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 Green Eggs and HAM: Tetration for ALL bases, real and complex, now possible? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/29/2014, 04:07 AM So we can represent the tetration via $G(z) = \sum_{n=-\infty}^{\infty} a_n \left(\frac{1 + z}{1 - z}\right)^n$. where $a_n$ 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 $b_n(z) = \left(\frac{1 + z}{1 - z}\right)^n$. 100 coefficients then gives 32 places accuracy. sheldonison Long Time Fellow Posts: 668 Threads: 24 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 $G(z) = e^{z^2} \mathrm{tet}(z)$, then .... we can represent the tetration via $G(z) = \sum_{n=-\infty}^{\infty} a_n \left(\frac{1 + z}{1 - z}\right)^n$. where $a_n$ 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 $b_n(z) = \left(\frac{1 + z}{1 - z}\right)^n$. 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)/(1-x)=-2 for x=3. But for the inverse, (x-1)/(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 mike3 Long Time Fellow 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 $G(z) = e^{z^2} \mathrm{tet}(z)$, then .... we can represent the tetration via $G(z) = \sum_{n=-\infty}^{\infty} a_n \left(\frac{1 + z}{1 - z}\right)^n$. where $a_n$ 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 $b_n(z) = \left(\frac{1 + z}{1 - z}\right)^n$. 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)/(1-x)=-2 for x=3. But for the inverse, (x-1)/(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 well-behaved at infinity, this function has a bad, bad singularity there. But a Fourier expansion is a real-analytic 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. mike3 Long Time Fellow 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 15-16 digits of accuracy. Here's the coefficients for the tetrational base $e$, modulated by the Gaussian as shown: (The small imaginary part is a round off error): Code:a[0] = 0.397875391058100595418111 + 2.9850441245110140462 E-61*I a[1] = 0.416189154885943649018174 + 1.9195842352032945454 E-60*I a[2] = 0.145193802913168035746894 - 6.153556682827798856 E-61*I a[3] = -0.0329201654231789117932072 - 6.609222044925913430 E-61*I a[4] = -0.0146936116956140366967053 - 1.0607130827070990655 E-60*I a[5] = 0.0110622948810356852131289 + 0.E-61*I a[6] = -0.00172833502092149930384961 + 3.994678610137473554 E-61*I a[7] = -0.00180835508937351785635017 + 1.2661662572059834638 E-60*I a[8] = 0.00157515772698227975979782 - 6.186108626681812204 E-61*I a[9] = -0.000535236409811992349076167 + 8.966021985857690207 E-61*I a[10] = -9.77675994121522083519057 E-5 + 1.3849254092325444506 E-60*I a[11] = 0.000250010664600388723240219 - 9.227946283399708211 E-61*I a[12] = -0.000170812756598325659089718 + 0.E-61*I a[13] = 5.73706487998321205152364 E-5 + 0.E-61*I a[14] = 1.16033585501542046248279 E-5 - 5.400813015342964984 E-62*I a[15] = -3.22628720257092790720677 E-5 + 6.273421451219000834 E-61*I a[16] = 2.59666125133106224062854 E-5 - 4.462573445252261917 E-61*I a[17] = -1.23572147925053013371470 E-5 - 1.4900556311994949756 E-60*I a[18] = 1.71404012816780638286111 E-6 + 0.E-61*I a[19] = 3.39074764592592092593282 E-6 + 5.546244708549837783 E-61*I a[20] = -4.23330953435767752463413 E-6 - 1.4593180859561573286 E-60*I a[21] = 2.99761918164106298894254 E-6 + 5.859022626312693552 E-61*I a[22] = -1.35468891893702642331490 E-6 + 2.0183108005574934644 E-60*I a[23] = 1.38465637192648226018902 E-7 + 2.0525080575792498574 E-60*I a[24] = 4.59321733662540982796999 E-7 - 9.606564334553191880 