Alternate solution of tetration for "convergent" bases discovered

[mike3]

... not always easy to prove things like analytic , size of radius , existance , uniqueness etc but at least explain your method decently.

i might not be the smartest person on the planet but a better explaination is sometimes really really required.

a typical thing is something in the line of :

" we start with an intial guess depending on base q and then ... "
" ... we take an integral and do as in ' link ' ( containing similar intial guesses ) ..." " ... and repeat the process k times where k is depending on q ... " " and then we estimate the contour integral of ... " " ... and repeat till it converges .. "

First off, the method is not well-proven. It's more conjectural at this point. But anyway, this is the best explanation I can give.

Let \( g(z) \) be a periodic, analytic function with period \( P \). Then there should exist a Fourier expansion (perhaps needing a shift or translation depending on location of singularities)

\( g(z) = \sum_{n=-\infty}^{\infty} a_n e^{\frac{2 \pi}{P} i nz} \).

Now we create the "continuum sum":

\( \sum_{n=0}^{z-1} g(n) = \sum_{n=-\infty}^{\infty} \frac{a_n}{e^{\frac{2 \pi}{P} i n} - 1} \left(e^{\frac{2 \pi}{P} i nz} - 1\right) \)

where the term at \( n = 0 \) is interpreted as \( a_0 z \).

So, if \( f(z) \) is some non-periodic complex function, we may be able to define a continuum sum for this function as follows. Let \( f_k(z) \) be a sequence of periodic, analytic functions that converges pointwise to \( f(z) \), i.e. \( \lim_{k \rightarrow \infty} f_k(z) = f(z) \). Then we define

\( \sum_{n=0}^{z-1} f(n) = \lim_{k \rightarrow \infty} \sum_{n=0}^{z-1} f_k(n) \),

provided that this limit exists. The hypothesis (not yet proven!) is that this limit is the same regardless of the sequence of periodic approximations.

Now that we have a way to define continuum sums, we start with an initial guess \( F_0(z) \) to our tetrational function \( \mathrm{tet}_b(z) \), and then apply the iteration formula

\( F_{k+1}(z) = F'(0) \int_{-1}^{z} \log(b)^w \exp_b\left(\sum_{n=0}^{w-1} F_k(n)\right) dw \)

and take the convergent as \( k \rightarrow \infty \). If we do not know \( F'(0) \), we replace the multiplication by \( F'(0) \) with multiplication by a normalization constant so that \( F(0) = 1 \). That is,

\( U(z) = \int_{-1}^{z} \log(b)^w \exp_b\left(\sum_{n=0}^{w-1} F_k(n)\right) dw \)
\( F_{k+1}(z) = \frac{U(z)}{U(0)} \).

Now the adaptation of this theory to construct a numerical algorithm is a little more complicated. I could post that, if you'd like it, in the Computation forum. But the above is the basic outline of the method.

If you have any difficulty with the above, please indicate which parts confuse you, and I'll try to explain it better.

the continuum sum is clear to me.

and trying the continuum sum on some four estimate makes sense too.

but nowhere do i see what fourier series is used as approximation of base 2.33 + 1.28i

i might have missed this or that , but im sure a newbie cant follow this way and i guess we dont what that do we ?

clarity is at the heart of mathematical intentions ...

So you want the numerical algorithm, then?

quote :

" gradually stepping back the base, using the already-calculated solution as an initial guess "

thats what i meant.

stepping back the base ... is that math terminology ? is it clear to anyone ?

It means you take the already-calculated solution as an initial guess for solving a base a little bit closer to the target one, then you take that as an initial guess for a base a little closer still, and so on.

you say it is different then regular , and later on you mention its the same on the real line.

Not sure what you mean. I don't seem to have mentioned it is the same as the regular at the real line -- it isn't.

EDITED: I left out the fact the method needs the derivative at 0, just put that in.