SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I – FALL 2006 (1) (2) (3) (4) FindR the following using the definition of the Riemann integral: 0 (a) −3 (2x + 1)dx Consider the partition Pn = {x0 = −3, x1 = −3 + 1 2 , x2 = −3 + , ... n n 3n − 1 −1 = , x3n = 0} n n of the interval [−3, 0]. Observe that for all k ∈ {1, 2, ..., 3n}, . . . , x3n−1 = −3 + inf {2x + 1 : xk−1 ≤ x ≤ xk } = 2xk−1 + 1 k−1 = 2 −3 + +1 n 2 (k − 1) = −5 + , and similarly n 2k sup {2x + 1 : xk−1 ≤ x ≤ xk } = −5 + . n Thus, letting f (x) = 2x + 1, 3n X 1 2 (k − 1) L (f, Pn ) = −5 + n n k=1 −5 2 (3n − 1) 3n (3n) + 2 = n n 2 3 = −6 − . n Similarly, 3n X 1 2k U (f, Pn ) = −5 + n n k=1 −5 2 3n (3n + 1) = (3n) + 2 n n 2 3 = −6 + . n 3 Thus, −6− n = L (f, Pn ) ≤ L (f ) ≤ U (f ) ≤ U (f, Pn ) = −6+ n3 . By the Order Limit Theorem and the algebraic limit theorem (with n → ∞), we have −6 ≤ L (f ) ≤ U (f ) ≤ −6, so Z 0 L (f ) = U (f ) = (2x + 1)dx = −6. −3 1 2 REAL ANALYSIS I – FALL 2006 (b) R1 (3 − x2 ) dx Consider the partition −2 Pn = {x0 = −2, x1 = −2 + 1 2 k , x2 = −2 + , ..., xk = −2 + , n n n 3n − 1 1 = 1 − , x3n = 1} n n 2 0 of the interval [−2, 1]. Observe that (3 − x ) = −2x, so the function f (x) = 3 − x2 is decreasing for x ≥ 0 and increasing for x ≤ 0. Then, for all k ∈ {1, 2, ..., 3n}, ( 2 if 1 ≤ k ≤ 2n 3 − −2 + nk 2 , and sup 3 − x : xk−1 ≤ x ≤ xk = k−1 2 3 − −2 + n if 2n + 1 ≤ k ≤ 3n ( 2 if 1 ≤ k ≤ 2n 3 − −2 + k−1 2 n inf 3 − x : xk−1 ≤ x ≤ xk = k 2 3 − −2 + n if 2n + 1 ≤ k ≤ 3n . . . , x3n−1 = −2 + Thus, 2 ! 2 ! 2n 3n X X 1 k k−1 1 L (f, Pn ) = 3 − −2 + + 3 − −2 + n n n n k=1 k=2n+1 3n 2n X X 4 1 2 4 1 1 1 2 − + 2k − 3k + − + 2 (k − 1) − 3 (k − 1) = n n n n n n k=2n+1 k=1 1 4 2n (2n + 1) 1 (4n + 1) 2n (2n + 1) = − (2n) + 2 − 3 n n 2 n 6 1 4 (3n − 1) (3n) (2n − 1) (2n) + − (3n − 2n) + 2 − n n 2 2 1 (6n − 1) (3n − 1) (3n) (4n − 1) (2n − 1) (2n) − − 3 n 6 6 1 5 − = 6+ 2n 2n2 Similarly, 2 ! 2 ! 2n 3n X X 1 k−1 1 k U (f, Pn ) = 3 − −2 + + 3 − −2 + n n n n k=1 k=2n+1 2n 3n X X 1 4 1 1 4 1 2 2 − + 2k − 3k = − + 2 (k − 1) − 3 (k − 1) + n n n n n n k=1 k=2n+1 1 4 (2n − 1) (2n) 1 (4n − 1) (2n − 1) (2n) = − (2n) + 2 − 3 n n 2 n 6 1 4 (3n) (3n + 1) (2n) (2n + 1) + − (3n − 2n) + 2 − n n 2 2 1 (6n + 1) (3n) (3n + 1) (4n + 1) (2n) (2n + 1) − 3 − n 6 6 5 1 = 6− − 2 2n 2n 5 5 Thus, 6 + 2n − 2n1 2 = L (f, Pn ) ≤ L (f ) ≤ U (f ) ≤ U (f, Pn ) = 6 − 2n − 2n1 2 . By the Order Limit Theorem and the algebraic limit theorem (with n → ∞), we have SAMPLE QUESTIONS FOR FINAL EXAM 3 6 ≤ L (f ) ≤ U (f ) ≤ 6, so Z 1 3 − x2 dx = 6. L (f ) = U (f ) = −2 (c) R1 |2x| dx −3 Consider the partition Pn = {x0 = −3, x1 = −3 + 1 2 , x2 = −3 + , ... n n . . . , x4n−1 = −3 + 4n − 1 , x4n = 1} n of the interval [−3, 1]. Letting ψ (x) = |2x|, observe that −2xk if k ≤ 3n (because |2x| is decreasing here) inf {ψ (x) : x ∈ [xk−1 , xk ]} = 2xk−1 if k > 3n (because |2x| is increasing here) −2 −3 + nk if k ≤ 3n = . if k > 3n 2 −3 + k−1 n Thus, 4n X k−1 1 k 1 2 −3 + −2 −3 + + L (ψ, Pn ) = n n n n k=3n+1 k=1 3n 4n X X 6 2k −6 2 (k − 1) = − + + n n2 n n2 k=1 k=3n+1 ! 4n X 6 2 3n (3n + 1) 2 −6 = (3n) − 2 (n) + 2 (k − 1) . + n n 2 n n k=3n+1 3n X Note that 4n X (k − 1) = k=3n+1 4n−1 X k= k=3n 4n−1 X k=1 k− 3n−1 X k k=1 (4n − 1) 4n (3n − 1) 3n − 2 2 7 2 1 = n − n, 2 2 = so 6 2 L (ψ, Pn ) = (3n) − 2 n n 4 = 10 − n 3n (3n + 1) 2 −6 2 + (n) + 2 n n 7 2 1 n − n 2 2 Similarly, sup {ψ (x) : x ∈ [xk−1 , xk ]} = −2 −3 + k−1 n 2 −3 + nk if k ≤ 3n if k > 3n , 4 REAL ANALYSIS I – FALL 2006 so 3n X 4n X k−1 1 k 1 U (ψ, Pn ) = + −2 −3 + 2 −3 + n n k=3n+1 n n k=1 4n 3n X X −6 2k 6 2 (k − 1) − + + 2 = 2 n n n n k=3n+1 k=1 ! 