插值法
许多实际问题都用函数
y
=
f
(
x
)
y=f(x)
y=f(x)来表示某种内在规律的数量关系,其中相当一部分函数是通过实验或观测得到的,虽然
f
(
x
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f(x)
f(x)在某个区间
[
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[a,b]
[a,b]上是存在的,其中一些是连续的，但是有相当的一部分却只能给出
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[a,b]
[a,b]上一系列点
x
i
x_i
xi的函数值
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,
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y_i=f(x_i),(i=0,1,\cdots,n)
yi=f(xi),(i=0,1,⋯,n),这相当于是一张函数表.有的函数虽然有表达式,但是由于计算复杂,使用起来很不方便,通常也是造一个函数表,但是由于一些情况下,需要求出不在表上的函数值.
因此,我们可以根据给定的函数表做一个既能反应
f
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x
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f(x)
f(x)特性又便于计算的简单函数
P
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P(x)
P(x)用
P
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P(x)
P(x)近似
f
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f(x)
f(x).通常选一类较简单的函数(如代数多项式或分段代数多项式)作为
P
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x
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P(x)
P(x),并使
P
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f
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P(x_i)=f(x_i)
P(xi)=f(xi)对
i
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0
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1
,
⋯
,
n
i=0,1,\cdots,n
i=0,1,⋯,n成立.这样确定的
P
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P(x)
P(x)就是插值函数.
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设函数 y = f ( x ) y=f(x) y=f(x)在区间 [ a , b ] [a,b] [a,b]上有定义,且已知在点 a ≤ x 0 < x 1 < ⋯ < x n ≤ b a \le x_0 <x_1< \cdots < x_n \le b a≤x0<x1<⋯<xn≤b上的值 y 0 , y 1 , ⋯ , y n y_0,y_1,\cdots,y_n y0,y1,⋯,yn,若存在一简单函数 P ( x ) P(x) P(x),使
P ( x i ) = y i , ( i = 0 , 1 , ⋯ , x n ) P(x_i)=y_i,(i=0,1,\cdots,x_n) P(xi)=yi,(i=0,1,⋯,xn)
成立,就称 P ( x ) P(x) P(x)为 f ( x ) f(x) f(x)的插值函数,点 x 0 , x 1 , ⋯ , x n x_0,x_1,\cdots,x_n x0,x1,⋯,xn称为插值节点,包含插值节点的区间 [ a , b ] [a,b] [a,b]称为插值空间,求插值函数 P ( x ) P(x) P(x)的方法称为插值法.若 P ( x ) P(x) P(x)是次数不超过 n n n的代数多项式,即
P ( x ) = a 0 + a 1 x + ⋯ + a n x n P(x)=a_0+a_1x+\cdots+a_nx^n P(x)=a0+a1x+⋯+anxn,
其中 a i a_i ai为实数,就称 P ( x ) P(x) P(x)为插值多项式,相应的插值法称为多项式插值.若 P ( x ) P(x) P(x)为分段的多项式,就称为分段插值.
拉格朗日插值
首先
n
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1
n=1
n=1的情况下,假定区间
[
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[x_k,x_{k+1}]
[xk,xk+1]及端点函数值
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y_k=f(x_k),y_{k+1}=f(x_{k+1})
yk=f(xk),yk+1=f(xk+1),要求线性插值多项式
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1
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L_1(x)
L1(x),使之满足
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L_1(x_k)=y_k,L_1(x_{k+1})=y_{k+1}
L1(xk)=yk,L1(xk+1)=yk+1.
y
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1
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x
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y=L_1(x)
y=L1(x)的几何意义就是通过
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,
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(x_k,y_k),(x_{k+1},y_{k+1})
(xk,yk),(xk+1,yk+1)两点的直线.由几何意义直接给出
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L_1(x)=y_k+\frac{y_{k+1}-y_k}{x_{k+1}-x_k}(x-x_k)
L1(x)=yk+xk+1−xkyk+1−yk(x−xk) (点斜式),
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L_1(x)=\frac{x_{k+1}-x}{x_{k+1}-x_k}y_k+\frac{x-x_k}{x_{k+1}-x_k}y_{k+1}
L1(x)=xk+1−xkxk+1−xyk+xk+1−xkx−xkyk+1(两点式).
