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| author | Opheliar99 <> | 2010-07-04 02:17:53 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 02:17:53 +0000 |
| commit | 300fc6a5d057f6cc642ff3f3131952e6a45828c2 (patch) | |
| tree | b65e9ba2462393a56c160001173a3b42b6b0c826 /Problem Set 2.page | |
| parent | f68b8a554c2620e2b7f40a6498a3cd778940cc34 (diff) | |
| download | afterklein-wiki-300fc6a5d057f6cc642ff3f3131952e6a45828c2.tar.gz afterklein-wiki-300fc6a5d057f6cc642ff3f3131952e6a45828c2.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 329265b..518dccb 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -39,7 +39,7 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ # Solutions 2. Since -$\sin(x) = \frac{e^{ix}-e^{-ix}}{2}$, +$\sin(x) = \frac{\exp^{ix}-e^{-ix}}{2}$, $\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^4}{16}$, $ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$ |