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**
