Estimates for sums over primes

Transcription

1 8 Estimates for sums over primes Let 8 Principles of the method S = n N fnλn If f is monotonic, then we can estimate S by using the Prime Number Theorem and integration by parts If f is multiplicative, then we can gain information concerning S by studying the properties of the associated Dirichlet series fnn s This has already been especially successful when f is of the form fn = χnn s We now introduce an entirely different method that is most successful when f is far from being multiplicative Let P = p N p Vinogradov 937 had the idea of writing f + fp = N<p N n N n,p = fn = t P t N µt r N/t frt If we can demonstrate that there is considerable cancellation the inner sum on the right, then we can obtain a non-trivial estimate for the left hand side However, when t is near N in size, one expects to have little cancellation, and indeed when N/2 < t N the sum has only one term, and hence no cancellation at all Hence the terms on the right must be rearranged before satisfactory estimates can be derived This approach, known as Vinogradov s method for prime number sums, is rather complicated, but Vaughan 977 devised a much simpler variant Vaughan s version of Vinogradov s method, which we now describe Our first step involves expressing Λn as a linear combination of several other arithmetic functions Put 8 F s = m U Λnn s, Gs = µdd s Clearly 82 ζ ζ s = F s ζsf sgs ζ sgs + ζ ζ s F s ζsgs 67

2 68 Estimates for sums over primes for σ > By calculating the Dirichlet series coefficients of the four Dirichlet series on the right hand side, we deduce that 83 Λn = a n + a 2 n + a 3 n + a 4 n where { Λn if n U, a n = 0 if n > U, a 2 n = Λmµd, a 3 n = mdr=n m U hd=n a 4 n = mk=n m>v k> µd log h, Λn d k We multiply 83 through by fn and sum to see that µd S = S + S 2 + S 3 + S 4 where S i = n N fna i n We generally estimate S trivially, but the other terms require individual treatment We note that 84 S 2 = t UV bt r N/t frt where bt = md=t m U µdλm Since bt m t Λm = log t log UV, it follows that 85 S 2 log UV t UV r N/t frt

3 8 Principles of the method 69 As for S 3, we find that 86 S 3 = µd fdh log h = µd h N/d N = µd fdh dw w w h N/d log N max fdh w w h N/d h N/d h dw fdh w Let ck = d k Since ck = 0 for < k V, it follows that S 4 = U<m N/V Λm µd V <k N/m Suppose that M = M, N, f is defined so that ckfmk 87 b m k N/m c k fmk M b m 2 /2 c k 2 /2 for arbitrary complex numbers b m and c k By cutting the interval U m N/V into log N subintervals of the form M < m 2M, we deduce that S 4 log N max M U M N/V Λm 2 /2 ck 2 /2 Here the sum over m is ψ2m log 2M M log 2M Since ck dk for all k, we deduce by 23 that the sum over k is NM log N 3 Hence 88 S 4 N /2 log N 3 max U M N/V M We interrupt our development at this point in order to assess the situation For purposes of discussion, in this paragraph only, we assume that fn for all n The bound S N is trivial, and if f is oscillatory we hope to show that S = on Trivially S U, so S poses no problem provided that U = on In 85 the trivial bound would be that S 2 log UV t UV N t Nlog UV 2