4n X 6 2 (3n − 1) 3n −6 2 = (3n) − 2 + (n) + 2 k , n n 2 n n k=3n+1 and 4n X k=3n+1 k = 4n X k=1 k− 3n X k=1 k= 4n (4n + 1) 3n (3n + 1) − 2 2 7 2 1 = n + n, 2 2 which implies 2 (3n − 1) 3n 2 7 2 1 6 −6 (3n) − 2 (n) + 2 n + n U (ψ, Pn ) = + n n 2 n n 2 2 4 = 10 + . n Since L (ψ, Pn ) ≤ L (ψ) ≤ U (ψ) ≤ U (ψ, Pn ), we have 4 4 10 − ≤ L (ψ) ≤ U (ψ) ≤ 10 + n n for all n ∈ N, so that we can use the order limit theorem as n → ∞ to conclude that the function ψ is integrable and that Z 1 L (ψ) = U (ψ) = |2x| dx = 10. −3 R2 (5) Using the definition of Riemann integral, find −2 ψ (x) dx if it exists, if 0 if x 6= 32 ψ (x) = 7 if x = 23 For k ∈ N, consider the partition Pk = −2, 23 − k1 , 23 + k1 , 2 of the interval [−2, 2]. Then 2 1 2 1 inf ψ (x) : x ∈ − , + = 0, and 3 k 3 k 2 1 2 1 sup ψ (x) : x ∈ − , + = 7; 3 k 3 k the infima and suprema of ψ over the other intervals are zero. Thus, L (f, Pk ) = 0 + 0 + 0 = 0 14 2 U (f, Pk ) = 0 + 7 · +0= . k k Since L (f, Pk ) ≤ L (f ) ≤ U (f ) ≤ U (f, Pk ), we have 14 0 ≤ L (f ) ≤ U (f ) ≤ k SAMPLE QUESTIONS FOR FINAL EXAM 5 for R 2 all k ∈ N. Using the order limit theorem as k → ∞, we have that L (f ) = U (f ) = ψ (x) dx = 0. −2 2 (6) Prove using the definition that lim x −2x+3 = 2. x x→3 Proof: For any ε > 0, let δ = min {ε, 1} > 0. If 0 < |x − 3| < δ, then 1 < x − 1 < 3 and |x − 3| < ε, which implies x |x − 3| < ε , so that x−1 (x − 3) (x − 1) < ε, or x 2 x − 2x + 3 < ε. − 2 x (7) Using the definition, prove that each function below is continuous on its domain. 1 (a) g (x) = 2x+1 ε 2 1 Proof: For any ε > 0, if c 6= −1 , let δ = min |2c + 1| , |2c + 1| > 0. If 2 4 4 |x − c| < δ, then 1 1 − |2c + 1| < x − c < |2c + 1| , so 4 4 1 1 2c − |2c + 1| < 2x < 2c + |2c + 1| , or 2 2 1 1 2c + 1 − |2c + 1| < 2x + 1 < 2c + 1 + |2c + 1| , so that 2 2 1 |2x + 1| > |2c + 1| > 0, or |2c + 1| < 2 |2x + 1| 2 We also have that |x − c| < δ implies ε |c − x| < |2c + 1|2 , which implies 4 1 |c − x| < ε (2x + 1) (2c + 1) , so that 2 2 (c − x) (2x + 1) (2c + 1) < ε, or 1 1 2x + 1 − 2c + 1 < ε. Thus, g is continuous at any c such that c 6= −1 . 2 √ (b) h (x) = 4x + 3 Proof: Let c be any element of − 34 , ∞ , the domain of h. 2 Case 1: If c = − 43 , then for any ε > 0, let δ = ε4 > 0. If |x − c| = x + 43 < δ and x ≥ − 34 , then 3 4 x + < ε2 , so that 4 √ 3 4x + 3 = h (x) − h − < ε. 4 Thus, h is continuous at − 34 . 6 REAL ANALYSIS I – FALL 2006 3√ 3 , then for any ε > 0, if δ = min ε 4c + 3, (4c + 3) > 0 and if Case 2: If c > −3 4 8 16 |x − c| < δ, then 4x + 3 = 4 (x − c) + 3 + 4c 3 > −4 (4c + 3) + 3 + 4c 16 1 = (4c + 3) , 4 so that √ 1√ 4c + 3 < 4x + 3. 