Taylor公式

单变量函数的微分

张瑞
中国科学技术大学数学科学学院

Taylor公式

若$f(x)$在$x_0$处可导，则在$x_0$附近，有

\[f(x)=f(x_0)+f'(x_0)(x-x_0)+o(x-x_0),\quad x\to x_0 \]

即，在$x_0$附近，可以用1次多项式近似$f(x)$。

类似地，若$f(x)$在$x_0$附近有$n$阶导数，是否可以用$n$次多项式来近似$f(x)$？

记n次多项式为

\[p_n(x)=a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots+a_n(x-x_0)^n \]

则

\[p^{(k)}_n(x_0)=k! a_k \]

若$f^{(k)}(x_0)=p_n^{(k)}(x_0), k=0,1,\cdots,n$，则有$a_k=\frac{f^{(k)}(x_0)}{k!}$。

定义 1.
称

\[T_n(x)=f(x_0)+f'(x_0)(x-x_0)+\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \]

为函数$f(x)$在点$x_0$的n阶Taylor多项式。

Taylor定理

定理 1. (Lagrange余项)
 $f(x)$在区间$I$上有$n+1$阶导数，$x_0\in I$。则$\forall x\in I$，有

\[\begin{aligned} f(x)=f(x_0)&+f'(x_0)(x-x_0)+\dfrac{f''(x_0)}{2!}(x-x_0)^2+\cdots \\ &+\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x) \end{aligned} \]

其中$R_n(x)=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}$称为Lagrange型余项， $\xi$在$x$与$x_0$之间。

证： 记

\[P_n(x)=f(x_0)+f'(x_0)(x-x_0)+\cdots +\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n \]

则$P_n(x)$是一个$n$次多项式。记$R_n(x)=f(x)-P_n(x)$，则有

\[R_n(x_0)=R'_n(x_0)=\cdots=R^{(n)}_n(x_0)=0 \]

这样，

\[\begin{aligned} &\dfrac{R_n(x)}{(x-x_0)^{n+1}}=\dfrac{R_n(x)-R_n(x_0)}{(x-x_0)^{n+1}n-0^{n+1}}\\ =&\dfrac{R'_n(\xi_1)}{(n+1)(\xi_1-x_0)^n}=\dfrac{R'_n(\xi_1)-R'_n(x_0)}{(n+1)(\xi_1-x_0)^n-0} \\ =&\dfrac{R''_n(\xi_2)}{(n+1)n(\xi_2-x_0)^{n-1}}=\dfrac{R''_n(\xi_2)-R''_n(x_0)}{(n+1)n(\xi_2-x_0)^{n-1}-0} \\ \end{aligned} \]

以此类推，

\[\begin{aligned} \dfrac{R_n(x)}{(x-x_0)^{n+1}}=\dfrac{R^{(n+1)}_n(\xi_{n+1})}{(n+1)!} \end{aligned} \]

而

\[R^{(n+1)}_n(\xi_{n+1})=f^{(n+1)}(\xi_{n+1})-0 \]

所以

\[R_n(x)=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}, \xi\in(x_0,x) \]

Peano余项

定理 2. (Peano余项)
 $f(x)$在$x_0$有直到$n$阶的导数，在含$x_0$的区间$I$内有直到$n-1$阶的导数，则

\[\begin{aligned} f(x)=f(x_0)&+f'(x_0)(x-x_0)+\cdots \\ &+\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x) \\ \end{aligned} \]

其中，

\[R_n(x)=o((x-x_0)^n), x \to x_0 \]

