# Tetration Forum

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I have been reading post in forum for two weeks. Now I feel I was too young, too simple, and naive for tetration.
I need somebody to help me clarifiy the elementary knowledge for tetration.

My understanding of the tetration is:

Quote:Code definition:
$\infty^*$=ComplexInfinity (infinite magnitude, undetermined complex phase)
Not consider the branch cut:
$tet_b(slog_b(z))=z$
$slog_b(tet_b(z))=z$
$tet_{sroot_h(z)}(h)=z$

and:
$tet_{conj(b)}(conj(z))=tet_b(z)$
$slog_{conj(b)}(conj(z))=slog_b(z)$
$sroot_h(conj(z))=sroot_h(z)$

and:
$tet_0(0)=1, tet_0(1)=0, tet_0(\infty^*)$is oscillates  infinitely, but maybe 0 and 1 are different branch of the infinite iterated exponential.
$tet_1(0)=1, tet_1(1)=1, tet_1(\infty^*)=1$
in other bases:
$tet_b(\infty^*)=\frac{\mathrm{W_{cut}}(-\ln{z})}{-\ln{z}},cut\in\mathbb{Z}$
$sroot_{\infty^*}(z)=({\frac{1}{z}})^{-\frac{1}{z}}$

tetration, super-root and super-logarithm is infinitely differentiable. (but I wasn't find code take the derivative...)

If the bases is hyperbolic, there is only one "regular" super-function. If the bases is parabolic, will have at least 2 "regular" super-function.(Leau-Fatou-flower)
The branch cut for super-function is infinitely.
fatou.gp will use all "regular" super-function to refactoring tetration.

bases regions for tetration:
$base=\pm\infty$, Andrew Robbins
base=0, not supported
$base\in(0,e^{-e})$, unknown
$base\in[e^{-e},1)$, Koenig, no code
base=1, Andrew Robbins
$base\in(1,e^{e^{-1}})$, Koenig, fatou.gp
$base=e^{e^{-1}}$, Ecalle, fatou.gp
$Arg(base)\in({\frac{14\pi}{30}},{\frac{21\pi}{30}})\wedge({\frac{42\pi}{30}},{\frac{47\pi}{30}}),\left| base \right|<1.76.$ill-region for Fatou.
other, Fatou, fatou.gp
First you need to edit your first some formulas, so that one can read it - you can not include line breakes via <br/> in a formula.
(07/09/2019, 09:26 PM)bo198214 Wrote: [ -> ]First you need to edit your first some formulas, so that one can read it - you can not include line breakes via <br/> in a formula.

Does it look better now?
(07/01/2019, 12:11 PM)Ember Edison Wrote: [ -> ]I have been reading post in forum for two weeks. Now I feel I was too young, too simple, and naive for tetration.
I need somebody to help me clarifiy the elementary knowledge for tetration.

My understanding of the tetration is:

Quote:Code definition:
$\infty^*$=ComplexInfinity (infinite magnitude, undetermined complex phase)
Not consider the branch cut:
$tet_b(slog_b(z))=z$
$slog_b(tet_b(z))=z$
$tet_{sroot_h(z)}(h)=z$

and:
$tet_{conj(b)}(conj(z))=tet_b(z)$
$slog_{conj(b)}(conj(z))=slog_b(z)$
$sroot_h(conj(z))=sroot_h(z)$

and:
$tet_0(0)=1, tet_0(1)=0, tet_0(\infty^*)$is oscillates  infinitely, but maybe 0 and 1 are different branch of the infinite iterated exponential.
$tet_1(0)=1, tet_1(1)=1, tet_1(\infty^*)=1$
in other bases:
$tet_b(\infty^*)=\frac{\mathrm{W_{cut}}(-\ln{z})}{-\ln{z}},cut\in\mathbb{Z}$
$sroot_{\infty^*}(z)=({\frac{1}{z}})^{-\frac{1}{z}}$

tetration, super-root and super-logarithm is infinitely differentiable. (but I wasn't find code take the derivative...)

If the bases is hyperbolic, there is only one "regular" super-function. If the bases is parabolic, will have at least 2 "regular" super-function.(Leau-Fatou-flower)
The branch cut for super-function is infinitely.
fatou.gp will use all "regular" super-function to refactoring tetration.

bases regions for tetration:
$base=\pm\infty$, Andrew Robbins
base=0, not supported
$base\in(0,e^{-e})$, unknown
$base\in[e^{-e},1)$, Koenig, no code
base=1, Andrew Robbins
$base\in(1,e^{e^{-1}})$, Koenig, fatou.gp
$base=e^{e^{-1}}$, Ecalle, fatou.gp
$Arg(base)\in({\frac{14\pi}{30}},{\frac{21\pi}{30}})\wedge({\frac{42\pi}{30}},{\frac{47\pi}{30}}),\left| base \right|<1.76.$ill-region for Fatou.
other, Fatou, fatou.gp
It looks better.  I would like to post more, but time is limited.  For sroot, all of the extensions need to be in the same analytic family.

