数值分析读书笔记（5）数值逼近问题(I)----插值极其数值计算

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最新推荐文章于 2020-11-11 23:39:59 发布

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分类专栏：数值分析文章标签： Numeric Calculation

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本文介绍了数值逼近的基本概念，重点讲解了Lagrange插值、Newton插值和Hermit插值的方法及其误差分析，探讨了如何有效构造插值多项式以逼近给定函数。

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数值分析读书笔记（5）数值逼近问题(I)—-插值极其数值计算

给出一般性的插值概念

给定 $f(x),x \in [a,b]$ ,已知它在n+1个互异的节点 $x_{0},x_{1},\dots$ 上的函数值为 $y_{0},y_{1},\dots$
目的即寻求 $\varphi (x)$ ,使得 $\varphi(x_{i})=f(x_{i})=y_{i}$
令所有的 $\varphi(x)$ 组成 $\Phi$ ,通常 $\Phi$ 是有限维线性空间,记

$Φ = s p a n {φ i (x)} n i = 0$ $\Phi=span\{\varphi_{i}(x)\}_{i=0}^{n}$
其中 $\varphi_{i}(x)$ 为一组基, 于是有 $φ (x) = \sum i = 0 n a i φ i (x)$ $\varphi(x)=\sum_{i=0}^{n}{a_{i}\varphi_{i}(x)}$
故我们可以利用序列 $\{a_{i}\}_{i=0}^{n}$ 来确定 $\varphi(x)$ , 这里的 $\varphi(x)$ 就是插值函数

通过概念我们可以看出来,目的就是让插值函数去接近给定的函数

1.关于多项式插值

当给定插值函数是多项式函数的时候, 我们可以产生一种插值的方案, 下面介绍一下Lagrange插值

令 $P(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n}=\sum_{i=0}^{n}{a_{i}x^{i}}$
由于
$P (x i) = y i, i = 0, 1, 2, \dots, n$ $P(x_{i})=y_{i},i=0,1,2,\dots,n$
$⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ a 0 + a 1 x 0 + \dots + a n x n 0 ⋮ a 0 + a 1 x n + \dots + a n x n n = y 0 = y 0$ $\left\{ \begin{aligned} a_{0}+a_{1}x_{0}+\cdots+a_{n}x^{n}_{0}&=y_{0}\\\vdots\\a_{0}+a_{1}x_{n}+\cdots+a_{n}x^{n}_{n}&=y_{0}\end{aligned}\right.$
得系数阵为 $A = ⎛ ⎝ ⎜ ⎜ 1 ⋮ 1 x 0 ⋮ x n \dots ⋱ \dots x n 0 ⋮ x n n ⎞ ⎠ ⎟ ⎟$ $A=\begin{pmatrix}1&x_{0}&\cdots&x_{0}^{n}\\\vdots&\vdots&\ddots&\vdots\\1&x_{n}&\cdots&x_{n}^{n}\end{pmatrix}$
由Vandermonde行列式的特性,我们可以知道 $| A | = \prod 0 \leq i < j \leq n (x j - x i)$ $\begin{vmatrix}A\end{vmatrix}=\prod_{0\leq i<j\leq n}(x_{j}-x_{i})$
若 $x_{i}$ 互异,则有唯一解
构造插值基函数 $l i (x j) = δ i j = {10 i = j i \neq j$ $l_{i}(x_{j})=\delta_{ij}=\left\{\begin{aligned} &1 && i=j\\&0&&i\neq j\end{aligned}\right.$
$i \neq j$ 时, $l_{i}(x_{j})=0$ ,故
$l i (x) = c i \prod k = 0, k \neq i n (x - x k)$ $l_{i}(x)=c_{i}\prod_{k=0,k\neq i}^{n}{(x-x_{k})}$
又对于 $c_{i}$ 由于 $l_{i}(x_{j})=1$ 当仅当 $i=j$ ,故
$l i (x i) = c i \prod k = 0, k \neq i n (x i - x k) = 1$ $l_{i}(x_{i})=c_{i}\prod_{k=0,k\neq i}^{n}{(x_{i}-x_{k})}=1$
所以可以得到 $c i = \prod k = 0, k \neq i n (x i - x k) - 1$ $c_{i}=\prod_{k=0,k\neq i}^{n}{(x_{i}-x_{k})}^{-1}$
我们将 $c_{i}$ 回代回去,则有
$l i (x) = \prod k = 0, k \neq i n ( x - x k ) ( x i - x k )$ $l_{i}(x)=\prod_{k=0,k\neq i}^{n}{\frac{(x-x_{k})}{(x_{i}-x_{k})}}$
令 $L n (x) = \sum i = 0 n y i l i (x) = \sum i = 0 n (y i \prod k = 0, k \neq i n ( x - x k ) ( x i - x k ))$ $L_{n}(x)=\sum_{i=0}^{n}y_{i}l_{i}(x)=\sum_{i=0}^{n} \left ( y_{i}\prod_{k=0,k\neq i}^{n}{\frac{(x-x_{k})}{(x_{i}-x_{k})}} \right)$
上式即为所求的Lagarange插值函数

