Editing Iterated function (section)

=== 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.