CONCLUSION 1 : conditions for nice tetration : second derivative !!

[tommy1729]

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

[tommy1729]

ok the word " flat " I used might be a bit misleading.

But I guess you know what I mean.

if 

f(x + g(x)) = f(x) at a given x  and the approximation there is f(x + g(x)) = f(x + ax + O(x^3) )

and not 

f(x + ax + O(x^2) )

then the words " flat " or linear " come to mind.
Especially when a is or is close to 0.


regards

tommy1729

[tommy1729]

Ok this idea does have issues and I guess I should adress them.

If you take the product formula for the derivative of tet(x) a bit too far , although that has logical justifications (!) , we run into this

tet(-1) = 0

tet(0) = 1

tet ' (0) = tet(0) * tet ' (-1).

But tet(0) = 1 , so

tet ' (0) = 1 * tet ' (-1) = tet ' (-1)

Since tet ' (0) = tet ' (-1) this means the function cannot be convex in [-1,0].

SO sufficiently-differentiable solutions must have an inflection point in [-1,0] !!


This means our unique zero of the second derivative is not so unique afterall.


Or maybe we should set

sexp ' ' ( - 1/2) = 0

to make the Riemann fans happy

I have to think about these issues.

Although these issues are very old ofcourse.

My apologies if I made things sound " completed " , they are not.

However  the remark about the theta function theta(x) has analogues here !

So theta(x) is not so free to choose as you might have believed !


regards

tommy1729

[tommy1729]

perhaps worth mentioning

This seems like a typical problem with f(0) = 1 type functions.

If we study 2 exp(x) iterations we do not get that issue.

Now when you think about it , 

iterations of 2 exp(x) are related to iterations of (e^2)^x.

So a base change.

So maybe for larger bases than e ....

Then again we want to include general ideas for all bases ...


regards

tommy1729

[tommy1729]

bottom line

be careful with your one periodic function.


regards

tommy1729

[tommy1729]

ok in the previous posts I considered 

Let 0 < x <  1

Then 

0 < tet(x-1) < x


and

tet ' (-1) < 1.

This leads to at least 3 inflection points if we make the logical claim tet ' (-1) = tet ' (0).



The opposite situation might be more appealing :

tet ' (-1) > 1
tet ' (0) > 1

tet ' (-1) = tet ' (0)

Now we get that the function tet(x-1) is sometimes above x and sometimes below.

We could have only ONE inflection point now !!

Maybe we should set the inflection point at 1/2.

Then again I do not like that for many reasons , 1/2 seems to high.

Maybe set the inflection point at the intersection of tet(x-1) with x ( for x-1 between -1 and 0 ).
And maybe make that intersection at 1/3 or 1/e.



Where do the tetration methods get their inflection points ?
Where do the tetration methods give a fixpoint of tet(x-1) ?

Let me know what you found or know.

Im going to think about it.

I might add stuff to the gaussian method and 2sinh methods about this and also some error terms.



regards

tommy1729

[tommy1729]

ok in the previous posts I considered 

Let 0 < x <  1

Then 

0 < tet(x-1) < x


and

tet ' (-1) < 1.

This leads to at least 3 inflection points if we make the logical claim tet ' (-1) = tet ' (0).



The opposite situation might be more appealing :

tet ' (-1) > 1
tet ' (0) > 1

tet ' (-1) = tet ' (0)

Now we get that the function tet(x-1) is sometimes above x and sometimes below.

We could have only ONE inflection point now !!

Maybe we should set the inflection point at 1/2.

Then again I do not like that for many reasons , 1/2 seems to high.

Maybe set the inflection point at the intersection of tet(x-1) with x ( for x-1 between -1 and 0 ).
And maybe make that intersection at 1/3 or 1/e.



Where do the tetration methods get their inflection points ?
Where do the tetration methods give a fixpoint of tet(x-1) ?

Let me know what you found or know.

Im going to think about it.

I might add stuff to the gaussian method and 2sinh methods about this and also some error terms.



regards

tommy1729

Now we also get that

d/dx  exp^[n](x) at x = 0 < d/dx  sexp(-1+n)

which makes sense.

exp(exp(x)) or any finite positive !!integer!! amount of exp iterates afterall is slower than tetration


For noninteger iterates then again it is different.

regards

tommy1729