Numerical evidence indicates that Jacobi polynomials with negative integer parameters satisfy the identity $$P_n^{(-m,-k)}(x)=\left(\frac{x-1}{2}\right)^m\left(\frac{1+x}{2}\right)^kP_{n-m-k}^{(m,k)}(x),$$ where $n\ge m+k$. How this identity can be proved?

I came across this supposed identity when comparing some linearisation relations for the product of two Laguerre polynomials from https://cds.cern.ch/record/1378854?ln=en (after correcting typos) and https://iopscience.iop.org/article/10.1088/0305-4470/18/9/022 but such a proof (assuming both papers are correct) is too complicated and I prefer a direct proof which then I can use to argue that the results of these two papers are equivalent to each other.

P.S. The identity can be proved also by using $$P_n^{(-m,\beta)}(x)=\frac{\Gamma(n+\beta+1)}{\Gamma(n+\beta+1-m)}\frac{(n-m)!}{n!}\left(\frac{x-1}{2}\right)^m\,P_{n-m}^{(m,\beta)}(x),$$ which can be found in the book G. Szego, Orthogonal Polynomials (formula (4.22.2)), in combination with the symmetry relation $$P_n^{(\alpha,\beta)}(x)=(-1)^n\,P_n^{(\beta,\alpha)}(-x).$$


1 Answer 1


The identity you want is (typo in second factor of proposer's formula) $$ P_n^{(-m,-k)}(x)=\Big(\frac{x-1}{2}\Big)^{m} \Big(\frac{x+1}{2}\Big)^{k} P_{n-m-k}^{(m,k)}(x). $$ Start with the ID (it's on the wiki) $$ \frac{P_n^{(a,b)}(x)}{(n+a)!(n+b)!}=\sum_s \Big( s!(n+a-s)!(b+s)!(n-s)! \Big)^{-1} \Big(\frac{x-1}{2}\Big)^{n-s} \Big(\frac{x+1}{2}\Big)^{s} $$ The conditions for this formula to be valid are as the proposer stated, $n\ge m+k.$ Since the factorials within the sum are in the denominator, we can extend the sum from $-\infty$ to $+\infty.$ The left-hand side of the top equation is $$ \frac{P_n^{(-m,-k)}(x)}{(n-m)!(n-k)!}=\sum_s \Big( s!(n-m-s)!(s-k)!(n-s)! \Big)^{-1} \Big(\frac{x-1}{2}\Big)^{n-s} \Big(\frac{x+1}{2}\Big)^{s} $$ and likewise $$ \frac{P_{n-m-k}^{(m,k)}(x)}{(n-m)!(n-k)!}=\sum_s \big(s!(n\!-\!k-\!s)!(k+s)!(n\!-\!m\!-\!k\!-\!s)! \big)^{-1} * $$ $$\big(\frac{x\!-\!1}{2}\big)^{n-m-k-s} \big(\frac{x\!+\!1}{2}\big)^{s}.$$

Shift the second sum by $s \to s - k$ and you will find the same sum as the previous. Algebra completes the proof.


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