# Tetration Forum

Full Version: Why bases 0<a<1 don't get love on the forum?
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I had being looking what had being done with bases lower than 1, and found nothing.

I didn't expected to get oscillating/damped periodic functions. They also seem to converge to $\vspace{25}{\lim_{x\to \infty}{^xa=c} \Leftrightarrow {c}^{\frac{1}{c}}=a}$

¿Does this hints that $\vspace{20}{^{i.x}a\,=\,^x{(\frac{1}{a})}$ ??

(04/13/2015, 04:26 AM)marraco Wrote: [ -> ]I had being looking what had being done with bases lower than 1, and found nothing.

I didn't expected to get oscillating/damped periodic functions. They also seem to converge to $\vspace{25}{\lim_{x\to \infty}{^xa=N_a} \Leftrightarrow {N_a}^{\frac{1}{N_a}}=a}$

¿Does this hints that $\vspace{20}{^{i.x}a\,=\,^x{(\frac{1}{a})}$ ??

Well there is at least one post; http://math.eretrandre.org/tetrationforu...0&pid=6748 tetration base $\exp(-e)\approx0.0660$

This is an example of a really difficult case, where you can't use Koenig's solution, since the multiplier at the fixed point is $\frac{\exp(\2\pi i)}{5}$, which is an indifferent case with a rational multiplier. Otherwise, using Koenig's solution is fairly straightforward, but of course Koenig's solution is never the same as the complex base solution you would get using both fixed points, if you started with tetration for a real base greater than $\eta=\exp(1/e)$, and then slowly changed the base in the complex plane, while avoiding the singularity at $\eta$.

I think Mike and I and Henryk are the only people on the forum that have any understanding of complex base tetration for $b<\eta$; not that we understand it that well. One of the interesting things is that for real bases; $b<\eta$; that it ($\text{Tet}_b(z);\;$ complex base tetration for base b) is not real valued! For bases close to eta, it is almost real valued, as the imaginary pseudo period for both fixed points gets arbitrarily large.
(04/13/2015, 03:27 PM)sheldonison Wrote: [ -> ]Well there is at least one post; http://math.eretrandre.org/tetrationforu...0&pid=6748 tetration base $\exp(-e)\approx0.0660$

That's were excel gets itchy, because it can't take logarithms of negative numbers. The polynomial gets too close to zero.
I get this

for bases between $\vspace{15}{0 it remains bounded between 0 and 1 (for x>0), 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.
(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
(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.
(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.
(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.
(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 $\theta(z+1)=-\theta(z)$.

[attachment=1190]
I had just installed Sage, which includes PariGP, but still are stuck on excel, because I need to learn PariGP.

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
(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.

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