Counting distinct zeros of the Riemann zeta-function

David W. Farmer

Submitted: December 1, 1994; Accepted: December 13, 1994.

Abstract. Bounds on the number of simple zeros of the derivatives of a function are used to give bounds on the number of distinct zeros of the function.

The Riemann ξ-function is defined by ξ(s) = H(s)ζ(s), where H(s) = \frac{1}{2}s(s−1)π^{− \frac{1}{2} s}Γ(\frac{1}{2}s) and ζ(s) is the Riemann ζ-function. The zeros of ξ(s) and its derivatives are all located in the critical strip 0 < σ < 1, where s = σ + it. Since H(s) is regular and nonzero for σ > 0, the nontrivial zeros of ζ(s) exactly correspond to those of ξ(s). Let ρ^{(j)} = β + iγ denote a zero of the j^{th} derivative ξ^{(j)}(s), and denote its multiplicity by m(γ). Define the following counting functions:

N^{(j)}(T) = ∑_{ρ^{(j)}=β+iγ} 1  zeros of ξ^{(j)}(σ + it) with 0 < t < T
N(T) = N^{(0)}(T)  zeros of ξ(σ + it) with 0 < t < T
N_{s}^{(j)}(T) = ∑_{{ρ^{(j)}=β+iγ \atop m(γ)=1}} 1  simple zeros of ξ^{(j)}(σ + it) with 0 < t < T

N_{s, \frac{1}{2}}^{(j)} (T) = ∑_{{ρ^{(j)}= \frac{1}{2} +iγ \atop m(γ)=1}} 1  simple zeros of ξ^{(j)}(\frac{1}{2} + it) with 0 < t < T

M_{r}(T) = ∑_{{ρ^{(0)}=β+iγ \atop m(γ)=r}} 1  zeros of ξ(σ + it) of multiplicity r with 0 < t < T

M_{≤r}(T) = ∑_{{ρ^{(0)}=β+iγ \atop m(γ)≤r}} 1  zeros of ξ(σ + it) of multiplicity ≤ r with 0 < t < T

where all sums are over 0 < γ < T, and zeros are counted according to their multiplicity. It is well known that N^{(j)}(T) ∼ \frac{1}{2π} T log T. Let

α_{j} = lim inf_{T →∞} \frac{N_{s, \frac{1}{2}}^{(j)} (T)}{N^{(j)}(T)}. β_{j} = lim inf_{T →∞} \frac{N_{s}^{(j)}(T)}{N^{(j)}(T)}.

Thus, β_{j} is the proportion of zeros of ξ^{(j)}(s) which are simple, and α_{j} is the proportion which are simple and on the critical line. The best currently available bounds are α_{0} > 0.40219, α_{1} > 0.79874, α_{2} > 0.93469, α_{3} > 0.9673, α_{4} > 0.98006, and α_{5} > 0.9863. These bounds were obtained by combining Theorem 2 of [C2] with the methods of [C1]. Trivially, β_{j} ≥ α_{j}.

1991 Mathematical Subject Classification: 05A20, 11M26 Let N_{d}(T) be the number of distinct zeros of ξ(s) in the region 0 < t < T. That is,

N_{d}(T) = ∑_{n=1}^{∞} \frac{M_{n}(T)}{n}.  (1)

It is conjectured that all of the zeros of ξ(s) are distinct: N_{d}(T) = N(T), or equivalently, all of the zeros are simple: N_{s}^{(0)}(T) = N(T). From the bound on α_{0} we have N_{s}^{(0)}(T) > κ N(T), with κ = 0.40219. We will use the bounds on β_{j} to obtain the following

Theorem. For T sufficiently large,

N_{d}(T) > k N(T),

with k = 0.63952 . . .. Furthermore, given the bounds on β_{j}, this result is best possible.

