Posts: 368
Threads: 44
Joined: Sep 2009
So we can represent the tetration via
.
where are the coefficients of the Fourier series for the wrapped unit circle, and the fraction is just the inverse Moebius mapping taking the imaginary axis to the circle. That is, the choice of basis functions for the HAM is given by
.
100 coefficients then gives 32 places accuracy.
Posts: 641
Threads: 22
Joined: Oct 2008
06/30/2014, 11:17 PM
(This post was last modified: 06/30/2014, 11:36 PM by sheldonison.)
(06/29/2014, 04:07 AM)mike3 Wrote: If we let , then .... we can represent the tetration via
.
where are the coefficients of the Fourier series for the wrapped unit circle, and the fraction is just the inverse Moebius mapping taking the imaginary axis to the circle. That is, the choice of basis functions for the HAM is given by
.
100 coefficients then gives 32 places accuracy.
Mike, it sounds very promising. I like the Gaussian scaling, and the equations; it seems like a novel promising approach, to combine all of these different ideas with Kouznetsov's Cauchy integral. I'm struggling with the Laurent series on a unit circle, representing G(i oo) to G(+i oo), where the inside of the circle is either the left or right half of the complex plane. Since the function is represented as a Laurent series, I guess it doesn't matter which is whiich (inside/outside unit circle <=> left/right half complex plane).
So where are the nearest singularities to the unit Laurent series circle? There would be sexp(2,3,4.....) And also sexp(oo). But doesn't real(oo) also get mapped to the unit circle boundary too? I'm also confused about which half of the complex plane is inside the circle, and which half is outside... (1+x)/(1x)=2 for x=3. But for the inverse, (x1)/(1+x), then (1/3) would map to 2. I think real(oo) is mapped to 1 or 1, depending on whether we use the Mobius or its inverse. So how did the singularity for real(oo), which must be on the unit circle, get cancelled out so that it doesn't mess up the Laurent series? Even if the real(oo) gets cancelled out, sexp(4)~=oo so it may as well be a singularity and sexp(4) is mapped to either x=0.6 or x=5/3. I'm just trying to figure out the radius of convergence of this approach.
 Sheldon
Posts: 368
Threads: 44
Joined: Sep 2009
07/01/2014, 12:34 AM
(This post was last modified: 07/01/2014, 12:41 AM by mike3.)
(06/30/2014, 11:17 PM)sheldonison Wrote: (06/29/2014, 04:07 AM)mike3 Wrote: If we let , then .... we can represent the tetration via
.
where are the coefficients of the Fourier series for the wrapped unit circle, and the fraction is just the inverse Moebius mapping taking the imaginary axis to the circle. That is, the choice of basis functions for the HAM is given by
.
100 coefficients then gives 32 places accuracy.
Mike, it sounds very promising. I like the Gaussian scaling, and the equations; it seems like a novel promising approach, to combine all of these different ideas with Kouznetsov's Cauchy integral. I'm struggling with the Laurent series on a unit circle, representing G(i oo) to G(+i oo), where the inside of the circle is either the left or right half of the complex plane. Since the function is represented as a Laurent series, I guess it doesn't matter which is whiich (inside/outside unit circle <=> left/right half complex plane).
So where are the nearest singularities to the unit Laurent series circle? There would be sexp(2,3,4.....) And also sexp(oo). But doesn't real(oo) also get mapped to the unit circle boundary too? I'm also confused about which half of the complex plane is inside the circle, and which half is outside... (1+x)/(1x)=2 for x=3. But for the inverse, (x1)/(1+x), then (1/3) would map to 2. I think real(oo) is mapped to 1 or 1, depending on whether we use the Mobius or its inverse. So how did the singularity for real(oo), which must be on the unit circle, get cancelled out so that it doesn't mess up the Laurent series? Even if the real(oo) gets cancelled out, sexp(4)~=oo so it may as well be a singularity and sexp(4) is mapped to either x=0.6 or x=5/3. I'm just trying to figure out the radius of convergence of this approach.
