(12/14/2009, 01:33 PM)bo198214 Wrote:(12/08/2009, 10:32 AM)mike3 Wrote: So far, I've found 2 types of transseries representation such that if the series converges, the continuum sum does as well (proof given earlier here):
I was a bit puzzled why \( e^{e^x} \) converges via transseries while \( e^{x^2} \) does not converge. Actually \( e^{\frac{x^2}{2}} \) must also converge as we have \( \frac{x^2}{2}<e^x \), and indeed it also converges! But it seems that this already the limit, i.e. every series \( e^{c x^2} \) for \( c>\frac{1}{2} \) does not converge. I guess that this is the same for \( e^{\frac{x^n}{n!} \), \( n>1 \), while Faulhaber always works for \( n=1 \), i.e. \( e^{c x} \). Do you have any idea how to sum \( e^{x^2} \)?
Hi,
I've been reading on this forum for a while, and just now I've signed up.
I have been exploring the sum operator, what you guys call "continuum sum" for a couple of years as a hobby.
I think I have an idea how to sum \( e^{x^2} \).
Recall that \( e^{x^2} = \lim_{n \to \infty} \left( 1 + \frac{x^2}{n} \right)^n \)
Now calculate the polynomial sum of \( \left( 1 + \frac{x^2}{n} \right)^n \) (using Faulhaber's formula).
This seems to convergent to a nice function.

