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Why bases 0<a<1 don't get love on the forum?
#11
(04/14/2015, 08:49 PM)marraco 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.
Maybe; first figure out the bases in between...

Quote:The derivative must converge to a periodic and alternating Diract Delta, multiplied by c₁-c₂, and the surface of each "Diract" must be constant on the positive axis.

That's because there is at least one point in each period with value c₁ and c₂.

Cool stuff! I think the next step is to figure out the Taylor series using some of the techniques we've developed for Tetration. Here, I spent a few minutes generating a four term series approximation using a variant of Andrew's slog technique. This approximation is good between z=-1, and z=0. Taking the ; the 2nd derivative is continuous for a piecemeal approximation. The other boundary requirements are: tet(0)=1; tet(-1)=0; tet'(0)=0; tet'(-1)=0;

   
The tet(z) function is real valued for z>-3.9... surprisingly remaining real valued after the logarithmic singularity at -2. I expect there is a unique solution for the complex plane, which can be developed using Kneser's Riemann mapping and the equivalent mapping techniques we've invented ... which should be really interesting too! We expect this function to approach two of the function's other complex fixed points, and its conjugate, as gets arbitrarily large, positive or negative.
- Sheldon
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#12
Did anybody tried tetrations with negative bases, are there some special intervals here (like we have e^-e and e^(1/e)) on positive axis side?
So far I see that there should be at least 4 differnt tetration classes:
z < 0
e^-e > z > 0
e^(1/e) > z > e^-e
z > e^-e

And special cases:
e^-e
e^(1/e)
and maybe 0?
Is there any general summary of tetriation behavior in all those cases?

Best Regards.
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#13
Plot exp(-exp(-x)) then you see why its not so popular.
It has a Nice nonparabolic fix yet it is not nice.
Notice the limits at - , + infinity.

Regards

tommy1729
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#14
I have a conjecture I haven't been able to prove. I think the Fractional Calculus approach I took for hyper operators in between 1 and eta will work for bases inbetween

To be explicit:



This is of course if there exists an exponentially bounded solution. Since these bases do not induce a periodic tetration there requires more thought into how to solve this case.

Constructing the Koenigs function of produces a function where . Using regular Koenig iteration we can construct a tiny area about the fixed point which exists. If we can induce the Koenigs function such that is a well defined function for and large enough so that this function is defined for all we can extend this to all in the immediate basin (including 1) using Fractional Calculus and similar techniques that I used here. This would give , and then using more tricks can be recovered.

The trick is of course ensuring that defined for in some can be analytically continued to the right half plane, and then further is appropriately bounded. This would require analytically continuing the koenig function and as well its inverse.

Unless someone knows if there exists a solution such that for with , I see no other way of making this method work.

However there is excitement, because certain holomorphic functions with a multiplier that is not real positive and between 0 and 1 can be complex iterated using fractional calculus.


To give further motivation of this technique, consider the function for
We can iterate this using fractional calculus and for in the immediate basin of attraction, z in the right half plane, is analytic in and consequently the iterate must converge in the Fractional calculus transform for in an epsilon neighbourhood of (0,1). The suspicion is that it will converge for all on suitable . I'm betting the exponential function is one of these suitable functions.
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#15
Conjecture: For all bases between , there are a pair of complex conjugate fixed points. Each of these fixed points has a Koenig solution, that can be used to generate a complex valued superfunction. Such a Koenig solution is entire, and never takes on the value of 0. The conjecture is the pair of Koenig complex superfunctions can be Kneser (Riemann) mapped (or equivalently) theta mapped, to a Tetration solution that is real valued between the singularity at -2, and infinity. Such a solution would be like tetration for bases> analytic if or and at the real axis, for z>-2, with singularities at -2, -3, and other values <-3.

For tetration bases , there is a primary real valued fixed point, with a negative multiplier at the fixed point. For bases the fixed point is repelling, but bifurcates to an attracting two-cycle. marraco and I have posted plausible real valued tetration solutions with Tet(0)=1 for base b=0.01. For bases> the fixed point is attracting, and the solution will slowly oscillate towards the fixed point, exponentially scaling as it does so. The initial seed for such a solution would be generated in much the same way as the earlier post for b=0.01; I haven't done the Kneser/theta mapping yet, from the complex pair of fixed points.