4 620 Estimates for sums over primes Thus in order to get a bound that is on we only need to demonstrate a modest amount of cancellation in the sum over r in 85, and even this only on average over t We note, however, that there will be little or no cancellation if the inner sum has very few terms a single term is the worst case For this reason it will be necessary to choose the parameters U and V so that UV is considerably smaller than N Similar remarks apply to 86 where the situation is even more favorable since the range of d in 86 is shorter than that of t in 85 To obtain a trivial bound for M we first observe that c k fmk b m c k b m k N/m By Cauchy s inequality, this in turn is M N/M /2 b m 2 /2 c k 2 /2 Thus the bound M N /2 is trivial By inserting this in 88 we deduce that S 4 Nlog N 3 trivially That is, we will be able to show that S 4 = on if we can obtain a bound for M that is only a power of a logarithm smaller than trivial In summary, it seems that we have not dug ourselves into too deep a hole, and that we can expect to show that S = on whenever we can derive estimates that are only moderately better than trivial We note, however, that if f were to be unimodular and totally multiplicative, then we might obtain nontrivial estimates for S 2 and S 3, but no nontrivial estimate for M can hold because of the possibility that b m = fm and c k = fk Despite this observation, we shall find in Chapters 20 and 25 that we can still use our present approach when we average over several multiplicative functions f i In order to estimate M, we first observe that by Cauchy s inequality the left hand side of 87 is Here the second sum over m is 89 = j N/M b m 2 /2 c j c k k N/m c k fmk 2 /2 fmjfmk By the arithmetic-geometric mean inequality we know that c j c k 2 c j c k 2 Thus the above is 80 c k 2 fmjfmk 2 j N/M M m 2M

5 Thus 8 M 8 Principles of the method 62 c k 2 max fmjfmk 2 max and so by 88 we conclude that j N/M j N/M M m 2M fmjfmk 2 /2, 82 S 4 N /2 log N 3 max U M N/V max j N/M fmjfmk 2 /2 Clearly our bound 85 for S 2 becomes better when UV is reduced On the other hand, our bound above for S 4 becomes better when U and V are increased In practice, we choose the parameters to optimize these bounds Our strategy for bounding S 4 may be inferior, for two reasons In the first place, we need to bound the double sum on the left hand side of 87 not for arbitrary b m and c k but only in the special case that b m = Λm and c k = ck Secondly, the double sum on the left hand side of 87 is a linear function of the b m, and is also linear in the c k Such an expression is known as a bilinear form, and in Appendix F we develop a general theory concerning bounds for bilinear forms Indeed, we could have passed directly from 87 to 8 simply by appealing to Corollary F4 Although we have taken a more elementary route, the general theory offers some insights From Theorem F we see that from 87 up to 89 we have thrown nothing away In 89 we again have a bilinear form, but this time the coefficient matrix is not only square, but Hermitian as well, and hence normal Thus by Corollary F the problem is to determine or estimate the spectral radius of this matrix In passing from 89 to 80 we have in effect derived a bound for this spectral radius, but our bound may be considerably larger than the truth An expression of the form 83 b m fmk m W X k N/m is known as a Type I sum Thus S 2 and S 3 are Type I sums An expression of the form 84 b m c k fmk Y m Z k N/m

6 622 Estimates for sums over primes is known as a Type II sum Thus S 4 is a Type II sum In some situations, the estimate we derive can be improved by playing the following trick: We take a Type I sum such as S 2 and write it as 85 m W b m X k N/mfmk = m Y b m = S I + S II, X k N/m fmk + Y <m W b m X k N/m fmk say Here S I is treated as a Type I sum, but the estimate is better because Y is smaller than W, and we treat S II as a Type II sum 8 Exercises Suppose that M = M, N, V, f is defined so that 86 /2 c k fmk M b m 2 b m V <k N/m for arbitrary complex numbers b m and c k a Show that 87 S 4 N /2 log N 3 max U M N/V M b Deduce that 88 M c Conclude that max V < V <j N/M fmjfmk V < 2 /2, c k 2 /2 89 S 4 N /2 log N 3 max U M N/V max V < V <j N/M M m 2M fmjfmk 2 /2 2 Let S 2 be defined as in 84, and write S 2 = bt frt + t V = S I + S II, r N/t V <t UV bt r N/t frt

