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regular sexp: curve near h=-2 (h=-2 + eps*I)
#1
Just to have a nice picture for the vizualization of that unusual thing "complex heights" I made the following graph with our beloved base b=sqrt(2), where the imaginary part of the height parameter increases from 0 to 2 Pi /log(log(2)).
The latter means simply to rotate the value of the scröder-function for some real h, say, if we get for x0=1 the schröder-value s(x0)= s(1) = -0.632, then we use the rotation of this value in the complex plane by a parameter k in the following way s(1)(k) = -0.632 * exp(i*2*Pi*k/64) for k=0..64 and plug that value into the inverse schröder-function to get the iterate of desired complex height.
This gives ellipses for -2<real(h)< inf , and if the imaginary part of the h- value is not zero we can even plot for real(h) = -3, real(h) = -4 , ... real(h)-> - inf
This gives the following curves for some h with real heights and k=0..64

   

Then I was interested in the behaviour of the curve where the real part is -2. Does it diverge to imaginary +- infinity? Does it converge to real(-inf)? Or will the two parts of the curve converge to parallels of the x-axis (=constant imaginary part)?
I found the result surprising: not only, that it seems, they converge to an imaginary part of +-c where c = Pi/log(2), but also, that in the plot with exponentially scaled x-axis the stepwidth approximates to a constant value.
The "exponentially scaled" x-axis means simply, that the (negative) x-coordinates go into the exponent in b^x, so for x-> -inf I get b^x->0 instead.
In the following graph I took the very small stepwidth of k/2^32 for the imaginary part of h - and whatever small stepwidth I use, it seems always, that the curve approximates that parallels to the x-axis with distant of c.

   

Hmm. I'm not familiar with things like Riemann-sphere, but I remember something like if we take the plot from the euclidean plane to the surface of a sphere, then the infinities meet at one point. Would such translation be useful here too?

Gottfried
Gottfried Helms, Kassel
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#2
Didnt Jay D already show similar pictures here?
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#3
(03/09/2010, 10:33 PM)bo198214 Wrote: Didnt Jay D already show similar pictures here?

Yes - thanks! Only that they are much more informative and much much nicer... Smile
I liked them very much, when i saw them first, but wouldn't have realized the relation to my own question in focus...

Especially that one, of which mine is just a rough&small part:
[Image: attachment.php?aid=143]

It seems, that my rescaling of the x-axis to exp(real(f(k))), which maps all on the way to the negative infinity to a finite open interval - to the left. To have the same to the right I'd do a quadratic scaling - something like exp(- real(f(k))^2 ) or so... let's see...



----

Hey Lord, won't you buy me ... a new directory structure for our forum ... some treeview were nice...
Gottfried Helms, Kassel
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