Solutions to Problem Sheet 4

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1 Solutions to Problem Sheet 4 ) From Theorem 4.0 we have valid for Res > 0. i) Deduce that for Res >. ζ(s) = + s s [u] ζ(s) = s u +sdu ote the integral contains [u] in place of. ii) Deduce that for 0 < Res <. ζ(s) = s 0 u +sdu, ote how the integral runs from 0 and not. iii) Deduce from (59) that for real σ > 0,σ we have In particular, ζ(σ) < 0 for 0 < σ <. σ < ζ(σ) < σ σ. In Theorem.5 we had such bounds but only for σ >. Hint for Part iii) Use 0 <. Solution i) From (59) we have ζ(s) = s s s for Res > 0, For Res > we have u [ u +sdu = du u s u = s s These combine to give the stated result. u+sdu, (59) u [u] u +s du, ] = s. 58

2 ii) For Res < we see that 0 [ ] u +sdu = u u s 0 u +sdu = = s 0 s. (60) Then with both Res > 0 and Res < the result (59) gives ζ(s) = s s s = s 0 = s 0 u +sdu u +sdu s u +sdu. iii) Using (59) with s = σ > 0 real we get for σ, ζ(σ) = + σ σ having used 0, as an upper bound. u+sdu by (60) u+σdu + σ = σ σ, For a lower bound use < to get ζ(σ) + [ σ σ du u σ = + u+σ σ σ σ = σ. ] 2) Prove that for all θ. Deduce that 5+8cosθ+4cos2θ+cos3θ 0, (6) ζ 5 (σ) ζ(σ +it) 8 ζ(σ +2it) 4 ζ(σ +3it). Thus the results in Lemmas 4.20 and 4.2 are not the only ones of their type. ote that (6) has a property in common with Lemma 4.20, namely the polynomials are zero when θ = π. 59

3 See additional notes on the web site showing how to construct non-negative sums of cosines such as (6) containing cosnθ for any n. Solution Recall that sin 2 θ+cos 2 θ = and cos(α+β) = cosαcosβ sinαsinβ, sin(α+β) = cosαsinβ +sinαsinβ. These two identities are used to show the well-known Perhaps less well-known is cos2θ = cos 2 θ sin 2 θ = 2cos 2 θ, sin2θ = 2sinθcosθ. cos3θ = cos(2θ+θ) = cos2θcosθ sin2θsinθ = ( 2cos 2 θ ) cosθ (2sinθcosθ)sinθ = ( 2cos 2 θ ) cosθ (2cosθ) ( cos 2 θ ) = 4cos 3 θ 3cosθ. The above results can be substituted in to give 5+8cosθ+4cos2θ+cos3θ = 5+8cosθ+4 ( 2cos 2 θ ) + ( 4cos 3 θ 3cosθ ) = 4cos 3 θ+8cos 2 θ+5cosθ+. Write x in place of cosθ. Looking at 4x 3 +8x 2 +5x+ we see, by substituting in small x, that the polynomial has a root at x =, i.e. a factor of x+, and in fact 4x 3 +8x 2 +5x+ = (x+) ( 4x 2 +4x+ ) by observation. Therefore = (x+)(2x+) 2, 5+8cosθ+4cos2θ+cos3θ = (+cosθ)(+2cosθ) 2 0, since cosθ. As in the notes 60

4 for Res >. Also log ζ(s) = Re p m= mp ms Re e itmlogp = Re pms p mσ = cos( mtlogp) p mσ = cos(θ m,t,p) p mσ, say, where θ m,t,p = mtlogp. Thus log ( ζ 5 (σ) ζ(σ +it) 8 ζ(σ +2it) 4 ζ(σ +3it) ) = 5logζ(σ)+8log ζ(σ +it) +4log ζ(σ +2it) +log ζ(σ +3it) = p m= (5+8cos(θ m,t,p )+4cos(2θ m,t,p )+cos(3θ m,t,p )) mp mσ 0, by (6). Exponentiate to get the stated result. 3) In the last Chapter we had Euler s Summation formula: Lemma Let f be differentiable on [, ). Then f (n) = f (u)du+f ()+ n= for any integer 2. i) Apply this with f (u) = /u to deduce that f (u)du. n= n = log + du, (62) u2 for integer. Explain why this can be written as n= n = log + ( ) u du+o. 2 6

5 This gives a second proof of a result already seen in a Theorem Chapter 3, proved by Partial Summation, where the constant u 2 du was called Euler s constant, γ. ii) From Theorem 4.0 deduce that lim s ( ζ(s) s ) = γ. Solution i) If f (u) = /u then for integer, n= n = du u ++ f (u)du. as required. Write this as = log + u 2 du n= n = log + u du+ 2 u du. 2 The integral over the tail can be bounded as u du 2 du u = 2, which is the error in the stated result. ii) Let s in (7) from the notes to deduce ( lim ζ(s) ) = s s by part i. du = γ, u2 4) You cannot put s = into Theorem 4.9: n n = + s s + s s s du u s+, 62

6 because of the s on the denominator. Instead what is the limit as s, of these two terms with s in their denominator, i.e. ( ) lim s s + s? s In this way give an alternative proof of Question 3i. Solution Most simply done by applying L Hopital s rule, so ( ) lim s s + s s = lim s s s = lim s e ( s)log s (log)e ( s)log = lim, by L Hopital s rule, s = log. The result (62) from the last question follows. 5) Apply Euler s Summation with to show that n= logn n = 2 log2 f (u) = logu u, (logu ) du+o u 2 ( log This gives an alternative proof of a result asked for in a level 4 question on Sheet 3. In that question, though, the constant was given as E (x) dx x 2 where logn = n x x logydy +E (x). It may be thought that the interpretation of the constant in this question is more enlightening. Solution From f (n) = n= f (u)du+f ()+ 63 f (u)du ).

