Math 341 Summer 2016 Midterm Exam 2 Solutions. 1. Complete the definitions of the following words or phrases:

Transcription

1 Math 34 Summer 06 Midterm Exam Solutions. Complete the definitions of the following words or phrases: (a) A sequence (a n ) is called a Cauchy sequence if and only if for every ɛ > 0, there exists and N N such that a n a m < ɛ whenever m, n N. (b) A sequence (a n ) converges to a real number a if and only if for every ɛ > 0 there exists an N N such that a n a < ɛ whenever n N. (c) [Reminder: We use the notation V ɛ (a) = {x R: x a < ɛ} = (a ɛ, a + ɛ), and this set is called the ɛ-neighborhood of a.] A set O is open if and only if for all points a O, there exists an ɛ-neighborhood V ɛ (a) O. (d) A point x is a limit point of a set A if and only if every ɛ-neighborhood V ɛ (x) intersects the set A as some point other than x. (e) A point a A is called an isolated point of A if and only if it is not a limit point of A. (f) A set F R is closed if and only if if it contains all of it s limit points. (g) If A R the closure of A is the set Ā = A L, where L is the set of all limit points of A.

2 . Provide an example of each or explain why the request is impossible. (a) Two function f and g, neither of which is continuous at 0 but such that f(x)g(x) and f(x) + g(x) are continuous at 0. Example. Let f(x) = { 0 if x < 0 if x 0 and g(x) = { if x < 0, 0 if x 0. Then f(x)g(x) = 0 and f(x) + g(x) = for all x. Constant functions are continuous at 0. (b) A function f(x) continuous at 0 and g(x) not continuous at 0 such that f(x)+g(x) is continuous at 0. Disproof. This statement is impossible. If f(x) is continuous at 0 and if f(x) + g(x) is continuous at 0, then the Algebraic Continuity Theorem implies that g(x) = [f(x) + g(x)] f(x) must be continuous at 0, contrary to the statement. (c) A function f(x) continuous at 0 and a g(x) not continuous at 0 such that f(x)g(x) is continuous at 0. { 0 if x < 0, Example. Let f(x) = 0 be the constant function 0. Let g(x) = if x 0. Then g(x) is not continuous at 0, but the constant function f(x)g(x) = 0 is continuous at 0. (d) A function f(x) not continuous at 0 such that f(x) + f(x) Example. Let f(x) = { if x 0 if x = 0 and then is continuous at 0. { f(x) = if x 0 if x = 0. The function f has been defined to be discontinuous at 0 and continuous at all x 0. The reciprocal /f(x) is also discontinuous at 0 and continuous at all x 0. The constant function f(x) + = 5 is continuous at 0. f(x) (e) A function f(x) not continuous at 0 such that [f(x)] 3 is continuous at 0. Disproof. The function g(x) = x /3 is continuous at 0. If [f(x)] 3 were also continuous at 0, then the composition of continuous functions g([f(x)] 3 ) = ([f(x)] 3 ) /3 = f(x) would also be continuous at zero. So, no such f(x) exists.

3 3. For each of the following statements, circle True or False. No justification is necessary. For any set A R, (Ā)c is open. True False A set A is closed if and only if A = Ā. True False If A is a bounded set, then s = sup A is a limit point of A. True False An open set that contains every rational number must necessarily be all of R. True False An arbitrary intersection of compact sets is compact. True False If F F F 3 is a nested sequence of nonempty closed sets, then the intersection n=f n. True False The notation V δ (c) in the textbook denotes the interval (c δ, c + δ). True False The set Cantor set is compact. True False Any finite set is compact. True False The set S = { : n N} is compact. True False n 3

4 4. (Limit of a sequence) Suppose lim n x n =. Directly use the definition of the limit of a convergent sequence to show that lim n (x3 n 7x n + 8) =. You should not use the Algebraic Limit Theorem in your argument. [Note: If you get a better score on this question than on Midterm question 4, it will raise the score on Midterm. This question is not optional. You still need to answer it for this exam.] Proof. Let ɛ > 0 be given. Consider the following calculation (x 3 n 7x n + 8) = x 3 n 7x 6 = x n x + x 3 (Since x n = is a root factor using long division.) Since lim x n =, there exists N N such that x n < whenever n N. This implies < x n < 3 whenever x N. Thus for n N, applying the triangle inequality gives (x n 7x n + 8) = x n ( x n + x n + 3) x n ( ) = 3 x n. Since lim x n =, there exists N such that whenever n N. (x n 7x n + 8) < ɛ 3 Thus, if we set N = max(n, N ), then we have This proves that (x n 7x n + 8) since n N 3 x n since n N < 3 ɛ 3 = ɛ. lim n (x3 n 7x n + 8) =. 4

