# Tetration Forum

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Xorter and I have been exchanging some emails.  Here's the question.  Start with tetration, base e.  If you want to extend to analytic tetration, download fatou.gp from this website.  Fatou was recently updated to do some really really cool stuff, but that's for another post.
sexpinit(exp(1));
sexp(-1)=0
sexp(0)=1
sexp(1)=e

Now onto Pentation.  The latest version of fatou.gp also has built in support for pentation for real bases, but lets start with the integer values for Pentation.
genpent(exp(1));
pent(-2)=-1 /* slog(pent(-1))=slog(0)=-1 */
pent(-1)=0  /* slog(pent(0))=slog(1)=0 */
pent(0)=1 /* pent(0,1) are by definition 1 and the base=e */
pent(1)=e

We don't yet have analytic hexation, but if it existed ....
hex(-3)=-2 /* pent^{-1}(hext(-2))=pent^{-1}(-1)=-2 */
hex(-2)=-1 /* pent^{-1}(hext(-1))=pent^{-1}(0)=-1 */
hex(-1)=0 /* pent^{-1}(hext(0))=pent^{-1}(0)=-1 */
hex(0)=1 /* hex(0,1) are by definition, 1 and the base=e  */
hex(1)=e

If one uses the simple linear extension between hex<-2,-1>, then there would be a singularity somewhere near hex(-3.86) since hex(-2.86)~=-1.85, and the fixed point for tetration base(e) is ~=-1.85.  Near that value, hex(z) would go to minus infinity.  One could imagine generating an analytic hexation function by using a bipolar theta mapping using the two primary complex conjugate fixed points of pentation.  This would require a fair amount of work.
-2.2597543772948619217694227344490 +/- 1.3844243840414798988939409459371*I

So does this pattern really hold? would hept(-4,-3,-2,-1,0,1)=(-3,-2,-1,0,1,e)?  Could it be analytic? Hexation probably has a real fixed point....

Here is what hexation base(e) would look like from <-3.85,1.5> with the linear approximation on <-2,-1>.  I also show the line y=x for reference.
[attachment=1282]
[attachment=1283]
Hey Sheldon. Bo proved this somewhere on here. It's actually pretty straight forward to prove that ANY analytic hyper-operator necessarily satisfies $f(-k) = 1-k$ for $0 \le k \le n-1$ for $n$ the rank of the hyper-operator. (exponentiation is rank 1, tetration is rank 2, pentation is rank 3, hexation is rank 4, so on and so forth.)
(08/21/2017, 08:05 PM)JmsNxn Wrote: [ -> ]Hey Sheldon. Bo proved this somewhere on here. It's actually pretty straight forward to prove that ANY analytic hyper-operator necessarily satisfies $f(-k) = 1-k$ for $0 \le k \le n-1$ for $n$ the rank of the hyper-operator. (exponentiation is rank 1, tetration is rank 2, pentation is rank 3, hexation is rank 4, so on and so forth.)

Yeah, that makes sense.  I hadn't seen Henryk's proof.  I'm still working on the finishing touches for analytic Hexation, with a theta mapping.
I'm waiting for an inductive proof that gets you from $e \uparrow^n x$ to $e \uparrow^{n+1} x$.
(08/22/2017, 10:38 PM)JmsNxn Wrote: [ -> ]I'm waiting for an inductive proof that gets you from $e \uparrow^n x$ to $e \uparrow^{n+1} x$.

I've been playing around with these functions, but work and life have been busy.  I got hexation with a theta mapping working, which is pretty cool all by itself.   Its actually the inverse of hexation, with roughly 30 decimal digits accuracy for real bases.   Here's a quick picture showing the algorithm, where we sample n points around a circle between the pentation conjugate fixed points; -2.25975437729486 +/- 1.38442438404148*I, and solve for the hexation Abel function on that circle.  On the left, a theta mapping with 60 samples, and 8 theta samples gives 11 decimal digits of precision, with precision of around 30 decimal digits with 240 samples.  On the right, without a theta mapping we are required to extend the sample points all the way to the fixed points.  With 256 samples, we only get 7 decimal digits of precision.  So a theta mapping is pretty important.  The yellow sample points are pairing z with pent(z), and the brown sample points pair z with pent^{-1}(z).  The pink and green points on the bigger circle involve the theta mapping and the Schroder function from the complex fixed point, generated using the 8 green and pink samples in the smaller circle.  Using the sample points in the picture, we setup a linear systems of equations and solve it, much like the routine I use for the slog matrix solution in the latest version of fatou.gp

[attachment=1296]

Then we can show this solution between these two conjugate fixed points is unique.  Of course, there may be other fixed points for pentation depending on the base.  Which sort of brings me to a sort of obvious observation.  This pentation and hexation function doesn't seem particularly unique.  There are other real fixed points for pentation for some bases, and there are other complex conjugate fixed points too.  It turns out tetration has complex conjugate fixed points too, so maybe we could have a used a theta mapping to generate an alternative pentation solution as well.  So, we have defined a family of functions that is not particularly unique, but allows analytic superfunctions to at least septation and probably octation.
(1) pentation defined using leftmost real fixed point of tetration;
(2) hexation defined using the leftmost complex conjugate fixed points of pentation from (1)
(3) heptation is defined using the leftmost real fixed point of hexation.
(4) ocation defined using the leftmost complex conjugate fixed points of heptation.
.....

