03/27/2010, 10:16 PM
(01/30/2009, 09:14 AM)Gottfried Wrote: just for my own exercise I've looked at the iteration of f(x) = 1/(1+x).
The interesting thing about fractional linear function is indeed that it does not depend on the fixed point. Particularly interesting does this become with non-real fixed points, like for example:
\( f(z)=\frac{z-1}{z+1} \)
This function has two non-real fixed points: i and -i.
The regular iterations at both fixed points coincide (which is only the case with linear fractions, and we know for example that this is not the case for \( f(z)=e^z \)).
Particularly it is real.
There is a different way to compute the regular iteration.
These linear fractions have the interesting property that they can be represented by its 2x2 matrices, in this case:
\( \begin{pmatrix}1 &-1\\1 &1\end{pmatrix} \)
composition of two linear fractions corresponds to multiplication of their matrices.
And hence we can use here also matrix powers to obtain the regular iteration.
Without making the calculations too explicit, I give the result here:
\( f^{\circ u}(z)=\frac{\cos(\frac{\pi}{4}u)z-\sin(\frac{\pi}{4}u) }{\sin(\frac{\pi}{4}u)z+\cos(\frac{\pi}{4}u)} \)
To optically verify the iteration, I give the graphs for \( u=0\dots 1 \).
If you wonder where this \( \frac{\pi}{4} \) factor comes from, its the angle of the eigenvalue(s) of the matrix, which are in this case \( 1+i \) and \( 1-i \).
