Editing Iterated function (section)

==Fractional iterates and flows, and negative iterates==
[[File:TrivFctRootExm svg.svg|thumb|{{color|#20b080|''g'': '''R'''→'''R'''}} is a trivial functional 5th root of {{color|#901070|2=''f'': '''R'''<sup>+</sup>→'''R'''<sup>+</sup>, ''f''(''x'') = sin(''x'')}}. The computation of ''f''({{frac|π|6}}) = {{frac|1|2}} = ''g''<sup>5</sup>({{frac|π|6}}) is shown.]]
The notion {{math|''f''{{i sup|1/''n''}}}} must be used with care when the equation {{math|1=''g''<sup>''n''</sup>(''x'') = ''f''(''x'')}} has multiple solutions, which is normally the case, as in [[Functional square root|Babbage's equation]] of the functional roots of the identity map. For example, for {{math|1=''n'' = 2}} and {{math|1=''f''(''x'') = 4''x'' − 6}}, both {{math|1=''g''(''x'') = 6 − 2''x''}} and {{math|1=''g''(''x'') = 2''x'' − 2}} are solutions; so the expression {{math|''f''<sup> 1/2</sup>(''x'')}} does not denote a unique function, just as numbers have multiple algebraic roots. A trivial root of ''f'' can always be obtained if ''f''{{'}}s domain can be extended sufficiently, cf. picture. The roots chosen are normally the ones belonging to the orbit under study.

Fractional iteration of a function can be defined: for instance, a [[functional square root|half iterate]] of a function {{mvar|f}} is a function {{mvar|g}} such that {{math|1=''g''(''g''(''x'')) = ''f''(''x'')}}.<ref>{{cite web |work=MathOverflow |title=Finding f such that f(f(x))=g(x) given g |url=https://mathoverflow.net/q/66538 }}</ref> This function {{math|''g''(''x'')}} can be written using the index notation as {{math|''f''<sup> 1/2</sup>(''x'')}} . Similarly, {{math|''f''<sup> 1/3</sup>(''x'')}} is the function defined such that {{math|1=''f''<sup>1/3</sup>(''f''<sup>1/3</sup>(''f''<sup>1/3</sup>(''x''))) = ''f''(''x'')}}, while {{math|''f''{{i sup|2/3}}(''x'')}} may be defined as equal to {{math|''f''{{i sup| 1/3}}(''f''{{i sup|1/3}}(''x''))}}, and so forth, all based on the principle, mentioned earlier, that {{math|1=''f''<sup> ''m''</sup> ○ ''f''<sup> ''n''</sup> = ''f''<sup> ''m'' + ''n''</sup>}}. This idea can be generalized so that the iteration count {{mvar|n}} becomes a '''continuous parameter''', a sort of continuous "time" of a continuous [[Orbit (dynamics)|orbit]].<ref>{{cite journal |first1=R. |last1=Aldrovandi |first2=L. P. |last2=Freitas |title=Continuous Iteration of Dynamical Maps |journal=J. Math. Phys. |volume=39 |issue=10 |pages=5324 |year=1998 |doi=10.1063/1.532574 |arxiv=physics/9712026 |bibcode=1998JMP....39.5324A |hdl=11449/65519 |s2cid=119675869 |hdl-access=free }}</ref><ref>{{cite journal |first1=G. |last1=Berkolaiko |first2=S. |last2=Rabinovich |first3=S. |last3=Havlin |title=Analysis of Carleman Representation of Analytical Recursions |journal=J. Math. Anal. Appl. |volume=224 |pages=81–90 |year=1998 |doi=10.1006/jmaa.1998.5986 |doi-access=free }}</ref>

In such cases, one refers to the system as a [[flow (mathematics)|flow]] (cf. section on [[#Conjugacy|conjugacy]] below.)

