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?

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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.

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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.

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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

$j$-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 2

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 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;
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

Answer 4

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:

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