这里为了运算记录方便, 记

ω n + 1 (x) = \prod k = 0 n (x - x k)

$\omega_{n+1}(x)=\prod_{k=0}^{n}(x-x_{k})$
则有

ω' n + 1 (x i) = \prod k = 0, k \neq i n (x i - x k), ω n + 1 ( x ) x - x i = \prod k = 0, k \neq i n (x - x k)

$\omega'_{n+1}(x_i)=\prod_{k=0,k\neq i}^{n}{(x_{i}-x_{k})},\frac{\omega_{n+1}(x)}{x-x_{i}}=\prod_{k=0,k\neq i}^{n}{(x-x_{k})}$
故

L n (x) = \sum i = 0 n (y i ω n + 1 ( x ) ( x - x i ) ω ' n + 1 ( x i ))

$L_{n}(x)=\sum_{i=0}^{n}\left({y_{i}\frac{\omega_{n+1}(x)}{(x-x_{i})\omega'_{n+1}(x_{i})}}\right)$

下面继续讨论Lagrange插值的误差,引入误差余项

R (x) \equiv f (x) - L n (x)

$R(x)\equiv f(x)-L_{n}(x)$
我们引入一个辅助函数
令

F (t) = f (t) - L n (t) - K (x) ω n + 1 (t), t \in [a, b], x \neq x i, i = 0, 1, 2, \dots

$F(t)=f(t)-L_{n}(t)-K(x)\omega_{n+1} (t) , t \in [a,b] , x \neq x_{i} , i=0, 1, 2, \dots$
这里面对于辅助函数的构造,其中末尾一项是保证当x等于节点中的一个时,误差为0
其中

K (x) ω n + 1 (x) = f (x) - L n (x)

$K(x)\omega_{n+1}(x)=f(x)-L_{n}(x)$

t = x 0, x 1, \dots, x n

$t=x_{0}, x_{1},\dots,x_{n}$ 是辅助函数的n+2个相异的零点

注意到Rolle定理

$φ (x), x \in [a, b], φ (a) = φ (b) = 0, 其中 φ (x) 满足连续等条件, 则必有 ξ \in [a, b], 使得 φ' (ξ) = 0$ $\varphi (x) , x \in [a ,b] , \varphi (a) =\varphi (b) = 0, 其中\varphi (x) 满足连续等条件, 则必有 \xi \in [a,b],使得\varphi '(\xi) = 0$

F (t) = f (t) - L n (t) - K (x) ω n + 1 (t), 对 F (t), F' (t), . . . . 应 用 R o l l e 定 理

$F(t) = f(t) - L_{n}(t) - K(x)\omega _{n+1}(t),对F(t),F'(t),....应用Rolle定理$

必 有 ξ \in [a, b], 使 得 F (n + 1) (ξ) = 0, 故 f (n +!) (ξ) - K (x) (n + 1)! = 0 故 K (x) = f ( n + 1 ) ( x ) ( n + 1 ) ! 从 而 R n (x) = f ( n + 1 ) ( x ) ( n + 1 ) ! ω n +! (x)

$必有\xi \in [a,b] , 使得F^{(n+1)}(\xi) =0 ,故f^{(n+!)}(\xi)-K(x)(n+1)!=0\\故K(x)=\frac{f^{(n+1)}(x)}{(n+1)!}\\从而 R_{n}(x)=\frac{f^{(n+1)}(x)}{(n+1)!}\omega_{n+!}(x)$

