Tetration series for integer exponent. Can you find the pattern? marraco Fellow   Posts: 100 Threads: 12 Joined: Apr 2011 01/30/2016, 09:06 PM (This post was last modified: 02/12/2016, 11:06 PM by marraco.) $^0(1+x) \,=\,$${\color{Red} 1}$$+ 0+ 0+0...$ $^1(1+x) \,=\,$${\color{Red} 1}+ x$$+ 0+0+0...$ $^2(1+x) \,=\,$${\color{Red} 1}+ x+ x^2$$+\frac {1}{2}x^{3}+\frac {1}{3}x^{4}+\frac {1}{12}x^{5}+\frac {3}{40}x^{6}+\frac {-1}{120}x^{7}+\frac {59}{2520}x^{8}+\frac {-71}{5040}x^{9}+\frac {131}{10080}x^{10}+\frac {-53}{5040}x^{11}+O(x^{12})$ $^3(1+x) \,=\,$${\color{Red} 1}+x+x^2 +\frac {3}{2}x^{3}$$+\frac {4}{3}x^{4}+\frac {3}{2}x^{5}+\frac {53}{40}x^{6}+\frac {233}{180}x^{7}+\frac {5627}{5040}x^{8}+\frac {2501}{2520}x^{9}+\frac {8399}{10080}x^{10}+\frac {34871}{50400}x^{11}+O(x^{12})$ $^4(1+x) \,=\,$${\color{Red} 1}+x+x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}$$+3x^{5}+\frac {163}{40}x^{6}+\frac {1861}{360}x^{7}+\frac {33641}{5040}x^{8}+\frac {8363}{1008}x^{9}+\frac {22391}{2160}x^{10}+\frac {7589}{600}x^{11}+O(x^{12})$ $^5(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}$$+\frac {243}{40}x^{6}+\frac {3421}{360}x^{7}+\frac {71861}{5040}x^{8}+\frac {54371}{2520}x^{9}+\frac {69281}{2160}x^{10}+\frac {7200983}{151200}x^{11}+O(x^{12})$ $^6(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}$$+\frac {4321}{360}x^{7}+\frac {102941}{5040}x^{8}+\frac {85871}{2520}x^{9}+\frac {61333}{1080}x^{10}+\frac {886763}{9450}x^{11}+O(x^{12})$ $^7(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}$$+\frac {118061}{5040}x^{8}+\frac {106661}{2520}x^{9}+\frac {81583}{1080}x^{10}+\frac {10169449}{75600}x^{11}+O(x^{12})$ $^8(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}$$+\frac {115481}{2520}x^{9}+\frac {93013}{1080}x^{10}+\frac {12169699}{75600}x^{11}+O(x^{12})$ $^9(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}+\frac {118001}{2520}x^{9}$$+\frac {97333}{1080}x^{10}+\frac {13165099}{75600}x^{11}+O(x^{12})$ $^{10}(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}+\frac {118001}{2520}x^{9}+\frac {98413}{1080}x^{10}$$+\frac {13505299}{75600}x^{11}+O(x^{12})$ with each integer power of n, the tetration $\\[15pt] {^n(x+1)}$ converges, one extra coefficient (highlighted in red), to the Taylor series of $\\[15pt] {\frac {lambertW(-ln(x+1))}{ln(x+1)} }$, which gives the known asymptotic limit for $\\[15pt] {e^{-e}<(x+1)tolerance,break,maxConvergedCoeff=n); )} /*finding the coefficients for taylor series around x=1*/ f(x)=sum(n=1,maxConvergedCoeff,Tcoeff_at_r[n]*(x-(r-1))^(n-1)); Tcoeff_at_1=vector(maxConvergedCoeff,n,polcoeff(f(x),n-1)); cj=vector(maxConvergedCoeff-1,n,Tcoeff_at_1[n+1]*(n)!) bj=vector(length(cj),n,sum(j=1,n,stirling(n,j,2)*cj[j])) Unfortunately, even using a large precision (it took more than one day running on an i7 920 processor), I only got 4 derivatives. The other have stability problems, and the ones I got are not very precise. Anyways, these are the closest numbers I got for bj of \( \\[15pt] {^{1.5}(x+1)}$: Code:%38 = [1.000152704474621237210741951681248733595287259524430467296, 2.861320494145353285475346429548911736678675196129994723805, -13.6475978846314412502384523327046630578, 756.92186972158813546692266102007258863] Those should be "number of forests with height 1/2" for 1,2,3 and 4 nodes. The first number is the more accurate, but not for much. It should be 1, and the rest loses many digits of accuracy each one. So, for 2 nodes, there should be 2.86 forests of 1/2 height. But that number is surely a rough approximation. It may well be 2.5 instead of 2.86, and I find dubious that 3 nodes have a negative number of forests. I have no idea on how to improve precision. Increasing the decimals in Pari/GP does nothing. Maybe using a more accurate numerical difference equation? I used finite difference of order 40. I have the result, but I do not yet know how to get it. sheldonison Long Time Fellow    Posts: 684 Threads: 24 Joined: Oct 2008 02/14/2016, 03:25 AM (This post was last modified: 02/14/2016, 03:30 AM by