Sketch Solutions for Some

Exercises in Calculus for the Ambitious

T. W. Körner

Introduction

When I was young, I used to be surprised when the answer in the back of the book was wrong. I could not believe that the wise and gifted people who wrote textbooks could possibly make mistakes. I am no longer surprised.

Here are what I believe to be sketch solutions to the bulk of the exercises. By the nature of things, they cannot be guaranteed free of error. I would appreciate the opportunity to remedy problems. Please tell me of any errors, unbridgeable gaps, misnumberings etc. I welcome suggestions for additions.

ALL COMMENTS GRATEFULLY RECEIVED. If you can, please use L^{A}TEX 2ε or its relatives for mathematics. If

not, please use plain text. My e-mail is twk@dpmms.cam.ac.uk. You may safely assume that I am both lazy and stupid, so that a message saying ‘Presumably you have already realised the mistake in Exercise Z’ is less useful than one which says ‘I think you have made a mistake in Exercise Z because you have have assumed that the sum is necessarily larger than the integral. One way round this problem is to assume that f is decreasing.’

It may be easiest to navigate this document by using the table of contents which follow on the next few pages. To avoid disappointment, observe that those exercises marked ⋆ have no solution given.

Contents

Introduction 2 Exercise 1.2.1 8 Exercise 1.2.2 9 Exercise 1.2.3 10 Exercise 1.2.4 11 Exercise 1.2.5 12 Exercise 1.2.7 13 Exercise 1.2.8 14 Exercise 1.2.9 15 Exercise 1.2.10 16 Exercise 1.3.1⋆ 16 Exercise 1.3.2 17 Exercise 1.3.4⋆ 17 Exercise 1.4.1⋆ 17 Exercise 1.4.2 18 Exercise 1.4.4⋆ 18 Exercise 1.4.5 19 Exercise 1.4.7 21 Exercise 1.4.8 22 Exercise 1.4.9 23 Exercise 1.5.1 24 Exercise 1.5.2⋆ 24 Exercise 1.5.3 25 Exercise 2.2.1 26 Exercise 2.2.2⋆ 26 Exercise 2.2.3 27 Exercise 2.2.4 28 Exercise 2.2.5 29 Exercise 2.2.6 30 Exercise 2.2.7 31 Exercise 2.2.8 33 Exercise 2.2.9 34 Exercise 2.2.10 35 Exercise 2.2.11 36 Exercise 2.2.12 39 Exercise 2.3.1 41 Exercise 2.3.2 42 Exercise 2.3.3 43 Exercise 2.3.4 44 Exercise 2.3.5 45 Exercise 2.4.1 46 Exercise 2.4.2 47 Exercise 2.4.3 48 Exercise 2.5.1 49 Exercise 2.5.2 50 Exercise 2.5.3 51 Exercise 2.5.4⋆ 51 Exercise 2.5.5 52 Exercise 2.5.6 53 Exercise 2.5.7 54 Exercise 2.5.8 55 Exercise 2.5.9 56 Exercise 2.5.10 57 Exercise 2.5.11 58 Exercise 2.5.12⋆ 58 Exercise 2.5.13 59 Exercise 2.5.14 60 Exercise 2.5.15 61 Exercise 2.6.1 62 Exercise 2.6.2 63 Exercise 2.6.3 64 Exercise 2.6.4 65 Exercise 2.6.5 66 Exercise 2.6.6 67 Exercise 3.1.1 69 Exercise 3.1.2⋆ 69 Exercise 3.1.3 70 Exercise 3.1.4 71 Exercise 3.1.5 72 Exercise 3.1.6 73 Exercise 3.1.7 74 Exercise 3.2.1 75 Exercise 3.2.2 76 Exercise 3.2.3 77 Exercise 3.2.4 78 Exercise 3.2.5 79 Exercise 3.2.6 80 Exercise 3.2.7 81 Exercise 3.2.8 82 Exercise 3.2.9⋆ 82 Exercise 3.2.10 83 Exercise 3.2.11 84 Exercise 3.2.12 86 Exercise 3.2.13 88 Exercise 3.3.1 90 Exercise 3.3.2 91 Exercise 3.3.3 92 Exercise 3.3.4 93 Exercise 3.3.5 94 Exercise 3.3.6 95 Exercise 3.3.7 96 Exercise 3.3.8 97 Exercise 3.3.9 98 Exercise 4.1.1 99 Exercise 4.1.2 100 Exercise 4.1.3 101 Exercise 4.1.4 102 Exercise 4.1.5 103 Exercise 4.2.1 104 Exercise 4.2.2 105 Exercise 4.3.1 108 Exercise 4.3.2 109 Exercise 4.3.3 110 Exercise 4.3.4 111 Exercise 4.3.5 113 Exercise 5.1.1⋆ 113 Exercise 5.1.2 114 Exercise 5.1.3 115 Exercise 5.1.4 116 Exercise 5.1.5 117 Exercise 5.1.6 118 Exercise 5.1.7 119 Exercise 5.2.1⋆ 119 Exercise 5.2.2 120 Exercise 5.2.3 121 Exercise 5.2.4 122 Exercise 5.3.1 123 Exercise 5.3.2 124 Exercise 5.3.3 125 Exercise 5.3.4 126 Exercise 6.1.1 127 Exercise 6.2.1 128 Exercise 6.2.2 129 Exercise 6.2.3 130 Exercise 6.2.4 131 Exercise 6.2.5 132 Exercise 6.2.6 133 Exercise 6.3.1 134 Exercise 6.3.2 135 Exercise 6.3.3 136 Exercise 6.3.4 137 Exercise 6.3.5 138 Exercise 6.3.6⋆ 138 Exercise 6.3.7⋆ 138 Exercise 6.3.8 139 Exercise 6.3.9⋆ 139 Exercise 6.3.10 140 Exercise 6.3.11 141 Exercise 6.3.12 142 Exercise 6.3.13 143 Exercise 6.3.14 144 Exercise 6.3.15 145 Exercise 7.1.1 146 Exercise 7.1.2 147 Exercise 7.1.3 148 Exercise 7.1.4 149 Exercise 7.1.5 150 Exercise 7.2.1 151 Exercise 7.2.2 152 Exercise 7.2.3 153 Exercise 7.2.4 154 Exercise 7.2.5 155 Exercise 7.2.6 156 Exercise 7.2.7 157 Exercise 7.3.1 158 Exercise 7.3.2 159 Exercise 7.3.3 160 Exercise 7.3.4 161 Exercise 7.3.5 162 Exercise 7.3.6 163 Exercise 8.1.1 164 Exercise 8.1.3 165 Exercise 8.1.5 166 Exercise 8.2.1 167 Exercise 8.2.2 168 Exercise 8.2.3 169 Exercise 8.2.4 170 Exercise 8.2.5 171 Exercise 8.2.6⋆ 171 Exercise 8.2.7 172 Exercise 8.2.8 173 Exercise 8.3.1 174 Exercise 8.3.2⋆ 174 Exercise 9.2.1 175 Exercise 9.2.2 176 Exercise 9.2.3 177 Exercise 9.2.4⋆ 177 Exercise 9.3.1 178 Exercise 9.3.2⋆ 178 Exercise 9.3.3⋆ 178 Exercise 9.3.4 179 Exercise 9.3.5 180 Exercise 9.3.6 181 Exercise 9.3.7 182 Exercise 10.1.1 183 Exercise 10.1.2 184 Exercise 10.1.3 185 Exercise 10.1.4 186 Exercise 10.1.5 187 Exercise 10.1.6 188 Exercise 10.1.7 189 Exercise 10.1.8 190 Exercise 10.2.1 191 Exercise 10.2.2 192 Exercise 10.2.3 193 Exercise 10.2.4 195 Exercise 11.3.2 197 Exercise 11.3.3 198 Exercise 11.3.4 199 Exercise 11.4.1 200 Exercise 11.5.1 201 Exercise 11.5.2 202

Exercise 1.2.1

We have

20 019 × 299 987 = (20 000 + 19) × (300 000 − 13)
= 20 000 × 300 000 + 19 × 300 000 − 13 × 20 000 − 19 × 13

so, to zeroth order,

20 019 × 299 987 ≈ 20 000 × 300 000 = 6 000 000 000

and, to first order,

20 019 × 299 987 = (20 000 + 19)

≈ 20 000 × 300 000 + 19 × 300 000 − 13 × 20, 000 = 6 000 000 000 + 5700 000 − 260 000 = 6 005 440 000.

The exact answer is 20 019 × 299 987 = 6 005 439 753.

Exercise 1.2.2

By the binomial theorem (see Exercise 5.3.3) or direct calculation,

(x + δx)^{3} = x^{3} + 3x^{2}δx + 3x(δx)^{2} + (δx)^{3} ≈ x^{3} + 3x^{2}δx. Alternatively, apply the product rule,

(a + δa) × (b + δb) ≈ a × b + a × δb + b × δa

to first order, twice.

By the binomial theorem, induction or repeated use of the product rule,

(x + δx)^{n} ≈ x^{n} + nx^{n−1}δx

to the first order.

Exercise 1.2.3

(a + δa) − (b + δb) = a − b + (δa − δb)

exactly and thus certainly to first order.

Exercise 1.2.4

With the notation given,

a = 1 000 000, v = 1 000, δa = 3

so

δv ≈ \frac{1}{2\sqrt{a}} × δa = \frac{3}{2 000}.

Thus

\sqrt{1 000 003} = v + δv ≈ 1000.0015.

in agreement with my calculator to these decimal places (and actually rather more).

Exercise 1.2.5

Let us write

v = a^{1/3} and v + δv = (a + δa)^{1/3}.

If δa is small in magnitude compared to a, then a + δa will be close to a and so v + δv will be close to v. In other words, δv will be small in magnitude compared to v. Thus, working to first order,

a + δa = (v + δv)^{3} ≈ v^{3} + 3v^{2} × δv = a + 3(a^{1/3})^{2} × δv.

Subtracting a from both sides and rearranging, we get

δa ≈ 3a^{2/3} × δv,

(where a^{2/3} = (a^{1/3})^{2}) so

δv ≈ \frac{1}{3a^{2/3}} × δa

to the first order.

Take

a = 1 000 000, δa = 3.

To first order

(1 000 003)^{1/3} = v + δv ≈ 100 + \frac{δa}{3a^{2/3}} = 100 + \frac{1}{10 000} = 100.0001

in agreement with my calculator.

(Working to more figures it gets 100.0000999999000001666663333341 . . ..)

Exercise 1.2.7

Since u = s − a, we have

u = (b + c − a)/2

and, by the addition rule,

δu = (δb + δc − δa)/2.

Similar formulae hold for v and w, whilst

δs = (δa + δb + δc)/2.

Thus

δA ≈\frac{\sqrt{(a + b + c)(a + b − c)(a − b + c)(−a + b + c)}}{8}

× (\frac{δa + δb + δc}{a + b + c} + \frac{−δa + δb + δc}{−a + b + c} + \frac{δa − δb + δc}{a − b + c} + \frac{δa + δb − δc}{a + b − c})

to first order.

If we observe that

u + v + w = 3s − a − b − c = s,

we obtain

δs = δu + δv + δw

and

δA ≈ \sqrt{(u + v + w)uvw} × (\frac{δu + δv + δw}{u + v + w} + \frac{δu}{u} + \frac{δv}{v} + \frac{δw}{w})

to first order.

Exercise 1.2.8

If S = b^{2} − 4c and S + δS = (b + δb)^{2} − 4(c + δc), then, to first order,

δS = 2bδb − 4δc

so, if

u = \frac{−b + \sqrt{b^{2} − 4c}}{2},

u + δu = \frac{−(b + δb) + \sqrt{(b + δb)^{2} − 4(c + δc)}}{2},

we have

δu = \frac{1}{2} (−δb + \frac{δS}{2\sqrt{S}} )

= \frac{1}{2} (−δb + \frac{bδb − 2δc}{\sqrt{b^{2} − 4c}}) .

Similarly,

\frac{−(b + δb) − \sqrt{(b + δb)^{2} − 4(c + δc)}}{2}

= \frac{−b − \sqrt{b^{2} − 4c}}{2} − \frac{1}{2} (δb + \frac{bδb − 2δc}{\sqrt{b^{2} − 4c}})

to first order.

Exercise 1.2.9

Let R be the radius of the moon in meters. The intended length is 2π(R + 1), so the actual length is 10 + 2π(R + 1) and the radius of the ring will be

\frac{10 + 2π(R + 1)}{2π} = \frac{10}{2π} + 1 + R.

The extra height is 10/2π ≈ 1.6 meters.

Exercise 1.2.10

Using the formula (x + y)(x − y) = x^{2} − y^{2}, with x = 1 and 100y the percentage rise, we see that, after one rise and fall, Pumpkin shares have fallen by .01%. Since .01 is very small compared with 100, we expect an overall fall of roughly 40 × .01 = .4 percent (binomial theorem or Exercise 1.2.2). The value is essentially unchanged.

Melon shares will be worth 1^{2} − .4^{2} = .84 times their original value, that is to say 84% of their original value, a substantial change.

Exercise 1.3.2

(i) Let x ≥ y ≥ 0. Then

|x| + |y| = x + y = |x + y|,
|x| + | − y| = x + y ≥ x − y = |x + (−y)|,
| − x| + |y| = x + y ≥ x − y = |(−x) + y|,
| − x| + | − y| = x + y = | − (x + y)| = |(−x) + (−y)|.

(ii) Let x, y ≥ 0. Then

|x||y| = xy = |x||y|,
|x|| − y| = xy = |x(−y)|,
| − x||y| = xy = |(−x)y|,
| − x|| − y| = xy = |(−x)(−y)|.

(iii) By (ii),

|a − b| + |b − c| ≥ |(a − b) + (b − c)| = |a − c|.

(iv) We have

|3 − 2| + |2 − 1| = 2 = |3 − 1|,

but

|2 − 3| + |2 − 1| = 2 > 1 = |2 − 1|.

Exercise 1.3.4⋆

Exercise 1.4.2

We have

150 = f(32) − f(31) = A,

so

f(31 + \frac{2}{3}) = f(31) + 2A/3 = f(31) + 100

as required.

Exercise 1.4.5

(i) g(x + h) = g(x) = g(x) + 0h + o(h), so g′(x) = 0. (ii) If g(t) = a, then

u′(t) = g′(t)f(t) + g(t)f′(t) = 0 × f(t) + a × f′(t) = af′(t).

(iii) g_{1}(x + h) = x + h = g(x) + 1 × h + o(h), so g′_{1}(x) = 1. (iv) g′_{2}(t) = g_{1}(t)g′_{1}(t) + g′_{1}(t)g_{1}(t) = t × 1 + 1 × t = 2t.

(v) g′_{3}(t) = g_{1}(t)g′_{2}(t) + g′_{1}(t)g_{2}(t) = 2t^{2} + t^{2} = 3t^{2}. (vi) If g′_{n−1}(t) = (n − 1)g_{n−2}(t), then, since g_{n}(t) = g_{1}(t)g_{n−1}(t),

g′_{n}(t) = g_{1}(t)g′_{n−1}(t)+g′_{1}(t)g_{n−1}(t) = (n−1)g_{n−1}(t)+1×g_{n−1}(t) = ng_{n−1}(t), so the result follows by induction.

(vii) Write u(t) = 1. We have u(t) = g_{n}(t)g_{−n}(t) and

0 = u′(t) = g′_{n}(t)g_{−n}(t) + g_{n}(t)g′_{−n}(t)
= ng_{n−1}(t)g_{−n}(t) + g_{n}(t)g′_{−n}(t)
= ng_{−1}(t) + g_{n}(t)g′_{−n}(t)

so

g_{n}(t)g′_{−n}(t) = −ng_{−1}(t)

and

g′_{−n}(t) = −\frac{n}{t^{n+1}} .

Combining our results for positive and negative integers, we get

g′_{n}(t) = ng_{n−1}(t)

for all integers n ≠ 0.

By (i), g′_{0}(t) = 0. (viii) Using the rule (f + g)′ = f′ + g′,

P′(t) = na_{n}t^{n−1} + (n − 1)a_{n−1}t^{n−2} + … + a_{1}.

(ix) h(t) = g_{−1} ∘ f(t) so

h′(t) = f′(t)g′_{−1} ∘ f(t) = −\frac{f′(t)}{f(t)^{2}} .

