1+x 1 + α+x. x = 2(α x2 ) 1+x

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1 Math 2030 Homework 6 Solutios # [Problem 5] For coveiece we let α lim sup a ad β lim sup b. Without loss of geerality let us assume that α β. If α the by assumptio β < so i this case α + β. By Theorem 3.7 there exists N N such that b < β + for all N. O the other had sice α implies a ) for all M R there exists N N such that a M β for all N. Therefore for all max{n N } we get a + b M β + β + M. Cosequetly a + b ) ad hece lim sup a + b α + β. If β the by assumptio α > so α + β ad there is othig to prove i this case. Fially if < α β < the for all ɛ > 0 there exists N N 2 N such that a α + ɛ/2 for all N b β + ɛ/2 for all N 2. Thus for all max{n N 2 } there holds a + b α + β + ɛ. Now let a k + b k ) be a subsequece of a + b ) covergig to lim sup a + b ). Choosig k 0 such that k0 > max{n N 2 } we see that a k + b k α + β + ɛ for all k k 0. Lettig k yields lim sup a + b ) α + β + ɛ. Sice ɛ is arbitrary we get the desired iequality. [Problem 7] We first make some geeral observatios about the sequece. For coveiece we let fx) α+x +x. Note that fx) α α + x α x). I particular whe x is positive we have fx) < α if ad oly if x > α. Moreover if we iterate the fuctio f twice ad compare the result with x we get ffx)) x α + α+x +x + α+x +x x 2α x2 ) + α + 2x. Thus for x > 0 we have ffx)) > x if ad oly if x < α. ) & 2) We will prove parts ) ad 2) simultaeously. Sice x > α by the first observatio above we have x 2 fx ) < α. Suppose we ve prove that x 2 > α ad x 2 < α for some the agai by the same observatio we get x2+ fx 2 ) > α ad x 2+2 fx 2+ ) < α. Hece by iductio the odd terms are always larger tha α whereas the eve terms are always smaller. The by the secod observatio above we get x 2+ ffx 2 )) < x 2 sice x 2 > α. Similarly we have x 2+2 ffx 2 )) > x 2 sice x 2 < α. 3) Sice x 2 ) is decreasig ad bouded below by α we kow that it coverges to some limit x > 0. Moreover passig to limits i the equatio x 2+ x 2 2α x2 2 ) + α + 2x 2 we get 0 2α x2 ) +α+2x which implies that x α sice x is positive. Similarly we ca show that x 2 ) coverges to α. Thus for all ɛ > 0 there exists some N N 2 N

2 2 such that x 2 α < ɛ for all N x 2 α < ɛ for all N 2. Thus x α < ɛ for all 2 max{n N 2 } ad we are doe. [Problem 20] For all ɛ > 0 sice p ) is a Cauchy sequece there exists N N such that dp p m ) < ɛ 2 for all m N. For all N sice the subsequece p k ) coverges to p there must exist k N such that k > N ad dp k p) < ɛ 2 ad thus Hece p ) coverges to p i X. dp p) dp p k ) + dp k p) < ɛ 2 + ɛ 2 ɛ. #2 ) Let x ) be a Cauchy sequece i 0 ] with respect to d. The sice 0 < x for all we observe that x x m x x m x x m dx x m ). Hece x ) is a Cauchy sequece with respect to the stadard metric as well ad by the completeess of [0 ] with the stadard metric there exists x [0 ] such that lim x x 0. Next we claim that x 0 for if ot the we would have lim x 0. For all x 0 ] ad R > 0 choose N N such that 0 x < x/ + xr) for all N. The for all N we have dx x) x x > R implyig that the sequece x ) is ot bouded with respect to d cotradictig the assumptio that it is Cauchy. Hece x 0 ] ad it remais to show that x ) x with respect to d but this follows directly from Theorem 3.3. Specifically sice x ) x i the stadard metric ad x > 0 we have lim x x 0 ad therefore x ) x with respect to d. To see that [ ) is icomplete with respect to d we cosider the sequece x. The for all N N we have dx x m ) m 2 for all m N N which implies that x ) is a Cauchy sequece with respect to d. coverge with respect to d to some α [ ) the we would have lim α 0 ad thus /α 0 which is clearly impossible. If x ) were to

