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08/17/2007, 09:13 PM
(This post was last modified: 08/17/2007, 09:14 PM by bo198214.)
What really would interest me is a numerical comparison (graphing) of your solution, of Daniel's solution (development at the fixed point of \( b^x=x \) for \( b<\eta \)) and of Jay's solution (but perhaps it nees still some explanation).
But at least for the first two we have all numerical methods at hand.
It simply needs a volunteer (and in the moment my time becomes rare.) ...
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If by my solution you mean \( {\ }^{y}(e^{1/e}) \) through parabolic iteration of \( e^x1 \) and by Daniel's you mean \( {\ }^{y}(b^{1/b}) \) through hyperbolic iteration of \( b^x1 \), then yes, I'd be happy to provide some highquality graphics and numerical tables. I still need to reread Jay's solution, since it seems dependent on the superlogarithmic constant which I don't understand yet. Also you brought up an interesting point that may prevent comparison of all methods, since some methods work over nonoverlapping intervals.
Tetration can be divided into several intervals in terms of what values of x in \( {}^{y}x \) are valid for that definition: the parabolic point \( e^{1/e} \), the hyperbolic interval \( (e^{e}, e^{1/e}) \), and I suppose would you could call the elliptic? interval \( (e^{1/e}, \infty) \), which is what my definition of the superlogarithm is valid for. These are all closely tied to the interval of convergence of the infinite hyperpower \( {}^{\infty}x \) which was found almost 230 years ago. My how time has passed...
Andrew Robbins
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I've gotten SAGE working and I'm currently writing a tetration package for an arbitrary base greater than eta (though it's most stable for bases in a smaller range, probably b<20 or so, depending on precision). I'm using Andrew's SAGE code above to generate the FMA or whatever you would call it, but the rest of the code is based on my cheta function and change of base formula.
I'm about halfway done.
~ Jay Daniel Fox
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08/17/2007, 11:45 PM
(This post was last modified: 08/17/2007, 11:48 PM by bo198214.)
andydude Wrote:If by my solution you mean \( {\ }^{y}(e^{1/e}) \) through parabolic iteration of \( e^x1 \) No, I meant the piecewise infinite differentiable definition of slog.
Isnt it defined for arbirtrary bases? (I think you wrote for base greater 1.)
So if you have the slog you have also the "sexp".
And I would compare this with what comes out from Daniel's solution for the hyperbolic case (there is then no problem with the convergence radius of the series).
Quote:and by Daniel's you mean \( {\ }^{y}(b^{1/b}) \) through hyperbolic iteration of \( b^x1 \)
Rather \( b^xx_0 \), i.e. to develop \( b^x \) at the (lower) fixed point \( b^{x_0}=x_0 \) which is given (if I remember correctly) by
\( x_0=\frac{W(\log(b))}{\log(b)} \) and then take hyperbolic iterations \( {}^xb=\exp^{\circ x}_b(1.0) \).
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bo198214 Wrote:andydude Wrote:If by my solution you mean \( {\ }^{y}(e^{1/e}) \) through parabolic iteration of \( e^x1 \) No, I meant the piecewise infinite differentiable definition of slog.
Isnt it defined for arbirtrary bases? (I think you wrote for base greater 1.) No, the slog function is only defined for bases greater than 1. For bases between 0 and 1, the tetration function is not onetoone, so its inverse is not a function, because there are multiple values. It's much like saying that y=x^2 does not have an inverse function. y=sqrt(x) only covers part of the domain of the original function.
Also, there's a question about whether you consider the slog function to only apply to the inverse of the function of iterated exponentials/logarithms from 1. For b=2, for example, the domain of slog is negative infinity to 2. However, you can perform iterated exponentials/logarithms from any real number as a starting point, so you could also include the graph for x>4 and the corridor between 2 and 4.
~ Jay Daniel Fox
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08/18/2007, 08:19 AM
(This post was last modified: 08/18/2007, 08:20 AM by bo198214.)
jaydfox Wrote:No, the slog function is only defined for bases greater than 1. For bases between 0 and 1, the tetration function is not onetoone, so its inverse is not a function, because there are multiple values. It's much like saying that y=x^2 does not have an inverse function. y=sqrt(x) only covers part of the domain of the original function. Surely, by "arbitrary" I meant: also for \( b<\eta \).
I was referring to
andydude Wrote:and I suppose would you could call the elliptic? interval \( (e^{1/e}, \infty) \), which is what my definition of the superlogarithm is valid for however in his pdf he defined it for bases \( >1 \).