以上是关于Lagrange插值的介绍,针对Lagrange插值,节点个数的增加或者减少的时候,插值基函数需要变动,为了解决这一问题,我们引入Newton插值

$N n (x) = a 0 + a 1 (x - x 0) + \dots + a n (x - x 0) (x - x 1) \dots (x - x n) a l s o d e f i n e N n - 1 (x) = a 0 + a 1 (x - x 0) + \dots + a n - 1 (x - x 0) (x - x 1) \dots (x - x n - 1) F o r x i (i = 0, 1, \dots, n - 1) N n - 1 (x) = N n (x) = y i S o N n (x) - N n - 1 (x) = c (x - x 0) (x - x 1) \dots (x - x n - 1) W h e n x = x n N n (x n) - N n - 1 (x n) = y n - N n - 1 (x n) = c \prod i = 0 n - 1 (x - x i) W e c a n g e t c = [y n - N n - 1 (x)] \prod i = 0 n - 1 (x - x i) - 1 P a y a t t e n t i o n t o t h a t N n - 1 (x) = \sum i = 0 n - 1 y i l i (x n) = \sum i = 0 n - 1 (y i \prod k = 0, k \neq i n - 1 (x n - x k x i - x k)) S o c = y n \prod i = 0 n - 1 (x n - x i) - 1 - N n - 1 (x n) \prod i = 0 n - 1 (x n - x i) - 1 N n - 1 (x n) \prod i = 0 n - 1 (x n - x i) - 1 = \sum i = 0 n - 1 (y i \prod k = 0, k \neq i n - 1 (x n - x k x i - x k)) / \prod i = 0 n - 1 (x n - x i) S o N n - 1 (x n) \prod i = 0 n - 1 (x n - x i) - 1 = - \sum i = 0 n - 1 (y i \prod k = 0, k \neq i n (x i - x k) - 1) W e c a n g e t c = y n \prod i = 0 n - 1 (x n - x i) - 1 + \sum i = 1 n - 1 (y i \prod k = 0, k \neq i n (x i - x k) - 1) B e c a u s e o f t h a t W e g e t . N n (x) - N n - 1 (x) = c \prod i = 0 n - 1 (x - x i) = [\sum i = 1 n - 1 (y i \prod k = 0, k \neq i n (x i - x k) - 1)] \sum i = 0 n - 1 (x - x i)$ $N_{n}(x) = a_{0}+a_{1}(x-x_{0}) + \cdots + a_{n}(x-x_{0})(x-x_{1})\cdots(x-x_{n})\\also\ define \quad N_{n-1}(x)= a_{0}+a_{1}(x-x_{0}) + \cdots + a_{n-1}(x-x_{0})(x-x_{1})\cdots(x-x_{n-1})\\For \ \ x_{i} \ (i=0,1,\cdots,n-1) \ \ \ N_{n-1}(x)=N_{n}(x)=y_{i} \\ So \ \ N_{n}(x)-N_{n-1}(x)=c(x-x_{0})(x-x_{1})\cdots(x-x_{n-1}) \\ When \ \ x=x_{n} \\ N_{n}(x_{n})-N_{n-1}(x_{n})=y_{n}-N_{n-1}(x_{n})=c\prod _{i=0}^{n-1}(x-x_{i}) \\ We \ can \ get \ c=[y_{n}-N_{n-1}(x)]\prod _{i=0}^{n-1}(x-x_{i})^{-1} \\ Pay \ attention \ to \ that\\N_{n-1}(x)=\sum _{i=0}^{n-1}{y_{i}l_{i}(x_{n})}=\sum_{i=0}^{n-1}\left({y_{i}\prod_{k=0,k\neq i}^{n-1}\left(\frac{x_{n}-x_{k}}{x_{i}-x_{k}}\right)}\right)\\So \ \ c=y_{n}\prod _{i=0}^{n-1}(x_{n}-x_{i})^{-1} - N_{n-1}(x_{n})\prod_{i=0}^{n-1}(x_{n}-x_{i})^{-1}\\N_{n-1}(x_{n})\prod_{i=0}^{n-1}(x_{n}-x_{i})^{-1} =\sum_{i=0}^{n-1}\left({y_{i}\prod_{k=0,k\neq i}^{n-1}\left(\frac{x_{n}-x_{k}}{x_{i}-x_{k}}\right)}\right) / \prod_{i=0}^{n-1}(x_{n}-x_{i}) \\ So \\N_{n-1}(x_{n})\prod_{i=0}^{n-1}(x_{n}-x_{i})^{-1} =- \sum _{i=0}^{n-1}\left(y_{i}\prod_{k=0,k\neq i}^{n}{(x_{i}-x_{k})^{-1}} \right)\\ We \ can \ get \ \ \ c=y_{n}\prod_{i=0}^{n-1}{(x_{n}-x_{i})^{-1}}+\sum_{i=1}^{n-1}{\left(y_{i}\prod_{k=0,k\neq i}^{n}{(x_{i}-x_{k})^{-1}} \right)} \\ Because \ of \ that \\ We \ get. \ \ \ N_{n}(x)-N_{n-1}(x)=c\prod _{i=0}^{n-1}(x-x_{i})= \left[\sum_{i=1}^{n-1}{\left(y_{i}\prod_{k=0,k\neq i}^{n}{(x_{i}-x_{k})^{-1}} \right)} \right] \sum_{i=0}^{n-1}(x-x_{i})$

