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| author | Opheliar99 <> | 2010-07-04 02:20:32 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 02:20:32 +0000 |
| commit | 1d9ba4013d87ddeb4a1404f5c9812b2ebf13c417 (patch) | |
| tree | ded5222a324b28e1d5c1ec2a5a7bbfbdb1386a58 | |
| parent | bd1a5eb91d0f8c5adfc4e3d71cb6ff992616713e (diff) | |
| download | afterklein-wiki-1d9ba4013d87ddeb4a1404f5c9812b2ebf13c417.tar.gz afterklein-wiki-1d9ba4013d87ddeb4a1404f5c9812b2ebf13c417.zip | |
posted solutions of 2 and 3 in pset2
| -rw-r--r-- | Problem Set 2.page | 5 |
1 files changed, 3 insertions, 2 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index d001972..5d48dd5 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -40,8 +40,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, -$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^{4}}{16}$, -$ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$ +$\int_0^{2\pi} \sin^4(x) dx = \frac{{\left e^{ix}-e^{-ix} \right}^{4}}{16}$, + +$ = \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 |