2 Further, again if |x − c| < δ, 3√ 4 |x − c| < ε 4c + 3, so that 2 √ √ |4x − 4c| < ε 4x + 3 + 4c + 3 , or √ √ |(4x + 3) − (4c + 3)| < ε 4x + 3 + 4c + 3 . Then |(4x + 3) − (4c + 3)| < ε, or √ √ 4x + 3 + 4c + 3 √ √ 4x + 3 − 4c + 3 < ε. Thus, h is continuous at c. (8) Prove or disprove: If φ : [0, 1] → R is continuous, and if (xn ) is a Cauchy sequence such that xn ∈ [0, 1] for all n ∈ N, then (φ (xn )) is Cauchy. True: same proof as 10b below. √ (9) Using the definiton, prove that the derivative of x is 2√1 x . Observe that if x > 0, √ √ y− x 1 √ lim = lim √ y→x y→x y−x y+ x 1 = √ 2 x by the algebraic limit theorem. (10) Prove that the derivative of x1 is −1 , using the definition. x2 1 Proof: Let f (x) = x . If x and y are in the domain of f , that is x, y ∈ R \ {0}, then 1 y − 1 x x−y yx −1 . y−x y−x yx We may now take the limit as y → x of the expression above, using the algebraic limit theorem and the continuity of g (t) = −1 (algebraic continuity theorem). Thus, the limit tx exists and is 1 − x1 −1 −1 −1 y 0 f (x) = lim = lim = = 2. y→x y − x y→x yx x·x x (11) Assuming Rolle’s Theorem, prove the Mean Value Theorem. (a) Hint: Apply Rolle’s Theorem to the function g (x) = f (x) − f (b)−f x. b−a (12) Determine if the each series below converges or diverges, and prove your answer. = = SAMPLE QUESTIONS FOR FINAL EXAM (a) ∞ P n=2 7 √ 1 n3 −n Observe that for n ≥ 2, n2 ≥ 4, so that n3 ≥ 4n, or 1 −n ≥ − n3 . 4 Thus, for n ≥ 2, 0≤ √ Since ∞ P n=2 1 √2 3/2 3n = √2 3 1 1 2 1 ≤q = √ 3/2 . n3 − n 3n n3 − 41 n3 ∞ P n=2 1 n3/2 converges by the p-series test (p = ∞ X n=2 (b) √ 3 2 > 1), 1 n3 − n converges by the direct comparison test. ∞ P (−1)n √ n=4 n2 +3n+1 1 < n1 for all n ∈ N, since Observe that this series is alternating. Since 0 < √n2 +3n+1 1 lim n1 = 0, the Squeeze Theorem implies that lim √n2 +3n+1 = 0. Next, observe that for all n ≥ 4, 1 q (n + 1)2 + 3 (n + 1) + 1 ≥√ 1 , n2 + 3n + 1 so the absolute values of the terms of the series are decreasing. By the alternating series test, the series converges. (13) True or False. Prove your response (you may refer to theorems). (a) If φ : (0, 1] → R is continuous, and if (xn ) is a Cauchy sequence such that xn ∈ (0, 1] for all n ∈ N, then (φ (xn )) is Cauchy. False: For example, let φ (x) = x1 (which is continuous on (0, 1] by the algebraic properties of continuous functions), and consider the sequence (xn ) defined by xn = 1 ∈ (0, 1]. Then (xn ) converges to 0 and is thus Cauchy, but φ (xn ) = n, and so n (h (xn )) diverges and is therefore not Cauchy. (b) If φ : [0, 1] → R is continuous, and if (xn ) is a Cauchy sequence such that xn ∈ (0, 1) for all n ∈ N, then (φ (xn )) is Cauchy. True. Since [0, 1] is closed, (xn ) converges to a point x ∈ [0, 1]. Then, since φ is continuous, (φ (xn )) converges to φ (x). Thus, (φ (xn )) is Cauchy, since it is a convergent sequence. (c) Every bounded function on [a, b] is continuous on [a, b]. False: For example, let a = 0, b = 1, and consider the function q : [0, 1] → R defined by 1 if x = 21 q (x) = 0 otherwise Then |q (x)| ≤ 1 for all x ∈ [0, 1], so q is bounded. However, q is not continuous at 1 ; even though 12 is a limit point of [0, 1], lim1 q (x) = 0 6= 1 = q (0). 2 x→ 2 8 REAL ANALYSIS I – FALL 2006 (d) Every bounded function on [a, b] is integrable on [a, b]. False: For example, if 1 if x ∈ Q g (x) = 0 otherwise (e) (f) (g) (h) (i) then g is bounded ( 0 ≤ g (x) ≤ 1 ∀x ), but each upper sum is (b − a), and each lower sum is zero, so the function g is not Riemann integrable. If lim xn = x is a convergent sequence, then the set {xn | n ∈ N} ∪ {x} is compact. True: Since (xn ) converges, the set {xn | n ∈ N} is bounded, say |xn | ≤ B for all n ∈ N. Then every element of S = {xn | n ∈ N}∪{x} is bounded in absolute value by max {B, |x|}. Because lim xn = x, every neighborhood of x contains all but