Solutios to Problem Sheet ) Use Theorem. to rove that loglog for all real 3. This is a versio of Theorem. with the iteger N relaced by the real. Hit Give 3 let N = [], the largest iteger. The, imortatly, for a sum over ay terms, a = N a. Solutio Give 3 let N = [], the largest iteger. So N 3 ad N < N +. The =, N because there is o iteger betwee N ad. We ca ow aly Theorem., loglogn +) loglog, sice N + >. N ) Show that for every iteger k there eist a cosecutive sequece of k comosite itegers i.e. o-rimes). Solutio k +)!+, k +)!+3, k +)!+4,..., k +)!+k +). 3) I the roof of Theorem. it is show that logn +) logn +, for iteger N. Prove that for all real. log N log+
Solutio Let N = [] whe = N, thus log logn +) logn + < log+. The idea glbf t)b a) [a,b] b a f t)dt lubf t)b a), 5) [a,b] has lots of alicatios, though it sometimes eeds a little tweakig. 4) Let f be a fuctio itegrable o [M,N]. Prove that for N > M, i) if f is decreasig M f t)dt+f N) ii) if f is icreasig M f t)dt+f M) f ) M N M f ) M N M f t)dt+f M). f t)dt+f N), These two arts ca be summed u by sayig that if f is mootoic the mif M),f N)) f ) M N M f t)dt maf M),f N)). Hit For Part i follow closely the roof of logn +) N logn +. give i the lectures, though the result here i Part i imlies logn + N N 3 logn +.
Solutio i) Sice f is decreasig we have + f t)dt f ) f t)dt. Sum the left had iequality over M N, ad sum the right had oe over M + N. Thus we get the stated result. ii) Sice f is icreasig we have f t)dt f ) + f t)dt. Sum the left had iequality over M + N, ad sum the right had oe over M N. Thus we get the stated result. 5) Prove that for N N logn N + log educe that for such N, e N ) N N N! < N e log N +)logn N + log. e ) ) N N. e Hit It might hel to ote that N log = N log. Solutio Start from log = 0 so log = N The by Questio 4 i we get logtdt+log The stated result follows from N N log log. logtdt+logn. logtdt = N logn N log ). 4
If you caot remember, you itegrate logt by arts). The result o N! follows from logn! = log. N 6) i) Prove that for θ > 0, θ N θ M N N θ M θ θ θ M θ. ii)youcaotutθ = itoartibecauseofthe θ ithedeomiator but what is the limit N θ M θ lim? θ θ Solutio i) Aly the result i Questio 4 ii) with f ) = / θ to get M dt t + θ N θ M N N θ M A straightforward itegratio gives the stated result. dt t + θ M θ. ii) The most direct aroach is to aly L Hoital s Rule. I leave this to the studet. The Otherwise, erhas ote that N θ M θ θ is, i fact, the defiitio of = M θn/m) θ. θ N/M) θ lim θ θ ) θ d N = d dθ M dθ e θ)logn/m) θ= θ= = logn/m) e θ)logn/m) θ= = logn/m). 5
The the Product Rule for limits gives lim θ N θ M θ θ = limm θ θ θ= θ= = logn/m) lim θ N/M) θ θ θ= 7) i) Use 5) to show that logy < y for all y >. ii) By a aroriate choice of y i art i show that for all we have log < / for >. iii) educe that for all ε > 0, we have log ε ε. This meas that the logarithm of grows slower tha ay ower of. Here f ) ε g) meas that there eists a costat C = Cε), deedig o ε, such that f ) < Cε)g). Solutio i) For y >, logy = y dt t < y = y. ii) Choose y = / i logy < y < y to get log / < / which rearrages to the required result. iii)letε > 0begive. BytheArchimedeaProertyfortherealumbers R, there eists a iteger such that / < ε. The by art ii log < / ε. The iteger deeds o ε hece log ε ε. 8) A imortat task throughout this course is the estimatig of itegrals. The first is logy) l, with iteger l. A alicatio of 5) gives log) l logy) l log) l. 6) i) Which of these bouds do you thik reresets more truly the rate of growth of the itegral as? 6
Hit Itegrate by arts. ii) Aly a tweak ad slit the itegral at : logy) = l logy) + l Aly the uer boud i 5) to both arts to show logy) l l logy) l. log) l. 7) Hit You may eed to make use of Questio 4 i the itegral over [, ]. iii) Show, usig itegratio by arts a umber of times ad the Part ii, that for m we have m ) logy) = l +j)! l l )! log l+j +O l,m log l+m+ j=0 Solutio i) Itegratio by arts shows that [ ] logy) = y l logy) l +l logy) l+. If we assume 7) for ow, the itegral o the right is of size /log) l+ which is smaller tha the /log) l that comes from the first term. Thus we would thik that the lower boud i 6) most truly reresets the growth of the itegral. ii) Firstly logy) l log) l From Questio 4 we have log < l /l) which rearrages as l)l log) l. Thus logy) l log) l l)l log) l log) l. 7
