1 1

¤ dx.

n+1 x

n

Hence or otherwise show that if we write

n

1

’ log n

Tn =

r

r=1

468 A COMPANION TO ANALYSIS

we have Tn+1 ¤ Tn for all n ≥ 1. Show also that 1 ≥ Tn ≥ 0. Deduce that

Tn tends to a limit γ (Euler™s constant) with 1 ≥ γ ≥ 0. [It is an indication

of how little we know about speci¬c real numbers that, after three centuries,

we still do not know whether γ is irrational. Hardy is said to have o¬ered

his chair to anyone who could decide this.]

(ii) By considering T2n ’ Tn , show that

1111111

1’ + ’ + ’ + ’ · · · = log 2

2345678

(iii) By considering T4n ’ 2 T2n ’ 1 Tn , show that

1

2

11111 3

’ + + ’ + · · · = log 2

1+

32574 2

This famous example is due to Dirichlet. It gives a speci¬c example where

rearranging a non-absolutely convergent sum changes its value.

(iv) Show that

11 1 1 1

+ + ··· + ’ log n ’ γ,

24 2n 2 2

11 1 1 1

1 + + + ··· + ’ log n ’ γ + log 2

35 2n + 1 2 2

as n ’ ∞.

Deduce that

1 1 11 1 1 1 1

’ ’ ’ ··· ’ + ··· + ’ ...,

1+ + ... + +

2p ’ 1 2 4 4p ’ 1

3 2q 2p 2p + 2

where p positive terms alternate with q negative terms, has sum log 2 +

(1/2) log(p/q).

(v) Use the ideas of (iv) to show how the series may be arranged to

converge to any desired sum. (Be careful, convergence of a subsequence of

sums does not imply convergence of the whole sequence.)

Exercise K.73. [5.3, P] We work in C.

(i) Review the proof of Abel™s lemma in Exercise 5.2.6. Show that, if

| N aj | ¤ K for all N and »j is a decreasing sequence of positive terms

j=0

with »j ’ 0 as j ’ ∞, then ∞ »j aj converges and

j=0

∞

»j aj ¤ K sup »j .

j≥0

j=0

469

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

∞

(ii) Suppose that j=0 bj converges. Show that, given > 0, we can ¬nd

an N ( ) such that

∞

bj xj ¤

j=M

whenever M ≥ N ( ) and x is a real number with 0 ¤ x < 1. (This result

was also obtained in Exercise K.61 (vii) but the suggested proof ran along

di¬erent lines. It worth mastering both proofs.)

(iii) By considering the equation

∞ M ’1 ∞

bj xj = bj xj + bj xj ,

j=0 j=0 j=M

∞

or otherwise, show that, if j=0 bj converges, then

∞ ∞

bj xj ’ bj

j=0 j=0

when x ’ 1 through real values of x with x < 1.

(iv) Use the result of Exercise 5.4.4 together with part (iii) to prove the

following result. Let aj and bj be sequences of complex numbers and write

n

cn = an’j bj .

j=0

∞ ∞ ∞

Then, if all the three sums aj , j=0 bj and j=0 cj converge (not nec-

j=0

essarily absolutely), we have

∞ ∞ ∞

aj bj = cj .

j=0 j=0 j=0

(v) By choosing aj = bj = (’1)j j ’1/2 , or otherwise, show that, if cn is

de¬ned as in (iv), the two sums ∞ aj and ∞ bj may converge and yet

j=0 j=0

∞

j=0 cj diverge.

Exercise K.74. [5.3, P] This exercise is fairly easy but is included to show

that the simple picture painted in Theorem 4.6.19 of a ˜disc of convergence™

for power series fails in more general contexts. More speci¬cally we shall

deal with the absolute convergence of ∞ ∞

with cn,m ∈ R

nm

m=0 cn,m x y

n=0

[n, m ≥ 0] and x, y ∈ R. (We shall deal with absolute convergence only

470 A COMPANION TO ANALYSIS

because as we saw in Section 5.3 this means that the order in which we sum

terms does not matter.)

(a) Show that, if there exists a δ > 0 such that ∞ ∞ nm

m=0 cn,m x y

n=0

converges absolutely for |x|, |y| ¤ δ then there exists a ρ > 0 and an K > 0

such that

|cn,m | ¤ Kρn+m for all n, m ≥ 0.

(b) Identify the set

∞ ∞

E = {x, y ∈ R :2

|cn,m xn y m | converges}

n=0 m=0

in the following cases.

(i) cn,m = 1 for all n, m ≥ 0.

(ii) cn,m = n+m for all n, m ≥ 0.

n

(iii) c2n,2m = n+m for all n, m ≥ 0 and cn,m = 0 otherwise.

n

(iv) cn,m = (n!m!)’1 for all n, m ≥ 0.

(v) cn,m = n!m! for all n, m ≥ 0.

(vi) cn,n = 1 for all n ≥ 0. and cn,m = 0 otherwise.

(c) Let E be as in (i). Show that if (x0 , y0 ) ∈ E then (x, y) ∈ E whenever

|x| ¤ |x0 | and |y| ¤ |y0 |. Conclude that E is the union of rectangles of the

form [’u, u] — [’v, v].

Exercise K.75. [5.3, T!] Part (ii) of this fairly easy question proves a ver-

sion the so called ˜Monotone Convergence Theorem™. Suppose that aj,n ∈ R

for all j, n ≥ 1, that aj,n+1 ≥ aj,n for all j, n ≥ 1, and aj,n ’ aj as n ’ ∞

for all j ≥ 1 .

