# Prime Secret: ζ(s)

Riemann intuitively found the Zeta Function ζ(s), but couldn’t prove it. Computer ‘tested’ it correct up to billion numbers.

$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots$

Or equivalently (see note *)

$\frac {1}{\zeta(s)} =(1-\frac{1}{2^{s}})(1-\frac{1}{3^{s}})(1-\frac{1}{5^{s}})(1-\frac{1}{p^{s}})\dots$

ζ(1) = Harmonic series (Pythagorean music notes) -> diverge to infinity
(See note #)

ζ(2) = Π²/6 [Euler]

ζ(3) = not Rational number.

1. The Riemann Hypothesis:
All non-trivial zeros of the zeta function have real part one-half.

ie ζ(s)= 0 where s= ½ + bi

Trivial zeroes are s= {- even Z}:
s(-2) = 0 =s(-4) =s(-6) =s(-8)…

You might ask why Re(s)=1/2 has to do with Prime number ?

There is another Prime Number Theorem (PNT) conjectured by Gauss and proved by Hadamard and Poussin:

π(Ν) ~ N / log N
ε = π(Ν) – N / log N
The error ε hides in the Riemann Zeta Function’s non-trivial zeroes, which all lie on the Critical line = 1/2 :

All non-trivial zeroes of ζ(s) are in Complex number between ]0,1[ along real line x=1/2

2. David Hilbert:

If I were to awaken after 500 yrs, my 1st question would be: Has Riemann been proven?’

It will be proven in future by a young man. ‘uncorrupted’ by today’s math.

Note (*):

$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots = \sum \frac {1}{n^{s}}$ …[1]

$\frac {1}{2^{s}}\zeta(s) = \frac{1}{2^{s}}(1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots)$

$\frac {1}{2^{s}}\zeta(s) = \frac {1}{2^{s}}+ \frac{1}{4^{s}} + \frac{1}{6^{s}} + \frac{1}{8^s} +\dots$ … [2]

[1]-[2]:

$(1- \frac{1}{2^{s}})\zeta(s) = 1+ \frac{1}{3^{s}} + \frac{1}{5^{s}} + \dots + \frac{1}{p^{s}} +\dots$

$\text {Repeat with} (1-\frac{1}{3^s}) \text { both sides:}$

$(1- \frac{1}{3^{s}})(1- \frac{1}{2^{s}})\zeta(s) = 1+ \frac{1}{5^{s}} + \frac{1}{7^{s}} + \dots + \frac{1}{p^{s}} +\dots$

Finally,

$(1- \frac{1}{p^{s}}) \dots (1- \frac{1}{5^{s}})(1- \frac{1}{3^{s}})(1- \frac{1}{2^{s}})\zeta(s) = 1$

Or

$\zeta(s) = \prod \frac {1} {1- \frac{1}{p^{s}}}= \sum \frac {1}{n^{s}}$

Note #:
$\zeta(s) = \prod \frac {1} {1- \frac{1}{p^{s}}}= \sum \frac {1}{n^{s}}$

Let s=1
RHS: Harmonic series diverge to infinity
LHS:
$\prod \frac {1}{1- \frac{1}{p}}= \prod \frac{p}{p-1}$
Diverge to infinity => there are infinitely many primes p

English: Zero-free region for the Riemann_zeta_function (Photo credit: Wikipedia)