(12/23/2014, 11:31 PM)tommy1729 Wrote: ....function
for which every real iterate
" works " is found , though it might not be as nontrivial as mick hoped. ( not saying a nontrivial case cannot exist ).
...
 = x - 2^t / x)
 + 1))
=>
with  + 1)
=> })
=> )
Solve .. => )
Thus :
 = x - 2(x-t) = x - 2x + t = -x + t)
....
Hey Tommy,
Not sure I understood all of that ... But it inspired me to consider the following sequence of functions
...
Does g(x) converge, and is it a solution of interest to Mick? If g(x) converges, and it is analytic, then it has a Taylor/Laurent series....
Update:, by brute force, using a lot of computer cycles to estimate the limit, and then turn the coefficents it back into a fraction with power's of 2's... I get the following Laurent series, as the function that Mick might be looking for.
It would probably be normally expressed as
update2:
This would be compactly expressed via the Abel function as:
And then we get:
Finally, Mick's desired function in closed form would be as follows. With a little algebra, we generate all of the fractional iterates of g(z) as well. Then, using Mick's notation we have the desired g(z,t) function, which has all fractional iterates defined as:
=\; \frac{1}{f(1/z)}\; = \; sqrt{z^2-2t}\; = \; z -<br />
\frac{t}{z} -<br />
\frac{t^2}{2z^{3}} -<br />
\frac{t^3}{2z^{5}} -<br />
\frac{5t^4}{8z^{7}} -<br />
\frac{7t^5}{8z^{9}} -<br />
\frac{21t^6}{16z^{11}} - ... \;\;\;)
for t=1, this is the same as the Laurent series above