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| author | Opheliar99 <> | 2010-07-04 04:28:53 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 04:28:53 +0000 |
| commit | 0774a65c9cdbd4992b972eefff31b2da83c2c40b (patch) | |
| tree | 12c18ae1f8b0b77b8237bf08c2447940052f5d3b /Problem Set 2.page | |
| parent | 60d31ba77b860e5f788fb97fb87d44043288a27e (diff) | |
| download | afterklein-wiki-0774a65c9cdbd4992b972eefff31b2da83c2c40b.tar.gz afterklein-wiki-0774a65c9cdbd4992b972eefff31b2da83c2c40b.zip | |
posted solutions of 2 and 3 in pset2
Diffstat (limited to 'Problem Set 2.page')
| -rw-r--r-- | Problem Set 2.page | 1 |
1 files changed, 0 insertions, 1 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index de38296..fd21bf0 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -40,7 +40,6 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, - $\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$ $= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. |