题目：
\(\text { Bernstein定理: } 设 f 在 (a, b) 内无穷阶可导, 且各阶导数均只取正值. \\证明: 对 \forall x_{0} \in(a, b), \exists r> 0, \mathrm{~s} . \mathrm{t}. 当 x \in\left[x_{0}-r, x_{0}+r\right] \subset(a, b) 时,有 \\ f(x)=\sum_{k=0}^{\infty} \frac{f^{(k)}\left(x_{0}\right)}{k !}\left(x-x_{0}\right)^{k} .\)
证明 ：
$\forall x_{0} \in(a, b)$, 取 \(0<\delta<\frac{1}{2} \min \left\{x_{0}-a, b-x_{0}\right\}\). 对 \(\forall \bar{x}_{0} \in\left(x_{0}-\delta, x_{0}+\delta\right)\), 在 \(\left[\bar{x}_{0}-\delta\right.,\left.\bar{x}_{0}+\delta\right]\) 中作有限 Taylor 展开, 有
\[f(x)=\sum_{k=0}^{n} \frac{f^{(k)}\left(\bar{x}_{0}\right)}{k !}\left(x-\bar{x}_{0}\right)^{k}+\frac{f^{(n+1)}(\bar{\xi})}{(n+1) !}\left(x-\bar{x}_{0}\right)^{n+1},\]其中 $\bar{\xi}$ 介于 \(\bar{x}_{0}\) 与 $x$ 之间, \(\xi \in(a, b)\) . 因 $f$ 在 \(\left[x_{0}-2 \delta, x_{0}+2 \delta\right]\) 中连续, 故有界. 设 \(|f(x)| \leqslant M, \forall x \in\left[x_{0}-2 \delta, x_{0}+2 \delta\right] .\) 取 $x=\bar{x}_{0}+\delta$, 则由 $f^{(j)}(x)>0(j \in \mathrm{N})$ 及上述 Taylor 展开得到
\[0<\frac{\frac{f^{(k)}\left(\bar{x}_{0}\right)}{k !} \delta^{t} \leqslant f\left(\bar{x}_{0}+\delta\right)-f\left(\bar{x}_{0}\right) \leqslant\left|f\left(\bar{x}_{0}+\delta\right)\right|+\left|f\left(\bar{x}_{0}\right)\right| \leqslant M+M=2 M,}{\frac{f^{(k)}\left(\bar{x}_{0}\right)}{k !} \leqslant \frac{2 M}{\delta^{k}}, \quad k \in \mathrm{N} .}\]于是, 再取 $0<r<\delta$, 则 $\left[x_{0}-r, x_{0}+r\right] \subset\left(x_{0}-\delta, x_{0}+\delta\right) \subset(a, b)$. 作 Taylor 展开有
\[f(x)=\sum_{k=0}^{n} \frac{f^{(k)}\left(x_{0}\right)}{k !}\left(x-x_{0}\right)^{n}+\frac{f^{(n+1)}(\xi)}{(n+1) !}\left(x-x_{0}\right)^{n+1}, \quad x \in\left[x_{0}-r, x_{0}+r\right],\]其中 $\xi$ 介于 $x_{0}$ 与 $x$ 之间, $\xi \in\left(x_{0}-r, x_{0}+r\right) \subset\left(x_{0}-\delta, x_{0}+\delta\right)$, 因此
\[\left|\frac{f^{(n+1)}(\xi)}{(n+1) !}\left(x-x_{0}\right)^{n+1}\right| \leqslant \frac{2 M}{\delta^{n+1}} r^{n+1}=2 M\left(\frac{r}{\delta}\right)^{n+1} \rightarrow 0 \quad(n \rightarrow+\infty) .\]由此, 当 $x \in\left[x_{0}-r, x_{0}+r\right]$ 时, 有
\[\begin{aligned} \sum_{k=0}^{\infty} \frac{f^{(k)}\left(x_{0}\right)}{k !}\left(x-x_{0}\right)^{k} &=\lim _{n \rightarrow+\infty} \sum_{k=0}^{n} \frac{f^{(k)}\left(x_{0}\right)}{k !}\left(x-x_{0}\right)^{k} \\&=\lim _{n \rightarrow+\infty}\left[f(x)-\frac{f^{(n \mid 1)}(\xi)}{(n+1) !}\left(x-x_{0}\right)^{n+1}\right] \\ &=f(x)-0=f(x) . \end{aligned}\]