Some elementary explicit bounds for two mollifications of the Moebius function
12 Pages
English
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Some elementary explicit bounds for two mollifications of the Moebius function

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12 Pages
English

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Some elementary explicit bounds for two mollifications of the Moebius function O. Ramare March 8, 2012 Abstract We prove that the sum ∑ { d≤x, (d,r)=1 µ(d)/d1+? is bounded by 1 + ?, uniformly in x ≥ 1, r and ? > 0. We prove a similar estimate for the quantity ∑ { d≤x, (d,r)=1 µ(d) Log(x/d)/d1+?. When ? = 0, r varies between 1 and a hundred, and x is below a million, this sum is non-negative and this raises the question as to whether it is non-negative for every x. 1 Introduction and results Our first result is the following: Theorem 1.1. When r ≥ 1 and ? ≥ 0, we have ? ? ? ? ∑ d≤x, (d,r)=1 µ(d) d1+? ? ? ? ? ≤ 1 + ?. This Lemma generalizes the estimate of [5, Lemme 10.2] which corre- sponds to the case ? = 0. This generalization is not straightforward at all and requires a change of proof. The case ? = 0 and r = 1 is classical.

  • follows readily

  • ??1 ?

  • explicit estimates

  • also ≤

  • very erratical

  • when ?

  • moebius function

  • useful when applying

  • lowest lower bound


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1

Some elementary explicit bounds for two
mollifications of the Moebius function

O.Ramar´

March 8, 2012

Abstract
P
1+ε
We prove that the sumµ(d)/dis bounded by 1+ε,
d≤x,
(d,r)=1
uniformly inx≥1,randε >prove a similar estimate for the0. We
P
1+ε
quantityµ(d) Log(x/d)/d. Whenε= 0,rvaries between
d≤x,
(d,r)=1
1 and a hundred, andxis below a million, this sum is non-negative and
this raises the question as to whether it is non-negativefor everyx.

Introduction and results

Our first result is the following:
Theorem 1.1.Whenr≥1andε≥0, we have
X
µ(d)
≤1 +ε.