The unproved Riemann hypothesis states that all nontrivial zeros of the Riemann zeta function lie along a line in the complex plane with real part of one half. With a JavaScript implementation of this function, such as that of Math, one can easily visualize how this conjecture is possible.

The zeta function is defined by an infinite summation and an equivalent product:

$\zeta \left(z\right)=\sum _{n=1}^{\infty }\frac{1}{n^z}=\prod _{\mathrm{primes}}\frac{1}{1-\frac{1}{p^z}}$

Their equivalence is verified by expanding the fractions on the right-hand side in geometric series, and recognizing that the resulting inverse products are all possible factorizations of integers.

The zeta function has no zeros on the right half-plane $Rez>1$ . If it did, one of the factors in the product definition would have to be zero, but each factor is proportional to

$p^z=p^{x}exp(iylnp)=p^x[cos(ylnp)+isin(ylnp\left)\right]$

The power of a positive integer for $x>1$ is always greater than one, and since sine and cosine are never simultaneously zero, this expression cannot be zero.

The Riemann zeta function satisfies the functional equation

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

In the left half-plane $Rez<0$ this equation becomes

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

It was just established that the final factor is never zero on this domain. The first two factors are never zero for the same reason as the factor above, and the gamma function is always greater than zero by definition. One might think there could be a zero as $z\to -\infty$ , but the gamma function grows faster than any power and prevents this. That leaves only the zeros of the sine function at negative even integers:

$\zeta (-2n)=0,n=1,2,3,\dots$

These are known as the trivial zeros. Given a relation between the zeta function and Bernoulli numbers,

$\zeta (-2n)=-\frac{B_{2n+1}}{2n+1},n>0$

the existence of these zeros is equivalent to the fact that odd Bernoulli numbers other than $B_1$ are identically zero.

From the direct functional equation, it initially appears that there would be zeros at the origin and positive even integers. The first does not occur because the zeta function has a simple pole with residue one at $z=1$ , so that at the origin one has

$\zeta \left(0\right)\approx \frac{1}{\pi }\times \frac{\pi z}{2}\times \frac{1}{(1-z)-1}=-\frac{1}{2}$

There are no zeros at positive integers because a pole of the gamma function balances the zero of the sine function. It is not difficult to show this, but it is simpler to just set $z=1-2n$ and rearrange:

$\begin{array}{l}\zeta \left(2n\right)=(-1{)}^n{2}^{2n-1}{\pi }^{2n}\frac{\zeta (1-2n)}{\Gamma \left(2n\right)}\\ \phantom{\zeta \left(2n\right)}=(-1{)}^{n+1}\frac{(2\pi {)}^{2n}}{2\left(2n\right)!}{B}_{2n}=\frac{(2\pi {)}^{2n}}{2\left(2n\right)!}\left|B_{2n}\right|\end{array}$

This expression was known to Euler.

If there are no zeros for $Rez>1$ and only trivial zeros for $Rez<0$ , then the interesting zeros are in the critical strip $0<Rez<1$ . This is can be explored by visualizing cross-sections of the real and imaginary parts of the function along the imaginary length of the strip. Here is how that appears, with the real part in blue and imaginary in red:

A zero of a complex function means that both real and imaginary parts are zero at the same time, which in this case is the same point on the real axis. Manipulating the imaginary part by holding down a cursor key allows one to locate these zeros to two decimal places. These values are approximately

±14.13, ±21.02, ±25.01, ±30.42, ±32.94, ±37.59, ±40.92, ±43.33, ±48.01, ±49.77

The value necessarily occur in conjugate pairs, since the zeta function has purely real coefficients in its expansion. They also all occur at the same point on the real axis, in agreement with the Riemann hypothesis. Spooky cool!

This interactive graphic constitutes a visual proof of the Riemann hypothesis in the given portion of the critical strip. One can easily see that there are no other zeros than those on the critical line $z=\frac{1}{2}+iy$ . For much of the imaginary length of the strip, one or both parts of the function are far from the real axis, and as such far from a zero of the complex function.

The functional equation above is useful for relating the two halves of the complex plane. It can be put in a symmetrical form thusly:

$\begin{array}{c}\zeta \left(z\right)=2^z{\pi }^{z-1}\frac{\pi }{\Gamma \left(\frac{z}{2}\right)\Gamma (1-\frac{z}{2})}\Gamma (1-z)\zeta (1-z)\\ \Gamma \left(\frac{z}{2}\right){\pi }^{-z/2}\zeta \left(z\right)=2^{z+1}{\pi }^{1/2}\frac{\Gamma (-z)}{\Gamma (-\frac{z}{2})}{\pi }^{-(1-z)/2}\zeta (1-z)\\ \Gamma \left(\frac{z}{2}\right){\pi }^{-z/2}\zeta \left(z\right)=\Gamma \left(\frac{1-z}{2}\right){\pi }^{-(1-z)/2}\zeta (1-z)\end{array}$

In the first step the sine is replaced using a reflection formula for the gamma function. In the second factors are rearranged and arguments of the remaining gamma functions reduced by one. The third uses the Legendre duplication to simplify.

This symmetrical form makes clear that the combination on the left-hand side is invariant under the replacement $z\to 1-z$ . The combination of functions has simple poles at $z=0$ and $z=1$ , one each from the gamma and zeta functions, both with residue one. These poles can be removed by multiplying both sides of the equation by $\frac{z(z-1)}{2}$ , since the numerator is zero at both poles. This combination is invariant under the same replacement. The denominator is presumably chosen to match the argument of the gamma function on the left-hand side.

Thus if one defines the function

$\xi \left(z\right)=\frac{z(z-1)}{2}\Gamma \left(\frac{z}{2}\right){\pi }^{-z/2}\zeta \left(z\right)$

the functional equation assumes the extremely simple form

$\xi \left(z\right)=\xi (1-z)$

Evaluating the value of this function at the two removed poles is instructive and verifies the functional equation for these points:

$\begin{array}{l}\xi \left(0\right)\approx -\frac{z}{2}\times \frac{2}{z}\times \zeta \left(0\right)=\frac{1}{2}\\ \xi \left(1\right)\approx \frac{z-1}{2}\times \Gamma \left(\frac{1}{2}\right)\times \frac{1}{\sqrt{\pi }}\times \frac{1}{z-1}=\frac{1}{2}\end{array}$

This function has the additional feature that it is pure real on the critical line

$\xi (\frac{1}{2}+iy)=\xi (\frac{1}{2}-iy)=\overline{\xi (\frac{1}{2}+iy)}$

and for this reason is useful for finding zeros numerically along this line. Unfortunately the function becomes small in absolute value fairly quickly, so a plot along the critical line is not interesting:

What is more interesting is to repeat the interactive graphic above, replacing the zeta function with this Riemann xi function. Here is how that appears, again with the real part in blue and imaginary in red:

The imaginary part of this function is in general quite linear across the critical strip, naturally passing through the critical line. The real part of the function appears to be tangent to the real axis at zeros. And once again one can easily see that there are no other zeros than those on the critical line.

Uploaded 2023.09.15 — 2023.12.27 analyticphysics.com