sheldonison.) (02/13/2016, 06:07 AM)marraco Wrote: ... I have no idea on how to improve precision. Increasing the decimals in Pari/GP does nothing. Maybe using a more accurate numerical difference equation? I used finite difference of order 40. See http://math.eretrandre.org/tetrationforu...729&page=2 post#15, In that post, there is a Taylor series for the 1st derivative of sexp_b(z) developed around b=2, and a Taylor series for sexp_b(-0.5), also developed around b=2. At the time I posted that, I also had developed Taylor series for the first 50 or so derivatives in the neighborhood of b=2. The key is to treat sexp_b(z) as analytic for the base b in the complex plane, in a circle around b=2. Then sample a set of points around a unit circle, and use Cauchy/Fourier which will give really good approximations to the derivatives. There is a mild singularity at b=eta which barely effects results at all, and a much much much more severe singularity at b=1. There is also reduced precision on the Shell Thron boundary. As I recall, the Taylor series posted were accurate to >25 decimal digits; using a relatively small number of sample points. I can post some pari-gp code that would work with fatou.gp to recreate that effort, if Maracco is interested. Of course, these results will be numeric, not a series. - Sheldon - Sheldon Gottfried Ultimate Fellow     Posts: 901 Threads: 130 Joined: Aug 2007 02/14/2016, 03:43 AM (This post was last modified: 02/14/2016, 03:46 AM by Gottfried.) (01/30/2016, 09:06 PM)marraco Wrote: $^0(1+x) \,=\,$${\color{Red} 1}$$+ 0+ 0+0...$ $^1(1+x) \,=\,$${\color{Red} 1}+ x$$+ 0+0+0...$ $^2(1+x) \,=\,$${\color{Red} 1}+ x+ x^2$$+\frac {1}{2}x^{3}+\frac {1}{3}x^{4}+\frac {1}{12}x^{5}+\frac {3}{40}x^{6}+\frac {-1}{120}x^{7}+\frac {59}{2520}x^{8}+\frac {-71}{5040}x^{9}+\frac {131}{10080}x^{10}+\frac {-53}{5040}x^{11}+O(x^{12})$ $^3(1+x) \,=\,$${\color{Red} 1}+x+x^2 +\frac {3}{2}x^{3}$$+\frac {4}{3}x^{4}+\frac {3}{2}x^{5}+\frac {53}{40}x^{6}+\frac {233}{180}x^{7}+\frac {5627}{5040}x^{8}+\frac {2501}{2520}x^{9}+\frac {8399}{10080}x^{10}+\frac {34871}{50400}x^{11}+O(x^{12})$Perhaps this is interesting for you: http://go.helms-net.de/math/tetdocs/Pasc...trated.pdf The tetration of the Pascalmatrix give that coefficients by matrix-exponentiation, and from this might possibly result smoother formulae for the expression of the single coefficients. Gottfried Gottfried Helms, Kassel marraco Fellow   Posts: 100 Threads: 12 Joined: Apr 2011 02/14/2016, 05:08 AM (02/14/2016, 03:25 AM)sheldonison Wrote: (02/13/2016, 06:07 AM)marraco Wrote: ... I have no idea on how to improve precision. Increasing the decimals in Pari/GP does nothing. Maybe using a more accurate numerical difference equation? I used finite difference of order 40. See http://math.eretrandre.org/tetrationforu...729&page=2 post#15, In that post, there is a Taylor series for the 1st derivative of sexp_b(z) developed around b=2, and a Taylor series for sexp_b(-0.5), also developed around b=2. At the time I posted that, I also had developed Taylor series for the first 50 or so derivatives in the neighborhood of b=2. The key is to treat sexp_b(z) as analytic for the base b in the complex plane, in a circle around b=2. Then sample a set of points around a unit circle, and use Cauchy/Fourier which will give really good approximations to the derivatives. There is a mild singularity at b=eta which barely effects results at all, and a much much much more severe singularity at b=1. There is also reduced precision on the Shell Thron boundary. As I recall, the Taylor series posted were accurate to >25 decimal digits; using a relatively small number of sample points. I can post some pari-gp code that would work with fatou.gp to recreate that effort, if Maracco is interested. Of course, these results will be numeric, not a series. - Sheldon As I understand, those are series for the variable in the exponent, but we need series for the variable in the base; a series for sexp_{x+r}(1.5). I mean not the series for $\\[15pt] {^xb}$, but for $\\[15pt] {^{1.5}(x+r)}$. (I write r instead of 1). That's why I used your knesser.gp to calculate points for different bases separated by n*dx steps. knesser gave me better precission than fatou. I don't know if I should had configured some parameter in fatou to get more digits. I also used a step dx=1e-15. When I tried a smaller dx=1e-20 precision was destroyed, so the error may be numerical instability inherent to the differentiation scheme. (02/14/2016, 03:43 AM)Gottfried Wrote: (01/30/2016, 09:06 PM)marraco Wrote: $^0(1+x) \,=\,$${\color{Red} 1}$$+ 0+ 0+0...