The singularity, I believe, is not canceled: as a complex function the wrapped function is not holomorphic there. It's a pole, therefore any vicinity of that singularity maps to a neighborhood of (complex) infinity, and since the modified tetrational is not wellbehaved at infinity, this function has a bad, bad singularity there. But a Fourier expansion is a realanalytic thing: in particular, it doesn't have to care about the complex plane, but only about the values on the circle itself (you can have a Fourier series for a square wave, after all, which isn't even continuous, much less analytic or extensible to the complex plane!). And the integral equation need not know about the plane either, only the imaginary axis: once you have that, then since the tetration is holomorphic, this will be extensible to the whole complex plane again. So if the representation only converges at the imaginary axis, that's no problem  the limit is the tetrational, and thus extensible off the axis even if everything up to that does not converge in the rest of the plane.
Posts: 368
Threads: 44
Joined: Sep 2009
07/01/2014, 10:07 AM
(This post was last modified: 07/01/2014, 11:44 AM by mike3.)
Bummer... did a more accurate measurement on the convergence  looks like 100 coefficients in each direction will only give 1516 digits of accuracy.
Here's the coefficients for the tetrational base , modulated by the Gaussian as shown:
(The small imaginary part is a round off error):
Code: a[0] = 0.397875391058100595418111 + 2.9850441245110140462 E61*I
a[1] = 0.416189154885943649018174 + 1.9195842352032945454 E60*I
a[2] = 0.145193802913168035746894  6.153556682827798856 E61*I
a[3] = 0.0329201654231789117932072  6.609222044925913430 E61*I
a[4] = 0.0146936116956140366967053  1.0607130827070990655 E60*I
a[5] = 0.0110622948810356852131289 + 0.E61*I
a[6] = 0.00172833502092149930384961 + 3.994678610137473554 E61*I
a[7] = 0.00180835508937351785635017 + 1.2661662572059834638 E60*I
a[8] = 0.00157515772698227975979782  6.186108626681812204 E61*I
a[9] = 0.000535236409811992349076167 + 8.966021985857690207 E61*I
a[10] = 9.77675994121522083519057 E5 + 1.3849254092325444506 E60*I
a[11] = 0.000250010664600388723240219  9.227946283399708211 E61*I
a[12] = 0.000170812756598325659089718 + 0.E61*I
a[13] = 5.73706487998321205152364 E5 + 0.E61*I
a[14] = 1.16033585501542046248279 E5  5.400813015342964984 E62*I
a[15] = 3.22628720257092790720677 E5 + 6.273421451219000834 E61*I
a[16] = 2.59666125133106224062854 E5  4.462573445252261917 E61*I
a[17] = 1.23572147925053013371470 E5  1.4900556311994949756 E60*I
a[18] = 1.71404012816780638286111 E6 + 0.E61*I
a[19] = 3.39074764592592092593282 E6 + 5.546244708549837783 E61*I
a[20] = 4.23330953435767752463413 E6  1.4593180859561573286 E60*I
a[21] = 2.99761918164106298894254 E6 + 5.859022626312693552 E61*I
a[22] = 1.35468891893702642331490 E6 + 2.0183108005574934644 E60*I
a[23] = 1.38465637192648226018902 E7 + 2.0525080575792498574 E60*I
a[24] = 4.59321733662540982796999 E7  9.606564334553191880 E61*I
a[25] = 5.76687416451818575771329 E7  1.1160783086457494800 E60*I
a[26] = 4.39224799220836982766874 E7  1.4513595207362045963 E60*I
a[27] = 2.31697536804857357892810 E7 + 4.134703860459057680 E61*I
a[28] = 5.92620019777023837219732 E8 + 6.385664986253645420 E61*I
a[29] = 4.20898986292613309771500 E8  1.2450292354861060336 E60*I
