Assignment 5: Solutions

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1 McGill Uiversity Departmet of Mathematics ad Statistics MATH 54 Aalysis, Fall 05 Assigmet 5: Solutios. Let y be a ubouded sequece of positive umbers satisfyig y + > y for all N. Let x be aother sequece ad suppose that the it exists. Prove that x + x y + y x y = x + x y + y. Hit: You may use the Problem 3 o Assigmet 4. Usig the above result, prove that for ay p N the followig holds: a b c p + p + + p p+ = p + p + + p p = p + 3 p + + p p+ = p Solutio: Let x + x y + y = L. The, for ay ɛ > 0, N N such that for N, x + x L y + y < ɛ L ɛ < x + x < L + ɛ. y + y Usig that y is a strictly icreasig sequece, L ɛy + y < x + x < L + ɛy + y, N Summig from some 0 N to some m 0 +, m m m L ɛ y + y < x + x < L + ɛ y + y. = 0 = 0 = 0 Sice we have a telescopig sum, the previous equatio is equivalet to: L ɛy m y 0 < x m x 0 < L + ɛy m y 0

2 Sice y is ubouded, we ca choose m large eough so that y m is positive; the we may divide by y m without chagig the sides of the iequalities: L ɛ y 0 < x m x 0 < L + ɛ y 0 y m y m y m y m Sice y is ubouded, taig the it as m gives us: x m L ɛ L + ɛ. m y m Sice ɛ is a arbitrary positive umber, we have after replacig the dummy m by : as we wated to show. x y = L = x + x y + y a We wat to compute if it exists. Expadig, + p + p+ p+ = p + p = p + p+ = + p + p+ p+ p p p = + p = + p+ = p Taig the it as, + p + p+ p+ = b Let By the theorem above, p + p + + p p+ = The p + p + + p p = = p p = = p p+ p x + x = + p + p+ + p+ y + y [ + p p. ] Usig biomial theorem, the above expads ito [ p + p p + ] [ p p = p p + [ p p + ] p p = p p + p =3 = x y. ] p+

3 Cacellig the p terms ad regroupig the p terms, we get [ ] p p+! p!! p + p = Dividig top ad bottom by p, we get p p+! p!! p [ p p + p = + p = p [ p + p = p p p =3 p+ ]. p + p =3 + ]. + Sice each -depedet term has a egative expoet, we will get i the it as, p p p+! p!! p = p p+pp! p!! p = =. Sice x + x y + y =, it follows from our theorem above that p + p + + p p c Wat to compute = x y =. + + p + p+ p+ = + 3 p + p+ p+. Expadig, + 3 p + p+ p+ = p p + p = p + p+ = p 3 p = p p + p = + p+ = p 3. Taig the it as, + 3 p p + p+ = p+. By the theorem above, p + 3 p + + p p+ = p.. Let x ad y be two sequeces defied recursively as follows: x = a 0, y = b 0, x + = x y, y + = x + y, 3

4 Prove that the sequeces x ad y are coverget ad that x = y. Solutio: Lemma. For ay s, t R, Proof. It is clear that Expadig, st s + t. s t 0 s st + t 0 s + t st I particular, if s = u ad t = v, the for ay u, v 0, The, for ay, u + v uv. x + = x y, y + = x + y Sice x, y 0, x, y 0 for by defiitio. We ca therefore use our lemma to coclude that for, x y. The for, x + = x y x = x, y + = x + y y = y. Thus, x + is a icreasig sequece ad y + is a decreasig sequece. Sice x + y + y for 3, we iductively coclude x + y 3. Thus, x + is a mootoically icreasig sequece bouded above, ad coverges by the mootoe covergece theorem. Sice y + 0 for 0, y + is a mootoically decreasig sequece bouded below ad coverges by the mootoe covergece theorem. Lettig L be the it of x +, ad thus of x, ad oticig: we have: 3. Prove that for all N. y = x + x y = x y x, = L L = L = x. + < l + < Solutio: 4