We present two methods for determining lower bounds for N_{d}(T). These methods employ combinatorial arguments involving the β_{j}. We note that the added information that α_{j} detects zeros on the critical line is of no use in improving our result. Everything below is phrased in terms of the Riemann ξ-function, but the manipulations work equally well for any function such that it and all of its derivatives have the same number of zeros. We write f(T) ≳ g(T) for f(T) ≥ g(T)+o(N(T)) as T → ∞. For example, N_{s}^{(j)}(T) ≳ β_{j}N(T) means N_{s}^{(j)}(T) ≥ (β_{j} + o(1)) N(T) as T → ∞.

The first method starts with the following inequality of Conrey, Ghosh, and Gonek [CGG]. A simple counting argument yields

N_{d}(T) ≥ ∑_{r=1}^{R} \frac{M_{≤r}(T)}{r(r + 1)} + \frac{M_{≤R+1}(T)}{R + 1}.  (2)

To obtain lower bounds for M_{≤r}(T) we note that if ρ is a zero of ξ(s) of order m ≥ n + 2 then ρ is a zero of order m − n ≥ 2m/(n + 2) ≥ 2 for ξ^{(n)}(s). Thus,

N_{s}^{(n)}(T) ≤ N(T) − \frac{2}{n + 2}(N(T) − M_{≤n+1}(T)),

which gives

M_{≤n}(T) ≳ (\frac{β_{n−1}(n + 1) − n + 1}{2}) N(T).  (3)

The bounds for α_{j} now give: M_{≤1}(T) ≳ 0.40219N(T), M_{≤2}(T) ≳ 0.69812N(T), M_{≤3}(T) ≳ 0.86938N(T), M_{≤4}(T) ≳ 0.91825N(T), M_{≤5}(T) ≳ 0.94019N(T), and M_{≤6}(T) ≳ 0.9520N(T). Inserting these bounds into inequality (2) with R = 5 gives N_{d}(T) ≳ 0.62583N(T). We note that the lower bounds for M_{≤n}(T) are best possible in the sense that, for each n separately, equality could hold in (3). However, inequality (3) is not simultaneously sharp for all n, and this possibility imparts some weakness to the result. A lower bound for N_{d}(T) was calculated in [CGG] in a spirit similar to the above computation, but it was mistakenly assumed that M_{≤n}(T) ≳ β_{n−1}N(T), rendering their bound invalid.

Our second method eliminates the loss inherent in the first method. We start with this

the electronic journal of combinatorics 2 (1995), #R1 3

Lemma. In the notation above,

N_{s}^{(n)}(T) ≤ ∑_{j=1}^{n+1} M_{j}(T) + n ∑_{j=n+2}^{∞} \frac{M_{j}(T)}{j}.

Proof. Suppose ρ is a zero of order j for ξ(s). If j ≥ n + 2 then ρ is a zero of order j − n for ξ^{(n)}(s), so ξ^{(n)}(s) has at least ∑_{j=n+2}^{∞} \frac{(j − n)M_{j}(T)}{j} zeros of order ≥ 2. Thus,

N_{s}^{(n)}(T) ≤ N^{(n)}(T) − ∑_{j=n+2}^{∞} \frac{(j − n)M_{j}(T)}{j}

= ∑_{j=0}^{∞} M_{j}(T) − ∑_{j=n+2}^{∞} \frac{(j − n)M_{j}(T)}{j}

= ∑_{j=1}^{n+1} M_{j}(T) + n ∑_{j=n+2}^{∞} \frac{M_{j}(T)}{j},

as claimed.

Combining the Lemma with (1) we get

N_{s}^{(n)}(T) ≤ nN_{d}(T) + n ∑_{j=1}^{n+1} (\frac{1}{n} − \frac{1}{j})M_{j}(T).  (4)

Let I_{n} denote the inequality (4). Then, in the obvious notation, a straightforward calculation finds that the inequality

I_{J} + ∑_{n=1}^{J−1} 2^{J−n−1}I_{n}

is equivalent to

(2^{J} − 1)N_{d}(T) + ∑_{n=1}^{J+1} \frac{M_{n}(T)}{n} ≥ 2^{J−1}M_{1}(T) + N_{s}^{(J)}(T) + ∑_{n=1}^{J−1} 2^{J−n−1}N_{s}^{(n)}(T).  (5)

This implies

N_{d}(T) ≥ 2^{−J} (2^{J−1}N_{s}^{(0)}(T) + N_{s}^{(J)}(T) + ∑_{n=1}^{J−1} 2^{J−n−1}N_{s}^{(n)}(T))
≳ 2^{−J} (2^{J−1}β_{0} + β_{J} + ∑_{n=1}^{J−1} 2^{J−n−1}β_{n})N(T).  (6)

Choose J = 5 and use the trivial inequality β_{j} ≥ α_{j} and the bounds for α_{j} to obtain the Theorem.

