12
$\begingroup$

I found this approximation of which an earlier version I posted in the chat room:

$$7 \pi -\text{Log}\left[\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi \right] = 14.13472514154629716253329494571302508888...$$

The first non trivial zeta zero: $$14.13472514173469379045725198356247027078$$

Can you improve on the formula above?


Edit 2.9.2012

Based on the comments below I would like to explain how I reasoned:

Any Taylor series evaluated at $x=1$ is convergent for variants of it when multiplied element wise with rows in this matrix:

$$\begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$

Many Taylor series have the second row as part of its coefficients. That is: $$(1,-1,1,-1,1,-1,1,-1,1,-1,...)$$

Such Taylor series are for example $\log 2$, $\sqrt 2$, $\cos 1$, $\sin 1$. The reason for the convergence of such series and divisibility defined variants of thereof, seems to be that in the matrix above, a period sums to zero.

The simplest Dirichlet series that sums to zero and is not a an element wise multiplication of two other Dirichlet series, is the first row:

$$\frac{0}{1}+\frac{0}{2}+\frac{0}{3}+\frac{0}{4}+\frac{0}{5}+... = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

This suggests that one should try to find an expression for a such sequence.

The definition of a number raised to a complex number is:

$$n^{(a+ib)} = n^{a}(\cos (b \log (n))+i\sin (b \log (n)))$$

and the Riemann zeta function is:

$$\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+...$$

where $s$ is a complex number.

Here I then made a mistake. I started studying the equation: $$\cos (\log (n)) = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ in order to get something similar to the Dirichlet series with numerators equal to the all zeros sequence in expression $(1)$ above. But if I understand correctly this would be the same as seeking the undefined sequence:

$$\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+$$

After that I just guessed that by combining values from the solutions to equation $(2)$ one could possibly find an expression for the zeta zeros.


Edit 23.12.2012: For what it is worth. Here is how the actual calculation went:

The first Riemann zeta zero is:

$$\Im(\rho _1)$$ $$=14.1347251417346937904572519836$$

A number close to the first Riemann zeta zero is:

$$\frac{9 \pi }{2}$$ $$=14.1371669411540695730818952248$$

That number can be split up into:

$$\frac{9 \pi }{2} = 7 \pi -\log \left(e^{\frac{5 \pi }{2}}\right)$$

To see what is missing within the logarithm I added an $x$ and solved the equation:

$$\text{Solve}\left[N\left[7 \pi -\log \left(x+e^{\frac{5 \pi }{2}}\right),30\right]=N[\Im(\rho _1),30],x\right]$$

This gives the solution:

$$\{\{x\to 6.297688980465813720589098\}\}$$

which is close to:

$$2\pi = 6.28318530717958647692528676656...$$

Substituting $x$ with $2\pi$:

$$7 \pi -\log \left(e^{\frac{5 \pi }{2}}+2 \pi \right)$$

which is closer:

$$=14.1347307583914370155699744066$$

Some small number seems to be missing, the second harmonic number could be it:

$$7 \pi -\log \left(e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

which again is closer:

$$=14.1347272795405950845865949010$$

Multiplying the added number with $\frac{3}{2}$

$$7 \pi -\log \left(\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

closer still:

$$=14.1347255401197125097619679160$$

continuing the trick with similar numbers:

$$7 \pi -\log \left(\frac{5}{2} e^{-\frac{1}{2} (5 \pi )}+\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

works:

$$=14.1347251642841507747886817861$$

and once more:

$$7 \pi -\log \left(\frac{7}{2} e^{-\frac{1}{2} (7 \pi )}+\frac{5}{2} e^{-\frac{1}{2} (5 \pi )}+\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

it works:

$$=14.1347251415462971625332949457$$

but then I can't get further.


Edit: 5.11.2013:

$$\frac{\sqrt{\frac{\Im(\rho _1)}{\pi }+\frac{1}{2}}}{\sqrt{5}}=0.999922272089659461895288929782$$

$$\frac{\Im(\rho _1)}{\pi }+\frac{1}{2}=4.99922275110473484848654142318$$

$\rho _1$ = first riemann zeta zero = 14.134725141734693790457...

