已知函数 ${f_0}\left(x\right) = \dfrac{\sin x}{x}$ $\left(x > 0\right)$，设 ${f_n}\left(x\right)$ 为 ${f_{n - 1}}\left(x\right)$ 的导数，$n \in {{\mathbb{N}}^ * }$．
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【出处】
2014年高考江苏卷
【标注】
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求 $2{f_1}\left( {\dfrac{{\mathrm \pi} }{2}} \right) + \dfrac{{\mathrm \pi} }{2}{f_2}\left( {\dfrac{{\mathrm \pi} }{2}} \right)$ 的值；标注答案$-1$解析由已知，并根据导数的运算可得\[{f_1}\left(x\right) = {f'_0}\left(x\right) = {\left( {\dfrac{\sin x}{x}} \right)^\prime } = \dfrac{\cos x}{x} - \dfrac{\sin x}{x^2},\]于是\[\begin{split}{f_2}\left(x\right) = {f'_1}\left(x\right) &= {\left( {\dfrac{\cos x}{x}} \right)^\prime } - {\left( {\dfrac{\sin x}{x^2}} \right)^\prime } \\&= - \dfrac{\sin x}{x} - \dfrac{2\cos x}{x^2} + \dfrac{2\sin x}{x^3},\end{split}\]所以\[{f_1}\left(\dfrac{{\mathrm{\mathrm \pi} } }{2}\right) = - \dfrac{4}{{{{\mathrm{\mathrm \pi} } ^2}}},{f_2}\left(\dfrac{{\mathrm{\mathrm \pi} } }{2}\right) = - \dfrac{2}{{\mathrm{\mathrm \pi} } } + \dfrac{16}{{{{\mathrm{\mathrm \pi} } ^3}}},\]故\[2{f_1}\left(\dfrac{{\mathrm{\mathrm \pi} } }{2}\right) + \dfrac{{\mathrm{\mathrm \pi} } }{2}{f_2}\left(\dfrac{{\mathrm{\mathrm \pi} } }{2}\right) = - 1.\]
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证明：对任意的 $n \in {{\mathbb{N}}^ * }$，等式 $\left| {n{f_{n - 1}}\left( {\dfrac{{\mathrm \pi} }{4}} \right) + \dfrac{\mathrm \pi} {4}{f_n}\left( {\dfrac{{\mathrm \pi} }{4}} \right)} \right| = \dfrac{\sqrt 2 }{2}$ 都成立．标注答案略解析由已知，得 $x{f_0}\left(x\right) = \sin x $，等式两边分别对 $ x $ 求导，得\[{f_0}\left(x\right) + x{f'_0}\left(x\right) = \cos x,\]即\[{f_0}\left(x\right) + x{f_1}\left(x\right) = \cos x = \sin \left(x + \dfrac{{\mathrm{\mathrm \pi} } }{2}\right),\]类似可得\[2{f_1}\left(x\right) + x{f_2}\left(x\right) = - \sin x = \sin \left(x + {\mathrm{\mathrm \pi} } \right),\]\[3{f_2}\left(x\right) + x{f_3}\left(x\right) = - \cos x = \sin \left(x + \dfrac{{3{\mathrm{\mathrm \pi} } }}{2}\right),\]\[4{f_3}\left(x\right) + x{f_4}\left(x\right) = \sin x = \sin \left(x + 2{\mathrm{\mathrm \pi} } \right).\]下面用数学归纳法证明等式 $n{f_{n - 1}}\left(x\right) + x{f_n}\left(x\right) = \sin \left(x + \dfrac{{n{\mathrm{\mathrm \pi} } }}{2}\right)$ 对所有的 $n \in {{\mathbb{N}}^*}$ 都成立．
（i）当 $ n=1 $ 时，由上可知等式成立．
（ii）假设当 $ n=k $ 时等式成立，即\[k{f_{k - 1}}\left(x\right) + x{f_k}\left(x\right) = \sin \left(x + \dfrac{{k{\mathrm{\mathrm \pi} } }}{2}\right).\]根据导数的运算可得\[\begin{split}&\left[k{f_{k - 1}}\left(x\right) + x{f_k}\left(x\right)\right]' \\=& k{f'_{k - 1}}\left(x\right) + {f_k}\left(x\right) + x{f'_k}\left(x\right) \\=& \left(k + 1\right){f_k}\left(x\right) +x {f_{k + 1}}\left(x\right),\end{split}\]\[\begin{split}&\left[\sin \left(x + \dfrac{{k{\mathrm{\mathrm \pi} } }}{2}\right)\right]' \\=& \cos \left(x + \dfrac{{k{\mathrm{\mathrm \pi} } }}{2}\right) \cdot \left(x + \dfrac{{k{\mathrm{\mathrm \pi} } }}{2}\right)' \\=& \sin \left[x + \dfrac{{\left(k + 1\right){\mathrm{\mathrm \pi} } }}{2}\right],\end{split}\]所以\[\left(k + 1\right){f_k}\left(x\right) + x{f_{k + 1}}\left(x\right) = \sin \left[x + \dfrac{{\left(k + 1\right){\mathrm{\mathrm \pi} } }}{2}\right] .\]所以当 $ n=k+1 $ 时，等式也成立．
综合（i），（ii）可知等式 $n{f_{n - 1}}\left(x\right) + x{f_n}\left(x\right) = \sin \left(x + \dfrac{{n{\mathrm{\mathrm \pi} } }}{2}\right)$ 对所有的 $n \in {{\mathbb{N}}^*}$ 都成立．
令 $x = \dfrac{{\mathrm{\mathrm \pi} } }{4}$，可得\[n{f_{n - 1}}\left(\dfrac{{\mathrm{\mathrm \pi} } }{4}\right) + \dfrac{{\mathrm{\mathrm \pi} } }{4}{f_n}\left(\dfrac{{\mathrm{\mathrm \pi} } }{4}\right) = \sin \left(\dfrac{{\mathrm{\mathrm \pi} } }{4} + \dfrac{{n{\mathrm{\mathrm \pi} } }}{2}\right) \left(n \in {{\mathbb{N}}^*} \right) .\]因为 $ \sin \left(\dfrac{{\mathrm{\mathrm \pi} } }{4} + \dfrac{{n{\mathrm{\mathrm \pi} } }}{2}\right) =\pm \dfrac {\sqrt 2}2\left(n \in {{\mathbb{N}}^*} \right) $，所以\[\left| {n{f_{n - 1}}\left(\dfrac{{\mathrm{\mathrm \pi} } }{4}\right) + \dfrac{{\mathrm{\mathrm \pi} } }{4}{f_n}\left(\dfrac{{\mathrm{\mathrm \pi} } }{4}\right)} \right| = \dfrac{\sqrt 2 }{2} \left(n \in {{\mathbb{N}}^*} \right) .\]
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