From 4cb4aca40267f73499ad6669620381b44d68ebf3 Mon Sep 17 00:00:00 2001
From: Opheliar99 <>
Date: Sun, 4 Jul 2010 04:38:05 +0000
Subject: posted solutions of 2 and 3 in pset2

---
 Problem Set 2.page | 12 ++++++++++++
 1 file changed, 12 insertions(+)

(limited to 'Problem Set 2.page')

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 63e60ad..6eb1452 100644
--- a/Problem Set 2.page	
+++ b/Problem Set 2.page	
@@ -62,6 +62,18 @@ $\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times < f_0, f >$
 $= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
 
 
+3. Since
+$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
+
+$\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$
+
+$a_m = \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx$
+
+$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx$
+
+$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx$.
+
+
 ## Cardinality
 
 Cardinality of the natural numbers (countable): $\mathbf{N}$,$\mathbf{Z}$  
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