泰勒级数与递归的柯西中值

若 $f(x)$ 在一段区间上连续，则

$$\begin{array}{rl}f(x)& =f(x_0)+f^{\prime }(\xi )(x-x_0)\end{array}$$

若 $f(x)$ 在区间上无穷可微，则

$$\begin{array}{rl}f(x)& =f(x_0)+f^{\prime }({\xi }_1)(x-x_0)\\ & =f(x_0)+(f^{\prime }(x_0)+f^{\prime \prime }({\xi }_2)({\xi }_1-x_0))(x-x_0)\\ & =f(x_0)+f^{\prime }(x_0)(x-x_0)+f^{\prime \prime }({\xi }_2)({\xi }_1-x_0)(x-x_0)\\ & =f(x_0)+f^{\prime }(x_0)(x-x_0)+f^{\prime \prime }(x_0)({\xi }_1-x_0)(x-x_0)+f^{\prime \prime }({\xi }_3)({\xi }_2-x_0)({\xi }_1-x_0)(x-x_0)\\ & =f(x_0)+f^{\prime }(x_0)(x-x_0)+f^{\prime \prime }(x_0)({\xi }_1-x_0)(x-x_0)+f^{\prime \prime }(x_0)({\xi }_2-x_0)({\xi }_1-x_0)(x-x_0)+f^{\prime \prime }({\xi }_4)({\xi }_3-x_0)({\xi }_2-x_0)({\xi }_1-x_0)(x-x_0)\\ & =...\end{array}$$

看起来与泰勒展开很像，但是仍有差距，只能说上面的本意是好的，都是下面的人执行歪了

拉格朗日余项

出于简便考虑，这里全部取 $x_0=0$

$$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{(3)}(0)}{3!}{x}^3+\frac{f^{(4)}(\xi )}{4!}{x}^4$$

其中 $\frac{f^{(4)}(\xi )}{3!}{x}^3$ 称为拉格朗日余项，记为 $R_4(x)$

下面证明上式，构造 $F(x)$：

$$F(x)=f(x)-(f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{(3)}(0)}{3!}{x}^3)$$

则 $F(x)$ 有性质 $F(0)=F^{\prime }(0)=F^{\prime \prime }(0)=F^{(3)}(0)=0$，且对于更高阶导，$F^{(4)}(x)=f^{(4)}(x)$

由柯西中值定理：

$$\begin{array}{rl}F(x)& =F(x)-F(0)\\ & =x^4\frac{F(x)-F(0)}{x^4-0}\\ & =x^4\frac{F^{\prime }({\xi }_1)}{4{\xi }_1^3}\\ & =x^4\frac{1}{4}\frac{F^{\prime }({\xi }_1)-F^{\prime }(0)}{{\xi }_1^3-0}\\ & =x^4\frac{1}{4}\frac{F^{\prime \prime }({\xi }_2)}{3{\xi }_2^2}\\ & ...\\ & =x^4\frac{1}{4}\frac{1}{3}\frac{1}{2}\frac{F^{(4)}({\xi }_4)}{{\xi }_4^0}\\ & =\frac{x^4}{4!}{F}^{(4)}({\xi }_4)\\ & =\frac{x^4}{4!}{f}^{(4)}({\xi }_4)=R_4(x)\end{array}$$

从拉格朗日余项到皮亚诺余项

对拉格朗日余项取极限是显然的

$$\underset{x\to 0}{\lim }\frac{f(x)}{x^3}=\underset{x\to 0}{\lim }\frac{x}{4!}{f}^{(4)}({\xi }_4)=0$$

显式递归

$$\begin{array}{rl}f(x)& =f(x_0)+(x-x_0)\frac{f(x)-f(x_0)}{x-x_0}\\ & =f(x_0)+(x-x_0)f^{\prime }({\xi }_1)\\ & =f(x_0)+(x-x_0)f^{\prime }({\xi }_1)\end{array}$$