Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I think it may help mathworld.wolfram.com/Riemann-SiegelFormula.html –  Joe Berg Apr 20 '12 at 13:34
5  
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). –  J. M. Apr 20 '12 at 13:37
1  
As Ginger mentions, one uses the Riemann-Siegel formula numerically to compute the nontrivial zeroes (there are no known closed forms for them). –  J. M. Apr 20 '12 at 13:39

3 Answers 3

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.

share|improve this answer
3  
In particular: $$\zeta(s)=2(2\pi)^{s-1}\sin\frac{\pi s}{2}\Gamma(1-s)\zeta(1-s)$$ Replace $s$ in both sides with a negative even integer and observe... –  J. M. Apr 20 '12 at 13:46
1  
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. –  plusepsilon.de Apr 20 '12 at 13:50
    
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... –  J. M. Apr 20 '12 at 13:53
1  
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. –  Thomas Andrews Apr 20 '12 at 13:53
1  
Sorry, I was only focusing on the last question. The OP seems really to have difficulties about the notion of analytic continuation. –  plusepsilon.de Apr 20 '12 at 14:08

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

share|improve this answer

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<Complex> 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.

    /// <summary>
    /// Calculates the converged point for a Dirichlet series expansion.
    /// </summary>
    /// <param name="t">imaginary part of s. The first zero is at 14.134725</param>
    /// <param name="numberOfTerms">Use a higher number to find more accurate convergence.</param>
    /// <returns></returns>
    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;
    }
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.