08/04/2014, 11:49 PM
(This post was last modified: 08/05/2014, 01:41 AM by sheldonison.)
(08/03/2014, 11:38 PM)tommy1729 Wrote: \( f^{o n}(x) \) grows like \( exp(a^{2^n - 1} ln(x)^{2^n} + C) \).
Much much slower then exp or even its half-iterate.
And that for a function that is similarly defined. (second derivative = Original => second difference = Original )
Thats what I meant by counterintuitive.
I noticed the resemblance of the plots with the fake half-iterate exp too.
For the zero's Im convinced they are all real.
I made some numerical approximations, and for big enough values of x, using Jay's Taylor series, I get that \( \ln(f(x)) \approx \frac{(\ln(x))^2}{2\ln(2)} \), which match the equations in Gottfried's link. If ln(x)=1E100, then ln(f(x))~=7.213E199, ln(f(f(x)))~=3.753E399
So then \( \ln(f(f(x))) \approx (\ln(x))^4 \). And yeah, that's growing much slower than
\( \exp^{0.5}(\exp^{0.5}(x))=\exp(x) \;\; \exp(x) \gt \exp((\ln(x))^4) \), if x is big enough.
This is despite many similarities in the complex plane graphs of f(x) and f(f(x)) as compared with the entire asymptotic to \( \exp^{0.5}(x) \).
- Sheldon

