• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Alternate solution of tetration for "convergent" bases discovered mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/13/2010, 11:30 AM (This post was last modified: 09/13/2010, 09:16 PM by mike3.) Hi. I think I may have discovered an alternate solution, not the regular iteration(!), for the tetration of the bases in the "convergent" or "Shell-Thron Region" (STR) of the complex plane, that resembles the "holomorphic in the right halfplane and decays to two fixed points at imag infinity" solution for bases outside that region. After all, this would make sense, no -- there doesn't seem to be any reason there should be a "bump!" when we hit the STR boundary that causes us to lose that behavior, after all. Indeed, I'd expect the only "bumps" to be at the points 0, 1, and perhaps $e^{1/e}$, which would be singularities or branch points. Consider the complex base $b = 2.33 + 1.28i$, of which such a solution was constructed via the continuum sum method with a Fourier approximation (have had no luck using the Cauchy integral at complex bases). See here: http://math.eretrandre.org/tetrationforu...hp?tid=437 The graph at the complex plane looks like: (scale is +/-10 on both axes) The limits at $\pm i \infty$ are $L_{+} \approx 0.45506976073281089070317940138591656146 + 0.88874773514083174495912637427337221333i$ and $L_{-} \approx -0.31786548580783041558330120589997441842 - 1.6391492649644773520573981723012379287i$ (where the +/- indicates positive/negative imaginary infinity, respectively). Using Newton's method to approximate fixed points of exponential and gradually stepping the base backward parallel to the real axis (using the last fixpoint as a guess to find the one for the "stepped-back" base), we get that for $b = 1.33 + 1.28i$, which is in the STR, that this solution should approach $L_{+} \approx 0.45506976073281089070317940138591656146 + 0.88874773514083174495912637427337221333i$ and $L_{-} \approx -0.31786548580783041558330120589997441842 - 1.6391492649644773520573981723012379287i$. And the most astonishing thing is that this solution actually appears to exist! An experiment with the same continuum-sum Fourier method, taking the assumption the solution is holomorphic in the right half-plane (implementing the Fourier expansion along the imaginary axis is equivalent to this and must exclude the regular as it has singularities that are at minimum arbitrarily close to the axis, which'd mean a Fourier expansion of zero convergence width), and then gradually stepping back the base, using the already-calculated solution as an initial guess, we get this:     This certainly seems to make more sense as opposed to the "regular" solution, which looks like this:     Note how it lacks the fractal zones and is not holomorphic in the right half-plane (note the cut lines and singularities) -- features the tetrational outside the STR had. I'd hypothesize that this "alternative" solution is what the analytic continuation of the "Cauchy"/"Kneser" tetrational in the base to the STR would look like. Also, notice the strange branchpoints and cutlines in the left halfplane. The cuts and points are probably too complicated to give a simple description. If tetration ever is developed to the point where it could be a "real" special function, it'd probably be the most exotic. I believe this solution of tetration will not be real at $1 < b < e^{1/e}$, and $b = e^{1/e}$ will be a branch point, along with $b = 0$ and $b = 1$ (suggest cut at $b \in (-\infty, e^{1/e}]$.). However, due to the fractal nature, it'll likely satisfy $\tet_b(\slog_b(z)) = z$ for all $b$ but the singularities. Oddly enough, I can't seem to get the construction via this sum method to work for bases nearing the negative real axis for some reason or too far in to the left half-plane -- not sure why. But I don't see why a function which has similar imaginary-asymptotic properties can't exist there. Regular was appealing, since it gave real-at-the-real-axis solutions for $1 < b < e^{1/e}$, but appears to have a natural boundary at the STR boundary, and so is worthless for creating a full-scale tetrational on the overall complex plane. Puzzling observations: why does Carleman-power seem to give regular at $1 < b \le e^{1/e}$ and Cauchy-like when $b > e^{1/e}$??? And why does continuum sum seem to yield regular, too, when $b$ is on the real axis in that range? What is going on here? What is this strange "affinity" these two seem to have for each other? