# Tetration Forum

Full Version: Semi-exp and the geometric derivative. A criterion.
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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