ 0$)? Or is it false at all?

Thanks!

Here are some thoughts.

Suppose $ HH^Tv \equal{} \lambda v$ for some vector $ v \neq 0$, and $ \lambda \neq 0$.  Multiply both sides by $ H^T$ on the left, and you get $ H^THH^Tv \equal{} \lambda H^Tv$, which shows that $ H^Tv$ is an eigenvector of $ H^TH$ with eigenvalue $ \lambda$.  So every non-zero eigenvalue of $ HH^T$ is also an eigenvalue of $ H^TH$.  The other direction works in the same way.  This proof doesn't make any assumptions about which field of scalars we are using, so it still applies when our field of scalars is a finite field.

The proof I know of the second assertion does not generalize so well.  Suppose our field of scalars is $ \mathbb{R}$.  Then the nullspace of $ H$ is equal to the nullspace of $ H^TH$, because $ Hx \equal{} 0 \implies H^THx \equal{} 0 \implies x^TH^THx \equal{} 0 \implies \parallel{}Hx\parallel{}^2 \equal{} 0 \implies Hx \equal{} 0$.  The multiplicity of $ 0$ as an eigenvalue of $ H^TH$ is the dimension of the nullspace of $ H^TH$ which is the dimension of the nullspace of $ H$.  But this proof uses the fact that our field of scalars is $ \mathbb{R}$.

The first generalizes to any pair $ A,B$ rather than just transposes.

That second part is false in finite characteristic- consider $ H\equal{}\begin{bmatrix}1&1\\1&1\end{bmatrix}$ in characteristic 2. It breaks down whenever a sum of squares can be zero.

[quote="jmerry"]That second part is false in finite characteristic- consider $ H \equal{} \begin{bmatrix}1 & 1 \\
1 & 1\end{bmatrix}$ in characteristic 2. It breaks down whenever a sum of squares can be zero.[/quote]Exactly the example I've been struggling on. We have a symmetric matrix,  but the algebraic multiplicity of zero (2) is lower than the geometric multiplicity (1).

Are any results known concerning this? Is there a way we can still find out the geometric multiplicity in some way (other than explicitly computing it), over characteristic $ p$?


Euler
graph theory
combinatorics unsolved
combinatorics

Let G be a graph. Show that:

(a) if each degree in G is even, then G has no cut edge.
(b) if G is a k-regular bipartite graph with k>=2, then G has no cut edge.

A cut edge of G is an edge e such that w(G-e) > w(G), where w(G) denotes the number of components of G.

(a) is easy because if each degree of G is even then each component of G is an Euler cicle. So obviously removing only one edge won't change the number of components.

(b) is also easy: assuming the contrary gives that one less than a multiple of $k$ is equal to a multiple of $k$.


function
advanced fields
advanced fields unsolved

Let $ f_n$ be a sequence of functions in $ L^1(0,1)$ such that $ f_n(x) \to \zeta(x)$ pointwise a.e. and $ f_n \to \xi \in L^1(0,1)$ weakly. Show that $ \zeta \equal{} \xi$ a.e..

Try to use Lusin and Egoroff to make life simpler.


[quote="nebula42"][quote="bpms"][quote="nebula42"]We have a $5 \times 6 \times 7$ rectangular prism made up of unit cubes, and we paint all of the outside faces of the rectangular prism.

a) How many unit cubes have exactly one face painted?
b) How many unit cubes have exactly two faces painted?
c) How many unit cubes have exactly three faces painted?
d) How many unit cubes have no faces painted?[/quote]
[hide="a"]The ones with one face are not on an edge, but on the surface. This is the same as subtracting the edges which is the same as subtracting two which is the surface area of a 3x4x5 prism which is 94.[/hide]
[hide="b"]This is the ones that are on edges but not corners. This is 6+8+10=24[/hide]
[hide="c"]This is the corners is 8[/hide]
[hide="d"]This is the ones on the "inside". If you cut off all of the cubes that are on the surface you eliminate all the painted cubes and have a 3x4x5 unpainted cube with [b]60[/b] unit cubes...[/hide][/quote]
You got parts a, c, and d correct  :) , but check your work for part b again.  :wink:[/quote]
I think I see what I did wrong...
[hide="b"]I only counted half of the edges so it is now 48.[/hide]

Correct.  :)
Here's an extension/generalization:
Do the problem the same way as before, but now, instead of a $5 \times 6 \times 7$ prism, we have an $l \times w \times h$ prism (assuming $l,w,h \geq 2$).


search
geometry
3D geometry
group theory
abstract algebra
Galois Theory
superior algebra

These two users inspired me to create my own challenge. It doesnt look that hard, but I didnt work on it.

[b]Problem:[/b] Let $ E=\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},...,\sqrt{p_{n}})$ where $ p_{1},p_{2},...,p_{n}$ are distinct primes. Prove that $ \mbox{Gal}(E/\mathbb{Q}) \simeq \underbrace{\mathbb{Z}_{2}\times ... \times \mathbb{Z}_{2}}_{n}$.

-oo- has done it here:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1058707505&t=153532

in the link it was shown that $ [E : \mathbb{Q}] = 2^{n}$. obviously, $ E$ is the splitting field of $ (X^{2}-p_{1})*...*(X^{2}-p_{n})$, so that $ E$ is galois. now the galois group has exactly $ 2^{n}$ elements and each has to perform $ \sqrt{p_{i}}\mapsto \pm \sqrt{p_{i}}$. thus, every choice of signs occurs and this yields an isomorphism as desired.

instead of distinct primes we may use coprime squarefree integers.

If we replace square roots by cube-roots, I think we have [E:Q]=3^i.
But the extension is obviously not normal.

Yes, you get this by adjoining a third root of unity first and then using Kummer theory (by this, you get it for any roots as long as no bad things happen).

What kind of bad things? As long as we take kth roots with k odd, no bad things will occur, right?

[quote="Abel"]What kind of bad things? As long as we take kth roo
ts with k odd, no bad things will occur, right?[/quote]
Well, it seems the extension isn't normal in general. Show how can we talk about the Galois group?

Never talked about any group, did I?
But we can talk about groups if we attach the root of 1.
If we attach qth roots of i distinct rationals  and the qth root of unity to Q (where q is prime), the group is
 [C(q)^i]C(q-1)    (semi-direct product)
where C(q)^i stands for C(q)X...XC(q) (i times)
This group has exactly q(q^i -1/q-1) subgroups of index q. So, if we attach the qth root of a new rational (new, here, meaning not a product/ quotient of the other i rationals), we see that it is not contained in F and hence, not contained in E.


AMC
USA(J)MO
USAMO
number theory theorems
number theory

Theorems in Number Theory that I might need for USAMO??? Help would be greatly appreciated.

1. Please don't post the same topic in more than one place.

2. No one is going to give you a comprehensive answer. There's just way too many things that can appear on Olympiads.

3. It's not going to help you just knowing theorems. You could know every theorem found in every textbook, and you still may not be able to solve USAMO problems, because that's not what they are about. The best way to prepare for an Olympiad is to work on past problems or work on an Olympiad book.


calculus
integration
logarithms
trigonometry
function
complex analysis

Hi, I know how to evaluate this integral by three different ways.  However, now I'm trying to evaluate this integral by the method of Gauss's Mean Value Theorem.  I got stuck attempting this.  I was wondering if anyone can help me out.  Thanks!

Question:  Using Gauss's Mean Value Theorem, prove that $ \displaystyle\int_0^{\pi}\ln\sin\theta d\theta \equal{} \minus{} \pi\ln 2$.

Consider the harmonic function $ \ln(|1\minus{}z|)$ on the unit disk.

By the way, is there any way to do this by the normal substitution $ z \equal{} \exp i\theta$, $ \sin \theta \equal{} \frac{1}{2i}\Big(z \minus{} \frac{1}{z}\Big)$?


