Residue公式和Lagrange插值恒等式的关系

Lagrange插值恒等式适用的时候. 也就是n+1个插值互不相同的时候.

只要\deg f \le \deg P - 2并且P是一个无重根的多项式, 这时f(x) / P(x)在很大的地方积分一圈就有了0 = \sum \textrm{Res}_{x_i}(f(x)/P(x)) = f(x_i)/P'(x_i). 最后一个等号是因为如果f(x)有一个一阶零点而g(x)解析那么\textrm{Res}_{\alpha}(g(z)/f(z)) = g(\alpha)/f'(\alpha).

这样一来, 让P(x) = (x-\xi)\prod_{i=0}^n{(x-x_i)}, 就有了
\displaystyle 0 = \sum_{k=0}^n\frac{1}{(x_k - \xi)\prod_{j\ne k}{(x_k - x_j)}} f(x_k) + \frac{f(\xi)}{\prod_{k=0}^n(\xi - x_k)}.
整理一下就得到了Lagrange插值恒等式\displaystyle f(\xi) = \sum_{k=0}^n{\frac{\prod_{j\ne k}{(\xi - x_j)}}{\prod_{j\ne k}{(x_k - x_j)}} f(x_k)}.

\xi = x_k这种退化的时候随便验证一下就完了.

最后, 这个方法应该可以推广到Hermite插值, 并且终于可以写出Hermite插值的公式了. 我就不在这里算了.

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