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Equations for Kneser sexp algorithm
#9
(08/12/2010, 04:43 AM)sheldonison Wrote: ... the sexp(z) is about 100-200x more accurate than the RiemannCircle function from which it is generated.
correction The RiemannCircle(z) function is 100-200x more accurate than the sexp(z) from which it is generated -- however this data does include the renormalize update in the code, which improved convergence. In this case, all of the improvement appears to come from the Riemann mapping. But, as you'll see farther down, there is another alternative without the renormalize step where more of the improvement comes from the sexp(z) step!

Here is a table, showing how binary bits precision improves with each iteration. The initial seed has an accuracy of about 4 bits binary. The accuracy is for the lowest term in the Taylor series. The next term is about the same, for both Taylor series, and accuracy for higher order terms is more accurate (at least for awhile). I could plot it. Anyway, the point is the Riemann series is 7 or 8 bits more accurate than the sexp(z) series from which it is generated. The sexp(z) series has roughly the same accuracy as the Riemann function from which it is generated.
Riemann -8.41108211297959 bits
sexp(z) -9.11708328076014 bits
Riemann -15.1121338561072 bits
sexp(z) -15.1316170140021 bits
Riemann -22.4179232976446 bits
sexp(z) -22.4897567784208 bits
Riemann -30.0155349055434 bits
sexp(z) -30.1087094975908 bits
Riemann -37.7126066320302 bits
sexp(z) -37.8144609513923 bits
Riemann -45.4478768923269 bits
sexp(z) -45.5533186457503 bits
Riemann -53.2001314165620 bits
sexp(z) -53.3071463323265 bits
Riemann -60.9582476200045 bits
sexp(z) -61.0659125192666 bits
Riemann -68.7204813782899 bits
sexp(z) -68.8284373037337 bits
Riemann -76.4826669744581 bits
sexp(z) -76.5907565309966 bits
Riemann -84.2469508431293 bits
sexp(z) -84.3550685136293 bits
Riemann -92.0030319102508 bits
sexp(z) -92.1112561892320 bits

And here is a second table, without the renormalization step. The renormalize step is after the sexp(z) series is generated. It multiplies the sexp(z) series by whatever constant is required to force sexp(0.5)=exp(sexp(-0.5)). After 13 iterations, accuracy is at 69 bits, (as opposed to 92 bits). Here's what's interesting to me. This is a more elegant loop, without the somewhat ugly renormalize step. This leads to a more accurate sexp(z) series, but leads to less improvement overall, in the Riemann mapping, with a little singularity since sexp(0.5)<>exp(sexp(-0.5).

Overall, this approach took 17 or 18 iterations until convergence stopped at 92 bits, as opposed to 13 iterations. So, anyway, that's the data. Both routines are working to improve the results together.

Riemann -8.41108211297959
sexp(z) -9.11708328076014
Riemann -13.5881879738804
sexp(z) -17.1061334888054
Riemann -18.8839530119948
sexp(z) -21.4385286084149
Riemann -24.1419288537565
sexp(z) -26.8704129838725
Riemann -29.4092009637620
sexp(z) -32.1372761148881
Riemann -34.6784789286960
sexp(z) -37.4290603754555
Riemann -39.9512192183627
sexp(z) -42.7181894098719
Riemann -45.2270233243075
sexp(z) -48.0041816783103
Riemann -50.5061031443367
sexp(z) -53.3029348163809
Riemann -55.7895507636155
sexp(z) -58.6076934808030
Riemann -61.0771911987248
sexp(z) -63.9108585125742
Riemann -66.3693959721491
sexp(z) -69.2206441976381
Riemann -71.6653883035104
sexp(z) -74.5285659987165
Riemann -76.9672004092254
sexp(z) -79.8616209389427
Riemann -82.2744841556731
sexp(z) -85.1839830576327
Riemann -87.5851241316643
sexp(z) -90.5153257252998
Riemann -92.8193397830429
(convergence stops improving here)

Some other comments. The sexp(z) isn't even a superfunction. Its an approximation, with its accuracy as a superfunction decaying as the radius increases. Its most important property is that sexp(z) is real valued at the real axis.

Contrast that with the RiemannCircle function which is guaranteed by construction to be a true superfunction, with f(z+1)=exp(f(z)) for all z with imag(z)>=idelta. At idelta, it is the best superfunction matching sexp(z). But we only kept terms from the Taylor series, not the Laurent series, so it is not equal to sexp(z). It has guarenteed convergence for all imag(z)>=idelta.
- Sheldon
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Messages In This Thread
RE: Equations for Kneser sexp algorithm - by sheldonison - 08/13/2010, 07:39 PM

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