09/19/2017, 09:45 PM
The so-called geometric derivative from " non-newtonian calculus " : basically just f*(x) = exp( ln ' (f(x)) ) = exp( f ' (x) /f(x) ) plays a role for a criterion for the half-iterate of exp ( semi-exp ).
Notice that (b^x)* = b.
So in a sense it gives the average base locally.
Since semi-exp grows slower than any b^x for b>1 , it makes sense that the base Goes down towards 1 as x grows.
And this decrease should be smooth.
So the semi-exp should satisfy
For x > 1 and n a nonnegative integer.
S ' (x) > 1
S " (x) > 0
x < S(x) < exp(x)
Sign ( D^n S*(x) ) = (-1)^n
These 4 conditions may reduce to 3 or 2, I have not looked into it.
Notice I did not include analytic !
Also I did not say this gives uniqueness.
But I think these conditions are intresting.
And I think They have a solution.
I particular I conjecture that the semi-exp computed with my 2sinh method satisfies all 4 conditions !!
I tried to Find a proof but with no succes.
I mentioned all this before ( apart from the conjecture about my 2sinh method ), and I might not be the first, but Im bringing this back in the spotlight.
Your ideas ?
Regards
Tommy1729
Notice that (b^x)* = b.
So in a sense it gives the average base locally.
Since semi-exp grows slower than any b^x for b>1 , it makes sense that the base Goes down towards 1 as x grows.
And this decrease should be smooth.
So the semi-exp should satisfy
For x > 1 and n a nonnegative integer.
S ' (x) > 1
S " (x) > 0
x < S(x) < exp(x)
Sign ( D^n S*(x) ) = (-1)^n
These 4 conditions may reduce to 3 or 2, I have not looked into it.
Notice I did not include analytic !
Also I did not say this gives uniqueness.
But I think these conditions are intresting.
And I think They have a solution.
I particular I conjecture that the semi-exp computed with my 2sinh method satisfies all 4 conditions !!
I tried to Find a proof but with no succes.
I mentioned all this before ( apart from the conjecture about my 2sinh method ), and I might not be the first, but Im bringing this back in the spotlight.
Your ideas ?
Regards
Tommy1729