(x) 0 = f′(t)h(t) + f(t)h′(t), so

h′(t) = −\frac{f′(t)h(t)}{f(t)} = −\frac{f′(t)}{f(t)^{2}} .

(x) By the product rule and the quotient rule,

h′(t) = \frac{f′(t)}{g(t)} − \frac{f(t)g′(t)}{g(t)^{2}}.

Exercise 1.4.7

(i) If h(t) = t^{p}, then g_{1/p} = h^{−1}(t), so

g′_{1/p}(t) = (h^{−1})′(t) = \frac{1}{h′(h^{−1}(t))} = \frac{1}{p(h^{−1}(t))^{p−1}} = \frac{1}{pt^{(p−1)/p}} = \frac{1}{pt}g_{1/p}(t).

(ii) By the product rule (or applying the function of a function rule to u ∘ g_{1/p} with u(t) = t^{q}), we have

g′_{q/p}(t) = q(g_{1/p}(t))^{q−1}g′_{1/p}(t) = \frac{q}{p}g_{(q/p)−1}(t).

Exercise 1.4.8

(i) We know that

a′(x) = 0, S′(x) = \frac{1}{2\sqrt{x}}

and, by the addition rule,

u′(x) = 2x.

We have c(x) = S(u(x)), so, by the function of a function rule,

c′(x) = u′(x)S′(u(x)) = \frac{x}{\sqrt{1 + x^{2}}} .

By the addition rule,

h′(x) = a′(x) + c′(x) = \frac{x}{\sqrt{1 + x^{2}}} .

We also know that

b′(x) = \frac{1}{3x^{2/3}}

so, by the addition rule,

g′(x) = a′(x) + b′(x) = \frac{1}{3x^{2/3}} .

The quotient rule now gives

f′(x) = \frac{g′(x)}{h(x)} − \frac{g(x)h′(x)}{h(x)^{2}}

= \frac{1}{3x^{2/3}(1 + \sqrt{1 + x^{2}})} − \frac{x(1 + x^{1/3})}{(1 + \sqrt{1 + x^{2}})^{2}\sqrt{1 + x^{2}}} .

Exercise 1.4.9

If h > 0,

m(h) − m(0) = h = h + o(h),

so a = 1.

If h < 0,

m(h) − m(0) = −h = −h + o(h),

so b = −1.

Exercise 1.5.1

There are many ways of explaining this. One way (and there is no claim that this is better than any other) is to observe that

∠ABX = 2r − ∠CBX

(where r is a right angle), so

tan(2r − θ) = \frac{sin(2r − θ)}{cos(2r − θ)} = \frac{cos(r − θ)}{− sin(r − θ)}

= \frac{− sin(−θ)}{− cos(−θ)} = \frac{sin(−θ)}{cos(−θ)}

= −\frac{sin θ}{cos θ} = − tan θ.

Exercise 1.5.3

(i) sin(x+4r) = − cos(x+3r) = − sin(x+2r) = − cos(x+r) = sin x and

cos(x + 4r) = sin(x + 3r) = sin(x − r) = cos x.

Since sin 0 = − sin(−0) = − sin 0, we have sin 0 = 0. Finally,

cos(−x) = sin(−x + r) = − sin(x − r) = cos x.

(ii) We have

cos(u + v) = sin(u + v + r) = sin(u + r) cos v + cos(u + r) sin v

= cos u cos v − sin u sin v.

(iii) We have

1 = sin r = cos 0 = cos(u + (−u)) = cos u cos(−u) − sin u sin(−u)

= (cos u)^{2} + (sin u)^{2}. (iv) We have

sin 2x = sin(x + x) = sin x cos x + sin x cos x = 2 sin x cos x, and

cos 2x = cos(x + x) = cos x cos x − sin x sin x
= (cos x)^{2} − (sin x)^{2} = (1 − (sin x)^{2}) − (sin x)^{2}
= 1 − 2(sin x)^{2} = 1 − 2(1 − (cos x)^{2}) = 2(cos x)^{2} − 1.

(v) We have

tan(u + v) = \frac{sin(u + v)}{cos(u + v)}
= \frac{sin u cos v + cos u sin v}{cos u cos v − sin u sin v}
= \frac{tan u + tan v}{1 − tan u tan v} .

Setting u = v = x, we obtain tan 2x = \frac{2 tan x}{1 − (tan x)^{2}} .

Exercise 2.2.1

N(f) = N_{1}(f) + N_{2}(f) + … + N_{n}(f),
N(g) = N_{1}(g) + N_{2}(g) + … + N_{n}(g),
N(f + g) = N_{1}(f + g) + N_{2}(f + g) + … + N_{n}(f + g).

Now observe that

f(t) ≥ N_{r}(f)s and g(t) ≥ N_{r}(g)s

so

f(t) + g(t) ≥ (N_{r}(f) + N_{r}(g))s

whenever a + (r − 1)s ≤ t ≤ a + rs, so N_{r}(f + g) ≥ N_{r}(f) + N_{r}(g).

Exercise 2.2.3

(i) We have

g(t) = \begin{cases}f(t) ≥ 0 & if f(t) ≥ 0, \\ 0 ≥ 0 & otherwise \end{cases}

and

h(t) = \begin{cases}0 ≥ 0 & if f(t) ≥ 0, \\ −f(t) ≥ 0 & otherwise. \end{cases}

whilst

g(t) − h(t) = \begin{cases}f(t) + 0 & if f(t) ≥ 0, \\ 0 − (−f(t)) & otherwise \end{cases}
= f(t)

as required.

(ii) g(t)−h(t) = |f(t)|−(|f(t)|−f(t)) = f(t) and, since |f(t)| ≥ f(t), we have g(t), h(t) ≥ 0.

(iii) We have g(t) − h(t) = (M + f(t)) − M = f(t),

g(t) = M + f(t) ≥ M − M ≥ 0

and g(t) = M ≥ 0.

Exercise 2.2.4

We have G(t), H(t) ≥ 0 and G(t) − H(t) = F (t), so

I_{a}^{b}F (t) dt = ∫_{a}^{b} G(t) dt − ∫_{a}^{b} H(t) dt

= ∫_{a}^{b} F (t) dt − 0 = ∫_{a}^{b} F (t) dt.

Exercise 2.2.5

Choosing g and h so that g(t), h(t) ≥ 0 and f(t) = g(t) − h(t), we have −f(t) = h(t) − g(t). Thus

∫_{a}^{b} ( − f(t)) dt = ∫_{a}^{b} h(t) dt − ∫_{a}^{b} g(t) dt

= − (∫_{a}^{b} g(t) dt − ∫_{a}^{b} h(t) dt)

= − ∫_{a}^{b} f(t) dt.

Exercise 2.2.6

Let f(t) = f_{1}(t) − f_{2}(t), g(t) = g_{1}(t) − g_{2}(t) with f_{j}(t), g_{j}(t) ≥ 0 [j = 1, 2]. We have

f(t) + g(t) = (f_{1}(t) + g_{1}(t)) − (f_{2}(t) + g_{2}(t))

and

(f_{1}(t) + g_{1}(t)), (f_{2}(t) + g_{2}(t)) ≥ 0,

so

∫_{a}^{b} f(t) + g(t) dt = ∫_{a}^{b}(f_{1}(t) + g_{1}(t)) dt − ∫_{a}^{b}(f_{2}(t) + g_{2}(t)) dt
= (∫_{a}^{b} f_{1}(t) dt + ∫_{a}^{b} g_{1}(t) dt) − (∫_{a}^{b} f_{2}(t) dt + ∫_{a}^{b} g_{2}(t) dt)
= (∫_{a}^{b} f_{1}(t) dt − ∫_{a}^{b} f_{2}(t) dt) + (∫_{a}^{b} g_{1}(t) dt − ∫_{a}^{b} g_{2}(t) dt)
= ∫_{a}^{b} f(t) dt + ∫_{a}^{b} g(t) dt.

Using the first part of the question and the result of Exercise 2.2.5, we have

∫_{a}^{b} f(t) − g(t) dt = ∫_{a}^{b} f(t) + (−g(t)) dt
= ∫_{a}^{b} f(t) dt + ∫_{a}^{b}(−g(t)) dt
= ∫_{a}^{b} f(t) dt − ∫_{a}^{b} g(t) dt.

Exercise 2.2.7

(i) We have

∫_{a}^{b} pf(t) dt = ∫_{a}^{b} \underbrace{f(t) + f(t) + … + f(t)}_{p} dt

= \underbrace{∫_{a}^{b} f(t) dt + ∫_{a}^{b} f(t) dt + … + ∫_{a}^{b} f(t) dt}_{p}

= p ∫_{a}^{b} f(t) dt.

If p = 0, we have ∫_{a}^{b} 0 × f(t) dt = ∫_{a}^{b} 0 dt = 0 = 0 × ∫_{a}^{b} f(t) dt.

(ii) By (i), we have

q ∫_{a}^{b} \frac{p}{q}f(t) dt = ∫_{a}^{b} q \frac{p}{q}f(t) dt
= ∫_{a}^{b} pf(t) dt
= p ∫_{a}^{b} f(t) dt,

so

∫_{a}^{b} \frac{p}{q}f(t) dt = \frac{p}{q} ∫_{a}^{b} f(t) dt.

(iii) Since

\frac{p}{q}f(t) ≤ uf(t) ≤ \frac{p + 1}{q}f(t),

the inequality rule for integrals yields

\frac{p}{q} ∫_{a}^{b} f(t) dt ≤ ∫_{a}^{b} uf(t) dt ≤ \frac{p + 1}{q} ∫_{a}^{b} f(t) dt.

Thus

(u − \frac{1}{q}) ∫_{a}^{b} f(t) dt ≤ ∫_{a}^{b} uf(t) dt ≤ (u + \frac{1}{q}) ∫_{a}^{b} f(t) dt.

Since we can make q as large as we please, ∫_{a}^{b} uf(t) dt = u ∫_{a}^{b} f(t) dt.

(iv) We can write g(t) = g_{1}(t) − g_{2}(t) with g_{1}(t), g_{2}(t) ≥ 0. Thus, if u ≥ 0, part (iii) yields

∫_{a}^{b} ug(t) dt = ∫_{a}^{b} ug_{1}(t) − ug_{2}(t) dt

= ∫_{a}^{b} ug_{1}(t) dt − ∫_{a}^{b} ug_{2}(t) dt

= u ∫_{a}^{b} g_{1}(t) dt − u ∫_{a}^{b} g_{2}(t) dt

= u (∫_{a}^{b} g_{1}(t) dt − ∫_{a}^{b} g_{2}(t) dt)

= u ∫_{a}^{b} g(t) dt.

Using Exercise 2.2.5, we deduce that

∫_{a}^{b}(−u)g(t) dt = ∫_{a}^{b} −(ug(t)) dt = − ∫_{a}^{b}(ug(t)) dt

= − (u ∫_{a}^{b} g(t) dt) = (−u) ∫_{a}^{b} g(t) dt.

The required result now follows.

Exercise 2.2.8

(i) We know the result for a ≤ b. If b ≤ a, then

∫_{a}^{b}(f(t) + g(t)) dt = − ∫_{b}^{a}(f(t) + g(t)) dt

= − (∫_{b}^{a} f(t) dt + ∫_{b}^{a} g(t) dt)

= − ∫_{b}^{a} f(t) dt − ∫_{b}^{a} g(t) dt

= ∫_{a}^{b} f(t) dt + ∫_{a}^{b} g(t) dt.

(ii) We know the result for a ≤ b. If b ≤ a, then ∫_{a}^{b} vf(t) dt = − ∫_{b}^{a} vf(t) dt = −v ∫_{b}^{a} f(t) dt = v ∫_{a}^{b} f(t) dt.

(iii) If A ≥ B, then −B ≥ −A.

Exercise 2.2.9

Write f(t) = f_{1}(t)−f_{2}(t) with f_{1} and f_{2} positive functions. We have

∫_{a}^{b} f(t) dt + ∫_{b}^{c} f(t) dt = (∫_{a}^{b} f_{1}(t) dt − ∫_{a}^{b} f_{2}(t) dt) + (∫_{b}^{c} f_{1}(t) dt − ∫_{b}^{c} f_{2}(t) dt)

= (∫_{a}^{b} f_{1}(t) dt + ∫_{b}^{c} f_{1}(t) dt) − (∫_{a}^{b} f_{2}(t) dt + ∫_{b}^{c} f_{2}(t) dt)

= ∫_{a}^{c} f_{1}(t) dt − ∫_{a}^{c} f_{2}(t) dt

= ∫_{a}^{c} f(t) dt.

Exercise 2.2.10

(i) We have

∫_{a}^{c}f(t) dt + ∫_{b}^{c} f(t) dt = − ∫_{c}^{a} f(t) dt − ∫_{b}^{c} f(t) dt

= − (∫_{b}^{c} f(t) dt + ∫_{c}^{a} f(t) dt) = − ∫_{b}^{a} f(t) dt = ∫_{a}^{b} f(t) dt.

(ii) We have

∫_{a}^{b} f(t) dt + ∫_{b}^{c} f(t) dt = ∫_{a}^{b} f(t) dt − ∫_{c}^{b} f(t) dt

= ∫_{a}^{b} f(t) dt − ∫_{c}^{a} f(t) dt − ∫_{a}^{b} f(t) dt = − ∫_{c}^{a} f(t) dt = ∫_{a}^{c} f(t) dt.

(iii) If a ≤ c ≤ b,

∫_{a}^{b} f(t) dt+∫_{b}^{c} f(t) dt = ∫_{a}^{c} f(t) dt+∫_{c}^{b} f(t) dt+∫_{b}^{c} f(t) dt = ∫_{a}^{c} f(t) dt.

We thus have the result for c ≤ b ≤ a, c ≤ a ≤ b, a ≤ c ≤ b and a ≤ b ≤ c. The remaining cases when b ≤ a follow as in (i).

(iv) We know the result for a ≤ b and k ≥ 0 (see note below), so, if k < 0,

∫_{a}^{b} k dt = − ∫_{a}^{b}(−k), dt = −(−k)(b − a) = k(b − a)

If b ≤ a, then

∫_{a}^{b} k dt = − ∫_{b}^{a} k dt = −k(b − a) = k(a − b).

(Note: So far as this book is concerned, the result for b ≥ a and k ≥ 0 is obvious. We could go one step further back and remark that if b > a and k > 0, our argument about splitting into squares shows that

∫_{a}^{b}(b − a) dt = (b − a)^{2}

and so

∫_{a}^{b} k dt = ∫_{a}^{b} \frac{k}{b − a}(b − a) dt = \frac{k}{b − a} ∫_{a}^{b}(b − a) dt = k(b − a).

However, this is all a bit hair splitting.)

Exercise 2.2.11

(i) We have

(r + 1)r − r(r − 1) = r((r + 1) − (r − 1)) = 2r.

Thus

1 + 2 + … + n = \frac{1}{2}((2 × 1 − 1 × 0) + …
+ ((r + 1)r − r(r − 1))
+ ((r + 2)(r + 1) − (r + 1)r) + …
+ ((n + 1)n − n(n − 1)))

= \frac{1}{2}((n + 1)n − 1 × 0) = \frac{1}{2}n(n + 1).

(ii) We have (r +1)r(r −1)−r(r −1)(r −2) = r(r −1)((r +1)−(r −2)) = 3r(r −1) Thus

1 × 0 + 2 × 1 + … + r(r − 1) + … + n(n − 1)

= \frac{1}{3}((2 × 1 × 0 − 2 × 0 × (−1)) + …

+ ((r + 1)r(r − 1) − r(r − 1)(r − 2))

+ ((r + 2)(r + 1)r − (r + 1)r(r − 1)) + …
+ ((n + 1)n(n − 1) − n(n − 1)(n − 2)))

= \frac{1}{3}((n + 1)n(n − 1) − 1 × 0 × (−1)) = \frac{1}{3}(n + 1)n(n − 1).

(iii) We have

1^{2} + 2^{2} + … + r^{2} + … + n^{2}

= (1 + 1 × 0) + … + (r(r − 1) + r) + … + (n(n − 1) + n) = (1 × 0 + 2 × 1 + … + r(r − 1) + … + n(n − 1))

+ (1 + 2 + … + n)

= \frac{1}{3}((n + 1)n(n − 1) + \frac{1}{2}n(n + 1)) = \frac{1}{6}n(n + 1)(2n + 1).

