Chapter 1 Additional Questions

Transcription

1 Chapter Additional Questions 8) Prove that n=3 n= n= converges if, and only if, σ >. nσ nlogn) σ converges if, and only if, σ >. 3) nlognloglogn) σ converges if, and only if, σ >. Can you see a pattern? Solution By Question 8 these series converge if, and only if, the integrals t σ, tlogt) σ and 3 tlogtloglogt) σ converge. But note, that if you substitute y = logt in the second integral and y = loglogt in the third they are all of the form c with c =, log or loglog3. Thus all three integrals, and thus all three sums, converge iff σ >. For the general pattern define the iterated logarithm by log r+ t = loglog r t) for r, log t = logt and, for simplicity, log 0 t = t. The subscript no longer refers to the base of the logarithm, these are all natural logarithms.) Then n n+ r=0 log rt ) log n t) σ converges iff σ >. dy y σ 9) i Using the observation that log = 0 improve the result of Question 4i to N logn N + log logn N +)logn N + log. n N 4

2 ii. Deduce that for such N, e ) N N N! < N e e ) ) N N. 4) e This is an increase on the lower bound of 0) by a factor of e/ and a decrease on the upper bound by a factor of e/ Alternatively the improvement can be seen in that the lower and upper bounds in 4) differ by a factor of N/ whilst in the earlier 0) the factor is N. Can we do better and have a factor that does not depend on N? See a later Problem Sheet. Solution i Start with the fact that log = 0 in which case logn = logn. n N Then by Question 3i we get N logt+log n N n N logn N logt+logn, since log t is an increasing function. The conclusion of the question then follows from N ii. The result on N! follows from logt = N logn N log ). N! = exp n N logn ). 0) Recall from notes the definition of the set N = {n : p n p N}. Then unique factorisation of integers justifies, for real s = σ >, the last equality in ζσ) = n σ n = ). 5) σ p σ n= n N This is a rather convoluted way of showing that ζσ) > 0 for σ > since this finite product can not be zero to be zero one of the factors would have to 5

3 be zero). Yet can 5) be generalised to complex s so that we can conclude that ζs) 0 for Res >? In the lectures we prove that ζs) ) p s n σ n/ N n N+ n σ σ )Nσ. 6) and deduced that ζs) = p p s ) 7) for Res >. From this we see, on multiplying both sides of 7) by a finite number of terms that, for any N >, p ) ζs) = ). 8) s p s p>n i) Prove that for any M > N, ) p s σ )N σ. N<p M Hint Write the product as a sum and use the same ideas that gave the bound 6). ii) Deduce that given s : Res > there exists N = N s) > such that p ) ζs) s. Hint Take a limit over M in Part i. iii) Deduce ζs) > p s ). Hint Perhaps use the triangle inequality in the form a > b a b. This is our generalisation of 5). The N depends on s but for a given s it is finite and this finite product is never zero and so ζs) > 0, i.e. ζs) 0 for Res >. 6

4 Solution i) By the unique factorisation of integers, N<p M p s ) = n p n N<p M n s = + n>n p n N<p M n s, having noticed that n = has no prime divisors and so the condition p n N < p M is satisfied. All other n have at least one prime divisor which must be larger than N in which case n must be larger than N. Therefore ) = p s N<p M n s n>n p n N<p M n>n p n N<p M n s = n>n p n N<p M n σ n>n n σ σ )N σ by the same argument that gave 6). ii) Since this result is true for all M > N we can let M when 8) tells us the limit of the product exists and even tells us what it is, in which case p ) ζs) s σ )N σ. So it suffices to choose any N > ) /σ ), σ to get p ) ζs) s 7

5 iii) The triangle inequality in the form a > b a b gives p ) ζs) p ) ζs) s s as required. =, ) i) By looking at an integral justify ) log < x x for 0 x < /. ii) Use this to prove a weaker form of Theorem.4, p > loglogn +). It may be weaker, but the proof is shorter.) Hint Part i gives p > log ) so p p > log n N Why? This latter sum over n is seen in the proof of Theorem.4 and so follow the steps found there. Solution i) The assumption 0 x < / implies that x > /. Thus by bounding an integral we get ) log = x t x < x. 9) x x n. 8

