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

f ( x ) = ∑ k = 0 n f ( k ) ( x 0 ) k ! ( x − x 0 ) k + R n ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + 1 2 ! f ′ ′ ( x 0 ) ( x − x 0 ) 2 + . . . + 1 n ! f ( k ) ( x 0 ) ( x − x 0 ) n + R n ( x ) \begin{aligned} f(x) &= \sum\limits^{n}_{k=0}\frac{f^{(k)}(x_{0})}{k!}(x-x_{0})^{k}+R_{n}(x) \\ &= f(x_{0})+f'(x_{0})(x-x_{0})+\frac{1}{2!}f''(x_{0})(x-x_{0})^{2}+...+\frac{1}{n!}f^{(k)}(x_{0})(x-x_{0})^{n}+R_{n}(x) \end{aligned} f(x)​=k=0∑n​k!f(k)(x0​)​(x−x0​)k+Rn​(x)=f(x0​)+f′(x0​)(x−x0​)+2!1​f′′(x0​)(x−x0​)2+...+n!1​f(k)(x0​)(x−x0​)n+Rn​(x)​

P n ( x ) = ∑ k = 0 n f ( k ) ( x 0 ) k ! ( x − x 0 ) k P_{n}(x)=\sum\limits^{n}_{k=0}\frac{f^{(k)}(x_{0})}{k!}(x-x_{0})^{k} Pn​(x)=k=0∑n​k!f(k)(x0​)​(x−x0​)k称为 f f f在 x = x 0 x=x_{0} x=x0​处的n次Taylor多项式

带Peano余项的Taylor公式

定义：

设 f ( x ) f(x) f(x)在 x 0 x_{0} x0​有n阶导数，则在 x 0 x_{0} x0​的邻域成立 f ( x ) = P n ( x ) + R n ( x ) f(x)=P_{n}(x)+R_{n}(x) f(x)=Pn​(x)+Rn​(x)

R n ( x ) = o ( ( x − x 0 ) n ) R_{n}(x)=o((x-x_{0})^{n}) Rn​(x)=o((x−x0​)n)称为 f f f在 x = x 0 x=x_{0} x=x0​处的Peano余项

注意： f ( x ) 在 f(x)在 f(x)在 x 0 x_{0} x0​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 ) R_{n}(x)=o((x-x_{0})^{n}) Rn​(x)=o((x−x0​)n)满足Taylor公式

由

R n ( x ) = f ( x ) − ∑ k = 0 n f ( k ) ( x 0 ) k ! ( x − x 0 ) k R_{n}(x)=f(x)-\sum\limits^{n}_{k=0}\frac{f^{(k)}(x_{0})}{k!}(x-x_{0})^{k} Rn​(x)=f(x)−k=0∑n​k!f(k)(x0​)​(x−x0​)k

R n ′ ( x ) = f ′ ( x ) − ∑ k = 1 n f ( k ) ( x 0 ) ( k − 1 ) ! ( x − x 0 ) k − 1 R'_{n}(x)=f'(x)-\sum\limits^{n}_{k=1}\frac{f^{(k)}(x_{0})}{(k-1)!}(x-x_{0})^{k-1} Rn′​(x)=f′(x)−k=1∑n​(k−1)!f(k)(x0​)​(x−x0​)k−1

R n ′ ′ ( x ) = f ′ ′ ( x ) − ∑ k = 2 n f ( k ) ( x 0 ) ( k − 2 ) ! ( x − x 0 ) k − 2 R''_{n}(x)=f''(x)-\sum\limits^{n}_{k=2}\frac{f^{(k)}(x_{0})}{(k-2)!}(x-x_{0})^{k-2} Rn′′​(x)=f′′(x)−k=2∑n​(k−2)!f(k)(x0​)​(x−x0​)k−2

