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 Does anyone have taylor series approximations for tetration and slog base e^(1/e)? JmsNxn Ultimate Fellow Posts: 895 Threads: 110 Joined: Dec 2010 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 semi-operators behave for bases less than or equal to $e^{1/e}$. 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 $log_{e^{1/e}}(x)$ 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 $0 \le q \le 1$ and $r:f(x) = f^{\alpha r}(x) = f^{[r]}(x)$: $x\, \{-q\}\, y = {\small (1-q):}\log_{b}({\small (q-1):}\log_{b}x+ {\small (q-1):}\log_{b}y)$ which behaves as addition $x\, \{1-q\}\, y = {\small (q-1):}\log_{b}({\small (1-q):}\log_{b}x + {\small (1-q):}\log_{b}y)$ which behaves as multiplication $x\, \{2-q\}\, y = {\small (q-1):}\log_{b}(y [{\small (1-q):}\log_{b}x])$ which behaves as exponentiation Thanks for reading this. Any help would be greatly appreciated. Gottfried Ultimate Fellow Posts: 851 Threads: 126 Joined: Aug 2007 05/20/2011, 07:34 AM (05/19/2011, 07:49 PM)JmsNxn Wrote: I ask because I want to observe how logarithmic semi-operators behave for bases less than or equal to $e^{1/e}$.Hi James, I'd like to help for this neat idea. However, after my comparision of the four interpolation-methods I do no more trust in the relevance of the "regular-tetration"-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 Cauchy-integral-method 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/GP-procedures for this. (in that case please contact me via email) Regards- Gottfried Gottfried Helms, Kassel JmsNxn Ultimate Fellow Posts: 895 Threads: 110 Joined: Dec 2010 05/20/2011, 06:34 PM Well I wouldn't be analyzing the Taylor series per se. I'd just like to see a graph of $f(t) = a\,\{t\}\,b$ for any appropriate constants a and b over $t\, \epsilon\, [-1, 2]$. 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. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/20/2011, 06:43 PM 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 Gottfried Ultimate Fellow Posts: 851 Threads: 126 Joined: Aug 2007 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 matrix-logarithm-method based on the exponentials recentered at e; g(h) = f(1-e,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 1.0991226259673585977E-10 -4.5760148165255342291E-11 1.8525282102101486163E-11 -7.2799477529346619758E-12 2.7731492194588571934E-12 -1.0228781132287377086E-12 3.6501818149236149850E-13 -1.2594229104590033680E-13 4.1994992731976094776E-14 -1.3528699883474457412E-14 4.2097881296159269770E-15 -1.2652080888516218381E-15 3.6723460072987803371E-16 -1.0294636636067164354E-16 2.7873275063956835675E-17 -7.2897324850428282565E-18 1.8417406660994510113E-18 -4.4956448474692462035E-19 1.0603756293758013064E-19 -2.4170833917511111844E-20 5.3253529443700961287E-21 -1.1342016411529572217E-21 2.3354793597373322700E-22 -4.6500987042706443614E-23 8.9536936179097868826E-24 -1.6674248602720150332E-24 3.0035886177128642416E-25 -5.2339146859694508036E-26 8.8235255818014501117E-27 -1.4391944413647261141E-27 2.2713605538391230710E-28 -3.4686799592348898270E-29 5.1259082336317069982E-30 -7.3302241660374222986E-31 1.0144000556887333339E-31which 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 unit-intervals of the iteration. The restriction to, say, 20 digits is because I've seen that ae^g(0.5) - g(1.5) ~ 1e-15 and possibly I need a significant extension of my matrices/powerseries to achieve more accuracy. @James: as promised I'll send the Pari/GP-procedures, 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/GP-script files library... Gottfried Helms, Kassel sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 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 semi-operators behave for bases less than or equal to $e^{1/e}$. 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 $log_{e^{1/e}}(x)$ 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 $0 \le q \le 1$ and $r:f(x) = f^{\alpha r}(x) = f^{[r]}(x)$: $x\, \{-q\}\, y = {\small (1-q):}\log_{b}({\small (q-1):}\log_{b}x+ {\small (q-1):}\log_{b}y)$ which behaves as addition $x\, \{1-q\}\, y = {\small (q-1):}\log_{b}({\small (1-q):}\log_{b}x + {\small (1-q):}\log_{b}y)$ which behaves as multiplication $x\, \{2-q\}\, y = {\small (q-1):}\log_{b}(y [{\small (1-q):}\log_{b}x])$ 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 $\eta = e^{1/e}$. 