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Ivars Wrote:Quote:
A complete analysis of such questions considering also the complex
roots involves the T function as shown by Hayes in ND Hayes "The roots of the equation and the cycles of the substitution
I think the tree function is T(x) = W(x).
I like T because it allows H to be defined very simply.
Andrew Robbins
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Well, this is approximately what I was thinking. If , then we are trying to solve and where right? OK, so I started with a series expansion, and this is easier if you deal with a single exponential rather than an iterated exponential, so I found the series expansions:
and with these expansions at hand, we notice that if , then , so these two expansions should be the same (this is not true, but for some reason I was thinking it was... It is true for but not for all x). If these two expansions are the same, then the first two coefficients should be equal, which means
Solving for in the secondtolast equation gives , which I suppose is what I was thinking would be a general function, but I don't think it works that way. Also, solving for in the secondtolast equation gives .
Andrew Robbins
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Just a proposal  it might be useful to plot the graphs not with xaxis = base b, but, assuming b=t^(1/t), u=log(t), with xaxis =u. It rescales the plot a bit, so the behaviour may be better visible by such a graph. (your opinion?)
Example:
Gottfried
Gottfried Helms, Kassel
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01/16/2008, 10:03 AM
(This post was last modified: 01/16/2008, 09:48 PM by Gottfried.)
andydude Wrote:Well, this is approximately what I was thinking. If , then we are trying to solve and where right? OK, so I started with a series expansion, and this is easier if you deal with a single exponential rather than an iterated exponential, so I found the series expansions: Hi Andrew 
I actually do not understand the job, that x0 and x1 shall do. But it gives a mental association to me, which may be helpful for your considerations, too.
From the eigensystemanalysis of decremented iterated exponentiation
where
and ) I get a bivariate polynomial in u and u^h, where h indicates the "height" (see below).
Since on the other hand, I found, that the iterated exponentiation
(with x=1 means our usual tetration)
can be converted into
by scaling and shifting of x we may use these polynomials also for Ttetration.
We have, using b=t^(1/t), (t complex for b>eta) the substitution x'=x/t1, and inverse substitution x"=(x+1)*t,
so we may apply the polynomials to x' and use the result y' by applying the inverse substitution y"
Since we have denominators of the powerseries for U() , which generate singularities for all complex u, where abs(u)=1 (which means for instance, b=e^(e^1) = eta,u=1, or b=(e^1)^e = beta , u=1) these cases come clearly out as special cases. However, by the construction of the polynomials, for integer heights h the denominators can be cancelled by the same factors in the numerators, and we have a powerseries with 0/0terms, or, when cancelled, surely an equivalent solution to the logarithmapproach (but I didn't check this so far).
Here is the powerseries for (sorry , when editing the formula I overlooked one occurence of f(), which should be rewritten as U())
which I assume is the "ultimate" description for the tetrationproblem (assuming continuation of tetration via powerseriesexpansion) from where other considerations about the behaviour of tetration can be derived (at least for real u, which means for bases b 0<b<eta)
Perhaps your x0,x1idea may be inserted into this polynomial and be resolved this way. If x0=t (the fixpoint of b), then x' = x/x01 in the above formula, and the terms of the powerseries for U(x) expand and mix then with binomial weights. (don't know, whether this agrees with your idea, however).
I can provide the matrix of polynomially coefficients at the powers of x, where one might then insert values for u and for u^h up to size 64x64. Unfortunately the polynomials for each term of the powerseries grow binomially c(r,2) with their rowindex r, so the size of this matrix is huge (200 MB in textformat): for 96 terms of the powerseries (which was my intended goal) one even had a polynomial of degree ~4400 in the 96'term and if I count the number of all occuring coefficients at powers of u individually I get about 10 million coefficients for that number of x^kpolynomials... urrgs...
Gottfried
Gottfried Helms, Kassel
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01/16/2008, 10:26 PM
(This post was last modified: 01/16/2008, 10:30 PM by Ivars.)
May be a stupid question:
1) For domain x<e^e can odd and even h(x) be looked upon as separate continuous or infinitely differentiable functions?
2) Would the same apply to superroots in their oscillating domain?
3) If we denote them h odd(x) , h even (x) , can we make a function:
f(x) = h odd (x) / h even (x)? or h even (x) * h odd (x)? And perhaps similar one for supperroots?
