[PDF] [PDF] Solutions to Homework 6- MAT319

[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

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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

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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 

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(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 

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Solutions to Homework 6- MAT319

November 10, 2008

1 Section 3.3

Exercise 1(# 4).Letx1= 1andxn+1=p2 +xn. Then limxn= 2

First we show thatxnis increasing by using an induction argument.x2=p3>1 =x1so the base case holds. Now suppose thatxn> xn1. Then

x n+ 2> xn1+ 2. Thenpx n+ 2>px n1+ 2. Thusxn+1> xn. Now we show thatxnis bounded above by 2. We use induction again:x1<2 so the base case holds. Ifxn<2, thenxn+1=p2 +xnL= limxn+1= limpx n+ 2 =p(limxn+ 2) =pL+ 2. SoL2L2 = 0. This equation has two solutions, namelyL= 2,L=1. Sincexn>0 for alln, we deduce thatL= 2. Exercise 2(# 7).Supposex1=a >0andxn+1=xn+ 1=xn. Thenxn diverges. We proceed by contradiction. Suppose thatxnconverges. Then writeL= limxn. ThenL= lim(xn+ 1=xn) = limxn+ 1=limxn, soL=L+ 1=L. Thus

0 = 1=L. Since this is impossible, we deduce thatxnmust diverge.

Exercise 3(# 10).snandtnare monotone and iflimsn= limtn, thenxn converges. First we show thatsn= supfxk:xngis decreasing. Notice thatsn= maxfxn;sn+1g. Thussn+1sn. Thussnis decreasing. Now we show that t n= inffxk:xngis increasing. Notice thattn= inffxn;tn+1g. Thus t ntn+1. sotnis a decreasing sequence. Now suppose thatsnandtnconverge. sincesnxntnand since limsn= limtn, by the squeeze theorem it follows thatxnconverges. Exercise 4(# 15).Calculatep5, correct to within5decimal places Following the example, setsn+1= 1=2(sn+5=sn), ands1= 5. Thens2= 3, s

3= 7=3,s4= 2:23809524,s5= 2:2360689. By the inequalitysnp(5)

(s2n5)=snwe see that s

5p5:0000018 (1)

which tells us thats5is correct up to 5 decimal places. 1

2 Section 3.4

Exercise 5(# 3).letfnbe the bonacci sequence and letxn=fn+1=fn.

Suppose thatL= limxn. What is the value ofL?

Recall thatfn+1=fn+fn1so thatxn=fn1=fn+1+1. Taking limits, we haveL= limfn1=fn+1+ 1 = 1=L+ 1. (Prove to yourself that limfn1=fnis the reciprocal of limfn=fn1.) SoL= 1=L+1 and thusL2L1 = 0. By the quadratic formula and the fact thatxn>0 we deduce thatL= 1=2 +p(5)=2.

Exercise 6(#8).

(a).lim(3n)1=2n= 1 Notice that (3n)1=2n= (3=2)1=2n(2n)1=2n. This is a subsequence (the even terms) of the sequence (3=2)1=nn1=n. Since limn1=n= 1 and lim(3=2)1=n= 1 it follows that the limit of the product is the product of the limits, thus the limit is 1. (b).lim(1 + 1=(2n))3n=e3=2 Notice that (1 + 1=(2n))3n= (1 + 1=(2n))2n(1 + 1=(2n))n= (1 + 1=(2n))2np(1 + 1=(2n))2n) But (1 + 1=(2n))2nis a subsequence (the even terms) of (1 + 1=n)nwhich converges toe. Thus lim(1 + 1=(2n))3n= lim(1 + 1=(2n))2np(1 + 1=(2n))2n)= (2) lim(1 + 1=(2n))2nlimp(1 + 1=(2n))2n) =eplim(1 + 1=(2n)2n= (3) epe=e3=2(4) Exercise 7(#9).Suppose every subsequence ofxnhas a subsequence that converges to0. Thenxnconverges to0. By considering the contrapositive statement, it suces to prove that ifxn does not converge to 0, then there exists a subsequencexnkthat does not converge to 0. Ifxndoes not converge to 0, it follows from the denition that there exists >0 so that for allN >0, there existsnk> Nso thatjxnkj> . So we recursively construct a subsequence ofxnthat does not converge to 0 as follows: choosen1>1 so thatjxn1j> . Then to ndxnk+1, choosenk+1> nk so thatjxnk+1j> . (Here we are choosingN=nkin the denition above). Then it follows thatnk+1nkso thatxnkis actually a subsequence, and by construction, we see thatxnkdoes not converge to 0.

3 Section 3.5

Exercise 8.#2

(a).Show directly thatn+1n is a Cauchy sequence. 2 Suppose >0. Then chooseNso that ifk > N, 1=k < =2. Then notice that, for anym;n > N j n+ 1n m+ 1m j=jmn+mnmnmn =jmnmn j m=mn+n=mn= (5)

1=n+ 1=m <2(=2) =:(6)

so the sequence is cauchy. (b).1 + 1=2! +:::+ 1=n!is a Cauchy sequence. As an exercise for yourself, prove that 1=n!<1=2nas long asn4. Then, by using induction onm, prove that

1=2n+ 1=2n+1+:::+ 1=2n+m1=2n1(7)

Now if >0 is given, chooseNso that ifn > Nthen 1=2n< . Then if n > m > N, we have j1 + 1=2! +:::+ 1=n!(1 + 1=2! +:::+ 1=m!)j= (8) j1=(m+ 1)! +:::+ 1=(m+n)!j j1=2m+1+:::+ 1=2m+nj(9)

1=2m(10)

so the sequence is Cauchy. Exercise 9(#5).Show thatlimjp(n+ 1)pnj= 0butpnis not a Cauchy sequence. We have already shown thatjpn+ 1pnjconverges to 0 in a previous homework assignment. To show it is not Cauchy, choose= 1=4, and suppose

Nis given. Choosen=N, and choosemso thatn2

pm <1=4 (we can do this since the sequence 1=2pmconverges, andnis a xed number.) Then pmpn=mnpm+pn mn2 pm =pm 2 n2 pm >(11) 1=2n2 pm >1=21=4 = 1=2 (12)

So the sequence isn't cauchy.

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