tpm
Apr2-07, 06:27 AM
If we have the Laplace transform:
\int_{0}^{\infty}dtf(t)exp(-st) = \sum_{n=0}^{\infty}(f(n)-f(n-1))\frac{exp(-sn)}{s}=g(s)
(We have used Abel sum formula on the right hand)
then making Z=exp(s) we find the Z-transform:
\sum_{n=0}^{\infty}(f(n)-f(n-1))Z^{-n}
which can be inverted to get:
2\pi i (f(n)-f(n-1))=\oint g(lnZ) (lnZ)Z^{n-1}
my quetion is if using 'Circle method' you can get an asymptotic expansion for: f(n)-f(n-1) n big, also another question if we have that:
f(n)-f(n-1) \sim h(n) then ??? f(n) \sim \int h(n)dn
\int_{0}^{\infty}dtf(t)exp(-st) = \sum_{n=0}^{\infty}(f(n)-f(n-1))\frac{exp(-sn)}{s}=g(s)
(We have used Abel sum formula on the right hand)
then making Z=exp(s) we find the Z-transform:
\sum_{n=0}^{\infty}(f(n)-f(n-1))Z^{-n}
which can be inverted to get:
2\pi i (f(n)-f(n-1))=\oint g(lnZ) (lnZ)Z^{n-1}
my quetion is if using 'Circle method' you can get an asymptotic expansion for: f(n)-f(n-1) n big, also another question if we have that:
f(n)-f(n-1) \sim h(n) then ??? f(n) \sim \int h(n)dn