Suppose $0 \leq a;b;c \leq1$.Prove that:
$a+b^{2}+c^{3}-ab-bc-ca\leq 1$

Let $f: [0,1]\to \mathbb R$, $f(b)=b^{2}-(a+c)b+c^{3}+a-ca-1$
So $f$ is convex therefore $f(b)\leq f(1)$ or $f(b)\leq f(0)$ $\forall b\in [0,1]$
So it's enough to prove that $f(1)\leq 0$ $\Leftrightarrow$ $c(c^{2}-a-1)\leq 0$ which it's obviously true , or $f(0)\leq 0$ $\Leftrightarrow$ $c^{3}+a-ca-1\leq 0$.
Let $g: [0,1]\to \mathbb R$ $g(c)=c^{3}-ca+a-1$ .Because $g$ is convex $\Rightarrow$ $g(c)\leq g(1)$ or $g(c)\leq g(0)$ ,so it's enough to prove that

$g(1)\leq 0$ $\Leftrightarrow$ $1+a-a-1\leq 0$ ,which it's true
or

$g(0)\leq 0$ $\Leftrightarrow$ $a-1\leq 0$

,true
and we are done


An ant walks around a square of insect repellent that is one half meter on a side, always remaining exactly 20 cm away from it. What is the total area enclosed by his path in sq. centimeters?

Well... kind of.

[hide="answer"]
It traces out a square with curved edges - the curves can be put together to make a circle with radius 20 and the lines can be put together to make a square of side length 50.

$\boxed{200+40\pi}$
[/hide]

200 plus 40 pie, r u serious?

dont u mean 2500 plus 400 pie?

Yeah, the answer is $\boxed{2500+400\pi}$

Holy crap I'm an idiot... I found the perimeter instead of the area.


[hide="#22"]For each positive integer $ n$, let $ S(n)$ denote the sum of the digits of $ n$.
For how many values of $ n$ is $ n+S(n)+S(S(n)) = 2007$?

$ (A)$ $ 1$     $ (B)$ $ 2$     $ (C)$ $ 3$     $ (D)$ $ 4$     $ (E)$ $ 5$[/hide]

[quote="vishalarul"][url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=161[/url]

Search "digits" by doing [Ctrl]+F. The seventh occurrence (I think) leads you to the solution.[/quote]

I dont like that solution.
I would just set limits first the max s(n) can be is 28 when n=1999, the max s(s(n)) can be is 10, so n has to be greater than 2007-28-10=1969, also the minimum s(n) can be is 2 and s(s(n)) is also 2 (note s(s(n)) cannot be 1, otherwise then s(n) has to be 10 but s(n) cannot be 10).  So it has to be lessthan 2007-2-2=2003.

Furthermore notice 2007 is 0 mod 9.  Also if n is k mod 9, then s(n) is also k mod 9, same with s(s(....s(n)))))..) is also k mod 9.   So since 2007 is 0 mod 9, then n must either be 0,3,or 6 mod 9.

Also note that any number from 1990-1999 cannot possible work  because the minimum of s(n) in that situation is 18 when n=1990.  But already 1990+18=2008 which is greater than 2007, so all other values also must be greater than 2007.

So we have n from [1969,1989],[2000,2003] to choose from and also n has to be 0,3,or 6 mod 9.

So we have a small case set of (1971,1974,1977,1980,1983,1986,1989,2001) to try and see which work.  Cleary only $ 1977,1980,1983,$ and $ 2001$ work.  So $ 4$ solutions (when you try those numbers out it helps to be quick at adding, only should takes like 2 mins max to try out all those numbers)


All positive integers whose digits add up to 11 are listed in increasing order:  $ 29, 38, 47, ...$. What is the eleventh number in that list?

continuing the list, we get

$ 29, 38, 47, 56, 65, 74, 83, 92, 119, 128, 137$

we find that $ 137$ is the 11th number in this list, so our answer is $ 137$

Well, all numbers with digits that add up to $ n$ will have a common difference of 9.  Since we know that $ 29$ is the first integer, you can easily find out the eleventh integer.

edit: Nvm.  That doesn't work because the digits add up to $ >10$.

Mohammed's way would work IF the sequence had a common difference. However, it does not, so listing the numbers is probably the only way.
PS you may think that the sequence has a common difference of 9, but not once it gets to 3 digits.


geometry
3D geometry
calculus
integration
vector
LaTeX
Gauss

There is a cube with each surface with surface charge density $ \sigma$ .If we remove one surface as shown in the figure,what will be the electric field in the centre of the cube?

For purposes of analysis we can consider each face of the cube as if it were a point charge.

From which we find the following model (in electrostatic units),

	 \[ \Phi\left(\mathbf{x}\right)\equal{}\sum_{i \equal{} 1}^{N} \frac{Q_i}{\left|\mathbf{x} \minus{}\mathbf{x}_i\right|}
\]

	\[ \mathbf{E}\left(\mathbf{x}\right) \equal{}\minus{}\nabla\Phi\left(\mathbf{x}\right)\equal{}\sum_{i\equal{}1}^{N}\frac{Q_i\left(\mathbf{x}\minus{}\mathbf{x}_i\right)}{\left|\mathbf{x}\minus{}\mathbf{x}_i\right|^3}
\]

For the cube $ \forall i\,Q_i \equal{} Q$

	 \[ \mathbf{E}\left(\mathbf{x}\right) \equal{}\sum_{i\equal{}1}^{N}\frac{Q\left(\mathbf{x}\minus{}\mathbf{x}_i\right)}{\left|\mathbf{x}\minus{}\mathbf{x}_i\right|^3}
\]

For purposes of simplification we shall take the center of the cube as $ \mathbf{x} \equal{} 0$ .

With all faces in place,

\[ \mathbf{E}\left(0\right) \equal{}\sum_{i\equal{}1}^{6}\frac{\minus{}Q\mathbf{x}_i}{\left|\mathbf{x}_i\right|^3}\equal{}0
\], because of symmetry.

Removing one face in accordance in accordance with the diagram,

	 \[ \mathbf{E}^{\prime}\left(0\right) \equal{}\sum_{i\equal{}1}^{5}\frac{\minus{}Q\mathbf{x}_i}{\left|\mathbf{x}_i\right|^3}\equal{}\frac{\minus{}Q\mathbf{x}_5}{\left|\mathbf{x}_5\right|^3}
\]

In this case the prime does not denote differentiation. It indicates this is a different field from the one above it. If the area of a cube face is $ w^2$ then $ Q \equal{}\sigma w^2$.

	 \[ \mathbf{x}_5 \equal{}\frac{1}{2}w\,\mathbf{\hat{u}}_5\Rightarrow \frac{\mathbf{x}_5}{\left|\mathbf{x}_5\right|^3} \equal{}\frac{4}{w^2}\,\mathbf{\hat{u}}_5
\]

	\[ \mathbf{E}^{\prime}\left(0\right) \equal{}\frac{\minus{}Q\mathbf{x}_5}{\left|\mathbf{x}_5\right|^3}\equal{}\minus{}\left(\sigma w^2\right)\left(\frac{4}{w^2}\,\mathbf{\hat{u}}_5\right)\equal{}\minus{}4\sigma\,\mathbf{\hat{u}}_5
\]

Finally, $ \mathbf{E}_{SI}\left(0\right)\equal{}k\mathbf{E}^{\prime}\left(0\right)\equal{}\frac{\mathbf{E}^{\prime}\left(0\right)}{4\pi\epsilon_0}$

	\[ \boxed{\therefore\mathbf{E}_{SI}\left(0\right)\equal{}\frac{\mathbf{E}^{\prime}\left(0\right)}{4\pi\epsilon_0}\equal{}\minus{}\frac{\sigma}{\pi\epsilon_0}\,\mathbf{\hat{u}}_5}
\]

NO,correct answer is ${ \frac{\sigma}{\epsilon_0}}$.

Observe that magnitude of electric field at the center is equal to that due to a single face at the same distance. The field due to a square plate can be computed by integration, I will try to think of an easier method.

this is cos the electric field due to the opposite faces get cancelled because they oppose themselves....
but since a side is missing so the answer i.e.. at center E is sigma/epsilon

[quote="Nihat"]NO,correct answer is ${ \frac {\sigma}{\epsilon_0}}$.[/quote]

The difference is due to the presence of different proportionality constants in Maxwell's equations in the two systems of units, SI vs. esu (CGS),

In SI units, $ \iint \vec{D}_{SI}\cdot\hat{n}\, da = \iint \epsilon_0\vec{E}_{SI}\cdot\hat{n}\, da = Q_{\text{net}}$.

In esu, $ \iint \vec{E}_{esu}\cdot\hat{n}\, da =  4\pi Q_{\text{net}}$

$ \therefore \iint \vec{E}_{esu}\cdot\hat{n}\, da =  4\pi\iint \epsilon_0\vec{E}_{SI}\cdot\hat{n}\, da$

$ \Rightarrow \vec{E}_{SI} = \frac{\vec{E}_{esu}}{4\pi\epsilon_0}$

Also, when I calculated the E-field, I calculated it based on the actual distance to the center of the cube from the cube face emitting the field, which is only half the length of the edge of the cube this gave my answer a factor of four that is not present if one simply stipulates that $ \epsilon E = \frac{Q}{A} =\sigma$.

The answer $ E =\frac{\sigma}{\epsilon_0}$ is the field in SI units (ignoring the factor of $ 4\pi$) at the location of the face that was removed; not at the center of the cube, that was specified in the problem.

i really doubt if that's the answerbecause the field due to rest is same as the the field due to removed part in magnitude....
which is upposedly $ \frac{\sigma}{\epsilon_{o}}$
this means even if we extend the side of the cube infinitely in all directions there is no net  contribution due to them

[quote="pardesi"]i really doubt if that's the answerbecause the field due to rest is same as the the field due to removed part in magnitude....
which is upposedly $ \frac {\sigma}{\epsilon_{o}}$
this means even if we extend the side of the cube infinitely in all directions there is no net  contribution due to them[/quote]

Let me take a somewhat different approach.

First, the recognition that opposing faces produce opposing fields is entirely correct. As a consequence the net electric field is due only to the face opposite the one removed.

Gauss’ law guarantees that $ \iint_S \vec{D}\cdot\hat{n}\, da \equal{} Q_{\text{net}}$ .

The Mean Value Theorem then tells us that:

	\[ \exists \mathbf{r^{\star}}\,\backepsilon\, \vec{D}\left(\mathbf{r^{\star}}\right)\cdot\hat{n}\equal{}\frac{Q_{\text{net}}\left(A_2\right)\minus{}Q_{\text{net}}\left(A_1\right)}{A_2 \minus{}A_1}
\]

One typically starts from zero area and integrates upward such that $ A_1 \equal{} 0$ .

Further, for an uniform charge distribution such as we have $ Q\left(A_i\right) \equal{} \sigma A_i$ .

Therefore $ Q_{\text{net}}\left(A_1\right) \equal{}\sigma A_1 \equal{} \sigma\cdot 0 \equal{} 0$.

Likewise, $ Q_{\text{net}}\left(A_2\right) \equal{}\sigma A_2$.

It follows then $ D\left(\mathbf{r^{\star}}\right) \equal{} \frac{\sigma A_2 \minus{} 0}{A_2 \minus{} 0} \equal{}\sigma$ .

(The vector arrow has been dropped intentionally because we are only concerned with the component of $ \vec{D}$ in the $ \hat{n}$ direction.)

From the definition of $ \vec{D}$, we know that $ \vec{D}\doteq \epsilon\vec{E}$.

In our case the dielectric is free space $ \Rightarrow \epsilon \to \epsilon_0$.

Therefore, $ \vec{D}\equal{}\epsilon_0\vec{E}\Rightarrow D \equal{} \epsilon_0 E$.

Hence $ \epsilon_0 E\left(\mathbf{r^{\star}}\right) \equal{} \sigma \Rightarrow E\left(\mathbf{r^{\star}}\right) \equal{} \frac{\sigma}{\epsilon_0}$.

My argument boils down to the fact that $ \mathbf{r^{\star}}$ is [b]not[/b] the location of the center of the cube.

The electrostatic field in SI units (in free space), due to a charge $ Q_i$ is given by:

	\[ \vec{E}\left(\mathbf{r}\right) \equal{} \frac{1}{4\pi\epsilon_0}\frac{Q_i}{\left|\mathbf{r}\minus{}\mathbf{r}_i\right|^2}\,\hat{u}_{\mathbf{r}\minus{}\mathbf{r}_i}
\]

Satisfy yourself that this is true; Google it for goodness sake. Just stop saying, “I don’t think that’s right because [i]the answer is supposed to be …[/i]”

Text books are sometimes wrong!  I don’t give a flying fig what the answer is supposed to be, and neither should you. It only matters what the answer actually is.

Applying the above formula to find $ \mathbf{r^{\star}}$,

	 \[ E\left(\mathbf{r^{\star}}\right) \equal{} \frac{1}{4\pi\epsilon_0}\frac{Q_{\text{net}}}{\left|\mathbf{r^{\star}}\minus{}\mathbf{r}_i\right|^2}\equal{}\frac{\sigma}{\epsilon_0}
\]

It follows that,

	\[ \frac{1}{4\pi}\frac{1}{\left|\mathbf{r^{\star}}\minus{}\mathbf{r}_i\right|^2}\equal{}\frac{\sigma}{Q_{\text{net}}}
\]

Re-writing this expression by evaluating the RHS,

	\[ \frac{1}{4\pi}\frac{1}{\left|\mathbf{r^{\star}}\minus{}\mathbf{r}_i\right|^2}\equal{}\frac{1}{A_2}
\]

Inverting the above expression,

	\[ 4\pi\left|\mathbf{r^{\star}}\minus{}\mathbf{r}_i\right|^2 \equal{} A_2 \equal{}w^2
\]

$ A_2 \equal{} w^2$ because $ A_2$ is the area of the face of a cube, i.e. a square.

This implies $ \left(r^{\star} \minus{} r_i\right) \equal{} \pm \sqrt{\frac{w^2}{4\pi}} \equal{} \pm \frac{w}{2\sqrt{\pi}}$.

$ \Rightarrow r^{\star} \equal{} r_i \pm \frac{w}{2\sqrt{\pi}}$

$ r_i$ is the location of the charged face.

With respect to the center of the cube (according to the diagram), $ r_i \equal{} \frac{w}{2}$

 $ r^{\star} \equal{} \frac{w}{2}\pm \frac{w}{2\sqrt{\pi}}$

We don’t care about the field outside the cube and therefore we can ignore the “+” solution.

$ \boxed{\therefore r^{\star} \equal{} \frac{w}{2}\minus{} \frac{w}{2\sqrt{\pi}}}$

[b]This is not the cube’s center.[/b]  If it were, we should have found $ r^{\star} \equal{} 0$.

[quote="pardesi"]please would u mind giving physical interprtations of ur proofs i am [b]bored[/b] of seeing ur $ \text{\LaTeX}$ skills each time there is a problem on electrostatics/dynamics :D[/quote]

No sweat. I will have to use some $ \text{\LaTeX}$ in order to show what I was thinking though.

To start with Gauss' law is a specific application of the concept of flux.

That is if you have some force field impinging on a surface, then the "flux" of that field across the boundary formed by said surface is equal to the surface integral of the dot product of the force field and the unit normal of the surface upon which the field impinges.

"Wow, that's nice," you say, "but what the [i]flux[/i] is flux?" From all I can tell, Gauss had an answer in search of a question. That is, he decided that the charge on a charged surface was a flux. He reasoned that there then had to be a force field, $ \vec{D}$, that was the source of the charge.  Gauss' law expressed as an integral equation is then:
\[ \iint_S \vec{D}\cdot\hat{n}\, da = Q
\]
You should notice that since an electric field is the force field emitted by an electric charge, Gauss' force field is proportional to the electric field. That is, $ \vec{D} = \epsilon\vec{E}$, where $ \epsilon$ is the constant of proportionality.

