Fourier Analysis, Stein and Shakarchi Chapter 8 Dirichlet s Theorem

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1 Fourier Aalysis, Stei ad Shakarchi Chapter 8 Dirichlet s Theorem Abstract Durig the course Aalysis II i NTU 208 Sprig, this solutio file is latexed by the teachig assistat Yug-Hsiag Huag with the discussios or help from the followig cotributors: Exercise 3-5 He-qig Huag; Exercise 7- Mighty Yeh; Exercise 0-???; Exercise -???; Exercise 2-???; Exercise 4-???; Exercise 5-???; Exercise 6-???; Problem -???; Problem 2-???; Problem 3-???; Problem 4-???; Exercises. Prove that there are ifiitely may primes by observig that there were oly fiitely may p,, p N,the N /p j = This is a simple cosequece of Theorem I the text we showed that there are ifiitely may primes of the form 4k+3 by a modificatio of Euclid s origial argumet. Oe ca easily adapt this techique to prove the similar result for primes of the form 3k + 2, ad for those of the form 6k d042200@tu.edu.tw

2 3. Usig the same map as Problem of Chapter 7 oe ca prove that if m ad are relatively prime, the Z (m Z ( is isomorphic to Z (m. For surjectivity (say, give (a, b Z (m Z (, oe has to verify k = bmx + ay Z (m where mx + y = (comes from Corollary.3. This ca be verified as follows: suppose ot, say there is a prime p k ad p m, the p a sice p y ad hece cotradicts to the fact a Z (m. 4. Let ϕ( deote the umber of positive itegers that are relatively prime to. Use the order of groups i the previous exercise, oe kows that if ad m are relatively prime, the ϕ(m = ϕ(ϕ(m. Moreover, oe ca give a formula for Euler phi-fuctio as follows: (a Calculate ϕ(p whe p is a prime by coutig the umber of elemets i Z (p. (b Give a formula for ϕ(p k whe p is a prime ad k by coutig the umber of elemets i Z (p k. (c Show that ( p i ϕ( = i where p i are the primes that divide. (a ϕ(p = p if p is a prime. (b Claim: ϕ(p k = p k p k for k. This ca be proved as follows: if p s, the s Z (p k. O the other had, if p s, sice p is a prime, s Z (p k. So ϕ(p k = p k p k, the order of Z(p k mius the umber of multiples of p that less tha p k. (c By the multiplicative property of ϕ ad (b, ϕ( = ϕ(p a p a ( p p a k k ( p k = k i= ( pi. 5. If is a positive iteger, show that p a 2 2 p a k k = ϕ(pa ϕ(p a 2 2 ϕ(p a k k = = d ϕ(d, where ϕ is the Euler phi-fuctio. [Hit: There are precisely ϕ(/d itegers m with gcd(m, = d.] Note that { i } { j } : i = d d : j d, gcd(d, j = =: d A d 2

3 ad {A d } d are pairwisely disjoit. So oe completes the proof by computig the cardiality of sets i both sides. 6. Write dow the characters of the groups Z (3, Z (4, Z (5, Z (6, ad Z (8. (a Which oes are real, or complex? (b Which oes are eve,or odd? otherwise. (A character is eve if χ( =, ad odd Sice Z (3, Z (4, ad Z (6 are all = Z(2 = {0, }, their characters cotai the trivial oe ad the oe χ(0 =, χ( = oly, both are real ad eve. For Z (5 = Z(4 = {0,, 2, 3}. The characters are χ j (k = e 2πi j 4 k (j, k = 0,, 2, 3. So χ 0, χ 2 are real. χ, χ 3 are complex. Oly χ 0 is eve. For Z (8 = Z(2 Z(2 = {(0, 0, (, 0, (0,, (, }. Because of (, 0 + (, 0 = (0, 0, χ((, 0 = ± for each character χ. Same for (0, ad (,. So every character is real. Note that A + B = C for {A, B, C} = {(, 0, (0,, (, }, so appears twice or ever appears i the values that each character takes at {A, B, C}. Hece the eve character are the trivial oe ad the oe χ((, = χ((0, 0 = ad χ((, 0 = χ((0, = 7. Recall that for z <, ( log = z k z k. k We have see that e log ( z = z. (a Show that if w = /( z, the z < if ad oly if Re(w > /2. (b Show that if Re(w > /2 ad w = ρe iϕ with ρ > 0, ϕ < π, the log w = log ρ + iϕ. [Hit: If e ζ = w, the the real part of ζ is uiquely determied ad its imagiary part is determied modulo 2π.] Remark. (a is the Möbius trasformatio. (a ca be proved by brutal computatios ad Arithmetic-Geometric Meas iequality. (b As hit, e log ρ+iϕ = ρe iϕ = w = z for some z < from (a. The e log ρ+iϕ = z = elog ( z = e log w. 3

