Self tetraroot constructed via Newton series interpolation

[mike3]

Hi.

I thought of a novel approach to aid in the "divination" of what the tetration function for real heights is: the self-tetraroot function. This function is defined so that \( ^x \mathrm{selftetroot(x)} = x \). It is analogous to the self-root function \( x^{1/x} \), only with tetration instead of exponentiation. It gives the base of the tetrational with a given fixed point.

It can be evaluated at the integers using numerical root-finding methods, and when the points are plotted, it reveals a scatter that slowly, but regularly (i.e. no noticeable "bumpiness") decays to a fixed value about 1.444782 as \( x \rightarrow \infty \).

This slow regular decay made me wonder if it would be a good candidate for interpolation via the Newton series. Newton series, when applied to tetrationals with bases in \( 1 < b \le e^{1/e} \), yields the regular iteration.

The Newton series for a function \( f(x) \) at a point \( a \) is

\( f(x) = \sum_{n=0}^{\infty} \frac{[\Delta^n f](a)}{n!} (x)_n \)

where \( (x)_n \) is the falling factorial and \( \Delta \) is the unit forward difference operator.

Doing this for \( a = 1 \) gives the following graph. Convergence is dog slow -- I needed 300 or so terms and over 100 decimals of accuracy just to get error that I'm confident is less than \( 10^{-3} \). This is very bad. Is there some way to accelerate the convergence of Newton series?

   

As you can see, it sort of looks like the graph of the self-root. I superimposed both graphs:

   

The maximum is at about \( x = 3.089 \) or so (maybe \( 3.0885 \) if you want to push it as this approximation may be good to 5 places -- may -- it's so dog-slow, where it reaches a value of \( 1.635 \) (maybe \( 1.6353 \)). I presume this is the base where "pentation" goes from convergent to explosive growth, and so is to pentation what \( e^{1/e} \) is to tetration. Flipping the self-tetroot along \( y = x \) and taking the lower branch would then yield a graph of \( x \uparrow \uparrow \uparrow \infty \).

For \( x = 1.5 \), I get a tetra-self-root of \( 1.389 \) (\( 1.390 \) rounded from \( 1.3897 \) whose last digit may be right or near). Plugging this into the regular iteration to try and compute \( ^{1.5} 1.389 \) suggests whatever "fractional iteration" this is creating may agree with it, but this is a dodgy bet with just 4 decimals. Yet if it does agree, then perhaps, since this function shows no "bumps" when the base crosses \( e^{1/e} \), maybe the regular iteration really doesn't have a natural boundary after all. But again, 4 decimals, 5 at best, geez... far too little to even bet seriously... If we could compute more, we might be able to get a better idea if it agrees with regular, and if it also agrees with the Cauchy integral (for the bases from \( e^{1/e} \) to 1.635...), hence helping to "divine" if these are really "good" methods to use and if they are actually analytic continuations of each other (meaning that the STB is not a natural boundary of the regular iteration.).

What do you think of this function? Especially the similarity between the shape of its graph and that of the selfroot. One could almost imagine a continuous spectrum of similar functions in between them -- "fractional-rank hyperoperations", anyone?

[bo198214]

What do you think of this function? Especially the similarity between the shape of its graph and that of the selfroot. One could almost imagine a continuous spectrum of similar functions in between them -- "fractional-rank hyperoperations", anyone?

That indeed looks interesting (Andrew posted this self tetra root already somewhere on the forum and I proved somewhere that the limit is indeed e^(1/e)).
However I would expect to have a functional equation for the tetra self root.
As well as to know how you derive tetration from the tetra self root.

[mike3]

What do you think of this function? Especially the similarity between the shape of its graph and that of the selfroot. One could almost imagine a continuous spectrum of similar functions in between them -- "fractional-rank hyperoperations", anyone?

That indeed looks interesting (Andrew posted this self tetra root already somewhere on the forum and I proved somewhere that the limit is indeed e^(1/e)).
However I would expect to have a functional equation for the tetra self root.
As well as to know how you derive tetration from the tetra self root.

Yes, a functional equation would be good, but I'm not sure of one. As for deriving tetration, you can't really, except that evaluating this at some fractional number will give you the value of a specific base \( b \in [1, K) \) (where \( K = 1.6353... \)) -- i.e. the value of the function, tetrated to a certain tower, i.e. the point of evaluation, and the value of such tetration, also the point of evaluation (this means the point of evaluation is a fixed point of the tetrational of the base given by the result of evaluation). For bases \( b > L \), where \( L \) is the limit of the function at \( +\infty \) (which you are saying is \( e^{1/e} \) -- if that's so then it must converge to it at a hideously slow rate: do you know what the big-\( O \) of \( |\mathrm{selftetroot}(n) - e^{1/e}| \) is as \( n \rightarrow \infty \)?), this procedure yields fractional tetration values at two points.

If the fractional tetration values this yields agree with regular iteration for \( b \in (1, e^{1/e}] \), that may suggest that regular iteration does not have a natural boundary and can be analytically continued to the wider complex plane. If not, then the regular may still have such a boundary, and I'd be curious what this "superior-to-regular-iteration" solution for \( b \in (1, e^{1/e}] \) is that it is "pointing at" with the values for the fractional towers that are fixed points.