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I have heard that for improving the error term in the Prime Number Theorem, we need better and better estimates on the zero free region. I have also heard that the best possible error term comes from assuming the Riemann Hypothesis. But can anyone explain intuitively why it is the case?

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Intuitively, we can reduce the error term for the prime number theorem to a statement about how far to the left we can take a line before we hit a zero of the Riemann zeta function. To try to illustrate this, let's go through a bird's eye proof of the prime number theorem.

The proof relies on understanding the analytic properties of the Riemann zeta function $\zeta(s)$. In particular, the classical approach to proving the Prime Number Theorem is to understand the growth of $\Lambda(n)$, the von Mongoldt function, which is $\log p$ if $n = p^k$ for $p$ a prime and $0$ otherwise.

One can show that $\displaystyle \psi(x) = \sum_{n \leq x} \Lambda(n)$ behaves like $\pi(x) \log x$, and so proving the Prime Number Theorem is akin to proving that $\psi(x) \sim x$. The relation to $\zeta(s)$ is the fact that $$ -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n \geq 1} \frac{\Lambda(n)}{n^s},$$ so the von Mongoldt function appears as the coefficients of a Dirichlet series whose analytic information is contained in the Riemann zeta function. To get that information, you perform an integral transform whose effect is to pull out the first $x$ coefficients. This is the Mellin transform $$ \psi(x) = \frac{1}{2\pi i} \int_{2 - i \infty}^{2 + i\infty} \left( - \frac{\zeta'(s)}{\zeta(s)} \right)\frac{x^s}{s} ds,$$ for $x$ non-integers.

Now it boils down to using the residue theorem and understanding convergence. We'll consider the former and waive the latter.

As we move the line of integration to the left, we pick up residues from the zeroes of $\zeta(s)$. The largest zero is at $s = 1$, which has the effect of contributing an $x$. For each additional zero of $\zeta(s)$, which I'll denote by $\rho$, we get additional contribution. In total, we would get $$ \psi(x) = x - \sum_{\rho} \frac{x^\rho}{\rho} - \frac{\zeta'(0)}{\zeta(0)} - \frac{1}{2} \log( 1 - x^{-2}).\tag{1}$$

So if there is a zero of real part $\rho$, then there is a contribution to the growth of $\pi(x)\log x$ of order $x^\rho$. The Riemann hypothesis states that the only zero of real greater than $1/2$ is from $1$, which means that the secondary growth terms contribute only things on the order of $x^{1/2}$. In particular, we would have that $\lvert\pi(x)\log x - x\rvert \ll x^{1/2 + \epsilon}$ for any $\epsilon > 0$, and where the $\epsilon$ comes from the fact that we are summing very many zeroes (and the sum contributes something like logarithmic growth).

The big idea is from equation $(1)$ above.Having zeroes with large real part give larger "error" terms in the prime number theorem.

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    $\begingroup$ thanks a lot for this nice answer $\endgroup$ – happymath Dec 27 '14 at 4:26
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It's hard to capture an intuition, but the authors of this book make a great point of considering the zeta zeros as the "spectrum of the primes", meaning that the distribution of the primes is encoded in the zeta zeros.

More specifically, the more closely the zeros stay to the critical line, the more evenly distributed they are. And the RH says that the primes are as evenly distributed as we could expect it - the $\frac{1}{2}$ in the real part of the zeros directly translates to the exponent $O(x^{1/2+\epsilon})$ in the error term. Think of it that way: We want to push the error bound closer and closer to the critical line, but we can't push it past any zero. So the best we could hope for is to have it an epsilon from the critical line. In the meantime, our best error terms rely on estimates of the zero free regions.

If you need a different angle, I wrote an article explaining how the RH lets us interpret primes as the random toss of a fair coin.

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  • $\begingroup$ thanks a lot for the article $\endgroup$ – happymath Dec 27 '14 at 4:53

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