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Wikipedia states that

The Riemann zeta function $\zeta(s)$ is defined for all complex numbers $s \neq 1$. It has zeros at the negative even integers (i.e. at $s = −2, −4, −6, ...)$. These are called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, and states that: The real part of any non-trivial zero of the Riemann zeta function is $\frac{1}{2}$.

What does it mean to say that $\zeta(s)$ has a $\text{trivial}$ zero and a $\text{non-trivial}$ zero. I know that $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$ what wikipedia claims it that $\zeta(-2) = \sum_{n=1}^{\infty} n^{2} = 0$ which looks absurd.

My question is can somebody show me how to calculate a zero for the $\zeta$ function.

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    I think it may help http://mathworld.wolfram.com/Riemann-SiegelFormula.html2012-04-20
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    The series is not applicable for $\Re(s)\leq 1$; one uses a different formula (an *analytic continuation*, if you will) of the $\zeta$ function (so yes, it does look absurd until you consider the extension of the function to the rest of the complex plane).2012-04-20
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    As Ginger mentions, one uses the Riemann-Siegel formula *numerically* to compute the nontrivial zeroes (there are no known closed forms for them).2012-04-20

3 Answers 3

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You are going to need a bit of knowledge about complex analysis before you can really follow the answer, but if you start with a function defined as a series, it is frequently possible to extend that function to a much larger part of the complex plane.

For example, if you define $f(x)=1+x+x^2+x^3+...$ then $f$ can be extended to $\mathbb C\setminus \{1\}$ as $g(x)=\frac{1}{1-x}$. Clearly, it is "absurd" to say that $f(2)=-1$, but $g(2)=-1$ makes sense.

The Riemann zeta function is initially defined as a series, but it can be "analytically extended" to $\mathbb C\setminus \{1\}$. The details of this really require complex analysis.

Calculating the non-trivial zeroes of the Riemann zeta function is a whole entire field of mathematics.

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    [In particular](http://dlmf.nist.gov/25.4.E2): $$\zeta(s)=2(2\pi)^{s-1}\sin\frac{\pi s}{2}\Gamma(1-s)\zeta(1-s)$$ Replace $s$ in both sides with a negative even integer and observe...2012-04-20
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    Analytic continuation for $\Re s>0$ does not really require so much knowledge other than integrals, and good notion of convergence. Analytic continuation to $s \neq 1$ requires only the Poisson summation formula, not really complex analysis either.2012-04-20
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    I don't know about the integrals late_learner is referring to, but the treatment I'm accustomed to for continuing to $\Re\,s > 0$ is to consider the related Dirichlet $\eta$ function...2012-04-20
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    Perhaps, but the whole notion of analytic continuations - What is analytic? Is it distinct? Why would we want this type of continuation? - really require beginning complex analysis.2012-04-20
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    Precisely and the relation $\eta(s)$ with $\eta(s^{-1})$ is a special case of Poisson summation formula.2012-04-20
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    Sorry, I was only focusing on the last question. The OP seems really to have difficulties about the notion of analytic continuation.2012-04-20
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    There is an even simpler example of analytic continuation: when we are first taught about $x^a$, we only consider positive integer $a$ at first. Then we figure out what it means for $a=0$ and $a$ a negative integer. Then we figure out what it means for $a\in\mathbb Q$ (algebra), and then $a\in\mathbb R$ (calculus/real analysis) and then finally $a\in\mathbb C$ (complex analysis)...2012-04-20
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    @J.M. this example does not fit here and is a wrong oversimplyfication. The one Thomas Anrews gives, is perfectly fine. Analytic continutation considers the question, whether a function usually given as series/integral makes sense outside a range of convergence, and not simply about artificially changing the domain of definition.2012-04-21
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    I never once thought of these "changes" you speak of as "artificial", @late_learner. Clearly, OP has trouble about the concept of having a function make sense beyond its original definition/domain, and I offered up the simplest example I have. We will have to agree to disagree on this matter, apparently.2012-04-21
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    Analytic continuation is nevertheless a well-defined concept (see e.g. wiki), but certainly the adjective "artificial" wasn't carefully chosen either;)2012-04-22
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Copied from Wikipedia:

For all $s\in\mathbb{C}\setminus\{1\}$ the integral relation $$\zeta(s) = \frac{2^{s-1}}{s-1}-2^s\!\int_0^{\infty}\!\!\!\frac{\sin(s\arctan t)}{(1+t^2)^\frac{s}{2}(\mathrm{e}^{\pi\,t}+1)}\,\mathrm{d}t,$$ holds true, which may be used for a numerical evaluation of the Zeta-function. http://mo.mathematik.uni-stuttgart.de/kurse/kurs5/seite19.html

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Here's an extension method in c# to calculate the zeroes for Re(s) > 0. However, it is not very efficient for large values of t. Note, .5 in the calculation is the Zeta(1/2+it). Try any other number and you will not get a zero.

Also, one could easily modify the function to return an IEnumerable and the user could create a query/filter against each term in the infinite sum. I found it interesting to plot each term on a graph and watch it converge in the plane. The zeroes are where the graph comes back to the origin. The Complex type is found in the System.Numerics namespace.

    /// 
    /// Calculates the converged point for a Dirichlet series expansion.
    /// 
    /// imaginary part of s. The first zero is at 14.134725
    /// Use a higher number to find more accurate convergence.
    /// 
    public static Complex CalcZetaZero(this double t, int numberOfTerms)
    {
        var range = Enumerable.Range(1, numberOfTerms);
        var zetaZero = Complex.Zero;

        foreach (int n in range)
        {
            var direction = n % 2 == 0 ? Math.PI : 0;
            var newTerm = Complex.Exp(new Complex(-Math.Log(n) * .5, -Math.Log(n) * t + direction));
            zetaZero += newTerm;
        }

        return zetaZero;
    }