E-61*I a[25] = -5.76687416451818575771329 E-7 - 1.1160783086457494800 E-60*I a[26] = 4.39224799220836982766874 E-7 - 1.4513595207362045963 E-60*I a[27] = -2.31697536804857357892810 E-7 + 4.134703860459057680 E-61*I a[28] = 5.92620019777023837219732 E-8 + 6.385664986253645420 E-61*I a[29] = 4.20898986292613309771500 E-8 - 1.2450292354861060336 E-60*I a[30] = -7.84364193301053660743543 E-8 + 4.587242661830708352 E-61*I a[31] = 7.27398375979498645162490 E-8 + 3.782960000987940630 E-61*I a[32] = -4.85109609270205489457946 E-8 + 7.224531536018421310 E-61*I a[33] = 2.23861456942567905105987 E-8 + 2.2410168612528228352 E-60*I a[34] = -2.86459520282896688975806 E-9 + 5.425108648749196066 E-61*I a[35] = -7.85379007070825266641794 E-9 - 1.6805656882674358601 E-60*I a[36] = 1.11868036739744560753676 E-8 - 1.5987419640240867624 E-61*I a[37] = -9.88486022637174088592919 E-9 + 1.2657073222455788301 E-60*I a[38] = 6.57973079390423678632267 E-9 + 1.0896547868791202914 E-61*I a[39] = -3.12685133933201448329449 E-9 + 1.2245179540360111058 E-60*I a[40] = 5.00793151317478543886855 E-10 - 1.0847905575555328089 E-60*I a[41] = 1.01921311820734388923147 E-9 - 1.9705505216136802586 E-61*I a[42] = -1.57486101968126885772701 E-9 + 0.E-61*I a[43] = 1.48382725528176917888905 E-9 + 5.800591992630454416 E-61*I a[44] = -1.07278266135766583611448 E-9 + 2.3673656365209571122 E-60*I a[45] = 5.91393342126508110476373 E-10 + 2.2624948620853793066 E-60*I a[46] = -1.88086564992327356679738 E-10 + 0.E-61*I a[47] = -7.73625553845716638140121 E-11 - 5.988289000338183422 E-61*I a[48] = 2.05846212745759269838244 E-10 - 1.4353069532990601708 E-60*I a[49] = -2.29938243046527653804108 E-10 + 0.E-61*I a[50] = 1.91340971688285908262136 E-10 + 1.0000753744597080188 E-60*I a[51] = -1.26965919854134121032240 E-10 + 3.354005847534243129 E-61*I a[52] = 6.27241433951610870046252 E-11 + 2.6680656443800172726 E-61*I a[53] = -1.27554698490738113179862 E-11 + 3.651119749002290437 E-61*I a[54] = -1.83236230992450431858269 E-11 + 0.E-61*I a[55] = 3.20921606979202761627920 E-11 + 2.5966855895604019986 E-61*I a[56] = -3.31681014233249216857662 E-11 - 9.166436613722241994 E-61*I a[57] = 2.68392089286889387832932 E-11 + 1.7914157461000540476 E-60*I a[58] = -1.76318602279710468751399 E-11 + 1.3166943389636563108 E-60*I a[59] = 8.69284889345558056935594 E-12 + 6.080470517003548930 E-61*I a[60] = -1.74232704665226971623959 E-12 - 2.4339713208887516650 E-61*I a[61] = -2.65334914006790198704635 E-12 - 1.6245329612991390394 E-60*I a[62] = 4.69145775573836397670701 E-12 - 1.5498501075859591586 E-60*I a[63] = -4.95935435400045178020460 E-12 - 8.557056786077598276 E-61*I a[64] = 4.14672723475903838162245 E-12 + 1.1072489386005317462 E-60*I a[65] = -2.86377465983331342292356 E-12 + 1.4859188838587007500 E-60*I a[66] = 1.55478646551640234218994 E-12 + 0.E-61*I a[67] = -4.82139005441705749124474 E-13 - 8.889365529047513239 E-61*I a[68] = -2.47143106719271718406506 E-13 - 9.196594103143828457 E-61*I a[69] = 6.36114497738363193531291 E-13 - 2.1486380323392822628 E-61*I a[70] = -7.50874402464770287965930 E-13 + 1.6320674706080463846 E-60*I a[71] = 6.81586835289400046205588 E-13 + 1.6223001498193322858 E-60*I a[72] = -5.15514528195414641348527 E-13 + 2.1512095618115911032 E-60*I a[73] = 3.22004369448179887030840 E-13 + 2.0167120121818167768 E-60*I a[74] = -1.47127859276329591059006 E-13 + 1.3530384902365274826 E-60*I a[75] = 1.48619138606160662654643 E-14 - 2.2107123486648895330 E-60*I a[76] = 6.81806683696976514667217 E-14 - 