Zeros of the Zeta FunctionDate: 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 http://mathforum.org/dr.math/ |
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