Solutions to Problem Sheet 4

Transcription

1 Solutios to Problem Sheet 4. a Evaluate. b By Partial Summatio deduce = ζ2 2 x2 +Oxlogx. Hit Fid f such that / = d f d. Solutio a Recall σ = j. The defiitio of covolutio ca be give i more tha oe way, so = jd = d d, d d which is the ormal defiitio of σ. But more useful to us is = dj = d d, so = d. d d d The Covolutio Method the gives = = d x = x d x d d = d d x o iterchagig summatios d [ x = d d] x d d +O d x d +O 2 d d x 22

2 The first sum here coverges so complete it to ifiity, = x d +Ologx 2 d 2 d= d>x = x ζ2+o = xζ2+ologx. +Ologx x Hece = ζ2x+ologx. b = = x x = x = x t = x ζ2x+ologx dt dt ζ2t+ologt dt = x ζ2x+ologx ζ2 2 x2 +Oxlogx = ζ2 2 x2 +Oxlogx. 2. O Problem Sheet 3 it was show that Q k = µ k where Q k is the characteristic fuctio of the k-free itegers while µ k a = if a = m k for some iteger m, 0 otherwise. Use the Covolutio Method to show that Q k = x ζk +O x /k, 23

3 for k 2. Solutio Q k = a x µ k a b x/a = x k a a xµ a +O = x m m xµm +O k k = x µm +O m k m k x = x m= m x /k µm µm m k m k m>x /k +O x /k. The secod sum here is µm m k m>x /k m>x /k m k x /k k, by Example 4., ad so Q k = x ζk +O +O x /k. x /k This is the required result. 3. Look back i the otes to recall how the result = logx+o 24

4 was improved to = logx+γ +O. 5 x Use the same method to improve Corollary 4. to for some costat C. Q 2 = logx+c +O ζ2 x /2 Ufortuately,thisdoes tdirectlyleadtoaimprovemetfor 2ω, just as 5 does t directly lead to a improvemet for d. Solutio Write Theorem 4.2 as Q 2 = ζ2 x+ex where E x = O x /2. Lookig back at the proof of Corollary 4. we fid that Partial Summatio gives Q 2 = x dt Q 2 + Q 2 x t 2 = x ζ2 x+ex + t, dt ζ2 t+et t 2 = ζ2 logx+ ζ2 + Ex x + Et t dt. 2 The itegral here is coverget ad so we complete it to ifiity ad estimate the tail ed as Ex t /2 dt = O x t 2 x t dt = O 2 x t 3/2dt = O. x /2 This boud also suffices as a boud for Ex/x. Thus Q 2 = logx+c +O, ζ2 x /2 25

5 with C = ζ2 + Et t dt Prove by iductio that for all l 2. d l = l! xlogl x+o xlog l 2 x 6 HitAssumig6usepartialsummatiotoprovearesultfor d l/. The use d l+ = d l to get a result for d l+. Solutio The result for l = 2 was proved i lectures. Iductio works because of the iductive defiitio d l+ = d l. Assume the result holds for l = k. For l = k + we start with the observatio that d l+ = d l. The the Covolutio Method gives l+ = d d l = x l d +O = x = x d l +O d l d l +O xlog l x, 7 by the iductive hypothesis o the error term. 26

6 O the mai term we apply Partial Summatio d l = x dt d l + d l x t 2 = x t l! xlogl x+o xlog l 2 x + + l! tlogl t+o tlog l 2 t dt t 2 = l! logl x+o log l 2 x + l! logl x+o log l x = l! logl x+o log l x. Substituted ito 7 this gives d l+ = l! xlogl x+o xlog l x, ad so the required result holds for k = l+. Hece by iductio 6 holds for all l a Prove that b Prove that 2 = ζ2logx+o. logx φ logx. Hit for part b use a result from Problem Sheet 4 which combies ad φ. 27

7 Solutio a Partial summatio o Questio a gives 2 = x + t dt t 2 = x ζ2x+ologx + x = ζ2logx+o. ζ2t+ologt dt t 2 You could also use Partial Summatio o the result from Questio??. b From Problem Sheet 4 we fid Thus 2 6 π φ φ π Hece the result follows from Part a. 6. I Problem Sheet 4 the characteristic fuctio, q 2, of square-full umbers was defied. So o prime powers qp a = 0 if a =, if a 2. It was show there that Recall that Therefore = ζ3s ζ6s = ζ3s ζ23s = where m= h = q 2 s = Q 2 s Q 2 m m 3s = = ζ2s ζ3s ζ6s. 9 = ζs ζ2s m= Q 2 m m 3 s = { Q2 m if = m 3 0 otherwise. 28 = =m 3 20 Q 2 m s = = h 3,

8 Thus 9 ca be writte as D q2 s = ζ2s ζ3s ζ6s = D sqsd h s = D sq h s, where sq = if is a square, 0 otherwise. This suggests that q 2 = sq h i Prove that q 2 = sq h, by showig that the two sides agree o all prime powers. ii Use the Compositio Method to prove q 2 = x /2ζ3/2 ζ3 +O x /3. Note there are more square-full umbers tha squarese.g. 8 is squarefull but ot square. The coefficiet here is approximately , so we might say that there are just over twice as may square-full itegers as squares Solutio iweeedolyshowequalityoprimepowers. Notefirstthathp a = if a = 0 or 3, 0 otherwise. The sq hp a = sq p b hp c b+c=a = { sqp a if a 2 sqp a +sqp a 3 if a 3., Fora 3,exactlyoeofaada 3iseveadsosqp a +sqp a 3 = for all such a. Hece sq hp a = if a 2, whe a =, i.e. sq hp a = q 2 p a. 29

9 ii q 2 = a x = a x ha b x/a ha 2 x/a sqb = x /2 +O a xha a = x /2 2 m +O m m xq 3 3 = x /2 Q 2 m +O Q m 3/2 2 m 2 m x /3 m x /3 Complete the first sum to ifiity with a error i doig so of Q 2 m m 3/2 m 3/2 x m>x /3 m>x /3 /2, /3 by Example 4.. The error i 2 is x /3. Hece q 2 = x /2 a= Q 2 a +O +O x /3, a 3/2 x /6 which leads to the stated result because of 20 with s = 3/2. 7. If f = g the covolutio method starts with f = [ x ga. 22 a] a x Use this equality to show that for Euler s phi fuctio φ ad itegral N we have 30

10 a N a N [ ] N φa = a 2 N N+. Solutio From 22 we have [ ] N φa = a where f = φ. Yet φ = µ j ad thus f = µ j N f = µ j by associativity = δ j by Möbius Iversio = j sice δ is the idetity. Hece a N [ ] N φa = j = = a 2 N N+. N N 8. i. Recall d = xlogx+ox. Prove, by usig Partial Summatio o this result, that d = 2 x2 logx+o x 2. ii. I a previous Questio Sheet you showed that σ φ = j j. Use this to show that σ φ = 2 x2 logx+o x 2. Hit make use of part i. 3

11 Solutio The method of Partial Summatio gives = d d x x = x = x d d d dt d dt t = x xlogx+ox = x 2 logx+o x 2 [ 2 t2 logt tlogt+ot dt ] x + 2 havig itegrated by parts. This last itegral is absorbed ito the error ad we get the required result. tdt ii. Note that j j = d jdj = d d d d = d = d. 32