# Tetration Forum

Full Version: Exploring Pentation - Base e
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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:
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:

[attachment=172]

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 came to a similar conclusion in this thread, only using a linear approximation, and our results seem to agree, especially on the specific value: $x{\uparrow}{\uparrow}{\uparrow}-2 = -1$.

Andrew Robbins
Yeah, as I was thinking about extending pentation to hexation, I began to visualize the alternative horizontal and vertical asymptotes, but I wasn't sure.

Note that, with a real fixed point of tetration, we can find a continuous pentation with either an Abel solution or a regular solution. Hopefully they'll agree with each other, but I think that depends on whether the real fixed point at -1.85 is the closest to the origin.
Dear Jaydfox!
Concerning:

jaydfox Wrote: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.
....................................

I am very happy to see that you are approaching this problem exactly like (KAR and myself) we did (perhaps with less precision) in a progress report we posted to the NKS Forum on 25-07-2006, copy attached (NKS Forum III - Final).

In fact, in the case of pentation (to the base b), we may indicate it as:
y = b-penta-x, or y = b ยง x , or y = b [5] x (GFR-KAR conventions), or:
y = b ||| x, according to your conventions (sorry for the arrows, they are .... up!).

As a matter of fact, in that occasion and for the particular case of of b = e, we have shown that pentation is definable also for (integer) hyperexponents x < 0 and that this generates an asymptotic behaviour for x -> -oo. See the attachment, Section 4, pages 8 and 9, formulas 13 to 15.

We noticed that two near successive points of the plot in that area must always be linked by y(x-1) = sln y(x), where sln is the "natural" slog (base e), and by y(x+1) = sexpn y(x), where sexpn is the "natural" tetration operator (base e). We then concluded that the asymptotic value of y could be immediately found by putting:
sexpn(y) = slog(y)

This means that we can find the asymptotic value of y, for x -> -oo, at the intersection of sln(x) with sexpn(x) and we called "Sigma" this numerical value. We got, with our first approximations:
Sigma = -1.84140566043697..
But, very probably, your numerical value is better.

I understand that this was also your conjecture, which was verified through your calculations. Could I have confirmation of the most precise value of Sigma obtainable via more formal and precise calculations, for example using the Andrew's excellent slog approximation? Or... else? I (together with KAR) would be very happy if you could kindly produce that.

Thank you very much in advance.

GFR

(Annex attached on 2-02-2008. Previous attachment missing. Sorry)
Hello,

This has a fascinating intutive appeal, especially the appearance of even negative integers-just like trivial 0 of Riemann zeta function.

What are the few next values on other axis (- 3, ..., - 5, ... ) and how accurate they seem to be? Meaning is e.g. -1,85.. really close to asymptotic value or it can be - 1,9..

Are there any analytical means to get these values, for any base- does there exist such base? Hoping it will be e^(pi/2) , of course, but may be some other nice value.

Ivars
Ivars Wrote:Hello,

This has a fascinating intutive appeal, especially the appearance of even negative integers-just like trivial 0 of Riemann zeta function.

What are the few next values on other axis (- 3, ..., - 5, ... ) and how accurate they seem to be? Meaning is e.g. -1,85.. really close to asymptotic value or it can be - 1,9..

Are there any analytical means to get these values, for any base- does there exist such base? Hoping it will be e^(pi/2) , of course, but may be some other nice value.

Ivars

Just repeating the question...
I am asking because of this sum:

=1/1,85035452902718^8-1/(2*1,8503545290271^7+1/(3*1,8503545290271^6-1/(4*1,8503545290271^5+1/(5*1,8503545290271^4=0,007297583=1/137,0316766

which makes sense to me.

Ivars Fabriciuss
Please see the NKS Forum III attachment of my previous posting. Sorry. It was missing.

GFR
Dear GFR,

That is great paper. Visionary. Also easy to understand, as it omits non-essential things.
Could You send to me all Yours papers You have available which You think may give insights in hyperoperations (insights, not definitions based on limits or sets) . I feel fine plugging in infinities and h-s/g- s directly in my head.

I would like to learn to work with them directly, starting from exponentation, tetration, pentation.

I would like to "integrate" (or differentiate) e.g pentation to obtain "slower" operation (or faster). I would like to do hyperoperation math with objects that they can deal with without focusing on rigor but developing intuition"already in that level. I think it is more important. Luckily, Euler did not have time to write down his intuitive results in this direction ( he must have had )