finitely many terms of the sequence (xn ). Thus, if y ∈ R {x}, then if we let ε = 12 |y − x|, Vε (y) ∩ Vε (x) = ∅, so that Vε (y) {y} can contain at most a finite number of terms {xn1 , ..., xnk } of the sequence. Thus, if εe = min {ε, |y − xn1 | , ..., |y − xnk |} > 0, we have that Vεe (y) {y} = ∅. Thus, y is not a limit point of S. We have shown that no element other than x is a limit point of S, so S is closed. Since S is closed and bounded, S is compact, by the Heine-Borel Theorem. Any bounded set that contains all of its limit points is compact. True: Since such a set would be closed and bounded, the Heine-Borel Theorem implies that the set is compact. Any set that contains all of its limit points is compact. False: For example, N is closed but not bounded, and so it contains all of its limit points (there are none) but is not compact. The set {x ∈ R | x3 − 3x6 + 17x11 − 7x ≤ 100} is closed. True: Call this set A; observe that A = f −1 ( (−∞, 100] ), where f is the function defined by f (x) = x3 − 3x6 + 17x11 − 7x. The function f is continuous because it is a polynomial, a sum of products of j (x) = x and constants, and the algebraic properties of continuous functions then imply f is continuous. The set (−∞, 100] is closed, and its complement (100, ∞) is open. Observe that B = R \ A=f −1 ( (100, ∞) ) is open because it is the continuous inverse image of an open set. Thus, A = R \ B is closed. If f is continuous on (−∞, 1)∪(2, ∞), then it is also continuous on (−∞, 1]∪ [2, ∞). False: Let 1 if x ∈ [1, 2] f (x) = 0 otherwise The function is constant and thus continuous on (−∞, 1) ∪ (2, ∞), but f is not continuous at 1 or at 2. For instance, the definition of continuous does not hold at 1 for ε ≤ 21 , because for any δ > 0, |f (x) − f (1)| = 1 > 12 for x = 1 − 2δ , even though |x − 1| = 2δ < δ. (j) If the function φ : R → R satisfies the intermediate value property, then it is continuous. False: For example, let sin x1 if x 6= 0 φ (x) = 0 if x = 0 Then φ is not continuous at 0, because φ ((−δ, δ)) = [−1, 1] for every δ > 0 (so the definition of continuous cannot hold at 0 with ε = 12 ). On the other hand, φ satisfies the intermediate value property (IVP). Since φ is continuous away from 0, SAMPLE QUESTIONS FOR FINAL EXAM 9 it automatically satisfies the IVP for intervals contained in R \ {0}. Next, if [a, b] is any interval containing zero, φ ([a, b]) = φ ((a, b)) = [−1, 1], so there are (an infinite number of) points x in (a, b) such that φ (x) is any possible R 1 intermediate value. (k) If f is nonnegative and integrable on [0, 1] and if 0 f = 0, then f (x) = 0 for every x ∈ (0, 1). False: Let 1 if x = 12 f (x) = 0 otherwise Then this function is nonnegative, and it is not identically zero on (0, 1). However, observe that every lower sum will be zero for this function (since every interval contains points x where f (x) = 0. Next, for k > 1, consider the partition Pk defined by 1 1 1 1 Pk = 0, − k , + k , 1 . 