Saturday, 24 November 2012

cv.complex variables - Inversion of Laurent series

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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