(iv) Observe that

\frac{r^{2}}{n^{2}} ≤ t^{2} ≤ \frac{(r + 1)^{2}}{n^{2}}

for r/n ≤ t ≤ (r + 1)/n and so

\frac{r^{2}}{n^{3}} = ∫_{r/n}^{(r+1)/n} \frac{r^{2}}{n^{2}} dt ≤ ∫_{r/n}^{(r+1)/n} t^{2} dt ≤ ∫_{r/n}^{(r+1)/n} \frac{(r + 1)^{2}}{n^{2}} dt = \frac{(r + 1)^{2}}{n^{3}} .

(v) Thus, adding,

\frac{0^{2}}{n^{3}} + … + \frac{r^{2}}{n^{3}} + … + \frac{(n − 1)^{2}}{n^{3}}

≤ ∫_{0}^{1/n} t^{2} dt + … + ∫_{r/n}^{(r+1)/n} t^{2} dt + … + ∫_{(n−1)/n}^{1} t^{2} dt

≤ \frac{1^{2}}{n^{3}} + … + \frac{(r + 1)^{2}}{n^{3}} + … + \frac{n^{2}}{n^{3}} ,

so

\frac{0^{2} + 1^{2} + … + (n − 1)^{2}}{n^{3}} ≤ ∫_{0}^{1} t^{2} dt ≤ \frac{1^{2} + 2^{2} + … + n^{2}}{n^{3}}

and, by (iii),

\frac{(1 − n^{−1})(1 − \frac{1}{2}n^{−1})}{3} ≤ ∫_{0}^{1} t^{2} dt ≤ \frac{(1 + n^{−1})(1 + \frac{1}{2}n^{−1})}{3}

Since we can make n as large as we please,

∫_{0}^{1} t^{2} dt = \frac{1}{3}.

(iv) Observe that

\frac{r^{2}a^{2}}{n^{2}} ≤ t^{2} ≤ \frac{(r + 1)^{2}a^{2}}{n^{2}}

for ra/n ≤ t ≤ (r + 1)a/n and so

\frac{r^{2}a^{3}}{n^{3}} = ∫_{ra/n}^{(r+1)a/n} \frac{r^{2}a^{2}}{n^{2}} dt ≤ ∫_{r/n}^{(r+1)a/n} t^{2} dt ≤ ∫_{ra/n}^{(r+1)a/n} \frac{(r + 1)^{2}a^{2}}{n^{2}} dt = \frac{(r + 1)^{2}a^{3}}{n^{3}} .

Thus, adding,

\frac{(0^{2} + 1^{2} + … + (n − 1)^{2})a^{3}}{n^{3}} ≤ ∫_{0}^{a} t^{2} dt ≤ \frac{(1^{2} + 2^{2} + … + n^{2})a^{3}}{n^{3}}

and, by (iii),

a^{3} \frac{(1 − n^{−1})(1 − \frac{1}{2}n^{−1})}{3} ≤ ∫_{0}^{a} t^{2} dt ≤ a^{3} \frac{(1 + n^{−1})(1 + \frac{1}{2}n^{−1})}{3}.

Since we can make n as large as we please,

∫_{0}^{a} t^{2} dt = a^{3} \frac{1}{3}.

We observe (using symmetry) that, if a > 0

∫_{0}^{−a} t^{2} dt = − ∫_{−a}^{0} t^{2} dt = − ∫_{0}^{a} t^{2} dt = −\frac{a^{3}}{3} = \frac{(−a)^{3}}{3} .

Thus (since the result for a = 0 is trivial)

∫_{0}^{a} t^{2} dt = \frac{a^{3}}{3}

for all values of a whether positive, negative or zero.

Thus

∫_{a}^{b} t^{2} dt = ∫_{0}^{b} t^{2} dt − ∫_{0}^{a} t^{2} dt = \frac{b^{3} − a^{3}}{3} .

Exercise 2.2.12

(i) Observe that

(r + 1)r(r − 1)(r − 2) … (r − k + 1) − r(r − 1)(r − 2) … (r − k)
= r(r − 1)(r − 2) … (r − k + 1) × (r + 1 − (r − k))
= (k + 1)u_{r},

so, adding, we have (k + 1)(u_{1} + u_{2} + … + u_{n}) = (n + 1)n(n − 1)(n − 2) … (n − k + 1). (ii) The result is true when k = 0, since

1^{0} + 2^{0} + … + n^{0} = n = \frac{n^{0+1}}{0 + 1}.

Suppose it is true for all k ≤ K − 1. Then, writing

u_{r} = (r + 1)r(r − 1)(r − 2) … (r − K + 1)

and

s_{p} = 1^{p} + 2^{p} + … + n^{p},

we have

r^{K} = u_{r} + a_{0}1^{0} + a_{1}r + a_{2}r^{2} + … + a_{K−1}r^{K−1}

and so

1^{K} + 2^{K} + … + n^{K} = (u_{1} + u_{2} + … + u_{n}) + (a_{0}s_{0} + a_{1}s_{1} + a_{2}s_{2} + … + a_{K−1}s_{k−1}

= (n + 1)n(n − 1)(n − 2) … (n − k + 1) + a_{0} \frac{n^{1}}{1} + a_{1} \frac{n^{2}}{2} + a_{K−1} \frac{n^{K}}{K}

+ a_{0}P_{0}(n) + a_{1}P_{1}(n) + a_{K−1}P_{K−1}(n)

= \frac{n^{K+1}}{K + 1} + P_{K}(n)

where P_{K} is some polynomial of degree at most K. The result now follows by induction.

(iii) Observe that, if a > 0,

\frac{a^{k}r^{k}}{n^{k}} ≤ t^{k} ≤ \frac{a^{k}(r + 1)^{k}}{n^{k}}

for ar/n ≤ t ≤ a(r + 1)/n and so

\frac{a^{k+1}r^{k}}{n^{k+1}} = ∫_{ar/n}^{a(r+1)/n} \frac{a^{k}r^{k}}{n^{k}} dt ≤ ∫_{ar/n}^{(ar+1)/n} t^{k} dt ≤ ∫_{ar/n}^{a(r+1)/n} \frac{a^{k}(r + 1)^{k}}{n^{k}} dt = \frac{a^{k+1}(r + 1)^{k}}{n^{k+1}} .

Thus, adding,

a^{k+1} (\frac{0^{k}}{n^{k+1}} + … + \frac{r^{k}}{n^{k+1}} + … + \frac{(n − 1)^{k}}{n^{k+1}} )

≤ ∫_{0}^{a/n} t^{k} dt + … + ∫_{ar/n}^{a(r+1)/n} t^{k} dt + … + ∫_{a(n−1)/n}^{a} t^{k} dt

≤ a^{k+1} (\frac{1^{k}}{n^{k+1}} + … + \frac{(r + 1)^{k}}{n^{k+1}} + … + \frac{n^{k}}{n^{k+1}} ) ,

so

a^{k+1} \frac{0^{k} + 1^{k} + … + (n − 1)^{k}}{n^{k+1}} ≤ ∫_{0}^{a} t^{k} dt ≤ a^{k+1} \frac{1^{k} + k^{k} + … + n^{k}}{n^{k+1}}

Thus

a^{k+1} \frac{(n − 1)^{k+1}/(k + 1) + P_{k}(n − 1)}{n^{k+1}} ≤ ∫_{0}^{a} t^{k} dt ≤ a^{k+1} \frac{n^{k+1}/(k + 1) + P_{k}(n)}{n^{k+1}},

i.e.

a^{k+1} \frac{1}{k + 1} + Q_{1}(1/n) ≤ ∫_{0}^{1} t^{k} dt ≤ a^{k+1} \frac{1}{k + 1} + Q_{2}(1/n)

where Q_{1} and Q_{2} are polynomials with zero constant term.

Since we can make n as large as we please,

∫_{0}^{a} t^{k} dt = \frac{a^{k+1}}{k + 1}.

Using symmetry or antisymmetry

∫_{0}^{−a} t^{k} dt = − ∫_{−a}^{0} t^{k} dt = (−1)^{k+1} ∫_{0}^{a} t^{k} dt = \frac{(−a)^{k+1}}{k + 1} .

Thus

∫_{0}^{a} t^{k} dt = \frac{a^{k+1}}{k + 1}

regardless of the sign of a and so

∫_{a}^{b} t^{k} dt = ∫_{0}^{b} t^{k} dt − ∫_{0}^{a} t^{k} dt = \frac{b^{k} − a^{k}}{3}

for all a and b.

Exercise 2.3.1

(Only part (ii).)

If δt > 0 and f(s) is negative for t ≤ s ≤ t + δt, then

∫_{t}^{t+δt} f(x) dx = − ∫_{t}^{t+δt}(−f(x)) dx

= −(−f(t)δt + o(δt))

= f(t)δt + o(δt).

Exercise 2.3.2

We have

f(t) − u ≤ f(x) ≤ f(t) + u

whenever |x − t| ≤ v. Thus, if 0 ≤ δt ≤ v, we have

∫_{t−δt}^{t}(f(t) − u) dx ≤ ∫_{t−δt}^{t} f(x) dx ≤ ∫_{t−δt}^{t}(f(t) + u) dx so that

(f(t) − u)δt ≤ ∫_{t−δt}^{t} f(x) dx ≤ (f(t) + u)δt

and, multiplying through by −1, (f(t) − u)(−δt) ≥ ∫_{t}^{t−δt} f(x) dx ≥ (f(t) + u)(−δt).

Exercise 2.3.3

If

g(x) = (F (x) − G(x)) − (F (a) − G(a)),

then

g′(x) = F′(x) − G′(x) − 0 = f(x) − f(x) = 0,

so

g(x) = c

a constant, whence G(x) = F (x) + c for all x.

Conversely, if F′(x) = f(x) and G(x) = F (x) + c, then G′(x) = F′(x) + 0 = f(x).

Exercise 2.3.4

(i) If

F (t) = a_{0}t + a_{1} \frac{t^{2}}{2} + a_{2} \frac{t^{3}}{3} + … + a_{n} \frac{t^{n+1}}{n + 1},

then

F′(t) = a_{0} + a_{1}t + a_{2}t^{2} + … + a_{n}t^{n}.

Thus

∫_{0}^{x}(a_{0} + a_{1}t + a_{2}t^{2} + … + a_{n}t^{n}) dt = [F (t)]_{0}^{x}

= a_{0}x + a_{1} \frac{x^{2}}{2} + a_{2} \frac{x^{3}}{3} + … + a_{n} \frac{x^{n+1}}{n + 1}.

(ii) Write k = p/q. If

F (t) = \frac{t^{k+1}}{k + 1},

then

F′(t) = t^{k}.

Thus

∫_{a}^{b} t^{k} dt = [F (t)]_{a}^{b} = \frac{1}{k + 1}(b^{k+1} − a^{k+1}).

Exercise 2.3.5

(i) Observe that

\frac{r^{a}}{n^{a}} ≤ t^{a} ≤ \frac{(r + 1)^{a}}{n^{a}}

for r/n ≤ t ≤ (r + 1)/n and so

\frac{r^{a}}{n^{a+1}} = ∫_{r/n}^{(r+1)/n} \frac{r^{a}}{n^{a}} dt ≤ ∫_{r/n}^{(r+1)/n} t^{a} dt = ∫_{r/n}^{(r+1)/n} \frac{(r + 1)^{a}}{n^{a}} dt ≤ \frac{(r + 1)^{a}}{n^{a+1}} .

Thus, adding,

\frac{0^{a}}{n^{a+1}} + … +\frac{r^{a}}{n^{a+1}} + … + \frac{(n − 1)^{a}}{n^{a+1}}

≤ ∫_{0}^{1/n} t^{a} dt + … + ∫_{r/n}^{(r+1)/n} t^{a} dt + … + ∫_{(n−1)/n}^{1} t^{a} dt

≤ \frac{1^{a}}{n^{a+1}} + … + \frac{(r + 1)^{a}}{n^{a+1}} + … + \frac{n^{a}}{n^{a+1}} ,

so

\frac{0^{a} + 1^{a} + … + (n − 1)^{a}}{n^{a+1}} ≤ ∫_{0}^{1} t^{a} dt ≤ \frac{1^{a} + 2^{a} + … + n^{a}}{n^{a+1}} .

(ii) Since

∫_{0}^{1} t^{a} dt = [\frac{t^{a+1}}{a + 1}]_{0}^{1} = \frac{1}{a + 1},

this gives us

0^{a} + 1^{a} + … + (n − 1)^{a} ≤ \frac{n^{a+1}}{a + 1} ≤ 1^{a} + 2^{a} + … + n^{a}.

(iii) Thus

\frac{n^{a+1}}{a + 1} ≤ 1^{a} + 2^{a} + … + n^{a} ≤ \frac{n^{a+1}}{a + 1} + n^{α}

and

0 ≤ \frac{1^{a} + 2^{a} + … + n^{a}}{n^{a+1}} − \frac{1}{a + 1} ≤ \frac{1}{n}.

Exercise 2.4.1

Observe that b^{3} − a^{3} = (b − a)(a^{2} + ab + b^{2}) = (b − a)((a + b/2)^{2} + 3b^{2}/4) > 0

if b > a. However, f′(t) = 3t^{2}, so f′(0) = 0.

Exercise 2.4.2

If g(t) = f(t) − Mt, then g′(t) = f′(t) − M ≤ 0 so g is decreasing and g(b) − g(a) ≤ 0. In other words,

(f(b) − Mb) − (f(a) − Ma) ≤ 0,

so

f(b) − f(a) ≤ M(b − a).

If h(t) = Mt + f(t), then h′(t) = M + f′(t) ≥ 0, so h is increasing and h(b) − h(a) ≥ 0. In other words,

(Mb + f(b)) − (Ma + f(a)) ≥ 0

so

f(b) − f(a) ≥ −M(b − a).

Exercise 2.4.3

By the definition of differentiation, given v > 0, we can find a u > 0 with u ≤ (t − s)/2 such that

|f(y + h) − f(y) − f′(y)h| ≤ v|h|

whenever |h| ≤ u. Since f′(y) > 0, we can take v = f′(y)/2. Now

f(s) ≥ f(y + u) = (f(y) + f′(y)u) + (f(y + u) − f(y) − f′(y)u)

≥ (f(y) + f′(y)u) − |f(y + u) − f(y) − f′(y)u|

≥ (f(y) + f′(y)u) − \frac{f′(y)}{2}u = f(y) + \frac{f′(y)u}{2}

> f(y) ≥ f(t) as stated.

Exercise 2.5.1

(i) If s < 0, then Supergirl will run a longer distance and swim a longer distance than if she had taken s = 0.

(ii) We have

f(s) = time running + time swimming

= \frac{distance run}{u} + \frac{distance swum}{v}

= \frac{s}{u} + \frac{\sqrt{(a − s)^{2} + b^{2}}}{v} .

Exercise 2.5.2

(i) f(s) = h(s) + g(s)/v with h(s) = s/u, g(s) = G(H(s)) where G(s) = \sqrt{s} and H(s) = (a − s)^{2} + b^{2}.

Now H′(s) = −2(s − a),

g′(s) = H′(s)G′(H(s)) = −2(s − a) × \frac{1}{2\sqrt{H(s)}} = −\frac{s − a}{\sqrt{(a − s)^{2} + b^{2}}}

and

f′(s) = \frac{g′(s)}{v} + h′(s) = \frac{1}{u} − \frac{1}{v} \frac{a − s}{((a − s)^{2} + b^{2})^{1/2}} .

(ii) We have

f(s) = \frac{1}{u} − \frac{F (s)}{vg(s)},

with g as in (i) and F (s) = a − s, so

f′(s) = −\frac{1}{v} (\frac{F′(s)}{g(s)} − \frac{F (s)g′(s)}{g(s)^{2}})

= \frac{1}{v} (\frac{1}{((a − s)^{2} + b^{2})^{1/2}} − \frac{(a − s)^{2}}{((a − s)^{2} + b^{2})^{3/2}} )

= \frac{1}{v} \frac{b^{2}}{((a − s)^{2} + b^{2})^{3/2}} > 0.

Exercise 2.5.3

(i) Since

a ≤ (a^{2} + b^{2})^{1/2},

we have

\frac{1}{u} − \frac{1}{v} \frac{a}{(a^{2} + b^{2})^{1/2}} ≥ 0

whenever v ≥ u.

