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 Green Eggs and HAM: Tetration for ALL bases, real and complex, now possible? sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 11/23/2013, 12:58 PM (11/22/2013, 11:30 PM)mike3 Wrote: .... where $b_n$ are properly-chosen "basis functions". I have thought about the Kneser-mapping solution (i.e. regular iteration warped with theta mapping) as a possible set of basis functions... but the problem is this only covers half of the plane (as given, the upper half-plane), and the Cauchy equations require both halves of the plane. .... Do you, perhaps, have any ideas as to how this could be done? The form of solution need not converge on the entire plane, only on and perhaps near the imaginary axis. You could try mapping the two unit infinite strip to the unit circle using, $f(z) = \frac{4}{\pi}\tan^{-1}(z)$ Here, f(z) maps the unit circle to an infinite strip, from -1 to +1. The left side of the unit circle is mapped to -1+iz, and the right side of the circle is mapped to 1+iz, where iz varies from $+/-\Im\infty$ I don't know if that would help or not, or whether the singularity at +/-I would be fairly mild, or not, given that the tetration solution also converges to a fixed point at $+/-\Im\infty$. - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/23/2013, 10:53 PM (This post was last modified: 11/23/2013, 11:14 PM by mike3.) (11/23/2013, 12:58 PM)sheldonison Wrote: (11/22/2013, 11:30 PM)mike3 Wrote: .... where $b_n$ are properly-chosen "basis functions". I have thought about the Kneser-mapping solution (i.e. regular iteration warped with theta mapping) as a possible set of basis functions... but the problem is this only covers half of the plane (as given, the upper half-plane), and the Cauchy equations require both halves of the plane. .... Do you, perhaps, have any ideas as to how this could be done? The form of solution need not converge on the entire plane, only on and perhaps near the imaginary axis. You could try mapping the two unit infinite strip to the unit circle using, $f(z) = \frac{4}{\pi}\tan^{-1}(z)$ Here, f(z) maps the unit circle to an infinite strip, from -1 to +1. The left side of the unit circle is mapped to -1+iz, and the right side of the circle is mapped to 1+iz, where iz varies from $+/-\Im\infty$ I don't know if that would help or not, or whether the singularity at +/-I would be fairly mild, or not, given that the tetration solution also converges to a fixed point at $+/-\Im\infty$. - Sheldon I'm not sure how that'd be useful, since what I need is a representation of the tetrational at the imaginary axis, that is, at $it$ where $t$ goes between $\pm \infty$, not $-1 + it$ or $1 + it$ (these are obtained by exp/log of the function's values at the imaginary axis within the integral equation under the integral sign, while on the other side of the equation (without the integral) is the function on the imaginary axis). So I don't see how this makes for a set of basis functions for the function on the imaginary axis. Edit: Although, I suppose that could work since you could just drop an "exp" or a "log" on the side of the integral equation that wants values at the imaginary axis. However, now that side of the integral equation can only be valid for input to the function in half the range of what the right side uses, namely the half of the unit circle we use to retrieve the values at the imaginary axis. So I'm not sure this'll work, as the integral equation, as specified, is for the whole function. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/23/2013, 11:37 PM (This post was last modified: 11/23/2013, 11:49 PM by mike3.) However, the rescaled function $f(z) = \frac{\frac{4}{\pi} \tan^{-1}(z) + 1}{2} = \frac{2}{\pi} \tan^{-1}(z) + \frac{1}{2}$ might work. This maps the right half of the u.c. to $1 + ix$ and the left half to $ix$, thus allowing us to create a basis for the function at the imaginary axis. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/24/2013, 01:14 AM (This post was last modified: 11/24/2013, 01:15 AM by mike3.) This shows $\mathrm{tet}(f(e^{it}))$ for your mapping (base-$e$ tetration), with $t$ going from $-\pi$ to $\pi$:     There is a small corner in the real part (red curve). This may slow down convergence of a Fourier series. The rescaled mapping has a much nastier spike, however, and so doesn't seem very useful sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 11/24/2013, 03:24 PM (11/24/2013, 01:14 AM)mike3 Wrote: .... There is a small corner in the real part (red curve). This may slow down convergence of a Fourier series. The rescaled mapping has a much nastier spike, however, and so doesn't seem very useful Yeah, the discontinuity in the derivative means that very large numbers