Tommy's Gaussian method.

[tommy1729]

You really don't have to go too much into depth in choosing your branch of logarithm. The principal branch is good enough if you add a \( \rho \) function.

If you write,

\(
\text{Tom}(s) = \Omega_{j=1}^\infty e^{A(s-j)z}\,\bullet z\\
\)

And,

\(
\text{Tom}_A(s) = A(s)\text{Tom}(s) = \log \text{Tom}(s+1)\\
\)

And construct a sequence \( \rho_n(s) \) where,

\(
\rho^{n+1}(s) = \log(1+\frac{\rho^n(s+1)}{\text{Tom}_A(s+1)}) + \log A(s+1)\\
\)

where \( \frac{\rho^n(s+1)}{\text{Tom}_A(s+1)} \) is very small for large \( \Re s \). So you are effectively calculating a \( \log(1+\Delta) \) for \( \Delta \) small, as opposed to a \( \log(X) \) where \( X \) is large. The branching won't be an issue at all.

Where then,

\(
\text{tet}_{\text{Tom}}(s + x_0) = \text{Tom}_A(s) + \rho(s)\\
\)

---

Remember Tommy that,

\(
\lim_{\Re(s) \to \infty} \text{Tom}_A(s) \to \infty\\
\)

Even though there are dips to zero this is still the asymptotic behaviour.

I take those dips very seriously.

If s is close to the real line that might work.

But for Im(s) substantially high things are more complicated I think.

For real s we can take the principal branch , in fact we must.
And then ofcourse by analytic continuation we do not need to bother with the other branches.

This indeed suggests we might not need not to look at other branches but for a proof I was pessimistic.

Let me put it like this ;

suppose we have tetration tet(s)

ln (ln ( tet(s+2) )) = tet(s) ONLY works with the correct branches.

example tet(s) = 1 + 400 pi i for some s.

regards

tommy1729

Absolutely Tommy!

But the branches are chosen by,

\(
\log \beta (s+1) = \beta(s) + \text{err}(s)\\
\)

So which ever branch satisfies this equation is the correct branch.

So, when you make a sequence of error functions \( \lim_{n\to\infty} \tau^n(s) \); given by,

\(
\tau^{n+1}(s) = \log(1+\frac{\tau^n(s+1)}{\beta(s+1)}) + \text{err}(s)\\
\)

you are choosing these logarithms already; because they're the only ones which satisfy the equation,

\(
\lim_{\Re(s) \to \infty} \tau^n(s) = 0\\
\)


If you were to right it the naive way,

\(
\lim_{n\to\infty} \log^{\circ n} \beta(s+n)\\
\)

Then, yes, choosing the branch is necessary. But if you write it the former way; there's no such problem.


As to the dips to zero; they produce the most trouble when trying to numerically evaluate. But in the math; they don't pose that much of a problem because we know that \( \lim_{\Re(s) \to \infty} \beta(s) = \infty \).

I want/wanted to prove boundedness without any assumptions such as analytic , continu , smooth etc.

IN A NUMERICAL WAY !!

If we start from the real line and assume analytic then the logs do not need branches.

the definion log( analytic f(z) ) = integral f ' (z)/f(z) for any f(z) (not = 0) will do.

or the taylor series does the work too.

the taylor of exp(exp(ln(ln(x)))) or ln(ln(exp(exp(x)))) is simply x without worrying about taking branches.

but clearly ln(exp(x)) does not equal x if we consider taking the principal value of ln ... since exp is periodic.
( a function of a periodic function is periodic normally ) 

so the taylor " automatically " picks the correct branches.

But we cannot assume analytic without proving it first !! A critic would call that circular logic !

Also I do not want to use statistical or intuitive arguments , because that is not a formal proof.

So I considered a more or less worst-case scenario from the perspective of numerical math.
( ingnoring decimal precision ofcourse )

That worst case numerical method shows a clear boundary.

that boundary can then be used to construct stronger proofs such as convergeance.
( fixed point methods or similar methods as the bounds )

It is well known that if a sequence of analytic functions f1 f2 f3 ... converges , then that limit is also analytic !!

And thus we would have a formal proof that the gaussian method is analytic near the positive real line.

I know you understand the difference between analytic and numerically very well.
But just to be clear ( to all ) I felt the need to reply this.

I know you are already convinced it is analytic but we need to get formal.



regards

tommy1729