= Hermite interpolation error formula
{c}
{title2=$f(x)-H(x)=\frac{f^{(2n+2)}(\xi)}{(2n+2)!}\prod_{i=0}^n(x-x_i)^2$}
Let $x_0<\cdots<x_n$ be distinct <interpolation nodes>, let $f\in C^{2n+2}[x_0,x_n]$, and let $H$ be the <Hermite interpolating polynomial> matching the values and first <derivatives> of $f$ at all nodes. For each $x$ in the interval there is an interior point $\xi$ for which the displayed formula holds. At a node the formula is immediate. Otherwise put $e=f-H$, $\pi(t)=\prod_i(t-x_i)$ and $\varphi(t)=e(t)-e(x)\pi(t)^2/\pi(x)^2$. This function vanishes at the $n+1$ nodes and at $x$, while its <derivative> vanishes at every node. <Rolle's theorem> gives $n+1$ additional zeros of its <derivative> between successive zeros of $\varphi$. Repeating the theorem therefore gives $\varphi^{(2n+2)}(\xi)=0$. Since $H^{(2n+2)}=0$ and $(\pi^2)^{(2n+2)}=(2n+2)!$, rearrangement yields the formula. The extra point $x$ is a zero of $\varphi$, not necessarily of its <derivative>.
For two nodes $a,b$, this gives the uniform bound $\|f-H\|_\infty\le\|f^{(4)}\|_\infty(b-a)^4/384$, since $(x-a)^2(b-x)^2\le(b-a)^4/16$. The squared nodal factors encode simultaneous matching of values and <derivatives>.
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