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authorOpheliar99 <>2010-07-04 04:38:05 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:38:05 +0000
commit4cb4aca40267f73499ad6669620381b44d68ebf3 (patch)
tree7b1e816bb71c5909e3f991255e21dce393cf39bc /Problem Set 2.page
parentd406b157e8e83ef13288d6bce42e1ecf6f1f4cb6 (diff)
downloadafterklein-wiki-4cb4aca40267f73499ad6669620381b44d68ebf3.tar.gz
afterklein-wiki-4cb4aca40267f73499ad6669620381b44d68ebf3.zip
posted solutions of 2 and 3 in pset2
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@@ -62,6 +62,18 @@ $\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times < f_0, f >$
$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
+3. Since
+$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
+
+$\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$
+
+$a_m = \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx$
+
+$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx$
+
+$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx$.
+
+
## Cardinality
Cardinality of the natural numbers (countable): $\mathbf{N}$,$\mathbf{Z}$