7 we get n= logn n = = 2 log2 + logu u du+0+ logu du u 2 logu du u 2 logu du. u 2 On completing the integral to the error in doing so is (logu ) du u 2 logu u du+ du 2 u log 2, by an estimate in a question on Sheet. This is the error in the stated result. 6) The Riemann zeta function has a Laurent Expansion at s =. This is a Taylor series, though finite negative powers are allowed, so ζ(s) = s + c k (s ) k, for s close to, for some coefficients c k, k 0. The /(s ) comes from the simple pole at s = with residue. The result of Question 3 can be rephrased as saying that c 0 = γ, Euler s constant. Prove that the next coefficient in the Laurent expansion satisfies Hint First justify c = and then look at Corollary 4.2. k=0 (logu ) du. u 2 ( ) c = lim ζ (s)+ s (s ) 2, Solution Differentiate the Laurent expansion to give ζ (s) = (s ) 2 +c + 64 c k (s ) k, k=2

8 in which case ( ) c = lim ζ (s)+ s (s ) 2. From Corollary 4.2 in the notes we have ζ (s)+ ( s) 2 = s (logu ) du. u +s Let s to get the result. It is allowable to take the limit inside the integral since the integral converges uniformly for Res δ for an δ > 0. 7) Prove Theorem 4.26, for σ and t > 2 we have Hint Estimate each term in (33) : ζ (s) = n= ζ (σ +it) (logt+7/4) 2. logn s log n s s s (s ) 2 I (s)+si 2 (s), where I (s) = u s+du and I 2(s) = logu du. u s+ Proof Again, given t 2 choose = [t], so 2. Again we estimate each term in (33) separately. So logn ( n s ) logn log log (log +). n σ n n= n= n= By the idea in the last proof s log s log t 2 log and For the remainder terms, firstly I (s) s (s ) 2 t 2 4. u σ+du = σ σ 2. 65

9 Then I (s) using integration by parts. Thus logu log uσ+du σ + σ σ 2 σ si (s) s (log +) 2(log +), σσ the bound s /σ σ 2 having already been seen in the proof of Theorem Collecting these individual bounds we get ζ (σ +it) log log + 4 ( logt+ 7 4) 2. 8) Results in the lectures concern the size of the Riemann zeta function for Res (or just to the left of Res = for the Level 4 students). In this question go to the line Res = /2. Prove that for t 4. ζ(/2+it) 4t /2 + Hint Follow the proof of Theorem 4.25, again making use of Theorem Solution In Theorem 4.24 with t 4 given choose = [t], so 4 and t < +. Then n = + dt + /2 n/2 t = /2 2/2. n s n= n= n=2 Also Finally s s = /2 /2+it /2 t /2. r (s) σ +it σ σ +t/σ /2 +2( +) /2 = 2 /2 + 3 /2. 66

10 Combining we find that for s = /2+it, ζ(s) n s + s s + r (s) since t and 4. n= ( 2 /2 ) + /2 +2/2 + 3 /2 4t /2 +, 9) Assume that the Dirichlet Series n= converges at s 0 C. Prove that the series converges in the half plane to the right of s 0, i.e. for all s with Res > Res 0. Deduce that the Riemann zeta function diverges for all Res <. Hint For the first part show that n n s = s s 0 n n s n s 0 (s s 0) n s 0 n t dt t s s 0+. For the deduction concerning divergence of the Riemann zeta function use proof by contradiction, the contradiction coming from the known fact that n= /nη diverges for any real η <. Solution Partial Summation gives, for any, n = n s n s 0 n s s 0 n = s s 0 n a n n (s s 0) s 0 n s 0 n t dt t s s 0+. In this we let. The first sum n /n s 0 converges by assumption while s s 0 = 0 σ σ 0 67

11 since σ σ 0 = Res Res 0 > 0. In the integral the series is bounded since convergent and so the integral is dt t σ σ 0+, which converges, again since σ σ 0 > 0. Hence the Dirichlet series converges. Assume for contradiction that there exists s 0 with Res 0 < for which the Dirichlet series ζ(s 0 ) converges. Choose σ = (Res 0 +)/2, so Res 0 < σ <. Then Res 0 < σ means that the point σ is in the half plane to the right of s 0 and so, by the Theorem above, ζ(σ) converges. Yet σ is real and σ < hence, as is well-known from first year analysis, ζ(σ) = diverges. This contradiction means that the last assumption is false, hence ζ(s) diverges for all s with Res <. n= n σ 68

ζ (s) = s 1 s {u} [u] ζ (s) = s 0 u 1+sdu, {u} Note how the integral runs from 0 and not 1.

ζ (s) = s 1 s {u} [u] ζ (s) = s 0 u 1+sdu, {u} Note how the integral runs from 0 and not 1. Problem Sheet 3. From Theorem 3. we have ζ (s) = + s s {u} u+sdu, (45) valid for Res > 0. i) Deduce that for Res >. [u] ζ (s) = s u +sdu ote the integral contains [u] in place of {u}. ii) Deduce that for