Switching gears momentarily, Coulomb discovered that electric charges exert a force on one another that is analogous to Newton's force of gravity.

Notice, Coulomb's force is given by: $ \vec{F} = k \frac {Q_1Q_2}{R^2}$.

$ R$, as with gravity, is the distance between the centers of $ Q_1$ and $ Q_2$.  Implying that $ \vec{E} = k\frac {Q_1}{R^2}\, \hat{u}_{R}$ is an electric field like $ \vec{g} = - G\frac {m_1}{R^2}\, \hat{u}_{R}$ is a gravity field.

If we substitute Coulomb's electric field into Gauss' integral we obtain,
\[ \iint_S \epsilon k\frac {Q}{R^2}\,\hat{u}_{R}\,\cdot\hat{n}\,da = Q
\]
The integration is simplest when $ \hat{u}_{R}\cdot\hat{n} = 1$, everywhere.

This situation occurs only when the surface over which one integrates is sphere. If the sphere has a fixed radius $ R$ then $ da = R^2\, d\Omega$, where $ \Omega$ is a solid angle in steradian units and
\[ \int_{0}^{4\pi}\epsilon k\frac {Q}{R^2}\, R^2\,d\Omega = Q
\]
Therefore $ 4\pi\epsilon k Q = Q$.

From which we get Coulomb's constant, $ k = \frac {1}{4\pi\epsilon}$.

From this point, the mean value theorem can be applied to this problem in two different ways:

1) $ \iint \epsilon\vec{E}\cdot\hat{n}\, da = \epsilon E_{ave}A = Q$

von Kleist (1745) discovered that charge stored in a Leyden jar was proportional to the potential applied to it. $ Q\propto V\,\Rightarrow Q = CV$

Noting that $ E_{ave} = \frac {\Delta V}{\Delta x}\to \epsilon\frac {\Delta V}{\Delta x}A = CV$.

If we take the bottom of the "Leyden jar" as ground ($ V_{\text{bottom}} = 0$) then $ \Delta V = V$; and then if we let $ \Delta x = \ell$, we have: $ \epsilon A\frac {V}{\ell} = CV$, which gives us[i]...drum roll, please...[/i] $ C = \frac {\epsilon A}{\ell}$, the formula for capacitance.

2) The second way to apply the mean value theorem is the way I did earlier.

If $ \int_{a}^{b} f^{\prime}\left(x\right)\,dx = f\left(b\right) - f\left(a\right)$ then there is a $ c\in\left[a,b\right]\,\backepsilon\,f^{\prime}\left(c\right) = \frac {f\left(b\right) - f\left(a\right)}{b - a}$

I applied the above form to the Gauss integral, where $ E\left(r^{\star}\right) = \frac {\sigma}{\epsilon_0}$.    The $ r^{\star}$ is the point "$ c$," if you will, in the interval $ \left[ - \frac {w}{2},\frac {w}{2}\right]$

If you stop and think about it for a second, the E-field, cannot just be $ \frac {\sigma}{\epsilon_0}$ because this is a constant. Meaning that the E-field would be the same everywhere in space. Such a situation is physically absurd.

I'll wrap this up by explaining a little about what I did that to obtain my answer of $ E\left(0\right) = \frac {\sigma}{\pi\epsilon_0}$. The original problem was to find the resultant E-field at the center of the cube.  I started by calculating the Coulomb field at the center of the cube. I knew that since the surface charge density was uniform then $ Q_i = \sigma w^2$. Also $ R^2 = \left|\mathbf{r} - \mathbf{r}_i\right|^2$. Recalling that $ R$ is the distance between the centers of two charges that are exerting force on one another, $ \mathbf{r}_i$ is the position vector of the "pusher" charge, $ Q_i$, that creates the E-field, and $ \mathbf{r}$ is the position vector of the (virtual) charge that is pushed upon, also known as the "field point."

I fixed the origin of the coordinate system to be coincident with the center of the cube so that $ \mathbf{r} = 0$; the field point is the origin. Next, I noted that if   the center was the origin then $ Q_i$, the net charge on the face of the cube opposite the one removed is located at $ \mathbf{r}_i = \frac {w}{2}\,\hat{u}_{\mathbf{r}}\to R^2 = \left|0 - \frac {w}{2}\right|^2 = \frac {w^2}{4}$. Combining all of this, I obtained $ E\left(0\right) = 4\sigma$ in electrostatic units, which I knew most people reading what I had written wouldn't recognize, so I multiplied by the Coulomb constant -- which we discovered earlier is $ k = \frac {1}{4\pi\epsilon_0}$ -- to obtain $ E\left(0\right) = \frac {\sigma}{\pi\epsilon_0}$.

[b]A word about electrostatic units:[/b] As an undergraduate student I learned that Maxwell's equations were:
\[ \begin{array}{lc} \nabla\cdot\vec{D} = \rho & \nabla\times\vec{E} = - \frac {\partial \vec{B}}{\partial t} \\
\nabla\cdot\vec{B} = 0 & \nabla\times\vec{H} = \vec{J} + \frac {\partial\vec{D}}{\partial t}\end{array}
\]
Later, as a graduate student, I re-learned Maxwell's equations as:
\[ \begin{array}{lc} \nabla\cdot\vec{E} = 4\pi\rho &{ \nabla\times\vec{E}=-\frac{1}{c}\frac{\partial\vec{H}}{\partial t}} \\
\nabla\cdot\vec{H} = 0 & \nabla\times\vec{H}=\frac{4\pi}{c}\vec{J}+\frac{1}{c}\frac{\partial\vec{E}}{\partial t}\end{array}
\]
This second version of Maxwell's equations has some special advantages for solving electrostatic and magnetostatic problems of this kind, because in esu (electrostatic units) $ \epsilon_0 = \mu_0 = k = 1$.

[quote="pardesi"]so what's ur answer and can u please stop giving the definition of flux nth time....it uses lots of space...just come straight to the point...just do what's necessary to solving the problme :huh:[/quote]

I have solved it three different times in this thread, should I do it a fourth time?

The answer is: $ E\left(\text{cube center}\right) \equal{} \frac{\sigma}{\pi\epsilon_0}$.

Apparently Nihat's text book gives the answer as: $ \frac{\sigma}{\epsilon_0}$ and while that is the average flux density over the surface of a single charged face, it is not the strength of the E-field at the center of the cube. It is equivalent to the strength of the E-field at a distance $ \frac{w}{2}\minus{}\frac{w}{2\sqrt{\pi}}$ from the center of the cube.

[quote="pardesi"][quote="Dr. No"]

The answer $ E \equal{} \frac {\sigma}{\epsilon_0}$ is the field in SI units (ignoring the factor of $ 4\pi$) at the location of the face that was removed; not at the center of the cube, that was specified in the problem.[/quote]
This was i was opposing...this answer is impossible according to my previous argument.....
...and can u please have less of the mathematical theory behind the proof and directly come to the point....and i don't understand 99.99% of the symbols u use....
it almost looks like u r talking to urself while writing the proofs it' highly unclear :)[/quote]

Sorry, I will try to use less math ... that sounds a bit odd on a math site ... but nonetheless I'll make an effort for your sake. I want to be clear, but I also want to be thorough. I was under the impression that in a discussion of "Advanced Physics" that the symbols I use would be familiar to most since they are those commonly used in discussing the topics to which I post.

If it sounds like I'm talking to myself, maybe I am, because the chains of the equations that I write and the order I write them in tend to follow my train of thought rather closely.

[quote="Immanuel Bonfils"][quote="Dr. No"]For purposes of analysis we can consider each face of the cube as if it were a point charge. [/quote]

It's always possible to reduce a charge distribution to a point charge (kind of an image); the problem is to find it's position...