5 5. (Limit of a sequence) Suppose lim n x n = 3. Directly use the definition of the limit of a convergent sequence to show that lim n x n ( x n ) = 6. You should not use the Algebraic Limit Theorem in your argument. [Note: If you get a better score on this question than on Midterm question 0, it will raise the score on Midterm. This question is not optional. You still need to answer it for this exam.] Proof. Let ɛ > 0 be given. We will show there exists N N such that ( x n ( x n ) < ɛ 6) whenever n N. Consider the following calculation: ( x n ( x n ) = 6 + x n ( x n ) 6) 6x n ( x n ) = x n x n 6 6 x n x n = x n 3 x n +. 6 x n x n Since lim x n = 3, there exists N N such that x n 3 < ɛ = whenever n N. So, for n N, we have < x n 3 < < 3 x n < < x n < 4 < x n < 3 Then for n N, we have ( x n ( x n ) ) = x n 3 x n x n x n x n (4 + ) 6 = x n. Since lim x n = 3, there exists N N such that x n 3 < ɛ whenever n N. Now, set N = max(n, N ). If n N, then we have ( x n ( x n ) ) (since n N ) x n 6 This proves that if lim x n = 3, then lim n x n( x n) = 6. (since n N ) < ɛ < ɛ. There is nothing particulary special about the number ɛ = here. The critical thing was to use a value of ɛ that prevents the denominator from being too small. Since lim x n = 3, we need to ensure that for large enough n, both x n and x n are bounded below by a positive constant. In this exercise, any choice of ɛ satisfying 0 < ɛ < would suffice. The choice ɛ = was merely convenient because it gave simple calculations. In problems of this type, avoid automatically using the number. You need to think about what choices of ɛ will cause the denominator to be bounded below by a positive number. For example, if the problem had been to prove lim n x = /3, you would need to choose an ɛ n(5/ x n) satisfying 0 < ɛ < /. So, to be concrete, you might decide to use the number ɛ = /4 in this case. 5

6 6. (Cauchy sequence) Suppose x n is a Cauchy sequence and 3 < x n < 0 for all n N. By directly using the definition of a Cauchy sequence, show that the sequence (b n ), where b n = x n + x n 5, is also a Cauchy sequence. Your argument should not use Cauchy s Criterion. Proof. Let ɛ > 0 be given. Since (x n ) is a Cauchy sequence, there exists N N such that x n x m < ɛ whenever m, n N. We will show that this choice of N suffices to show that (b n ) is also a Cauchy sequence. First, note that since 3 < x n < 0 for all n N, if follows that x n 5 > 5 for all n N. Now assume n, m N, then b n b m = x n + x n 5 x m + x m 5 = (x n + )(x m 5) (x m + )(x n 5) (x n 5)(x m 5) = x n x m + x m 5x n 30) (x m x n + x n 5x m 30) (x n 5)(x m 5) 7 x n x m = x n 5 x m 5 7 x n x m 5 5 7ɛ 5 < ɛ. This proves that (b n ) is a Cauchy sequence. 6

7 7. (ɛ-δ definition of limit) By directly using the ɛ-δ definition of the limit of a function, show that x 3 lim x x + = 8 5. Your argument should not rely on the Algebraic Limit Theorem. Proof. Let ɛ > 0 be given. We will show that there exists δ > 0 such that x 3 x < ɛ whenever 0 < x < δ. Let δ = min(, ɛ/). Note that if x 3 < δ, then < x < 3 and so x < 3. Now assume x 3 < δ, then x 3 x = 5x 3 8(x + ) 5(x + ) = 5x 3 8x 8 5(x + ) = x 5x + x x + x (5 x + x + 4) 5 x ( ) 5 = x ɛ < = ɛ. This proves that lim x x 3 x + =

8 8. Let {O λ : λ Λ} be a collection of open sets. Prove that is an open set. λ Λ Proof. Let x λ Λ O λ. Then x O λ for some λ Λ. Since O λ is open there exists an epsilon neighborhood V ɛ (x) O λ. But then O λ V ɛ (x) O λ λ Λ O λ. This proves that λ Λ O λ is open. 8