If a function is bijective (and so possesses an inverse function), then negative iterates correspond to function inverses and their compositions. For example, {{math|''f''<sup> −1</sup>(''x'')}} is the normal inverse of {{mvar|f}}, while {{math|''f''<sup> −2</sup>(''x'')}} is the inverse composed with itself, i.e. {{math|1=''f''<sup> −2</sup>(''x'') = ''f''<sup> −1</sup>(''f''<sup> −1</sup>(''x''))}}. Fractional negative iterates are defined analogously to fractional positive ones; for example, {{math|''f''<sup> −1/2</sup>(''x'')}} is defined such that {{math|1=''f''<sup> −1/2</sup>(''f''<sup> −1/2</sup>(''x'')) = ''f''<sup> −1</sup>(''x'')}}, or, equivalently, such that {{math|1=''f''<sup> −1/2</sup>(''f''<sup> 1/2</sup>(''x'')) = ''f''<sup> 0</sup>(''x'') = ''x''}}.

=== 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.<ref>{{cite web |title=Tetration.org |url=https://tetration.org/index.php/Fractional_Iteration }}</ref>

# First determine a fixed point for the function such that {{math|1=''f''(''a'') = ''a''}}.
# Define {{math|1=''f'' <sup>''n''</sup>(''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.
# Expand {{math|''f''<sup>''n''</sup>(''x'')}} around the fixed point ''a'' as a [[Taylor series]], <math display="block">
f^n(x) = f^n(a) + (x-a)\left.\frac{d}{dx}f^n(x)\right|_{x=a} + \frac{(x-a)^2}2\left.\frac{d^2}{dx^2}f^n(x)\right|_{x=a} +\cdots
</math>
# Expand out <math display="block">
f^n(x) = f^n(a) + (x-a) f'(a)f'(f(a))f'(f^2(a))\cdots f'(f^{n-1}(a)) + \cdots
</math>
# Substitute in for {{math|1=''f{{i sup|k}}''(''a'') = ''a''}}, for any ''k'', <math display="block">
f^n(x) = 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
</math>
# Make use of the [[geometric progression]] to simplify terms, <math display="block">
f^n(x) = 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
</math> There is a special case when {{math|1=''f'' '(a) = 1}}, <math display="block">
f^n(x) = x + \frac{(x-a)^2}2(n f''(a))+ \frac{(x-a)^3}6\left(\frac{3}{2}n(n-1) f''(a)^2 + n f'''(a)\right)+\cdots
</math>
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 {{math|''f''(''x'') {{=}} ''Cx'' + ''D''}} gives the fixed point {{math|''a'' {{=}} ''D''/(1 − ''C'')}}, so the above formula terminates to just
<math display="block">
f^n(x)=\frac{D}{1-C} + \left(x-\frac{D}{1-C}\right)C^n=C^nx+\frac{1-C^n}{1-C}D ~,
</math>
which is trivial to check.

====Example 2====
Find the value of <math>\sqrt{2}^{ \sqrt{2}^{\sqrt{2}^{\cdots}} }</math> where this is done ''n'' times (and possibly the interpolated values when ''n'' is not an integer). We have {{math|1=''f''(''x'') = {{sqrt|2}}<sup>''x''</sup>}}. A fixed point is {{math|1=''a'' = ''f''(2) = 2}}.

So set {{math|1=''x'' = 1}} and {{math|''f'' <sup>''n''</sup> (1)}} expanded around the fixed point value of 2 is then an infinite series,
<math display="block">
\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
</math>
which, taking just the first three terms, is correct to the first decimal place when ''n'' is positive. Also see [[Tetration]]: {{math|1=''f'' <sup>''n''</sup>(1) = <sup>''n''</sup>{{sqrt|2}}}}. Using the other fixed point {{math|1=''a'' = ''f''(4) {{=}} 4}} causes the series to diverge.

For {{math|1= ''n'' = −1}}, the series computes the inverse function {{sfrac|2|ln ''x''|ln 2}}.

====Example 3====
With the function {{math|''f''(''x'') {{=}} ''x''<sup>''b''</sup>}}, expand around the fixed point 1 to get the series
<math display="block">
f^n(x) = 1 + b^n(x-1) + \frac{1}2b^{n}(b^n-1)(x-1)^2 + \frac{1}{3!}b^n (b^n-1)(b^n-2)(x-1)^3 + \cdots ~,
</math>
which is simply the Taylor series of ''x''<sup>(''b''<sup>''n''</sup> )</sup> expanded around 1.