$ $^1(1+x) \,=\,$${\color{Red} 1}+ x$$+ 0+0+0...$ $^2(1+x) \,=\,$${\color{Red} 1}+ x+ x^2$$+\frac {1}{2}x^{3}+\frac {1}{3}x^{4}+\frac {1}{12}x^{5}+\frac {3}{40}x^{6}+\frac {-1}{120}x^{7}+\frac {59}{2520}x^{8}+\frac {-71}{5040}x^{9}+\frac {131}{10080}x^{10}+\frac {-53}{5040}x^{11}+O(x^{12})$ $^3(1+x) \,=\,$${\color{Red} 1}+x+x^2 +\frac {3}{2}x^{3}$$+\frac {4}{3}x^{4}+\frac {3}{2}x^{5}+\frac {53}{40}x^{6}+\frac {233}{180}x^{7}+\frac {5627}{5040}x^{8}+\frac {2501}{2520}x^{9}+\frac {8399}{10080}x^{10}+\frac {34871}{50400}x^{11}+O(x^{12})$Perhaps this is interesting for you: http://go.helms-net.de/math/tetdocs/Pasc...trated.pdf The tetration of the Pascalmatrix give that coefficients by matrix-exponentiation, and from this might possibly result smoother formulae for the expression of the single coefficients.That's great. You found that the tetration of the pascal matrix produces the bj sequences. If we can iterate logaritms maybe we can figure what a tree of negative height is. I wonder if we can generate the tetrated pascal with a series, and if that series matches the series for some real base. That base should be important. If just we could calculate $\\[15pt] {^{1.5}P}$, we would get the bj for trees of 1/2 height. I have the result, but I do not yet know how to get it. sheldonison Long Time Fellow    Posts: 684 Threads: 24 Joined: Oct 2008 02/14/2016, 07:55 AM (This post was last modified: 02/14/2016, 04:14 PM by sheldonison.) sheldon Wrote:See http://math.eretrandre.org/tetrationforu...729&page=2 post#15, In that post, there is a Taylor series for the 1st derivative of sexp_b(z) developed around b=2, and a Taylor series for sexp_b(-0.5), also developed around b=2. At the time I posted that, I also had developed Taylor series for the first 50 or so derivatives in the neighborhood of b=2. (02/14/2016, 05:08 AM)marraco Wrote: As I understand, those are series for the variable in the exponent, but we need series for the variable in the base; a series for sexp_{x+r}(1.5). I mean not the series for $\\[15pt] {^xb}$, but for $\\[15pt] {^{1.5}(x+r)}$. (I write r instead of 1). That's why I used your knesser.gp to calculate points for different bases separated by n*dx steps... Those series are for the base not the exponent! That's why they needed to be generated with tetcomplex rather than kneser; because getting such amazingly good analytic convergence for such a series requires complex base tetration, with results for n bases, equally spaced around a circle centered on base=2. fatou.gp (with small code updates) could also generate such results. To evaluate those two referenced Taylor series, substitute x=(b-2), to evaluate the Taylor series. This is probably off topic, but bases between 1 and eta=exp(1/e) aren't the same function as Kneser tetration for real bases>eta. Also, when you develop the iterated exponential using regular iteration from the attracting fixed point for baseseta because it is a completely different function than Kneser tetration. - Sheldon MorgothV8 Junior Fellow  Posts: 18 Threads: 6 Joined: Dec 2012 02/14/2016, 06:25 PM I'm not that good on math, but.... I can help as a programmer. Maybe precision problems are due to usage of "fixed" floating points (I mean fixed bit numbers). There are libraries that can work with "big number" or "arbitrary precision". For example GMP (this is a C/C++ library) gives non limited (limited by memory) number representation, so