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10
  • $\begingroup$ You might want to add some explanations about this exact-approximate formula (the one with $\pi$ and exponentials). Is it composed of the first terms of an explicit series? $\endgroup$
    – Did
    Sep 2, 2012 at 17:01
  • $\begingroup$ Interesting, but $2171956\over153661$ is closer. I mean, unless the pattern lurking in your exression leads to a convergent sequence (which seems is not the case), this looks quite fruitless to me. $\endgroup$ Sep 2, 2012 at 17:02
  • 2
    $\begingroup$ Agreed, unless you can come up with a "reason," this could easily just be a coincidence. $\endgroup$ Sep 2, 2012 at 17:06
  • 1
    $\begingroup$ I fail to see what the Edit section explains. $\endgroup$
    – Did
    Sep 2, 2012 at 18:15
  • 1
    $\begingroup$ I can improve on yours by adding 0.0000000002 $\endgroup$
    – fretty
    Dec 23, 2012 at 12:08

6 Answers 6

5
$\begingroup$

Can you improve on the formula above? -- Yes: $$ \text{Log}\left[\frac{2}{3} e^{-5 \pi /2}+\frac{e^{7 \pi }}{\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi }\right] = 14.1347251417343... $$

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2
$\begingroup$

Let $n=1$, then the formula where $\vartheta (t)$ is the Riemann Siegel theta function and $\text{sgn}$ is the sign function:

$$ 14.13472514173469...=\int _0^{16}\frac{1}{2} \left(1-\text{sgn}\left(\frac{\vartheta (t)+\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)}{\pi }-n+\frac{3}{2}\right)\right)dt$$

gives the first Riemann zeta zero.

Mathematica:

Clear[n, t, gamma];
gamma = 15; 
Quiet[
 Do[gamma = 
   N[NIntegrate[(1/2)*(1 - 
        Sign[(RiemannSiegelTheta[t] + Im[Log[Zeta[I*t + 1/2]]])/Pi - 
          n + 3/2]), {t, 0, gamma + 15}, PrecisionGoal -> 45, 
     MaxRecursion -> 220, WorkingPrecision -> 50], 40]; 
  Print[gamma], {n, 1, 1}]]

14.13472514173469379045725198356247027078

which gives the same result as:

N[Im[ZetaZero[1]], 40]

14.13472514173469379045725198356247027078

Update 16.8.2017:

The integral below might be solvable through Fourier series approximations of the Floor and Sign functions, and repeated integration by parts, since there is no function in the denominators of the power series for the Fourier series.

$n=1$ $$14.1347251417346937904572\text{...}=\int _0^{16}\frac{1}{2} \left(1-\text{sgn}\left(\left(\left\lfloor \frac{\vartheta (t)}{\pi }+1\right\rfloor +\frac{1}{2} \left(-1+\text{sgn}\left(\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)\right)\right)-n+\frac{3}{2}\right)\right)dt$$

The convergence of the symbolic solution is probably not great though, since the powers in the power series cause the complexity of the symbolic expression to grow rapidly.

Mathematica program:

