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| author | luccul <luccul@gmail.com> | 2010-07-04 19:54:36 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 19:54:36 +0000 |
| commit | 3f8b5e12eb198dd28960cd8b725c315ab69733bf (patch) | |
| tree | 0b19952870729631bb1d5903a05f4130deba09f6 | |
| parent | 7e25667dea286052527e96bfc43f12aa08f7fd3b (diff) | |
| download | afterklein-wiki-3f8b5e12eb198dd28960cd8b725c315ab69733bf.tar.gz afterklein-wiki-3f8b5e12eb198dd28960cd8b725c315ab69733bf.zip | |
more formatting
| -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 b4a7e19..b1b82a5 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -67,7 +67,7 @@ $$\begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-im &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx \\ -&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx. \\ +&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx. \end{array} $$ Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$, |