…

R n ( n − 1 ) ( x ) = f ( n − 1 ) ( x ) − ∑ k = n − 1 n f ( k ) ( x 0 ) ( k − ( n − 1 ) ) ! ( x − x 0 ) k − ( n − 1 ) R^{(n-1)}_{n}(x)=f^{(n-1)}(x)-\sum\limits^{n}_{k=n-1}\frac{f^{(k)}(x_{0})}{(k-(n-1))!}(x-x_{0})^{k-(n-1)} Rn(n−1)​(x)=f(n−1)(x)−k=n−1∑n​(k−(n−1))!f(k)(x0​)​(x−x0​)k−(n−1)

可得：

R n ( x 0 ) = f ( x 0 ) − f ( x 0 ) 0 ! ( x 0 − x 0 ) 0 = 0 R_{n}(x_{0})=f(x_{0})-\frac{f(x_{0})}{0!}(x_{0}-x_{0})^{0}=0 Rn​(x0​)=f(x0​)−0!f(x0​)​(x0​−x0​)0=0

R n ′ ( x 0 ) = f ′ ( x 0 ) − f ′ ( x 0 ) 0 ! ( x 0 − x 0 ) 0 = 0 R'_{n}(x_{0})=f'(x_{0})-\frac{f'(x_{0})}{0!}(x_{0}-x_{0})^{0}=0 Rn′​(x0​)=f′(x0​)−0!f′(x0​)​(x0​−x0​)0=0

R n ′ ′ ( x 0 ) = f ′ ′ ( x 0 ) − f ′ ′ ( x 0 ) 0 ! ( x 0 − x 0 ) 0 = 0 R''_{n}(x_{0})=f''(x_{0})-\frac{f''(x_{0})}{0!}(x_{0}-x_{0})^{0}=0 Rn′′​(x0​)=f′′(x0​)−0!f′′(x0​)​(x0​−x0​)0=0

…

R n ( n − 1 ) ( x 0 ) = f ′ ( x 0 ) − f ( n − 1 ) ( x 0 ) 0 ! ( x 0 − x 0 ) 0 = 0 R^{(n-1)}_{n}(x_{0})=f'(x_{0})-\frac{f^{(n-1)}(x_{0})}{0!}(x_{0}-x_{0})^{0}=0 Rn(n−1)​(x0​)=f′(x0​)−0!f(n−1)(x0​)​(x0​−x0​)0=0

又因为

R n ( n − 1 ) ( x ) = f ( n − 1 ) ( x ) − ∑ k = n − 1 n 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 ) \begin{aligned} R^{(n-1)}_{n}(x) &= f^{(n-1)}(x)-\sum\limits^{n}_{k=n-1}\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{aligned} Rn(n−1)​(x)​=f(n−1)(x)−k=n−1∑n​(k−(n−1))!f(k)(x0​)​(x−x0​)k−(n−1)=f(n−1)(x)−f(n−1)(x0​)−f(n)(x0​)(x−x0​)​

所以

lim ⁡ x → x 0 R n ( x ) ( x − x 0 ) n = 0 0 lim ⁡ x → x 0 R n ′ ( x ) n ( x − x 0 ) n − 1 = 0 0 . . . = 0 0 lim ⁡ x → x 0 R n ( n − 1 ) ( x ) n ( n − 1 ) . . . 2 ( x − x 0 ) \lim\limits_{x\rarr x_{0}}\frac{R_{n}(x)}{(x-x_{0})^{n}} \overset{\frac{0}{0}}{=} \lim\limits_{x\rarr x_{0}}\frac{R'_{n}(x)}{n(x-x_{0})^{n-1}} \overset{\frac{0}{0}}{=} ... \overset{\frac{0}{0}}{=} \lim\limits_{x\rarr x_{0}}\frac{R^{(n-1)}_{n}(x)}{n(n-1)...2(x-x_{0})} x→x0​lim​(x−x0​)nRn​(x)​=00​x→x0​lim​n(x−x0​)n−1Rn′​(x)​=00​...=00​x→x0​lim​n(n−1)...2(x−x0​)Rn(n−1)​(x)​

