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authorOpheliar99 <>2010-07-04 04:01:23 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:01:23 +0000
commit9cefc50326bfca41e5844321ead27af9f81020ba (patch)
treea65c530e303221c43fc246e321deeb140753bc17 /Problem Set 2.page
parentffd68f8715387b7c68cf07f7a27485d64b1e50e7 (diff)
downloadafterklein-wiki-9cefc50326bfca41e5844321ead27af9f81020ba.tar.gz
afterklein-wiki-9cefc50326bfca41e5844321ead27af9f81020ba.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