Discriminant of $\alpha P(u) + (z-u) P'(u)$

Question

I'm trying to find a “closed form” of $\textrm{Discriminant}_u(f(u))$, where $f(u) := \alpha P(u) + (z-u) P'(u)$.

Here $P(u)$ is a monic polynomial of degree $d > 1$ with $u\in\mathbb{C}$, $\alpha$ is a real parameter, and $f(u)$ is related to the zero distribution of a generalization of the Legendre polynomials. (When $\alpha = d$, then $f(u)$ is called the polar derivative of $P$ with pole $z$, which has been studied by e.g. Laguerre, Pólya and Szegő.) By “closed form” I mean something other than the standard product formulas of the differences of the roots of $f$ (and $f'$), the determinant of the standard Sylvester matrix, and so on.

While this seems to be a difficult problem (which I've been stuck on longer than I'd like to admit), there are a number of reasons that make me think it can be attacked successfully. For example:

$(1)$: If $P(u) = u^3 + bu^2 + cu + \kappa$, then $\mathrm{Discriminant}_u(f(u)) =$

$36(b^2-3c)z^4 +\left[48b\left(b^2-3c\right)-4\left(2b^3-9bc+27 \kappa\right)\alpha\right]z^3+\left[8\left(b^2-3c\right)\left(2b^2+3 c\right)-4\left(4b^4-21b^2c+18c^2+27b\kappa\right)\alpha+4\left(b^2-3 c\right)^2\alpha^2\right]z^2+\left[16bc\left(b^2-3c\right)+4\left(-4 b^3c+15bc^2+18b^2\kappa-81c\kappa\right)\alpha+4\left(b^2-3c\right)(bc-9\kappa)\alpha^2\right]z +4c^2\left(b^2-3c\right)+4\left(-3b^2c^2+10c^3+8b^3\kappa-27bc\kappa\right)\alpha+\left(13b^2c^2-48c^3-48b^3\kappa+198bc\kappa-243\kappa^2\right)\alpha^2+6\left(-b^2c^2+4c^3+4b^3\kappa-18bc\kappa+27\kappa^2\right)\alpha^3-\left(-b^2c^2+4c^3+4b^3\kappa-18bc\kappa+27\kappa^2\right)\alpha^4$.

Notice that

$\textrm{Resultant}_u(P(u),P'(u)) = -b^2c^2+4c^3+4b^3\kappa-18bc\kappa+27\kappa^2,$ $\textrm{Resultant}_u(P(u),P''(u)) = -8\left(2b^3-9bc+27\kappa\right),$ and $\textrm{Resultant}_u(P'(u),P''(u)) = -12\left(b^2-3c\right),$

so the coefficients of the $\alpha^n$ terms above appear to be resultant-like objects involving various orders of the derivative of $P$. Furthermore, there is apparently no $\alpha^3$ term in the coefficient of $z$. Additionally, we have that

$\textrm{Discriminant}_u(dP(u) + (z-u) P'(u)) = 4\left(b^2-3c\right)z^2+4(bc-9\kappa)z+4\left(c^2-3b\kappa\right),$

and we can see that the coefficient of $z$ in the discriminant of the polar derivative appears in the coefficient of $\alpha^2$ in the $z$ term above. Similarly,

$\textrm{Discriminant}_u(0\cdot P(u)+(z-u)P'(u)) = 4(b^2-3c)(P'(z))^2 = 36\left(b^2-3c\right)z^4+48b\left(b^2-3c\right)z^3+8\left(b^2-3 c\right)\left(2 b^2+3c\right)z^2+16bc\left(b^2-3c\right)z+4c^2\left(b^2-3c\right)$

and we see that the coefficient of $z^2$ here is constant term in the coefficient of $z^2$ above. Where do the remaining coefficients come from?

$(2)$: Polynomial division of $f(u)$ by $f'(u)$ yields a linear quotient in $u$ (and in $z$) with an interesting symmetry. Explicitly, standard properties of discriminants and resultants give that

$\mathrm{Discriminant}_u(f(u)) = e^{\pi i d(3d-7)/2} (\alpha - d) d^2 \cdot \mathrm{Resultant}_u(f'(u), r(u))$,

where $r(u) := f(u) - \dfrac{f'(u)}{d^2}\cdot\left(a_{d-1} + du + \dfrac{a_{d-1} + dz}{\alpha - d}\right)$ is a polynomial of degree $d-2$ and $a_{d-1}$ is the coefficient of $u^{d-1}$ in $P(u)$.

$(3)$: Terry Tao found an interesting connection to the Laguerre separation theorem (Theorem 4 in the text). As a result, this problem is possibly more manageable if $f(u)$ is rewritten as

$f(u) = (u-z)^{\alpha-1} g'(\Phi_I(u)),$

where $g(u) := u^{\alpha} P(\Phi(u)),\,\Phi(u) := z+u^{-1}$ and $\Phi_I(u) := \Phi^{-1}(u) = (u-z)^{-1}.$

Thus, if we introduce the function $h(u) := u^{1-\alpha}\,g'(u)$, we can (if we exclude the singularities) write

$f(u) = h(\Phi_I(u)).$

Now, it follows from standard properties of discriminants and resultants that

$\mathrm{Discriminant}_u(f(u)) = \mathrm{Discriminant}_u(h(\Phi_I(u))$

$=(-1)^{d(d-1)/2}\,(\alpha-d)^{-1}\,\mathrm{Resultant}_u(h(\Phi_I(u)),h'(\Phi_I(u))\,\Phi_I'(u))$

$=(-1)^{d(d-1)/2}\,(\alpha-d)^{-1}\,\mathrm{Resultant}_u(h(\Phi_I(u)),h'(\Phi_I(u)))\cdot\mathrm{Resultant}_u\left(h(\Phi(u)),-(u-z)^{-2}\right)$

$=\dfrac{i^{d(d+1)}}{(\alpha-d)(\alpha\,P(z))^2}\,\mathrm{Resultant}_u(h(\Phi_I(u)),h'(\Phi_I(u))).$

I see a few possible approaches from this. For example:

2.1: Rewrite

$h'(u) = u^{-\alpha}\left((1-\alpha)g'(u) + ug''(u)\right)$

in a more convenient form so that the resultant can be factored again.

