[PDF] [PDF] MATH201 INTRODUCTION TO ANALYSIS Worksheet for week 6

[PDF] 447 HOMEWORK SET 6 1 33 3) Let x1 ≥ 2 and x n+1 := 1 + √ xn

satisfies x =1+ √ x − 1 ⇔ x − 1 = √ x − 1 ⇔ x ∈ {1,2} ⇒ x = 2, since (xn) is bounded below by 2 0 4) Let x1 := 1 and xn+1 := √ 2 + xn,n ∈ N Show (xn) 

[PDF] 1 Section 33 - 1, 2, 3, 4

Show that (xn) is bounded and monotone Find the limit Proof First, let's show that it is monotone (decreasing) Note that x1 = 8 > x2 = 

[PDF] Practice Problems 2: Convergence of sequences and monotone

Let xn = (−1)n for all n ∈ N Show that the sequence (xn) does not converge 3 Let A be (a) x1 = 2 and xn+1 = 2 − 1 xnfor n ∈ N (b) x1 = √ 2 and xn+1 = √

[PDF] Solutions to Homework 6- MAT319 - Stony Brook Mathematics

10 nov 2008 · 1 Section 3 3 Exercise 1 (# 4) Let x1 = 1 and xn+1 = √ 2 + xn Then lim xn = 2 Following the example, set sn+1 = 1/2(sn +5/sn), and s1 = 5

[PDF] Solutions to Homework 6- MAT319

10 nov 2008 · 1 Section 3 3 Exercise 1 (# 4) Let x1 = 1 and xn+1 = √ 2 + xn Then lim xn = 2 Following the example, set sn+1 = 1/2(sn +5/sn), and s1 = 5

[PDF] Solutions to Homework 7- MAT319

17 nov 2008 · Let x1 = 2 and xn+1 =2+1/xn Then xn is contractive, and lim xn =1+ √ 2 First let us show xn is contractive

[PDF] MA 101 (Mathematics I) Hints/Solutions for Practice Problem Set - 2

(xn) such that the sequence (yn) is convergent, where yn = xn + 1 n Let f(x) = { ( x − 1)2 sin 1 (x−1)2 if 1 < x ≤ 2, 0 if x = 1 Clearly f : [1,2] → R is differentiable 

[PDF] MATH201 INTRODUCTION TO ANALYSIS Worksheet for week 6

Lecture sessions 1/2 Name: Tutorial Section: Student ID: 1 Suppose that (xn) is a convergent sequence and (yn + 1 n ) Let lim xn = x We know x1 ≥ 2, if xk ≥ 2, then xk+1 = 1+ √ xk − 1 ≥ 1+ √ 2 − 1 = 2 Hence xn ≥ 2 for all n ∈ N by 

[PDF] Homework 3 232 234 242 - Purdue Math

(b) =⇒ limxn+1 = lim 1 4−xn = 1 4−lim xn ⇒ x = 1 4−x ⇒ x = 2 − √3 Let x1 = √2, and xn+1 = √2xn ∀n ≥ 1 Claim: If 0 < xn < 2 ⇒ 0 < xn < xn+1 < 2 

[PDF] Math 242: Principles of Analysis Fall 2016 Homework 3 Part B

Show that the sequence x1 = 1, xn+1 = xn + 1 xn diverges (Hint: Suppose it did converge ) Solution Suppose xn converges Let a = lim xn Clearly xn > 0 for all 

[PDF] let x1=1 and xn+1=3xn^2

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MATH201 INTRODUCTION TO ANALYSIS

Worksheet for week 6: Sequences and series

Lecture sessions 1/2

Name:Tutorial Section:

Student ID:

1. Suppose that (xn) is a convergent sequence and (yn+1n

) is a sequence such that for anyε >0 there exists N?Nsuch that|xn-yn|< εfor alln >N. Prove that the sequence (yn+1n ) also converges. [3]

Proof. Let limxn=x. We have|yn+1n

We know for anyε >0, there existsN?Nsuch that|yn-xn|< εfor alln > N. Since limxn=x and lim 1n = 0, we have, for the sameε >0 as above, integersMandQ, such that|xn-x|< εfor all n > Mand|1n -0|=|1n |< εfor alln > P. So we obtain for anyε >0, there exitsQ?N, such that |yn+1n -x|< ε+ε+ε= 3εfor alln > Q, where Q can be chosen as max(N,M,P). Therefore (yn+1n converges.2. Letx1≥2 andxn+1= 1+⎷x n-1 forn?N. Show that (xn) is monotone decreasing and bounded below, and find its limit.[4] k-1 =xk+1. Hence (xn) is monotone decreasing. We knowx1≥2, ifxk≥2, thenxk+1= 1+⎷x k-1≥1+⎷2-1 = 2. Hencexn≥2 for alln?Nby induction. Therefore (xn) is bounded below. Since limxn+1= 1+lim⎷x n-1, sox:= limxnsatisfiesx= 1 +⎷x-1. Then we havex= 1 orx= 2. Sincex= 1 is impossible, we have

x= 2.3. LetAbe an infinite subset ofRthat is bounded above and letu= sup(A). Show that there exists an

increasing sequence (xn) withxn?Afor alln?Nsuch thatu= limxn. [3] Proof. We distinguish two cases. Ifu?A, then we simple choosexn=ufor eachn?Nto obtain the sequence (xn). Ifu??A, then sinceu= sup(A), so by Lemma 2.3.4 we have for everyε >0, there exists

ansε?Asuch thatu-ε < sε. SinceAis not an empty set, so we can find an integern1and such that

u-1n

1is not an upper bound, and an elementx1ofAsuch thatu-1n

1< x1. We now choose an integer

n

2withn2> n1such thatx1< u-1n

2(this is always possible since we can always find an integern2such

that 1n

2< u-x1. Note thatu-x1>0). Nowu-1n

2is not an upper bound ofA, so we can find an

elementx2ofAsuch thatu-1n x

2< u-1n

3. Then becauseu-1n

3is not an upper bound, so we can findx3?Awithu-1n

3< x3. Hence

x

Awhich is bounded above. The Monotone Convergence Theorem now implies limxn=u.4. (optional) Let (xn) be a sequence inRthat is bounded. For eachn?N, we letsn= sup{xk:k≥n}.

Prove that (sn) is monotone and convergent. The limit limsnis called thelimit superiorof (xn).

that (sn) is monotone decreasing. Because (xn) is bounded and in particular bounded below, so there exists

a real numberMsuch thatxn≥Mfor alln?N. Hence inf{(xn)} ≥M. But then s n= sup{xk:k≥n} ≥inf{xk:k≥n} ≥inf{(xk)} ≥M, holds for eachn?N. We see that (sn) is monotone decreasing. We conclude by monotone convergence theorem (Theorem 3.3.2) that (sn) is convergent.quotesdbs_dbs14.pdfusesText_20