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authorOpheliar99 <>2010-07-04 04:43:25 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:43:25 +0000
commit703b678fb47b34894a45a7a96d9fdf2768fecdad (patch)
treebe3630855652e3ab10a14f04d24ad31a1cfcb6c4 /Problem Set 2.page
parent4cb4aca40267f73499ad6669620381b44d68ebf3 (diff)
downloadafterklein-wiki-703b678fb47b34894a45a7a96d9fdf2768fecdad.tar.gz
afterklein-wiki-703b678fb47b34894a45a7a96d9fdf2768fecdad.zip
posted solutions of 2 and 3 in pset2
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+++ b/Problem Set 2.page
@@ -73,6 +73,12 @@ $= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx
$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx$.
+Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$,
+
+$m = 2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$,
+$m = 0 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \sqrt{2\pi}/2$,
+$m = -2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$,
+
## Cardinality