Since Supergirl can swim faster than she runs and a straight line is the path of shortest length, it is clear that she should dive in at once.

(ii) This is just the observation that

\frac{v}{u} = \frac{a − s_{0}}{((a − s_{0})^{2} + b^{2})^{1/2}}

and so the line joining (a, b) and (a − s_{0}, 0) is at angle θ_{0} with

sin θ_{0} = \frac{a − s_{0}}{((a − s_{0})^{2} + b^{2})^{1/2}} = \frac{v}{u}.

(iii) As s decreases, running along the bank for a fixed time becomes less effective in closing the distance to Superman.

Exercise 2.5.5

a is local maximum if and only if there is a u > 0 with b − a ≥ u such that f(a) ≥ f(s) for a ≤ s ≤ a + u or, equivalently, for all s with a ≤ s ≤ b, |s − a| ≤ u.

b is local maximum if and only if there is a u > 0 with b − a ≥ u such that f(b) ≥ f(s) for b − u ≤ s ≤ b or, equivalently, for all s with a ≤ s ≤ b, |s − b| ≤ u.

If a < t < b, then t is a local maximum if and only if we can find a u > 0 with u ≤ t − a, b − t such that f(s) ≤ f(t) whenever t − u ≤ s ≤ t + u or, equivalently, for all s with a ≤ s ≤ b, |s − t| ≤ u.

Exercise 2.5.6

(Parts (ii) and (iii) only.)

(ii) If f(s) ≥ f(t) for all t ∈ [a, b], then, automatically, f(s) ≥ f(t) for all t ∈ [a, b] and |t − s| ≤ 1.

(iii) We say that f has a local minimum at s with a ≤ s ≤ b if we can find a u > 0 such that, whenever a ≤ t ≤ b and |t − s| ≤ u, we have f(s) ≤ f(t).

Exercise 2.5.7

We give two arguments.

First argument

f(c + δt) = f(c) + f′(c)δt + o(δt).

If f′(c) < 0, then f(c + δt) < f(c) when δt is strictly positive and sufficiently small, so f does not have a minimum at c. If f′(c) > 0, then f(c + δt) < f(c) when δt is strictly negative and sufficiently small (in absolute value), so f does not have a minimum at c. Thus, if f attains a local minimum at c, we must have f′(c) = 0.

Second argument If f attains a local minimum at c, then −f attains a local maximum, so −f′(c) = (−f)′(c) = 0 and f′(c) = 0.

Exercise 2.5.8

We have

f(a + δt) = f(a) + f′(a)δt + o(δt),

so, if f′(a) > 0, then f(a + δt) > f(a) when δt is strictly positive and sufficiently small, and so f does not have a maximum at a. Thus, if f has a local maximum at a, we have f′(a) ≤ 0.

We have

f(b + δt) = f(b) + f′(b)δt + o(δt)

so, if f′(b) < 0, then f(b + δt) > f(b) when δt is strictly negative and sufficiently small, so f does not have a maximum at b. Thus, if f has a local maximum at b, we have f′(b) ≥ 0.

Exercise 2.5.9

This is just a repeat of Exercise 2.4.1. Observe that b^{3} − a^{3} = (b − a)(a^{2} + ab + b^{2}) = (b − a)((a + b/2)^{2} + 3b^{2}/4) > 0

if b > a, so f is everywhere increasing and can have neither a local maximum nor a local minimum.

However, f′(t) = 3t^{2}, so f′(0) = 0.

Exercise 2.5.10

If c is an interior point with f′(c) = 0 and we can find a u > 0 such that f′(t) ≤ 0 for c − u ≤ t ≤ c and f′(t) ≥ 0 for c ≤ t ≤ c + u, then we know that f(t) decreases as t runs from c − u to c and increases as t runs from c to c + u. Thus f attains a local minimum at c.

(We could obtain the same result by observing that f has a local minimum at c if and only if −f has a local maximum at c.)

Exercise 2.5.11

If we can find a u > 0 such that f′(t) ≤ 0 for a ≤ t ≤ a + u, then f is decreasing between a and a + u so f(a) ≥ f(t) for a ≤ t ≤ a + u and a is a local maximum.

If we can find a u > 0 such that f′(t) ≥ 0 for b − u ≤ t ≤ b, then f is increasing between b − u and b so f(b) ≥ f(t) for b − u ≤ t ≤ b and b is a local maximum.

Exercise 2.5.13

If j + k is odd and 1 ≤ j ≤ n, then f′(t) > 0 for x_{j−1} < t < x_{j}, so f(t) is strictly increasing as t runs from x_{j−1} to x_{j} and f′(t) < 0 for x_{j} < t < x_{j+1}, so f(t) is strictly decreasing as t runs from x_{j} to x_{j+1}. Thus f has a local maximum at x_{j} and no local maxima or minima at s with x_{j−1} < s < x_{j}. or with x_{j} < s < x_{j+1}.

If k = 1, then f′(t) < 0 for x_{0} < t < x_{1} so f(t) is strictly decreasing as t runs from x_{0} to x_{1}. Thus f has a local maximum at x_{0} and no local maxima or minima at s with x_{0} < s < x_{1}.

If k = 0, then f′(t) > 0 for x_{0} < t < x_{1}, so f(t) is strictly increasing as t runs from x_{0} to x_{1}. Thus f has a local minimum at x_{0} and no local maxima or minima at s with x_{0} < s < x_{1}.

The remaining cases are dealt with similarly.

Exercise 2.5.14

Observe that, by the statement proved in Exercise 2.5.13,

e_{j} = (−1)^{j+k+1},

so

\frac{1}{2}e_{0} + e_{1} + e_{2} + … + e_{n−1} + e_{n} + \frac{1}{2}e_{n+1}

= (−1)^{k+1}(\frac{1}{2}(−1)^{0} + (−1)^{1} + (−1)^{2} + … + (−1)^{n} + \frac{1}{2}(−1)^{n+1})

= \begin{cases}(−1)^{k+1}(\frac{1}{2} + 0 − \frac{1}{2}) = 0 & if n is even, \\ (−1)^{k+1}(\frac{1}{2} − 1 + \frac{1}{2}) = 0 & if n is odd. \end{cases}

Exercise 2.5.15

(i) f does not satisfy the conditions since f′ has infinitely many zeros. (ii) f does not satisfy the conditions since f′ does not change sign

as it passes through its zero.

Exercise 2.6.1

We have

f(s) = time running + time swimming

= \frac{distance run}{u} + \frac{distance swum}{v}

= \frac{1}{u}\sqrt{(a − s)^{2} + b^{2}} + \frac{1}{v}\sqrt{(c − s)^{2} + d^{2}}.

(ii) When |s| ≥ 2|a|, 2|c|, then |a − s|, |c − s| ≥ |s|/2 and so \sqrt{(a − s)^{2} + b^{2}}, \sqrt{(c − s)^{2} + d^{2}} ≥ |s|/2.

Exercise 2.6.2

If A(s) = a − s, then A′(s) = −1.

If B(s) = (a − s)^{2} = A(s)^{2}, then the function of a function rule gives B′(s) = 2A′(s)A(s) = −2(a − s).

If C(s) = (a − s)^{2} + b^{2} = B(s) + b^{2}, then the addition rule gives C′(s) = B′(s) + 0 = −2(a − s).

If D(s) = \sqrt{(a − s)^{2} + b^{2}} = \sqrt{C(s)}, then the function of a function rule gives

D′(s) = \frac{C′(s)}{2\sqrt{C(s)}} = −\frac{a − s}{\sqrt{(a − s)^{2} + b^{2}}} .

The result now follows.

Exercise 2.6.3

The time taken is

f(x) = \frac{1}{u}(\sqrt{(a − x)^{2} + b^{2}} + \sqrt{(p − x)^{2} + q^{2}}).

We observe that f(x) is large when |x| is large.

Now

f′(x) = \frac{−1}{u} (\frac{a − x}{\sqrt{(a − x)^{2} + b^{2}}} + \frac{p − x}{\sqrt{(p − x)^{2} + q^{2}}} ) .

We observe that f′(x) < 0 when x is large and negative and f′(x) > 0 when x is large and positive.

If

g(x) = \frac{a − x}{\sqrt{(a − x)^{2} + b^{2}}} ,

then, by the product rule,

g′(x) = \frac{−1}{\sqrt{(a − x)^{2} + b^{2}}} + \frac{(a − x)^{2}}{((a − x)^{2} + b^{2})^{3/2}} = \frac{−b^{2}}{((a − x)^{2} + b^{2})^{3/2}} ,

so

f′′(x) = \frac{1}{u} (\frac{b^{2}}{((a − x)^{2} + b^{2})^{3/2}} + \frac{q^{2}}{((p − x)^{2} + q^{2})^{3/2}} ) > 0.

Thus f′ is strictly increasing.

It follows that f′ has a unique zero at z say and this z gives a minimum. Let X = (z, 0).

Since

\frac{z − a}{\sqrt{(z − x)^{2} + b^{2}}} = \frac{p − z}{\sqrt{(p − z)^{2} + q^{2}}} ,

elementary trigonometry shows that that AX and XC make equal angles with the brook.

Exercise 2.6.4

If one side has length x, the adjacent sides have length 2a − x. Thus

f(x) = x(2a − x) = 2ax − x^{2}.

Since f′(x) = 2(a − x) we have f′(a) = 0, f′(x) > 0 for 0 ≤ x < a and f′(x) < 0 for a < x ≤ 2a. Thus f(x) has a global maximum at x = a and nowhere else.

In other words, among all rectangles of given perimeter, the squares have greatest area.

Remark Completing the square gives a non-calculus proof. Observe that

f(x) = x(2a − x) = −(a − x)^{2} + a^{2} ≤ a^{2}

with equality when x = a.

Exercise 2.6.5

In the notation of Example 1.2.6,

b = 2s − a − c,

so the area of the triangle is

\sqrt{s(s − a)(s − b)(s − c)} = \sqrt{s(s − c)(s − a)(a + c − s)}.

Thus the area is maximised when f(s) = (s−a)(a+c−s) is maximised. Since

f(a) = −a^{2} + (2s − c)a − s^{2} + sc,

we have f′(a) = −2a + (2s − c), f′(s − \frac{1}{2}c) = 0, f′(a) > 0 for 0 ≤ a < s − \frac{1}{2}c and f′(a) < 0 for s − \frac{1}{2}c < a ≤ 2s − c. Thus f(a) has a global maximum at a = s − \frac{1}{2}c (i.e. when a = b) and nowhere else. Thus, among the triangles with fixed perimeter 2s and one side of fixed length c, the triangles with the other two sides of equal length have greatest area.

Now consider such triangles and allow c to vary. The area of such triangles is

\sqrt{s(s − c)(\frac{1}{2}c)^{2}} = \frac{1}{2} \sqrt{s(s − c)c^{2}}.

Thus the area is maximised when g(c) = (s − c)c^{2} is maximised.

Using the product rule

g′(c) = −c^{2} + 2(s − c)c = (2s − 3c)c.

Thus g′(\frac{2}{3}s) = 0, g′(c) > 0 for 0 ≤ c < \frac{2}{3}s and g′(c) < 0 for \frac{2}{3}s < c ≤ s. Thus g(c) has a global minimum at c = \frac{2}{3}s (i.e. when a = b = c) and nowhere else.

Thus, among the triangles with fixed perimeter, the equilateral triangles have greatest area.

[Here again we could have used completing the square rather than differentiation.]

Exercise 2.6.6

(i) We have

f(x) = \frac{1}{(n − 1)^{n−1}} x(na − x)^{n−1}

and

f′(x) = \frac{1}{(n − 1)^{n−1}} ((na − x)^{n−1} − (n − 1)x(na − x)^{n−2})

= \frac{(na − x)^{n−2}}{(n − 1)^{n−1}} (na − x − (n − 1)x)

= n\frac{(na − x)^{n−2}}{(n − 1)^{n−1}} (a − x).

Thus f′(a) = 0, f′(x) > 0 for 0 ≤ x < a and f′(x) < 0 for a < x ≤ na. It follows that f(x) has a global maximum at x = a and nowhere else.

(ii) If n = 2, then f(x) is the area of a rectangle sides x, (2a − x) (i.e. perimeter 4a).

(iii) By (i) with n = 2, if z is fixed and x + y + z = 3a, x, y, z ≥ 0, then

xy ≤ (\frac{3a − z}{2} )^{2}

with equality only if x = y = (3a − z)/2.

Thus if x + y + z = 3a, x, y, z ≥ 0, (i) with n = 3, yields

xyz ≤ z (\frac{3a − z}{2} )^{2} ≤ a^{3}

with equality only if x = y = z = a.

(iv) Let P (n) be the proposition that, if the x_{j} ≥ 0 and

x_{1} + x_{2} + … + x_{n} = na,

then

x_{1}x_{2} … x_{n} ≤ a^{n}

with equality only if x_{1} = x_{2} = … = x_{n}. We know that P (2) is true.

If P (n − 1) is true, then if x_{j} ≥ 0 and

x_{1} + x_{2} + … + x_{n} = na,

the inductive hypothesis tells us that

x_{1}x_{2} … x_{n−1}x_{n} ≤ x_{n} (\frac{na − x_{n}}{n − 1} )^{n−1}

with equality only if x_{1} = x_{2} = … = x_{n−1} and part (i) tells us that

(\frac{na − x_{n}}{n − 1} )^{n−1} ≤ a^{n}

with equality only if x_{n} = a so

x_{1}x_{2} … x_{n−1}x_{n} ≤ a^{n}

with equality only if x_{1} = x_{2} = … = x_{n−1} = x_{n} = a Thus P (n) is true. The full result follows by induction.

(v) Setting x_{1} + x_{2} + … + x_{n} = na, we obtain

\frac{x_{1} + x_{2} + … + x_{n}}{n} ≤ (x_{1}x_{2} … x_{n})^{1/n}

with equality only if

x_{1} = x_{2} = … = x_{n}.

Exercise 3.1.1

If g(t) = 1 + t^{2}, then g′(t) = 2t. If G(t) = 1/g(t) the quotient rule gives

G′(t) = −\frac{g′(t)}{g(t)^{2}} = −\frac{2t}{(1 + t^{2})^{2}} .

Since F (t) = (−1/2)G(t), we have

F′(t) = \frac{t}{(1 + t^{2})^{2}} .

Exercise 3.1.3

If a ≥ 1, log a is the area under y = f(x) between x = 1 and x = a. If 0 < a ≤ 1, log a is minus the area under y = f(x) between x = a

and x = 1.

Exercise 3.1.4

(i) log 1 + log 1 = log 1, so log 1 = 0. log a + log a^{−1} = log aa^{−1} = log 1 = 0, so log a = − log(1/a). (ii) We have

log 1 = ∫_{1}^{1} \frac{1}{t} dt = 0

since ∫_{a}^{a} f(t) dt = 0 automatically.

Let f(x) = G(x) = 1/x in the ‘change of variable formula’ on page 19. Then

log(1/a) = ∫_{1}^{1/a} \frac{1}{t} dt = ∫_{G(1)}^{G(a)} f(y) dy

= ∫_{1}^{a} f(G(x))G′(x) dx = ∫_{1}^{a} x\frac{−1}{x^{2}} dx

= − ∫_{1}^{a} \frac{1}{x} dx = − log a.

Exercise 3.1.5

Note that log′ x = 1/x > 0, so log is strictly increasing.

(i) We have

log x = ∫_{1}^{x} \frac{1}{t} dt ≥ ∫_{1}^{x} \frac{1}{2} dt = \frac{x − 1}{2}.

(ii) We have

2^{−1} ≤ t^{−1}

for 1 ≤ t ≤ 2, so

\frac{1}{2} = ∫_{1}^{2} \frac{1}{2} dt ≤ ∫_{1}^{2} \frac{1}{t} dt = log 2

(iii) log 2^{n} = log 2 + log 2^{n−1} = … = n log 2 ≥ \frac{n}{2}, so log 2^{n} > K whenever n ≥ 2K.

(iv) log 2^{−n} = − log 2^{n} ≤ −K whenever n ≥ 2K.

Exercise 3.1.6

By previous results, we know that, given K, we can find an Q such that

log x ≥ K

when x ≥ Q.

Now we can find a L such that

log x ≥ Q

for x ≥ L.

Thus log log x ≥ log Q ≥ M when x ≥ L, Experiment shows that K = 10^{24} will do but K = 10^{23} will not. (In the next section we see that log log exp exp 5 = 5.)

