lokofer
Aug29-06, 02:29 AM
Let be a divergent series:
\sum _{n=0}^{\infty} a(n) (1)
then if you "had" that f(x)= \sum _{n=0}^{\infty} \frac{a(n)}{n!}x^{n}
You could obtain the "sum" of the series (1) as S= \int_{0}^{\infty}dte^{-t}f(t) 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 f(t)=O(e^{Mt}) M>0, but what happens if f(t) grows faster than any positive exponential?.. :cry:
\sum _{n=0}^{\infty} a(n) (1)
then if you "had" that f(x)= \sum _{n=0}^{\infty} \frac{a(n)}{n!}x^{n}
You could obtain the "sum" of the series (1) as S= \int_{0}^{\infty}dte^{-t}f(t) 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 f(t)=O(e^{Mt}) M>0, but what happens if f(t) grows faster than any positive exponential?.. :cry: