02/22/2023, 10:37 PM

I have been thinking about tetration for a while.

Good ideas, bad ideas, crazy ideas , guessing and confused.

Conjectures, proofs, disproofs, examples.

Now conditions or uniqueness criterions that make tetration " nice " deserve special attention.

Some ideas have gained popularity while others lost them.

And having multiple properties was often desired but not reached or not proven to be true.

So instead of combining many old ideas , (desired) properties and desired criterions I will talk about them seperately.

Im talking about tetration base e here.

Also ofcourse exp^[r](x) for real r > 0 , maps real x to a subset of reals.

The reason I post this thread Is because I became convinced that ONE extra condition RARELY seen is VERY important.

That extra condition saved me from thinking in wrong ways and pessimism.

That extra condition made things logical again when I arrived at weird conclusions.

Now some conditions may overlap and such, I will not go into that here.

So here we go :

** Conditions **

For real x :

basics :

sexp(x) = tet(x)

sexp(-2) = - oo

sexp(-1) = 0

sexp(0) = 1

sexp(1) = e

sexp(x+1) = exp( sexp(x) )

extra :

For real x > -2

d/dx sexp(x) > 0

For real x and real r > 0 :

d/dx exp^[r](x) > 0

(d/dx)^2 exp^[r](x) > 0

***

Now come the important extra ones :

Let 0 < x < 1

Then

0 < sexp(x-1) < x

(d/dx)^2 sexp(x) at x equal to -1 :

sexp ' ' (-1) = 0.

This is very important !!

sexp " (-1) = 0

This makes tetration grow in a logical sense : not sudden acceleration or deceleration.

So the main question is : which solutions have this property sexp " (-1) = 0 ??

Lets investigate further.

We know that sexp ' (x) = sexp(x) * sexp(x-1) * sexp(x-2) * ... * sexp ' (s)

this product cannot go beyond x - n < -2.

In fact not beyond x - n = -1 since sexp(-1) = 0 !!

So we have the continuum sum , well product actually :

for x > 0 :

sexp ' (x) = CP sexp(y) , y from 0 to x.

Now ofcourse the previously mentioned

Let 0 < x < 1

Then

0 < sexp(x-1) < x

( together with sexp ' (x) > 0 )

implies that

0 < sexp ' (-1) < 1

And ofcourse the second derivative also relates to the product and the first derivative.

Going into all the details and consequences would require 1000 posts but let me just make one thing clear :

sexp " (-1) = 0

where -1 is the only 0 in te interval [-2,+oo] is very interesting.

This is the full extra condition this topic was started for.

Let sexp(x + theta(x) ) be an alternative solution to tetration where theta (x) is the typical C^oo one-periodic function.

Now

sexp " (-1) = 0 implies that

sexp " (x + theta(x)) = 0 at x = -1.

So

D^2 sexp (x + theta(x)) =

d/dx [ sexp ' (x + theta(x)) ( 1 + theta ' (x) ) ]

=

sexp " (x + theta(x)) (1 + theta ' (x))^2 + sexp ' (x + theta(x)) theta " (x)

=

sexp " (-1) (1 + theta ' (x))^2 + sexp ' (-1) theta " (x)

=

0 * (1 + theta ' (x))^2 + sexp ' (-1) theta " (x)

=

sexp ' (-1) theta " (x)

=

sexp ' (-1) theta " (-1).

Now since we need to have that the sexp(x + theta(x)) also satisfies tet " (-1) = 0 ;

sexp ' (-1) theta " (-1) = 0

We know sexp ' (-1) =/= 0

so

theta " (-1) = 0

This is periodic so

theta " (-1) = theta " (0) = theta " (1) = ... = 0.

This implies that our theta is very flat near the integer iterations of 0, 1 , e , ...

So our new functions is locally very similar to the old one !!

This implies that

D^2 sexp (x + theta(x)) =

d/dx [ sexp ' (x + theta(x)) ( 1 + theta ' (x) ) ]

=

sexp " (x + theta(x)) (1 + theta ' (x))^2

FOR x = -1, 0 , 1 , 2 , 3 , ...

Ofcourse this has many implications on the complex plane , the continuum sum ( indefinite sum ) and many related tetration topics.

Long story short :

sexp " (-1) = 0

where -1 is the only 0 in te interval [-2,+oo]

FTW

Does the gaussian have this property ?

The 2sinh method ?

Walkers method ?

your own method ??

Ow ofcourse I assumed continu first and second derivatives so the functions are at least C^3.

Also I would like to remind that all the higher derivatives (d/dx)^n sexp(x) for x > -2 can not all be positive due to the log singularities tetration has.

Also I think I can modify functions to satisfy the condition.

