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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asked Dec 2, 2009 at 0:28
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2 Answers
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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.)
answered Dec 2, 2009 at 0:40
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$\begingroup$ 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! $\endgroup$ Commented Dec 2, 2009 at 1:18
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Examples of compositional inversion of significant Laurent series are given in the MO-Qs "Combinatorics for the action of Virasoro / Kac–Schwarz operators: partition polynomials of free probability theory" and "Infinite dimensional involutions: infinitely large sets of multivariate polynomials self-inverse under self-substitution".
answered Sep 18 at 23:57
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$\begingroup$ "On Emergent Geometry of the Gromov-Witten Theory of Quintic Calabi-Yau Threefold" and "Quantum Deformation Theory of the Airy Curve and Mirror Symmetry of a Point" by Jian Zhou contain such Laurent series. $\endgroup$ Commented Sep 23 at 14:33