仅供大致参考，有许多定义存在不严谨的地方；不同学校的考察重点自然是不同的
第二章 插值法
拉格朗日插值
P n ( x ) = L n ( x ) = ∑ i = 0 n f ( x i ) l i ( x ) P_{n}(x)=L_{n}(x)=\sum_{i=0}^{n} f\left(x_{i}\right) l_{i}(x) Pn(x)=Ln(x)=i=0∑nf(xi)li(x) 其中拉格朗日基函数 { l i ( x ) } i = 0 n \left\{l_{i}(x)\right\}_{i=0}^{n} { li(x)}i=0n定义如下： l i ( x ) = ∏ k = 0 k ≠ i n x − x k x i − x k = ω n + 1 ( x ) ( x − x i ) ω n + 1 ′ ( x i ) l_{i}(x)=\prod_{k=0 \atop k \neq i}^{n} \frac{x-x_{k}}{x_{i}-x_{k}}=\frac{\omega_{n+1(x)}}{\left(x-x_{i}\right) \omega_{n+1}^{\prime}\left(x_{i}\right)} li(x)=k=ik=0∏nxi−xkx−xk=(x−xi)ωn+1′(xi)ωn+1(x) 该基函数有如下性质(可由余项推得)： ∑ i = 0 n x i k l i ( x ) = x k , k = 0 , 1 , … , n \sum_{i=0}^{n} x_{i}^{k} l_{i}(\mathrm{x})=x^{k}, \quad k=0,1, \ldots, n i=0∑nxikli(x)=xk,k=0,1,…,n
牛顿插值
P n ( x ) = N n ( x ) = ∑ i = 0 n f [ x 0 , x 1 , ⋯ , x i ] ω i ( x ) P_{n}(x)=N_{n}(x)=\sum_{i=0}^{n} f\left[x_{0}, x_{1}, \cdots, x_{i}\right] \omega_{i}(x) Pn(x)=Nn(x)=i=0∑nf[x0,x1,⋯,xi]ωi(x) 其中 ω 0 ( x ) ≡ 1 , ω i ( x ) = ∏ k = 0 i − 1 ( x − x k ) \omega_{0}(x) \equiv 1, \omega_{i}(x)=\prod_{k=0}^{i-1}\left(x-x_{k}\right) ω0(x)≡1,ωi(x)=k=0∏i−1(x−xk)
插值余项
记原函数(被插值函数)为 f ( x ) f(x) f(x)，插值函数作为原函数的一个近似，为 P ( x ) P(x) P(x)，则有差值余项为： R ( x ) = def f ( x ) − P ( x ) R(x) \stackrel{\text { def }}{=} f(x)-P(x) R(x)= def f(x)−P(x) 事实上插值余项也可以看做是一种截断误差。
ω n + 1 ( x ) \omega_{n+1}(x) ωn+1(x)与 ω n ′ ( x j ) \omega_{n}'(x_j) ωn′(xj)
ω n + 1 ( x ) = ( x − x 0 ) ( x − x 1 ) … ( x − x n ) \omega_{n+1}(x)=\left(x-x_{0}\right)\left(x-x_{1}\right) \ldots\left(x-x_{n}\right) ωn+1(x)=(x−x0)(x−x1)…(x−xn) ω n + 1 ′ ( x j ) = ( x j − x 0 ) ⋯ ( x j − x j − 1 ) ( x j − x j + 1 ) ⋯ ( x j − x n ) \omega_{n+1}'(x_j)=\left(x_{j}-x_{0}\right) \cdots\left(x_{j}-x_{j-1}\right)\left(x_{j}-x_{j+1}\right) \cdots\left(x_{j}-x_{n}\right) ωn+1′(xj)=(xj−x0)⋯(xj−xj−1)(xj−xj+1)⋯(xj−xn)
插值余项的估计
注： M n + 1 = max a ⩽ x ⩽ b ∣ f ( n + 1 ) ( x ) ∣ M_{n+1}=\max _{a \leqslant x \leqslant b}\left|f^{(n+1)}(x)\right| Mn+1=maxa⩽x⩽b∣∣f(n+1)(x)∣∣为 ∣ f ( n + 1 ) ( x ) ∣ \left|f^{(n+1)}(x)\right| ∣∣f(n+1)(x)∣∣在 [ a , b ] [a,b] [a,b]的上界
考虑被插值函数 f ( x ) ∈ C n + 1 [ a , b ] f(x) \in C^{n+1}[a, b] f(x)∈Cn+1[a,b]，可得插值余项为： R n ( x ) = f ( x ) − L n ( x ) = f ( n + 1 ) ( ξ x ) ( n + 1 ) ! ω n + 1 ( x ) R_{n}(x)=f(x)-L_{n}(x)=\frac{f^{(n+1)}\left(\xi_{x}\right)}{(n+1) !} \omega_{n+1}(x) Rn(x)=f(x)−Ln(x)=(n+1)!f(n+1)(ξx)ωn+1(x) 插值余项的截断误差界为： ∣ R n ( x ) ∣ ⩽ M n + 1 ( n + 1 ) ! ∣ ω n + 1 ( x ) ∣ \left|R_{n}(x)\right| \leqslant \frac{M_{n+1}}{(n+1) !}\left|\omega_{n+1}(x)\right| ∣Rn(x)∣⩽(n+1)!Mn+1∣ωn+1(x)∣
