I know that GeneratingFunction can be used to compute the generating function $\sum_{n=0}^\infty a_n x^n$ of a sequence $(a_n)_n$ via
GeneratingFunction[a[n],n,x]
I also know that polynomial (in $n$) coefficients of the general term $a_n$ of the sequence result in an expression involving derivatives of the generating function, e.g.
GeneratingFunction[n a[n],n,x]
(* x (GeneratingFunction^(0,0,1))[a[n],n,x] *)
This does not seem to happen with rational (in $n$) coefficients, which would lead to indefinite integrals of the generating function. For example I would expect the input
GeneratingFunction[a[n]/(n + 1), n, x]
to result in
(*1/x Integrate[GeneratingFunction[a[n],n,y],{y,0,x}]*)
or
(*1/x GeneratingFunction^(0,0,-1))[a[n],n,x]*)
but this does not seem to be the case.
I thus have the following question: Is it possible to get Mathematica to compute the generating function of a linear combination of lagged values of $a_n$ with rational (in $n$) coefficients in terms of derivatives and integrals of the generating function of the sequence $(a_n)_n$?
GeneratingFunction[a[n]/(n + 1), n, x] === 1/x Integrate[GeneratingFunction[a[n], n, y], {y, 0, x}]
isFalse
. Are you sure that it lands there ? $\endgroup$ – Sejwal Sep 2 '13 at 18:18GeneratingFunction^(0,0,1)
is not recognized byM
, what do you mean by this ? $\endgroup$ – Sejwal Sep 2 '13 at 19:11GeneratingFunction[n a[n],n,x]
evaluates tox GeneratingFunction[(1+n)a[1+n], n, x]
. Which version of Mma are you using? (I'm using 9.) $\endgroup$ – Andrew Sep 18 '13 at 0:08