The answer is  $ \;\;\; {\sigma \over {6\;\epsilon_0}}\;$ and I'm afraid integration is the only way to get it. But it's not to hard ( it takes some geometry...)[/quote]

Go back and re-read the problem, son. The charge density on each face of the cube is $ \sigma$. Further, this problem has nothing necessarily to do with an image charge model -- we are not considering the field inside an otherwise uncharged cube due to a charge distribution outside of it. You are to determine the electric field inside the cube, at its center, due to the effect of removing one of the charged faces.

If you are still confident of your answer, the best way to prove it is to let us see your integral.

I said [b]kind[/b] of image and it was clear i wasn't meaning the image method.

In a Brazilian book: Problemas de Física   -   Eletricidade  -  Eletrostática , by Eduardo Wilner,
 there is a solved problem of the field due to a square in a generic point of it axis (perpendicular to its plane) given

\[ {\sigma \over {\pi\epsilon_0}}\,arc\,sin\,{1 \over {1\,+{4z^2 \over {a^2}}}}\].

In our case $ \; z\;=\;{a \over2} \;$, and we get the result above.

[quote="Immanuel Bonfils"]I said [b]kind[/b] of image and it was clear i wasn't meaning the image method.

In a Brazilian book: Problemas de Física   -   Eletricidade  -  Eletrostática , by Eduardo Wilner,
 there is a solved problem of the field due to a square in a generic point of it axis (perpendicular to its plane) given
\[ {\sigma \over {\pi\epsilon_0}}\,arc\,sin\,{1 \over {1\, + {4z^2 \over {a^2}}}
}\]
.

In our case $ \; z\; = \;{a \over2} \;$, and we get the result above.[/quote]

I have been considering your post for several hours. I agree with you that given that there is a distribution of charge on the charged face that an integral solution is needed. Further, I have no reason to doubt Professor Wilner solved his problem correctly. But it seems to me in looking at the geometry of the problem, that perhaps your statement of Dr. Wilner's solution may be missing a radical.

I tried differentiating the expression that you gave. It appears to be twice as large as needed, which one would expect if there is a missing $ \frac{1}{2}$ power (commensurate with a missing radical). Also one seldom sees the hypotenuse of any triangle expressed without a radical; this is further reason to suspect that one may be missing.

If one places a radical in your expression, where presumably it should have been,

\[ {\sigma \over {\pi\epsilon_0}}\,arc\,sin\,{1 \over \sqrt{{1\, + {4z^2 \over {a^2}}}}
}\]

then evaluation at $ z=\frac{a}{2}$ yields $ \frac{\sigma}{4\epsilon_0}$

The important point in this exercise however is not to prove how well one can integrate, but that the electric field at the center of the cube is not trivially equal to $ \frac{\sigma}{\epsilon_0}$.

one thing's for sure the answer is not $ \frac{\sigma}{\epsilon_{0}}$

it shld be man........
coz the flux due to the opp faces cancel each other.....
since along one of the axes,one side is misiing....
THe flux shld be only sigma/epsilon(not).........

[quote="pardesi"]i really doubt if that's the answerbecause the field due to rest is same as the the field due to removed part in magnitude....
which is upposedly $ \frac {\sigma}{\epsilon_{o}}$
this means even if we extend the side of the cube infinitely in all directions there is no net  contribution due to them[/quote]
Ranga go thru this post of mine about why it can't be so

ya i think what i said is right...(tell me properly -i don understand)... :D  :D  :D

i mean if due to a square sheet itself the field is $ \frac{\sigma}{\epsilon_{0}}$ ..and also due to an infinite plat the same ...then by suerposition the rest parts of the infinte plate are useless that is they contricute 0 field and that is surely wrong

yes man..,..
do u mean to say that there is  a flaw in the gauss law....

[quote="Dr. No"]I tried differentiating the expression that you gave.[/quote]

Be aware: it's a definite integral. Wat was your variable?

As for [quote="Dr. No"]The important point in this exercise however is not to prove how well one can integrate, but that the electric field at the center of the cube is not trivially equal to $ \frac{\sigma}{\epsilon_0}$.[/quote] Pardesi already gave a perfect answer: The field due a finite square can't exceed that due a infinite plane.

[quote="pardesi"]i mean if due to a square sheet itself the field is $ \frac{\sigma}{\epsilon_{0}}$ ..and also due to an infinite plat the same ..[/quote]
Just to reenforce your argument and to an correct expression, that due the plane (or infinte "plate") is half that value.

sorry just gotta bit carried away :rotfl:

Sorry for the mistake for answer.The correct answer is  $ \frac {\sigma}{6\epsilon_0}$.

can u explain to me pardesi how the answer is sigma/6*epsilon.

yes me working on it what i said was the answer is less than $ \frac{\sigma}{2 \epsilon_{0}}$

i still feel the answer is wrong
Why??
beacuse the field due to the other remaining sides is equal in magnitud eto the field due to the removed side.
Then the question is equivalen to finding the field at a distance $ \frac{a}{2}$ from the centre of that side where $ a$ is the edge length of the cube and if this does not depends on $ a$ this means no matter how close or far u r the field is the same whcih seems absurd for this finite object

When [b]a[/b] decreases you get closer to a square that also decreases (with a constant sigma, it's charge decreases).

[quote="Immanuel Bonfils"]The field is proportional to the charge and to the inverse of  $ \;a^2\;$.
But the charge is proprotional to the square area, that is, $ \;a^2\;$...[/quote]
yes that's true...but i still somehow can't satisfy myself...it isn't that obvious to me
I will try giving a proof,u try too

I already did it:
[quote="Imannuel Bonfils"]In a Brazilian book: Problemas de Física - Eletricidade - Eletrostática , by Eduardo Wilner,
there is a solved problem of the field due to a square in a generic point of it axis (perpendicular to its plane) given

$ {\sigma \over {\pi\epsilon_0}}\,arc\,sin\,{1 \over {1\,+{4z^2 \over {a^2}}}}$.

In our case$ \; z\;=\;{a \over2} \;$, and we get the result above.[/quote]

The only way to resolve this once and for all is to do the integration one’s self. In that spirit, for a continuous distribution of charge such as we have,

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{1}{4\pi\epsilon_0}\sum_{\parallel{}\Delta Q_i\parallel{}\to 0}\frac{\Delta Q_i}{\left|\mathbf{x}\minus{}\mathbf{x}_i\right|^2}\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}\doteq\frac{1}{4\pi\epsilon_0}\int\frac{d Q_i}{\left|\mathbf{x}\minus{}\mathbf{x}_i\right|^2}\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

	\[ \begin{array}{cc} \mathbf{x} \equal{}(0,0,z)\mbox{;} & \mathbf{x}_i \equal{} (x_i,y_i,0)\end{array}
\]

	\[ dQ_i \equal{} \sigma dA_i \equal{} \sigma dx_i\,dy_i
\]

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{1}{4\pi\epsilon_0}\iint\frac{\sigma dx_i\,dy_i}{\left(x_{i}^{2} \plus{} y_{i}^{2} \plus{} z^2\right)}\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

The above integral is rather hideous but can be substantially simplified with a coordinate transformation. In spherical coordinates:

	\[ \left(\begin{array}{c} x_i \\ y_i \\ z \end{array}\right) \equal{} \left(\begin{array}{c}R\sin\theta\cos\phi \\ R\sin\theta\sin\phi \\ R\cos\theta \end{array}\right)
\]

	\[ x_{i}^{2} \plus{} y_{i}^{2} \plus{} z^{2} \equal{} \left(R\sin\theta\cos\phi\right)^2 \plus{}\left(R\sin\theta\sin\phi\right)^2 \plus{}\left(R\cos\theta\right)^2 \equal{}R^2
\]

	\[ dx_i\,dy_i \equal{} \frac{\partial\left(x_i,y_i\right)}{\partial\left(R,\phi\right)}\,dR\,d\phi\equal{}R\sin^{2}\theta\,dR\,d\phi
\]