(i) Show that, given any > 0 and any M ≥ 1, we can ¬nd an N ( , M )

such that

∞ M

aj,n ≥ aj ’

j=1 j=1

for all n ≥ N ( , M ). Show also, that, if ∞ aj converges, we have ∞ aj ≥

j=1 j=1

∞

aj,n for all n.

j=1

∞

(ii) Deduce that, under the hypotheses of this exercise, j=1 aj,n con-

∞ ∞

aj,n ’ A as n ’ ∞ if and only if

verges for each n and j=1 aj

j=1

converges with value A.

Exercise K.76. [5.3, T!, ‘ ] In this question we work in C. If ∞ an z n

n=0

∞ n

converges to f (z), say, for all |z| < δ and n=0 bn z converges to g(z), say,

471

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

for all |z| < δ [δ, δ > 0] it is natural to ask if f (g(z)) = f —¦ g(z) can be

written as a power series in z for some values of z.

(i) Show that formal manipulation suggests that

∞ ∞

?

cn z n with cn =

f —¦ g(z) = ar bm(1) bm(2) . . . bm(r) .

n=0 r=0 m(1)+m(2)+···+m(r)=n

The rest of this question is devoted to showing that this equality is indeed

true provided that

∞

|bn z n | < δ

n=0

where δ is the radius of convergence of ∞ an z n . The exercise is only worth

n=0

doing if you treat it as an exercise in rigour. Since the case z = 0 is fairly

trivial we shall assume that z = 0.

(ii) Let us de¬ne cN r and CN r by equating coe¬cients of powers of w in

the equations

n n

∞ ∞

N N N N

cN r w r = bm w m CN r w r = |bm |wm

|an |

an and .

r=0 n=0 m=0 r=0 n=0 m=0

(Note that cN r = CN r = 0 if r is large.)

Show that CN q |z|r ¤ ∞ |an |κq where κ = ∞ |bn z n |. By using the

n=0 n=0

fact that an increasing sequence bounded above converges show that CN q ’

Cq for some Cr as N ’ ∞. Use the monotone convergence theorem (Ex-

ercise K.75) to show that ∞ |an | m(1)+m(2)+···+m(r)=n |bm(1) bm(2) . . . bm(r) |

n=0

converges to Cn . Deduce that ∞ ar m(1)+m(2)+···+m(r)=n bm(1) bm(2) . . . bm(r)

r=0

converges and now use the dominated convergence theorem (Lemma 5.3.3)

to show that

∞

cN n ’ ar bm(1) bm(2) . . . bm(r) .

r=0 m(1)+m(2)+···+m(r)=n

(iii) By further applications of the monotone and dominated convergence

theorems to both sides of the equations

n n

∞ ∞

N N N N

CN r |z|r = |bm z|m cN r z r = bm z m

|an | and an

r=0 n=0 m=0 r=0 n=0 m=0

472 A COMPANION TO ANALYSIS

show that the two sides of the following equation converge and are equal.

n

∞ ∞

N

m

cn z n

an bm w =

n=0 m=0 n=0

where cn = ∞ br m(1)+m(2)+···+m(r)=n bm(1) bm(2) . . . bm(r) .

r=0

(iv) Let us say that a function G : C ’ C is ˜locally expandable in power

series at z0 ™ if we can ¬nd an · > 0 and Aj ∈ C such that ∞ Aj z j has

j=0

∞

radius of convergence at least · and G(z) = j=0 Aj (z ’ z0 ) for all z ∈ C

j

with |z ’ z0 | < ·. Show that if G : C ’ C is locally expandable in power

series at z0 and F : C ’ C is locally expandable in power series at G(z0 ) then

F —¦ G is locally expandable in power series at z0 . [In more advanced work

this result is usually proved by a much more indirect route which reduces it

to a trivial consequence of other theorems.]

Exercise K.77. (The Wallis formula.) [5.4, P] Set

π/2

sinn x dx.

In =

0

n

In’1 for n ≥ 0.

(i) Show that In+1 =

n+1

(ii) Show that In+1 ¤ In ¤ In’1 and deduce that In+1 /In ’ 1 as n ’ ∞.

(iii) By computing I0 and I1 directly, ¬nd I2n+1 and I2n for all n ≥ 0.

(iv) By applying (ii) and (iii), show that

n

4k 2 π

’ as n ’ ∞.

4k 2 ’ 1 2

k=1

Exercise K.78. (In¬nite products.) [5.4, T]

(i) Prove that, if x ≥ 0, then exp x ≥ 1 + x.

(ii) Suppose a1 , a2 , . . . , an ∈ C. Show that

n n

(1 + aj ) ¤ (1 + |aj |)

j=1 j=1

and

n n

(1 + aj ) ’ 1 ¤ (1 + |aj |) ’ 1.

j=1 j=1

473

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

(iii) Suppose a1 , a2 , . . . , an , . . . ∈ C and 1 ¤ n ¤ m. Show that

n m n m

(1 + aj ) ’ (1 + aj ) ¤ exp |aj | |aj | ’1 .

exp

j=1 j=1 j=1 j=n+1

(iv) Show that, if ∞ aj converges absolutely, then n

j=1 (1 + aj ) tends

j=1

to a limit A, say. We write

∞

A= (1 + aj ).

j=1

(v) Suppose that ∞ aj converges absolutely, and, in addition that an =

j=1

’1 for all n. By considering bn such that

(1 + bn )(1 + an ) = 1,

or otherwise, show that ∞ (1 + aj ) = 0.

j=1

(vi) Suppose aj > 0 and ∞ aj converges. If we write Rn = ∞ j=n+1 aj ,

j=1

∞

show, by considering an appropriate in¬nite product, that j=1 aj /Rj di-

verges.

