Thread Rating:
  • 2 Vote(s) - 4 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Searching for an asymptotic to exp[0.5]
I think I'm very close to the ideal definition for the entire half iterate, using Kneser's sexp(z). This version should also be robust enough to pin down all the zeros as well, for the Weierstrass infinite product form.
(05/29/2014, 11:09 PM)sheldonison Wrote: Well, version V works a lot lot better, but it I was hoping for something with more theoretical power. So there's this discontinuity for Kneser's half iterate at the negative real axis. Why not get rid of it, with some sort of mapping that still converges to the half iterate as real(z) increases? Well, I got rid of most of it; not all. Here halfk refer's to Kneser's half iterate...

Here is the algorithm for the version V half iterate approximation.

The new version, call it version VI, says as |z| gets arbitrarily large, there is some optimal number of terms, k, for the following series, where k gets arbitrarily large as |z| gets arbitrarily large, and where the error term gets arbitrarily small.

The error term is and we choose k to give the smallest error term. For various values of k, the error term will decrease for awhile, before hitting a minimum, and then start increasing. The previous algorithm, half(v), arbitrarily used k=1, which is pretty darn accurate all by itself.

Now, take the Cauchy circle integral at radius r=|z|, to define all of the Taylor series coefficients for the entire half iterate. The conjecture is that as z gets arbitrarily large, all of the Taylor series coefficients will converge, and in fact, the Taylor series I posted for version V, is pretty much the same as you would get for version VI.

For example, if we were interested in calculating the 100th Taylor series coefficient for the previous version V, we would have used a radius~=exp(24.72). Version V has an imaginary offset of -6iE143, at negative number, -exp(24.72). At the positive number exp(24.72), the value is ~=2E334, so the 100th derivative can already be calculated accurate to 190 decimal digits using version V, in spite of the large imaginary error term. But version VI allows calculate all of the first 100 derivatives to a theoretical precision of 1000 decimal digits at this radius. For version VI, at a radius of exp(24.72), the ideal value for k to use is k=7, and the imaginary offset at -exp(24.72) shrinks to -3iE-1039, which is a truly tiny number. This extra precision using k=7 allows calculating all of the first 100 Taylor series coefficients accurate to nearly a 1000 digits. Not that version V was all that inaccurate to begin with; the lowest precision Taylor series coeffient for version V was a0, which was accurate to around 14 or 15 decimal digits.

Actually, the function has a Laurent series, but we throw away the Laurent coefficients as unhelpful for the entire half iterate approximation. The Laurent series needs to be taken into account when comparing the half(VI) approximation with the entire function generated using the Taylor series generated from that approximation.

Also, as a practical matter, it is difficult to calculate the lower Taylor series coefficients at the larger radii, due to precision errors. But if infinite precision were used, they would converge (This is a conjecture), so that VI defines all of the Taylor series coefficients for the entire half iterate asymptotic exactly.
- Sheldon

Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 07/12/2014, 07:46 PM

Possibly Related Threads...
Thread Author Replies Views Last Post
  Merged fixpoints of 2 iterates ? Asymptotic ? [2019] tommy1729 1 1,135 09/10/2019, 11:28 AM
Last Post: sheldonison
  Another asymptotic development, similar to 2sinh method JmsNxn 0 2,808 07/05/2011, 06:34 PM
Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)