📝 题目
例 2 求函数 $f\left( x\right) = \sin x$ 和 $g\left( x\right) = \cos x$ 的麦克劳林公式.
💡 答案与解析
解 我们有
$$ {f}^{\left( k\right) }\left( x\right) = \sin \left( {x + \frac{k\pi }{2}}\right) ,\;{f}^{\left( 2k\right) }\left( 0\right) = 0, $$
$$ {f}^{\left( 2k + 1\right) }\left( 0\right) = {\left( -1\right) }^{k},\;k = 0,1,2,\cdots ; $$
$$ {g}^{\left( k\right) }\left( x\right) = \cos \left( {x + \frac{k\pi }{2}}\right) ,\;{g}^{\left( 2k\right) }\left( 0\right) = {\left( -1\right) }^{k}, $$
$$ {g}^{\left( 2k + 1\right) }\left( 0\right) = 0,\;k = 0,1,2,\cdots . $$
于是
$$ \sin x = x - \frac{{x}^{3}}{3!} + \frac{{x}^{5}}{5!} - \cdots + {\left( -1\right) }^{n - 1}\frac{{x}^{{2n} - 1}}{\left( {{2n} - 1}\right) !} + o\left( {x}^{2n}\right) , $$
$$ \cos x = 1 - \frac{{x}^{2}}{2!} + \frac{{x}^{4}}{4!} - \cdots + {\left( -1\right) }^{n}\frac{{x}^{2n}}{\left( {2n}\right) !} + o\left( {x}^{{2n} + 1}\right) . $$