# Calculating the Zeroes of the Riemann-Zeta function

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.

• I think it may help mathworld.wolfram.com/Riemann-SiegelFormula.html – 89085731 Apr 20 '12 at 13:34
• 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. isn't a mathematician Apr 20 '12 at 13:37
• 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. isn't a mathematician Apr 20 '12 at 13:39

## 4 Answers

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.

• 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. isn't a mathematician Apr 20 '12 at 13:46
• 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. – Marc Palm 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. isn't a mathematician Apr 20 '12 at 13:53
• 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
• Sorry, I was only focusing on the last question. The OP seems really to have difficulties about the notion of analytic continuation. – Marc Palm 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

For the first question, what are the "trivial" and "non-trivial" zeros of the Riemann zeta function? These are terms used to describe 1) the zeros of the Riemann zeta function that occur at regular intervals to infinity (trivial) and those that do not (non-trivial). Take any function, not just the zeta function. For instance the sine function $$sin(s)$$. Anytime $$sin(s)=0$$, then that's a zero of the sine function. All the sine zeros would be considered "trivial" by comparison, as they occur at regular intervals: $$sin(\pi n)$$, anytime the argument is a natural number $$(0, 1, 2, 3, ...)$$ multiplied by $$\pi$$.

The trivial zeros of the Riemann zeta function occur at $$s=-2n$$, so for natural numbers $$n>0$$, one gets a zero at $$\zeta(-2)$$, $$\zeta(-4)$$, $$\zeta(-6)$$, etc.. So rather trivial.

The non-trivial are more complicated. Assuming you are familiar with complex arithmetic, the first one is $$\zeta(1/2 + i 14.134725141734...)$$, where $$i$$ is the imaginary number. The next non-trivial zero is its complex conjugate $$\zeta(1/2 - i 14.134725141734...)$$ and the next is $$\zeta(1/2 + i 21.022039638771...)$$, and likewise its complex conjugate $$\zeta(1/2 - i 21.022039638771...)$$, and they go on and on to infinity, getting closer and closer together, but never exhibiting any real clear pattern. The reason they are more complicated is due to the fact that no one knows much about the imaginary parts, almost nothing at all. It is known that all the non-trivial zeros whose imaginary part is up into somewhere greater than $$3*10^9$$ each have a real part equal to one half, as seen with the first couple examples. The Riemann hypothesis states that they all have a real part one half.

The second question...

The infinite series representation you reference cannot calculate any of the zeros of the Riemann zeta function, so that's probably why it doesn't make a lot of sense. Fortunately, there are other representations of the function available. Here's a straight forward way to calculate the non-trivial zeros with what you already have.

Multiply the infinite series you used in your post by $$(-1)^{1-s}/(1-2^{1-s})$$ in order to "alternate" the function. This gives $$\sum_{n=1}^{infinity} (-1)^{1-n}/(n^s(1-2^{1-s}))=\zeta(s), Re(s)>0,$$ which does converge for the non-trivial zeros. You can apply one of the non-trivial zeros I provided above and you will see that by summing to infinity the negative values begin to negate the positive. This representation is sometimes called the Dirichlet (alternating) zeta function.

For calculating the trivial zeros, one will need to use a functional representation of the zeta function or the Abel-Plana representation, which you can read up more at https://en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula .

Here's that representation, which will also work for the non-trivial zeros: $$\zeta(s)=1/(s-1)+1/2+2\int_0^{infinity} sin(s* arctan(t))/((e^{2 \pi t }-1) (1+t^2)^{s/2}) dt$$

While the Abel-Plana formula is powerful, it's a bit lengthy. I hope the sum I provided for the non-trivial zeros helps.

• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – Leucippus Feb 20 '20 at 19:50
• I have edited it with a clear answer...I hope. – Jeff Feb 20 '20 at 22:18
• $\sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ converges for $\Re(s) >0$ not for $s=-2$. Moreover the interesting part is how do we prove that a zero has real exactly $1/2$ and how do we prove that there are no other non-trivial zeros up to height $T$ – reuns Feb 21 '20 at 3:47
• Correct, it does not converge for s=-2, so I provided the Abel-Plana representation as well. jamie's highlighted question just asks how to calculate "a" zero. I provided two methods, the first calculates for Re(s)>0 and the second calculates both types of zeros (and of course every other argument s except s=1). The first representation is most similar to where jamie began. – Jeff Feb 21 '20 at 12:50
• Do you know how to prove that there is a zero near $1/2 + i 14.13$ and that it has real part exactly $1/2$ ? (you can do it from the evaluation of $\eta(s)=\sum_n (-1)^{n+1}n^{-s}$, but you have to know the functional equation) – reuns Feb 23 '20 at 6:54

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;
}