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 Why bases 00), and converges to c, were c is the solution of $\vspace{15}{a^{a^c}}\,=\,c$, which seems to have 2 roots, c₁ and c₂, with $\vspace{15}{a^{c_1}\,=\,c_2}$ $\vspace{15}{a^{c_2}\,=\,c_1}$ $\vspace{15}{{c_1}^{c_1}\,=\,{c_2^{c_2}\,=\,{a^{c_1.c_2}}$ $\vspace{15}{a\,=\,{c_1}^{\frac{1}{c_2}}\,={c_2}^{\frac{1}{c_1}}}$ I suspect that this relation is the key to solve tetration equations: $\vspace{15}{a^{c_1}\,=\,a^{a^{c_2}}\,=\,a^{a^{a^{c_1}}}\,=\,...$ Here is tetration base a=0.01: c₁ = 0,941488369 c₂ = 0,013092521 The negative axis probably converges to a real function akin to a cosine. I need something better than excel. Gottfried Ultimate Fellow Posts: 767 Threads: 119 Joined: Aug 2007 04/13/2015, 08:38 PM (04/13/2015, 08:01 PM)marraco Wrote: The negative axis probably converges to a real function akin to a cosine. I need something better than excel.Pari/GP? And I've made a utility "PariTTY" with which it is easy to copy results to Excel to make graphics or to evaluate something more. Pari/GP can compute to arbitrary precision, work with complex values, matrices, vectors, formal power series. Gottfried Gottfried Helms, Kassel sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 04/13/2015, 09:57 PM (This post was last modified: 04/13/2015, 11:23 PM by sheldonison.) (04/13/2015, 08:01 PM)marraco Wrote: ... Here is tetration base a=0.01: c₁ = 0,941488369 c₂ = 0,013092521 ... The negative axis probably converges to a real function akin to a cosine. .... I need something better than excel. Pari-gp is what you want. Let's assume you're only interested in Koenig's solution as opposed to the much more complicated complex base tetration solution. For your base b=0.1, you should be able to find a real valued fixed point, plus two complex conjugate repelling fixed points. For Koenig's solution, you'll only need the real valued fixed point, which is attracting for b=0.1. That's a good place to start. Figure out how the function behaves in the neighborhood of the fixed point, and what its periodicity is.... The problem with base exp(-e), is that the periodicity is 2, which is a really nasty case since it turns out there is no Koenig's solution... If you'll notice from the link, it took me 8 months to find a conjectured complex base solution, from my first post, to the post with the Taylor series for the complex base tetration solution. But anyway, base b=0.1, find the fixed point, and find the multiplier $\lambda$ at the fixed point, and from that the periodicity$=\frac{2\pi i}{\ln(\lambda)}$; and that's a pretty darn good start, assuming you ever get that far .... The multiplier $\lambda$ is defined where $b^L=L$ and $b^{L+\delta} \approx L + \lambda\delta$ So then there is a formal Koenig solution that has $S(\lambda z) = b^{S(z)}\;\;\;\exp_b^{\circ n} = S(\lambda^n)$. From that, you should be able to generate graphs, or a Taylor series, or whatever you like. - Sheldon marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 04/14/2015, 01:43 AM (This post was last modified: 04/14/2015, 09:31 AM by marraco.) (04/13/2015, 09:57 PM)sheldonison Wrote: But anyway, base b=0.1, find the fixed point, and find the multiplier $\lambda$ at the fixed point, and from that the periodicity$=\frac{2\pi i}{\ln(\lambda)}$; and that's a pretty darn good start, assuming you ever get that far .... The multiplier $\lambda$ is defined where $b^L=L$ and $b^{L+\delta} \approx L + \lambda\delta$ So then there is a formal Koenig solution that has $S(\lambda z) = b^{S(z)}\;\;\;\exp_b^{\circ n} = S(\lambda^n)$. From that, you should be able to generate graphs, or a Taylor series, or whatever you like. I had being reading about the Koenigs function, and I have not a clear understanding of it. My function is $\vspace{15}{y(x)\,=\,^xb\,=\,^x0.01}$ Do you mean finding a value $\vspace{15}{x_0}$ such that $\vspace{15}{0.01^{x_0}=\,x_0}$? (x₀ would be the fixed point of $\vspace{15}{b^x}$) Now, I need to find a function h(x), such that $\vspace{20}{h(0.01^x)=\lambda . h(x)}$, where $\vspace{20}{\lambda=\frac{\mathrm{d} (0.01^x)}{\mathrm{d} x}|_{x=x_0}}$ I read that λ is the derivative of $\vspace{15}{\frac{\mathrm{d} (b^x)}{\mathrm{d} x}|}$$\vspace{5}{_{x={\color{Red} 0}}}$, but you mean $\vspace{15}{\frac{\mathrm{d} (b^x)}{\mathrm{d} x}|}$$\vspace{0}{_{x={\color{Red} x_0}}}$ (that makes more sense to me). If I find h(x), then b˟ would be an eigenvector of h(x), and λ the eigenvalue of b˟. Do $\vspace{15}{S(z)=h^{-1}(z)}$ ?? Is correct that $\vspace{15}{^nb\,=\,h^{-1}(\lambda^n)}$ ?? Edit: Can't be correct. I can't make it work. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 04/14/2015, 02:36 PM (This post was last modified: 04/14/2015, 02:38 PM by sheldonison.) (04/14/2015, 01:43 AM)marraco Wrote: Do $\vspace{15}{S(z)=h^{-1}(z)}$ ?? Is correct that $\vspace{15}{^nb\,=\,h^{-1}(\lambda^n)}$ ?? Edit: Can't be correct. I can't make it work.Yes. For b=0.01; the primary fixed point, which is repelling is $L\approx0.277987425;\;\;\; \lambda\approx -1.28017940$ ${0.01}^{\left( z+0.277987425 \right)} \; ~= \; L -1.28017940 \cdot z +2.94772200\cdot z^2 -4.52492050\cdot z^3 ...