# Proof that a finite series expansion of $f(X)$ at $\alpha$ exists iff $Q(X)$ is a power of $(X-\alpha)$, in $f(X)=\frac{P(X)}{Q(X)}$

I'm working through Gouvea's P-adic numbers book, and early on they give the problem

Write $$f(X)=\frac{P(X)}{Q(X)}$$ in lowest terms, so that $$P(X)$$ and $$Q(X)$$ have no common zeros. Show that the expansion of $$f(X)$$ in powers of $$(X-\alpha)$$ is finite if and only if $$Q(X)=(X-\alpha)^m$$ for some $$m\geq 0$$

The solution/hint given is

"if the expansion is finite, it will certainly become a polynomial after we multiply by $$(X-\alpha)^m$$, where $$m$$ is the biggest exponnent appearing in a denominator.

From what I read so far, we should get a finite series in the event where we have a "pole", for example, the series expansion of $$\frac{X}{X-1}$$ at the point $$\alpha=1$$, gives us the series $$(X-1)^{-1}+1$$. But i'm not sure how to prove that we are assure a finite series expansion where the denomiator polynomial can be expressed as a power of $$(X-\alpha)$$ and I'm also unsure of going in the other direction.

Thanks

Assume that the expansion is finite, i.e,

$$f(x) = \sum_{n=-k}^m a_i (x-\alpha)^i$$

where we assume $$k>0$$, and we let $$a_{-k}=0$$ if needed (I assume it only for convenience. Also note that if the sequence actually start with $$n\geq 0$$, $$f$$ is a polynomial and there's nothing to prove). Note that after multiplying by $$(x-\alpha)^k$$ we get:

$$(x-\alpha)^kf(x)=\sum_{n=-k}^m a_i (x-\alpha)^{i+k}=\sum_{n=0}^{m+k}a_i (x-\alpha)^{i+k}$$

RHS is a sum of polynomials, hence a polynomial $$g(x)$$, so, we may write $$(x-\alpha)^k f(x)=g(x)$$ which gives $$f(x)=g(x)/(x-\alpha)^k$$. After canceling factors, this is still the form of the rational function $$f$$ (maybe $$k$$ will be changed).

The other way around, if $$f(x)=g(x)/(x-\alpha)^k$$ for some $$k\geq0$$, where $$g$$ is a polynomial, note that $$g$$ can be written as $$g(x)=g((x-\alpha)+\alpha)$$. Now if we write $$g$$ explicitly as $$g(x)=\sum_{i=0}^m b_i x^i$$, we get-

$$g((x-\alpha)+\alpha)=g(x)=\sum_{i=0}^m b_i ((x-\alpha)+\alpha)^i$$

using the binomial theorem we get that $$g(x)=\sum_{i=0}^m b_i' (x-\alpha)^i$$ ($$b_i'$$ are the modified coefficients; we can calculate them explicitly, but this is not interesting). This gives $$f(x)=\frac{g(x)}{(x-\alpha)^k}=\sum_{i=0}^m b_i' (x-\alpha)^{i-k}$$

And so the expansion is finite.