6 ii) Looking within the proof of Theorem.4 we see loglogn +) log n N log n N = n n log ) p by 9). Thus we get the stated result. p, ) A function f is convex on [a,b] iff fa+tb a)) fa)+tfb) fa)) for all 0 t. That is, the graph for f between x = a and x = b lies below the chord joining the points a,fa)) and b,fb)) i) Prove that if f is convex then b a ft) b a)fb)+fa)). ii) Prove that /t is concave on R + = {x R : x > 0}. Deduce that ) x log x x x) for 0 < x <. What change does this lead to in Theorem.4? 9

7 Solution i) Change variable t = a+yb a) to get b a ft) = b a) b a) 0 0 fa+yb a))dy fa)+yfb) fa)))dy since f is convex = b a) fa)+ ) fb) fa)) = b a)fb)+fa)). ii) For f t) = /t on R + take any interval [a,b] with b > a > 0. The condition fa+tb a)) fa)+tfb) fa)) becomes a+tb a) a +t b ). a All steps are now reversible i.e. ) a+tb a) a +t b ) a Multiply up ab b+ta b))a+tb a)) = b+ta b). ab = ab+taa b)+bb a)) t a b) = ab+ta b) t a b) = ab+t t)a b), which is a valid inequality since t t) 0 for 0 t. 30

8 For the logarithm ) log x = x t x)) + ) x by part i = x + x x) = x+ = x+ x x). Within the proof of Theorem.4 this gives p > loglogn +). x x) x ) 3) For σ > the Riemann zeta function converges absolutely and so ζσ) N n= for all N, while from Theorem. we have ζσ) = p nσ, 30) p σ ) 3) for such σ. Assume that there are only finitely many primes and use 30) and 3) to obtain a contradiction. Hint For each prime p the factor of the Euler Product is continuous, i.e. lim ) = ), σ σ 0 p σ p σ 0 for all σ 0 R. From second year analysis the finite product or sum) of functions continuous at a point, is continuous at that point, i.e. for two functions f and g if lim σ σ0 fσ) = fσ 0 ) and lim σ σ0 gσ) = gσ 0 ) then lim σ σ 0 fσ)+gσ)) = fσ 0 )+gσ 0 ) 3

9 and lim σ σ 0 fσ)gσ)) = fσ 0 )gσ 0 ). By repeated application these results hold for finitely many terms in the sum or product. Use these facts in your solution. For comparison the infinite sum or product of functions continuous at a point, are not necessarily continuous at that point. Solution If there are only finitely many primes then lim ζσ) = lim σ + σ + p p σ ) = p = p lim p ) σ + σ allowable since a finite product, p), which is finite. Yet from 30) we have lim ζσ) lim σ + σ + N n= n σ = = N n= N n= lim σ + n n σ allowable since a finite sum, logn +) by Corollary.), for each N. This last inequality contradicts the assertion that lim σ + ζσ) is finite. This contradiction means that the last assumption is false, hence there are infinitely many primes. 3

10 4) Prove that Theorem.4, or more precisely Question above, implies ) ) p σ e log, σ for < σ < +/log3. A weaker version of Theorem.3. p Hint Given σ, truncate the sum at x, to be chosen. Get the sum into a one of summing /p, not /p σ. BUT, to simply say p σ p throws away too much information. Instead write p σ = p ) σ p p ) σ, x since p x. Finally use Question and then choose x in terms of σ. Solution With x to be chosen, p σ p p x p = σ p x p ) σ p p p x ) σ = x x σ p, p x since p x for every term in the summation. Thus p loglogx ), σ xσ p by Question. Finally, look at the answer and, to get log/σ )) in place of the loglogx found here, choose logx = /σ ), for which x σ = e σ )logx = e. Note that in Question we require x 3, which becomes /σ ) log3, and rearranges to the given σ < +/log3. 33