02/23/2023, 12:49 PM
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
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