= lim ⁡ x → x 0 1 n ! ⋅ [ f ( n − 1 ) ( x ) − f ( n − 1 ) ( x 0 ) x − x 0 − f ( n ) ( x 0 ) ] = 0 =\lim\limits_{x\rarr x_{0}}\frac{1}{n!}\cdot[\frac{f^{(n-1)}(x)-f^{(n-1)}(x_{0})}{x-x_{0}}-f^{(n)}(x_{0})]=0 =x→x0​lim​n!1​⋅[x−x0​f(n−1)(x)−f(n−1)(x0​)​−f(n)(x0​)]=0

所以 R n ( x ) = o ( ( x − x 0 ) n ) R_{n}(x)=o((x-x_{0})^{n}) Rn​(x)=o((x−x0​)n)

带Lagrange余项的Taylor公式

定义：

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

设 x 0 ∈ [ a , , b ] x_{0}\in[a,,b] x0​∈[a,,b]为一定点，则对任意 x ∈ [ a , , b ] x\in[a,,b] x∈[a,,b]，

有 f ( x ) = P n ( x ) + R n ( x ) f(x)=P_{n}(x)+R_{n}(x) f(x)=Pn​(x)+Rn​(x)

R n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) n + 1 R_{n}(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_{0})^{n+1} Rn​(x)=(n+1)!f(n+1)(ξ)​(x−x0​)n+1称为 f f f在 x = x 0 x=x_{0} x=x0​处的Lagrange余项，其中 ξ \xi ξ在 x x x和 x 0 x_{0} x0​之间

证明：

即证 R n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) n + 1 R_{n}(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_{0})^{n+1} Rn​(x)=(n+1)!f(n+1)(ξ)​(x−x0​)n+1满足Taylor公式

不妨设 x > x 0 x>x_{0} x>x0​

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

则 G ( x ) = f ( x ) − ∑ k = 0 n f ( k ) ( x ) k ! ( x − x ) k = f ( x ) − f ( x ) = 0 G(x)=f(x)-\sum\limits^{n}_{k=0}\frac{f^{(k)}(x)}{k!}(x-x)^{k}=f(x)-f(x)=0 G(x)=f(x)−k=0∑n​k!f(k)(x)​(x−x)k=f(x)−f(x)=0

     G ( x 0 ) = f ( x 0 ) − ∑ k = 0 n f ( k ) ( x 0 ) k ! ( x − x 0 ) k = f ( x ) − f ( x ) = R n ( x ) G(x_{0})=f(x_{0})-\sum\limits^{n}_{k=0}\frac{f^{(k)}(x_{0})}{k!}(x-x_{0})^{k}=f(x)-f(x)=R_n(x) G(x0​)=f(x0​)−k=0∑n​k!f(k)(x0​)​(x−x0​)k=f(x)−f(x)=Rn​(x)

令 H ( t ) = ( x − t ) n + 1 H(t)=(x-t)^{n+1} H(t)=(x−t)n+1

则 H ( x ) = 0 H(x)=0 H(x)=0

     H ( x 0 ) = ( x − x 0 ) n + 1 H(x_{0})=(x-x_{0})^{n+1} H(x0​)=(x−x0​)n+1

G ( t ) , H ( t ) G(t),H(t) G(t),H(t)在 [ x 0 , x ] [x_{0},x] [x0​,x]连续，在 ( x 0 , x ) (x_{0},x) (x0​,x)可导

G ′ ( t ) = − ∑ k = 0 n 1 k ! [ f ( k ) ( t ) ( x − t ) k ] ′ = − ∑ k = 0 n f ( k + 1 ) ( t ) k ! ( x − t ) k + ∑ k = 1 n f ( k ) ( t ) ( k − 1 ) ! ( x − t ) k − 1 = − ∑ k = 1 n + 1 f ( k ) ( t ) ( k − 1 ) ! ( x − t ) k − 1 + ∑ k = 1 n f ( k ) ( t ) ( k − 1 ) ! ( x − t ) k − 1 = − f ( n + 1 ) ( t ) n ! ( x − t ) n \begin{aligned} G'(t) &= -\sum\limits^{n}_{k=0}\frac{1}{k!}[f^{(k)}(t)(x-t)^{k}]' \\ &= -\sum\limits^{n}_{k=0}\frac{f^{(k+1)}(t)}{k!}(x-t)^{k}+\sum\limits^{n}_{k=1}\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1} \\ &= -\sum\limits^{n+1}_{k=1}\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}+\sum\limits^{n}_{k=1}\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1} \\ &= -\frac{f^{(n+1)}(t)}{n!}(x-t)^{n} \end{aligned} G′(t)​=−k=0∑n​k!1​[f(k)(t)(x−t)k]′=−k=0∑n​k!f(k+1)(t)​(x−t)k+k=1∑n​(k−1)!f(k)(t)​(x−t)k−1=−k=1∑n+1​(k−1)!f(k)(t)​(x−t)k−1+k=1∑n​(k−1)!f(k)(t)​(x−t)k−1=−n!f(n+1)(t)​(x−t)n​