2 2 2 2 Then 1 1 1 2 1 − − +1 k +0 = 21−k . U (f, Pk ) = 0 2 2k 2 2 2k Since L (f, Pk ) ≤ L (f ) ≤ U (f ) ≤ U (f, Pk ), and since lim U (f, Pk ) = 0 and k→∞ L (f, Pk ) = 0, by the algebraic R 1 limit theorem, we have 0 ≤ L (f ) ≤ U (f ) ≤ 0, so that f is integrable and 0 f = 0. (l) If f and g are two integrable functions on [−2, 1], then Z 1 Z 1 Z 1 |g| . |f | − |f − g| ≥ −2 −2 −2 True: Z 1 Z |f − g| ≥ −2 1 ||f | − |g|| Z |f | − |g| = −2 Z 1 1 −2 −2 ≥ Z 1 |f | − −2 |g| . (m) If A is bounded, then inf A is a limit point of A. False: For example, the set A = {1, 2} has no limit points, but inf A = 1. (n) The function 0 if t is rational Bubba (t) = t if t is irrational is continuous at 0. True: Given ε > 0, if |t − 0| < δ = ε, then |Bubba (t) − Bubba (0)| = |Bubba (t) − 0| 0 if t is rational = |t| if t is irrational < ε. Rb (o) If a f (x) dx = 0, f is continuous, and f (x) ≤ 0 for all x ∈ [a, b], then f (x) = 0 for all x ∈ [a, b]. True: Suppose that f (x) is not always 0; that is, suppose that f (c) < 0 for some c ∈ [a, b]. Then, since f is continuous, letting ε = −f2(c) > 0, there exists δ > 0 such 10 REAL ANALYSIS I – FALL 2006 that |x − c| < δ and x ∈ [a, b] implies that |f (x) − f (c)| < −f2(c) , so f (x) − f (c) < −f (c) or f (x) < f (c) . Thus, 2 2 Z b Z Z Z 0 = f (x) dx = f (x) dx + f (x) dx ≤ f (x) dx a [a,b]Ic Ic Ic Z f (c) f (c) δ ≤ dx = < 0, 2 2 2 Ic where Ic ⊂ [a, b] is an interval of length 2δ that includes c as an endpoint. This is a contradiction, so f (x) = 0 for all x ∈ [a, b]. (p) Every real number is a limit point of the set of real numbers. True, since every real number x is a limit of the sequence (an ) defined by an = x + n1 for all n ∈ N. (14) Suppose that F is a differentiable function on R, and suppose that F (0) = 1, F (2) = −1, and F (4) = 0. Prove that there is a real number x ∈ (0, 4) such that F 0 (x) = 13 . (0) Proof: By the Mean Value Theorem, there exists c ∈ (0, 2) such that F 0 (c) = F (2)−F = 2 F (4)−F (2) 1 0 −1, and also there exists b ∈ (2, 4) such that F (b) = = 2 . By the Darboux 2 Theorem, F 0 satisfies the Intermediate Value Property, so since c < b and F 0 (c) = −1 < 1 1 < = F 0 (b) , 3 2 there exists x ∈ (c, b) ⊂ (0, 4) such that F 0 (x) = 13 . (15) Let p : [0, 1] → R be a continuous function. Which of the following must be true? (Justify your responses.) (a) There exists a constant a > 0 such that |p (x) − p (y)| < a for all x, y ∈ [0, 1]. True: By the Extreme Value Theorem, a continous function achieves its maximum and minimum on a compact set, so since [0, 1] is compact, in particular p is bounded, say |p (t)| ≤ M for all t ∈ [0, 1]. Then |p (x) − p (y)| ≤ |p (x)| + |−p (y)| ≤ 2M . So a = 2M + 1 works. (b) There exists a constant b > 0 such that |p (x) − p (y)| < 1 for all x, y ∈ [0, 1] that satisfy |x − y| < b. True: Since a continuous function on a compact set is uniformly continuous, p is uniformly continuous on [0, 1], which implies the statement above with ε = 1 > 0. (c) There exists a constant c > 0 such that |p (x) − p (y)| < c |x − y| for all x, y ∈ [0, 1]. |p (x) − p (y)| False: You can construct a function that is continuous on [0, 1] so that |x − y| is unbounded. For example, let x sin x1 if 0 < x ≤ 1 p (x) = 0 if x = 0 Observe that p is continuous on (0, 1], since the functions defined by g1 (x) = sin (x) , g2 (x) = x are continuous, and by the fact that the compositions and algebraic combinations of continuous functions are continuous. Observe that 0 is a limit point of [0, 1], and since for x > 0, −x ≤ x sin x1 ≤ x and lim+ x = 0, the Squeeze x→0 Theorem implies that 1 lim+ x sin = 0. x→0 x SAMPLE QUESTIONS FOR FINAL EXAM 11 Thus, p is continuous at 0 as well. On the other hand, for any n ∈ N, we may let 2 2 