10/06/2007, 09:18 AM
Even if we choose a function that is completely symmetric at 
The regular iterations at both fixed points dont coincide. They have the difference:
Details: If we have a function)
that is symmetric at the y-Axis then we can make a function )
out of it, which is symmetric at the straight line =-x)
by the following procedure:
\circ (\text{id}-f)^{-1})
.
This roughly corresponds to rotating the function graph by 45 degrees anticlockwise. The property of being symmetric at can be expressed by =x)
. Directly translated it means mirror the function at the y-Axis then mirror it at 
(function inversion) and then mirror it at the x-Axis. The result of these three mirrorings is a mirroring at and this should not change anything. With some arithmetic you can indeed verify that .
The current graph resulted from letting=\frac{1}{4}x^2-\frac{1}{4})
and presents 
with the fixed points -1 and 1.
The regular iterations at both fixed points dont coincide. They have the difference:
Details: If we have a function
This roughly corresponds to rotating the function graph by 45 degrees anticlockwise. The property of being symmetric at
The current graph resulted from letting