(ii) Such a phenomenon can not take place for the reals. Suppose that

we are given intervals J1 , J2 , . . . in R of total length 1 ’ where 1 > > 0.

Show that we can ¬nd open intervals I1 , I2 , . . . of total length less than

1 ’ /2 such that Ir ⊇ Jr for each r. Now set Kj = [0, 1] \ j Ir . By r=1

applying Exercise 4.3.8 which tells that the intersection of bounded, closed,

nested non-empty sets in R is itself non-empty, show that ∞ Kj = … and

r=1

∞

deduce that r=1 Jr [0, 1]. We can not put a quart into a pint pot8 .

(iii) It could be argued that the example above does not completely rule

out a theory of integration for well behaved functions f : Q ’ Q. To show

that no such theory exists consider the function f : Q ’ Q is given by

if x2 < 2,

f (x) = 1

f (x) = 0 otherwise,

2

Examine how the the procedure for de¬ning an integral f (x) dx by means

0

of upper and lower sums and integrals breaks down

Exercise K.116. (First integral mean value theorem.) [8.3, T] (i)

Suppose F : [a, b] ’ R is continuous. Show that, if supt∈[a,b] F (t) ≥ » ≥

inf t∈[a,b] F (t), there exists a c ∈ [a, b] with F (c) = ».

(ii) Suppose that w : [a, b] ’ R is continuous and non-negative on [a, b].

If f : [a, b] ’ R is continuous, show that there exists a c ∈ [a, b] with

b b

f (t)w(t) dt = f (c) w(t) dt.

a a

(This is a pretty result, but in the view of the present author, resembles

the mean value theorem in being a mainly decorative extension of simpler

inequalities.)

(iii) Show that (ii) may fail if we do not demand w positive. Show that

it also holds if we demand w everywhere negative.

8

Or a litre into a half litre bottle.

491

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.117. [8.3, T] The real function f : [a, b] ’ R is strictly increas-

ing and continuous with inverse function g. Give a geometric interpretation

of the equality

b f (b)

g(y) dy = bf (b) ’ af (a)

f (x) dx +

a f (a)

and prove it by using upper and lower Riemann sums.

Suppose p an q are positive real numbers with p’1 + q ’1 = 1. By using

the idea above with f (x) = xp’1 , a = 0 and b = X show that, if X, Y > 0

then

Xp Y q

≥ XY.

+

p q

For which values of X and Y does this inequality become an equality?

Exercise K.118. [8.3, H] Historically and pedagogically the integrability

of continuous functions is always linked with the proof that a continuous

function on a closed set is uniformly bounded. The following exercise shows

that this is not the only path.

(i) Reread Exercise 8.2.17 (i). Show that, under the assumptions of that

exercise,

— — —

I[a,c] (f |[a,c] ) + I[c,b] (f |[c,b] ) = I[a,b] (f |[a,b] ) and

I—[a,c] (f |[a,c] ) + I—[c,b] (f |[c,b] ) = I—[a,b] (f |[a,b] )

—

where I[a,c] denotes the upper integral on [a, c] and so on.

(ii) Let f : [a, b] ’ R be continuous. Suppose, if possible, that f is not

integrable. Explain why this means that there is a κ > 0 such that

—

I[a,b] (f |[a,b] ) ’ I—[a,b] (f |[a,b] ) ≥ κ(b ’ a).

(iii) Suppose that c0 = (a + b)/2. Show that at least one of the following

two statements must be true.

—

I[a,c0 ] (f |[a,c0 ] ) ’ I—[a,c0 ] (f |[a,c0 ] ) ≥ κ(c0 ’ a) and, or

—

I[c0 ,b] (f |[c0 ,b] ) ’ I—[c0 ,b] (f |[c0 ,b] ) ≥ κ(b ’ c0 ).

Use this remark as the basis for a lion hunting argument. Use the continuity

of f to obtain a contradiction and deduce Theorem 8.3.1.

