Real Analysis Exam 2017, questions and answers PDF

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MAT00005C UNIVERSITY OF YORK BA, BSc and MMath Examinations 2017 MATHEMATICS Real Analysis Time Allowed: 3 hours. Answer ALL questions. There are four questions, each worth 50 marks. Your final mark for the paper will be calculated out of a possible 200 marks. Please write your answers in ink; penci...

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MAT00005C UNIVERSITY OF YORK

BA, BSc and MMath Examinations 2017 MATHEMATICS Real Analysis Time Allowed: 3 hours. Answer ALL questions. There are four questions, each worth 50 marks. Your final mark for the paper will be calculated out of a possible 200 marks. Please write your answers in ink; pencil is acceptable for graphs and diagrams. Do not use red ink. The marking scheme shown on each question is indicative only. Calculators may not be used.

Page 1 (of 4)

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MAT00005C 1 (of 4).

(a) Use the Archimedean property to establish that   2 1 n −1 : n∈N = . sup 2 2n 2

[10]

(b) Define what it means for a sequence of real numbers (an )n∈N to converge to a limit a ∈ R as n → ∞. [5] (c) Suppose that three sequences (an ), (bn ) and (cn ) satisfy • an ≤ bn ≤ cn for all n, and • an → a and cn → a as n → ∞.

Prove that bn → a as n → ∞.

[15]

(d) Consider the sequence (an ) given by a0 = 2 ,

an+1 =

1 2an , + an 3

n ≥ 0.

By first of all showing that an2 > 3 for all n ≥ 0, prove that an converges as n → ∞, and find the value of its limit. [20]

2 (of 4).

(a) By considering the sequence of partial sums, establish the range of x ∈ R for which the following series converges, and calculate its limit when it does converge: ∞ X xj . [10] j=1

(b) Determine whether or not each of the following statements is true or false. If false, give a counterexample; if true, provide a proof. (The numbers aj and bj are positive in each case.) P∞ aj converges. (i) If aj+1 /aj < 1 for all j ≥ 1 then j=1 P∞ P∞ bj con(ii) If limj→∞ (aj − bj ) = 0 and j=1 aj converges, then j=1 verges. P P∞ 2 aj converges then ∞ (iii) If j=1 j=1 aj converges. P∞ P∞ 2 (iv) If j=1 aj converges then j=1 aj converges. [20]

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MAT00005C

(c) Determine whether the following series converge or diverge. If you make use of any convergence tests, be sure to show that the relevant conditions are satisfied.  ∞ ∞  ∞ X X X (2j )! 1 1 (−1)j+1 j √ (i) √ − ; (iii) . ; (ii) 2 2j j! j − 2j + 2 j j+1 j=1 j=1 j=1 [20]

3 (of 4).

(a) Define what it means for a function f : R → R to be continuous at a point c ∈ R. [5] (b) Show that the function defined by ( 3 f (x) = 0 is not continuous at any c ∈ N.

x∈N x∈ /N [10]

(c) Define what it means for a function f : R → R to be differentiable at a point c ∈ R. Show that the function f (x) = 3x2 satisfies this definition for all c ∈ R. [15] (d) Give an example of a function that is continuous everywhere, but for which there exists at least one point where the derivative is not defined. (You are not required to prove anything here.) [5] (e) Let f : R → R be differentiable everywhere, and let the function g : R → R be defined by g(x) = f 2 (x). Show directly (i.e. without using the chain or product rules) that g is differentiable everywhere, and find its derivative. [15]

Page 3 (of 4)

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MAT00005C 4 (of 4).

(a) State the Mean Value Theorem.

[5]

(b) Suppose that f : [a, b] → R and g : [a, b] → R are both continuous on [a, b] and differentiable on (a, b). By applying the Mean Value Theorem to (f − g), or otherwise, prove that f ′ = g ′ if and only if f and g differ by a constant. (You may use without justification the fact that the derivative of a constant function is zero.) [15] (c) Define what it means for a function F : [a, b] → R to be a primitive for a function f : [a, b] → R. [5] (d) State the Fundamental Theorem of Calculus.

