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.

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Stanley Riley
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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 0 < Res <. {u} ζ (s) = s 0 u +sdu, ote how the integral runs from 0 and not. iii) Deduce from (45) that for real σ > 0, σ we have σ < ζ (σ) < In particular, ζ (σ) < 0 for 0 < σ <. σ σ. In Theorem.5 we had such bounds but only for σ >. Hint for Part iii) Use 0 {u} <. 55

2 2. Let a n C be a sequence of coefficients and set A (x) = n x a n. i) Use Partial Summation to prove n= a n n = A () + s s s A (t) dt t+s, (46) ii) Assume that there exists a constant C > 0 such that A (x) C for all x >. Prove that the Dirichlet Series converges and satisfies F (s) = n= a n for all Res > 0. F (s) C s σ 3. i) Prove, using the previous question, that the Dirichlet Series converges for all Res > 0. F (s) = ( ) n+ n= ii) For the Dirichlet series F (s) defined in Part i, prove that ( F (s) = ) ζ (s) 2 s for Res >. ote that we can now define ζ (s) for Res > 0, s, by ζ (s) = ( ) F (s). (47) 2 s In this way we have a continuation of ζ (s) to the larger half plane Res > 0. 56

3 Hint: For Part ii consider the partial sums 2 n= ( ) n+ and 2 n=, expressing each as sums over even and odd integers. Combine and then let. 4. i) Prove that for all θ cosθ + 4 cos 2θ + cos 3θ 0, (48) ii) Deduce that ζ 5 (σ) ζ (σ + it) 8 ζ (σ + 2it) 4 ζ (σ + 3it). Thus the results in Lemmas 3.7 and 3.8 are not the only ones of their type. ote that (48) has a property in common with Lemma 3.7, namely the polynomials are zero when θ = π. 5. Recall n ( ) n = log + γ + O, (49) where γ is Euler s constant. From Theorem 3. deduce that lim s ( ζ (s) s ) = γ. Hint Go back to the lecture notes to find expressions for γ. 57

4 6. You cannot put s = into Theorem 3.0: n n = + s s + s s s {u} du u s+, 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 (49). 7. Apply Euler s Summation with to show that n= log n n = 2 log2 f (u) = log u u, {u} (log u ) du + O u 2 ( log ), for integers. 8. 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 5 can be rephrased as saying that c 0 = γ, Euler s constant. Prove that the next coefficient in the Laurent expansioatisfies c = {u} 58 k=0 (log u ) du. u 2

5 Hint First justify ( ) c = lim ζ (s) + s (s ) 2, and then look at Corollary Prove Theorem 3.25, for σ and t > 2 we have Hint Estimate each term in (28) : ζ (σ + it) (log t + 7/4) 2. ζ (s) = n= log n s log s s (s ) 2 I (s) + si 2 (s), where I (s) = {u} u s+du and I 2 (s) = {u} log u du. u s+ 0. Results in the lectures concern the size of the Riemann zeta function for Res. In this question we go to the line Res = /2. Prove that for t 4. ζ (/2 + it) 4t /2 + Hint Follow the proof of Theorem 3.22, again making use of Theorem

6 . 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 a n n = s s s 0 n a n a n n (s s 0) s 0 a n 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 <. 60

7 Level 4 Problems Ch 3 2. i) Let a n C be a sequence of coefficients and set A (x) = n x a n. Prove that n=+ a n () = A + s ns s n= A (t) dt. ts+ ii) Assume there exists a constant C > 0 such that A (x) < C for all x. Let a n a n F (s) = and F (s) =. n= Show that F (s) F (s) C ( + s ), σ σ for all. Thus deduce that for any δ > 0 and T > 0 the Dirichlet series for F (s) converges uniformly in the semi-infinite rectangle {s = σ + it : σ δ, t < T }. Aside This means that the Dirichlet series ( ) n n= of Question 3 is holomorphic for Res > 0 in which case (47) gives an analytic continuation of ζ (s) to Re s > Generalise Question 3, replacing by any ζ C satisfying ζ = along with ζ. i) Prove that ζ n 2 ζ n= 6

8 for all. ii) Deduce that n= ζ n converges uniformly for all Res δ, for all δ > 0. iii) Deduce that as long as θ 2πk for any k Z, the Dirichlet Series converges for all Res > 0. n= sin nθ 4. i) Apply Euler s Summation formula with l Z, to show that f (u) = logl u u, n= log l n n = l + logl+ {u} logl u (log u l) du+o u 2 ( log l ). ii) Apply Euler s Summation with to show that f (u) = logl u u s, ζ (l) l! (s) + = ( )l+ l+ ( s) for Res > 0, s. {u} logl u (s log u l) du. u +s 62

9 5. The result of question 4 can be written as n log l n n = l + logl+ + C l + O ( log l ), for some constant C l. Prove that C l = ( ) l l!c l, where c l is the l-th coefficient in the Laurent expansion for ζ (s) around s =. Hint Look at the Laurent expansion as given in Question 8. How can c l be expressed as the l-th derivative of ζ (s)? How can Question 4 then be used? 63

10 Additional Questions Ch3. 6. Generalise Question 0 and show that for t 4 and /2 σ <. ζ (σ + it) t σ σ + 2t σ +, 7. For t > 4 prove that ζ (/2 + it) 4t /2 log t + 4t /2 + 2 log t + 25/8. Hint Either differentiate (25) and (26) or use (28). Then estimate each term. You may estimate some terms differently to how I do in the solutioo you may end with different coefficients, though you should be able to get the coefficient of the leading term, t /2 log t, to be 4 as shown. 64

Solutions to Problem Sheet 4

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