5 It was show i class that + < e < + + for all N. Taig logarithms yields [ l + ] = l + [ < < l + ] + = + l + The left part of this iequality implies that l + < l + which is what we had to show. 4. Prove that the sequece coverges. for all N. Combiig both iequalities yields + < whereas the right part implies that + < l + < x = l, N, Remar: The it of this sequece is called the Euler-Mascheroi costat; its umerical value is It is curretly uow whether this costat is ratioal or irratioal. Solutio: We start by showig that x is decreasig; we will use the estimates from questio 3. x x + = l + l + + = + l + l + = l = + + l + > 0 by questio 4 i.e. x is decreasig. We will show ext that x is bouded below by 0 i.e. that x 0 for all N. l = l l = l l + l l + + l l = l + l + + l = l + + l l + < by questio 3. Thus x = l > = > 0 Thus x is decreasig ad bouded below ad is thus coverget. 5

6 5. Prove that = l. Solutio: Let x := y := l + + l l + + ad z := The, by questio 3, we have that x < y < z for all N. For y we obtai: [ y = l = l Note that this a telescopig product! [ ] ] = l = l Furthermore, z = x + = x +. Combiig these results yields x < l < x + ad thus l < x < l. Sice l = l = l, it follows from the squeeze theorem that x coverges to l. 6. Let x = Prove that the sequece x coverges. Solutio: x + = , N = x + + > x 6

7 The sequece x is thus icreasig. We show ext that x is bouded above by e, usig estimates obtaied i questio 3 as well as the formula for the sum of a fiite geometric series recall that a + ar + ar + + ar = a r = a r r. [ lx = l = l + By questio 3, l + <. Thus = l ] + + l + lx < = where the right-had side is a fiite geometric series with a = ad r =. Thus lx < = The formula = ca also be proved by iductio. Cosequetly lx < which meas that x < e for all N. The sequece x is thus icreasig ad bouded above ad thus coverges. 7. Let x, x > 0 be a coverget sequece, the x x x 3 x = x Solutio: Let L x ad let y x x x 3 x There are two possibilities. First sceario is that L = 0. I this case, for a give ε > 0, there exists N N such that for all > N, Thus, 0 < x < ε y = x x N x = c x N + x c ε N 7 = = a N c ε N ε ε

8 where c x x N all N, ad a N c ε N. Sice a Thus, choosig N > max {N, N }, we have for all N. Thus, y = 0. a < y < ε = ε =, there exists N N such that for I the secod sceario, L > 0. I this case, by Theorem 3..3 b i Bartle ad Sherbert, we ca equivaletly show that y L = Thus, defiig z = y L, our goal becomes to show that z =. We ow examie the expressio for z : z = y x = x x x L = x x x = x L L x L x L Fixig ε > 0 Without loss of geerality, ε < we may choose N N large eough so that x L < ε L for all N. I particular, we have x > L L ε for N. Thus, x x z > L L = c ε xn L L ε L N where c = N N i= x i L ε N > 0 Oce agai, we shall use the fact that c Thus, choosig N > max {N, N }, we have that z > c = to choose N large eough so that > ε ε ε = ε + 4 > ε 8

9 for N. I a aalogous maer, oe ca esure that N is suitably large eough so that for N, z < + ε it is a exercise to show this rigorously. Summarizig, for a give ε > 0, we have foud N N, such that for all N. Equivaletly stated, This proves the claim. ε < z < ε z < ε N 8. Let x, x > 0 be a sequece such that the it exists. Prove that x + L = x x = L. Usig this result, prove that! = e Solutio: Let y be a sequece defied by y = x y = x for x The, oe ca easily verify that y > 0 for all N ad that y = L We may thus use the result from Exercise 7 to coclude that y y y = L However, we see that y y y = x x x x3 x x = x x 9

10 ad thus x = L. To prove the secod statemet i the problem, we defie the sequece x by x!. We the examie the ratio of cosecutive terms i the sequece, x + = + +! x +! = + +! +! + = = + Thus, by our defiitio of the umber e, we have x + = e which, usig the first result of this problem, implies that x =! = e. 9. Let x = be a positive subadditive sequece. That is, for ay, m N, we have 0 x +m x + x m Show that x exists ad x { = if x } : N Solutio: We let L if { x : N }. We must clearly have L x N To show covergece, we let ε > 0 be give ad choose N, such that a < L + ε 0

