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**

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**