Recent content by ReyChiquito

Well, first of all, it would be nice if someone tell me why the TeX doesn't work. Second of all, i got it, so nevermind. \int \Omega \omega no \int? nice...

Is it just me or nobody can see the TeX?

Let x \in \mathbb{R}^n and u_0>0, \qquad \int\limits_\Omega u_0(x) dx =1, \qquad E(t)=\int\limits_\Omega u(x,t)u_0(x)dx Im having trouble proving the following inequality \int\limits_\Omega \frac{u_0(x)}{(1+u(x,t))^2}dx \ge \dfrac{1}{(1+E)^2}. \qquad \hbox{(1)} I know i have to use Jensen's...

i don't understand the question, but i know this much \begin{array}{l} a_0(x)=\int_a^x F(t)dt\\ \\ a_1(x)=\int_a^x \int_a^{x_1} F(t) dt dx_1\\ \\ a_2(x)=2\int_a^x \int_a^{x_2} \int_a^{x_1} F(t)dtdx_1dx_2\\ \vdots\\ a_n(x)=n! \int_a^x\int_a^{x_{n-1}}\dots\int_a^{x_1}F(t)dt dx_1\dots dx_{n-1}...

\iint\limits_\Omega \int\limits_{\|\vec{X}-\vec{Y}\|=\epsilon} \mathop{\nabla}\limits_X all right! :approve:

thx mate... I am still not convinced though... ill get it right someday :)

\int \atop \|\vec{X}-\vec{Y}\|=\epsilon \nabla \atop X there has to be a better way

quick q. instead of having something like this \int_{\|\vec{X}-\vec{Y}\|=\epsilon} can i make the \|\vec{X}-\vec{Y}\|=\epsilon} right beneath the \int sign? the same here, instead of \nabla_X i want the X beneath the nabla sign thx :)

good point SEE KIDS what can happen if you don't read the whole question? *blush* :-p

i read the q. wrong :P but sure it works using the fundamental theorem of calculus

\int x^ndx=\frac{x^{n+1}}{n+1}+C

then, does your IF works? The only problem i see here is that the equation for \mu (x,y) as i wrote it has a unique solution (in this case). If the IF is not unique, then neither the solution to the DE right?

the integrating factor \mu (x,y) must satisfy the equation P\mu_y-Q\mu_x+(P_y-Q_x)\mu=0

how about learning some physics? classical mechanics for starters...

Brown is great too