11 Note that this is always possible by the defiitio of L. Now for each, we have = p +r for some p N ad r {0,..., }. By iductively applyig the subadditivity coditio, we fid that dividig by, we fid that x = x p +r x p + x r p x + x r for all. x p x + x r p + r Now, sice r {0,..., }, we have that x + x r < L + ε + x r x r max {x,..., x } M Thus, choosig N N such that N > max {, M ε N, we have }, equatios?? ad?? imply that for all L x L + ε ad sice ε was a arbitrary positive umber, we have completed the proof. 0. Let x be a bouded sequece ad for each N let s = sup{x : } ad S = if{s }. Show that there exists a subsequece of x that coverges to S. Solutio: First ote that for N s + = sup{x : + } = sup{x +, x, x,... } = sup{x + } {x, x,... } = max{x +, sup{x, x,... }} sup{x, x,... } = s. Sice x is bouded, M R such that x M for all N. By defiitio, we have that s x M. Therefore, sice N was arbitrary, s N is bouded below by M ad decreasig ad we coclude that S := if s = s. We will iductively costruct a subsequece x N which coverges to S. The first elemet Simply tae x = x. The -th elemet > Assume we costructed x. We wat to fid x such that x S < /. Sice, S = m s m, M N such that s m S < / for ay m M. I particular, we ca fid M > such that s M S < /. Now sice s M = sup M x, M > such that s M x < s M + /. The, x S x s M + s M S < / + / = /. Sice the sequece x N satisfies x S < / for all, by the Archimedea property, ɛ > 0, Kɛ N such that /Kɛ < ɛ x S < ɛ Kɛ ad we coclude x S as.

12 . Let L R. The set L is said to be ope if for ay x L there exists ɛ > 0 such that x ɛ, x + ɛ L. The set L is said to be closed if its complemet L c = {x R : x / L} is ope. a Prove that L is closed if ad oly if for ay coverget sequece x with x L, the it x = x = x is also a elemet of L. b Let x be a bouded sequece. A poit x R is called a accumulatio poit of x if there exists a subsequece x of x such that x = x. We deote by L the set of all accumulatio poits of x. By the Bolzao-Weierstraß Theorem, the set L is o-empty. Prove that L is a bouded closed set. c Let x be a bouded sequece, let L be as i part b ad let S be as i problem. Prove that S = sup L. Solutio: a Let L be a closed set. Let x N be covergig sequece with x L ad x = x. The, i particular, ɛ > 0, N such that x x < ɛ. Therefore, ɛ > 0, ca fid x L such that x x ɛ, x + ɛ. Therefore, ɛ > 0, x ɛ, x + ɛ L c. Equivaletly, ɛ > 0 such that x ɛ, x + ɛ L c. Sice L c is ope, we must coclude that x / L c ad therefore that x L. Let L be a set that is ot closed. The L c is ot ope, which implies that x L c such that N, x, x + L c x, x + L. Hece, N, x L such that x x, x +. We have costructed a sequece x N i L with the property that N, x x <, which implies that x = x where x / L. Therefore, it is ot true that for ay covergig sequece x N with x L, the it x = x is also a elemet of L. Remar: A lot of people did well o the first implicatio. For the secod, a lot of people have cofused L is ot closed ad L is ope : they do ot mea the same thig. Some sets are both ope ad closed clope ad some sets are either closed or ope. Also ote that closed [resp. ope] sets are ot ecessarily closed [resp. ope] itervals ad that i geeral, x ɛ, x + ɛ L c does ot imply x ɛ, x + ɛ L. b Let l L be a accumulatio poit. The there is a subsequece x N such that x = l. I particular, K N such that x K l <. This implies write the details l < + x K. Sice x N is a bouded sequece, M R such that M x for all N ad i particular, M x K. Therefore, l < + M. Sice l L was arbitrary, we coclude L is bouded by M +. Let l m m N be a coverget sequece with l m L ad l = m l m. We wat to costruct a subsequece x Mj,KMj j N of x N that coverges to l. Let x M,KM = x. For j >, sice l = m l m, Mj N such that l Mj l < /j. Sice l Mj L, we ca fid a subsequece x Mj, N such that x Mj, = l Mj. I particular, KMj N such that x Mj,KMj l Mj < /j ad Mj,KMj > Mj,KMj. This defies a subsequece x Mj,KMj j N of x N satisfyig x Mj,KMj l x Mj,KMj l Mj + l Mj l < /j + /j = /j.

