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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 $a\left(n\right)$.

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

$\mathrm{rec}≔\left\{a\left(0\right)=1\,a\left(1\right)=1\,a\left(n\right)=na\left(n-1\right)+a\left(n-2\right)\right\}\:$

$b≔\mathrm{borel}\left(\mathrm{rec}\,a\left(n\right)\right)$

$b≔\left\{-a\left(n\right)+\left(-n^2-3n-2\right)a\left(n+1\right)+\left(n^2+3n+2\right)a\left(n+2\right)\,a\left(0\right)=1\,a\left(1\right)=1\right\}$

$\mathrm{invborel}\left(b\,a\left(n\right)\right)$

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

$\mathrm{deq}≔\mathrm{rectodiffeq}\left(\mathrm{rec}\,a\left(n\right)\,f\left(x\right)\right)\:$

$\mathrm{newdeq}≔\mathrm{borel}\left(\mathrm{deq}\,f\left(x\right)\,\mathrm{diffeq}\right)$

$\mathrm{newdeq}≔\left\{-f\left(x\right)-2\frac{\ⅆ}{\ⅆx}f\left(x\right)+\left(1-x\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}f\left(x\right)\right)\,f\left(0\right)=1\,\mathrm{D}\left(f\right)\left(0\right)=1\right\}$

$\mathrm{diffeqtorec}\left(\mathrm{newdeq}\,f\left(x\right)\,a\left(n\right)\right)$

$\left\{-a\left(n\right)+\left(-n^2-3n-2\right)a\left(n+1\right)+\left(n^2+3n+2\right)a\left(n+2\right)\,a\left(0\right)=1\,a\left(1\right)=1\right\}$