称为Peano型余项。

证： $R_n(x_0)=0, R_n^{(k)}(x_0)=0, k=1,2,\cdots,n$，由L'Hostpital法则，

\[\begin{aligned} &\lim_{x\to x_0}\dfrac{R_n(x)}{(x-x_0)^n}=\lim_{x\to x_0}\dfrac{R'_n(x)}{n(x-x_0)^{n-1}} = \cdots \\ =&\lim_{x\to x_0}\dfrac{R^{(n-1)}_n(x)}{n(n-1)\cdots2(x-x_0)} \\ =&\lim_{x\to x_0}\dfrac{f^{(n-1)}(x)-f^{(n-1)}(x_0)-f^{(n)}(x_0)(x-x_0)}{n(n-1)\cdots2(x-x_0)} \\ =&\dfrac{1}{n!}\lim_{x\to x_0}\left(\dfrac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{(x-x_0)}-f^{(n)}(x_0) \right)\\ =&(f^{(n)}(x_0)-f^{(n)}(x_0))=0 \end{aligned} \]

Maclaurin公式

定义 2.
取$x_0=0$，称为Maclaurin公式。

\[\begin{aligned} f(x)=f(0) +& f'(0)x+\dfrac{f''(0)}{2!}x^2+ \cdots \\ +&\dfrac{f^{(n)}(0)}{n!}x^n +\dfrac{f^{(n+1)}(\theta x)} {(n+1)!} x^{n+1} , \theta\in(0,1) \end{aligned} \]

初等函数的Maclaurin展开

1. $e^x$

2. $\sin(x)$

3. $\cos(x)$

4. $(1+x)^\alpha$

5. $\ln(1+x)$

\[e^x=1+\dfrac{1}{1}x+\dfrac{1}{2!}x^2+\cdots+\dfrac{1}{n!}x^n+R^n(x) \]

其中

\[R_n(x)=\dfrac{x^{n+1}}{(n+1)!}e^{\theta x}, \theta\in(0,1) \]
\[\sin(x)=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+\cdots+(-1)^{m-1}\dfrac{1}{(2m-1)!}x^{2m-1}+R_{2m} \]

其中

\[\begin{aligned} R_{2m}=&\dfrac{1}{(2m+1)!}x^{2m+1}\sin(\theta x+\dfrac{2m+1}{2}\pi) \\ =& (-1)^m\dfrac{x^{2m+1}}{(2m+1)!}\cos(\theta x) , \theta\in(0,1) \end{aligned} \]
\[\cos(x)=1-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4-\dfrac{1}{6!}x^6+\cdots+(-1)^{m-1}\dfrac{1}{(2m-2)!}x^{2m-2}+R_{2m-1} \]

其中

\[R_{2m-1}=(-1)^m\dfrac{x^{2m}}{(2m)!}\cos(\theta x) , \theta\in(0,1) \]
\[\begin{aligned} (1+x)^\alpha=1+\alpha x+& \dfrac{\alpha(\alpha-1)}{2!}x^2+\cdots \\ +& \dfrac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n+R_{n} \end{aligned} \]

其中

\[R_{n}=\dfrac{\alpha(\alpha-1)\cdots(\alpha-n)}{(n+1)!}(1+\theta x)^{\alpha-n-1}x^{n+1} , \theta\in(0,1) \]
\[\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\cdots +(-1)^{n-1}\dfrac{x^n}{n}+R_{n} \]

