03/01/2012, 03:47 PM
(This post was last modified: 03/01/2012, 03:49 PM by sheldonison.)

(03/01/2012, 12:04 PM)Kouznetsov Wrote: Beautiful job, Sheldon!Thanks Dimitrii. Yes, my evaluations confirm such a result, but with caveat. Consider only the solutions I plotted for imag(b)>0, rotating counter clockwise, in a unit half circle with radius=0.25 around base eta=exp(1/e), and concatenate them with their imaginary conjugate solutions, rotating clockwise around base eta. This generates a full 360 degree unit circle with radius=0.25. Now, we have two different solutions for base=, depending on whether or not the solution was generated rotating clockwise, or rotating counter clockwise. One solution is the complex conjugate of the other solution. But in fact, these two solutions are nearly identical! In fact, using the default precision for my program, \p 28, gives results accurate to only about 12 decimal digits. The imaginary component of the solution is for all intents and purposes, zero. This is somewhat analogous to the paper that you and Henryk authored, about the multiple solutions to base=sqrt(2).

If you put points uniformly at the circle around exp(1/e),

then with few points, you can get a good precision evaluating the derivatives of tet_b(z) with respect to b by the Cauchi contour integral. So. you can plot, for example,

d tet_b(z) / db,

d^2 tet_b(z) / db^2,

as functions of z, and make conclusion about the behavior in vicinity of b=exp(1/e).

I try to guess the result:

As b approaches exp(1/e), the quasiperiods become large; so, all the cut lines at the z plane are far away from zero (and out of the field of your pics). In this sense, in vicinity of b=exp(1/e), there should be no cutlines at all, so, b=exp(1/e) is regular point for the most of values of z. Do your evaluations allow to confirm or to refute such a guess?

So we need to use much higher precision to see what the resulting taylor series really is, so we can know how large the discontinuity is at the real axis. So we switch to \p 67, and regenerate the result, which is now accurate to nearly 32 decimal digits. Now, I display 32 decimal digits of the first 32 taylor series terms, and we can clearly see the imaginary component of the taylor series is not equal to zero, but it is small. This implies that a cauchy integral for the taylor series coefficients at this radius would be accurate to about 12 or so decimal digits, which is what I reported in our earlier email. If the radius is smaller, than the imaginary component gets arbitrarily small, as the period goes to infinity as the base approaches eta.

But, as I have shown, there are valid solutions continuing around the unit circle, past Pi in either direction! These solutions may become more and chaotic as we rotate around the unit circle. More investigation is required.

Code:

`base = eta-0.25 = 1.1946678610097661336583391085964`

a0= 1.0000000000000000000000000000002 - 2.5045231329042313021714830574624 E-33 *I

a1= 0.38496815270132386352241683306460 + 2.9620626097717511899678016191509 E-15*I

a2= -0.30638546356948817486767367830749 - 2.6869898290903338543289980018355 E-12*I

a3= 0.17137553259854622103441222382774 + 4.2539626191909924099462858637538 E-12*I

a4= -0.078842783289222721712390323634160 + 5.2707372816265845032792519650466 E-12*I

a5= 0.033397834492629539091982856016043 - 1.1834996632986949273365960313094 E-11*I

a6= -0.014062642712262839852834627395487 - 1.0368427952696666700362295909949 E-12*I

a7= 0.0060726830996950253025451542948419 + 1.1849869471152561434791390649593 E-11*I

a8= -0.0026850556844815823250514434924228 - 3.7602323779180141105383026598257 E-12*I

a9= 0.0012040486094043950995855470134962 - 5.3158941236931785524929546755397 E-12*I

a10= -0.00054474814804614731040426857600308 + 3.1944313543928145461027272093334 E-12*I

a11= 0.00024832392231804542064225683981716 + 1.1076037258793794680406808795498 E-12*I

a12= -0.00011400411390499149013484464217955 - 1.2224736525685550155452450286042 E-12*I

a13= 0.000052675540606501631486599712359474 - 3.1538787710070020583821885985944 E-14*I

a14= -0.000024475226952919505782311190678321 + 2.7875704395526464117303918113594 E-13*I

a15= 0.000011427590750796733085407589555046 - 4.2995304405247787558561937922391 E-14*I

a16= -0.0000053584356003719399501796824446649 - 4.1002066255892877710071670543552 E-14*I

a17= 0.0000025221448337440401171565499390582 + 1.2948108547854285868732829804724 E-14*I

a18= -0.0000011911750001530884957050568945427 + 3.7704568003050785102096244717384 E-15*I

a19= 0.00000056429134007987082478685580515917 - 2.1444364924137420108000782393165 E-15*I

a20= -0.00000026805421022158351352476965081551 - 1.4207109662028524318097129012894 E-16*I

a21= 0.00000012764981127354399472472848135343 + 2.4160910253209088209640235836972 E-16*I

a22= -0.000000060925322495377065126875790768965 - 1.6669522319047266527277839758739 E-17*I

a23= 0.000000029138684091882210545890916875488 - 1.9324014251813970993635668172572 E-17*I

a24= -0.000000013962439484810850056781055681481 + 3.6832378817727104363013010565609 E-18*I

a25= 0.0000000067020194575469197752230803984840 + 1.0388483526899569925433126202680 E-18*I

a26= -0.0000000032221401102863250739766134720131 - 3.8394830991863341982500368103067 E-19*I

a27= 0.0000000015514056726555442749481756970557 - 2.4554286735248275372944881826322 E-20*I

a28= -0.00000000074800071493876431692501570839246 + 2.6920551709726843973680186854492 E-20*I

a29= 0.00000000036110428764854932891958471091099 - 1.7058757358747842185335268010642 E-21*I

a30= -1.7453389675757007922473601336416 E-10 - 1.2990416120980836858358608407113 E-21*I

a31= 8.4451935940173048361164518224634 E-11 + 2.4640693918496310042195760772726 E-22*I

Quote:Is your evaluation for fixed b and various z faster than evaluation for fixed z and varable b?

For a given base, the evaluation of sexp(z) is nearly instantaneous once you have generated the taylor series for the sexp(z) function and the two theta(z) functions, and the two superfunctions. Results are a little slower in the complex plane, for imag(z)>0.7i or <-0.7i, since the algorithm switches to using the superfunction, and the theta(z) function needs to be evluated before the superfunction can be evaluated, and then a few iterations of exp(z) or log(z) are usually required.

It takes much much more time to regenerate these six taylor series for a different base, than it does to regenerate the solution for a different value of z, for the same base.

- Sheldon