H ′ ( t ) = − ( n + 1 ) ( x − t ) n H'(t)=-(n+1)(x-t)^{n} H′(t)=−(n+1)(x−t)n

R n ( x ) ( x − x 0 ) n + 1 = G ( x 0 ) H ( x 0 ) = G ( x 0 ) − G ( x ) H ( x 0 ) − H ( x ) = G ′ ( ξ ) H ′ ( ξ ) = f n + 1 ( ξ ) ( n + 1 ) ! ( x 0 < ξ < x ) \begin{aligned} \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'(\xi)}{H'(\xi)} \\ &= \frac{f^{n+1}(\xi)}{(n+1)!} & (x_{0}<\xi<x)\end{aligned} (x−x0​)n+1Rn​(x)​​=H(x0​)G(x0​)​=H(x0​)−H(x)G(x0​)−G(x)​=H′(ξ)G′(ξ)​=(n+1)!fn+1(ξ)​​(x0​<ξ<x)​

所以 R n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) n + 1 R_{n}(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_{0})^{n+1} Rn​(x)=(n+1)!f(n+1)(ξ)​(x−x0​)n+1，其中 ξ \xi ξ在 x x x和 x 0 x_{0} x0​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 p_{n}=a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n} pn​=a0​+a1​x+a2​x2+...+an​xn至少需要n+1个条件

目标：对于 [ a , b ] [a,b] [a,b]上的函数 f ( x ) f(x) f(x)，找到n+1个条件来确定 f ( x ) f(x) f(x)的插值多项式 p n ( x ) p_{n}(x) pn​(x)

插值多项式的定义

已知 f ( x ) f(x) f(x)在 ( a , b ) (a,b) (a,b)内的m+1个互异点 x 0 , x 1 , . . . x m x_{0},x_{1},...x_{m} x0​,x1​,...xm​上的函数值和若干阶导数 f ( j ) ( x ) f^{(j)}(x) f(j)(x)，其中 i = 0 , 1 , 2 , . . . , m i=0,1,2,...,m i=0,1,2,...,m， j = 0 , 1 , . . . , m a x ( n i ) − 1 j=0,1,...,max(n_{i})-1 j=0,1,...,max(ni​)−1。

n i n_{i} ni​表示 f ( x ) f(x) f(x)在点 x i x_{i} xi​处已知的导数值的个数， m j m_{j} mj​表示已知 j j j阶导数值的点的个数， ∑ j = 0 n + 1 m j = ∑ i = 0 m n i = n + 1 \sum\limits_{j=0}^{n+1}m_{j}=\sum\limits_{i=0}^{m}n_{i}=n+1 j=0∑n+1​mj​=i=0∑m​ni​=n+1。

若存在一个n次多项式 P n ( x ) P_{n}(x) Pn​(x)满足所有插值条件 p n ( j ) ( x i ) = f ( j ) ( x i ) p^{(j)}_{n}(x_{i})=f^{(j)}(x_{i}) pn(j)​(xi​)=f(j)(xi​)，则称 p n ( x ) p_{n}(x) pn​(x)是 f ( x ) f(x) f(x)关于插值结点 x 0 , x 1 , . . . , x m x_{0},x_{1},...,x_{m} x0​,x1​,...,xm​的n次插值多项式

