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05/19/2011, 07:49 PM
(This post was last modified: 05/19/2011, 07:55 PM by JmsNxn.)
I ask because I want to observe how logarithmic semioperators behave for bases less than or equal to . I haven't a clue where I might going about getting these. I think it was Sheldon who posted coefficients for me before, but they were base 2, and the graphs didn't look pretty; so I just wonder if more erratic bases will give different results. I know that has a fix point at x = e, so I wonder if that might change anything and might shift the hump that appears with base 2.
If anybody wonders what I'm talking about it's and :
which behaves as addition
which behaves as multiplication
which behaves as exponentiation
Thanks for reading this. Any help would be greatly appreciated.
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(05/19/2011, 07:49 PM)JmsNxn Wrote: I ask because I want to observe how logarithmic semioperators behave for bases less than or equal to . Hi James,
I'd like to help for this neat idea. However, after my comparision of the four interpolationmethods I do no more trust in the relevance of the "regulartetration"powerseries as I used them up to now  so I do not know whether you want possibly waste your time with analysis of that powerseries. Unfortunately I still do not have a clue how the Dmitri Kouznetzov Cauchyintegralmethod works at all so I also cannot provide powerseries of that presumably better method.
But if it is just for the fun of exploration/exercise you can well have my Pari/GPprocedures for this. (in that case please contact me via email)
Regards
Gottfried
Gottfried Helms, Kassel
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Well I wouldn't be analyzing the Taylor series per se. I'd just like to see a graph of for any appropriate constants a and b over . My understanding is that the differences between each version of tetration are minute and probably wouldn't affect an approximate graph (It should at least be enough to verify that there are no humps).
I'll give you an email then.
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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
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05/21/2011, 02:14 PM
(This post was last modified: 05/22/2011, 01:10 AM by Gottfried.)
Just for an impression. With the matrixlogarithmmethod based on the exponentials recentered at e; g(h) = f(1e,h)+e depending on the iteration height h I get the following coefficients for g(h)
Code: 1.0
0.61109545377165144382
0.23170261447676551100
0.091781287662054951830
0.037564921705011073397
0.015773722201726337117
0.0067614637772198390363
0.0029477476181950503707
0.0013032939674486044656
0.00058306881272407925943
0.00026347551443499556924
0.00012007976722060208750
0.000055130237268268005131
0.000025472518304042709786
0.000011834707545807564518
0.0000055251010640897040596
0.0000025902729494681159146
0.0000012187271454481672018
0.00000057508317717804681085
0.00000027193027228176594270
0.00000012870902697527732596
0.000000060889266619581435803
0.000000028734742331292897512
0.000000013494371318392422045
0.0000000062882996391919038758
0.0000000028984843383423711421
0.0000000013171193093747536382
0.00000000058812247376504184749
0.00000000025724643591308081238
1.0991226259673585977E10
4.5760148165255342291E11
1.8525282102101486163E11
7.2799477529346619758E12
2.7731492194588571934E12
1.0228781132287377086E12
3.6501818149236149850E13
1.2594229104590033680E13
4.1994992731976094776E14
1.3528699883474457412E14
4.2097881296159269770E15
1.2652080888516218381E15
3.6723460072987803371E16
1.0294636636067164354E16
2.7873275063956835675E17
7.2897324850428282565E18
1.8417406660994510113E18
4.4956448474692462035E19
1.0603756293758013064E19
2.4170833917511111844E20
5.3253529443700961287E21
1.1342016411529572217E21
2.3354793597373322700E22
4.6500987042706443614E23
8.9536936179097868826E24
1.6674248602720150332E24
3.0035886177128642416E25
5.2339146859694508036E26
8.8235255818014501117E27
1.4391944413647261141E27
2.2713605538391230710E28
3.4686799592348898270E29
5.1259082336317069982E30
7.3302241660374222986E31
1.0144000556887333339E31
which means
Code: g(h)= 1.0 + 0.611... h  0.2317... h^2 + 0.09178... h^3 + ...
gives the the h'th iteration to base ae=exp(exp(1))~ 1.4446... beginning at x0=1 in the version of the "regular tetration". This is valid for 20 digits accuracy for at least 0<=h<=2, so two unitintervals of the iteration. The restriction to, say, 20 digits is because I've seen that ae^g(0.5)  g(1.5) ~ 1e15 and possibly I need a significant extension of my matrices/powerseries to achieve more accuracy.
@James: as promised I'll send the Pari/GPprocedures, so you may extend the precision to higher degree yourself, but give me some time to harvest the relevant script procedures from the the jungle of my Pari/GPscript files library...
Gottfried Helms, Kassel
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05/21/2011, 10:55 PM
(This post was last modified: 05/23/2011, 08:11 PM by sheldonison.)
(05/19/2011, 07:49 PM)JmsNxn Wrote: I ask because I want to observe how logarithmic semioperators behave for bases less than or equal to . I haven't a clue where I might going about getting these. I think it was Sheldon who posted coefficients for me before, but they were base 2, and the graphs didn't look pretty; so I just wonder if more erratic bases will give different results. I know that has a fix point at x = e, so I wonder if that might change anything and might shift the hump that appears with base 2.
