方企勤 第二章 一元函数微分学 第12题
📝 题目
例 12 设 $y = {\left( \arcsin x\right) }^{2}$ .
(1) 求证: $\left( {1 - {x}^{2}}\right) {y}^{\prime \prime } - x{y}^{\prime } = 2$ ; (2) 求 ${y}^{\left( n\right) }\left( 0\right)$ .
💡 答案解析
证 $\left( 1\right) {y}^{\prime } = 2\frac{\arcsin x}{\sqrt{1 - {x}^{2}}} \Rightarrow \left( {1 - {x}^{2}}\right) {\left( {y}^{\prime }\right) }^{2} = {4y}$
$$ \overset{\text{ 对 }}{ \Rightarrow }2{y}^{\prime }{y}^{\prime \prime }\left( {1 - {x}^{2}}\right) - {2x}{\left( {y}^{\prime }\right) }^{2} = 4{y}^{\prime }, $$
化简即得 $\left( {1 - {x}^{2}}\right) {y}^{\prime \prime } - x{y}^{\prime } = 2$ .
解 (2) 显然 $y\left( 0\right) = 0$ ,由第 (1) 小题知 ${y}^{\prime }\left( 0\right) = 0,{y}^{\prime \prime }\left( 0\right) = 2$ . 为了求 ${y}^{\left( n\right) }\left( 0\right)$ ,我们对第 (1) 小题所证的方程,两边求 $n$ 阶导数,得
$$ \left( {1 - {x}^{2}}\right) {y}^{\left( n + 2\right) } - {2nx}{y}^{\left( n + 1\right) } - n\left( {n - 1}\right) {y}^{\left( n\right) } - x{y}^{\left( n + 1\right) } - n{y}^{\left( n\right) } = 0, $$
化简得
$$ \left( {1 - {x}^{2}}\right) {y}^{\left( n + 2\right) } - \left( {{2n} + 1}\right) x{y}^{\left( n + 1\right) } - {n}^{2}{y}^{\left( n\right) } = 0\;\left( {n \geq 1}\right) . $$
由此,令 $x = 0$ ,得 ${y}^{\left( n + 2\right) }\left( 0\right) = {n}^{2}{y}^{\left( n\right) }\left( 0\right) \left( {n \geq 1}\right)$ . 这是 ${y}^{\left( n\right) }\left( 0\right)$ 的递推公式, 根据这个公式, 有
$$ {y}^{\prime }\left( 0\right) = 0 \Rightarrow {y}^{\left( 2n + 1\right) }\left( 0\right) = 0\;\left( {n = 0,1,2,\cdots }\right) ; $$
$$ {y}^{\prime \prime }\left( 0\right) = 2 \Rightarrow {y}^{\left( 2n\right) }\left( 0\right) = 2{\left\lbrack \left( 2n - 2\right) !!\right\rbrack }^{2}\;\left( {n = 1,2,\cdots }\right) . $$