Why are the trivial zeros negative even integers?

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Discussion Overview

The discussion centers around the nature of trivial zeros of the Riemann zeta function, specifically why they are negative even integers. Participants explore the implications of substituting these values into the zeta function and the conditions under which the function is defined.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions how substituting a trivial zero, such as -2, into the zeta function results in a series that equals zero.
  • Another participant points out that the formula provided is only valid for complex s with real parts greater than 1, suggesting a limitation in the initial approach.
  • A subsequent reply raises the question of whether the Riemann Hypothesis relies on the functional equation of the zeta function.
  • Additional participants recommend exploring analytic continuation and the role of Bernoulli numbers and the function \xi in this context.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as multiple competing views and questions about the implications of the zeta function's properties remain unresolved.

Contextual Notes

There are limitations regarding the assumptions made about the validity of the zeta function at certain values and the dependence on definitions related to analytic continuation.

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[tex]\varsigma(s) = \sum^{\infty}_{n=1}n^{-s}[/tex]

If you substitute a trivial zero, let's say -2. Wouldn't it be

[tex]\varsigma(s) = \sum^{\infty}_{n=1} = 1^2 + 2^2 + 3^2 + 4^2 + . . .[/tex]

How would this series be equals to zero?

Thanks
 
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So that means that the Riemann Hypothesis is based on the functional equation instead?
 
& the bernoulli numbers, and the function [itex]\xi[/itex]
 

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