If anybody wonders what I'm talking about it's and :
which behaves as addition
which behaves as multiplication
which behaves as exponentiation
Thanks for reading this. Any help would be greatly appreciated. Hey James,
Hey Andy, haven't heard from you in awhile! I've been laying low for awhile, trying to learn more background math so I can talk more intelligently on these forums, and maybe eventually write a paper....
I'll have to take a pass on James's larger problem, but I think I can get you a series for base . Jaydfox (Jay Daniels) called the upper superfunction for this base cheta(z). For all real(z), cheta(z) is > e, and increasing, and cheta(z) is entire. Henryk has made numerous other posts on this function, with references to earlier work, which is mathematically equivalent to iterating exp(z)1. I believe Gottfried may have posted results for the lower superfunction base eta. I'm using Jay's suggestion to normalize cheta(0)=2*e. First off a caveat: I feel like I have some understanding of all of the other bases for tetratation. Even though cheta(z) was the very first super exponential base I explored, there many things I don't understand cheta(z). Unlike all other bases, it has no periodicity or pseudo periodicity. I'm using Newton interpolation algorithm here, that seems to work quite well. Before I knew how to calculate sexp(z) for any other bases, I was very curious about cheta(z), and cheta(z) in the complex plane, and this is the algorithm I used to investigate it.
I center the Newton polynomial at cheta(95) with 25 sample points on either side. So that the Newton_polynomial(0)=cheta(95). This gives a consistent well behaved polynomial, whose 50th coefficient is 5E101, with consistent accurate values for values of cheta in the neighborhood of cheta(95), with an accuracy radius of at least 10 units, in either direction.
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
Now, for the inverse function I am also centering the series around z=2*e. So, you need to substitute y=z2*e, and this is the Taylor series for the inverse function, .
 Sheldon
Code: 0 2.8394844317361184366323169886 E46
1 0.84951664340115092455978946179380
2 0.29299658208028424957861348941347
3 0.10509962628439971957233240144673
4 0.038157724678352862884934417549581
5 0.013925154552107876079062165565924
6 0.0050952225123435103030773323796416
7 0.0018671879930152709206427694717335
8 0.00068490296133904125688872118488805
9 0.00025139034517175714464979472515850
10 0.000092313189599831207400315327952637
11 0.000033909513426122719199099928773847
12 0.000012459097214278183406543059927027
13 0.0000045786159697733865794175565753500
14 0.0000016828557401554335864965747890457
15 0.00000061860247032573763467393855008516
16 0.00000022741483226518418491094851146220
17 0.000000083610506876659067250809903046197
18 0.000000030741996153004058819109207794717
19 0.000000011303887549718411176045726840015
20 0.0000000041566598052944547999518543402356
21 0.0000000015285478957891868574437546094353
22 0.00000000056212008538080452772480392355221
23 2.0672482156588402950705541420527 E10
24 7.6027015127751986795782560107348 E11
25 2.7961060314772058262329305454437 E11
26 1.0283679197507576516530461271023 E11
27 3.7822616925408613984109447810316 E12
28 1.3911114500866907702507009276973 E12
29 5.1165688309305946581953915277227 E13
30 1.8819219526223455358781434824655 E13
31 6.9219700998269053341103265344978 E14
32 2.5460249489325270551084541622129 E14
33 9.3648317462180677670443574299830 E15
34 3.4446198869227001149385448196462 E15
35 1.2670282251343045884635356804874 E15
36 4.6605227237301411241081317439108 E16
37 1.7142969554422764221773785470395 E16
38 6.3058018366578812057806434170850 E17
39 2.3195151116741095091654975044205 E17
40 8.5321114627106034923460832031476 E18
41 3.1384707379328624159531298493402 E18
42 1.1544675657668811615413235021795 E18
43 4.2466584398146493466987617123853 E19
44 1.5621212453791332783782145294831 E19
45 5.7462410024361900551890874499463 E20
46 2.1137543536539174110135441193169 E20
47 7.7754704178615174823047812681216 E21
48 2.8602251382786823028314838146849 E21
49 1.0521437215593346136619106291665 E21
50 3.8703580282475950589132824159248 E22
51 1.4237323972688916255261870725321 E22
52 5.2372906974743051783073917784620 E23
53 1.9265754323196651962839735628669 E23
54 7.0870634936127301616271498413745 E24
55 2.6070391010756576820822103151750 E24
56 9.5902434635083516682157243528384 E25
57 3.5278697393657245345167652687742 E25
58 1.2977655109484285361881296652688 E25
59 4.7739806502062836993834657752311 E26
60 1.7561668807408021340395405357838 E26
61 6.4602835243675049421151226060374 E27
62 2.3765010449423382541883982332437 E27
63 8.7422867197032152849307293297797 E28
64 3.2159749628188831946088316129999 E28
65 1.1830437957766415405064649862914 E28
66 4.3520062549876002351263488987314 E29
67 1.6009534383143544576118352603138 E29
68 5.8893636455378353194841863940980 E30
69 2.1664990350170076098431454251477 E30
70 7.9698301848454300290272438589004 E31
71 2.9318385870247681422532261029951 E31
72 1.0785280589301633949989045005489 E31
73 3.9675574722314662059345458319316 E32
74 1.4595378961050141920140942761514 E32
75 5.3691789767797473462281237059548 E33
76 1.9751529348915986487843860664041 E33
77 7.2659750890193416485885457517150 E34
78 2.6729288396177802100060770186772 E34
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(05/19/2011, 07:49 PM)JmsNxn Wrote: I ask because I want to observe how logarithmic semioperators behave for bases less than or equal to . I haven't a clue where I might going about getting these.
