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 Base 'Enigma' iterative exponential, tetrational and pentational Cherrina_Pixie Junior Fellow Posts: 6 Threads: 3 Joined: Jun 2011 06/29/2011, 08:39 PM Code:base          1.63532449671527639934534 fixed point   1.63698995729105702725471 + 1.52462061222409266551551*I ... pentation base        1.63532449671527639934534 pentation(-0.5)       0.540797083552851488750873 sexp fixed point      -1.64087257571659334856123 sexp slope at fixed   4.80600575430175169638439 pentation period      4.00236960853189042690462*I pentation singularity -1.64567803532871618956796 + 2.00118480426594521345231*I pentation precision, via sexp(pent(-0.5))-pent(0.5)                       -8.68755408487736716409823 E-22 complex sexp Taylor series centered at 3.0885322718067176544821807826411 sexp base, sexp(upfixed)=upfixed 1.6353244967152763993453446183062 init;loop iterations required    7 upfixed, parabolic fixed point   3.0885322718067176544821807826411 sexp'(upfixed) base B            1.0000000000000000000000000000000 sexp(upfixed)-upfixed error      9.7027744501722372002703206199650 E-33 ? sexp(3.000) %210 = 3.0022195105134555312059775955040 ? pent(2.000) %184 = 2.0088543076992014631570864956219Pentation code from http://math.eretrandre.org/tetrationforu...34#pid5334 Let $\nu$ 'enigma' (Greek αίνιγμα, aínigma) be the base at which the tetrational has a parabolic fixed point, as shown above. This is analogous to $\eta = e^{1/e}$, the base at which the exponential has a parabolic fixed point. The interesting things about this base are that several values shown above are close to $\nu$, often to within one part in one hundred or less, and that some others come to near-integers. Ignoring signs, the real part of the fixed point of the exponential, the lower fixed point of the tetrational (sexp) and the real part of the pentation singularity differ from $\nu$ by at most 0.010456, roughly one part in 95. Also, the pentation period, the upper fixed point of the tetrational and the fixed point of the pentational aren't so far off from 4.00, 3.00 and 2.00, respectively. Furthermore, base $e$ tetration and pentation give values close to $\nu$: Code:? exp(1/2) %42 = 1.6487212707001281468486507878142 ? sexp(1/2) %43 = 1.6463542337511945809719240315921 ? pent(1/2) %44 = 1.6323247404360631184869762532583 ? pent(-3) %45 = -1.6363583542860289796292230421033 Are these all 'coincidencies' or is there some good reason for the minute differences between these values? Any reason for it seems rather obscure, it's like a puzzle, a mystery to figure out. That is why I call that number, the 'enigma constant'. Perhaps one can use base $\nu$, instead of base $\eta$ or $e$ to evaluate non-integer iterations of exponential. There's no real fixed point of base $\nu$ exponential so the half-iterate and such should be real-analytic on the entire domain         $f_1(x) = \nu^x, \ f_2(x) = \exp_{\nu}^{\circ 1/2}(x), \ g(x) = \textrm{sexp}_\nu(x)$     $h(x) = \textrm{pent}_\nu(x)$ bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/29/2011, 09:13 PM For naming see also our Wiki Article about the critical base. sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 07/01/2011, 10:47 PM (This post was last modified: 07/01/2011, 10:51 PM by sheldonison.) (06/29/2011, 08:39 PM)Cherrina_Pixie Wrote: ... Let $\nu$ 'enigma' (Greek αίνιγμα, aínigma) be the base at which the tetrational has a parabolic fixed point, as shown above. This is analogous to $\eta = e^{1/e}$, the base at which the exponential has a parabolic fixed point. .... Are these all 'coincidencies' or is there some good reason for the minute differences between these values? Any reason for it seems rather obscure, it's like a puzzle, a mystery to figure out. That is why I call that number, the 'enigma constant'. Perhaps one can use base $\nu$, instead of base $\eta$ or $e$ to evaluate non-integer iterations of exponential. There's no real fixed point of base $\nu$ exponential so the half-iterate and such should be real-analytic on the entire domain $f_1(x) = \nu^x, \ f_2(x) = \exp_{\nu}^{\circ 1/2}(x), \ g(x) = \textrm{sexp}_\nu(x)$ $h(x) = \textrm{pent}_\nu(x)$Cherrina, Would you use the sexp(z) for the tetra-euler base, or the pent(z) for the tetra-euler base? b~1.6353245. In the pentation.gpp code, the genpen function uses the lower repelling fixed point (-1.6409), not the upper parabolic fixed point which is ~3.0885323. The