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 regular tetration base sqrt(2) : an interesting(?) constant 2.76432104 Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 06/22/2013, 12:52 PM (This post was last modified: 06/22/2013, 01:15 PM by Gottfried.) In the regular tetration we use the Schröder-function for the linearization of the fractional heights-computation beginning at some x0. Experimenting with it, say tetration with the base $b=\sqrt 2$ we have a lower fixpoint $t_0=2$. Here the values of the Schröder-function for $x \gt t_0$ are positive and that of $x \lt t_0$ are negative. Since the effect of changing sign of the Schröder-value is the same as using an imaginary component to the height-parameter, we can say, that iterations from the region above the fixpoint $t_0$ down to that below that fixpoint can be achieved by an imaginary height - so, in some sense, the imaginary oversteps the infinite iteration-height. But this allows to define a pairwise relation between x-values, whose Schröder-values have opposite signs. So $x=1$(below the fixpoint) has the negative schröder-value from $x=2.46791405...$ (above the fixpoint). Let's call the two related points "duals" of each other. One can find that the dual of $x=- \infty$ for base $b=\sqrt 2$ is about $x_w \approx 2.76432104000012572327981201783...$ Does someone "know" this value and knows (or with some seriousness guesses) more properties of this value? For instance, what is then the dual of that $\log_b(x_w) \approx 2.93385035151$ : is this $\log_b(-\infty)$ ? Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 06/22/2013, 08:43 PM (This post was last modified: 06/22/2013, 08:44 PM by tommy1729.) Dear Gottfried It may be my fault since Im not so familiar with the terms you use but I fail to completely understand your question or even method. For instance I do not know what you mean by "linearization" here. For me it is a term used for differential equations or the truncated Taylor series a_0 + a_1 x for an analytic function. Are you using Koenigs function to solve the Schroeder equation ? Or is this one of those matrix methods ? How is the connection to uniqueness , existance , multiple solutions and fixpoints ? Also I do not know what all this imaginary stuff is. Imaginary component ? Imaginary height ? Imaginary oversteps the infinite iteration height ? What is meant with " dual of x = -oo " ? It may be my fault , but I think you need to be more explicit if you want to speak to a large audience and be understood. And even if it has been explained before you should have added at least a link imho. Maybe it helps even yourself if you write it out completely. Im - perhaps surprising - convinced that this question has an answer. Most things have an answer in math. Regards tommy1729 Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 06/22/2013, 09:55 PM (This post was last modified: 06/22/2013, 10:51 PM by Gottfried.) Well, for the case it is needed, here is some more explanation (see a more general remark at the end). In the "regular tetration" (this is that method, where we use the exponential series wich is recentered around a fixpoint, say the lower (attracting) fixpoint $t_0$ ) we realize the tetration to fractional heights via the "Schröder"-function (see wikipedia), say $\hspace{48} s_0= S (x - t_0)$ where $S(\cdot)$ denotes the Schröder function. After that we calculate the "height"-parameter, say "h" into it, where "h" goes into the exponent of the log of the fixpoint: $\hspace{48} r_0 = s_0 \cdot \log(t_0)^h$ Then we use the inverse Schröder-function to find the value $x_h$ which is the (fractional) h'th iterate "from" $x_0$ : $\hspace{48} x_h = S^{\circ -1} (r_0) + t_0$ Now if we let the "height" h equal zero, thus no iteration, but just change the sign of $r_0 = -s_0$ then we get a "dual" $\tilde x_0$ , where if $x_0$ is between 2 and 4, then the dual is below 2, and if $x_0$ is below 2 then its "dual" is between 2 and 4. This is what I meant with "dual" or "a pair of related numbers". This simple idea reduces to the formula: $\hspace{48} \tilde x_0 = S^{\circ -1} (-s_0) + t_0 = S^{\circ -1} ( - S(x_0 - t_0)) + t_0$ But the effect of changing sign in $s_0$ is alternatively reachable, if we simply introduce an imaginary value for the height-parameter, since $\hspace{48} -s_0 = s_0 \cdot \ln(t_0) ^{i \pi \over \ln \ln t_0}$ Now, starting at some $x_0$ between 2 and 4 we can infinitely iterate and at most approach 2, but we can never arrive with any real height a value below 2 by any number of iterations. We might say, that 2 is the infinite iteration from