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 tetration exp(z)-1+k sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 01/31/2015, 02:45 AM (This post was last modified: 02/02/2015, 05:26 AM by sheldonison.) (01/27/2015, 12:28 AM)sheldonison Wrote: where k is a small perturbation constant in the neighborhood of zero... The idea is that each positive value of k corresponds directly to tetration for some real base greater than eta. One idea is that also, there is some similar function to f(x), which has the same x^2 multiplier, and which has a well defined superfunction which can be expressed in a closed form in terms of tan(bx); see previous mathstack question. First a quick note about what's interesting about this alternative form for tetration, for $f_k^{\circ z}(0)$ where $f_k(z)=k+\exp(z)-1$. Let us call this superfunction $s_k(z)= f_k^{\circ z}(0)$ First off, if $k=\ln(ln(b))+1\;\;$ then $s_k(z) = \ln(b)\cdot \text{sexp}\left(z+\text{slog}(e)-2\right)\;+\;\ln\left(\ln(b)\right)\;$ with some algebra So for k>0 $s_k(z)$ is directly equivalent to Tetration base b. Secondly, the inflection point occurs between zero and one. By definition, $s_k(0)=0\;$ which gives a nice equation showing the derivative at zero and one are equivalent. $\frac{d}{dz}s_k(0) = \frac{d}{dz}(1)\;\;$ trivially proven by using the chain rule since $s_k(z+1)=\exp(s_k(z))-1+k\;\;\frac{d}{dz}s_k(z+1)=\exp(s_k(z))\cdot \frac{d}{dz}s_k(z)\;\;s_k(0)=0$ So, this particular point on the tetration curve is the place where the linear approximation piecemeal approximation works best, since a linear approximation between zero and one gives rise to a continuous and once differentiable approximation for the superfunction. The inflection point for s(z) will occur approximately midway between zero and one. This is exactly equivalent to the piecemeal approximation for tetration base(e), where sexp(z)=1+z, between -1 and 0, which is then extended... So I've always found this part of the Tetration curve which contains the inflection point fascinating. Since real Tetration for bases>exp(1/e) has complex fixed points, perhaps this section of the sexp(z) curve is the part of the curve from which we can build up an analytic tetration function, for small positive values of k. 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, working with $f_k(z)=k+\exp(z)-1$ is a lot like working with $g_k(z)=k+\frac{2z}{2-z}\;\;$ and $g_k^{\circ z}(0)$ has an exact closed form solution... Perhaps there is a quasiconformal mapping from $g_k^{\circ z}(0)$ to $s_k(z)$? - 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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