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| author | Opheliar99 <> | 2010-07-04 02:19:10 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 02:19:10 +0000 |
| commit | bd1a5eb91d0f8c5adfc4e3d71cb6ff992616713e (patch) | |
| tree | b8f50806bd4d0783719e943ba09323320dc9bf6c | |
| parent | 300fc6a5d057f6cc642ff3f3131952e6a45828c2 (diff) | |
| download | afterklein-wiki-bd1a5eb91d0f8c5adfc4e3d71cb6ff992616713e.tar.gz afterklein-wiki-bd1a5eb91d0f8c5adfc4e3d71cb6ff992616713e.zip | |
posted solutions of 2 and 3 in pset2
| -rw-r--r-- | Problem Set 2.page | 4 |
1 files changed, 2 insertions, 2 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 518dccb..d001972 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -39,8 +39,8 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ # Solutions 2. Since -$\sin(x) = \frac{\exp^{ix}-e^{-ix}}{2}$, -$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^4}{16}$, +$\sin x = \frac{e^{ix}-e^{-ix}}{2}$, +$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^{4}}{16}$, $ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$ If we express any periodic function $f(x)$ as |