From f68b8a554c2620e2b7f40a6498a3cd778940cc34 Mon Sep 17 00:00:00 2001
From: Opheliar99 <>
Date: Sun, 4 Jul 2010 02:16:06 +0000
Subject: posted solutions of 2 and 3 in pset2

---
 Problem Set 2.page | 7 +++----
 1 file changed, 3 insertions(+), 4 deletions(-)

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 6fec599..329265b 100644
--- a/Problem Set 2.page	
+++ b/Problem Set 2.page	
@@ -39,10 +39,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
 # Solutions
 
 2. Since
-$sin(x) = \frac{e^{ix}-e^{-ix}}{2}$,
+$\sin(x) = \frac{e^{ix}-e^{-ix}}{2}$,
 $\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^4}{16}$,
-
-$ = \frac{e^{i4x}+e^{-4ix}-4 e^{i2x} -4 e^{-i2x}+6}{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 
 
@@ -57,7 +56,7 @@ $a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $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>$
 
-$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
+$ = \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
 
 
 ## Cardinality
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