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 by the following procedure:

.

This roughly corresponds to rotating the function graph by 45 degrees anticlockwise. The property of being symmetric at can be expressed by . 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 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 that is symmetric at the y-Axis then we can make a function out of it, which is symmetric at the straight line by the following procedure:

.

This roughly corresponds to rotating the function graph by 45 degrees anticlockwise. The property of being symmetric at can be expressed by . 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 and presents with the fixed points -1 and 1.