The Riemann Hypothesis and Claude

Anthropic announced that their latest model has made progress on the Riemann hypothesis. I’ll give enough background on the hypothesis to understand the result and its importance (but not the argument). Much of this is based on Terence Tao’s blog and also this video.

Consider the infinite sum $\sum^\infty_{n=1}n^{-s}$ for complex numbers $s$ with $\text{Re}(s)>1$ which Riemann wrote about in 1859 in a short note. This series can be analytically continued to a function $\zeta(s)$ whose domain is all of $\mathbb{C}$. That is to say, the analytic continuation function agrees with the original on the smaller domain and is also complex differentiable everywhere. With these two properties, it turns out that the analytic continuation is unique. Well, almost; for $\zeta$, it has a pole at $s=1$.

As an example, $\zeta(-1)=-\dfrac{1}{12}$. If $\text{Re}(s)>1$, then $\zeta(s)\cdot\dfrac{1}{2^s} = \sum_{n=2k}n^{-s}$. That is, the sum is now over all the positive even integers. Thus, $\zeta(s) - \zeta(s)\cdot\dfrac{1}{2^s} = \zeta(s)(1-\dfrac{1}{2^s})$ is the sum but without any of the terms where $n$ is even. Similarly, $\zeta(s)(1-\dfrac{1}{2^s})\dfrac{1}{3^s}$ is a sum where $n$ is a multiple of 3 but not of 2. Then subtract: $\zeta(s)(1-\dfrac{1}{2^s}) - \zeta(s)(1-\dfrac{1}{2^s})\dfrac{1}{3^s} = \zeta(s)(1-\dfrac{1}{2^s})(1-\dfrac{1}{3^s})$. This equals the sum $\sum n^{-s}$ but $n$ can neither be a multiple of 2 or 3. If we continue on in this manner for all primes, we find $\zeta(s)\prod_p(1-\dfrac{1}{p^s})$ is equal to a sum over all the $n$ which have no prime factors which is just 1. So we see that for $\text{Re}(s)>1$, then $\zeta(s) = \prod_p (1-p^{-s})^{-1}$ for primes $p$. So we see that the function does have some relationship with primes.

Interestingly, there is also a kind of symmetry when taking complex conjugates. That is, $\zeta(\bar{s})=\overline{\zeta(s)}$. In particular, if $s$ is a zero, then so is $\bar{s}$.

Question: How do the zeros of $\zeta$ relate to the distribution of primes?

Let $\pi(x)$ be the prime counting function which tells you how many primes are less than or equal to a real number $x$. We’ll typically be interested in $x$ being an integer since that’s where the function could jump. So we’ll write $\pi(n)$. Gauss proposed that for large $n$, $\pi(n) \approx \dfrac{n}{\ln n}$. Later, he improved his proposal to $\pi(n) \approx \text{Li}(n)$ where $\text{Li}(n)=\int^n_2\dfrac{dx}{\ln x}$, the logarithmic integral. It has since been shown that $\pi(x)$ is approximately $\text{Li}(x)$ with some error terms for large $x$. This is called the prime number theorem and the Riemann Hypothesis would imply how large the error terms are.

In fact, there is a way to explicitly write $\pi$ in terms of the nontrivial zeros of $\zeta$ and $\text{Li}$ (and some other terms). As it turns out, all the negative even integers are zeros of $\zeta$. Remember, we analytically continued to the left half plane so the series formula doesn’t apply to this left side. These zeros are called trivial. It has been shown that all the other zeros of $\zeta$ lie in the critical strip where the real part of $s$ is between 0 and 1. The critical line is where the real part is $1/2$ and the Riemann Hypothesis is that all of the nontrivial zeros lie on the critical line. In 1951, Wiener proved that the prime number theorem is equivalent to there being no nontrivial zeros with real part equal to 1.

In 1914, Hardy showed that there are infinitely zeros on the critical line. Later, Selberg in 1942 showed something about the asymptotic ratio of nontrivial zeros. That is, let $T>0$ be a height at which we cut off the critical strip. So we have a compact set with finitely many zeros. If there were infinitely, because the function is analytic and the zeros accumulate, you’d have the constant function 0. Anyways, we can count the number of zeros on the critical line in this compact set divided by the total number of zeros in the compact set. This gives some number $L(T)$. Selberg showed that for large $T$, $L(T)>\kappa>0$.

