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The rectodiffeq(eqns, u(n), f(z)) command converts a linear recurrence into a differential equation.

Let f be the generating function associated with the sequence u(n) where $f\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}u\left(n\right)z^n$. The rectodiffeq function returns a linear differential equation with polynomial coefficients satisfied by f.

The input syntax is the same as for rsolve . The first argument is a single recurrence relation or a set containing one recurrence relation and boundary conditions. The recurrence relation is linear in the variable u, with polynomial coefficients in n. The terms of the sequence appearing in the relation should be of the form $u\left(n+k\right)$, with k as an integer.

The function returns either a single differential equation, or a set containing a differential equation and initial conditions.

$\mathrm{with}\left(\mathrm{gfun}\right)\:$

$\mathrm{deq}≔\mathrm{rectodiffeq}\left(\left\{\left(5n+10\right)u\left(n\right)+au\left(n+1\right)-u\left(n+2\right)\,u\left(0\right)=0\,u\left(1\right)=0\right\}\,u\left(n\right)\,f\left(t\right)\right)$

$\mathrm{deq}≔\left(at+10t^2-1\right)f\left(t\right)+5t^3\left(\frac{\ⅆ}{\ⅆt}f\left(t\right)\right)$

$\mathrm{diffeqtorec}\left(\mathrm{deq}\,f\left(t\right)\,u\left(n\right)\right)$

$\left(5n+10\right)u\left(n\right)+au\left(n+1\right)-u\left(n+2\right)$

$\mathrm{deq}≔\mathrm{rectodiffeq}\left(\left(n-10\right)u\left(n+1\right)-u\left(n\right)\,u\left(n\right)\,y\left(z\right)\right)$

$\mathrm{deq}≔\left(-z-11\right)y\left(z\right)+z\left(\frac{\ⅆ}{\ⅆz}y\left(z\right)\right)$

$\mathrm{dsolve}\left(\mathrm{deq}\,y\left(z\right)\right)$

$y\left(z\right)=\mathrm{c\_\_1}{\ⅇ}^z{z}^{11}$