05/10/2014, 12:14 PM
(This post was last modified: 05/10/2014, 07:41 PM by sheldonison.)

(05/08/2014, 04:25 PM)sheldonison Wrote:(05/07/2014, 12:22 PM)tommy1729 Wrote: Im searching for an asymptotic to exp[0.5](x)....Emperical testing suggests that an "entire" pseudo half iterate is very likely possible, with all positive Taylor series coefficients at z=0, and a probable growth value of 0.5, as defined by the "growth" equation. .... Each derivative is bounded to a maximum value by a particular value of half(z)....still working on how to formalize the definition of the conjectured Taylor series.

- Sheldon

Here, for a_n, factorial is extended to the reals with the gamma(n+1) function. I do not know what the limiting equation for a_n is as n gets arbitrarily large. These values were generated numerically. With 150 Taylor series terms, this series can accurately generate the half iterate for numbers up to 10^12.

Code:

`a_1= 1.289368074687`

a_2= 2.685542084449

a_3= 4.481892104368

a_4= 6.478743767633

a_5= 8.617330003873

a_6= 10.86883764614

a_7= 13.21549097355

a_8= 15.64480318533

a_9= 18.14765340676

a_10= 20.71664375389

a_11= 23.34605172725

a_12= 26.03107295368

a_13= 28.76759185629

a_14= 31.55218923626

a_15= 34.38183207053

a_16= 37.25395409452

a_17= 40.16617725981

a_18= 43.11650203072

a_19= 46.10306787913

a_20= 49.12423329725

a_21= 52.17844415943

a_22= 55.26438661508

a_23= 58.38072977033

a_24= 61.52640919181

a_25= 64.70028943211

a_26= 67.90141159249

a_27= 71.12888751647

a_28= 74.38183008946

a_29= 77.65944293902

a_30= 80.96100674887

a_31= 84.28582227410

a_32= 87.63323120070

a_33= 91.00261369743

a_34= 94.39338557884

a_35= 97.80499516680

a_36= 101.2369199152

a_37= 104.6886902658

a_38= 108.1598101190

a_39= 111.6498571303

a_40= 115.1583810162

a_41= 118.6850011605

a_42= 122.2293283987

a_43= 125.7909699804

a_44= 129.3696195111

a_45= 132.9648897794

a_46= 136.5764573911

a_47= 140.2040557334

a_48= 143.8473645074

a_49= 147.5060942647

a_50= 151.1799685187

a_51= 154.8686996662

a_52= 158.5720762206

a_53= 162.2898267064

a_54= 166.0217152588

a_55= 169.7675306959

a_56= 173.5270093037

a_57= 177.2999786775

a_58= 181.0862092231

a_59= 184.8855000278

a_60= 188.6976766463

a_61= 192.5225306431

a_62= 196.3598778123

a_63= 200.2095425885

a_64= 204.0713710077

a_65= 207.9451779799

a_66= 211.8308191885

a_67= 215.7281180480

a_68= 219.6369408364

a_69= 223.5570997849

a_70= 227.4885071866

a_71= 231.4309843808

a_72= 235.3843931661

a_73= 239.3486192792

a_74= 243.3235140576

a_75= 247.3089677325

a_76= 251.3048375095

a_77= 255.3110228135

a_78= 259.3273875110

a_79= 263.3538324564

a_80= 267.3902471867

a_81= 271.4365024525

a_82= 275.4924928105

a_83= 279.5581471603

a_84= 283.6333119804

a_85= 287.7179392030

a_86= 291.8118947335

a_87= 295.9151023881

a_88= 300.0274504396

a_89= 304.1488634944

a_90= 308.2792519888

a_91= 312.4185117624

a_92= 316.5665730911

a_93= 320.7233354589

a_94= 324.8887317451

a_95= 329.0626821729

a_96= 333.2451078801

a_97= 337.4359118041

a_98= 341.6350372412

a_99= 345.8424068745

a_100= 350.0579336860

a_101= 354.2815588969

a_102= 358.5132133215

a_103= 362.7528106159

a_104= 367.0003160828

a_105= 371.2556293444

a_106= 375.5186854508

a_107= 379.7894500858

a_108= 384.0678314842

a_109= 388.3537801607

a_110= 392.6472189586

a_111= 396.9481223641

a_112= 401.2563957391

a_113= 405.5719986392

a_114= 409.8948585083

a_115= 414.2249480026

a_116= 418.5621839981

a_117= 422.9065248225

a_118= 427.2579174535

a_119= 431.6162930147

a_120= 435.9816335167

a_121= 440.3538537394

a_122= 444.7329047543

a_123= 449.1187668605

a_124= 453.5113755327