there is no problem with numbers like 10^(10^10) etc. And MPFR library does similar thing for floating point numbers ..... I've written some tetration programs using those two, and they're much more precise, but at the horrible cost of memory and time. Fuji GSW690III Nikon D3, Nikkors 14-24/2.8, 24/1.4, 35/2, 50/1.4, 85/1.4, 135/2, 80-200/2.8 marraco Fellow   Posts: 100 Threads: 12 Joined: Apr 2011 02/14/2016, 07:31 PM (This post was last modified: 02/14/2016, 07:42 PM by marraco.) (02/14/2016, 07:55 AM)sheldonison Wrote: Those series are for the base not the exponent! That's why they needed to be generated with tetcomplex rather than kneser; because getting such amazingly good analytic convergence for such a series requires complex base tetration, with results for n bases, equally spaced around a circle centered on base=2. fatou.gp (with small code updates) could also generate such results. To evaluate those two referenced Taylor series, substitute x=(b-2), to evaluate the Taylor series. ...I'm royally confused now. the coefficients in the last code tag of this post are the Taylor series coefficients of the function $\\[15pt] {f_2(x)\,=\,^{-0.5}(x+2)}$ ? Then it should be $\\[15pt] {f_{2\,(0)}\,=\, ^{-0.5}(2)=0.544764121459557...}$ $\\[15pt] {f_{2\,(1)}\,=\, ^{-0.5}(3)=0.486648923935738...}$ $\\[15pt] {f_{2\,(2)}\,=\, ^{-0.5}(4)=0.457213238216139...}$ but this code gives Code:\p 100 /*coefficients of Taylor around 2 of ^{-0.5}x*/ a=[0.5447641214595567339801218858257244685854,-0.09026490293475114180982800726025252487179,0.05334642698935378617403396491528890594804,-0.03638190492562309183765608353362070821840,0.02665589484943122254265742189263438424835,-0.02047608577133435850738520805893632252252,0.01628939391559684527389871185757624228228,-0.01331802035638468229849633176805710250959,0.01113080347039454404398917618932270486539,-0.009471945601741301301799666960159500493414,0.008181870472918983418952481797363865140773,-0.007156971109633091475785436209879906176635,0.006327698270413005651257016844418549882893,-0.005646005057506155565996841622134687059648,0.005077852297548008377590502397935756807242,-0.004598579564709383679003395147264261288216,0.004189960720720897899813797031566114828076,-0.003838281581489355718364575037825080832826,0.003533055202798846826754080155559369484869,-0.003266144873505376098605478730250127897586,0.003031153706186763516973507099991793317656,-0.002822992160610217843850530938287773588695,0.002637566583362648226391359841543536218954,-0.002471551499838161939220088380143251371568,0.002322220812129558330686955182616565785365,-0.002187321052201244713039707812416923885954,0.002064975080105974232286030081964966694861,-0.001953608108640999822645409586956590179027,0.001851890298539576237491912671987439471748,-0.001758691790434811807511081574014212915353,0.001673047168806168273884584564550319035826,-0.001594127148975512837191984637091772279620,0.001521215846034519175841930450931000214014,-0.001453692394304569109742037483974061901089,0.001391015984731342378663288972983370893753,-0.001332713607717317327426366538502721958881,0.001278369952559800466677607874323180950794,-0.001227619037446880124253911053592339307366,0.001180137236855264191223682764928143626889,-0.001135637444029102801867724419804092337925,0.001093864160641107689052187995385697434389,-0.001054589347847405322195327651270883669674,0.001017608905751088252280461602251417505390,-0.0009827396740070429313553656082778187553405,0.0009498168665858747234148639733725614346048,-0.0009186918698079533934930814875314534697736,0.0008892303455966866573662840017051026789547,-0.0008613105921955727892242590292733293823957,0.0008348221228916550711398042195621478636400,-0.0008096644300085595816921514966352297536765,0.0007857459069005338594524734325749675459648,-0.0007629829051481069438849922812669102687654,0.0007412989078245228555308404776565708380003,-0.0007206238027261275601524070871984237792707,0.0007008932419630012770449395923516096144246,-0.0006820480763866325438818551382445843370545,0.0006640