# Inversion of Laurent series

For a power series $f(z) = \sum_{i=0}^{\infty} a_i z^i$ with $a_1$ nonzero, Lagrange's inversion formula gives an explicit way to compute the Taylor coefficients of the inverse function.

Is there any analogous formula for Laurent series?

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## 1 Answer

The Lagrange inversion formula is meant to give you the Taylor series expansion of $f^{-1}$ at the point $f(0)$. If $f$ has a Laurent series instead, then it means that $f(0) = \infty$ and that $f$ is meromorphic. The Taylor series at $\infty$ of $f^{-1}$ then doesn't particularly mean anything unless you change to a different coordinate patch on the Riemann sphere, for instance $\zeta = 1/z$. So you can first switch to the function $1/f$, which has a usual Taylor series, and then use the standard Lagrange inversion formula for $(1/f)^{-1}$.

(If I have understood the question correctly. Maybe this answer is too straightforward to address the real question.)

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Actually, that's just what I needed. Thanks. I'm embarrassed that I wasn't able to figure out such a straightforward thing on my own! –  Kevin H. Lin Dec 2 '09 at 1:18
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