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| -rw-r--r-- | Problem Set 2.page | 12 |
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diff --git a/Problem Set 2.page b/Problem Set 2.page index a26852a..523ce40 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -73,12 +73,24 @@ $= \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 $n \neq 0$, $m = 2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$, + $m = 0 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \sqrt{2\pi}/2$, + $m = -2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$, +Then, + +$\sum |a_n|^2 = {\sqrt{2\pi}/4}^2 + {- \sqrt{2\pi}/2}^2 + {\sqrt{2\pi}/4}^2 = \frac{3 \pi}{4}$ +And, it was shown in Prob 2 that $\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$. +Therefore, + +$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$ + + ## Cardinality |