- #1
lokofer
- 106
- 0
Borel "resummation"...
Let be a divergent series:
[tex] \sum _{n=0}^{\infty} a(n) [/tex] (1)
then if you "had" that [tex] f(x)= \sum _{n=0}^{\infty} \frac{a(n)}{n!}x^{n} [/tex]
You could obtain the "sum" of the series (1) as [tex] S= \int_{0}^{\infty}dte^{-t}f(t) [/tex] in case the integral converges...
- Yes that's "beatiful" the problem is ..what happens if the coefficients a(n) are complicate?..then how can you obtain the sum of the series?...
- By the way i think that Borel resummation can be applied if [tex] f(t)=O(e^{Mt}) [/tex] M>0, but what happens if f(t) grows faster than any positive exponential?..
Let be a divergent series:
[tex] \sum _{n=0}^{\infty} a(n) [/tex] (1)
then if you "had" that [tex] f(x)= \sum _{n=0}^{\infty} \frac{a(n)}{n!}x^{n} [/tex]
You could obtain the "sum" of the series (1) as [tex] S= \int_{0}^{\infty}dte^{-t}f(t) [/tex] in case the integral converges...
- Yes that's "beatiful" the problem is ..what happens if the coefficients a(n) are complicate?..then how can you obtain the sum of the series?...
- By the way i think that Borel resummation can be applied if [tex] f(t)=O(e^{Mt}) [/tex] M>0, but what happens if f(t) grows faster than any positive exponential?..