$ $S(z) = h^{-1}(z)=L + z + a_2 z^2 + a_3 z^3 + ... a_n\cdot z^n$ There is a formal solution for a2, a3 etc in terms of the Taylor series of the desired function at its fixed point. I get a2~=1.00982628 ... Normally, we would use $f(z)=S(\lambda^z)$ for the superfunction, but since lambda is a negative number, perhaps one could use $f(z)=S\left( (-\lambda)^z \cdot \sin(\pi z)\right)$ to generate the exponentially increasing pseudo 2-periodic solution the Op is looking for. - Sheldon sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 04/14/2015, 04:48 PM (This post was last modified: 04/15/2015, 03:23 PM by sheldonison.) (04/14/2015, 02:36 PM)sheldonison Wrote: $L\approx0.277987425;\;\;\; \lambda\approx -1.28017940$ ... Normally, we would use $f(z)=S(\lambda^z)$ for the superfunction, but since lambda is a negative number, perhaps one could use $f(z)=S\left( (-\lambda)^z \cdot \sin(\pi z)\right)$ to generate the exponentially increasing pseudo 2-periodic solution the Op is looking for. $S(z) = \sum_{n = 0}^{\infty} a_n \cdot z^n\;\;\; S(\lambda z) = {0.01}^{S(z)}\;\;\; S(0)=L \;\;\; {0.01}^L=L \;\;\; \lambda \approx -1.28017940081259$ Code:a0=   0.277987424809561 a1=   1 a2=   1.00982628479736 a3=  -1.74658670929184 a4=  -3.98120395207580 a5=   3.59453988520567 a6=   13.8485552663038 a7=  -6.48959552733049 a8=  -44.4747907113725 a9=   7.91461618658992$\text{superfunction}(z)=S\left( (-\lambda)^z \cdot \sin(\pi z)\right)$ And here is a graph, showing an analytic super-function for ${0.01}^z$ the Op might be interested in, from -10 to 10, where it starts out oscillating around the primary fixed point, and then converges to the two cycle that the Op noted. Of course, this isn't tetration; Tet(-2) is by convention a logarithmic singularity. And this function has no uniqueness, I could have just as easily used cosine instead of sine, or any of other 2-cyclic function with $\theta(z+1)=-\theta(z)$.     - Sheldon marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 04/14/2015, 08:17 PM (This post was last modified: 04/14/2015, 08:51 PM by marraco.) I had just installed Sage, which includes PariGP, but still are stuck on excel, because I need to learn PariGP. Thanks for your help. I still chewing it. Since λ is negative, that means that his powers take complex values, then h(x) should be complex, and that's a mess with excel. So, as practice, I tried with base b=2^⁽½⁾, and still trying to make it work. The fixed point is 2, and λ=0,693147181 The blue line should be the tetration base b=1,414213562, and should be matching the cyan line (which I got with Excel). (04/14/2015, 04:48 PM)sheldonison Wrote: [/code] $\text{superfunction}(z)=S\left( (-\lambda)^z \cdot \sin(\pi z)\right)$ And here is a graph, showing an analytic super-function for ${0.01}^z$ the Op might be interested in, from -10 to 10, where it starts out oscillating around the primary fixed point, and then converges to the two cycle that the Op noted. Of course, this isn't tetration; Tet(-2) is by convention a logarithmic singularity. And this function has no uniqueness, I could have just as easily used sine or cosine, or an infinite number of other 1-cyclic functions. Maybe that curve is inverted? That way, it would match my own solution, and would give ⁰b≈1 and ⁻¹b≈0 I got that solution adjusting the coefficients with Excel's Solver, for a Taylor series with 12 coefficients, (expanded around 0), subject to these restrictions: a₀=1 ⁻¹b=0 ⁰b=1 $^xb=b^{\,{^{x-1}b}}$ I got these coefficients: Code:a0    1 a1    -1,339425732123610E+00 a2    -3,822141746305190E+00 a3    4,850222934263920E+00 a4    5,354079775248650E+00 a5    -6,056488564613110E+00 a6    -2,938727974303330E+00 a7    2,550689528770280E+00 a8    4,116553795338360E-01 a9    1,989418887288240E-06 a10    1,291559624345350E-04 a11    -1,519921425061600E-07 a12    1,409996370575610E-08 a13    1,348473734742680E-10 a14    1,133187395821660E-11 a12    -3,860717643983480E-09 marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 04/14/2015, 08:49 PM (This post was last modified: 04/15/2015, 04:50 AM by marraco.) (04/14/2015, 04:48 PM)sheldonison Wrote: I conjecture that tetration base 0 should be a discontinuous function, alternating between the values 0 and 1, with period 2, and as the base approach zero, the negative axis turns into a real, continuous function. The derivative must converge to a periodic and alternating Dirac Delta, multiplied by c₁-c₂, and the surface of each "Dirac" must be constant on the positive axis. That's because there is at least one point in each period with value c₁ and c₂. « Next Oldest | Next Newest »

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