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? enter image description here

or is it this? enter image description here or perhaps it was this guy: enter image description here

Do you see what I'm saying? Where's the y=f(x) versions that make those plots of zeta going to zero along re=1/2? I feel like knowing how to plot that would help me understand the Riemann hypothesis a little better.

  • 1
    $\begingroup$ 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 $\Re(s) > 0,1-2^{1-s}\ne 0$. $\endgroup$
    – reuns
    Oct 8, 2019 at 6:59
  • $\begingroup$ @reuns are you saying that's the equation to use? And if it is one of the inf series would you just compute some large partial sum to get outputs? $\endgroup$
    – user273872
    Oct 8, 2019 at 9:22
  • $\begingroup$ 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. $\endgroup$
    – reuns
    Oct 9, 2019 at 9:27
  • $\begingroup$ I take that as a yea? (every algorithm to eval bla bla part of your comment sounds like "yes it's partial sums and yes it's hard" but I don't for sure understand you please clarify) btw I just made a new question cause of this: math.stackexchange.com/questions/3389209/… it seems like you could answer both questions. $\endgroup$
    – user273872
    Oct 11, 2019 at 6:39

1 Answer 1


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.


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