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 tetration exp(z)-1+k sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 01/27/2015, 12:28 AM (This post was last modified: 01/27/2015, 03:06 PM by sheldonison.) where k is a small perturbation constant in the neighborhood of zero, what if we develop the superfunction of $f(z)=\exp(z)-1+k= k + z + \frac{z^2}{2} + \frac{z^3}{6} + .... + \frac{z^n}{n!}\;\;\; k= \ln(\ln(b))+1\;\;$ where b is the corresponding tetration base to k I plan on developing these ideas over the next several weeks or months ... feel free to contribute or critique. In particular, for a value of k, can we develop the Fatou coordinate/Abel function, and the $f^{[\circ z]}(0)$ superfunction? Perhaps involving quasi conformal mappings and parabolic implosion, and perturbed fatou coordinates, which unfortunately, I don't understand that much. The math is very hard, and some of the best research is in French. k=0 corresponds to eta, which is the singularity point since zero is z fixed point. 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. The goal would be to develop an Abel/Fatou function for arbitrary values of k... Here is a graph of the superexponential corresponding to k=0.1, from -5 to 7. The inflection point is always about halfway between 0 and 1, and the superexponential is developed at z=0. k=0.1 corresponds to tetration b=1.501657. Where f(0)=0, this corresponds to where the $\text{sexp}(z)=\ln(\ln(e))$     Another idea is to develop an asymptotic formal series for $f^{[\circ z]}(0)$ the superfunction of f in tems of the pertubation value of k; I posted some preliminary results for such a formal series below. I think both ideas have promise. I see the possibility for a more rigorous definition of complex base tetration coming out of these ideas. So far, I got a pretty good asymptotic (non-converging) series for the first few derivatives of tetration base e, at z=-1. This would be plugging in k=1. For smaller values of k, correspondign to real bases and complex bases closer to eta, the asymptotic series converges better before diverging. Anyway, I'm not thrilled with the series since it is an asymptotic non-converging series... Code:a0= 0 a1= k + 1/12*k^2 + 1/120*k^3 + 1/5040*k^4 - 1/10080*k^5 + 1/332640*k^6 -323/86486400*k^7 a2= -1/4*k^2  - 1/16*k^3 - 1/90*k^4  - 11/12096*k^5 + 19/1209600*k^6 + 683/79833600*k^7 a3= 1/6*k^2 + 1/8*k^3 + 11/240*k^4 + 1957/181440*k^5 + 5191/3628800*k^6 + 2791/19958400*k^7 a4= -5/48*k^3 - 23/288*k^4 - 121/3456*k^5 - 28303/2903040*k^6 -31687/17418240*k^7 a5= 1/30*k^3 + 13/180*k^4 + 247/4320*k^5 + 39913/1451520*k^6 + 151307/17418240*k^7 a6= -49/1440*k^4 - 469/8640*k^5 - 15053/345600*k^6 - 274711/12441600*k^7 a7= 17/2520*k^4 + 1861/60480*k^5 + 241/5600*k^6 + 1498627/43545600*k^7 a8= - 79/8064*k^5 - 983/35840*k^6 - 204851/5806080*k^7 substituting in k=1, we get a reasonably good approximation for sexp_e at z=-1 a0=   0 a1=   1.09176514457764 a2=  -0.324496239678531 a3=   0.349856276054193 a4=  -0.230607971873163 a5=   0.198915562077454 a6=  -0.153946357381687 a7=   0.114967367541152 a8=  -0.0725062348434744 For comparison the "correct" values of the Taylor series coefficients at z=-1 are: a0=   0 a1=   1.09176735125832 a2=  -0.324494761735110 a3=   0.349836269767157 a4=  -0.230854426837443 a5=   0.201330212284523 a6=  -0.164352165253219 a7=   0.142836335724573 a8=  -0.124694993215245 - Sheldon MphLee Fellow Posts: 127 Threads: 10 Joined: May 2013 01/27/2015, 11:18 AM There is a reason (maybe is a standard) behind the use of the letter "o" in $f^{o n}$ instead of the "\circ" sign $f^{\circ n}$? MathStackExchange account:MphLee sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 01/27/2015, 03:07 PM (01/27/2015, 11:18 AM)MphLee Wrote: There is a reason (maybe is a standard) behind the use of the letter "o" in $f^{o n}$ instead of the "\circ" sign $f^{\circ n}$?Thanks; I should have been using "\circ n" and not "on". - Sheldon tommy1729 Ultimate Fellow Posts: 1,419 Threads: 345 Joined: Feb 2009 01/30/2015, 07:20 AM In your example you say f(0)=0 but f(0)=0,1. So no fixpoint ! Regards