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

[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