a[30] = 7.84364193301053660743543 E8 + 4.587242661830708352 E61*I
a[31] = 7.27398375979498645162490 E8 + 3.782960000987940630 E61*I
a[32] = 4.85109609270205489457946 E8 + 7.224531536018421310 E61*I
a[33] = 2.23861456942567905105987 E8 + 2.2410168612528228352 E60*I
a[34] = 2.86459520282896688975806 E9 + 5.425108648749196066 E61*I
a[35] = 7.85379007070825266641794 E9  1.6805656882674358601 E60*I
a[36] = 1.11868036739744560753676 E8  1.5987419640240867624 E61*I
a[37] = 9.88486022637174088592919 E9 + 1.2657073222455788301 E60*I
a[38] = 6.57973079390423678632267 E9 + 1.0896547868791202914 E61*I
a[39] = 3.12685133933201448329449 E9 + 1.2245179540360111058 E60*I
a[40] = 5.00793151317478543886855 E10  1.0847905575555328089 E60*I
a[41] = 1.01921311820734388923147 E9  1.9705505216136802586 E61*I
a[42] = 1.57486101968126885772701 E9 + 0.E61*I
a[43] = 1.48382725528176917888905 E9 + 5.800591992630454416 E61*I
a[44] = 1.07278266135766583611448 E9 + 2.3673656365209571122 E60*I
a[45] = 5.91393342126508110476373 E10 + 2.2624948620853793066 E60*I
a[46] = 1.88086564992327356679738 E10 + 0.E61*I
a[47] = 7.73625553845716638140121 E11  5.988289000338183422 E61*I
a[48] = 2.05846212745759269838244 E10  1.4353069532990601708 E60*I
a[49] = 2.29938243046527653804108 E10 + 0.E61*I
a[50] = 1.91340971688285908262136 E10 + 1.0000753744597080188 E60*I
a[51] = 1.26965919854134121032240 E10 + 3.354005847534243129 E61*I
a[52] = 6.27241433951610870046252 E11 + 2.6680656443800172726 E61*I
a[53] = 1.27554698490738113179862 E11 + 3.651119749002290437 E61*I
a[54] = 1.83236230992450431858269 E11 + 0.E61*I
a[55] = 3.20921606979202761627920 E11 + 2.5966855895604019986 E61*I
a[56] = 3.31681014233249216857662 E11  9.166436613722241994 E61*I
a[57] = 2.68392089286889387832932 E11 + 1.7914157461000540476 E60*I
a[58] = 1.76318602279710468751399 E11 + 1.3166943389636563108 E60*I
a[59] = 8.69284889345558056935594 E12 + 6.080470517003548930 E61*I
a[60] = 1.74232704665226971623959 E12  2.4339713208887516650 E61*I
a[61] = 2.65334914006790198704635 E12  1.6245329612991390394 E60*I
a[62] = 4.69145775573836397670701 E12  1.5498501075859591586 E60*I
a[63] = 4.95935435400045178020460 E12  8.557056786077598276 E61*I
a[64] = 4.14672723475903838162245 E12 + 1.1072489386005317462 E60*I
a[65] = 2.86377465983331342292356 E12 + 1.4859188838587007500 E60*I
a[66] = 1.55478646551640234218994 E12 + 0.E61*I
a[67] = 4.82139005441705749124474 E13  8.889365529047513239 E61*I
a[68] = 2.47143106719271718406506 E13  9.196594103143828457 E61*I
a[69] = 6.36114497738363193531291 E13  2.1486380323392822628 E61*I
a[70] = 7.50874402464770287965930 E13 + 1.6320674706080463846 E60*I
a[71] = 6.81586835289400046205588 E13 + 1.6223001498193322858 E60*I
a[72] = 5.15514528195414641348527 E13 + 2.1512095618115911032 E60*I
a[73] = 3.22004369448179887030840 E13 + 2.0167120121818167768 E60*I
a[74] = 1.47127859276329591059006 E13 + 1.3530384902365274826 E60*I
a[75] = 1.48619138606160662654643 E14  2.2107123486648895330 E60*I
a[76] = 6.81806683696976514667217 E14  1.5956928302738256176 E60*I
a[77] = 1.06835153965874312282071 E13  1.8502631798810804800 E60*I
a[78] = 1.11855715496354125502501 E13  2.3301484139823512510 E61*I
a[79] = 9.56074174059497298207089 E14 + 3.459308371959962772 E60*I
a[80] = 6.92406912140772493555385 E14 + 2.0177144019872742998 E60*I
a[81] = 4.12344666488088741457145 E14 + 1.4027704083224675329 E60*I
