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Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - Printable Version +- Tetration Forum (https://tetrationforum.org) +-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3) +--- Thread: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? (/showthread.php?tid=633) Pages:
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RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - bo198214 - 05/24/2011 (05/24/2011, 12:12 AM)sheldonison Wrote: 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. You mean, you take the interpolation polynomial at the integer-points of sexp? Yeah true, that also yields regular iteration. I think we discussed this a lot with Ansus - the Newton (interpolation) formula. RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - Gottfried - 05/24/2011 (05/23/2011, 08:42 PM)sheldonison Wrote: (...)Well, now I'm surprised. I've get that coefficients -however accurate only up to 6 digits, but maybe they converge if I use higher precision - by the most simple eigen-decomposition of the 32x32 carleman matrix. While the formal h'th powers of the carleman-matrices occur if I raise the diagonalmatrix of the eigenvalues to the h'th power, Pari/GP is able to convert this to a powerseries in x, if I enter the indeterminate x instead of a explicite h-value for the powers. (Note that this is the fourth method in my short treatize on "four methods of interpolation") Well, Sheldon's method seem to allow much more precision and I do not see yet, how I could reproduce this by simply increasing the Pari/GP-resources in decimal precision and matrix-size. Gottfried RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - JmsNxn - 05/26/2011 (05/20/2011, 06:43 PM)andydude Wrote: 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... Your close to what I'm talking about. I'll write it out in full math format \( f(t) = a\, \{t\}\, b = \left\{ \begin{array}{c l} \exp_\eta^{\alpha t}(\exp_\eta^{\alpha-t}(a) + \exp_\eta^{\alpha -t}(b)) & t \in [-1,1]\\ \exp_\eta^{\alpha t-1}(b * \exp_\eta^{\alpha 1-t}(a)) & t \in [1,2]\\ \end{array} \right. \) which maps the growth of addition to multiplication to exponentiation. I hope that base eta will behave better than base 2 and base e behaved. RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - JmsNxn - 06/02/2011 (05/21/2011, 10:55 PM)sheldonison Wrote: 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. I hate to be a bit of a dunce but: \( \text{sexp}_\eta(z) = \sum_{n=0}^{\infty} a_n (z-2e)^n \) is the correct formula for the first series correct? I only ask because this is the code I'm using (and I've also tried\( \text{sexp}_\eta(z) = \sum_{n=0}^{\infty} a_n (z+2e)^n \)) and neither seem to converge anywhere? The only convergence I do get, is \( \text{cheta}(0) = 2e \) when I remove the plus/minus 2e. RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - sheldonison - 06/02/2011 (06/02/2011, 09:37 PM)JmsNxn Wrote: I hate to be a bit of a dunce but:Hey James, Hopefully, this helps. The correct formula is \( \text{sexp}_\eta(z) = \sum_{n=0}^{\infty} a_n (z^n) \) Cheta(z) is the upper entire superfunction of eta, which grows superexponentially as z increases. Unfortunately, how to define cheta(0) is a somewhat arbitrary, since cheta(z) is always bigger than e, as z goes to minus infinity. Jay suggested defining cheta(0)=2e. Then \( \text{cheta}(1)=\eta^{2e}=e^2 \), and cheta(2)=e^e, which seems like a reasonable choice for how to define cheta(0). Centered around 0, the series I posted will generate these values. More recently, Henryk and Dimitrii have written a paper where the upper \( \text{SuperFunction}_\eta(0)=3 \). Perhaps there will eventually be a reason to pick a definitive value, but that hasn't happened yet. There is also a lower superfunction for base eta, that I usually refer to as \( \text{sexp}_\eta(z) \), since \( \text{sexp}_\eta(0)=1 \), \( \text{sexp}_\eta(-1)=0 \), and there is a singularity at \( \text{sexp}_\eta(-2) \). \( \text{sexp}_\eta(z) \) does not grow superexponentially, but converges towards e as z increases. I recently posted the Taylor series for that function, centered at z=0, here: http://math.eretrandre.org/tetrationforum/showthread.php?mode=threaded&tid=634&pid=5817#pid5817 - Sheldon RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - JmsNxn - 