x= ,y= , and so x, y ∈ [0, 1], but (4n + 1) π (4n − 1) π 2 (4n + 1) π 2 (4n − 1) π sin − sin (4n + 1) π |p (x) − p (y)| 2 (4n − 1) π 2 = 2 2 |x − y| − (4n + 1) π (4n − 1) π 2 2 + 2 (4n − 1) + 2 (4n + 1) (4n + 1) π (4n − 1) π = = = 4n, 2 2 2 (4n + 1) − 2 (4n − 1) − (4n − 1) π (4n + 1) π which is unbounded as n → ∞. (16) What facts are impicitly used to make the following deductions? x+2 < 4x + 1 y y 1 > : assuming x + 2 and 4x + 1 are nonzero, x+2 4x + 1 x+2 assuming and 4x + 1 are the same sign, and using y 1 the fact that r (x) = is a strictly decreasing function. x 2 y 1 1 =⇒ > : assuming first that 2 > x+2 4x + 1 (4x + 1) 1 y and are both positive and using the fact x+2 4x + 1 that s (x) = x2 is strictly increasing on (0, ∞) ; then 1 assuming that is greater than 1 and using 4x + 1 the fact that x2 > x if x > 1. =⇒ (17) Consider the function g defined by 0 if x = 0 g (x) = 1 2 x cos x2 if x ∈ [−π, 0) ∪ (0, π] Is this function bounded? Is this function continuous at zero? ... differentiable at zero? ... Is g 0 (x) continous at zero? Provide justification. 2 2 Answer: The function g is bounded, because|g (x)| ≤ x ≤ π for all x in the domain of x2 cos 1 − 0 |x|2 x2 g. If ε > 0, and if |x| < δ = ε, then < ε. Thus, g is differentiable ≤ x−0 |x| at zero with g 0 (0) = 0, and as a result g is also continuous at zero. Observe that by the algebraic properties of derivatives, if x 6= 0, g is differentiable at x, and 1 x2 1 0 g (x) = 2x cos − 2 sin x2 x3 x2 1 2 1 = 2x cos − sin . 2 x x x2 12 REAL ANALYSIS I – FALL 2006 Observe that ! r π π 2 π + 2nπ − 2 + 2nπ sin + 2nπ g = pπ cos 2 2 2 + 2nπ 2 r √ π + 2nπ ≤ − 2 2π n1/2 , = −2 2 √ 1 and xn = p π approaches zero and − 2 2π n1/2 decreases without bound as n + 2nπ 2 increases, so lim g 0 (xn ) = −∞. We conclude by the sequential criterion that g 0 (x) has no limit as x → 0 . Thus, g 0 is not continous at 0. (18) Consider the function 1 if t is rational G (t) = et if t is irrational 0 1 pπ + 2nπ 2 Find the set S of real numbers at which G is continuous. Justify your response. Solution: G is continuous only at 0. Observe that since 0 ≤ |G (t) − 1| ≤ |et − 1|, and since lim |et − 1| = 0, the Squeeze Theorem implies that lim |G (t) − 1| = 0. Thus t→0 t→0 lim (G (t) − 1) = 0, so that lim G (t) = 1 = G (0), which implies that G is continuous at t→0 t→0 0. On the other hand, if t 6= 0, then for ε = 21 |1 − et | > 0 and for any δ > 0, by the density of Q in R and of (RQ) in R, there are irrational numbers x and rational numbers y such that |x − t| < δ and |y − t| < δ so that max {|G (x) − G (t)| , |G (y) − G (t)|} ≥ min {|1 − et | , |1 − ex | , |1 − ey |}. The latter quantity is strictly greater than ε = 12 |1 − et | if δ is sufficiently small, showing that G is not continuous at t. (19) Suppose that (fn ) is a sequence of bounded, real-valued functions on R. If 3 |fk+1 (x)| ≤ |fk (x)| for all x ∈ R and all k ∈ N, prove that ∞ X fn (x) n=1 converges absolutely and uniformly on R. (x) Proof: (Note: Ratio Test does not work, because you do not know that lim fk+1 fk (x) exists. Plus it only gives you pointwise convergence.) 