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{1}{4\pi\epsilon_0}\iint\frac{\sigma R\sin^2\theta}{R^2}\,dR\,d\phi\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{1}{4\pi\epsilon_0}\iint\frac{\sigma \sin^2\theta}{R}\,dR\,d\phi\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

Now, since the charged surface is flat $ \theta$ is fixed, we have:

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{1}{4\pi\epsilon_0}\iint\frac{\sigma \sin\theta}{R}\,d\left(R\sin\theta\right)\,d\phi\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

Transforming the spherical coordinates to cylindrical coordinates,

	\[ \left(\begin{array}{c}r \\ z\end{array}\right) \equal{} \left(\begin{array}{c}R\sin\theta \\ R\cos\theta\end{array}\right)
\]

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{1}{4\pi\epsilon_0}\iint\frac{\sigma \sin\theta}{\sqrt{\left(R\sin\theta\right)^2 \plus{} \left(R\cos\theta\right)^2}}\,d\left(R\sin\theta\right)\,d\phi\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{1}{4\pi\epsilon_0}\iint\frac{\sigma \sin\theta}{\sqrt{r^2 \plus{} z^2}}\,dr\,d\phi\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

Now, $ \theta \equal{} \frac{\pi}{2}\to \sin\theta \equal{} 1$.

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{\sigma}{4\pi\epsilon_0}\int_{\minus{}\frac{\pi}{4}}^{\frac{\pi}{4}}\int_{\frac{\minus{}a}{2}}^{\frac{a}{2}}\frac{dr\,d\phi}{\sqrt{r^2 \plus{} z^2}}\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{\sigma}{8\epsilon_0}\int_{\frac{\minus{}a}{2}}^{\frac{a}{2}}\frac{dr}{\sqrt{r^2 \plus{} z^2}}\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

	\[ \mathbf{E}\left(\mathbf{x}\right)\equal{}\frac{\sigma}{8\epsilon_0}\int_{\frac{\minus{}a}{2}}^{\frac{a}{2}}\frac{dr/z}{\sqrt{\left(\frac{r}{z}\right)^2 \plus{} 1}}\,\hat{u}_{\mathbf{x}\minus{}\mathbf{x}_i}
\]

	\[ \boxed{\therefore\mathbf{E}\left(\mathbf{x}\equal{}\mathbf{z}\right)\equal{}\frac{\sigma}{4\epsilon_0}\sinh^{\minus{}1}\left(\frac{a}{2z}\right)\,\hat{u}_{\mathbf{z}\minus{}\mathbf{x}_i}}
\]

[quote="pardesi"][quote="Immanuel Bonfils"]I already did it:
[quote="Imannuel Bonfils"]In a Brazilian book: Problemas de Física - Eletricidade - Eletrostática , by Eduardo Wilner,
there is a solved problem of the field due to a square in a generic point of it axis (perpendicular to its plane) given

$ {\sigma \over {\pi\epsilon_0}}\,arc\,sin\,{1 \over {1\, + {4z^2 \over {a^2}}}}$.

In our case$ \; z\; = \;{a \over2} \;$, and we get the result above.[/quote][/quote]
ya you did :)[/quote]

As I tried to point out in an earlier post, Immanuel's answer cannot be correct.

If the anti-derivative of $ f\left(u\right)$ is $ \sin^{-1} u + C$ then $ f\left(u\right)$ = $ \frac{1}{\sqrt{1-u^2}}$. No amount of finagling I can imagine would ever give us an integrand of this form for the problem at hand.

[quote="Dr. No"]
$ \mathbf{E}\left(\mathbf{x}\right)=\frac{\sigma}{8\epsilon_0}\int_{\frac{-a}{2}}^{\frac{a}{2}}\frac{dr/z}{\sqrt{\left(\frac{r}{z}\right)^2 + 1}}\,\hat{u}_{\mathbf{x}-\mathbf{x}_i}$

$ \boxed{\therefore\mathbf{E}\left(\mathbf{x}=\mathbf{z}\right)=\frac{\sigma}{4\epsilon_0}\sinh^{-1}\left(\frac{a}{2z}\right)\,\hat{u}_{\mathbf{z}-\mathbf{x}_i}}$[/quote]

$ \;\hat{u}_{\mathbf{z}-\mathbf{x}_i\;}$ is not a constant in the integration...

[quote="Immanuel Bonfils"][quote="Dr. No"]
$ \mathbf{E}\left(\mathbf{x}\right) = \frac {\sigma}{8\epsilon_0}\int_{\frac { - a}{2}}^{\frac {a}{2}}\frac {dr/z}{\sqrt {\left(\frac {r}{z}\right)^2 + 1}}\,\hat{u}_{\mathbf{x} - \mathbf{x}_i}$

$ \boxed{\therefore\mathbf{E}\left(\mathbf{x} = \mathbf{z}\right) = \frac {\sigma}{4\epsilon_0}\sinh^{ - 1}\left(\frac {a}{2z}\right)\,\hat{u}_{\mathbf{z} - \mathbf{x}_i}}$[/quote]

$ \;\hat{u}_{\mathbf{z} - \mathbf{x}_i\;}$ is not a constant in the integration...[/quote]

True ... I hadn't noticed and its not entirely trivial; let's see what we get,

	\[ \mathbf{x}-\mathbf{x}_i =\left[\begin{array}{c} 0 \\ 0 \\ z\end{array}\right] -\left[\begin{array}{c}x_i \\ y_i \\ 0\end{array}\right]=\left[\begin{array}{r}-x_i\\-y_i\\z\end{array}\right]
\]

	\[ \mathbf{x}-\mathbf{x}_i =\left[\begin{array}{c}-x_i\\-y_i\\z\end{array}\right]=\left[\begin{array}{c}-R\cos\psi\cos\phi\\-R\cos\psi\sin\phi\\R\sin\psi\end{array}\right]
\]
	\[ \left|\mathbf{x}-\mathbf{x}_i\right| = R
\]
	\[ \hat{u}_{\mathbf{x}-\mathbf{x}_i} =\frac{\mathbf{x}-\mathbf{x}_i}{\left|\mathbf{x}-\mathbf{x}_i\right|}=\left[\begin{array}{c}-\cos\psi\cos\phi\\-\cos\psi\sin\phi\\\sin\psi\end{array}\right]
\]

	\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\hat{u}_{\mathbf{x}-\mathbf{x}_i}\,d\phi =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left[\begin{array}{c}-\cos\psi\cos\phi\\-\cos\psi\sin\phi\\\sin\psi\end{array}\right]\,d\phi=\left[\begin{array}{c}-\sqrt{2}\cos\psi\\0\\\frac{\pi}{2}\sin\psi\end{array}\right]
\]

	\[ \psi =\tan^{-1}\left(\frac{2z}{a}\right)
\]

	\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\hat{u}_{\mathbf{x}-\mathbf{x}_i}\,d\phi =\left[\begin{array}{c}-\sqrt{2}\cos\left(\tan^{-1}\left(\frac{2z}{a}\right)\right)\\0\\\frac{\pi}{2}\sin\left(\tan^{-1}\left(\frac{2z}{a}\right)\right)\end{array}\right]
\]

Therefore,

	\[ \mathbf{E}\left(\mathbf{x} =\mathbf{z}\right)=\frac{\sigma}{2\pi\epsilon_0}\sinh^{-1}\left(\frac{a}{2z}\right)
\left[\begin{array}{c}-\sqrt{2}\cos\left(\tan^{-1}\left(\frac{2z}{a}\right)\right)\\0\\\frac{\pi}{2}\sin\left(\tan^{-1}\left(\frac{2z}{a}\right)\right)\end{array}\right]
\]

Oh my god, all this argumentation is funny.

I'll try to resolve this issue using simple notation and relatively simple math.  From my calculations, I am [b]100%[/b] sure that the answer is $ \frac{\sigma}{6\epsilon_0}$.