Tommy1729 sheldonison Long Time Fellow Posts: 663 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 tommy1729 Ultimate Fellow Posts: 1,419 Threads: 345 Joined: Feb 2009 01/31/2015, 09:18 AM Your third equation has typo's. I know what u intended. Regards Tommy1729 "Never underestimate THE master" Mick sheldonison Long Time Fellow Posts: 663 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 tommy1729 Ultimate Fellow Posts: 1,419 Threads: 345 Joined: Feb 2009 02/01/2015, 11:06 PM (01/31/2015, 02:45 AM)sheldonison Wrote: [quote='sheldonison' pid='7609' dateline='1422314910'] where k is a small perturbation constant in the neighborhood of zero... First off, if $k=\ln(ln(b))+1\;\;$ then $s_k(z) = \ln(b)\cdot \text{sexp}\left(\text{slog}(e)-2\right)\;+\;\ln\left(\ln(b)\right)\;$ with some algebraDespite the edit this equation is still wrong. Look , your "z" is on the LHS but not on the RHS. Quote:$\frac{d}{dz}s_k(z) = \frac{d}{dz}s_k(z+1)\;\;$ trivially proven by using the chain rule since $s_k(z+1)=\exp(s_k(z))-1+k$ This is also wrong. sexp'(x+1) = sexp(x+1) sexp'(x). If you say f ' (x) = f ' (x+1) then basicly you say f ' (x) is periodic. Quote:So, this particular point on the tetration curve is the place where the linear approximation piecemeal approximation works best, since ... Considering these mistakes or miscommunications I have trouble following and as a reader am not seduced to do so. Maybe Im harsch but if you ever want to write a paper you will loose the intrest of people after page 1 by these things , if it even gets accepted. As for my opinion , well , considering what I wrote its clear. Also the idea or post seems unfinished and is therefore hard to Judge. Nothing new under the sun till now ? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,419 Threads: 345 Joined: Feb 2009 02/01/2015, 11:16 PM (02/01/2015, 05:57 AM)sheldonison Wrote: 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. There are many nonlinear approximations that give a continuous first and second derivative. Why is this preferred ? Is there a uniqueness condition ? What is the advantage of this over just fitting the derivatives at the connection points ? ( Like A Robbins rediscovered ) regards tommy1729 sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 02/02/2015, 05:43 AM (This post was last modified: 02/02/2015, 04:27 PM by sheldonison.) (02/01/2015, 11:16 PM)tommy1729 Wrote: (02/01/2015, 05:57 AM)sheldonison Wrote: 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. There are many nonlinear approximations that give a continuous first and second derivative. Why is this preferred ? Is there a uniqueness condition ? ok, fixed the typos. I'm still inventing this post on the fly; as I said in my first post, "I plan on developing these ideas over the next several weeks or months ..." The next step would be to look at the Abel function of $g(z) = k+\frac{2z}{2-z} \;$, as well as to look at how the superfunction of g(z) changes as k rotates around 0 in the complex plane, and compare it to the closely related Abel function of $\exp(z)-1+k$. The long term goal which this post may lead to would be to put some kind of theoretical base behind this complex plane tetration post. For tetration for real bases>eta, we have Kneser's Riemann mapping, and I can show that the $z+\theta(z)$ mapping is equivalent. But for complex base tetration, there is no Riemann mapping; Henryk's post discusses the Abel function for complex perturbations. So the conjecture is that there is a quasi-conformal mapping between the Abel function of g(z), and the Abel function of $\exp(z)-1+k$, and that the pretty pictures and pari-gp code for complex base tetration is computing the quasi-conformal mapping between the tangent superfunction, and complex base tetration. It would be nice to be able to say if these bipolar theta mappings converge, than they are computing the complex Abel function. More later when I have time. These ideas are still under construction, and it takes me time to think it through. - Sheldon « Next Oldest | Next Newest »

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