MATH20101 Exam, Real Analysis part

MATH20101 Exam, Real Analysis part

January 2016

A1

i. Prove, by verifying the appropriate definition, that a. lim x→1 x 3_{− 4x}2_{+ 4x + 4
= 5.} b. lim x→+∞ x2_{+ 2x} x2_{+ x − 2} = 1.

ii. 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 limx→ah (x) = L and limx→ag (x) = L. Prove that limx→af (x) = L.

Solution A1

i. a. Rough work Assume 0 < |x − 1| < δ, where δ is to be found. Consider |f (x) − L| =   x3− 4x2+ 4x + 4
− 5   =  x3− 4x2+ 4x − 1   = |x − 1|x2− 3x + 1   < δ x2− 3x + 1  .

Assume δ ≤ 1 when |x − 1| < δ ≤ 1 expands out as 0 < x < 2. Next use the triangle inequality:

 x2− 3x + 1  ≤ |x| 2 + 3 |x| + 1 ≤ 11, since |x| < 2.

Note problems occur if students simply put x = 0 or x = 2 into x2_{− 3x + 1}

and choose the largest absolute value, since the maximum occurs at x = 3/2. Then

|f (x) − L| < 11δ

Proof Let ε > 0 be given. Choose δ = min (1, ε/11) and assume 0 < |x − 1| < δ. Consider   x3− 4x2+ 4x + 4
− 5  < δ  x2− 3x + 1  ,

as seen in the rough work. Also seen there is |x − 1| < δ ≤ 1 implies |x| < 2, and x2− 3x + 1 ≤ |x|2+ 3 |x| + 1 ≤ 11. Thus   x3− 4x2+ 4x + 4
− 5  < 11δ ≤ 11  ε 11  = ε. Hence we have verified the definition of

lim

x→1 x

3_{− 4x}2_{+ 4x + 4
= 5.}

Many such examples on problem sheets[7 marks; all but 1 for getting an allowable δ]

b. Rough work Perhaps observe that lim x→+∞ x2_{+ 2x} x2_{+ x − 2} = lim_x→+∞ x (x + 2) (x − 1) (x + 2) = limx→+∞ x (x − 1). With X > 0 to be chosen assume x > X. Consider

    x x − 1− 1     = 1 x − 1 < 1 X − 1,

which we demand is ≤ ε, i.e. X > 1 + 1/ε. End of Rough work

Proof Let ε > 0 be given. Choose X = 1 + 1/ε and assume x > X. Then, by the argument above,

|f (x) − L| =     x2_{+ 2x} x2_{+ x − 2}− 1     < 1 X − 1 = ε.

Let ε > 0 be given.

Then limx→ah (x) = L implies ∃δ1 > 0 such that

0 < |x − a| < δ1⇒ |h (x) − L| < ε ⇒ L − ε < h (x) < L + ε.

In particular

0 < |x − a| < δ1 ⇒ L − ε < h (x) . (2)

Similarly limx→ag (x) = L implies ∃δ2> 0 such that

0 < |x − a| < δ2⇒ |g (x) − L| < ε ⇒ L − ε < g (x) < L + ε.

In particular

0 < |x − a| < δ2 ⇒ g (x) < L + ε. (3)

Let δ = min (δ0, δ1, δ2) > 0. Assume 0 < |x − a| < δ. For such x, (1), (2)

and (3) hold. Then

L − ε < h (x) ≤ f (x) ≤ g (x) < L + ε, ie.

L − ε < f (x) < L + ε, equivalently |f (x) − L| < ε. Hence we have verified the definition that limx→af (x) = L.

A2

x4

1 + x4

is differentiable on R, and find the derivative.

iii. Assume that f : A → R is differentiable at a ∈ R. Prove, using the laws of limits, that

 1 f ′ (a) = −f ′_(a) f2_(a),

provided that f (a) 6= 0.

(You may assume that if f is differentiable at a and f (a) 6= 0 then f (x) 6= 0 in some neighbourhood of a.)

Solution A2

i Start from − |y| ≤ y ≤ |y| for all y ∈ R. With y = x cos (1/x) , in which case |y| ≤ |x| since |cos θ| ≤ 1 for all θ ∈ R, this gives

− |x| ≤ x cos 1 x

 ≤ |x|

for all x 6= 0. Let x → 0 using the Sandwich Rule along with limx→0|x| = 0. Example on Problem Sheets[4 marks]

iii Let a ∈ R be given. Consider x4

4 −

a4

Let x → a and use the Product and Quotient Rules for Limits to deduce, first, that lim x→a x4 1 + x4 − a4 1 + a4 x − a

exists, and so x4_{/ (1 + x}4_{) is differentiable at a. Since a was arbitrary the}

function is differentiable on R. Secondly, the value of the limit, and thus the value of the derivative at x = a is

4a3

(1 + a4₎2.

Plenty of such examples on problem sheet[9 marks]

iv. We are told that f is differentiable at a so by assumption in the question, f is non-zero in some deleted neighbourhood of a. For such x the reciprocal 1/f (x) is well-defined and we can consider

1 f (x) − 1 f (a) x − a = 1 f (x)− 1 f (a) x − a = − 1 f (x) f (a) f (x) − f (a) x − a . Let x → a to get  1 f ′ (a) = −f ′_(a) f2_(a),

having used the Quotient Rule for limits, allowable since f (a) 6= 0.