1 Observe that |f2 (x)| ≤ 13 |f1 (x)|, and in general if |fk (x)| ≤ 3k−2 |f1 (x)|, then |fk+1 (x)| ≤ 1 1 |fk (x)| ≤ 3k−1 |f1 (x)|. By induction, 3 |fn (x)| ≤ 1 3n−2 |f1 (x)| for all n ≥ 2, and the equation is also valid for n = 1. Since f1 is bounded on R, there exists M > 0 such that |f1 (x)| ≤ M for all x ∈ R, so that |fn (x)| ≤ 1 3n−2 M P 1 for all x ∈ R.PSince M is a convergent geometric series, the Weierstrass M -test 3n−2 implies that fn (x) converges absolutely and uniformly on R. (20) Prove that the Taylor series for f (x) = sin (x) that is centered at zero converges to sin (x) at each x ∈ R. SAMPLE QUESTIONS FOR FINAL EXAM Proof: If pk (x) = k P n=0 f (n) (0) n x n! 13 is the k th partial sum of the Taylor series of sin (x), the (n+1) Lagrange Remainder Theorem implies that for each x ∈ R, sin (x)−pn (x) = f (n+1)!(c) xn+1 , d d where c is a number between 0 and x. Observe that since sin (x) dx (n+1) = cos (x), dx cos (x) = − sin (x), |sin (x)| ≤ 1, and |cos (x)| ≤ 1, we have that f (t) ≤ 1 for all t ∈ R. Then (n+1) f (c) |x|n+1 n+1 0 ≤ |sin (x) − pn (x)| = x ≤ . (n + 1)! (n + 1)! |x|n+2 (n + 1)! = (n + 2)! |x|n+1 n=1 |x| , which has a limit of zero as n → ∞. By the ratio test, the series converges, so by n+2 Next, for fixed x ∈ R, consider the series ∞ P |x|n+1 . (n+1)! Observe that for any n+1 |x| = 0, so by the Squeeze Theorem and the the divergence criterion for series, lim (n+1)! n→∞ inequality above, lim |sin (x) − pn (x)| = 0, n→∞ so the Taylor series for sin (x) converges to sin (x) for all x ∈ R. (21) Prove using basic principles that a sequence of continuous functions that converges uniformly converges to a continuous function. Proof: Suppose (fn ) is a sequence of continuous functions that converges uniformly to a function f on A ⊂ R. Let c be any element of A. By the definition of uniform convergence, ε given any ε > 0, we may choose N ∈ N such that for all n ≥ N , |fn (x) − f (x)| < for 3 all x ∈ A. Next, for that specific N , the continuity of fN implies that we may choose ε δ > 0 so that if x ∈ A and |x − c| < δ then |fN (x) − fN (c)| < . Then if |x − c| < δ, we 3 have |f (x) − f (c)| = |f (x) − fN (x) + fN (x) − fN (c) + fN (c) − f (c)| ≤ |f (x) − fN (x)| + |fN (x) − fN (c)| + |fN (c) − f (c)| ε ε ε < + + = ε. 3 3 3 Thus f is continuous at c. (22) Given an example of a sequence of continuous functions that converges but does not converge to a continous function. (Justify your response.) For each n ∈ N, let αn : [0, 1] → R be defined by αn (x) = xn ; all of these functions are continuous by the algebraic continuity theorem. Then let 0 if 0 ≤ x < 1 β (x) = lim αn (x) = 1 if x = 1 n→∞ Observe that lim− β (x) = 0 6= 1 = β (1), so β is not continuous at 1. x→1 (23) Let B be a bounded, nonempty set of real numbers, and let b be the least upper bound of B. If b is not an element of B, which of the following is necessarily true? Justify your responses. (a) B is closed. False: for example, if B = (0, 1), b = 1 ∈ / B. (b) B is not open. False: see above. (c) b is a limit point of B. 14 REAL ANALYSIS I – FALL 2006 True. By the sup lemma, given any ε > 0, there is an x ∈ B such that b − ε < x ≤ b, but since b ∈ / B we also have b − ε < x < b. Thus, every ε-neighborhood of b contains a point x ∈ B {b} = B; thus, b is a limit point of B. (d) No sequence in B converges to b. False. By part (c), b is a limit point of B, so there exists a sequence of points in B {b} = B that converges to b. (e) There is an open interval containing b that contains no points of B. False. By part (c), b is a limit point of B. Every open interval