Calculating the Zeroes of the Riemann-Zeta function

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

Regarding your question on how to calculate zeros of $\zeta(s)$...

The algorithm below will converge to a nontrivial zero of $\zeta(s)$ along the line $s=1/2+t \in \mathbb{C}$.

$\mathbf{Newton-Raphson}$ $\mathbf{Algorithm}$ $\mathbf{for}$ $\zeta(s)$: Given an initial $t_{k} \in \mathbb{R}$, iterative solutions $t_{k+1}$ converge to nontrivial zeros of $\zeta(s)$,

$$t_{k+1}=t_{k}-\frac{2 i}{\frac{16 i t_{k}}{1+4t_{k}^2}+\log_{e}(\pi) - \psi(1/4+i t_{k}/2)-\frac{2 \zeta’(1/2+i t_k)}{\zeta(1/2+i t_k)}}.$$

Note that $\psi(s)$ is the di-gamma function and that $t_k$ is the imaginary part of the root $s=1/2+i t_k$, with $\zeta(1/2+it_k)\approx 0$.

Answer 4

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 $3\cdot10^9$ 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 $$\sum_{n=1}^{\infty} (-1)^{1-n}/(n^s(1-2^{1-s}))=\zeta(s),\quad \operatorname{Re}(s)>0,$$ 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: $$\zeta(s)=1/(s-1)+1/2+2\int_0^{\infty} \sin(s\cdot \arctan(t))/((e^{2 \pi t }-1) (1+t^2)^{s/2}) dt$$

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