parabolic fixed point would generate a similar but different pentation function for this base. Also, presumably one might be able to generate another pentation function going to infinity, from the parabolic fixed point. I haven't generated numerical approximations for either of those functions. - Sheldon Cherrina_Pixie Junior Fellow Posts: 6 Threads: 3 Joined: Jun 2011 07/02/2011, 02:00 AM (07/01/2011, 10:47 PM)sheldonison Wrote: Cherrina, Would you use the sexp(z) for the tetra-euler base, or the pent(z) for the tetra-euler base? b~1.6353245. In the pentation.gpp code, the genpen function uses the lower repelling fixed point (-1.6409), not the upper parabolic fixed point which is ~3.0885323. The parabolic fixed point would generate a similar but different pentation function for this base. Also, presumably one might be able to generate another pentation function going to infinity, from the parabolic fixed point. I haven't generated numerical approximations for either of those functions. - Sheldon The function sexp(z) is used for non-integer iterates of the exponential, which for the 'tetra-euler' base has no real fixed point. The non-integer iterates of exponential can then be used to construct operators between addition and multiplication along with corresponding means. The function pent(z) can generate non-integer iterates of sexp(z) but they can't be analytic at the upper fixed point, so I just used the regular iteration at the lower fixed point of sexp(z). There is also a possible function that rises to infinity based on iteration of sexp(z) but it wouldn't be 'pentation' in the sense that $S(0) \not = 1$ much like the 'cheta' function. Besides, I'm not totally sure about how to generate an analytic superfunction from a parabolic fixed point... From what I have read there is no way to construct an analytic non-integer iterate of a function that is analytic at a parabolic fixed point, is that correct? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/02/2011, 07:13 AM (This post was last modified: 07/02/2011, 07:39 AM by bo198214.) (07/02/2011, 02:00 AM)Cherrina_Pixie Wrote: Besides, I'm not totally sure about how to generate an analytic superfunction from a parabolic fixed point... From what I have read there is no way to construct an analytic non-integer iterate of a function that is analytic at a parabolic fixed point, is that correct? Only the fixpoint is not analytic (though it is infinitely times differentiable). For more details see my explanation on mathoverflow. So generally we have 2 or more superfunctions (not even counting the constant superfunction, with value of the fixpoint). I recently proved a rather simple formula to calculate the non-integer iterates: $f^t(x) = \lim_{n\to\infty} f^{-n}(f^n(x) - ta f^n(x)^{p})$ if $f(x)=x-ax^{p}+o(x^{p}),x\to 0$ and $a,x>0$, $p\ge 2$. For example if $f$ can be devloped into a powerseries $f(x)=x-ax^{p} + c_{p+1}x^{p+1}+\dots$. If $a<0$ and/or $x<0$ it possibly does not converge anymore. You can use the identity $f^t(x)=(f^{-1})^{-t}(x)$ to obtain a convergent version. If the fixpoint is not 0 but $x_0$ then one considers $g(x)=f(x+x_0)-x_0$, the iterates of $f$ are then: $f^t(x)=g^t(x-x_0)+x_0$. Originally I developed this formula to also have a direct formula for the superfuntion $x\mapsto f^x(x_0)$, while the in my mathoverflow-answer mentioned last (quickly converging) formula for the Abel function of exp(x)-1 is not directly invertible opposing for example Lévy's formula. Unfortunately it seems my formula is also not very quickly converging. You can improve the initial exactness, when changing the $\tau(x)=x-tax^p$ in the formula $f^{-n}(\tau(f^n(x)))$. $\tau$ is a truncation of the iteration $f^t$ of the formal powerseries of $f$. Though the whole formal powerseries of $f^t$ has zero convergence radius, you can use any truncation $\tau$ in the above formula, and it will converge. With these non-converging powerseries there is a trick: If you truncate in the right moment, then you may very close to the real value. This moment is usually the point where the indexed convergence radius $r_n = 1/\sqrt[n]{|c_n|}$ is largest. There is a recursive formula to calculate the coefficients of that formal iteration. I wrote a sage-package formal_powerseries.py, which can be found here, which can calculate these iterates of formal powerseries. « Next Oldest | Next Newest »

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