(Quite ridiculous how misunderstandings reach its maximum in the communication between Jay and me ...)
Quote:Also, there's a question about whether you consider the slog function to only apply to the inverse of the function of iterated exponentials/logarithms from 1. For b=2, for example, the domain of slog is negative infinity to 2.
Dont understand this, \( t\mapsto \exp_2^{\circ t}(1.0) \) maps \( (2,\infty) \) to \( (\infty,\infty) \), so the slog is defined on \( (\infty,\infty) \)?
Quote: However, you can perform iterated exponentials/logarithms from any real number as a starting point, so you could also include the graph for x>4 and the corridor between 2 and 4.
If you use the fixed point method however for base \( <\eta \) you have to specify which fixed point you use in case you start with \( x_0 \) between these fixed points.
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I just wanted to clarify that my superlogarithm solution only works for \( x > e^{1/e} \) even though my pdf says \( x > 1 \). I think that the reason for this is that tetration for bases between 1 and \( e^{1/e} \) has an upper limit (due to the convergence of \( {}^{\infty}x \)), and thus the superlogarithm has a limited domain over which it is defined. This upper limit of tetration also means that there will be a singularity on the boundary of this domain, which might explain why I have had such poor results as \( x \rightarrow e^{1/e} \) with experiments with my method.
When I first discovered the superlogarithm solution, the only singularities in the matrix equation (rather than the series expansion) seemed to be less than 1, this is what led me to the requirement \( x > 1 \), which I later learned to be not strict enough. Sorry. Hope this helps.
For an abstraction of my superlogarithm method, see http://math.eretrandre.org/tetrationforu...181#pid181
Andrew Robbins
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08/18/2007, 11:59 AM
(This post was last modified: 08/18/2007, 11:59 AM by bo198214.)
andydude Wrote:I just wanted to clarify that my superlogarithm solution only works for \( x > e^{1/e} \) even though my pdf says \( x > 1 \). I think that the reason for this is that tetration for bases between 1 and \( e^{1/e} \) has an upper limit (due to the convergence of \( {}^{\infty}x \)), and thus the superlogarithm has a limited domain over which it is defined. Are you sure, what goes wrong for bases \( b<\eta \)?
I mean it is clear that the domain of the superlog is bounded above by the lower fixed point of \( b^x=x \), which I will call \( \beta(b) \) and which is \( \beta(b)=W(\log(b))/(\log(b)) \).
So \( \text{slog}_b(x) \) is allowed only for \( x<\beta(b) \) because \( \lim_{n\to\infty}{}^nb=\beta(b) \).
What goes wrong for those arguments with your superlog?
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08/18/2007, 03:49 PM
(This post was last modified: 08/18/2007, 03:49 PM by jaydfox.)
bo198214 Wrote:Quote:Also, there's a question about whether you consider the slog function to only apply to the inverse of the function of iterated exponentials/logarithms from 1. For b=2, for example, the domain of slog is negative infinity to 2.
Dont understand this, \( t\mapsto \exp_2^{\circ t}(1.0) \) maps \( (2,\infty) \) to \( (\infty,\infty) \), so the slog is defined on \( (\infty,\infty) \)? Argh, sorry, I meant \( \sqrt{2} \). \( t\mapsto \exp_{\sqrt{2}}^{\circ t}(1.0) \) maps \( (2,\infty) \) to \( (\infty,2) \), so the slog is defined on \( (\infty,2) \)
~ Jay Daniel Fox
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08/19/2007, 12:50 AM
(This post was last modified: 08/19/2007, 01:14 PM by bo198214.)
The iterated exp(x)1 seems to work nicely. It takes a minute or two to generate all the helper functions for a 35term power series, but then I'm able to calculate 505 points for the interval [248/128, 256/128], step size 1/128, in just over 59 seconds. This was sufficient to produce the following graph of \( {}^{x} e \) in 1 minute:
By the way, I'm using a radius of 0.005, 35 terms. I could take the number of terms out to 50 pretty easily, and still get increased precision with each new term. As it is, precision seems to be about 6570 decimal digits.
By the way, sorry for the extremely generic appearance of the graph. This was my first attempt at plotting in SAGE, so I haven't tried changing the tick spacings on the axes, adding gridlines and labels, etc. I also plan to add graphs of the derivatives (approximated with secants).
~ Jay Daniel Fox