Exercise K.79. [5.4, T, ‘ ] (i) Suppose that ± > 0. By expanding into

geometric series and multiplying those series show that

(1 ’ 2’± )’1 (1 ’ 3’± )’1 = n’± ,

n∈S(2,3)

where S(2, 3) is the set of strictly positive integers with only 2 and 3 as factors

and we sum over the set S(2, 3) in order of increasing size of the elements.

(ii) Find a similar expression for N (1 ’ p± )’1 where p1 , p2 , . . . , pN

j

j=1

are the ¬rst N primes. If ± > 1 use dominated convergence (Lemma 5.3.3)

to show that

∞

1 1

=

1 ’ p’± n±

n=1

p∈P

where P is the set of primes and we sum over P in order of increasing size

of the elements.

(iii) By considering carefully what happens as ± ’ 1 from above, show

that

1

’∞

1 ’ p’1

p∈P, p¤N

474 A COMPANION TO ANALYSIS

as N ’ ∞. Deduce that

1

diverges.

p

p∈P

(iv) Part (iii) and its proof are due to Euler. It shows that, not merely are

there an in¬nity of primes, but that, in some sense, they are quite common.

Show that if M (n) is the number of primes between 2n and 2n+1 then the

sequence 2’βn M (n) is unbounded whenever β < 1.

Exercise K.80. [5.4, P, ‘] Suppose that an is real and 0 ¤ an < 1 for all

n ≥ 1. Suppose further that an ’ 0 as n ’ ∞. Let the numbers Pn be

de¬ned by taking P0 = 1 and

if Pn’1 ¤ 1,

(1 + an )Pn’1

Pn =

(1 ’ an )Pn’1 if Pn’1 > 1,

for all n ≥ 1. Show that Pn tends to a limit l, say.

What, if anything, can you say about l if ∞ an diverges? What, if

n=1

anything, can you say about l in general? Prove your answers.

n j

(1 + z 2 )

Exercise K.81. [5.4, P] By considering the partial products j=1

and using the dominated convergence theorem, show that

∞ ∞

2j

z k = (1 ’ z)’1

(1 + z ) =

j=1 k=0

for all z ∈ C with |z| < 1.

Exercise K.82. (Vieta™s formula.) [5.5, T] (See Exercise K.78 for the

de¬nition of an in¬nite product if you really need it.) Show that if x ∈ R

and x = 0 then

n

’n

cos(2’j x) = 2’n sin x.

sin(2 x)

j=0

Deduce that

∞

sin x

cos(2’j x) = .

x

j=0

Does the result hold if x ∈ C and x = 0?

475

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

By setting x = π/2 obtain Vieta™s formula (probably the ¬rst published

in¬nite product)

2 1 11 1 11 11 1

= + + + ...

π 2 22 2 22 22 2

∞

2

un where u2 = (1 + un )/2

or (in a more reasonable notation) = n+1

n=1

π

and u1 = 2’1/2 .

Exercise K.83. [5.6, P, G] We say that a function f : A ’ B is injective

if f (a) = f (a ) implies a = a . We say that a function f : A ’ B is surjective

if, given b ∈ B, we can ¬nd an a ∈ A such that f (a) = b. If f is both injective

and surjective we say that f is bijective.

(a) Consider the functions f : A ’ B, g : B ’ C and their composition

g —¦ f : A ’ C given by g —¦ f (a) = g(f (a)). Prove the following results.

(i) If f and g are surjective, then so is g —¦ f .

(ii) If f and g are injective, then so is g —¦ f .

(iii) If g —¦ f is injective, then so is f .

(iv) If g —¦ f is surjective, then so is g.

(b) Give an example where g —¦ f is injective and surjective but f is not

surjective and g is not injective.

(c) If any of your proofs of parts (i) to (iv) of (a) involved contradiction,

reprove them without such arguments5 .

(d) Have you given the simplest possible example in (b)? (If you feel that

this is not a proper question, let us ask instead for the smallest possible sets

A and B.)

Exercise K.84. [5.6, P, G] Show that an injective function f : [0, 1] ’

[0, 1] is continuous if and only if it is strictly monotonic (that is strictly

increasing or strictly decreasing). Is the result true if we replace ˜injective™

by ˜surjective™ ? Give a proof or a counterexample.

Exercise K.85. (Multiplication before Napier.) [5.6, T, G]

(i) The ancient Greek astronomers drew up tables of chords where

chord ± = 2 sin(±/2).

Prove that

chord θ chord(π ’ φ) = chord(θ + φ) + chord(θ ’ φ).

5

Conway refers to arguments of the form ˜Assume A is true but B is false. Since A

implies B it follows that B is true. This contradicts our original assumption. Thus A

implies B.™ as ˜absurd absurdums™.

476 A COMPANION TO ANALYSIS

If 0 < x < 2 and 0 < y < 2, show that the following procedure computes xy

using only addition, subtraction and table look up.

(a) Find θ and φ so that x = chord θ, y = chord(π ’ φ).

(b) Compute θ + φ and θ ’ φ.

(c) Find chord(θ + φ) and chord(θ ’ φ).

(d) Compute chord(θ + φ) + chord(θ ’ φ).

(ii) In fact we can multiply positive numbers using only addition, sub-

traction, division by 4 and a table of squares. Why? (Do think about this

for a little before looking at Exercise 1.2.6 (iii) for a solution.)