2.2: Interpret $\mathrm{Resultant}_u(h(\Phi_I(u)),h'(\Phi_I(u)))$ in terms of a change of variables, or a transformed discriminant.

$(4)$: The coefficients of $\alpha^i z^j$ (for various $i,\,j$) in $\mathrm{Discriminant}_u(f(u))$ are possibly related to “resultants” of fractional and sometimes negative (in the differintegral sense) orders of the derivative of $P$. For example, if $P(u) = u^3 + b u^2 + c u + \kappa$, then the coefficient of $\alpha^2 z^0$ in $\mathrm{Discriminant}_u(f(u))$ (which is one of the four coefficients not divisible by any of the three resultants in the example in $(2)$ above) is approximately $b^2 c^2 - 3.6923 c^3 - 3.6923 b^3 \kappa + 15.2308 b c \kappa - 18.6923 \kappa^2$ after normalization.

Compare this to the following determinant of a “Sylvester matrix” inspired by fractional orders $D_1$ and $D_2$ of the derivative of $P(u)$:

$\lim_{D_1\to 33/100,\,D_2\to 166/100}\left| \begin{array}{cccccc} \frac{6}{\Gamma \left(4-D_1\right)} & \frac{2 b}{\Gamma \left(3-D_1\right)} & \frac{c}{\Gamma \left(2-D_1\right)} & \frac{\kappa}{\Gamma \left(1-D_1\right)} & 0 & 0 \\ 0 & \frac{6}{\Gamma \left(4-D_1\right)} & \frac{2 b}{\Gamma \left(3-D_1\right)} & \frac{c}{\Gamma \left(2-D_1\right)} & \frac{\kappa}{\Gamma \left(1-D_1\right)} & 0 \\ 0 & 0 & \frac{6}{\Gamma \left(4-D_1\right)} & \frac{2 b}{\Gamma \left(3-D_1\right)} & \frac{c}{\Gamma \left(2-D_1\right)} & \frac{\kappa}{\Gamma \left(1-D_1\right)} \\ \frac{6}{\Gamma \left(4-D_2\right)} & \frac{2 b}{\Gamma \left(3-D_2\right)} & \frac{c}{\Gamma \left(2-D_2\right)} & \frac{\kappa}{\Gamma \left(1-D_2\right)} & 0 & 0 \\ 0 & \frac{6}{\Gamma \left(4-D_2\right)} & \frac{2 b}{\Gamma \left(3-D_2\right)} & \frac{c}{\Gamma \left(2-D_2\right)} & \frac{\kappa}{\Gamma \left(1-D_2\right)} & 0 \\ 0 & 0 & \frac{6}{\Gamma \left(4-D_2\right)} & \frac{2 b}{\Gamma \left(3-D_2\right)} & \frac{c}{\Gamma \left(2-D_2\right)} & \frac{\kappa}{\Gamma \left(1-D_2\right)} \\ \end{array} \right|,$

which after normalization is approximately

$b^2 c^2 - 3.79847 c^3 - 3.59317 b^3 \kappa + 15.2455 b c \kappa - 18.7119 \kappa^2.$

$(5)$: If $P(u) = u^3 + bu^2 + cu + \kappa$, then $\mathrm{Discriminant}_u(f(u)) =$

$36\left(b^2-3c\right)z^4 + \left[-8b^3(\alpha-6)+36bc(\alpha-4)-108\kappa \alpha\right]z^3 + 4\left[-3 b^2 c (\alpha (2\alpha-7)+2)+b^4 (\alpha-2)^2-27b \kappa \alpha+9 c^2 ((\alpha-2)\alpha-2)\right]z^2 + 4\left[b^3 c (\alpha-2)^2-9b^2\kappa\alpha(\alpha-2)-3bc^2(\alpha-1)(\alpha-4)+27c \kappa\alpha(\alpha-3)\right]z + b^2 c^2 \left(\alpha^2-3 \alpha+2\right)^2 - 4 c^3 (\alpha-1)^3(\alpha-3) - 4 b^3 \kappa\alpha(\alpha-2)^3+18 b c \kappa \alpha(\alpha-1)(\alpha-2)(\alpha-3)-27\kappa ^2 \alpha^2 (\alpha-3)^2.$

Here the coefficient of $z^4$ equals $9\cdot\mathrm{Discriminant}_u(P'(u)).$ Additionally, the constant term looks strikingly similar to $\mathrm{Discriminant}_u(P(u)) = b^2c^2 - 4c^3 - 4b^3\kappa + 18bc\kappa - 27\kappa^2$, while the coefficient of $z^3$ resembles $\mathrm{Resultant}_u(P(u),P''(u))/2 = -8b^3 + 36bc - 108\kappa.$

Inspired by the above, if we replace the factors that contain $\alpha$ with $1$ in the coefficient of $z$, this modified coefficient is equal to $4(b^2-3c)(bc-9\kappa) = (bc-9\kappa)\,\mathrm{Discriminant}_u(P'(u)).$

In other words, it appears as if the coefficients of $z^k$ are (in some sense) transformed resultant-like objects that involve $P(u)$ and its derivatives.

Unfortunately, this is as far as I've gotten with the above approaches. Suggestions or other ideas to evaluate this discriminant would be greatly appreciated.