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 09/13/2010, 11:20 PM (This post was last modified: 09/13/2010, 11:22 PM by tommy1729.) for starters i like this forum and its members but im kinda tired of people thinking that they are clear about their methods. sorry , but apart from the classical methods like koenigs function , carleman matrix etc , there are not so many methods well defined and well explained. i know its not always easy to prove things like analytic , size of radius , existance , uniqueness etc but at least explain your method decently. i might not be the smartest person on the planet but a better explaination is sometimes really really required. a typical thing is something in the line of : " we start with an intial guess depending on base q and then ... " " ... we take an integral and do as in ' link ' ( containing similar intial guesses ) ..." " ... and repeat the process k times where k is depending on q ... " " and then we estimate the contour integral of ... " " ... and repeat till it converges .. " continuing with a " im not sure about this and that and how to do this and that " but then adding a nice picture in the next post ?? also i have nothing again C code and similar , but i want to see the math , C code is not my language. although i of course see use of that for software computations and plots. ( i must say Bo did an excellent job explaining Kneser partially and andrews slog completely but many others are still vague , integral methods e.g. undefined estimates and guesses mean little to me ! ) this is a perfect example of what i meant. suddenly there is a new method for base 2.33 + 1.28 i. no tex ! just a link. to another vaguelue described method. the continuum sum is clear to me. and trying the continuum sum on some four estimate makes sense too. but nowhere do i see what fourier series is used as approximation of base 2.33 + 1.28i i might have missed this or that , but im sure a newbie cant follow this way and i guess we dont what that do we ? clarity is at the heart of mathematical intentions ... quote : " gradually stepping back the base, using the already-calculated solution as an initial guess " thats what i meant. stepping back the base ... is that math terminology ? is it clear to anyone ? maybe but i doubt it. 1 unclear thing and whole method is unclear. this is nothing personal mike. its a general remark. although i dont understand how you got that pic. initial guess .. stepping back ... further you seem to contradict yourself : if your method is analytic : you say it is different then regular , and later on you mention its the same on the real line. then it cant be analytic ! those 3 exclude eachother. sorry for being so critical. but i waited to long to say this. tommy1729 sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 09/13/2010, 11:45 PM (This post was last modified: 09/14/2010, 01:24 AM by sheldonison.) (09/13/2010, 11:30 AM)mike3 Wrote: Hi. I think I may have discovered an alternate solution, not the regular iteration(!)When you compare to the regular iteration solution, are you developing the regular solution from a fixed point at -real infinity? $b = 2.33 + 1.28i$ Generating the "regular" solution, I get fixed point= 0.45507 + 0.88875i, which matches one of your two fixed points! The period=3.98+0.23*I. But wouldn't the solution generated from the fixed point (if repelling) be an entire solution? Here is a parametric plot, from z=-100 to z=0, at the real axis. The function is nicely developing from the fixed point.     