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  1. Say I have a generating function $f(z) = \sum_{n=0}^\infty a_nz^n$. And I want a generating function for $g(z) = \sum_{n=0}^\infty n!a_nz^n$ for the same $a_n$. How do I modify $f(z)$?

EDIT: In my case $f(z) = \frac{1}{(1-z)(2-e^z)}$.

  1. Are there any results on evaluating coefficients of generating functions modulo primes?

Thanks.

user75619
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  • If $g(z)$ is the ordinary generating function for some series, then $f(z)$ is the exponential generating function for that series. I don't believe that there is some general method you can use to transform one in the other. – orlp Oct 12 '17 at 20:10
  • As an example, $f(x) = \frac{2}{\sqrt5 }\exp(\frac{1}{2}x)\sinh(\frac{\sqrt 5}{2}x)$ gives $g(x) = \dfrac{x}{1 - x - x^2}$. – orlp Oct 12 '17 at 20:14
  • I have $f(z) = \frac{1}{(1-z)(2-e^z)}$. – user75619 Oct 12 '17 at 20:16
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    Turns out that there is such a way, and it's called the Borel transform. I believe that's what you're looking for. – orlp Oct 12 '17 at 20:24

1 Answers1

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Regarding 1, we have (this is the Borel transform):

$$g(z) = \sum_{n=0}^\infty n!a_nz^n$$

$$g(z) = \sum_{n=0}^\infty n!a_n \frac{1}{n!}\left(\int_0^\infty e^{-t}t^ndt\right)z^n$$

$$g(z) = \int_0^\infty e^{-t}\sum_{n=0}^\infty a_n (tz)^ndt$$ $$g(z) = \int_0^\infty e^{-t}f(tz)dt$$

So in your case:

$$g(z) = \int_0^\infty \frac{e^{-t}}{(1-tz)(2-e^{tz})}dt$$

orlp
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  • Thanks. Interesting theoretically, doesn't seem to lead to any manageable result though... :) But I guess there's simply no such result. – user75619 Oct 12 '17 at 21:06
  • @reuns To be honest, I haven't got the slightest clue. This isn't particularly my strong point (quite new to this). – orlp Oct 12 '17 at 21:22
  • Forget about my comment. Note $g(z) = \sum_{n=0}^\infty n! a_n z^n$ converges for one $z \ne 0$ implies $f(z) = \sum_{n=0}^\infty a_n z^n$ is entire, but for your integral $g(z) = \int_0^\infty e^{-t}f(tz)dt$ to converge (and be holomorphic) on a disk we don't need $f$ to be entire. – reuns Oct 12 '17 at 21:41