From e5e3a0620c1bee420a836df7255f3e453262056d Mon Sep 17 00:00:00 2001
From: Opheliar99 <>
Date: Sun, 4 Jul 2010 02:23:33 +0000
Subject: posted solutions of 2 and 3 in pset2

---
 Problem Set 2.page | 5 ++---
 1 file changed, 2 insertions(+), 3 deletions(-)

(limited to 'Problem Set 2.page')

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 02a8f48..7918c1a 100644
--- a/Problem Set 2.page	
+++ b/Problem Set 2.page	
@@ -40,9 +40,8 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
 
 2. Since
 $\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
-$ \sin^4(x) dx = \frac{{( e^{ix}-e^{-ix})}^{4}}{16}$,
-
-$ \= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$.
+$ \sin^4 x = \frac{{( e^{ix}-e^{-ix})}^{4}}{16}$,
+$ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$.
 
 If we express any periodic function $f(x)$ as 
 
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