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討論串[中學] 二階遞回式
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上式(1)A1 = A0 = 1 (筆誤). 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
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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
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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) =
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