some formatting for solution 2

1 files 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 $f(x) = \sin^4(x)$,