Taylor公式和插值多项式

Taylor公式和插值多项式

笔记总结自：复旦大学-陈纪修-《数学分析》课程-第5章第3节-Taylor公式和插值多项式

一、Taylor公式

$\begin{array}{rl}f(x)& =\sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k+R_n(x)\\ & =f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{1}{2!}{f}^{\prime \prime }(x_0)(x-x_0{)}^2+...+\frac{1}{n!}{f}^{(k)}(x_0)(x-x_0{)}^n+R_n(x)\end{array}$

$P_n(x)=\sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$称为$f$在$x=x_0$处的n次Taylor多项式

带Peano余项的Taylor公式

定义：

设$f(x)$在$x_0$有n阶导数，则在$x_0$的邻域成立$f(x)=P_n(x)+R_n(x)$

$R_n(x)=o((x-x_0{)}^n)$称为$f$在$x=x_0$处的Peano余项

注意：$f(x)在$$x_0$处n阶可导，n-1阶连续，n阶不一定连续

证明：

即证$R_n(x)=o((x-x_0{)}^n)$满足Taylor公式

由

$R_n(x)=f(x)-\sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$

$R_n^{\prime }(x)=f^{\prime }(x)-\sum _{k=1}^n\frac{f^{(k)}(x_0)}{(k-1)!}(x-x_0{)}^{k-1}$

$R_n^{\prime \prime }(x)=f^{\prime \prime }(x)-\sum _{k=2}^n\frac{f^{(k)}(x_0)}{(k-2)!}(x-x_0{)}^{k-2}$

…

$R_n^{(n-1)}(x)=f^{(n-1)}(x)-\sum _{k=n-1}^n\frac{f^{(k)}(x_0)}{(k-(n-1))!}(x-x_0{)}^{k-(n-1)}$

可得：

$R_n(x_0)=f(x_0)-\frac{f(x_0)}{0!}(x_0-x_0{)}^0=0$

$R_n^{\prime }(x_0)=f^{\prime }(x_0)-\frac{f^{\prime }(x_0)}{0!}(x_0-x_0{)}^0=0$

$R_n^{\prime \prime }(x_0)=f^{\prime \prime }(x_0)-\frac{f^{\prime \prime }(x_0)}{0!}(x_0-x_0{)}^0=0$

…

$R_n^{(n-1)}(x_0)=f^{\prime }(x_0)-\frac{f^{(n-1)}(x_0)}{0!}(x_0-x_0{)}^0=0$

又因为

$\begin{array}{rl}{R}_n^{(n-1)}(x)& =f^{(n-1)}(x)-\sum _{k=n-1}^n\frac{f^{(k)}(x_0)}{(k-(n-1))!}(x-x_0{)}^{k-(n-1)}\\ & =f^{(n-1)}(x)-f^{(n-1)}(x_0)-f^{(n)}(x_0)(x-x_0)\end{array}$

所以

$\underset{x\to x_0}{\lim }\frac{R_n(x)}{(x-x_0{)}^n}\stackrel{\frac{0}{0}}{=}\underset{x\to x_0}{\lim }\frac{R_n^{\prime }(x)}{n(x-x_0{)}^{n-1}}\stackrel{\frac{0}{0}}{=}...\stackrel{\frac{0}{0}}{=}\underset{x\to x_0}{\lim }\frac{R_n^{(n-1)}(x)}{n(n-1)...2(x-x_0)}$

$=\underset{x\to x_0}{\lim }\frac{1}{n!}\cdot [\frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0}-f^{(n)}(x_0)]=0$

所以$R_n(x)=o((x-x_0{)}^n)$

带Lagrange余项的Taylor公式

定义：

$f(x)$在$[a,b]$有n阶连续导数，在$(a,b)$有n+1阶导数。

设$x_0\in [a,,b]$为一定点，则对任意$x\in [a,,b]$，

有$f(x)=P_n(x)+R_n(x)$

$R_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}$称为$f$在$x=x_0$处的Lagrange余项，其中$\xi$在$x$和$x_0$之间

证明：

即证$R_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}$满足Taylor公式

不妨设$x>x_0$

令$G(t)=f(x)-\sum _{k=0}^n\frac{f^{(k)}(t)}{k!}(x-t{)}^k$

则$G(x)=f(x)-\sum _{k=0}^n\frac{f^{(k)}(x)}{k!}(x-x{)}^k=f(x)-f(x)=0$

    $G(x_0)=f(x_0)-\sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k=f(x)-f(x)=R_n(x)$

令$H(t)=(x-t{)}^{n+1}$

则$H(x)=0$

    $H(x_0)=(x-x_0{)}^{n+1}$

$G(t),H(t)$在$[x_0,x]$连续，在$(x_0,x)$可导

$\begin{array}{rl}{G}^{\prime }(t)& =-\sum _{k=0}^n\frac{1}{k!}[f^{(k)}(t)(x-t{)}^k{]}^{\prime }\\ & =-\sum _{k=0}^n\frac{f^{(k+1)}(t)}{k!}(x-t{)}^k+\sum _{k=1}^n\frac{f^{(k)}(t)}{(k-1)!}(x-t{)}^{k-1}\\ & =-\sum _{k=1}^{n+1}\frac{f^{(k)}(t)}{(k-1)!}(x-t{)}^{k-1}+\sum _{k=1}^n\frac{f^{(k)}(t)}{(k-1)!}(x-t{)}^{k-1}\\ & =-\frac{f^{(n+1)}(t)}{n!}(x-t{)}^n\end{array}$

