Computing a rational function at a point in terms of a uniformising parameter I am not quite sure how to ask this precisely, but vaguely I would like to know how difficult it is to write a function on an algebraic curve at a point $P$ as a power series of a uniformising parameter (a function with order 1 at $P$). For example, is it practically possible, and if so would it in general require using a computer.
I should clarify that I am trying to do this to find the residue at $P$ and hence I am trying to write this power series with coefficients only in the field that I am working over. It is not clear to me if this is possible, or if the coefficents have to be allowed to be functions which are invertible at $P$.
I imagine this may be contextual, so in particular, I am trying to compute this on hyperelliptic curves where the affine part is defined by $y^2 = f(x)$ for some polynomial $f(x)$ of degree $2g+1$.
I am struggling to even write $x$ as a power series of the uniformising parameter $\frac{y}{x^{g+1}}$ and the branch point above $\infty$.
 A: It is practically possible, and I would use a computer, although I've done it by hand.
I use the approach described by Bliss in "Algebraic Functions", which is to
postulate a series expansion, substitute it into your minimal polynomial,
expand, and set everything equal to zero.  The Newton polygon is a cleaver and useful
tool for figuring out what powers you need.  Once you've got $x$ and $y$ expanded
out in terms of some $t$, then you can figure out the expansion for any
rational function.  If that rational function is itself a uniformising
variable, then by inverting the series expansion, you get $t$ as a series expansion
in terms of your new uniformising variable, and can convert everything else.
Here's a specific example, using Maxima.  I'll let $y^2 = x^3-1$
and compute the expansion for $x$ in terms of $t = \frac{y}{x^2}$ at $\infty$.
First, I define a helper function to pull out the smallest term
in a series expansion, to keep the machine from burying me in algebra:

mincoeff(p, t) := coeff(p, t, lopow(p, t)) * t^lopow(p,t);

Next, I know that $x$ has a second order pole at infinity (Newton polygon),
so I postulate my series expansion:

x: sum(c[i]*t^i,i,-2,6)

$$c_{6}\,t^6+c_{5}\,t^5+c_{4}\,t^4+c_{3}\,t^3+c_{2}\,t^2+c_{1}\,t+{{c
 _{-1}}\over{t}}+{{c_{-2}}\over{t^2}}+c_{0}$$
I set $y=x^2 t$, expand my minimal polynomial, and look at the smallest term:

mincoeff(expand(x^4 * t^2 - x^3 +1), t)

$${{c_{-2}^4-c_{-2}^3}\over{t^6}}$$
I've got four choices, it seems!  (and three of them are zero)
Looking at 
$t$, we see that it is zero not only at infinity, but also at the
three places where $y=0$ and $x$ is a cube root of unity.  Those
are the three places where we'll get a zero coefficient for our $t^{-2}$
term, so we want the one where $c_{-2}$ is 1.

mincoeff(ev(expand(x^4 * t^2 - x^3 +1), c[-2]=1), t)

$${{c_{-1}}\over{t^5}}$$
Only one choice here, and it's $c_{-1}=0$.  Continuing on for several steps...

mincoeff(ev(expand(x^4 * t^2 - x^3 +1), c[-2]=1, c[-1]=0, c[0]=0, c[1]=0, c[2]=0, c[3]=0), t)
  $$c_{4}+1$$

Our next non-zero term is $c_{4}=-1$, and I'll stop here, with my expansion at:
$$x = t^{-2} - t^{4} + \cdots$$
