In this post we consider the Ramanujan tau function $\tau(n)$, see the Wikipedia Ramanujan tau function, and we consider partial sums of its corresponding Dirichlet series (see for example the article Tau Dirichlet Series from the encyclopedia Wolfram MathWorld)

$$\varphi_N(s):=\sum_{n=1}^N\frac{\tau(n)}{n^s},\tag{1}$$ where thus $N>1$ is a positive integer, and $s=x+iy$ denotes the complex variable.

Question 1. How do you determine and calculate an approximation of a zero of $$\varphi_3(s)=\sum_{n=1}^3\frac{\tau(n)}{n^s}?$$ Many thanks.

I'm inspired in section 4 from [1], thus I know that at least one can to deduce these equations $$0=1-24\cdot 2^{-x}\cos(y \log 2)+252\cdot 3^{-x}\cos(y\log 3)$$ and

$$0=-24\cdot 2^{-x}\sin(y \log 2)+252\cdot 3^{-x}\sin(y\log 3),$$

but I don't know how get an approximation for one of those zeros $s=x+iy$ or if this is a good way.

Question 2. Inspired in section 3 I would like to know if it is possible to state or conjecture a region for which the partial sum of the Ramanujan's zeta function $\varphi_N(s)$, for $N$ large enough, doesn't vanish (see the quoted statement due to Montgomery in page 23). Many thanks.

If some of previous questions are in the literature feel free to refer it, and I try to search and read the answers to my questions from the literature. I hope that both questions have mathematical sense.


[1] Peter Borwein, Greg Fee, Ron Ferguson and Alexa van der Waall, Zeros of Partial Sums of the Riemann Zeta Function, Experimental Mathematics, Vol. 16 (2007), No. 1, A K Peters, Ltd.

  • $\begingroup$ Why are we calling a zeta function phi? $\endgroup$ Nov 4, 2019 at 11:18
  • $\begingroup$ Many thanks for your comment, I haven't a concise motivation for it. If you provide other notation I can edit my post @GerryMyerson Or feel free to do the edit $\endgroup$
    – user142929
    Nov 4, 2019 at 11:22
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    $\begingroup$ If the function appears in the literature, and if it's called zeta in the literature even though it gets denoted by phi, then I'm not going to argue with precedent and provide alternative notation. $\endgroup$ Nov 4, 2019 at 11:25
  • 1
    $\begingroup$ You start from any reasonable value of $s$ and do a stupid Newton iteration. In this way you will find as many zeros as you want (but of course you do not prove anything about the region). For instance $$Z=4.8045238352373374953993503610565632974 + 2.1946196473041207514471882037135248165 i $$ is an approximate value for a zero. Also, when $N=\infty$, GRH asserts that all zeros have real part equal to $6$. $\endgroup$ Nov 4, 2019 at 14:08
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    $\begingroup$ I have made some further computations of zeros of $\varphi_3$ and their real part is always less than $6$ (in the range I computed, between $4.3$ and $5.7$). Maybe this is easy to prove ? $\endgroup$ Nov 4, 2019 at 18:19

1 Answer 1


Here is how you can find reasonable answers using Mathematica (or WolframAlpha).

To see the first two zeros:

ContourPlot[ Log[Abs[1 - 24/2^(sig + I t) + 252/3^(sig + I t)]], {sig, 2, 7}, {t, 0, 10}]

Making a contour plot of the Log of the absolute value (instead of just the absolute value) is a useful trick.

If you examine this plot:

Plot[Abs[24/2^sig] + Abs[252/3^sig], {sig, 5, 10}]

you will see that it is less than 1 for sig>5.64 . Therefore, phi_3 has no zeros with real part larger than 5.64. The explanation is that if z = x + i y with x and y real and b is real, then Abs[b^z] = b^x . Thus, for sig>5.64 the function has strictly positive real part, and so is nonzero.

There will be zeros arbitrarily close to the prohibited region.

  • $\begingroup$ Thanks you very much, I'm going to check these. $\endgroup$
    – user142929
    Feb 10, 2020 at 14:35

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