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Zeros of the Zeta Function

Date: 08/09/99 at 15:36:04
From: Xi Wang
Subject: Zeros of the Zeta Function

Hello Dr. Math,

I recently read in a book that the only real zeros to the zeta 
function are -2, -4, -6, -8, etc. However, since:

     zeta(x) = 1 + 1/2^x + 1/3^x + 1/4^x + ...

How can any of the zeros work? It doesn't matter which one of them you 
plug in, the answer is obviously not zero.

Thanks in advance for your help, 
Xi Wang

Date: 08/09/99 at 16:10:33
From: Doctor Tom
Subject: Re: Zeros of the Zeta Function

To understand what's going on, you have to understand enough of 
complex analysis to know what "analytic continuation" means.

Unlike functions of a real variable, differentiable functions of a 
complex variable are not flexible - once you know the values of the 
function on any infinite set of points, that function can be extended 
in a unique way to a certain function.

The expression for zeta(x) that you gave above converges for positive 
x, so it does define an infinite set of values of the function. This 
function defined on the positive reals can be extended in a unique way 
to most of the complex plane, and that extended function has a set of 
zeroes outside the range of convergence of the series you gave above.

Here's an example that's very easy to understand. You've probably 
learned that:

     f(x) = 1 + x + x^2 + x^3 + ... = 1/(1-x)

But if x is greater than 1 (say, for example, x = 2), the series above 
doesn't converge, it gives:

     f(2) = 1 + 2 + 4 + 8 + ...

But the expression 1/(1-x) makes perfect sense for x = 2. The function 
1/(1-x) is the analytic continuation of the series:

     1 + x + x^2 + x^3 + ...

- Doctor Tom, The Math Forum
Associated Topics:
College Analysis

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