I would like to find the roots of the polynomial sequence given by a recurrence relation as follows:


$V\_0(x) = 1-a^2$


$V\_1(x) = 1-a^2 - x$


$V\_{k \geq 2}(x) = (1+a^2 - x)V\_{k-1}(x) - a^2V\_{k-2}(x)$


I know that $a \in (0,1)$.


Another formulation of the problem could be to find the eigenvalues of the tridiagonal symmetric matrix $C\_n$


$C\_n = \left[ \begin{matrix}
1-a^2 & -a\sqrt{1-a^2} & & \\ 
-a\sqrt{1-a^2} & 1+a^2 & -a & \\ 
 & -a & 1+a^2 & \ddots \\ 
 & & \ddots & \ddots
\end{matrix} \right]$


Since the matrix is symmetric, the roots are going to be real. I also know that the matrices $C\_n$ are positive definite.


This is almost like a Chebyshev recursion, but a little bit perturbed.


**A bonus question**: I have some vague memories about the proof of the roots of Chebyshev polynomials that involved an argument that the roots of the consecutive polynomials in the sequence separate each other, and from this, with some additional tools the roots were derived. If someone can point to a location where I can find that proof, it would be really helpful.


**Edit after the solutions: Based on the trigonometric equations derived by Pietro Mejer, I suspect that there is no closed formula that describes the roots.**



Finding roots of a recursively given polynomial sequence / eigenvalues of an almost Toeplitz matix
For $k\in\mathbb{N}$ and $0<a<1$ your polynomial $V\_k$ has indeed $k$ distinct real roots in the interval 
$\big( (1-a)^2, (1+a)^2 \big)$ ; these can be written 
$$x\_j:=1+a^2-2a\cos(t\_j)$$
where $ t\_1<\dots < t\_k $ are the $k=\lceil k/2\rceil+\lfloor k/2\rfloor$ solutions in the interval $(0,\pi)$ to either
$$\cos\Big(\frac{k+1}{2}t\Big)=a\cos\Big(\frac{k-1}{2}t\Big)$$
 or  

$$\sin\Big(\frac{k+1}{2}t\Big)=a\sin\Big(\frac{k-1}{2}t\Big)\, .$$


To see it, write $x:=1+a^2-2az$ and introduce the polynomials $P\_k(z):=a^{-k}V\_k(x)$. This way $P\_k$
satisfy the Chebyshev polynomials linear recurrence, $$P\_{k+2}=2zP\_{k+1} - P\_k\, ,$$ with initial conditions $$P\_0=1-a^2$$ $$P\_1=2z-2a\, .$$
Solving the linear recurrence one finds a representation
$$P\_k(z):=\big(u^{k+1}-au^k-au+1 \big)\big(u^{k+1}-au^k+au-1\big)\frac{u^{-k}}{u^2-1}=$$


$$\frac{\Big(u^{\frac{k+1}{2}}+u^{-\frac{k+1}{2}}-a(u^{\frac{k-1}{2}}+u^{-\frac{k-1}{2}}) \Big)\Big(u^{\frac{k+1}{2}}-u^{-\frac{k+1}{2}}-a(u^{\frac{k-1}{2}}-u^{-\frac{k-1}{2}}) \Big)}{u-u^{-1}}\, ,$$


where $u:=z+i\sqrt{1-z^2}$. 


If we write $u:=\exp(it)$ and $z=\cos(t)$ we get


$$P\_k\big(\cos(t)\big)=\frac{2\bigg(\cos\Big(\frac{k+1}{2}t\Big)-a\cos\Big(\frac{k-1}{2}t\Big)\bigg)\bigg(\sin\Big(\frac{k+1}{2}t\Big)-a\sin\Big(\frac{k-1}{2}t\Big)\bigg)}{\sin(t)}\, .$$


It is easy to see that the first factor has $\lceil k/2\rceil$ zeros in $(0,\pi)$, while the other has $\lfloor k/2\rfloor$ zeros, and of course they have no common zeros, as $|a|\neq 1$. Since $\cos$ is bijective on $(0,\pi)$, these $k$ values $t\_1\dots t\_k$ correspond to distinct roots $z\_j:=\cos(t\_j)$ of $P\_k(z)$, thus all of them. 
$$\*$$
Note that for $a=0$ the roots of $P\_k$ are $\cos(t\_j)$ with $\cos\Big(\frac{k+1}{2}t\_j\Big)\sin\Big(\frac{k+1}{2}t\_j\Big)=0$, that is $\sin\Big((k+1)t\_j\Big)=0$, which is what we expect since $P\_k(z)=U\_k(z)$ for $a=0$, whose roots are $z\_j(0)=\cos\big(\pi j/(k+1)\big)$ for $1\le j\le k$.


For $a=1$ one finds $P\_k(z)=2(z-1)U\_{k-1}(z)$, with roots $z\_j(1)=\cos\big(\pi (j-1)/k\big)$ for $1\le j\le k$.


Also, if I'm not wrong, it's easy to see that the $z\_j=z\_j(a)$ are increasing wrto the parameter $a\in[0,1]$, so that $z\_j(0)\le z\_j(a)\le z\_j(1)$ for all $j$ and $a$. 


