Hi Sheldon -
thanks for the hint to the tanh-function. I'll give it a try!
No, the "dual" is just the iterate with an appropriate (purely) imaginary height (which -in the case we work using such a function - changes the sign of the schröder-function-value), where the result is again real. if \( K(h,x) \) would denote the Kneser-method iteration from x using height h, then in our case with base = sqrt(2), it should be \( h=\pi i/ \log( \log(2)) \) . It's similar to take the purely imaginary with a multiple of pi as exponent of the exponential-function to arrive at the negative part of the number line .
And the result should be real and near the "dual" taken by the regular iteration.
Gottfried
thanks for the hint to the tanh-function. I'll give it a try!
(06/24/2013, 04:09 PM)sheldonison Wrote:(06/23/2013, 10:20 PM)Gottfried Wrote: The last point (the only one which I cannot answer myself, perhaps you can look at it): Can we look at your Kneser-method what the dual of, say x_0 = 1 or x_0 = 0 or x_0 = -infty were? I think we need only the appropriate imaginary iteration height to compute the respectively duals. Levenstein- numbers? ;-)What is the definition/equation for a Kneser solution dual? The Kneser solution is not periodic. Would the dual of -infinity, which is sexp(-2) be sexp(2)?
- Sheldon
No, the "dual" is just the iterate with an appropriate (purely) imaginary height (which -in the case we work using such a function - changes the sign of the schröder-function-value), where the result is again real. if \( K(h,x) \) would denote the Kneser-method iteration from x using height h, then in our case with base = sqrt(2), it should be \( h=\pi i/ \log( \log(2)) \) . It's similar to take the purely imaginary with a multiple of pi as exponent of the exponential-function to arrive at the negative part of the number line .
And the result should be real and near the "dual" taken by the regular iteration.
Gottfried
Gottfried Helms, Kassel