492 A COMPANION TO ANALYSIS

Exercise K.119. [8.3, H] For the reasons sketched in Section 9.3 there it

is hard to extend the change of variable formula given in Theorem 8.3.13 for

one-dimensional integrals to many dimensions without a fundamental rethink

about the nature of area. But, whether we face or evade this issue, the proof

of Theorem 8.3.13 is essentially one-dimensional. In this exercise we give a

lion hunting argument which could be extended to higher dimensions.

(i) We use the notation and hypotheses of Theorem 8.3.13. Suppose, if

possible, that the theorem is false. Explain why this means that there is a

κ > 0 such that

g(d) d

f (s) ds ’ f (g(x))g (x) dx ≥ κ(d ’ c).

g(c) c

(ii) Suppose that e0 = (c + d)/2. Show that at least one of the following

two statements must be true.

g(e0 ) e0

f (s) ds ’ f (g(x))g (x) dx ≥ κ(e0 ’ c) and, or

g(c) c

g(d) d

f (s) ds ’ f (g(x))g (x) dx ≥ κ(d ’ e0 ).

g(e0 ) e0

Use this remark as the basis for a lion hunting argument and deduce Theo-

rem 8.3.13 from the resulting contradiction.

Exercise K.120. [8.3, H, ‘ ] (i) Prove Lemma 8.3.18 (the formula for in-

tegration by parts) by direct calculation in the special case G(x) = Ax + B,

f (x) = Kx + L by direct calculation.

(ii) Prove Lemma 8.3.18 by lion hunting along the lines of Exercise K.119.

Exercise K.121. [8.3, H, ‘ ] Let f ; [a, b] ’ R be di¬erentiable (with one

sided derivatives at the end points). Use lion hunting to prove the theorem

due to Darboux which states that, if f is Riemann integrable then

b

f (t) dt = f (b) ’ f (a).

a

[This is a considerable generalisation of Exercise 8.3.12 but is not as ¬nal

a result as it looks at ¬rst sight since it need not be true that f is Riemann

integrable.]

493

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.122. [8.3, T] (i) Show directly, using uniform continuity and

an argument along the lines of our proof of Theorem 8.3.1, that, if f : [a, b] ’

R is continuous, then

N ’1 b

b’a

f (a + j(b ’ a)/N ) ’ f (t) dt

N a

j=0

as N ’ ∞. (The general result, which holds for all Riemann integrable

functions, was given in Exercise K.113 (v) but the special case is easier to

prove.)

(ii) It is natural to ask whether we cannot de¬ne the integral of a con-

tinuous function directly in this manner, without going through upper and

lower sums. Here we show one way in which it it can be done. We shall

suppose f : [a, b] ’ R a continuous function and write

N ’1

b’a

f (a + j(b ’ a)/N ).

SN =

N j=0

As might be expected, our main tool will be uniform continuity.

Show that, given > 0, we can ¬nd an N0 ( ) such that, if N ≥ N0 ( ) and

P ≥ 1, then |SN ’SN P | < . Show that if M, N ≥ N0 ( ), then |SN ’SM | < 2 .

Deduce that SN tends to a limit as N ’ ∞. We can de¬ne this limit to be

b

f (t) dt.

a

(iii) Explain brie¬‚y why the new de¬nition gives the same value for the

integral as the old.

The objections to this procedure are that it obscures the geometric idea

of area, that many branches of pure and applied mathematics deal with

functions like the Heaviside function which, though well behaved, are not

continuous, that it gives a special status to points of the form a + j(b ’ a)/N

and that it does not generalise well. (These objections, although strong,

are not, however, conclusive. The ideas sketched here are frequently used in

obtaining various generalisations of our simple integral. An example of this

is given in Exercise K.137.)

Exercise K.123. [8.3, T] Let f : [’1, 1] ’ R be de¬ned by f (1/n) = 1

for all integers n ≥ 1 and f (t) = 0, otherwise. By ¬nding appropriate

dissections, show that f is Riemann integrable. If

t

F (t) = f (x) dx,

0

¬nd F and show that F is everywhere di¬erentiable. Observe that F (0) =

f (0). Is f continuous at 0? (Prove your answer.)

