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Searching for an asymptotic to exp[0.5]
#43
(07/01/2014, 10:10 PM)tommy1729 Wrote: ...
That settles the case if that assumption t_n > n^A is true.

However Im not sure if sheldon or anyone else has a proof of that assumption WITHOUT assumptions about the Hadamard product for the fake exp^[1/2] ... to avoid circular reasoning.

It seems the attention is now 100% at the zero's as far as product expansions are concerned.
edit fixed lots of typos, getting back into "half iterate" mode; sorry. The zeros of the asymptotic of the half iterate of sexp(z) are on the negative real axis, approximately where, for negative values of z, using the Kneser half iterate, where we follow the branch counterclockwise around and above the fixed point, towards the negative real axis, where it is complex valued instead of real valued, and where the real part of that number is zero.
approximate zeros of the entire half iterate, for negative values of z
halfk(z)=exp(halfk(log(z)))
log(-z)=log(|z|)+pi i
if Re(exp(z))=0, then Im(z)=(n+0.5) pi i, this is for n>=0

So understanding the zeros of the entire half iterate of the negative axis behaves turns out to be equivalent to understanding how the half iterate of (log(|z|)+pi i) behaves. This approximation gets arbitrarily accurate as z gets larger, based on empirical results. Also, using this branch of the Kneser half iterate for the negative real axis, we can say that the entire half iterate
only for negative values of z

This ratio also gets arbitrarily accurate, approaching arbitrarily close to 2. This might be justifiable/proveable based on the algorithm used for version V, my best entire half iterate approximation.

Here is a sample calculation of the approximate fourth zero of the entire half iterate, compared with the actual zero of the entire asymptotic. The approximation says we want imag(halfk(log(z)))=3.5pi, where halfk is the Kneser half iterate. This is true for z~=-43.661481, where halfk(log(-43.66148105371))=halfk(3.776466272986+pi i)=
3.390550517856 + 10.99557428756i, which has imag part=3.5pi. The actual third "zero" of the entire half iterate is at z=-43.661461, just a tiny bit away from this approximation.

For the 10th zero, the approximation gives z=-2623.212352003, which is accurate to 1.5E-14. I will repost an updated list of these zeros; the old list in post#28, uses a less accurate asymptotic half iterate.
- Sheldon
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RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 07/01/2014, 11:41 PM

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