MATH20101 Real Analysis, Exam Solutions and Feedback. 2013\14

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1 MATH200 Real Analysis, Exam Solutions and Feedback. 203\4 A. i. Prove by verifying the appropriate definition that ( 2x 3 + x ) = 7. x 2 ii. By using the Rules for its evaluate a) b) x 2 x + x 2 + x 2 2x 2 + 7x + 6, x 2 + x 2 2x 2 + 7x + 6. iii. Prove the Sandwich Rule: Suppose that f, g and h are three functions for which h (x) f (x) g (x) for all x in some deleted neighbourhood of a R. Assume further that x a h (x) = L and x a g (x) = L. Prove that x a f (x) = L. Solution A i. Let ε > 0 be given. Choose δ = min (, ε/33). Assume 0 < x ( 2) < δ. Next Consider 2x 3 + x ( 7) = x + 2 2x 2 3x + 6 < δ 2x 2 3x < x + 2 < δ 3 < x < < 2x 2 3x + 6 < 33 Thus 2x 3 + x ( 7) ( ε ) < 33δ 33 = ε. 33 Hence we have verified the definition of x 2 (2x 3 + x 2 + 5) = 7. Similar to problem sheet example [8 marks]

2 ii. a) x 2 x 2 + x 2 2x 2 + 7x + 6 = x 2 (x + 2) (x ) (x + 2) (2x + 3) = x 2 (x ) (2x + 3) = x 2 (x ) x 2 (2x + 3) by the Quotient Rule = 3 = 3. b) x + x 2 + x 2 2x 2 + 7x + 6 = + 2 x x 2 x x x 2 = ( x + + ) 2 ( x x 2 x ) 6 x x 2 [2 marks] by the Quotient Rule = 2. [2 marks] iii In symbols the assumption of the question can be written as we are told there exists δ 0 > 0 such that Let ε > 0 be given. 0 < x a < δ 0 h (x) f (x) g (x) Then x a h (x) = L implies δ > 0 such that 0 < x a < δ h (x) L < ε/3. () Similarly x a g (x) = L implies δ 2 > 0 such that 0 < x a < δ 2 g (x) L < ε/3. (2) Let δ = min (δ 0, δ, δ 2 ). Assume 0 < x a < δ. For such x both () and (2) hold. 2

3 Consider f (x) L = f (x) h (x) + h (x) L (3) f (x) h (x) + h (x) L by the triangle inequality. Next h (x) f (x) g (x) 0 f (x) h (x) g (x) h (x). So f (x) h (x) g (x) h (x) = g (x) L + L h (x) g (x) L + L h (x) again by the triangle inequality. Inserted into (3) gives f (x) L g (x) L + 2 h (x) L < ε/3 + 2ε/3 by () and (2) = ε Hence we have verified the definition that x a f (x) = L. Feedback on Question. Bookwork [8 marks] i. Students failed to tell me that ε > 0 was given. And having chosen δ many forgot the lower bound in assume 0 < x ( 2) < δ. Many failed to correctly deduce 3 < x < from x + 2 <. Even with the correct interval 3 < x < many got the wrong upper bound on the quadratic 2x 2 3x+6. In this case the largest (absolute) value of the quadratic does not occur at the largest value of x. Students often simply put x = 3 and x = into the quadratic and wrote i.e. 33 <. 33 < 2x 2 3x + 6 <, The easiest way to find an upper bound is by using the triangle inequality in 2x 2 3x + 6 2x 2 + 3x + 6 3

4 Next 3 < x < implies that x < 3 which, when put in the bound above gives 2x 2 + 3x This is, in fact, the best possible upper bound since 33 is the value taken by the polynomial at x = 3. Do not waste time in the exam by finding the best possible bound (by, for example, completing the square), just find an upper bound. ii) a. You cannot write x 2 x 2 + x 2 2x 2 + 7x + 6 = x 2 x 2 + x 2 x 2 2x 2 + 7x + 6 since the it on the denominator is zero. I did not want you to use L Hopital s Rule for this is not one of the Limit Rules (Sum, Product or Quotient Rule). Unfortunately so many of you used it that I had to be generous and allow it. But even if you applied L Hopital s Rule you would still have to apply the Quotient Rule and you needed to have told me you had used it. iii). So many students forgot δ 0 that I had to be generous and not deduct marks for this. A common problem is to say x a h (x) = L means ε > 0 δ > 0 such that 0 < x a < δ h (x) L < ε/3. And x a g (x) = L means ε > 0 δ 2 > 0 such that 0 < x a < δ 2 g (x) L < ε/3. Next choose δ = min (δ, δ 2 ). But if you do this, what ε does δ depend on? Compare what is written here with the answer above. A number of students finished the proof in a different way to mine: Let ε > 0 be given. Then δ > 0 such that 0 < x a < δ h (x) L < ε L ε < h (x) Also δ 2 > 0 such that 0 < x a < δ 2 g (x) L < ε g (x) < L + ε Let δ = min (δ 0, δ, δ 2 ). Assume 0 < x a < δ. For such x i.e. f (x) L < ε as required. L ε < h (x) f (x) g (x) < L + ε 4