494 A COMPANION TO ANALYSIS

Exercise K.124. [8.3, T] (i) Use the relation

(r + 1)r(r ’ 1) ’ r(r ’ 1)(r ’ 2)

r(r ’ 1) =

3

to ¬nd n r(r ’ 1). Use a similar relation to ¬nd n r and deduce the

r=1 r=1

n 2

value of r=1 r .

(ii) By using the method of part (i), ¬nd n r3 . Show that

r=1

n

rm = (m + 1)’1 nm+1 + Pm (n)

r=1

where Pm is a polynomial of degree at most m.

(iii) Use dissections of the form

D = {0, 1/n, 2/n, . . . , 1}

to show that

1

1

xm dx =

m+1

0

for any positive integer m.

a b

(iv) Use the result of (iii)9 to compute 0 xm dx and a xm dx for all values

of a and b. Obtain the same result by using a version of the fundamental

theorem of the calculus.

(v) Use dissections of the form

D = {0, 1/n, 2/n, . . . , 1}

to show that

1

ex dx = e ’ 1.

0

Obtain the same result by using a version of the fundamental theorem of the

calculus.

(vi) Use dissections of the form

D = {brn , brn’1 , brn’2 , . . . , b}

9

This method is due to Wallis who built on earlier, more geometric, ideas and represents

one of the high spots of analysis before the discovery of the fundamental theorem of

the calculus united the theories of di¬erentiation and integration. Wallis was appointed

professor at the ¬ercely royalist university of Oxford as a reward for breaking codes for

the parliamentary (that is, anti-royalist) side in the English civil war.

495

Please send corrections however trivial to twk@dpmms.cam.ac.uk

with 0 < r and r n = a/b to compute

b

xm dx

a

for any positive integer m.

Exercise K.125. (Numerical integration.) [8.3, T] We saw in Exer-

N ’1

cise K.122 (i) that, if f : [a, b] ’ R is continuous, then b’a j=0 f (a + j(b ’

N

b

a)/N ) is a good approximation to a f (t) dt. In this exercise we shall see

that, if we know that f is better behaved, then we can get better approxi-

mations. The calculations give a good example of the use of global Taylor

theorems such as Theorem 7.1.2.

(i) Show that, if g is linear, then

1

g(t) dt = g(’1) + g(1).

’1

(ii) Suppose that g : R ’ R is twice di¬erentiable with |g (t)| ¤ K for

all t ∈ [’1, 1]. Explain why |g(t) ’ g(0) ’ g (0)t| ¤ Kt2 /2 for all t ∈ [’1, 1]

and deduce that

1

4K

g(t) dt ’ g(’1) + g(1) ¤ .

3

’1

(iii) Suppose that f : R ’ R is twice di¬erentiable with |f (t)| ¤ K for

all t ∈ [a, b]. If N is a strictly positive integer and N h = b ’ a, show that

a+rh

Kh3

f (a + (r ’ 1)h) + f (a + rh) h

f (t) dt ’ ¤

2 6

a+(r’1)h

for all integers r with 1 ¤ r ¤ N . Let

h

f (a) + 2f (a + h) + 2f (a + 2h) + · · · + 2f (a + (N ’ 1)h) + f (b)

Th f =

2

Show that

b

K(b ’ a)h2

f (t) dt ’ Th f ¤ .

6

a

b

We call the approximation a f (t) dt ≈ Th f . the trapezium rule. Speaking

informally, we can say that ˜the error in the trapezium rule decreases like the

square of the step length h™.

496 A COMPANION TO ANALYSIS

(iv) Let a = 0, b = π, N h = π with N a strictly positive integer and let

F (t) = sin2 (N t). Show that

b

F (t) dt ’ Th F ≥ A sup |F (t)|(b ’ a)h2

t∈[a,b]

a

where A is a strictly positive constant. Conclude that the bound in can

not be substantially improved. (In fact, it can be halved by careful thought

about worst cases.)

