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| author | luccul <luccul@gmail.com> | 2010-07-04 19:53:39 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 19:53:39 +0000 |
| commit | 7e25667dea286052527e96bfc43f12aa08f7fd3b (patch) | |
| tree | f79d30e98a89faa314b9fc8f68bb01bb41a892e9 | |
| parent | 1db3e56bce88b7d8e99cd18a69a714c817bd5e7d (diff) | |
| download | afterklein-wiki-7e25667dea286052527e96bfc43f12aa08f7fd3b.tar.gz afterklein-wiki-7e25667dea286052527e96bfc43f12aa08f7fd3b.zip | |
more formatting/editing
| -rw-r--r-- | Problem Set 2.page | 27 |
1 files changed, 11 insertions, 16 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 5ba4b65..b4a7e19 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -61,34 +61,29 @@ $$ \begin{array}{ccl} \int_0^{2\pi} \sin^4(x) dx = <1, f> &=& \sqrt{2\pi} \times &=& \frac{3 \pi}{4} \end{array} $$ -3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, +3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, $\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$, and therefore, -$\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$ +$$\begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx \\ -$a_m = \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx$ - -$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx$ - -$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx$. +&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx \\ +&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx. \\ Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$, -$m = 2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$, - -$m = 0 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \sqrt{2\pi}/2$, +$$m = 2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4,$$ -$m = -2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$, +$$m = 0 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \frac{\sqrt{2\pi}{2},$$ -Then, +$$m = -2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \frac{\sqrt{2\pi}}{4},$$ -$\sum |a_n|^2 = {(\sqrt{2\pi}/4)}^2 + {(- \sqrt{2\pi}/2)}^2 + {(\sqrt{2\pi}/4)}^2 = \frac{3 \pi}{4}$. +It follows that -And, it was shown in Prob 2 that $\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$. +$$\sum |a_n|^2 = {(\sqrt{2\pi}/4)}^2 + {(- \sqrt{2\pi}/2)}^2 + {(\sqrt{2\pi}/4)}^2 = \frac{3 \pi}{4}$$. -Therefore, +It was shown in Prob 2 that $\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$, so therefore, -$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$ +$$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$$ |