# Recurrence relation for Jacobi polynomials with negative parameter $\beta$

The Jacobi polynomials $P^{(\alpha,\beta)}_n(z)$ have the following recurrence relation in $n$ (see e.g. here):

$2n(n+\alpha+\beta)(2n+\alpha+\beta-2)P^{(\alpha,\beta)}_n(z)$ $=(2n+\alpha+\beta-1)\Big((2n+\alpha+\beta)(2n+\alpha+\beta-2)z+\alpha^2-\beta^2\Big)P^{(\alpha,\beta)}_{n-1}(z)$ $-2(n+\alpha-1)(n+\beta-1)(2n+\alpha+\beta)P^{(\alpha,\beta)}_{n-2}$ .

I'm trying to use this to write a program for calculating the polynomial. However for some of the parameter values I need (e.g. $n=2$, $\alpha=4$ and $\beta=-6$) the relation is not valid, since the factor in front of $P^{(\alpha,\beta)}_n(z)$ is zero.

Does anyone know if there is a different relation valid in these cases? Or otherwise have ideas on how I can deal with these cases without completely changing the program (e.g. not using recurrence relations)?

## 1 Answer

In my own library implementation $\alpha+\beta$ shall not be a negative integer. Many other implementations assume $\alpha, \beta > -1,$ because for these values the orthogonality relation holds. If you really want these unusual ranges for $\alpha, \,\beta$ you can use relations to the Gauss hypergeometric function ${_2}F_{1}$. If (as in your case) $\alpha$ is no negative integer there is $$P^{(\alpha,\beta)}_n(z) =\frac{(\alpha+1)_n}{n!} {_2}F_{1}\left(-n,1+\alpha+\beta +n; 1+\alpha; \frac{1-z}{2}\right)$$ which will give for example $$P^{(4,-6)}_2\left(\frac{1}{2}\right)=\frac{217}{16}$$ A general formula for $n \in \mathbb{N}$ you can find at the Wolfram function site: $$P^{(\alpha,\beta)}_n(z) = \frac{1}{n!}\sum_{k=0}^{n}\frac{(-n)_k (\alpha+\beta +n+1)_k (\alpha+k+1)_{n-k}}{k!}\left(\frac{1-z}{2}\right)^k$$ With your parameters the polynomial is (if I did not make any mistakes) $$P^{(4,-6)}_2\left(z\right)=\frac{49}{4}+\frac{5}{2}z+\frac{1}{4}z^2$$

• Thanks! Btw if I understand correctly it seems the orthogonality relation holds universally: ac.els-cdn.com/S0377042701005891/… Commented Oct 30, 2013 at 14:38
• I must thank you. I had never heard of Sobolev orthogonality! Commented Oct 30, 2013 at 14:49