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[函数] Lagrange插值的极限

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hbghlyj Posted at 2023-5-6 22:40:11 |Read mode
Last edited by hbghlyj at 2023-5-7 02:20:00存在唯一的≤2次多项式$p$使得\begin{cases}p(0)=a\\p'(0)=b\\p(1)=c\end{cases}设$p(x)=a+bx+kx^2,$
代入$p(1)=c$得$k=c-(a+b).$
即$p(x)=a+bx+(c-a-b)x^2.$
存在唯一的≤3次多项式$p$使得\begin{cases}p(0)=a\\p'(0)=b\\p(1)=c\\p'(1)=d\end{cases}相当于同余方程组\begin{cases}p(x)\equiv a+bx\pmod{x^2}\\p(x)\equiv c+d(x-1)\pmod{(x-1)^2}\end{cases}一般的问题: $n$次多项式$p$, 给定$p$图象上的$m\;(1≤m≤n+1)$个点 给定$p'$图象上的$n-m+1$个点, 这样的$p$唯一存在吗

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 Author| hbghlyj Posted at 2023-5-6 22:51:04
Last edited by hbghlyj at 2023-5-7 11:31:00Confluent Vandermonde matrices
As described before, a Vandermonde matrix describes the linear algebra interpolation problem of finding the coefficients of a polynomial $ p(x) $ of degree $ n-1 $ based on the values $ p(\alpha _{1}),\,...,\,p(\alpha _{n}) $, where $ \alpha _{1},\,...,\,\alpha _{n} $ are distinct points. If $ \alpha _{i} $ are not distinct, then this problem does not have a unique solution (which is reflected by the fact that the corresponding Vandermonde matrix is singular). However, if we give the values of the derivatives at the repeated points, then the problem can have a unique solution. For example, the problem
\begin{cases}p(0)=a\\p'(0)=b\\p(1)=c\end{cases}where $ p $ is a polynomial of degree ≤ 2, has a unique solution for all $ a,b,c $. In general, suppose that $ \alpha _{1},\alpha _{2},...,\alpha _{n} $ are (not necessarily distinct) numbers, and suppose for ease of notation that equal values come in continuous sequences in the list. That is
$$ \alpha _{1}=\cdots =\alpha _{m_{1}},\ \alpha _{m_{1}+1}=\cdots =\alpha _{m_{2}},\ \ldots ,\ \alpha _{m_{k-1}+1}=\cdots =\alpha _{m_{k}} $$
where $ m_{k}=n, $ $ m_{1}<m_{2}<\cdots <m_{k}, $ and $ \alpha _{m_{1}},\ldots ,\alpha _{m_{k}} $ are distinct. Then the corresponding interpolation problem is
\begin{cases}p(\alpha _{1})=\beta _{1},&p'(\alpha _{1})=\beta _{2},&\ldots ,&p^{(m_{1}-1)}(\alpha _{1})=\beta _{m_{1}}\\p(\alpha _{m_{1}+1})=\beta _{m_{1}+1},&p'(\alpha _{m_{1}+1})=\beta _{m_{1}+2},&\ldots ,&p^{(m_{2}-m_{1}-1)}(\alpha _{m_{2}})=\beta _{m_{2}}\\\qquad \vdots \\p(\alpha _{m_{k-1}+1})=\beta _{m_{k-1}+1},&p'(\alpha _{m_{k-1}+2})=\beta _{m_{k-1}+2},&\ldots ,&p^{(m_{k}-m_{k-1}-1)}(\alpha _{m_{k}})=\beta _{m_{k}}\end{cases}
And the corresponding matrix for this problem is called a confluent Vandermonde matrices. In our case (which is the general case, up to permuting the rows of the matrix) the formula for it is given as follows: if $ 1\leq i,j\leq n $, then $ m_{\ell }<i\leq m_{\ell +1} $ for some (unique) $ 0\leq \ell \leq k-1 $ (we consider $ m_{0}=0 $). Then, we have\begin{equation}\label1 V_{i,j}={\begin{cases}0,&{\text{if }}j<i-m_{\ell };\\[6pt]{\dfrac {(j-1)!}{(j-(i-m_{\ell }))!}}\alpha _{i}^{j-(i-m_{\ell })},&{\text{if }}j\geq i-m_{\ell }.\end{cases}} \end{equation}
This generalization of the Vandermonde matrix makes it non-singular (such that there exists a unique solution to the system of equations) while retaining most properties of the Vandermonde matrix. Its rows are derivatives (of some order) of the original Vandermonde rows.

Another way to receive this formula is to let some of the $ \alpha _{i} $'s go arbitrarily close to each other. For example, if $ \alpha _{1}=\alpha _{2} $, then letting $ \alpha _{2}\to \alpha _{1} $ in the original Vandermonde matrix, the difference between the first and second rows yields the corresponding row in the confluent Vandermonde matrix. This allows us to link the generalized interpolation problem (given value and derivatives on a point) to the original case where all points are distinct: Being given $ p(\alpha ),p'(\alpha ) $ is similar to being given $ p(\alpha ),p(\alpha +\varepsilon ) $ where $ \varepsilon $ is very small.

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 Author| hbghlyj Posted at 2023-5-7 18:32:41
请问式\eqref{1}如何得到?2#没有详细写

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