📝 题目
例 4 求函数 $f\left( x\right) = {\left( 1 + x\right) }^{a}$ 的麦克劳林公式.
💡 答案与解析
解 计算导数得
$$ {f}^{\left( k\right) }\left( x\right) = \alpha \left( {\alpha - 1}\right) \cdots \left( {\alpha - k + 1}\right) {\left( 1 + x\right) }^{\alpha - k}, $$
$$ {f}^{\left( k\right) }\left( 0\right) = \alpha \left( {\alpha - 1}\right) \cdots \left( {\alpha - k + 1}\right) \;\left( {k = 1,2,\cdots }\right) . $$
于是
$$ {\left( 1 + x\right) }^{\alpha } = 1 + {\alpha x} + \frac{\alpha \left( {\alpha - 1}\right) }{2}{x}^{2} + \cdots $$
$$ + \frac{\alpha \left( {\alpha - 1}\right) \cdots \left( {\alpha - n + 1}\right) }{n!}{x}^{n} + o\left( {x}^{n}\right) . $$
我们引入记号
$$ \left( \begin{array}{l} \alpha \\ k \end{array}\right) = \frac{\alpha \left( {\alpha - 1}\right) \cdots \left( {\alpha - k + 1}\right) }{k!}. $$
用这记号可以把 ${\left( 1 + x\right) }^{a}$ 的麦克劳林公式写成更紧凑的形式:
$$ {\left( 1 + x\right) }^{a} = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} \alpha \\ k \end{array}\right) {x}^{k} + o\left( {x}^{n}\right) . $$
为了求函数 $\arctan x$ 和 $\arcsin x$ 的麦克劳林公式,我们将要用到以下引理.
引理 3 设函数 $f\left( x\right)$ 在 $U\left( {0,\eta }\right)$ 有定义,在 0 点 $n$ 次可导,如果
$$ {f}^{\prime }\left( x\right) = {A}_{0}^{\prime } + {A}_{1}^{\prime }x + \cdots + {A}_{n - 1}^{\prime }{x}^{n - 1} + o\left( {x}^{n - 1}\right) $$
$$ \left( {{A}_{0}^{\prime },{A}_{1}^{\prime },\cdots ,{A}_{n - 1}^{\prime } \in \mathbb{R}}\right) , $$
那么
$$ f\left( x\right) = f\left( 0\right) + {A}_{0}^{\prime }x + \frac{{A}_{1}^{\prime }}{2}{x}^{2} + \cdots + \frac{{A}_{n - 1}^{\prime }}{n}{x}^{n} + o\left( {x}^{n}\right) . $$
证明 在所给的条件下, 应该有
$$ {f}^{\prime }\left( x\right) = {f}^{\prime }\left( 0\right) + {f}^{\prime \prime }\left( 0\right) x + \frac{{f}^{\prime \prime \prime }\left( 0\right) }{2!}{x}^{2} + \cdots $$
$$ + \frac{{f}^{\left( n\right) }\left( 0\right) }{\left( {n - 1}\right) !}{x}^{n - 1} + o\left( {x}^{n - 1}\right) . $$
因为函数 ${f}^{\prime }\left( x\right)$ 的麦克劳林公式是唯一的,所以必须有
$$ {f}^{\prime }\left( 0\right) = {A}_{0}^{\prime },{f}^{\prime \prime }\left( 0\right) = {A}_{1}^{\prime },{f}^{\prime \prime \prime }\left( 0\right) = 2{A}_{2}^{\prime },\cdots $$
$$ {f}^{\left( n\right) }\left( 0\right) = \left( {n - 1}\right) !{A}_{n - 1}^{\prime }. $$
于是, 根据定理 1 , 我们得到
$$ f\left( x\right) = f\left( 0\right) + {f}^{\prime }\left( 0\right) x + \frac{{f}^{\prime \prime }\left( 0\right) }{2}{x}^{2} + \cdots + \frac{{f}^{\left( n\right) }\left( 0\right) }{n!}{x}^{n} + o\left( {x}^{n}\right) $$
$$ = f\left( 0\right) + {A}_{0}^{\prime }x + \frac{{A}_{1}^{\prime }}{2}{x}^{2} + \cdots + \frac{{A}_{n - 1}^{\prime }}{n}{x}^{n} + o\left( {x}^{n}\right) . $$