Real-analytic tetration uniqueness criterion?
#1
Hi.

I was wondering about this. it appears that the Kneser tetrational satisfies the following real-analytic property on \( (-2, \infty) \):

\( \mathrm{tet}^{(2n)}(x) \) is strict-monotone increasing for \( n \ge 0 \), \( n \in \mathbb{Z} \)
\( \mathrm{tet}^{(2n+1)}(x) \) is positive for \( n \ge 0 \), \( n \in \mathbb{Z} \)

and

\( \mathrm{tet}^{(2n+1)}(x) \) is convex for \( n \ge 0 \), \( n \in \mathbb{Z} \).

(the notation denotes differentiation)

These are equivalent. Easy theorems from calculus concerning monotonicity, convexity, and derivatives and integrals shows that 1 and 2 imply each other and that 3 implies 1.

But here's the thing: could this be a uniqueness criterion for tetration? (here, I'm thinking of base \( e \)) I tried some numerical tests of Kneser's tetration solution, warping it with small \( \theta(z) \) 1-cyclic warping mappings (specifically \( \theta(z) = \frac{\sin(2\pi z)}{K} \) and \( \theta(z) = \frac{\sin(2\pi z - \pi) + 1}{K} \)) with amplitudes of down to \( 10^{-5} \) (which is \( K = 10^5 \)) and the criterion seems to fail if the derivative is high enough (for amplitudes of the given magnitude, at around the 32nd derivative). The derivative at which it fails seems to increase rapidly as \( K \) shrinks, so I'm not sure if it is singular, meaning there is a range of \( K \) for which the condition is satisfied and hence uniqueness is not obtained, or whether or not it will eventually fail no matter how small the 1-cyclic wobble is, which would mean this should provide a uniqueness condition when combined with, perhaps, analyticity or maybe even just smoothness and the usual functional equations.

What do you think?
#2
Im very convinced its not unique.

Many reasons.

Asymptotics , interpolation , " fake function theory " ,...

In fact I posted similar conjectures years ago and came to reconsider them.

Hence I posted " TPID 16 " and a related uniqueness claim for exp^[1/2].

See :

http://math.eretrandre.org/tetrationforu...hp?tid=881

http://math.eretrandre.org/tetrationforu...hp?tid=879

( were the suggestion in post 2 is considered solid now )

Somewhat different :

http://math.eretrandre.org/tetrationforu...hp?tid=842

---
Here I was a bit upset because I posted these things first

http://math.eretrandre.org/tetrationforu...hp?tid=503
post 13.
---

http://math.eretrandre.org/tetrationforu...hp?tid=474

http://math.eretrandre.org/tetrationforu...hp?tid=484

and probably more posts.
( I currently posted more than 13% of this forum so I dont remember them all so well )

So the idea is far from completely new.
But its intresting.
I hope you get more responses than I did.

As a remark : Notice this type of uniqueness was not sufficient for a unique real-analytic gamma function in the past.
Gamma also grows fast and satisfies a functional equation !
See the Bohr-Möllerup and Wielandt theorems.

If you want to replace convex with log convex , probably better to replace with arc2sinh or so ... which then again resembles my conjectures.
Imho the lenght idea is also underrated although I owe some credit to gottfried.

A few members here have already proved analogues of the Wielandt for tetration.

regards

tommy1729
#3
I am not convinced by the "2sinh"-stuff. A composition with inverse of 2sinh superfunction may not even be analytic (I think sheldonison did some work on related stuff involving tetrational base changes -- and maybe even 2sinh, I don't remember), and converges to a linear function plus a small 1-cyclic wobble. A small, 1-cyclic oscillation doesn't seem too good for "convexity"-related purposes -- something that wiggles a little about a line is not quite convex.

Also, with more testing it looks like that this method of repeated differentiation eventually teases out even tiny 1-cyclic wobbles applied to the gamma function, i.e. \( \Gamma(x) \theta(x) \) where \( \theta(x) \) is a small-amplitude 1-cyclic wobble with \( \theta(0) = 1 \). Just that it seems with tetration, you need to differentiate it a whole lot more (probably because the growth rate is soo much faster than the gamma function's). 32 derivatives was more than enough to tease out a wobble of amplitude \( 10^{-12} \).
#4
I assume mike3 meant :

\( \Gamma(\theta(x)) \)

As for the numerical testing , did mike test for x > 0 or x >-2 or a Taylor series at some x ? Or x > some C ?

