If we have the Laplace transform:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \int_{0}^{\infty}dtf(t)exp(-st) = \sum_{n=0}^{\infty}(f(n)-f(n-1))\frac{exp(-sn)}{s}=g(s) [/tex]

(We have used Abel sum formula on the right hand)

then making Z=exp(s) we find the Z-transform:

[tex] \sum_{n=0}^{\infty}(f(n)-f(n-1))Z^{-n} [/tex]

which can be inverted to get:

[tex] 2\pi i (f(n)-f(n-1))=\oint g(lnZ) (lnZ)Z^{n-1} [/tex]

my quetion is if using 'Circle method' you can get an asymptotic expansion for: [tex] f(n)-f(n-1) [/tex] n big, also another question if we have that:

[tex] f(n)-f(n-1) \sim h(n) [/tex] then ??? [tex] f(n) \sim \int h(n)dn [/tex]

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# Circle method for Laplace transform.

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