均差(差商)
称 f [ x 0 , x k ] = f ( x k ) − f ( x 0 ) x k − x 0 f\left[x_{0}, x_{k}\right]=\frac{f\left(x_{k}\right)-f\left(x_{0}\right)}{x_{k}-x_{0}} f[x0,xk]=xk−x0f(xk)−f(x0)为函数 f ( x ) f(x) f(x)关于点 x 0 x_{0} x0， x k x_{k} xk的一阶均差(一种看起来像斜率的东西)， f [ x 0 , x 1 , x k ] = f [ x 0 , x k ] − f [ x 0 , x 1 ] x k − x 1 f[x_{0},\left.x_{1}, x_{k}\right]=\frac{f\left[x_{0}, x_{k}\right]-f\left[x_{0}, x_{1}\right]}{x_{k}-x_{1}} f[x0,x1,xk]=xk−x1f[x0,xk]−f[x0,x1]称为 f ( x ) f(x) f(x)的二阶均差。一般地，有： f [ x 0 , x 1 , ⋯ , x k ] = f [ x 0 , ⋯ , x k − 2 , x k ] − f [ x 0 , x 1 , ⋯ , x k − 1 ] x k − x k − 1 f\left[x_{0}, x_{1}, \cdots, x_{k}\right]=\frac{f\left[x_{0}, \cdots, x_{k-2}, x_{k}\right]-f\left[x_{0}, x_{1}, \cdots, x_{k-1}\right]}{x_{k}-x_{k-1}} f[x0,x1,⋯,xk]=xk−xk−1f[x0,⋯,xk−2,xk]−f[x0,x1,⋯,xk−1] 为 f ( x ) f(x) f(x)的 k k k阶均差。
均差的性质
n阶均差可表示为函数值 f ( x 0 ) , f ( x 1 ) , . . . , f ( x n ) f(x_0),f(x_1),...,f(x_n) f(x0),f(x1),...,f(xn)的线性组合： f [ x 0 , x 1 , ⋯ , x n ] = ∑ j = 0 k f ( x j ) ( x j − x 0 ) ⋯ ( x j − x j − 1 ) ( x j − x j + 1 ) ⋯ ( x j − x n ) \left.f[x_{0}, x_{1}, \cdots, x_{n}\right]=\sum_{j=0}^{k} \frac{f\left(x_{j}\right)}{\left(x_{j}-x_{0}\right) \cdots\left(x_{j}-x_{j-1}\right)\left(x_{j}-x_{j+1}\right) \cdots\left(x_{j}-x_{n}\right)} f[x0,x1,⋯,xn]=j=0∑k(xj−x0)⋯(xj−xj−1)(xj−xj+1)⋯(xj−xn)f(xj) 更常见的是下面这种形式： f [ x 0 , x 1 , ⋯ , x n ] = ∑ j = 0 n f ( x j ) ω n ′ ( x j ) f\left[x_{0}, x_{1}, \cdots, x_{n}\right]=\sum_{j=0}^{n} \frac{f\left(x_{j}\right)}{\omega_{n}^{\prime}\left(x_{j}\right)} f[x0,x1,⋯,xn]=j=0∑nωn′(xj)f(xj)若 f ( x ) f(x) f(x)在 [ a , b ] [a,b] [a,b]上存在 n n n阶导数，且节点 x 0 ， x 1 ， . . . ， x n ∈ [ a , b ] x_{0}，x_{1}，...，x_{n} \in [a,b] x0，x1，...，xn∈[a,b]，则 n n n阶均差与导数的关系为： f [ x 0 , x 1 , ⋯ , x n ⌋ = f ( n ) ( ξ ) n ! , ξ ∈ [ a , b ] f\left[x_{0}, x_{1}, \cdots, x_{n}\right\rfloor=\frac{f^{(n)}(\xi)}{n !}, \quad \xi \in[a, b] f[x0,x1,⋯,xn⌋=n!f(n)(ξ),ξ∈[a,b]
差分
令 x k x_k xk处的函数值为 f k = f ( x k ) f_k=f(x_k) fk=f(xk)，称 Δ f k = f k + 1 − f k \Delta f_k=f_{k+1}-f_k Δfk=fk+1−fk为 x k x_k xk处以 h h h为步长的一阶(向前)差分。类似地， Δ 2 f k = Δ f k + 1 − Δ f k \Delta^2 f_k=\Delta f_{k+1}-\Delta f_k Δ2fk=Δfk+1−Δfk为 x k x_k xk处的二阶差分。一般地，记 Δ n f k = Δ n − 1 f k + 1 − Δ n − 1 f k \Delta^n f_k=\Delta^{n-1} f_{k+1}-\Delta^{n-1} f_k Δnfk=Δn−1fk+1−Δn−1fk 为 x k x_k xk处的 n n n阶差分。
三次样条插值
