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(05/24/2011, 12:12 AM)sheldonison Wrote: I'll describe what I did to generate the Taylor series for sexp(z). Its not fancy or anything  its just what worked for me to investigate base eta, with its parabolic convergence. I use parigp, to generate an interpolating polynomial. Only I center the polynomial around sexp(100), where the sexp(z) function is already converging towards e, and is fairly well behaved. Then I generate 25 points on either side of sexp(100), using 67 digits of precision.
You mean, you take the interpolation polynomial at the integerpoints of sexp?
Yeah true, that also yields regular iteration.
I think we discussed this a lot with Ansus  the Newton (interpolation) formula.
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(05/23/2011, 08:42 PM)sheldonison Wrote: (...)
Code: a0 = 0
a1 = 1.661129667441415
a2 = 1.137387400487982
a3 = 0.841151615164940
a4 = 0.657512962174043
a5 = 0.535494578310460
a6 = 0.449853109363909
a7 = 0.387026076215351
a8 = 0.339240627153272
a9 = 0.301798047541097
Well, now I'm surprised. I've get that coefficients however accurate only up to 6 digits, but maybe they converge if I use higher precision  by the most simple eigendecomposition of the 32x32 carleman matrix.
While the formal h'th powers of the carlemanmatrices occur if I raise the diagonalmatrix of the eigenvalues to the h'th power, Pari/GP is able to convert this to a powerseries in x, if I enter the indeterminate x instead of a explicite hvalue for the powers. (Note that this is the fourth method in my short treatize on "four methods of interpolation")
Well, Sheldon's method seem to allow much more precision and I do not see yet, how I could reproduce this by simply increasing the Pari/GPresources in decimal precision and matrixsize.
Gottfried
Gottfried Helms, Kassel
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(05/20/2011, 06:43 PM)andydude Wrote: That is a really interesting question. First of all, that is one of the bases for which the series expansion of (\exp_b^t(x)) in terms of x is relatively simple with "nice" coefficients, but substituting x=1 (which I think you are talking about) gives a function of t which is not strictly a power series, which makes finding that power series more difficult. Anyways, I believe I have done this before, but I don't have access to my notes right, now, so let me get back to you later today...
Andrew Robbins
Your close to what I'm talking about. I'll write it out in full math format
which maps the growth of addition to multiplication to exponentiation. I hope that base eta will behave better than base 2 and base e behaved.
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06/02/2011, 09:37 PM
(This post was last modified: 06/02/2011, 09:48 PM by JmsNxn.)
(05/21/2011, 10:55 PM)sheldonison Wrote: For the results I'm posting here centered at cheta(0)=2e, I iterate the exponent of that function 95 times, to make a unit circle in the complex plane centered around cheta(0), from which a taylor series can be generated. It appears to work; I've haven't posted it before. Initialized to 67 digits accuracy in parigp, the algorithm seems to give results with nearly 50 decimal digits of accuracy. Here is the Taylor Series. a0=2e, printed to 32 digits.
Code: 0 5.4365636569180904707205749427053
1 1.1771399745582020467487064927981
2 0.47791083712959936964236746127117
3 0.18626062152494972692276478391796
4 0.070474191198539960880465202693624
5 0.026056306225434063913977558720610
6 0.0094541495787515083484748872855356
7 0.0033764647774015865179387607261247
8 0.0011895908149927411979137386055855
9 0.00041416349743994006206357899506395
10 0.00014268359371573690572984247219736
11 0.000048694763765091835931424063371768
12 0.000016477512451260383444394568944931
13 0.0000055326597652388183384746557130853
14 0.0000018445541337171731425492507600409
15 0.00000061095142258861804599507950586002
16 0.00000020113633929013309964268387743384
17 0.000000065845717087468591004558852969906
18 0.000000021442747870947309095492187967455
19 0.0000000069485439464512255857882560746267
20 0.0000000022412832385662916992460615895339
21 0.00000000071978893862885391677345614556987
22 2.3020973206030329181894361145544 E10
23 7.3341040297826856350206259498267 E11
24 2.3278852998967291568233733165642 E11
25 7.3628505815581778431734554753314 E12
26 2.3209857992934250244177110110812 E12
27 7.2930204918450243443324949177246 E13
28 2.2846097982451633980559339079833 E13
29 7.1358033041146574466639840152247 E14
30 2.2225543653988499920641676567938 E14
31 6.9038181736676583445161044386747 E15
32 2.1389425595138842758935272382177 E15
33 6.6103445086593382475520541375449 E16
34 2.0379986212392242975901900360689 E16
35 6.2686731343734042989649316156714 E17
36 1.9238630883507697992098052847583 E17
37 5.8915865956656546031878586526468 E18
38 1.8004488782209858871936565169317 E18
39 5.4909672937838092357031986642134 E19
40 1.6713366863796135261320082560915 E19
41 5.0775249729439841733736059243593 E20
42 1.5397061803790039357741194269061 E20
43 4.6606303423839124073481231546437 E21
44 1.4082983164868122075336349130200 E21
45 4.2482389646419479670428084576358 E22
46 1.2794035465663755925461443200159 E22
47 3.8468885494278543290882325049087 E23
48 1.1548705835256123872218006706955 E23
49 3.4617530598681323751620914079008 E24
50 1.0361306324224779753428626830104 E24
51 3.0967381932148655440014679355037 E25
52 9.2423246088852631003347605140667 E26
53 2.7546045204596969212253367181553 E26
54 8.1988428443303828566587256203397 E27
55 2.4371065951483711807719660406315 E27
56 7.2349910531853903620035897727885 E28
57 2.1451384654610549641820852967050 E28
58 6.3524081182337135717494628619063 E29
59 1.8788780939059811381889624923360 E29
60 5.5506897778611357646729108553389 E30
61 1.6379251020900380753695185487417 E30
62 4.8278087421120487722463748941123 E31
63 1.4214279502607189609788849783519 E31
64 4.1804970061336688633898121429008 E32
65 1.2281981008321929711060382301444 E32
66 3.6045845774315208753223996337347 E33
67 1.0568098942930644804288412600335 E33
68 3.0952921594935758380591422397535 E34
I hate to be a bit of a dunce but:
is the correct formula for the first series correct? I only ask because this is the code I'm using (and I've also tried ) and neither seem to converge anywhere?
