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Ben Weiss
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Let's look at $\zeta(s)$ for some large $\sigma$ (the real part of $s$). We can bound the function by $\int_1^\infty \frac{dx}{x^\sigma} + 1$, which is $\frac{\sigma}{\sigma-1}.$ So all these values in any $A_m$ ($m > 1$) can be bounded by a half-plane. This definitely doesn't pin down where they are, but does give a nice bound on where they are not.

Let's look at $\zeta(s)$ for some large $\sigma$ (the real part of $s$). We can bound the function by $\int_1^\infty \frac{dx}{x^\sigma} + 1$, which is $\frac{\sigma}{\sigma-1}.$ So all these values in any $A_m$ can be bounded by a half-plane.

Let's look at $\zeta(s)$ for some large $\sigma$ (the real part of $s$). We can bound the function by $\int_1^\infty \frac{dx}{x^\sigma} + 1$, which is $\frac{\sigma}{\sigma-1}.$ So all these values in any $A_m$ ($m > 1$) can be bounded by a half-plane. This definitely doesn't pin down where they are, but does give a nice bound on where they are not.

Source Link
Ben Weiss
  • 1.6k
  • 1
  • 17
  • 34

Let's look at $\zeta(s)$ for some large $\sigma$ (the real part of $s$). We can bound the function by $\int_1^\infty \frac{dx}{x^\sigma} + 1$, which is $\frac{\sigma}{\sigma-1}.$ So all these values in any $A_m$ can be bounded by a half-plane.