Coefficient of Taylor expansion of rational function?

Question

We know that given a rational number of the from $p/q$ if we expand it in the decimal base that is an expansion in power of 10 and we know that coefficient is repeating after finite step. In other words, we can say that we need only finitely many coefficients to describe the rest. Let $F(z)$ be a rational function that is $F(z)=\frac{p(z)}{q(z)}$. I take the Taylor series expansion of $F(z)$ at the point away from the pole (maybe in a local coordinate around the point). Say $$F(z)\mid_z=a \sum t_k (z-a)^k $$ My question is there any finite condition on the coefficient of $t_k$ ? That is we need only finitely many $t_k$'s to determine the rest? If so what will be the proof?

Answer 1

For simplicity of notation, we can consider $a = 0$. Then we have $$\frac {P(z)}{Q(z)} = \sum_{n=0}^\infty c_nz^n$$ or $$P(z) = \sum_{n=0}^\infty c_nz^nQ(z)$$ if $Q(z) = \sum_{n=0}^k b_nz^n$, the coefficient of $z^n$ on the RH side for sufficiently high $n$ will be given by $$\sum_{j=0}^k b_jc_{n-j}$$

When $n$ is greater than the degree of $P$, this will be $0$. So we can solve (assuming $b_0 \ne 0$) to get $$c_n = -\sum_{j-1}^k \frac{b_j}{b_0}c_{n-j}$$

So, yes, rational functions have a recursion formula for their Taylor series coefficients, which holds for $n$ greater than the degrees of both $P$ and $Q$.