The Riemann hypothesis: Equivalents of the hypothesis

One reason that the Riemann hypothesis is so important to number theorists is that, as we’ve noted, it implies the smallest possible error estimate in the prime number theorem. In other words, it gives as much information as possible about the distribution of prime numbers. In fact, the condition

|π(x) – Li(x)| = O(x1/2 log(x))

has been proven equivalent to the Riemann hypothesis. Knowing this about the distribution of primes can be used to derive the best possible estimates for many other number theoretic quantites as well.

Because of the importance of the hypothesis, it is often helpful to have other equivalent formulations available. So we’ll just mention some of them now. It’s fairly easy to guess at some equivalent statements. For example, think of functions that can be defined in terms of ζ(s) which would have singularities if ζ(s) = 0 for some s with Re(s) > 1/2. Thus the Riemann hypothesis is true if and only if 1/ζ(s) is holomorphic for Re(s) > 1/2.

Since ζ(s) has a simple pole at s=1, (s-1)ζ(s) is holomorphic there. So its logarithm, log(s-1) + log(ζ(s)), is too. Except, perhaps, right at s=1, but in a neighborhood of s=1, the derivative of that is also holomorphic, and equals (s-1)-1 + ζ′(s)/ζ(s). ζ′(s)/ζ(s) is the “logarithmic derivative” of ζ(s), (log ζ(s))′. Using power series representations, it’s clear

ζ′(s)/ζ(s) + (s-1)-1

is holomorphic at s=1, and so we conclude it is holomorphic in the region {s ∈ ℂ: Re(s) > 1/2} if and only if the Riemann hypothesis is true. Equivalently, log((s-1)ζ(s)) is holomorphic in the same region. (Remark: the logarithmic derivative ζ′(s)/ζ(s) has at most a simple pole at zeros of ζ(s), even if some of those zeros are multiple – although none have been found that are.)

We worked a bit with ζ′(s)/ζ(s) in connection with the function ψ(x). When all the gory details are processed, it turns out that two things are true:

If for some α < 1 one has

ψ(x) = x + O(xα) as x → ∞

then all zeros ρ of ζ(s) in the critical strip have Re(ρ) ≤ α

and

If all zeros of ζ(s) are contained in the strip {s ∈ ℂ: 1-θ ≤ Re(s) ≤ θ} for some θ < 1, then

ψ(x) = x + O(xθ (log(x))2) as x → ∞

Since ζ(s) = 0 for infinitely many s with Re(s) = 1/2, then by the first statement, α = 1/2 is the smallest exponent we can have such that ψ(x) = x + O(xα). In other words, no matter what, we can never get any substantially better estimate of ψ(x) than ψ(x) = x + O(x1/2). If that is true for ψ(x), then we have also π(x) = Li(x) + O(x1/2).

On the other hand, if the Riemann hypothesis is correct, we can apply the second statement with θ = 1/2 to conclude that in fact ψ(x) = x + O(x1/2 (log(x))2) and hence π(x) = Li(x) + O(x1/2 log(x)). Conversely, by the first statement, this estimate implies the Riemann hypothesis.

In other words, the Riemann hypothesis is equivalent to the estimate π(x) = Li(x) + O(x1/2 log(x)), and no better estimate of this sort is possible.

Unfortunately, we must point out that no results have actually been proven of the form that all zeros of ζ(s) must have Re(ρ) ≤ α for some α < 1. Even now, the best result we have for a zero-free region is Hadamard’s, which illustrates how little actual progress on the Riemann hypothesis has been made in almost 100 years.

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