## A mean value theorem

Christine Riedtmann asked me about the proof of the general mean value theorem mentioned in the very first entry of this blog. Let me recall the statement. In what follows ${I}$ is an interval and ${{x_0, \ldots, x_n}}$ are (pairwise distinct) points on ${I}$.

Theorem 1 Let ${{f_0, \ldots, f_n}}$ be ${n}$ times differentiable functions on ${I}$. Then $\displaystyle \det \bigl[f_i(x_j)\bigr]= \det \left[\frac{f_i^{(j)}(\tau_{ij})}{j!}\right]V(x_0, \ldots, x_n)$

with some ${{\tau_{ij}\in I}}$.

Here ${V(x_0, \ldots, x_n)}$ is the Vandermonde determinant.

The proof is based on the more special Schwarz mean value theorem, also mentioned in that entry. It is more convenient for us to state the Schwarz theorem as follows.

Theorem 2 (Schwarz) Let ${f}$ be an ${n}$ times differentiable function on ${I}$. Denote by ${P(t)}$ the Lagrange interpolation polynomial for ${f}$ at the points ${{x_0, \ldots, x_n}}$ (the polynomial of degree at most~ ${n}$ taking value ${f(x_j)}$ at ${x_j}$ for ${j=0,\ldots,n}$.) Then for some ${{\tau \in I}}$ we have ${{f^{(n)}(\tau)=P^{(n)}(=P^{(n)}(\tau))}}$.

In this form the Schwarz theorem is immediate: the “auxiliary function” ${{f(t)-P(t)}}$ vanishes at ${{x_0, \ldots, x_n}}$, and by the Rolle theorem its ${n}$-th derivative vanishes somewhere on ${I}$.

To prove Theorem 1 we consider the interpolation polynomials ${P_{ij}(t)}$ of ${f_i}$ at ${{x_0, \ldots, x_j}}$. Let ${{t_0, \ldots, t_n}}$ be independent indeterminates. Elementary transformations of determinants show that $\displaystyle \det\bigl[P_{ij}(t_j)\bigr] =\frac{\prod_{0\le k< \ell\le n}(t_\ell-x_k) }{V(x_0, \ldots, x_n)}\det\bigl[f_i(x_j)\bigr].$

Applying to the both sides the differential operator $\displaystyle \frac{1}{0!1!\cdots n!}\,\frac{\partial^{0+1+\cdots+n}}{\partial t_0^0\partial t_1^1\cdots \partial t_n^n},$

we obtain $\displaystyle \det \left[\frac{P_{ij}^{(j)}}{j!}\right]=\frac{1 }{V(x_0, \ldots, x_n)}\det\bigl[f_i(x_j)\bigr].$

Now the Schwarz theorem implies that ${{P_{ij}^{(j)}= f_i^{(j)}(\tau_{ij})}}$ with a suitably chosen ${\tau_{ij}}$.