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author | Opheliar99 <> | 2010-07-04 04:38:05 +0000 |
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committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 04:38:05 +0000 |
commit | 4cb4aca40267f73499ad6669620381b44d68ebf3 (patch) | |
tree | 7b1e816bb71c5909e3f991255e21dce393cf39bc | |
parent | d406b157e8e83ef13288d6bce42e1ecf6f1f4cb6 (diff) | |
download | afterklein-wiki-4cb4aca40267f73499ad6669620381b44d68ebf3.tar.gz afterklein-wiki-4cb4aca40267f73499ad6669620381b44d68ebf3.zip |
posted solutions of 2 and 3 in pset2
-rw-r--r-- | Problem Set 2.page | 12 |
1 files changed, 12 insertions, 0 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 63e60ad..6eb1452 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -62,6 +62,18 @@ $\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times < f_0, f >$ $= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$ +3. Since +$\sin x = \frac{e^{ix}-e^{-ix}}{2}$, + +$\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$ + +$a_m = \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx$ + +$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx$ + +$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx$. + + ## Cardinality Cardinality of the natural numbers (countable): $\mathbf{N}$,$\mathbf{Z}$ |