Is the Riemann Zeta Function of 0 Infinity or -1/2?

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Discussion Overview

The discussion revolves around the value of the Riemann zeta function at zero, specifically whether it is -1/2 or infinity. Participants explore the definitions and properties of the Riemann zeta function, particularly in relation to its analytic continuation and behavior for values less than or equal to one.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that the Riemann zeta function at zero is -1/2, while questioning if it could also be considered infinity based on the series representation.
  • Others clarify that the Riemann zeta function is defined for s > 1, and for s ≤ 1, it is defined through analytic continuation.
  • A participant mentions that the sum has infinitely many continuous extensions to the real line but only one analytic extension to the complex plane.
  • Another participant specifies that the extension is meromorphic due to a pole at s = 1.
  • There is a request for a formula for s < 1, indicating interest in further representations of the zeta function.
  • Participants reference external sources for additional information on the Riemann zeta function.

Areas of Agreement / Disagreement

Participants express differing views on the value of the Riemann zeta function at zero, with some supporting -1/2 and others suggesting infinity. The discussion includes clarifications about definitions and extensions, but no consensus is reached on the value itself.

Contextual Notes

Limitations include the dependence on the definitions of the Riemann zeta function and the nature of its analytic continuation. The discussion does not resolve the mathematical implications of these definitions.

dimension10
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The riemann zeta function of 0 is supposed to be -1/2. But isn't it infinity?

1/1^0 +1/2^0+1/3^0+1/4^0+...
1/1+1/1+1/1+1/1+...
1+1+1+1+...
infinity

But many sources claim that it is -1/2
 
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dimension10 said:
The riemann zeta function of 0 is supposed to be -1/2. But isn't it infinity?

1/1^0 +1/2^0+1/3^0+1/4^0+...
1/1+1/1+1/1+1/1+...
1+1+1+1+...
infinity

But many sources claim that it is -1/2

Yes, you are correct, but the definition af the Riemann-zeta function as

[tex]\zeta (s)=\sum_{n=1}^{+\infty}{\frac{1}{n^s}}[/tex]

is only defined as s>1 (or in the complex case, as Re(s)>1). The Riemann-zeta function for [itex]s\leq 1[/itex] is defined as the "analytic continuation" of the above function. That is: the unique function that looks most like [itex]\zeta(s),s>1[/itex].
 
micromass said:
is only defined as s>1 (or in the complex case, as Re(s)>1). The Riemann-zeta function for [itex]s\leq 1[/itex] is defined as the "analytic continuation" of the above function. That is: the unique function that looks most like [itex]\zeta(s),s>1[/itex].

The sum has infinitely many continuous extensions to the real line, but only one analytical extension to the complex plane.
 
disregardthat said:
The sum has infinitely many continuous extensions to the real line, but only one analytical extension to the complex plane.

Well, a meromorphic extension really, since there is a pole in 1.
 
micromass said:
Well, a meromorphic extension really, since there is a pole in 1.

You are right about that :redface:
 
micromass said:
Yes, you are correct, but the definition af the Riemann-zeta function as

[tex]\zeta (s)=\sum_{n=1}^{+\infty}{\frac{1}{n^s}}[/tex]

is only defined as s>1 (or in the complex case, as Re(s)>1). The Riemann-zeta function for [itex]s\leq 1[/itex] is defined as the "analytic continuation" of the above function. That is: the unique function that looks most like [itex]\zeta(s),s>1[/itex].

Thanks, but is there a formula for s<1?
 

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