1.5956928302738256176 E-60*I a[77] = -1.06835153965874312282071 E-13 - 1.8502631798810804800 E-60*I a[78] = 1.11855715496354125502501 E-13 - 2.3301484139823512510 E-61*I a[79] = -9.56074174059497298207089 E-14 + 3.459308371959962772 E-60*I a[80] = 6.92406912140772493555385 E-14 + 2.0177144019872742998 E-60*I a[81] = -4.12344666488088741457145 E-14 + 1.4027704083224675329 E-60*I a[82] = 1.69989528696582883423590 E-14 - 1.1939093710245961698 E-60*I a[83] = 8.28557333333947259946431 E-16 - 1.2812244721545093977 E-60*I a[84] = -1.17301959482471538616070 E-14 + 0.E-61*I a[85] = 1.65548785804244879058777 E-14 + 9.250786505007335062 E-61*I a[86] = -1.68354953604888918779494 E-14 + 1.6947032525180505029 E-60*I a[87] = 1.42635476691793834697834 E-14 + 4.586704609864632169 E-61*I a[88] = -1.03462501346013446634608 E-14 - 5.956702009493176365 E-61*I a[89] = 6.23014521174691018831580 E-15 - 1.8923225574097490690 E-60*I a[90] = -2.65386727870540686082214 E-15 - 9.457280254834658315 E-61*I a[91] = -1.36743824456341755390483 E-17 - 9.198759367435836521 E-61*I a[92] = 1.68991171106908868158372 E-15 + 0.E-61*I a[93] = -2.48151945854288445675137 E-15 + 2.559961278697615104 E-60*I a[94] = 2.59472725640198465329043 E-15 + 1.2416813791761960441 E-60*I a[95] = -2.26454587544363434925095 E-15 + 8.848369631381160165 E-61*I a[96] = 1.70679875149646918218844 E-15 - 7.455253132690566890 E-61*I a[97] = -1.09166406110655826382257 E-15 - 1.2893324994708104573 E-60*I a[98] = 5.34526532821100880705770 E-16 + 0.E-61*I a[99] = -9.88572418981299229968335 E-17 + 0.E-61*I a[100] = -1.93977952794815338960238 E-16 - 1.1932106382924937024 E-60*I and the negative-degree coefficients: Code:a[-1] = 0.118714366068275491139864 + 1.2835397909662120082 E-60*I a[-2] = -0.0369421953926547148001681 - 9.773327445418275734 E-61*I a[-3] = -0.00955787685160441768978072 - 2.3683477764127354318 E-60*I a[-4] = 0.0105346633896152201891680 - 4.750906105176736143 E-61*I a[-5] = -0.00278544093753364182911849 - 6.812575863861415539 E-61*I a[-6] = -0.00113746278912202521777083 + 1.3525989300396415025 E-60*I a[-7] = 0.00149606845332274563487211 + 5.187478321895119788 E-61*I a[-8] = -0.000692053329891718522941987 + 5.685336628751597375 E-61*I a[-9] = 3.72699540037986232050485 E-5 - 1.6031405792330813983 E-60*I a[-10] = 0.000202673430359728381444716 + 0.E-61*I a[-11] = -0.000181155346083026835684956 + 1.5710306168750673432 E-61*I a[-12] = 8.31323708325276189154988 E-5 + 0.E-61*I a[-13] = -6.60938888491505880235531 E-6 - 4.790275268171344574 E-61*I a[-14] = -2.60140940617459515381288 E-5 - 1.1626175538036606982 E-60*I a[-15] = 2.74318412498564493421101 E-5 - 7.947392844057401237 E-61*I a[-16] = -1.62573052848227757666247 E-5 - 7.764172327980947514 E-61*I a[-17] = 4.87293045103624148056718 E-6 + 1.3446294372037274322 E-60*I a[-18] = 1.91116681343650782771301 E-6 - 4.089456297645928344 E-61*I a[-19] = -4.09267008579672159051381 E-6 + 4.741043558227001726 E-61*I a[-20] = 3.49558406867503292406780 E-6 + 2.0642426247743434862 E-60*I a[-21] = -1.93628188950229441011984 E-6 + 1.8313309409859436603 E-60*I a[-22] = 5.32401910689934060147273 E-7 + 0.E-61*I a[-23] = 2.99062098588705940667515 E-7 - 9.676901700219074217 E-61*I a[-24] = -5.82707454307740529747510 E-7 - 1.0218183086733718353 E-60*I a[-25] = 5.20338395105059715584950 E-7 + 4.753551074903132911 E-61*I a[-26] = -3.19579982184098489760597 E-7 - 1.4905070317308039177 E-61*I a[-27] = 1.20122734827712474542618 E-7 + 3.0131320758905032910 E-61*I a[-28] = 1.47054770875714223143751 