The Convolution Method of Evaluating Sums of Multiplicative Functions

Transcription

1 The Convolution Method of Evaluating Sums of Multiplicative Functions Ernie Croot May 26, 2004 Introduction In this document we consider the problem of estimating q S ) ϕq), and S 2 ) ϕq). I know of several simple methods for estimating S ) to within an error of about O log A ) for some A > 0); however, to get the error all the way down to Olog ) ) requires some work, and in the next section, we will give such an estimate, namely S ) +Olog ) ). It may be possible to give some clever, ad hoc proof of an estimate this sharp, but the proof we present will use a general approach that I like to call the convolution method. It is possible to use much more sophisticated methods, such as what are called contour methods, but we will not concern ourselves with these in this course. Before we embark on our proof of our estimate for S ), we show how to use such an estimate to produce one for S 2 ). To do this, we will require the following simple theorem: Theorem Abel Summation Formula) Suppose that fn) is supported on the positive integers and that gt) is a function supported on the reals

2 having a first derivative for t. For each real number x, define F x) n x fn), and Gx) n x gn). Note here that F x) Gx) 0 for all x < because the sums are empty). We have the following identity fn)gn) gx)f x) g t)f t)dt. n x Proof. One way to prove this identity is to observe that the right-hand-side is linear in F t); so, it suffices to prove that both sides hold for when fn) is zero for all n x, except for a single value n x where fn). In this case, the left-hand-side will just equal gn); and, the right-hand-side will be gx) g t)f t)dt gx) g t)dt gx) gx) gn)) gn). n So, both sides match, and the theorem follows. Another way to prove the identity, and which is more general and easier to remember, is to use Riemann-Stiljes integration and integration-by-parts: We have that fn)gn) gt)df t) n x ɛ gt)f t) x ɛ gx)f x) ɛ g t)f t)dt g t)f t)dt. Now, using this theorem, we show how to recover S 2 ) from S ): If we let fn) n/ϕn) for n square-free and 0 if n is not square-free, and let gt) /t, then S 2 ) fq)gq) S ) + + o) + log + O). 2 S t) dt t 2 t + Olog t) t) dt t 2

3 2 The Convolution Trick, and Estimation of S ) Given a multiplicative function, one way of evaluating sums involving this function is to use the convolution method. Basically, we want to find two functions un) and vn) so that for q square-free, q/ϕq) u v)q). One such pair of functions is vn) and un) { /ϕn), if n square free; 0, otherwise. These functions are obviously multiplicative; and so, to check that q/ϕq) u v)q) for square-free q, it suffices to prove that it holds for q prime: For this case, we have q ϕq) + q d q which is just what we wanted. Now, then, expanding S ) gives S ) d ud)vn/d) u v)q), d q ϕd) 2. Treatment of the Inner Sum ϕd), d q µ 2 q). ) The inner sum in this expansion of S ) also will require an application of the convolution trick: We may write µ 2 q) e 2 q µe). 2) Let us now see that this formula holds: The right hand side is a convolution α )q), where αn) equals 0 if n is not a square, and equals µm) for n m 2. So, since convolutions of multiplicative functions is multiplicative, 3

4 to verify that the formula holds, it suffices to prove it for q equal to a prime power. It is easy to check that for, α )p a ) 0 for a 2, and equals for a 0,. So, the formula 2) holds. Thus, the inner sum in ) is µe) µe) d q e 2 q e 2 e 2 µe) [d, e 2 ] d q, e 2 q + O ), 3) where [d, e 2 ] is the lcm of d and e 2. The error term here comes from the fact that there are /[d, e 2 ] + O) integers that are divisible by d and by e 2. Now, extending this last sum to all e 2, we get that the last line of 3) is µe) [d, e 2 ] + O ). e 2 Here we got an additional error term besides the one in 3)) when we added the terms with e 2 >. This last sum can be expressed as an Euler product as follows: e 2 µe) [d, e 2 ] d d d µe) e 2 p e p d p d dζ2) e 2,e) p p ) ) p 2 ) p ) ) p 2 2 p + p), where ζ2) n n 2 p 2 ). 4

5 Resuming our estimation of S ), we find that S ) ζ2) d dϕd) Now, one can show that for a square-free d, dϕd) + p) + p) + Olog ) ). ) p2 O. d 2 Thus, we may extend the range of summation in our expression for S ) to all d, and produce only a small error: We have S ) ζ2) ζ2) ζ2) ζ2) d + p 2 p 2 + Olog ) ) ) + Olog ) ) + Olog ) ). + Olog ) ) 5