* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Advanced Calculus
Survey
Document related concepts
Infinitesimal wikipedia , lookup
Vincent's theorem wikipedia , lookup
Large numbers wikipedia , lookup
Law of large numbers wikipedia , lookup
Non-standard calculus wikipedia , lookup
Central limit theorem wikipedia , lookup
Non-standard analysis wikipedia , lookup
Real number wikipedia , lookup
Hyperreal number wikipedia , lookup
Collatz conjecture wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
Transcript
Advanced Calculus Unit XIX SEQUENCES OF REAL NUMBERS – PART III Objectives From this session a learner is expected to achieve the following Introduce the concept of monotone sequences Study that a bounded monotonic sequence is convergent Learn that monotone convergence theorem helps in finding a sequence of real numbers that converges to the positive square root of a positive real number. Construct a sequence that converges to the transcendental number e. Contents 1. Introduction 2. Montone Sequences 3. The Calculation of Square Roots 4. Euler Number 1. Introduction In the previous session (Sequences of Real Numbers -Part II) we have seen that every convergent sequence is bounded, but the converse is not true. In this session, we search for a condition which ensures the convergence of a bounded sequence; and end up with the Monotone Convergence Theorem that establishes the existence of the limit of a bounded monotone sequence. We begin with the definition of monotone sequence and discuss associated results and examples. We will see that Monotone Convergence Theorem is helpful in finding limit of certain seqeunces. It helps us in finding a sequence of real numbers that converges to the positive square root of a positive real number. A discussion on Euler Number will be made in this session. 2. Montone Sequences 1 Definition Let X ( xn ) be a sequence of real numbers. We say that the sequence X is increasing (or monotonically increasing) if it satisfies the inequalities x1 x2 xn xn 1 We say that X is decreasing (or monotonically decreasing)if it satisfies the inequalities x1 x2 xn xn 1 We say that X is monotone if it is either increasing, or it is decreasing. The following sequences are increasing: (1, 2, 3, 4, , n, ), ( a, a 2 , a 3 , , an , (1, 3, 5, 7, ), if a 1 . ) The following sequences are decreasing: 1, 1 , 1 , 2 3 (b, b2 , b3 , 1 1 1, 3 , 2 , 3 , 1 , , 3 , bn , 1 n1 3 , , ) if 0 b 1. The following sequences are not monotone: (1, 1, 1, , (1)n , (1, 2, 3, ), , (1)n n, ). The following sequences are not monotone, but they are ultimately monotone: (5, 3, 2, 4, 1, 2, 3, 4, ), 3, 8, 1, 1 , 1 , 1 , 2 3 4 . Example 1 Show that the sequence ( xn ) where xn 1 1 1 n n 1 n 2 1 , n , 2n 1 is monotonically decreasing. Solution We note that for n , xn1 1 1 n 1 n 2 1 2(n 1) 1 1 1 n 1 n 2 1 1 1 2n 1 2 n 2 n 1 Hence xn1 xn 1 1 1 1 0. 2n 2n 1 n 2n(2n 1) Therefore, xn 1 xn for n and hence the given sequence ( xn ) is monotonically decreasing. Theorem 1 (Monotone Convergence Theorem) A monotone sequence of real numbers is convergent if and only if it is bounded. Also, (a) If X ( xn ) is a bounded increasing sequence, then 2 lim( xn ) sup{xn : n } (b) If Y ( yn ) is a bounded decreasing sequence, then lim( yn ) inf{ yn : n } . Proof. First of all we note that a convergent sequence is bounded. Hence, in particular, a bounded monotone sequence of real numbers is also convergent. Conversely, let X be a bounded monotone sequence. Then X is either increasing or decreasing. We first treat the case where X ( xn ) is a bounded increasing sequence. By (a) hypothesis, there exists a real number M such that xn M for all n . Thus the subset {xn : n } of the set of real numbers is bounded above. Hence by the Supremum Property of real numbers, supremum of the set {xn : n } exists. Let the supremum be x* sup{xn : n } . Claim x* lim( xn ). If 0 is given, then x * is not an upper bound for the set {xn : n } , and hence there exists a natural number K K ( ) such that x K is a member of the set xn : n } and x * xK . But since ( xn ) is an increasing sequence it follows that x * xK xn x * for all n K . Therefore it follows that x xn x for all n K , i.e., xn x for all n K i.e., | xn x*| for all n K . Since 0 is arbitrary, we have ( xn ) converges to x * . (b) If Y ( yn ) is a bounded decreasing sequence, then it is clear that X Y ( yn ) is a bounded increasing sequence. We have seen in part (a) that lim X sup{ yn : n } . On the other hand, lim X limY ; and also, we have sup{ yn : n } inf{ yn : n } . Therefore lim Y lim X sup{ yn : n } inf{ yn : n } . This completes the proof. Remark The above theorem gives us a way of calculating the limit of the sequence provided we can evaluate the supremum in case (a), or the infimum in case (b). Example 2 lim 1 0 . n Solution Obviously the sequence 1 1 1 , , 1, 2 3 n is decreasing and bounded (1 is a bound for the sequence). Hence by part (b) of Theorem 1, 3 1 1 lim : n . inf n n …(1) Clearly 0 is a lower bound for the set 1 : n , and also 0 is the infimum of the set n 1 1 : n ; hence (1) implies lim 0. n n Example 3 Examine the convergence of the sequence ( xn ) , where 1 1 n 1 n 2 xn 1 nn for n . Solution It can be seen that xn1 xn 1 0 2(2n 1)(n 1) and hence ( xn ) is monotonically increasing. Obviously the sequence ( xn ) is bounded below by 1 . 2 Also, 1 1 n 1 n 2 xn 1 n 1 nn 1 n n times n 1. n Hence the sequence ( xn ) is bounded above by 1. We conclude that the sequence is a bounded monotone sequence and hence is convergent. Example 4 Examine the convergence of the sequence ( xn ) , where xn 1 1 1 2 3 1 for n n . Solution Since xn1 xn 1 xn , we see that ( xn ) is an increasing sequence. Hence by the n 1 Monotone convergence Theorem the question of whether the sequence is convergent or not is reduced to the question of whether the sequence is bounded or not. We note that x2n 1 1 1 1 2 3 4 1 1 1 1 2 4 4 1 n1 2 1 1 n 2 1 2n 1 2n 4 1 1 1 2 2 1 n . 2 1 2 Hence the sequence ( xn ) is unbounded, and therefore by Monotone convergence Theorem ( xn ) is divergent. Example 5 Let X ( xn ) be defined inductively by x1 1, xn 1 14 (2 xn 3) for n 1 . Show that lim X 3 . 2 Solution Direct calculation shows that x2 5 . Hence we have x1 x2 2 . We show, by induction, 4 that xn 2 for all n . We have already noted that this is true for n 1, 2 . If xk 2 holds for some k , then xk 1 14 (2 xk 3) 14 (4 3) 74 2 , so that xk 1 2 . Therefore xn 2 for all n . We now show, by induction, that xn xn 1 for all n . The truth of this assertion has been verified for n 1 . Now suppose that xk xk 1 for some k ; then 2 xk 3 2 xk 1 3 , and hence it follows that xk 1 14 (2 xk 3) 14 (2 xk 1 3) xk 2 . Thus xk xk 1 implies that xk 1 xk 2 . Therefore xn xn 1 for all n . We have shown that the sequence X ( xn ) is increasing and bounded above by 2. It follows from the Monotone convergence Theorem that the sequence X converges to a limit that is at most 2. In this case it is not so easy to evaluate lim( xn ) by calculating sup{xn : n } . Hence we proceed as follows: Let x lim( xn ). Then limit of the 1-tail ( xn 1 ) of X also has the limit x . Since xn 1 14 (2 xn 3) for all n , we have the limit of the sequence ( xn 1 ) and the limit of the sequence 14 (2xn 3) are the same. Hence, we obtain x 14 (2 x 3) , Hence it follows that x 3 . . 2 Example 6 Let a and b be such that a 0 and b a. Let ( xn ) be the sequence of real numbers defined by x1 a, xn 1 ab 2 xn2 a 1 for n . Show that lim( xn ) b. 5 Solution We first prove that ( xn ) is bounded above by b. i.e., we have to show that xn b for n . We prove this by Principle of Mathemtical Induction. The result is true for n 1, becuase x1 a b. As induction argument, assume the result is true for n k . i.e., suppose xk b. Then ab2 xk2 xk2 b2 2 xk 1 b b 0, since xk2 b2 0. a 1 a 1 2 2 Hence x k 1 b, and this shows that xn b for n . Hence ( xn ) is bounded above by b. Also, for n xn1 2 xn2 ab2 xn2 a(b2 xn2 ) 2 xn 0, a 1 a 1 since b xn , and a 0. Hence x n21 