For the function $\dfrac{f(x)}{g(x)}$, we have, $\left(\dfrac{f}{g}\right)' = \dfrac{gf'-fg'} {g^2}$.

We can write the numerator as $W(g,f) = \left|\begin{matrix} g & f \\ g' & f'\end{matrix}\right|$ which is called Wronskian.

I wonder why determinant appears in the numerator ? Is there any mathematical relationship between derivative of $\dfrac fg$ and determinant that gives us $W(g,f)$ right away ?

  • 2
    $\begingroup$ At the very least, they both vanish iff $f$ and $g$ are linearly dependent (under some mild conditions on $f$ and $g$). $\endgroup$
    – anomaly
    Jun 30 at 14:39

If $f$ is $g$ times a constant $c$ then the quotient is $c$ and has derivative zero and the two columns of the Wronskian are linearly dependent, the left column equalling the right column times $c$, and thus the Wronskian has determinant $0$. This immediately suggests a relation.

One can make this a calculation-free proof that the determinant of the Wronskian appears in the numerator by using the ring of dual numbers. If $f$ and $g$ are elements of $\mathbb C[x]/(x^2)$ with $g$ invertible, then $\frac{f}{g}$ has derivative $0$ (at $x=0$) if and only if it is a constant, in which case the determinant of the Wronskian vanishes, so the determinant of the Wronskian must divide $\left( \frac{f}{g} \right)'$.

  • 7
    $\begingroup$ This does not answer the question. The fact that the Wronskian is the numerator is obvious and does not need new proofs. The question is what is a connection between $W(f,g)$ and $(f/g)'$ that makes it true. An answer can be something like "$W(f,g)$ and $(f/g)'g^2$ measure the same <<geometric>> object". $\endgroup$
    – markvs
    Jun 30 at 2:47
  • 2
    $\begingroup$ @MarkSapir The question specifically asked why the determinant appears in the numerator. $\endgroup$
    – Will Sawin
    Jun 30 at 2:49
  • 8
    $\begingroup$ @WillSavin: Exactly. And you do not answer that question. You gave new proofs of a trivial fact. $\endgroup$
    – markvs
    Jun 30 at 3:05

With the risk of making a simple thing confusing, here's another point of view.

Two functions $f,g$ on an interval $I$ give a map to the projective line $I \dashrightarrow \mathbb P^1$ by $x\mapsto [f(x):g(x)]$ (some indeterminacies can occur if both functions vanish at some point). You can ask: when does this map have a critical point?

If you choose the chart $[y:1]$ on $\mathbb P^1$ you have to look at the vanishing of the derivative of $f/g$.

If you lift the map to $I\to \mathbb R^2$ via $x\mapsto (f(x),g(x))$ then you get a curve in the plane, with tangent vector $(f'(x),g'(x))$. The map to $\mathbb P^1$ has a critical point precisely when the vectors $(f(x),g(x))$ and $(f'(x),g'(x))$ are proportional, so you get the Wronskian.

The book of Ovsienko and Tabachnikov "Projective Differential Geometry, Old and New" contains a wealth of additional information.


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