From 3f8b5e12eb198dd28960cd8b725c315ab69733bf Mon Sep 17 00:00:00 2001 From: luccul Date: Sun, 4 Jul 2010 19:54:36 +0000 Subject: more formatting --- Problem Set 2.page | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/Problem Set 2.page b/Problem Set 2.page index b4a7e19..b1b82a5 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -67,7 +67,7 @@ $$\begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-im &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx \\ -&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx. \\ +&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx. \end{array} $$ Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$, -- cgit v1.2.3