这里引入差商(Difference Quotient)的概念

$L e t . f [x 0, \dots, x n] = \sum i = 1 n - 1 (y i \prod k = 0, k \neq i n (x i - x k) - 1) N n (x) = f [x 0] + f [x 0, x 1] (x 0 - x 1) + \dots + f [x 0, x 1, \dots, x n] \prod i = 0 n - 1 (x - x i) T h e D i f f e r e n c e Q u o t i e n t i s l i k e t h a t f [x i, x j] = f ( x i ) - f ( x j ) x i - x j, f [x i, x j, x k] = f [ x i , x j ] - f [ x j , x k ] x i - x k$ $Let. \ f[x_{0},\cdots,x_{n}]=\sum_{i=1}^{n-1}{\left(y_{i}\prod_{k=0,k\neq i}^{n}{(x_{i}-x_{k})^{-1}} \right)} \\ N_{n}(x)=f[x_{0}]+f[x_{0},x_{1}](x_{0}-x_{1})+\cdots + f[x_{0},x_{1},\cdots , x_{n}]\prod _{i=0}^{n-1}(x-x_{i})\\ The \ Difference \ Quotient \ is \ like \ that \\f[x_{i},x_{j}]=\frac{f(x_i)-f(x_{j})}{x_{i}-x_{j}}\ , \ \ f[x_{i},x_{j},x_{k}]=\frac{f[x_{i},x_{j}]-f[x_{j},x_{k}]}{x_{i}-x_{k}}$

我们可以利用这里的差商的概念写出Newton插值公式

N n (x) = f (x 0) + f [x 0, x 1] (x 0 - x 1) + \dots + f [x 0, x 1, \dots, x n] (x - x 0) (x - x 1) \dots (x - x n - 1)

$N_{n}(x)=f(x_{0})+f[x_{0},x_{1}](x_{0}-x_{1})+\cdots + f[x_{0},x_{1},\cdots,x_{n}](x-x_{0})(x-x_{1})\cdots (x-x_{n-1})$

其实Newton插值公式和Lagrange插值公式其实本质上是一样的,只不过是书写的方式不同,但是这样的不同的书写方式在实际操作中带来了很大的便利,当需要增加一个插值点的时候,只需要在原插值多项式的后面再添加一个新的项就可以了

有时候我们不但要求插值函数P(x)在节点处的函数值与被插值函数f(x)的值相等,而且要求在节点处的导数值也相等,这就引出了了一种新的插值方案Hermit插值

G i v e n f (x), a n d n + 1 d i f f e r e n t p o i n t s x 0, x 1, \dots, x n L e t f (x i) = y i f' (x i) = y' i W e c a n c o n s t r u c t a p o l y n o m i a l H 2 n + 1 (x i) = y i H' 2 n + 1 (x i) = y' i

$Given \ f(x) , \ and \ n+1 \ different\ points\ x_{0},x_{1},\cdots , x_{n}\\Let \ f(x_i)=y_i\ \ \ \ f'(x_i)=y_i'\\We \ can\ construct\ a\ polynomial \\H_{2n+1}(x_i)=y_i \ \ \ \ H'_{2n+1}(x_i)=y_i'$