So the problem reduces to finding the electric field at some distance above the center of a square plate of side length $ s$ and uniform charge density $ \sigma$.  Call this distance $ a$, and center the square at the origin.  For a charge $ dq$ at coordinates $ x, y$ on the plate, after some simple similar right-triangle ratios, we find that the $ z$-component (vertical component) of the electric field due to it at the height $ a$ above the plate is $ E_z$ where

\[ \frac{\sqrt{a^2\plus{}x^2\plus{}y^2}}{a} \equal{} \frac{\frac{dq}{4\pi\epsilon_0(a^2\plus{}x^2\plus{}y^2)}}{E_z} \implies E_z \equal{} \frac{adq}{4\pi\epsilon_0(a^2\plus{}x^2\plus{}y^2)^{3/2}}.\]

We only need to consider $ E_z$ for each $ dq$ since it is the only component that doesn't cancel.  The total electric field is given by

\[ E \equal{} \int_{\mbox{square}} E_z\]

\[ dq \equal{} \sigma dxdy \implies E \equal{} \frac{a\sigma}{4\pi\epsilon_0}\int_{\minus{}s/2}^{s/2}\int_{\minus{}s/2}^{s/2}\frac{1}{(a^2\plus{}x^2\plus{}y^2)^{3/2}} dxdy\]

Now, after not much effort, we find that $ \int \frac{1}{(a^2\plus{}x^2)^{3/2}} dx \equal{} \frac{x}{a^2\sqrt{a^2\plus{}x^2}}$, which reduces the above integral to:

\[ \frac{a\sigma}{4\pi\epsilon_0}\int_{\minus{}s/2}^{s/2}\frac{s}{(a^2\plus{}y^2)\sqrt{a^2\plus{}y^2\plus{}\frac{s^2}{4}}}dy\]

After a bit more effort, we can evaluate this integral as follows:

\[ E \equal{} \frac{a\sigma}{4\pi\epsilon_0}\left[\frac{2}{a}\arctan\left(\frac{sy}{a\sqrt{4a^2\plus{}4y^2\plus{}s^2}}\right)\right]_{\minus{}s/2}^{s/2} \]
\[ \equal{} \frac{\sigma}{\pi\epsilon_0}\arctan\left(\frac{s^2}{2a\sqrt{4a^2\plus{}2s^2}}\right).\]

Plugging in $ a \equal{} s/2$ to solve our original problem, we have

\[ E \equal{} \frac{\sigma}{\pi\epsilon_0}\arctan\left(\frac{s^2}{s\sqrt{3s^2}}\right) \equal{} \frac{\sigma}{\pi\epsilon_0}\arctan\left(\frac{1}{\sqrt{3}}\right) \equal{} \frac{\sigma}{6\epsilon_0}.\]

You would arrive at answers like $ \frac{\sigma}{\epsilon_0}$ and $ \frac{\sigma}{2\epsilon_0}$ by using Gauss's law and ignoring edge effects.

Can't fault your logic -- wish I could, but I can't. You solved it.


Let [b]a[/b],[b]b[/b] be positive integers, and let [b]a*[/b] be [u](Sum of digits of a)^2[/u], and b respectively.

Find all pairs (a,b) for which the following holds:

[b][i] a* = b [/i][/b]  and [b][i] b* = a [/i][/b]

Problem difficulty: [i]Easy[/i]

Fun problem.

[hide="Solution"]Denote by $ s(n)$ the sum of the digits of $ n$ squared. . Note that this problem is equivalent to finding all $ a$ such that $ s(s(a)) \equal{} a$.

First, note that if $ n$ has 4 or more digits, $ s(n)$ is at most $ 36^2 \equal{} 1296$. $ n \equal{} s(s(n))$ is therefore at most $ 27^2 \equal{} 729$, an impossibility.

Now, note that for any 3 digit number less than or equal to some integer $ m$, the greatest possible value of $ s(x)$ is $ \overline{a99}$, where $ a$ is less than or equal to the hundreds digit of $ m$.

So $ n$ has at most 3 digits. Therefore, $ s(n) \leq 729$, so $ n \equal{} s(s(n)) \leq 576$.

Since $ n \leq 576$, $ s(n) \leq 484$, so $ n \equal{} s(s(n)) \leq 441$. Since $ s(n) \leq 441$ as well, it must be that the sum of the digits of $ n$ is at most 21. Note that as a consequence, $ n < 400$.

Now, in order for $ n$ to be the square of the sum of the digits of something, $ n$ must be a perfect square. We only need to test the squres between 1 and 361 inclusive. 1 works, 81 works, 169 works, and 256 works.

I think that's all. [/hide]

[quote="Zhero"]Fun problem.

[hide="Solution"]Denote by $ s(n)$ the sum of the digits of $ n$ squared. . Note that this problem is equivalent to finding all $ a$ such that $ s(s(a)) \equal{} a$.
 [/hide][/quote]
dude the conditions are $ (s(s(a)^2))^2 \equal{} a$ how is that $ s(s(a))$? (Its not)
Reread the question  :wink:
your bounding solution will work the same though I think

Bleh, I used bad notation. I defined $ s(n)$ to be the square of the sum of the digits of $ n$.


linear algebra
matrix
vector
function
integration
calculus
linear algebra unsolved

I found this problem and found it quite interesting.

Let $A$ be an $n \times n$ real matrix, and let $f(t)$ be a T-Period vector function. Suppose that the homogeneous system $\vec{x}' = A\vec{x}$ admits no nonzero T-periodic solutions. Prove that the nohomogeneous system $\vec{x}' = A\vec{x} + f(t)$ admits a unique T-periodic solution.

If $k\in\mathbb{Z},$ then $\dfrac{2\pi ik}T$ is not an eigenvalue of $A.$

We know this because if $Av=\dfrac{2\pi ik}Tv$ then $y(t)=e^{2\pi ikt/T}v$ is a $T$-periodic solution of $y'=Ay,$ and our hypothesis is that there are no such solutions.

Now, take Fourier series. We can do vector-valued Fourier series with vector-valued coefficients.

If $f$ is a $T$-periodic vector-valued function, then define

$\hat{f}(k)=\dfrac1T\int_0^Te^{-2\pi ikt/T}f(t)\,dt$.

For nice enough functions (and we'll assume that everything in this problem is "nice enough" - $C^1$ ought to be sufficient),

$f(x)=\sum\limits_{k=-\infty}^{\infty}e^{2\pi ikt/T}\hat{f}(k).$

Now apply the Fourier transform to the differential equation. Note that $\widehat{f'}(k)=\dfrac{2\pi ik}T\hat{f}(k).$

$\dfrac{2\pi ik}T\hat{x}(k)=A\hat{x}(k)+\hat{f}(x)$

$\left(\dfrac{2\pi ik}TI-A\right)\hat{x}(k)=\hat{f}(k)$

Since we have established that $\dfrac{2\pi ik}TI-A$ is invertible, we find a unique solution for $\hat{x}(k)$ and hence a unique solution for $x(t).$

Applying the method of integrating factors to the (system)
$x'=Ax+f(t)$, we have $x(t)=e^{At}\int e^{-At}f(t)\,dt$
This gives
$x(t+T)-x(t)$
$=e^{At+AT}\left(\int_?^{t+T} e^{-As}f(s)\,ds-\int_?^t e^{-As}f(s)\,ds\right)$ $+e^{At+AT}\int_?^t e^{-As}f(s)\,ds-e^{At}\int_?^t e^{-As}f(s)\,ds$
$=e^{At+AT}\int_t^{t+T} e^{-As}f(s)\,ds+\left(e^{AT}-1\right)\int_?^t e^{-As}f(s)\,ds$
$=e^{AT}\int_0^Te^{-As}f(s+t)\,ds+\left(e^{AT}-1\right)x(t)$
1 is of course the identity here, in whatever algebra is relevant. By hypothesis, $e^{AT}-1$ is invertible; a singular matrix would give periodic solutions to the homogeneous equation.

