Analytic Number Theory Solutions

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1 Aalytic Number Theory Solutios Sea Li Corell Uiversity Ja. 03 Itrouctio This ocumet is a work-i-progress solutio maual for Tom Apostol s Itrouctio to Aalytic Number Theory. The solutios were worke out primarily for my learig of the subject, as Corell Uiversity curretly oes ot offer a aalytic umber theory course at either the uergrauate or grauate level. However, this ocumet is public a available for use by ayoe. If you are a stuet usig this ocumet for a course, I recomme that you first try work out the problems by yourself or i a group. My math ocumets are store o a math blog at 3 Averages of Arithmetical Fuctios 3.. Use Euler s summatio formula to euce the followig for. a log log log + A + O, where A is a costat. Proof. Note that log t log t t t t.

2 By Euler s summatio formula we have log log t t + t log log + O. t [t] log t t t [t] log t t t + t + log [] t [t] log t t Note that the remaier results from the fact that [] is boue by a 0. Now t [t] log t t t which are boue by t [t] t t t [t] log t t, t t b a t [t] t t t [t] log t t t t t log t t t so that the term coverges a is equal to a costat, call it A. Similarly for the log t t [t] t term, we have a t t [t] log t t Hece log t t [t] t t O O, we have log log t [t] t t t log t t log t log + t. t a sice O + O log log + A + O log loglog + B + O, where B is a costat. log log Proof. The erivative is. log t t log t + log t t log t.

3 By Euler s summatio formula we have log t t log t t [t] + log t t log t t + [] log loglog loglog + t [t] + log t t log t t t [t] + log t t log t t + O. log Now the first itegral term is boue a hece a costat as t [t] + log t t log t t a the seco itegral term follows t [t] + log t t log t t + log t t log t t log, + log t t log t t log O. log Hece the sum is equal to loglog + B + O. log log 3.. If prove that log + C log + O, where C is Euler s costat. 3

4 Proof. Let q. The usig Theorem 3.a a Eercise 3.a, we have q q, q q q { log log + C + O { log + C + O } } { log + C + O log + C log + C + O log A + O log + C log + O. } log log + C log 3.3. If a α < 0, α, prove that α α log α + ζα + O α. 4

5 Proof. Let q. Usig Theorem 3.b, we have α q, q α q α α q α q { } α α α α + ζα + O α { } α log + C + O α { } α {ζα + α + ζα + O α + O α } α α α log + C α α + O α + α α ζα + ζα + ζαo α + O α α α α log + ζα + O α 3.4. If prove that: a [ µ ] + O log. ζ Proof. Let g N u, so that g a G g []. By Theorems 3.0 a 3.7, we have [ µ µ N u ] ϕ I b µ [ ] + Olog. ζ 6 π + O log ζ + O log. 5

6 Proof. Note that [ ] + O + O. Thus by part a a Theorem 3., we have [ ] µ { [ ] + O [ ] µ + O + O log + O 3.5. If prove that: a ϕ ζ + Olog. ζ [ ] µ +. } [ µ ] Proof. Recall from the proof of Eercise 3.4a that [ µ ϕ I. ] Sice I, rearragig gives ϕ µ b ϕ µ [ ]. [ ] +. [ µ ] Proof. Let g, the G g []. By Theorem 3.0, we have µ [ ] N µ u µ µ ϕ. 6

7 3.6. If prove that ϕ ζ log + C ζ A + O log, where C is Euler s costat a A µ log. Proof. Note that N is completely multiplicative, so ϕ N µ N N µ N. The by Theorem 3.0 a 3.a, we have Now ϕ N µ N µ { 6 π + O ζ log + µ log { log + C + O } } { log + C + O log C ζ + O µ log } µ log µ log. > µ log. The sum from to ifiity is clearly fiite, a the other sum satisfies µ log log log O, > > hece ϕ ζ log + C ζ A + O log. 7

