MATH202 Introduction to Analysis (2007 Fall and 2008 Spring) Tutorial Note #12

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1 MATH0 Introduction to Analysis (007 Fall and 008 Spring) Tutorial Note #1 Limit (Part ) Recurrence Relation: Type 1: Monotone Sequence (Increasing/ Decreasing sequence) Theorem 1: Monotone Sequence Theorem f {x n } is increasing and bounded from above, then {x n } converges. Similarly, if {x n } is decreasing and bounded from below, then {x n } converges. Example 1 Given a sequence y n such that Show {y n } is convergent. By computing first -4 terms, we see that y 1 = and y 1 n+1 = y n + n y = , y = , y 4 = also see y n is bounded from by 1 (or any number greater than 1) 17, we suspect y n is increasing. And we (Step 1: Show y n is increasing, i.e. y n+1 y n ) By induction, for n = 1, y = > 1 = y 1. Assume y k+1 > y k, for n = k + 1, y k+ = y k n > y k+1 which completes the proof. (Step : Show y n is bounded from above by 1) Note that y n = 1 n +1 + y n 1 = 1 n y n = 1 n +1 + y n =..= 1 n Note that y n = n +1 < n < = = 1. So y n is bounded from above by 1. (Step ) Applying Monotone Sequence Theorem, we see y n converges.

2 ========================================================================= Exercise 1: Let a sequence {x n } which x 1 = 0 and x n+1 = sin +x n for n = 1,,,. Show {x n } converges. (One can use Newton s Method to approximate the limit) ========================================================================= Type : Intertwining Sequence Theorem : (Nested Interval Theorem) If I n = [a n, b n ] such that I 1 I I., then n=1 I n = [a, b] where a = lim n a n lim n b n = b, Furthermore, if lim n (b n a n ) = 0 (i.e. a = b), then exactly one number Theorem : (Intertwining Sequence Theorem) If {x m } and {x m 1 } converge to x, then {x n } converges to x. n=1 I n contains Remark: In normal situation, we use first theorem to show lim n x n 1 and lim n x n exists. Then we show lim n x n 1 = lim n x n. Finally, use the second theorem to conclude the sequence converges. Example A sequence x n is defined as follows: x 1 = 1, x n+1 = + 1 x n Show x n converges and compute the limit. Solution: By computing 7 more terms, we get x =, x = 5, x 4 =., x 5 = 1.909, x 6 =.048, x 7 = 1.977, x 8 =.01, we see the sequence is intertwining type. By plotting the points on the real line, X X X X------X X------X X x 1 x x 5 x 7 x 8 x 6 x 4 x Let I 1 = x 1, x, I = x, x 4, I = x 5, x 6,., I k = [x k 1, x k ] We need to show I k I k+1, i.e. x k 1 x k+1 x k+ x k (We can try to prove by induction) For n = 1, it is true that x 1 x x 4 x Assume x k 1 x k+1 x k+ x k

3 x k 1 x k+1 x k+ x k 1 x k 1 x k+1 x k+ x k 1 + x k 1 + x k+1 + x k+ + x k (Multiply both side by and then add 1) x k x k+ x k+ x k+1 1 x k+1 x k+ x k+ x k 1 + x k+1 + x k+ + x k+ + x k x k+ x k+4 x k+ x k+1 which completes our induction. Therefore by nested interval theorem, we see lim k x k and lim k x k 1 exists, let lim k x k = a and lim k x k 1 = b, from the recurrence relation x k+1 = x k + 1 (By putting n = k), take k, we get a = b + 1, also x k+ = x k (By putting n = k + 1). Take k, we get b = + 1. Solving equations, we get a = b = a Hence by intertwining Sequence Theorem, {x n } converges and lim n x n =. (One more example is given in Example 7 of Tutorial Note #11) Exercise In Example, solve a = + 1 and b = + 1 to get a = b = b a Exercise Given a sequence which defined as x 1 = and x n+1 = x n Show {x n } converges and find lim n x n More difficult sequence Example Given a sequence defined by x 1 = 1.5, x =, x n+ = 4x n Determine whether x 1, x, x, converges or not. In case the sequence converges, find the limit also. By testing a few terms, we see x = 1.44, x 4 = 1.710, x 5 = 1.404, x 6 = 1.566, x 7 = 1.78, x 8 = 1.48, x 9 = We see the sequence does not belong one of the above types.

4 However when we look at the sequence separately, we look at the sequence {x 1, x, x 5, } and {x, x 4, x 6, }. We see each of them is decreasing sequence (monotone sequence) X X X X----X-----X------X X X x 9 x 7 x 5 x x 8 x 1 x 6 x 4 x So our proof is as follows: Step 1: Show {x 1, x, x 5, } converges and find the limit --Show the sequence is decreasing and bounded from below by?? Step : Show {x, x 4, x 6,. } converges and find the limit --Show the sequence is decreasing and bounded from below by?? Step : If both limit (in step 1 and step ) are the same, we can conclude the sequence converges by intertwining sequence theorem. Solution: (Step 1) First, we show x n 1 is bounded from below by 1 For n = 1, x 1 = 1.5 > 1 Assume x k 1 > 1, then x k+1 = 4x k 1 + By induction, x n 1 > (bounded from below) > = 7 > 1 Second, we show x n 1 is decreasing (i.e. x n+1 < x n 1 ). We use induction to show this. For n = 1, x = 1.44 < 1.5 = x 1, assume x k+1 < x k 1, Consider x k+ x k+1 = 4x k+1 = 4x k+1 + 4xk+1 = 4x k +1 4x k 1 4x k 1 4x k+1 4xk 1 = 4 x k +1 x k 1 4x k 1 < 0 (Note > 0). ( ) + 4x k 1 (*Note: We have used the formula: a b = a b (a + ab + b )) Therefore, x n 1 is decreasing sequence. So by monotone theorem, {x n 1 } converges, similar argument also shows {x n } Converges (left as exercise). Now we write lim n x n 1 = a and lim n x n = b. From x k+1 = 4x k 1, let k, we get a 4a + = 0 a 1 (a + a

5 ) = 0 a = 1 and a a = 0 a = We get b = ± 1. So we get a = 1+ 1, similarly Therefore, since a = b, so by intertwining sequence theorem, the sequence {x n } converges and lim n x n = Exercise In Example, show the sequence {x n } converges.. Exercise 5 (006 Fall Final) Given a sequence {x n } defined as x 1 =, x = 4, x n+ = 10x n 9 Show {x n } converges and compute the limit Appendix: In fact, the intertwining sequence theorem can be extended to three or more subsequences. Theorem 4: (Extension of Intertwining sequence theorem, subsequences) Suppose {x n }, {x n+1 }, {x n+ } converges to x, then {x n } converges to x Proof: for any ε > 0, We would like to show there exists N such that n > N, x n x < ε Since lim n x n = x, there exist M 1, such that n > M 1, x n x < ε Similarly there exists M, such that n > M, x n+1 x < ε There exists M, such that n > M, x n+ x < ε Pick N = max {M 1, M, M }, then for any n > N, x n x = x n x or x n+1 x or x n+ x, in any case, we must have x n x < ε. Therefore by definition of limit, {x n } converges to x So one can make use the theorem and use the similar method as in Example to do the following: Try to do it if you have time Exercise 6a Given a sequence, defined as x 1 = 1, x =, x = and x n+ = 7x n Show x n converges and compute the limit. Of course, you may try to extend the theorem further to k subsequences. k = 4,5,6, Exercise 6b (Try to write the proof by yourselves) Suppose {x kn }, {x kn +1 },...{x kn + k 1 } converges to x, then {x n } converges to x.