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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.

Thanks for reading.

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  • $\begingroup$ 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 $\endgroup$
    – Will Jagy
    Commented Nov 22, 2011 at 6:35

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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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There is a general technique for doing this, found in expositions dealing with transseries. One example is my own...

Transseries for beginners, Real Analysis Exchange 35 (2010) 253--310

see Problem 4.2.

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  • $\begingroup$ Great article. I use these objects all the time but never saw this formal treatment before. It's good to know that all the manipulations I assumed were valid are in fact valid. $\endgroup$
    – Brendan McKay
    Commented Nov 23, 2011 at 1:08
  • $\begingroup$ I was not aware of the existence of transeries. I increasingly find myself staring at these types of expansions. So thank you for introducing these to me, the article looks very interesting (I'll need some more time to digest it thoroughly). $\endgroup$
    – AUK1939
    Commented Nov 24, 2011 at 5:03

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