4) what will happen with that function h odd(x)/h even(x) at x=0
And what will happen with h even (x) * h odd (x) at point x=0?
The values will be 0, but how can it be?
5) If we could than divide
h odd (x)/h even(x) at x=0 which is 0 with
h even (x) * h odd (x) at x=0 which is 0 again then we have
(h odd (0) /h even (0) ) / ((h even (0) * h odd (0)) = 1/h^2 even(0) = 1/1= (0/0)
If we do it other way around, we get h^2 even (0)/1 = 1/1= 0/0
For superroots, it will be odd function squared that will lead to 0/0. Obviosly that seems not to be true, since 0/0 is undefined, so we need to differentiate away the 0es but it seems to me we can not do it even by infinite application of lHopitals rule.
I have a feeling that somehow we need an extra dimension to resolve this. But may be this is just another miss
Excuse me.
Ivars
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Ivars Wrote:1) For domain x<e^e can odd and even h(x) be looked upon as separate continuous or infinitely differentiable functions?
2) Would the same apply to superroots in their oscillating domain?
My guessing is that what you called heven and hodd (the lower and upper limit of the "off limit" area) are the lower and upper branch of the same twovalued "function". The application of this idea to the superroots would imply the inversion of that "function", giving, this time arespectable onevalued function
To investigate that, I think it would be necessary to use all (so to say) the real and complex branches of the Lambert "function".
But this is my guessing!
GFR
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Here is Andrews' observation from the thread tetration of h(0) :
I do not know how to copy Tex so I copy the link:
http://math.eretrandre.org/tetrationforu...825#pid825
Quote: Each one of these definitions gives a different answer as x approaches 0. The strange thing is that all 3 definitions are equivalent for e^e<x<e^1/e.
Andrew Robbins
When I was speaking in previous post about the missing dimension, I was thinking about rotating plane ( or rotation in plane) perpendicular the graphs of GFR and parallel to y axis.
Then I have not even a guess, but just a vision of these 3 definitions of h(x) forming a double helix (odd/even) with a thread (x^1/x) in the midlle in region e^e<x<e^(1/e). The projection on GFR graph is just a single line, but projection on that perpendicular plane is 3 points at each x.
When bifurcation happens x< e^e, we get 2 opening spirals rotating left (odd?) and right (even?) in that plane and I am not sure what happens with x^(1/x) central thread thereit may tend to become a rotating connection between these spirals of that circle.
At x=0 we may have a unit circle and x^(1/x) a rotating diameter or radius?
That perpendicular plane is necesserily complex plane.
Would that be possible to somehow then connect with Gotfrieds spider like imaginary zeros x>e^1/e of so that his complexplane is situated in this construction perpendicularly x but behind x> e^1/e? With all the branches.
Then we would be able to connect somehow that region above e^1/e on that complex plane with the region x(0, 1) in other complex plane I proposed via region in the middleby moving this plane as a crossection of h(x) in 3 dimensions.
May be this would not work without involving superroots as well. Would that not allow to understand the phenomena of bifurcation as such better? And phase transitions?
Ivars
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01/17/2008, 11:34 PM
(This post was last modified: 01/18/2008, 09:25 AM by andydude.)
@Ivars: First of all those are not my definitions, they are Knoebel's. His paper, called "Exponentials Reiterated" is an amazing exposition of the infinitely iterated exponential. It covers almost all aspects of it, and mentions many aspects of tetration as well.
@Gottfried: Your graph confuses me, why is the 1tick for b=... and the 1tick for f=... not aligned? and why do you have X's for points, do you have any options for boxes? diamonds? circles? no dots? That would make the graph much nicer to look at. But anyway, I got the general idea of the graph, I just think it could use some work. I'm not sure what the benefit of such a changeofvariables would be, unless it would simplify the series for b^b^x.
@Gottfried: As for your series expansion, I really, really like your presentation of the series for (b^x1), it shows patterns in the coefficients, where I never saw patterns before Good job.
About the role of x1 and x2, the role is as follows. The whole point of is to parameterize the two branches of oddH, and evenH as described by Knoebel. Individually, both x0 and x1 satisfy this equation, but when viewed together, they satisfy:
 and
 ,
as this represents the fact that exponentials oscilate between two values when . The values x0 and x1 are the two values that the iterated exponential oscilates between. That is their role.