由两点式可以看出,
L
1
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L_1(x)
L1(x)是由两个线性函数
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l_k(x)=\frac{x-x_{k+1}}{x_k-x_{k+1}},l_{k+1}(x)=\frac{x-x_k}{x_{k+1}-x_k}
lk(x)=xk−xk+1x−xk+1,lk+1(x)=xk+1−xkx−xk的线性组合得到,其系数分别为
y
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y_k,y_{k+1}
yk,yk+1
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L_1(x)=y_kl_k(x)+y_{k+1}l_{k+1}(x)
L1(x)=yklk(x)+yk+1lk+1(x).其中函数
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,
l
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1
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l_k(x),l_{k+1}(x)
lk(x),lk+1(x)为线性插值基函数
当
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2
n=2
n=2的情况.假定插值节点为
x
k
−
1
,
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x_{k-1},x_k,x_{k+1}
xk−1,xk,xk+1,其二次插值多项式
L
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L_2(x)
L2(x)使其满足
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L_2(x_j)=y_j,(j=k-1,k,k+1)
L2(xj)=yj,(j=k−1,k,k+1),此时
y
=
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2
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x
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y=L_2(x)
y=L2(x)在几何意义上是通过
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,
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,
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∗
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(x_{k-1},y_{k-1}),(x_k,y_k),(x_{k+1*},y_{k+1})
(xk−1,yk−1),(xk,yk),(xk+1∗,yk+1)三点的抛物线
该节点满足:
{
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\begin{cases} l_{k-1}(x_{k-1})=1,l_{k-1}(x_j)=0,(j=k,k+1) \\ l_k(x_k)=1,l_k(x_j)=0,(j=k-1,k+1) \\ l_{k+1}(x_{k+1})=1,l_{k+1}(x_j)=0,(j=k-1,k) \end{cases}
⎩⎪⎨⎪⎧lk−1(xk−1)=1,lk−1(xj)=0,(j=k,k+1)lk(xk)=1,lk(xj)=0,(j=k−1,k+1)lk+1(xk+1)=1,lk+1(xj)=0,(j=k−1,k),由它的两个零点有:
l
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1
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=
A
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l_{k-1}(x)=A(x-x_k)(x-x_{k+1})
lk−1(x)=A(x−xk)(x−xk+1),其中
A
A
A为待定系数,由条件
l
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=
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⟹
A
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l_{k-1}(x_{k-1})=1 \Longrightarrow A=\frac{1}{(x_{k-1}-x_k)(x_{k-1}-x_{k+1})}
lk−1(xk−1)=1⟹A=(xk−1−xk)(xk−1−xk+1)1所以有:
{
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\begin{cases} l_{k-1}(x)=\frac{(x-x_k)(x-x_{k+1})}{(x_{k-1}-x_k)(x_{k-1}-x_{k+1})} \\ l_k(x)=\frac{(x-x_{k-1})(x-x_{k+1})}{(x_k-x_{k-1})(x_k-x_{k+1})} \\ l_{k+1}(x)=\frac{(x-x_{k-1})(x-x_k)}{(x_{k+1}-x_{k-1})(x_{k+1}-x_k)} \end{cases}
⎩⎪⎨⎪⎧lk−1(x)=(xk−1−xk)(xk−1−xk+1)(x−xk)(x−xk+1)lk(x)=(xk−xk−1)(xk−xk+1)(x−xk−1)(x−xk+1)lk+1(x)=(xk+1−xk−1)(xk+1−xk)(x−xk−1)(x−xk)
