Let $\log(1),\log(2),\log(3),\log(4)...\log(n)$ be approximated by fractions generated by the truncated sums:

$$\text{log1}=\sum_{n=0}^k \frac{0}{(1 n+1)^c}=0$$ $$\text{log2}=\sum _{n=0}^k \left(\frac{1}{(2 n+1)^c}-\frac{1}{(2 n+2)^c}\right)=\frac{1}{2}$$ $$\text{log3}=\sum _{n=0}^k \left(\frac{1}{(3 n+1)^c}+\frac{1}{(3 n+2)^c}-\frac{2}{(3 n+3)^c}\right)=\frac{5}{6}$$ $$\text{log4}=\sum _{n=0}^k \left(\frac{1}{(4 n+1)^c}+\frac{1}{(4 n+2)^c}+\frac{1}{(4 n+3)^c}-\frac{3}{(4 n+4)^c}\right)=\frac{13}{12}$$

Letting $k \rightarrow \infty$ makes the sums above converge to the precise values of $\log(n)$ by their Dirichlet generating function:

$$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)$$

The alternating series which is valid in the critical strip can be written as, and is set to zero:


The truncated alternating series:

$$\sum_{n=1}^{n=4}(-1)^{(n+1)}\frac{1}{\left(e^{\log(n)}\right)^s}=0$$ is solved for $s$ in Mathematica 8.0.1 with the program:

Clear[log1, log2, log3, log4, n, k, s];
c = 1;
k = 0;
log1 = Sum[0/(1*n + 1)^c, {n, 0, k}];
log2 = Sum[1/(2*n + 1)^c - 1/(2*n + 2)^c, {n, 0, k}];
log3 = Sum[1/(3*n + 1)^c + 1/(3*n + 2)^c - 2/(3*n + 3)^c, {n, 0, k}];
log4 = Sum[
   1/(4*n + 1)^c + 1/(4*n + 2)^c + 1/(4*n + 3)^c - 3/(4*n + 4)^c, {n, 
    0, k}];
$MaxRootDegree = 1000;
   1/(E^(log1))^s - 1/(E^(log2))^s + 1/(E^(log3))^s - 
     1/(E^(log4))^s == 0, s]];

which gives the output:

$$\left\{s\to -4 i \pi +\log \left(\text{Root}\left[\text{$\#$1}^{10}-3 \text{$\#$1}^9+3 \text{$\#$1}^8+23 \text{$\#$1}^7+40 \text{$\#$1}^6 \\-2 \text{$\#$1}^5+42 \text{$\#$1}^4+12 \text{$\#$1}^2+1\&,10\right]\right)\right\} \label{1}\tag{$*$}$$

which has the form:

$$\left\{s\to -\text{integer } i \pi +\log \left(\text{polynomial root}\right)\right\}$$

Now the following number also has a similar form:

$$7 \pi -\text{Log}\left[\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi \right] = \\ 14.1347251415...$$ $$14.1347251417...$$ with the value of the actual first Riemann zeta zero Im[ZetaZero[1]] right below it for comparison. This can't be all coincidence.


How does Mathematica arrive at the output in \eqref{1}$?

In the edit at 19.7.2021 in https://mathematica.stackexchange.com/q/63541/328 I made a guess at the solution for a simpler case which is not the alternating series, but one that also has $1/(e^{\text{fraction}})^s$ as its terms.

  • 5
    $\begingroup$ I don't know what $12\#1^2$ means. I don't know what $1\&,10$ means. It's not surprising that you can approximate any number you like by a multiple of $\pi$ plus the log of a polynomial. You have a lot of degrees of freedom there. $\endgroup$ Jul 20, 2021 at 2:32
  • $\begingroup$ I have to admit I don't know either what the precise meaning of the hashtag followed by the number "1" and the ampersand are. But the hashtag serves as the variable $x$ in any normal polynomial which can be seen from this Mathematica program: x = N[Root[1 + 12 #1^2 + 42 #1^4 - 2 #1^5 + 40 #1^6 + 23 #1^7 + 3 #1^8 - 3 #1^9 + #1^10 &, 10], 100]; 1 + 12 x^2 + 42 x^4 - 2 x^5 + 40 x^6 + 23 x^7 + 3 x^8 - 3 x^9 + x^10 which outputs 0 to 93 decimal places: 0.*10^-93 + 0.*10^-93 I The number $10$ at the end means the 10-th root of that polynomial. And yes there are many degrees of freedom. $\endgroup$ Jul 20, 2021 at 2:43
  • $\begingroup$ in your Mathematica code you set $k=0$ and then sum from $n=0$ to $k$; what is the point of that? $\endgroup$ Jul 20, 2021 at 6:23
  • $\begingroup$ Setting $k=1$ instead of $k=0$, makes the computation time take significantly longer. Or at least I have not seen the answer on my computer within a few minutes. The smaller the fractions the faster the computation. The fractions for $k=1$ would be: ${0, \frac{7}{12}, \frac{19}{20}, \frac{341}{280}}$ $\endgroup$ Jul 20, 2021 at 15:31
  • 1
    $\begingroup$ I don't have approximations for the other Riemann zeta zeros. The starting point was the observation that the first Riemann zeta zero is close to $\frac{9 \pi }{2}$ and the equality $\frac{9 \pi }{2}=7 \pi -\log \left(e^{\frac{5 \pi }{2}}\right)$ from there on I put an $x$ inside the logarithm and solved for $x$ and observed that the number is close to $2 \pi$ and from that I guessed the rest. There is an explanation here: math.stackexchange.com/q/190080/8530 below "Edit 23.12.2012". $\endgroup$ Jul 21, 2021 at 10:31

1 Answer 1


The actual question asked here is a Mathematica question, so this is not really the right site for it, but here goes. Suppressing the extraneous notation in $c$ and $k$, the expression in $s$ is $$ -e^{-13 s/12}+e^{-5 s/6}-e^{-s/2}+1 $$ which is a polynomial of degree 13 in $x=e^{-s/12}$: $$ - x^{13} + x^{10} - x^6+1=-(x-1)(x^2+x+1)(x^{10}+x^3+1) $$ The Mathematica code 'Solve' tries to find roots of this polynomial in $x$, and the $\log$ expressions will then solve for $s$. (I suspect the OP is showing output from a different choice of $c$ and/or $k$, because already I'm seeing something different.). The first two factors are solvable by radicals of course. The OP has asked for the 'last' solution. To avoid introducing new variables which may conflict with user defined variables, it is merely expressed as Root[1#^10+1#^3+1&,10]. Here 1# is just the variable name (Mathematica can do this in more than one variable - the others would be 2#, 3#, etc.). The & at the end is just a way of saying'Think of this as a function of 1#' - a 'pure function' in the language of the Mathematica documentation . The 10 indicated the last root.

Addendum: the -integer $\pi i$ in the title is misleading of course. the questions is not about the zeros of $\zeta(s)$, but the imaginary parts of the zeros, which are real numbers. The $\exp(\text{integer}\pi i)$ is just $\pm 1$, and Mathematica is just trying to give the most general form of the solution when one uses Log to solve for $s$.


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