borel - Maple Help

gfun

 borel
 compute the Borel transform of a generating function

 Calling Sequence borel(expr, a(n), t)

Parameters

 expr - linear recurrence with polynomial coefficients a - name; recurrence name n - name; index of the recurrence a t - (optional) 'diffeq'; specify as a linear differential equation

Description

 • The borel(expr, a(n)) command computes the Borel transform of a generating function.
 • If $a\left(n\right),n=0..\mathrm{\infty }$ is the sequence of numbers defined by the recurrence expr, the borel function computes the recurrence for the numbers $\frac{a\left(n\right)}{n!}$.
 • If $a\left(n\right),n=0..\mathrm{\infty }$ is the sequence of numbers defined by the recurrence expr, the procedure computes the recurrence for the numbers $\frac{a\left(n\right)}{n!}$.
 • If t is specified as 'diffeq', expr is considered as a linear differential equation with polynomial coefficients for the function $a\left(n\right)$. In this case, the function returns a linear differential equation satisfied by the Borel transform of $a\left(n\right)$.

Examples

 > $\mathrm{with}\left(\mathrm{gfun}\right):$
 > $\mathrm{rec}≔\left\{a\left(n\right)=na\left(n-1\right)+a\left(n-2\right),a\left(0\right)=1,a\left(1\right)=1\right\}:$
 > $b≔\mathrm{borel}\left(\mathrm{rec},a\left(n\right)\right)$
 ${b}{≔}\left\{{-}{a}{}\left({n}\right){+}\left({-}{{n}}^{{2}}{-}{3}{}{n}{-}{2}\right){}{a}{}\left({n}{+}{1}\right){+}\left({{n}}^{{2}}{+}{3}{}{n}{+}{2}\right){}{a}{}\left({n}{+}{2}\right){,}{a}{}\left({0}\right){=}{1}{,}{a}{}\left({1}\right){=}{1}\right\}$ (1)

The invborel command is the inverse command.

 > $\mathrm{invborel}\left(b,a\left(n\right)\right)$
 $\left\{{-}{a}{}\left({n}\right){+}\left({-}{n}{-}{2}\right){}{a}{}\left({n}{+}{1}\right){+}{a}{}\left({n}{+}{2}\right){,}{a}{}\left({0}\right){=}{1}{,}{a}{}\left({1}\right){=}{1}\right\}$ (2)

You can also perform Borel transforms on the corresponding differential equations.

 > $\mathrm{deq}≔\mathrm{rectodiffeq}\left(\mathrm{rec},a\left(n\right),f\left(x\right)\right):$
 > $\mathrm{newdeq}≔\mathrm{borel}\left(\mathrm{deq},f\left(x\right),\mathrm{diffeq}\right)$
 ${\mathrm{newdeq}}{≔}\left\{{-}{f}{}\left({x}\right){-}{2}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({x}\right){+}\left({1}{-}{x}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({x}\right)\right){,}{f}{}\left({0}\right){=}{1}{,}{\mathrm{D}}{}\left({f}\right){}\left({0}\right){=}{1}\right\}$ (3)
 > $\mathrm{diffeqtorec}\left(\mathrm{newdeq},f\left(x\right),a\left(n\right)\right)$
 $\left\{{-}{a}{}\left({n}\right){+}\left({-}{{n}}^{{2}}{-}{3}{}{n}{-}{2}\right){}{a}{}\left({n}{+}{1}\right){+}\left({{n}}^{{2}}{+}{3}{}{n}{+}{2}\right){}{a}{}\left({n}{+}{2}\right){,}{a}{}\left({0}\right){=}{1}{,}{a}{}\left({1}\right){=}{1}\right\}$ (4)