Clear[n, t, gamma]
gamma = 15;
Quiet[Do[gamma = 
   N[NIntegrate[(1/2)*(1 - 
        Sign[((Floor[
              RiemannSiegelTheta[t]/Pi + 
               1]) + (Sign[Im[Zeta[1/2 + I*t]]] - 1)/2) - n + 
          3/2]), {t, 0, gamma + 16}, PrecisionGoal -> 45, 
     MaxRecursion -> 220, WorkingPrecision -> 50], 40];
  Print[gamma], {n, 1, 10}]]
$\endgroup$
5
  • $\begingroup$ How did you come up with such a formula? $\endgroup$
    – Alphonse
    May 24, 2017 at 18:36
  • 1
    $\begingroup$ @Alphonse See this answer by Raymond Manzoni: math.stackexchange.com/a/442686/8530 Manzoni's step function /zeta zero counting function, is the same as the last formula on page 128 in the book "Riemann' zeta function, by H.M. Edwards." The zeta zero counting function has been know for at least since the year 1912, maybe even longer back to Riemann, I don't know the history behind it. What I did was to wrap the counting function with the sign function and compute the numerical integral for between 0 and 16, a simple rectangle. Solving the symbolical integral would be more valuable. $\endgroup$ May 25, 2017 at 6:43
  • 2
    $\begingroup$ The length of the rectangle is equal to the first zeta zero = 14.1347251417... and the height of the rectangle is equal to 1. The rectangle starts at 0 and ends at the first zeta zero. The rectangle thereby falls within the interval 0 to 16, but the upper integration limit 16 is arbitrary, (although if one wants to set it closer to the first zeta zero, then one needs to use the mathematica command MinRecursion->12). So one can see with the bare eyes that the integral must be the first zeta zero. Be aware that I used the Quiet command in Mathematica which silences the error messages. $\endgroup$ May 25, 2017 at 6:56
  • 1
    $\begingroup$ The upper integration limit is arbitrary as long as it is greater than the zeta zero to be computed. For the second zeta zero 21.022... the integration limit would have to be 24 or more. This formula works for the first 126 zeta zeros and a few more later on. It probably has to do with the sign of the derivative at the zeta zeros, I don't know for sure. $\endgroup$ May 25, 2017 at 7:07
  • 1
    $\begingroup$ @MatsGranvik it probably stops working after n=126 cause thats the first 'bad' Gram point where the argument is not on the principal branch, math.stackexchange.com/questions/2885232/… $\endgroup$
    – crow
    Apr 7, 2019 at 22:05
0
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Let $h(s,n)$ be:

$$h(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-2}}{(n-2)!}\zeta (c)^{n-2} \sum _{k=1}^{n-1} \frac{(-1)^{k-1} \binom{n-2}{k-1}}{\zeta ((c-1) (k-1)+s)}$$

and let $g(s,n)$ be:

$$g(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-1}}{(n-1)!} \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}$$

Conjecture:

The ratio $$\rho_s = i s+\lim\limits_{n \rightarrow \infty}\frac{h(i s,n)}{g(i s,n)}$$ appears to converge to the nearest Riemann zeta zero.

For $s=15$ we get: $0.5 +14.1347 i$

The plot of real part which starts off at the trivial zero $-2$ and then tends close to $1/2$ except at singularities. The Gram points appear to be a subset of the singularities.

Real and imaginary part of computed ratios

The second plot is the imaginary part of the output (ratios) which has heights close to imaginary parts of Riemann zeta zeros.

(*start*)
(*Mathematica program for the plots*)
Clear[n, k, s, c, z, f, g];
n = 11;
ss = 40;
h[s_] = Limit[((-1)^(n - 2) Zeta[
      c]^(n - 2) Sum[(-1)^(k - 1)*
        Binomial[n - 2, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, 
        n - 1}]/(n - 2)!), c -> 1];
g[s_] = Limit[((-1)^(n - 1) Zeta[
      c]^(n - 1) Sum[(-1)^(k - 1)*
        Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, 
        n}]/(n - 1)!), c -> 1];
Monitor[b = Table[s*I + h[s*N[I]]/g[s*N[I]], {s, 0, ss, 1/10}];, s*10]
ListLinePlot[Re[b], DataRange -> {0, ss}]
ListLinePlot[Im[b], DataRange -> {0, ss}]
(*end*)

(*start*)
(*Mathematica program for the first non trivial zeta zero*)
Clear[n, k, s, c, z, f, g];
n = 12;
h[s_] = Limit[((-1)^(n - 2) Zeta[
      c]^(n - 2) Sum[(-1)^(k - 1)*
        Binomial[n - 2, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, 
        n - 1}]/(n - 2)!), c -> 1];
g[s_] = Limit[((-1)^(n - 1) Zeta[
      c]^(n - 1) Sum[(-1)^(k - 1)*
        Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, 
        n}]/(n - 1)!), c -> 1];
s = 15;
s*I + h[s*N[I]]/g[s*N[I]]
(*end*)

