(09/08/2022, 06:03 PM)bo198214 Wrote: Just to test my sage-lib:
let \(a(x)=-x+x^4+x^5\) then \(f(x)=a(a(x))=x-2x^5+O(x^6)\), \({\rm valit}[f]=4\) and we have again the 4 4th roots:
\begin{align}
f^{\circ \frac{1}{4}|1}&=x - \frac{1}{2} x^{5} - x^{7} + \frac{1}{4} x^{8} - \frac{5}{8} x^{9} + \frac{3}{2} x^{10} - \frac{17}{2} x^{11} - \frac{17}{16} x^{12} - \frac{209}{16} x^{13} + 22 x^{14} - \frac{429}{8} x^{15} + \frac{147}{8} x^{16} - \frac{31917}{128} x^{17} + \frac{2407}{16} x^{18} - \frac{4057}{8} x^{19}+O(x^20)\\
f^{\circ \frac{1}{4}|2}&=i x + 2 i x^{3} + \left(-\frac{1}{2} i + \frac{1}{2}\right) x^{4} + \frac{11}{2} i x^{5} + \left(-3 i + 4\right) x^{6} + \left(-\frac{15}{2} i - 1\right) x^{7} + \left(-16 i + \frac{101}{4}\right) x^{8} + \left(-\frac{1513}{8} i - 12\right) x^{9} + \left(\frac{43}{4} i + 101\right) x^{10} + \left(-\frac{18907}{12} i - \frac{211}{2}\right) x^{11} + \left(\frac{12329}{16} i + \frac{669}{8}\right) x^{12} + \left(-\frac{406943}{48} i - \frac{1119}{2}\right) x^{13} + \left(\frac{52139}{6} i - \frac{40595}{12}\right) x^{14} + \left(-\frac{2215209}{80} i - \frac{1901}{2}\right) x^{15} + \left(\frac{2798725}{48} i - \frac{1946047}{48}\right) x^{16} + \left(\frac{25537919}{640} i + \frac{130513}{6}\right) x^{17} + \left(\frac{7666655}{32} i - \frac{6096887}{20}\right) x^{18} + \left(\frac{3365536581}{2240} i + \frac{3905543}\\{12}\right) x^{19}+O(x^{20})\\
f^{\circ \frac{1}{4}|3}&=- x + x^{4} + \frac{1}{2} x^{5} - x^{7} + \frac{9}{4} x^{8} + \frac{5}{8} x^{9} + \frac{11}{2} x^{10} - \frac{21}{2} x^{11} + \frac{133}{16} x^{12} - \frac{367}{16} x^{13} + \frac{187}{2} x^{14} - \frac{699}{8} x^{15} + \frac{699}{4} x^{16} - \frac{82707}{128} x^{17} + \frac{17417}{16} x^{18} - \frac{11755}{8} x^{19}+O(x^{20})\\
f^{\circ \frac{1}{4}|4}&=-i x - 2 i x^{3} + \left(\frac{1}{2} i + \frac{1}{2}\right) x^{4} - \frac{11}{2} i x^{5} + \left(3 i + 4\right) x^{6} + \left(\frac{15}{2} i - 1\right) x^{7} + \left(16 i + \frac{101}{4}\right) x^{8} + \left(\frac{1513}{8} i - 12\right) x^{9} + \left(-\frac{43}{4} i + 101\right) x^{10} + \left(\frac{18907}{12} i - \frac{211}{2}\right) x^{11} + \left(-\frac{12329}{16} i + \frac{669}{8}\right) x^{12} + \left(\frac{406943}{48} i - \frac{1119}{2}\right) x^{13} + \left(-\frac{52139}{6} i - \frac{40595}{12}\right) x^{14} + \left(\frac{2215209}{80} i - \frac{1901}{2}\right) x^{15} + \left(-\frac{2798725}{48} i - \frac{1946047}{48}\right) x^{16} + \left(-\frac{25537919}{640} i + \frac{130513}{6}\right) x^{17} + \left(-\frac{7666655}{32} i - \frac{6096887}{20}\right) x^{18} + \left(-\frac{3365536581}{2240} i + \frac{3905543}{12}\right) x^{19}+O(x^{20})\\
a^{\circ \frac{1}{2}|1} &= f^{\circ \frac{1}{4}|2}\\
a^{\circ \frac{1}{2}|2} &= f^{\circ \frac{1}{4}|4}
\end{align}
this is completely logical , almost trivial :
near fixpoint 0 , for small x and n > 2 , we have
f(x) = a x + b x^n + O(x^(n+1))
Notice the other fixed points are always a nonzero distance away from 0.
so if say a is 2 we are nearly solving
y = 2 x.
and hence the hyperbolic method works by the approximation y = 2x near 0.
AND we then get a unique solution.
Now if a = 1 we get a different situation
f(x) = x + b x^n + ...
so we are nearly solving
y = x + b x^n
or
y = x ( 1 + b x^(n-1) )
y and x are close to 0 and close to eachother so
we are basicly solving
y = ( 1 + b x^(n-1) )
or
0 = 1 + b x^(n-1)
so we get (n-1) local solutions near the fixpoint 0.
and we can use those truncated (n-1) local solutions as asymptotics in our limit formulas for continu iterations , in the same way as we did with hyperbolic.
***
confused ?
ok look at it geometrically
x + x^4 + ... = x ( 1 + x^3 + ...) = id(x) + x^4 + ...
SO near the point 0 we get 3 solutions or said differently , 3 branches for the functional inverse.
so we have a riemann surface ( for the functional inverse ) with 3 layers near 0.
if it is analytic at 0 OR!! near 0 then by analytic continuation and the principles of riemann surfaces , the riemann surface will REMAIN to have 3 layers.
So for t element of ]0,1] we must also have 3 solutions for
f^[t](x) for x near 0.
***
yes your examples had -1 instead of 1 but its the same principle.
The higher terms are irrelevant because they vanish as x gets close to 0.
To avoid confusion i mean they vanish for a given analytic near 0 case f(x), I did mean vanish for the divergeant taylor or non-existant taylor of fractional derivates near 0. AND that is all that is required.
***
keep in mind , i said n > 2.
because n - 1 is 1 when n = 2 but that is invalid and absurd because then we get another linear.
Also keep in mind that I am considering x in the direction re(x) > 0.
This matters for the amount of petals.
for x = -1 we might not get solutions for half-iterates agreeing with x = 0,1. but at x = -1 there might be a petal that works.
This suggests - by symmetry * re(x) > 0 or re(x) < 0 * - that the number of petals when n > 2 is equal to 2*(n-1).
***
im wondering if real-analytic f(x) such that
f(x) = x + b x^n + b_2 x^(n+1) + ...
where b_i are nonnegative.
and g(g(x)) = f(x) such that g(x) is the real half-iterate for x > 0 but not analytic at 0 ,
gives rise to the taylor coefficients g_m of g(x) growing as (m!)^(n-1) IFF some growth condition is given for b_i.
regards
tommy1729