2
$\begingroup$

Clausen’s identity for Legendre polynomials has the form (see, for example, A generating function of the squares of Legendre polynomials, by Wadim Zudilin: https://arxiv.org/abs/1210.2493) $$P_n(\cos{\theta})^2=\sum_{k=0}^n\frac{(-1)^k}{2^{2k}}\binom{n}{k}\binom{n+k}{n}\binom{2k}{k}\sin^{2k}{\theta}.$$ Do the analogous identity exist for the square of associated Legendre polynomials $[P_n^l(\cos{\theta})]^2$?

$\endgroup$
3
$\begingroup$

There is a similar formula $$ \small{\left(P_n^m(\cos\theta)\right)^2=(\sin{\theta})^{2m}\frac{(m+n)!}{(n-m)!}\sum_{k=0}^{n-m}\frac{(-1)^k}{4^{k+m}}\binom{n+m}{k+2m}\binom{n+k+m}{n+m}\binom{2k+2m}{k+m}\sin^{2k}{\theta}} $$ obtained from the following representation for associated Legendre polynomials $$ P_n^m(z)=(-1)^m \left(\frac{1-z}{1+z}\right)^{\frac{m}{2}} \frac{(m+n)!}{m! (n-m)!} \, _2F_1\left(-n,n+1;m+1;\frac{1-z}{2}\right) $$ and the formula (6.1) from the paper W.N. Bailey, Some Theorems Concerning Products of Hypergeometric Series, Proc. London Math. Soc. 38, 377–384 (1935): $$ {}_2F_1(a,b;c;x){}_2F_1(a,b;a+b-c+1;x)={}_4F_3\left({a,b,\tfrac{a+b}2,\tfrac{1+a+b}2\atop a+b,c,a+b-c+1};4x(1-x)\right). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.