Given

$$\begin{aligned} \varphi^{-1} (f(\varphi(w))) = w^2 = g(w) \\ \end{aligned}$$

What is $f(z)$ in terms of $g$ ?

Let

$$\begin{aligned} z &= \varphi(w)) \\ \varphi^{-1}(z) &= w \\ \varphi^{-1} (f(\varphi(w))) &= g(w) \\ f(\varphi(w)) &= \varphi(g(w)) \\ \therefore \,\, f(z) &= \varphi(g(\varphi^{-1}(z))) \\ \end{aligned}$$

Why is $\displaystyle \varphi^{-1} \circ f \circ \varphi$ interesting ?

It's rather easy to show that the function $\displaystyle p(z) = \varphi^{-1} \circ f \circ \varphi$ behaves the same as $\displaystyle f(z)$ under iteration. In other words,

$$\begin{aligned} p^n = \varphi^{-1} \circ f^n \circ \varphi \end{aligned}$$

So instead of studying $p$, which can be a bit more complicated, the problem can be reduced to the study of $f$; or vice-versa.

In general, given any quadratic polynomials in $\mathbb{C}$

$$\begin{aligned} p(z) = az^2 + bz + d \\ \end{aligned}$$

Let

$$\begin{aligned} f(z) &= z^2 + c \\ c &= ad + \frac{b}{2} - (\frac{b}{2})^2 \\ \varphi(z) &= az + \frac{b}{2} \\ \end{aligned}$$

It can be readily verified that

$$\begin{aligned} p &= \varphi^{-1} \circ f \circ \varphi \\ f &= \varphi \circ p \circ \varphi^{-1} \\ \end{aligned}$$

Furthermore,

$$\begin{aligned} p^n &= \varphi^{-1} \circ f^n \circ \varphi \\ f^n &= \varphi \circ p^n \circ \varphi^{-1} \\ \end{aligned}$$

More info at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 1:18 AM