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@@ -40,7 +40,7 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
2. Since
$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
-$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$,
+$ \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}$.
If we express any periodic function $f(x)$ as