例如：

	x 0 x_{0} x0​	x 1 x_{1} x1​	x 2 x_{2} x2​	x 3 x_{3} x3​	m j m_{j} mj​
j = 0 j=0 j=0	f ( x 0 ) f(x_{0}) f(x0​)	f ( x 1 ) f(x_{1}) f(x1​)	f ( x 2 ) f(x_{2}) f(x2​)	f ( x 3 ) f(x_{3}) f(x3​)	m 0 = 4 m_{0}=4 m0​=4
j = 1 j=1 j=1	f ′ ( x 0 ) f'(x_{0}) f′(x0​)	—	f ′ ( x 2 ) f'(x_{2}) f′(x2​)	f ′ ( x 3 ) f'(x_{3}) f′(x3​)	m 1 = 3 m_{1}=3 m1​=3
j = 2 j=2 j=2	f ′ ′ ( x 0 ) f''(x_{0}) f′′(x0​)	—	f ′ ′ ( x 2 ) f''(x_{2}) f′′(x2​)	f ′ ′ ( x 3 ) f''(x_{3}) f′′(x3​)	m 2 = 3 m_{2}=3 m2​=3
j = 3 j=3 j=3	—	—	f ′ ′ ′ ( x 2 ) f'''(x_{2}) f′′′(x2​)	—	m 3 = 1 m_{3}=1 m3​=1
n i n_{i} ni​	n 0 = 3 n_{0}=3 n0​=3	n 1 = 1 n_{1}=1 n1​=1	n 2 = 4 n_{2}=4 n2​=4	n 3 = 3 n_{3}=3 n3​=3	n + 1 = 11 n+1=11 n+1=11

插值多项式余项定理

定义：

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

R n ( x ) = f ( x ) − p n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ∏ i = 0 m ( x − x i ) n i R_{n}(x)=f(x)-p_{n}(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod\limits^{m}_{i=0}(x-x_{i})^{n_{i}} Rn​(x)=f(x)−pn​(x)=(n+1)!f(n+1)(ξ)​i=0∏m​(x−xi​)ni​

这里 ξ \xi ξ是介于 x m i n = m i n ( x 0 , x 1 , . . . , x m , x ) x_{min}=min(x_{0},x_{1},...,x_{m},x) xmin​=min(x0​,x1​,...,xm​,x)和 x m a x = m a x ( x 0 , x 1 , . . . , x m , x ) x_{max}=max(x_{0},x_{1},...,x_{m},x) xmax​=max(x0​,x1​,...,xm​,x)之间的一个数（一般依赖于 x x x）

证明：

令 x ≠ x i x\neq x_{i} x​=xi​

令n+1次多项式 w n + 1 ( t ) = ∏ i = 0 m ( t − x i ) n i w_{n+1}(t)=\prod\limits^{m}_{i=0}(t-x_{i})^{n_{i}} wn+1​(t)=i=0∏m​(t−xi​)ni​

则 w n + 1 ( j ) ( x i ) = 0 , ( i = 0 , 1 , 2 , . . . , m j = 0 , 1 , . . . , m a x ( n i ) − 1 ) w^{(j)}_{n+1}(x_{i})=0, \left(\begin{array}{l} i=0,1,2,...,m \\ j=0,1,...,max(n_{i})-1\end{array}\right) wn+1(j)​(xi​)=0,(i=0,1,2,...,mj=0,1,...,max(ni​)−1​)

令 φ ( t ) = f ( t ) − p n ( t ) − w n + 1 ( t ) w n + 1 ( x ) ( f ( x ) − p n ( x ) ) \varphi(t)=f(t)-p_{n}(t)-\frac{w_{n+1}(t)}{w_{n+1}(x)}(f(x)-p_{n}(x)) φ(t)=f(t)−pn​(t)−wn+1​(x)wn+1​(t)​(f(x)−pn​(x))