Exercise 3.1.7

Observe that

f(x) = \frac{1}{2a}(log(a + x) − log(a − x))

so, using the function of a function rule,

f′(x) = \frac{1}{2a} (\frac{1}{a + x} + \frac{1}{a − x})

= \frac{1}{2a} × \frac{(a − x) + (a + x)}{a^{2} − x^{2}}

= \frac{1}{a^{2} − x^{2}} .

Exercise 3.2.1

(i) If a, b > 0, then

E(log a + log b) = E(log ab) = ab = E(log a)E(log b),

so, setting a = E(x), b = E(y), we have

E(x + y) = E(x)E(y).

(ii) E(0) = E(log 1) = 1, so E(x)E(−x) = E(x − x) = E(0) = 1 and 1/E(x) = E(−x).

(iii) There are lots of ways. Probably best to look at definition, but we could also note that E(x) = E(x/2)^{2} ≥ 0 and (by (ii)) E(x) ≠ 0.

(iv) E′(x) = E(x) > 0, so E is strictly increasing.

(v) If x > log a, then E(x) > E(log a) = a. If log b > x > 0, then b = E(log b) > log x.

Exercise 3.2.2

If f(x) = x^{−n} exp x, then, using the product rule,

f′(x) = −nx^{−n−1} exp x + x^{−n} exp x = x^{−n−1}(x − n) exp x > 0 when x > n, so f(x) = x^{−n} exp x is an increasing function of x for x ≥ n.

Exercise 3.2.3

Let

f(y) = ay^{2} log \frac{1}{y} = −ay^{2} log y.

Then, using the product rule,

f′(y) = −a (2y log y + \frac{y^{2}}{y} ) = −ay(2 log y + 1)

so f′(exp(−1/2)) = 0, f′(y) > 0 for 0 < y < exp(−1/2) and f′(y) < 0 for exp(−1/2) < y. Thus f(y) has a unique maximum at y = exp(−1/2).

Exercise 3.2.4

(i) Observe that

n log x = \underbrace{log x + log x + … + log x}_{n}

= log(\underbrace{x × x × … × x}_{n})

= log x^{n}

for n ≥ 1.

We have also

0 log x = 0 = log 1 = log x^{0}.

(ii) If n ≥ 0 then

(−n) log x = −(n log x) = − log x^{n} = log(1/x^{n}) = log x^{−n}. Thus n log x = log x^{n} for all integers n.

(iii) Using part (ii),

q log x^{p/q} = log(x^{p/q})^{q} = log x^{p} = p log x

and so

\frac{p}{q} log x = log x^{p/q}.

Thus

exp (\frac{p}{q} log x) = exp(log x^{p/q}) = x^{p/q}.

Exercise 3.2.5

(i) Observe that

x^{a+b} = exp ((a + b) log x) = exp(a log x + b log x)

= exp(a log x) exp(b log x) = x^{a}x^{b}.

(ii) Observe that

(xy)^{a} = exp (a log(xy)) = exp(a log x + a log y)

= exp(a log x) exp(a log y) = x^{a}y^{a}.

(iii) Observe that

log(x^{a})^{b} = b log x^{a} = ba log x = ab log x = log x^{ab},

so that

(x^{a})^{b} = exp log(x^{a})^{b} = exp log x^{ab} = x^{ab}.

(iv) e^{a} = exp(a log e) = exp a.

(v) If f(t) = t^{a}, then

f(t) = exp(a log t)

and, using the function of a function rule,

f′(t) = \frac{a}{t} exp(a log t) = at^{−1}t^{a} = at^{a−1}.

Exercise 3.2.6

Observe that, if x ≥ exp(exp(exp 4)) then (since log is an increasing function)

log(log(log x)) ≥ log(log(log(exp(exp(exp 4)))))

= log(log(exp(exp 4))) = log(exp 4) = 4

Similarly if x < exp(exp(exp 4)), then log log log x < 4 (or, if x is small, log log log x is not even defined).

My calculator gives

exp exp 4 > 5 × 10^{23}

so

exp exp exp 4 > exp(5 × 10^{23}) = e^{5×10^{23}}
= (e^{5})^{10^{23}} ≥ 100^{10^{23}} = 10^{2×10^{23}}.

Thus M has at least 2 × 10^{23} figures.

70 years contain roughly 2.2 × 10^{9} seconds. Thus M is far too large to be written down in a lifetime.

Exercise 3.2.7

We have g(t) = f(h(t)) with f(t) = exp t, h(t) = (log a)t, so g′(t) = h′(t)f′(h(t)) = (log a)f(h(t)) = (log a)a^{t}.

Exercise 3.2.8

If g(t) = t^{1/t}, then g(t) = f(h(t)) with f(t) = exp t, h(t) = t^{−1}(log t), so

g′(t) = h′(t)f′(h(t)) = (t^{−2} − t^{−2} log t)f(h(t)) = t^{−2}(1 − log t)t^{1/t}.

Thus g′(e) = 0, g′(y) > 0 for 0 < y < e and g′(y) < 0 for y > e. Thus g(t) is strictly increasing as t increases up to a maximum at t = e and thereafter is strictly decreasing.

Thus, if g(a) = g(b) with a < b, we have a < e < b. Since 2 < e < 3, we must have a ≤ 2, b ≥ 3.

If n^{m} = m^{n} with n > m, then m log n = n log m and g(n) = g(m). Thus m is a positive integer with m ≤ 2. By inspection m = 1 is not a solution, but m = 2 is a solution (2^{4} = 4^{2}). Thus there is exactly one solution.

Exercise 3.2.10

Let f(t) = log t and g(t) = t. We have

∫_{1}^{n} log t dt = ∫_{1}^{n} f(t)g′(t) dt

= [f(t)g(t)]_{1}^{n} − ∫_{1}^{n} f′(t)g(t) dt

= [t log t]_{1}^{n} − ∫_{1}^{n} \frac{1}{t} × t dt

= n log n − ∫_{1}^{n} 1 dt = n log n − (n − 1).

Exercise 3.2.11

(i) Observe that

∫_{1}^{n+1} log t dt = ∫_{1}^{n} log t dt + ∫_{n}^{n+1} log t dt

≤ ∫_{1}^{n} log t dt + ∫_{n}^{n+1} log(n + 1) dt = ∫_{1}^{n} log t dt + log(n + 1).

(ii) Recall that

log n! ≤ ∫_{1}^{n+1} log t dt ≤ log(n + 1)!

so

∫_{1}^{n} log t dt ≤ log n!

and, using (i),

∫_{1}^{n} log t dt ≤ log n! ≤ ∫_{1}^{n+1} log t dt ≤ ∫_{1}^{n} log t dt + log(n + 1).

Exercise 3.2.10 now gives (n log n) − (n − 1) ≤ log n! ≤ (n log n) − (n − 1) + log(n + 1).

(iii) I get

6908.76 ≤ log 1000! ≤ 6913.37.

(iv) Observe that, between 1 and 1000 inclusive, there are

200 integers divisible by 5
40 integers divisible by 5^{2}
8 integers divisible by 5^{3}
1 integer divisible by 5^{4}
0 integers divisible by 5^{5}.

Thus 1000! is divisible by 5^{N} with

N = 200 + 40 + 8 + 1 = 249,

but not by 5^{N+1}.

Between 1 and 1000 inclusive, there are 500 integers divisible by 2, so 1000! is divisible by 2^{500}. Thus 1000! is divisible by 10^{249}, but not by 10^{250}.

Thus 1000! ends in exactly 249 zeros.

(v) We have

\frac{log 1000!}{log 10} ≈ 2957

so 1000! has about 2957 digits.

(vi) Applying exp to the inequality in (ii) yields n^{n}e^{1−n} ≤ n! ≤ (n + 1)n^{n}e^{1−n}.

Exercise 3.2.12

(i) It is plausible that using the value of f at a mid-point rather than one of the end points will give a better idea of the behaviour of f on a given interval.

(ii) If

f(t) = 2 log r − ( log(r − t) + log(r + t)),

then

f′(t) = 0 − (−\frac{1}{r − t} + \frac{1}{r + t}) = \frac{1}{r − t} − \frac{1}{r + t}

and

f′′(t) = \frac{1}{(r − t)^{2}} + \frac{1}{(r + t)^{2}} ≥ 0

for 0 ≤ t ≤ 1/2.

Thus f′(t) is increasing as t increases from 0 to 1/2 so, since f′(0) = 0, f′(t) ≥ 0 for 0 ≤ t ≤ 1/2. Thus

0 ≤ f′(t) ≤ f′(1/2)

and, since f(0) = 0, the mean value inequality gives

0 ≤ f(t) ≤ \frac{1}{2} (\frac{1}{r − \frac{1}{2}} − \frac{1}{r + \frac{1}{2}} )

for 0 ≤ t ≤ 1/2.

(iii) Now

∫_{0}^{1/2} f(t) dt = ∫_{0}^{1/2} 2 log r − ( log(r − t) + log(r + t)) dt
= log r − ∫_{0}^{1/2} log(r − t) dt − ∫_{0}^{1/2} log(r + t) dt
= log r − ∫_{r−1/2}^{r} log t dt − ∫_{r}^{r+1/2} log t dt

= log r − ∫_{r−1/2}^{r+1/2} log t dt

and, by (ii),

0 ≤ ∫_{0}^{1/2} f(t) dt ≤ ∫_{0}^{1/2} \frac{1}{2} (\frac{1}{r − \frac{1}{2}} − \frac{1}{r + \frac{1}{2}} ) dt = \frac{1}{2} (\frac{1}{2r − 1} − \frac{1}{2r + 1}) .

Thus

0 ≤ log r − ∫_{r−1/2}^{r+1/2} log t dt ≤ \frac{1}{2} (\frac{1}{2r − 1} − \frac{1}{2r + 1})

(iv) Adding these inequalities yields

0 ≤ log n! − ∫_{1/2}^{n+1/2} log t dt ≤ \frac{1}{2}.

Now the argument of Exercise 3.2.10 shows that

∫_{1/2}^{n+1/2} log t dt = (n + \frac{1}{2}) log(n + \frac{1}{2}) − \frac{1}{2} log \frac{1}{2} − n,

so (n+\frac{1}{2}) log(n+\frac{1}{2})−\frac{1}{2} log \frac{1}{2}−n ≤ log n! ≤ (n+\frac{1}{2}) log(n+\frac{1}{2})−\frac{1}{2} log \frac{1}{2}−n+\frac{1}{2} and

2^{−1/2}e^{−n}(n + \frac{1}{2})^{(n+ \frac{1}{2} )} ≤ n! ≤ e^{1/2}2^{−1/2}e^{−n}(n + \frac{1}{2})^{(n+ \frac{1}{2} )}.

(v) The same arguments give

0 ≤ log n! − log m! − ∫_{m−1/2}^{n+1/2} log t dt ≤ \frac{1}{(m − 1/2)(m + 1/2)}

so

(n + \frac{1}{2}) log(n + \frac{1}{2}) − (m − \frac{1}{2}) log(m − \frac{1}{2}) + m − n

≤ log n! − log m!

≤ (n + \frac{1}{2}) log(n + \frac{1}{2}) − (m − \frac{1}{2}) log(m − \frac{1}{2}) + m − n + \frac{1}{m − \frac{1}{2}}

and m!e^{m−n}(n+\frac{1}{2})^{(n+ \frac{1}{2} )}(m−\frac{1}{2})^{−(m− \frac{1}{2} )} ≤ n! ≤ e^{1/(m− \frac{1}{2} )}m!e^{m−n}(n+\frac{1}{2})^{(n+ \frac{1}{2} )}(m−\frac{1}{2})^{−(m− \frac{1}{2} )}.

Exercise 3.2.13

(i) If 0 ≥ a > −1, then

(r − 1)^{a} ≥ t^{a} ≥ r^{a}

for r − 1 ≤ t ≤ r, so

∫_{r−1}^{r}(r − 1)^{a} dt ≥ ∫_{r−1}^{r} t^{a} dt ≥ ∫_{r−1}^{r} r^{a} dt

and so

(r − 1)^{a} ≥ ∫_{r−1}^{r} t^{a} dt ≥ r^{a}.

Summing these inequalities, we obtain

1^{a} + 2^{a} + … + (n − 1)^{a} ≥ ∫_{1}^{n} t^{a} dt ≥ 2^{a} + … + n^{a}

so

1^{a} + 2^{a} + … + (n − 1)^{a} ≥ \frac{n^{a+1} − 1}{a + 1} ≥ 2^{a} + … + n^{a}

and

|(1^{a} + 2^{a} + … + n^{a}) − \frac{n^{a+1}}{a + 1}| ≤ \frac{1}{a + 1}.

(ii) Similarly,

(r − 1)^{−1} ≥ ∫_{r−1}^{r} t^{−1} dt ≥ r^{−1}.

so

1^{−1} + 2^{−1} + … + (n − 1)^{−1} ≥ ∫_{1}^{n} t^{−1} dt ≥ 2^{−1} + … + n^{−1}

and

1^{−1} + 2^{−1} + … + (n − 1)^{−1} ≥ log n ≥ 2^{−1} + … + n^{−1}

whence

|(1^{−1} + 2^{−1} + … + n^{−1}) − log n| ≤ 1.

(iii) If a ≥ 0,

(r − 1)^{a} ≤ t^{a} ≤ r^{a}

for r − 1 ≤ t ≤ r so ∫_{r−1}^{r}(r − 1)^{a} dt ≤ ∫_{r−1}^{r} t^{a} dt ≤ ∫_{r−1}^{r} r^{a} dt

and so

(r − 1)^{a} ≤ ∫_{r−1}^{r} t^{a} dt ≤ r^{a}.

Summing these inequalities, we obtain

1^{a} + 2^{a} + … + (n − 1)^{a} ≤ ∫_{1}^{n} t^{a} dt ≤ 2^{a} + … + n^{a} so

1^{a} + 2^{a} + … + (n − 1)^{a} ≤ \frac{n^{a+1} − 1}{a + 1} ≤ 2^{a} + … + n^{a}

and

|(1^{a} + 2^{a} + … + n^{a}) − \frac{n^{a+1}}{a + 1}| ≤ n^{a}.

(iv) If −1 > a, the result of (i) remains true, but the inequality does not improve as we increase n.

Exercise 3.3.1

If an angle is θ in the A system, then it is k_{B}k_{A}^{−1}θ in the B system. Thus, taking θ = k_{A}t, we have

sin_{A} k_{A}t = sin_{B} k_{B}t.

Exercise 3.3.2

(i) We have an isosceles triangle with base angle T_{N} and side 1, so altitude cos_{R} T_{N}/2 and base 2 sin_{R} T_{N}/2, so area

\frac{1}{2} × 2 sin_{R}(T_{N}/2) cos_{R}(T_{N}/2) = \frac{1}{2} × sin_{R} T_{N}.

(ii) We have N equal angles adding up to 4 right angles so T_{N} = 4k_{R}/N and the total area

P_{N} = \frac{N}{2} × sin_{R}(4k_{R}/N).

(iii) We have

sin_{R} δt = sin′_{R}(0)δt + o(δt) = δt + o(δt).

so

P_{N} = \frac{N}{2} × (\frac{4k_{R}}{N} + o (\frac{4k_{R}}{N} )) = 2k_{R} + o(1)

(where o(1) represents a quantity which becomes arbitrarily small provided N is large enough).

Thus P_{N} gets arbitrarily close to 2k_{R} as N increases.

(iv) The area of our polygon becomes arbitrarily close to that of our circle i.e. arbitrarily close to π. Thus 2k_{R} = π.

Exercise 3.3.3

(i) We have already seen that X_{0}X_{1} has length

2 sin(T_{N}/2) = 2 sin(2k_{R}/N),

so the perimeter of the polygon is

2N sin(2k_{R}/N).

For large N, the perimeter is close to the length of the circle, so

2π ≈ 2N sin(2k_{R}/N) ≈ 2N × \frac{2k_{R}}{N} ≈ 4k_{R},

with the approximation improving as N gets larger. Thus

2π = 4k_{R}

and k_{R} = π/2.

(ii) There are several ways of seeing this. One is to place points X_{0}, X_{1}, … X_{N} at equal distances along the arc. Then

length arc ≈ N × length line segment X_{0}X_{1}
= N(2a sin (θ/(2N)) ≈ N(2a(θ/(2N)) = aθ

The approximation improves as N increases, so the length of the arc of the circle is aθ.

