From 3797a85d15f62fc6a469a055d6d7c25bdec9aece Mon Sep 17 00:00:00 2001
From: Opheliar99 <>
Date: Sun, 4 Jul 2010 02:11:39 +0000
Subject: posted solutions of 2 and 3 in pset2

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

diff --git a/Problem Set 2.page b/Problem Set 2.page
index cc84acb..05c4885 100644
--- a/Problem Set 2.page	
+++ b/Problem Set 2.page	
@@ -54,8 +54,7 @@ $a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \tim
 
 
 Since $a_m = < f_m, f >$,
-
-$\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times $<f_0, f>$$
+$ \int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times <f_0, f>$
 
 $= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
 
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