I have an asymptotic series expansion that I would like to invert of the form

$ t_2 z + t_3 z^2 + t_4 z^3 + \frac{A_1}{z} + \frac{A_2}{z^2} + ... + \frac{A_9}{z^9} + \mathcal{O}(z^{-10}) = w$

i.e. I would like to be able to write $z$ solutions in the form

$ z_{1,2,3} = (\frac{w}{t_4})^{1/3} \big(...\big)$

where the terms in the brackets are a power series with coefficients that are functions of the coefficients of my original equation.

I have tried using InverseFunction and InverseSeries but to no avail. Is this inversion possible in mathematica?


1 Answer 1


You should always provide a code block that can be copied into Mathematica! At any rate, here is your series:

expr = Sum[Subscript[t,i]z^(i-1),{i,2,4}] + Sum[Subscript[A,i]z^-i,{i,9}] //TeXForm

$\frac{A_9}{z^9}+\frac{A_8}{z^8}+\frac{A_7}{z^7}+\frac{A_6}{z ^6}+\frac{A_5}{z^5}+\frac{A_4}{z^4}+\frac{A_3}{z^3}+\frac {A_2}{z^2}+\frac{A_1}{z}+t_4 z^3+t_3 z^2+t_2 z$

The inversion of this series is quite large with symbolic coefficients, so I won't invert the order 9 series. You can invert it as follows:

InverseSeries @ Series[expr, {z, Infinity, 1}]//TeXForm

$\frac{\sqrt[3]{z}}{\sqrt[3]{t_4}}-\frac{t_3}{3 t_4}+\frac{\left(\frac{t_3^2}{9 t_4^{4/3}}-\frac{t_2}{3 \sqrt[3]{t_4}}\right) \sqrt[3]{\frac{1}{z}}}{\sqrt[3]{t_4}}+\frac{\left(\frac{t _2 t_3}{9 t_4}-\frac{2 t_3^3}{81 t_4^2}\right) \left(\frac{1}{z}\right)^{2/3}}{\sqrt[3]{t_4}}-\frac{A_1} {3 z}+O\left(\left(\frac{1}{z}\right)^{4/3}\right)$


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