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In searching for various choices for the interpolation of exponential-towers to fractional heights (aka tetration) I came to the following type of function: $$ f_b(x) =\left[ \frac {t_0}{2} + \sum_{k=1}^{\infty} v_k \cdot (x-t_0)^k \right] + \left[ \frac{t_1}{2} + \sum_{k=1}^{\infty} w_k \cdot (x-t_1)^k \right]$$ which is an analytical formula for the doubly-infinite alternating series of the exponential-towers of consecutive heights (for $ ...+\log_b(\log_b(x))-\log_b(x) + x -b^x+b^{b^x}-...$ where $b$ is inside the range $]1 \ldots e^{\frac 1e}[$ and $t_0,t_1$ are the fixpoints, and the $x$ is in the range between the fixpoints - for what this information might be worth).

To find some smooth translation depending on another parameter I would like to have an inverse for that function, say I have some value for $f(x)$ and want to determine $x$; until now I apply Newton-iteration for this, but that is not much satisfying.

So perhaps there might be one clever idea how one could proceed to determine an inverse by some analytical formula.

[update]: After @Helge's comment I should perhaps add, that $f_b(x)$ seems to have no real fixpoint in the admissible range for $x$ and so the path, which is otherways the most straightforward one in more general problems, seems to be not viable here [/update]

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  • $\begingroup$ Here's a general strategy. 1) Write as a power series around a single point $t$. 2) Invert this power series. There are "nice" expressions for 2). For 1) it depends on how you choose $t$. $\endgroup$
    – Helge
    Jun 11, 2013 at 17:13
  • $\begingroup$ @Helge: this seems to be impossible. To have a correct expression for the two-way-infinite series of iterated exponentials (towards the lower fixpoint $t_0$) and iterated logarithms (towards the upper fixpoint $t_1$) it was just essential to have the two power series developed around the respective fixpoint. Only then the power series can be made convergent (in case this is required by additional integer-iterations of $b^x$ resp. $\log_b(x)$ towards the resp. fixpoint) $\endgroup$ Jun 11, 2013 at 18:03

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