A conjugacy problem

mark

Let [imath]f(x) = x^2 + 1/4[/imath], let [imath]g(x) = x^2+2 x+1/4[/imath], and let [imath]\varphi(x) = x+1[/imath].

• Show that [imath]\varphi[/imath] conjugates [imath]f[/imath] to [imath]g[/imath], i.e. [imath]f\circ \varphi = \varphi \circ g[/imath]
• Suppose I tell you that [imath]f[/imath] has a neutral fixed point at [imath]x=1/2[/imath]. How can you use this information to find a neutral fixed point for [imath]g[/imath]?

theoldernoah

First we must show that [imath] f \circ \varphi = \varphi \circ g [/imath]:

[dmath] f(\varphi(x))= (x+1)^2 + \frac{1}{4} = x^2+2x+1 + \frac{1}{4} = x^2 +2x +\frac{5}{4} [/dmath]

and...

[dmath] \varphi(g(x))= \left(x^2+2x+\frac{1}{4}\right)+1 = x^2+2x+\frac{5}{4}, [/dmath]

therefore [imath]\varphi[/imath] conjugates [imath] f [/imath] to [imath]g[/imath].

Now we must find a neutral fixed point for [imath]g[/imath] given that [imath]f[/imath] has a neutral fixed point at [imath]x=\frac{1}{2}[/imath].

We may need [imath] \varphi^{-1}[/imath], which is simply [imath] x-1 [/imath].

If we then evaluate [imath] \varphi^{-1}(\frac{1}{2})= \frac{1}{2}-1=-\frac{1}{2} [/imath] and then plug our answer into [imath]g'(x)=|2x+2|[/imath] we should get one by the definition of neutral fixed points.

[dmath] g'(-\frac{1}{2})=2\left(-\frac{1}{2}\right) +2 = -1+2=1.[/dmath]

Thus [imath]-\frac{1}{2}[/imath] is a neutral fixed point for [imath]g[/imath].