E-8 + 0.E-61*I a[-29] = -7.63225621367399383010778 E-8 - 5.968455892835055058 E-61*I a[-30] = 8.36791727655117218101898 E-8 + 0.E-61*I a[-31] = -6.24246309431715664962425 E-8 + 1.1255404619245750633 E-60*I a[-32] = 3.34462516932598301110728 E-8 - 2.6583464337186196360 E-61*I a[-33] = -9.10688232594550776584457 E-9 + 9.540970786115621652 E-61*I a[-34] = -5.92779372572677067492525 E-9 - 1.9650874852508057502 E-61*I a[-35] = 1.20323850083452019291928 E-8 - 6.483548289837628729 E-61*I a[-36] = -1.19093229700253504587441 E-8 + 4.035975234800842194 E-61*I a[-37] = 8.63559964962720920997923 E-9 - 1.3145152429715187371 E-60*I a[-38] = -4.62051640065370451908076 E-9 + 0.E-61*I a[-39] = 1.28561146893290583816355 E-9 - 5.283791543417498821 E-61*I a[-40] = 8.22863985770012259991888 E-10 - 1.0457773727539229825 E-60*I a[-41] = -1.74394101069019941430219 E-9 + 2.2447070902738456334 E-60*I a[-42] = 1.80338785172728153254588 E-9 + 9.996570091213026754 E-61*I a[-43] = -1.38972254762024517621378 E-9 + 9.388465677448883675 E-61*I a[-44] = 8.25238360883481060964940 E-10 - 1.3255208419053970184 E-60*I a[-45] = -3.17325318526524320237867 E-10 - 1.0123287544259977184 E-60*I a[-46] = -3.76888806832179544950598 E-11 - 1.6683173843410761278 E-60*I a[-47] = 2.25171487351773085674629 E-10 + 2.2320233303164905300 E-61*I a[-48] = -2.77091696418487305763715 E-10 + 1.2075997623732215851 E-60*I a[-49] = 2.42538238284964521710153 E-10 + 1.7521139906568798183 E-60*I a[-50] = -1.68206588381801979019770 E-10 + 9.312100328125669223 E-61*I a[-51] = 8.86778193399147529408203 E-11 - 4.800925697999852935 E-61*I a[-52] = -2.40760544223348638787067 E-11 - 1.1254977818362073943 E-60*I a[-53] = -1.78663899312889338420433 E-11 - 5.830601395922755530 E-61*I a[-54] = 3.78952392006332329241532 E-11 + 3.504390548186142454 E-61*I a[-55] = -4.12278752589250636158352 E-11 + 3.0596921797453011188 E-61*I a[-56] = 3.43949992116145399144005 E-11 + 8.751393358481279748 E-61*I a[-57] = -2.32358512986154833240055 E-11 + 2.3083903193421552646 E-61*I a[-58] = 1.19532569663009671002094 E-11 - 9.758346281421433379 E-61*I a[-59] = -2.94639468338014118784073 E-12 - 3.420748699541095055 E-61*I a[-60] = -2.89822830020187414794661 E-12 - 1.0298137458562810148 E-60*I a[-61] = 5.72765224984859659596865 E-12 + 4.283326808727990317 E-61*I a[-62] = -6.24100353769722057596258 E-12 + 5.811818892310327139 E-61*I a[-63] = 5.30699853047606608204884 E-12 - 2.5395251271869018626 E-61*I a[-64] = -3.71501500370105790077365 E-12 + 9.418938714617855370 E-61*I a[-65] = 2.05112778246914099072157 E-12 + 9.408049915823806533 E-61*I a[-66] = -6.68854036779409633895353 E-13 - 1.0120200114991650599 E-60*I a[-67] = -2.81257305556733932148009 E-13 - 9.543496085046087047 E-61*I a[-68] = 7.94868549839678126659465 E-13 - 1.0528864539925234357 E-60*I a[-69] = -9.52953113039993652956814 E-13 - 3.585756009068627262 E-61*I a[-70] = 8.70430522912296317987430 E-13 + 8.007512533533776157 E-61*I a[-71] = -6.60146252637862386950784 E-13 + 6.589232225946199111 E-61*I a[-72] = 4.12498258220042912909922 E-13 + 1.3357319612143506715 E-60*I a[-73] = -1.87880303840560995974459 E-13 + 0.E-61*I a[-74] = 1.79009589906311932561769 E-14 + 0.E-61*I a[-75] = 8.85750482258138756415336 E-14 - 2.1753355395551205912 E-60*I a[-76] = -1.37721223883570899336952 E-13 - 1.9114560326653858132 E-60*I a[-77] = 1.43496588961083972036201 E-13 + 1.3386303332392950945 E-60*I a[-78] = -1.21978846122519073841780 E-13 + 2.1862102205662737612 E-60*I a[-79] = 8.76469854442097685439913 E-14 + 