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| author | Opheliar99 <> | 2010-07-04 02:23:33 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 02:23:33 +0000 |
| commit | e5e3a0620c1bee420a836df7255f3e453262056d (patch) | |
| tree | 57a3358ed377642a235847829e5a722567689bd0 /Problem Set 2.page | |
| parent | 413b07d78a4b1f0bd292ff982e061862b003e9cf (diff) | |
| download | afterklein-wiki-e5e3a0620c1bee420a836df7255f3e453262056d.tar.gz afterklein-wiki-e5e3a0620c1bee420a836df7255f3e453262056d.zip | |
posted solutions of 2 and 3 in pset2
Diffstat (limited to 'Problem Set 2.page')
| -rw-r--r-- | Problem Set 2.page | 5 |
1 files changed, 2 insertions, 3 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 02a8f48..7918c1a 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -40,9 +40,8 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, -$ \sin^4(x) 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}$. +$ \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 |