Now, the leap of faith: Define $y(t)=\left(1-e^{AT}\right)^{-1}e^{AT}\int_0^T e^{-As}f(s+t)\,ds$.
It is clear that $y$ is periodic with period $T$; we need to show that it satisfies the equation, and that no other periodic function of period $T$ does.
By algebra on the definition and a change of variables,
$y(t)=\left(e^{-AT}-1\right)^{-1}e^{At}\int_t^{t+T} e^{-As}f(s)\,ds$.
Then
$y'(t)=\left(e^{-AT}-1\right)^{-1}Ae^{At}\int_t^{t+T} e^{-As}f(s)\,ds +\left(e^{-AT}-1\right)^{-1}e^{At}\left(e^{-A(t+T)}f(t+T)-e^{-At}f(t)\right)$
$y'(t)=Ay(t)+\left(e^{-AT}-1\right)^{-1}e^{At}\left(e^{-At}e^{-AT}-e^{-At}\right) f(t)$
$y'(t)=Ay(t)+\left(e^{-AT}-1\right)^{-1}\left(e^{-AT}-1\right)f(t)=Ay(t)+f(t)$

Now that we know $y$ is a solution, uniqueness is simple. The difference of two functions with period $T$ has period $T$, and the difference of two solutions is a solution of the homogeneous equation. Since the only solution of the homogeneous equation with period $T$ is 0, $y$ is the unique solution of the nonhomogeneous equation with period $T$ and we are done.

[I really don't know why the integrals are displaying that way]

[After spending a very long time composing this post, I was posted under just a few seconds before I finished- it came after my last preview]


[quote="firefox"]4) Let x be an integer one less than a multiple of 24. Prove that if a and b are positive integers such that ab=24, then a+b is a multiple of 24.[/quote]
4. where does x come in?

sorry, mistyped the problem. It's fixed now.

[quote="firefox"]1) Show that 4^(3x+1) + 2^(3x+1) + 1 is divisible by 7.

2) Determine all integral solutions of a^2 + b^2 + c^2 = (a^2)(b^2)

3) If n is a positive integer greater than 1 such that 2^n + n^2 is prime, show that n  :equiv: 3 (mod 6)

4) Let x be an integer one less than a multiple of 24. Prove that if a and b are positive integers such that ab=x, then a+b is a multiple of 24.[/quote]
3) if 2^n+n^2 was prime, n must be odd.now we will show that n is a multiple of 3.  the pattern for 2^3n in modulo 10 is 8, 4,2, 8,4,2....  Since n is odd and divisible by 3, you get the units digit as 7,3,1,7,3,1... This fits.
If it was not divisible by 3, it would be either 1 mod 3 or 2 mod 3.  after a quick guess and check, their units digits dont correspond.

4)logically, a and b must both be odd.   a must be -1(mod 24) and b must be 1(mod 24).  this is proved because x must be =1(mod 24) and it is clear that that is the only way.

Hi Scrambled,

    [quote]4)logically, a and b must both be odd. a must be -1(mod 24) and b must be 1(mod 24). this is proved because x must be =1(mod 24) and it is clear that that is the only way.[/quote]

I guess there is some error in your reasoning. First, you say x must be 1 (mod 24), but actually x is -1 (mod 24) as it is 1 less than a multiple of 24. Further, a is not always -1 (mod 24) and neither is b always 1 (mod 24). eg, x = 95 (= 96 - 1) = 5 * 19, so a can be 5 and b can be 19.

A rather long attempt at (4):

Since $ab = x \equiv -1 \pmod{24}$, $a$ and $b$ are odd. Moreover, $3$ doesn't divide $a$ or $b$; nor does $8$. Also, $ab$ is of the form $3k+2$ and also of the form $8k+7$.

First Note : out of $a$ and $b$, one is of the form $3k+1$ and the other is of the form $3k+2$.
   Because, if $a$ and $b$ both are of the form $3k+1$, then $ab$ is of the form $3l+1$
          and, if $a$ and $b$ both are of the form $3k+2$, then $ab$ is of the form $3l+1$.

Thus, $a+b$ is of the form $3k+3$ and hence is a multiple of $3$. ---------- (1)

Second Note : out of $a$ and $b$, one is of the form $8k+1$ and the other is of the form $8k+7$; or one is of the form $8k+3$ and the other is of the form $8k+5$
     Simliar to the previous argument, one can check that other possible forms of $a$ and $b$ imply that $ab$ is not of the form $8k+7$.

Thus, $a+b$ is of the form $8k+8$ and hence is a multiple of $8$. ----------- (2)

From (1) and (2), $a+b$ is a multiple of $24$.

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Sambit


induction
vector
number theory
least common multiple
modular arithmetic

hello friends, i have a very hard question which is from a  tst test. any help is appreciated.

Let $P$ be the set of all points in $\mathbb{R}^{n}$ with rational  coordinates.  For $A,B\in P$, one can move from $A$ to $B$ if the distance $AB$ is 1.  Prove that every point in $P$ can be reached from everyother point in $P$ by a finite sequence of moves if and only if $n\geq 5$.

Note that if any two points in $P$ can be connected for a given $n$, then the same is true for any larger $n$.  Thus it suffices to show that this is not the cas for $n=4$ and that it is the case for $n=5$.
First suppose $n=4$.  We say that a rational number is a  2-integer if it can be written as p/q with p and q integers and q odd.  we will show that for any point $(x_{1},x_{2},x_{3},x_{4})$ in $P$  reachable from the origin, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$ is a 2-integer.  Thus for example, (1/2,0,0,0) will not be reachable.
We proceed by induction on the min number of steps required to reach the point.  Obviously the claim holds for the origin.  Now suppose the claim holds for $(x_{1},x_{2},x_{3},x_{4})$ and $(y_{1},y_{2},y_{3},y_{4})$ is a unit  vector with rational components.  Write $x_{i}=\frac{p_{i}}{q_{i}}$ and $\sum x_{i}^{2}=\frac{p}{q}$ in lowest terms; then $\frac{p}{q}$ is a 2-integer and $q$ is odd.  we also notice that $q_{i}$ must all be odd or must all be twice an odd number.  If the lcm, d, of the $q_{i}$ is not odd, then $\sum (\frac{p_{i}d}{q_{i}})^{2}=\frac{d^{2}p}{q}$ is an integer.  Since $d$ is even and $q$ is odd, the right side is $\equiv 0 \pmod{4}$, while the left side is the sum of termswhich are 0 or 1, mod 4, and not all of the former.  Hence, they are all of the latter form, so the $q_{i}$ are all dividble by exactly the same power of $2$.  Moreover, the left side is now $\equiv 4 \pmod{8}$, so $d$ is not divisible by 4.
Likewise, if we write $y_{i}=\frac{v_{i}}{w_{i}}$, in lowest terms, the $w_{i}$ are either all odd or all twice an odd number.  Let $e$ be the lcm of $w_{i}$.  Then
$\sum(x_{i}+y_{i})^{2}=\sum x_{i}^{2}+\sum y_{i}^{2}+\frac{2}{de}\sum (dx_{i})(ey_{i}).$
the first two terms on RHS are 2 integers by the inductive hypotheisis.  Furthermore, the third sum is the sum of 4 odd integers, multiplied by 2 and divided by an integer not divisibly by 8.  Therefore it is also a 2-integer.  SInce thesum of 2-integers is a 2-integer, wehave that $\sum(x_{i}+y_{i})^{2}$ is a 2-ingteger, and induciton is complete.
Now suppose $n\geq 5$, in fact, it suffices to consider $n=5$.  The key fact is that any positive integer can be written as the sum of four perfect squares, this is commonly known as  [i]Legendre's Theorem[/i].  From this it follows that any rational number is the sum of four rational prfect squares.
It suffices to show that one can get from the origin to the point $(a,0,0,0)$.  To do so, find a solution to:
$(\frac{a}{2})^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=1$ in rational numbers $x_{1},x_{2},x_{3},x_{4}$, and then step from the origin to $(\frac{a}{2},0,0,0)$ then to $(a,0,0,0)$.
so we are done :)