若函数 S ( x ) ∈ C 2 [ a , b ] S(x)\in C^{2}[a,b] S(x)∈C2[a,b]，且在每个子区间 [ x k , x k + 1 ] [x_k,x_{k+1}] [xk,xk+1](k=0,1,…,n-1)上是三次多项式，则称 S ( x ) S(x) S(x)是关于节点 a = x 0 < x 1 < . . . < x n = b a=x_0<x_1<...<x_n=b a=x0<x1<...<xn=b的三次样条函数。若进一步有 S ( x k ) = f ( x k ) ( k = 0 , 1 , . . . , n ) S(x_k)=f(x_k)(k=0,1,...,n) S(xk)=f(xk)(k=0,1,...,n)，则称 S ( x ) S(x) S(x)是 f ( x ) f(x) f(x)关于 { x i } i = 0 n \left\{x_{i}\right\}_{i=0}^{n} { xi}i=0n的三次样条插值函数。相应的插值形式如下： S ( x ) = m k ( x k + 1 − x ) 3 6 h k + M k + 1 ( x − x k ) 3 6 h k + ( f ( x k ) − M k h k 2 6 ) x k + 1 − x h k + ( f ( x k + 1 ) − M k + 1 h k 2 6 ) x − x k h k x ∈ [ x k , x k + 1 ] , k = 0 , 1 , ⋯ , n − 1 \begin{aligned} S(x) &=m_{k} \frac{\left(x_{k+1}-x\right)^{3}}{6 h_{k}}+M_{k+1} \frac{\left(x-x_{k}\right)^{3}}{6 h_{k}} \\ &+\left(f\left(x_{k}\right)-\frac{M_{k} h_{k}^{2}}{6}\right) \frac{x_{k+1}-x}{h_{k}}+\left(f\left(x_{k+1}\right)-\frac{M_{k+1} h_{k}^{2}}{6}\right) \frac{x-x_{k}}{h_{k}} \\ x & \in\left[x_{k}, x_{k+1}\right], k=0,1, \cdots, n-1 \end{aligned} S(x)x=mk6hk(xk+1−x)3+Mk+16hk(x−xk)3+(f(xk)−6Mkhk2)hkxk+1−x+(f(xk+1)−6Mk+1hk2)hkx−xk∈[xk,xk+1],k=0,1,⋯,n−1 其中 h k = x h + 1 − x k h_k=x_{h+1}-x_{k} hk=xh+1−xk
例1
设 x i ( i = 0 , 1 , 2 , 3 , 4 , 5 ) x_i(i=0,1,2,3,4,5) xi(i=0,1,2,3,4,5)是互异结点， l i ( x ) ( i = 0 , 1 , 2 , 3 , 4 , 5 ) l_i(x)(i=0,1,2,3,4,5) li(x)(i=0,1,2,3,4,5)是对应的5次拉格朗日插值基函数，求 ∑ i = 0 5 x i 5 l i ( 0 ) \sum_{i=0}^{5} x_{i}^{5} l_{i}(0) ∑i=05xi5li(0)
解 根据拉格朗日基函数的性质，有 ∑ i = 0 n x i k l i ( x ) = x k , k = 0 , 1 , … , n \sum_{i=0}^{n} x_{i}^{k} l_{i}(\mathrm{x})=x^{k}, \quad k=0,1, \ldots, n i=0∑nxikli(x)=xk,k=0,1,…,n 这里的话 n = 5 n=5 n=5， x = 0 x=0 x=0，因此 ∑ i = 0 5 x i 5 l i ( 0 ) = 0 5 = 0 \sum_{i=0}^{5} x_{i}^{5} l_{i}(0)=0^5=0 ∑i=05xi5li(0)=05=0
例2
设函数 f ( x ) = e − 2 x f(x)=e^{-2x} f(x)=e−2x， L 2 ( x ) L_2(x) L2(x)是以0,1,2为节点的 f ( x ) f(x) f(x)的二次拉格朗日插值多项式，求其余项
解 无论是拉格朗日插值还是别的插值，其余项都可以用上面那个余项公式求得，即 R n ( x ) = f ( n + 1 ) ( ξ x ) ( n + 1 ) ! ω n + 1 ( x ) = − 4 e − 2 ξ 3 x ( x − 1 ) ( x − 2 ) R_{n}(x)=\frac{f^{(n+1)}\left(\xi_{x}\right)}{(n+1) !} \omega_{n+1}(x) = -\frac{4 e^{-2 \xi}}{3} x(x-1)(x-2) Rn(x)=(n+1)!f(n+1)(ξx)ωn+1(x)=−34e−2ξx(x−1)(x−2)
例3
多项式 f ( x ) = x 3 + x + 1 f(x)=x^3+x+1 f(x)=x3+x+1，求差商 f [ 0 , 1 , 2 ] f[0,1,2] f[0,1,2]
解 根据均差可以表示为函数值的线性组合，有 f [ 0 , 1 , 2 ] = f ( 0 ) ( 0 − 1 ) ( 0 − 2 ) + f ( 1 ) ( 1 − 0 ) ( 1 − 2 ) + f ( 2 ) ( 2 − 0 ) ( 2 − 1 ) = 3 f[0,1,2]=\frac{f(0)}{(0-1)(0-2)} + \frac{f(1)}{(1-0)(1-2)} + \frac{f(2)}{(2-0)(2-1)} = 3 f[0,1,2]=(0−1)(0−2)f(0)+(1−0)(1−2)f(1)+(2−0)(2−1)f(2)=3
例4