a_125= 457.9106702550

a_126= 462.3166165089

a_127= 466.7291677505

a_128= 471.1482630381

a_129= 475.5738880517

a_130= 480.0059669786

a_131= 484.4444725900

a_132= 488.8893596420

a_133= 493.3405854525

a_134= 497.7981092466

a_135= 502.2618743290

a_136= 506.7318543022

a_137= 511.2080081242

a_138= 515.6902967113

a_139= 520.1786803772

a_140= 524.6731047065

a_141= 529.1735644829

a_142= 533.6799885996

a_143= 538.1923555703

a_144= 542.7106288953

a_145= 547.2347567376

a_146= 551.7647258028

a_147= 556.3004650155

a_148= 560.8419393211

a_149= 565.3890426629

a_150= 569.9416519995

One can compare this to Tommy's hypothetical candidate (quote below), and see that the (n^2)! in the denominator is growing much quicker than necessary, as compared to the empirical results. But one can also see that the denominator in the Taylor series coefficients for this asymptotic half iterate grow much faster than for the exp(z). The conjecture is as n gets arbitrarily large, for any z0>1, the slog(f^n(z0))/n~=0.5, and therefore this would be an entire function with half exponential growth.

tommy1729 Wrote:f1(z) = sum z^n/(n^2)!

The asymptotic half iterate function is defined such that all Taylor series coefficients are positive, and that the function is always less than but approaching the Kneser half iterate, for real(z)>0. The construction for the Taylor series I used is a somewhat complicated two stage process; I'll post more later. But the first stage is to note that if the Taylor series terms are all positive, than no individual Taylor series term can be larger than the desired sum, so we require that, . This gives an upper bounds for the Taylor series coefficients. This function is always bigger than the Kneser half(z) function. The second stage is to scale down the terms by observing that for each value of half(z), one particular Taylor series term is largest contributer to the Taylor series sum. That determines for each term how much to scale that term down by. I think the resulting equation will always be less than the Kneser half iterate, but the ratio of this function over the Kneser half iterate will approach arbitrarily close to 1. Because the function is bounded to the right by the Kneser half iterate, we can safely say that this assymptotic half iterate is entire, assuming the construction works for arbitrarily large values of z and arbitrarily large Taylor series terms.

Here are some example of calculations using this assymptotic half Taylor series, as compared to the Kneser half iterate. It would probably make sense to set a_0 of the half iterate to sexp(-0.5), which is the half iterate of 0. I will half to generate a complex plan plot for this half iterate...

z, assymptotic_half, Kneser_half

0 0 0.4985632879411

1 1.126644950749 1.646354233751

10 58.93202104249 61.48617436731

100 187646.5930113 192708.5721853

1000 425414280682.2 432750850493.0

10000 9.638915213265 E21 9.750966938073 E21

100000 5.362748331798 E37 5.406412389290 E37

1000000 3.362567348729 E60 3.382539228002 E60

10000000 2.187210706560 E92 2.196854946875 E92

100000000 2.935957885769 E135 2.945782901678 E135

1000000000 3.788233214763 E192 3.798003781412 E192

10000000000 5.577154174589 E266 5.588492690694 E266

100000000000 3.101249943705 E361 3.106297055696 E361

1000000000000 6.614359301415 E480 6.622925643007 E480

One obvious questions from the Taylor series result, that I can't answer, because I have no idea how fast these functions grow as x goes to infinity, relative to exponentiaton. What is the "growth" of a functions like these, which should grow slower than exponentiation, but faster than any polynomial?