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)
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 =
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 = √
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
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
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
(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
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
(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
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