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 tetration base conversion, and sexp/slog limit equations sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 03/03/2009, 07:27 PM (This post was last modified: 03/04/2009, 02:42 PM by sheldonison.) bo198214 Wrote:Are you sure that it is 1-periodic? I mean it is well known that $f^{-1}(g(x))-x=\theta(x)$ must be 1-periodic for two superexponentials f and g. This implies that $g(x)=f(\theta(x)+x)$ But $f(\theta(x)+x)-f(x)$ does not look 1-periodic? $f(\theta(x+1)+x+1)-f(x+1)=\exp(f(\theta(x)+x)))-\exp(f(x))\neq f(\theta(x)+x)-f(x)$ mostlyYou are correct, but it turns out not to matter that much. First, the two bases I'm comparing are the same. One is the ideal $\text{sexp}_b$, for b a little bigger than $\eta$, and the other is $\text{sexp}_b$ converted from base e to base b. So slog(sexp(x))-x would have to be 1-cyclic. Since the base is approaching $\eta$, and since I'm comparing sexp in the critical section, which is very linear, so I can get away with a short cut of just subtracting the two sexp functions, since I don't have an slog function in my spreadsheet. But for an sexp with a linear approximation over the critical section, its not that big a deal. One of the two waves is a line segment, and the other is a line segment plus a sinusoid with an amplitude of 0.0004 times the slope of the line segment. The difference is too small to be seen in the rough graphs I made. Mostly, I'm looking for a curve fit to a sine wave for the 1-periodic transfer function, as opposed to a 1-cyclic wave with higher order terms. If it is a sine wave, then there is hope for calculating it theoretically, as opposed to empirically. Once we have the delta sinusoid, in theory the sine wave can be applied to the critical section for a base approaching $\eta$, and then used to generate sexp for another base with all positive odd derivatives. I realize that this is pretty hypothetical, and that the only thing I'm basing this on is how good a curve fit there is between the graph, and an ideal sine wave, but an ideal sine wave makes the math a lot cleaner. Maybe Dimitrii could verify my results, by graphing $\text{slog}_b(\text{sexp}_e(x))-x$ for b approaching $\eta$. If I'm right, the result will approach a perfect sine wave, as b approaches $\eta$. The amplitude of the sine wave will converge on about 0.0004 as the conversion base approaches approaches $\eta$. new prediction: its even possible that all base conversions for Dimitrii's extension of sexp to real numbers will converge on 1-periodic sine waves, but surely someone would've noticed this already. Its all possible that it will be a sine wave in the sexp domain, (at the critical section, as the limit approaches e^(1/e)), but only an approximate sine wave in the slog domain. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 03/09/2009, 06:34 PM (This post was last modified: 03/10/2009, 03:51 PM by sheldonison.) $\theta(x) = \text{slog}_{v2}(\text{sexp}_{v1}(x))-x$ Back to characterizing the $\theta$ for the published sexp base e equation (version v1), compared to base 1.45 approximated with a 3rd order slog (version v2). At one point I thought the theta might be a relatively simple sin(2*pi*x+phase) type of function, but there are definitely higher harmonics, although the second harmonic sin(4*pi*x) is about 120 times smaller than the main harmonic. Given everything I know so far, the proposal is the following limit, using a linear approximation for the critical section for slog/sexp b $\text{sexp}_c(x) = \lim_{b \to \eta^+}\text{ } \lim_{n \to \infty} \text{log}_c^{\circ n}(\text{sexp}_b (x + \text{slog}_b(\text{sexp}_c(n)))$ converges, and all of its derivatives are continuous. For each base=c, there's a 1-cyclic $\theta$ function, linking this sexp definition with the traditional sexp function. Unless $\theta$ is badly behaved, the $\text{sexp}_{v2}(x)=\text{sexp}_{v1}(x+\theta(x))$ will also be continuous for all of its derivatives, and possibly analytic. This isn't new, Dimitrii points out on his wiki that the theta transfer function allows constructing other solutions of tetration with a reduced range of convergence in the complex plane, due to singularities caused by theta. The only advantage of this particular alternate sexp definition is that it has constant base conversions between any two bases. But the addition of the theta function introduces a wobble, that becomes more apparent in the higher derivatives; the odd derivatives are no longer positive for all x>-2. Finally, for smaller values of bases, the magnitude of the 1-cyclic $\theta$ function, generated by comparing the two sexp functions for the same base, gets arbitrarily small as the base=c approaches e^(1/e). This is also critical to the convergence of the limit equation. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 07/31/2009, 06:55 PM (This post was last modified: 08/01/2009, 10:37 AM by sheldonison.) Henryk sent me a nice note asking me whether I wanted to contribute to the co-authored paper. I don't think I'm up to the task, mathematically. But a few months ago, I did spend some effort trying to analyze the sexp base change equations in the complex plane, converting to base(e). Perhaps some of you may be interested. What I found, heuristically, is that for values of x with imag>=1 the limit equation for different values of n seems to give different non-converging values, and I don't think the equation converges for any values with imag>0. And that is really about as far as I got. Now, I want to describe some of the equations I used in the analysis. The first thing I did was start with this equation, that I was using all along. In these equations, sexp_e is referring to a definition of sexp with a base conversion constant, as described earlier in this post. $\text{sexp}_e(x) = \lim_{b \to \eta^+}\text{ } \lim_{n \to \infty} \text{log}_e^{\circ n}(\text{sexp}_b (x + \text{slog}_b(\text{sexp}_e(n)))$ After reading the posts about sexp upper, I made the following change, which should be identical at the real axis. Here the limit as b approaches $\eta^+$ is changed to $\eta \text{.upper}$. The advantage is that eta.upper is complete, and defined everywhere in the complex plane, and now there is only one limit in the equation. Also, once again, I think this turns out to be the approach Jay had in mind! $\text{sexp}_e(x) = \lim_{n \to \infty} \text{log}_e^{\circ n}(\text{sexp}_{\eta\text{.upper}} (x + \text{slog}_{\eta\text{.upper}}(\text{sexp}_e(n)))$ Further simplification comes from the fact that this is really just a constant, equal to the base conversion constant from eta.upper to base e. $\text{const}= \lim_{n \to \infty} \text{slog}_{\eta\text{.upper}}(\text{sexp}_e(n))-n$ Plugging this back in, we get the following fairly clean equation. $\text{sexp}_e(x) = \lim_{n \to \infty} \text{log}_e^{\circ n}(\text{sexp}_{\eta\text{.upper}} (x + n + \text{const}))$ This equation converges nicely at the real axis, and it does so for relatively small values of n. In fact, convergence is "super-exponential" as n increases, and the number of digits of accuracy quickly becomes larger than the number of atoms in the universe. But as soon as you consider complex values things get much messier. My analysis skills are pushed beyond their limits in iterating the logarithms for the complex valued function for $\eta \text{.upper}$. And as far as I can tell, it doesn't converge, period. That's because the complex argument for x means sexp_eta.upper no longer increases to arbitrarily large numbers. In fact, for values with imag=1, the sexp function for base eta.upper reaches a maximum value with real=~6.3, and then a minimum around real=~-6.6. After that, the real portion slowly grows towards e as x goes to infinity. Taking the iterated natural logarithms of eta.upper(i=1) doesn't converge. In other words, solving for a strip of the sexp(x, i=1), I find that sexp(x+1)<>e^sexp(x), so the result isn't converging. At best, it may pretend to converge to the fixed point of sexp_e(x), but even if that was the case that it was converging towards the fixed point of sexp(e), then at the strip boundaries, the derivative is discontinous, d/dx sexp(x+1)<> d/dx e^sexp(x). Again, that's as far as I got. For smaller values of the imag, it the real portion does grow for awhile, but as soon as the imaginary portion catches up, the results become chaotic, and I couldn't see how the iterated logarithms could converge. So I gave up. It converges so nicely at the real number line, perhaps it is non-analytic or has a taylor series convergence radius of 0. - Sheldon Levenstein sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 08/01/2009, 10:32 AM (This post was last modified: 08/01/2009, 03:17 PM by sheldonison.) I'm adding a graphs of sexp eta.upper, at the real axis and at imaginary=1. This is the same as Jay's cheta function. The function was shifted so that $\text{sexp}_{\eta.