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authorOpheliar99 <>2010-07-04 04:09:15 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:09:15 +0000
commitaffb42424db9b15740145cd7ba1f323b082126b5 (patch)
treeb63b6aef6dd24a7ec310bc43fa388e8fad2cc962
parentfeffa8323b8f71ff31fe0042ec541bf91d6bab58 (diff)
downloadafterklein-wiki-affb42424db9b15740145cd7ba1f323b082126b5.tar.gz
afterklein-wiki-affb42424db9b15740145cd7ba1f323b082126b5.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 aadb1f6..3f1a214 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