5 A2. i. a) State the Intermediate Value Theorem. b) Prove that cos (πx) = 0 + x2 has at least two solutions with x [, ]. ii. By verifying the definition show that + x 2, is differentiable on R and find its derivative. Any results on its you use should be stated carefully. iii) Prove that if f is differentiable on (a, b) with a local maximum at c (a, b) then f (c) = 0. iv) a) State carefully Rolle s Theorem. b) Prove that has exactly one solution in [0, ] cos (πx) = 0 + x2 Solution A2 i. a) The Intermediate Value Theorem: Suppose that f is a function continuous on a closed interval [a, b]. For all γ between f (a) and f (b) there exist c : a c b for which f (c) = γ. Bookwork [2marks] b) Let f (x) = cos (πx). + x2 The trick is to look at f at 0 as well as and. f ( ) = 2, f (0) = 5, f () = 2. Because of the sign changes we have, by the Intermediate Value Theorem with γ = 0 applied to both [, 0] and [0, ], two zeros in [, ]. [2 marks] 5

6 ii. Let a R be given. Consider +x 2 +a 2 x a = = a 2 x 2 ( + x 2 ) ( + a 2 ) (x a) (a x) (a + x) ( + x 2 ) ( + a 2 ) (x a) (a + x) = ( + x 2 ) ( + a 2 ). Let x a when the Quotient Rules for its gives d dx + x 2 = 2a x=a ( + a 2 ) 2. Thus the function is differentiable. True for all a R implies it is differentiable on R. Hence d dx + x = 2x 2 ( + x 2 ) 2. Similar to problem sheet example [5 marks] iii. Assuming that f has a local maximum at c then f (x) f (c) for all x in some open neighbourhood of c. That is, there exists δ > 0 such that if x c < δ then f (x) f (c). Consider first such x satisfying x < c. Then f (x) f (c) 0 and x c < 0. The quotient of two negative numbers is positive so f (x) f (c) x c 0. By assumption f is differentiable at c so the it of (f (x) f (c)) / (x c) as x c exists which implies the two one sided its exits. Thus f (x) f (c) x c x c 0. Consider next x satisfying x > c. Then f (x) f (c) 0 and x c > 0. The quotient of a negative number by a positive is negative so f (x) f (c) x c 0. 6

7 Hence f (x) f (c) 0. x c+ x c Finally, since f is differentiable the two one-sided its not only exist but are equal. This can only be if both are zero. Hence f (a) = 0. Bookwork [6 marks] iv. Rolle s Theorem If f is continuous on [a, b], differentiable on (a, b) and f (a) = f (b) then there exists c (a, b) for which f (c) = 0. Bookwork [2 marks] Assume that f (x) = 2/ ( + x 2 ) + 3 cos (πx) has two zeros in [0, ] then there exists c (0, ) for which f (c) = 0. Yet f 4 (x) = ( + x 2 2 x 3π (sin πx) < 0, ) since both terms are < 0 for 0 < x <. This contradiction gives the required result. [2 marks] Feedback on Question 2. i. A number of students gave me the statements of Rolle s Theorem or the Mean Value Theorem instead of the Intermediate Value Theorem. The correct statement for the Intermediate Value Theorem starts as γ..., c.... The order of the quantifiers is vital and I saw all variants on this, i.e. γ..., c..., γ..., c... and γ..., c.... The second one here was the most common mistake. It is incorrect to say that f (a) γ f (b) for it might be that f (a) > f (b). But again I had to be generous and not deduct a mark for this. iii. In the definition of local maximum so many students forgot what local means here that I had to be generous and not deduct marks for not using δ in the definition. The order of the steps in the proof that f (a) = 0 is extremely important. You must start by saying that the assumption in the question implies the two one-sided its exist. For only then can you calculate them. iv It is important in the application of Rolle s Theorem that the c found satisfies c (0, ), i.e. c 0 since f (x) = 0 when x = 0 and for our contradiction we require f (x) < 0. 7

8 A3 i. Suppose that f and g are two functions differentiable at a R. Prove the Product Rule for differentiation, namely that using the Rules for Limits. (fg) (a) = f (a) g (a) + f (a) g (a), (4) ii) Give an example of two functions f and g not differentiable at a point a R for which the product fg is differentiable at a. iii) Calculate the Taylor Polynomial T 5,0 ( ln ( + x) + x ). Solution A3 i. Consider (fg) (x) (fg) (a) x a = = f (x) g (x) f (a) g (a) x a f (x) g (x) f (x) g (a) + f (x) g (a) f (a) g (a) x a = f (x) g (x) g (a) x a + g (a) f (x) f (a) x a Now let x a and use the it rules to get the required result. ii For example, f (x) = g (x) = x at a = 0. iii. Let f (x) = ln ( + x) + x. Bookwork [6 marks] unseen [2 marks] 8

