Re: [微積] 幾個不簡單的微積分

看板Math作者ntust661 (XDeutesh)時間13年前 (2011/08/15 22:24)推噓4(4推 0噓 3→)

留言7則, 2人參與討論串2/3 (看更多)

y※ 引述《light0617 (EDWIN)》之銘言： : 有幾個問題想請教板上神人 : 1. : 假設V(x)=(x^2-1)^n : (1-x^2)V'(x)+2nxV(x)=0 : 左右兩邊微分n+1次 : (n+2) (n+1) (n) : (1-x^2)V (x)-2xV (x)+n(n+1)V (x)=0 : (n) : n(n+1)V (x)這一項是怎麼出來的一直為不出來@@ : 萊不妮茲微分 [n+1] ((1-x^2)V'(x)) = ... 2 [n+1] n+1 2 [n] n+1 2 [n] 2 [n+1] (1-x ) V'(x) + C (1-x ) V''(x) ... C ((1-x )') V' + (1-x)V' 1 n n+1 [n] n+1 [n+1] 0 + 0 + ... + C (-2)V + C (-2x)V n-1 n --------------------------- 2 [n+2] + (1-x ) V --------------------(1) ┌───────────────────────┐ │ [n] [n+1] 2 [n+2] │ │ -(n+1)n V -2x(n+1) V + (1-x ) V ├──→ (1) └───────────────────────┘ (n+1)! n+1 1 => ──── = C = ── (n+1)n 2!(n-1)! n-1 2 (n+1)! n+1 => ──── = C = (n+1) 1!(n!) n [n+1] (2nxV(x)) = ... 同上 n+1 [n] [n+1] [n] [n+1] C 2n V + 2nxV = (2n)(n+1) V + 2nx V n ------------------------------(2) 再相加就知道嚕^^ 2.Pn(x)=Σ_(k=0)^[n/2]-1)^k ] *2^(-n)((2n-2k)?n)(n?k)x^(n-2k) : PS:此為legendre方程式的解 : WHY Pn'(-x)=(-1)^(n+1)Pn'(x)??? : 解答是先算Pn'(x)再把-x代入變成Pn'(-x) : 但我是從Pn(-x)去微分結果是(-1)^n 答案就不一樣了 : ? ? 一直不解.... : 請各問大大幫小弟解答~ -- -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.122.217.41

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Frobenius

08/15 23:27, , 1^F

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light0617

08/16 00:21, , 2^F

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light0617

08/16 00:21, , 3^F

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