= Solution
Write $\omega(t)=\prod_{i=0}^n(t-x_i)$. If $x$ is an <interpolation node>, the <polynomial interpolation error> is zero. Otherwise set
$$
K=\frac{f(x)-p(x)}{\omega(x)},\qquad F(t)=f(t)-p(t)-K\omega(t).
$$
The function $F$ has $n+2$ distinct <zeros>: the $n+1$ interpolation nodes and $x$. Repeated <Rolle's theorem> therefore gives a point $\xi$ strictly between the smallest and largest of these points with $F^{(n+1)}(\xi)=0$. Since $p^{(n+1)}=0$ and $\omega^{(n+1)}=(n+1)!$, this yields the <polynomial interpolation error>
$$
\boxed{f(x)-p(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n(x-x_i).}
$$
All differentiations are justified by $f\in C^{n+1}[a,b]$. At an interpolation node the same displayed equality holds with any $\xi\in[a,b]$, because the product is zero.
Back to article page