4 8. Let ζ deote the zeta fuctio defied for s >. (a Compare ζ(s with x s dx to show that (b Prove as a cosequece that p ζ(s = s + O( as s +. ( p = log + O( as s +. s s (a Use mea-value theorem, oe has ζ(s x = s = + s x dx = s (b is a cosequece of (a ad the fact log ζ(s = p = + s x dx s s. s+ = +O( proved i Propositio.. p s 9. Let χ 0 deote the trivial Dirichlet character mod q, ad p,, p k the distict prime divisors of q. Recall that L(s, χ 0 = ( p s ( p s ζ(s, ad show as a cosequece L(s, χ 0 = ϕ(q q + O( as s + s k Note that, by Exercise 8 ad mea-value theorem to f(s = k ( p s j, L(s, χ 0 = k [ k ( p s j ζ(s = ( p s = O(s ( s + O( + ϕ(q q j k ] ( p j ζ(s + ϕ(q q s + O(. ζ(s 0. Show that if l is relatively prime to q, the p l p = ( s ϕ(q log + O( as s +. s This is a quatitative versio of Dirichlet s Theorem. 4

5 . Use the characters for Z (3, Z (4, Z (5, ad Z (6 to verify directly that L(, χ 0 for all o-trivial Dirichlet characters modulo q whe q = 3, 4, 5, ad 6. [Hit: Cosider i each case the appropriate alteratig series.] 2. Suppose χ is real ad o-trivial; assumig the theorem that L(, χ 0, show directly that L(, χ > 0. [Hit: Use the product formula for L(s, χ.] 3. Let {a } = be a sequece of complex umbers such that a = a m if = m mod q. Show that the series coverges if ad oly if q = a = 0. [Hit: Summatio by parts.] = a Let A j = j k= a k with covetio that A 0 = 0. Recall that N = a = N = A ( + + A N N + The periodicity implies that ([x] is the floor fuctio of x. A N = A q [ N q ] + O(. So the secod term is always bouded. A q = 0. Moreover, the first term coverges if ad oly if 4. The series F (θ = =0 e iθ, for θ < π, coverges for every θ ad is the Fourier series of the fuctio defied o [ π, π] by F (0 = 0 ad i( π θ if π θ < 0 F (θ = i(π θ if 0 < θ π, ad exteded by periodicity (period 2π to all of R (see Exercise 8 i Chapter 2. 5

6 Show also that if θ 0 mod 2π, the the series E(θ = = e iθ coverges, ad that E(θ = ( 2 log + i 2 2 cos θ 2 F (θ 5. To sum the series = a / with a = a m if = m mod q ad q = a = 0, proceed as follows. (a Defie A(m = q a ζ m where ζ = e 2πi/q = Note that A(q = 0. With the otatio of the previous exercise, prove that = a = q [Hit: Use Fourier iversio o Z(q.] q m= A(mE(2πm/q. (b If {a m } is odd, (a m = a m for m Z, observe that a 0 = a q = 0 ad show that A(m = <q/2 (c Still assumig that {a m } is odd, show that = a = 2q q m= a (ζ m ζ m. A(mF (2πm/q. [Hit: Defie Ã(m = q = a ζ m ad apply the Fourier iversio formula.] 6. Use the previous exercises to show that π 3 3 = , which is L(, χ for the o-trivial (odd Dirichlet character modulo 3. 6

7 2 Problems. Here are other series that ca be summed by the methods i (a For the otrivial Dirichlet character modulo 6, L(, χ equals π 2 3 = , (b If χ is the odd Dirichlet character modulo 8, the L(, χ equals π 2 2 = , (c For a odd Dirichlet character modulo 7, L(, χ equals π 7 = , (d For a eve Dirichlet character modulo 8, L(, χ equals log( = , (e For a eve Dirichlet character modulo 5, L(, χ equals 2 5 log ( + 5 = , 2. Let d(k deote the umber of positive divisors of k. (a Show that if k = p a p a is the prime factorizatio of k, the d(k = (a + (a +. Although Theorem 3.2 shows that o average d(k is of the order of log k, prove that the followig o the basis of (a: (b d(k = 2 for ifiitely may k. (c For ay positive iteger N, there is a costat c > 0 so that d(k c(log k N for ifiitely may k. [Hit: Let p, p N be N distict primes, ad cosider k of the form (p p 2 p N m for m =, 2,.] (a(b are easy. (c 7

8 3. Show that if p is relatively prime to q, the χ ( χ(p ( g, = p s p fs where g = ϕ(q/f, ad f is the order of p i Z (q (that is, the smallest for which p mod q. Here the product is take over all Dirichlet characters modulo q. 4. Prove as a cosequece of the previous problem that L(s, χ = χ where a 0, ad the product is over all Dirichlet characters modulo q. a s, 8