其中

\[R_{n}=(-1)^n\dfrac{x^{n+1}}{(n+1)(1+\theta x)^{n+1}} , \theta\in(0,1) \]
\[\begin{aligned} e^x=&1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^n}{n!}+o(x^n) \\ \ln(1+x)=& x-\dfrac{x^2}{2}+\cdots+(-1)^{n-1}\dfrac{x^n}{n}+o(x^n) \\ \sin(x)=& x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots+(-1)^{n-1}\dfrac{x^{2n-1}}{(2n-1)!}+o(x^{2n}) \\ \cos(x)=&1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots+(-1)^{n}\dfrac{x^{2n}}{(2n)!}+o(x^{2n+1}) \\ \arctan x=&x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+\cdots+(-1)^{n-1}\dfrac{x^{2n-1}}{2n-1}+o(x^{2n}) \\ \arcsin x=&x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+\cdots+\dfrac{(2n-3)!!}{(2n-2)!!(2n-1)}x^{2n-1}+o(x^{2n}) \end{aligned} \]
\[\begin{aligned} (1+x)^m=& 1+mx+\dfrac{m(m-1)}{2!}x^2+\cdots \\ &+\dfrac{m(m-1)\cdots(m-n+1)}{n!}x^n+o(x^n) \\ \dfrac{1}{1+x}=&1-x+x^2-\cdots+(-1)^nx^n+o(x^n) \\ \dfrac{1}{(1+x)^2}=&1-2x+3x^3-\cdots+(-1)^n(n+1)x^n+o(x^n) \\ \sqrt{1+x}=&1+\displaystyle\frac12 x-\displaystyle\frac18 x^2+\cdots+(-1)^{n-1}\dfrac{(2n-3)!!}{(2n)!!}x^n+o(x^n) \\ \dfrac{1}{\sqrt{1+x}}=&1-\displaystyle\frac12 x+\dfrac{3}{8}x^2+\cdots+(-1)^n\dfrac{(2n-1)!!}{(2n)!!}x^n+o(x^n) \end{aligned} \]

例 1. Maclaurin展开 $e^{-x^4}$

例 2. Maclaurin展开 $\cos^2(x)$

例 3. Maclaurin展开 $\tan(x)$ 到$x^5$项

例 4. Maclaurin展开 $\sin(\sin(x))$到$x^3$项

例 5. Maclaurin展开 $\sqrt{\cos(2x)}$到$x^2$项

例 6. Maclaurin展开 $\sqrt{\cos(2x)}\sqrt[3]{\cos(3x)}$到$x^2$项

例 7. Maclaurin展开 $e^{\cos x}$到$x^8$项

例 8. $f(x)=\frac1{2x^2-1}$，求$f^{(2018)}(0)$。

例 9. 在$x=2$处展开$\frac1x$

1.

\[\begin{aligned} e^{-x^4}=&1+(-x^4)+\dfrac{(-x^4)^2}{2!}+\cdots+\dfrac{(-x^4)^n}{n!}+o((-x^4)^n) \\ =&1-x^4+\dfrac{x^8}{2!}+\cdots+\dfrac{(-1)^nx^{4n}}{n!}+o(x^{4n}) \end{aligned} \]

2.

\[\begin{aligned} \cos^2(x)=&\frac12+\frac12\cos(x) \\ =&\frac12+\frac12(1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}+\cdots \\ &+\dfrac{(-1)^{n-1}(2x)^{2n-2}}{(2n-2)!}+o((2x)^{2n-1})) \\ =&1-\dfrac{2}{2!}x^2+\dfrac{2^3}{4!}x^4+\cdots+\dfrac{(-1)^{n-1}2^{2n-3}}{(2n-2)!}x^{2n-2}+o(x^{2n-1}) \end{aligned} \]

3. $\tan(x)$展开到$x^5$项

\[\begin{aligned} \sin(x)=&x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+o(x^6) \\ \cos(x)=&1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+o(x^5)=1-\Delta \end{aligned} \]
\[\begin{aligned} \tan(x)=&\dfrac{\sin(x)}{\cos x}=(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+o(x^6))\dfrac{1}{1-\Delta} \\ =&(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+o(x^6))\cdot(1+\Delta+\Delta^2+o(x^4)) \\ =& (x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+o(x^6))\cdot (1+\dfrac{x^2}{2}+\dfrac{5x^4}{24}+o(x^4)) \\ =&x+\dfrac{x^3}{3}+\dfrac{2}{15}x^5+o(x^5) \end{aligned} \]

4.

