07/28/2021, 12:02 AM
Again I want to give some more details about my thoughts.
considering 2 things.
First a note on functional inverse.
I estimated the bounds of ln ln ... a^b^...(A)
but let B = ln ln ... a^b^...(A)
then A = ln_ln(...)(...ln_ln(b)(ln_ln(a)(exp(exp...exp(B))))))
notice how this gives similar bounds by the same method !!
so the function is bounded and its inverse is also bounded !!
That is a powerful thing !!
Also This is a nice alternative way of looking at it , that might convince ppl who were perhaps confused or arguing " chaos as objection " for one them.
***
The second thing is this thought experiment :
Let abs(s) < sqrt(2)
Then how does ln^[n](exp^[n](s) + C) behave ?
It turns out for real s this converges nice and any real C > 0 this converges fast.
However when s or C are complex things are slightly different !
This is hard to test numerically because of overflow or even with calculus.
But it is clear that when C goes to zero ( as function of n ) FAST ENOUGH , then it converges to s.
so how fast is fast ??
Well for abs(s) < sqrt(2) and sufficiently large n ( usually 2,3 or 4 will already do ) :
ln^[n](exp^[n](s) + exp(- n^2) ) = s + o(exp(- n^2)) + o(exp(- n^2)) i
(if we take the correct branches).
Now this looks familiar not ??
This strenghtens my previous ideas and bound arguments ofcourse.
regards
tommy1729
considering 2 things.
First a note on functional inverse.
I estimated the bounds of ln ln ... a^b^...(A)
but let B = ln ln ... a^b^...(A)
then A = ln_ln(...)(...ln_ln(b)(ln_ln(a)(exp(exp...exp(B))))))
notice how this gives similar bounds by the same method !!
so the function is bounded and its inverse is also bounded !!
That is a powerful thing !!
Also This is a nice alternative way of looking at it , that might convince ppl who were perhaps confused or arguing " chaos as objection " for one them.
***
The second thing is this thought experiment :
Let abs(s) < sqrt(2)
Then how does ln^[n](exp^[n](s) + C) behave ?
It turns out for real s this converges nice and any real C > 0 this converges fast.
However when s or C are complex things are slightly different !
This is hard to test numerically because of overflow or even with calculus.
But it is clear that when C goes to zero ( as function of n ) FAST ENOUGH , then it converges to s.
so how fast is fast ??
Well for abs(s) < sqrt(2) and sufficiently large n ( usually 2,3 or 4 will already do ) :
ln^[n](exp^[n](s) + exp(- n^2) ) = s + o(exp(- n^2)) + o(exp(- n^2)) i
(if we take the correct branches).
Now this looks familiar not ??
This strenghtens my previous ideas and bound arguments ofcourse.
regards
tommy1729