If $\kappa = 1$, then the Riemann Hypothesis is proved. Mathematicians have worked to improve lower bounds on $\kappa$. Prior to Anthropic’s announcement, it was known that $\kappa\geq 5/12$. Anthropic is claiming that $\kappa \geq 8/12 = 2/3$. I can give no comments on the proof as I know very little number theory. But it is very interesting to see Claude (the Anthropic model) write a paper like a human would. For example, it wrote an acknowledgement section, thanking one of Anthropic’s employees for the support and encouragement.

Chebyshev $\psi$ Function

At any rate, I would like to show how the nontrivial zeros of $\zeta$ are related to a different function $\psi$ which is still related to primes. This $\psi$ is often preferable to work with over $\pi$. Above, we have that for real part greater than 1, $\zeta(s) = \prod_p (1-p^{-s})^{-1}$. Taking the log, we have $\ln \zeta(s) = -\sum_p \ln(1-p^{-s})$. The Taylor series for $\ln(1-x) = -\sum_{k=1}^\infty \dfrac{x^k}{k}$ so the expression is now $\sum_p \sum_{k=1}^\infty \dfrac{p^{-sk}}{k}$. If we differentiate the LHS, this will become $\dfrac{\zeta’(s)}{\zeta(s)}$ and the RHS will, after some algebra, be $-\sum_p \sum_{k=1}^\infty\dfrac{\ln p}{p^{sk}}$. One could write out this double sum like a 2D grid. The rows will be indexed by a prime $p$ and across the row, we see that the numerator will be $\ln p$ while the denominator will have powers of $p$. For example, $\ln(2)/2^s + \ln(2)/4^s + \ln(2)/8^s+…$

Let’s define $\Lambda(n)= \begin{cases}\ln p & n=p^k \ 0 & \text{otherwise} \end{cases}$ to make it easier to express. With this, we now have $-\dfrac{\zeta’(s)}{\zeta(s)} = \sum_{n=p^k} \dfrac{\Lambda(n)}{n^s}$ where the $p$ are primes. The LHS is entirely about the zeta function, the right is a sum over powers of prime. Because $\Lambda(n)$ is 0 when $n$ is not a power of prime, we can also just have the sum run over all natural numbers $n$.

This $\Lambda(n)$ can be used to define the Chebyshev psi function $\psi(x) = \sum_{n \leq x}\Lambda(n)$. This is like a counting function which jumps everytime we encounter a power of a prime (though the jump is not uniform). Okay, so now we have $\sum^\infty_{n=1}\dfrac{\Lambda(n)}{n^s} = -\dfrac{\zeta’(s)}{\zeta(s)}$. If we multiply both sides by $\dfrac{x^s}{2\pi i s}$, the LHS becomes $\dfrac{1}{2\pi i} \sum^\infty_{n=1}\dfrac{\Lambda(n)(x/n)^s}{s}$. Next, integrate both sides over the vertical line $\gamma$ with real part 2 (think of this as a circle of infinite radius). This line $\gamma$ is in the region where the series $\sum^\infty_{n=1}n^{-s}$ converges; we can swap the integral and sum: $\sum^\infty_{n=1}\Lambda(n) \cdot \dfrac{1}{2\pi i} \int_\gamma \dfrac{(x/n)^s}{s}\, ds$.

The Cauchy Residue theorem shows that the integral is 0 when $x<n$ and is $2\pi i$ when $x >n$. This is great because then, the LHS is simply the step function:

The RHS is defined everywhere other than $s=1$, not just when the real part is greater than 1. The Cauchy Residue Theorem tells us that the integral is just the sum of the residues of the integrand. So we just need to find the poles of the function which are at $s=0$ and all the zeros of $\zeta$ and at the pole of $\zeta$ as around there, $\zeta’(s)$ also approaches infinity.

Near $s=0$, the function looks like $\dfrac{\zeta’(0)}{\zeta(0)}\cdot \dfrac{1}{s}$ so $\dfrac{\zeta’(0)}{\zeta(0)} = \ln(2\pi)$ is the residue. For $s=1$, the residue is $-x$. At the trivial zeros $s=-2k$, one can find them to be $-\dfrac{x^{-2k}}{2k}$. The sum of all of those, by a Taylor series, is $\dfrac{1}{2}\ln(1-x^{-2})$. Let $\rho$ represent nontrivial zeros. The sum of the residues for all of those is $\sum_\rho \dfrac{x^\rho}{\rho}$. So in summary,

If we assume the Riemann Hypothesis, then the $\rho = \dfrac{1}{2}\pm i\gamma$. Introducing and then expanding exponentials gives some sines and cosines. We find that $\psi(x) \approx x- 2\sqrt{x}\sum_{\gamma > 0} \dfrac{\sin(\gamma \ln x)}{\gamma}-\ln(2\pi)$ if the Riemann Hypothesis is true.