more formatting

1 files changed, 6 insertions, 10 deletions

diff --git a/Problem Set 2.page b/Problem Set 2.page
index c114851..5ba4b65 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -38,12 +38,10 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
 
 # Solutions
 
-2. Since
-$$\sin x = \frac{e^{ix}-e^{-ix}}{2}$$,
+2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
 
 $$\begin{array}{ccl} \sin^4 x &=& \frac{{( e^{ix}-e^{-ix} )}^4}{16} \\
-  &=& \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. \end{array} $$
-
+  &=& \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 
 
@@ -58,14 +56,12 @@ $$\begin{array}{rcl} a_{-4} = a_{4} &=& \sqrt{2\pi} \times 1/16, \\
 
 Since $a_m = < f_m, f >$ and setting $f(x) = \sin^4(x)$,
 
-$\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times < f_0, f >$
-
-$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
-
+$$ \begin{array}{ccl} \int_0^{2\pi} \sin^4(x) dx = <1, f> &=& \sqrt{2\pi} \times < f_0, f > \\
+&=& \sqrt{2\pi} \times a_0 \\ 
+ &=& \frac{3 \pi}{4} \end{array} $$
 
 
-3. Since
-$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
+3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
 
 $\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$