Quadratic function
Iteration

To iterate a function f (x) = a x 2 + b x + c {\displaystyle f(x)=ax^{2}+bx+c}, one applies the function repeatedly, using the output from one iteration as the input to the next.

One cannot always deduce the analytic form of f (n) (x) {\displaystyle f^{(n)}(x)}, which means the n iteration of f (x) {\displaystyle f(x)}. (The superscript can be extended to negative numbers, referring to the iteration of the inverse of f (x) {\displaystyle f(x)} if the inverse exists.) But there are some analytically tractable cases.

For example, for the iterative equation

one has

where

So by induction,

can be obtained, where g (n) (x) {\displaystyle g^{(n)}(x)} can be easily computed as

Finally, we have

as the solution.

See Topological conjugacy for more detail about the relationship between f and g. And see Complex quadratic polynomial for the chaotic behavior in the general iteration.

The logistic map

with parameter 2< r <4 can be solved in certain cases, one of which is chaotic and one of which is not. In the chaotic case r =4 the solution is

where the initial condition parameter θ {\displaystyle \theta } is given by θ = 1 π sin − 1 (x 0 1 / 2) {\displaystyle \theta ={\tfrac {1}{\pi }}\sin ^{-1}(x_{0}^{1/2})}. For rational θ {\displaystyle \theta }, after a finite number of iterations x n {\displaystyle x_{n}} maps into a periodic sequence. But almost all θ {\displaystyle \theta } are irrational, and, for irrational θ {\displaystyle \theta }, x n {\displaystyle x_{n}} never repeats itself – it is non-periodic and exhibits sensitive dependence on initial conditions, so it is said to be chaotic.

The solution of the logistic map when r =2 is

x n = 1 2 − 1 2 (1 − 2 x 0) 2 n {\displaystyle x_{n}={\frac {1}{2}}-{\frac {1}{2}}(1-2x_{0})^{2^{n}}}

for x 0 ∈ [ 0, 1) {\displaystyle x_{0}\in [0,1)}. Since (1 − 2 x 0) ∈ (− 1, 1) {\displaystyle (1-2x_{0})\in (-1,1)} for any value of x 0 {\displaystyle x_{0}} other than the unstable fixed point 0, the term (1 − 2 x 0) 2 n {\displaystyle (1-2x_{0})^{2^{n}}} goes to 0 as n goes to infinity, so x n {\displaystyle x_{n}} goes to the stable fixed point 1 2. {\displaystyle {\tfrac {1}{2}}.}
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