If we just want to multiply two numbers together, the methods of (i)

and (ii) are not much longer than using logarithms but, if we want to multi-

ply several numbers together, logarithms are much more convenient. (Check

this statement by considering how you would multiply 10 numbers together

by the various methods.) Logarithms also provide an easy method of ¬nd-

ing ±th roots. (Describe it.) We may say that logarithms are more useful

computational tools because they rely on isomorphism rather than clever

tricks.

[This question was suggested by the beautiful discussion in [39]. If the reader

wants to know more about the historical context of logarithms, Phillips™ book

is the place to start.]

Exercise K.86. [5.6, P, S] Arrange the functions

1/2 3 1/2

1 1 1 1

, log , exp log , exp

x x x x

as f1 (x), f2 (x), f3 (x), f4 (x) in such a way that fr (x)/fr+1 (x) ’ 0 as x ’ 0

through positive values of x. Justify your answer.

Exercise K.87. [5.6, P] We shall give an number of alternative treatments

of the logarithm. This is one I like lees than some of the others.

Let x0 = x > 0. The sequence of real strictly positive numbers xn is

de¬ned by the recurrence relation

x2 = x n .

n+1

Prove that the two expressions 2n (xn ’ 1) and 2n (1 ’ 1/xn ) both tend to the

same limit as n ’ ∞. Taking this limit as the de¬nition of log x, prove that,

for real, strictly positive x and y,

(i) log 1 = 0,

(ii) log xy = log x + log y,

(iii) log 1/x = ’ log x,

(iv) log x increases with x.

477

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

Exercise K.88. [5.7, P, S] Let f : R ’ R be a strictly positive function.

1/·

f (x + ·)

tends to a limit l, say, as · ’ 0 if and only if f is

Show that

f (x)

di¬erentiable at x. If l exists, give its value in terms of f (x) and f (x).

Exercise K.89. [5.7, S] (This is really just a remark.) I have been unable

to trace the story told in Exercise 5.7.10 to a sure source. However, one of

Pierce™s papers6 does refer to the ˜mysterious formula™

√

i’i = ( e)2π

(which he makes still more mysterious by using a notation of his own inven-

tion). By remarking that eiπ/2 = e’3iπ/2 , show that, even if we interpret this

equation formally, the left hand side of his equation is ambiguous. How did

we prevent this ambiguity in part (ii) of Exercise 5.7.10?

Exercise K.90. (Functional equations.) [5.7, T]

(i) Suppose f : R ’ R is a function such that

f (x + y) = f (x) + f (y)

for all x, y ∈ R. Let f (1) = a.

(a) Find f (n) for n a strictly positive integer.

(b) Find f (n) for n an integer.

(c) Find f (1/n) for n a non-zero integer.

(d) Find f (x) for x a rational.

Now suppose that f is continuous. Show that f (x) = ax for all x ∈ R.

(ii) Suppose ± : Rn ’ Rm is a continuous function such that

±(x + y) = ±(x) + ±(y)

for all x, y ∈ Rn . Show that ± is a linear map.

(iii) Suppose g : (0, ∞) ’ (0, ∞) is a continuous function such that

g(xy) = g(x)g(y)

for all x, y ∈ (0, ∞). Find the general form of g.

(iv) Suppose g : R \ {0} ’ R \ {0} is a continuous function such that

g(xy) = g(x)g(y)

for all x, y ∈ R \ {0}. Find the general form of g.

6

Linear Associative Algebra, Vol 4 of the American Journal of Mathematics, see

page 101.

478 A COMPANION TO ANALYSIS

Exercise K.91. [5.7, T, ‘ ] In this question we seek to characterise the

continuous homomorphisms χ from the real numbers R under addition to

the group S 1 = {» ∈ C : |»| = 1} under multiplication. In other words we

want to ¬nd all continuous functions χ : R ’ S 1 such that

χ(x + y) = χ(x)χ(y)

for all x, y ∈ R.

(i) For each t ∈ T let θ(t) be the unique solution of

χ(t) = exp iθ(t)

with ’π < θ(t) ¤ π. Explain why we cannot establish the equation

?

θ(t) = 2θ(t/2)

for all t but we can ¬nd a δ > 0 such that

θ(t) = 2θ(t/2)

for all |t| < δ.

(ii) If θ(δ/2) = γ and » = 2δ ’1 γ establish carefully that χ(t) = exp i»t

for all t. Conclude that the continuous homomorphisms χ from R to S 1 are

precisely the functions of the form χ(t) = exp i»t for some real ».

(iii) Find the continuous homomorphisms from the group S 1 to itself.

Exercise K.92. [5.7, T, ‘ ] (i) Suppose f : R ’ R is a function such that

f (2t) = 2f (t)

for all t ∈ R. Show that f (0) = 0.

(ii) Let pj be the jth prime. Show that we can ¬nd a function F : R ’ R

such that

F (2t) = 2F (t)

for all t ∈ R and F (p’1 ) = j. Show that F is not continuous at 0.

j

(iii) Suppose f : R ’ R is a function such that

f (2t) = 2f (t)

for all t ∈ R and, in addition, f is di¬erentiable at 0. Show that f (t) = At

for all t ∈ R and some A ∈ R.

479

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

(iv) Show that we can ¬nd a function F : R ’ R such that

F (2t) = 2F (t)

for all t ∈ R and F is continuous at 0 but not di¬erentiable there.

(v) Suppose u : R ’ R is a function such that

u(2t) = u(t)2

for all t ∈ R and, in addition, u is di¬erentiable at 0. Show that either

u(t) = 0 for all t ∈ R or there exists an ± > 0 such that u(t) = ± t for all

t ∈ R.