[5]

(e) Let f : [a, b] → R be continuous and G : [a, b] → R be any primitive for f . Prove that Z b f = G(b) − G(a) . a

(You may find the result of part (b) helpful here.) R1 Hence, or otherwise, find the Riemann integral 0 f of the function f : [0, 1] → R defined by f (x) = x3 . [20]

Page 4 (of 4)

End of examination.

SOLUTIONS 1.

MAT00005C

(a) First show that 1/2 is an upper bound for the set: n2 n2 − 1 1 ≤ = . 2n2 2 2n2

✞ ✝5

☎

Marks ✆

To see that 1/2 is the least upper bound, we need to show that for any ε > 0 we can find n ∈ N such that n2 − 1 1 > − ε. 2n2 2 Rearranging this inequality, we see that we need to find n ∈ N such that 1 n>√ , 2ε and the existence of such an n is guaranteed by the Archimedean ✞ prop-☎ erty. ✝5 Marks ✆ TD2 (b) an converges to a as n tends to infinity, if for any ε > 0 there✞ exists a☎ natural number Nε such that if n > Nε then |an − a| < ε. ✝5 Marks ✆ TD1 (c)

(a)

(a) (c) Given ε > 0 there exist Nε , Nε ∈ N such that if n > N ✞ε then☎ (c) |an − a| < ε and if n > Nε then |cn − a| < ε. ✝5 Marks ✆ (c) (a) If n > Nε := max{Nε , Nε } then both of these inequalities are true, and we have

a − ε < an < a + ε ;

a − ε < cn < a + ε.

✞

☎

✝5 Marks ✆

Now use the fact that an ≤ bn ≤ cn to deduce that a − ε < a n ≤ bn ≤ c n < a + ε so in particular a − ε < bn < a + ε

which is equivalent to |bn − a| < ε. This shows that bn → a, as✞required.☎ ✝5 Marks ✆ TD2 (d) We first of all show that an2 > 3 for all n. This is clearly true for n = 0. Suppose that it is true for some value of n ≥ 0. Then  2  2an + 3 2 2 a n+1 − 3 = −3 3an 4a4 − 15an2 + 9 (4an2 − 3)(a2n − 3) = n = , 9a2n 9an2 5

SOLUTIONS

MAT00005C ✞

and this is > 0 thanks to our inductive assumption.

☎

✝5 Marks ✆

We know that our sequence is bounded below: if we can show it is decreasing then it follows by the Principle of Bounded Monotone Con-☎ ✞ vergence that it will converge. 5 Marks ✝ ✆ But an 3 − a2n 1 < 0, − an+1 − an = = 3an an 3 ✞ ✝5

and so the sequence is indeed decreasing.

☎

Marks ✆

Since the sequence converges, we know that a = limn→∞ an = limn→∞ an+1. Thus a satisfies 2a 1 + a= a 3 √ with solutions a = ± √3. It’s clear that an ≥ 0 for all n, and so ✞ it must☎ be the case that a = 3. ✝5 Marks ✆ TD2 Remarks. Part (a) is similar to seen examples; parts (b) and (c) are bookwork; part (d) is an unseen example, but is similar to examples seen during the module. ✞ ✝Total:

2.

