From 802c90d926a3ffc36dec7b4c792a36419eb6f7aa Mon Sep 17 00:00:00 2001 From: luccul Date: Tue, 6 Jul 2010 06:02:58 +0000 Subject: more formatting --- ClassJuly5.page | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'ClassJuly5.page') diff --git a/ClassJuly5.page b/ClassJuly5.page index fa72182..ff7cfb3 100644 --- a/ClassJuly5.page +++ b/ClassJuly5.page @@ -58,7 +58,7 @@ $$ \lambda = \frac{\pi n}{L} $$ So, the most general solution we can generate in this manner is: $$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-(\frac{\pi n}{L})^2t} \sin(\frac{\pi n x}{L}) $$ -We would like to assert that any solution takes this form. One way to prove this assertion would be to show that any function $f:[0,L] \to \R$ satisfying $f(0) = f(L) = 0$ has a unique ``[Fourier sine expansion](http://mathworld.wolfram.com/FourierSineSeries.html)'': +We would like to assert that any solution takes this form. One way to prove this assertion would be to show that any function $f:[0,L] \to \mathbb{R}$ satisfying $f(0) = f(L) = 0$ has a unique ``[Fourier sine expansion](http://mathworld.wolfram.com/FourierSineSeries.html)'': $$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$ One could then allow the coefficients $c_n$ to vary with $t$ and apply the same method of solution that we used in the case of periodic boundary conditions. -- cgit v1.2.3