Exercise K.93. [6.1, P] We work in R3 with the usual inner product. Con-

sider the function f : R3 ’ R3 given by

x

f (x) = for x = 0

x

and f (0) = 0. Show that f is di¬erentiable except at 0 and

h x

’ x, h

Df (x)h = 3

x x

Verify that Df (x)h is orthogonal to x and explain geometrically why this is

the case.

Exercise K.94. (The Cauchy-Riemann equations.) [6.1, T] (This ex-

ercise is really the ¬rst theorem in a course on complex variable but it can

do the reader no harm to think about it in advance.)

We say that a function f : C ’ C is complex di¬erentiable at z0 if there

exists a f (z0 ) ∈ C such that

f (z0 + h) ’ f (z0 )

’ f (z0 ) as |h| ’ 0.

h

Observe that C can be considered as the vector space R2 and so we we can

write

f (x + iy) = u(x, y) + iv(x, y)

with x, y, u and v real, obtaining a function F : R2 ’ R2 given by

x u(x, y)

F = .

y v(x, y)

480 A COMPANION TO ANALYSIS

Show that the following statements are equivalent

(i) f is complex di¬erentiable at z0 .

x0

(ii) F is di¬erentiable at and its Jacobian matrix at this point is

y0

given by

‚u ‚u

cos θ ’ sin θ

‚x ‚y

=»

‚v ‚v

sin θ cos θ

‚x ‚y

with » real and » ≥ 0.

x0

(iii) F is di¬erentiable at and its derivative at this point is the

y0

composition of a dilation and a rotation.

x0

(iv) F is di¬erentiable at and its partial derivatives at this point

y0

satisfy the Cauchy-Riemann conditions

‚u ‚v ‚v ‚u

=’ .

= ,

‚x ‚y ‚x ‚y

Exercise K.95. [6.2, P] Let f : R2 ’ R be di¬erentiable and let g(x) =

f (x, c ’ x) where c constant. Show that g : R ’ R is di¬erentiable and ¬nd

its derivative

(i) directly from the de¬nition of di¬erentiability,

and also

(ii) by using the chain rule.

Deduce that if f,1 = f,2 throughout R then f (x, y) = h(x + y) for some

di¬erentiable function h.

Exercise K.96. [6.2, P] Consider a function f : R ’ R. State and prove

a necessary and su¬cient condition for there to be a continuous function

F : R2 ’ R with

f (x) ’ f (y)

F (x, y) =

x’y

whenever x = y.

Prove that, if f is twice di¬erentiable, then F is everywhere di¬erentiable.

Exercise K.97. [6.2, T] (The results of this question are due to Euler.) Let

± be a real number. We say that a function f : Rm \{0} ’ R is homogeneous

of degree ± if

f (»x) = »± f (x) †

481

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

for all » > 0 and all x = 0.

(i) By di¬erentiating both sides of † and choosing a particular value of

», show that, if f : Rm \ {0} ’ R is a di¬erentiable function which is

homogeneous of degree ±, then

m

††

xj f,j (x) = ±f (x)

j=1

for all x = 0.

(ii) Suppose conversely, that f : Rm \ {0} ’ R is a di¬erentiable function

satisfying ††. By setting up a di¬erential equation for the function v de¬ned

by v(») = f (»x), or otherwise, show that f is homogeneous of degree ±.

Exercise K.98. [6.2, T] (i) Suppose that ± : R2 ’ R2 is a linear map and

that its matrix with respect to the standard basis is

ab

.

cd

Show how ± may be calculated. (There is no particular point in carrying

through the algebra in detail.)

(ii) Suppose that ± : Rm ’ Rp is a linear map and that its matrix with

respect to the standard basis is A = (aij ). Show how ± might be calculated.

(If you know about Lagrange multipliers this gives you an opportunity to

apply them.)

We discuss this problem further in Exercises K.99 to K.101.

Exercise K.99. [6.2, T] A linear map ± : Rm ’ Rm is called symmetric if

±(x), y = x, ±(y)

for all x, y ∈ Rm . Show that ± : Rm ’ Rm is symmetric if and only if its

matrix (aij ) with respect to the standard bases is symmetric, that is aij = aji

for all i and j.

In algebra courses (and, indeed, in Exercise K.30) it is shown that the

linear map ± : Rm ’ Rm is symmetric if and only if it has m orthonormal

eigenvectors. Show that a symmetric linear map ± : Rm ’ Rm has operator

norm equal to the largest absolute value of its eigenvalues, that is to say,

± = max{|»| : » an eigenvalue of ±}. (†)

By considering the linear map ± : R2 ’ R2 with matrix

01

A= ,

00

or otherwise, show that the equality (†) may fail if ± is not symmetric.

482 A COMPANION TO ANALYSIS

Exercise K.100. [6.2, T, ‘ ] Question K.99 tells us that the operator norm

of a symmetric linear map ± : Rm ’ Rm is the largest absolute value of its

eigenvalues but does not tell us how to ¬nd this value.

(i) Our ¬rst thought might be to look at the roots of the characteris-

tic polynomial. Examine the kind of calculations involved when m = 50.

(Actually matters are even worse than they look at ¬rst sight.)

(ii) Suppose ± has eigenvalues »j with associated orthonormal eigenvec-

tors ej . Choose an x0 ∈ V and de¬ne yk = xk ’1 xk (if xk = 0 the process

terminates) and xk+1 = ±(yk ). If »1 > |»j | [2 ¤ j ¤ m] show that, except in

a special case to be speci¬ed, the process does not terminate and

x k ’ » 1 , y k ’ e1 ’ 0

as k ’ ∞.

