[NT] more zeta stuff for the fans

[JmsNxn]

It can still make sense if you consider the zero's for every truncated sum.

So zero's outside analytic continuation can make sense.

And a summability method for that integral might help.

I remember as a kid , I somewhat rediscovered the Riemann Hypothesis.

I was looking for zero's of the zeta function , being aware it had a pole.

At that time I did not know complex analysis or even got officially introduced to complex numbers.

I also missed the whole number theory thing.

So i just wanted to find zero's of a function that looked nice, without knowing analytic continuation , calculus , formal complex numbers or number theory.

And I did.

With the truncated sum for the zeta and some more tricks I rediscovered that the zero's were on the critical line.

The intuition I had was that every function apart from of the form exp(..) and gamma(..) had a zero.
This probably followed from rediscovering of the complex numbers.

Therefore I said all those functions had zero's.

I considered a function were the zero was hard to find and the function easy to define.
so ended with zeta.
probably inspired by exp because they did not have a 0.

So I tested my ideas and found critical zero's with 123lotus or whatever that predeccessor of excel was called.

I did not tell anyone because I thought I had something new and special lol.


regards

tommy1729

Okay, Tommy

That's all well and good. But, sorry. That doesn't explain the situation at hand, not at all.

Guessing:

\[
\sum_{n=1}^M p(n)n^{-s} = A_M(s)\\
\]

And saying zeroes of \(A_M(s)\) are within \(1 \le \Re(s) \le 2\) has no statement on the zeroes of \(\lim_{M\to\infty} A_M(s) = A(s)\). Especially when your formula of \(A(s)\) only exists for \(\Re(s) > 2\). The zeroes as you observe \(M\to \infty\)--will obviously be in \(1 \le \Re(s) \le 2\); because \(A(s)\) is non-zero for \(\Re(s) > 2\)....

I'm sorry, Tommy; this doesn't make sense. None of it does.