# Fractals/Iterations in the complex plane/fractional-iterations

Fractional iteration doesn't have a unique value. For example the square root of 4 is 2 or -2.[1][2]

# Definition

Fractional iteration ${\displaystyle \phi }$  is called an ${\displaystyle {\frac {m}{n}}}$ -iterate of ${\displaystyle f}$  iff [3]

 ${\displaystyle \phi ^{n}(x)=f^{m}(x)}$


for instance, a half iterate ( = functional square root) of a function f is a function g such that

g(g(x)) = f(x).


This function g(x) can be written using the index notation as

 f ½(x) .


Similarly,

 f ⅓(x)


is the function defined such that f(f(f(x))) = f(x), while f(x) may be defined equal to f(f(x)), and so forth, all based on the principle, mentioned earlier, that f mf n = f m + n. This idea can be generalized so that the iteration count n becomes a continuous parameter, a sort of continuous "time" of a continuous orbit.

In such cases, one refers to the system as a flow, specified by Schröder's equation.

Negative iterates correspond to function inverses and their compositions. For example, f −1(x) is the normal inverse of Template:Mvar, while f −2(x) is the inverse composed with itself, i.e. f −2(x) = f −1(f −1(x)). Fractional negative iterates are defined analogously to fractional positive ones; for example, f −½(x) is defined such that f − ½(f −½(x)) = f −1(x), or, equivalently, such that f −½(f ½(x)) = f 0(x) = x.

The notion f1/n must be used with care when the equation gn(x) = f(x) has multiple solutions, which is normally the case, as in Babbage's equation of the functional roots of the identity map. For example, for n = 2 and f(x) = 4x−6, both g(x) = 6−2x and g(x) = 2x−2 are solutions; so the expression f ½(x) doesn't denote a unique function, just as algebraic roots of numbers are multiple. The issue is quite similar to division by zero. The roots chosen are normally the ones belonging to the orbit under study.

## Question. It is possible to iterate a function non-integer times?

• Given a differentiable function with a formal power series, and its power series can be used as a means to study its iterates and interpolate between them, and in some cases, this

interpolation also produces an differentiable function as well.

# Examples

## Solutions of g(g(z) = z^2+c

Solutions :

• the closed-form solution
• expansion

### c=0

Find a continous function [4]

 ${\displaystyle g:\mathbb {C} \to \mathbb {C} }$


satisfying

 ${\displaystyle g(g(z))=z^{2}}$


for all ${\displaystyle z\in \mathbb {C} }$ .

In other words find fractional iteration ${\displaystyle g(z)=f^{\frac {1}{2}}(z)}$  from functional equation

 ${\displaystyle g(g(z))=f(z)=z^{2}}$


Lets try :

 ${\displaystyle g(z)=z^{s}}$


so

 ${\displaystyle g(g(z))=(z^{s})^{s}=z^{s*s}}$


then

 ${\displaystyle s^{2}=2}$


and

 ${\displaystyle s={\sqrt {2}}=1.414\ 213\ 562\ 373\ 095\ 048\ 801\ 688\ 724\ 209\ \ldots }$


Square root of 2 is irrational number.

 ${\displaystyle f^{\frac {1}{2}}(z)=g(z)=z^{s}=z^{\sqrt {2}}}$


using "powers via logarithm"

 ${\displaystyle z^{s}=(e^{\ln z})^{s}=e^{s\cdot \ln z}}$


for each real number s.

 ${\displaystyle \ln(a+b\imath )=\ln \left(|z|\exp ^{\imath \varphi }\right)=\ln |z|+\ln \exp ^{\imath \varphi }=\ln |z|+\imath \varphi }$


### c = −2 , description by G A Edgar

https://people.math.osu.edu/edgar.2/preprints/trans_frac/fractional.pdf "Fractional Iteration of Series and Transseries" by G. A. Edgar .

" 6. Julia Example

As an example we will consider fractional iterates for the function

${\displaystyle M(x)=x^{2}+c}$

near ${\displaystyle x=+\infty }$

Of course, positive integer iterates of this function are used for construction of Julia sets or the Mandelbrot set. For the theory of real transseries to be applicable, we must restrict to real values c. But once we have nice formulas, they can be investigated for general complex c.