\text{upper}}(0)=4$. I started with f(x) is approximately e*(1-(1/2x)), as x goes to -infinity, and optimized further by replacing x with a polynomial. I got this from the http://en.citizendium.org/wiki/Tetration web page. Anyway, taking the iterated $\tex{log}_e$ of $\text{sexp}_{\eta.\text{upper}}$ converges nicely at the real axis due to super exponential growth, but may not converge in the complex plane. At imaginary=1, the $\text{sexp}_{\eta.\text{upper}}$ function gradually grows towards a real value of "e", as x goes to (imag=1, real=infinity). sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 12/22/2009, 11:39 PM (This post was last modified: 12/23/2009, 03:46 AM by sheldonison.) In continuing to ponder the base change properties of tetration, I have come across an interesting function, that is analytic, but seems like it must converge to a non-analytic function. Consider the function f(x), defined such that $\text{sexp}_e(f(x)+x) = \text{sexp}_{\eta.\text{upper}}(x)$ I chose sexp_e, since Dimitrii's base e super exponential taylor series has been published. And super $\eta$.upper is exactly the same as Jay's "cheta" function, which is a well defined analytic super exponential, with base $\eta=exp^{(1/e)}$, and which has previously been explored in many posts, especially http://math.eretrandre.org/tetrationforu...hp?tid=333. Jay suggests choosing cheta(0)=2e. Since $\text{cheta}(x)=\eta^{\text{cheta}(x-1)}$, this leads to cheta(1)=e*e, and cheta(2)=e^e. Cheta(x) has super exponential growth as x increases in the positive reals. For negative values of x, cheta(x) assymptotically approaches e. Continuing on, solving for f(x) $f(x)= \text{slog_e}(\text{cheta}(x)) - x$ Here we use the inverse of Dimitrii's base(e) super exponential, which is an analytic function. Since cheta(x) is analytic, then f(x) must also be an analytic function. Here is the graph of f(x), for real values of x. F(x) is an interesting analytic function. On the left, it converges to the line f(x)=-x+1. On the right, it converges to a 1-cyclic function. By definition $\text{sexp_e}(f(x)+x) = \text{cheta}(x)$ So f(x) converts from $\text{sexp}_e(x)$ to cheta(x) or $\text{sexp_{\eta.upper}}(x)$. As x increases past 4, f(x) is converging to a one-cyclic function. You can see the detail of the one cyclic function in the graph. It averages around -0.5843. The convergence is super-exponential, with the value of f(5.0) converging to 42 decimal digits, and the value of f(6.0) converging to something like 10^42 decimal digits. f(2.0) = 0.000000000000 = slog_e(sexp(2.0))-2.0 = slog_e(e^e)-2.0 = 0 f(2.5) = -0.267810330237 = slog_e(sexp(2.5))-2.0 = slog_e(35.32)-2.5 f(3.0) = -0.455343068531 = slog_e( 263.734)-3.0 f(3.5) = -0.554032644495 = slog_e( 439911)-3.5 f(4.0) = -0.583032469819 = slog_e( 1.3686e+042)-4.0 f(4.5) = -0.584037761409 = slog_e( (439911)/e)-3.5 f(5.0) = -0.584589240445 = slog_e((1.3686e+42/e)-4.0 f(5.5) = -0.584037958805 = slog_e(( 439911)/e-1)-5.5 f(6.0) = -0.584589240445 = slog_e((1.37e+42/e-1)-6.0 f(6.5) = -0.584037958805 = f(5.5) Back to the base change function Since we have $\text{cheta}(x)=\text{sexp_e}(f(x)+x)$, then we can convert from cheta(x) to the sexp_e_baseconv(x), by iterating ln of cheta(x). I'm using the equations for the sexp.base.change_e(x) from earlier in this post. Define the small constant $k=\lim_{n \to \infty} \text{slog}_{\eta\text{.upper}}(\text{sexp_e(n))-n$ $\text{sexp.basechange}_e(x) = \text{ } \lim_{n \to \infty} \ln^{\circ n}(\text{cheta(x+k+n))$ But here, cheta(x) can be replaced with $\text{sexp_e}(f(x)+x)$ $\text{sexp.basechange}_e(x) = \text{ } \lim_{n \to \infty} \ln^{\circ n}(\text{sexp_e}(f(x+k+n)+x+k+n))$ We can substitute and simplify, getting rid of the iterated ln since $\text{ } \lim_{n \to \infty} \ln^{\circ n}(\text{sexp_e}(g(x)+n))=\text{sexp_e}(g(x))$ $\text{sexp.basechange}_e(x) = \text{ } \lim_{n \to \infty} (\text{sexp_e}(f(x+k+n)+x+k))$ Next, we define $\theta(x)=\lim_{n \to \infty}f(x+n)$, which is just the 1-cyclic limit of f(x), and plug it in, now with all limits removed. $\text{sexp.basechange}_e(x) = (\text{sexp_e}(\theta(x+k)+x+k))$ Next question, is the 1-cyclic function $\theta(x)$ analytic? Previous arguments made by Henryk and Jay suggest that it cannot be, since the the base.change version of sexp_e seems to be only defined at the real axis. But this new equation is the same as the base change version of sexp_e! We have succeeded in removing all the iterated logarithms from the base change function. But this time, the base change version of sexp_e is being generated by sexp_e combined with $\theta(x)$, which is the limiting behavior of the graph of f(x), an analytic function. The other version was generated by iterated logarithms of cheta(x). It makes sense that infinitely iterated logarithms might not be analytic. But f(x) has been shown to be a well defined analytic function. And since f(x) quickly converges to $\theta(x)$, it seems like $\theta(x)$ should be analytic. Is it analytic? For a quick review, the base change versions of super exponentials is defined at the real axis, and infinitly differentiable. But so far, it cannot be defined on the complex plane. The nice thing about the base change version of super exponentials is that the value of "k" converges to a base change constant for increasing values of x, for any two bases a, and b. converges to a base conversion constant as x grows. $k = \lim_{x\to \infty} \text{slog}_a(\text{sexp}_b(x))-x$ then the value of k is a constant for increasing values of x. sexp_a(x+k) = sexp_b(x) Henryk and Jay showed that the base change version of sexp has an infinite number of singularities, arbitrarily close to any point on the real axis. So the radius of convergence of the taylor series would be zero, or, as I understand it, the function isn't analytic. And that means $\theta(x)$ isn't analytic either, because if $\theta(x)$ was analytic, then sexp_e(x+$\theta(x)$) would also be analytic. But $\theta(x)$ is merely the limiting behavior of f(x), for values of x>=6, where f(x) is a 1-cyclic sinusoid. Can an analytic function converge in this way, to a non-analytic function? For me, the base change $\theta(x)$ is a source of continued fascination. Why is it that for any particular value of x, there is a value of k such that the equation sexp_a(x+n)=sexp_b(x+n+k) holds for integer values of n, but not for fractional or real values of n? For cheta(x+n), why are there some values of x where cheta(x) seems "larger" than it should be compared with cheta(x-0.5), at least as compared with other sexp bases? I'm also still attempting to make some sense of the base change function on the complex plane, and may eventually post some graphs of cheta, on the complex plane, along wth some thoughts on how the base change version of sexp might be defined on the complex plane. I've also made many graphs of patterns of the amplitude and phase of the $\theta(x)$ function, that I may eventually post, which show a predictable amplitude and phase. I would also like to look at $\theta(x)$ for different bases, and compare the fourier series for different versions of $\theta(x)$. Unfortunately, so far, any real progress I make seems to be made at a glacially slow pace... Hopefully, others also find this subject interesting. - Sheldon Levenstein mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/25/2009, 08:51 PM (This post was last modified: 12/26/2009, 01:55 AM by mike3.) I think I might have an idea about what's going on here. This shows the result of graphing $\mathrm{tet}(z)$ in the complex plane. The x-scale (real part) runs from 2 to 10 and the y-scale (imag part) runs from -8i to 8i. This function was obtained via the Cauchy integral (Kouznetsov's construction):     This, now, is the result of graphing $\check{\eta}(z)$ over the same parameter range, obtained via the regular iteration in the "repelling" fashion (the limit formula converges dog slowly, by fortunately we don't need much precision to get a good graph). Both have value $e^e$ at 2:     There should be more detail in the white regions than is seen here (especially toward the right, the leftmost bits are correct), indeed, the behavior there is extremely complicated, but I can't get that, as the computer overflows when iterating the recurrence equation to draw the graph. So I set a "bail out" to suppress this, yet that has the effect of losing detail. But still, these graphs can be useful in examining the behavior of the functions involved. As one can easily see, they can't even