05/25/2014, 03:00 PM
(This post was last modified: 05/25/2014, 03:40 PM by sheldonison.)

(05/23/2014, 10:53 PM)tommy1729 Wrote: I anticipated your approximation for the zeros of the asymptotic function, in terms of Kneser's half exponential.A proof requires an analytic function, with a definition. Right now, all I have is a series of constructive algorithms to generate an asymptotic half Taylor series, along with some estimates for the error terms. Perhaps one can formalize a definition in terms of how well f(f(z)) converges to exp(z), or to the Kneser half iterate. The best algorithm is Version IV, which uses a Cauchy circle integral approximation of a function with a fairly large discontinuity at the negative real axis, which causes us to carefully choose where to evaluate the Cauchy integral to minimize the numeric impact of the jump discontinuity at the negative real axis. A picture showing the discontinuity at the negative real axis is at the end of this post.

The Weierstrass product form does not only need a computation but also a proof ofcourse. Too illustrate why, I gave those counterintuitive functions as examples.

this a_n is the most accurate entire asymptotic Taylor series! b_n are the coefficients of the Kneser Taylor series at h_n, and for the definite integral, x=h_n

Now from the Kneser function itself, one can generate a Weiestrass zero approximation half iterate, which represents yet another version, it appears to be not quiet as good as version IV, but this is primarily because even though the approximation for the zeros, as compared to version IV, is very good, for the larger zeros, is is not for the smaller zeros. This approximation gives the 10th zero as -2623.212352003 as compared with -2623.216046490, for version IV, from my previous post. I am having trouble formalizing why this approximation for the zeros would approximately match the zeros of the version IV approximation, even though it does quite well. This approximation says the zeros are where the real part of the Kneser half iterate at the negative axis goes to zero.

equation used for zeros of a Weiestrass half iterate approximation

The basic problem is how to remove the singularities and the the discontinuity at the negative real axis for the Kneser half iterate. Below, I show the Kneser half iterate, from -10 to 10, with grids every 2 units, singularities at L and L* with cutpoints on a circle of abs(L), extended on the real axis to minus infinity; this is different than the normal way to draw the function with the cutpoints extended so the function is real valued at the real axis. Relatively speaking, this version of the Kneser half iterate is much larger in absolute magnitude at the positive reals, than at the negative reals, which is what the approximations are all based on. Version IV does "Cacuhy type integrals" on this function at carefully chosen radii to minimize the discontinuity. The Weiestrass approximation takes the negative reals of this function as the zeros of an entire function. Meanwhile, I have a new version I am experimenting with that will greatly reduce the discontinuity at the negative real axis, that may allow approximating both the Weiestrass zeros, and the error term for the convergence to the Kneser half iterate... but there will still be a discontinuity cut point at |L|; still working.