\[\begin{aligned} \sin(\sin(x))=&\sin x-\dfrac{\sin^3x}{3!}+o(x^4) \\ =&\left(x-\dfrac{x^3}{3!}+o(x^4)\right)-\dfrac{1}{3!}\left(x-\dfrac{x^3}{3!}+o(x^4)\right)^3+o(x^4) \\ =& x-\dfrac{2x^3}{3!}+o(x^4) \end{aligned} \]
\[\begin{aligned} \sqrt{\cos(2x)}=&\sqrt{1-\frac12(2x)^2+o(x^2)} \\ =& 1+\frac12(-2x^2+o(x^2))-\frac18(-2x^2+o(x^2))^2+o((-2x^2)^2)\\ =&1-x^2+o(x^2) \end{aligned} \]
\[\begin{aligned} \sqrt[3]{\cos(3x)}=&\sqrt[3]{1-\frac12(3x)^2+o(x^2)} \\ =& 1+\frac13(-\frac92x^2+o(x^2))+o((-\frac92x^2)^2)\\ =&1-\frac32x^2+o(x^2) \end{aligned} \]

应用例子

求极限

例 10. $\displaystyle\lim_{x\to0+}\dfrac{a^x+a^{-x}-2}{x^2} , a>0$

例 11. $\displaystyle\lim_{x\to\infty}(x-x^2\ln(1+\dfrac{1}{x}))$

1.

\[\begin{aligned} a^x+a^{-x}-2= & 1+x\ln a+\dfrac{x^2}{2}\ln^2a+o(x^2) \\ &+1-x\ln a+\dfrac{x^2}{2}\ln^2a+o(x^2)-2 \\ =&x^2\ln^2a+o(x^2) \end{aligned} \]

2.

\[\begin{aligned} x-x^2\ln(1+\dfrac{1}{x})=&x-x^2(\dfrac{1}{x}-\displaystyle\frac1{2x^2}+\displaystyle\frac1{3x^3}+o(\displaystyle\frac1{x^3})) \\ =&\frac12-\displaystyle\frac1{3x}+o(\displaystyle\frac1 x) \end{aligned} \]

所以

\[\lim_{x\to\infty}(x-x^2\ln(1+\dfrac{1}{x}))=\frac12 \]

例 12. $\displaystyle\lim_{x\to\infty}(x(x+1)\ln(1+\dfrac{1}{x})-x)$

\[\begin{aligned} &\lim_{x\to\infty}(x(x+1)\ln(1+\dfrac{1}{x})-x)\\ =&\lim_{x\to\infty}(x^2\ln(1+\dfrac{1}{x})-x+x\ln(1+\displaystyle\frac1 x)) \\ =&\frac12 \end{aligned} \]

例 13. $\displaystyle\lim_{n\to+\infty}(n(n+1)\ln(1+\dfrac{1}{n})-n)$

例 14. $\displaystyle\lim_{n\to+\infty}\dfrac{(1+\dfrac{1}{n})^{n(n+1)}}{e^n}$

如何知道Taylor展开的需要的阶？看看如下形式的极限，

\[\lim_{x\to0}\dfrac{a_nx^n+o(x^n)}{b_mx^m+o(x^m)}=? \]

(1) $m=n$,

\[\lim_{x\to0}\dfrac{a_nx^n+o(x^n)}{b_mx^m+o(x^m)} =\lim_{x\to0}\dfrac{a_n+o(1)}{b_m+o(1)} \]

(2) $m>n$

\[\lim_{x\to0}\dfrac{a_nx^n+o(x^n)}{b_mx^m+o(x^m)} =\lim_{x\to0}\dfrac{a_n+o(1)}{b_mx^{m-n}+o(x^{m-n})} \]

(3) $m<n$

\[\lim_{x\to0}\dfrac{a_nx^n+o(x^n)}{b_mx^m+o(x^m)} =\lim_{x\to0}\dfrac{a_nx^{n-m}+o(x^{n-m})}{b_m+o(1)} \]

