Solutios to Problem Sheet ) Use Theorem.4 to prove that p log for all real x 3. This is a versio of Theorem.4 with the iteger N replaced by the real x. Hit Give x 3 let N = [x], the largest iteger x. The, importatly, the sets of itegers { : x} ad { : N} are equal. Therefore sums ad products) over both sets are equal, i.e. for ay terms a, x a = N a. Solutio Give x 3 let N = [x], the largest iteger x. So N 3 ad N x < N +. The p = p, p N because there is o iteger betwee N ad x. We ca ow apply Theorem.4, loglogn +) log, p sice N + > x. p N 2) Corollary.2 states that for iteger N. i. Prove that for N. logn +) N logn + N N logn +, 8) logn + ii. Why is the lower boud i Part i better tha that i 8)? iii. Prove that x + 6
for all real x. Solutio i N = N + N logn + N, by the lower boud i 8) with N replaced by N. ii. logn +) = + dt N t = dt N+ t + dt N t = logn + + N dt t logn + N + N dt iii Let N = [x] whe thus = logn + N. x logn +) = x N, logn + < +. The idea b a)glbf t) [a,b] b a f t)dt b a)lubf t), 9) [a,b] has lots of applicatios, ad oe of the most importat is see i the ext questio. Whe ca a sum be replaced by a itegral? 3) Boudig a Sum by a Itegral. Let f be a fuctio itegrable o [,N]. Prove that for itegers N >, 7
i) if f is icreasig ii) if f is decreasig ft)dt+f) N ft)dt+fn) N f) f) ft)dt+fn), ft)dt+f). These two parts ca be summed up by sayig that if f is mootoic the mif),fn)) for all itegers N >. Hit Apply 9). N Solutio i) If f is icreasig the f) ft)dt f) ft)dt maxf),fn)), + ft)dt. Sum the left had iequality over + N, ad sum the right had oe over N. Thus we get the stated result. ii) If f is decreasig the + ft)dt f) ft)dt. Sum the left had iequality over N, ad sum the right had oe over + N. Thus we get the stated result. 4) i. Use Questio 3 to prove that for itegers N N logn N ) N log N +)logn N ) ii educe that e ) N N N! en e ) N N, 20) e 8
for N. This is a weak result, there is a factor of N differece betwee the upper ad lower bouds. Which boud is closer to N!? Perhaps it lies somewhere i the middle? See later questios for the aswers. Solutio i. From Questio 3i logtdt+log Itegratio by parts gives Combiig gives stated result. ii. N! = explogn!) = exp N log logtdt = N logn N ). log Take the expoetial of Part i gives logtdt+logn, ) N N ) = exp log. expn logn N )) N! expn +)logn N )). For the lower boud = = expn logn N )) = exp logn N) exp N)exp = N N e N e = e ) N N. e N. The upper boud is idetical other tha a extra factor of explogn) = 5) i) Use Questio 3 to prove that for σ > 0, σ N σ N for all itegers N >. N σ σ σ σ σ, 2) ii) You caot substitute σ = ito part i because of the σ i the deomiator but what is the limit lim σ N σ σ? σ 9
What does the result 2) become uder the limit σ? Solutio i) Apply the result i Questio 3ii) with the decreasig fuctio f) = / σ to get dt t σ + N σ N N σ dt t σ + σ. A straightforward itegratio gives the stated result. ii) The most direct approach is to apply L Hopital s Rule. I leave this to the studet. Otherwise, perhaps ote that The N σ σ σ = σn/) σ. σ N/) σ lim σ σ is, i fact, the defiitio of the derivative ) σ d N = d dσ σ= dσ e σ)logn/) σ= = logn/) e σ)logn/) σ= = logn/). The the Product Rule for limits gives N σ σ lim σ σ = lim σ σ σ= σ= = logn/). lim σ Thus uder the limit of σ 2) becomes N ) N log. N 20 N/) σ σ σ=
6) Show that ) < p, 22) for all real x 3. Thus deduce that lim ) = 0. x p I which case we say that the ifiite product diverges. Hit Replace x by a iteger, look at the iverse of the product, ad use ideas ad results from the proof of Theorem.4. Later i the course we will show that, with a appropriate costat c, we have ) c p as x, where f x) gx) as x meas lim x f x)/gx) =. Solutio Let N = [x], so N x < N + ad ) = p p N p sice there are o itegers betwee x ad N. Cosider the iverse, whe p N ) = p N as see i the proof of Theorem.4. As also see i that proof N {,2,...,N} so logn +) >. N N Combiig all three displayed results we get ) < p. 2, ),
7) Let πx) =, the umber of primes less tha equal to x. The ifiitude of primes is equivalet to lim x πx) = Prove that πx) c for some costat c > 0. Justify each step i the followig argumet. With > a costat to be chose πx) < e 2, < p p 2 log log )) for a appropriate choice of. What is that choice ad why? Solutio With > a costat to be chose, we throw away some terms to get a lower boud: πx) = <. as The, we rearrage the lower coditio i the summatio, )/ < p p. This lower boud o ca be used as < < p = < p. 22
Write this sum as a differece of two ad replace the secod by the summatio over all itegers 2 /, p = p p < Usig Questios ad 2iii gives log log p p 2 )).. You ca see ow why we eeded results o sums over p x ad x with real x rather tha p N ad N for iteger N. This justifies Questios ad 2iii.) The log log x factors cacel, leavig πx) log. The costat is still to be chose. Try to choose it to maximise log )/. Though it is of iterest to derive a result o πx) = from the weighted sum /p, later i the course we will show that πx) > Cx/, for some costat C > 0, which is a far larger lower boud. 23