If his conjecture means for x > -2 then it is quite a strong idea ,
but Im not sure about existance since going from f(-1.9) to f(-0.9) is not likely convex.

SO I assume for x > 0.

Right ?

regards

tommy1729
#5
(06/09/2014, 10:44 AM)tommy1729 Wrote: I assume mike3 meant :

\( \Gamma(\theta(x)) \)

As for the numerical testing , did mike test for x > 0 or x >-2 or a Taylor series at some x ? Or x > some C ?

If his conjecture means for x > -2 then it is quite a strong idea ,
but Im not sure about existance since going from 0 to 1 is not convex.

SO I assume for x > 0.

Right ?

regards

tommy1729

Um, it should be \( \Gamma(x) \theta(x) \). \( \Gamma(\theta(x)) \) is a 1-periodic function itself and not at all a solution of the Gamma function equations!

Convexity is over odd derivatives (for tetration), and even derivatives (for gamma function).
#6
IMPORTANT!

I just was running some more tests and I've discovered that these results may not be trustworthy. Apparently, increasing the precision on sheldonison's Kneser PARI/GP program does not seem to make it necessarily generate more than 64 digits of the tetrational. So results requiring high precision seem to be suspect. I'll need to figure out how to get more digits out of the code before trying again. I had done a "\p 128" and it only would display 64 digits -- upon "\p 128"ing again to force it to cough up more digits, I found all the succeeding digits after the initial 64 were different from those I got for "\p 256", suggesting it is not getting beyond 64 digits.

I'm not sure how to get the program to give a correct result with more than 64 digits -- sheldonison?
#7
(06/09/2014, 10:50 AM)mike3 Wrote:
(06/09/2014, 10:44 AM)tommy1729 Wrote: I assume mike3 meant :

\( \Gamma(\theta(x)) \)

As for the numerical testing , did mike test for x > 0 or x >-2 or a Taylor series at some x ? Or x > some C ?

If his conjecture means for x > -2 then it is quite a strong idea ,
but Im not sure about existance since going from 0 to 1 is not convex.

SO I assume for x > 0.

Right ?

regards

tommy1729

Um, it should be \( \Gamma(x) \theta(x) \). \( \Gamma(\theta(x)) \) is a 1-periodic function itself and not at all a solution of the Gamma function equations!

Convexity is over odd derivatives (for tetration), and even derivatives (for gamma function).

Sorry I meant to say \( \Gamma(x+\theta(x)) \).

Afterall you consider \( sexp(x+\theta(x)) \) right ?

Or are you talking about \( sexp(x)\theta(x) \) ??
( that would not make sense )

Anyway for all clarity , I like the idea and wish it was true.
But I fear not.

You tested theta function that have \( |\theta'(x)| > 0 \) for x=0,1. I think that is the mistake.

When I tried periodic functions I took functions that satisfied
\( |D^m \theta(x)| = 0 \) for all 24 >= m >= 0 and all x = n/6 for any integer n.

I did however take x > 1 I think.

regards

tommy1729
#8
@Tommy1729:

The Gamma function is different from tetration. To make an alternative tetration, use

\( \mathrm{tet}^{*}(z) = \mathrm{tet}(z + \theta(z)) \)

where \( \theta(z) \) is a 1-cyclic function with \( \theta(0) = 0 \).

To make an alternative "Gamma function", use

\( \Gamma^{*}(z) = \Gamma(z) \theta(z) \).

where \( \theta(z) \) is a 1-cyclic function with \( \theta(0) = 1 \).

This is because the functional equation for tetration is

\( \mathrm{tet}(z + 1) = \exp(\mathrm{tet}(z)) \)

whereas that for the Gamma function is

\( \Gamma(z + 1) = z \Gamma(z) \).

Take the ratio of two solutions of this equation, and you will see it is 1-periodic. Thus, a 1-periodic multiplication factor (unlike for tetration, where you need composition!) will convert one "Gamma-like function" into another.

