I've somewhat reached a natural stopping point in my experimenting with the natural slog for base e. There's more to do, but I'm at a point of diminishing returns and want to do something else, hoping to get inspiration.

I've decided to move on to extending the continuous tetration solution to a continuous pentation solution. The first thing we need to know is what the fixed points are. Hyperbolic fixed points tell us where logarithmic singularities will be in the inverse function (the penta-logarithm, or whatever it's called). The location of the closest such fixed point tells us what the radius of convergence of the power series will be, which we can use as a rough validation tool for any power series we might try to derive, e.g., by an Abel matrix solution.

For base e, the first fixed point I've identified is at about -1.85. This can be seen trivially to exist by looking at the graph of tetration for base e. Without looking at the graph, we know that \( \exp_e^{\circ {\small -2}}(1) = -\infty \) and \( \exp_e^{\circ {\small-1}}(1) = 0 \) Therefore, somewhere in that interval, we must have a crossing. And we can also tell that the fixed point will be repelling under tetration, because the slope at the crossing will be greater than 1.

The quick and dirty way to find the fixed point is to take iterated superlogarithms. As it turns out, this is also how we can extend pentation to negative iterations. I'll use a triple arrow to notate pentation, though I suppose that \( \mathrm{sexp}_e^{\circ n}(1) \) would work as well.

We know that \( e\uparrow\uparrow\uparrow0=1 \), and \( e\uparrow\uparrow\uparrow-1=0 \). But we can find \( e\uparrow\uparrow\uparrow-2 \) by finding \( \mathrm{slog}_e(0) \), which is -1. Then we can find \( e\uparrow\uparrow\uparrow-3 \) by finding \( \mathrm{slog}_e(-1) \). This will quickly take us outside the radius of convergence, so in order to get maximum accuracy, we'll find \( \mathrm{slog}_e\left(\exp_e(-1)\right)-1 \).

Using my 1200-term accelerated solution, the first few iterations give us the following:

\( e\uparrow\uparrow\uparrow0=1 \)

\( e\uparrow\uparrow\uparrow-1=\mathrm{slog}_e(1)=0 \)

\( e\uparrow\uparrow\uparrow-2=\mathrm{slog}_e(0)=-1 \)

\( e\uparrow\uparrow\uparrow-3=\mathrm{slog}_e(-1)=-1.636358354286028979629049436 \)

\( e\uparrow\uparrow\uparrow-4=\mathrm{slog}_e(-1.636358354286028979629049436)=-1.813170483098635639971748853 \)

And so on... Taken to similar precision, the fixed point is -1.850354529027181418483437788.

Going in the forward direction for iteration:

\( e\uparrow\uparrow\uparrow1=\mathrm{sexp}_e(1)=2.718281828459045235360287471 \)

\( e\uparrow\uparrow\uparrow2=\mathrm{sexp}_e(2.718281828459045235360287471)=2075.968335058065833574141757 \)

And so on... Obviously, the next iteration is beyond the scope of scientific notation.

In table form, the integer pentations of e, from -20 to 2:

Plotted, we get the following for integer pentations, noting that the second pentation is at about 2,076, well off the top of this graph:

Note that if we flip this graph about the line y=x, we'll see the pentalog. There will be a logarithmic singularity at about x=1.850354529. We can calculate the base of the logarithm by dividing the differences of two consecutive pairs of integer pentates. Going out to -100 iterations, This yields a value of about 6.460671295681839390208370083.

We can also get the value by considering the slog and the reciprocal of its derivative at -1.850354529... This is outside the radius of convergence, so we can't simply take the derivative of the power series I developed at 0. However, we can get there using the Abel functional definition of the slog:

\( \mathrm{slog}(z) = \mathrm{slog}\left(\exp(z)\right)-1 \)

\(

\begin{eqnarray}

D_z \left[\mathrm{slog}(z)\right]

& = & D_z \left[\mathrm{slog}\left(\exp(z)\right)-1\right] \\

\mathrm{slog}^{'}(z) & = & \mathrm{slog}^{'}\left(\exp(z)\right)\exp(z) \\

\end{eqnarray}

\)

This evaluates to 0.1547826772534266617145246066. The reciprocal is 6.460671295681839390208370083, which matches the value I previously calculated by comparing successive negative iterates.

We now have the location and base of the logarithmic singularity. The only potential problem is if there are closer singularities in the complex plane, meaning there are other fixed points of the slog near the origin (which at a glance I doubt). But I'll cross that bridge if and when I get there.

I've decided to move on to extending the continuous tetration solution to a continuous pentation solution. The first thing we need to know is what the fixed points are. Hyperbolic fixed points tell us where logarithmic singularities will be in the inverse function (the penta-logarithm, or whatever it's called). The location of the closest such fixed point tells us what the radius of convergence of the power series will be, which we can use as a rough validation tool for any power series we might try to derive, e.g., by an Abel matrix solution.