338550679383137667663988829019940730,-0.0006468003814946490331294868382872885969872,0.0006303013193831178111678644623963748370560,-0.0006144938420376038744680011450272121474617,0.0005993383200741946673108758684076009577129,-0.0005847980430851115584178527733490186488280,0.0005708389714760195065962019338923580583096,-0.0005574295152845303003325133006371836953863,0.0005445403373002463827629604348689289744143,-0.0005321441782716469402918017155794455019026,0.0005202157024181592772390603384936993523783,-0.0005087313618820156701812678182987572351923,0.0004976692791697413648392248040346423784153,-0.0004870091470646722871348673605764087543724,0.0004767321459596961918548061350965022366495,-0.0004668208790873150944172435667760750155470,0.0004572593267416065803546985077871002284578,-0.0004480328213309398865880156803821028719749,0.0004391280460191844112860195774985849996584,-0.0004305330608687577109307488887504560310821,0.0004222373618726151852083171686555860996115,-0.0004142319801615488633254732943499557223787,0.0004065096311401751693759190311085411363661,-0.0003990649265288728503867805273227150792807,0.0003918946665218884447060007230972331199284,-0.0003849982348510417760694970639142865634935,0.0003783781269217428875539830317674205408416,-0.0003720406509700054967447945428476416910439,0.0003659968551918890293717641839010994444982,-0.0003602637511201750326035146298122448667960,0.0003548659266520138706765657962715630833711,-0.0003498376730740476659615080146624803134097,0.0003452257919088022517001222379805220095853,-0.0003410933031098378510324434621473247433045,0.0003375243510812609314744505648626776451409,-0.0003346307060211903759381304723940593677783,0.0003325603945037557882292885310966766766416,-0.0003315091777419398624384857779237117250571,0.0003317358460153907833479388101334190239424,-0.0003335826371421790840947235619676003126446,0.0003375025483293688889372157683210232309158,-0.0003440959391986930633575807516000693209207,0.0003541596811015852018702302245010936694298,-0.0003687532792553722994412436565871821184048,0.0003892879974033369355506985563538084697749,-0.0004176472122482091920832394022681079546693,0.0004563492419325118494433069912923916819317,-0.0005087680413650084606119114211931555096770,0.0005794328703416784956152178590887291116773,-0.0006744359202315312364732927380874961661109,0.0008019877695049482375578463387186199880001,-0.0009731755956181394060092634776613016580966,0.00120299993096644682942613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/*Taylor series around 2 of ^{-0.5}x*/ f2(x)=sum(n=0,150,a[n+1]/n!*x^n)` $\\[15pt] {f_{2\,(0)}\,=\, 0.5447641214595567...}$ $\\[15pt] {f_{2\,(1)}\,=\, 0.4760690431769200...}$ $\\[15pt] {f_{2\,(2)}\,=\, 0.4358972773362133...}$ ... and why textcomplex.gp "generates tetration for arbitrary bases" instead of "arbitrary exponents"? Sounds like the same as knesser and fatou (the base is constant, and the exponent is the variable) I have the result, but I do not yet know how to get it. marraco Fellow   Posts: 100 Threads: 12 Joined: Apr 2011 02/14/2016, 07:41 PM (This post was last modified: 02/14/2016, 07:54 PM by marraco.) (02/14/2016, 06:25 PM)MorgothV8 Wrote: I'm not that good on math, but.... I can help as a programmer. Maybe precision problems are due to usage of "fixed" floating points (I mean fixed bit numbers).I'm not even a mathematician. Just an engineer. Pari/GP uses arbitrary precision, and I used like 300 or 400 decimals. I also inverted the function for the calculation of the derivatives and verified that there is no loss of information. ...but beyond the precision used, the method of finite differences has inherent error. That may be the cause, or knesser may need more precision (with the one I choose, it took more than a day to run, but it may not be enough), or there is a bug on my posted code (ask for clarification if you want to check it), or all of those... I have the result, but I do not yet know how to get it. « Next Oldest | Next Newest »

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