(a) Let sn =

Pn

j=1

☎

50 Marks ✆

xj . Then sn = x + x2 + . . . xn =

x − xn+1 . 1−x

✞ ✝5

☎

Marks ✆

Taking the limit as n → ∞, we see that this converges if and ✞ only if☎ |x| < 1, in which case the infinite series has value x/(1 − x). ✝5 Marks ✆ TD2 (b) (i) False: e.g. aj = 1/j gives the harmonic series, which we✞ know is☎ divergent. ✝5 Marks ✆ P∞ aj converges, and (ii) False: e.g. let aj = 1/j 2 and bj = 1/j. Then j=1 P∞ 2 bj diverges. limj→∞(aj − bj ) = limj→∞ (1/j − 1/j) = 0, but j=1 ✞ ☎ 5 Marks ✝ ✆ P∞ aj converges then we know that aj → 0. So there (iii) True: if j=1 exists N ∈ N such that aj < 1 for all j > N , and hence aj2 < aj P∞ 2 aj converges. for all j > N . By the comparison test the series j=1 6

SOLUTIONS

MAT00005C ✞

☎

✝5 Marks ✆

(iv) False: if aj = 1/j then

P∞

j=1

a2j converges but

P∞

j=1

aj diverges. ✞ ☎ ✝5 Marks ✆

P∞ (−1)j+1aj where aj is a non-negative de(c) (i) This takes the form j=1 creasing sequence tending to zero. (A little work is needed to see that it’s decreasing, but nothing substantial!) So this converges by☎ ✞ the Leibniz alternating series test. ✝7 Marks ✆ √ P∞ aj where aj = bj − bj+1 with bj = 1/ j , (ii) This takes the form j=1 i.e. this is a telescopic series. So n X j=1

aj = b1 − bn+1 = 1 − √

1 →1 n+1

as n → ∞. Thus the series converges (with limit 1).

✞ ✝7

☎

Marks ✆

(iii) We use the ratio test:      (2j + 1)(2j + 2)  aj+1   (2j + 2)! 2j j!  =  = (2j + 1) .  aj   (2j )! 2j+1(j + 1)!  = 2(j + 1)

Since this tends to ∞ as j → ∞, the ratio test tells us✞that this☎ series diverges. ✝6 Marks ✆

TD2 (any part) Remarks. Part (a) is straightforward; part (b) requires some thought, but not a lot of writing, and should test how well students really understand the concept of series convergence; part (c) requires use of convergence tests which students have seen used in many previous examples. ✞ ✝Total:

3.

☎

50 Marks ✆

(a) We say a function f : R → R is continuous at c ∈ R if limx→c f (x) = f (c). In other words, for any ε > 0 there exists δ > 0 such that |x✞− c| < δ☎ implies that |f (x) − f (c)| < ε. ✝5 Marks ✆ TD1 NB. Students may obtain full marks for this part by correctly making either of the statements above.

7

SOLUTIONS

MAT00005C

(b) Fix c ∈ N. There are a number of ways to show that f is not continuous at c, and full marks may be obtained for any of these. Here are two possibilities: • Consider the sequence xn = c + 1/2n; we have xn → c as n → ∞, with xn 6= c for all n, and f (xn ) = 0 for all n. Thus limn→∞ f (xn ) = 0, and since this disagrees with f (c) = 3 we see that f is not continuous at c. • Alternatively, let ε = 1 (any value of ε < 3 will do). For any choice of δ > 0 there will exist points x ∈ (c − δ, c + δ) for which x ∈ / N, and so for which f (x) = 0. That is, there exists ε > 0 such that for all δ > 0 there are points x with |x − c| < δ but |f (x) − f (c)| = 3 > ε. So f is not continuous at c. ✞ ✝10

TD3

☎

Marks ✆

(c) We say a function f : R → R is differentiable at c ∈ R whenever the limit f (x) − f (c) f (x + h) − f (x) lim = lim x→c h h→0 x−c ✞

exists. TD1

✝5

☎

Marks ✆

With f (x) = 3x2 we have f (x + h) − f (x) 6xh + 3h2 3(x + h)2 − 3x2 = lim = 6x . = lim h→0 h→0 h→0 h h h ✞ ☎ ✝6 Marks ✆ Since this limit exists for any x ∈ R it follows that f is differentiable ✞ ☎ everywhere. ✝4 Marks ✆ TD3 lim