The only convergence I do get, is when I remove the plus/minus 2e.
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06/02/2011, 10:27 PM
(This post was last modified: 06/02/2011, 11:53 PM by sheldonison.)
(06/02/2011, 09:37 PM)JmsNxn Wrote: I hate to be a bit of a dunce but:
is the correct formula for the first series correct? I only ask because this is the code I'm using (and I've also tried) and neither seem to converge anywhere?
The only convergence I do get, is when I remove the plus/minus 2e. Hey James,
Hopefully, this helps. The correct formula is
Cheta(z) is the upper entire superfunction of eta, which grows superexponentially as z increases. Unfortunately, how to define cheta(0) is a somewhat arbitrary, since cheta(z) is always bigger than e, as z goes to minus infinity. Jay suggested defining cheta(0)=2e. Then , and cheta(2)=e^e, which seems like a reasonable choice for how to define cheta(0). Centered around 0, the series I posted will generate these values. More recently, Henryk and Dimitrii have written a paper where the upper . Perhaps there will eventually be a reason to pick a definitive value, but that hasn't happened yet.
There is also a lower superfunction for base eta, that I usually refer to as , since , , and there is a singularity at . does not grow superexponentially, but converges towards e as z increases. I recently posted the Taylor series for that function, centered at z=0, here:
http://math.eretrandre.org/tetrationforu...17#pid5817
 Sheldon
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06/02/2011, 11:40 PM
(This post was last modified: 06/02/2011, 11:41 PM by JmsNxn.)
(06/02/2011, 10:27 PM)sheldonison Wrote: There is also a lower superfunction for base eta, that I usually refer to as , since , , and there is a singularity at . does not grow superexponentially, but converges towards e as z increases. I recently posted the Taylor series for that function, centered at z=0, here:
http://math.eretrandre.org/tetrationforu...17#pid5817
 Sheldon
This is more what I was looking for, thanks. And another question, do you have a similar taylor series for ? That would be the inverse of the lower super function.
(06/02/2011, 10:27 PM)sheldonison Wrote: Jay suggested defining cheta(0)=2e. Then , and cheta(2)=e^e, which seems like a reasonable choice for how to define cheta(0).
Woah! have you ever thought to consider that since , , and that the cheta function maps the growth of , where is a hyper operator of x order (0 is addition, 1 is multiplication etc etc); or mathematically speaking, another conjecture:
, which should at least be true over domain [0, 2]. To see if its universally true would be very difficult, though.
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06/03/2011, 12:48 AM
(This post was last modified: 06/03/2011, 12:51 AM by sheldonison.)
(06/02/2011, 11:40 PM)JmsNxn Wrote: This is more what I was looking for, thanks. And another question, do you have a similar taylor series for ? That would be the inverse of the lower super function.
......
