$\lim_{n \to \infty}n^2\left(4-2\pi+n^2\int_{x_n}^2P_n(x)dx\right)$

Question

Let $a, b$ and $c$ be complex numbers such that $abc=1$.

Denote $\Omega(n)=$ $\begin{vmatrix}b+n^3c & n(c-b) & n^2(b-c) \\ n^2(c-a) & c+n^3a & n(a-c) \\ n(b-a) & n^2(a-b) & a+n^3b\end{vmatrix}$

Now, Consider the sequence of polynomials $P_0(x)=2$, $P_1(x)=x$ and $P_n(x)=xP_{n-1}(x)-P_{n-2}(x)$ for $n\geq2$. Let $x_n$ be the greatest zero of $P_n$ in the interval $|x|\leq2$.

If $$\lim_{n \to \infty}\big(\sqrt[3]{\Omega(\sqrt[3]n)}-1\big)^2\left(4-2\pi+\big(\sqrt[3]{\Omega(\sqrt[3]n)}-1\big)^2\int_{x_n}^2P_n(x)dx\right)$$ $$=p\pi-q-\frac{\pi^3}{r}$$where $p, q, r \in\mathbb Z^{+}$, such that $r$ is a two digit number, compute $p+q+r$.

My Attempt at the Solution:

$$\Omega(n)=abc(n^9+3n^6+3n^3+1)$$ $$=n^9+3n^6+3n^3+1$$ Therefore, $$\sqrt[3]{\Omega(\sqrt[3]n)}-1=n$$ The limit reduces to

$$\lim_{n \to \infty}n^2\left(4-2\pi+n^2\int_{x_n}^2P_n(x)dx\right)$$

I am not able to take my solution any further, as I am not that well versed in polynomial recurrence relations.

Answer 1

Consider the reduction-of-order equivalent recurrence relation

$$\mathbf{p}_n\equiv\begin{pmatrix}P_n \\ P_{n-1}\end{pmatrix} = \begin{pmatrix}x & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}P_{n-1} \\ P_{n-2}\end{pmatrix} \equiv \mathbf{T}\mathbf{p}_{n-1} = \mathbf{T}^{n-1}\mathbf{p}_1$$

This reduces to an eigenvalue problem. The eigenvalues of $\mathbf{T}$ are

$$\lambda_{\pm}(x) = \frac{x\pm i\sqrt{4-x^2}}{2}$$

on the interval $|x|\leq 2$, with corresponding eigenvectors

$$\mathbf{v}_{\pm}(x) = \begin{pmatrix}\lambda_{\pm}(x) \\ 1\end{pmatrix}$$

Notice that

$$\mathbf{v}_+ + \mathbf{v}_- = \begin{pmatrix}x \\ 2\end{pmatrix} = \mathbf{p}_1$$

thus

$$\mathbf{p}_n = \lambda_+^{n-1}\mathbf{v}_+ + \lambda_-^{n-1}\mathbf{v}_-$$

and therefore

$$P_n(x) = \left(\frac{x+i\sqrt{4-x^2}}{2}\right)^n + \left(\frac{x-i\sqrt{4-x^2}}{2}\right)^n$$

Equivalently we could say let $x=2\cos\theta$ and

$$P_n(2\cos\theta) = e^{in\theta} + e^{-in\theta} = 2\cos( n\theta)$$

which means the integral can be rewritten as

$$\int_0^{\frac{\pi}{2n}}4\cos(n\theta)\sin\theta\:d\theta \approx \int_0^{\frac{\pi}{2n}}4\theta\cos(n\theta) - \frac{2}{3}\theta^3\cos(n\theta)\:d\theta + O(n^{-6}) $$

$$= \frac{2\pi-4}{n^2} + \frac{2\pi-4}{n^4} - \frac{\pi^3}{12n^4} + O(n^{-6})$$

And the limit simplifies to

$$\require{cancel}\lim_{n\to\infty}n^2\left(\cancel{4-2\pi}+\cancel{2\pi-4}-\frac{4-2\pi}{n^2}-\frac{\pi^3}{12n^2}+O(n^{-4})\right) = \boxed{-\frac{\pi^3}{12} + (2\pi -4)}$$

Answer 2

A closed form of $P_n(x)$ is not hard to find. In its shortest form, $$P_n(2\cos t)=2\cos nt$$ is easy to show using induction on $n$. Then clearly $x_n=2\cos\frac\pi{2n}$ and, for $n>1$, $$\int_{x_n}^2 P_n(x)\,dx=\int_0^{\frac\pi{2n}}2\cos nt\cdot2\sin t\,dt=\frac4{n^2-1}\left(n\sin\frac\pi{2n}-1\right).$$ Now the limit is easily evaluated (e.g. via the power series for $\sin$) as $\color{blue}{2\pi-4-\dfrac{\pi^3}{12}}$.