A3.

i Given a function f whose first n derivatives exist at a ∈ R define the Taylor polynomial Tn,af (x) of degree n at a.

ii Prove that if the first n + 1 derivatives of f exist on an open interval I then d dtTn,tf (x) = (x − t)n n! f (n+1)_(t) for all x, t ∈ I. iii. Calculate T5,0  sin x 1 + x 

Solution A3

ii. For t between a and x. d dtTn,tf (x) = d dt  f (t) + f′_{(t) (x − t) +} f ′′_(t) 2! (x − t) 2 + ... +f (n)_(t) n! (x − t) n  = f′_{(t) +}_f′′ (t) (x − t) − f′_(t)₊ +1 2! f ′′′ (t) (x − t)2− 2f′′(t) (x − t)
+ .... ... ... + 1 f(n+1)_{(t) (x − t)}n − nf(n)_{(t) (x − t)}n−1
.

iii A suggested manner of calculation is to let f (x) = sin x/ (1 + x) and multiply up so (1 + x) f (x) = sin x. Differentiate repeatedly

(1 + x) f (x) = sin x f (x) + (1 + x) f(1)(x) = cos x 2f(1)(x) + (1 + x) f(2)(x) = − sin x 3f(2)_{(x) + (1 + x) f}(3)_{(x) = − cos x} 4f(3)(x) + (1 + x) f(4)(x) = sin x 5f(4)(x) + (1 + x) f(5)(x) = cos x. Put x = 0 to get f (0) = 0 f (0) + f(1)(0) = 1 2f(1)_{(0) + f}(2)_{(0) = 0} 3f(2)(0) + f(3)(0) = −1 4f(3)(0) + f(4)(0) = 0 5f(4)(0) + f(5)(0) = 1. So f (0) = 0, f(1)_{(0) = 1, f}(2)_{(0) = −2, f}(3)_{(0) = −1 + 6 = 5, f}(4)_{(0) =} −20 and f(5)_{(0) = 1 + 5 × 20 = 101. Thus} T5,0  sin x 1 + x  = x − x2+5 6x 3₋5 6x 4₊101 120x 5_. Bookwork[12 marks]

A4

i. Let f : [a, b] → R be a bounded function and let P = {xi: 0 ≤ i ≤ n}

be a partition of [a, b] .

a. Define the Upper Sum U (P, f ) and Lower Sum L (P, f ) , explaining the meaning of all terms.

b. Show that if y ∈ [a, b] and y /∈ P then L (P ∪ {y} , f ) ≥ L (P, f ) . Hint If A ⊆ B ⊆ R then glbA ≥ glbB.

ii Let f : [1, 2] → R, x 7→ x (x + 1) and, for every n ≥ 1, define Pn =  1 + i n : 0 ≤ i ≤ n  . a. Prove that U (Pn, f ) = 23n2_{+ 12n + 1} 6n2

(You may assume thatPn

i=1i = n (n + 1) /2 and

Pn

i=1i2= n (n + 1) (2n + 1) /6.)

b. Prove, by verifying the definition, that f is integrable over [1, 2] and find the value of the integral.

(You may assume that L (Pn, f ) = (23n2− 12n + 1) /6n2.)

Solution A4

i. a. Assume f is a bounded function on the interval [a, b]. The Upper Sum is U (P, f ) = n X i=1 Mi(xi− xi−1)

Definitions and thus considered easy marks[4 marks]

b. Given y ∈ [a, b] there exists 1 ≤ j ≤ n such that xj−1< y < xj. Then

L (P ∪ {y} , f ) =X i6=j mi(xi− xi−1) + glbf [x_j−1,y] (y − xj−1) + glbf [y,xj] (xj− y) (4)

We then use the idea that f (x) may take smaller values on [xj−1, xi] than

on any of the subsets [xj−1, y] or [y, xj] so

glbf [x_j−1,y] ≥ glbf [x_j−1,xj] = mj and glbf [y,xj] ≥ glbf [x_j−1,xj] = mj.

To relate this to the given hint A = {f (x) : x ∈ [xj−1, y]} and B =

{f (x) : x ∈ [xj−1, xj]} for the first inequality.

Hence RHS(4) is ≥ X i6=j mi(xi− xi−1) + mj(y − xj−1) + mj(xj− y) = X i6=j

mi(xi− xi−1) + mj{(y − xj−1) + (xj− y)}

= X

i6=j

mi(xi− xi−1) + mj(xj− xj−1) = L (P, f ) .

That is, L (P ∪ {y} , f ) ≥ L (P, f ).

One of the few proofs in the Integration Chapter[6 marks]

ii. a. On [1, 2] the function f (x) = x (1 + x) is increasing and so takes it maximum at the right hand of each sub-interval [xi−1, xi]. The width of each

sub-interval is 1/n. Thus U (Pn, f ) = n X i=1  1 + i n   2 + i n   1 n  = 1 n n X i=1  2 + 3 ni + 1 n2i 2  = 1 n  2n + 3 n n (n + 1) 2 + 1 n2 n (n + 1) (2n + 1) 6  = 1 n  12n2 6n + 9n (n + 1) 6n + (n + 1) (2n + 1) 6n  = 1 6n2 12n 2_{+ 9n}2_{+ 9n + 2n}2_{+ 3n + 1}
= 1 6n2 23n 2_{+ 12n + 1
.}

An exercise in arithmetic summation[8 marks]

b. For every n ≥ 1 we have L (Pn, f ) ≤ Z 2 1 f ≤ Z 2 1 f ≤ U (Pn, f ) .

So, by the assumption given, 23n2_{− 12n + 1} 6n2 ≤ Z 2 1 f ≤ Z 2 1 f ≤ 23n 2_{+ 12n + 1} 6n2 .

Let n → ∞ when the expressions at both ends converge to 23/6 and so we must have equality in the centre, i.e. R2

1f =

R2

1f , and so the function in

integrable over [1, 2]. The common value, 23/6 , is therefore the value of the integral.