Previously, we visualised the Riemann Zeta series $\sum 1/n^s$ in its domain $\sigma > 1$ and saw that its shape strongly suggested the function should continue to the left of $\sigma=1$.
Slides are here [pdf], and a video is here [youtube].
Series And Functions
Before we try to extend the Riemann Zeta series to the left of $\sigma=1$ we should first understand how it might even be possible.
Let's take a fresh look at the well known Taylor series expansion of $f(x)=(1-x)^{-1}$ developed around $x=0$.
$$S_{0}=1+x+x^{2}+x^{3}+\ldots$$
The series $S_{0}$ is only valid for $|x|<1$, but the function $f(x)$ is defined for all $x$ except $x=1$. This apparent discrepancy requires some clarification.
That series $S_{0}$ is just one representation of the function, valid for some of that function's domain, specifically $|x|<1$. We can use the standard method for working out Taylor series to find a different representation of $f(x)$ valid outside $|x|<1$. For example, the following series $S_{3}$ is developed around $x=3$, and is valid for $1<x<5$.
$$S_{3} = -\frac{1}{2} + \frac{1}{4}(x-3) - \frac{1}{8}(x-3)^{2} + \frac{1}{16}(x-3)^{3} - \ldots$$
So the series $S_{0}$ and $S_{3}$ both represent $f(x)=(1-x)^{-1}$ but over different parts of its domain. This clarifies the distinction between a function, and any series which represent it in parts of its domains.
Perhaps the series $\sum1/n^{s}$ only gives us a partial view of a much richer function that encodes information about the primes. Could that function be represented by a different series over a different domain?
A New Series
Let's write out the familiar series for $\zeta(s)$.
$$\zeta(s)=\sum\frac{1}{n^{s}}=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\ldots$$
An alternating version of the zeta function is called the eta function $\eta(s)$.
$$\eta(s)=\sum\frac{(-1)^{n+1}}{n^{s}}=1-\frac{1}{2^{s}}+\frac{1}{3^{s}}-\frac{1}{4^{s}}+\frac{1}{5^{s}}-\ldots$$
This is a Dirichlet series which, as explained in a previous post, converges for $\sigma>0$. If we could express $\zeta(s)$ in terms of $\eta(s)$, we would have a new series for the Riemann Zeta function that extends to the left of $\sigma=1$, even if only as far as $\sigma>0$.
Looking at the difference $\zeta(s)-\eta(s)$, we can see a pattern to exploit.
$$\begin{align}\zeta(s)-\eta(s)&=\frac{2}{2^{s}}+\frac{2}{4^{s}}+\frac{2}{6^{s}}+\ldots\\ \\&=\frac{2}{2^{s}}\left(1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\ldots\right)\\ \\&=2^{1-s}\zeta(s)\end{align}$$
Isolating $\zeta(s)$ gives us a new series that is valid in the larger domain $\sigma>0$.
$$\boxed{\zeta(s)=\frac{1}{1-2^{1-s}}\sum\frac{(-1)^{n+1}}{n^{s}}}$$
The denominator $(1-2^{1-s})$ is zero at $s=1+0i$, and provides $\zeta(s)$ with its divergence at that point.
Visualising The New Series
Any enthusiastic mathematician would be impatient to visualise this new series. The plot below shows a contour plot of $\ln\left|\zeta(s)\right|$, this time evaluated using the new series.
Our previous intuition was justified, the surface does continue smoothly to the left of $\sigma=1$. In fact the contours suggest the function should again continue smoothly even further into $\sigma<0$.
Even more interesting is the appearance of zeros, all of which seem to be on the line $s=1/2+it$.
These zeros are critically important, but we'll have to continue our journey to see why.
Animated 3D View
The following animation shows the the logarithm of the magnitude of this extended Riemann Zeta function from different angles to better illustrate its shape.