I've generated the first pentation, hexation, and heptation, accurate to 25-30 decimal digits.  I've also generated the complex conjugate fixed points for heptation as well, which is often enough to show I can generate a nicely converging theta mapping.  So octation also appears plausible; and octation would have real valued fixed points...  Here is a graph of base(e) exp, sexp, pentation, hexation, and septation!  Octation would probably behave somewhat like hexation.  The pattern is pretty cool, with pentation and heptation tending towards their real valued fixed points as real(z) gets more negative, whereas tetration and hexation have a logarithmic singularity.
[attachment=1299]

Let us assume that there is an infinite sequence of these functions.  Now, for base 2, its easy to show that the sequence gets arbitrarily large since the Knuth uparrow function is well defined for integer valued bases.  But Knuth's uparrow only gives us integer values for integer bases and tells us nothing about real valued bases.

So lets make some observation.
Tetration or iterated exponentiation is bounded for bases<=exp(1/e), since there is a real valued upper fixed point.
For pentation Nuinho observed that for bases<= ~1.635 sexp(z) has an upper fixed point, and therefore pentation is bounded for these bases.
For hexation, I observe that for bases<= ~1.738, pent(z) has an upper fixed point, and therefore hexation is bounded for these bases
For heptation, I observe that for bases<=1.799, hex(z) has an upper fixed point, and therefore heptation is bounded for these bases
For octation, I observe that for bases<=1.839, hept(z) has an upper fixed point, and therefore octation is bounded for these bases, assuming one could generation analytic octation ...

Perhaps for all bases<2, the iterated function sequence eventually becomes bounded?
There is a result posted on here about how the "eta constants" converge to 2 and the "euler constants" converge to 4. The n'th eta constant is the sup of the  x'th n'th hyperoperator root of x. And the "euler constants" are the actual values x_n such that x_n'th n'th hyperoperator root of x_n = n'th eta constant. I think from this we can show that anything less than 2 does not grow unbounded, (just observe the meaning of the n'th hyperoperator root). Any base less than the n'th eta constant has a bounded hyperoperator for N>n-1.

We get something similar with the bounded hyperoperators as what you're talking about. It's a little messier but essentially there exists $\mathbb{Z}^n$  n-2'th hyperoperators. They satisfy the following really weird recursion

$\alpha \uparrow^n_{a_1a_2...a_n} (\alpha \uparrow^{n+1}_{a_1a_2...a_n b} z) = \alpha \uparrow^{n+1}_{a_1a_2...a_n b} (z+1)$

for ANY arbitrary b, and $a_1,...,a_n \in \mathbb{Z}$ fixed. The first n numbers have to equal for this recursion to work though, but the last number can be anything. Essentially, everytime we go up a hyperoperator, we get another $\mathbb{Z}$ amount of iterates.

I haven't actually proved this, but it's really straightforward.

Firstly, there exists $\mathbb{Z}$ exponentiations. namely $f(z) = \,e^{2 \pi i a_1 z}\alpha^z$ for all integer a_1. Then we fix an a_1 and there exists a kind of fixed point that is unique (its uniqueness and what qualifies it is a little hard to explain, which is the weakest point of my argument). It's geometrically attracting and hence can be iterated. Now the iterate $g(z,\xi)$ is conjugate to SOME exponential $e^{2 \pi i a_2 z} \lambda^z$ for $\lambda$ the multiplier of the above fixed point. This generates $\mathbb{Z} \times \mathbb{Z}$ tetration functions g(z,1). (they interpolate the natural power towers, send to complex values and are bounded on SOME half plane of the complex plane that includes $\mathbb{R}^+$, and satisfy the recursion $f(g(z)) = g(z+1)$. This allows us to find another FIXED point and it's unique because of Schwarz' lemma. Rinse and repeat, and we get $\mathbb{Z} \times \mathbb{Z}\times \mathbb{Z}$ pentations. Then $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}\times\mathbb{Z}$ hexations, so on and so forth.

