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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

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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.

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