Roots of Recursive Polynomials

Question

Define a series of polynomials as such: $$p_0(x)=1,$$ $$p_1(x)=x,\text{and}$$ $$p_n(x)=xp_{n-1}(x)-p_{n-2}(x)\forall n\in\left\{2,3,4,\dots\right\}.$$ I am interested in the roots of these polynomials. Let $S_k=\left\{t:p_k(t)=0\right\}.$ Then, we have that $$S_0=\emptyset,$$ $$S_1=\left\{0\right\},$$ $$S_2=\left\{-1,1\right\},$$ $$S_3=\left\{-\sqrt{2},0,\sqrt{2}\right\},$$ $$S_4=\left\{-\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2},\frac{\sqrt{5}-1}{2},\frac{1+\sqrt{5}}{2}\right\},$$ $$S_5=\left\{-\sqrt{3},-1,0,1,\sqrt{3}\right\},$$ $$S_6,$$ $$\text{and}$$ $$S_7=\left\{-\sqrt{2+\sqrt{2}},-\sqrt{2},-\sqrt{2-\sqrt{2}},0,\sqrt{2-\sqrt{2}},\sqrt{2},\sqrt{2+\sqrt{2}}\right\}.$$ As you can see, most of these roots seem to have "nice" closed forms. This is why I am interested in these roots. I know that $p_m$ is a polynomial of degree $m.$ So, $S_m$ has at most $m$ elements. If we could prove that $p_m$ never has repeated roots, then we could say that $S_m$ has exactly $m$ elements. Is there anything else that can be said about the roots, or am I being too ambitious?

Answer 1

The polynomials $U_n(x) = p_n(2x)$ satisfy $$ U_0(x) = 1 \\ U_1(x) = 2x \\ U_{n}(x) = p_n(2x) = 2x p_{n-1}(2x) - p_{n-2}(2x) = 2x U_{n-1}(x) - U_{n-2}(x) $$ and the solution of that recurrence is well-known: $U_n$ are the Chebyshev polynomials of the second kind and their roots are $$ x_k = \cos\left( \frac{k \pi}{n+1}\right) \, , \, k = 1, \ldots, n \, . $$ So the roots of your polynomials $p_n$ are $$ x_k = 2 \cos\left( \frac{k \pi}{n+1}\right) \, , \, k = 1, \ldots, n \, . $$ Example: The roots of $p_5$ are $$ 2 \cos\left( \frac{\pi}{6}\right) = \sqrt 3 \, ,\\ 2 \cos\left( \frac{2\pi}{6}\right) = 1 \, ,\\ 2 \cos\left( \frac{3\pi}{6}\right) = 0 \, ,\\ 2 \cos\left( \frac{4\pi}{6}\right) = -1 \, ,\\ 2 \cos\left( \frac{5\pi}{6}\right) = -\sqrt 3 \, .\\ $$