MA2108 Mathematical Analysis I AY 2009/2010 Sem 2 NATIONAL UNIVERSITY OF SINGAPORE MATHEMATICS SOCIETY PAST YEAR PAPER SOLUTIONS solutions prepared by Wei Boyan, Tay Jun Jie MA2108 Mathematical Analysis I AY 2009/2010 Sem 2 Question 1 (a) Let P (n) : xn 6 2. When n = 1, x1 = 1 6 2, so P (1) is true. Suppose P (k) is true, thus xk 6 2. Then xn+1 = 51 (xk 2 + 6) 6 15 (4 + 6) = 2. By Principle of Mathematical Induction, P (n) is true for all n ∈ N. (b) Claim: xn is increasing. Proof: 1 xn+1 − xn = (x2n − 5xn + 6) 5 1 = (xn − 2)(xn − 3) 5 Since xn 6 2, we have xn+1 > xn . Therefore xn is increasing. By Monotone Convergence Theorem, xn is converges. Let x be the limit of xn . 1 1 2 (xn + 6) = (x2 + 6) 5 5 (x − 2)(x − 3) = 0 x = lim xn+1 = lim n→∞ n→∞ Thus x = 2 or x = 3. Since xn 6 2, we obtain x = 2. we conclude that xn is convergent and its limit is 2. Question 2 (a) (i) Firstly, observe that 1 √ 2n+ n+1 is a decreasing sequence of strictly positive terms with lim n→∞ 2n + 1 √ n+1 = 0. Therefore the series converges by Alternating Series Test. (ii) 2 n1 n2 1 6n ρ = lim n 1 + n→∞ 3 3n 1 1 1 1 3n 1 3n lim 1 + = lim n n lim n n lim 1 + n→∞ n→∞ n→∞ 3 n→∞ 3n 3n 1 = e2 > 1 3 Therefore the series diverges by Root Test. NUS Math LaTeXify Proj Team Page: 1 of 6 NUS Mathematics Society MA2108 Mathematical Analysis I (b) Observe that 1 (2n−1)(2n+1) ∞ X n=1 1 = 12 ( 2n−1 − 1 2n+1 ). AY 2009/2010 Sem 2 Thus 1 1 1 1 1 1 − + − + ··· + − 3 3 5 2n − 1 2n + 1 1 1 = lim 1− n→∞ 2 2n + 1 1 = 2 1 1 = lim (2n − 1)(2n + 1) n→∞ 2 (c) Since the an , bn > 0 for all n ∈ N, an bn > 0 for all n ∈ N. In addition, bn → 0 as ρ = lim n→∞ Since P an converges, P P bn converges. an bn = lim bn = 0. n→∞ an an bn converges by Limit Comparison Test. Question 3 (a) Given ε > 0, choose δ = min Since 0 < |x| < 16 , we have (b) 1 3 6 , 20 ε . Suppose 0 < |x − 0| < δ, 2 (2x + 1)(x − 2) 2x + 3x + 2 = 3x + 1 3x + 1 |x| |2x + 3| = |3x + 1| δ |2x + 3| < |3x + 1| |2x+3| |3x+1| 20 3 . Then, (2x + 1)(x − 2) δ |2x + 3| + 2 < 3x + 1 |3x + 1| 20 6 δ 3 =ε 6 (i) Let f (x) = (x2 + x + 1) sin( λ3 ). Let xn = yn → 0. 3 (2n+1)π , yn = 3 2nπ . Then xn 6= 0, xn → 0, yn 6= 0 and lim f (yn ) = 0 n→∞ lim f (xn ) 6= 0 n→∞ Therefore limn→∞ f (x) does not exist by the Divergent Criterion. (ii) 6 6 6 −1< 6 x x x x x 6 3− < 63 ∵x>0 2 2 x Since limx→0+ 3 − x 2 = limx→0+ 3 = 3, limx→0+ NUS Math LaTeXify Proj Team x 2 Page: 2 of 6 6 x = 3 by Squeeze Theorem. NUS Mathematics Society MA2108 Mathematical Analysis I AY 2009/2010 Sem 2 Question 4 Let ε > 0 be given. Since limx→a g(x) = 0, ∃δ > 0 