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authorOpheliar99 <>2010-07-04 02:22:35 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 02:22:35 +0000
commit413b07d78a4b1f0bd292ff982e061862b003e9cf (patch)
treee4edf6130593a46a6e6912de7f84ba314ff6ecfd
parent3ceb2d9901fb49cb0a12828dc2fc45e410ca886f (diff)
downloadafterklein-wiki-413b07d78a4b1f0bd292ff982e061862b003e9cf.tar.gz
afterklein-wiki-413b07d78a4b1f0bd292ff982e061862b003e9cf.zip
posted solutions of 2 and 3 in pset2
-rw-r--r--Problem Set 2.page4
1 files changed, 2 insertions, 2 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page
index 4811096..02a8f48 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -40,9 +40,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
2. Since
$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
-$\int_0^{2\pi} \sin^4(x) dx = \frac{{( e^{ix}-e^{-ix})}^{4}}{16}$,
+$ \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}$.
+$ \= \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