Here, the graph continues from z=0 to z=8.4, with the beginning of superexponential growth seen. This doesn't really look like any of your solutions..... I also tried your second base=1.33+1.28i. and it also seems to work nicely via regular tetration, with a steeper "slope". By the way, the color graphics in your post are really nice! Is there a way to generate the "png" color graphics file from pari-gp?     mike3 Wrote:.... Consider the complex base $b = 2.33 + 1.28i$, of which such a solution was constructed via the continuum sum method with a Fourier approximation (have had no luck using the Cauchy integral at complex bases)..... The parametric graph at the complex plane from z==100 to z=0 looks like:So, this solution is no longer periodic in the complex plane. I wonder if you're developing something analogous to using the other fixed attracting point (assuming the other fixed point is attracting, and there is an attracting fixed point). Or else its somehow equivalent to a Schwarz reflection, with some sort of 1-cyclic transformation of the regular sexp solution, that creates the second fixed point, along with singularities for f(z)=log(0), log(log(0)), log(log(log(0))) .... - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/14/2010, 01:50 AM (This post was last modified: 09/14/2010, 09:31 PM by mike3.) (09/13/2010, 11:20 PM)tommy1729 Wrote: ... not always easy to prove things like analytic , size of radius , existance , uniqueness etc but at least explain your method decently. i might not be the smartest person on the planet but a better explaination is sometimes really really required. a typical thing is something in the line of : " we start with an intial guess depending on base q and then ... " " ... we take an integral and do as in ' link ' ( containing similar intial guesses ) ..." " ... and repeat the process k times where k is depending on q ... " " and then we estimate the contour integral of ... " " ... and repeat till it converges .. " First off, the method is not well-proven. It's more conjectural at this point. But anyway, this is the best explanation I can give. Let $g(z)$ be a periodic, analytic function with period $P$. Then there should exist a Fourier expansion (perhaps needing a shift or translation depending on location of singularities) $g(z) = \sum_{n=-\infty}^{\infty} a_n e^{\frac{2 \pi}{P} i nz}$. Now we create the "continuum sum": $\sum_{n=0}^{z-1} g(n) = \sum_{n=-\infty}^{\infty} \frac{a_n}{e^{\frac{2 \pi}{P} i n} - 1} \left(e^{\frac{2 \pi}{P} i nz} - 1\right)$ where the term at $n = 0$ is interpreted as $a_0 z$. So, if $f(z)$ is some non-periodic complex function, we may be able to define a continuum sum for this function as follows. Let $f_k(z)$ be a sequence of periodic, analytic functions that converges pointwise to $f(z)$, i.e. $\lim_{k \rightarrow \infty} f_k(z) = f(z)$. Then we define $\sum_{n=0}^{z-1} f(n) = \lim_{k \rightarrow \infty} \sum_{n=0}^{z-1} f_k(n)$, provided that this limit exists. The hypothesis (not yet proven!) is that this limit is the same regardless of the sequence of periodic approximations. Now that we have a way to define continuum sums, we start with an initial guess $F_0(z)$ to our tetrational function $\mathrm{tet}_b(z)$, and then apply the iteration formula $F_{k+1}(z) = F'(0) \int_{-1}^{z} \log(b)^w \exp_b\left(\sum_{n=0}^{w-1} F_k(n)\right) dw$ and take the convergent as $k \rightarrow \infty$. If we do not know $F'(0)$, we replace the multiplication by $F'(0)$ with multiplication by a normalization constant so that $F(0) = 1$. That is, $U(z) = \int_{-1}^{z} \log(b)^w \exp_b\left(\sum_{n=0}^{w-1} F_k(n)\right) dw$ $F_{k+1}(z) = \frac{U(z)}{U(0)}$. Now the adaptation of this theory to construct a numerical algorithm is a little more complicated. I could post that, if you'd like it, in the Computation forum. But the above is the basic outline of the method. If you have any difficulty with the above, please indicate which parts confuse you, and I'll try to explain it better. (09/13/2010, 11:20 PM)tommy1729 Wrote: the continuum sum is clear to me. and trying the continuum sum on some four estimate makes sense too. but nowhere do i see what fourier series is used as approximation of base 2.33 + 1.28i i might have missed this or that , but im sure a newbie cant follow this way and i guess we dont what that do we ? clarity is at the heart of mathematical intentions ... So you want the numerical algorithm, then? (09/13/2010, 11:20 PM)tommy1729 Wrote: quote : " gradually stepping back the base, using the already-calculated solution as an initial guess " thats what i meant. stepping back the base ... is that math terminology ? is it clear to anyone ? It means you take the already-calculated solution as an initial guess for solving a base a little bit closer to the target one, then you take that as an initial guess for a base a little closer still, and so on. (09/13/2010, 11:20 PM)tommy1729 Wrote: you say it is different then regular , and later on you mention its the same on the real line. Not sure what you mean. I don't seem to have mentioned it is the same as the regular at the real line -- it isn't. EDITED: I left out the fact the method needs the derivative at 0, just put that in. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/14/2010, 01:53 AM (This post was last modified: 09/14/2010, 03:04 AM by mike3.) @sheldonison: Um, I'm not comparing base $2.33 + 1.28i$ to regular, rather base $1.33 + 1.28i$ to its regular iteration at the real attracting fixed point (so-called "regular tetration"). tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 09/14/2010, 12:26 PM (09/14/2010, 01:50 AM)mike3 Wrote: Now that we have a way to define continuum sums, we start with an initial guess $F_0(z)$ to our tetrational function $\tet_b(z)$, and then apply the iteration formula $F_{k+1}(z) = \int_{-1}^{z} \log(b)^w \exp_b\left(\sum_{n=0}^{w-1} F_k(n)\right) dw$ and take the convergent as $k \rightarrow \infty$. Now the adaptation of this theory to construct a numerical algorithm is a little more complicated. I could post that, if you'd like it, in the Computation forum. But the above is the basic outline of the method. thanks thats a lot better. 1) is that similar or equal to my " using the sum hoping for convergence " thread ? 2) i assume the intial guess needs to be a taylor , laurent or coo fourier series , in other words analytic. and also map R+ -> R+ , but then we have to start from an already tetration solution ? so what is the advantage of this sum method solution ? afterall it is just a 1periodic wave transform of another sexp/slog. what are its properties ? thanks for the reply. ( btw i believe in the continuum sum ) tommy1729 sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 09/14/2010, 02:18 PM (This post was last modified: 09/14/2010, 07:58 PM by sheldonison.) (09/14/2010, 01:53 AM)mike3 Wrote: @sheldonison: Um, I'm not comparing base $2.33 + 1.28i$ to regular, rather base $1.33 + 1.28i$ to its regular iteration at the real attracting fixed point (so-called "regular tetration").Mike, Thanks for the clarifications. Again, love those png color plots, and I would really appreciate it if you have a link to how to create them ..... (ideally from pari-gp). Now, onto B=1.33+1.28i. The fixed repelling point is L1 = -1.13500 - 1.98958i with a corresponding period of -3.6051+2.4470i. $\text{period}(B)=2Pi*I/(L*\log(B)+\log(\log(B)))$ The fixed attracting point (which would be your third graph, developed by iterating log(log(log(z))) starting from near the attracting fixed point) is L2 = 0.57937+0.64723i with a corresponding period of 3.58745 - 0.32994i This matches your third graph, which has approximately five iterations over 20 unit lengths. Now your second graph might appear to be a combination of these two periods, with a period corresponding to L2 on the left, and a period corresponding to L1 on the right. This seems difficult to imagine to me update, seems like it might, (or might not) work just fine, but it will take one or two posts to explain. Suppose the regular super function(z) is developed from the repelling fixed point by regular iteration, from L1. RegularSuper(z) is entire. Call your function f(z). Since your function approximates the periodicity of the Regular Super function on the right, would this equation hold, where $\theta(z)$ is a 1-cyclic function? $f(z)=\text{RegularSuper}(z+\theta(z))$ edit question Mike, where are the singularities in your second graph? Are they organized in one line? - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 09/14/2010, 07:41 PM (09/14/2010, 02:18 PM)sheldonison Wrote: Now, onto B=1.33+1.28i. The fixed repelling point is L1 = -1.13500 - 1.98958i with a corresponding period of -3.6051+2.4470i. $\text{period}(B)=2Pi*I/(L*\log(B)+\log(\log(B)))$ its seems familiar , so forgive me for not having my day , but period of what ? a kind of sexp with base B i assume. which sexp ? computed how ? i wanted to ask about the formula , but i think it follows when i see the answer to the