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**

[leon]

The iterates tan(expression + P) and all others that stack variables into functionless categories are a good example that proves both methods are better than the iterates that don't have like umpteen sums in the formulas.Analytic tetration would be missing all of the steps and procedures if more fundamental tetration. And sexp is like Schroeder-Cauchy inversion and inflection more than it operates on P or reconciles the inequality without composite function logic or super function strings. 

This makes analytically tetrating a bicomposite derivative bigger than two TRI-composites and derivatives a convergent and divergent enough approach to integrate variables more than sums of decomposed functions or super functions, and can give three riemanns sums per derivative that are bigger than another three or two that make the hyper-infinite derivation and functioning possible or not (like a simultaneous hyper-operator

So if this was a good technique what would you say about it vs the newest pentations and higher order hyperoperations? And the lower? Using two at the same time can beat some higher order that could diverge or differ more than derivative functions and methods

About the other variable, the derivatives add up and make two magic rieman sums that are bigger than 2 more derivative ones. (  So how does tetrating or hyperooerating so fast you could be doing two hyperoperations at the same time (but aren't) show two Riemannian sums that differ enough from 2 others? So, one hyo with a bico <bicomposite> or two tricos or one quadco (quadcomoosite) or two of these at the same time? In/outside an anything? Anymore more derivative than x gets Cauchy for gaussian bucks and super function bicos

Can cauchys and riemanns that make xs that can operate on it octocomposite function use two riemmans or two super functions to operate or hyperoperate on two tricos one tricos or one bico where everything isn't derivative from two bigger riemanns?

The way to say this is if a function and it's antiderivatives can Cauchy a bigger function it can make a riemanns sum that beats two tricos.

See the two Riemann sums bigger than the other two in and out of everything more derivative than x  A Schroeder equation is a backwards Cauchy on big enough things so check the ariemann sums like a quadco or to bicos and it will beat the trico like a Schroeder equation too Riemann sums bigger than the other two in and out of everything more derivative than x

[tommy1729]

The iterates tan(expression + P) and all others that stack variables into functionless categories are a good example that proves both methods are better than the iterates that don't have like umpteen sums in the formulas.Analytic tetration would be missing all of the steps and procedures if more fundamental tetration. And sexp is like Schroeder-Cauchy inversion and inflection more than it operates on P or reconciles the inequality without composite function logic or super function strings. 

This makes analytically tetrating a bicomposite derivative bigger than two TRI-composites and derivatives a convergent and divergent enough approach to integrate variables more than sums of decomposed functions or super functions, and can give three riemanns sums per derivative that are bigger than another three or two that make the hyper-infinite derivation and functioning possible or not (like a simultaneous hyper-operator

So if this was a good technique what would you say about it vs the newest pentations and higher order hyperoperations? And the lower? Using two at the same time can beat some higher order that could diverge or differ more than derivative functions and methods

About the other variable, the derivatives add up and make two magic rieman sums that are bigger than 2 more derivative ones. (  So how does tetrating or hyperooerating so fast you could be doing two hyperoperations at the same time (but aren't) show two Riemannian sums that differ enough from 2 others? So, one hyo with a bico <bicomposite> or two tricos or one quadco (quadcomoosite) or two of these at the same time? In/outside an anything? Anymore more derivative than x gets Cauchy for gaussian bucks and super function bicos

Can cauchys and riemanns that make xs that can operate on it octocomposite function use two riemmans or two super functions to operate or hyperoperate on two tricos one tricos or one bico where everything isn't derivative from two bigger riemanns?

The way to say this is if a function and it's antiderivatives can Cauchy a bigger function it can make a riemanns sum that beats two tricos.

See the two Riemann sums bigger than the other two in and out of everything more derivative than x  A Schroeder equation is a backwards Cauchy on big enough things so check the ariemann sums like a quadco or to bicos and it will beat the trico like a Schroeder equation too Riemann sums bigger than the other two in and out of everything more derivative than x

euh what ?

regards

tommy1729

[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

[leon]

I was saying that you need to slope the function to heteilromorphically meet x like in the middle then to get bigger and could use an tan(derivative) that goes to infinity to decrease the nonfunctional area slope and asymptote region to determine a variable P like x this would make the function simple and tell you that x gets greater as it goes to infinity as P goes to smaller infinity

The sexp and derivative of sexp have similar enough slopes this time to say that about P and backderive in the sets of p and P and x like if you knew where x and p we're going

That should be easier to see since the derivative x is a locus

[tommy1729]

Ok here is an idea for uniqueness or at least putting solutions in category.

f(x) is an analytic function on [0,oo] such as analytic semiexp(x) or tet(x), so the idea is general.

More specific

Let f^{n} mean n th derivative of f.

say f(z) satisfies 

f(0) = 0

f(1) = 1

f(x+1) = exp(f(x))

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


then 

TR( f(x) ) = integral from 1 to infinity  sum_(n>=0)  (f^{n}(x))^(-2) dx

LET Min TR( f(x) , c ) = G©

where the minimum is for a fixed c , and f(x) is picked to get a minimum value.  

Then we wonder what G© is as a function of c.

And how many analytic solutions f(x) exist with the same G© value.

We can further wonder what value of c makes G© at minimum.


Notice that TR can diverge for non tetration like functions.

Adding a weight might help there.


the details for systematic adding weights needs to be investigated.


regards

tommy1729

[leon]

Ok here is an idea for uniqueness or at least putting solutions in category.

f(x) is an analytic function on [0,oo] such as analytic semiexp(x) or tet(x), so the idea is general.

More specific

Let f^{n} mean n th derivative of f.

say f(z) satisfies 

f(0) = 0

f(1) = 1

f(x+1) = exp(f(x))

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


then 

TR( f(x) ) = integral from 1 to infinity  sum_(n>=0)  (f^{n}(x))^(-2) dx

LET Min TR( f(x) , c ) = G©

where the minimum is for a fixed c , and f(x) is picked to get a minimum value.  

Then we wonder what G© is as a function of c.

And how many analytic solutions f(x) exist with the same G© value.

We can further wonder what value of c makes G© at minimum.


Notice that TR can diverge for non tetration like functions.

Adding a weight might help there.


the details for systematic adding weights needs to be investigated.


regards

tommy1729

Good; how functional the system looks because tetration makes big things look small 
Not bad but not tremendous; how slow the variables transpose in a mental equation (with two other things in mind)
overall: a good way to look at it. Helps to find out how much stuff there is more rapidly and easily