From 465bf280a3f16938d7423d7cd1cc4fafe8a6cec8 Mon Sep 17 00:00:00 2001
From: Opheliar99 <>
Date: Sun, 4 Jul 2010 04:55:55 +0000
Subject: posted solutions of 2 and 3 in pset2

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

diff --git a/Problem Set 2.page b/Problem Set 2.page
index a26852a..523ce40 100644
--- a/Problem Set 2.page	
+++ b/Problem Set 2.page	
@@ -73,12 +73,24 @@ $= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx
 
 $= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx$.
 
+
 Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$,
 
 $m = 2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$,
+
 $m = 0 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \sqrt{2\pi}/2$,
+
 $m = -2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$,
 
+Then,
+
+$\sum |a_n|^2 = {\sqrt{2\pi}/4}^2 + {- \sqrt{2\pi}/2}^2 + {\sqrt{2\pi}/4}^2 = \frac{3 \pi}{4}$
+And, it was shown in Prob 2 that $\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$.
+Therefore, 
+
+$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$
+
+
 
 ## Cardinality
 
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