9 Then f () (x) = ln ( + x) 2 ( + x) ( + x) 2 f (2) 3 ( + x) (x) = 3 + 2ln ( + x) ( + x) 3 f (3) (x) = ( + x) 4 6ln ( + x) ( + x) 4 f (4) (x) = 50 ( + x) ln ( + x) ( + x) 5 f (5) (x) = 274 ( + x) 6 20ln ( + x) ( + x) 6 So f (0) = 0, f () (0) =, f (2) (0) = 3, f (3) (0) =, f (4) (0) = 50 and f (5) (0) = 274. Thus T 5,0 ( ln ( + x) + x ) = 0 + x 3 x2 2! + x3 3! 50x4 4! + 274x5 5! = x 3 2 x2 + 6 x x x5 On problem sheet [2 marks] Feedback on Question 3 i. As proof that students never read this feedback, we see the same problems with this question as we have in previous years. And in previous years I have given the same feedback that I m going to give now. Too many students started by looking at the right hand side of (4). So f (a) g (a) + f g (x) g (a) f (x) f (a) (a) g (a) = f (a) + g (a) x a x a x a x a f (a) (g (x) g (a)) + (f (x) f (a)) g (a) =, x a x a 9

10 by the Sum Rule. But what happens in f (a) (g (x) g (a))+(f (x) f (a)) g (a)? There is no cancellation, and no way of getting to f (x) g (x) f (a) g (a), which you need to see for the right hand side of (4). The most common way out of this difficulty was to make a s look like x s and x s look like a s and then discover some mysterious cancellation. Time was taken when marking to look out for this. ii. Be careful. You might choose { if x > 0 f (x) = g (x) = if x < 0 for these are not differentiable at x = 0. But fg (x), which is for all x 0, is still not differentiable at x = 0 since it is not defined there. iii. This was done well. Many students started with ( + x) f (x) = ln ( + x), differentiated 5 times and found the f (j) (0) by substitution. 0

11 A4 i. State carefully the Cauchy Mean Value Theorem. Show that if f is continuous on [a, b] and differentiable on (a, b) and f (x) 0 for all x (a, b) then there exists c (a, b) such that f (b) = f (a) + e f(b) f(c) e f(a) f(c). (Hint apply the Cauchy Mean Value Theorem with ln g (x) = f (x).) ii. a. Let f : [, 2] R, x x 2 + x + and, for every n, define the partition P n = { + in } : 0 i n of [, 2]. Show that the Upper Sum is U (P n, f) = 29n2 + 2n + 6n 2. Find a similar expression for L (P n, f). (You may assume that n i= i = n(n+) 2 and n i= i2 = n(n+)(2n+) 6.) b. Prove, by verifying the definition, that f is integrable over [, 2] and find the value of the integral. Solution A4 i. Cauchy Mean Value Theorem: If f and g are continuous on [a, b], differentiable on (a, b) and g (x) 0 for all x (a, b) then there exists c (a, b) for which f (b) f (a) g (b) g (a) = f (c) g (c). Bookwork [2 marks] If ln g (x) = f (x) then g (x) = e f(x) and g (x) = f (x) e f(x). The assumption of f (x) 0 implies g (x) 0 as required for the Cauchy Mean Value Theorem. Thus there exists c (a, b) for which f (b) f (a) e f(b) e f(a) = f (c) f (c) e f(c) = e f(c), which rearranges to the stated result. On Problem Sheet[3 marks]

12 ii. U (P n, f) = n M i (x i x i ) = n i= n i= M i since x i x i = /n for all i. Here M i = sup x i x x i f (x) = x 2 i + x i + = since f is increasing ( ( + i ) 2 ( + + i ) ) + n n Then n M i (x i x i ) = n i= = n = i2 n i n + 3. n ( i 2 n + 3 i ) 2 n + 3 i= ( n 2 = 29n2 + 2n + 6n 2. ( ) n (n + ) (2n + ) + 3 ( ) n (n + ) 6 n 2 ) + 3n [8 marks] For the lower sum L (P n, f) = n m i (x i x i ) = n i= n m i, i= where m i = inf f (x) = x 2 i + x i + = M i. x i x x i 2

13 So L (P n, f) = n m i = n M i = n n n n i= i= i=0 ( n ) = M i + M 0 M n n i= M i = U (P n, f) + (3 7) n = 29n2 + 2n + 4 6n 2 n = 29n2 2n + 6n 2 [3 marks] b) From the theory we have L (P n, f) for all n. Thus f f U (P n, f) 29n 2 2n + 6n 2 f f 29n2 + 2n + 6n 2 for all n. Let n to get This implies 29 6 f f = f and so f in integrable over [, 2]. Secondly, the common value, 29/6, is the integral of f over [, 2]. Bookwork [4 marks] f, Feedback on Question 4 i. The most common error was not knowing which is the CAUCHY mean value theorem and even when it was stated correctly, the conditions were not usually verified when applying it to the second part of the question. 3

14 ii. The calculations of the Upper and Lower sums are exercises in being careful. Too often I saw a page long calculation of ever increasingly complex expressions which suddenly resolve themselves into the expression given in the question. Time was taken when marking to spot initial error. 4

Part 2 Continuous functions and their properties

Part 2 Continuous functions and their properties 2.1 Definition Definition A function f is continuous at a R if, and only if, that is lim f (x) = f (a), x a ε > 0, δ > 0, x, x a < δ f (x) f (a) < ε. Notice