There are lots of questions on this site asking how to find a Laurent expansion for functions involving rational functions, exponential functions and some common friends. I have encountered a situation where I need to understand the residues of a function which is defined implicitly by a functional equation, and I have never seen anything like this tackled in any literature. Maybe because it is too hard, maybe because it is too general, maybe it has not been considered. But my first question is if anyone knows any references for this general kind of problem.
Now I'll elaborate on my specific situation. I have a function $G= G(z)$ defined locally in the complex plane by the equation $G^3 + 1/G^3 = P$, where $P = P(z)$ is a polynomial of degree $3$. Around a given non-branch point of this equation, there is a meromorphic solution to this equation given by analytic continuation. Essentially, $G$ is defined implicitly by a degree $6$ polynomial, and for any nearby point, one can find the $6$ solutions to this equation, and pick the one that makes the function continuous. It is then easy to show that the resulting function is meromorphic. Or if you like, you can think of $G$ as living on a Riemann surface.
In any case, I want to calculate the series expansion of $G$ near the branch points of the equation.
If this is possible, I'd like to go further. I am studying $G$ in a $2-$parameter family, by upgrading $P$ from a single cubic polynomial to $P(z;a,b)$. I want to show that a certain integral involving a messy combination of $G$ and $P$ is non-zero by cranking out a series expansion of the integrand at each point and getting estimates for the residues as a function of $a,b$. My guess is that if I understood how to get a Laurent expansion from an implicit equation, it would only be more bookkeeping to do this for my two parameter family.
With that context and motivation out of the way, my specific $P(z;a,b) = i/ \sqrt 2 \big( 10 + az + bz^2 + 27 z^3\big)$. Branch points can be located by a trick: by squaring both sides and subtracting $4$, we obtain $G^6 - 2 - 1/G^6 = P^2-4$, but the left side of this is nothing more than $(G^3 - 1/G^3)^2$. By defining $y:= G^3 - 1/G^3$ (and accepting on faith that this is kosher), I can write $y^2 = P^2 - 4$, and we see that the branch points are the zeroes of $P^2-4$.
Thanks for any pointers.
The method outlined below works when you are centered at the branch point(s). Some preliminary steps are needed to restate the problem, using various minor algebraic substitutions.
Let $w= g+g^{-1}$ so that $g^3+ g^{-3} =w^3 - 3w$. Then you want to solve $w^3- 3w= p(z)$ near a branch point where the derivative $3w^2-3=0;$ i,e. the two branch point cases are those critical points of $w$ and corresponding critical values for $p$ listed below:
(i) $(w,p)= (1,-2)$ or
(ii) $(w,p)=(-1,2)$.
In case (i) write $w= 1 +u$ and $p= -2+q$ where $u$ and $q(z)$ are small. By algebra obtain $ u^2 + \frac{1}{3}u^3 =\frac{1}{3}q(z)$ and expand $\frac{1}{3}q(z)= a_1 z+ a_2 z^2 + a_3 z^3+\ldots$ to see that we expect $u$ to be a power series in $\sqrt{z}$; that is, a Puiseux series series.
(An entertaining exposition of the history and applications of such expansions is in V. I. Arnold's little book Huygens and Barrow, Newton and Hooke: Pioneers in Mathematical Analysis and Catastrophe Theory from Evolvements to Quasicrystals)
This motivates setting $t^2=z$ and solving $ u^2 (1+ \frac{1}{3}u) =a_1t^2+ a_2 t^4 + \ldots$ as an ordinary power series in $t$.
Substitute $ u= \sqrt{a_1} t v$ into preceding and cancel the power $t^2$ across both sides to obtain $ v^2( 1+ c tv) = 1 + b_2 t^2 +\ldots$
So $v(t)= \sqrt{\frac{ 1+ b_2 t^2+ \ldots}{ 1+ c t v(t)}}$. Here the numerator is an explicit known power series in $t$, and we seek a power series for $v(t)$ whose initial value at $t=0$ is $v_0=1$. You can interpret this as a fixed-point equation for $v(t)$. My hunch is that the recursive algorithm
$v_{k+1}(t) = \sqrt{ \frac{1+ b_2 t^2+ \ldots}{ 1+ c t v_k(t)}}$ whose initial term $v_0=1$ should generate the explicit expansion you desire (using e.g Mathematica). This ought to work; there may be numerically more efficient ways however.
Case (ii) is essentially similar of course.
After having solved for $w$, one then needs to solve for $g$, but that is "merely" a matter of solving a quadratic equation.