4
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ I think this is equation 2.32 of On Multivariate Interpolation --- upon taking the determinant of both sides. $\endgroup$ – Carlo Beenakker Aug 2 '17 at 8:07
  • 1
    $\begingroup$ @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 $a$. $\endgroup$ – Abdelmalek Abdesselam Aug 2 '17 at 13:31
  • 2
    $\begingroup$ @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 $1!2!\ldots N!$. $\endgroup$ – Abdelmalek Abdesselam Aug 2 '17 at 13:33
  • $\begingroup$ @Abdelmalek. I agree. This is actually the product which I found. I mistakenly wrote it as $(N!)!$. I fix it. $\endgroup$ – Denis Serre Aug 2 '17 at 13:34
  • 3
    $\begingroup$ 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 $\gamma'(c)$. $\endgroup$ – Paata Ivanishvili Aug 2 '17 at 22:52
6
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ indeed, it only holds in the limit $a,b\rightarrow c$. $\endgroup$ – Carlo Beenakker Aug 2 '17 at 21:21
  • $\begingroup$ @Christian: good catch. It might still be worth finding an integral formula for LHS divided by Vandermonde of points and combinatorial factor. $\endgroup$ – Abdelmalek Abdesselam Aug 2 '17 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.