※ 引述《sainteyes (立其)》之銘言：
: Qm(x)={(x^2-1)^m}^(m)->指微分m次 (m=0,1,2,3.....)
: ^^^
: Prove:
: Qm"(x)(1-x^2)+Qm'(x)(-2x)+m(m+1)Qm(x)=0
: The anwer as following
: {(1-x^2)(x^2-1)^m}^(m+2) + {(x^2-1)^m+1}^(m+2)=0
: ^^^^^^^^^^^^^^^^ ^^^^^^^^^^
: ()內相乘後一負一正
: Qm(x)={(x^2-1)^m}^(m) => Qm'(x)={(x^2-1)^m}^(m+1) Qm"(x)=......^(m+2)-----(*)
: 上式 =>
: (1-x^2){(x^2-1)^m}^(m+2)+{C(m+2) take 1}(-2x){(x^2-1)^m}^(m+1)+{C(m+2) take 2}*
: ^^^^^^^^^^^^^
: 表組合C X 取 Y
: (-2){(x^2-1)^m}^(m+2)+{(m+1)(2x)(x^2-1)^m}^(m+1)=0
: From (*) (1-x^2)Qm"(x)+m+2(-2x)Qm'(x)-(m^2+3m+2)Qm(x)+(m+1)Qm'(x)(2x)+Qm(x)(2m
: ^2+4m+2)=0
: 化簡
: => Qm"(x)(1-x^2)+Qm'(x)(-2x)+m(m+1)Qm(x)=0------(proved)
: It is a very interesting problem with concise anwer,right?
mine:
令U=(x^2-1)^m ~~~>Qm(x)=U^(m) U'=m[(x^2-1)^m-1]2x 左右同乘x^2-1還原
=> (x^2-1)U'=2mxU 左右各微m+1次
(x^2-1)U^(m+2)+(C m+1 take 1)(2x)U^(m+1)+(C m+1 take 2)2U^(m)
=2m{xU^(m+1)+(C m+1 take 1)U^(m)} 移向整理(至右邊)
(1-x^2)Qm''(x)-2xQm'(x)+m(m+1)Qm(x)=0 proved ^^~
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