证明 n n n阶均差有下列性质：若 F ( x ) = f ( x ) + g ( x ) F(x)=f(x)+g(x) F(x)=f(x)+g(x)，则： F [ x 0 , x 1 , ⋯ x n ] = f [ x 0 , x 1 , ⋯ , x n ] + g [ x 0 , x 1 , ⋯ , x n ] F\left[x_{0}, x_{1}, \cdots x_{n}\right]=f\left[x_{0}, x_{1}, \cdots, x_{n}\right]+g\left[x_{0}, x_{1}, \cdots, x_{n}\right] F[x0,x1,⋯xn]=f[x0,x1,⋯,xn]+g[x0,x1,⋯,xn] 解 F [ x 0 , x 1 , ⋯ , x n ] = ∑ j = 0 n F ( x j ) ω n ′ ( x j ) = ∑ j = 0 n f ( x j ) + g ( x j ) ω n ′ ( x j ) = ∑ j = 0 n f ( x j ) ω n ′ ( x j ) + ∑ j = 0 n g ( x j ) ω n ′ ( x j ) = f [ x 0 , x 1 , ⋯ , x n ] + g [ x 0 , x 1 , ⋯ , x n ] . \begin{aligned} F\left[x_{0}, x_{1}, \cdots, x_{n}\right] &=\sum_{j=0}^{n} \frac{F\left(x_{j}\right)}{\omega_{n}^{\prime}\left(x_{j}\right)}=\sum_{j=0}^{n} \frac{f\left(x_{j}\right)+g\left(x_{j}\right)}{\omega_{n}^{\prime}\left(x_{j}\right)} \\ &=\sum_{j=0}^{n} \frac{f\left(x_{j}\right)}{\omega_{n}^{\prime}\left(x_{j}\right)}+\sum_{j=0}^{n} \frac{g\left(x_{j}\right)}{\omega_{n}^{\prime}\left(x_{j}\right)} \\ &=f\left[x_{0}, x_{1}, \cdots, x_{n}\right]+g\left[x_{0}, x_{1}, \cdots, x_{n}\right] . \end{aligned} F[x0,x1,⋯,xn]=j=0∑nωn′(xj)F(xj)=j=0∑nωn′(xj)f(xj)+g(xj)=j=0∑nωn′(xj)f(xj)+j=0∑nωn′(xj)g(xj)=f[x0,x1,⋯,xn]+g[x0,x1,⋯,xn].
例5
设 f ( x ) ∈ C 2 [ a , b ] f(x) \in C^{2}[a, b] f(x)∈C2[a,b]且 f ( a ) = f ( b ) = 0 f(a)=f(b)=0 f(a)=f(b)=0，求证： max a ⩽ x ⩽ b ∣ f ( x ) ∣ ⩽ 1 8 ( b − a ) 2 max a ⩽ x ⩽ b ∣ f ′ ′ ( x ) ∣ \max _{a \leqslant x \leqslant b}|f(x)| \leqslant \frac{1}{8}(b-a)^{2} \max _{a \leqslant x \leqslant b}\left|f^{\prime\prime}(x)\right| a⩽x⩽bmax∣f(x)∣⩽81(b−a)2a⩽x⩽bmax∣f′′(x)∣
注 f ( x ) ∈ C 2 [ a , b ] f(x) \in C^{2}[a, b] f(x)∈C2[a,b]表示 f ( x ) f(x) f(x)在闭区间 [ a , b ] [a,b] [a,b]上有连续二阶导数
解 首先构造一个 f ( x ) f(x) f(x)的线性插值： L 1 ( x ) = f ( a ) x − b a − b + f ( b ) x − a b − a ≡ 0 L_{1}(x)=f(a) \frac{x-b}{a-b}+f(b) \frac{x-a}{b-a} \equiv 0 L1(x)=f(a)a−bx−b+f(b)b−ax−a≡0 现在，我们有了原函数与插值函数，可以计算差值余项如下： ∣ f ( x ) − L 1 ( x ) ∣ ⩽ M 2 2 ! ∣ ω 2 ( x ) ∣ \left|f(x)-L_{1}(x)\right| \leqslant \frac{M_{2}}{2!}\left|\omega_{2}(x)\right| ∣f(x)−L1(x)∣⩽2!M2∣ω2(x)∣ 其中 ω 2 ( x ) = ( x − a ) ( x − b ) \omega_{2}(x)=\left(x-a\right)\left(x-b\right) ω2(x)=(x−a)(x−b)；考虑 f ′ ′ ( x ) f^{\prime \prime}(x) f′′(x)在 ξ \xi ξ处取得最大值，即 M 2 = f ′ ′ ( ξ ) M_{2}=f^{\prime \prime}(\xi) M2=f′′(ξ)，有： ∣ f ( x ) − L 1 ( x ) ∣ = ∣ 1 2 ! f ′ ′ ( ξ ) ( x − a ) ( x − b ) ∣ ξ ∈ ( a , b ) ⩽ 1 2 max a ⩽ x ⩽ b ∣ f ′ ′ ( x ) ∣ max a ⩽ x ⩽ b ∣ ( x − a ) ( x − b ) ∣ \begin{aligned} \left|f(x)-L_{1}(x)\right| &=\left|\frac{1}{2 !