7 8 Principles of the method 623 say Show that 820 S II N /2 log N 2 max V M UV M 82 3 Let M denote the best constant in the bilinear form inequality 87 By appealing to an apropriate result from Appendix F, or otherwise, show also that if fn for all n, then maxm /2, N/M /2 Hence our method, as presently constituted, never gives an upper bound better than N 3/4 when f is unimodular 4 Linnik 96 Let d k n = card{n,, n k : n n 2 n k = n, n i > } Show that Λn K log n = k d k n/k if K log n/ log 2 k= 5 Montgomery & Vaughan 98 Let Gs be defined as in 8 From the identity ζs = 2Gs Gs2 ζs + ζs Gs ζsgs, or otherwise, show that where µn = a 0 n + a n + a 2 n { 2µn n V, a 0 n = 0 n > V, a n = µdµe, dem=n e V a 2 n = 6 Show that if V N, then dk=n d>v k>v µd e k e V µe N µnfn = T 0 + T + T 2 n=

8 624 Estimates for sums over primes where V 822 T = 2 µnfn, n= 823 T 2 = fmn, m V 2 b m n N/m b m = de=m d,e V µdµe, and 824 T 3 = V <m N/V V <n N/m µmc n fmn, c n = d n µd 7 With the T i defined as above, show that 825 T 0 n V fn, 826 T dr r V 2 and k N/r frk, 827 T 2 N /2 log N 5/2 max V M N/V max j N/M fmjfmk /2 2 Applications We begin with an historically important example, which will be invaluable in Chapter 2 Theorem 8 For N 2, let 828 Sα = If a, q = and α a/q /q 2, then N Λnenα 829 Sα Nq /2 + N 4/5 + N /2 q /2 log N 4 n=

9 Here eθ = e 2πiθ is the complex exponential with period 82 Applications 625 Proof By the formula for a segment of a geometric series we see that if β is not an integer, then N enβ = n= en + β eβ eβ enβ/2 e Nβ/2 = en + β/2 eβ/2 e β/2 sin πnβ = en + β/2 sin πβ But sin πβ 2 β where β denotes the distance from β to the nearest integer, β = min n Z β n, so 830 Thus 83 0<t T max w N enβ min n= w r N/t ertα N, 0<t T 2 β N min t, tα To estimate the right hand side, we write t = hq + r and sum over 0 h T/q and r q Let δ = α a/q We consider first the case in which h = 0 and r q/2 Since δ /q 2, rα differs from ra/q by at most /2q But ra/q /q for these r, and hence rα ra/q Consequently r q/2 rα r q/2 ra/q r q/2 q r q log 2q For all other terms we have hq + r h + q Thus it suffices to estimate 832 q 0 h T /q r= min N h + q, hqα + ra/q + rδ For any given h, the q points hqα + ra/q + rδ are uniformly within /q of the equallyspaced points hqα + ra/q Thus if hqα + ra/q + rδ < /q, then hqα + ra/q < 2/q, and this holds for at most 4 values of r For all other r, the numbers hqα + ra/q + rδ are comparable to the numbers r/q for 0 < r < q Hence the double sum 832 is 0 h T /q N h + q + q log 2q N log 2T/q + T log 2q + q log 2q q

10 626 Estimates for sums over primes That is, we have shown that 833 0<t T By 85 we deduce that Similarly, from 86 we see that By 82 and 830 we find that S 4 N /2 log N 3 Here the sum over j is M + 0<j N/M N min t, N/q + T + q log 2T q tα S 2 N/q + UV + qlog 2qUV 2 S 3 N/q + V + qlog 2qV N 2 max U M N/V max min M, j N/M /2 j kα min M, M + N min jα j, jα 0<j N/M since M N/j for j N/M Thus by a further application of 833 we deduce that S 4 Nq /2 + NU /2 + NV /2 + N /2 q /2 log 2qN 4 By taking U = V = N 2/5 we deduce that Sα Nq /2 + N 4/5 + N /2 q /2 log 2qN 4 To complete the argument it suffices to note that we may assume that q N, since otherwise the estimate 829 is weaker than the trivial estimate Sα N 82 Exercises Show that if α a/q /q 2 and a, q =, then µnenα Nlog N 3 Nq /2 + N 4/5+ε + N /2 q /2 log N 3 n N 2 Show that if q is a positive integer, then for any integer c, ec/q = d q d c φq/d χ mod q/d τχχc/d