06/02/2011 (06/02/2011, 10:27 PM)sheldonison Wrote: There is also a lower superfunction for base eta, that I usually refer to as \( \text{sexp}_\eta(z) \), since \( \text{sexp}_\eta(0)=1 \), \( \text{sexp}_\eta(-1)=0 \), and there is a singularity at \( \text{sexp}_\eta(-2) \). \( \text{sexp}_\eta(z) \) does not grow superexponentially, but converges towards e as z increases. I recently posted the Taylor series for that function, centered at z=0, here: This is more what I was looking for, thanks. And another question, do you have a similar taylor series for \( \text{slog}_\eta(z) \)? That would be the inverse of the lower super function. (06/02/2011, 10:27 PM)sheldonison Wrote: Jay suggested defining cheta(0)=2e. Then \( \text{cheta}(1)=\eta^{2e}=e^2 \), and cheta(2)=e^e, which seems like a reasonable choice for how to define cheta(0). Woah! have you ever thought to consider that since \( \text{cheta}(0) = 2*e = e + e \), \( \text{cheta}(1) = e^2 = e*e \), and \( \text{cheta}(2) = \text{sexp}_e(2) = e^e \) that the cheta function maps the growth of \( e\, \{x\}\, e \), where \( \{x\} \) is a hyper operator of x order (0 is addition, 1 is multiplication etc etc); or mathematically speaking, another conjecture: \( \text{cheta}(x)\, =\, e\, \{x\}\, e\, =\, e\,\{x+1\}\,2 \), which should at least be true over domain [0, 2]. To see if its universally true would be very difficult, though. RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - sheldonison - 06/03/2011 (06/02/2011, 11:40 PM)JmsNxn Wrote: This is more what I was looking for, thanks. And another question, do you have a similar taylor series for \( \text{slog}_\eta(z) \)? That would be the inverse of the lower super function.It is an interesting sequence. Here is the series, centered at z=1, \( \text{slog}_\eta(z) = \sum_{n=0}^{\infty} a_n (z-1)^n \) Code: a0= 0RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - JmsNxn - 06/05/2011 Thank you very much for the series approximations sheldon, but sadly the humps still occur in base \( \eta \). I'm wondering now if there is a better base to work with or if it's smarter to dump the idea of logarithmic semi-operators altogether, as they seem to be a poor extension of ackerman function to domain real. The only interesting thing I have to report is that: just like \( \lim_{p\to -\infty} \text{cheta}(p) = e \), \( \lim_{p\to -\infty} e\, \{p\}\, e = \ln^{\alpha -p}(2 \exp^{\alpha -p}(e)) = e \) or that \( \lim_{p\to -\infty} \text{cheta}(p) = e\, \{p\}\, e = e \) where \( \{p\} \) is a logarithmic semi operator. So my question was, what's the radius of convergence for the cheta series you gave me, and whats the recurrence relation so that I can produce the full \( \text{cheta}(x) \) function. I just want to test some values. for ex: if \( \text{cheta}(-1) = e + \ln(2) \) then I think we have something, but if it doesn't, oh well. RE: Does anyone have taylor series approximations for tetration and slog base e^(1/e)? - sheldonison - 06/05/2011 (06/05/2011, 06:24 PM)JmsNxn Wrote: Thank you very much for the series approximations sheldon, but sadly the humps still occur in base \( \eta \). I'm wondering now if there is a better base to work with or if it's smarter to dump the idea of logarithmic semi-operators altogether, as they seem to be a poor extension of ackerman function to domain real.Cheta(-1)=e*(log(2)+1). cheta(z-1)=\( \log_\eta(\text{cheta}(z))=e*\log(\text{cheta}(z)) \) cheta(z+1)=\( \exp_\eta(\text{cheta}(z))=\exp(\text{cheta}(z)/e) \) Cheta(z) is entire, but the series convergence for a finite number of terms is limited by how close we are to the region of superexponential growth, and how many terms are used. It probably has an effective radius of convergence of about 2 with the number of terms I posted. If you want more convergence as real(z) increases or decreases, use the recurrence relation below. If you want more convergence as imag(z) increases, one way to get that is to generate the series for cheta(z-100), centered at -100. For cheta/sexpeta, I actually use Newton Polynomial interpolation, centered around -100 for cheta, and +100 for sexpeta, with a 50 term polynomial, which has pretty good convergence out to a radius of 25 (~28 digits), and is accurate to 50 digits within a unit radius. I also recently included cheta/sexpeta support in my latest kneser.gp program. - Sheldon |