Section01c_中值定理 (3)

Th4 Taylor中值定理

导入

$$\begin{split} &\lim_{x\to 0}\frac{x-\sin x}{x^{3}}\neq\frac{x-x}{x^{3}}\\ \text{Conjection}\ &\exists\ k, \sin x=x+ kx^{3} + o(x^{3}) \\ & \lim_{x\to 0}\frac{x-\sin x}{x^{3}} = \frac{x-x-kx^{3}-o(x^{3})}{x^{3}} = -k \end{split}$$

定理

• 条件：$f(x)$在$x=x_{0}$的领域内$n+1$阶可导
• 结论：$f(x) = P_{n}(x)+R_{n}(x)$
  • $\displaystyle{P_{n}(x) = f(x_{0})+f'(x_{0})(x-x_{0})+\frac{f''(x_{0})}{2!}(x-x_{0})^{2}+\cdots+\frac{f^{(n)}(x_{0})}{n!}(x-x_{0})^{n}}$
  • $\displaystyle{R_{n}(x) = \begin{cases}\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_{0})^{n+1}&\text{拉格朗日型}\\o((x-x_{0})^{n})&\text{皮亚诺型}\end{cases}}$
• 特殊情况：若$x_{0}=0$ $$\begin{split} P_{n}(x) = & f(0) +f'(0)x+\frac{f''(0)}{2!}x^{2}+\cdots+\frac{f^{(n)}(0)}{n!}x^{n} \\ = & f(0)+\sum_{i=1}^{n}\frac{f^{(i)}(0)}{i!}x^{i} \end{split}$$

需记 $x_{0}=0$时的Taylor展开

1. $e^{x} = 1 + x +\frac{x^{2}}{2!}+\cdots+\frac{x^{n}}{n!}+o(x^n)$
2. $\sin x = x - \frac{1}{3!}x^{3}+\frac{1}{5!}x^{5} - \cdots+\frac{(-1)^{n}}{(2n+1)!}x^{2n+1}+o(x^{2n+1})$
3. $\cos x = 1 - \frac{1}{2!}x^{2} + \frac{1}{4!}x^{4}-\cdots+\frac{(-1)^{n}}{2n!}x^{2n}+(x^{2n})$
4. $\frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \cdots+x^{n} + o(x^{n})$
5. $\frac{1}{1+x} = 1 -x+x^{2}-x^{3}+\cdots-(-1)^{n}x^{n}+o(x^n)$
6. $\ln(1+x)= x - x^{2}+x^{3}+\cdots +(-1)^{n-1}x^{n}+o(x^{n})$
7. $(1+x)^{a} = 1 + ax + \frac{a(a-1)}{2!}x^{2}+\cdots+\frac{a!}{(a-n)!n!}x^{n}+o(x^{n})$
8. $\arctan x = x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots+\frac{(-1)n}{2n+1}x^{2n+1}+o(x^{2n+1})$
   • Proof $$\begin{array}{ll} \because & (\arctan x)' =\frac{1}{1 + x^{2}} \text{ and } \frac{1}{1+x} = 1-x+x^{2}-x^{3}+\cdots+(-1)^{n}x^{n}+o(x^{n}) \\ \therefore & \frac{1}{1+x^{2}} = 1-x^{2} +x^{4} -x^{6} + \cdots +(-1)^{n}x^{2n} +o(x^{2n}) \\ \because & \arctan x = \int \frac{1}{1+x^{2}}\cdot dx \\ \therefore & \arctan x = \int (1-x^{2}+x^{4}-x^{6}+\cdots +(-1)^{n}x^{2n})\cdot dx + R_{n}(x) \\ & = x-\frac{x^{3}}{3} + \frac{x^{5}}{5}-\frac{x^{7}}{7} +\cdots +\frac{(-1)^{n}x^{2n+1}}{2n+1} +o(x^{2n+1}) \end{array}$$

例题

• 例1 $\displaystyle{\lim_{x\to 0}\frac{e^{-\frac{x^{2}}{2}}-1+\frac{x^{2}}{2}}{x^{3}\sin x}}$ $$\begin{array}{ll} \because & e^{x} = 1 + x + \frac{1}{2!}x^{2} + o(x^{2}) \\ \therefore & e^{- \frac{x^{2}}{2}} = 1-\frac{x^{2}}{2}+\frac{x^{4}}{2^{2}2!}+o(x^{4}) \\ & \text{原式} = \displaystyle{\lim_{x\to 0}\frac{1-\frac{x^{2}}{2}+\frac{1}{8}x^{4}-1+\frac{x^{2}}{2}+o(x^{4})}{x^{4}}}\\ & = \displaystyle{\frac{1}{8}+\lim_{x\to 0}\frac{o(x^{4})}{x^{4}} = \frac{1}{8}} \end{array}$$

• 例2 $\displaystyle{\lim_{x\to 0}\frac{\sqrt{1+x}+\sqrt{1-x}-2}{x^{2}}}$ $$\begin{array}{ll} \because & (1+x)^{a} = 1 + a x +\frac{a(a-1)}{2!}x^{2} + o(x^{2}) \\ \therefore & \sqrt{1+x} = 1+\frac{1}{2}x + \frac{\frac{1}{2}(-\frac{1}{2})}{2} x^{2} + o(x^{2}) \\ & \sqrt{1-x} = 1-\frac{1}{2}x +\frac{\frac{1}{2}(-\frac{1}{2})}{2}x^{2} + o(x^{2}) \\ \therefore & \text{原式} = \displaystyle{\lim_{x\to 0}\frac{2-\frac{1}{4}x^{2}+o(x^{2})-2}{x^{2}}} = -\frac{1}{4} \\ \end{array}$$