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 Solving tetration for base 0 < b < e^-e mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/12/2009, 02:00 AM Hi. I was wondering about the possibility of extension of the tetrational function $^{z} b$ to real and complex towers z (also called "heights", but I like "tower" because it rhymes with "power") for exotic bases in the range $0 < b < e^{-e}$, as well as at the point $b = e^{-e}$. Most extensions to tetration seem to concentrate on bases greater than 1, however I'm interested in the range of bases less than 1 due to the exotic behavior of their integer tetration. It seems that the tetration for bases $e^{-e} < b < e^{1/e}$ can be defined via the regular iteration technique, and for $b < 1$ it yields complex values. The technique also seems to be applicable to all bases inside the convergent region of the complex plane (the "kidney-bean" shaped area of the "tetration fractal"), and when using the principal branch of the logarithm and Lambert function we get a "principal branch" of the tetrational in that region of bases with a branch cut going to the left from the base b = 1, where it is continuous from above. The bases on the interval $e^{-e} < b < 1$ lie along this cut line. When tetration of these bases to real towers is plotted, it forms a spiral pattern. However, once one gets below $e^{-e}$, there is no more attracting fixed point and $^n b$ for natural numbers n does not settle down as $n \rightarrow \infty$, but instead oscillates between two values (solutions of $b^{b^z} = z$ -- I'm not sure if there exists a closed-form solution for them like how is a closed form for the fixed points of the exponential via Lambert's function, though), i.e. it converges to a 2-cycle. Thus the regular method does not seem to be useful for extending to these bases (regular iteration down into the repelling fixed point is possible but it yields a function that is entire on the complex plane, and that is not good, as I'd expect tetration to have branch points/singularities at z = -2, -3, -4, -5, ...). Yet I'm really curious to know what would happen if this tetration were extended to real and complex values of z, as it doesn't seem much has been done with it before, and it's a sort of "last frontier" of the real bases. The behavior for $e^{-e} < b < 1$ is interesting enough, but those bases still converge on a fixed point. I was curious about Ansus' formula and also Kouznetsov's Cauchy integral method, though it is possible (but by no means certain) that there may be additional singularities on the z-plane (additional to those at z = -2, -3, -4, -5, ...), as the bases in $e^{-e} < b < 1$ have additional singularities on the z-plane (in both the left and right half-planes) due to complex periodicity with period $\frac{2\pi i}{\log(\log(F))}$ where $F = -\frac{W(-\log(b))}{\log(b)}$ is the principal fixed point of the exponential), so choosing a good countour for the latter may be tricky, plus there are uncertainties in what the asymptotic behavior should be. And by Ansus' formula, I mean the formula from the continuous product, given by $\mathrm{tet}'_b(x) = \log(b)^{x} \mathrm{tet}'_b(0) \prod_{k=0}^{x-1} \mathrm{tet}_b(k)$. which can also be written as an iterative formula with an integral: $\mathrm{tet}_b(x) = \mathrm{tet}'_b(0) \int_{-1}^{x} \log(b)^x \prod_{x} f(x+1) dx$. However I haven't had much luck in getting the iterative integral formula to stabilize/converge. For base e and an initial function (f(x) = 1) given as a truncated Taylor series it seemed to converge for a bit then diverged again (I could post the code if you want). The thread these formulas come from is here: http://math.eretrandre.org/tetrationforu...273&page=1 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/12/2009, 06:56 AM Gottfried considered this case already here. He provided pictures of the regular iteration (base 0.04) at the repelling fixed point. If we however rather want to consider iteration at the attracting 2-cycle, the following comes to my mind: At this base range the limits $\lim_{n\to\infty} f^{\circ 2n}(x_0) = p_1$ and $\lim_{n\to\infty} f^{\circ 2n+1}(x_0)= p_2 = f(p_1)$ exist ($f(x)=b^x$). So I would consider the regular iteration of $g(x)=f^{\circ 2}(x)=b^{b^x}$ and then just always take the half of the iteration number $f^{\circ t}(x)=g^{\circ t/2}(x)$. I will carry that out perhaps in the next post. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/12/2009, 07:20 AM (09/12/2009, 06:56 AM)bo198214 Wrote: So I would consider the regular iteration of $g(x)=f^{\circ 2}(x)=b^{b^x}$ and then just always take the half of the iteration number $f^{\circ t}(x)=g^{\circ t/2}(x)$. The interesting thing is that $g$ not only has the two attracting fixed points $p_1$ and $p_2$ but also a repelling fixed point in between. Which is attractive for $g^{-1}$ in the range $(p_1,p_2)$.     