1.2990126947682685171 E-60*I a[-80] = -5.14705216926262643814896 E-14 - 1.0018008419977926074 E-60*I a[-81] = 2.03979604111418701986726 E-14 - 1.3233368591067467349 E-60*I a[-82] = 2.24192775496443330325499 E-15 - 1.4996906318551638389 E-60*I a[-83] = -1.58712754588968736068534 E-14 + 8.426085523305792649 E-61*I a[-84] = 2.16693093743371355272937 E-14 + 1.5376892740692408496 E-60*I a[-85] = -2.16763197278933496352268 E-14 + 1.4770850422580782269 E-60*I a[-86] = 1.81095267077642830630561 E-14 + 0.E-61*I a[-87] = -1.29217532221333541104969 E-14 - 9.882759084708351006 E-61*I a[-88] = 7.57976874806336744272561 E-15 - 1.5877929675157080387 E-60*I a[-89] = -3.01138071696685183820669 E-15 - 1.8255317833205893866 E-60*I a[-90] = -3.37380814768684205008675 E-16 + 1.2143230804412328186 E-60*I a[-91] = 2.38833050705337079031841 E-15 + 2.570873151632171509 E-60*I a[-92] = -3.30173234356011298634007 E-15 + 1.2549002395433393497 E-60*I a[-93] = 3.35871548378515198055450 E-15 + 7.207925042196062712 E-61*I a[-94] = -2.86994211237359841277488 E-15 + 0.E-61*I a[-95] = 2.11494906498676096740399 E-15 - 1.4620541430877854840 E-60*I a[-96] = -1.30987593100428238834775 E-15 - 8.398699513278232340 E-61*I a[-97] = 5.97641773826568741126775 E-16 + 0.E-61*I a[-98] = -5.33936990807526121723619 E-17 + 0.E-61*I a[-99] = -3.01549958759416732988085 E-16 - 1.0449032549761315903 E-60*I a[-100] = 4.82723205179637784396592 E-16 + 1.1553428412097541002 E-60*I sheldonison Long Time Fellow Posts: 668 Threads: 24 Joined: Oct 2008 07/01/2014, 02:25 PM (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 15-16 digits of accuracy. So, I tried generating a Laurent series for $f=\exp((\frac{z-1}{z+1})^2)$, 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 tommy1729 Ultimate Fellow Posts: 1,472 Threads: 353 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 15-16 digits of accuracy. So, I tried generating a Laurent series for $f=\exp((\frac{z-1}{z+1})^2)$, 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 double-periodicity for sexp. regards tommy1729 mike3 Long Time Fellow 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 15-16 digits of accuracy. So, I tried generating a Laurent series for $f=\exp((\frac{z-1}{z+1})^2)$, 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 $G(z) = \sum_{n=0}^{\infty} a_n e^{\epsilon_n (z - z_n)^2}$, where $\epsilon_n > 0$ are real parameters, and $z_n \in i\mathbb{R}$ 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 ($\epsilon_n$ and $z_n$) so as to optimize the convergence now. The trouble is doing the arithmetic on these series. Addition is simple enough (assuming same $\epsilon_n$ and $z_n$ between the two series), but multiplication seems a little trickier. tommy1729 Ultimate Fellow Posts: 1,472 Threads: 353 Joined: Feb 2009 07/02/2014, 10:13 PM (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 15-16 digits of accuracy. So, I tried generating a Laurent series for $f=\exp((\frac{z-1}{z+1})^2)$, 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 $G(z) = \sum_{n=0}^{\infty} a_n e^{\epsilon_n (z - z_n)^2}$, where $\epsilon_n > 0$ are real parameters, and $z_n \in i\mathbb{R}$ 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 ($\epsilon_n$ and $z_n$) so as to optimize the convergence now. The trouble is doing the arithmetic on these series. Addition is simple enough (assuming same $\epsilon_n$ and $z_n$ 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 « Next Oldest | Next Newest »

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