of terms would be required in the fourier series. It doesn't seem like a terribly useful idea. With 8,192 terms, I was able to get accuracy to 1E-7 or so. Mapping an infinite strip to a unit circle seemed like a good idea, but in practice, not usable. - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/28/2013, 06:53 AM I've been experimenting with some other functions and mappings for this purpose. There do exist more complicated mappings which map the unit circle to the imaginary axis twice, with strong divergence to imaginary infinity at their singularities, so that the wrapped function will more quickly approach the fixed point and so be smoother. One such function is: $f(z) = \tanh\left(\frac{\pi}{4} (x - 1/x)\right)$ (unit circle -> imag axis). The inverse, giving the basis functions as powers of this, is $b(z) = f^{-1}(z) = \frac{1}{2} \left( \frac{4}{\pi} \tanh^{-1}(z) + \sqrt{\frac{16}{\pi^2} \tanh^{-1}(z)^2 + 4} \right)$. And the basis functions are $b_n(z) = b(z)^n$. The resulting mapping of the tetrational to base $e$, $\mathrm{tet}(f(e^{it}))$, looks like (red real, green imag):     We can see this is much smoother and has no nasty corners. When we use this to get the Fourier series, however, we find that both positive and negative degree terms are required (i.e. the sum in terms of $b_n(z)$ I gave must go from $-\infty$ to $\infty$ instead of from $0$ to $\infty$), but it is significantly more accurate than the original mapping. In particular, with 203 terms (that's 101 positive and 101 negative degree terms plus the constant term), I get an error on the order of somewhat more than 10^-11 for the tetrational for imaginary-axis inputs close to 0. With 461 terms (230 positive, 230 negative), I get accuracy of around 10^-18 for close to 0, 10^-17 for inputs around $20i$. Clearly, this is very much improved, however I'm left wondering if it's possible a still better set of basis functions exists. While this is no use for getting an analytic approximation out of the HAM since it cannot be integrated exactly (the HAM was originally conceived, actually, as a method to get analytic, i.e. as formulas, approximations to the solutions of nonlinear problems, even though it can be operated as a numerical algorithm as well), we are interested in numerical approximations, and so this should still suffice for that. But it would be interesting if a simple enough basis function existed, that enabled the obtaining of an analytic approximation, since it might help in the finding of a series formula for tetration. sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 06/24/2014, 10:57 PM (This post was last modified: 06/24/2014, 11:02 PM by sheldonison.) How about looking for a function that maps L to 1, and L* to -1? Call this first function "t", since its just a temporary step. $t(z) = \frac{(z-\Re(L))}{i\Im(L)}$ Now, if t(z)=+/-1, then 1-t(z)^2=0, so 1-t(L)^2=0 and 1-t(L*)^2=0. So this might be your function, which maps L to 0 and L* to 0, and does so in a nice analytic function. $f(z) = 1-t(z)^2 = (\frac{(z-\Re(L))}{i\Im(L)})^2-1$ $f(z) \approx 0.5592217758698x^2 - 0.3558121306015x - 1.056597524339$ Here is a graph of f(sexp(z)) at z=0, from -5i to +5i For all bases with a pair of complex fixed points, the function should exponentially decay to 0 as z goes to $\pm \Im \infty$. I would think f(z) would have an infinite convolution/Fourier representation.     This does not work for real bases, b<=exp(1/e). - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/25/2014, 06:10 AM (This post was last modified: 06/25/2014, 06:10 AM by mike3.) (06/24/2014, 10:57 PM)sheldonison Wrote: How about looking for a function that maps L to 1, and L* to -1? Call this first function "t", since its just a temporary step. $t(z) = \frac{(z-\Re(L))}{i\Im(L)}$ Now, if t(z)=+/-1, then 1-t(z)^2=0, so 1-t(L)^2=0 and 1-t(L*)^2=0. So this might be your function, which maps L to 0 and L* to 0, and does so in a nice analytic function. $f(z) = 1-t(z)^2 = (\frac{(z-\Re(L))}{i\Im(L)})^2-1$ $f(z) \approx 0.5592217758698x^2 - 0.3558121306015x - 1.056597524339$ Here is a graph of f(sexp(z)) at z=0, from -5i to +5i For all bases with a pair of complex fixed points, the function should exponentially decay to 0 as z goes to $\pm \Im \infty$. I would think f(z) would have an infinite convolution/Fourier representation. This does not work for real bases, b<=exp(1/e). Hmm. It might work. However the function will not be complex-periodic, and therefore will not have a Fourier/exponential series expansion. It needs the periodicity in order to work. sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 06/25/2014, 08:21 AM (This post was last modified: 06/25/2014, 08:25 AM by sheldonison.) (06/25/2014, 06:10 AM)mike3 Wrote: Hmm. It