Anyways, I found a way to construct a power series for this function that expresses the coefficients. Lets call it since its a function that gives b as an output, and it takes x as an input, such that for all x in [0, 1].
Starting with the definition of this function, and differentiating:
And we now have a differential equation, or in other words, a closed form for the derivative. Many cases in which we have a closed form for the derivative, we can find a series expansion around (almost) any point. So for example, we know the closed form of the derivative of the Lambert W function, which is one way its coefficients can be found (this is the hard way). Anyways, to get back to my point. Using the secondtolast equation (which allows either side to be 0, the last equation prevents the denominator from being 0), We can start with an "unknown" series expansion for and solve for its coefficients to satisfy this differential equation. Doing this at , we get (factoring out to simplify things):
With this series expansion, we can then test to see if we have a "correct" function, by testing its derivatives, rather than continually guessing. Maybe all we have to do to find the right function is look it up on OEIS, maybe its already there...
Andrew Robbins
PS. I love finding new series
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01/17/2008, 11:56 PM
(This post was last modified: 01/18/2008, 12:28 AM by GFR.)
Hi IVARS!
A small preliminary comment (during the weekend I shall supply more) to your interesting "vision":
Ivars Wrote:.............
When I was speaking in previous post about the missing dimension, I was thinking about rotating plane (or rotation in plane) perpendicular the graphs of GFR and parallel to y axis.
Then I have not even a guess, but just a vision of these 3 definitions of h(x) forming a double helix (odd/even) with a thread (x^1/x) in the midlle in region e^e<x<e^(1/e). The projection on GFR graph is just a single line, but projection on that perpendicular plane is 3 points at each x.
When bifurcation happens x< e^e, we get 2 opening spirals rotating left (odd?) and right (even?) in that plane and I am not sure what happens with x^(1/x) central thread thereit may tend to become a rotating connection between these spirals of that circle.
Ivars
Actually, three dimensions are already there:
 y is the dimensionn [h(b)] where the tetration "heights" are measured ("vertical" in the traditional, so to say, representation);
 b is the dimension allowing us to define the "bases" (horizontal, in the same convention);
 x is the dimension perpendicular to y and b (a cross section perpendicular of the yb plane, perpendicular to the worksheet, in the same conventional representation).
On the yx cross sction planes (allow me to say so) the plots of the tetrational functions are, indeed (in my opinion, of course) oscillating lines plotted in a 2dimensional plane, limited by maxima of y defined by the even integer values of x (0,2,4,6,...) and minima of y defined by their odd (1,3,5,...) integer values. These plots show, in the available approximated simulations, a decreasing, but not vanishing, oscillating behaviour in the region 0 < b < e^(e). In fact, their asymptotical shapes, for x > + oo, seem to be describable as "permanent oscillations" around an asypmptotic axis, defined by:
y = plog(log/b x)(log/b x).
Therefore, I agree with you (IVARS) thet we need an extra dimension, but I think that it should be represented by the imaginary axis itself. In this case, it will be the "fourth dimension" of the diagram.
I also agree with your "spirallylike vision", but I guess that the (deformed) spirals (for any fixed b) should be ... unique. What we "see" in the yx plane is (I guess) the "real projection" of the spiral, at any fixed b. On the contrary, what we see in the "traditional" yb diagram (the upper and lower h's) is (always in my guessing dream) the projection of the spiral on the yb plane, for x > *oo. The upper and lower y, in the "Forbidden Zone for Integer Superexponents" should show the upper and lower maximum elongations around the center of the spiral, for any constamt b and for x > +oo.
Sorry for the ... approximated description and thank you for your kind attention.
GFR
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Ivars, I think I know what you might be talking about, or at least, I have something that may be described in similar words. This is a 3D plot of the function and a 3D plot of "zero". The interesting thing here, is that the intersection of the two is what we are trying to parameterize. You can see the intersection of the two in this plot:
What is also interesting about this plot is that is shows (in some sense), why the outer "legs" should be attracting periodic points, and the middle is a repeling fixed point. It shows this because the slope/derivative is in the opposite direction fot the middle than it is for the other two branches/legs.
Andrew Robbins
