The Wronskian determinant I have a general question about the Wronskian (the determinant). Suppose you have $n$ functions $f_i(x)$ and you compute the Wronskian $W$ and find that it is non-zero everywhere except for a countable number of points. For instance, say $W=x$. Clearly, then, $W=0$ only if $x=0$. What exactly does this imply? 
Thanks for any replies
 A: This is actually a deep question. The answer depends on the set of functions $f_i(x)$ you're dealing with. But the general answer is that Wronskians actually count certain things. Here's one possible reference.
For linear differential equations, by Abel's theorem the wronskian of all linearly independent solutions is either zero everywhere or never zero. This is related to the fact that you can always solve linear ode's uniquely.
Suppose $f_i(x)$ are complex valued polynomials, let $P$ be the vector space that's spanned by these polynomials. An incredible result due to Mukhin, Tarasov and Varchenko is that if the Wronskian of these polynomials has only real roots, then $P$ has a basis in terms of real polynomials. This is related to other areas like the Shapiro conjecture for Grassmanian's. 
Here's an example of this kind of stuff: take the wronskian of $f,g$. You get $W(f,g)(x)=f(x)g'(x)-f'(x)g(x)$. Pick $f,g$ to be any linearly independent solutions to $\frac{d^4y}{dx^4}=y$.
Now look at the fourth derivative of $W(f,g)$. Some manipulation will give: 
$$\frac{d^4}{dx^4}W(f,g)(x)=2W(f'',g'')(x)+2W(f,g)(x).$$
Now imagine you draw four lines in projective three dimensions. This just means lines have no preferred direction. Also make sure that no two lines are parallel. It turns out that the first $2$ in the expansion counts the number of lines you can draw that will simultaneously intersect all four lines. The second $2$ is beyond the scope of this answer (it counts the number of degree-1 morphisms from the 4-pointed projective line to the Grassmanian of the four projective lines). This is all generally known as "Schubert Calculus." Whether or not Wronskians have zeros will imply certain facts about these coefficients that come out differentiation. It turns out that the class of functions that the Wronskian acts on makes a difference in such counting arguments.