How must your answer be modi¬ed if ’»1 > |»j | [2 ¤ j ¤ m].

(iii) Now suppose simply that ± = 0 and |»1 | ≥ |»j | [2 ¤ j ¤ m]. Show

that it remains true that, except in a special case to be speci¬ed, the process

does not terminate and

xk ’ |»1 |.

as k ’ ∞.

If ± = 0, all the »j are positive and »1 ≥ »j [2 ¤ j ¤ m] show that,

except in a special case to be speci¬ed, the process does not terminate and

xk tends to a limit. Show that this limit is an eigenvector but (unless »1 > »j

[2 ¤ j ¤ m]) this eigenvector need not be a multiple of e1 .

(iv) We thus have an algorithm for computing the largest absolute value

of the eigenvalues of ± by applying the iterative procedure described in (ii).

The reader may raise various objections.

(a) ˜The randomly chosen x0 ∈ V may be such that the process fails.™

Suppose that V = R50 and you are use your favourite computer and your

favourite random number generator. Estimate the probability that a ran-

domly chosen x0 will be such that the process fails. Recalling that computers

use ¬nite precision arithmetic discuss the e¬ect of errors on the process.

(b) ˜We give an iterative procedure rather than an exact formula.™ Re-

member that if V = R50 we are seeking a root of a polynomial of degree 50.

How do you propose to ¬nd an exact formula for such an object?

(c) ˜We may have to make many iterations.™ Estimate the number of

operations required for each iteration if the matrix A associated with ± is

given and conclude that, unless m is very large, each iteration will be so fast

that the number of iterations hardly matters.

483

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

(v) The method is (usually) quite e¬ective for ¬nding the largest (abso-

lute value) of the eigenvalues. Sometimes it is less e¬ective in ¬nding the

associated eigenvector. Why should this be?

(vi) Discuss how you might go about ¬nding the second largest (absolute

value) of the eigenvalues.

(vi) Suppose that ± is a linear map ± : Cm ’ Cm whose eigenvalues

satisfy |»1 | > |»j | [2 ¤ j ¤ m]. Show that the method outlined above will

continue to work.

(vii) Let e1 and e2 be linearly independent vectors in R2 with e1 ’ e2

small. Let ± : R2 ’ R2 be the linear map such that ±e1 = e1 , ±e2 = ’e2

Set

’1

x = e 1 ’ e2 (e1 ’ e2 ).

and examine the the behaviour of ±r x.

(viii) Explain why our algorithm may fail for an ± : Cm ’ Cm whose

eigenvalues merely satisfy |»1 | ≥ |»j | [2 ¤ j ¤ m].

(vii) Explain why the method of (vii) may prove very slow even if |»1 | >

|»j | [2 ¤ j ¤ m]. Why did we not have the problem in the real symmetric

case?

Exercise K.101. [6.2, T, ‘ ] We now consider general linear maps ± :

Rm ’ Rm . In this general case mathematicians remain very interested in

¬nding the largest absolute value of its eigenvalues. (Some of the reasons

for this are set out in Exercises K.282 to K.286.) However, they are much

less interested in ¬nding ± which, as I have stressed, is mainly used as a

convenient theoretical tool.

If you really need to obtain ± , this exercise justi¬es one method of

going about it. Recall that ±— is the linear map ±— : Rm ’ Rm whose matrix

A— = (bij ) with respect to the standard bases is given by bij = aji for all i

and j (where A = (aij ) is the matrix of ± with respect to the same basis).

(i) Show that

±x, y = x, ±— y

for all x, y ∈ Rm .

(ii) Explain why

x, ±— ±x = ±x 2

for all x ∈ Rm . Deduce that

±— ± ≥ ± 2

484 A COMPANION TO ANALYSIS

and so ±— ≥ ± .

(iv) Show that, in fact, ±— = ± and

±— ± = ± 2 .

Show that ±— ± is a symmetric linear map all of whose eigenvalues are positive.

(v) Use the results of (ii) and Exercise K.100 to give a method of obtaining

±.

(vi) Estimate the number of operations required if the matrix A associated

with ± is given. What step requires the bulk of the calculations?

(vii) Find A— A in the special case

11

A= .

02

What are the eigenvalues of A and AA— ? What is A ?

Exercise K.102. [6.3, P] We work in R2 . Let

¯

B = {x : x < 1}, B = {x : x ¤ 1} and ‚B = {x : x = 1}.

¯

Suppose that f : B ’ R is a continuous function which is di¬erentiable on B.

Show that, if f (x) = 0 for all x ∈ ‚B, there exists a c ∈ B with Df (c) = 0.

¯

Construct a continuous function g : B ’ R which is di¬erentiable on B

but such that we cannot ¬nd a linear map ± : R2 ’ R such that f ’ ± is

constant on ‚B. [Hint: This is easy.]

Discuss the foregoing results in the light of Rolle™s theorem (Theorem 4.4.4)

and the proof of the one dimensional mean value theorem given in Exer-

cise 4.4.5.

Exercise K.103. [7.1, P, S] Suppose that q1 , q2 , . . . is a strictly increasing

∞ 1

sequence of strictly positive integers such that n=1 qn diverges. Show that

∞ ∞

1 1 1

log 1 ’ log 1 ’

diverges but + converges.

qn qn qn

n=1 n=1

Exercise K.104. [7.3, P] (This question has low theoretical interest but

tests your power of clear organisation.) Find the maximum of

ax2 + bx + c

on the interval [0, 10] where a, b and c are real constants.

