(2022年中科大创新班)设
$$
f_n\left( x \right) =1-x+\frac{x^2}{2!}+\cdots +\frac{\left( -x \right) ^n}{n!},n\in \mathbb{N}^\ast,x\in \mathbb{R}.
$$
(1)证明:方程$f_{2n-1}(x)=0$有唯一实数解,记作$a_n$.
(2)数列$\{a_n\}$是否为单调数列?请证明你的结论.
解法一. (1)当$x\leqslant 0$时,显然$f_n(x)>0$,故只需考虑$x>0$的情况即可.
令
$$
g_{2n-1}\left( x \right) =e^{x}f_{2n-1}\left( x \right) =e^{x}\left( 1-x+\frac{x^2}{2!}+\cdots -\frac{x^{2n-1}}{\left( 2n-1 \right) !} \right),
$$
则
$$
g'_{2n-1}\left( x \right) =-e^x\frac{x^{2n-1}}{\left( 2n-1 \right) !}.
$$
当$x>0$时,
$$
g'_{2n-1}\left( x \right) =-e^x\frac{x^{2n-1}}{\left( 2n-1 \right) !}<0.
$$
故$g_{2n-1}\left( x \right)$在$(0,+\infty)$上单调递减.
又$g_{2n-1}\left( 0 \right)=1>0$,而
$$
g_{2n-1}\left( x \right) =e^x\sum_{k=1}^n{\left[ \frac{x^{2k-2}}{\left( 2k-2 \right) !}-\frac{x^{2k-1}}{\left( 2k-1 \right) !} \right]}=e^x\sum_{k=1}^n{\frac{x^{2k-2}}{\left( 2k-1 \right) !}\left( 2k-1-x \right)},
$$
故
$$
g_{2n-1}\left( 2n-1 \right) =e^x\sum_{k=1}^n{\frac{\left( 2n-1 \right) ^{2k-2}}{\left( 2k-1 \right) !}\left( 2k-2n \right)}<0.
$$
由零点存在定理可知方程$g_{2n-1}(x)=0$在$(0,2n-1)$上有唯一正数解$a_n$.即$f_{2n-1}(x)=0$在$(0,2n-1)$上有唯一正数解$a_n$.
(2)注意到
$$
g_{2n+1}\left( x \right) =g_{2n-1}\left( x \right) +e^x\left( \frac{x^{2n}}{\left( 2n \right) !}-\frac{x^{2n+1}}{\left( 2n+1 \right) !} \right) ,
$$
则
$$\begin{aligned}
g_{2n+1}\left( a_n \right) &=g_{2n-1}\left( a_n \right) +e^x\left( \frac{a_{n}^{2n}}{\left( 2n \right) !}-\frac{a_{n}^{2n+1}}{\left( 2n+1 \right) !} \right)\\
&=e^x\left( \frac{a_{n}^{2n}}{\left( 2n \right) !}-\frac{a_{n}^{2n+1}}{\left( 2n+1 \right) !} \right) =e^x\left( \frac{a_{n}^{2n}}{\left( 2n+1 \right) !}\left( 2n+1-a_n \right) \right)\\
&>0=g_{2n+1}\left( a_{n+1} \right) ,\\
\end{aligned}$$
再由函数$g_{2n+1}\left( x\right)$单调递减可知$a_n< a_{n+1}$,故数列$\{a_n\}$单调递增.
解法二. (1)当$x\leqslant 0$时,显然$f_n(x)>0$,故只需考虑$x>0$的情况即可.
由于
$$
f_{2n-1}\left( x \right) =1-x+\frac{x^2}{2!}+\cdots -\frac{x^{2n-1}}{\left( 2n-1 \right) !},
$$
则
$$
f'_{2n-1}\left( x \right) =-1+x-\frac{x^2}{2!}+\cdots -\frac{x^{2n-2}}{\left( 2n-2 \right) !}=-f_{2n-2}\left( x \right).
$$
同理可得$f'_{2n}(x)=-f'_{2n-1}$.
下面用数学归纳法证明:函数$f_{2n-1}(x)$单调递减,有唯一正零点$a_n\in (0,2n-1)$,而$f_{2n-2}\left( x \right)>0$.
(a)当$n=1$时,函数$f_1(x)=1-x$单调递减,有唯一正零点$x=a_1=1$,由$f'_2(x)=-f_1(x)$可知$f_2(x)$在$(-\infty,1]$上递减,在$(1,+\infty)$上递增,故$\displaystyle f_2(x)\geqslant f_2(1)=\frac{1}{2}>0$.
(b)当$n=k$时,假设函数$f_{2k-1}(x)$单调递减,有唯一正零点$a_ k\in (0,2k-1)$,而$f_{2k-2}\left( x \right)>0$.
由$f'_{2k}(x)=-f_{2k-1}\left( x \right)$可知$f_{2k}(x)$在$(-\infty,a_k)$上单调递减,在$[a_k,+\infty)$上单调递增,故
$$
f_{2k}(x)\geqslant f_{2k}\left( a_k \right) =f_{2k-1}\left( a_k \right) +\frac{a_{k}^{2k}}{\left( 2k \right) !}=\frac{a_{k}^{2k}}{\left( 2k \right) !}>0.
$$
再由$f'_{2k+1}(x)=-f_{2k}\left( x \right)$可知$f_{2k+1}(x)$在$\mathbb{R}$上单调递减.
而$f_{2k+1}\left( 0 \right) =1>0$,再由
$$
f_{2k+1}\left( x \right) =\sum_{m=0}^k{\left[ \frac{x^{2m}}{\left( 2m \right) !}-\frac{x^{2m+1}}{\left( 2m+1 \right) !} \right]}=\sum_{m=0}^k{\frac{x^{2m}}{\left( 2m+1 \right) !}\left( 2m+1-x \right)}
$$
可知
$$
f_{2k+1}\left( 2k+1 \right) =\sum_{m=0}^k{\frac{\left( 2k+1 \right) ^{2m}}{\left( 2m+1 \right) !}\left( 2m-2k \right)}<0.
$$
由零点存在定理可知方程$f_{2k+1}(x)=0$在$(0,2k+1)$上有唯一正零点$a_{k+1}$.
故$n=k+1$时结论也成立.
综上所述,对任意正整数$n$,函数$f_{2n-1}(x)$单调递减,有唯一正零点$a_n\in (0,2n-1)$,而$f_{2n-2}\left( x \right)>0$.即方程$f_{2n-1}(x)=0$有唯一实数解.
(2)注意到
$$
f_{2n+1}\left( x \right) =f_{2n-1}\left( x \right) +\frac{x^{2n}}{\left( 2n \right) !}-\frac{x^{2n+1}}{\left( 2n+1 \right) !},
$$
则
$$\begin{aligned}
f_{2n+1}\left( a_n \right) &=f_{2n-1}\left( a_n \right) +\frac{a_{n}^{2n}}{\left( 2n \right) !}-\frac{a_{n}^{2n+1}}{\left( 2n+1 \right) !}
\\
&=\frac{a_{n}^{2n}}{\left( 2n \right) !}-\frac{a_{n}^{2n+1}}{\left( 2n+1 \right) !}=\frac{a_{n}^{2n}}{\left( 2n+1 \right) !}\left( 2n+1-a_n \right)
\\
&>0=f_{2n+1}\left( a_{n+1} \right),
\end{aligned}$$
再由函数$f_{2n+1}\left( x\right)$单调递减可知$a_n< a_{n+1}$,故数列$\{a_n\}$单调递增.