} f^{\prime \prime}(\xi)(x-a)(x-b)\right| \quad \xi \in(a, b) \\ & \leqslant \frac{1}{2} \max _{a \leqslant x \leqslant b}\left|f^{\prime \prime}(x)\right| \max _{a \leqslant x \leqslant b}|(x-a)(x-b)| \\ \end{aligned} ∣f(x)−L1(x)∣=∣∣∣∣2!1f′′(ξ)(x−a)(x−b)∣∣∣∣ξ∈(a,b)⩽21a⩽x⩽bmax∣f′′(x)∣a⩽x⩽bmax∣(x−a)(x−b)∣ 又 max a ⩽ x ⩽ b ∣ ( x − a ) ( x − b ) ∣ = 1 4 ( b − a ) 2 \max _{a \leqslant x \leqslant b}|(x-a)(x-b)|=\frac{1}{4}(b-a)^{2} maxa⩽x⩽b∣(x−a)(x−b)∣=41(b−a)2(二次函数极值)，因此有 max a ⩽ x ⩽ b ∣ f ( x ) ∣ ⩽ 1 8 ( b − a ) 2 max a ⩽ x ⩽ b ∣ f ′ ′ ( x ) ∣ \max _{a \leqslant x \leqslant b}|f(x)| \leqslant \frac{1}{8}(b-a)^{2} \max _{a \leqslant x \leqslant b}\left|f^{\prime\prime}(x)\right| maxa⩽x⩽b∣f(x)∣⩽81(b−a)2maxa⩽x⩽b∣f′′(x)∣
例6
证明 Δ ( f k g k ) = f k Δ g k + g k + 1 Δ f k \Delta\left(f_{k} g_{k}\right)=f_{k} \Delta g_{k}+g_{k+1} \Delta f_{k} Δ(fkgk)=fkΔgk+gk+1Δfk
解 Δ ( f k g k ) = f k + 1 g k + 1 − f k g k = f k + 1 g k + 1 − f k g k + 1 + f k g k + 1 − f k g k = ( f k + 1 − f k ) g k + 1 + f k ( g k + 1 − g k ) = g k + 1 Δ f k + f k Δ g k . \begin{aligned} \Delta\left(f_{k} g_{k}\right) &=f_{k+1} g_{k+1}-f_{k} g_{k} \\ &=f_{k+1} g_{k+1}-f_{k} g_{k+1}+f_{k} g_{k+1}-f_{k} g_{k} \\ &=\left(f_{k+1}-f_{k}\right) g_{k+1}+f_{k}\left(g_{k+1}-g_{k}\right) \\ &=g_{k+1} \Delta f_{k}+f_{k} \Delta g_{k} . \end{aligned} Δ(fkgk)=fk+1gk+1−fkgk=fk+1gk+1−fkgk+1+fkgk+1−fkgk=(fk+1−fk)gk+1+fk(gk+1−gk)=gk+1Δfk+fkΔgk.
例7
证明 ∑ k = 0 n − 1 f k Δ g k = f n g n − f 0 g 0 − ∑ k = 0 n − 1 g k + 1 Δ f k \sum_{k=0}^{n-1} f_{k} \Delta g_{k}=f_{n} g_{n}-f_{0} g_{0}-\sum_{k=0}^{n-1} g_{k+1} \Delta f_{k} ∑k=0n−1fkΔgk=fngn−f0g0−∑k=0n−1gk+1Δfk
解 ∑ k = 0 n − 1 f k Δ g k + ∑ k = 0 n − 1 g k + 1 Δ f k = ∑ k = 0 n − 1 ( f k Δ g k + g k + 1 Δ f k ) = ∑ k = 0 n − 1 ( f k + 1 g k + 1 − f k g k ) = f n g n − f 0 g 0 \begin{aligned} \sum_{k=0}^{n-1} f_{k} \Delta g_{k}+\sum_{k=0}^{n-1} g_{k+1} \Delta f_{k} &=\sum_{k=0}^{n-1}\left(f_{k} \Delta g_{k}+g_{k+1} \Delta f_{k}\right) \\ &=\sum_{k=0}^{n-1}\left(f_{k+1} g_{k+1}-f_{k} g_{k}\right) \\ &=f_{n} g_{n}-f_{0} g_{0} \end{aligned} k=0∑n−1fkΔgk+k=0∑n−1gk+1Δfk=k=0∑n−1(fkΔgk+gk+1Δfk)=k=0∑n−1(fk+1gk+1−fkgk)=fngn−f0g0 因此 ∑ k = 0 n − 1 f k Δ g k = f n g n − f 0 g 0 − ∑ k = 0 n − 1 g k + 1 Δ f k \sum_{k=0}^{n-1} f_{k} \Delta g_{k}=f_{n} g_{n}-f_{0} g_{0}-\sum_{k=0}^{n-1} g_{k+1} \Delta f_{k} ∑k=0n−1fkΔgk=fngn−f0g0−∑k=0n−1gk+1Δfk
例8