例 15. $\displaystyle\lim_{x\to0}\dfrac{1-\cos x\cos2x}{1-\cos x}$

例 16. $\displaystyle\lim_{x\to0}\dfrac{1-\cos x\sqrt{\cos2x}\sqrt[3]{\cos3x}}{x^2}$

例 17. $\displaystyle\lim_{x\to0}(\dfrac{1}{x}-\displaystyle\frac1{\sin x})$

例 18. $\displaystyle\lim_{x\to0}\frac{\arcsin(x)-\arctan x}{\sin x-\tan x}$

3.

\[1-\cos x=1-(1-\frac{x^2}2+o(x^2))=\frac{x^2}2+o(x^2) \]

则把分子也展开到$x^2$项

\[\begin{aligned} 1-\cos x\cos 2x=&1-(1-\frac{x^2}2+o(x^2))(1-\frac{(2x)^2}2+o((2x)^2)) \\ =&\frac{5}2x^2+o(x^2) \end{aligned} \]

4.

\[\begin{aligned} \displaystyle\lim_{x\to0}(\dfrac{1}{x}-\displaystyle\frac1{\sin x}) =&\lim_{x\to0}\dfrac{\sin x-x}{x \sin x} \\ =&\lim_{x\to0}\dfrac{\sin x-x}{x^2} \\ =&\lim_{x\to0}\dfrac{-\dfrac{x^3}{3}+o(x^3)}{x^2} \\ =&0 \end{aligned} \]

例 19. 设$f(x)$有二阶导数连续，且 $f(0)=f'(0)=0$, $f''(0)=6$。求

\[\lim_{x\to0}\frac{f(\sin^2x)}{x^4} \]

例 20. $f(x)$在$a$处二阶可导，求

\[\lim_{h\to0}\dfrac{f(a+2h)-3f(a)+2f(a-h)}{h^2} \]

证明

例 21. [复习题] $f(x)$在$x_0$的某邻域内有连续的$n+1$阶导数，

\[f(x_0+h)=f(x_0)+hf'(x_0)+\cdots+\dfrac{f^{(n)}(x_0+\theta h)}{n!}h^n, \theta\in(0,1) \]

且$f^{(n+1)}(x)\neq0$，证明：

\[\lim_{h\to0}\theta=\dfrac{1}{n+1} \]

1.

\[\begin{aligned} f(x_0+h)=f(x_0)+&hf'(x_0)+\cdots \\ +&\dfrac{f^{(n)}(x_0)}{n!}h^n+\dfrac{f^{(n+1)}(x_0)}{(n+1)!}h^{n+1}+o(h^{n+1}) \end{aligned} \]

比较可得，

\[\dfrac{h^n}{n!}f^{(n)}(x+\theta h)=\dfrac{h^n}{n!}f^{(n)}(x)+\dfrac{h^{n+1}}{(n+1)!}f^{(n+1)}(x)+o(h^{n+1}) \]

则有

\[f^{(n)}(x+\theta h)=f^{(n)}(x)+\dfrac{h}{n+1}f^{(n+1)}(x)+o(h) \]
\[\theta\dfrac{f^{(n)}(x+\theta h)-f^{(n)}(x)}{\theta h}=\dfrac{1}{n+1}f^{(n+1)}(x)+o(1) \]

令$h\to0$

\[\lim_{h\to0}\theta\lim_{h}\dfrac{f^{(n)}(x+\theta h)-f^{(n)}(x)}{\theta h}=\dfrac{1}{n+1}f^{(n+1)}(x) \]

即有

\[\lim_{h\to0}\theta=\dfrac{1}{n+1} \]

例 22. $f(x)$在$[0,2]$有连续的三阶导数，$f(0)=1$， $f(2)=2$，$f'(1)=0$，证明： $\exists \xi\in(0,2)$，满足

\[f'''(\xi)=3 \]

例 23. $f(x)$在$[a,b]$上有连续的二阶导数，$f'(a)=f'(b)=0$，证明： $\exists \xi \in(a,b)$， s.t.