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 5E-101, 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 pari-gp, 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 E-10 23 7.3341040297826856350206259498267 E-11 24 2.3278852998967291568233733165642 E-11 25 7.3628505815581778431734554753314 E-12 26 2.3209857992934250244177110110812 E-12 27 7.2930204918450243443324949177246 E-13 28 2.2846097982451633980559339079833 E-13 29 7.1358033041146574466639840152247 E-14 30 2.2225543653988499920641676567938 E-14 31 6.9038181736676583445161044386747 E-15 32 2.1389425595138842758935272382177 E-15 33 6.6103445086593382475520541375449 E-16 34 2.0379986212392242975901900360689 E-16 35 6.2686731343734042989649316156714 E-17 36 1.9238630883507697992098052847583 E-17 37 5.8915865956656546031878586526468 E-18 38 1.8004488782209858871936565169317 E-18 39 5.4909672937838092357031986642134 E-19 40 1.6713366863796135261320082560915 E-19 41 5.0775249729439841733736059243593 E-20 42 1.5397061803790039357741194269061 E-20 43 4.6606303423839124073481231546437 E-21 44 1.4082983164868122075336349130200 E-21 45 4.2482389646419479670428084576358 E-22 46 1.2794035465663755925461443200159 E-22 47 3.8468885494278543290882325049087 E-23 48 1.1548705835256123872218006706955 E-23 49 3.4617530598681323751620914079008 E-24 50 1.0361306324224779753428626830104 E-24 51 3.0967381932148655440014679355037 E-25 52 9.2423246088852631003347605140667 E-26 53 2.7546045204596969212253367181553 E-26 54 8.1988428443303828566587256203397 E-27 55 2.4371065951483711807719660406315 E-27 56 7.2349910531853903620035897727885 E-28 57 2.1451384654610549641820852967050 E-28 58 6.3524081182337135717494628619063 E-29 59 1.8788780939059811381889624923360 E-29 60 5.5506897778611357646729108553389 E-30 61 1.6379251020900380753695185487417 E-30 62 4.8278087421120487722463748941123 E-31 63 1.4214279502607189609788849783519 E-31 64 4.1804970061336688633898121429008 E-32 65 1.2281981008321929711060382301444 E-32 66 3.6045845774315208753223996337347 E-33 67 1.0568098942930644804288412600335 E-33 68 3.0952921594935758380591422397535 E-34 Now, for the inverse function I am also centering the series around z=2*e. So, you need to substitute y=z-2*e, and this is the Taylor series for the inverse function, $\text{cheta}^{-1}(z-2e)$. - Sheldon Code:0 -2.8394844317361184366323169886 E-46 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 E-10 24 -7.6027015127751986795782560107348 E-11 25 2.7961060314772058262329305454437 E-11 26 -1.0283679197507576516530461271023 E-11 27 3.7822616925408613984109447810316 E-12 28 -1.3911114500866907702507009276973 E-12 29 5.1165688309305946581953915277227 E-13 30 -1.8819219526223455358781434824655 E-13 31 6.9219700998269053341103265344978 E-14 32 -2.5460249489325270551084541622129 E-14 33 9.3648317462180677670443574299830 E-15 34 -3.4446198869227001149385448196462 E-15 35 1.2670282251343045884635356804874 E-15 36 -4.6605227237301411241081317439108 E-16 37 1.7142969554422764221773785470395 E-16 38 -6.3058018366578812057806434170850 E-17 39 2.3195151116741095091654975044205 E-17 40 -8.5321114627106034923460832031476 E-18 41 3.1384707379328624159531298493402 E-18 42 -1.1544675657668811615413235021795 E-18 43 4.2466584398146493466987617123853 E-19 44 -1.5621212453791332783782145294831 E-19 45 5.7462410024361900551890874499463 E-20 46 -2.1137543536539174110135441193169 E-20 47 7.7754704178615174823047812681216 E-21 48 -2.8602251382786823028314838146849 E-21 49 1.0521437215593346136619106291665 E-21 50 -3.8703580282475950589132824159248 E-22 51 1.4237323972688916255261870725321 E-22 52 -5.2372906974743051783073917784620 E-23 53 1.9265754323196651962839735628669 E-23 54 -7.0870634936127301616271498413745 E-24 55 2.6070391010756576820822103151750 E-24 56 -9.5902434635083516682157243528384 E-25 57 3.5278697393657245345167652687742 E-25 58 -1.2977655109484285361881296652688 E-25 59 4.7739806502062836993834657752311 E-26 60 -1.7561668807408021340395405357838 E-26 61 6.4602835243675049421151226060374 E-27 62 -2.3765010449423382541883982332437 E-27 63 8.7422867197032152849307293297797 E-28 64 -3.2159749628188831946088316129999 E-28 65 1.1830437957766415405064649862914 E-28 66 -4.3520062549876002351263488987314 E-29 67 1.6009534383143544576118352603138 E-29 68 -5.8893636455378353194841863940980 E-30 69 2.1664990350170076098431454251477 