Subject: asymptotic expansion of the error function
From: Dan
Date: Wed, 25 Apr 2012 18:10:45 +0100 (BST)
On Wed, 25 Apr 2012, Raymond Toy wrote:
> Too bad this doesn't work:
>
> taylor(erfc(z),z,inf,4)
It can't, because erfc(1/y) has an essential singularity at y = 0, and
therefore no series representation there in powers of y alone. The
asymptotic series is allowed to evade that by containing an explicit
exponential factor.