485

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

Exercise K.105. (Routh™s rule.) [7.3, T] It is Routh™s misfortune to be

known chie¬‚y as the man who beat Maxwell in the Cambridge mathematics

examinations, but he was an able mathematician and teacher7 . Part (v) of

this question gives his method for determining if a real symmetric matrix

A = (aij )1¤i,j¤n is positive de¬nite.

(i) We said that A is positive de¬nite if all its eigenvalues are strictly posi-

tive. It is more usual to say that A is positive de¬nite if n n

j=1 xi aij xj > 0

i=1

for all x = 0. Show that the two de¬nitions are equivalent. [Think about

the map x ’ xT Ax (with x as a column vector).]

(ii) By choosing a particular x, show that, if A is positive de¬nite, then

a11 > 0.

(iii) If a11 = 0, show that (whether A is positive de¬nite or not)

n n n n

2

xi aij xj = a11 y1 + xi bij xj

i=1 j=1 i=2 j=2

where y1 = x1 + a’1 a12 x2 + a’1 a13 x3 + · · · + a’1 a1n xn , and

11 11 11

aij a11 ’ a1i a1j

a1i a1j

bij = aij ’ = .

a11 a11

Hence, or otherwise, show carefully that A is positive de¬nite if and only if

a11 > 0 and the matrix B = (bij )2¤i,j¤n is positive de¬nite. (Of course, we

must perform the very simple veri¬cation that B is a real symmetric matrix.)

The matrix B is called the Schur complement of a11 in A.

(iv) By considering the e¬ect of row operations of the form

[new ith row] = [old ith row] ’ ai1 a’1 [old 1st row]

11

on A, or otherwise, show that det A = a11 det B.

(v) Use (iii), ideas based on (iv) and induction on n to prove Routh™s

rule:“ The real symmetric matrix A is positive de¬nite if and only if the

determinants of the n leading minors

«

a11 a12 a13

a11 a12

, det a21 a22 a23 , . . . , det A

a11 = det(a11 ), det

a21 a22

a31 a32 a33

are all strictly positive.

(vi) Show that A is negative de¬nite if and only if ’A is positive de¬nite.

7

He was one of the two greatest Cambridge coaches (people who tried to make students

do slightly better in mathematics examinations than they should, an occupation which I

have always felt to be fairly harmless, if not terribly useful).

486 A COMPANION TO ANALYSIS

Figure K.2: Shortest road system for vertices of a square

Exercise K.106. [7.3, P] (i) Let A(t) be a 2 — 2 real symmetric matrix

with

a(t) b(t)

A(t) =

b(t) c(t)

Suppose that a, b, c : R ’ R are continuous. Show that, if A(t) has

eigenvalues »1 (t) and »2 (t) with »1 (t) ≥ »2 (t), then »1 and »2 are continuous.

Show that, if A(0) is positive de¬nite and A(1) is negative de¬nite, then there

exists an ± ∈ (0, 1) with A(±) singular. Show, more strongly, that, either

we can ¬nd ±1 and ±2 with 1 > ±1 > ±2 > 0 and A(±1 ), A(±2 ) singular, or

there exists an ± ∈ (0, 1) with A(±) the zero matrix.

(ii) State and prove a result along the lines of Exercise 7.3.19 (ii).

(iii) Find a 2 — 2 real matrix B(t) = (bij (t))1¤i,j¤2 such that the entries

bij : R ’ R are continuous, B(0) = I and B(π) = ’I but B(t) is nowhere

singular. [Think rotations.]

Exercise K.107. [7.3, M!] Four towns lie on the vertices A, B, C, D of

a square. Find the shortest total length of the system of roads shown in

Figure K.2 where the diagram is symmetric about lines through the centres

of opposite sides of the square. By formal or informal arguments, show that

this arrangement gives the shortest total length of a system of roads joining

all four towns. (You should, at least, convince yourself of this.) Is the

arrangement unique?

The interesting point here is that the most highly symmetric road systems

do not give the best answer.

Exercise K.108. [7.3, P] We continue with the notation of Exercise 7.3.12.

Show that, if g is continuously di¬erentiable, then f is di¬erentiable except

at (0, 0) (there are various ways of doing this but, whichever one you choose,

be careful), and has directional derivatives in all directions at (0, 0). For

which functions g is f di¬erentiable at (0, 0)?

487

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

Exercise K.109. [7.3, P] (i) Let a > b > 0 and de¬ne f : R2 ’ R by

f (x, y) = (y ’ ax2 )(y ’ bx2 ).

Sketch the set E = {(x, y) : f (x, y) > 0}. Show that f has a minimum

at (0, 0) along each straight line through the origin. (Formally, if ± and β

are real numbers which are not both zero, the function g : R ’ R given by

g(t) = f (±t, βt) has a strict local minimum at 0.) Show, however, that f has

no minimum at (0, 0). Does f have any minima?

(ii) Suppose F : R2 ’ R is twice continuously di¬erentiable with a non-

singular Hessian at (0, 0). Show that F has a minimum at (0, 0) if and only if

it has a minimum at (0, 0) along each straight line through the origin. Why

is this result consistent with part (i)?

[Like many examples in this subject, part (i) is due to Peano.]

Exercise K.110. [8.2, P] (i) Let F be the function given in Exercise 8.2.23.

By considering F ’ 1 , or otherwise, show that if a bounded function f :

2

[a, b] ’ R is not Riemann integrable, it does not follow that |f | is not

Riemann integrable. If a bounded function f : [a, b] ’ R is such that |f |

is not Riemann integrable, does it follow that f is not Riemann integrable?

Give a proof or counterexample.

