As I mentioned, for integer values, we can solve exactly. The inverse function \( \mu_e^{\small -1}(x) \) can be solved for integer values, as shown below:
\( \begin{eqnarray}
\mu_e^{\small -1}(-3) & \approx & 1.6882141580329708209627762045
\\
\mu_e^{\small -1}(-2) & \approx & 1.8030232398553705179704668746
\\
\mu_e^{\small -1}(-1) & \approx & 2.0404716999485556015209199031
\\
\mu_e^{\small -1}(0) & \approx &
2.7182818284590452353602874713 = e
\\
\mu_e^{\small -1}(1) & \approx & 7.514088623758291009709030684
\\
\mu_e^{\small -1}(2) & \approx & 3814263.950501952154234481426
\end{eqnarray}
\)
For base 2, we have:
\( \begin{eqnarray}
\mu_2^{\small -1}(-2) & \approx & 1.678120055270763107817891
\\
\mu_2^{\small -1}(-1) & \approx & 1.78467418460380830116416988
\\
\mu_2^{\small -1}(0) & = & 2 \\
\mu_2^{\small -1}(1) & \approx & 2.5861097400293629228959
\\
\mu_2^{\small -1}(2) & \approx & 6.121264365801078940145846
\end{eqnarray}
\)
For base 10:
\( \begin{eqnarray}
\mu_{10}^{\small -1}(-2) & \approx & 2.09007667990965647973914
\\
mu_{10}^{\small -1}(-1) & \approx & 2.894243914558514321612672 \\
\mu_{10}^{\small -1}(0) & = & 10 \\
\mu_{10}^{\small -1}(1) & \approx & 9999999990.00000000043429448210153789
\end{eqnarray}
\)
Much like the tetration function itself, it would seem that this function, whatever it is, is iteratively definable for integer steps, but who knows how to interpolate in between? With an exact solution for a particular base, we may get an answer.
One careful observation as x decreases to negative infinity:
\( \lim_{x \to -\infty} \mu_e^{\small -1}(x) = \eta \)
Also note that it would appear to grow as quickly as tetration itself as we move into positive territory. At a glance, we can tell that for \( \mu_e^{\small -1}(x) \), for values of x > 2, it should be almost exactly \( {}^{(x+1)} e \), if you only look at the order of magnitude. More succintly, we can state:
\( ln \left(\mu_e^{\small -1}(x)\right) \approx {}^{x} e \)
I'm curious to see this function graphed, once we've put together the tools to solve the various bases. We're almost there. This graph would appear to grow hyperexponentially to the right, but asymptotically approaches eta to the left. In some ways, I bet this graph looks more interesting than the tetration graph, because the curvature should always be positive.
For tetration, remember, there's an inflection point. To the right of the inflection point, the graph becomes more and more hyperexponential, i.e., repeated tetrations of the (for intuive purposes) approximately linear critical interval. While to the left of the inflection point, the graph becomes more and more hyperlogarithmic (repeated logarithms of the critical interval).
\( \begin{eqnarray}
\mu_e^{\small -1}(-3) & \approx & 1.6882141580329708209627762045
\\
\mu_e^{\small -1}(-2) & \approx & 1.8030232398553705179704668746
\\
\mu_e^{\small -1}(-1) & \approx & 2.0404716999485556015209199031
\\
\mu_e^{\small -1}(0) & \approx &
2.7182818284590452353602874713 = e
\\
\mu_e^{\small -1}(1) & \approx & 7.514088623758291009709030684
\\
\mu_e^{\small -1}(2) & \approx & 3814263.950501952154234481426
\end{eqnarray}
\)
For base 2, we have:
\( \begin{eqnarray}
\mu_2^{\small -1}(-2) & \approx & 1.678120055270763107817891
\\
\mu_2^{\small -1}(-1) & \approx & 1.78467418460380830116416988
\\
\mu_2^{\small -1}(0) & = & 2 \\
\mu_2^{\small -1}(1) & \approx & 2.5861097400293629228959
\\
\mu_2^{\small -1}(2) & \approx & 6.121264365801078940145846
\end{eqnarray}
\)
For base 10:
\( \begin{eqnarray}
\mu_{10}^{\small -1}(-2) & \approx & 2.09007667990965647973914
\\
mu_{10}^{\small -1}(-1) & \approx & 2.894243914558514321612672 \\
\mu_{10}^{\small -1}(0) & = & 10 \\
\mu_{10}^{\small -1}(1) & \approx & 9999999990.00000000043429448210153789
\end{eqnarray}
\)
Much like the tetration function itself, it would seem that this function, whatever it is, is iteratively definable for integer steps, but who knows how to interpolate in between? With an exact solution for a particular base, we may get an answer.
One careful observation as x decreases to negative infinity:
\( \lim_{x \to -\infty} \mu_e^{\small -1}(x) = \eta \)
Also note that it would appear to grow as quickly as tetration itself as we move into positive territory. At a glance, we can tell that for \( \mu_e^{\small -1}(x) \), for values of x > 2, it should be almost exactly \( {}^{(x+1)} e \), if you only look at the order of magnitude. More succintly, we can state:
\( ln \left(\mu_e^{\small -1}(x)\right) \approx {}^{x} e \)
I'm curious to see this function graphed, once we've put together the tools to solve the various bases. We're almost there. This graph would appear to grow hyperexponentially to the right, but asymptotically approaches eta to the left. In some ways, I bet this graph looks more interesting than the tetration graph, because the curvature should always be positive.
For tetration, remember, there's an inflection point. To the right of the inflection point, the graph becomes more and more hyperexponential, i.e., repeated tetrations of the (for intuive purposes) approximately linear critical interval. While to the left of the inflection point, the graph becomes more and more hyperlogarithmic (repeated logarithms of the critical interval).
