eta as branchpoint of tetrational
#3
(06/02/2011, 12:51 PM)bo198214 Wrote: Lets see what happens, with the fixpoints \( M^\pm(b) \) when moving on the circle \( b=\beta(t)=\eta+ (\eta-\sqrt{2})e^{\pi i t} \) for \( -1<t <1 \) in the next post.

Ok, this happens when moving the base in the lower halfplane \( t\in (-1,0) \). The blue curve is the movment of the upper fixpoint \( M^+(\beta(t)) \) and the red one is the lower fixpoint \( M^-(\beta(t)) \). At t=0 we have the two complex fixpoints with real part around 2.5.
   

And this happens when moving the base in the upper halfplane \( t\in (0,1) \):
   

The funny thing is, the curves indeed seem to lie on a circle with center 3 and radius 1.
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Messages In This Thread
eta as branchpoint of tetrational - by mike3 - 06/02/2011, 01:55 AM
RE: eta as branchpoint of tetrational - by mike3 - 06/03/2011, 10:57 PM
RE: eta as branchpoint of tetrational - by mike3 - 06/04/2011, 09:08 AM
RE: eta as branchpoint of tetrational - by mike3 - 06/04/2011, 09:50 AM

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