(v) Show that, if g is constant, then

1

g(t) dt = 2g(’1).

’1

By imitating the earlier parts of this exercise show that if f : R ’ R is once

di¬erentiable with |f (t)| ¤ K for all t ∈ [a, b], N is a strictly positive integer,

N h = b ’ a and we write

Sh f = ’h f (a) + f (a + h) + 2f (a + 2h) + · · · + f (a + (N ’ 1)ha

b

f (t) dt ’ Sh f ¤ K(b ’ a)h. Speaking informally, we can say that

then a

b

˜the error when we use the approximation rule a f (t) dt ≈ Sh f decreases

like the step length h™. Show that we can not improve substantially on the

bound obtained.

(vi) Show that, if g is a cubic (that is to say, a polynomial of degree 3),

then

1

g(’1) + 4g(0) + g(1)

g(t) dt = .

3

’1

(vi) Show that, if g is a cubic (that is to say, a polynomial of degree 3),

then

1

g(’1) + 4g(0) + g(1)

g(t) dt = .

3

’1

Suppose that f : R ’ R is four times di¬erentiable with |f (4) (t)| ¤ K for

all t ∈ [a, b]. If N is a strictly positive even integer and N h = b ’ a, let us

write

h

Qh f = f (a) + 4f (a + h) + 2f (a + 2h) + 4f (a + 3h) + 2f (a + 4h) + . . .

3

+ 2f (a + (N ’ 2)h) + 4f (a + (N ’ 1)h) + f (b)

497

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Show that

b

K(b ’ a)h4

f (t) dt ’ Qh f ¤ .

90

a

b

We call the approximation a f (t) dt ≈ Qh f Simpson™s rule. Speaking in-

formally, we can say that ˜the error in the Simpson™s rule decreases like the

fourth power of the step length h™. Show that the bound in can not be

substantially improved. (Again, it can be halved by careful thought about

worst cases.)

(The reader may be tempted to go on and consider more and more compli-

cated rules along these lines. However such rules involve assumptions about

the behaviour of higher derivatives which are often unrealistic in practice.)

Exercise K.126. [8.3, T] The object of this exercise is to de¬ne the loga-

rithm and the real exponential function, so no properties of those functions

should be used. You should quote all theorems that you use, paying partic-

ular attention to those on integration.

x

We set l(x) = 1 1 dt.

t

(i) Explain why l : (0, ∞) ’ R is a well de¬ned function.

(ii) Use the change of variable formula for integrals to show that

xy

1

dt = l(y)

t

x

whenever x, y > 0. Deduce that l(xy) = l(x) + l(y).

(iii) Show that l is everywhere di¬erentiable with l (x) = 1/x.

(iv) Show that l is a strictly increasing function.

(v) Show that l(x) ’ ∞ as x ’ ∞.

(vi) Show that l : (0, ∞) ’ R is a bijective function.

(vii) In Section 5.6 we de¬ned the logarithm as the inverse of the expo-

nential function e : R ’ (0, ∞). Turn this procedure on its head by de¬ning

e as the inverse of l. Derive the main properties of e, taking particular care

to quote those theorems on inverse functions that you require. When you

have ¬nished, glance through section 5.4 to see if you have proved all the

properties of the (real) exponential given there.

(viii) By expanding (1 + t)’1 as a geometric series and integrating, obtain

a Taylor series for log(1 + x), giving the range over which your argument is

valid.

(ix) Use (viii) to ¬nd the Taylor series of log (1 + x)/(1 ’ x) , giving the

range over which your argument is valid. Show that

3 5

y’1 y’1 y’1

1 1

log y = 2 + + + ...

y+1 3 y+1 5 y+1

498 A COMPANION TO ANALYSIS

for all y > 0.

Exercise K.127. [8.3, P, S, ‘ ] Use the Taylor series for log(1 + x) and

some result on products like Exercise 5.4.4 to obtain

∞

(’1)n

2