A mysterious expression from a discriminant

Question

I recently investigated $\textrm{Discriminant}_u(\omega P(u) +(z-u)P'(u))$, where $P(u) := u^3 + au + b$ and $\omega$ is a real parameter (with $\omega\in(0,1)\cup(1,3)$) associated with the order of a fractional derivative. (This is a special case of the problem previously discussed here.) In doing so, I came across the following expression (plus some additional terms) that looks quite striking, but I haven't been able to place it:

$$\frac{3a(b-z^3)}{(\omega-1)\,\lambda} + \frac{z}{\lambda^3}\left(a^2\lambda\left(1-\frac{1}{(\omega-1)^2}\right)-3bz\right)$$

Here $\lambda:= \sqrt[3]{\omega-3}.$ I'm wondering if anyone has seen this expression, or something sufficiently similar to it, in any other context.

Due to a natural change of variables used to arrive at the above expression (which reduces the constant part of the above discriminant to the discriminant of the polynomial $\widetilde{P}(u)$ below), the expression should possibly be thought of as the “image” of the polynomial

$$\widetilde{P}(u) := u^3 + \frac{a}{(\omega-1)\sqrt[3]{\omega-3}}u + \frac{b}{\omega(\omega-3)}$$

under some discriminant-like “function”.

Does the expression above look familiar, or is it perhaps some determinant involving the coefficients in $\widetilde{P}(u)$, and $z$? Alternatively, do you see some way to attack $\textrm{Discriminant}_u(\omega P(u) +(z-u)P'(u))$ directly?

Answer 1

It might be illuminating to write the discriminant as the resultant of your polynomial with its derivative. If I've done the algebra correctly (but please check), the discriminant that you're computing is $$ \operatorname{Resultant}\Bigl( (\omega -3)u^3+3z u^2 + a(\omega -1)u+(az+b\omega ),\, 3(\omega -3)u^2+6z u+a(\omega -1) \Bigr). $$ Then subtracting $\frac13u$ times the second polynomial from the first gives the resultant of two quadratics, $$ \operatorname{Resultant}\left(z u^2 + \frac23a(\omega -1)u+(az+b\omega ),\, 3(\omega -3)u^2+6z u+a(\omega -1)\right). $$