<s>[INST] 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. 
 [/INST] 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,
$$P_n=2^{-n}[\frac{x+\sqrt{x^2-4}}{2\sqrt{x^2-4}}(x+\sqrt{x^2-4})^n+\frac{\sqrt{x^2-4}-x}{2\sqrt{x^2-4}}(x-\sqrt{x^2-4})^n]$$
</s>