(d) The function f (x) = |x| is continuous on R but not differentiable ✞ at zero.☎ ✝5 Marks ✆ (e) Fix some c ∈ R. We need to consider f 2 (x) − f 2 (c) g(x) − g(c) = lim x→c x→c x−c x−c (f (x) − f (c))(f (x) + f (c)) = lim x→c x−c f (x) − f (c) = lim lim (f (x) + f (c)) , x→c x→c x−c lim

✞ ✝5

☎

Marks ✆

by the algebra of limits. The first of these limits is equal to f✞′ (c) since☎ we know that f is differentiable at c. ✝5 Marks ✆ 8

SOLUTIONS

MAT00005C

Since f is differentiable at c, it is continuous at c, and so the second limit is equal to 2f (c). We conclude that g is differentiable at c, with ✞

g ′ (c) = 2f (c)f ′ (c) .

✝5

☎

Marks ✆

TD3 Remarks. Part (a) is bookwork; part (b) is unseen but similar to seen examples; part (c) is a mixture of bookwork and a straightforward example; part (d) should be simple - this example has been seen a number of times in the module; part (e) is unseen. ✞

✝Total:

4.

☎

50 Marks ✆

(a) Let f : [a, b] → R be continuous, and differentiable on (a, b). Then there exists c ∈ (a, b) for which f ′ (c) =

f (b) − f (a) . b−a

✞

☎

✝5 Marks ✆

(b) First suppose that f = g +k for some constant k. Since g is differentiable we know that the derivative of the RHS exists and is given by (g + k)′ = g ′ + k ′ = g ′ (since the derivative of a constant function is zero). Thus☎ ✞ ′ ′ f =g. ✝5 Marks ✆ Now suppose instead that f ′ = g ′ . Choose any two points x, y satisfying a ≤ x < y ≤ b. The MVT applied to the function (f − g) says that there exists c ∈ (x, y) for which (f − g )′ (c) =

(f − g)(y) − (f − g)(x) . y−x

But since f ′ = g ′ the left hand side of this equation is zero, and we obtain (f − g)(y) = (f − g )(x) .

✞

☎

✝5 Marks ✆

Since this is true for all x, y ∈ [a, b] we see that (f − g) is constant on☎ ✞ [a, b], and thus f = g + k for some constant k, as required. ✝5 Marks ✆ TD3 (c) A continuous function F : [a, b] → R which is differentiable on ✞ (a, b) is☎ ′ called a primitive for f : [a, b] → R if f (x) = F (x) on (a, b). ✝5 Marks ✆ TD1 Rx (d) The FTC says that if f : [a, b] → R is continuous and F (x) = a f , then F (x) is differentiable on (a, b) with F ′ (x) = f (x). (In other✞words, if☎ Rx f : [a, b] → R is continuous then a f is a primitive for f .) ✝5 Marks ✆ 9

SOLUTIONS

MAT00005C

Rx (e) The FTC says that F (x) = a f is a primitive for f , i.e. that✞F ′ (x) =☎ f (x). If G is also a primitive for f then G′ = F ′ . ✝5 ✞ Marks ✆ ☎ By part (b), this means that F and G differ by a constant. ✝5 Marks ✆ Therefore Z b f = F (b) = F (b) − F (a) = G(b) − G(a) a

✞

☎

✞

☎

as required. R ✝5 Marks ✆ 1 To evaluate 0 x3 it suffices by the above result to find a primitive for f . By observation, the function G(x) = x4 /4 is a primitive, and so Z

0

1

f = G(1) − G(0) = 1/4 .

✝5 Marks ✆

TD3 Remarks. Part (a) is bookwork; part (b) has been seen in a related example, but not in exactly this setting; parts (c) and (d) are bookwork; part (e) requires putting together a number of (seen) concepts and applying the result to a simple example. ✞ ✝Total:

10

☎

50 Marks ✆...