8 3.7. I a later chapter we will prove that µ α /ζα if α >. Assumig this, prove that for a α >, α, we have ϕ α α ζ α + ζα ζα + O α log. Proof. Note that N α is completely multiplicative, so Also, ote that α ϕ N α µ N N α µ N α. µ α µ α > µ α, where the first sum is assume to be /ζα a the seco sum satisfies µ α α O α. > > The by Theorem 3.0 a 3.b, we have ϕ α N α µ N α µ α α α + ζα + O α α µ α + ζα + O α ζα + O α α ζ α + ζα ζα + O α log If α a prove that ϕ α α ζ + O α log. α Proof. The proof is the same as for Eercise 3.7 ecept that ue to the coitio o α, we use Theorem 3. istea of Theorem 3.b I a later chapter we will prove that the ifiite prouct p p, etee over all primes, coverges to the value /ζ 6/π. Assumig this result, prove that 8

9 a σ < ϕ < π σ 6 if. Proof. Let i api i. The from the formula o p.39 a usig the fact that σ is multiplicative, we have σ i p ai+ i p i. Now we also have Diviig gives ϕ i p i p i. /ϕ σ/ i i i i p i p ai+ i p ai+ i p ai+ i p a i+ i p ai+ i p ai+ i The prouct is clearly miimize whe, i.e. there are o prime factors. To maimize the prouct, we ee to have as may prime factors as possible a miimize the a i. Sice a i by our costructio of the prime factorizatio, we have that the maimum is This proves both iequalities. p p ζ π 6. b If prove that ϕ O. Proof. From part a we kow that ϕ O σ, 9

10 thus it suffices to show σ O. I fact, we show the asymptotic behavior σ σ ζ + Olog. Let q, the q / [ ] + O + O + ζ + O ζ + Olog. + Olog 3.0. If prove that Olog. ϕ Proof. I Eercise 3.9a we showe that for, Hece it suffices to show σ Olog. Let q, the σ q / q q / q log + C + O σ < ϕ < π 6 log + C + O log log + C + O log + ζ + O + O log ζ log + ζc + O. σ. 0

11 Thus σ Olog a we are oe. 3.. Let ϕ µ /. a Prove that ϕ is multiplicative a that ϕ p + p. Proof. Let m a be positive itegers such that m,. The ϕ mϕ m µ µ m µ ϕ m, m m hece ϕ is multiplicative. Pluggig i prime powers shows µ + /p + /p. p p k b Prove that ϕ µσ where the sum is over the ivisors of for which. Proof. Sice ϕ is multiplicative it suffices to show this for prime powers p k. The prouct efiitio i part a yiels ϕ p k p k p + p p k + p k. The formula i part b yiels µσ σp k σp k pk+ p k p p k p p p k p + p k + p k, so that the two efiitios are the same. c Prove that ϕ µs, where S σk, k

12 the use Theorem 3.4 to euce that, for, ϕ ζ ζ4 + O log. As i Eercise 3.7, you may assume the result µ α /ζα for α >. Proof. The proof of the first statemet follows easily from usig Theorem 3.0 o the result i part b. The seco statemet ca be show as follows: ϕ µs ζ µ ζ + O log µ 4 + O log µ ζ ζ4 + O log. µ log 3.. For real s > 0 a iteger k fi a asymptotic formula for the partial sums s,k with a error term that tes to 0 as. Be sure to iclue the case s. Proof. Note that, k imply that a k share o commo prime factors. Hece, for each prime factor p of k, we ee to subtract from /s those,p /s. For eample, if k the we have, s s s s s s / s / s.

13 Now we ee to be careful whe there are multiple prime factors. For eample, if k 6, the whe we subtract the multiples of a multiples of 3, we have ee up subtractig the multiples of 6 twice, so we ee to a back the multiples of 6. The fuctio that takes o at, at primes, at the prouct of two primes, etc. is the Möbius fuctio, µ. Hece we may write the sum as,k s k µ s / s. Now we ee to cosier the two cases s a s. If s the we have µ µ log log + C + O k / k log k µ + C k µ k µ log + O, where k µ/ a k µ log / are costats base o k. If s, k µ s / s s + ζs + O s s s s µ s k s µ s + ζs k k µ s + O s, where k µ/ a k µ/s are costats base o k a s For each real the symbol [] eotes the greatest iteger. Eercises 3 through 6 escribe some properties of the greatest-iteger fuctio. I these eercises a y eote real umbers, eotes a iteger Prove each of the followig statemets. a If k + y where k is a iteger a 0 y <, the k []. Proof. y k is a iteger, a sice 0 y <, k is also the greatest iteger. b [ + ] [] +. Proof. Let k + y, where k is a iteger a 0 y <. The by part a, we have [ + ] [y + k + ] k + [] +. { [] if [], c [ ] [] if []. 3

