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### Zeros of the Zeta Function

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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.

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
http://mathforum.org/dr.math/
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