Suppose that we have the following general Laurent series with a pole of order 1 at $x=0$: $$y(x)=\frac{1}{x}+\sum_{n=0}^{\infty}a_{n}x^{n}$$

How can I then compute the Laurent series expansion of a new function $u(t)$ given by $$u(t)=y(x),\quad t=x^3$$ in terms of $t$ (around $t=0$)?


If the substitution were in the other direction (ie $x=t^3$) I'd say to just substitute and use the uniqueness of the Laurent series.

But now it's different. You still can substitute, but don't get a Laurent series out of it. What you get is something like:

$$u(t) = {1\over x} + \sum a_n t^{n/3} = {1\over t} + \sum a_{3n}t^n + t^{1/3}\sum a_{3n+1}t^n + t^{2/3}\sum a_{3n+2}t^{n}$$

here you've got a problem since $t^{1/3}$ and $t^{2/3}$ are not uniquely determined, but the result of a Laurent series is. If we call the last sums $f$ and $g$ (which are analytic around zero) we can see that they have to have special relation in order for $u$ to be unique. For example if we go one turn around the origin we would have:

$$\xi(t) f(t) + \xi^2(t)g(t) = \xi(t) e^{2\pi i/3}f(t) + \xi^2(t)e^{-2\pi i/3} g(t)$$

where $\xi(t)$ is the principal cubic root. This gives the following equation

$$\xi(t)(1 - e^{2\pi i/3}) f(t) = -\xi^2(t)(1-e^{-2\pi i/3})g(t)$$

$$f(t) = -\xi(t){1-e^{2\pi i/3}\over1-e^{-2\pi i/3}}g(t)$$

Now we see that due to analytic continuations we are required to have:

$$f(t) = -t^{1/3}{1-e^{2\pi i/3}\over1-e^{-2\pi i/3}}g(t)$$

which is not unique unless $f=g=0$

Bottom line is that iff you can do the substitution in the Laurent series and still get a Laurent series there's a Laurentseries after the substitution.

  • $\begingroup$ Dear @skyking, Thank you for the detailed explanation. $\endgroup$ – Jack Oct 18 '15 at 9:17

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