What is the formula for the first Riemann zeta zero?

Question

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...

Answer 1

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

Answer 2

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}]]

Answer 3

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.

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;
sI + h[sN[I]]/g[sN[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

Answer 4

$j$-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*)

Answer 5

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:

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

Answer 6

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