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本帖最后由 hbghlyj 于 2023-1-10 15:41 编辑 lect1_Lagrange.pdf
$p_n(x)$是$f(x)$在$n+1$个点$x_k,k=0,1,\cdots,n$的Lagrange插值多项式(所以$p_n(x)$的次数$≤n$),
设$e(x) \stackrel{\text { def }}{=} f(x)-p_{n}(x)$,则存在与$x$有关的点$\xi\in(x_0,x_n)$使得
$$e(x) =\left(x-x_{0}\right)\left(x-x_{1}\right) \cdots\left(x-x_{n}\right) \frac{f^{(n+1)}(\xi)}{(n+1) !}$$
证明:
当$x∈\{x_k:k=0,1,\cdots,n\}$时,$e(x)=x-x_k=0$,等式成立.下面设$x≠x_k,k=0,1,\cdots,n$.
设$\pi(t) \stackrel{\text { def }}{=}\left(t-x_{0}\right)\left(t-x_{1}\right) \cdots\left(t-x_{n}\right)$, 则$\pi(x)\ne0$.
设$\phi(t) \stackrel{\text { def }}{=} e(t)-\frac{e(x)}{\pi(x)} \pi(t)$
注意到$\phi(t)$在$t=x$和$t=x_k, k = 0, 1, \cdots , n$共$n+2$个点处都为0,
由罗尔定理, 存在 $ξ \in (x_0, x_n)$ 使 $\phi^{(n+1)}(\xi)=0$.
下面计算$\phi^{(n+1)}(t)$:
因为$\pi(t)$是$n+1$次首一多项式,所以$\pi^{(n+1)}(t)=(n+1)!$
因为$p_n(x)$的次数$≤n$,所以$p_n^{(n+1)}(t)=0$,所以$e_n^{(n+1)}(t)=f^{(n+1)}(t)$.
$$\phi^{(n+1)}(t)=e^{(n+1)}(t)-\frac{e(x)}{\pi(x)} \pi^{(n+1)}(t)=f^{(n+1)}(t)-\frac{e(x)}{\pi(x)}(n+1) !$$
由$\phi^{(n+1)}(\xi)=0$得,
$$e(x)=\pi(x)\frac{f^{(n+1)}(\xi)}{(n+1) !}$$
即$$e(x) =\left(x-x_{0}\right)\left(x-x_{1}\right) \cdots\left(x-x_{n}\right) \frac{f^{(n+1)}(\xi)}{(n+1) !}$$
Example: $f(x)=\log (1+x)$ on $[0,1]$. Here, $\left|f^{(n+1)}(\xi)\right|=n ! /(1+\xi)^{n+1}<n !$ on $(0,1)$. So $|e(x)|<|\pi(x)| n ! /(n+1) ! \leq 1 /(n+1)$ since $\left|x-x_k\right| \leq 1$ for each $x, x_k, k=0,1, \ldots, n$, in $[0,1] \Longrightarrow|\pi(x)| \leq 1$. This is probably pessimistic for many $x$, e.g. for $x=\frac{1}{2}, \pi\left(\frac{1}{2}\right) \leq 2^{-(n+1)}$ as $\left|\frac{1}{2}-x_k\right| \leq \frac{1}{2}$.
This shows the important fact that the error can be large at the end points when samples $\left\{x_k\right\}$ are equispaced points, an effect known as the "Runge phenomena" (Carl Runge, 1901), which we return to in lecture 4 .
demonstrations.wolfram.com/RungesPhenomenon/ |
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