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| author | Opheliar99 <> | 2010-07-04 04:02:12 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 04:02:12 +0000 |
| commit | a7bd395cd869fd41f40131024ff2b4d8063a6f15 (patch) | |
| tree | 529306786939b18383d252d3d529df02705c6bef | |
| parent | 9cefc50326bfca41e5844321ead27af9f81020ba (diff) | |
| download | afterklein-wiki-a7bd395cd869fd41f40131024ff2b4d8063a6f15.tar.gz afterklein-wiki-a7bd395cd869fd41f40131024ff2b4d8063a6f15.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 51968f8..abb923b 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 |