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 Road testing Ansus' continuum product formula mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/13/2009, 10:55 AM Hi. On the thread http://math.eretrandre.org/tetrationforu...273&page=3 the following integral+continuum product iterative formula was mentioned for tetration to base e: $f(x)=f'(0)\int_{-1}^x \prod _x f(x+1)dx$ So I put together a Pari/GP code to road-test it numerically, to see if maybe it would converge on something. I use the formula $\prod_x f(x) = \exp\left(\sum_x \log(f(x))\right)$ to convert the product to a sum (which also has the nice effect of changing the x+1 to x), and then the Faulhaber's formula (expressed using Bernoulli polynomials/coefficients) is used to do the continuous sum on a set of Taylor coefficients. The Bell polynomials can be used to compute the exponential exp of the series. When I run the code, I notice something strange. For a small number of terms in the test array (I think around 15), it seems to converge and give like 4-5 digits of accuracy. For a larger number (say 28 or more), however, it starts converging but then it blows up.... Not sure if this is a problem with the code, or what... That's what I want to find out. As ultimately I'd like to be able to explore the general formula for other bases, including the really intriguing case of $0 < b < e^{-e}$... Code:/* Continuous sum transformation. */ bernoulli(n) = if(n==0, 1, (-1)^(n+1) * n * zeta(1 - n)); contsumcoef(series, k) = if(k==0, 0, k!*sum(n=1,matsize(series)[2], series[n]/(n!)*binomial(n, k)*bernoulli(n-k))); sumoperator(series) = { local(v=series); for(i=1,matsize(series)[2],v[i] = contsumcoef(series, i-1)); return(v); } tsum(series, x) = sum(k=0,matsize(series)[2]-1,series[k+1]/k! * x^k); /* Compute exponential of a Taylor series. */ bellcomplete(coefs) = {             local(n=matsize(coefs)[2]);             local(M=matrix(n,n));             for(i=1,n,\                 for(j=1,n,\                     M[i, j] = 0;                     if(j-i+1 > 0, M[i, j] = binomial(n-i, j-i)*coefs[j-i+1]);                     if(j-i+1 == 0, M[i, j] = -1);                 );             );             return(matdet(M)); } expoperator(series) = {            local(v = series);            for(i=1,matsize(series)[2]-1,                v[i+1] = bellcomplete(vector(i,n,series[n+1])));            v[1] = 1;            return(v*exp(series[1])); } /* Compute integral of a Taylor series at lower bound a. */ intoperator(series, a) = {            local(v = series);            for(i=1,matsize(series)[2]-1,                v[i+1] = series[i]);            v[1] = -sum(i=2,matsize(series)[2],v[i]*(a^(i-1))/((i-1)!));            return(v); } /* Tetration iteration */ tetiter(f0, series) = f0*intoperator(expoperator(sumoperator(series)), -1); To use the code, run tetiter several times over a vector containing coefficients for an initial guess (all zeroes will work), and then use tsum to evaluate it at a given x-value. f0 should be set equal to a numerical value for the derivative at 0 (see the original thread for the value I used). « Next Oldest | Next Newest »

 Messages In This Thread Road testing Ansus' continuum product formula - by mike3 - 09/13/2009, 10:55 AM RE: Road testing Ansus' continuum product formula - by bo198214 - 09/17/2009, 09:35 AM RE: Road testing Ansus' continuum product formula - by bo198214 - 09/17/2009, 02:22 PM RE: Road testing Ansus' continuum product formula - by mike3 - 09/17/2009, 08:37 PM RE: Road testing Ansus' continuum product formula - by Gottfried - 09/17/2009, 10:35 PM the summation problem, references - by bo198214 - 09/21/2009, 02:29 PM RE: the summation problem, references - by mike3 - 09/21/2009, 08:06 PM RE: the summation problem, references - by bo198214 - 09/22/2009, 01:02 AM RE: the summation problem, references - by mike3 - 09/22/2009, 05:27 AM RE: the summation problem, references - by mike3 - 09/22/2009, 05:30 AM RE: the summation problem, references - by bo198214 - 09/22/2009, 07:39 AM RE: Road testing Ansus' continuum product formula - by Gottfried - 09/22/2009, 12:48 PM RE: Road testing Ansus' continuum product formula - by mike3 - 09/22/2009, 08:58 PM

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