For example, consider b=0.1, which has an attracting fixed point , and a pair of repelling complex conjugate fixed points . Here is an initial seed, a 4th degree polynomial approximation roughly accurate between tet(-1) and tet(0), with boundary conditions tet(-1)=0, tet(0)=1, tet'(-1)=0, tet'(0)=0, and the with a piecemeal approximation having a continuous 2nd derivative:



Then this is what it looks like, at the real axis from -3.76 to 10. As
   

And here is the function at from - Notice that the function would go to the secondary fixed point of as . For , it would go to the conjugate fixed point, . So this graph is why I think there is a unique Kneser mapping to generate this Tetration function, with the boundary conditions that tet(-1)=0, tet(0)=1, tet'(-1)=0, tet'(0)=0, (approaching Koenig's solution)

The algorithm just sketched could work for all bases .
   

I also have some bonus questions. Is it possible that we derive the requirement that tet'(-1)=0 from the requirement that the Riemann mapping go to the pair of conjugate fixed points in the complex plane? btw, if Tet'(-1)=0, then it is trivial to show that tet'(0)=0, from the definition of the derivative of exp(f(x))

Also, this function is different than Tetration ,http://math.eretrandre.org/tetrationforu...0&pid=6748. Mike and I think that the function at this link is actually the Tetration function that is analytically connected to real base tetration for b>eta. What happens if we start with this new Tetration function, for 0<b<1, and slowly move the base in the complex plane, avoiding the singularities at 1 and eta, back to the real axis for base(e)? Does this new function have a natural analytic boundary someplace, perhaps at the Shell Thron boundary or something like that???? That's a wild guess, btw. For example returning to base(e), I think you wind up with tetration from the secondary fixed point~=2.0623+ 7.589i, in the upper half of the complex plane, along with 0.318-1.337i in the lower half of the complex plane, so this could be challenging... see the somewhat related post tetration from secondary fixed point.
- Sheldon
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#16
What is the justification to demand that the derivative at zero must be zero?

Every time I tried to develop a Taylor series of tetration, I found that there is ever a degree of liberty, in choosing the derivative at 0.

This was part of one of my failed attempts.

marraco Wrote:Let suppose that there exist a series for convergent in a radius equal or larger than 1.

Since °a=1, the first coefficient must be a₀=1:




Let's call p to the sum of even terms, and q to the sum of odd ones



Since we know the values of at -1 and 1:


If we need to make a linear systems of equations to solve the coefficients, this would be the first one.


Looking at the series, is evident that the functions generated by the coefficients p, P(x) and q, Q(x) would be calculable from this way:





Those functions may be related to an analogous decomposition of e^x into sines and cosines.
[Image: x8RXtVL.png?1]
[Image: 54RSKT9.png?1]
[Image: wgtELKc.png?1]

We also know this differential equation:



The derivative at 0 is

Then:



The derivative at x=1 is




But, P series only has even powers of x, and Q only odd, so:


Then

Which is a system of 2 equations with 3 unknown. If I try to add anything else, I ever get one extra unknown.
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#17
(04/17/2015, 02:50 AM)marraco Wrote: What is the justification to demand that the derivative at zero must be zero?

Every time I tried to develop a Taylor series of tetration, I found that there is ever a degree of liberty, in choosing the derivative at 0.

This was part of one of my failed attempts.

marraco Wrote:Let suppose that there exist a series for convergent in a radius equal or larger than 1.

Since °a=1, the first coefficient must be a₀=1:




Let's call p to the sum of even terms, and q to the sum of odd ones





Since we know the values of at -1 and 1:


If we need to make a linear systems of equations to solve the coefficients, this would be the first one.

I also find it interesting because



Looking at the series, is evident that the functions generated by the coefficients p, P(x) and q, Q(x) would be calculable from this way:





Those functions may be related to an analogous decomposition of e^x into sines and cosines.
[Image: x8RXtVL.png?1]
[Image: 54RSKT9.png?1]
[Image: wgtELKc.png?1]

We also know this differential equation:



The derivative at 0 is

Then:



The derivative at x=1 is




But, P series only has even powers of x, and Q only odd, so:


Then

Which is a system of 2 equations with 3 unknown. If I try to add anything else, I ever get one extra unknown.