Exercise 3.3.4

We will need the following remark. If

f(x) = \sqrt{1 − x},

then f(0) = 1 and

f′(x) = −\frac{1}{2} × \frac{1}{\sqrt{1 − x}}

so f′(0) = −1/2 and

\sqrt{1 − δx} = f(δx) = f(0) + f′(0)δx = 1 − \frac{1}{2}δx

to first order.

Let O be the centre of the earth, A the top of the lighthouse, B the crow’s nest and X the point where the straight line AB touches the surface of the earth. Writing |AX| for the length of AX, we have (since ∠OXA is a right angle)

|OA|^{2} = |OX|^{2} + |XA|^{2}

and (taking R to be radius of the earth)

|AX| = \sqrt{(R + h)^{2} − R^{2}} = \sqrt{2Rh − h^{2}}

= 2^{1/2}R^{1/2}h^{1/2}\sqrt{1 − h/(2R)} = 2^{1/2}R^{1/2}h^{1/2}(1 − h/(4R) + o(h/R)).

Similarly

|BX| = 2^{1/2}R^{1/2}H^{1/2}(1 − h/(4r) + o(H/R))

so

|AB| = |AX| + |BX| = 2^{1/2}R^{1/2}(h^{1/2} + H^{1/2})

to the zeroth order and so (since h/R and H/R very small indeed) the distance required is 2^{1/2}R^{1/2}(h^{1/2} + H^{1/2}).

(ii) Observe that quadrupling the height of the towers only doubles the viewing distance (and so only halves the number of towers required) so it will be cheaper to build fairly short towers rather than very tall ones.

Exercise 3.3.5

(i) Let g(t) = sin t, h(t) = t + π/2 so that cos t = g(h(t)). By the function of a function rule,

cos′ t = h′(t)g′(h(t)) = cos(t + π/2) = − sin t.

(ii) We have tan t = sin t/ cos t, so, by the product and quotient rules,

tan′ t = \frac{sin′ t}{cos t} − \frac{sin t cos′ t}{(cos t)^{2}} = \frac{cos t}{cos t} − \frac{−(sin t)^{2}}{(cos t)^{2}}

= 1 + \frac{(sin t)^{2}}{(cos t)^{2}} = \frac{(cos t)^{2} + (sin t)^{2}}{(cos t)^{2}} = \frac{1}{(cos t)^{2}}

= (sec t)^{2}.

Since cot t = − tan(t + π/2), we have

cot′ t = −( sec(t + π/2))^{2} = −(cosec t)^{2}.

Since cosec t = 1/ sin t, the quotient rule gives

cosec′(t) = −\frac{sin′ t}{(sin t)^{2}} = −\frac{cos t}{(sin t)^{2}} = − cot t cosec t.

Since sec t = cosec(t + π/2), we have

sec′ t = − cot(t + π/2) cosec(t + π/2) = tan t sec t.

Exercise 3.3.6

The time taken is

f(θ) = time running + time swimming

= \frac{distance run}{u} + \frac{distance swum}{v}

= \frac{distance run}{u} + \frac{2a sin ((π − θ)/2)}{v}

= \frac{aθ}{u} + \frac{2a cos(θ/2)}{v}.

Now

f′(θ) = \frac{a}{u} − \frac{a}{v} sin θ/2

so f′(θ) decreases from \frac{a}{u} to \frac{a}{u} − \frac{a}{v} as θ goes from 0 to π.

If u ≤ v, then f′(θ) ≥ 0 for all 0 ≤ θ ≤ π and f increases as θ increases, so she should dive in at once.

If u > v, then there is a θ_{0} such that f′(θ) > 0 for 0 ≤ θ < θ_{0} and f′(θ) < 0 for θ_{0} < θ ≤ π. Thus f attains its global maximum at an end point. If u > πv, she should run all the way, if u < πv, she should swim all the way. If u = πv, she should do one or the other.

Exercise 3.3.7

(i) sin′ x = cos x > 0 for −π/2 < x < π/2, so sin x is strictly increasing as x runs from −π/2 to π/2.

Since −π/2 ≤ sin^{−1} y ≤ π/2,

cos(sin^{−1} y) ≥ 0

so, since

( cos(sin^{−1} y))^{2} = 1 − ( sin(sin^{−1} y))^{2} = 1 − y^{2},

we have

cos(sin^{−1} y) = \sqrt{1 − y^{2}}.

By the inverse function rule,

(sin^{−1})′(y) = \frac{1}{sin′(sin^{−1} y)} = \frac{1}{cos(sin^{−1} y)} = \frac{1}{\sqrt{1 − y^{2}}} .

(ii) cos′ x = − sin x < 0 for 0 < x < π, so cos x is a strictly decreasing as x runs from 0 to π.

Since 0 ≤ cos^{−1} y ≤ π/2, it follows that

sin(cos^{−1} y) ≥ 0

so, since

( sin(cos^{−1} y))^{2} = 1 − ( cos(cos^{−1} y))^{2} = 1 − y^{2},

we have

sin(cos^{−1} y) = \sqrt{1 − y^{2}}.

By the inverse function rule,

(cos^{−1})′(y) = \frac{1}{cos′(cos^{−1} y)} = \frac{−1}{sin(cos^{−1} y)} = \frac{−1}{\sqrt{1 − y^{2}}} .

(iii) tan′(x) = cosec^{2} x, so tan x is a strictly increasing as x runs from −π/2 to π/2.

We observe that

\frac{1}{1 + (tan x)^{2}} = \frac{(cos x)^{2}}{(cos x)^{2} + (sin x)^{2}} = (cos x)^{2}

so that

\frac{1}{1 + y^{2}} = \frac{1}{1 + ( tan(tan^{−1} y))^{2}} = ( cos(tan^{−1} y))^{2}.

By the inverse function rule,

(tan^{−1})′(y) = \frac{1}{tan′(tan^{−1} y)} = ( cos(tan^{−1} y))^{2} = \frac{1}{1 + y^{2}} .

Exercise 3.3.8

(Parts (i) to (iii) only.)

(i) We have

cosh(−x) = \frac{exp(−x) + exp(x)}{2} = cosh x

and

sinh(−x) = \frac{exp(−x) − exp(x)}{2} = − sinh x.

(ii) We have

cosh′(x) = \frac{exp(x) − exp(−x)}{2} = sinh x

and

sinh′(x) = \frac{exp(x) + exp(−x)}{2} = cosh x.

(iii) sinh′ x = cosh x ≥ 1/2 > 0 for all x, so sinh x is strictly increasing as x increases.

cosh′ x = sinh x > 0 for all x > 0, so cosh x is strictly increasing as x increases from 0.

Exercise 3.3.9

(i) If f(t) = m^{−1} sin mt, then f′(t) = cos mt. Thus

∫_{a}^{b} cos mt dt = ∫_{a}^{b} f′(t) dt = [f(t)]_{a}^{b} = \frac{sin mb − sin ma}{m}.

(ii) We have

cos(u + v) + cos(u − v) = cos u cos v − sin u sin v + cos u cos v + sin u sin v
= 2 cos u cos v.

(iii) We have, using (ii),

∫_{−π}^{π} cos nx cos mx dx = \frac{1}{2} (∫_{−π}^{π} cos(n + m)x dx + ∫_{−π}^{π} cos(n − m)x dx)

By (i),

∫_{−π}^{π} cos qt dt = \frac{sin(qπ) − sin(−qπ)}{q} = \frac{0 − 0}{q} = 0

whenever q is a non-zero integer.

Thus if n ≠ m,

∫_{−π}^{π} cos nx cos mx dx = 0.

If n = m ≠ 0,

∫_{−π}^{π} cos nx cos mx dx = \frac{1}{2} (∫_{−π}^{π} cos(2nx) dx + ∫_{−π}^{π} 1 dx) = \frac{1}{2}(0+2π) = π

and, if n = m = 0, ∫_{−π}^{π} cos nx cos mx dx = ∫_{−π}^{π} 1 dx = 2π.

Exercise 4.1.1

y(t) = −\frac{gt^{2}}{2} ,

so

\frac{y((N + 1)T ) − y(NT)}{y((M + 1)T ) − y(MT)} = \frac{((N + 1)T )^{2} − (NT)^{2}}{((M + 1)T )^{2} − (MT)^{2}}

= \frac{(N + 1)^{2} − N^{2}}{(M + 1)^{2} − M^{2}}

= \frac{2N + 1}{2M + 1}.

Exercise 4.1.2

(i) If

y′′(t) = P (t),

then, writing

z(t) = y′(t) − (a_{0}t + \frac{a_{1}}{2} t^{2} + … + \frac{a_{n}}{n + 1}t^{n+1}) ,

we have

z′(t) = 0,

so z(t) = c_{1} for some constant c_{1} and, writing

w(t) = y(t) − (c_{1}t + a_{0} \frac{t^{2}}{2 × 1} + a_{1} \frac{t^{3}}{3 × 2} + … + a_{n} \frac{t^{n+2}}{(n + 2) × (n + 1)}) ,

we have

w′(t) = z(t) − c_{1} = 0,

so w(t) = c_{0} for some constant c_{0} and

y(t) = c_{0} + c_{1}t + a_{0} \frac{t^{2}}{2 × 1} + a_{1} \frac{t^{3}}{3 × 2} + … + a_{n} \frac{t^{n+2}}{(n + 2) × (n + 1)}.

(ii) If

y′′′(t) = P (t),

then, by part (i),

y′(t) = c_{1} + 2c_{2}t + a_{0} \frac{t^{2}}{2 × 1} + a_{1} \frac{t^{3}}{3 × 2} + … + a_{n} \frac{t^{n+2}}{(n + 2) × (n + 1)},

for some constants c_{1}, c_{2}.

Thus writing

u(t) = y(t) − (c_{1}t + c_{2}t^{2} + a_{0} \frac{t^{3}}{3 × 2 × 1} +a_{1} \frac{t^{4}}{4 × 3 × 2}

+ … + a_{n} \frac{t^{n+3}}{(n + 3) × (n + 2) × (n + 1)})

we have

u′(t) = 0,

so u(t) = c_{0} for some constant c_{0} and

y(t) = c_{0} + c_{1}t + c_{2}t^{2} + a_{0} \frac{t^{3}}{3 × 2 × 1}

+ a_{1} \frac{t^{4}}{4 × 3 × 2} + … + a_{n} \frac{t^{n+3}}{(n + 3) × (n + 2) × (n + 1)}.

Exercise 4.1.3

tan θ = \frac{sin θ}{cos θ} = \frac{V sin θ}{V cos θ} = \frac{v}{u}.

Exercise 4.1.4

Completion of the square with a > 0 gives

−as^{2} + bs = −(a^{1/2}s − b/(2a^{1/2}))^{2} + b^{2}/(4a),

so

y = − (\frac{g^{1/2} sec θ}{2^{1/2}V}x − \frac{V sin θ}{2^{1/2}g^{1/2}} )^{2} + \frac{V^{2} sin^{2} θ}{2g}.

Since squares are positive, the maximum of −(a^{1/2}s − b/(2a^{1/2}))^{2} + b^{2}/(4a) occurs when

a^{1/2}s − b/(2a^{1/2}) = 0

(i.e. s = b/(2a)) and is b^{2}/(4a).

Thus the highest point on the trajectory is V^{2} sin^{2} /(2g) above its initial height and the projectile travelled a horizontal distance

x = tan θ × \frac{V^{2}}{g sec^{2} θ} = cos θ sin θ × \frac{V^{2}}{g} = \frac{V^{2}}{g} sin 2θ.

(This is, as we shall see, half the total range.)

If V sin θ < 0, we must be firing from the top of a cliff and the particle is at highest point when it is fired.

Exercise 4.1.5

y = Y − b = CX^{2} − b = C(x + a)^{2}.

If C ≠ 0, Y = CX^{2} is a parabola, so y = C(x + A)^{2} + B is a parabola with respect to different axes. Thus equation ⋆⋆ is the equation of a parabola.

Exercise 4.2.1

Set f(t) = A exp t, g(t) = −kt. If u(t) = f(g(t)), the function of a function rule yields

u′(t) = g′(t)f′(g(t)) = −ku(t).

If w(t) = (exp kt)u(t), the product rule gives w′(t) = k(exp kt)u(t) + (exp kt)u′(t) = (exp kt)(u′(t) + ku(t)) = 0.

Exercise 4.2.2

(i) Using the fundamental theorem of the calculus,

f′(t) = −a exp(−at)g(t) + exp(−at)g′(t) − exp(−at)h(t)
= exp(−at)(g′(t) − ag(t) − h(t)) = 0,

so f(t) = A a constant and so

A = exp(−at)g(t) − ∫_{0}^{t} exp(−ax)h(x) dx,

whence

g(t) = A exp at + (exp at) ∫_{0}^{t} exp(−ax)h(x) dx.

(ii) If h(t) = C, we have

g(t) = A exp at + exp at ∫_{0}^{t} C exp(−ax) dx

= A exp at + [−\frac{C}{a} exp(−ax)]_{0}^{t} exp at

= A exp at + \frac{C}{a} (1 − exp(−at)) exp at

= B exp at − \frac{C}{a}

for some constant B.

If a = 0, the fundamental theorem of the calculus tells us that

g(t) = Ct + A

for some constant A.

If h(t) = C exp bt for all t and a ≠ b, then

g(t) = A exp at + (exp at) ∫_{0}^{t} C exp(−ax) exp(bx) dx

= A exp at + exp at ∫_{0}^{t} C exp ((b − a)x) dx

= A exp at + [\frac{C}{b − a} exp ((b − a)x)]_{0}^{t} exp at

= A exp at + \frac{C}{b − a}( exp ((b − a)t) − 1) exp at

= B exp at + \frac{C}{b − a} exp bt

for some constant B.

If h(t) = C exp at for all t,

g(t) = A exp at + (exp at) ∫_{0}^{t} C exp(−ax) exp(ax) dx
= A exp at + exp at ∫_{0}^{t} C dx
= A exp at + Ct exp at

for some constant A.

(iii) We have u′(t)−bu(t) = (g′′(t)−ag′(t))−b(g′(t)−ag(t)) = g′′(t)+pg′(t)+qg(t) = 0. Thus

u(t) = C exp bt

for some constant C.

It follows that

g′(t) − ag(t) = C exp bt

and, by (ii),

g(t) = A exp at + \frac{C}{b − a} exp bt = A exp at + B exp bt

for some constants A and B.

(iv) We have u′(t)−au(t) = (g′′(t)−ag′(t))−a(g′(t)−ag(t)) = g′′(t)−2ag′(t)+a^{2}g(t) = 0. Thus

u(t) = B exp at

for some constant B, so

g′(t) − ag(t) = B exp at

and

g(t) = (A + Bt) exp at

for some constant A.

(v) We have u′(t)−bu(t) = (g′′(t)−ag′(t))−b(g′(t)−ag(t)) = g′′(t)+pg′(t)+qg(t) = 1. Thus

u(t) = C exp bt − \frac{1}{b}

for some constant C, so

g′(t) − ag(t) = C exp bt − \frac{1}{b}.

Thus

g(t) = A exp at + exp at ∫_{0}^{t} exp(−ax)(C exp bx − b^{−1}) dx

= A exp at + B exp bt + \frac{1}{ab}

= A exp at + B exp bt + \frac{1}{q}

for some constants A and B.

(vi) Let u(t) = g′(t) − ag(t) and v(t) = u′(t) − bu(t). Then

v′(t) − cv(t) = (u′′(t) − bu′(t)) − c(u′(t) − bu(t))

= (g′′′(t) − ag′′(t)) − b(g′′(t) − ag′(t)) − c(g′′(t) − ag′(t)) + cb(g′(t) − ag(t)) = g′′′(t) + pg′′(t) + qg′(t) + rg(t) = 0. Thus v(t) = A_{1} exp(ct), so u(t) = B_{1} exp(bt) + B_{2} exp(ct) and, using

the formula of (i) again,

g(t) = C_{1} exp(at) + C_{2} exp(bt) + C_{3} exp(ct)

where A_{1}, B_{i}, C_{j} are constants.

Exercise 4.3.1

Whatever f we choose, f′(x)^{2} ≥ 0 > −1 − x^{2}.

Exercise 4.3.2

If u(t) = 0, then

u′(t) = 0 = 3 × 0^{2/3} = 3 × u(t)^{2/3}.