above. forgive me if i ask this again or if it has been explained before. sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 09/14/2010, 10:30 PM (09/14/2010, 07:41 PM)tommy1729 Wrote: (09/14/2010, 02:18 PM)sheldonison Wrote: Now, onto B=1.33+1.28i. The fixed repelling point is L1 = -1.13500 - 1.98958i with a corresponding period of -3.6051+2.4470i. $\text{period}(B)=2Pi*I/(L*\log(B)+\log(\log(B)))$ its seems familiar , so forgive me for not having my day , but period of what ? a kind of sexp with base B i assume. which sexp ? computed how ? i wanted to ask about the formula , but i think it follows when i see the answer to the above. forgive me if i ask this again or if it has been explained before.The period for the regular superfunction, is a somewhat complicated mess. As z->-infinity, the behavior of the regular superfunction is approximated by an exponential. This is the formula I use. $\text{RegularSuper}_{B}(z) = \lim_{n \to \infty} B^{[n](L + {(L\times\ln(B))}^{z-n})}$ The next step, is to figure out what the periodicity is, taking into account because L is a fixed point, then B^L=L. As z -> -infinity, $\text{RegularSuper}(z) = L+{(L\times\log(B))}^z$ $\text{Period}=2Pi*I/\log(L*\log(B))$ $\text{Period}=2Pi*i/(\log(L) + \log(\log(B)))$ substitute: $L=B^L$, $\log(L)=\log(B^L)$, $\log(L)=L\times\log(B)$ $\text{Period}=2Pi*i/(L\times\log(B) + \log(\log(B)))$ mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/14/2010, 11:04 PM (This post was last modified: 09/14/2010, 11:08 PM by mike3.) (09/14/2010, 12:26 PM)tommy1729 Wrote: (09/14/2010, 01:50 AM)mike3 Wrote: Now that we have a way to define continuum sums, we start with an initial guess $F_0(z)$ to our tetrational function $\tet_b(z)$, and then apply the iteration formula $F_{k+1}(z) = \int_{-1}^{z} \log(b)^w \exp_b\left(\sum_{n=0}^{w-1} F_k(n)\right) dw$ and take the convergent as $k \rightarrow \infty$. Now the adaptation of this theory to construct a numerical algorithm is a little more complicated. I could post that, if you'd like it, in the Computation forum. But the above is the basic outline of the method. thanks thats a lot better. Yes. Though I accidentally left out a detail -- see the post again, I updated it. (09/14/2010, 12:26 PM)tommy1729 Wrote: 1) is that similar or equal to my " using the sum hoping for convergence " thread ? It should be. (09/14/2010, 12:26 PM)tommy1729 Wrote: 2) i assume the intial guess needs to be a taylor , laurent or coo fourier series , in other words analytic. and also map R+ -> R+ , but then we have to start from an already tetration solution ? In the abstract, it's just a holomorphic (multi-)function. But for the numerical algorithm, we'd use a (truncated) Fourier series. (09/14/2010, 12:26 PM)tommy1729 Wrote: so what is the advantage of this sum method solution ? afterall it is just a 1periodic wave transform of another sexp/slog. Biggest advantage seems to be that it gives the widest range of bases I've seen for any tetration method. (09/14/2010, 12:26 PM)tommy1729 Wrote: what are its properties ? Not sure exactly what you'd want to know. It seems to be similar to the "Cauchy integral" for bases outside and in some parts of the STR, but the weird part is that when it is used for $1 < b < e^{1/e}$, it gives the attracting regular iteration. ??? « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Special tetration bases Daniel 0 113 12/13/2019, 02:55 AM Last Post: Daniel tommy's simple solution ln^[n](2sinh^[n+x](z)) tommy1729 1 2,506 01/17/2017, 07:21 AM Last Post: sheldonison Why bases 0 2 tommy1729 0 1,797 04/18/2015, 12:24 PM Last Post: tommy1729 on constructing hyper operations for bases > eta JmsNxn 1 2,713 04/08/2015, 09:18 PM Last Post: marraco Further observations on fractional calc solution to tetration JmsNxn 13 13,935 06/05/2014, 08:54 PM Last Post: tommy1729 An alternate power series representation for ln(x) JmsNxn 7 13,720 05/09/2011, 01:02 AM Last Post: JmsNxn Seeking a solution to a problem Qoppa/ssqrtQoppa=2 0 2,071 01/13/2011, 01:15 AM Last Post: Qoppa/ssqrtQoppa=2 my accepted bases tommy1729 0 2,197 08/29/2010, 07:38 PM Last Post: tommy1729 [Regular tetration] bases arbitrarily near eta Gottfried 0 2,830 08/22/2010, 09:01 AM Last Post: Gottfried

Users browsing this thread: 1 Guest(s)