Kneser analytic solution can be extended to complex bases, and creates a family of complex base solutions, but
such solutions have singularities at bases like base=0,1, eta so one cannot talk about analytic base=0 or base=1 or base=exp(1/e) for Kneser.  One cannot use the Koenig/Schroeder solutions or the Ecalle solutions in the Kneser construct for an sroot family.
For $b>1$ there are the following cases:
[attachment=1374][attachment=1373][attachment=1372]

Actually, I also wanted to show the half iterate in the picture. However it took too long time, so I just explain, how it works.

For the hyperbolic case we have to fixpoints. For each fixpoint there exists an analytic regular solution for fractional iterates that is analytic at that fixpoint. And this solution is not analytic at the other fixpoint. This insight was causing the Bummer thread.

For the parabolic case, which you can imagine as moving the fixpoints together into one fixpoint, there are also two fractional iterates (from left and right) that are (I think its called) asymptotically analytic at the fixpoint.

For the elliptic case, there are two conjugate complex fixpoints and there is a uniqueness criterion for the Abel function, which then can be used to calculate the fractional iterates. The fractional iterates are not analytic at both fixpoints.

Somehow for me its strange that in the hyperbolic and parabolic case there are always two "regular" solutions, while in the elliptic case there is only one "right" solution. Does this indicate that the fractional iterates have are not analytic with respect to the basis b, in $b=e^{1/e}$?
(07/11/2019, 04:57 PM)sheldonison Wrote: [ -> ]
(07/01/2019, 12:11 PM)Ember Edison Wrote: [ -> ]I have been reading post in forum for two weeks. Now I feel I was too young, too simple, and naive for tetration.
I need somebody to help me clarifiy the elementary knowledge for tetration.

My understanding of the tetration is:

Quote:Code definition:
$\infty^*$=ComplexInfinity (infinite magnitude, undetermined complex phase)
Not consider the branch cut:
$tet_b(slog_b(z))=z$
$slog_b(tet_b(z))=z$
$tet_{sroot_h(z)}(h)=z$

and:
$tet_{conj(b)}(conj(z))=tet_b(z)$
$slog_{conj(b)}(conj(z))=slog_b(z)$
$sroot_h(conj(z))=sroot_h(z)$

and:
$tet_0(0)=1, tet_0(1)=0, tet_0(\infty^*)$is oscillates  infinitely, but maybe 0 and 1 are different branch of the infinite iterated exponential.
$tet_1(0)=1, tet_1(1)=1, tet_1(\infty^*)=1$
in other bases:
$tet_b(\infty^*)=\frac{\mathrm{W_{cut}}(-\ln{z})}{-\ln{z}},cut\in\mathbb{Z}$
$sroot_{\infty^*}(z)=({\frac{1}{z}})^{-\frac{1}{z}}$

tetration, super-root and super-logarithm is infinitely differentiable. (but I wasn't find code take the derivative...)

If the bases is hyperbolic, there is only one "regular" super-function. If the bases is parabolic, will have at least 2 "regular" super-function.(Leau-Fatou-flower)
The branch cut for super-function is infinitely.
fatou.gp will use all "regular" super-function to refactoring tetration.

bases regions for tetration:
$base=\pm\infty$, Andrew Robbins
base=0, not supported
$base\in(0,e^{-e})$, unknown
$base\in[e^{-e},1)$, Koenig, no code
base=1, Andrew Robbins
$base\in(1,e^{e^{-1}})$, Koenig, fatou.gp
$base=e^{e^{-1}}$, Ecalle, fatou.gp
$Arg(base)\in({\frac{14\pi}{30}},{\frac{21\pi}{30}})\wedge({\frac{42\pi}{30}},{\frac{47\pi}{30}}),\left| base \right|<1.76.$ill-region for Fatou.
other, Fatou, fatou.gp
It looks better.  I would like to post more, but time is limited.  For sroot, all of the extensions need to be in the same analytic family.

Kneser analytic solution can be extended to complex bases, and creates a family of complex base solutions, but
such solutions have singularities at bases like base=0,1, eta so one cannot talk about analytic base=0 or base=1 or base=exp(1/e) for Kneser.  One cannot use the Koenig/Schroeder solutions or the Ecalle solutions in the Kneser construct for an sroot family.

