gfun[rectodiffeq] - convert a linear recurrence into a differential equation

Calling Sequence

rectodiffeq(eqns, u,n, f,z, <homogeneous=bool>, <operator=name>)

eqns - a single equation or a set of equations

u,n - the name and index of the recurrence

f,z - the name and variable of the function

homogeneous - (optional) boolean requesting homogeneous output; the default is false

operator - (optional) name used verbatim as the differential operator

Let f be the generating function associated to the sequence (u(n)): f(z)=sum(u(n)*z^n,n=0..infinity). The procedure outputs a linear differential equation with polynomial coefficients verified by f.
The input syntax is the same as for rsolve: the first argument should be a single recurrence relation or a set containing one recurrence relation and boundary conditions. The recurrence relation should be linear in the variable u, with polynomial coefficients in n. The terms of the sequence appearing in the relation should be of the form u(n+k), with k an integer.
The output is either a single differential equation, or a set containing a differential equation and initial conditions.
With operator=name, initial conditions are not computed. Homogeneous output is a polynomial in the supplied name and inhomogeneous output has the form L(name)=-q.

> with(gfun):
deq:=rectodiffeq({(5*n+10)*u(n)+a*u(n+1)-u(n+2),u(0)=0,u(1)=0},u(n),f(t));

\[\left(a t +10 t^{2}-1\right) f \! \left(t \right)+5 t^{3} \left(\frac{d}{d t}f \! \left(t \right)\right)\]

> diffeqtorec(deq,f(t),u(n));

\[\left(5 n +10\right) u \! \left(n \right)+a u \! \left(n +1\right)-u \! \left(n +2\right)\]

> deq:=rectodiffeq((n-10)*u(n+1)-u(n),u(n),y(z));

\[\left(-z -11\right) y \! \left(z \right)+z \left(\frac{d}{d z}y \! \left(z \right)\right)\]

> dsolve(deq,y(z));

\[y \! \left(z \right) = c_{1} {\mathrm e}^{z} z^{11}\]