Secodly logy) l log ) l l log) l. Add these two bouds to get the required result. iii) Itegratig by arts m+ times gives logy) l = m j=0 l +j)! l )! [ t log l+j t ] + l+m)! l )! logy) l+m+. By art i the last term i O l,m /log) l+m+), ad all the costats that arise from the t = i the earlier itegrals ca be absorbed ito this boud. 9) Aother itegral we will meet i the course is the tail ed i.e. from to ) logy) l, 8) y with iteger l. i) Use log < l /l), which follows from Questio 4, to show that logy) l y l)l /. o you thik this uer boud truly reresets the size of the itegral? ii) We caot directly aly 5) to 8) because the iterval of itegratio is ifiite. Istead tweak as i Questio 5 but this time slit the itegral at : logy) l y = logy) l y + l logy). y Now remove the log factors from both itegrals, leavig a itegratio over a ower of y i both cases. Thus show that logy) l log) l y l. 8
Solutio i) The boud is immediate after the substitutio. It does ot rereset the true growth of the itegral, as ca be see by artial itegratio, logy) l y = log)l logy) l +l 9) y If we assume the coclusio of this questio the the itegral o the right of 9) is of size log) l / which is smaller tha the first term log) l / which thus reresets the true growth of the itegral. ii) Firstly logy) l y log )) l y llog)l. Secodly, use log < l /l) as i art i to get Add to get the required result. l logy) y l) l l)l y y ) = l) l. / 0) Show that for all 3. Thus deduce that lim ) log, 0) ) = 0. Hit Relace by a iteger, look at the iverse of the roduct, ad use ideas ad results from the roof of Theorem.. Later i the course we will show, for level 4 studets, that with a aroriate costat c we have ) c log as. Here f ) g) as meas lim f )/g) =.) 9
Solutio Let N = [], so N < N + ad ) = N N sice there are o itegers betwee ad N. Cosider the iverse, whe ) =, as see i the roof of Theorem.. As also see i that roof N {,,...,N} so logn +) > log. N N N Combiig all 3 dislayed results we get ) log. ) For σ > the Riema zeta fuctio coverges absolutely ad so ζσ) N = for all N, while from Theorem.8 we have ) σ ), σ, ) ζσ) = for such σ. Assume that there are oly fiitely may rimes ad use these two facts to obtai a cotradictio. Hit For each rime the factor of the Euler Product is cotiuous, i.e. lim ) = ), σ σ 0 σ σ 0 for all σ 0 R. From secod year aalysis the fiite roduct or sum) of fuctios cotiuous at a oit, is cotiuous at that oit, i.e. for two fuctios f ad g if lim σ σ0 f σ) = f σ 0 ) ad lim σ σ0 gσ) = gσ 0 ) the lim f σ)+gσ)) = f σ 0 )+gσ 0 ) σ σ 0 0
ad lim f σ)gσ)) = f σ 0 )gσ 0 ). σ σ 0 By reeated alicatio these results hold for fiitely may terms i the sum or roduct. Use these facts i your solutio. Note that the ifiite sum or roduct of fuctios cotiuous at a oit, are ot ecessarily cotiuous at that oit. Solutio If there are oly fiitely may rimes the lim ζσ) = lim σ + σ + σ ) = = lim ) σ + σ allowable sice a fiite roduct, ), which is fiite. Yet from ) we have lim ζσ) lim σ + σ + N = σ = = N = N = lim σ + σ allowable sice a fiite sum, logn +) by4),foreachn.thislastiequalitycotradictstheassertiothatlim σ + ζσ) is fiite. This cotradictio meas that the last assumtio is false, hece there are ifiitely may rimes.
) Let π) = be the umber of rimes less tha equal to. The ifiitude of rimes is equivalet to lim π) = Prove that π) clog for some costat c > 0. Justify each ste i the followig argumet. With > a costat to be chose log π) log < log < log log e 3 log. log loglog log ) log )) + Solutio With > a costat to be chose, we throw away some terms to get a lower boud: π) = log <. The, we rearrage the lower coditio i the summatio, log/ < as log. This lower boud o ca be used as log < log < log = log log < Writethissumasadiffereceoftwoadrelacethesecodbythesummatio over all itegers log/, log = log log < log log log..
Usig Questios ad 3 gives log loglog The log log factors cacel, leavig log ) log )) +. log log. Thecostatisstilltobechose. Trytochooseittomaimiselog )/. Though it is of iterest to derive a result o π) = from the weighted sum /, later i the course we will show that π) > C/log, for some costat C > 0, which is a far larger lower boud, 3) Prove that coverges for σ > 0. m= m+) σ m Show that the Euler roduct ) + s ) coverges for Res > 0. Solutio As i 8) we have m+) σ m m σ+. N N m +σ du u N σ =. +σ σ This itegral coverges as N sice σ > 0. Thus /m+σ coverges for such σ. Therefore, by the comariso test for series, This series m= m+) σ m 3
coverges for σ > 0. As see i the otes the Euler roduct will coverge for those s for which s ) coverges. Yet s ) σ ) = m= m+) σ m relacig the sum over rimes by oe over all itegers, ad the chagig to a sum over m =. The result o Euler Product tha follows from the first art of this questio. 4