Two types of tetration : sexp ' > or < 1.

[tommy1729]

The chain rule implies for analytic tetration :

sexp ' (x) = D exp( sexp(x-1) ) = sexp(x) sexp '(x-1)

LET

f(0) = 0 , f(x+ 1) = exp(f(x)).

then by the above we have

f '(0) = f '(1) !!

You can check with limits that

f ' (0) = f ' (-1) f(0)  is consistant AND free to choose.

Let f(h) = c h + d h^2

Then f(-1 + h) = ln( c h + d h^2) 

taking derivatives 

f ' (-1 + h) = ln© + ln(h) + O(h^2)

So 

f ' (0) = f ' (-1) f(0)  is consistant AND free to choose.

***

So 

sexp ' (x) = D exp( sexp(x-1) ) = sexp(x) sexp '(x-1)

holds for the full real range as desired.

and

f '(0) = f '(1) = c

where real c is free to choose.


You guessed it, I am considering this c as fundamental and cruxial and make a fuss about it.

But for some good reasons.

I have 2 methods that I feel are important.

https://tetrationforum.org/showthread.php?tid=1750

call it  the 3/5 method

and 

https://tetrationforum.org/showthread.php?tid=1339

; the gaussian method.


(lemma)
As is typical for iterations, for positive integers a : 

t^[a](  b( t^[-a](x) )  ) 

somewhat resembles the structure of b(x)

(end lemma)

Taking that into account it seems that 



the gaussian method implies 0 < c < 1.

If you look at how the gaussian applies to exp^[v](1) it seems to imply 

lim (exp^[v](1) - 1)/v

for small v equals the derivative c.

and since erf(x) is such at flat function it seems to imply 0 < c < 1. 

On the other hand, applying to lemma to the 3/5 method it seems to suggest

1 < c < e - 1

I assume c = 8/5 (or very close) here.

( notice 1 + 3/5 = 8/5 and 8/5 < e-1 )

 
***

So this seems to prove that the two methods are very different.

Notice neither c equals 0.

And neither c equals 1 , so a smooth transition from one to the other seems problematic.

***

This also seems to implies that the gaussian method does not satisfy the semi-group isomorphism.

***

So what property do we prefer ?

Well we want that

semiexp(semiexp(x)) = exp(x)

and we want 

D_h exp^[h](1) > 1.

This seems the case for the 3/5 method.

We also want

D semiexp(x) >  1 for x > 1.

This last inequality leads me to 

semiexp(x) < (exp(x) + x)/2 

for x > 1.



Together with the typical

semiexp ' (x) > 0 for all real x

and

semiexp " (x) > 0 for all x > 1


I think this can be achieved by the 3/5 method.


***

I start to wonder when variants of the 3/5 method converge to the same solution.

Lets say they use the same helper function with also derivative 8/5 at 0.

And lets say both of them give a solution semiexp that satisfies the above ( such as sexp ' (x) > 1 for x > 1 )

And lets say such 2 helper functions have a half iterate that both stay close to eachother within ...

within what ?? 

---

To analyse this ...

We could consider the third derivative.

Or make sharper bounds than (exp(x) + x)/2



I will stop here for now.


However I conjecture


sexp( slog(x) + w ) = < w exp(x) + (1 - w) x

for x > 1 and 1 > w > 0



regards

tommy1729

**edited**

It seems a good idea to consider the slight generalization 

f(x) = a b^x + c

for a > 0 , b > 1 ,  c >= 0

In particular when f(0) > 1 and/or  f ' (0) > 1.

Now as you well know 

exp(x) - 1 is related to tetration base eta ( e^(1/e) )

as a small example.

The idea is that the super of f(x) does not have the annoying case above with the repeated derivatives values.

So we could demand or investigate bernstein function solutions ( all derivatives > 0 ) for the superfunctions at a larger domain/range. 

For instance

f(x) = 2 exp(x) 

F(0) = 0

F(x+1) = 2 exp(F(x))

and then 

D^n F(x) >= 1

for x > 0


I need to think more about it, but I wanted to share this.

Notice

2 exp(x) - 2 exp(- 3/5 x) can be used as replacement for the 3/5 method.

and the same idea for the gaussian method.


regards

tommy1729