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 Tetration series for integer exponent. Can you find the pattern? marraco Fellow   Posts: 100 Threads: 12 Joined: Apr 2011 02/13/2016, 06:07 AM (This post was last modified: 02/13/2016, 06:55 PM by marraco.) (01/30/2016, 09:06 PM)marraco Wrote: According of OEIS, the sequence for seems to be number of "forests" with n nodes and height at most z-1 (OEIS A210725) Following this, I had been trying to figure what a forest of fractional height is. should correspond to forests with height =. I tried various definitions for counting forests of half height, but none seems to work. Them I attempted to reverse engineer it by getting the coefficients first, and then trying to figure how to get them. My plan was to use Sheldonison's code to calculate the values of at various points, and numerically obtaining the coefficients of his Taylor series by applying finite differences. The problem is that for we need his Taylor series developed around x=1, and both knesser.gp and fatou.gp do not converge here. So, I planned to get the Taylor series around r, and with it, calculate back the Taylor around 1. I tried an r=1.01. calculated n (41) points of using knesser.gp, and from those points estimated the derivatives, and the bj values. This is the Pari/GP code I used. Code:\r kneser.gp \p 400 z=1.5 /*fractional exponent*/ r=1.01; /*place where derivatives will be calculated. Initially, coefficientes of Taylor developped around x=r*/ dx=1e-15; num=20; /*number of derivatives to calculate*/ T=vector(1+2*num,i,0); /*vector where tetrations are stored*/ /*^z(x+r) at various points separated by dx*/ {for(n=1,1+2*num,   init(r+(-1-num+n)*dx);   T[n]=real(sexp(z)); );} /*finding the coefficients for taylor series around x=r*/ /*using finite differences of order 'num'*/ m=matrix(1+2*num,1+2*num,r,c,((-(1+num)+r)*dx)^(c-1)); Tcoeff_at_r=m^-1*T~ /*indentification of the converged coefficients*/ /*calculating ^{1.5}1=1+tolerance*/ tolerance=abs(2*r); maxConvergedCoeff=0; {T1=0; for(n=1,num,     T1=T1+Tcoeff_at_r[n]*(-(r-1))^(n-1);     if(abs(T1)>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 : 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. « Next Oldest | Next Newest »

 Messages In This Thread Tetration series for integer exponent. Can you find the pattern? - by marraco - 01/30/2016, 09:06 PM RE: Tetration series for integer exponent. Can you find the pattern? - by Gottfried - 02/14/2016, 03:43 AM

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