Re: [其他] fourier series
※ 引述《a2626 (Lucas)》之銘言:
: 想請問一下,x^2的an我算到最後會
: 是4/(n^2pi^2)*cosnpix然後積分的
: 上下限是-pi到pi,怎麼帶進去都是0
: 阿@@!一直沒辦法算出4/(n^2pi^2)(-1)^n
: 手機發文排版很亂,不好意思
∫ x^2 * cos(nx) dx
= x^2 sin(nx)/n - ∫2 x sin(nx)/n dx
= x^2 sin(nx)/n + 2 x cos(nx)/n^2 - ∫2 cos(nx)/n^2 dx
下限-π 上限π
= 4 π (-1)^n / n^2
n > 0:
a_n = (1/π)∫ x^2 * cos(nx) dx
= 4 (-1)^n / n^2
n = 0:
π
a_0 = (1/(2π))∫ x^2 dx = [1/(6π)]2π^3 = π^2 / 3
-π
--
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※ 編輯: Honor1984 (61.228.130.168), 04/17/2014 21:17:25
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