a[82] = 1.69989528696582883423590 E14  1.1939093710245961698 E60*I
a[83] = 8.28557333333947259946431 E16  1.2812244721545093977 E60*I
a[84] = 1.17301959482471538616070 E14 + 0.E61*I
a[85] = 1.65548785804244879058777 E14 + 9.250786505007335062 E61*I
a[86] = 1.68354953604888918779494 E14 + 1.6947032525180505029 E60*I
a[87] = 1.42635476691793834697834 E14 + 4.586704609864632169 E61*I
a[88] = 1.03462501346013446634608 E14  5.956702009493176365 E61*I
a[89] = 6.23014521174691018831580 E15  1.8923225574097490690 E60*I
a[90] = 2.65386727870540686082214 E15  9.457280254834658315 E61*I
a[91] = 1.36743824456341755390483 E17  9.198759367435836521 E61*I
a[92] = 1.68991171106908868158372 E15 + 0.E61*I
a[93] = 2.48151945854288445675137 E15 + 2.559961278697615104 E60*I
a[94] = 2.59472725640198465329043 E15 + 1.2416813791761960441 E60*I
a[95] = 2.26454587544363434925095 E15 + 8.848369631381160165 E61*I
a[96] = 1.70679875149646918218844 E15  7.455253132690566890 E61*I
a[97] = 1.09166406110655826382257 E15  1.2893324994708104573 E60*I
a[98] = 5.34526532821100880705770 E16 + 0.E61*I
a[99] = 9.88572418981299229968335 E17 + 0.E61*I
a[100] = 1.93977952794815338960238 E16  1.1932106382924937024 E60*I
and the negativedegree coefficients:
Code: a[1] = 0.118714366068275491139864 + 1.2835397909662120082 E60*I
a[2] = 0.0369421953926547148001681  9.773327445418275734 E61*I
a[3] = 0.00955787685160441768978072  2.3683477764127354318 E60*I
a[4] = 0.0105346633896152201891680  4.750906105176736143 E61*I
a[5] = 0.00278544093753364182911849  6.812575863861415539 E61*I
a[6] = 0.00113746278912202521777083 + 1.3525989300396415025 E60*I
a[7] = 0.00149606845332274563487211 + 5.187478321895119788 E61*I
a[8] = 0.000692053329891718522941987 + 5.685336628751597375 E61*I
a[9] = 3.72699540037986232050485 E5  1.6031405792330813983 E60*I
a[10] = 0.000202673430359728381444716 + 0.E61*I
a[11] = 0.000181155346083026835684956 + 1.5710306168750673432 E61*I
a[12] = 8.31323708325276189154988 E5 + 0.E61*I
a[13] = 6.60938888491505880235531 E6  4.790275268171344574 E61*I
a[14] = 2.60140940617459515381288 E5  1.1626175538036606982 E60*I
a[15] = 2.74318412498564493421101 E5  7.947392844057401237 E61*I
a[16] = 1.62573052848227757666247 E5  7.764172327980947514 E61*I
a[17] = 4.87293045103624148056718 E6 + 1.3446294372037274322 E60*I
a[18] = 1.91116681343650782771301 E6  4.089456297645928344 E61*I
a[19] = 4.09267008579672159051381 E6 + 4.741043558227001726 E61*I
a[20] = 3.49558406867503292406780 E6 + 2.0642426247743434862 E60*I
a[21] = 1.93628188950229441011984 E6 + 1.8313309409859436603 E60*I
a[22] = 5.32401910689934060147273 E7 + 0.E61*I
a[23] = 2.99062098588705940667515 E7  9.676901700219074217 E61*I
a[24] = 5.82707454307740529747510 E7  1.0218183086733718353 E60*I
a[25] = 5.20338395105059715584950 E7 + 4.753551074903132911 E61*I
a[26] = 3.19579982184098489760597 E7  1.4905070317308039177 E61*I
a[27] = 1.20122734827712474542618 E7 + 3.0131320758905032910 E61*I
a[28] = 1.47054770875714223143751 E8 + 0.E61*I
a[29] = 7.63225621367399383010778 E8  5.968455892835055058 E61*I
a[30] = 8.36791727655117218101898 E8 + 0.E61*I
a[31] = 6.24246309431715664962425 E8 + 1.1255404619245750633 E60*I
a[32] = 3.34462516932598301110728 E8  2.6583464337186196360 E61*I
a[33] = 9.10688232594550776584457 E9 + 9.540970786115621652 E61*I
