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authorOpheliar99 <>2010-07-04 02:19:10 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 02:19:10 +0000
commitbd1a5eb91d0f8c5adfc4e3d71cb6ff992616713e (patch)
treeb8f50806bd4d0783719e943ba09323320dc9bf6c
parent300fc6a5d057f6cc642ff3f3131952e6a45828c2 (diff)
downloadafterklein-wiki-bd1a5eb91d0f8c5adfc4e3d71cb6ff992616713e.tar.gz
afterklein-wiki-bd1a5eb91d0f8c5adfc4e3d71cb6ff992616713e.zip
posted solutions of 2 and 3 in pset2
-rw-r--r--Problem Set 2.page4
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diff --git a/Problem Set 2.page b/Problem Set 2.page
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+++ b/Problem Set 2.page
@@ -39,8 +39,8 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
# Solutions
2. Since
-$\sin(x) = \frac{\exp^{ix}-e^{-ix}}{2}$,
-$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^4}{16}$,
+$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
+$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^{4}}{16}$,
$ = \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