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