Clear[n, k, s, c];
n = 7;
s = N[14*I];
s - n*Limit[
   1/Zeta[c]*
    Sum[(-1)^(k - 1)*
       Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, 
       n}]/
     Sum[(-1)^(k - 1)*
       Binomial[n, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n + 1}], 
   c -> 1]

For $n=7$ and $s=14i$:

$$0.5 + 14.1347i = s-n \left(\lim_{c\to 1} \, \frac{\sum _{k=1}^{n} \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}\right)$$

The conjecture is that as $n \rightarrow \infty$ the limit above converges to the Riemann zeta zero nearest to $s$.

Derivation:

Clear[s, c, A]
A0 = 1/Zeta[s];
Limit[Zeta[c] A0 - Zeta[c]/Zeta[-1 + c + s], c -> 1];

A1 = Zeta[c]/Zeta[-0 + 0 c + s] - Zeta[c]/Zeta[-1 + 1 c + s];
A2 = Zeta[c]/Zeta[-1 + 1 c + s] - Zeta[c]/Zeta[-2 + 2 c + s];
A3 = Zeta[c]/Zeta[-2 + 2 c + s] - Zeta[c]/Zeta[-3 + 3 c + s];
A4 = Zeta[c]/Zeta[-3 + 3 c + s] - Zeta[c]/Zeta[-4 + 4 c + s];
A5 = Zeta[c]/Zeta[-4 + 4 c + s] - Zeta[c]/Zeta[-5 + 5 c + s];

B1 = ReplaceAll[A1, Zeta[-1 + 1 c + s] -> 1/A2];
B2 = ReplaceAll[B1, Zeta[-0 + 0 c + s] -> 1/A1];

C1 = ReplaceAll[B2, Zeta[-2 + 2 c + s] -> 1/A3];
C2 = ReplaceAll[C1, Zeta[-1 + 1 c + s] -> 1/A2];
C3 = ReplaceAll[C2, Zeta[-0 + 0 c + s] -> 1/A1];

D1 = ReplaceAll[C3, Zeta[-3 + 3 c + s] -> 1/A4];
D2 = ReplaceAll[D1, Zeta[-2 + 2 c + s] -> 1/A3];
D3 = ReplaceAll[D2, Zeta[-1 + 1 c + s] -> 1/A2];
D4 = ReplaceAll[D3, Zeta[-0 + 0 c + s] -> 1/A1];

E1 = ReplaceAll[D4, Zeta[-4 + 4 c + s] -> 1/A5];
E2 = ReplaceAll[E1, Zeta[-3 + 3 c + s] -> 1/A4];
E3 = ReplaceAll[E2, Zeta[-2 + 2 c + s] -> 1/A3];
E4 = ReplaceAll[E3, Zeta[-1 + 1 c + s] -> 1/A2];
E5 = ReplaceAll[E4, Zeta[-0 + 0 c + s] -> 1/A1];

FullSimplify[A0]
FullSimplify[A1]
FullSimplify[B2]
FullSimplify[C3]
FullSimplify[D4]
FullSimplify[E5]

B1 = ReplaceAll[A1, Zeta[-1 + 1 c + s] -> 1/A2] means:
B1 equals the result of: "In A1 replace all Zeta[-1 + 1 c + s] with 1/A2"

FullSimplify[A0] $$\frac{1}{\zeta (s)}$$ FullSimplify[A1] $$\zeta (c) \left(\frac{1}{\zeta (s)}-\frac{1}{\zeta (c+s-1)}\right)$$ FullSimplify[A2] $$\zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right)$$ FullSimplify[A3] $$\zeta (c)^3 \left(\frac{1}{\zeta (s)}-\frac{3}{\zeta (c+s-1)}+\frac{3}{\zeta (2 c+s-2)}-\frac{1}{\zeta (3 c+s-3)}\right)$$ FullSimplify[A4] $$\zeta (c)^4 \left(\frac{1}{\zeta (s)}-\frac{4}{\zeta (c+s-1)}+\frac{6}{\zeta (2 c+s-2)}-\frac{4}{\zeta (3 c+s-3)}+\frac{1}{\zeta (4 c+s-4)}\right)$$ FullSimplify[A5] $$\zeta (c)^5 \left(\frac{1}{\zeta (s)}-\frac{5}{\zeta (c+s-1)}+\frac{10}{\zeta (2 c+s-2)}-\frac{10}{\zeta (3 c+s-3)}+\frac{5}{\zeta (4 c+s-4)}-\frac{1}{\zeta (5 c+s-5)}\right)$$