I think makes sense.

Regards

Tommy1729
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#18
(04/17/2015, 02:50 AM)marraco Wrote: What is the justification to demand that the derivative at zero must be zero?

Every time I tried to develop a Taylor series of tetration, I found that there is ever a degree of liberty, in choosing the derivative at 0.

I could have been more clear; that boundary condition, tet'(-1,0)=0, seemed to work to get a seed from which a Kneser bipolar mapping seems plausible, based on the behavior going smoothly to the pair of conjugate fixed points, in a path to -infinity, from just above and below the real axis. My bonus question, was pretty much the same as your question, "Is it possible that we derive the requirement that tet'(-1)=0 from the requirement that the kneser mapping go to the pair of conjugate fixed points in the complex plane?". edit perhaps tet'(0)=0 is not a requirement for a Kneser solution. Then there may be an infinite number of Kneser solutions with different derivatives... don't know.

edit#2 the solution isn't upside down; For in the lower half of the complex plane, and in the upper half of the complex plane The lower half of the complex plane should be somewhat like the lower half of http://math.eretrandre.org/tetrationforu...0&pid=6748 post#10 and in the upper half, will be its conjugate.

By the way, Andrew's slog won't work, since the Abel function has singularities where sexp'(z)=0. The Kouznetsov's algorithm might work, though from previous experience with b=exp(-e), the function doesn't behave that nicely on the path to the fixed point. The algorithm most likely to converge is the Kneser/theta mapping algorithm. So, now that I have this initial piecemeal seed, the next step would be to write a program that would iterate, generating the Kneser/theta mapping, from the initial seed, and the two complex valued Koenig's solution. This would let us iterate and generate results of arbitrarily high precision. Its not a trivial undertaking, and will require several days of work; probably more like weeks. At the end, running the program will take less than 30 seconds for arbitrary bases between 0 and 1, and then you can instantaneously get very accurate result, and can calculate the function anywhere in the complex plane, so you can make pretty complex plane graphs too. As imag(z) increases, or decreases, the function will go to the complex conjugate pair of fixed points, and will converge to one of the Koenigs solution's in the upper half of the complex plane, and the conjugate Koenig's solution in the lower half of the complex plane ...

Anyway, assume the Kneser/theta mapping converges, then the conjectured uniqueness criteria is that the solution will be analytic in the upper half of the complex plane and the lower half of the complex plane, and at the real axis for z>-2, and converging to the two complex valued Koenig's solutions in the upper/lower halves of the complex plane. The uniqueness criteria and proofs that Henryk has written rely on the slog/Abel function (and the sexp function) both being analytic on a sickle, between the two fixed points. But since the slog/Abel function has singularities where sexp'(z)=0 ... I don't know how one might apply that proof technique to the solution at hand.
- Sheldon
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#19
For bases b > 2 i think the following makes sense :

For x > 0 :
F " (x) > 0
F ' (x) > 0
And
F ' (0) = b/2

What do you think ?

Regards

Tommy1729
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#20
(04/17/2015, 08:26 AM)tommy1729 Wrote: I think makes sense.

Regards

Tommy1729

(04/17/2015, 09:15 PM)tommy1729 Wrote: For bases b > 2 i think the following makes sense :

For x > 0 :
F " (x) > F ' (x)/2
F ' (x) > 0
And
F ' (0) = b/2

What do you think ?

Seems arbitrary to me. Do you care to expand?

There is probably some value for a₁ that provides some benefit. For example, some values makes (which is right bracket tetration). Other values make .

I don't like a₁=0, because it makes most bases look weird.
For example, this is tetration base , when a₁=0 (up), and when there is no restriction in a₁ (down; It gives a₁=0,6118~~~~~)

[Image: 4b7mbuR.png?1]

I find the smoothed curve more appealing, because I suspect that flow in pipes is governed by tetration.

This is a semi empiric "Moody diagram" of flow in pipes. To me, it resembles tetration. When flow turns fully turbulent, it converges to a constant value.

[Image: iajnDJO.png?1]
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