Means and Hermite Interpolation

Alan Horwitz

Published 2007-11-30, updated 2008-05-20Version 2

Let $m_{2}<m_{1}$ be two given nonnegative integers with $n=m_{1}+m_{2}+1$. For suitably differentiable $f$, we let $P,Q\in \pi_{n}$ be the Hermite polynomial interpolants to $f$ which satisfy $P^{(j)}(a)=f^{(j)}(a),j=0,1,...,m_{1}$ and $P^{(j)}(b)=f^{(j)}(b),j=0,1,...,m_{2},$ $Q^{(j)}(a)=f^{(j)}(a),j=0,1,...,m_{2}$ and $Q^{(j)}(b)=f^{(j)}(b),j=0,1,...,m_{1}$. Suppose that $f\in C^{n+2}(I)$ with $f^{(n+1)}(x)\neq 0$ for $x\in (a,b)$. If $m_{1}-m_{2}$ is even, then there is a unique $x_{0},a<x_{0}<b,$ such that $P(x_{0})=Q(x_{0})$. If $m_{1}-m_{2}$ is odd, then there is a unique $x_{0},a<x_{0}<b,$ such that $f(x_{0})=\tfrac{1}{2}(P(x_{0})+Q(x_{0})) $. $x_{0}$ defines a strict, symmetric mean, which we denote by $M_{f,m_{1},m_{2}}(a,b)$. We prove various properties of these means. In particular, we show that $f(x)=x^{m_{1}+m_{2}+2}$ yields the arithmetic mean, $f(x)=x^{-1}$ yields the harmonic mean, and $f(x)=x^{(m_{1}+m_{2}+1)/2}$ yields the geometric mean.