Difference between revisions of "Iteration of fractional linear maps"
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$L_{1,2}=-\frac{d-a}{2c} \pm \sqrt{\left(\frac{d-a}{2c}\right)^2 + \frac{b}{c}}$, or for later use: | $L_{1,2}=-\frac{d-a}{2c} \pm \sqrt{\left(\frac{d-a}{2c}\right)^2 + \frac{b}{c}}$, or for later use: | ||
− | $L_{1,2}=\ | + | $L_{1,2}=\Big(a_2-d \pm \underbrace{\sqrt{1-\frac{ad-bc}{a_2^2}}}_{=:q}\Big)/c$, where $a_2=\frac{a+d}{2}$. |
== Derivative at the fixpoints == | == Derivative at the fixpoints == |
Revision as of 13:32, 1 May 2013
Our concern here are functions of the form $f(z)=\frac{az+b}{cz+d}$ with $c\neq 0$.
Fixpoints
In the case $c\neq 0$, the fixpoints are given by
$L_{1,2}=-\frac{d-a}{2c} \pm \sqrt{\left(\frac{d-a}{2c}\right)^2 + \frac{b}{c}}$, or for later use:
$L_{1,2}=\Big(a_2-d \pm \underbrace{\sqrt{1-\frac{ad-bc}{a_2^2}}}_{=:q}\Big)/c$, where $a_2=\frac{a+d}{2}$.
Derivative at the fixpoints
The derivative of the fractional linear map $f$ is:
$f'(z)=\frac{ad-bz}{(cz+d)^2}$
If we plug in $z=L_1$ we get:
$f'(L_1) = \frac{ad-bz}{(a_2(1+q))^2} = \frac{ad-bz}{a_2^2}/\left(1+q\right)^2 = \frac{1-q^2}{\left(1+q\right)^2} = \frac{1-q}{1+q}$
$f'(L_2) = \frac{1+q}{1-q} = 1/f'(L_1)$.
Schröder coordinate at the fixpoints
$\sigma_k(f(z))=f'(L_k) \sigma(z)$, $\sigma_k$ is unique up to a multiplicative constant.
$\sigma_1(z) = \frac{z-L_1}{z-L_2}$.
$\sigma_1\left(\frac{az+b}{cz+d}\right) = \frac{(a-cL_2)z + b-dL_2}{(a-cL_1)z + b -dL_2} = \frac{a_2(1-q)z+}{a_2(1+q)z + }$.