Graph for b=0.01 with p1=0.941482102273016 and p2=0.0130925205079953. So why not do regular iteration at that inbetween fixed point? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/12/2009, 07:40 AM (09/12/2009, 07:20 AM)bo198214 Wrote: (09/12/2009, 06:56 AM)bo198214 Wrote: So I would consider the regular iteration of $g(x)=f^{\circ 2}(x)=b^{b^x}$ and then just always take the half of the iteration number $f^{\circ t}(x)=g^{\circ t/2}(x)$. The interesting thing is that $g$ not only has the two attracting fixed points $p_1$ and $p_2$ but also a repelling fixed point in between. Which is attractive for $g^{-1}$ in the range $(p_1,p_2)$. [attachment=551] Graph for b=0.01 with p1=0.941482102273016 and p2=0.0130925205079953. So why not do regular iteration at that inbetween fixed point? Because it's a repelling point, so the resulting superfunction will still be entire, and I want to try and find one with the singularities, etc., as that is more like what is expected from "tetration". bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/12/2009, 07:47 AM (This post was last modified: 09/12/2009, 07:50 AM by bo198214.) (09/12/2009, 07:40 AM)mike3 Wrote: Because it's a repelling point, so the resulting superfunction will still be entire, and I want to try and find one with the singularities, etc., as that is more like what is expected from "tetration". The method does anyway not work, even if we take the regular iteration at one of the attracting fixed points. Because the regular iteration returns a strictly increasing function, particularly $g^{\circ 1/2}$ is strictly increasing and hence $g^{\circ 1/2} \neq f$ because $f$ is strictly decreasing. Its a bit like you want to know what $(-3)^t$ is. Then you can not just consider $((-3)^2)^{t/2}=3^t$ as $3^t\neq (-3)^t$. PS: Quoting policy is to only quote what is necessary for your reply. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/12/2009, 08:07 AM Hmm. However in the emails, you mentioned the use of a "multiplier" that works similar to the derivative at the fixed point but for a cycle. How does one employ this? Oh yeah, and sorry about the "big" quote. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/12/2009, 08:35 AM (This post was last modified: 09/12/2009, 08:40 AM by bo198214.) (09/12/2009, 08:07 AM)mike3 Wrote: Hmm. However in the emails, you mentioned the use of a "multiplier" that works similar to the derivative at the fixed point but for a cycle. Generally the multiplier of a cycle $p_1,\dots,p_n$ (i.e. $p_{k+1}=f(p_k)$ and $f(p_n)=p_1$) is defined as: $f'(p_1)\cdot f'(p_2)\dots f'(p_n)$. This is equal to the multiplier of $f^{\circ n}$ at any point of the cycle (by the chain rule). Example n=2 $f(f(x))'=f'(f(x))\cdot f'(x)$. If you now plug in $x=p_1$ you get $f'(p_2)\cdot f'(p_1)$ and if you plug in $x=p_2$ you get the same result $f'(p_1)\cdot f'(p_2)$. If you would depict the tangents at the left and right fixed point in the graph before, they would be parallel. But this approach to consider $f^{\circ 2}$ already failed. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/12/2009, 08:50 AM So it would seem that the regular iteration is not applicable to this case... (as I suspected) What about the other methods, e.g. the continuum-product formula? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/12/2009, 03:48 PM (09/12/2009, 08:50 AM)mike3 Wrote: So it would seem that the regular iteration is not applicable to this case... (as I suspected) What about the other methods, e.g. the continuum-product formula? There are two other methods. Intuitive slog and matrix power tetration. The intuitive slog seems not applicable to bases $\le 1$ (see Andrew's post about the base range of the intuitive slog). Then I tried the matrix power method for base 0.04. It converges, however the result is not a superexponential. I plot the sexp on (0,1) (lower curve) and b^sexp(x) on (-1,0). They should be the same, but differ remarkably. The curve is nearly the same for matrix size 30 and matrix size 40.     . mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/12/2009, 09:04 PM (This post was last modified: 09/12/2009, 09:05 PM by mike3.) So then since neither of those worked, it seems all we're left with is the Ansus formula and the Cauchy integral (but determining the correct contours and asymptotic behavior, now that's the rub...). I'm also curious: What about $b = e^{-e}$ exactly? You said it converges slowly, but how do you iterate it at all? 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