若 f ( x ) = a 0 + a 1 x + ⋯ + a n − 1 x n − 1 + a n x n f(x)=a_{0}+a_{1} x+\cdots+a_{n-1} x^{n-1}+a_{n} x^{n} f(x)=a0+a1x+⋯+an−1xn−1+anxn有 n n n个不同实根 x 1 , x 2 , . . . , x n x_1,x_2,...,x_n x1,x2,...,xn，证明： ∑ j = 1 n x j k f ′ ( x j ) = { 0 , 0 ⩽ k ⩽ n − 2 ; 1 a n , k = n − 1 \sum_{j=1}^{n} \frac{x_{j}^{k}}{f^{\prime}\left(x_{j}\right)}=\left\{\begin{array}{l} 0, \quad 0 \leqslant k \leqslant n-2 ; \\ \frac{1}{a_{n}}, k=n-1 \end{array}\right. j=1∑nf′(xj)xjk={
0,0⩽k⩽n−2;an1,k=n−1
解 根据 f ( x ) f(x) f(x)的互异实根，可以得到： f ( x ) = a n ( x − x 1 ) ⋯ ( x − x n ) = a n ω n ( x ) f(x)=a_{n}\left(x-x_{1}\right) \cdots\left(x-x_{n}\right)=a_{n} \omega_{n}(x) f(x)=an(x−x1)⋯(x−xn)=anωn(x) 这样我们就可以更方便地把 ∑ j = 1 n x j k f ′ ( x j ) = ∑ j = 1 n x j k a n ω n ′ ( x j ) = 1 a n ∑ j = 1 n x j k ω n ′ ( x j ) \sum_{j=1}^{n} \frac{x_{j}^{k}}{f^{\prime}\left(x_{j}\right)}=\sum_{j=1}^{n} \frac{x_{j}^{k}}{a_{n} \omega_{n}^{\prime}\left(x_{j}\right)}=\frac{1}{a_{n}} \sum_{j=1}^{n} \frac{x_{j}^{k}}{\omega_{n}^{\prime}\left(x_{j}\right)} j=1∑nf′(xj)xjk=j=1∑nanωn′(xj)xjk=an1j=1∑nωn′(xj)xjk记 g ( x ) = x k g(x)=x^k g(x)=xk，利用差商的函数值表达形式，有： ∑ j = 1 n x j k f ′ ( x j ) = 1 a n ∑ j = 1 n g ( x j ) ω n ′ ( x j ) = 1 a n g [ x 1 , x 2 , ⋯ , x n ] \sum_{j=1}^{n} \frac{x_{j}^{k}}{f^{\prime}\left(x_{j}\right)}=\frac{1}{a_{n}} \sum_{j=1}^{n} \frac{g\left(x_{j}\right)}{\omega_{n}^{\prime}\left(x_{j}\right)}=\frac{1}{a_{n}} g\left[x_{1}, x_{2}, \cdots, x_{n}\right] j=1∑nf′(xj)xjk=an1j=1∑nωn′(xj)g(xj)=an1g[x1,x2,⋯,xn] 再根据差商与导数的关系，得到： ∑ j = 0 n x j k f ′ ( x j ) = { 0 , 0 ⩽ k ⩽ n − 2 1 a n g ( n − 1 ) ( n − 1 ) ! ( n − 1 a n , k = n − 1 \sum_{j=0}^{n} \frac{x_{j}^{k}}{f^{\prime}\left(x_{j}\right)}=\left\{\begin{aligned} 0, & 0 \leqslant k \leqslant n-2 \\ \frac{1}{a_{n}} \frac{g^{(n-1)}(n-1) !}{\left(n-\frac{1}{a_{n}}\right.}, & k=n-1 \end{aligned}\right. j=0∑nf′(xj)xjk=⎩⎪⎪⎨⎪⎪⎧0,an1(n−an1g(n−1)(n−1)!,0⩽k⩽n−2k=n−1
例9
给定数据表如下 x j 0.25 0.30 0.39 0.45 0.53 y j 0.5000 0.5477 0.6245 0.6708 0.7280 \begin{array}{|c|c|c|c|c|c|} \hline x_{j} & 0.25 & 0.30 & 0.39 & 0.45 & 0.53 \\ \hline y_{j} & 0.5000 & 0.5477 & 0.6245 & 0.6708 & 0.7280 \\ \hline \end{array} xjyj0.250.50000.300.54770.390.62450.450.67080.530.7280 试求三次样条插值 S ( x ) S(x) S(x)，并满足条件 S ′ ′ ( 0.25 ) = S ′ ′ ( 0.53 ) = 0 S^{\prime \prime}(0.25)=S^{\prime \prime}(0.53)=0 S′′(0.25)=S′′(0.53)=0。
注 两端的二阶导数已知， S ′ ′ ( x 0 ) = S ′ ′ ( x n ) = 0 S^{\prime \prime}(x_0)=S^{\prime \prime}(x_n)=0 S′′(x0)=S′′(xn)=0，属于一种自然边界条件。
