From ba5b7c03e8bc4ed4f57c52f278152c3fa0c74dd1 Mon Sep 17 00:00:00 2001
From: luccul <luccul@gmail.com>
Date: Sun, 4 Jul 2010 19:48:13 +0000
Subject: some formatting for solution 2

---
 Problem Set 2.page | 15 ++++++++-------
 1 file changed, 8 insertions(+), 7 deletions(-)

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 218287f..c114851 100644
--- a/Problem Set 2.page	
+++ b/Problem Set 2.page	
@@ -39,20 +39,21 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
 # Solutions
 
 2. Since
-$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
+$$\sin x = \frac{e^{ix}-e^{-ix}}{2}$$,
 
-$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$
-$= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$.
+$$\begin{array}{ccl} \sin^4 x &=& \frac{{( e^{ix}-e^{-ix} )}^4}{16} \\
+  &=& \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. \end{array} $$
 
 
 If we express any periodic function $f(x)$ as 
 
-$f(x) = \sum a_n f_n(x)$, where $f_n(x) = \frac{e^{inx}}{\sqrt{2\pi}}$ and $f_0(x) = \frac{1}{\sqrt{2\pi}}$,
-
+$$f(x) = \sum a_n f_n(x),$$
 
-The Fourier coefficients for the above functions are:
+where $f_n(x) = \frac{e^{inx}}{\sqrt{2\pi}}$ and $f_0(x) = \frac{1}{\sqrt{2\pi}}$, then the Fourier coefficients for the above functions are:
 
-$a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \times 4/16$, $a_0 = \sqrt{2\pi} \times 6/16$
+$$\begin{array}{rcl} a_{-4} = a_{4} &=& \sqrt{2\pi} \times 1/16, \\
+ a_{-2} = a_{2} &=& - \sqrt{2\pi} \times 4/16 \\
+ a_0 &=& \sqrt{2\pi} \times 6/16 \end{array} $$
 
 
 Since $a_m = < f_m, f >$ and setting $f(x) = \sin^4(x)$,
-- 
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