# A Taylor formula for a Vandermonde-like determinant

Let $f_0,\ldots,f_N$ be smooth functions over an interval $I\subset{\mathbb R}$. Let $x_0,\ldots,x_N\in I$ be given, and form the Vandermonde-like determinant $$\Delta_N=\det((f_i(x_j)))_{0\le i,j\le N}.$$ I should bet that there exists an $a\in I$ such that $$\Delta_N=\frac1{0!1!\cdots N!}\,V_N(\vec x)W_N(\vec f)(a),$$ where $V_N$ is the Vandermonde determinant $\prod_{i<j}(x_j-x_i)$ and $W_N$ is the Wronskian $\det((f_i^{(j)}))_{0\le i,j\le N}$.

This must be a well-known Taylor-like formula. I should appreciate an accurate reference, or a sketch of proof.

Edit. After Christian R.'s answer, who shows by a counter-example that the above claim is wrong, let me suggest a weaker property, that $$\left|\Delta_N\right|\le\frac1{0!1!\cdots N!}\,\left|V_N(\vec x)\right|\cdot\|W_N(\vec f)\|_{L^\infty(I)}.$$ After all, if $W_N(\vec f)\equiv0$, then the functions $f_j$ are linearly dependent and $\Delta_N$ vanishes too.

• I think this is equation 2.32 of On Multivariate Interpolation --- upon taking the determinant of both sides. Aug 2, 2017 at 8:07
• @Carlo: Eq. 2.32 does not seem to me to answer the OP's question which is more about a kind of mean value theorem with appearance of this mysterious $a$. Aug 2, 2017 at 13:31
• @Denis: are you sure about $(N!)!$ ? This combinatorial factor does not look right to me. Usually with Wronskians the kind of things that show up are products of factorials like $1!2!\ldots N!$. Aug 2, 2017 at 13:33
• @Abdelmalek. I agree. This is actually the product which I found. I mistakenly wrote it as $(N!)!$. I fix it. Aug 2, 2017 at 13:34
• It seems to me that a slightly different identity holds in 3D. Let $\gamma(t) = (f_{1}(t), f_{2}(t), f_{3}(t))$. For each $a\leq b$ there exists $c \in [a,b]$ such that $\det(\gamma'(a), \gamma'(b), \gamma(a)-\gamma(b))=\frac{(a-b)^{4}}{12}W(\gamma'(c))$, where $W(\gamma'(c))$ is the Wronskian of $\gamma'(c)$. Aug 2, 2017 at 22:52

This isn't working. Let's look at the simplest case, $N=1$ (but the example works in general). Then the identity you are hoping for becomes $$\det \begin{pmatrix} f(a) & g(a) \\ f(b) & g(b) \end{pmatrix} = (b-a)\det \begin{pmatrix} f(c) & g(c) \\ f'(c) & g'(c) \end{pmatrix} ,$$ for some $a\le c\le b$. However, taking $f(a)=f(b)=0$ won't force the Wronskian to be zero anywhere: for example, $f=\sin\pi\frac{x-a}{b-a}$, $g=\cos\pi\frac{x-a}{b-a}$ will make $W$ equal to a non-zero constant.
• indeed, it only holds in the limit $a,b\rightarrow c$. Aug 2, 2017 at 21:21