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| author | Opheliar99 <> | 2010-07-04 04:12:05 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 04:12:05 +0000 |
| commit | 14244b4506f772541be6b2289b2ace8651d0e7f1 (patch) | |
| tree | 7f850e8089b6db44b6c54681580a63c7d217553f /Problem Set 2.page | |
| parent | f5cb3cc89c0eb29347cdcb95dc6543f46af095d9 (diff) | |
| download | afterklein-wiki-14244b4506f772541be6b2289b2ace8651d0e7f1.tar.gz afterklein-wiki-14244b4506f772541be6b2289b2ace8651d0e7f1.zip | |
posted solutions of 2 and 3 in pset2
Diffstat (limited to 'Problem Set 2.page')
| -rw-r--r-- | Problem Set 2.page | 4 |
1 files changed, 3 insertions, 1 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index be65cd5..2d5a503 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -40,7 +40,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, -$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$, + + +$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$ $= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. If we express any periodic function $f(x)$ as |