be translated to fit each other, not even in the asymptotic. Though they seem like they're asymptotic on the real line, on the complex plane, they're very different, which may be why the constructed "base change tetrational" is not analytic (Taylor series has 0 convergence radius). Smooth functions, analytic nowhere, are well known, and it is also well known that in Real analysis, a sequence of real-analytic functions can converge pointwise to a non-analytic function, just look at Fourier series, where the convergent may not even be continuous! In addition, the limits at imaginary infinity are different. For the tetration, we have $\lim_{z \rightarrow \mathrm{i} \infty}\ \mathrm{tet}(z) = -W_{-1}(-1)$ $\lim_{z \rightarrow -\mathrm{i} \infty}\ \mathrm{tet}(z) = -W_{0}(-1)$ while for the cheta function, we have $\lim_{z \rightarrow \mathrm{i} \infty}\ \check{\eta}(z) =\ \lim_{z \rightarrow -\mathrm{i} \infty}\ \check{\eta}(z) = e$. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 12/26/2009, 01:44 AM (This post was last modified: 12/26/2009, 02:11 AM by sheldonison.) (12/25/2009, 08:51 PM)mike3 Wrote: I think I might have an idea about what's going on here. This shows the result of graphing $\mathrm{tet}(z)$ in the complex plane. The x-scale (real part) runs from 2 to 10 and the y-scale (imag part) runs from -8i to 8i. This function was obtained via the Cauchy integral (Kouznetsov's construction): .... As one can easily see, they can't even be translated to fit each other, not even in the asymptotic. Though they seem like they're asymptotic on the real line, on the complex plane, they're very different, which may be why the constructed "base change tetrational" is not analytic (Taylor series has 0 convergence radius). Smooth functions, analytic nowhere, are well known, and it is also well known that in Real analysis, a sequence of real-analytic functions can converge pointwise to a non-analytic function, just look at Fourier series, where the convergent may not even be continuous!Great plots, thanks. Do you use "mathematica"? The only tools I'm using are excel, and "perl". I like the "eta" plot. I've plotted some of the n*e*pi contour lines of eta in the complex plane, which all iterate to i=0 contour lines, and which seem to be the defining features of the eta graph, along with the real=e contour lines. I'll probably post some time. The two functions, sexp_eta, and sexp_e, seem like they should be assymptotic on real number line, but its a ringing pattern. And, as you point out, in the complex plane, they're completely different functions. Also, I now realize that the $\theta(x)$ to which f(x) converges to is probably infinitly differentiable, but not be analytic. If you shift sexp_eta so that its super-exponential growth lines up with sexp_e, by taking sexp_eta(x+0.584) and sexp_e(x), the two graphs will intersect each other an infinite number of times, as they both super-exponentially climb to infinity, each on a slightly different pattern. - Shel mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/26/2009, 01:54 AM (This post was last modified: 12/26/2009, 01:58 AM by mike3.) (12/26/2009, 01:44 AM)sheldonison Wrote: Great plots, thanks. Do you use "mathematica"? The only tools I'm using are excel, and "perl". I like the "eta" plot. I've plotted some of the n*e*pi contour lines of eta in the complex plane, which all iterate to i=0 contour lines, and which seem to be the defining features of the eta graph, along with the real=e contour lines. I'll probably post some time. The two functions seem like they should be assymptotic on real number line, but its a ringing pattern. If you shift them so that they're super-exponential growth lines up, by taking sexp_eta(x+0.584) and sexp_e(x), the two graphs will intersect each other an infinite number of times, as they both super-exponentially climb to infinity, each on a slightly different pattern. - Shel The calculation was done using the Pari/GP package, with a hand written graphing code. The approximation for tet was obtained from the article on the tetration at the Citizendium and from Kouznetsov's papers, both of which list coefficients. The papers also describe the methods for obtaining them. I don't have the money to get programs like Mathematica, which costs like 1 to 2 thousand bucks or something. And that's interesting about the ringing, it confirms the lack of a viable asymptotic here even without looking at the complex graph. So it would seem that the "cheta" method is not a good method for extending tetration. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 12/27/2009, 06:53 AM (This post was last modified: 12/27/2009, 06:57 AM by sheldonison.) (12/26/2009, 01:54 AM)mike3 Wrote: The calculation was done using the Pari/GP package, .... And that's interesting about the ringing, it confirms the lack of a viable asymptotic here even without looking at the complex graph. So it would seem that the "cheta" method is not a good method for extending tetration.I'll have to try out that Pari/GP package. I agree, the "cheta" method is not a good method for extending tetration, and I also agree that $\theta(x)$ is not analytic. But it tells us something about tetration, that fact that tetration for two different bases can be made to ring. There is an approximate sexp base conversion constant for large numbers, but the ringing means it is only approximate. The ringing for base e, and base eta, it is 0.08% from min to max. For cheta, and e, (or any other two sexp bases greater than $\eta$) there is a 1-cyclic conversion factor, such that as integer "n" increases, $\text{sexp}_e(x+n+\theta(x)) = \text{cheta}(x+n)$ for sexp_e, cheta $\theta(x)$ varies in the range -0.58432+/-0.0004 And there's still a couple of more interesting things I'm trying to figure out. But actually, the only bases for which I know how to generate reasonably accurate slog/sexp are base eta, and base e. Also, I can generate pretty good approximations for sexp for bases between eta and maybe 1.6 or so. It would help a lot if I had a Taylor series, for example for base 2, to verify some of the patterns I'm seeing. The phase and amplitude patterns hold for bases between eta, and 1.6, and also for base e. I'm sure someone must have a link to Andy's slog solution Taylor series results... - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/31/2009, 11:45 PM (12/27/2009, 06:53 AM)sheldonison Wrote: (12/26/2009, 01:54 AM)mike3 Wrote: The calculation was done using the Pari/GP package, .... And that's interesting about the ringing, it confirms the lack of a viable asymptotic here even without looking at the complex graph. So it would seem that the "cheta" method is not a good method for extending tetration.I'll have to try out that Pari/GP package. I agree, the "cheta" method is not a good method for extending tetration, and I also agree that $\theta(x)$ is not analytic. But it tells us something about tetration, that fact that tetration for two different bases can be made to ring. There is an approximate sexp base conversion constant for large numbers, but the ringing means it is only approximate. The ringing for base e, and base eta, it is 0.08% from min to max. For cheta, and e, (or any other two sexp bases greater than $\eta$) there is a 1-cyclic conversion factor, such that as integer "n" increases, $\text{sexp}_e(x+n+\theta(x)) = \text{cheta}(x+n)$ for sexp_e, cheta $\theta(x)$ varies in the range -0.58432+/-0.0004 And there's still a couple of more interesting things I'm trying to figure out. But actually, the only bases for which I know how to generate reasonably accurate slog/sexp are base eta, and base e. Also, I can generate pretty good approximations for sexp for bases between eta and maybe 1.6 or so. It would help a lot if I had a Taylor series, for example for base 2, to verify some of the patterns I'm seeing. The phase and amplitude patterns hold for bases between eta, and 1.6, and also for base e. I'm sure someone must have a link to Andy's slog solution Taylor series results... - Sheldon The graph I gave for $\mathrm{tet}(z)$ was done via the Cauchy integral. It should be possible also to use the Cauchy integral at other bases greater than $\eta$. I'll see if I could try one for $\mathrm{tet}_2(z)$ to get a graph and Taylor series approximation. I do wonder though, even if $\check{\eta}(z)$ cannot be used to approximate $\mathrm{tet}(z)$, whether it is still possible that maybe $\mathrm{tet}_{b_1}(z)$ and $\mathrm{tet}_{b_2}(z)$ can be used to approximate each other, for real $b_1$ and $b_2$ greater than $\eta$. However, the fine detail in that fractal thingy in the graph makes it seem questionable. « Next Oldest | Next Newest »

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