Fractional Composite of Functions

Solution 1:

I put a number of relevant articles at http://zakuski.utsa.edu/~jagy/other.html I tell people to start with the obituary of Baker.

The short version is this:

• if your function has no fixpoints, you can mostly manage this by Kneser's method.
• If there is just one fixpoint,
  • and the modulus of the derivative there is not $1,$ fairly easy by Schroder's equation.
  • However, if the derivative there is $1,$ it is a big song and dance invented by Ecalle in the 1970's. I have only second-hand accounts, but enough to apply, see https://mathoverflow.net/questions/45608/formal-power-series-convergence including my own answer.

Solution 2:

Functional square roots are relatively easy to find through Functional Conjugacy specifications of generic iterated functions; find the general iterate, and set n=1/2 in the general expression, such as the ones available and listed there.

For example, the functional square root of the nth Chebyshev polynomial $T_n$ is evidently $\cos (\sqrt{n} \arccos (x)) $ --- which, however, is not a polynomial, in general.

As Will Jagy already mentioned, in general, around a fixed point, say $f(0)=0$, the standard tried-and-true method is Schröder's equation, $\psi(f(x))=f'(0) ~\psi(x)$, a triumph of functional conjugacy, provided f ' there ≠1. Having found ψ, one has $f^{1/2}(x)=\psi^{-1}\left (\sqrt{f'(0)} ~ \psi(x)\right )$.

If it is =1, at the fixed point, however, all is not lost, and conjugacy still delivers: Curtright, Jin, & Zachos, JouPhys A Math-Th 44.40 (2011): 405205 illustrates how $f^n \circ g \circ f^{-n}$ will improve an initial approximant g rapidly and dramatically for large n.

(This is illustrated there for intuitive functions such as $f=\sin x$, blue, around the origin, where the half iterate is orange, the second iterate is red, and so on...)

Solution 3:

Just some simple examples, allowing much obvious/natural versions of a fractional iterate:

• Let $f(x)=x+a$ then $f(f(x))=x+2a$ , $f(f(f(x))) = x+3a$ and in general $f°^h(x)=x+h\cdot a$. Then $f°^{0.5} (x) = x+0.5 a$

• Let $f(x)=x \cdot a$ then $f(f(x))=x \cdot a^2$ , $f(f(f(x))) = x \cdot a^3$ and in general $f°^h(x)=x \cdot a^h $ . Then $f°^{0.5} (x) = x \cdot a^{0.5} $

• Let $f(x)=x ^ a$ then $f(f(x))=x ^{ a^2}$ , $f(f(f(x))) = x ^ { a^3}$ and in general $f°^h(x)=x ^{ a^h} $ . Then $f°^{0.5} (x) = x ^ {\sqrt {a}} $

If a function has a power series without constant term , then you can find a formal power series for a fractional iterate. If $f(x) = ax + O(x^2)$ and $0<a<1$ then sometimes that series for fractional iterates might have a nonzero radius of convergence. If the power series has a constant term, one can sometimes use conjugacy to find a formal power series. More on this using the method of Carleman-matrices, see wikipedia:Carleman-matrix. For the specific case of $f(x) = a^x$ see wikipedia:tetration also there is something in wikipedia:hyperoperations .
Also note that a collection of Q&A about this subject is in MSE (tag::tetration, tag::hyperoperation) and in MO tag::fractional-iteration.

A very involved discussion can be found at tetration-forum: http://math.eretrandre.org/tetrationforum and many more links can be found using google-search. In mathoverflow you can find this question with answers: MO1 , MO2 , MO3