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| author | Opheliar99 <> | 2010-07-04 04:12:42 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 04:12:42 +0000 |
| commit | f0c461bce6678793f61c2e0c7fafd8b1801129f6 (patch) | |
| tree | e0509520d752c0e14e94e3f82213f607bbf15b94 /Problem Set 2.page | |
| parent | 14244b4506f772541be6b2289b2ace8651d0e7f1 (diff) | |
| download | afterklein-wiki-f0c461bce6678793f61c2e0c7fafd8b1801129f6.tar.gz afterklein-wiki-f0c461bce6678793f61c2e0c7fafd8b1801129f6.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 2d5a503..667abad 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -41,7 +41,6 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, - $\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$ $= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. @@ -56,6 +55,7 @@ $a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \tim Since $a_m = < f_m, f >$, + $\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times < f_0, f >$ $= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$ |