我们这次要构造的多项式比起之前的lagrange多项式,多了导数值相等的条件,那我们就利用两组基函数来试着构造这一多项式

l 0 j (x i) = δ i j, l' 0 j (x i) = 0 l 1 j (x i) = 0, l' 1 j (x i) = δ i j S o w e g e t t h e s i m p l e p o l y n o m i a l : H 2 n + 1 (x) = \sum j = 0 n y j l 0 j (x) + \sum j = 0 n y' j l 1 j (x) W h e n i \neq j, l 0 j (x i) a n d l 0 j (x i)' a l w a y s e q u a l z e r o B e c a u s e o f t h a t, A s s u m e l 0 j (x) = (a x + b) l 2 j (x) {(a x j + b) l 2 j (x j) = 1 l j (x j) [a l j (x j) + 2 (a x j + b) l' j (x j)] = 0 (3) (4) A b o u t t h e s e c o n d f o r m u l a, t u r n e d f r o m " a l 2 j (x j) + 2 l j (x j) l' j (x j) (a x j + b) " T h a t m e a n s w e j u s t t o s o l v e t h e f o l l o w i n g : a x j + b = 1 a n d a + 2 (a x j + b) l' j (x j) = 0 W e g e t t h a t : a = - 2 l' j (x j), b = 1 + 2 x j l' j (x j) S i m i l a r l y l 1 j (x) = (c x + d) l j (x j) A n d w e a l s o g e t c = 1 a n d d = - x j S o w e c a n e a s i l y b u i l d H 2 n + 1 (x) T h e r e s u l t i s t h a t H 2 n + 1 (x) = \sum j = 0 n y j [2 (x j - x) l' j (x) + 1] l 2 j (x) + \sum j = 0 n y' j [(x - x j) l 2 j (x)] D o s o m e j o b a b o u t m a k i n g f o r m u l a n e a t W e g e t t h a t H 2 n + 1 (x) = \sum j = 0 n l 2 j (x) [y j + (x - x j) (y' j - 2 y j) l' j (x j)]

$l_{0j}(x_i)=\delta _{ij},\ l_{0j}'(x_{i})=0\\ l_{1j}(x_i)=0, \ l_{1j}'(x_i)=\delta_{ij}\\So\ we\ get\ the\ simple \ polynomial\ : \ \ H_{2n+1}(x)=\sum_{j=0}^{n}y_{j}l_{0j}(x)+\sum_{j=0}^{n}{y'_{j}l_{1j}(x)}\\When\ i\neq j\ , l_{0j}(x_i)\ and\ l_{0j}(x_i)' \ always\ equal\ zero\\ Because\ of\ that\ , Assume\ l_{0j}(x)=(ax+b)l_{j}^{2}(x) \\ \left\{{\begin{align}(ax_j+b)l_j^{2}(x_j)=1 \\ l_j(x_j)[al_j(x_j)+2(ax_j+b)l_j'(x_j)]=0\end{align}}\right.\\ About \ the \ second \ formula \ , \ turned \ from\ " al_{j}^{2}(x_j)+2l_{j}{(x_{j})}l_{j}'(x_{j})(ax_{j}+b) " \\ That\ means\ we\ just\ to\ solve\ the\ following: \\ax_{j}+b=1\ \ \ and\ \ \ a+2(ax_{j}+b)l_{j}'(x_j)=0\\ We\ get\ that\ :\ a=-2l_{j}'(x_{j}),\ \ b=1+2x_{j}l_{j}'(x_j)\\ Similarly\ l_{1j}(x)=(cx+d)l_{j}(x_{j}) \\ And\ we\ also\ get\ c=1\ and\ d=-x_{j}\\ So\ we\ can\ easily\ build\ H_{2n+1}(x)\\ The\ result\ is\ that\ H_{2n+1}(x)=\sum_{j=0}^{n}{y_{j}[2(x_{j}-x)l_{j}'(x)+1]l^{2}_{j}(x)+\sum_{j=0}^{n}y'_{j}[(x-x_{j})l_{j}^{2}(x)]}\\ Do\ some\ job\ about\ making\ formula\ neat \\ We\ get\ that\ H_{2n+1}(x)=\sum^{n}_{j=0}{l^{2}_{j}(x)[y_{j}+(x-x_{j})(y_{j}'-2y_j)l_{j}'(x_{j})]}$

这里我们需要提及的是,使用上述方法对各个节点进行插值的时候,很有可能在端点处产生一定程度的Runge现象,解决的手段可以使用分段线性插值构造出一系列的分段函数,对于分段线性插值,我们可以理解为对于多个划分的子区间进行Lagrange插值得到的一系列分段函数,当然分段插值也有非线性的,例如分段的二次插值,就是在划分的多个子区间上使用Lagrange2次插值.

这里由于某些教材的不同,可能介绍了Hermit三次插值的方案,在上述的公式中可以令n=1即可.