因为 f ( j ) ( x i ) = p n ( j ) ( x i ) f^{(j)}(x_{i})=p_{n}^{(j)}(x_{i}) f(j)(xi​)=pn(j)​(xi​)且 w n + 1 ( j ) ( x i ) = 0 w^{(j)}_{n+1}(x_{i})=0 wn+1(j)​(xi​)=0

所以 φ ( j ) ( x i ) = f ( j ) ( x i ) − p n ( j ) ( x i ) − w n + 1 ( j ) ( x i ) w n + 1 ( x ) ( f ( x ) − p n ( x ) ) = 0 , ( i = 0 , 1 , 2 , . . . , m j = 0 , 1 , . . . , m a x ( n i ) − 1 ) \varphi^{(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) φ(j)(xi​)=f(j)(xi​)−pn(j)​(xi​)−wn+1​(x)wn+1(j)​(xi​)​(f(x)−pn​(x))=0,(i=0,1,2,...,mj=0,1,...,max(ni​)−1​)

令上式中 j = 0 j=0 j=0，可得 φ ( x i ) = 0 \varphi(x_{i})=0 φ(xi​)=0

又因为 φ ( x ) = f ( x ) − p n ( x ) − w n + 1 ( x ) w n + 1 ( x ) ( f ( x ) − p n ( x ) ) = 0 \varphi(x)=f(x)-p_{n}(x)-\frac{w_{n+1}(x)}{w_{n+1}(x)}(f(x)-p_{n}(x))=0 φ(x)=f(x)−pn​(x)−wn+1​(x)wn+1​(x)​(f(x)−pn​(x))=0

所以 φ ( t ) \varphi(t) φ(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 m_{0}+1 m0​+1个

由Rolle定理：

φ ′ ( t ) \varphi'(t) φ′(t)的零点有 m 0 + m 1 m_{0}+m_{1} m0​+m1​个

φ ′ ′ ( t ) \varphi''(t) φ′′(t)的零点有 m 0 + m 1 + m 2 − 1 m_{0}+m_{1}+m_{2}-1 m0​+m1​+m2​−1个

…

φ ′ ( t ) \varphi'(t) φ′(t)的零点有 m 0 + m 1 + . . . + m n + 1 − n = 1 m_{0}+m_{1}+...+m_{n+1}-n=1 m0​+m1​+...+mn+1​−n=1个，设该零点为 ξ \xi ξ

则 ξ ∈ ( x m i n , x m a x ) , φ ( n + 1 ) ( ξ ) = 0 \xi\in(x_{min},x_{max}),\varphi^{(n+1)}(\xi)=0 ξ∈(xmin​,xmax​),φ(n+1)(ξ)=0

因为 p n ( x ) p_{n}(x) pn​(x)是n次多项式，所以其n+1阶导数为0

对 w n + 1 ( t ) = ∏ i = 0 m ( t − x i ) n i w_{n+1}(t)=\prod\limits^{m}_{i=0}(t-x_{i})^{n_{i}} wn+1​(t)=i=0∏m​(t−xi​)ni​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 t^{n+1} tn+1求n+1阶导数，得到 w n + 1 ( n + 1 ) ( t ) = ( n + 1 ) ! w_{n+1}^{(n+1)}(t)=(n+1)! wn+1(n+1)​(t)=(n+1)!

因此：

φ ( n + 1 ) ( ξ ) = f ( n + 1 ) ( x ) − ( n + 1 ) ! w n + 1 ( x ) ( f ( x ) − p n ( x ) ) = 0 \varphi^{(n+1)}(\xi)=f^{(n+1)}(x)-\frac{(n+1)!}{w_{n+1}(x)}(f(x)-p_{n}(x))=0 φ(n+1)(ξ)=f(n+1)(x)−wn+1​(x)(n+1)!​(f(x)−pn​(x))=0

⇒ R n ( x ) = f ( x ) − p n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ∏ i = 0 m ( x − x i ) n i \Rarr R_{n}(x)=f(x)-p_{n}(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod\limits^{m}_{i=0}(x-x_{i})^{n_{i}} ⇒Rn​(x)=f(x)−pn​(x)=(n+1)!f(n+1)(ξ)​i=0∏m​(x−xi​)ni​