From 14244b4506f772541be6b2289b2ace8651d0e7f1 Mon Sep 17 00:00:00 2001
From: Opheliar99 <>
Date: Sun, 4 Jul 2010 04:12:05 +0000
Subject: posted solutions of 2 and 3 in pset2

---
 Problem Set 2.page | 4 +++-
 1 file changed, 3 insertions(+), 1 deletion(-)

(limited to 'Problem Set 2.page')

diff --git a/Problem Set 2.page b/Problem Set 2.page
index be65cd5..2d5a503 100644
--- a/Problem Set 2.page	
+++ b/Problem Set 2.page	
@@ -40,7 +40,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
 
 2. Since
 $\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
-$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$,
+
+
+$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$
 $= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$.
 
 If we express any periodic function $f(x)$ as 
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