解 首先计算 h i h_i hi，有： h 0 = 0.30 − 0.25 = 0.05 h 1 = 0.39 − 0.30 = 0.09 h 2 = 0.45 − 0.39 = 0.06 h 3 = 0.53 − 0.45 = 0.08 \begin{array}{ll} h_{0}=0.30-0.25=0.05 & h_{1}=0.39-0.30=0.09 \\ h_{2}=0.45-0.39=0.06 & h_{3}=0.53-0.45=0.08 \end{array} h0=0.30−0.25=0.05h2=0.45−0.39=0.06h1=0.39−0.30=0.09h3=0.53−0.45=0.08 然后，根据 μ i = h i − 1 h i − 1 + h i , λ i = h i h i − 1 + h i \mu_{i}=\frac{h_{i-1}}{h_{i-1}+h_{i}}, \lambda_{i}=\frac{h_{i}}{h_{i-1}+h_{i}} μi=hi−1+hihi−1,λi=hi−1+hihi，可解得： μ 1 = 5 14 , λ 1 = 9 14 μ 2 = 3 5 , λ 2 = 2 5 μ 3 = 3 7 , λ 3 = 4 7 \begin{aligned} &\mu_{1}=\frac{5}{14}, \quad \lambda_{1}=\frac{9}{14} \\ &\mu_{2}=\frac{3}{5}, \quad \lambda_{2}=\frac{2}{5} \\ &\mu_{3}=\frac{3}{7}, \quad \lambda_{3}=\frac{4}{7} \end{aligned} μ1=145,λ1=149μ2=53,λ2=52μ3=73,λ3=74 现在已知二阶导数边界条件，有 M 0 = M 4 = 0 M_0=M_4=0 M0=M4=0，构建弯矩方程组： [ 2 9 14 0 3 5 2 2 5 0 3 7 2 ] [ M 1 M 2 M 3 ] = 6 [ − 0.719 3 − 0.544 0 − 0.405 0 ] \left[\begin{array}{ccc} 2 & \frac{9}{14} & 0 \\ \frac{3}{5} & 2 & \frac{2}{5} \\ 0 & \frac{3}{7} & 2 \end{array}\right]\left[\begin{array}{l} M_{1} \\ M_{2} \\ M_{3} \end{array}\right]=6\left[\begin{array}{l} -0.719 & 3 \\ -0.544 & 0 \\ -0.405 & 0 \end{array}\right] ⎣⎡25301492730522⎦⎤⎣⎡M1M2M3⎦⎤=6⎣⎡−0.719−0.544−0.405300⎦⎤ 利用追赶法，解得 M 1 ≈ − 1.8809 , M 2 ≈ − 0.8616 , M 3 ≈ − 1.0314 M_{1} \approx-1.8809, \quad M_{2} \approx-0.8616, \quad M_{3} \approx-1.0314 M1≈−1.8809,M2≈−0.8616,M3≈−1.0314，即 S ( x ) = { − 6.2697 x 3 + 4.7023 x 2 − 0.2059 x + 0.3555 , x ∈ [ 0.25 , 0.30 ] 1.8876 x 3 − 2.639 x 2 + 1.9966 x + 0.1353 , x ∈ [ 0.25 , 0.39 ] − 0.4689 x 3 + 0.1178 x 2 + 0.9213 x + 0.2751 , x ∈ [ 0.39 , 0.45 ] 2.1467 x 3 − 3.4132 x 2 + 2.5103 x + 0.0367 , x ∈ [ 0.45 , 0.53 ] S(x)=\left\{\begin{array}{cc} -6.2697 x^{3}+4.7023 x^{2}-0.2059 x+0.3555, x \in[0.25,0.30] \\ 1.8876 x^{3}-2.639 x^{2}+1.9966 x+0.1353, x \in[0.25,0.39]\\ -0.4689 x^{3}+0.1178 x^{2}+0.9213 x+0.2751, x \in[0.39,0.45] \\ 2.1467 x^{3}-3.4132 x^{2}+2.5103 x+0.0367, x \in[0.45,0.53] \end{array}\right. S(x)=⎩⎪⎪⎨⎪⎪⎧−6.2697x3+4.7023x2−0.2059x+0.3555,x∈[0.25,0.30]1.8876x3−2.639x2+1.9966x+0.1353,x∈[0.25,0.39]−0.4689x3+0.1178x2+0.9213x+0.2751,x∈[0.39,0.45]2.1467x3−3.4132x2+2.5103x+0.0367,x∈[0.45,0.53]
例10
在等距节点的二次拉格朗日插值多项式中，若函数值带有误差 ε i = f ( x i ) − f i , i = 0 , 1 , 2 {\varepsilon _i} = f({x_i}) - {f_i},i = 0,1,2 εi=f(xi)−fi,i=0,1,2，并且 ε = max { ∣ ε 1 ∣ , ∣ ε 2 ∣ , ∣ ε 3 ∣ } \varepsilon = \max \{ |{\varepsilon _1}|,|{\varepsilon _2}|,|{\varepsilon _3}|\} ε=max{ ∣ε1∣,∣ε2∣,∣ε3∣}，试证，用近似值 f i f_i fi进行运算后，相应的误差界为 5 4 ε \frac{5}{4} \varepsilon 45ε
