02/16/2023, 01:43 PM
All right, so \(C\) is not a constant. It is bounded by a constant, and that is all we're going to get. We can find a bound:
\[
|C(x)| \le C_0\\
\]
And that's about the best we're going to get. We can speed up some shit, but don't cross your fingers. We start with:
\[
C(x) = \frac{A(x)}{\log(x)}\\
\]
And the best we have is a constant bound as \(x\) grows. The \(\limsup_{x\to\infty}C(x)\) converges, and so does \(\liminf_{x\to \infty} C(x)\).
Here is \(C(x)\) from \(x=100\) up to \(x=1000\):
This means we have a strict display between each value; but not as good as we'd like.
I still believe we have meromorphy for \(\Re(s) > -1\); but it's gonna be harder than I thought......
\[
|C(x)| \le C_0\\
\]
And that's about the best we're going to get. We can speed up some shit, but don't cross your fingers. We start with:
\[
C(x) = \frac{A(x)}{\log(x)}\\
\]
And the best we have is a constant bound as \(x\) grows. The \(\limsup_{x\to\infty}C(x)\) converges, and so does \(\liminf_{x\to \infty} C(x)\).
Here is \(C(x)\) from \(x=100\) up to \(x=1000\):
This means we have a strict display between each value; but not as good as we'd like.
I still believe we have meromorphy for \(\Re(s) > -1\); but it's gonna be harder than I thought......