13 Usig the Archimedea property as i part a, we get that j x Mj,KMj = l ad we coclude l L. Remar: Oe has to be careful with otatio i order ot to give ew meaig to objects that were already defied. This might require a lot of idices. c We first eed to show that S is a upper boud of L. Let l L. The, there is a subsequece x N such that x = l. Sice, we have x s, which implies that x s, that is l S usig Problem. Sice l L was arbitrary, we coclude L is bouded above by S ad S sup L. By Problem, S is a accumulatio poit of x, i.e. S L, so that S sup L. We coclude S = sup L.. Usig the Cauchy Covergece Criterio, prove that the sequece is coverget. Solutio: x = Let ɛ > 0. By the Archimedea property, N N such that N > ɛ. Let, m N satisfy N ad m N. Without loss of geerality, m >. The, x m x = m < m m = m m = m < N < ɛ. Sice ɛ > 0 was arbitrary, we coclude x is a Cauchy sequece ad therefore coverget. 3

14 3. Defiitio: a sequece x has bouded variatio if there exists c > 0 such that for all N, x x + x x + + x x < c. Show that if a sequece has bouded variatio, the the sequece is covergig. Fid a example of a coverget sequece which does ot have bouded variatio. Solutio: We show the cotrapositive: if a sequece is ot covergig, the it does ot have bouded variatio. Assume x does ot coverge. The, x is ot Cauchy, i.e. ɛ 0 such that N N, N, mn N such that x mn x N ɛ 0. 3 Let c > 0 be arbitrary. The, by the Archimedea property, K N such that Kɛ 0 c.. By?? with N =, there exists, m N such that x m x ɛ 0 without loss of gereality m >.. By?? with N = m +, there exists m > m + > m > such that x m x ɛ 0.. K. By?? with N = mk +, we get mk > mk > mk > K such that x mk x K ɛ 0. The, c Kɛ 0 x m x + x m x + + x mk x K x x + x x 3 + x mk x mk. Write the details. Sice c > 0 was arbitrary, we coclude x does ot have bouded variatio. Cosider the sequece x = /. It is easy to show, usig the Archimedea property, that x coverges to 0. Note that write the details x x + x 3 x + + x x ad that there exists o c > 0 such that < c for all N see Example 3.3.3b i Bartle ad Sherbert, fourth editio. We coclude x does ot have bouded variatio. 4

15 4. Let x < x be arbitrary real umbers ad Fid the formula for x ad x. x = x 3 + x, >. 3 Solutio: The characteristic equatio is λ = λ/3 + /3, which has roots λ = /3 ad λ =. The th term is therefore give by x = /3 c + c. We have x = /3c + c ad x = 4/9c + c. Solvig the liear system write the details yields c = 9x x /0 ad c = x + 3x /5, so that x = /3 9x x /0 + x + 3x /5. Sice /3 0 as, we have /3 9x x /0 0 as ad x = /3 9x x /0 + x + 3x /5 x + 3x /5 as. 5

16 5. Let x > 0 ad x + = + x,. Show that x is a cotractive sequece ad fid x. Solutio: We first show by iductio that x > 0 for all N. Base case = is give. Assume x > 0. The, + x > 0 so that x + = +x > 0. Now, we have x + x + = + x + + x = + x x + + x + + x x x + = + x + + x < x x + 4 sice x + > 0 ad x > 0. This shows x is cotractive. Therefore, it is Cauchy ad it coverges. Let x = x. The, + x = + x + x = x = x + = = + x + x = + x + x x + x = x = ±. However, sice x > 0 for all N, we must have x 0 ad we coclude x = +. 6