containing b is an open set, so there exists ε > 0 so that Vε (b) is a subset of that interval. But every such Vε (b) contains a point of B {b} = B, so the open interval does as well. (24) Prove that every nonempty open set in R is a union of open intervals. Proof: If A is a nonempty open subset of R, then for every x ∈ A, there is an ε > 0 such that Vε (x) ⊂ A. Then consider [ B= Vε (x) . x∈A,V ε(x)⊂A The notation above implies that the union is taken over all possible x ∈ A and all possible ε > 0 such that Vε (x) ⊂ A. Then B is a union of open intervals, and y ∈ B, then y ∈ Vε (x) ⊂ A for some ε > 0 and some x ∈ A, so y ∈ A. Next, if y ∈ A, there exists ε > 0 such that y ∈ Vε (y) ⊂ A, so y ∈ B. Thus, A = B is a union of intervals. (25) Assuming that f and g are real-valued functions on R, negate the following statement: “For each s ∈ R, there exists r ∈ R such that if f (r) > 0, then g (r) > 0.” Negation: “There exists s ∈ R such that for all r ∈ R such that f (r) > 0, we have g (r) ≤ 0.” (26) Suppose that φ : R → R is a function such that lim φ(x) is a real number L x x→0 and φ (0) = 0. Which of the following are necessarily true? (Justify your responses.) (a) φ is differentiable at 0. φ (x) − φ (0) True: definition of the derivative: φ0 (0) = lim = L. x→0 x−0 (b) L = 0. False: For example, if φ (x) = x for all x ∈ R, then φ0 (0) = 1 = L. (c) lim φ (x) = 0. x→0 True: Differentiable implies continuous at zero, so since 0 is a limit point of the domain, lim φ (x) = φ (0) = 0. x→0 (27) Provide a definition for the statement lim f (x) = ∞. Then prove that if x→c 1 lim f (x) = ∞, then lim = 0. x→c x→c f (x) The statement lim f (x) = ∞ means that given M > 0, ∃ δ > 0 such that 0 < |x − c| < x→c δ and x ∈ (domain of f ) implies f (x) ≥ M . If this is true, then, given ε > 0, choose N ∈ N such that N1 < ε. Then choose δ as above with M = N . Then 0 < |x − c| < δ 1 1 and x ∈ (domain of f ) \ {0} implies that f (x) − 0 ≤ N1 < ε. Thus, lim = 0. x→c f (x) (28) Give an example of a function α : R → R that is uniformly continuous and an example of a function β : R → R that is not uniformly continuous. Solution: For example, let α (t) = t. Then, given any ε > 0, if |t − x| < δ = ε, then |α (t) − α (x)| = |t − x| < ε. Thus α is uniformly continuous. SAMPLE QUESTIONS FOR FINAL EXAM 15 2 Next, let β (x) that for every δ > 0, there exists n such that δ > n1 , or = xδ . Observe nδ > 1. Then n + 2 − n < δ, but 2 δ δ 2 = n+ β n + − β (n) − n 2 2 1 = nδ + δ 2 > 1, 4 so the definition of uniformly continuous does not hold for any ε < 1. (29) Be able to prove the product and quotient rules. OK, I’m able. (30) Prove that if f is differentiable on an interval with f 0 (x) 6= 1 on that interval, then f can have at most one fixed point. (Note: a fixed point of f is a number x such that f (x) = x.) Proof: Consider the function h (x) = f (x) − x. Then by hypothesis, h is the sum of differentiable functions and is thus differentiable on the interval, and h0 (x) 6= 0 on that interval. Suppose that h (a) = 0 and h (b) = 0 for some a and b in the interval (ie both a and b are fixed by f ). Then, by Rolle’s Theorem, if a 6= b, then there exists c between x and y such that h0 (c) = 0, which is a contradiction. Thus, there is at most one fixed point for f on that interval.

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