11 82 Applications Let Mx; χ, δ = n x χnµnenδ where χ is a Dirichlet character, x is real, and δ T Let A and B be given positive real numbers Show that if α = a/q + δ with a, q =, then n x µnenα = d q µd φq/d χ mod q/d τχχamx/d; χχ 0d, δ where χ 0d denotes the principal character modulo d 4 Let Mx; χ, δ be defined as in the preceding problem Show that if χ is a character modulo q and q log x A, then Mx; χ, δ + x δ xlog x B 5 Davenport 937a,b Show that if α a/q /q 2, a, q =, and q log x A, then 834 µnenα xlog x B n x By combining this with the result of Exercise, show that the above estimate holds uniformly in α 6 Show that the series µn n enα n= is uniformly convergent, and thus defines a continuous function on T Suppose that F n = d n fd for all n, and let sx denote the saw-tooth function with period, { {x} /2 x / Z, sx = 0 x Z By the Fourier series expansion of Lemma D, we see that d= fd d sdα = = d= n= fd d F n πn m= sin 2πmdα πm sin 2πnα,

12 628 Estimates for sums over primes by grouping together those pairs m, d for which md = n This is merely a formal argument, since we have not justified the reorganization of terms in passing from 835 to 936 In the next several exercises, we treat this issue in the interesting case that fd = µd 7 Let 837 S D α = d D µd d sdα a Let N be a parameter to be chosen later such that N > D, and let E K x be defined as in Lemma D Show that where b Show that S D α = π sin 2πα + T α + T 2 α T = π T 2 = d D T = D<d N n N/D µd d n N/d µd d E N/dα n D<d N/n µd d sin 2πndα, πn sin 2πndα c Use 834 to show that T log D B log N/D 2 d Explain why E K 0 = 0 e Show that if a, q = and q D, then T 2 a/q DN log 2q f Take N = Dlog D A, and deduce that 838 S D α = π sin 2πα + O log D B when α = a/q, a, q =, and q D 8 Let S D α be defined as in 837 a Show that S D α is piecewise linear with slope MD = d D µd and jump discontinuities at the Farey fractions of order D b Write x n x n,q= µn n = n x µn[x/n] + n,q= = Σ + Σ 2, µnx/n n x n,q=

13 82 Applications 629 say Show that Σ is the number of integers not exceeding x that are composed entirely of prime numbers that divide q Hence deduce that Σ x c Explain why Σ 2 x d Deduce that n x n,q= µn n 2 uniformly in x and q e Let S D α be defined as in 837 and let a/q denote a Farey fraction of order D Show that the jump discontinuity of S D α at α = a/q is d D q d µd d f Show that the above expression has absolute value not exceeding 2/q g Let R denote the set of numbers composed entirely of primes dividing q Show that { µn if n, q =, µn/d = 0 otherwise h Deduce that d n d R n x n,q= µn n = d d x d R m x/d µm m i By adapting the techniques developed in 7, show that if q x 2, then the number of members of R not exceeding x is x ε j Deduce that if q x, then n x n,q= µn n exp c log x k Davenport 937a,b Conclude that 838 holds uniformly in α 9 Montgomery & Vaughan 98; continued from Exercise 326 Let Sx, y = d y µdb {x/d 2 } where B u = u /2 is the first Bernoulli polynomial a By van der Corput s method, or otherwise, show that µdew/d 2 W /2 D 7/24 log D A d D

14 630 Estimates for sums over primes for D 3/4 W D 7/2 b Deduce that Sx, y for c Conclude that if RH is true, then the number Qx of squarefree numbers not exceeding x is Qx = 6 π 2 x + O x 2/64+ε