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authorOpheliar99 <>2010-07-04 02:16:06 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 02:16:06 +0000
commitf68b8a554c2620e2b7f40a6498a3cd778940cc34 (patch)
tree1185219cc1937495351cf3e6952103549da7626c
parent0c978dbfdade8624698d91f19245d0fecf5a3356 (diff)
downloadafterklein-wiki-f68b8a554c2620e2b7f40a6498a3cd778940cc34.tar.gz
afterklein-wiki-f68b8a554c2620e2b7f40a6498a3cd778940cc34.zip
posted solutions of 2 and 3 in pset2
-rw-r--r--Problem Set 2.page7
1 files changed, 3 insertions, 4 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page
index 6fec599..329265b 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -39,10 +39,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
# Solutions
2. Since
-$sin(x) = \frac{e^{ix}-e^{-ix}}{2}$,
+$\sin(x) = \frac{e^{ix}-e^{-ix}}{2}$,
$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^4}{16}$,
-
-$ = \frac{e^{i4x}+e^{-4ix}-4 e^{i2x} -4 e^{-i2x}+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
@@ -57,7 +56,7 @@ $a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \tim
Since $a_m = < f_m, f >$,
$ \int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times <f_0, f>$
-$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
+$ = \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
## Cardinality