# When may Function (meromorphic) be expanded as power series with coefficients of integers

Let F be meromorphic function,with what properties may it expanded as power series with coefficients of integers in such a form:

$$F=\sum_0^{\infty}a_i x^i,a_i\in \mathcal{N} \bigcup 0,\exists M \space a_i \leq M^i$$.

and when the coefficients consist of a sequence of computably enumerable relation.

If the question is ambiguous ,please tell me but please do not downvote it.

When may Function (meromorphic) be expanded as power series with coefficients of integers

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Without specifying what kind of properties you are asking for, this seems too open-ended for me. How about "F has the property that it may be expanded as a power series with integer coefficients"? Perfectly well-defined property. – Yemon Choi Aug 24 '11 at 10:08
I assume you mean on the whole complex plane. The power series coefficients come from C, so there are very few meromorphic functions with integer coefficients. Certainly 1/z^k * f(z) for f(z) = e(z), sin(z), cos(z), but it's hard to say something specific. Can you expound on your question? – Robert K Aug 24 '11 at 10:12
@Robert: A power series with integer coefficients can never converge on the whole complex plane, unless it is a polynomial. Indeed, the Cauchy–Hadamard theorem implies that a series with integer coefficients, infinitely many of which are nonzero, has radius of convergence at most $1$. – Emil Jeřábek Aug 24 '11 at 10:31
A sufficient condition is: Write the meromorphic function as quotient $f/g$ of homolomorphic functions. If the powerseries of $f$ and $g$ (around the origin) have integral coefficients and $g(0) = 1$ holds, than the powerseries of the meromorphic function will also have integral coefficients. – Ralph Aug 24 '11 at 10:52
@Robert: Can you please clarify what $1/z^k*f(z) ...$ means.For instance, $z^{-1}\cdot sin(z)$ doesn't epand with integral coefficients. – Ralph Aug 24 '11 at 11:04