xn2 so that x n 1 xn . Hence ( xn ) is monotonic increasing. Being monotonic increasing and bounded, the sequence ( xn ) is convergent. Let l lim( xn ). Then, also, lim( xn 1 ) l. Since x n21 ab2 xn2 , a 1 we have lim xn 1 2 ab2 lim xn2 n n a 1 , implies l2 ab 2 l 2 , a 1 implies al 2 l 2 ab 2 l 2 , implies 6 l 2 b2 , implies l b. i.e., lim( xn ) b. Example 7 Let Y ( yn ) be the sequence of real numbers defined by y1 1, yn 1 2 yn for n . Show that lim( yn ) 2. Solution Note that y1 1 and y2 2 ; hence 1 y1 y2 2 . We claim that the sequence Y is increasing and bounded above by 2. To show this we will show, by induction, that 1 yn yn 1 2 for all n . This fact has been verified for n 1 . Suppose that it is true for n k ; then 2 2 yk 2 yk 1 4 , and hence it follows that 1 2 yk 1 2 yk 2 yk 1 4 2. Noting that yk 2 2 yk 1 , the above implies 1 yk 1 yk 2 2. Hence the validity of the inequality 1 yk yk 1 2 , implies the validity of 1 yk 1 yk 2 2 . Therefore 1 yn yn 1 2 for all n . Since Y ( yn ) is a bounded increasing sequence, it follows from the Monotone convergence Theorem that it converges. Let the limit of Y be y . The relation yn 1 2 yn gives lim yn1 lim 2 yn which implies y 2y . Hence y 2 2 y which has the roots y 0, 2 . Since the terms of y ( yn ) all satisfy 1 yn 2 , it follows that we must have 1 y 2. Therefore y 2 . M2 3. The Calculation of Square Roots Now we show that Monotone Convergence Theorem helps us in finding a sequence of real numbers that converges to the positive square root of a positive real number. Let a 0 ; we will construct a sequence ( sn ) of real numbers that converges to a. Let s1 0 be arbitrary and define sn1 12 sn a for n . We now show that the s n sequence ( sn ) converges to a . 7 We first show that sn2 a for n 2 . Since sn satisfies the quadratic equation sn2 2sn1sn a 0 , this equation has a real root. Hence the discriminant 4sn21 4a must be nonnegative; that is sn21 a for n 1 . To see that ( sn ) is ultimately decreasing, we note that for n 2 we have 2 1 a 1 ( s a) sn sn1 sn sn n 0. 2 sn 2 sn Hence, sn 1 sn for all n 2 . It follows from the Monotone Convergence Theorem that s lim( sn ) exists. Moreover, it follows that the limit s must satisfy the relation 1 a s s . 2 s whence it follows that 1 1a s 2 2s s or a s or s2 a . s a . Thus For the purposes of calculation, it is often important to have an estimate of how rapidly the sequence ( sn ) converges to a . As above, we have a sn for all n 2 , whence it follows that a a sn . Thus we have sn 0 sn a sn 2 a ( sn a) sn sn Using this inequality we can calculate for n 2 . a to any desired degree of accuracy. Mod 3 4. Euler’s Number We now construct a sequence that converges to one of the most important transcendental numbers in mathematics. For this, let en 1 1 n n for n . We will now show that the sequence (en ) is bounded and increasing; hence it is convergent. If we apply the Binomial Theorem, we have 1 n 1 n(n 1) 1 n(n 1)(n 2) 1 en 1 1 2 3 1 n 2! 3! n n n n n(n 1) n! 2 1 1 . nn If we divide the powers of n into the terms in the numerators of the binomial coefficients, we get 8 1 1 1 1 2 1 1 1 2! n 3! n n en 1 1 1 n 1 . n 1 1 2 1 1 n! n n Similarly, we have en 1 1 1 1 1 1 1 2 1 1 1 2! n 1 3! n 1 n 1 1 n 1 n 1 1 1 2 1 1 n! n 1 n 1 1 1 2 1 1 (n 1)! n 1 n 1 1 n . n 1 Note that the expression for en contains n 1 terms, while that for en 1 contains n 2 terms. Moreover, each term appearing in en is less than or equal to the corresponding term in en 1 , and 2 e1 e2 en 1 en en 1 has one more positive term. Therefore we have , so that the terms of E are increasing. To show that the terms of the sequence (en ) are bounded above, we note that if p 1, 2, , n, then 1 p 1 1 1 . Moreover 2 p 1 p ! so that . Therefore, if n 1 , n p ! 