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| author | Opheliar99 <> | 2010-07-04 04:08:28 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 04:08:28 +0000 |
| commit | feffa8323b8f71ff31fe0042ec541bf91d6bab58 (patch) | |
| tree | 44861586887f1e14a1f4214671cd70431fb521de | |
| parent | a7bd395cd869fd41f40131024ff2b4d8063a6f15 (diff) | |
| download | afterklein-wiki-feffa8323b8f71ff31fe0042ec541bf91d6bab58.tar.gz afterklein-wiki-feffa8323b8f71ff31fe0042ec541bf91d6bab58.zip | |
posted solutions of 2 and 3 in pset2
| -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 abb923b..aadb1f6 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 $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, -$ \sin^4 x = \frac{{( e^{ix}-e^{-ix})}^{4}}{16}$, +$\sin^4 x = \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 |