New answers tagged orthogonal-polynomials
1
vote
Sum of squared hypergeometric polynomials
The identity can also be shown without divergences using generating functions, as suggested in the comment of @TimothyBudd above.
Using the generating functions
\begin{align}
\tag{1a}\label{eq:1a}
f(u,...
- 3,671
5
votes
Sum of squared hypergeometric polynomials
I think the proof below should work. Although it looks more complicated than the one of Fred Hucht, it doesn't involve divergent series.
I write
\begin{equation}\tag{1}S(t)=\sum_{m=1}^\infty\frac {t^m}...
- 2,306
9
votes
Sum of squared hypergeometric polynomials
The identity can be derived by differentiating both sides with respect to $u$:
The rhs
\begin{align}\tag{1}\label{eq:1}
R(u)=\frac 1 4-\frac 1 2 \log u
\end{align}
simply gives
\begin{align}\tag{2}\...
- 3,671
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