How to find the polynomial corresponding to positive roots of $f(x)$?

Question

Let $$P_0(x)=1$$

$$P_1(x)=x$$

$$P_n(x)=\frac{(x+\sqrt{x^2-4})^{n+1}-(x-\sqrt{x^2-4})^{n+1}}{2^{n+1}\sqrt{x^2-4}}; n\ge2$$

and $$f(x)=x P_{2n+1}(x)− P_{2}(x).P_{2n-2}(x)$$

It is given that roots of $f(x)$ are real and symmetric about origin that is $\alpha$ is a root of $f(x)$ iff $-\alpha$ too is a root. So $f(x)$ has some positive roots.

How to find the polynomial corresponding to positive roots of $f(x)$?

Answer 1

Setting $x=2\cosh u$ one rapidly sees that there are no roots $x>2$.

Set $x=2\cos(u)$ for $x\in[-2,2]$, then $\sqrt{x^2-4}=2i\sin x$, $x\pm\sqrt{x^2-4}=e^{\pm i u}$ and $$ P_n(x)=\frac{\sin((n+1)u)}{\sin u}. $$

Then $$ f(x)=\frac{\sin(2u)\sin((2n+2)u)-\sin(3u)\sin((2n-1)u)}{\sin^2(u)} $$ which can be transformed but not simplified using the cosine theorem $\sin A\sin B=\cos(A-B)-\cos(A+B)$.

---

Now if one had $$ f(x)=\begin{vmatrix}P_1(x)&P_2(x)\\P_{2n-2}(x)&P_{2n-1}(x)\end{vmatrix}=xP_{2n-1}(x)-P_2(x)P_{2n-2}(x) $$ then one would get \begin{align} f(x)&=\frac{\sin(2u)\sin((2n)u)-\sin(3u)\sin((2n-1)u)}{\sin^2(u)} \\&=\frac{\cos((2n-2)u)-\cos((2n+2)u)-\cos((2n-4)u)+\cos((2n+2)u)}{2\sin^2u} \\&=-\frac{\sin((2n-3)u)}{\sin u} \end{align}