We know that $p_1(x) = 0$ and $p_2(x) = x^2 - 1$. The recurrence is given by
$$ p_n(x) = xp_{n-1}(x) - p_{n-2}(x). $$
Is there a way to find a general formula for $p_n(x)$ that does not depend on the previous terms?
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1Yes, follow the methods here. – Dietrich Burde Oct 29 '17 at 17:53
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1For $n \ge 2$, $p_n(x) = (x^2-1)U_{n-2}(x/2)$ where $U_n(x)$ is the Chebyshev polynomial of second kind. – achille hui Oct 29 '17 at 18:04
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One can consider a solution of the difference equation in the for $p_{n}(x) = r^{n}(x)$ and determine that $r^{2} - x \, r + 1 = 0$ and has solutions $\{\alpha(x), \beta(x)\}$, where $2 \, \alpha(x) = x + \sqrt{x^2 - 4}$ and $2 \, \beta(x) = x - \sqrt{x^2 - 4}$. Now a solution for $p_{n}(x)$ takes the form $$p_{n}(x) = A \, \alpha^{n}(x) + B \, \beta^{n}(x).$$ Since $p_{1}(x) = 0$ then $B = - A (\alpha/\beta)$, or $$p_{n}(x) = \alpha(x) \, A \, \left( \alpha^{n-1}(x) - \beta^{n-1}(x) \right).$$ Since $p_{2}(x) = x^{2} - 1$ then $$p_{n}(x) = (x^{2} - 1) \, \left( \frac{\alpha^{n-1}(x) - \beta^{n-1}(x)}{\alpha(x) - \beta(x)} \right).$$
A generating function may be obtained for which $$\sum_{n=1}^{\infty} p_{n}(x) \, t^{n} = \frac{(x^{2}-1) \, t^{2}}{1-x t + t^{2}}.$$ Further properties may be obtained by use of the generating function.
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