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L_2(x)=y_{k-1}l_{k-1}(x)+y_kl_k(x)+y_{k+1}l_{k+1}(x)
L2(x)=yk−1lk−1(x)+yklk(x)+yk+1lk+1(x),带入上式得:
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L_2(x)=y_{k-1}\frac{(x-x_k)(x-x_{k+1})}{(x_{k-1}-x_k)(x_{k-1}-x_{k+1})}+y_k\frac{(x-x_{k-1})(x-x_{k+1})}{(x_k-x_{k-1})(x_{k}-x_{k+1})}+y_{k+1}\frac{(x-x_{k-1})(x-x_k)}{(x_{k+1}-x_{k-1})(x_{k+1}-x_k)}
L2(x)=yk−1(xk−1−xk)(xk−1−xk+1)(x−xk)(x−xk+1)+yk(xk−xk−1)(xk−xk+1)(x−xk−1)(x−xk+1)+yk+1(xk+1−xk−1)(xk+1−xk)(x−xk−1)(x−xk)
拉格朗日插值多项式
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若 n n n次多项式 l j ( x ) ( j = 0 , 1 , ⋯ , n ) l_j(x)(j=0,1,\cdots,n) lj(x)(j=0,1,⋯,n)在 n + 1 n+1 n+1个节点 x 0 < x 1 < ⋯ < x n x_0<x_1<\cdots<x_n x0<x1<⋯<xn满足条件
l j ( x k ) = { 1 , k = j ; 0 , k ≠ j . ( j , k = 0 , 1 , ⋯ , n ) l_j(x_k)= \begin{cases} 1,k=j; \\ 0,k \ne j. \end{cases}(j,k=0,1,\cdots,n) lj(xk)={1,k=j;0,k=j.(j,k=0,1,⋯,n)就称这 n + 1 n+1 n+1个 n n n次多项式 l 0 ( x ) , l 1 ( x ) , ⋯ , l n ( x ) l_0(x),l_1(x),\cdots,l_n(x) l0(x),l1(x),⋯,ln(x)为节点 x 0 , x 1 , ⋯ , x n x_0,x_1,\cdots,x_n x0,x1,⋯,xn上得 n n n次插值基函数.由此得到 n n n次插值得基函数:
l k ( x ) = ( x − x 0 ) ⋯ ( x − x k − 1 ) ( x − x k + 1 ) ⋯ ( x − x n ) ( x k − x 0 ) ⋯ ( x k − x k − 1 ) ( x k − x k + 1 ) ⋯ ( x k − x n ) , ( k = 0 , 1 , ⋯ , n ) l_k(x)=\frac{(x-x_0)\cdots(x-x_{k-1})(x-x_{k+1})\cdots(x-x_n)}{(x_k-x_0)\cdots(x_k-x_{k-1})(x_k-x_{k+1})\cdots(x_k-x_n)},(k=0,1,\cdots,n) lk(x)=(xk−x0)⋯(xk−xk−1)(xk−xk+1)⋯(xk−xn)(x−x0)⋯(x−xk−1)(x−xk+1)⋯(x−xn),(k=0,1,⋯,n)
插值多项式 L n ( x ) L_n(x) Ln(x)可表示为:
L n ( x ) = ∑ k = 0 n y k l k ( x ) L_n(x)=\sum_{k=0}^{n}y_kl_k(x) Ln(x)=∑k=0nyklk(x).
由 l k ( x ) l_k(x) lk(x)的定义知:
L n ( x j ) = ∑ k = 0 n y k l k ( x j ) = y j , ( j = 0 , 1 , ⋯ , n ) L_n(x_j)=\sum_{k=0}^{n}y_kl_k(x_j)=y_j,(j=0,1,\cdots,n) Ln(xj)=∑k=0nyklk(xj)=yj,(j=0,1,⋯,n),形如上式 L n ( x ) L_n(x) Ln(x)称为拉格朗日(Lagrange)插值多项式.
若引入记号
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\omega_{n+1}(x)=(x-x_0)(x-x_1)\cdots(x-x_n)
ωn+1(x)=(x−x0)(x−x1)⋯(x−xn),
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\omega^{\prime}_{n+1}(x_k)=(x_k-x_0)\cdots(x_k-x_{k-1})(x_k-x_{k+1})\cdots(x_k-x_n) \Longrightarrow L_n(x)=\sum_{k=0}^{n}y_k\frac{\omega_{n+1}(x)}{(x-x_k)\omega^{\prime}_{n+1}(x_k)}.
ωn+1′(xk)=(xk−x0)⋯(xk−xk−1)(xk−xk+1)⋯(xk−xn)⟹Ln(x)=∑k=0nyk(x−xk)ωn+1′(xk)ωn+1(x).
插值余项与误差估计
若在
[
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b
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[a,b]
[a,b]上用
L
n
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L_n(x)
Ln(x)近似
f
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f(x)
f(x),则其截断误差为
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R_n(x)=f(x)-L_n(x)
Rn(x)=f(x)−Ln(x),也称作插值多项式余项.
