Prove that largest root of $Q_k(x)$ is greater than that of $Q_j(x)$ for $k>j$. Consider the recurrence relation:
$P_0=1$
$P_1=x$
$P_n(x)=xP_{n-1}-P_{n-2}$;
1). What is closed form of $P_n$?
2). Let $k,j.\in\{1,2,...,n+1\}$ and $Q_k(x)=xP_{2n+1}-P_{k-1}.P_{2n+1-k}$. Then prove that largest root of $Q_k$ is greater than that of $Q_j$ for $k>j$. Another observation is taking $k,j$ to be even and $k>j$, smallest positive root of $Q_k$ is greater than that of $Q_j$. Without explicitly calculating the roots, is there any way to prove these observations. 
 A: The expression for $P_n$ in terms of $n$ is pretty ugly,
$$P_n=xP_{n-1}-P_{n-2}$$
Exact form for $P_n$ can be solved by using characteristic equation: 
It comes from guessing $P_n=ar^n$ might be an answer for $P_n$, you can search "recurrence relation" online to see how it works.
$ar^n=x(ar^{n-1})-ar^{n-2}$
$r^2=xr-1$
Here $x$ can be treated as a constant since it wont affect our result,
$$r=\frac{x\pm\sqrt{x^2-4}}2$$
So that means the general form of $P_n$ is:
$$P_n=A(\frac{x+\sqrt{x^2-4}}2)^n + B(\frac{x-\sqrt{x^2-4}}2)^n$$
A and B are constant (with respect to $n$) that  need to satisfy initial conditions.
$$P_0=A+B=1$$
And $$P_1= A\frac{x+\sqrt{x^2-4}}2 + B\frac{x-\sqrt{x^2-4}}2=x$$
So $$(A+B)\frac x2 +(A-B)\frac{\sqrt{x^2-4}}2=x$$
$$A-B=\frac{x}{ \sqrt{x^2-4}}$$
So $$A=\frac{\sqrt{x^2-4}+x}{2\sqrt{x^2-4}}$$ 
And
$$B=\frac{\sqrt{x^2-4}-x}{2\sqrt{x^2-4}}$$
Finally,
$$P_n=2^{-n}[\frac{x+\sqrt{x^2-4}}{2\sqrt{x^2-4}}(x+\sqrt{x^2-4})^n+\frac{\sqrt{x^2-4}-x}{2\sqrt{x^2-4}}(x-\sqrt{x^2-4})^n]$$
A: Claim:
$$
P_n(x)=\frac{\sin((n+1)\arccos(x/2))}{\sin\arccos(x/2)}
$$
This is obviously true for $n=0,1$ and (using $u=\arccos(x/2)$)
$$
P_{n+1}-P_{n-1}=\frac{\sin((n+2)u)-\sin(nu)}{\sin u}=\frac{2\sin((n+1)u)\cos u}{\sin u}=xP_n(x)
$$
so it satisfies the recursion equation.
For $|x|>2$ the formula modifies to
$$
P_n(x)=\frac{\sinh((n+1)\text{Arcosh}(x/2))}{\sinh(\text{Arcosh}(x/2))}
$$
by the usual connection of trigonometric and hyperbolic functions.

From this one observes that
$$
\det\begin{bmatrix}P_1&P_k\\P_{m+1}&P_{m+k}\end{bmatrix}
=xP_{m+k}-P_{k}P_{m+1}
\\=\frac{\sin(2u)\sin((m+k+1)u)-\sin((k+1)u)\sin((m+2)u)}{\sin^2u}
\\
=\frac{\cos(m+k-1)u-\cos(m+k+3)u-\cos(m-k+1)u+\cos(m+k+3)u}{2\sin^2u}
\\
=\frac{\sin(mu)\sin((k-1)u)}{\sin^2u}=P_{m-1}P_{k-2}
$$
This factorization allows statements on the root set, however the original determinant expression is incompatible with $Q_k$, and the $Q_k$ do not simplify that nicely. However, one gets close with the replacement $(m,k)\to(2n-k,k+1)$ which gives
$$
xP_{2n+1}-P_{k+1}P_{2n-k+1}=P_{2m-k-1}P_{k-1}
$$