such that |g(x)| < ε whenever 0 < |x − a| < δ. M Let δ1 = min(δ, h) > 0. If 0 < |x − a| < δ1 , then |f (x)g(x)| < M · ε = ε. M Therefore limx→a f (x)g(x) = 0. Question 5 Let a ∈ R, take a rational sequence (xn ) and an irrational sequence (yn ) such that xn → a, and yn → a. Then f (xn ) = −xn → −a f (yn ) = 3yn − 8 → 3a − 8. If f is continuous at x = a, then −a = 3a − 8 a = 2. It follows that if a 6= 2, then f is not continuous at x = a. At x = 2, given ε > 0, we choose δ = 3ε , then for |x − 2| < δ, we have | − x + 2| = |x − 2| < δ < ε |3x − 8 + 2| = 3|x − 2| < 3δ = ε Therefore, |f (x) − f (2)| < ε, so f is continuous at x = 2. Question 6 Let ε > 0, since f and g are uniformly continuous on R, there exists δ1 , δ2 > 0 such that ε 4 ε x, y ∈ R, |x − y| < δ2 ⇒ |g(x) − g(y)| < 4 x, y ∈ R, |x − y| < δ1 ⇒ |f (x) − f (y)| < Let δ = min(δ1 , δ2 ), then for x, y ∈ R, with |x − y| < δ, we have |F (x) − F (y)| = |f (x)g(x) − f (y)g(y)| = |f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)| 6 |f (x)g(x) − f (x)g(y)| + |f (x)g(y) − f (y)g(y)| = |f (x)| |g(x) − g(y)| + |g(y)| |f (x) − f (y)| ε ε < |f (x)| + |g(y)| 4 4 1ε ε 6 +2 24 4 5ε = <ε 8 Thus, F is also uniformly continuous. NUS Math LaTeXify Proj Team Page: 3 of 6 NUS Mathematics Society MA2108 Mathematical Analysis I AY 2009/2010 Sem 2 Question 7 (a) Let m = lim inf(yn ), M = lim sup(xn ) and ε > 0 be given. Thus ∃K ∈ N such that for n > K, m − ε < yn and xn < M + ε. Hence M − m > xn − yn for n ≥ K. Let x ∈ C(xn − yn ), so there exist subsequence (xnk − ynk ) such that xnk − ynk → x. Thus ∃K1 ∈ N such that |xnk − ynk − x| < ε whenever k ≥ K1 . xnk − ynk − ε < x < xnk − ynk + ε ∀k ≥ K1 Now, ∃K2 ∈ N such that K2 ≥ K1 and nk ≥ K whenever k ≥ K2 . Hence, x < xnk − ynk + ε < M − m + ε k ≥ K2 Therefore x < M − m + ε for all ε > 0, that is, x ≤ M − m. In conclusion, M − m is an upper bound of C(xn − yn ) and lim sup(xn − yn ) = sup C(xn − yn ) ≤ M − m. (b) (i) Since bn > 0 ∀n ∈ N, Sn > Sn−1 . Then Sn2 > Sn Sn−1 . Therefore, bn bn < Sn2 Sn Sn−1 Sn − Sn−1 = Sn Sn−1 1 1 = − Sn−1 Sn (ii) Let Tn = Pn bk k=1 S 2 , k then b1 1 1 1 1 1 1 + − + − + ··· + − 2 S1 S2 S2 S3 Sn−1 Sn S1 2 1 = − S1 Sn 2 < S1 P bn So (Tn ) is bounded, since Sbn2 > 0, (Tn ) is increasing. Therefore, ∞ n=1 S 2 is convergent. Tn < n n Question 8 (a) Let ε > 0 be given, ∃µ > 0 such that x > µ implies |f (x) − L| < ε Since limn→∞ xn = ∞, ∃K ∈ N such that n ≥ K implies xn > µ. Therefore, n ≥ K implies |f (xn ) − L| < ε Therefore, limn→∞ f (xn ) = L. (b) Let ε > 0 be given. By assumption, ∃M ∈ R such that |g(x) − g(x0 )| < NUS Math LaTeXify Proj Team ε whenever x, x0 > M. 3 Page: 4 of 6 NUS Mathematics Society MA2108 Mathematical Analysis I AY 2009/2010 Sem 2 Let (xn ) be a sequence in R such that xn → ∞. Now, ∃N ∈ N such that xn > M whenever n ≥ N . Hence ε |g(xn ) − g(xm )| < whenever n, m ≥ N. 3 That is, (g(xn )) is Cauchy and whence it converges to some L ∈ R. Let (yn ) be another sequence in R such that yn → ∞. By the above argument, g(yn ) → L0 for some L0 ∈ R. Now, ∃K1 ∈ N such that ε |g(xn ) − L| < whenever n ≥ K1 . 3 Similarly, ∃K2 ∈ N such that ε whenever m ≥ K2 . 3 |g(ym ) − L0 | < Lastly, ∃K3 ∈ N such that xn , ym > M whenever n, m ≥ K3 . Hence |g(xn ) − g(ym )| < ε whenever n, m ≥ K3 . 3 Let K = max{K1 , K2 , K3 }. If n, m ≥ K, |L − L0 | ≤ |L − g(xn )| + |g(xn ) − g(ym )| + |g(ym ) − L0 | ε ε ε < + + 3 3 3 =ε Thus |L − L0 | < ε for all ε > 0, that is, L = L0 . In conclusion, for every sequence (zn ) in R such that zn → ∞, the sequence (g(zn )) converges to L. Therefore limx→∞ g(x) = L. Question 9 P (a) Let a = min{x1 , . . . , xn } and b = max{x1 , . . . , xn }. If a = b, then n1 nk=1 f (xk ) = f (x1 ) and we are done. Suppose a < b, hence [a, b] ⊂ (0, 1) and f is continuous on [a, b]. By Extreme Value Theorem, ∃c, d ∈ [a, b] such that f (c) ≤ f (x) ≤ f (d) for all x ∈ [a, b]. f (c) ≤ f (xk ) ≤ f (d) ∀k = 1, . . . , n n 1X f (c) ≤ f (xk ) ≤ f (d) n k=1 P P P If f (c) = n1 nk=1 f (xk ) or f (d) = n1 nk=1 f (xk ) then we are done. Suppose f (c) < n1 nk=1 f (xk ) < f (d), applying Intermediate Value Theorem to f on [c, d] or [d, c], ∃e ∈ (c, d) or (d, c) such that P f (e) = n1 nk=1 f (xk ). (b) Firstly, ∃δ > 0 such that for all x, y ∈ [0, ∞), Now, since kδ 2 − |g(x) − g(y)| < 1 whenever |x − y| < δ. (k−1)δ < δ for all k ∈ N, we have 2 g kδ − g (k − 1)δ < 1 ∀k ∈ N 2 2 g kδ < 1 + g (k − 1)δ ∀k ∈ N 2 2 g kδ < k ∀k ∈ N 2 NUS Math LaTeXify Proj Team Page: 5 of 6 NUS Mathematics Society MA2108 Mathematical Analysis I AY 2009/2010 Sem 2 h S Let C = 2δ > 0. Now, k∈N (k−1)δ , kδ forms a partition for [0, ∞). Let x ∈ (0, ∞), then 2 2 h (m−1)δ mδ (m−1)δ x∈ , for some m ∈ N. Furthermore, x − < δ. If m = 1, then 2 2 2 |g(x) − g(0)| < 1 |g(x)| < 1 < 1 + Cx If m > 1, since (m−1)δ 2 ≤ x, we have 1 x ≤ 2 (m−1)δ . Therefore, (m − 1)δ g(x) − g <1 2 (m − 1)δ |g(x)| < 1 + g 2 1 |g(x)| < 1 + (m − 1) = 1 + (m − 1)x x 2 |g(x)| < 1 + (m − 1)x (m − 1)δ 2 |g(x)| < 1 + x = 1 + Cx δ In conclusion, |g(x)| < 1 + Cx for all x ∈ (0, ∞). NUS Math LaTeXify Proj Team Page: 6 of 6 NUS Mathematics Society
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