定理:设
f
n
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f^n(x)
fn(x)在
[
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[a,b]
[a,b]上连续,
f
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1
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(
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f^{(n+1)}(x)
f(n+1)(x)在
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(a,b)
(a,b)内存在节点
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<
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<
⋯
<
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a \le x_0<x_1<\cdots<x_n \le b
a≤x0<x1<⋯<xn≤b,
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L_n(x)
Ln(x)是满足条件$的插值多项式,则对任何的
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∈
[
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x \in [a,b]
x∈[a,b],插值余项
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!
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R_n(x)=f(x)-L_n(x)=\frac{f^{n+1}(\xi)}{(n+1)!}\omega_{n+1}(x)
Rn(x)=f(x)−Ln(x)=(n+1)!fn+1(ξ)ωn+1(x),这里
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\xi \in (a,b)
ξ∈(a,b)且依赖于
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证明:由给定条件知
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R_n(x)
Rn(x)在节点
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x_k(k=0,1,\cdots,n)
xk(k=0,1,⋯,n)上为0.即
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R_n(x_k)=0(k=0,1,\cdots,n)
Rn(xk)=0(k=0,1,⋯,n)于是
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⋯
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=
K
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ω
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1
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R_n(x)=K(x)(x-x_0)(x-x_1)\cdots(x-x_n)=K(x)\omega_{n+1}(x)
Rn(x)=K(x)(x−x0)(x−x1)⋯(x−xn)=K(x)ωn+1(x),其中
K
(
x
)
K(x)
K(x)是与
x
x
x有关的待定函数.
现把
x
x
x看成
[
a
,
b
]
[a,b]
[a,b]上的一个固定点,作函数
φ
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f
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L
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⋯
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\varphi(t)=f(t)-L_n(t)-K(x)(t-x_0)(t-x_1)\cdots(t-x_n)
φ(t)=f(t)−Ln(t)−K(x)(t−x0)(t−x1)⋯(t−xn),
φ
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t
)
\varphi(t)
φ(t)在点
x
0
,
x
1
,
⋯
,
x
n
x_0,x_1,\cdots,x_n
x0,x1,⋯,xn及
x
x
x处均为0.根据罗尔定理,
φ
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n
+
1
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(
ξ
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=
f
(
n
+
1
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(
ξ
)
−
(
n
+
1
)
!
K
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x
)
=
0
\varphi^{(n+1)}(\xi)=f^{(n+1)}(\xi)-(n+1)!K(x)=0
φ(n+1)(ξ)=f(n+1)(ξ)−(n+1)!K(x)=0
K
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x
)
=
f
(
n
+
1
)
(
ξ
)
(
n
+
1
)
!
,
ξ
∈
(
a
,
b
)
K(x)=\frac{f^{(n+1)(\xi)}}{(n+1)!},\xi \in(a,b)
K(x)=(n+1)!f(n+1)(ξ),ξ∈(a,b)且依赖于
x
x
x.
令
m
a
x
a
<
x
<
b
∣
f
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n
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1
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∣
=
M
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max_{a<x<b}|f^{(n+1)}(x)|=M_{n+1}
maxa<x<b∣f(n+1)(x)∣=Mn+1,插值多项式
L
n
(
x
)
L_n(x)
Ln(x)逼近
f
(
x
)
f(x)
f(x)的阶段误差限是
∣
R
n
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x
)
∣
≤
M
n
+
1
(
n
+
1
)
!