3 2) To avoid havig too may subscripts for a sequece s l p we will deote its terms by s) s2) ). Also we will oly preset the argumets for p < sice the case p follows from what we did i class. I fact l is exactly the same as BN; R).) Take a Cauchy sequece s ) i l p. We first otice that for all m ad k N by the defiitio of the p -orm we have s k) s m k) s s m p. Cosequetly for all k N the sequece s k)) N is Cauchy i R ad hece coverges to some limit which we deote by sk). To coclude we eed to show that this sequece s s) s2) ) we just defied lies i l p ad that lim s s p 0. To see that s l p ote that sice s ) is a Cauchy sequece i l p there exists some R > 0 such that s p R for all N. Thus for all N N N s k) p s p p R p. Recallig that sk) is the limit of s k)) N for all k we ifer that for every N N N N sk) p lim s k) p R p which implies that s l p ad i fact s p R. To see that lim s s p 0 take ay ɛ > 0; the sice s ) is a Cauchy sequece i l p there exists N N such that s s m p < ɛ/2 for all m N. I particular M s k) s m k) p < ɛ/2) p for all M N ad m N. Fixig m ad lettig ted to ifiity we get M sk) s m k) p ɛ/2) p for all M N m N. This implies s s m p ɛ/2 < ɛ for all m N ad we are doe. #3 ) By direct computatio we get fx) fy) α x y. + x) + y) Sice x y 0 ad < α < 2 we get fx) fy) α ) x y. 2) We prove this by iductio. Of course it holds for 0. Suppose x + x α ) x x for some 0 the we get x +2 x + fx + ) fx ) α ) x + x α ) + x x 0. Thus we are doe by iductio. 3) For all > m ote that x x m km x k+ x k α ) k x x 0 km m α ) m α ) x x 0 2 α α )m 2 α x x 0. 3

4 4 This implies that diam E N α )N x x 0 2 α. Sice α 2) the right-had side coverges to zero as N ad we coclude that x ) is a Cauchy sequece. Deotig the limit by x the by applyig Theorem 3.3 ad passig to the limit i the relatio x + α+x +x we see that x α+x +x which gives x α sice x 0. #4 ) Fix x [0 ] ad deote a + x ). We will prove that a ) is icreasig ad bouded from above. By the biomial theorem a x ) k x ) k ) k + ) x) x k ) k ) + x k ) k + + x k ) k ) a Thus a ) is icreasig. The above calculatios also tells us that a + x k ) 2 k ) 3 where i goig from the first lie to the secod we used the fact that 2 k for all k a fact that ca be verified easily by iductio. Sice a ) is icreasig ad bouded from above we coclude that it coverges i R. 2) It suffices to cosider oly those > N 0 2 max{ a b }. The agai by the biomial theorem we have + a + b ) 2 + ) a ) b k 2 + a ) k b k 2k + a ) b k k + a ) where i the last iequality we used the fact that / 2k / k for all k. We pause here to ote that by a calculatio similar to that i ) for all > N 0 we have + a ) N 0 + kn 0 + a N 0 N 0! N 0 N 0 + ) k

5 N 0 N 0 + a N 0 N 0! + a N 0 N 0! 2 k N 0 kn 0 + ) ) where the very last lie is a umber idepedet of. Hece the sequece + a is bouded say by some costat M. Pickig up where we left off i the previous strig of iequalities we get that for all > N 0 + a + b ) 2 + a ) M b k k M 2 k M. This clearly implies the desired result. 3) By part ) we have covergece for x [0 ]. ) ) Fix x 0 ] we prove by iductio that for all N N the sequece + Nx coverges. The case N is part ). Assume the claim for N the the followig sequece coverges as. + Nx ) + x By part 2) this implies that ) N + )x + + N+)x ) + Nx2 2. ) ) coverges as well completig the proof by iductio. Now sice every x > ca be writte as Ny for some N N ad y 0 ] we get the desired covergece for all x ). It remais to treat egative values of x. For this we ote that for all x < 0 + x ) + x ) x2 2 which coverges to as by part 2). Sice x ) coverges to some o-zero limit we coclude that + x ) coverges as well. 4) Take x y R ad ote that + x ) + y ) ) + x + y + xy ) 2 which coverges to expx + y) by part 2). Hece expx) expy) expx + y). To get the last iequality we use the biomial theorem agai to get + x ) x k ) x k k x k x x 3 x. 2k 5