See my exact solution for base e^(1/e) for background. Assuming the solution I provided is accepted as unique (i.e., correct), it should provide strength to the claim that the following solution is correct as well.

Let's start with the basics. Here, for brevity, assume that eta is e^(1/e). (Eta looks like an "n", but it's effectively a greek letter "e", and given its relationship to the constant e, I thought it would make a good symbol.)

\( \begin{eqnarray}

\eta^y

& = & (e^{\frac {\tiny 1}{e}})^y \\

\\[5pt]

\\

& = & e^{{\frac {\tiny 1}{e}}y}

\end{eqnarray} \)

Okay, so far, so good. Now, try this one on for size:

\( \begin{eqnarray}

\eta^{(e^y)}

& = & (e^{\frac {\tiny 1}{e}})^{(e^y)} \\

\\[5pt]

\\

& = & e^{\left({\frac {\tiny 1}{e}}\ \times\ {e^y}\right)} \\

\\[5pt]

\\

& = & e^{\left(e^{(y-1)}\right)}

\end{eqnarray} \)

Interesting... But is this useful? Well, let's go one more:

\( \begin{eqnarray}

{\Large \eta^{\left(e^{(e^y)}\right)}}

& = & \left(\large e^{\frac {\tiny 1}{e}}\right)^{\left(\large e^{(e^y)}\right)} \\

\\[10pt]

\\

& = & e^{\left({\frac {\tiny 1}{e}}\ \times\ {\large e^{\left(e^y\right)}}\right)} \\

\\[10pt]

\\

& = & e^{\large e^{\left( (e^y)-{\normalsize 1}\right)}} \\

\\[10pt]

\\

& = & e^{\large e^{\left( (e^y)\left(1-{\normalsize \delta}\right) \right)}},\text{ where } \delta\ =\ \frac{1}{e^y}

\end{eqnarray} \)

Hopefully it's apparent where I'm going with this. Let y be the mth tetration of e, with m=5 sufficient to exceed machine precision in all practical circumstances (when evaluating delta at m-1=4).

\( \begin{eqnarray}

{\Large \eta^{(^m e)}}

& = & {\Large \eta^{\Large \left(e^{(e^{(^{(m-2)} e)})}\right)}} \\

\\[10pt]

\\

& = & \left(\large e^{\frac {\tiny 1}{e}}\right)^{\left(\Large e^{(e^{(^{(m-2)} e)})}\right)} \\

\\[10pt]

\\

& = & e^{\left({\frac {\tiny 1}{e}}\ \times\ {\Large e^{\left(e^{(^{(m-2)} e)}\right)}}\right)} \\

\\[10pt]

\\

& = & {\Large e^{e^{\large \left( (e^{(^{(m-2)} e)})-{\normalsize 1}\right)}}} \\

\\[10pt]

\\

& = & {\Large e^{\large e^{\large \left({(^{(m-1)} e)\left(1-{\normalsize \delta}\right) }\right)}}},\text{ where } \delta\ =\ \frac{1}{(^{(m-1)} e)}

\end{eqnarray} \)

Fascinating. But still, is this useful? Well, for this, we need a new function, based on tetration of base eta, but which equals e at negative infinity, and which equals something greater than e at y = 0. Think of it as taking eta^^y just below its asymptote, all the way to infinity and beyond, wrapping around at negative infinity, but just above the asymptote instead of below. I call this new function eta_b-check, where b is a particular base we're interesting in. Here's how it looks, omitting the b for the assumed base e.

\( \begin{eqnarray}

{\Large ^{\normalsize(y+1)} \check \eta}

& = & {\Large \eta^{\left(^{y} \check \eta\right)}} \\

\\[10pt]

\\

\text{ or, alternatively:} \\

\\[10pt]

\\

{\Large log_{\eta}\left(^{\normalsize y} \check \eta \right)}

& = & {\Large ^{\normalsize (y-1)} \check \eta} \\

\\[10pt]

\\

{\Large ^{\normalsize (-\infty)} \check \eta} & = & {\Large e} \\

\\[10pt]

\\

{\Large ^{\normalsize y} \check \eta} & > & {\Large e},\text{ for all } y\ >\ -\infty

\end{eqnarray} \)

(The proof that there is a unique solution for eta-check is very similar to the one I gave for the tetration of eta. I can provide it in a separate post if required.)

As it turns out, we can even find the exact value of eta_b-check(0) with arbitrary precision, relative to any particular base. And I know that I'm not the first to find this function. At the very least, Peter Walker described its inverse, or something very similar, if I'm reading his paper correctly. So this function is not new. However, I haven't seen the following proof (Peter had something very close, so again, I can't take all the credit, if I can even take any at all.)