that $x_0$. We see by this that this concept of "imaginary height" (an imaginary value in the height-parameter h) allows to proceed not only to the value 2 but to values below 2. Thus I said with a sloppy expression: "imaginary height can overstep infinite height". Now, if we take the dual of, say $x_0 = 0$ this gives something above 2. If we iterate this one time towards 4 (which means with one negative height), we get that mentioned value of 2.764... . And its dual is then negative infinity (which itself is $x_0=0$ iterated one time with negative height - a thing which is not possible otherwise because of the occuring singularity). Remark: just to restate it again: "my matrix-method" is nothing else than the regular tetration. I came to this by the (accidentally) rediscovery of the concept of Carleman-matrices (see also wikipedia) not knowing that name, and where I was always working under the assumption of infinite size (no truncation), which has some sophisticated impact against a concept of truncated matrices for instance for the conception of fractional powers Gottfried Helms, Kassel sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 06/23/2013, 11:13 AM (This post was last modified: 06/23/2013, 11:19 AM by sheldonison.) (06/22/2013, 09:55 PM)Gottfried Wrote: .... $\hspace{48} s_0= S (x - t_0)$ where $S(\cdot)$ denotes the Schröder function.Hey Gottfried! So then, the Schröder function of -infinity for base sqrt(2) from the fixed point of 2 turns out to be $s_0 = \lim_{x\to -\infty} S (x - 2) \approx -1.31563054604637354754$. The significance of this value is that it is also the radius of convergence of the Taylor series for the inverse Schröder function, since -infinity is the nearest singularity, so the radius of convergence would be approximately 1.3156. Then we use the inverse Schröder function of $-s_0$ to get Gottfried's number. $z = S^{-1} (-s_0) + 2 \approx 2.7643210400001$ - Sheldon Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 06/23/2013, 10:20 PM (This post was last modified: 06/24/2013, 04:53 AM by Gottfried.) Hi Sheldon - just "in shortness" (06/23/2013, 11:13 AM)sheldonison Wrote: Then we use the inverse Schröder function of $-s_0$ to get Gottfried's number. $z = S^{-1} (-s_0) + 2 \approx 2.7643210400001$ Well, ;-) I think such a naming deserves at least one index, so let's index it with the related fixpoint... Unfortunately, that constant has no obvious relation to the asum- zero-height from the other thread - if there were some elegant relation: that were really great. Also whether there might be a relation to the same effect /constant using the upper/repelling fixpoint were an interesting thing: if we perform a handful of integer-height iterations towards the upper fixpoint and find the fixpoint_4-dual and if we could make this somehow consistent, then we had a nice relation over/connecting the whole real line. For the connection from the duals below 2, between 2 and 4 and above 4 I'd made a picture (and some mail here, around 07'2010) but where I did not yet understand fully the different fixpoint-implications. I used that dual for a "norming", setting one value between 2 and 4 as having height = 0, namely the dual of 1.     The last point (the only one which I cannot answer myself, perhaps you can look at it): Can we look at your Kneser-method what the dual of, say x_0 = 1 or x_0 = 0 or x_0 = -infty were? I think we need only the appropriate imaginary iteration height to compute the respectively duals. Levenstein- numbers? ;-) Gottfried Gottfried Helms, Kassel sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 06/24/2013, 04:09 PM (This post was last modified: 06/25/2013, 08:58 AM by sheldonison.) (06/23/2013, 10:20 PM)Gottfried Wrote: ... Unfortunately, that constant has no obvious relation to the asum- zero-height from the other thread - if there were some elegant relation: that were really great. Also whether there might be a relation to the same effect /constant using the upper/repelling fixpoint were an interesting thing: if we perform a handful of integer-height iterations towards the upper fixpoint and find the fixpoint_4-dual and if we could make this somehow consistent, then we had a nice relation over/connecting the whole real line. ...Hi Gottfried, It occurs to me that you might want to use a simpler iteration function to study the asum equations. In particular, the simplest iteration equation I know of is the following iteration function based on the tangent sum equation, which has symmetrical fixed points of +/-1. $f(x)=\frac{x+\tanh(1)}{1+x\tanh(1)}\approx \frac{x+1.5574}{1+1.5574x} \hspace{24} f^{o-1}(x)=\frac{x-\tanh(1)}{1-x\tanh(1)}$ $f^{oz}=\tanh(z) \hspace{24}$ the superfunction agrees with the Schröder function solution generated from both symmetrical fixed points! Generating the asum for this function is more straightforward, since the superfunction for this tangent sum equation is the remarkably simple hyperbolic tangent equation, which is symmetrical and valid everywhere in the complex plane, with simple poles at $\frac{\pi i}{2}+n\pi i$. I think the asum iteration equations are interesting, in that they allow one to generate another different analytic superfunction, but that in the complex plane, the resulting superfunction will never behave as well as the Schroder superfunction or the Kneser superfunction, since it will have additional singularities near imaginary Pi/2, even as x goes to +/- real infinity, whereas tanh(z) is very well behaved near +/- real infinity, where it converges to the +/-1 fixed points for all $\Im(z)$. It might be interesting to generate this asum superfunction for f(x) and compare it to the hyperbolic tangent. The asum of tanh would be an analytic 2-periodic function, with some maximum amplitude. Then the asum superfunction would be generated by taking the $\tanh$ of the inverse of this equation. $\frac{1}{\pi}\sin^{-1}\frac{\text{asum}(\tanh(z))}{\text{amplitude}}$. - Sheldon (06/23/2013, 10:20 PM)Gottfried Wrote: The last point (the only one which I cannot answer myself, perhaps you can look at it): Can we look at your Kneser-method what the dual of, say x_0 = 1 or x_0 = 0 or x_0 = -infty were? I think we need only the appropriate imaginary iteration height to compute the respectively duals. Levenstein- numbers? ;-)What is the definition/equation for a Kneser solution dual? The Kneser solution is not periodic. Would the dual of -infinity, which is sexp(-2) be sexp(2)? - Sheldon Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 06/25/2013, 10:45 AM (This post was last modified: 06/25/2013, 10:48 AM by Gottfried.) Hi Sheldon - thanks for the hint to the tanh-function. I'll give it a try! (06/24/2013, 04:09 PM)sheldonison Wrote: (06/23/2013, 10:20 PM)Gottfried Wrote: The last point (the only one which I cannot answer myself, perhaps you can look at it): Can we look at your Kneser-method what the dual of, say x_0 = 1 or x_0 = 0 or x_0 = -infty were? I think we need only the appropriate imaginary iteration height to compute the respectively duals. Levenstein- numbers? ;-)What is the definition/equation for a Kneser solution dual? The Kneser solution is not periodic. Would the dual of -infinity, which is sexp(-2) be sexp(2)? - Sheldon No, the "dual" is just the iterate with an appropriate (purely) imaginary height (which -in the case we work using such a function - changes the sign of the schröder-function-value), where the result is again real. if $K(h,x)$ would denote the Kneser-method iteration from x using height h, then in our case with base = sqrt(2), it should be $h=\pi i/ \log( \log(2))$ . It's similar to take the purely imaginary with a multiple of pi as exponent of the exponential-function to arrive at the negative part of the number line . And the result should be real and near the "dual" taken by the regular iteration. Gottfried Gottfried Helms, Kassel sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 06/25/2013, 01:37 PM (This post was last modified: 06/25/2013, 01:50 PM by sheldonison.) (06/25/2013, 10:45 AM)Gottfried Wrote: (06/24/2013, 04:09 PM)sheldonison Wrote: What is the definition/equation for a Kneser solution dual? The Kneser solution is not periodic. Would the dual of -infinity, which is sexp(-2) be sexp(2)? - Sheldon.... if $K(h,x)$ would denote the Kneser-method iteration from x using height h, then in our case with base = sqrt(2)... GottfriedThe Kneser complex conjugate solution is only defined for bases>e^(1/e), not for base=sqrt(2), where we normally use regular iteration instead. The Kneser solution is built on a modification of the complex Schroder function solution, but for the Kneser solution, S(0) is always singularity, so there would be no definition for a dual of 0. For the tangent angle sum equation, the Schröder function of the limit as z approaches real infinity is defined and the dual is $S^{-1} (-S (z - t_0)) + t_0 = 0$ which is exactly halfway between the two fixed points of -1 and 1. Hope you enjoy working with the alternating sum of the htan function, which is both 2-periodic, and Pi*I periodic, with singularities at Pi*I/2 and 1+Pi*I/2. - Sheldon « Next Oldest | Next Newest »

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