# 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.

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

• 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