14 Proof. If [] the statemet is trivial. Else, let k + y, where k is a iteger a 0 < y <. The [ ] [ k y] [ k + y] k []. [/] [[]/] if. Proof. Let k + y, where k is a iteger a 0 y <. The let k/ m + z, where m is a iteger a 0 z <. We have a [/] [k + y/] [k/ + y/] [m + z + y/], [[]/] [k/] m, so it remais to show that z + y/ <. Note that k/ m + z a implies z a/ for some a i 0,,,, Thus z /. A sice 0 y <, we have y/ < /. Hece z + y/ < / + /, a we are oe If 0 < y <, what are the possible values of [] [ y]? Proof. Let k + z, where k is a iteger a 0 z <. If z y the a if z < y the [] [ y] k [k + z y] k k 0, [] [ y] k [k + z y] k [k + y z] k k. Hece [] [ y] equals 0 or The umber {} [] is calle the fractioal part of. It satisfies the iequalities 0 {} <, with {} 0 if a oly if is a iteger. What are the possible values of {} + { }? Proof. If [] the the sum is trivially 0. Else, let k + y, where k is a iteger a 0 < y <. The {} y, a sice [ ] [], we have { } [] y +. Hece {} + { } y + y a Prove that [] [] is either 0 or. Proof. Let k + y, where k is a iteger a 0 y <. If y < / the [] [] k k 0, a if y / the k + y, where k <, so that [] [] k + k. 4

15 b Prove that [] + [y] [] + [y] + [ + y]. Proof. Let m + a a y + b, where m a are itegers a 0 a, b <. We cosier 4 cases: a, b < /, a < /, b /, a + b <, 3 a < /, b /, a + b, 4 a, b /. Combie with symmetry of a a b i cases a 3, these cover all cases. Case : [] + [y] m + [] + [y] + [ + y] m + + m +. Case : [] + [y] m + + > [] + [y] + [ + y] m + + m +. Case 3: []+[y] m++ []+[y]+[+y] m++m++. Case 4: [] + [y] m > [] + [y] + [ + y] m + + m + +. This shows the statemet hols i all cases Prove that [] + [ + ] [] a, more geerally, [ + k ] []. k0 Proof. Let m + a where m is a iteger a 0 a <. If a < the a if a, the [] + [ + ] m + m m [], [] + [ + ] m + m + m + []. More geerally, let j be the iteger i 0,,..., such that j/ a < j + /. The k0 [ + k ] j k0 [ + k ] + j jm + jm + m + j []. [ + k ] 5

16 3.8. Let f []. Prove that a euce that m f f k0 + k f + for all m a all real. Proof. By usig the previous eercise, 3.7, we have f k0 + k For the seco part, ote that so f k0 f + f k0 + k [ + k ] + [] f. [] + k f, + f +. Thus for each, we have f + f + f, a summig from to m results i m f + m f + f f m f sice f for all Give positive o itegers h a k, h, k, let a k /, b h /. 6