$H^{\prime }(t)=-(n+1)(x-t{)}^n$

$\begin{array}{rlr}\frac{R_n(x)}{(x-x_0{)}^{n+1}}& =\frac{G(x_0)}{H(x_0)}& \\ & =\frac{G(x_0)-G(x)}{H(x_0)-H(x)}& \\ & =\frac{G^{\prime }(\xi )}{H^{\prime }(\xi )}& \\ & =\frac{f^{n+1}(\xi )}{(n+1)!}& (x_0<\xi <x)\end{array}$

所以$R_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}$，其中$\xi$在$x$和$x_0$之间

二、插值多项式

已知：确定多项式$p_n=a_0+a_{1}x+a_2{x}^2+...+a_n{x}^n$至少需要n+1个条件

目标：对于$[a,b]$上的函数$f(x)$，找到n+1个条件来确定$f(x)$的插值多项式$p_n(x)$

插值多项式的定义

已知$f(x)$在$(a,b)$内的m+1个互异点$x_0,x_1,...x_m$上的函数值和若干阶导数$f^{(j)}(x)$，其中$i=0,1,2,...,m$，$j=0,1,...,max(n_i)-1$。

$n_i$表示$f(x)$在点$x_i$处已知的导数值的个数，$m_j$表示已知$j$阶导数值的点的个数，$\sum _{j=0}^{n+1}{m}_j=\sum _{i=0}^m{n}_i=n+1$。

若存在一个n次多项式$P_n(x)$满足所有插值条件$p_n^{(j)}(x_i)=f^{(j)}(x_i)$，则称$p_n(x)$是$f(x)$关于插值结点$x_0,x_1,...,x_m$的n次插值多项式

例如：

	$x_0$	$x_1$	$x_2$	$x_3$	$m_j$
$j=0$	$f(x_0)$	$f(x_1)$	$f(x_2)$	$f(x_3)$	$m_0=4$
$j=1$	$f^{\prime }(x_0)$	—	$f^{\prime }(x_2)$	$f^{\prime }(x_3)$	$m_1=3$
$j=2$	$f^{\prime \prime }(x_0)$	—	$f^{\prime \prime }(x_2)$	$f^{\prime \prime }(x_3)$	$m_2=3$
$j=3$	—	—	$f^{\prime \prime \prime }(x_2)$	—	$m_3=1$
$n_i$	$n_0=3$	$n_1=1$	$n_2=4$	$n_3=3$	$n+1=11$

插值多项式余项定理

定义：

设$f^{(n)}(x)$在$[a,b]$连续，$f^{(n+1)}(x)$在$(a,b)$存在，其插值多项式为$p_n(x)$。那么对任意$x\in (a,b)$，插值余项：

$R_n(x)=f(x)-p_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}\prod _{i=0}^m(x-x_i{)}^{n_i}$

这里$\xi$是介于$x_{min}=min(x_0,x_1,...,x_m,x)$和$x_{max}=max(x_0,x_1,...,x_m,x)$之间的一个数（一般依赖于$x$）

证明：

令$x\ne x_i$

令n+1次多项式$w_{n+1}(t)=\prod _{i=0}^m(t-x_i{)}^{n_i}$

则$w_{n+1}^{(j)}(x_i)=0,\left(\begin{array}{l}i=0,1,2,...,m\\ j=0,1,...,max(n_i)-1\end{array}\right)$

令$\phi (t)=f(t)-p_n(t)-\frac{w_{n+1}(t)}{w_{n+1}(x)}(f(x)-p_n(x))$

因为$f^{(j)}(x_i)=p_n^{(j)}(x_i)$且$w_{n+1}^{(j)}(x_i)=0$

所以${\phi }^{(j)}(x_i)=f^{(j)}(x_i)-p_n^{(j)}(x_i)-\frac{w_{n+1}^{(j)}(x_i)}{w_{n+1}(x)}(f(x)-p_n(x))=0,\left(\begin{array}{l}i=0,1,2,...,m\\ j=0,1,...,max(n_i)-1\end{array}\right)$

令上式中$j=0$，可得$\phi (x_i)=0$

又因为$\phi (x)=f(x)-p_n(x)-\frac{w_{n+1}(x)}{w_{n+1}(x)}(f(x)-p_n(x))=0$

所以$\phi (t)$的零点： { x 0 , x 1 , . . . , x m , x } \{x_{0},x_{1},...,x_{m},x\} { x0​,x1​,...,xm​,x}，共$m_0+1$个

由Rolle定理：

${\phi }^{\prime }(t)$的零点有$m_0+m_1$个

${\phi }^{\prime \prime }(t)$的零点有$m_0+m_1+m_2-1$个

…

${\phi }^{\prime }(t)$的零点有$m_0+m_1+...+m_{n+1}-n=1$个，设该零点为$\xi$

则$\xi \in (x_{min},x_{max}),{\phi }^{(n+1)}(\xi )=0$

因为$p_n(x)$是n次多项式，所以其n+1阶导数为0

对$w_{n+1}(t)=\prod _{i=0}^m(t-x_i{)}^{n_i}$是n+1次多项式，对它求n+1阶导数相当于对它的最高次项$t^{n+1}$求n+1阶导数，得到$w_{n+1}^{(n+1)}(t)=(n+1)!$

因此：

${\phi }^{(n+1)}(\xi )=f^{(n+1)}(x)-\frac{(n+1)!}{w_{n+1}(x)}(f(x)-p_n(x))=0$

$\Rightarrow R_n(x)=f(x)-p_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}\prod _{i=0}^m(x-x_i{)}^{n_i}$