\[|f''(\xi)|\geq\dfrac{4}{(b-a)^2}|f(b)-f(a)| \]

6. $f(0)$，$f(2)$在$1$处Taylor展开

7. $f(\dfrac{a+b}{2})$分别在$a$,$b$点展开，

定理 3.
设$[a,b]$上的函数$f(x)$，满足$f''(x)>0$，则

$\forall x_1,x_2,\cdots,x_n\in[a,b]$，有

\[\dfrac{f(x_1)+f(x_2)+\cdots+f(x_n)}{n}\geq f(\dfrac{x_1+x_2+\cdots+x_n}{n}) \]

更一般地，正数$\lambda_i>0, i=1,2,\cdots,n$，满足$ \displaystyle\sum_{i=1}^n \lambda_i=1$， 则$\forall x_1,x_2,\cdots,x_n\in[a,b]$，有

\[\begin{aligned} \lambda_1 f(x_1)+\lambda_2 f(x_2)&+\cdots+\lambda_n f(x_n) \\ & \geq f(\lambda_1 x_1+ \lambda_2 x_2+\cdots+\lambda_n x_n) \end{aligned} \]

证： 令$x_0=\dfrac{x_1+x_2+\cdots+x_n}{n}$，$f(x_i)$在$x_0$处Taylor展开，

\[\begin{aligned} f(x_i) =&f(x_0)+f'(x_0)(x_i-x_0)+\frac{f''(\xi_i)}{2!}(x_i-x_0)^2 \\ \geq& f(x_0)+f'(x_0)(x_i-x_0) \end{aligned} \]

则有

\[\begin{aligned} \sum_{i=1}^n f(x_i) \geq& nf(x_0)+\sum_{i=1}^n f'(x_0)(x_i-x_0) \\ = & n f(x_0)+f'(x_0)\left(\sum_{i=1}^n x_i-n x_0\right) \\ = & n f(x_0) \end{aligned} \]

近似计算

\[f(x_1)\approx f(x_0)+f'(x_0)\Delta x+\dfrac{f''(x_0)}{2!}\Delta x^2+\cdots+\dfrac{f^{(n)}}{n!}\Delta x^n \]

误差不超过

\[\dfrac{M|\Delta x|^{n+1}}{(n+1)!} \]

$M$为$|f^{(n+1)}|$的上界

例 24. 求近似值 $\log_{10}(11)=?$

例 25. 求近似值 $\sqrt 5=?$

\[\begin{aligned} \log_{10}11 =&1+\log_{10}(1+0.1) \\ =&1+\dfrac1{\ln(10)}\left(0.1-\dfrac{0.1^2}{2}+\dfrac{0.1^3}3-\cdots\right) \\ \approx& 1+\left(0.1-\dfrac12(0.1)^2+\dfrac13(0.1)^3-\dfrac14(0.1)^4\right)\dfrac1{\ln(10)} \end{aligned} \]
\[\begin{aligned} \sqrt5 =&\sqrt{4(1+\dfrac14)}=2\left(1+\dfrac14\right)^{\frac12} \\ \approx& 2\left(1+\dfrac12\dfrac14-\dfrac1{2!}\dfrac{1}{2^2}(\dfrac14)^2 +\dfrac1{3!}\dfrac{3}{2^3}(\dfrac14)^3\right) \\ =&2.2361 \end{aligned} \]

目录

本节读完

例 26. Thanks

26.