证 首先在真值 f ( x i ) f(x_i) f(xi)上进行插值，可以得到插值多项式： L 2 ( x ) = ( x − x 1 ) ( x − x 2 ) ( x 0 − x 1 ) ( x 0 − x 2 ) f ( x 0 ) + ( x − x 0 ) ( x − x 2 ) ( x 1 − x 0 ) ( x 1 − x 2 ) f ( x 1 ) + ( x − x 0 ) ( x − x 1 ) ( x 2 − x 0 ) ( x 2 − x 1 ) f ( x 2 ) {L_2}(x) = { {(x - {x_1})(x - {x_2})} \over {({x_0} - {x_1})({x_0} - {x_2})}}f({x_0}) + { {(x - {x_0})(x - {x_2})} \over {({x_1} - {x_0})({x_1} - {x_2})}}f({x_1}) + { {(x - {x_0})(x - {x_1})} \over {({x_2} - {x_0})({x_2} - {x_1})}}f({x_2}) L2(x)=(x0−x1)(x0−x2)(x−x1)(x−x2)f(x0)+(x1−x0)(x1−x2)(x−x0)(x−x2)f(x1)+(x2−x0)(x2−x1)(x−x0)(x−x1)f(x2) 既然是等距节点，我们可以先令 x 1 = x 0 + h x_1 = x_0 + h x1=x0+h， x 2 = x 0 + 2 h x_2 = x_0 + 2h x2=x0+2h，则有： L 2 ( x ) = ( x − x 1 ) ( x − x 2 ) 2 h 2 f ( x 0 ) − ( x − x 0 ) ( x − x 2 ) h 2 f ( x 1 ) + ( x − x 0 ) ( x − x 1 ) 2 h 2 f ( x 2 ) {L_2}(x) = { {(x - {x_1})(x - {x_2})} \over {2{h^2}}}f({x_0}) - { {(x - {x_0})(x - {x_2})} \over { {h^2}}}f({x_1}) + { {(x - {x_0})(x - {x_1})} \over {2{h^2}}}f({x_2}) L2(x)=2h2(x−x1)(x−x2)f(x0)−h2(x−x0)(x−x2)f(x1)+2h2(x−x0)(x−x1)f(x2) 即在非精确值 f i f_i fi上插值得到的插值多项式为 L 2 ∗ ( x ) {L_2^*}(x) L2∗(x)，题中"相应的误差界"指的就是要证： ∣ L 2 ( x ) − L 2 ∗ ( x ) ∣ ≤ 5 4 ε |{L_2}(x) - {L_2^*}(x)| \leq \frac{5}{4} \varepsilon ∣L2(x)−L2∗(x)∣≤45ε 而实际上： ∣ L 2 ( x ) − L 2 ∗ ( x ) ∣ = ∣ f ( x 0 ) − f 0 2 h 2 ( x − x 1 ) ( x − x 2 ) − f ( x 1 ) − f 1 h 2 ( x − x 0 ) ( x − x 2 ) + f ( x 2 ) − f 2 2 h 2 ( x − x 0 ) ( x − x 1 ) ∣ ≤ ∣ ε 2 h 2 ( x − x 1 ) ( x − x 2 ) ∣ + ∣ ε h 2 ( x − x 0 ) ( x − x 2 ) ∣ + ∣ ε 2 h 2 ( x − x 0 ) ( x − x 1 ) ∣ \begin{aligned} &\left|L_{2}(x)-L_{2}^{*}(x)\right| \\ &=\left|\frac{f\left(x_{0}\right)-f_{0}}{2 h^{2}}\left(x-x_{1}\right)\left(x-x_{2}\right)-\frac{f\left(x_{1}\right)-f_{1}}{h^{2}}\left(x-x_{0}\right)\left(x-x_{2}\right)+\frac{f\left(x_{2}\right)-f_{2}}{2 h^{2}}\left(x-x_{0}\right)\left(x-x_{1}\right)\right| \\ &\leq\left|\frac{\varepsilon}{2 h^{2}}\left(x-x_{1}\right)\left(x-x_{2}\right)\right|+\left|\frac{\varepsilon}{h^{2}}\left(x-x_{0}\right)\left(x-x_{2}\right)\right|+\left|\frac{\varepsilon}{2 h^{2}}\left(x-x_{0}\right)\left(x-x_{1}\right)\right| \end{aligned} ∣L2(x)−L2∗(x)∣=∣∣∣∣2h2f(x0)−f0(x−x1)(x−x2)−h2f(x1)−f1(x−x0)(x−x2)+2h2f(x2)−f2(x−x0)(x−x1)∣∣∣∣≤∣∣∣2h2ε(x−x1)(x−x2)∣∣∣+∣∣∣h2ε(x−x0)(x−x2)∣∣∣+∣∣∣2h2ε(x−x0)(x−x1)∣∣∣ 注意到 ( x − x 1 ) ( x − x 2 ) (x - {x_1})(x - {x_2}) (x−x1)(x−x2)等二式均为"二元一次方程"，容易求得其最大值各自为 h 2 / 4 h^2/4 h2/4， h 2 h^2 h2， h 2 / 4 h^2/4 h2/4，代回上式即可得到 ∣ L 2 ( x ) − L 2 ∗ ( x ) ∣ ≤ 5 4 ε |{L_2}(x) - {L_2^*}(x)| \leq \frac{5}{4} \varepsilon ∣L2(x)−L2∗(x)∣≤45ε。