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authorOpheliar99 <>2010-07-04 04:02:12 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:02:12 +0000
commita7bd395cd869fd41f40131024ff2b4d8063a6f15 (patch)
tree529306786939b18383d252d3d529df02705c6bef
parent9cefc50326bfca41e5844321ead27af9f81020ba (diff)
downloadafterklein-wiki-a7bd395cd869fd41f40131024ff2b4d8063a6f15.tar.gz
afterklein-wiki-a7bd395cd869fd41f40131024ff2b4d8063a6f15.zip
posted solutions of 2 and 3 in pset2
-rw-r--r--Problem Set 2.page2
1 files changed, 1 insertions, 1 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page
index 51968f8..abb923b 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -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