∣
ω
n
+
1
(
x
)
∣
|R_n(x)| \le \frac{M_{n+1}}{(n+1)!}|\omega_{n+1}(x)|
∣Rn(x)∣≤(n+1)!Mn+1∣ωn+1(x)∣.*
均差
将插值多项式表示为便于计算的形式有:
P
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+
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⋯
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P_n(x)=a_0+a_1(x-x_0)+a_2(x-x_0)(x-x_1)+\cdots+a_n(x-x_0)\cdots(x-x_{n-1})
Pn(x)=a0+a1(x−x0)+a2(x−x0)(x−x1)+⋯+an(x−x0)⋯(x−xn−1),其中
a
0
,
a
1
,
⋯
,
a
n
a_0,a_1,\cdots,a_n
a0,a1,⋯,an为待定系数,可由插值条件
P
n
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x
j
)
=
f
j
,
(
j
=
0
,
1
,
⋯
,
n
)
P_n(x_j)=f_j,(j=0,1,\cdots,n)
Pn(xj)=fj,(j=0,1,⋯,n)
当
x
=
x
0
x=x_0
x=x0时,
P
n
(
x
0
)
=
a
0
=
f
0
P_n(x_0)=a_0=f_0
Pn(x0)=a0=f0
当
x
=
x
1
x=x_1
x=x1时,
P
n
(
x
1
)
=
a
0
+
a
1
(
x
−
x
0
)
=
f
1
⟹
a
1
=
f
1
−
f
0
x
1
−
x
0
P_n(x_1)=a_0+a_1(x-x_0)=f_1 \Longrightarrow a_1=\frac{f_1-f_0}{x_1-x_0}
Pn(x1)=a0+a1(x−x0)=f1⟹a1=x1−x0f1−f0
当
x
=
x
2
x=x_2
x=x2时,
P
n
(
x
2
)
=
a
0
+
a
1
(
x
2
−
x
0
)
+
a
1
(
x
2
−
x
0
)
+
a
2
(
x
2
−
x
0
)
(
x
2
−
x
1
)
=
f
2
⟹
a
2
=
f
2
−
f
0
x
2
−
x
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−
f
1
−
f
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x
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−
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−
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P_n(x_2)=a_0+a_1(x_2-x_0)+a_1(x_2-x_0)+a_2(x_2-x_0)(x_2-x_1)=f_2 \Longrightarrow a_2=\frac{\frac{f_2-f_0}{x_2-x_0}-\frac{f_1-f_0}{x_1-x_0}}{x_2-x_1}
Pn(x2)=a0+a1(x2−x0)+a1(x2−x0)+a2(x2−x0)(x2−x1)=f2⟹a2=x2−x1x2−x0f2−f0−x1−x0f1−f0
-
称 f [ x 0 , x k ] = f ( x k ) − f ( x 0 ) x k − x 0 f[x_0,x_k]=\frac{f(x_k)-f(x_0)}{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 k x_0,x_k x0,xk的一阶均差. f [ x 0 , x 1 , x k ] = f [ x 0 , x k ] − f [ x 0 , x 1 ] x k − x 1 f[x_0,x_1,x_k]=\frac{f[x_0,x_k]-f[x_0,x_1]}{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,称
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[x_0,x_1,\cdots,x_k]=\frac{f[x_0,\cdots,x_{k-2},x_k]-f[x_0,x_1,\cdots,x_{k-1}]}{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(均差也称为差商).
| x k x_k xk | f ( x k ) f(x_k) f(xk) | 一阶均差 | 二阶均差 | 三阶均差 | 四阶均差 |
|---|---|---|---|---|---|
| x 0 x_0 x0 | f ( x 0 ) ‾ \underline{f(x_0)} f(x0) | ||||
| x 1 x_1 x1 | f ( x 1 ) f(x_1) f(x1) | f [ x 0 , x 1 ] ‾ \underline{f[x_0,x_1]} f[x0,x1] | |||
| x 2 x_2 x2 | f ( x 2 ) f(x_2) f(x2) | f [ x 1 , x 2 ] f[x_1,x_2] f[x1,x2] | f [ x 0 , x 1 , x 2 ] ‾ \underline{f[x_0,x_1,x_2]} f[x0,x1,x2] | ||
| x 3 x_3 x3 | f ( x 3 ) f(x_3) f(x3) | f [ x 2 , x 3 ] f[x_2,x_3] f[x2,x3] | f [ x 1 , x 2 , x 3 ] f[x_1,x_2,x_3] f[x1,x2,x3] | f [ x 0 , x 1 , x 2 , x 3 ] ‾ \underline{f[x_0,x_1,x_2,x_3]} f[x0,x1,x2,x3] | |
| x 4 x_4 x4 | f ( x 4 ) f(x_4) f(x4) | f [ x 3 , x 4 ] f[x_3,x_4] f[x3,x4] | f [ x 2 , x 3 , x 4 ] f[x_2,x_3,x_4] f[x2,x3,x4] | f [ x 1 , x 2 , x 3 , x 4 ] f[x_1,x_2,x_3,x_4] f[x1,x2,x3,x4] | f [ x 0 , x 1 , x 2 , x 3 , x 4 ] ‾ \underline{f[x_0,x_1,x_2,x_3,x_4]} f[x0,x1,x2,x3,x4] |