Also,

\( \Gamma((z + 1) + \theta(z + 1)) = \Gamma(z + \theta(z) + 1) = (z + \theta(z)) \Gamma(z + \theta(z)) \ne z \Gamma(z + \theta(z)) \).
#9
(06/09/2014, 12:49 PM)mike3 Wrote: @Tommy1729:

The Gamma function is different from tetration. To make an alternative tetration, use

\( \mathrm{tet}^{*}(z) = \mathrm{tet}(z + \theta(z)) \)

where \( \theta(z) \) is a 1-cyclic function with \( \theta(0) = 0 \).

To make an alternative "Gamma function", use

\( \Gamma^{*}(z) = \Gamma(z) \theta(z) \).

where \( \theta(z) \) is a 1-cyclic function with \( \theta(0) = 1 \).

This is because the functional equation for tetration is

\( \mathrm{tet}(z + 1) = \exp(\mathrm{tet}(z)) \)

whereas that for the Gamma function is

\( \Gamma(z + 1) = z \Gamma(z) \).

Take the ratio of two solutions of this equation, and you will see it is 1-periodic. Thus, a 1-periodic multiplication factor (unlike for tetration, where you need composition!) will convert one "Gamma-like function" into another.

Also,

\( \Gamma((z + 1) + \theta(z + 1)) = \Gamma(z + \theta(z) + 1) = (z + \theta(z)) \Gamma(z + \theta(z)) \ne z \Gamma(z + \theta(z)) \).

Yes of course you are right.
Sorry silly of me.

But the argument still remains ...

What is your opinion about the periodic functions that I tried ?
Did you try those ?

regards

tommy1729
#10
(06/09/2014, 12:20 PM)mike3 Wrote: IMPORTANT!

I just was running some more tests and I've discovered that these results may not be trustworthy. Apparently, increasing the precision on sheldonison's Kneser PARI/GP program does not seem to make it necessarily generate more than 64 digits of the tetrational. So results requiring high precision seem to be suspect. I'll need to figure out how to get more digits out of the code before trying again. I had done a "\p 128" and it only would display 64 digits -- upon "\p 128"ing again to force it to cough up more digits, I found all the succeeding digits after the initial 64 were different from those I got for "\p 256", suggesting it is not getting beyond 64 digits.

I'm not sure how to get the program to give a correct result with more than 64 digits -- sheldonison?

Hey Mike,

After changing the precision with "\p 134" or any large number, type in
"init(exp(1));" You will get approximately n/2 decimal digits for the results, limited by the Schroeder functions algorithm.
Code:
\p 134
init(exp(1));

This regenerates the tetration system with arbitrarily high precision numbers. The "init(exp(1));" after the default of "\p 67" takes 3 or 4 seconds. After "\p 134" takes about 30 seconds, and after "\p 221" takes 3-4 minutes. There s an out of memory (>4meg) error around "\p 230" or so. But I have used "\p 512" I think with 30 minutes computer time. The out of memory error is due to initializing my base eta approximation function, in the initcheta code. Here is a patch to remove that memory error.
Code:
initcheta() = {
/* automatically initialize cheta during program initialization */
  local(z,local);
  z=1.0;
  precis=precision(z);
  /* initialization for xcheta and xsexpeta required for cheta, sexpeta, invcheta, invsexpeta functions */
  /* initalizes to match precis, 50 digits/67 digits, this routine aims for 75% precision               */
  /* called from init(initbase) when program is loaded, and when init detects the precision has changed */
  if (precis<=180,
    chterms   = 2*(floor(precis/2)-8)+1;
,
    chterms   = 2*(floor(180/2)-8)+1;
);
  chdelta   = (chterms-1)*2;
  xcheta = genpoly(chterms,chdelta,0);
  invchetr = imag(cheta(-chdelta+(chterms-1)*I/2));
  invprecis = 10000.*(10^-chterms);
  /* invprecis = 100*abs(chetaerr(I+0.5,chdelta)); */
  /* invprecis = 1E-47; */
  chetadlt=cheta(-chdelta);
  xsexpeta = genpoly(chterms,chdelta,1);
  sxpetadlt=sexpeta(chdelta);
  return(0);
}
- Sheldon


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