[工數] Fourier series (O'Neil , 14.3 , P.6)
Fourier series (O'Neil , 14.3 , P.6)
f(x) = sin (2x) , -π≦ x ≦ π
Ans. f(x) = sin (2x) is its own fourier series
我的檢驗過程
f(x) is odd function ∴a0、an = 0
1 π
bn = ---- ∫ sin(2x) sin(nπx) dx
π -π
1 π
= ----∫ cos(2-nπ)x - cos(2+nπ)x dx
π 0
1 sin(2-nπ)x sin(2+nπ)x π
= ---- [ -------------- - -------------- ]
π 2-nπ 2+nπ 0
1 sin(2π-nπ^2) sin(2π+nπ^2)
= ---- [ ---------------- - ---------------- ]
π 2-nπ 2+nπ
-2sin (nπ^2) - nπsin(nπ^2) - 2sin (nπ^2) + nπsin(nπ^2)
= ---------------------------------------------------------------
π[ 4 - (nπ)^2]
4sin (nπ^2)
= ------------------
π[ 4 - (nπ)^2]
∞ sin (nπ^2)
∴ f = 4 Σ ------------------- sin(nπx) = sin(2x)
n=1 π[ (nπ)^2 - 4 ]
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※ 編輯: zirconium 來自: 220.141.205.181 (02/10 20:00)
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