# Inverting an asymptotic series

I have the first few terms of a series of the form,

$y(x)=\ln(x)+x+a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\cdots$.

Knowing that the inverse $x(y)$ exists, I am looking for method to write x in terms of y (at least the first few terms of the expansion). Does anybody know how I could achieve this?

Thanks to a mathematician much greater than I, I know that this is certainly possible in the case, the $x$ term is not present in the expansion of $y$ (i.e. $y(x)=\ln(x)+a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+O(\frac{1}{x^3})$). It turns out in this case $x(y)$ can be written as a series expansion in powers of $e^{-y}$. But I can't seem to be able adapt that method to handle this new case.

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Perhaps you could give a reference and more detail about the case that you can already do. Also, given your style, it seems you would have a better time on math.stackexchange.com/questions?sort=newest –  Will Jagy Nov 22 '11 at 6:35

Since you say that you only want the first few terms, one way you can do this type of thing is by making a contraction mapping. As $x\to\infty$, inspection shows $y\sim x$, so rewrite the equation as an assignment: $$x := y - \ln(x) - a_0-\frac{a_1}{x} - \frac{a_2}{x^2}-\cdots$$ The idea is that the right side is a more slowly varying function of $x$ than the left side.

Now start with the approximation $x=y$, and apply the assignment repeatedly, each time simplifying and pruning terms smaller than you need. After a finite number of steps it will converge to the precision you have been pruning to. Maple or Mathematica can handle it.

The first iteration makes $$y - \ln(y) - a_0 - \frac{a_1}{y} - \frac{a_2}{y^2}-\cdots$$

The second iteration makes $$y -\ln(y - \ln(y) - a_0 - \cdots) - a_0 - a_1/(y - \ln(y) - a_0 - \cdots)+\cdots$$ and you need to expose the smaller terms using $$\ln(y - \delta) = \ln(y) - \frac{\delta}{y} - \frac{\delta^2}{2y^2} - \cdots$$ and $$\frac{1 }{y-\delta } = \frac{1}{y}+\frac{\delta}{y^2}+ \frac{\delta^2}{y^3} + \cdots$$ and so on. You will get a series of terms with powers of $y$ in the denominators and powers of $\ln(y)$ in the numerators.

Another way is to solve your equation using Newton-Raphson iteration.

If you want a more formal method, with a chance at an expression for the general term, look at Lagrange inversion.

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