It turns out, there exists an exponent mu for eta-check such that:

\( {\Large ^{\mu} \check \eta\ =\ ^y e},\text{ for sufficiently large } y \)

I say sufficiently large, but really we're talking about a limit as we go to infinity. But beyond a certain point, mu increases pretty much linearly with y, with extreme precision (this has a very significant interpretation, which I'll get to if someone doesn't beat me to the punch). For y=4 or 5 or so, computer precision is not sufficient to tell the limited value of mu from the "approximate" value.

Anyway, here comes the fun part:

\( \begin{eqnarray}

{\Large ln^{\small(2)}\left(^{\normalsize(\mu)} \check \eta\right)}

& = & {\Large ln^{\small(2)}\left(^{\normalsize y} e\right)} \\

\\[10pt]

\\

& = & {\Large ^{\normalsize (y-2)} e} \end{eqnarray} \)

and

\( \begin{eqnarray}

{\Large ln^{\small(2)}\left(^{\normalsize(\mu+1)} \check \eta \right)}

& = & {\Large ln^{\small(2)}\left(\eta^{\left(^{\mu} \check \eta \right)} \right) } \\

\\[10pt]

\\

& = & {\Large ln^{\small(2)}\left(\eta^{\left(^{y} e\right)} \right) } \\

\\[10pt]

\\

& = & {\Large ln^{\small(2)}\left(e^{\large e^{\large \left({(^{(y-1)} e)\left(1-{\normalsize \delta}\right) }\right)}} \right) },\text{ where } \delta\ =\ \frac{1}{(^{(y-1)} e)} \\

\\[10pt]

\\

& = & {\Large {(^{\normalsize (y-1)} e)\left(1-{\normalsize \delta}\right)}},\text{ where } \delta\ =\ \frac{1}{(^{(y-1)} e)}

\end{eqnarray} \)

For sufficiently large y, this function is effectively exact (it is exact when you take the limit to infinity). Furthermore, notice that I swapped out the integer m for the real exponent y. Finally, notice that if eta-check is unique as I claim it is, then the tetration of base e that I just defined has a very strong claim on uniqueness. In other words, if some other method does not agree, I claim that this function is "correct", and the other function is displacing its exponents by some cyclic function.

Let's start with the basics. Here, for brevity, assume that eta is e^(1/e). (Eta looks like an "n", but it's effectively a greek letter "e", and given its relationship to the constant e, I thought it would make a good symbol.)

\( \begin{eqnarray}

\eta^y

& = & (e^{\frac {\tiny 1}{e}})^y \\

\\[5pt]

\\

& = & e^{{\frac {\tiny 1}{e}}y}

\end{eqnarray} \)

Okay, so far, so good. Now, try this one on for size:

\( \begin{eqnarray}

\eta^{(e^y)}

& = & (e^{\frac {\tiny 1}{e}})^{(e^y)} \\

\\[5pt]

\\

& = & e^{\left({\frac {\tiny 1}{e}}\ \times\ {e^y}\right)} \\

\\[5pt]

\\

& = & e^{\left(e^{(y-1)}\right)}

\end{eqnarray} \)

Interesting... But is this useful? Well, let's go one more:

\( \begin{eqnarray}

{\Large \eta^{\left(e^{(e^y)}\right)}}

& = & \left(\large e^{\frac {\tiny 1}{e}}\right)^{\left(\large e^{(e^y)}\right)} \\

\\[10pt]

\\

& = & e^{\left({\frac {\tiny 1}{e}}\ \times\ {\large e^{\left(e^y\right)}}\right)} \\

\\[10pt]

\\

& = & e^{\large e^{\left( (e^y)-{\normalsize 1}\right)}} \\

\\[10pt]

\\

& = & e^{\large e^{\left( (e^y)\left(1-{\normalsize \delta}\right) \right)}},\text{ where } \delta\ =\ \frac{1}{e^y}

\end{eqnarray} \)

Hopefully it's apparent where I'm going with this. Let y be the mth tetration of e, with m=5 sufficient to exceed machine precision in all practical circumstances (when evaluating delta at m-1=4).

