Riemann Hypothesis/The hypothesis

Theorem 1

\zeta (-2s)=0\forall s\in \mathbb {N}

Proof

Consider the functional equation for Zeta,

\zeta (s)=2^{s}\pi ^{s-1}\sin \left({\frac {\pi s}{2}}\right)\Gamma (1-s)\zeta (1-s)

Notice that for $\zeta (-2s)$ , the sine term evaluates to $\sin(-\pi s)$ which evaluates to 0 for all integers $s$ , hence $\zeta (-2s)=0$ for all natural $s$ $\blacksquare$ .

Definition 1

These zeroes are referred to as trivial zeroes. As a set,

\zeta _{t}=\{-2s:s\in \mathbb {N} \}

Zeroes that do not lie in this set are to be referred to as non-trivial zeroes.

Note 1

The above argument cannot be applied to $(s<1)\in \mathbb {Z}$ , as $\zeta (1)$ is a simple pole ( $s=0$ ), as are negative arguments of $\Gamma$ ( $s<0$ ).

Theorem 2

All non-trivial zeroes of $\zeta$ have a real part that lies in the interval $(0,1)$

Theorem 3: $\zeta (1+it)\neq 0\forall t\in \mathbb {R}$

Take the inequality,

\cos \theta \geq -1

\implies 2(1+\cos \theta )^{2}\geq 0

\implies 3+4\cos \theta +\cos 2\theta \geq 0

Using the definition of $\zeta$ deduced in an earlier chapter,

\zeta (s)=\prod _{p|{\text{prime}}}{\frac {1}{1-p^{-s}}}

Taking the log of both sides, using $\log \left(\prod f(x)\right)=\sum \log \left(f(x)\right)$

\log \zeta (s)=\sum _{p|{\text{prime}}}\log \left({\frac {1}{1-p^{-s}}}\right)=-\sum _{p|{\text{prime}}}\log(1-p^{-s})

Writing $\log$ as a power series,

\log \zeta (s)=\sum _{p|{\text{prime}}}\sum _{n=1}^{\infty }{\frac {p^{-sn}}{n}}

Substituting $s=\sigma +it$ ,

\log \zeta (\sigma +it)=\sum _{p|{\text{prime}}}\sum _{n=1}^{\infty }{\frac {p^{-n\sigma -nit}}{n}}=\sum _{p|{\text{prime}}}\sum _{n=1}^{\infty }{\frac {1}{p^{n\sigma }}}\exp(-nit\log p)

Taking the modulus of the argument,

\log |\zeta (\sigma +it)|=\sum _{p|{\text{prime}}}\sum _{n=1}^{\infty }{\frac {1}{p^{n\sigma }}}\Re \exp(-nit\log p)=\sum _{p|{\text{prime}}}\sum _{n=1}^{\infty }{\frac {1}{p^{n\sigma }}}\Re \left(\cos(nt\log p)-i\sin(nt\log p)\right)=\sum _{p|{\text{prime}}}\sum _{n=1}^{\infty }{\frac {1}{p^{n\sigma }}}\cos(nt\log p)

Substituting appropriate values,

3\log |\zeta (\sigma )|+4\log |\zeta (\sigma +it)|+\log |\zeta (\sigma +2it)|=\sum _{p|{\text{prime}}}\sum _{n=1}^{\infty }{\frac {3+4\cos(nt\log p)+\cos(2nt\log p)}{p^{n\sigma }}}

If one lets $nt\log p=\theta$ , it should become apparent that,

\sum _{p|{\text{prime}}}\sum _{n=1}^{\infty }{\frac {3+4\cos(nt\log p)+\cos(2nt\log p)}{p^{n\sigma }}}\geq 0

Clearly implying,

\log |\zeta ^{3}(\sigma )|+\log |\zeta ^{4}(\sigma +it)|+\log |\zeta (\sigma +2it)|\geq 0

Exponentiating both sides,

\zeta ^{3}(\sigma )\zeta ^{4}(\sigma +it)\zeta (\sigma +2it)\geq 1

Let's assume that $\zeta$ has a zero at $1+it_{0}$ , then,

\lim _{\sigma \to 1^{+}}\zeta ^{3}(\sigma )\zeta ^{4}(\sigma +it_{0})\zeta (\sigma +2it_{0})=0

As $\lim _{\sigma \to 1^{+}}\zeta (\sigma )$ gives a pole, and $\zeta (1+it_{0})$ gives a zero, contradicting the previously stated inequality, proving theorem 3 by contradiction $\blacksquare$ .