If v(t) = t^{3}, then v′(t) = 3t^{2} = 3 × v(t)^{2/3}.

Exercise 4.3.3

Observe that, if v(t) > V, then v′(t) < 0. Thus, if v(S) ≤ V + b, we have v(t) ≤ V + b for all t ≥ S and, if v(T) ≥ V + b, we have v(t) ≥ V + b for all t ≤ T.

Suppose that v(T) ≥ V + b. Then v(t) ≥ V + b, h(v(t)) ≥ h(V + b) and

v′(t) = g − h(v(t)) ≤ g − h(V + b) ≤ g − (g + a) = −a

for 0 ≤ t ≤ T. By the mean value inequality, this gives

v(0) − v(T) ≤ −aT

and

T ≤ (v(0) − V)/a.

Thus

v(t) ≤ V + b

for t ≥ (v(0) − V)/a.

Exercise 4.3.4

(i) Using Exercise 4.2.2, we have

x(t) = −A_{1} exp(−kt) + A_{2}
y(t) = −B_{1} exp(−kt) + B_{2} − gt/k

with A_{j}, B_{j} constants. We have

x′(t) = kA_{1} exp(−kt)
y′(t) = kB_{1} exp(−kt) − g/k.

Thus, taking t = 0,

A_{1} = u_{0}/k, B_{1} = (v_{0} + g/k)/k,
A_{2} = −A_{1} = −u_{0}/k, B_{2} = −B_{1} = −(v_{0} + g/k)/k.

We have

x(t) = \frac{u_{0}}{k} (1 − exp(−kt))
y(t) = \frac{1}{k} (v_{0} + \frac{g}{k} ) (1 − exp(−kt)) − \frac{gt}{k}.

(ii) When t is large, exp(−kt) ≈ 0 so, since

x′(t) = −u_{0} exp(−kt),
y′(t) = − (v_{0} + \frac{g}{k} ) exp(−kt) − \frac{g}{k} ,

we have

x′(t) ≈ 0
y′(t) ≈ −\frac{g}{k}

when t is large.

Observing that exp(−kt) > 0 for all t, we see that x(t) ≈ u_{0}/k for t large, but x(t) < u_{0}/k whenever t ≥ 0.

(iii) We have

y(t) = \frac{1}{k} (v_{0} + \frac{g}{k} ) (1 − exp(−kt)) − \frac{gt}{k} = (v_{0} + \frac{g}{k} ) \frac{x(t)}{u_{0}} − \frac{gt}{k}.

Now

exp(−kt) = \frac{kx(t)}{u_{0}} − 1,

so

−kt = log \frac{kx(t) − u_{0}}{u_{0}}

and

t = \frac{1}{k} log \frac{u_{0}}{u_{0} − kx(t)}.

Thus the equation of the path is

y = (\frac{c + v_{0}}{u_{0}} ) x − \frac{c}{k} log (\frac{u_{0}}{u_{0} − kx}) ,

where c = g/k.

Exercise 4.3.5

To first order, the weight of air in the column at height between x and x + δx is

−W′(x)δx = W (x) − W (x + δx) = KP (x)δx

and

W′(x) = −KP (x),

whence

P′(x) = −cP (x)

for some constants K and c.

We thus have

P (x) = A exp(−cx)

for some constant A and so P (x) = P (0) exp(−cx).

Exercise 5.1.2

x log (1 + \frac{a}{x}) = x log (\frac{a + x}{x}) = x(log(a + x) − log x).

Exercise 5.1.3

By the product rule,

f′(x) = (log(a + x) − log x) + x (\frac{1}{a + x} − \frac{1}{x})

= log(a + x) − log x + x (\frac{x − (a + x)}{x(a + x)})

= log(a + x) − log x − \frac{a}{a + x}.

Thus

f′′(x) = \frac{1}{a + x} − \frac{1}{x} + \frac{a}{(a + x)^{2}}

= \frac{x(a + x) − (a + x)^{2} + ax}{x(a + x)^{2}}

= \frac{−a(a + x) + ax}{x(a + x)^{2}}

= −\frac{a^{2}}{x(a + x)^{2}} < 0.

Exercise 5.1.4

We shall see that the argument we have given for a > 0 will, in fact, work for 0 > a. (However, we need a condition on x to make sure that 1 + a/x > 0.)

More specifically, we shall show that

g(x) = (1 + \frac{a}{x})^{x}

increases as x increases for x > max{0, −a}. This gives the result required.

Observe that it is sufficient to show that

f(x) = log g(x) = x log (1 + \frac{a}{x}) = x( log(x + a) − log x)

increases as x increases for x > max{0, −a}.

Now

f′(x) = (log(x + a) − log x) + x (\frac{1}{x + a} − \frac{1}{x})

= (log(x + a) − log x) − \frac{a}{x + a}

so

f′′(x) = \frac{1}{x + a} − \frac{1}{x} + \frac{a}{(a + x)^{2}}

= −\frac{a}{(a + x)x} + \frac{a}{(a + x)^{2}}

= −\frac{a^{2}}{(a + x)^{2}x} > 0

for x > max{0, −a}.

Since f′′(x) < 0 for x > max{0, −a}, we know that f′ is decreasing. Thus, if f′(x_{0}) = −u_{0} < 0 for some x_{0} > max{0, −a}, we will have

log (1 + \frac{a}{x}) − \frac{a}{a + x} = f′(x) ≤ −u_{0}

whenever x ≥ x_{0} and this is clearly false when x is very large. Thus f′(x) ≥ 0 whenever x > max{0, −a} and f(x) is increasing for x > max{0, −a}, as we hoped.

Exercise 5.1.5

We have

f(0) = log 1 + 0 = 0

and

f′(x) = \frac{1}{1 + x} − 1 = \frac{−x}{1 + x},

so

|f′(x)| ≤ \frac{|x|}{1/2} = 2|x|

for |x| ≤ 1/2, whence, by the mean value inequality,

|f(h)| = |f(h) − f(0)| ≤ 2|h||h| = 2h^{2}

for |h| ≤ 1/2.

Exercise 5.1.6

Since

0 < exp t = exp′ t = exp t ≤ exp(a + 1)

for a − 1 ≤ t ≤ a + 1, the mean value inequality gives

| exp(a + k) − exp a| ≤ exp(a + 1)|k|

for |k| ≤ 1.

Exercise 5.1.7

(i) The advantage of compounding is not large enough for the ordinary saver to press for it. The institution which offers the saving system prefers to keep its costs down and to tie the saver to a longer period.

For high rates of interest, compounding is much more advantageous. Institutions which lend at high interest generally have the whip hand over their borrowers and take every advantage of this.

(ii) We need to invest for a period of N years, where

(1 + c/100)^{N} ≈ 2

i.e.

N log(1 + c/100) ≈ log 2

i.e.

N ≈ \frac{log 2}{log(1 + c/100)}.

If c is small, then

log(1 + c/100) ≈ log′(1)c/100 = c/100

so

N ≈ \frac{100 log 2}{c} = \frac{69.3}{c}.

(iii) About 4.7 × 10^{25} by my calculations. But the question requires the institution to survive 2000 years, to be able to find the appropriate investments for 2000 years, not to spend the money and to keep other institutions’ hands off it. (The problem of inflation can be considered under the previous heads.)

Exercise 5.2.2

Set f(x) = (exp x − 1)/x.

We have

f′(x) = \frac{exp x}{x} − \frac{exp x − 1}{x^{2}} = \frac{x exp x − exp x + 1}{x^{2}} = \frac{g(x)}{x^{2}},

where

g(x) = x exp x − exp x + 1.

Now

g′(x) = x exp x > 0,

so g is strictly increasing for x > 0. Since g(0) = 0, we have g(x) > 0 for x > 0, so f′(x) > 0 for x > 0 and f is strictly increasing.

It follows that

\frac{exp kx − 1}{x} = k^{−1}f(x)

is strictly increasing and so f is increasing.

(ii) We have

L(b) = \frac{w}{T}f(T b) + c exp bT

so L is the sum of two strictly increasing functions and so strictly increasing.

If interest rates are higher, projects incur higher borrowing costs and must be more expensive.

Exercise 5.2.3

We have, whilst we are in debt, to first order,

h(t + δt) = h(t) − uδt + bh(t)δt

so, to first order,

h(t + δt) = h(t) + (bh(t) − u)δt

and

h′(t) = bh(t) − u.

Thus

h(t) = A exp(bt) + \frac{u}{b}

for some constant A. Taking t = 0, we see that A = L − u/b and

h(t) = (L − \frac{u}{b}) exp(bt) + \frac{u}{b}

whilst we are in debt. We pay of our debt at time x given by h(x) = 0, i.e.

(L − \frac{u}{b}) exp(bx) = −\frac{u}{b}

i.e.

exp(bx) = \frac{u}{u − bL},

so at time

\frac{1}{b} log (\frac{u}{u − bL}) .

Exercise 5.2.4

(i) We have

( exp(bT/2) − 1)^{2} = exp(bT) − 2 exp(bT/2) + 1,

so

(c+w/b)( exp(bT)−2 exp(bT/2))+ \frac{w}{b} = (c+w/b)( exp(bT/2)−1)^{2} −c

and our inequation becomes

( exp(bT/2) − 1)^{2} > \frac{c}{c + w/b} = \frac{cb}{cb + w}.

(ii) If T > 0,

( exp(bT/2) − 1)^{2} > 0,

so, if c = 0, we will always be better off starting from both ends.

The project takes half the time (so reducing interest charges) and the cost, not including interest, is unchanged.

(iii) If 4T > \frac{2}{b} log 2,

( exp(bT/2) − 1)^{2} > 1^{2} = 1 = \frac{cb}{cb} > \frac{cb}{cb + w},

so we should always dig from both ends.

If interest rates are very high it is cheaper to incur the cost and associated interest on two base camps, two tunnelling machines and two of everything else for half the time than on one for the whole time.

Exercise 5.3.1

\begin{matrix}1 & 2 & ∗ & ∗ \\ 1 & ∗ & 2 & ∗ \\ 1 & ∗ & ∗ & 2 \\ ∗ & 1 & 2 & ∗ \\ ∗ & 1 & ∗ & 2 \\ ∗ & ∗ & 1 & 2 \\ 2 & 1 & ∗ & ∗ \\ 2 & ∗ & 1 & ∗ \\ 2 & ∗ & ∗ & 1 \\ ∗ & 2 & 1 & ∗ \\ ∗ & 2 & ∗ & 1 \\ ∗ & ∗ & 2 & 1 \end{matrix}

and, if we cannot distinguish 1 and 2,

\begin{matrix}B & B & ∗ & ∗ \\ B & ∗ & B & ∗ \\ B & ∗ & ∗ & B \\ ∗ & B & B & ∗ \\ ∗ & B & ∗ & B \\ ∗ & ∗ & B & B \end{matrix}

Exercise 5.3.2

q_{0} = (1 − p)^{n} = 1 × 1 × (1 − p)^{n} = \binom{n}{0}p^{0}(1 − p)^{n−0}.

Exercise 5.3.3

The number ways in which we can place r indistinguishable balls in n holes is the same as the number of ways in which we can take r indistinguishable things from n places (taking at most one thing from any particular place). Thus, if we multiply out (x + y)^{n} in full, the term x^{n−r}y^{r} obtained by choosing y from r of the bracketed terms and

x from the remainder occurs \binom{n}{r} times. Thus

(x + y)^{n} = \binom{n}{0}x^{n} + \binom{n}{1}x^{n−1}y + \binom{n}{2}x^{n−2}y^{2} + … + \binom{n}{n}y^{n}.

Taking x = 1, we obtain

(1 + y)^{n} = \binom{n}{0} + \binom{n}{1}y + \binom{n}{2}y^{2} + … + \binom{n}{n}y^{n}.

Taking x = y = 1, we obtain

\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + … + \binom{n}{n} = 2^{n}

Taking x = 1, y = −1, we obtain

\binom{n}{0} − \binom{n}{1} + \binom{n}{2} − … + (−1)^{n}\binom{n}{n} = 0.

Exercise 5.3.4

(1 + x)^{n} = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^{2} + … + \binom{n}{n}x^{n},

so, differentiating,

n(1 + x)^{n−1} = 1 × \binom{n}{1} + 2 × \binom{n}{2}x + … + n × \binom{n}{n}x^{n−1}.

If p ≠ 1 and we set x = p/(1 − p), we get

n\frac{1}{(1 − p)^{n−1}} = 1 × \binom{n}{1} + 2 × \binom{n}{2}\frac{p}{(1 − p)} + … + n × \binom{n}{n}\frac{p^{n−1}}{(1 − p)^{n−1}} .

Multiplying both sides by p(1 − p)^{n−1} yields

np = 1 × \binom{n}{1}p(1 − p)^{n−1}

+ 2 × \binom{n}{2}p^{2}(1 − p)^{n−2} + … + n × \binom{n}{n}p^{n}

= E,

as required.

If p = 1 the result is immediate from the definition.

Exercise 6.1.1

We have

f(x) = sin x + 10^{−20} sin 10^{12}x ≈ sin x

and

f′(x) = cos x + 10^{−8} cos 10^{12}x ≈ cos x,

so both f and f′ can graphed on standard scales, but

f′′(x) = − sin x − 10^{2} sin 10^{12}x ≈ −10^{2} sin 10^{12}x,

so the scale for the y axis must be chosen so as to show y with |y| ≤ 10^{2} +1 and the scale for the x axis so that points a distance 10^{−13} apart are readily distinguishable.

Exercise 6.2.1

If u(t) = −t, then the function of a function rule gives

h′(t) = (g ∘ u)′(t) = u′(t)(g′ ∘ u)(t) = −g′(−t)

Repeating the argument (or reusing the result) r times gives h^{(r)}(t) = (−1)^{r}g^{(r)}(−t).

Thus, if |g^{(n)}(t)| ≤ M for −a ≤ t ≤ 0, we have |h^{(n)}(t)| ≤ M for 0 ≤ t ≤ a and

h(0) = h′(0) = h′′(0) = … = h^{(n−1)}(0) = 0,

so

|h(t)| ≤ M \frac{t^{n}}{n!}

whenever 0 ≤ t ≤ a. It follows that

|g(t)| ≤ M \frac{|t|^{n}}{n!}

whenever −a ≤ t ≤ 0.

Exercise 6.2.2

g^{(r)}(t) = f^{(r)}(t) − f^{(r)}(0) − \frac{f^{(r+1)}(0)(r + 1) × r × … × 2}{r!}t − …

− \frac{f^{(n−1)}(0)(n − 1) × (n − 2) … × (n − r − 1)}{(n − 1)!}t^{n−r},

for 0 ≤ r ≤ n − 1 and

g^{(n)}(t) = f^{(n)}(t),

so

g(0) = g′(0) = g′′(0) = … = g^{(n−1)}(0) = 0

and |g^{(n)}(t)| ≤ M whenever |t| ≤ a.

Exercise 6.2.3

This is just a translation.

Set f(t) = F (t − y), so f^{(r)}(t) = F^{(r)}(t − y). Then |f^{(n)}(s)| ≤ M whenever |s| ≤ a, so

|f(s) − f(0) − \frac{f′(0)}{1!} s − \frac{f′′(0)}{2!}s^{2} − … − \frac{f^{(n−1)}(0)}{(n − 1)!}s^{n−1}|

≤ M \frac{|s|^{n}}{n!} .

whenever |s| ≤ a and

|F (t) − F (y) − \frac{F′(y)}{1!} (t − y) − \frac{F′′(y)}{2!}(t − y)^{2} − … − \frac{F^{(n−1)}(t)}{(n − 1)!}(t − y)^{n−1}|

≤ M \frac{|t − y|^{n}}{n!} .

whenever |t − y| ≤ a.

Exercise 6.2.4

If

g(h) = a_{0} + a_{1}h + a_{2}h^{2} + … + a_{n−1}h^{n−1} + o(h^{n−1})

and

g(h) = b_{0} + b_{1}h + b_{2}h^{2} + … + b_{n−1}h^{n−1} + o(h^{n−1}),

then

0 = g(h) − g(h) = u_{0} + u_{1}h + u_{2}h^{2} + … + u_{n−1}h^{n−1} + o(h^{n−1}) where u_{j} = a_{j} − b_{j}.

