I am looking for inverse functions for the following family of functions:

$ \begin{aligned} f_0(z) &= z+e^z \\ f_1(z) &= ze^z \\ f_2(z) &= z^z \\ &\cdots \\ f_{n+1}(z) &= e^{\,f_n(\log(z))} \\ \end{aligned} $

Of course, we have $f_1^{-1}(z) = W(z)$ with $W$ being the Lambert W function.

We also know that:

$ \begin{aligned} f_0^{-1}(z) &= z - W(e^z) \\ f_2^{-1}(z) &= e^{\displaystyle W\big(\log(z)\big)} \\ \end{aligned} $

We can get a better sense for what is going on by defining a predecessor to $W$:

$V(z) = z - W(e^z)$

This gives us:

$ \begin{aligned} f_0^{-1}(z) &= V(z) \\ f_1^{-1}(z) &= \log(z) - V\big(\log(z)\big) \\ f_2^{-1}(z) &= e^{\displaystyle \log\big(\log(z)\big) - V\Big(\log\big(\log(z)\big)\Big)} \\ \end{aligned} $

If we write $E(z) = e^z$ and $L(z) = \log(z)$, I suspect that we have:

$\,f_{n+1}^{-1} = E^{n}\Big(L^{n+1}(z) - V\big(L^{n+1}(z)\big)\Big)$

But I have a hard time establishing it. Or my intuition is wrong and the generic inverse is something else altogether. I would be perfectly happy with a direct expression or a recurrence rule, but having both would be awesome. Also, for the time being, I am not interested in details about domains, codomains, and branches. These can be figured out later on.

Note: I suspect that the solution is trivial, but I keep running around in circles...

  • 1
    $\begingroup$ You have $f_{n+1}=E\circ f_{n} \circ L$, so the inverse would satisfy $f_{n+1}^{-1}=E\circ f_n^{-1} \circ L$. Then you can use induction from $f_0^{-1}= V$ to obtain $f_n^{-1}=E^n\circ V \circ L^n$. $\endgroup$ – nathan.j.mcdougall Jan 29 at 22:18
  • $\begingroup$ @nathan.j.mcdougall Of course! Thank you. $\endgroup$ – Ismael Ghalimi Jan 29 at 22:21

As per the comments, since $f_{n+1}=E\circ f_n\circ L$, the inverse must satisfy $$f_{n+1}^{-1}=L^{-1}\circ f_{n}^{-1}\circ E^{-1}=E\circ f_{n}^{-1}\circ L.$$ Induction gives $f_{n}^{-1}=E^n\circ f_0^{-1}\circ L^n$.

  • $\begingroup$ Thanks a lot for the clear answer. I gave you first-name-only credit on this notebook. I hope you don’t mind. If you do, I am more than happy to remove it. $\endgroup$ – Ismael Ghalimi Jan 29 at 22:38
  • $\begingroup$ There's no need, but thank you. $\endgroup$ – nathan.j.mcdougall Jan 29 at 22:39

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