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author | Opheliar99 <> | 2010-07-05 04:03:33 +0000 |
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committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-05 04:03:33 +0000 |
commit | 37bad950149205dbc8a52f1b88d3a11911305859 (patch) | |
tree | 983c023ee1c6915f913131396a0029bf8b8a0302 | |
parent | c96f9ee6c3d935947acfef30c2113e4695b668e4 (diff) | |
download | afterklein-wiki-37bad950149205dbc8a52f1b88d3a11911305859.tar.gz afterklein-wiki-37bad950149205dbc8a52f1b88d3a11911305859.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 d62036a..a6ed816 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -69,7 +69,7 @@ $$ \begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-i Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$, we have -$$a_2 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\ pi}/4,$$ +$$a_2 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \frac{\sqrt{2\pi}}{4},$$ $$a_0 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \frac{\sqrt{2\pi}}{2},$$ |