a[34] = 5.92779372572677067492525 E9  1.9650874852508057502 E61*I
a[35] = 1.20323850083452019291928 E8  6.483548289837628729 E61*I
a[36] = 1.19093229700253504587441 E8 + 4.035975234800842194 E61*I
a[37] = 8.63559964962720920997923 E9  1.3145152429715187371 E60*I
a[38] = 4.62051640065370451908076 E9 + 0.E61*I
a[39] = 1.28561146893290583816355 E9  5.283791543417498821 E61*I
a[40] = 8.22863985770012259991888 E10  1.0457773727539229825 E60*I
a[41] = 1.74394101069019941430219 E9 + 2.2447070902738456334 E60*I
a[42] = 1.80338785172728153254588 E9 + 9.996570091213026754 E61*I
a[43] = 1.38972254762024517621378 E9 + 9.388465677448883675 E61*I
a[44] = 8.25238360883481060964940 E10  1.3255208419053970184 E60*I
a[45] = 3.17325318526524320237867 E10  1.0123287544259977184 E60*I
a[46] = 3.76888806832179544950598 E11  1.6683173843410761278 E60*I
a[47] = 2.25171487351773085674629 E10 + 2.2320233303164905300 E61*I
a[48] = 2.77091696418487305763715 E10 + 1.2075997623732215851 E60*I
a[49] = 2.42538238284964521710153 E10 + 1.7521139906568798183 E60*I
a[50] = 1.68206588381801979019770 E10 + 9.312100328125669223 E61*I
a[51] = 8.86778193399147529408203 E11  4.800925697999852935 E61*I
a[52] = 2.40760544223348638787067 E11  1.1254977818362073943 E60*I
a[53] = 1.78663899312889338420433 E11  5.830601395922755530 E61*I
a[54] = 3.78952392006332329241532 E11 + 3.504390548186142454 E61*I
a[55] = 4.12278752589250636158352 E11 + 3.0596921797453011188 E61*I
a[56] = 3.43949992116145399144005 E11 + 8.751393358481279748 E61*I
a[57] = 2.32358512986154833240055 E11 + 2.3083903193421552646 E61*I
a[58] = 1.19532569663009671002094 E11  9.758346281421433379 E61*I
a[59] = 2.94639468338014118784073 E12  3.420748699541095055 E61*I
a[60] = 2.89822830020187414794661 E12  1.0298137458562810148 E60*I
a[61] = 5.72765224984859659596865 E12 + 4.283326808727990317 E61*I
a[62] = 6.24100353769722057596258 E12 + 5.811818892310327139 E61*I
a[63] = 5.30699853047606608204884 E12  2.5395251271869018626 E61*I
a[64] = 3.71501500370105790077365 E12 + 9.418938714617855370 E61*I
a[65] = 2.05112778246914099072157 E12 + 9.408049915823806533 E61*I
a[66] = 6.68854036779409633895353 E13  1.0120200114991650599 E60*I
a[67] = 2.81257305556733932148009 E13  9.543496085046087047 E61*I
a[68] = 7.94868549839678126659465 E13  1.0528864539925234357 E60*I
a[69] = 9.52953113039993652956814 E13  3.585756009068627262 E61*I
a[70] = 8.70430522912296317987430 E13 + 8.007512533533776157 E61*I
a[71] = 6.60146252637862386950784 E13 + 6.589232225946199111 E61*I
a[72] = 4.12498258220042912909922 E13 + 1.3357319612143506715 E60*I
a[73] = 1.87880303840560995974459 E13 + 0.E61*I
a[74] = 1.79009589906311932561769 E14 + 0.E61*I
a[75] = 8.85750482258138756415336 E14  2.1753355395551205912 E60*I
a[76] = 1.37721223883570899336952 E13  1.9114560326653858132 E60*I
a[77] = 1.43496588961083972036201 E13 + 1.3386303332392950945 E60*I
a[78] = 1.21978846122519073841780 E13 + 2.1862102205662737612 E60*I
a[79] = 8.76469854442097685439913 E14 + 1.2990126947682685171 E60*I
a[80] = 5.14705216926262643814896 E14  1.0018008419977926074 E60*I
a[81] = 2.03979604111418701986726 E14  1.3233368591067467349 E60*I
a[82] = 2.24192775496443330325499 E15  1.4996906318551638389 E60*I
a[83] = 1.58712754588968736068534 E14 + 8.426085523305792649 E61*I
a[84] = 2.16693093743371355272937 E14 + 1.5376892740692408496 E60*I