E-30 70 -7.9698301848454300290272438589004 E-31 71 2.9318385870247681422532261029951 E-31 72 -1.0785280589301633949989045005489 E-31 73 3.9675574722314662059345458319316 E-32 74 -1.4595378961050141920140942761514 E-32 75 5.3691789767797473462281237059548 E-33 76 -1.9751529348915986487843860664041 E-33 77 7.2659750890193416485885457517150 E-34 78 -2.6729288396177802100060770186772 E-34 bo198214 Administrator Posts: 1,493 Threads: 95 Joined: Aug 2007 05/23/2011, 07:04 PM (05/19/2011, 07:49 PM)JmsNxn Wrote: I ask because I want to observe how logarithmic semi-operators behave for bases less than or equal to $e^{1/e}$. I haven't a clue where I might going about getting these. Hmm, interesting question. The theory is that the regular fractional iterates of $f(x)=\eta^x$ ($\eta=e^{1/e}$) 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 $\alpha$, i.e. the inverse of the superfunction $\sigma(z)=f^{[z]}(z_0)$, as the following limit: $\alpha_\theta(z) = \lim_{n\to\infty} F_{\theta}(f^{[-\theta n]}(z)/e-1) + \theta n$ where $F_\theta(z)=\frac{1}{3}\log(\theta z) - \frac{2}{z}$ for $\theta z > e$, $\theta=\pm 1$. (a value of $\theta=-1$ indicates the function which is real on the left side of e and $\theta=+1$ indicates the function which is real on the right side of e.) Though this formula is usually designed to compute values of $\alpha_\theta$ numerically, it can also be used to compute the powerseries of $\alpha_\theta$, say at $z=0$. To compute this we see that we have to calculate the formal powerseries of $f^{[n]}$ (if we start with the case of $\theta=-1$) and further then apply logarithm and reciprocal to this powerseries. The composition of two powerseries $(g\circ h)(x)=g(h(x))$ is possible with rational terms of the coefficients of $g\circ h$ if h(0)=0. But $f(0)=\eta^0=1\neq 0$. Despite we can construct $f^{[n]}$ with only using compositions of powerseries $H_n$ which satisfy $H_n(0)=0$. This can be done with the following inductive definition: $H_1(x)=\eta^x - 1$ and $H_{n+1}(x) = (\eta[4]n)^{H_n(x)}-1$ Verify that always: $f^{[n]}(x)=(\eta[4](n-1))*(H_n(x)+1)$ for example: $f^{[2]}(x)]=\eta^{\eta^x} = \eta\eta^{\eta^x-1}=\eta*(\eta^{H_1(x)} - 1 + 1) = (\eta[4]1)(H_2(x)+1)$ We see that the constant term of the powerseries of $f^{[n]}$ is $\eta[4](n-1)$ and so it converges to $e$, hence the constant term of $f^{[n]}(x)/e-1$ is $\epsilon=\frac{\eta[4](n-1)}{e}-1$ and $\epsilon\neq 0$ (though $\epsilon\to 0$). Hence we can apply the logarithm powerseries which is $\log(x+\epsilon)=\log(\epsilon)+\log(1+x/\epsilon) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n \epsilon^n} x^{n}$ and also the reciprocal powerseries $\frac{1}{x+\epsilon} = \frac{1}{\epsilon} \sum_{n=0}^\infty \frac{(-1)^{n}}{\epsilon^n} x^n$. Ok, when one applies all this stuff to the calculation of $\alpha_\theta$ with $\theta=-1$, i.e. the Abel function which is real on $x, 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, 4.47e-6, 1.64e-6, 6.05e-7, 2.23e-7, 8.20e-8, 3.02e-8, 1.11e-8 it seems however that they are accurate only up to 3 digits. The corresponding super-function 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 sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/23/2011, 08:42 PM (05/23/2011, 07:04 PM)bo198214 Wrote: ....The corresponding super-function 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, 0.141I got my code working for the lower superfunction at $\eta$, or the sexp. Your results matches the Taylor series I get for $\text{sexp}_{\eta}(z-1)$, 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 bo198214 Administrator Posts: 1,493 Threads: 95 Joined: Aug 2007 05/23/2011, 09:53 PM (05/23/2011, 08:42 PM)sheldonison Wrote: Your results matches the Taylor series I get for $\text{sexp}_{\eta}(z-1)$, developed at -1, where sexp(-1)=0. Ya but your values are much more accurate as it seems, how do you obtain them? sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 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 $\text{sexp}_{\eta}(z-1)$, 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 pari-gp, 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 ~2E-102. 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(z-1). 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 « Next Oldest | Next Newest »

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