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