Hmm, interesting question.
The theory is that the regular fractional iterates of ( ) have 0 convergence radius around the development (and fix)point e. From the theory of Ă‰calle it follows that one can compute the regular Abel function , i.e. the inverse of the superfunction , as the following limit:
where
for ,
.
(a value of indicates the function which is real on the left side of e and indicates the function which is real on the right side of e.)
Though this formula is usually designed to compute values of numerically, it can also be used to compute the powerseries of , say at .
To compute this we see that we have to calculate the formal powerseries of (if we start with the case of ) and further then apply logarithm and reciprocal to this powerseries. The composition of two powerseries is possible with rational terms of the coefficients of if h(0)=0.
But . Despite we can construct with only using compositions of powerseries which satisfy .
This can be done with the following inductive definition:
and
Verify that always:
for example:
We see that the constant term of the powerseries of is and so it converges to , hence the constant term of is and (though ). Hence we can apply the logarithm
powerseries which is
and also the reciprocal powerseries
.
Ok, when one applies all this stuff to the calculation of with , i.e. the Abel function which is real on , I get the following coefficients for n=1000 iterations:
Code: 2.03, 0.602, 0.248, 0.0942, 0.0351, 0.0130, 0.00482, 0.00178, 0.000657, 0.000242, 0.0000893, 0.0000329, 0.0000121, 4.47e6, 1.64e6, 6.05e7, 2.23e7, 8.20e8, 3.02e8, 1.11e8
it seems however that they are accurate only up to 3 digits.
The corresponding superfunction developed at 2.03 has then the coefficients:
Code: 0, 1.66, 1.14, 0.839, 0.656, 0.534, 0.448, 0.385, 0.337, 0.300, 0.270, 0.245, 0.225, 0.207, 0.192, 0.179, 0.168, 0.158, 0.149, 0.141
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(05/23/2011, 07:04 PM)bo198214 Wrote: ....The corresponding superfunction developed at 2.03 has then the coefficients:
Code: 0, 1.66, 1.14, 0.839, 0.656, 0.534, 0.448, 0.385, 0.337, 0.300, 0.270, 0.245, 0.225, 0.207, 0.192, 0.179, 0.168, 0.158, 0.149, 0.141
I got my code working for the lower superfunction at , or the sexp. Your results matches the Taylor series I get for , developed at 1, where sexp(1)=0.
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
a10 = 0.271726518049431
a11 = 0.247071598485337
a12 = 0.226503399721030
a13 = 0.209089537153272
a14 = 0.194158863580830
a15 = 0.181216899324044
a16 = 0.169891743680233
a17 = 0.159898527811493
a18 = 0.151015469479131
a19 = 0.143067376977840
 Sheldon
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(05/23/2011, 08:42 PM)sheldonison Wrote: Your results matches the Taylor series I get for , developed at 1, where sexp(1)=0.
Ya but your values are much more accurate as it seems, how do you obtain them?
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05/24/2011, 12:12 AM
(This post was last modified: 05/24/2011, 03:52 AM by sheldonison.)
(05/23/2011, 09:53 PM)bo198214 Wrote: (05/23/2011, 08:42 PM)sheldonison Wrote: Your results matches the Taylor series I get for , developed at 1, where sexp(1)=0.
Ya but your values are much more accurate as it seems, how do you obtain them? 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.
Now I have a 50 term polynomial, centered around z=100, which seems to have nice convergence properties. The 50th series term of the interpolating polynomial is ~2E102. The error terms seem to be around 10^50 close to z=100, and increases to 10^33 as the radius increases to 20 units. I haven't done a theoretical interpolation error analysis, but I have graphed sexpeta(z)eta^sexpeta(z1).
The next step is akin to a discreet version of analytic continuation, where we pick a sample circle. I used a unit circle centered around z=100. I usually pick 200 evenly spaced points around the unit circle, and for each of those points, take the logarithm base eta 100 times. Now we have another unit circle, with 200 sample points, centered around z=0. Use the Cauchy integral to generate a Taylor series from those 200 sample points, and you get a Taylor series, centered around z=0. I have routines that will generate a Taylor series with any number of sample points and sample radius, sort of like a discreet version of analytic continuation.
 Sheldon
