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authorOpheliar99 <>2010-07-04 02:23:33 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 02:23:33 +0000
commite5e3a0620c1bee420a836df7255f3e453262056d (patch)
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parent413b07d78a4b1f0bd292ff982e061862b003e9cf (diff)
downloadafterklein-wiki-e5e3a0620c1bee420a836df7255f3e453262056d.tar.gz
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posted solutions of 2 and 3 in pset2
-rw-r--r--Problem Set 2.page5
1 files changed, 2 insertions, 3 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page
index 02a8f48..7918c1a 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -40,9 +40,8 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
2. Since
$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
-$ \sin^4(x) 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}$.
+$ \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