In the case c = −2 there is a closed form known,

${\displaystyle M^{[s]}=2cosh(2^{s}acosh(x/2))}$

[Of course, ${\displaystyle x^{2}-2=2cosh(2acosh(x/2))}$  is essentially the double-angle formula for cosines.]

And of course in the case c = 0 the closed form is ${\displaystyle M^{[s]}=x^{2^{s}}}$

For other values of c no closed form is known, and it is likely that there is none (but that must be explained).

.....

${\displaystyle M^{\frac {1}{2}}=2cosh({\sqrt {2}}\ acosh({\frac {x}{2}}))}$

"

## Some formulas for fractional iteration

One of several methods of finding a series formula for fractional iteration, making use of a fixed point, is as follows.

(1) First determine a fixed point for the function such that f(a)=a .

(2) Define f n(a)=a for all n belonging to the reals. This, in some ways, is the most natural extra condition to place upon the fractional iterates.

(3) Expand f n(x) around the fixed point a as a Taylor series,

${\displaystyle f^{n}(x)=f^{n}(a)+(x-a){\frac {d}{dx}}f^{n}(x)|_{x=a}+{\frac {(x-a)^{2}}{2!}}{\frac {d^{2}}{dx^{2}}}f^{n}(x)|_{x=a}+\cdots }$

(4) Expand out

${\displaystyle f^{n}\left(x\right)=f^{n}(a)+(x-a)f'(a)f'(f(a))f'(f^{2}(a))\cdots f'(f^{n-1}(a))+\cdots }$

(5) Substitute in for f k(a)= a, for any k,

${\displaystyle f^{n}\left(x\right)=a+(x-a)f'(a)^{n}+{\frac {(x-a)^{2}}{2!}}(f''(a)f'(a)^{n-1})\left(1+f'(a)+\cdots +f'(a)^{n-1}\right)+\cdots }$

(6) Make use of the geometric progression to simplify terms,

${\displaystyle f^{n}\left(x\right)=a+(x-a)f'(a)^{n}+{\frac {(x-a)^{2}}{2!}}(f''(a)f'(a)^{n-1}){\frac {f'(a)^{n}-1}{f'(a)-1}}+\cdots }$

(6b) There is a special case when f '(a)=1,

${\displaystyle f^{n}\left(x\right)=x+{\frac {(x-a)^{2}}{2!}}(nf''(a))+{\frac {(x-a)^{3}}{3!}}\left({\frac {3}{2}}n(n-1)f''(a)^{2}+nf'''(a)\right)+\cdots }$

(7) When n is not an integer, make use of the power formula y n = exp(n ln(y)).

This can be carried on indefinitely, although inefficiently, as the latter terms become increasingly complicated.

A more systematic procedure is outlined in the following section on Conjugacy.

### Example 1

For example, setting f(x) = Cx+D gives the fixed point a = D/(1-C), so the above formula terminates to just

${\displaystyle f^{n}(x)={\frac {D}{1-C}}+(x-{\frac {D}{1-C}})C^{n}=C^{n}x+{\frac {1-C^{n}}{1-C}}D~,}$

which is trivial to check.

### Example 2

Find the value of ${\displaystyle {\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{\cdots }}}}$  where this is done n times (and possibly the interpolated values when n is not an integer). We have f(x)=Template:Sqrtx. A fixed point is a=f(2)=2.

So set x=1 and f n (1) expanded around the fixed point value of 2 is then an infinite series,

${\displaystyle {\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{\cdots }}}=f^{n}(1)=2-(\ln 2)^{n}+{\frac {(\ln 2)^{n+1}((\ln 2)^{n}-1)}{4(\ln 2-1)}}-\cdots }$

which, taking just the first three terms, is correct to the first decimal place when n is positive—cf. Tetration: f n(1) = nTemplate:Sqrt . (Using the other fixed point a = f(4) = 4 causes the series to diverge.)

For n = −1, the series computes the inverse function, 2 lnx/ln2.

### Example 3

With the function f(x) = xb, expand around the fixed point 1 to get the series

${\displaystyle f^{n}(x)=1+b^{n}(x-1)+{\frac {1}{2!}}b^{n}(b^{n}-1)(x-1)^{2}+{\frac {1}{3!}}b^{n}(b^{n}-1)(b^{n}-2)(x-1)^{3}+\cdots ~,}$

which is simply the Taylor series of x(bn ) expanded around 1.