(ii) Let f, g : [a, b] ’ R be bounded functions. If both f and g are not

Riemann integrable, does it follow that f + g is not Riemann integrable? If

both f and g are not Riemann integrable, does it follow that f +g is Riemann

integrable? If f is Riemann integrable but g is not Riemann integrable, does

it follow that f +g is not Riemann integrable? If f is Riemann integrable but

g is not Riemann integrable, does it follow that f + g is Riemann integrable?

(Give proofs or counterexamples.)

(iii) Compose and answer similar questions for the product f g and for »f

where » is a non-zero real number.

Exercise K.111. [8.2, P] De¬ne f : [0, 1] ’ R by f (p/q) = 1/q when p

and q are coprime integers with 1 ¤ p < q and f (x) = 0 otherwise.

1

(i) Show that f is Riemann integrable and ¬nd 0 f (x) dx.

(ii) At which points, if any, is f continuous? Prove your answer.

(iii) At which points, if any, is f di¬erentiable? Prove your answer.

Use the ideas of this exercise to de¬ne a function g : R ’ R which is

unbounded on every interval [a, b] with b > a.

Exercise K.112. [8.2, P] Suppose that f : [0, 1] ’ R is Riemann inte-

grable. Show that, if we write gn (x) = f (xn ), then gn : [0, 1] ’ R is Riemann

488 A COMPANION TO ANALYSIS

integrable. State, with appropriate proof, which of the following conditions

imply that

1

f (xn ) dx ’ f (0)

0

as n ’ ∞.

(i) f is continuous on [0, 1].

(ii) f is continuous at 0.

(iii) No further condition on f .

Exercise K.113. [8.2, H] (i) Let

D = {x0 , x1 , . . . , xn } with a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xn = b

be a dissection of [a, b] and let > 0. Show that there exists a δ > 0

depending on and D such that, if

D = {y0 , y1 , . . . , ym } with a = y0 ¤ y1 ¤ y2 ¤ · · · ¤ ym = b

is a dissection of [a, b] with |yj ’ yj’1 | < δ for all j, then the total length

of those intervals [yk’1 , yk ] not completely contained in some [xi’1 , xi ] is less

than . More formally,

n

|yk ’ yk’1 | < .

i=1 [yk’1 ,yk ] [xi’1 ,xi ]=…

(ii) Show that, if f : [a, b] ’ R is a function with |f (t)| ¤ M for all t, D

is a dissection of [a, b] and > 0, we can ¬nd an · > 0 (depending on M ,

and D) such that if

D = {y0 , y1 , . . . , ym } with a = y0 ¤ y1 ¤ y2 ¤ · · · ¤ ym = b

is a dissection of [a, b] with |yj ’yj’1 | < · for all j, then S(f, D ) < S(f, D)+

and S(f, D) > s(f, D) ’ .

(iii) Show that a bounded function f : [a, b] ’ R is Riemann integrable

with integral I if and only if, given any > 0 we can ¬nd an · > 0 such that,

if

D = {x0 , x1 , . . . , xn } with a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xn = b

is a dissection of [a, b] with |xj ’ xj’1 | < · for all j, then

S(f, D) ’ s(f, D) < , and |S(f, D) ’ I| ¤ .

489

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(iv) Show that a bounded function f : [a, b] ’ R is Riemann integrable

with integral I if and only if, given any > 0, we can ¬nd a positive integer

N such that, if n ≥ N and

D = {x0 , x1 , . . . , xn } with xj = a + j(b ’ a)/N,

then

S(f, D) ’ s(f, D) < , and |S(f, D) ’ I| ¤ .

(v) Show that, if f : [a, b] ’ R is Riemann integrable, then

n’1 b

b’a

f (a + j(b ’ a)/n) ’ f (t) dt

n a

j=0

as N ’ ∞. Why is this result compatible with Dirichlet™s counterexample

(Exercise 8.2.23)?

Exercise K.114. [8.2, H] (This exercise is best treated informally.) If we

insist on considering the integral as the area under a curve, then our de¬nition

of the Riemann integral of a function which can be negative looks a bit odd.

We could restrict ourselves initially to positive bounded functions and then

extend to general functions in (at least) two ways.

(A) If f : [a, b] ’ R is bounded, we can write f (t) = f+ (t) ’ f’ (t) with

f+ and f’ positive. We set

b b b

f+ (t) dt ’

f (t) dt = f’ (t) dt

a a a

if the right hand side of the equation exists.

(B) If f : [a, b] ’ R is bounded we can write f (t) = fκ (t) ’ κ with fκ

positive and κ a real number. We set

b b

fκ (t) dt ’ κ(b ’ a).

f (t) dt =

a a

Run through the checks that you must make to see that these de¬nitions

work as expected. For example, if we use (B), we must check that the value

of the integral is independent of the value of κ chosen. In both cases we must

check that, if f and g are integrable, so is f g.

Explain why we get the same integrable functions and the same integrals

from the three de¬nitions considered (the one we actually used, (A) and (B)).

490 A COMPANION TO ANALYSIS

Exercise K.115. [8.2, H] No modern mathematician would expect any

reasonable theory of integration on the rationals for the following reason.

Since the rationals are countable we can enumerate the elements of [0, 1] © Q.

Enumerate them as x1 , x2 , . . . . Let Jr be an interval of length 2’r’1 con-

taining xr . We observe that the union of the Jr covers [0, 1] © Q but that the

total length of the intervals is only ∞ 2’r’1 = 1/2.

r=1

(i) Show that, given any > 0, we can ¬nd intervals J1 , J2 , . . . of total

length less than with ∞ Jr ⊇ Q.

r=1

(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