Finally, we show that our result is best possible. In other words, if our lower bounds for the β_{j} were actually equalities, then the lower bound given by (6) is sharp. We will accomplish this by showing that the M_{n}(T), the number of zeros of ξ(s) with multiplicity exactly n, can be assigned values which achieve the bounds on β_{j}, and which yield a value of N_{d}(T) which is arbitrarily close to the lower bound given by (6).

Suppose we have lower bounds for β_{j}, for 0 ≤ j ≤ J, and let K ≥ J + 2. Suppose we had the following four equalities:

M_{1}(T) = β_{0}N(T),
M_{K}(T) = \frac{K}{K−J} (1 − β_{J})N(T),

M_{J+1}(T) = \frac{J + 1}{2} (β_{J} − β_{J−1} − \frac{1 − β_{J}}{K − J} ) N(T),

and for 2 ≤ n ≤ J,

M_{n}(T) = \frac{n}{2} (\frac{3β_{n−1}}{2} − β_{n−2} − 2^{n−J−1}β_{J} − \frac{1 − β_{J}}{2^{J−n+1}(K − J)} − ∑_{j=n}^{J−1} 2^{n−j−2}β_{j})N(T)

and M_{j}(T) = 0 otherwise. Then ∑_{j=1}^{∞} M_{j}(T) = N(T) and for 0 ≤ n ≤ J we have

∑_{j=1}^{n+1} M_{j}(T) + n ∑_{j=n+2}^{∞} \frac{M_{j}(T)}{j} = β_{n}N(T),  (7)

and

∑_{n=1}^{∞} \frac{M_{n}(T)}{n} = 2^{−J}(2^{J−1}β_{0} + β_{J} + ∑_{n=1}^{J−1} 2^{J−n−1}β_{n})N(T) + \frac{(1 − β_{J})2^{−J}}{K − J} N(T).  (8)

Since the left side of (8) is N_{d}(T) and the second term on the right side can be made arbitrarily small by choosing K large, we conclude that (6) is sharp. There are two things left to check. The given values of M_{n}(T) must be positive when K is large. It is easy to check this for J = 5 and our lower bounds for β_{j}. And since we supposed that our bounds for β_{j} are sharp, we must show that N_{s}^{(j)}(T) = β_{j}N(T). To see this, note that, generically, the left side of (7) equals N_{s}^{(j)}(T). In other words, the zeros of the derivatives of a generic function are all simple, except for those which are “tied up” in high-order zeros of the original function.

By computing further values of α_{j}, enabling us to take a larger value of J in (6), we could improve the result slightly: this is due to a decrease in the loss in passing from (5) to (6). The bound M_{≤6}(T) ≳ 0.952N(T) implies that this improvement could increase the lower bound we obtained by at most 0.00021N(T).

References

[C1] J. B. Conrey, Zeros of derivatives of Riemann’s ξ-function on the critical line, II, J. Number Theory 17 (1983), 71-75.

[C2] J. B. Conrey, More than two fifths of the zeros of the Riemann zeta function are on the critical line, J. reine angew. Math. 399 (1989), 1-26.

[CGG] J. B. Conrey, A. Ghosh, and S. M. Gonek, Mean values of the Riemann zeta-function with application to distribution of zeros, Number Theory, Trace Formulas and Discrete Groups, (1989), 185-199.

[L] N. Levinson, More than one-third of the zeros of Riemann’s zeta-function are on σ = \frac{1}{2}, Adv. in Math., 13 (1974), 383-436.

Mathematical Sciences Research Institute 1000 Centennial Drive Berkeley, CA 94720 farmer@msri.org