This is all really roughly argued, more intuition than rigor. But I call it the "branches of hyper operators."
(09/03/2017, 11:52 PM)JmsNxn Wrote: [ -> ]There is a result posted on here about how the "eta constants" converge to 2 and the "euler constants" converge to 4. The n'th eta constant is the sup of the  x'th n'th hyperoperator root of x. And the "euler constants" are the actual values x_n such that x_n'th n'th hyperoperator root of x_n = n'th eta constant...

How interesting!  I haven't seen this conjecture before.  This is base 1.84, for which septation almost has an upper fixed point, that would be somewhere near 3.7  So octation for base=1.84 wouldn't quite be bounded, but oct(45)~=4.05, so it would take awhile before octation escapes; I can only calculate integer values for octation, until I do a theta mapping.  Also, notice that the solution for base(e) above, behaves nicer than base 1.84.

Why would do the "euler constants" converge to 4, as n gets arbitrarily large for the "nth" hyperoperator for bases approaching 2?
[attachment=1300]
(09/04/2017, 03:04 AM)sheldonison Wrote: [ -> ]
(09/03/2017, 11:52 PM)JmsNxn Wrote: [ -> ]There is a result posted on here about how the "eta constants" converge to 2 and the "euler constants" converge to 4. The n'th eta constant is the sup of the  x'th n'th hyperoperator root of x. And the "euler constants" are the actual values x_n such that x_n'th n'th hyperoperator root of x_n = n'th eta constant...

How interesting!  I haven't seen this conjecture before.  This is base 1.84, for which septation almost has an upper fixed point, that would be somewhere near 3.7  So octation for base=1.84 wouldn't quite be bounded, but oct(45)~=4.05, so it would take awhile before octation escapes; I can only calculate integer values for octation, until I do a theta mapping.  Why would do the "euler constants" converge to 4, as n gets arbitrarily large for the "nth" hyperoperator for bases approaching 2?

I'll be a bit clearer, I was in a rush before. I wish I could find the post (i think it was by jaydfox). I can't remember the finesse involved. This is all based off foggy memories, but I think it went something like this.

$h_n(x) \uparrow^n x = x$

then $h'_n(e_n) = 0$

and $h_n(e_n) = \eta_n$

then

$e_n \to 4$

$\eta_n \to 2$

But these constants are not unique (depend on which tetration, pentation, hexation, we're using), it simply follows that they converge to 4 or 2. (Maybe it was just a conjecture though...).

Again, I can't remember how to prove the following but, $b > \eta_n$ if and only if $b \uparrow^n x \to \infty$ as $x \to \infty$. Which are the sequence of constants you're talking about. I'm pretty sure they're equivalent. They are for $\uparrow^1$ and $\uparow^0 = \times$.

These two facts would claim if $b<2$ then $b \uparrow^n x$ is bounded as $n$ grows.

Again, I can't remember the full details, I may be forgetting something very important. Maybe it'll come to me in the morning, been a long day.
(08/07/2017, 07:33 PM)sheldonison Wrote: [ -> ]Now onto Pentation.  The latest version of fatou.gp also has built in support for pentation for real bases, but lets start with the integer values for Pentation.

Hi, Do you have any idea make pentinit() support complex base? The real base>eta is Kneser's solution in sexpinit(), like base<E^(-E) and complex base.

Take a step back, base>eta is not All real field.
(09/18/2019, 06:34 AM)Ember Edison Wrote: [ -> ]
(08/07/2017, 07:33 PM)sheldonison Wrote: [ -> ]Now onto Pentation.  The latest version of fatou.gp also has built in support for pentation for real bases, but lets start with the integer values for Pentation.

Hi, Do you have any idea make pentinit() support complex base? The real base>eta is Kneser's solution in sexpinit(), like base<E^(-E) and complex base.

Take a step back, base>eta is not All real field.

pentation, as implemented, relies on the Schroeder/Konig Abel function from the lower fixed point of tetration where tet(z)=z.  This is somewhere between tet(-2)=-infinity and tet(-1)=0.  Its a simpler solution than Kneser's solution; also less elegent and less well behaved.

Unfortunately, I'm too busy to write any code right now, and complex base pentation would seem to be a very large effort project.  The existing pentation code starts with an estimate of -1.97 and finds the fixed point iterating z=slog(z), and generates the Pentation solution, assuming it finds a repelling fixed point.  I think for complex bases it crashes when it tries to estimate the accuracy of the resulting Pentation solution, so it might not be too difficult for you to experiment with it.
n=0;while (z1<1,z1=sexp(z1);n++);
Code:
z1 = subst(pentz,x,pentr);
z2 = subst(pentz,x,pentr/lambdat);
n=0;while (z1<1,z1=sexp(z1);n++);
n++;while (n>0, z2=sexp(z2);n--;);
errz = -log(abs(z1-z2))/log(10);
...