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| author | Opheliar99 <> | 2010-07-04 02:21:52 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 02:21:52 +0000 |
| commit | 3ceb2d9901fb49cb0a12828dc2fc45e410ca886f (patch) | |
| tree | 1931048c53dd536296dd6d1b4a508e70383f114b /Problem Set 2.page | |
| parent | 32e82967016f48f091427bc4f57706aba5194d01 (diff) | |
| download | afterklein-wiki-3ceb2d9901fb49cb0a12828dc2fc45e410ca886f.tar.gz afterklein-wiki-3ceb2d9901fb49cb0a12828dc2fc45e410ca886f.zip | |
posted solutions of 2 and 3 in pset2
Diffstat (limited to 'Problem Set 2.page')
| -rw-r--r-- | Problem Set 2.page | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 1672b23..4811096 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -40,7 +40,7 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, -$\int_0^{2\pi} \sin^4(x) dx = \frac{(\left e^{ix}-e^{-ix})}^{4}}{16}$, +$\int_0^{2\pi} \sin^4(x) dx = \frac{{( e^{ix}-e^{-ix})}^{4}}{16}$, $ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. |