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 tetration exp(z)-1+k sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 02/01/2015, 05:57 AM (This post was last modified: 02/01/2015, 02:33 PM by sheldonison.) (01/31/2015, 02:45 AM)sheldonison Wrote: ....So all of that was background material to get ready for the next post; the next post will give the exact equations for a closed form superfunction which approaches $s_k(z)$ arbitrarily well as k gets gets closer to zero. In particular .... $g_k(z)=k+\frac{2z}{2-z}\;\;$ and $g_k^{\circ z}(0)$ has an exact closed form solution... The closed form solution for the tangent superfunction for $g(z) = k+\frac{2z}{2-z} \;$ is: $r =\sqrt{2k - \frac{k^2}{4}}\;\;\; s = \sqrt {\frac{k}{8-k}}\;\;\; r \cdot s = \frac{k}{2}$ $t = \arctan(s)$ the fixed points are: $\frac{k}{2} \pm i r\;\;$ and the period is $\frac{\pi}{2t}$ the tangent superfunction is: $g_k^{\circ z}(0) = r \cdot \tan(-t + 2tz) + \frac{k}{2}$ It is interesting to compare the graph of the tangent superfunction for k=0.1, shown in red, with the graph of supexponential for k=0.1 where $f_{0.1}(z)=\exp(z)-1+0.1\;$ shown in green. Between zero and one, the two superfunctions match each other to better than $1.4\cdot 10^{-6}$ with the maximum error occurring near the inflection point, which for the tangent superfunction is at exactly 0.5, and for the superexponential it is near 0.5     One reason for this is that the two functions match until the z^3 term, where they differ by $\frac{z^3}{12}$. As k approaches zero, both superfunctions are perturbations of the parabolic case $k+z+\frac{z^2}{2}$ $g_k(z)= k+\frac{2z}{2-z} = k + z + \frac{z^2}{2} + \frac{z^3}{4} + \frac{z^4}{8} + ...$ $f_k(z)=k + \exp(z) - 1 = k + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + ...$ The two functions have different periodicities, and fixed points, but those are also close. For the superexponential, the fixed points are $\approx 0.0332 \pm i0.4447$ and for g(z), the fixed points are $0.05 \pm i 0.4444$. The superfunction for g(z) is exactly real periodic, with a real period of $\approx 14.0203$, where as the sexp(z) function has a pseudo periodicity of $14.0498\pm i1.0484$ Another interesting experiment to try is to plug in k=1, and generate the superfunction at z=1, and compare the tangent superfunction to tetration, sexp(z). $g_1^{o z}(1) \approx 1 + 1.0927x + 0.29848x^2 + 0.27179x^3 + ...$ $\text{sexp}(z) \approx 1 + 1.0918x + 0.27148x^2 + 0.21245x^3 +...$ The tangent superfunction provides an excellent piecemeal definition of tetration, and when it replaces the linear approximation, the resulting sexp(z) approximation has a continuous first and second derivative. This works for all tetration bases. - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread tetration exp(z)-1+k - by sheldonison - 01/27/2015, 12:28 AM RE: tetration exp(z)-1+k - by MphLee - 01/27/2015, 11:18 AM RE: tetration exp(z)-1+k - by sheldonison - 01/27/2015, 03:07 PM RE: tetration exp(z)-1+k - by tommy1729 - 01/30/2015, 07:20 AM RE: tetration exp(z)-1+k - by sheldonison - 01/31/2015, 02:45 AM RE: tetration exp(z)-1+k - by sheldonison - 02/01/2015, 05:57 AM RE: tetration exp(z)-1+k - by tommy1729 - 02/01/2015, 11:16 PM RE: tetration exp(z)-1+k - by sheldonison - 02/02/2015, 05:43 AM RE: tetration exp(z)-1+k - by tommy1729 - 02/01/2015, 11:06 PM RE: tetration exp(z)-1+k - by tommy1729 - 01/31/2015, 09:18 AM

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