17 a Prove that a r [hr/k] + b r [kr/h] ab. Proof. Cosier the lie i the plae that passes through the cooriates 0, 0 a h, k. Sice h a k are relatively prime, this lie oes ot itersect ay lattice poits i the rectagle with corers, a h, k ecept for the poit h, k itself, call this rectagle R. The umber of eve lattice poits, i.e. the poits m, where m a are both eve, isie R is precisely ab. Note that the first sum couts the umber of eve lattice poits i the rectagle above the lie, a the seco sum couts the umber of eve lattice poits below the lie. Sice there are o eve lattice poits o the lie, the two sums combie to ab. b Obtai a correspoig result if h, k. Proof. Use the techique i part a. There are precisely / eve lattice poits i the rectagle which are ouble coute by the sums. Hece we have a r [ ] hr k + b r [ ] kr ab. h 3.0. If is a positive iteger prove that [ + + ] [ 4 + ]. Proof. Let k + a, where k a a are itegers satisfyig k [ ] a 0 a < k +. We cosier two cases: a < k a a k. I the first case, we have o the left sie [ + + ] [ k + a + k + a + ] k, as k + a + < k + k + 4 < k +, so that both a + are less tha k +. Hece the sum is less tha k + a the floor is k. O the other sie, we have [ ] as so that [ 4 + ] [ 4k + 4a + ] k + a + k + a + < k + k + 4 k +, k, k + a + < k + a hece k + a + < k +. Now for the seco case, a k. O the left sie we have [ + + ] [ k + a + k + a + ] k +, 7

18 as k + a + k + k + > k + k + 4 k +, a k + a is also greater tha this quatity if a > k; if a k the k + a k + a k O the right sie we have as a hece [ 4 + ] [ 4k + 4a + ] [ k + a + k + a + > k + k + 4 k +, k + a + > k +. ] k +, 3.. Determie all positive itegers such that [ ] ivies. Proof. Let k + a, where k a a are itegers satisfyig k [ ] a 0 a < k +. The k if a oly if k + mk where m 0,,. The if irectio is trivial. For the oly if irectio, we see that sice k k + a a k k, this implies k a, hece a mk. Now suppose m 3, the k + mk k + 3k k + k + k +, cotraictig the choice of k for [ ]. Hece m 0,,. 3.. If is a positive iteger, prove that [ ] [ [ ]] is iepeet of. Proof. Note that if we icrease by 5, both terms icrease by precisely 8, cacelig the effect. Thus to show that this ifferece has the same value for all, it suffices to show it has the same value for,,..., 5. This ca be show by ehaustio, resultig i 4 for all Prove that [ λ [ ] ]. Proof. Let s be the iicator fuctio for squares, amely { if is a square, s 0 otherwise. 8

19 Note that [ ] s, a by Theorem 3.0, [ λ ] λ s, so the two sies are equal Prove that [ ] [ ]. Proof. Oce agai let s be the iicator fuctio for squares, i.e. s if is a square a s 0 otherwise. Note that [ ] s. By Theorem 3.0, the left sie is [ ] u. s Rewritig the right sie a applyig Theorem 3.0 gives [ ] [ s ] u, s hece the two sies are equal Prove that a that k k [ ] k [ ] k 3 [ ] 4 [ 6 ]. Proof. For the first equatio, let m + r, where m [/] a r 0,. The left sie is k [ ] m k k + r + m mm + r + m m + rm, k0 9

20 a the right sie is [ ] [ 4m + 4rm + r ] 4 4 Sice r 0,, [r /4] 0, so the equatio hols. k ] [m + rm + r, 4 For the seco equatio, let 3m + r, where m [/3] a r 0,,. The left sie is [ ] m k 3 k + r + m 3m m + rm, 3 k0 a the right sie is [ ] [ 9m + 6mr 3m + r ] r 6 6 Sice r 0,,, [r r/3] 0, a the equatio hols. k [ 3m + mr m + r r If a,,..., 7 prove that there eists a iteger b epeig o a such that [ ] [ ] k + b. a 8a Proof. We show that b a works for a,,..., 7. As before, let am+r, where m [/a] a r 0,,..., a. The [ ] m k a k + r + m a k k0 am am + rm + m 4a m 4a m + 8arm + 8am. 8a Now let b a a write [ ] [ ] + b am + r + a, 8a 8a where the umerator epae out is 4a m 4a m + 8arm + 8am + 4r + 8r 4ar + 4 4a + a. To show a have the same value, we ee to show the umerator of mius the umerator of, call this ifferece, satisfies 0 < 8a. Note that 4r + 8r 4ar + 4 4a + a r + a 0. Now r a a a < 8 implies so we are oe. r + a a + a a < 8a, ]. 0