more formatting/editing

1 files changed, 5 insertions, 7 deletions

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 4fe7a1f..2637ab2 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -61,17 +61,15 @@ $$ \begin{array}{ccl} \int_0^{2\pi} \sin^4(x) dx = <1, f> &=& \sqrt{2\pi} \times
  &=& \frac{3 \pi}{4} \end{array} $$
 
 
-3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, $\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$, and therefore,
+3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, we have $\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$.  Therefore,
 
-$$\begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx \\
-
-&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx \\
-
-&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx.  \end{array} $$
+$$ \begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx \\
+  &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx \\
+  &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx.  \end{array} $$
 
 Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$, we have
 
-$$a_2 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4,$$
+$$a_2 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\ pi}/4,$$
 
 $$a_0 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \frac{\sqrt{2\pi}}{2},$$