例 27. $f(x)$在$\mathbb{R}$中满足$\displaystyle\lim_{x\to0}\dfrac{f(x)}{x}=1$，且$f''(x)>0$，证明：

\[f(x)\geq x, \forall x\in\mathbb{R} \]

2. 由题知$f(x)$是$x$等价无穷小。所以

\[f(0)=\lim_{x\to0}f(x)=0, f'(0)=\lim_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim_{x\to0}\dfrac{f(x)}{x}=1 \]

Taylor展开

\[\begin{aligned} f(x)=&f(0)+f'(0)x+\frac12 f''(\xi)x^2, \xi\in(0,x) \\ =&x+\frac12 f''(\xi)x^2 \end{aligned} \]

由$f''(x)>0$，则

\[f(x)>x, x\neq 0 \]

即有

\[f(x)\geq x, \forall x \]

例 28. $f(x)$在$(1,+\infty)$上定义，且$f'(x), f''(x)$存在。又$\displaystyle\lim_{x\to+\infty}f(x)=0$, $\displaystyle\lim_{x\to+\infty}f''(x)=0$， 证明：

\[\lim_{x\to+\infty}f'(x)=0 \]

例 29. $f(x)$在$[0,+\infty)$上定义，且有连续的二阶导数。又$\displaystyle\lim_{x\to+\infty}f(x)$存在, $f''(x)$在$[0,+\infty)$上有界， 证明：

\[\lim_{x\to+\infty}f'(x)=0 \]

4.

\[f(x+a)=f(x)+f'(x)a+\dfrac{f''(\xi)}{2!}a^2, \xi\in(x,x+a) \]

则

\[\lim_{x\to+\infty}f'(x)=\dfrac{1}{a}\left(f(x+a)-f(x)-\dfrac{f''(\xi)}{2}a^2\right)=0 \]

5. 由Taylor公式，$\forall h>0$

\[f(x+h)=f(x)+f'(x)h+\dfrac{f''(\xi)}{2!}h^2, \xi\in(x,x+h) \]

则

\[f'(x)=\dfrac{1}{h}\left(f(x+h)-f(x)-\dfrac{f''(\xi)}{2}h^2\right) \]

有

\[|f'(x)|\leq\dfrac{1}{h}\left|f(x+h)-f(x)\right|+\dfrac{|f''(\xi)|}{2}h , \forall h>0 \]

$\forall \epsilon>0$，由$f''(x)$有界，$\exists h>0$，使得$\dfrac{|f''(\xi)|}{2}h<\epsilon$。固定$h$，由$\displaystyle\lim_{x\to+\infty}f(x)=A$，则$\exists M>0$，当$x>M$时，

\[\dfrac{1}{h}|f(x+h)-f(x)|\leq\dfrac{1}{h}(|f(x+h)-A|+|f(x)-A|)\leq \epsilon \]

即有

\[|f'(x)|<2\epsilon, \forall x>M \]

例 30. $f(x)$在$[a,b]$上三次可导，证明：$\exists \xi\in(a,b)$，满足

\[f(b)=f(a)+f'(\frac{a+b}2)(b-a)+\dfrac{1}{24}f'''(\xi)(b-a)^3 \]

8.(待定常数法) 设$k$满足

\[f(b)=f(a)+f'(\frac{a+b}2)(b-a)+\dfrac{1}{24}k(b-a)^3 \]

则，问题变为$\exists \xi \in (a,b)$，满足 $k=f'''(\xi)$

令

\[g(x)=f(x)-\left(f(a)+f'(\frac{a+x}2)(x-a)+\dfrac{1}{24}k(x-a)^3\right) \]

则 $g(a)=g(b)=0$。

由Rolle，$\exists\xi\in(a,b)$，满足$g'(\xi)=0$，即

\[f'(\xi)-f'(\frac{\xi+a}2)-f''(\frac{a+\xi}2)\frac{\xi-a}2-\dfrac{k}{8}(\xi-a)^2=0 \]

与$f'(\xi)$在$\frac{\xi+a}2$处的Taylor展开比较

\[f'(\xi)=f'(\frac{\xi+a}2)+f''(\frac{a+\xi}2)\frac{\xi-a}2+\frac12 f'''(\xi)(\frac{\xi-a}2)^2 \]

即可得到结论