So Schroeder contribute 1 sroot, Kneser contribute 1 sroot, 3 Singularity contribute 3 sroot and "infinity" contribute 3 sroot(Complex analysis, Ordinal arithmetic and generalized continuum hypothesis).
It's fucking cool.
(07/11/2019, 08:16 PM)bo198214 Wrote: [ -> ]For $b>1$ there are the following cases:

Actually, I also wanted to show the half iterate in the picture. However it took too long time, so I just explain, how it works.

For the hyperbolic case we have to fixpoints. For each fixpoint there exists an analytic regular solution for fractional iterates that is analytic at that fixpoint. And this solution is not analytic at the other fixpoint. This insight was causing the Bummer thread.

For the parabolic case, which you can imagine as moving the fixpoints together into one fixpoint, there are also two fractional iterates (from left and right) that are (I think its called) asymptotically analytic at the fixpoint.

For the elliptic case, there are two conjugate complex fixpoints and there is a uniqueness criterion for the Abel function, which then can be used to calculate the fractional iterates. The fractional iterates are not analytic at both fixpoints.

Somehow for me its strange that in the hyperbolic and parabolic case there are always two "regular" solutions, while in the elliptic case there is only one "right" solution. Does this indicate that the fractional iterates have are not analytic with respect to the basis b, in $b=e^{1/e}$?

Can you introduce the base for 3 Singularity, 0, 1, eta?
Is all problems come from log?
Why andydude say he can evaluate the base-infinity? He didn't reply to my email.
(07/13/2019, 06:49 PM)Ember Edison Wrote: [ -> ]Can you introduce the base for 3 Singularity, 0, 1, eta?
Is all problems come from log?
Why andydude say he can evaluate the base-infinity? He didn't reply to my email.

iterating
$y\;\mapsto \eta^y;\;\;\;\eta=\exp(1/e)$
is congruent to iterating the cleaner problem with fixed point z=0
$z\;\mapsto \exp(z)-1;\;\;\; z=\frac{y}{e}-1;$

Now it turns out that there is a formal half iterate solution around 0 for exp(x)-1; but the formal equation doesn't converge!  See Will Jagy's thread on math overflow on the formal power series for f(f(x))=sin(x).  btw, I noticed a couple of our tetration forum folks on this thread, Daniel and Gottfried.

Here are the first 20 terms of the formal half iterate series for exp(x)-1.  half(half(x))~=exp(x)-1
Code:
{half=  x +x^ 2*  1/4 +x^ 3*  1/48 +x^ 4*  0 +x^ 5*  1/3840 +x^ 6* -7/92160 +x^ 7*  1/645120 +x^ 8*  53/3440640 +x^ 9* -281/30965760 +x^10* -1231/14863564800 +x^11*  87379/24222105600 +x^12* -13303471/7847962214400 +x^13* -54313201/40809403514880 +x^14*  10142361989/5713316492083200 +x^15*  2821265977/7617755322777600 +x^16* -10502027401553/5484783832399872000 +x^17*  1836446156249/5328075722902732800 +x^18*  2952828271088741/1220613711064989696000 +x^19* -1004826382596003137/680288708300220923904000 +x^20* -7006246797736924249/1943682023714916925440000 }
Even though the asymptotic series is eventually divergent, the half iterate itself is analytic (except at zero), and the series can be used to calculate the half iterate arbitrarily accurately!  If z is negative, iterate $z\;\mapsto \exp(z)-1$ to get close enough to zero before evaluating the half iterate series h, and then iterate $h\;\mapsto \ln(h+1)$ the same number of times, and you get the half iterate.

For the other petal of the Leau-fatou flower, do the iteration in reverse order. Iterate $z\;\mapsto \ln(z+1)$ and then calculate the half iterate h, and iterate $h\;\mapsto \exp(h)-1$ the same number of times.   So even though the half iterate series is not analytic at zero, it is useful.  Cool stuff, the parabolic case.

As far as Kneser and base eta, if the limit of Kneser half iterate were the base eta sexp as the base arbitrarily approaches close to eta, the problem would seem to be which petal?  The two petals are different so maybe not.  It is a singularity, where the fixed points switch from real valued to complex valued for bases>eta.  So, as Henryk pointed out, the formal half iterate family of functions is different for bases<eta then it is for bases>eta.  The Koenig's half iterate for bases<eta probably have the lower fixed point goes to the lower petal, and the upper fixed point goes to the upper petal.  Kneser can be analytically extended to bases<eta but it is no longer real valued, so it has a different half iterate.  Henryk's graph shows the change on either side of the parabolic fixed point.
Here I tried to make the difference between left/lower and right/upper regular iteration visible - by using a very extreme base close to 1:
[attachment=1378]
Ember, please do me a favour and don't quote the whole thing if you are only referring to some particular lines.

A half iterate of f is a function g, such that g(g(x))=f(x). You typically write it as $f^{\circ 1/2}(x)$. It's the superfunction at value 1/2.
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