( such as the gaussian )

regards

tommy1729

Good ideas, bad ideas, crazy ideas , guessing and confused.

Conjectures, proofs, disproofs, examples.

Now conditions or uniqueness criterions that make tetration " nice " deserve special attention.

Some ideas have gained popularity while others lost them.

And having multiple properties was often desired but not reached or not proven to be true.

So instead of combining many old ideas , (desired) properties and desired criterions I will talk about them seperately.

Im talking about tetration base e here.

Also ofcourse exp^[r](x) for real r > 0 , maps real x to a subset of reals.

The reason I post this thread Is because I became convinced that ONE extra condition RARELY seen is VERY important.

That extra condition saved me from thinking in wrong ways and pessimism.

That extra condition made things logical again when I arrived at weird conclusions.

Now some conditions may overlap and such, I will not go into that here.

So here we go :

** Conditions **

For real x :

basics :

sexp(x) = tet(x)

sexp(-2) = - oo

sexp(-1) = 0

sexp(0) = 1

sexp(1) = e

sexp(x+1) = exp( sexp(x) )

extra :

For real x > -2

d/dx sexp(x) > 0

For real x and real r > 0 :

d/dx exp^[r](x) > 0

(d/dx)^2 exp^[r](x) > 0

***

Now come the important extra ones :

Let 0 < x < 1

Then

0 < sexp(x-1) < x

(d/dx)^2 sexp(x) at x equal to -1 :

sexp ' ' (-1) = 0.

This is very important !!

sexp " (-1) = 0

This makes tetration grow in a logical sense : not sudden acceleration or deceleration.

So the main question is : which solutions have this property sexp " (-1) = 0 ??

Lets investigate further.

We know that sexp ' (x) = sexp(x) * sexp(x-1) * sexp(x-2) * ... * sexp ' (s)

this product cannot go beyond x - n < -2.

In fact not beyond x - n = -1 since sexp(-1) = 0 !!

So we have the continuum sum , well product actually :

for x > 0 :

sexp ' (x) = CP sexp(y) , y from 0 to x.

Now ofcourse the previously mentioned

Let 0 < x < 1

Then

0 < sexp(x-1) < x

( together with sexp ' (x) > 0 )

implies that

0 < sexp ' (-1) < 1

And ofcourse the second derivative also relates to the product and the first derivative.

Going into all the details and consequences would require 1000 posts but let me just make one thing clear :

sexp " (-1) = 0

where -1 is the only 0 in te interval [-2,+oo] is very interesting.

This is the full extra condition this topic was started for.

Let sexp(x + theta(x) ) be an alternative solution to tetration where theta (x) is the typical C^oo one-periodic function.

Now

sexp " (-1) = 0 implies that

sexp " (x + theta(x)) = 0 at x = -1.

So

D^2 sexp (x + theta(x)) =

d/dx [ sexp ' (x + theta(x)) ( 1 + theta ' (x) ) ]

=

sexp " (x + theta(x)) (1 + theta ' (x))^2 + sexp ' (x + theta(x)) theta " (x)

=

sexp " (-1) (1 + theta ' (x))^2 + sexp ' (-1) theta " (x)

=

0 * (1 + theta ' (x))^2 + sexp ' (-1) theta " (x)

=

sexp ' (-1) theta " (x)

=

sexp ' (-1) theta " (-1).

Now since we need to have that the sexp(x + theta(x)) also satisfies tet " (-1) = 0 ;

sexp ' (-1) theta " (-1) = 0

We know sexp ' (-1) =/= 0

so

theta " (-1) = 0

This is periodic so

theta " (-1) = theta " (0) = theta " (1) = ... = 0.

This implies that our theta is very flat near the integer iterations of 0, 1 , e , ...

So our new functions is locally very similar to the old one !!

This implies that

D^2 sexp (x + theta(x)) =

d/dx [ sexp ' (x + theta(x)) ( 1 + theta ' (x) ) ]

=

sexp " (x + theta(x)) (1 + theta ' (x))^2

FOR x = -1, 0 , 1 , 2 , 3 , ...

Ofcourse this has many implications on the complex plane , the continuum sum ( indefinite sum ) and many related tetration topics.

Long story short :

sexp " (-1) = 0

where -1 is the only 0 in te interval [-2,+oo]

FTW

Does the gaussian have this property ?

The 2sinh method ?

Walkers method ?

your own method ??

Ow ofcourse I assumed continu first and second derivatives so the functions are at least C^3.

Also I would like to remind that all the higher derivatives (d/dx)^n sexp(x) for x > -2 can not all be positive due to the log singularities tetration has.

Also I think I can modify functions to satisfy the condition.

( such as the gaussian )

regards

tommy1729