a[85] = 2.16763197278933496352268 E14 + 1.4770850422580782269 E60*I
a[86] = 1.81095267077642830630561 E14 + 0.E61*I
a[87] = 1.29217532221333541104969 E14  9.882759084708351006 E61*I
a[88] = 7.57976874806336744272561 E15  1.5877929675157080387 E60*I
a[89] = 3.01138071696685183820669 E15  1.8255317833205893866 E60*I
a[90] = 3.37380814768684205008675 E16 + 1.2143230804412328186 E60*I
a[91] = 2.38833050705337079031841 E15 + 2.570873151632171509 E60*I
a[92] = 3.30173234356011298634007 E15 + 1.2549002395433393497 E60*I
a[93] = 3.35871548378515198055450 E15 + 7.207925042196062712 E61*I
a[94] = 2.86994211237359841277488 E15 + 0.E61*I
a[95] = 2.11494906498676096740399 E15  1.4620541430877854840 E60*I
a[96] = 1.30987593100428238834775 E15  8.398699513278232340 E61*I
a[97] = 5.97641773826568741126775 E16 + 0.E61*I
a[98] = 5.33936990807526121723619 E17 + 0.E61*I
a[99] = 3.01549958759416732988085 E16  1.0449032549761315903 E60*I
a[100] = 4.82723205179637784396592 E16 + 1.1553428412097541002 E60*I
Posts: 641
Threads: 22
Joined: Oct 2008
(07/01/2014, 10:07 AM)mike3 Wrote: Bummer... did a more accurate measurement on the convergence  looks like 100 coefficients in each direction will only give 1516 digits of accuracy.
So, I tried generating a Laurent series for , wrapped around a unit circle, with similar results for the Laurent series. A Fourier/Laurent series on a unit circle will have the coefficients eventually decay according to the nearest singularity in the complex plane, a_n = r^n, or r^n, depending on the radius of that singularity. If the function is not analytic on the unit circle, then the first derivative with the discontinuity determines the rate of decay of the coefficients. I looked for an equation online, something like a_n~n^(k+1), but couldn't find it. Anyway, in this case, we have a Gaussian, and I'm pretty sure that all of the derivatives are continuous on the unit circle  and all of the derivatives go to zero at z=1, where we have the singularity. Hence, the coefficients apparently decay at an almost analytic rate, but not quite, but decay faster than n^k, where k is any integer.
 Sheldon
Posts: 1,372
Threads: 336
Joined: Feb 2009
07/01/2014, 11:37 PM
(This post was last modified: 07/01/2014, 11:41 PM by tommy1729.)
(07/01/2014, 02:25 PM)sheldonison Wrote: (07/01/2014, 10:07 AM)mike3 Wrote: Bummer... did a more accurate measurement on the convergence  looks like 100 coefficients in each direction will only give 1516 digits of accuracy.
So, I tried generating a Laurent series for , wrapped around a unit circle, with similar results for the Laurent series. A Fourier/Laurent series on a unit circle will have the coefficients eventually decay according to the nearest singularity in the complex plane, a_n = r^n, or r^n, depending on the radius of that singularity. If the function is not analytic on the unit circle, then the first derivative with the discontinuity determines the rate of decay of the coefficients. I looked for an equation online, something like a_n~n^(k+1), but couldn't find it. Anyway, in this case, we have a Gaussian, and I'm pretty sure that all of the derivatives are continuous on the unit circle  and all of the derivatives go to zero at z=1, where we have the singularity. Hence, the coefficients apparently decay at an almost analytic rate, but not quite, but decay faster than n^k, where k is any integer.
