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04/13/2015, 04:26 AM
(This post was last modified: 04/13/2015, 10:24 PM by marraco.)
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
¿Does this hints that ??
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04/13/2015, 03:27 PM
(This post was last modified: 04/13/2015, 05:07 PM by sheldonison.)
(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
¿Does this hints that ??
Well there is at least one post; http://math.eretrandre.org/tetrationforu...0&pid=6748 tetration base
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 , 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 , and then slowly changed the base in the complex plane, while avoiding the singularity at .
I think Mike and I and Henryk are the only people on the forum that have any understanding of complex base tetration for ; not that we understand it that well. One of the interesting things is that for real bases; ; that it ( 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.
 Sheldon
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04/13/2015, 08:01 PM
(This post was last modified: 04/13/2015, 09:43 PM by marraco.)
(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
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 it remains bounded between 0 and 1 (for x>0), and converges to c, were c is the solution of , which seems to have 2 roots, c₁ and c₂, with
I suspect that this relation is the key to solve tetration equations:
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.
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(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
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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.
Parigp 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 at the fixed point, and from that the periodicity ; and that's a pretty darn good start, assuming you ever get that far .... The multiplier is defined where and
So then there is a formal Koenig solution that has . From that, you should be able to generate graphs, or a Taylor series, or whatever you like.
 Sheldon
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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 at the fixed point, and from that the periodicity; and that's a pretty darn good start, assuming you ever get that far .... The multiplier is defined where and
So then there is a formal Koenig solution that has . 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
Do you mean finding a value such that ? (x₀ would be the fixed point of )
Now, I need to find a function h(x), such that , where
I read that λ is the derivative of , but you mean (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 ??
Is correct that ??
Edit: Can't be correct. I can't make it work.
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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 ??
Is correct that ??
Edit: Can't be correct. I can't make it work. Yes. For b=0.01; the primary fixed point, which is repelling is
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 for the superfunction, but since lambda is a negative number, perhaps one could use to generate the exponentially increasing pseudo 2periodic solution the Op is looking for.
 Sheldon
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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:
...
Normally, we would use for the superfunction, but since lambda is a negative number, perhaps one could use to generate the exponentially increasing pseudo 2periodic solution the Op is looking for.
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
And here is a graph, showing an analytic superfunction for 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 2cyclic function with .
 Sheldon
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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]
And here is a graph, showing an analytic superfunction for 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 1cyclic 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
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 4,116553795338360E01
a9 1,989418887288240E06
a10 1,291559624345350E04
a11 1,519921425061600E07
a12 1,409996370575610E08
a13 1,348473734742680E10
a14 1,133187395821660E11
a12 3,860717643983480E09
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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 c₁c₂, 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.