might work. However the function will not be complex-periodic, and therefore will not have a Fourier/exponential series expansion. It needs the periodicity in order to work.This one seems to converge pretty poorly; much slower than ideal. If you take $f(\text{sexp}(18\pi i))<\approx 10^{-32{$, and then wrap it around a unit circle, and then represent it with a Laurent series, then a 1000 term series from z^-500 to z^500 will be accurate to a little better than 10^-25. That's reasonably accurate, but it is also a very large number of Taylor/Laurent series terms. For comparison, the fast converging functions are like exp(-x^2), which is a perfect Gaussian. If you wrap it around a unit circle, so that f(-1)~-10^-32, then such a perfect Gaussian would require a little less than a 100 term Laurent series to be accurate to 32 decimal digits. So this function, needs 10x more terms and even then, it is less accurate -- not at all promising. Some background: I'm experimenting with Fourier/Laurent convolutions to calculate arbitrarily large Tetration Taylor series coefficients to arbitrary accuracy, and it works really well. But now I realize that it works best when the envelope function behaves like a Gaussian, which is often the case. But a Gaussian envelope converges to zero faster than this f(sexp(z*I)) function. - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/29/2014, 12:03 AM (This post was last modified: 06/29/2014, 02:25 AM by mike3.) (06/25/2014, 08:21 AM)sheldonison Wrote: (06/25/2014, 06:10 AM)mike3 Wrote: Hmm. It might work. However the function will not be complex-periodic, and therefore will not have a Fourier/exponential series expansion. It needs the periodicity in order to work.This one seems to converge pretty poorly; much slower than ideal. If you take $f(\text{sexp}(18\pi i))<\approx 10^{-32{$, and then wrap it around a unit circle, and then represent it with a Laurent series, then a 1000 term series from z^-500 to z^500 will be accurate to a little better than 10^-25. That's reasonably accurate, but it is also a very large number of Taylor/Laurent series terms. For comparison, the fast converging functions are like exp(-x^2), which is a perfect Gaussian. If you wrap it around a unit circle, so that f(-1)~-10^-32, then such a perfect Gaussian would require a little less than a 100 term Laurent series to be accurate to 32 decimal digits. So this function, needs 10x more terms and even then, it is less accurate -- not at all promising. Some background: I'm experimenting with Fourier/Laurent convolutions to calculate arbitrarily large Tetration Taylor series coefficients to arbitrary accuracy, and it works really well. But now I realize that it works best when the envelope function behaves like a Gaussian, which is often the case. But a Gaussian envelope converges to zero faster than this f(sexp(z*I)) function. Which makes one wonder: what about modifying the Cauchy integral equation so as to compute $e^{z^2} \mathrm{tet}(z)$ instead of $\mathrm{tet}(z)$ directly? Note that this function will still approach a limiting value at $\pm i\infty$, in particular, 0 at both ends! If we let $G(z) = e^{z^2} \mathrm{tet}(z)$, then $G(z+1) = e^{(z+1)^2} \mathrm{tet}(z+1) = e^{z^2 + 2z + 1} \exp(\mathrm{tet}(z)) = e^{z^2 + 2z + 1} \exp(e^{-z^2} G(z))$ $G(z-1) = e^{(z-1)^2} \mathrm{tet}(z-1) = e^{z^2 - 2z + 1} \log(\mathrm{tet}(z)) = e^{z^2 - 2z + 1} \log(e^{z^2} G(z))$. Then the Cauchy integral equation looks like $G_A(z) = \frac{1}{2\pi} \int_{-A}^{A} \frac{e^{-p^2 + 2ip + 1} \exp(e^{p^2} G(ip))}{1 + ip - z} dp - \frac{1}{2\pi} \int_{-A}^{A} \frac{e^{-p^2 - 2ip + 1} \log(e^{-p^2} G(ip))}{-1 + ip - z} dp$. where $G(z) = \lim_{A \rightarrow \infty} G_A(z)$. Note that unlike Kouznetsov's original Cauchy integral equation, there is no residual term $\mathcal{K}(z)$ because this function approaches 0 at $\pm i\infty$. This is just a theory, just a guess -- I have no idea if this will work or not. I am also concerned about the possibility of ambiguity because the fixed points are not specified in this equation and so you might wind up with the alternate fixed point solution instead. I suppose one could, perhaps, avoid that by adding a residual which is the Cauchy integral of the fixed point times the Gaussian, but that integral doesn't appear to be representable in terms of any standard functions. Or perhaps by trying to restrict it by some kind of restriction involving the branches of the $\log$ that appears there? Though the right initial guess should probably allow it to converge to the correct function, but finding a good initial guess for tetration seems tricky. But thought I'd toss it out there anyways... « Next Oldest | Next Newest »

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