📝 题目
例 5 将函数 $f\left( x\right) = \arctan \frac{2x}{1 - {x}^{2}}$ 在 $x = 0$ 点展开为幂级数.
💡 答案与解析
解法 $1{f}^{\prime }\left( x\right) = \frac{2}{1 + {x}^{2}} = 2\mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}{x}^{2n}\left( {\left| x\right| < 1}\right) ,f\left( 0\right) = 0$ , 因此
$$ f\left( x\right) = {\int }_{0}^{x}{f}^{\prime }\left( t\right) \mathrm{d}t = 2\mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}{\int }_{0}^{x}{t}^{2n}\mathrm{\;d}t $$
$$ = 2\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{\left( -1\right) }^{n}}{{2n} + 1}{x}^{{2n} + 1}\;\left( {\left| x\right| < 1}\right) . $$
解法 2 令 $t = \arctan x$ ,则当 $\left| x\right| < 1$ 时, $\left| t\right| < \pi /4$ ,于是
$$ f\left( x\right) = \arctan \frac{2\tan t}{1 - {\tan }^{2}t} = \arctan \left( {\tan {2t}}\right) $$
$$ = {2t} = 2\arctan x\;\left( {\left| x\right| < 1}\right) . $$
利用 $\arctan x$ 在 $x = 0$ 点展开的幂级数,即得
$$ f\left( x\right) = 2\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{\left( -1\right) }^{n}}{{2n} + 1}{x}^{{2n} + 1}\;\left( {\left| x\right| < 1}\right) . $$