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pentation and hexation
#1
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....
What about oct?  Oct(-5,-4,-3,-2,-1,0,1)=(-4,-3,-2,-1,0,e)?

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]
   
- Sheldon
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#2
Hey Sheldon. Bo proved this somewhere on here. It's actually pretty straight forward to prove that ANY analytic hyper-operator necessarily satisfies for for 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.)
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#3
(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 for for 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.
- Sheldon
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#4
I'm waiting for an inductive proof that gets you from to .
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