※ 引述《cutt1efish (喵喵)》之銘言：
: ※ 引述《mack (腦海裡依然記得妳)》之銘言：
: : Ai = (2n-1)/n * Ai-1 - Ai-2 = 2Ai-1 - (1/n)*Ai-1 - Ai-2
: : => (Ai - Ai-1) = (Ai-1 - Ai-2) - (1/n)*Ai-1
: : => (A2 - A1) = (A1 - A0) - (1/n)*A1
: : (A3 - A2) = (A2 - A1) - (1/n)*A2
: : ... = ...
: : +) (Am - Am-1) = (Am-1 - Am-2) - (1/n)*Am-1
: : ---------------------------------------------
: : (Am - Am-1) = -(1/n)*(A1 + A2 + ... + Am-1)
: : A1 + A2 + ... + Am-1算不出來 有大大可以解的嗎
: : 解出來後面就都可以解了
: Step 1: 引用樓上的:
: Am - Am-1 = -(1/n)*(A1 + A2 + ... + Am-1) ... (*)
: Step 2: 公式解
: 特徵方程式: x^2 - 2(1-1/2n)x + 1 = 0
: 令 1 - 1/2n = t ( 0 < t < 1)
: 兩根 x = t ±√(1-t^2) j ( where j = √(-1) )
: = cos(θ) ±j sin(θ) ( where cos(θ) = t, sin(θ) = √(1-t^2) )
: ( assume 0 < θ < π/2 )
: Ai = C1 * [cos(θ) + j sin(θ)]^i + C2 * [cos(θ) - j sin(θ)]^i
上式(1)
: A1 = A0 = 0 代入化簡解得
A1 = A0 = 1 (筆誤)
: Ai = cos[(i-1/2)θ] / cos[θ/2]
A0 = 1 so C1 + C2 = 1
A1 = 1 so C1 * [cos(θ) + j sin(θ)] + C2 * [cos(θ) - j sin(θ)] = 1
or j ( C1 - C2 ) sinθ = 1 - cosθ
or j ( C1 - C2 ) = tan(θ/2)
由(1)
Ai = ( C1 + C2 ) cos(iθ) + j ( C1 - C2 ) sin(iθ)
= cos(iθ) + tan(θ/2) sin(iθ)
= [ cos(iθ) cos(θ/2) + sin(iθ) sin(θ/2) ] / cos(θ/2)
= cos[(i-1/2)θ] / cos(θ/2)
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