📝 题目
例 13 设 ${P}_{n}\left( x\right) = \frac{1}{{2}^{n}n!}\frac{{\mathrm{d}}^{n}}{\mathrm{\;d}{x}^{n}}{\left( {x}^{2} - 1\right) }^{n}$ . 求证:
(1) ${P}_{n}\left( x\right)$ 的最高次项系数为 $\frac{\left( {2n}\right) !}{{2}^{n}{\left( n!\right) }^{2}}$ ;
(2) ${P}_{n}\left( 1\right) = 1,{P}_{n}\left( {-1}\right) = {\left( -1\right) }^{n}$ ;
(3) $\left( {{x}^{2} - 1}\right) {P}_{n}^{\prime \prime }\left( x\right) + {2x}{P}_{n}^{\prime }\left( x\right) - n\left( {n + 1}\right) {P}_{n}\left( x\right) = 0$ .
💡 答案与解析
证 (1) 因为 ${\left( {x}^{2} - 1\right) }^{n}$ 的最高次项为 ${2n}$ 次,所以 $\frac{{\mathrm{d}}^{n}}{\mathrm{\;d}{x}^{n}}{\left( {x}^{2} - 1\right) }^{n}$ 的最高次项系数为
$$ {2n}\left( {{2n} - 1}\right) \cdots \left( {n + 1}\right) = \frac{\left( {2n}\right) !}{n!}, $$
所以 ${P}_{n}\left( x\right)$ 的最高次项系数为 $\frac{\left( {2n}\right) !}{{2}^{n}{\left( n!\right) }^{2}}$ .
(2)按莱布尼茨公式:
$$ {P}_{n}\left( x\right) = \frac{1}{{2}^{n}n!}{\left\lbrack {\left( x - 1\right) }^{n}{\left( x + 1\right) }^{n}\right\rbrack }^{\left( n\right) } $$
$$ = \frac{1}{{2}^{n}n!}\left\{ {{\left( x - 1\right) }^{n}{\left\lbrack {\left( x + 1\right) }^{n}\right\rbrack }^{\left( n\right) } + {C}_{n}^{1}{\left\lbrack {\left( x - 1\right) }^{n}\right\rbrack }^{\prime }{\left\lbrack {\left( x + 1\right) }^{n}\right\rbrack }^{\left( n - 1\right) }}\right. $$
$$ + \cdots + {\left\lbrack {\left( x - 1\right) }^{n}\right\rbrack }^{\left( n\right) }{\left( x + 1\right) }^{n}\} $$
$$ = \frac{1}{{2}^{n}n!}\left\{ {n!{\left( x - 1\right) }^{n} + \cdots + n!{\left( x + 1\right) }^{n}}\right\} . $$
括弧中除首末两项外其余项均含有 $\left( {x + 1}\right) \left( {x - 1}\right)$ ,所以
$$ {P}_{n}\left( 1\right) = 1,\;{P}_{n}\left( {-1}\right) = {\left( -1\right) }^{n}. $$
(3) 令 $y = {\left( {x}^{2} - 1\right) }^{n}$ ,则有
$$ {y}^{\prime } = {2nx}{\left( {x}^{2} - 1\right) }^{n - 1} \Rightarrow \left( {{x}^{2} - 1}\right) {y}^{\prime } = {2nxy}. $$
在上式两端对 $x$ 求导 $n + 1$ 次,得
$$ \left( {{x}^{2} - 1}\right) {y}^{\left( n + 2\right) } + 2\left( {n + 1}\right) x{y}^{\left( n + 1\right) } + n\left( {n + 1}\right) {y}^{\left( n\right) } $$
$$ = {2nx}{y}^{\left( n + 1\right) } + {2n}\left( {n + 1}\right) {y}^{\left( n\right) }, $$
即
$$ \left( {{x}^{2} - 1}\right) {y}^{\left( n + 2\right) } + {2x}{y}^{\left( n + 1\right) } - n\left( {n + 1}\right) {y}^{\left( n\right) } = 0. $$
注意到 ${y}^{\left( n\right) } = {2}^{n} \cdot n!{P}_{n}\left( x\right)$ ,即得
$$ \left( {{x}^{2} - 1}\right) {P}_{n}^{\prime \prime }\left( x\right) + {2x}{P}_{n}^{\prime }\left( x\right) - n\left( {n + 1}\right) {P}_{n}\left( x\right) = 0. $$