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Figure 2.2.1 and Figure 2.5.1 show that the iterative behavior of the logistic family and the quadratic family are very similar. In a sense, they are identical. We make that notion precise in this section.


Let \(S\) and \(T\) be sets and suppose that \(f:S\to S\) and \(g:T\to T\text{.}\) We say that \(f\) is semi-conjugate to \(g\) if there is a surjective function \(\varphi:T\to S\) such that

\begin{equation*} f\circ \varphi = \varphi \circ g. \end{equation*}

The function \(\varphi\) is called a semi-conjugacy. In the case that \(\varphi\) is bijective, then we say that \(\varphi\) is a conjugacy and that \(f\) and \(g\) are conjugate.

A geo-symbolic way to remember the semi-conjugation formula is in the form of a commutative diagram:

\begin{equation*} \begin{array}[c]{ccc} T&\stackrel{g}{\longrightarrow}&T\\ \Big\downarrow\scriptstyle{\varphi}&&\Big\downarrow\scriptstyle{\varphi}\\ S&\stackrel{f}{\longrightarrow}&S \end{array} \end{equation*}

The formula states if you follow the arrows from the upper left to the lower right in either direction, you get the same result.

An immediate consequence of the definition of conjugacy is

\begin{equation*} f^2\circ \varphi = f\circ f\circ\varphi = f\circ\varphi\circ g = \varphi\circ g\circ g = \varphi\circ g^2 \end{equation*}

and, by induction

\begin{equation*} f^n\circ \varphi = \varphi\circ g^n. \end{equation*}

As a result, if \((t_i)\) is an orbit of \(g\text{,}\) then \((\varphi(t_i))\) is an orbit of \(f\text{.}\)

Generally, the nicer \(\varphi\) is, the closer the relationship between the dynamics of \(f\) and the dynamics of \(g\text{.}\) If \(\varphi\) is bijective, then the relationship is quite close. If \(\varphi\) is continuous with continuous inverse, then topological properties of the orbits will be preserved. If \(S\) and \(T\) are sets of real or complex numbers and \(\varphi(x)=ax+b\text{,}\) then an orbit of one function will be geometrically similar to an orbit of the other. The dynamical systems are truly identical, up to a scaling.


Show that \(f(x) = x^2-1\) is conjugate to \(g(x) = \frac{1}{2}{x^2}+2x-2\) via the conjugacy \(\varphi(x)=\frac{1}{2}x+1\text{.}\)


If you suspect that \(f\) is conjugate to \(g\) via a conjugacy of the form \(\varphi(x)=ax+b\text{,}\) then you can find that conjugacy by setting \(f(\varphi(x)) = \varphi(g(x))\text{.}\) If you compare coefficients, you should get a system of equations that you can solve for \(a\) and \(b\) yielding the conjugacy.


Find a conjugacy of the form \(\varphi(x)=ax+b\) from \(f(x)=x^2-2\) to \(g(x)=4x(1-x)\text{.}\)

Exercise Exercise 4 can be generalized. In fact, the quadratic family for \(-2\leq c \leq 1/4\) is identical to the logistic family for \(1 \leq \lambda \leq 4\text{.}\)


Show that \(f(x)=x^2+(2\lambda-\lambda^2)/4\) is conjugate to \(g(x)=\lambda x(1-x)\) via the conjugacy \(\varphi(x) = -\lambda x + \lambda/2\text{.}\)