Iterating at fixed points of b^x
#22
If we consider \( f(z)=b^z \) then \( \lim_{n\to\infty}f^{\circ n}(z) \) is either the lower fixed point of \( b^z \) or \( \infty \) if we start at a non fixed point \( z \).
This is because all fixed points of \( b^z \) are repelling except the lower real fixed point (which though only exists for \( b<\eta \)). All fixed points can be reached via iterating a branch of the logarithm except the lower real fixed point can be reached by iterating exponentiation.

Interestingly the upper real fixed point can be reached by iterating the logarithm even if we start at real \( z \) below the lower real fixed point.

So for the case \( 1<b<\eta \) either
\( f^{\circ n}(z)\to \infty \) or \( f^{\circ n}(z)\to a \) where \( a \) is the lower real fixed point.
However I am not sure in the moment for which area A the first case applies. Surely \( f^{\circ n}(z)\to\infty \) for each \( z>a_2 \) where \( a_2 \) is the upper fixed point and \( f^{\circ n}(z)\to a \) for all other real \( z \).

But for example \( f^{\circ n}(5+i)\to 2 \), but \( f^{\circ n}(5+0.1i)\to\infty \) for \( b=\sqrt{2} \). Jay?
Also we didnt consider yet the case of non-real \( b \).
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Messages In This Thread
Iterating at fixed points of b^x - by bo198214 - 09/08/2007, 10:02 AM
The fixed points of e^x - by bo198214 - 09/08/2007, 10:34 AM
The fixed points of b^x - by bo198214 - 09/08/2007, 11:36 AM
RE: Iterating at fixed points of b^x - by jaydfox - 09/12/2007, 06:23 AM
RE: Iterating at fixed points of b^x - by GFR - 10/03/2007, 11:03 PM
RE: Iterating at fixed points of b^x - by bo198214 - 10/04/2007, 01:30 PM
RE: Iterating at fixed points of b^x - by GFR - 01/31/2008, 03:07 PM

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