Consider the polynomial series defined by the following recursion formula:
$$ \begin{align} &\mathrm{P}_0 = 1 \\ &\mathrm{P}_1 = x-r \\ &\mathrm{P}_n = x\mathrm{P}_{n-1} - r\mathrm{P}_{n-2} \\ \end{align} \tag{1} $$
where $r \in \Bbb{R}_+$
What are the roots of $\mathrm{Q}_n(x)$, defined as
$$ \mathrm{Q}_n = \mathrm{P}_{n+1} - \mathrm{P}_{n} \tag{2} $$
Can a formula for these zeros be found? The recursion relation for $\mathrm{P}_n$ is similar to the definitions of Chebyshev polynomials and Fibonacci polynomials, so I suspect that a way to calculate their roots must exist as well - but I do not know what it is. Interestigly the numerical coefficients (without the various powers of the parameter $r$) are listed in OEIS sequence A108299.
The first few $\mathrm{Q}_n$ are
$$ \begin{align} & \mathrm{Q}_1 = \mathrm{P}_1 - \mathrm{P}_0 = x -(r+1) \\ & \mathrm{Q}_2 = \mathrm{P}_2 - \mathrm{P}_1 = x^2 -(r+1)x \\ & \mathrm{Q}_3 = \mathrm{P}_3 - \mathrm{P}_2 = x^3 -(r+1)x^2 - rx + r(r+1) \\ & \mathrm{Q}_4 = \mathrm{P}_4 - \mathrm{P}_3 = x^4 -(r+1)x^3 -2rx^2 +2r(r+1)x \\ & \mathrm{Q}_5 = \mathrm{P}_5 - \mathrm{P}_4 = x^5 -(r+1)x^4 -3rx^3 +3r(r+1)x^2 + r^2 x - r^2(r+1) \\ & \mathrm{Q}_6 = \mathrm{P}_6 - \mathrm{P}_5 = x^6 -(r+1)x^5 -4rx^4 +4r(r+1)x^3 + 3r^2 x^2 - 3r^2(r+1)x \\ & \mathrm{Q}_7 = \mathrm{P}_7 - \mathrm{P}_6 = x^7 -(r+1)x^6 -5rx^5 +5r(r+1)x^4 + 6r^2 x^3 - 6r^2(r+1)x^2 - r^3 x + r^3(r+1) \\ & \mathrm{Q}_8 = \mathrm{P}_8 - \mathrm{P}_7 = x^8 -(r+1)x^7 -6rx^6 +6r(r+1)x^5 +10r^2 x^4 -10r^2(r+1)x^3 -4r^3 x^2 + 4r^3(r+1)x \\ \end{align} $$
The magnitude of the numerical coefficients are listed in OEIS A011973 (the difference being that here they are doubled).
I also propose these conjectures, only one of which (#2) I have been able to prove:
- $r+1$ is a root of all $\mathrm{Q_n}$
- $0$ is a root for all $\mathrm{Q_n}$ of even order ($n$ even)
- if $\mathcal{R}$ is a root and $\mathcal{R} \neq r+1$ then $-\mathcal{R}$ is also a root.
- all roots of $\mathrm{Q}_n$ have multiplicity $1$ (are single roots).
This problem is related to solving the linear system I describe here.
I do not know how to tackle this problem generally, help will be greatly appreciated!
Define
$$ U_n(x) = r^{-n/2} (2\sqrt{r} x-r-1)^{-1} Q_{n+1}(2 \sqrt{r} x). $$
We see that $U_0(x) = 1$, $U_1(x) = 2x$, and
$$ U_n(x) = 2x U_{n-1}(x) - U_{n-2}(x), $$
so these $U_n$ are in fact Chebyshev polynomials of the second kind. Thus
$$ Q_n(x) = r^{n/2}(x-r-1) U_n(x/2\sqrt{r}). $$