For base e, the first fixed point I've identified is at about -1.85. This can be seen trivially to exist by looking at the graph of tetration for base e. Without looking at the graph, we know that \( \exp_e^{\circ {\small -2}}(1) = -\infty \) and \( \exp_e^{\circ {\small-1}}(1) = 0 \) Therefore, somewhere in that interval, we must have a crossing. And we can also tell that the fixed point will be repelling under tetration, because the slope at the crossing will be greater than 1.

The quick and dirty way to find the fixed point is to take iterated superlogarithms. As it turns out, this is also how we can extend pentation to negative iterations. I'll use a triple arrow to notate pentation, though I suppose that \( \mathrm{sexp}_e^{\circ n}(1) \) would work as well.

We know that \( e\uparrow\uparrow\uparrow0=1 \), and \( e\uparrow\uparrow\uparrow-1=0 \). But we can find \( e\uparrow\uparrow\uparrow-2 \) by finding \( \mathrm{slog}_e(0) \), which is -1. Then we can find \( e\uparrow\uparrow\uparrow-3 \) by finding \( \mathrm{slog}_e(-1) \). This will quickly take us outside the radius of convergence, so in order to get maximum accuracy, we'll find \( \mathrm{slog}_e\left(\exp_e(-1)\right)-1 \).

Using my 1200-term accelerated solution, the first few iterations give us the following:

\( e\uparrow\uparrow\uparrow0=1 \)

\( e\uparrow\uparrow\uparrow-1=\mathrm{slog}_e(1)=0 \)

\( e\uparrow\uparrow\uparrow-2=\mathrm{slog}_e(0)=-1 \)

\( e\uparrow\uparrow\uparrow-3=\mathrm{slog}_e(-1)=-1.636358354286028979629049436 \)

\( e\uparrow\uparrow\uparrow-4=\mathrm{slog}_e(-1.636358354286028979629049436)=-1.813170483098635639971748853 \)

And so on... Taken to similar precision, the fixed point is -1.850354529027181418483437788.

Going in the forward direction for iteration:

\( e\uparrow\uparrow\uparrow1=\mathrm{sexp}_e(1)=2.718281828459045235360287471 \)

\( e\uparrow\uparrow\uparrow2=\mathrm{sexp}_e(2.718281828459045235360287471)=2075.968335058065833574141757 \)

And so on... Obviously, the next iteration is beyond the scope of scientific notation.

In table form, the integer pentations of e, from -20 to 2:

Code:

`n | e penta n`

2 | 2075.968335058065833574141757

1 | 2.718281828459045235360287471

0 | 1.000000000000000000000000000

-1 | 0.000000000000000000000000000

-2 | -1.000000000000000000000000000

-3 | -1.636358354286028979629049436

-4 | -1.813170483098635639971748853

-5 | -1.844484246898395061868430374

-6 | -1.849443081393375287759562240

-7 | -1.850213384630118386703548774

-8 | -1.850332680687076371299817524

-9 | -1.850351147243492593015231122

-10 | -1.850354005584711078364293582

-11 | -1.850354448007332020493809851

-12 | -1.850354516486711680128769074

-13 | -1.850354527086133925479340624

-14 | -1.850354528726740890531493457

-15 | -1.850354528980678429204206706

-16 | -1.850354529019983561302333809

-17 | -1.850354529026067314878454466

-18 | -1.850354529027008974544720674

-19 | -1.850354529027154727148927025

-20 | -1.850354529027177287127222746

Plotted, we get the following for integer pentations, noting that the second pentation is at about 2,076, well off the top of this graph:

Note that if we flip this graph about the line y=x, we'll see the pentalog. There will be a logarithmic singularity at about x=1.850354529. We can calculate the base of the logarithm by dividing the differences of two consecutive pairs of integer pentates. Going out to -100 iterations, This yields a value of about 6.460671295681839390208370083.

We can also get the value by considering the slog and the reciprocal of its derivative at -1.850354529... This is outside the radius of convergence, so we can't simply take the derivative of the power series I developed at 0. However, we can get there using the Abel functional definition of the slog:

\( \mathrm{slog}(z) = \mathrm{slog}\left(\exp(z)\right)-1 \)

\(

\begin{eqnarray}

D_z \left[\mathrm{slog}(z)\right]

& = & D_z \left[\mathrm{slog}\left(\exp(z)\right)-1\right] \\

\mathrm{slog}^{'}(z) & = & \mathrm{slog}^{'}\left(\exp(z)\right)\exp(z) \\

\end{eqnarray}

\)

This evaluates to 0.1547826772534266617145246066. The reciprocal is 6.460671295681839390208370083, which matches the value I previously calculated by comparing successive negative iterates.

We now have the location and base of the logarithmic singularity. The only potential problem is if there are closer singularities in the complex plane, meaning there are other fixed points of the slog near the origin (which at a glance I doubt). But I'll cross that bridge if and when I get there.

~ Jay Daniel Fox