# 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 edit

Fractional iteration is called an -iterate of iff ^{[3]}

```
```

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*^{⅓}(*f*^{⅓}(*f*^{⅓}(*x*))) = *f*(*x*), while *f*^{ ⅔}(*x*) may be defined equal to *f*^{ ⅓}(*f*^{ ⅓}(*x*)), and so forth, all based on the principle, mentioned earlier, that *f*^{ m}○*f*^{ 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 f, 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 *f*^{1/n} must be used with care when the equation *g*^{n}(*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*) = 4*x*−6, both *g*(*x*) = 6−2*x* and *g*(*x*) = 2*x*−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? edit

"Short answer: it depends.

- 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.

- Given a function which is
**not continuous or differentiable**, it is possible to interpolate between iterates of the function, but there are many more possible ways of doing this." tetration forum FAQ by Henryk Trappman Andrew Robbins January 12, 2008

# Examples edit

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

Solutions :

- the closed-form solution
- expansion

### c=0 edit

Find a continous function ^{[4]}

```
```

satisfying

```
```

for all .

In other words find fractional iteration from functional equation

```
```

Lets try :

```
```

so

```
```

then

```
```

and

```
```

Square root of 2 is irrational number.

```
```

using "powers via logarithm"

```
```

for each real number *s*.

```
```

### c = −2 , description by G A Edgar edit

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

near

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 t**here is a **closed form** known,

[Of course, is essentially the double-angle formula for cosines.]

And of course in the case c = 0 the closed form is

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

.....

"

## Some formulas for fractional iteration edit

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,

**(4)** Expand out

**(5)** Substitute in for *f ^{k}(a)*=

*a*, for any

*k*,

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

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

**(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 edit

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

which is trivial to check.

### Example 2 edit

Find the value of where this is done *n* times (and possibly the interpolated values when *n* is not an integer). We have *f*(*x*)=Template:Sqrt^{x}. 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,

which, taking just the first three terms, is correct to the first decimal place when *n* is positive—cf. Tetration: *f* ^{n}(1) = ^{n}Template: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 ln*x*/ln2.

### Example 3 edit

With the function *f(x)* = *x*^{b}, expand around the fixed point 1 to get the series

which is simply the Taylor series of *x*^{(bn )} expanded around 1.

# See also edit

- of half-iterates of f(x) = x² + 1/4 by Gottfried Helms, 2012-10-10
- tetration forum : Iteration exercises: f(x)=x^2 - 0.5 ; Fixpoint-irritation...
- math.stackexchange question : half-iterate-of-x2c
- http://math.stackexchange.com/questions/911818/how-to-obtain-fx-if-it-is-known-that-ffx-x2x
- wikipedia : Carleman_matrix
- Neumann–Neumann methods
- w:Superfunction
- w:Infinite compositions of analytic functions
- math stackexchange: square-root-of-a-function-in-the-sense-of-composition: X^2+1