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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 $b_j$ sequence for $\vspace{15}{^z(x+1)}$ 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. $\vspace{25}{^{1+\frac{1}{n}}(x+1)}$ should correspond to forests with height =$\vspace{15}{\frac{1}{n}}$. 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$\vspace{15}{b_j}$ coefficients first, and then trying to figure how to get them. My plan was to use Sheldonison's code to calculate the values of $\vspace{25}{^{1+\frac{1}{2}}(x+1)}$ at various points, and numerically obtaining the coefficients of his Taylor series by applying finite differences. The problem is that for $\vspace{25}{^{1+\frac{1}{2}}(x+1)}$ 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 $\vspace{15}{^{1.5}(r+n.\Delta x)}$ 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 $\vspace{15}{^{1.5}(x+1)}$: 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. « 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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