Set $n=22$, use $1000$ decimal digits of $s=14i$, that is: $s=14.0000000000000000000000000000000000000000000000000000000...i$ with $1000$ zeros after the decimal point,
and set $c=1+1/10^{40}$ With those parameters compute the following formula:

$$s-\frac{n \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}$$

(*Mathematica*)
Clear[n, k, s, c];
n = 22;
s = N[14*I, 1000];
c = 1 + 1/10^40;
s - n*(1/Zeta[c]*
    Sum[(-1)^(k - 1)*
       Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n}]/
     Sum[(-1)^(k - 1)*
       Binomial[n, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n + 1}])

Output:

0.50000000000000000000000000869164952043539610585105410624219016910923
4350306635288663716698827334351521162266267711049536613660650 + 14.134725141734693790457251943361275230040269537603008106439999619334
167441157829302370040738141325840247264856336174742894610415300 I

which gives the first 25 decimal digits of the first Riemann zeta zero.

Related:
https://mathoverflow.net/a/368105/25104
https://math.stackexchange.com/a/3735702/8530
https://mathoverflow.net/q/368533/25104

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3
  • $\begingroup$ N[Log[-100/3 + Exp[14 + 15*Exp[-3/2*Pi]]], 12] 14.1347251416... $\endgroup$ Nov 20, 2021 at 16:36
  • $\begingroup$ Wolfram Alpha: ContinuedFraction[Re[N[Exp[-ZetaZero[1]/Im[ZetaZero[1]]*Pi/4], 50]]] {0, 1, 2, 4, 1, 16, 32, 1, 3, 4, 5, 2, 1, 1, 1, 1, 3, 1, 3, 5, 75, 1} which starts like powers of 2: {0, 1, 2, 4, 8, 16, 32, $\endgroup$ Jan 22, 2023 at 14:33
  • $\begingroup$ $s=14 i$ $$0.5 +14.1347 i=\lim_{n \rightarrow \infty} \left( \left[ 1- \left( \int _{-n}^{n} \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta(\tfrac{k}{n}+s)} \Bigg/ \int _{-n}^{n} \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta(\tfrac{k}{n}+s+\tfrac{1}{n})} \right) \right]^{-1} +\frac1n + s \right)$$ $\endgroup$ Feb 1, 2023 at 19:58
0
$\begingroup$

$j$-th zeta zero:

$$\rho _j=\frac{1}{2}+2 i \pi \exp (1) \exp \left(W\left(\frac{j-\frac{11}{8}}{\exp (1)}\right)\right)+\lim_{n\rightarrow \infty}\left(\frac{1}{1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+2 i \pi \exp (1) \exp \left(W\left(\frac{j-\frac{11}{8}}{\exp (1)}\right)\right)+\frac{1}{2}-\frac{1}{n}\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+2 i \pi \exp (1) \exp \left(W\left(\frac{j-\frac{11}{8}}{\exp (1)}\right)\right)+\frac{1}{2}\right)}}}\right)$$

$10$ first zeta zeros with the Franca-LeClair approximation:

(*Mathematica start*)
Clear[f, s, n];
nn = 10; (*nn=number of zeta zeros*)
n = 25;(*increase "n" for better precision*)
(*Franca LeClair approximation*)Monitor[
 z = Table[
   1/2 + I*2*Pi*Exp[1]*Exp[ProductLog[(j - N[11/8, 50])/Exp[1]]] + 
    1/(1 - Sum[((-1)^(k - 1)*Binomial[n - 1, k - 1])/
          Zeta[k/n + 1/2 + 
            I*2*Pi*Exp[1]*Exp[ProductLog[(j - N[11/8, 50])/Exp[1]]] - 
            1/n], {k, 1, n}]/
        Sum[((-1)^(k - 1)*Binomial[n - 1, k - 1])/
          Zeta[k/n + 1/2 + 
            I*2*Pi*Exp[1]*
             Exp[ProductLog[(j - N[11/8, 50])/Exp[1]]]], {k, 1, 
          n}]), {j, 1, nn}], j]
Zeta[z]
(*end*)
$\endgroup$
2
  • $\begingroup$ N[Arg[Mod[-2*ZetaZero[1], Pi/2]], 20] = 0.0085555555513109486408... $\endgroup$ Nov 24, 2021 at 19:33
  • $\begingroup$ Do[a = Sum[(-1)^n/n^(N[ZetaZero[j], 40]), {n, 1, 250000}]; Print[(Re[a]^2 + Im[a]^2)], {j, 1, 10}] $$9.999980008001570347769316372033*10^-7$$ $$9.99998001768694008135373598632*10^-7$$ $$9.99998002503156988695903834112*10^-7$$ $$9.99998003703670152046785002090*10^-7$$ $$9.99998004339847116423847131870*10^-7$$ $\endgroup$ Jul 30, 2022 at 19:29
0
$\begingroup$

log1,log2,log3,log4 in the program are rational numbers.

Mathematica 8.0.1:

(*start*)
Clear[log1, log2, log3, n, k, s, x]
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
s /. Last[
  Solve[(E^(log1))^s - (E^(log2))^s + (E^(log3))^s - (E^(log4))^s == 
    0, s]]
FullSimplify[%]
(*end*)

Output:

-6 I \[Pi] + 
 Log[Root[1 - 3 #1 + 3 #1^2 + 23 #1^3 + 40 #1^4 - 2 #1^5 + 42 #1^6 + 
     12 #1^8 + #1^10 &, 10]]

which in Latex is:

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

$\endgroup$
1
  • $\begingroup$ (*Mathematica*) N[ReplaceAll[Reduce[1/2 + (-1)^(x) Im[ZetaZero[-2]] == -2, x], C[1] -> 0], 20] x = 0.67777108520970950896 I $\endgroup$ Jul 26, 2023 at 19:26
0
$\begingroup$

It appears that we don't need analytic continuation to locate the first non-trivial Riemann zeta zero:

(*start*)
(*Mathematica 8.0.1*)
n = 100;(*set n=200 for more digits*)
s = (3/2 + 14*I);
s + 1/n + 
  1/(1 - Sum[(-1)^(k - 1)*
        Binomial[n - 1, k - 1]/HarmonicNumber[10^10000, s + k/n], {k, 
        1, n}]/Sum[(-1)^(k - 1)*
        Binomial[n - 1, k - 1]/
         HarmonicNumber[10^10000, s + k/n + 1/n], {k, 1, n}]);
N[%, n]
(*end*)

Output:

0.49999999999999999999999999999999999999999999999999999999999999999999\
821263876336076842104900392907 + 
 14.134725141734693790457251983562470270784257115699243175685567460149\
95997175784113908425467589217131 I

$s= 3/2 + 14i \\ M=10^{10000}$

$$\Re\lim_{n \rightarrow 100} \left( \left[ 1- \left( \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\sum_{m=1}^{m=M} 1/m^{\left(\tfrac{k}{n}+s\right)}} \Bigg/ \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\sum_{m=1}^{m=M} 1/m^{\left(\tfrac{k}{n}+s+\tfrac{1}{n}\right)}} \right) \right]^{-1} +\frac1n + s \right) = $$ 0.49999999999999999999999999999999999999999999999999999999999999999999\ 821263876336076842 + 14.134725141734693790457251983562470270784257115699243175685567460149\ 959971757841139084 I

$\endgroup$

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