2 p 1 then we have 2 en 1 1 1 1 2 22 1 n 1 2 . Being the finite sum of geometric progression with common ratio 1 1 2 22 1 2 n 1 1 1 2 n 1 1 , we have 2 1, and hence we deduce that 2 en 3 for all n . The Monotone Convergence Theorem implies that the sequence (en ) converges to a real number. Also, since 2 en 3 for all n , the limit of (en ) lies between 2 and 3. We define the number e to be the limit of the sequence (en ) . The limit of the sequence (en ) is the famous Euler number e, whose approximate value is 2.718281828459045 , which is taken as the base of the natural logarithm. Summary In this session, we have described Monotone Convergence Theorem and used it for finding limits of certain sequences. We have seen that it is helpful in finding a sequence of real numbers that converges to the positive square root of a positive real number. A discussion on Euler Number have been made in this session. Assignments 9 1 1. Let a 0 and let z1 0 . Define zn1 (a zn ) 2 for n . Show that ( zn ) converges and find the limit. 2. Let ( xn ) be a bounded sequence, and for each n let sn sup{xk : k n} and tn inf{xk : k n} . Prove that ( sn ) and (t n ) are convergent. Also prove that if lim( sn ) lim(tn ) , then ( xn ) is convergent. that is bounded above and let u sup A . Show there 3. Let A be an infinite subset of exists an increasing sequence ( xn ) with xn A for all n 4. Let xn : 12 12 1 2 1 for each n n2 such that u lim( xn ) . . Prove that ( xn ) is increasing and bounded, and hence converges. 5. Let x1 8 and xn1 1 xn 2 for n . Show that ( xn ) is bounded and monotone. 2 Find the limit. 6. Let x1 p , where p 0, and and xn 1 p xn for n . Find lim ( xn ). n QUIZ 1. Let x1 1 and xn1 2 1 for n 2 . Then pick the true statement. xn (a) ( xn ) is monotone, but not bounded. (b) ( xn ) is bounded, but not monotone. . (c) ( xn ) is bounded and monotone, but not convergent.. (d) ( xn ) is convergent. Ans. (d) ( xn ) is convergent. 2. Let x1 a 0 and xn1 xn 1 . Then pick the true statement. xn (a) ( xn ) is monotonically decreasing (b) ( xn ) is monotonically increasing (c) ( xn ) is monotonically decreasing and bounded below (d) none of the above. Ans. (b) ( xn ) is monotonically increasing 1 3. Limit of the sequence 1 n n 1 is _________ 10 (a) e (b) 1 e (c) e 2 (d) 1 e2 Ans. (a) e n 4. Limit of the sequence 1 1 is _________ n (a) e (b) 1 e (c) e 2 (d) 1 e2 Ans. (b) 1 e FAQ 1. State Supremum Property of real numbers. Answer: The Supremum Property (or Completeness Property) of is the following: Every nonempty set of real numbers that has an upper bound has a supremum in . 2. Define transcendental number. Answer: A transcendental number is a real or complex number, which is not a root of a nonconstant polynomial equation with rational coefficients. and e are transcendental numbers. 11 3. A careless assumption that a sequence is ‘convergent’ leads to aburd conclusions. Comment. Answer: The issue of convergence must not be ignored or casually assumed. The following example illustrates this: Consider the sequence ( xn ) defined by x1 1, xn 1 2 xn 1. Assuming the ‘convergence’ (actually wrong! The sequence is not convergent) with lim( xn ) x, we would obtain x 2x 1, so that x 1. Of course, this is absurd as always xn 1 and hence x must be 1. This absurdity is due to the wrong assumption that the sequence is convergent. Hence, it is required to examine the convergence of the sequence before finding its limit. Glossary Increasing Sequence: A sequence X ( xn ) of real numbers is increasing (or monotonically increasing) if it satisfies the inequalities x1 x2 Decreasing Sequence: xn xn 1 A sequence X ( xn ) of real numbers is decreasing (or monotonically decreasing) if it satisfies the inequalities x1 x2 xn xn 1 Monotone Sequence: A sequence X ( xn ) of real numbers is monotone if it is either increasing, or it is decreasing. References: 1. T. M. Apostol, Mathematical Analysis, Narosa Publishing House, New Delhi, 1985. 2. R. R. Goldberg, Real Analysis, Oxford & I.B.H. Publishing Co., New Delhi, 1970. 3. D. Soma Sundaram and B. Choudhary, A First Course in Mathematical Analysis, Narosa Publishing House, New Delhi, 1997. 4. Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, Wiley India Pvt. Ltd., New Delhi. 12