A Note on P\'Olya's Observation Concerning Liouville's Function

Home , Liouville function, Mertens conjecture

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ρ of ζ(s) in the upper half-plane. Ingham cast similar doubt on the Mertens conjecture M(n) √n, which was subsequently disproved in a remarkable tour| de force| ≤by Odlyzko and te Riele [11, 1985]. More recent results and improved bounds were given by Kotnik and te Riele [7, 2006]; see also Kotnik and van de Lune [6, 2004]. In view of Ingham’s results, it was no surprise when Haselgrove showed [2, 1958] that P´olya’s conjecture is false. He did not give an explicit counter-example, but his proof suggested that L(u) might be positive in the vicinity of u 1.8474 10361. Sherman Lehman [8, 1960]≈ gave an× algorithm for calculating L(n) similar to Meissel’s [10, 1885] formula for the prime-counting function π(x), and found the counter-example L(906180359) = +1. Tanaka [14, 1980] found the smallest counter-example L(n) = +1 for n = 906150257. Walter M. Lioen and Jan van de Lune [circa 1994] scanned the range n 2.5 1011 using a fast sieve, but found no counter-examples beyond≤ those× of Tanaka. More recently, Borwein, Ferguson and Mossinghoﬀ [1, 2008] showed that L(n) = +1160327 for n = 351753358289465. Humphries [3, 4] showed that, under certain plausible but unproved hypotheses (including the Riemann Hypothesis), there is a limiting logarithmic distribution of L(n)/√n, and numerical computations show that the logarithmic density of the set n N L(n) < 0 is approxi- mately 0.99988. Humphries’ approach followed{ ∈ | that of Rubinstein} and Sarnak [13], who investigated “Chebyshev’s bias” in prime “races”. Here we show in an elementary manner, and without any unproved hypotheses, that λ(n) is (in a certain sense) “negative on average”. To prove this, all that we need are some well-known facts about Mellin transforms, and the functional equation for the Jacobi theta function (which may be proved using Poisson summation). Our main result is:

Theorem 1. There exists a positive constant c such that for every (ﬁxed) N N ∈

∞ λ(n) c 1 = + + O(xN ) as x 0. enπx +1 −√x 2 ↓ Xn=1

Thus, a weighted mean of λ(n) , with positive weights initially close to a constant (1/2) and becoming{ } small for n 1/x, is negative for ≫ x < x0 and tends to as x 0. In the ﬁnal section−∞ we mention↓ some easy results on the M¨obius function µ(n) to contrast its behaviour with that of λ(n). POLYA’S´ OBSERVATION CONCERNING LIOUVILLE’S FUNCTION 3

2. Proof of Theorem 1 We prove Theorem 1 in three steps.

Step 1. For x> 0,

∞ λ(n) θ(x) 1 = φ(x)= − , enπx 1 2 Xn=1 − where ∞ 2 2 φ(x) := e−k πx, θ(x) := e−k πx. Xk=1 Xk∈Z Step 2. For x> 0,

∞ λ(n) = φ(x) 2φ(2x). enπx +1 − Xn=1 Step 3. Theorem 1 now follows from the functional equation 1 1 θ(x)= θ √x x for the Jacobi theta function θ(x).

Proof of Theorem 1.

(1) In the following, we assume that Re (s) > 1, so the Dirichlet series and integrals are absolutely convergent, and interchanging the orders of summation and integration is easy to justify. As mentioned above, it is well-known that

∞ −s λ(n) −1 1 p ζ(2s) = 1+ p−s = − = . ns 1 p−2s ζ(s) Xn=1 Yp Yp − Deﬁne ∞ λ(n) f(x) := , (x> 0). enx 1 Xn=1 − We will use the well known fact that if two suﬃciently well-behaved functions (such as ours below) have the same Mellin transform then the functions are equal. POLYA’S´ OBSERVATION CONCERNING LIOUVILLE’S FUNCTION 4

The Mellin transform of f(x) is

∞ ∞ ∞ λ(n) F (s) := f(x)xs−1 dx = xs−1 dx Z Z enx 1 0 0 Xn=1 − ∞ ∞ xs−1 ∞ λ(n) ∞ xs−1 = λ(n) dx = dx Z enx 1 ns × Z ex 1 Xn=1 0 − Xn=1 0 − ζ(2s) = ζ(s)Γ(s)= ζ(2s)Γ(s). ζ(s) × We also have

∞ ∞ ∞ x 2 φ xs−1 dx = e−n x xs−1 dx Z π Z 0 0 Xn=1 ∞ 1 ∞ = e−xxs−1 dx = ζ(2s)Γ(s), n2s × Z Xn=1 0 so the Mellin transforms of f(x) and of φ(x/π) are identical. Thus f(x)= φ(x/π). Replacing x by πx, we see that

∞ ∞ λ(n) 2 = e−k πx, enπx 1 Xn=1 − Xk=1 completing the proof of step (1). (2) Observe that 1 1 2 = , enπx +1 enπx 1 − e2nπx 1 − − so, from step (1),

∞ λ(n) = φ(x) 2φ(2x). enπx +1 − Xn=1

(3) Using the functional equation for θ(x), we easily ﬁnd that c 1 1 1 1 φ(x) 2φ(2x)= + + φ √2 φ − −√x 2 √x x − 2x with c =(√2 1)/2 > 0, proving our claim, since the “error” term is bounded by φ−(1/x)/√x exp( π/x)/√x = O(xN ) as x 0 (for any ﬁxed exponent N). ∼ − ↓ POLYA’S´ OBSERVATION CONCERNING LIOUVILLE’S FUNCTION 5

3. Remarks on the Mobius¨ function We give some further applications of the identity 1 1 2 ( ) = ∗ z +1 z 1 − z2 1 − − that we used (with z = enπx) in proving step (2) above. Lemma 2. For x < 1, we have | | ∞ xn µ(n) = x 2x2. xn +1 − Xn=1 Proof. Assume that x < 1. It is well known that | | ∞ xn µ(n) = x, 1 xn Xn=1 − in fact this “Lambert series” identity is equivalent to the Dirichlet series identity µ(n)/ns =1/ζ(s). Writing y =1/x, we have P ∞ µ(n) =1/y. yn 1 Xn=1 − If follows on taking z = yn in our identity ( ) that ∗ ∞ µ(n) ∞ µ(n) ∞ µ(n) = 2 = y−1 2y−2. yn +1 yn 1 − y2n 1 − Xn=1 Xn=1 − Xn=1 − Replacing y by 1/x gives the result. Corollary 3. ∞ µ(n) =0. 2n +1 Xn=1 Proof. Take x =1/2 in Lemma 2. If follows from Lemma 2 that ∞ xn lim µ(n) = 1, x↑1 xn +1 − Xn=1 so that one might say that in this sense µ(n) is negative on average. However, this is much weaker than what we showed in Theorem 1 for L(n), where the corresponding sum tends to . The “complex- analytic” reason for this diﬀerence is that ζ(2s)/ζ−∞(s) has a pole (with negative residue) at s =1/2, but 1/ζ(s) is regular at s = 1. POLYA’S´ OBSERVATION CONCERNING LIOUVILLE’S FUNCTION 6

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Mathematical Sciences Institute, Australian National University, Canberra, ACT 0200, Australia E-mail address: [email protected]

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