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