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How it looks (i.θ)ₐ
#17
(04/26/2015, 12:50 AM)marraco Wrote: I don't know on the complex field, but on the real line your code seems to produce very good results for many bases a<e^(e^-1)

For example, here is the comparison with results obtained with Excel for base a=1.3:
Maracco
I looked up my old kneser.gp code for bases<eta. The results are very accurate (32 decimal digits by default); wherever it doesn't blow up and always very accurate, out to about 0.15*I less than the imaginary Period2 of the real axis for bases<eta. If you want the entire complex plane, the workaround is trivial. Instead of sexp(z), use superf2(z) instead. Its really hard to explain what its doing, and why it blows up in the complex plan if imag(z) is large enough, but I'll try to explain it. It is actually generating a Kneser mapping from the repelling fixed point to the attracting fixed point! For b=sqrt(2), the repelling fixed point is 4 and the attracting fixed point is 2, so it is basically generating a Kneser mapping, starting from the repelling fixed point of 4; and then winding up exactly with the Koenig's solution for the attracting fixed point of 2! This was an interesting intellectual exercise at the time, and I suppose it is cool, but it also mostly causes confusion now, when you try to use sexp(20*I) or something like that and the theta approximation series causes an overflow. Anyway, because its a slightly different kind of Kneser mapping, the theta approximation no longer goes to a constant as imag(z) goes to infinity, unlike the theta approximation for bases>eta, and instead. The theta approximation for bases<eta is exactly like a Laurent series instead of a Taylor series, where a Laurent series blows up at some point as you approach the origin, which is at imaginary infinity.

But the workaround is trivial. Instead of sexp(z), use superf2(z). The results will be nearly exactly same every where sexp(z) doesn't blow up (at the real axis for example). And superf2(z) for bases<eta is accurate everywhere in the complex plane, since it just uses the Schroder/Koenig's solution.
- Sheldon
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Messages In This Thread
How it looks (i.θ)ₐ - by marraco - 04/18/2015, 11:20 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 02:40 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/19/2015, 08:40 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 11:19 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 02:35 AM
RE: How it looks (i.θ)ₐ - by Gottfried - 04/20/2015, 07:53 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 02:34 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 03:20 AM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/21/2015, 08:23 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 04:52 PM
RE: How it looks (i.θ)ₐ - by JmsNxn - 04/23/2015, 11:15 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/23/2015, 11:20 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/26/2015, 12:50 AM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/26/2015, 05:08 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 03:46 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 02:21 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/25/2015, 07:52 PM



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