\( \begin{eqnarray}

{\Large \eta^{(^m e)}}

& = & {\Large \eta^{\Large \left(e^{(e^{(^{(m-2)} e)})}\right)}} \\

\\[10pt]

\\

& = & \left(\large e^{\frac {\tiny 1}{e}}\right)^{\left(\Large e^{(e^{(^{(m-2)} e)})}\right)} \\

\\[10pt]

\\

& = & e^{\left({\frac {\tiny 1}{e}}\ \times\ {\Large e^{\left(e^{(^{(m-2)} e)}\right)}}\right)} \\

\\[10pt]

\\

& = & {\Large e^{e^{\large \left( (e^{(^{(m-2)} e)})-{\normalsize 1}\right)}}} \\

\\[10pt]

\\

& = & {\Large e^{\large e^{\large \left({(^{(m-1)} e)\left(1-{\normalsize \delta}\right) }\right)}}},\text{ where } \delta\ =\ \frac{1}{(^{(m-1)} e)}

\end{eqnarray} \)

Fascinating. But still, is this useful? Well, for this, we need a new function, based on tetration of base eta, but which equals e at negative infinity, and which equals something greater than e at y = 0. Think of it as taking eta^^y just below its asymptote, all the way to infinity and beyond, wrapping around at negative infinity, but just above the asymptote instead of below. I call this new function eta_b-check, where b is a particular base we're interesting in. Here's how it looks, omitting the b for the assumed base e.

\( \begin{eqnarray}

{\Large ^{\normalsize(y+1)} \check \eta}

& = & {\Large \eta^{\left(^{y} \check \eta\right)}} \\

\\[10pt]

\\

\text{ or, alternatively:} \\

\\[10pt]

\\

{\Large log_{\eta}\left(^{\normalsize y} \check \eta \right)}

& = & {\Large ^{\normalsize (y-1)} \check \eta} \\

\\[10pt]

\\

{\Large ^{\normalsize (-\infty)} \check \eta} & = & {\Large e} \\

\\[10pt]

\\

{\Large ^{\normalsize y} \check \eta} & > & {\Large e},\text{ for all } y\ >\ -\infty

\end{eqnarray} \)

(The proof that there is a unique solution for eta-check is very similar to the one I gave for the tetration of eta. I can provide it in a separate post if required.)

As it turns out, we can even find the exact value of eta_b-check(0) with arbitrary precision, relative to any particular base. And I know that I'm not the first to find this function. At the very least, Peter Walker described its inverse, or something very similar, if I'm reading his paper correctly. So this function is not new. However, I haven't seen the following proof (Peter had something very close, so again, I can't take all the credit, if I can even take any at all.)

It turns out, there exists an exponent mu for eta-check such that:

\( {\Large ^{\mu} \check \eta\ =\ ^y e},\text{ for sufficiently large } y \)

I say sufficiently large, but really we're talking about a limit as we go to infinity. But beyond a certain point, mu increases pretty much linearly with y, with extreme precision (this has a very significant interpretation, which I'll get to if someone doesn't beat me to the punch). For y=4 or 5 or so, computer precision is not sufficient to tell the limited value of mu from the "approximate" value.

Anyway, here comes the fun part:

\( \begin{eqnarray}

{\Large ln^{\small(2)}\left(^{\normalsize(\mu)} \check \eta\right)}

& = & {\Large ln^{\small(2)}\left(^{\normalsize y} e\right)} \\

\\[10pt]

\\

& = & {\Large ^{\normalsize (y-2)} e} \end{eqnarray} \)

and

\( \begin{eqnarray}

{\Large ln^{\small(2)}\left(^{\normalsize(\mu+1)} \check \eta \right)}

& = & {\Large ln^{\small(2)}\left(\eta^{\left(^{\mu} \check \eta \right)} \right) } \\

\\[10pt]

\\

& = & {\Large ln^{\small(2)}\left(\eta^{\left(^{y} e\right)} \right) } \\

\\[10pt]

\\

& = & {\Large ln^{\small(2)}\left(e^{\large e^{\large \left({(^{(y-1)} e)\left(1-{\normalsize \delta}\right) }\right)}} \right) },\text{ where } \delta\ =\ \frac{1}{(^{(y-1)} e)} \\

\\[10pt]

\\

& = & {\Large {(^{\normalsize (y-1)} e)\left(1-{\normalsize \delta}\right)}},\text{ where } \delta\ =\ \frac{1}{(^{(y-1)} e)}

\end{eqnarray} \)

For sufficiently large y, this function is effectively exact (it is exact when you take the limit to infinity). Furthermore, notice that I swapped out the integer m for the real exponent y. Finally, notice that if eta-check is unique as I claim it is, then the tetration of base e that I just defined has a very strong claim on uniqueness. In other words, if some other method does not agree, I claim that this function is "correct", and the other function is displacing its exponents by some cyclic function.