Woah! have you ever thought to consider that since , , and that the cheta function maps the growth of , where is a hyper operator of x order (0 is addition, 1 is multiplication etc etc); or mathematically speaking, another conjecture:
, which should at least be true over domain [0, 2]. To see if its universally true would be very difficult, though. It is an interesting sequence. Here is the series, centered at z=1,
Code: a0= 0
a1= 1.6364055628757310098612069643305
a2= 1.0153219515675015927934054348231
a3= 0.60179106451341218323473861841285
a4= 0.35339094138233716197130631267800
a5= 0.20678022805222642138569218148044
a6= 0.12077316515278617589978715489636
a7= 0.070466597739279935817347004921549
a8= 0.041088245566444697493192432669165
a9= 0.023947858691724371628412673016534
a10= 0.013953603992552690728627252149490
a11= 0.0081285329698961693398261557991908
a12= 0.0047344290157003070913140831631485
a13= 0.0027572046775250114750438564059555
a14= 0.0016055653508202650722548361996376
a15= 0.00093487396331912588479860322878510
a16= 0.00054431539381581329680504421391946
a17= 0.00031690239123736519247306265808875
a18= 0.00018449370377075905076738832103314
a19= 0.00010740430864046301276907787649912
a20= 0.000062524232227043156013995698181196
a21= 0.000036396822638059218270554447632005
a22= 0.000021186957652934110380738251130135
a23= 0.000012332895471487133422245016435244
a24= 0.0000071788340691456493380397479368256
a25= 0.0000041786510411693011327284124119868
a26= 0.0000024322734042405353969581364435823
a27= 0.0000014157397670132180924517455634643
a28= 0.00000082404283673548797945624589827040
a29= 0.00000047963619402863627513438889010921
a30= 0.00000027917101829902645542548058586310
a31= 0.00000016248949999299272203056483378299
a32= 0.000000094575181457179592966349170760430
a33= 0.000000055046061997819873417493220126620
a34= 0.000000032038543589826489735755462890716
a35= 0.000000018647341967554525035509582154381
a36= 0.000000010853228813698946831281796207325
a37= 0.0000000063168274147666049763200649142710
a38= 0.0000000036765226692784746720709040758475
a39= 0.0000000021398030836117500316624962487179
a40= 0.0000000012453998812027015780706541464576
a41= 0.0000000007248404052267904098654234587018
a42= 0.0000000004218661070727302103010309894460
a43= 0.0000000002455306059973809636590738454709
a44= 1.4290106986271220339032492794899 E10
a45= 8.3169529782673189838917026421766 E11
a46= 4.8405198310832728584491149379096 E11
a47= 2.8172073806752434116210055497734 E11
a48= 1.6396258116384352659166197756211 E11
a49= 9.5426686705080144354640068061238 E12
a50= 5.5538502152136607519127775538049 E12
a51= 3.2323452797056846816888732034200 E12
a52= 1.8812245466465630762639581584087 E12
a53= 1.0948708034538274092484245688059 E12
a54= 6.3721280701164076567624351207206 E13
a55= 3.7085619145590230253906096801015 E13
a56= 2.1583706178992941955046700446313 E13
a57= 1.2561629614172847097262503849877 E13
a58= 7.3108092357697884172557590560006 E14
a59= 4.2548519124264617256971973371109 E14
a60= 2.4762985306030106868904605580111 E14
a61= 1.4411896206100053414406968180449 E14
a62= 8.3876219415829221560909589871987 E15
a63= 4.8815324822646088477541099811656 E15
a64= 2.8410125546371166957177683205131 E15
a65= 1.6534450727661439551155006062390 E15
a66= 9.6229018805487868709553017018049 E16
a67= 5.6004383866369015723533210751934 E16
a68= 3.2594001769485484340769051661933 E16
a69= 1.8969376400216642523206535563610 E16
a70= 1.1039976379052598207387127395496 E16
a71= 6.4251454988114292160593966066487 E17
a72= 3.7393621961593764263505452461775 E17
a73= 2.1762653817172712533134127189169 E17
a74= 1.2665604966549427015232899034031 E17
a75= 7.3712271908237475668169947042325 E18
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06/05/2011, 06:24 PM
(This post was last modified: 06/05/2011, 08:02 PM by JmsNxn.)
Thank you very much for the series approximations sheldon, but sadly the humps still occur in base . I'm wondering now if there is a better base to work with or if it's smarter to dump the idea of logarithmic semioperators altogether, as they seem to be a poor extension of ackerman function to domain real.
The only interesting thing I have to report is that:
just like ,
or that where is a logarithmic semi operator.
So my question was, what's the radius of convergence for the cheta series you gave me, and whats the recurrence relation so that I can produce the full function. I just want to test some values. for ex: if then I think we have something, but if it doesn't, oh well.
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(06/05/2011, 06:24 PM)JmsNxn Wrote: Thank you very much for the series approximations sheldon, but sadly the humps still occur in base . I'm wondering now if there is a better base to work with or if it's smarter to dump the idea of logarithmic semioperators altogether, as they seem to be a poor extension of ackerman function to domain real.
The only interesting thing I have to report is that:
just like ,
or that where is a logarithmic semi operator.
So my question was, what's the radius of convergence for the cheta series you gave me, and whats the recurrence relation so that I can produce the full function. I just want to test some values. for ex: if then I think we have something, but if it doesn't, oh well. Cheta(1)=e*(log(2)+1).
cheta(z1)=
cheta(z+1)=
Cheta(z) is entire, but the series convergence for a finite number of terms is limited by how close we are to the region of superexponential growth, and how many terms are used. It probably has an effective radius of convergence of about 2 with the number of terms I posted. If you want more convergence as real(z) increases or decreases, use the recurrence relation below. If you want more convergence as imag(z) increases, one way to get that is to generate the series for cheta(z100), centered at 100. For cheta/sexpeta, I actually use Newton Polynomial interpolation, centered around 100 for cheta, and +100 for sexpeta, with a 50 term polynomial, which has pretty good convergence out to a radius of 25 (~28 digits), and is accurate to 50 digits within a unit radius. I also recently included cheta/sexpeta support in my latest kneser.gp program.
 Sheldon
