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authorOpheliar99 <>2010-07-04 04:08:28 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:08:28 +0000
commitfeffa8323b8f71ff31fe0042ec541bf91d6bab58 (patch)
tree44861586887f1e14a1f4214671cd70431fb521de /Problem Set 2.page
parenta7bd395cd869fd41f40131024ff2b4d8063a6f15 (diff)
downloadafterklein-wiki-feffa8323b8f71ff31fe0042ec541bf91d6bab58.tar.gz
afterklein-wiki-feffa8323b8f71ff31fe0042ec541bf91d6bab58.zip
posted solutions of 2 and 3 in pset2
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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