# Why are there so many equations for the Riemann zeta function and how do you go about calculating it when it times to actually crunch some numbers

For example if you look at this graph the real and imaginary parts along the critical line $$x$$ is plugged into what equation? It can't be the normal function: $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}$$ my understanding is that that is divergent in the critical strip. I know enough to know that you have to analytically continue the normal function, but reading all this different stuff there's all kinds a weird series and confusion. Perhaps I don't know enough to even ask my question, but for them to have plotted that graph up there don't they have some functions to plug x into? So for example was it this one?

• For $\Re(s) > 1$, $(1-2^{1-s})\zeta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ and the latter series converges and is analytic for $\Re(s) > 0$ thus we have found $\frac1{1-2^{1-s}} \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ the meromorphic continuation of $\zeta(s)$ to $\Re(s) > 0$ and the analytic continuation to $\Re(s) > 0,1-2^{1-s}\ne 0$. Oct 8, 2019 at 6:59
• Concretely $|(1-2^{1-s})\zeta(s)-\sum_{n=1}^{2N}(-1)^{n+1}n^{-s}|=|\sum_{n=N}^\infty (2n+1)-(2n+2)^{-s}|=|\sum_{n=N}^\infty \int_{2n+1}^{2n+2} s x^{-s-1}dx|$ $\le \sum_{n=N}^\infty \int_{2n}^{2n+2} |s| x^{-\Re(s)-1}dx= |s|\int_{2N}^\infty x^{-\Re(s)-1}dx=\frac{|s|}{\Re(s)}(2N)^{-\Re(s)}$ so larger is $|\Im(s)|$ and smaller is $\Re(s)$ larger you need $N$ to approximate $(1-2^{1-s})\zeta(s)$ with a finite sum. Every algorithm to evaluate $\zeta$ suffer from this kind of problem but some do better than others. Oct 9, 2019 at 9:27
As a random example, the Boost C++ libaries (v. 1.51.0) have a zeta() implementation. Suppose you wish to compute $$\zeta(s)$$ to within some (small) target precision. They start by, if necessary, using the reflection formula to ensure $$\mathrm{Re}(s) \geq 0$$. Then they choose the parameter $$n$$ so that $$\frac{1}{8^n |1 - 2^{1-s}|}$$ is smaller than your target precision. They use $$\zeta(s) = \frac{-1}{s^n(1-2^{1-s})} \sum_{j = 0}^{2n-1} \frac{e_j}{(j+1)^s} + \gamma(s) \text{,}$$ where $$e_j = (-1)^j \left(\sum_{k=0}^{j-n} \frac{n!}{k!(n-k)!} - 2^n \right) \text{.}$$ Then they compute and return $$\frac{-1}{s^n(1-2^{1-s})} \sum_{j = 0}^{2n-1} \frac{e_j}{(j+1)^s} \text{.}$$
Since it is known that $$|\gamma(s)| \leq \frac{1}{8^n |1 - 2^{1-s}|}$$, the difference between the computed term and the exact value of $$\zeta(s)$$ is smaller than your target precision.