Have you tried using lim n> oo (1+x/n)^n ~ exp(x) ?
Sorry if this is a bad idea.
Another idea Is using the parallelogram contour integration of G(z).
The parallelogram should be paralel to the pseudoperiod of sexp(z).
Since G(iz) = 0 in the real limit z > +oo we get that the integral around this paralellogram of G(z) = 0 ( since analytic).
And thus this integral reduces to 2 integrals over a line with slope equal to the speudoperiod.
Now by a staircase nesting of contour integrals we can set up the equations for the points on this line. Though that is complicated.
This relates to my recent idea and thread about pseudo doubleperiodicity for sexp.
regards
tommy1729
Posts: 368
Threads: 44
Joined: Sep 2009
07/02/2014, 07:56 AM
(This post was last modified: 07/02/2014, 08:00 AM by mike3.)
(07/01/2014, 02:25 PM)sheldonison Wrote: (07/01/2014, 10:07 AM)mike3 Wrote: Bummer... did a more accurate measurement on the convergence  looks like 100 coefficients in each direction will only give 1516 digits of accuracy.
So, I tried generating a Laurent series for , wrapped around a unit circle, with similar results for the Laurent series. A Fourier/Laurent series on a unit circle will have the coefficients eventually decay according to the nearest singularity in the complex plane, a_n = r^n, or r^n, depending on the radius of that singularity. If the function is not analytic on the unit circle, then the first derivative with the discontinuity determines the rate of decay of the coefficients. I looked for an equation online, something like a_n~n^(k+1), but couldn't find it. Anyway, in this case, we have a Gaussian, and I'm pretty sure that all of the derivatives are continuous on the unit circle  and all of the derivatives go to zero at z=1, where we have the singularity. Hence, the coefficients apparently decay at an almost analytic rate, but not quite, but decay faster than n^k, where k is any integer.
What if, perhaps, a better approach here would be to use Gaussians themselves as the set of basis functions? That is, write
,
where are real parameters, and are imaginary parameters? Now we have three, instead of one, coefficient sequences to deal with, but this might work. We have great freedom to tweak the various parameters ( and ) so as to optimize the convergence now.
The trouble is doing the arithmetic on these series. Addition is simple enough (assuming same and between the two series), but multiplication seems a little trickier.
Posts: 1,372
Threads: 336
Joined: Feb 2009
(07/02/2014, 07:56 AM)mike3 Wrote: (07/01/2014, 02:25 PM)sheldonison Wrote: (07/01/2014, 10:07 AM)mike3 Wrote: Bummer... did a more accurate measurement on the convergence  looks like 100 coefficients in each direction will only give 1516 digits of accuracy.
So, I tried generating a Laurent series for , wrapped around a unit circle, with similar results for the Laurent series. A Fourier/Laurent series on a unit circle will have the coefficients eventually decay according to the nearest singularity in the complex plane, a_n = r^n, or r^n, depending on the radius of that singularity. If the function is not analytic on the unit circle, then the first derivative with the discontinuity determines the rate of decay of the coefficients. I looked for an equation online, something like a_n~n^(k+1), but couldn't find it. Anyway, in this case, we have a Gaussian, and I'm pretty sure that all of the derivatives are continuous on the unit circle  and all of the derivatives go to zero at z=1, where we have the singularity. Hence, the coefficients apparently decay at an almost analytic rate, but not quite, but decay faster than n^k, where k is any integer.
What if, perhaps, a better approach here would be to use Gaussians themselves as the set of basis functions? That is, write
,
where are real parameters, and are imaginary parameters? Now we have three, instead of one, coefficient sequences to deal with, but this might work. We have great freedom to tweak the various parameters ( and ) so as to optimize the convergence now.
The trouble is doing the arithmetic on these series. Addition is simple enough (assuming same and between the two series), but multiplication seems a little trickier.
A generalized zeta type function of z^2 !?
This is more complicated than a general dirichlet series !!
Whatever you ask me , dont ask for the zero's
regards
tomm1729