9 9. Prove exactly one of the following two theorems. (Don t prove both.) Theorem. (Alternating Series Test) Let (a n ) be a sequence of positive terms such that a > a > a 3 > a 4 > and such that (a n ) 0. Prove that the series n= ( )n+ a n converges by showing that the sequence (s n ) of partial sums is a Cauchy sequence. s n = a a + a 3 + ( ) n+ a n Theorem. (Absolute Convergence Test). If the series n= a n converges, then the series n= a n converges as well. Proof of Alternating Series Test. Let ɛ > 0 be given. We must show that there exists an N N such that s m s n < ɛ whenever m, n N. Let m, n N. Assume n > m N. Then s m s n = a m+ a m+ + ± a n where the sum on the right hand side is alternating. Depending on whether the sum for s m s n has an even or odd number of terms, we have either or 0 (a m+ a m+ ) }{{} >0 = a m+ (a m+ a m+3 ) a m+ + + (a n a n ) }{{} >0 } {{ } >0 (a n a n ) a }{{} n >0 In either case, 0 (a m+ a m+ ) + + (a n a n ) + a n = a m+ (a m+ a m+3 ) (a n a n ) a m+. s m s n a m+. Since lim(a n ) = 0, there exists N N such that a k+ < ɛ whenever k N. If m, n N, then s m s n a m+ < ɛ. This proves that the sequence of partial sums of the alternating series is a Cauchy sequence. This sequence converges by Cauchy s Criterion. 9

10 Proof of Absolute Converges Test. Assume the series n= a n converges. By Cauchy s Criterion, there exists N N such that a m+ + a m+ + + a n < ɛ for all n > m N. By the triangle inequality, we then have a m+ + a m+ + + a n a m+ + a m+ + + a n < ɛ for all n > m N. This shows that the sequence of partial sums of n= is a Cauchy sequence, and by Cauchy s Criterion, the series n= a n converges. 0. Let f : R R be the function defined by { sin( ) if x 0, x f(x) = 0 if x = 0. Either prove that f is continuous at 0 or that f is not continuous at 0. [Hint: This problem can be solved using either the ɛ-δ characterization of continuity or the sequential characterization of continuity.] Solution. We will prove that f(x) is not continuous at x = 0. Let ɛ = / and let δ be any positive number. Then there exists k large enough such that x 0 = π(k + /) < δ. Then x 0 belongs to the δ-neighborhood V δ (0), but f(x 0 ) f(0) = 0 = > ɛ = /. This shows that f(x) is not continuous at x = 0. Solution. Consider the the two sequences (x n ) and (y n ) defined by x n = nπ and y n = (n + /)π. Then (x n ) 0 and (y n ) 0. However, f(x n ) = and f(y n ) = 0 for all n. So, lim f(x n) = 0 = lim f(y n ). n n By the sequential criterion for continuity, f is not continuous at x = 0. 0

11 . Prove exactly one of the following two theorems. (Don t prove both.) Theorem. (Composition of continuous functions) Let f : A R and g : B R be functions such that (i) f(a) = {f(x): x A} B, (ii) f is continuous at c A, (iii) g is continuous at f(c) B. Then g f is continuous at c. Theorem. If f : R R is continuous and B R is open, then is open. f (B) = {x R: f(x) B} Proof of Theorem (Solution ). Let ɛ > 0 be given. We must show that there exists δ > 0 such that x c < δ implies g(f(x)) g(f(c)) < ɛ. Since g is continuous at f(c), there exists α > 0 such that g(y) g(f(c)) < ɛ whenever y B and y g(f(c)) < α. Since f is continuous at c, there exists δ > 0 such that f(x) f(c) < α whenever x c < δ and x A. Then if x c < δ and x A, it follows that So, g f is continuous at c. g(f(x)) g(f(c)) < ɛ. [Note: Most attempted solutions starting off with Since f is continuous... wrong.] were Proof of Theorem (Solution ). Let (x n ) be any sequence in A such that lim x n = c. Since f is continuous at c, lim f(x n ) = f(c). Since ( f(x n ) ) is a sequence in B converging to f(c) and since g is continuous at g(c), it follows that lim g(f(x n )) = g(f(c)). By the sequential characterization of continuity, g f is continuous at c.

12 Proof of Theorem. If f (B) =, then f (B) is open. So, assume f (B). Let x 0 f (B). Then f(x 0 ) = y 0 B. Since B is open there exists an ɛ-neighborhood of y 0 such that V ɛ (y 0 ) B. Since f is continuous at x 0, there exists δ > 0 such that f(x) f(x 0 ) < ɛ, whenever x x 0 < δ. In other words, f(v δ (x 0 )) V ɛ (y 0 ) B. So, we have V δ (x 0 ) f (B). Since x 0 was arbitrary, f (B) is open.. Suppose f : R R is continuous at all points of R and suppose f(c) > 0. Prove that there exists some neighborhood V δ (c) such that f(x) > 0 for all x V δ (c). Proof. Assume f is continuous at c and f(c) > 0. From the definition of continuity, for the number ɛ = f(c) > 0, there exists δ > 0 such that f(x) f(c) < ɛ = f(c) whenever x c < δ. Thus if x V δ (c), then f(c) f(c) and so 0 < f(c) Thus, f(x) > 0 whenever x V δ (c). < f(x) < f(c) + f(c) < f(x) < 3f(c).