We need to prove that u_{j} = 0 for 0 ≤ j ≤ n − 1. But, if u_{0} = u_{1} = … = u_{r−1} = 0, we have

0 = u_{r}h^{r} + u_{r+1}h^{r+1} … + u_{n−1}h^{n−1} + o(h^{n−1}),

so

0 = u_{r} + u_{r+1}h … + u_{n−1}h^{n−r} + o(h^{n−r}),

whence

0 = u_{r} + o(h)

and u_{r} = 0. By repeating this argument n − 2 times, we get

u_{0} = u_{1} = … = u_{n−2} = 0

and

0 = u_{n−1}(h^{n−1}) + o(h^{n−1}),

so u_{n−1} = 0 and we are done.

To obtain the last part set c_{j} = a_{j} and b_{j} = F^{(j)}(y)/j!.

Exercise 6.2.5

If

f(a + h) = f(a) + Ah^{2} + o(h^{2})

and A < 0, then the o notation tells us that we can find a u > 0 such that

|f(a + h) − (f(a) + Ah^{2})| ≤ \frac{−A}{2}h^{2}

whenever |h| ≤ u. We thus have

f(a + h) − (f(a) + Ah^{2}) ≤ −\frac{A}{2}h^{2},

so

f(a + h) ≤ f(a) + \frac{A}{2}h^{2} = f(a) − \frac{|A|}{2}h^{2}

whenever |h| ≤ u. Thus f attains a local maximum at a.

Alternatively Look at −f.

Exercise 6.2.6

Let us suppose first that f^{(r)}(a) > 0. By the local Taylor’s theorem, we can find a u > 0 such that

|f(a + h) − (f(a) + \frac{f^{(r)}(a)}{r!}h^{r}))| ≤ \frac{f^{(r)}(a)}{2 × r!}|h|^{r}  ⋆

whenever |h| ≤ u.

If r is even, ⋆ tells us that

f(a + h) ≥ f(a) + \frac{f^{(r)}(a)}{2 × r!}|h|^{r}

whenever |h| ≤ u, so f attains a local minimum at a.

If r is odd, ⋆ tells us that

f(a + h) ≥ f(a) + \frac{f^{(r)}(a)}{2 × r!}|h|^{r}

for 0 ≤ h ≤ u, but

f(a + h) ≤ f(a) − \frac{f^{(r)}(a)}{2 × r!}|h|^{r}

for −u ≤ h ≤ 0, so we have neither a maximum nor a minimum.

Now suppose f^{r}(a) < 0. By looking at −f and applying our previous results we see that:-

If r is even, f attains a local minimum at a. If r is odd, we have neither a maximum nor a minimum.

Exercise 6.3.1

Observe that E^{(r)}(t) = E(t) and so

E^{(r)}(0) = E(0) = 1.

Now substitute into the formula of Taylor’s theorem.

Exercise 6.3.2

Since E is increasing,

0 < E^{(n)}(t) = E(t) ≤ E(0) = 1

for all t ≤ 0.

Exercise 6.3.3

Our error estimate shows that

| exp(1) − 2.71806| ≤ 0.006,

so exp(1) ≤ 2.8 and the error is less than

\frac{exp(1)}{6!} ≤ \frac{2.8}{720} ≤ 0.0039.

Exercise 6.3.4

(i) We have

u_{0} = 1
u_{1} = 1
u_{2} = 0.5
u_{3} ≈ 0.166666667
u_{4} ≈ 0.041666667
u_{5} ≈ 0.008333333
u_{6} ≈ 0.001388889
u_{7} ≈ 0.000198413
u_{8} ≈ 0.000024802
u_{9} ≈ 0.000002756
u_{10} ≈ 0.000000276
2.8 × u_{11} ≈ 0.000000070

giving

exp(1) ≈ u_{0} + u_{1} + u_{2} + … + u_{10} ≈ 2.71828180

with an error of less than about 0.00000007.

(ii) We take u_{j} = (10)^{−j}/j! giving

u_{0} = 1
u_{1} = 0.1
u_{2} = .05
u_{3} ≈ 0.001666667
u_{4} ≈ 0.000041667
u_{5} ≈ 0.000000833
2.8 × u_{6} ≈ 0.000000039

giving

exp(1/10) ≈ u_{0} + u_{1} + u_{2} + … + u_{5} ≈ 1.10517092

with an error of about 0.00000004.

(iii) Note that the error will now be no greater than

10^{−6}/6! ≈ 0.000000001.

We take u_{j} as in (ii).

exp(−1/10) ≈ u_{0} − u_{1} + u_{2} − … − u_{5} ≈ 0.90483742 correct to the number of figures given.

Exercise 6.3.5

By Taylor’s theorem,

|(1 + \frac{a}{N} + \frac{a^{2}}{2N^{2}} ) − exp(a/N)| ≤ \frac{exp 1}{3!} |\frac{a}{N} |^{3}

provided only that N > a.

Since log′(t) = 1/t, the mean value inequality tells us that

| log x − log y| ≤ 2|x − y|

whenever |x − 1|, |y − 1| ≤ 1/2. Thus

|log (1 + \frac{a}{N} + \frac{a^{2}}{2N^{2}} ) − \frac{a}{N} | ≤ K_{1} |\frac{a}{N} |^{3}

(where K_{1} is some constant) provided only that N is large enough. Thus

|log (1 + \frac{a}{N} + \frac{a^{2}}{2N^{2}} )^{N} − a| ≤ K_{1}a |\frac{a}{N} |^{2} ≤ K_{1}b |\frac{a}{N} |^{2}

By the mean value theorem

| exp x − exp a| ≤ exp(|a| + 1)|x − a|

for |x − a| ≤ 1. Thus, taking

x = log (1 + \frac{a}{N} + \frac{a^{2}}{2N^{2}} )

we see that there exists there exists an L (depending on a) with

|(1 + \frac{a}{N} + \frac{a^{2}}{2N^{2}} )^{N} − exp a| ≤ \frac{L}{N^{2}} ,

provided that N is large enough.

Exercise 6.3.6⋆

Exercise 6.3.8

Let u_{j} = 10^{j}/j!. Working to two decimal places.

u_{0} = 1
u_{1} = 10
u_{2} = 50
u_{3} = 166.67
u_{4} = 416.67
u_{5} = 833.33
u_{6} = 1388.89
u_{7} = 1984.13
u_{8} = 2480.16
u_{9} = 2755.73
u_{10} = 2755.73
u_{11} = 2505.21
u_{12} = 2087.68
u_{13} = 1605.90
u_{14} = 1147.07
u_{15} = 764.72
u_{16} = 477.95
u_{17} = 281.15
u_{18} = 156.19
u_{19} = 82.20
u_{20} = 41.10

Thus

exp 10 ≈ u_{0} + u_{1} = … + u_{20} ≈ 21991.48

with an error of at most about 41.1 Thus

22050 > exp 10 > 21950.

(If the reader feels that we should be more careful about the lower estimate she can observe that exp 10 ≥ u_{0}+u_{1}+…+u_{20}+u_{21} > 22000.)

Exercise 6.3.10

If we write u_{r} = E^{(r)}(0)(−100)^{r}/r! = (−100)^{r}/r! then, using Stirling’s inequality from Exercise 3.2.11, we have

100! ≤ 101 × 100^{100}e^{−99},

so

|u_{100}| ≥ \frac{e^{99}}{101} > 10^{40}

and working to 32 significant figures will allow errors of size 10^{8} and so certainly tell us nothing about exp(−100).

Exercise 6.3.11

exp 1 ≈ 2.718282
exp 2 ≈ 7.389056
exp 4 ≈ 5.459815 × 10^{1}
exp 8 ≈ 2.980958 × 10^{3}
exp 16 ≈ 8.886111 × 10^{6}
exp 32 ≈ 7.896296 × 10^{13}
exp 64 ≈ 6.235149 × 10^{27}

so

exp 100 = exp(64) × exp(32) × exp(4)

≈ 6.235149 × 7.896296 × 5.459815 × 10^{41} ≈ 2.688117 × 10^{2} × 10^{41} = 2.688117 × 10^{43},

so exp(−100) = 1/(exp 100) ≈ (1/2.688117) × 10^{−43} ≈ 3.720076 × 10^{−44} and exp(−100) = 3.270 × 10^{−44} to four significant figures.

Exercise 6.3.12

(i) We have

cos^{(4r)} t = cos t, cos^{(4r+1)} t = − sin t, cos^{(4r+2)} t = − cos t, cos^{(4r+3)} t = sin t.

Thus | cos^{(k)}(t)| ≤ 1 for all t and k whilst

cos^{(4r)} 0 = 1, cos^{(4r+1)} 0 = 0, cos^{(4r+2)} 0 = −1, cos^{(4r+3)} t = 0.

Thus

|cos t − (1 − \frac{t^{2}}{2!} + \frac{t^{4}}{4!} − … + (−1)^{r} \frac{t^{2r}}{(2r)!})| ≤ min \{\frac{|t|^{2r+1}}{(2r + 1)!}, \frac{|t|^{2r+2}}{(2r + 2)!}\}

Similarly,

sin^{(4r)} t = sin t, sin^{(4r+1)} t = cos t, sin^{(4r+2)} t = − sin t, sin^{(4r+3)} t = − cos t

and

|sin t − (t − \frac{t^{3}}{3!} − … + (−1)^{r} \frac{t^{2r+1}}{(2r + 1)!})| ≤ min \{\frac{|t|^{2r+2}}{(2r + 2)!}, \frac{|t|^{2r+3}}{(2r + 3)!}\} .

(ii) In particular,

|sin \frac{1}{10} − (10^{−1} − \frac{10^{−3}}{3!} )| ≤ \frac{10^{−5}}{5!} ≤ 10^{−7}

so, since

10^{−1} − \frac{10^{−3}}{3!} ≈ .0099833,

sin(1/10) = .009983

to six decimal places.

(iii) Since

\frac{|t|^{2r+4}}{(2r + 4)!} ≤ \frac{1}{4} \frac{|t|^{2r+2}}{(2r + 2)!}

for r ≥ 2|t| we can make the error term as small as we like. However, if |t| is large, our calculations will give sin t as the difference of two very large numbers and this is computationally unsound.

(iv) If we know π to high accuracy, we can compute x/2π and find an integer n such that |x−2πn| ≤ π. We can now compute y = x−2nπ to high accuracy and then sin x = sin y using Taylor’s theorem to compute sin y.

Exercise 6.3.13

(i) Unless the reader can think of something else, this seems to demand knowing the remainder when 2π is divided into 10^{100}. Presumably we would need to know π to at least 100 places of decimals and to carry out the calculation to at least 100 figure accuracy (in fact, slightly more).

(ii) Calculators do not work to this accuracy.

(iii) Computers can work to this accuracy, but this demands special programming techniques. If we replace tan 10^{100} by tan 10^{10^{10}}, things become really interesting.

Exercise 6.3.14

We have

sin_{D} t = sin_{R} \frac{2π}{360}t

so, using the function of a function rule, Taylor’s theorem yields

|sin t − (\frac{2π}{360}t − \frac{(2π)^{3}t^{3}}{3!(360)^{3}} − … + (−1)^{r} \frac{(2π)^{2r+1}t^{2r+1}}{(2r + 1)!(360)^{2r+1}} )|

≤ min \{\frac{(2π)^{2r+2}|t|^{2r+2}}{(2r + 2)!(360)^{2r+2}} , \frac{(2π)^{2r+3}|t|^{2r+3}}{(2r + 3)!(360)^{2r+3}} \} .

(We could also obtain the formula by simple substitution.) MORAL:- Use radians.

Exercise 6.3.15

We have

f^{(r)}(x) = m(m − 1) … (m − r + 1)(x + y)^{m−r}

and so

f^{(r)}(0) = m(m − 1) … (m − r + 1)y^{m−r}

for 0 ≤ r ≤ m, whilst

f^{(r)}(x) = 0

for r ≥ m + 1.

Thus |f^{(n)}(x)| = 0 ≤ 0 whenever n ≥ m + 1 and our Taylor series with error estimate yields the equality

f(x) = f(0) + \frac{f′(0)}{1!}x + \frac{f′′(0)}{2!}x^{2} + … + \frac{f^{(m)}(0)}{m!}x^{m}

when we take sufficiently many terms. Substituting in the values for f^{(r)}(0) already obtained we get

(x + y)^{m} = \binom{0}{0}y^{m} + \binom{m}{m − 1}y^{m−1}x + \binom{m}{2}y^{m−2}x^{2} + … + \binom{m}{m}x^{m}.

This is the binomial theorem.

Exercise 7.1.1

|f(x)| ≤ u if |x − b| ≤ u^{1/3}. Thus, for example, if we can compute f to six places of decimals but have no other information about b, we can only find b to two places of decimals.

Exercise 7.1.2

We have a_{j} < b_{j} and f(a_{j}) ≤ 0 ≤ f(b_{j}). Let c_{j} = (a_{j} +b_{j})/2. Either f(c_{j}) ≤ 0 and we set a_{j+1} = c_{j}, b_{j+1} = b_{j}

or 0 < f(c_{j}) and we set a_{j+1} = a_{j}, b_{j+1} = c_{j}.

Exercise 7.1.3

(i) We have a_{0} = 5/3, b_{0} = 8/3, so

f(a_{0}) < 0 < f(b_{0}).

c_{0} = 13/6, f(13/6) > 0, so

a_{1} = 5/3, b_{1} = 13/6.

c_{1} = 23/12, f(c_{1}) < 0, so

a_{2} = 23/12, b_{2} = 13/6.

c_{2} = 49/24, f(c_{2}) > 0, so

a_{3} = 23/12, b_{2} = 49/24.

c_{3} = 95/48, f(c_{3}) < 0, so

a_{4} = 95/48, b_{2} = 49/24 = 98/48.

(ii) a_{0} = 0, b_{0} = 2, so

f(a_{0}) < 0 < f(b_{0}).

c_{0} = 1, f(1) < 0, so

a_{1} = 1, b_{1} = 2.

c_{1} = 3/2, f(c_{1}) > 0, so

a_{2} = 1, b_{2} = 3/2.

c_{2} = 5/4, f(c_{2}) < 0, so

a_{3} = 5/4, b_{2} = 3/2.

c_{3} = 11/8, f(c_{3}) < 0, so a_{4} = 11/8, b_{2} = 3/2 = 12/8.

Exercise 7.1.4

Either look at previous argument applied to −f or write out as before:-

The method requires initial values a_{0}, b_{0} such that a_{0} < b_{0} and f(a_{0}) ≥ 0 ≥ f(b_{0}). Notice that this guarantees that there is a root between a_{0} and b_{0}.

Let c_{0} = (a_{0} + b_{0})/2. Either f(c_{0}) ≥ 0 and we set a_{1} = c_{0}, b_{1} = b_{0} or 0 > f(c_{0}) and we set a_{1} = a_{0}, b_{1} = c_{0}. In either case, f(a_{1}) ≥ 0 ≥ f(b_{1}) and b_{1} − a_{1} = (b_{0} − a_{0})/2, so we have halved the length of the interval in which we know there exists a root. We repeat the process as many times as we want, obtaining intervals with end points a_{j}, b_{j} such that f(a_{j}) ≥ 0 ≥ f(b_{j}) and b_{j} − a_{j} = 2^{−j}(b_{0} − a_{0}).

Exercise 7.1.5

2^{10} = 1024 ≈ 10^{3}

(indeed 2^{10} > 10^{3}) so doing the iteration 40 times produces an interval of size 2^{−40} ≈ 10^{−12}.

Exercise 7.2.1

(i) f′(x) = 2x, so

x_{j+1} = x_{j} − \frac{f(x_{j})}{f′(x_{j})} = x_{j} − \frac{x_{j}^{2} − 4}{2x_{j}} = \frac{1}{2}(x_{j} + 4x_{j}^{−1}).

Thus

x_{0} = 3
x_{1} ≈ 2.166666667
x_{2} ≈ 2.006410256
x_{3} ≈ 2.000010240
x_{4} ≈ 2.000000000

correct to the number of figures shown.

(ii) f′(x) = 2x, so

x_{j+1} = x_{j} − \frac{f(x_{j})}{f′(x_{j})} = x_{j} − \frac{x_{j}^{2} − 2}{2x_{j}} = \frac{1}{2}(x_{j} + 2x_{j}^{−1}).

Thus

x_{0} = 1
x_{1} = 1.5
x_{2} ≈ 1.416666667
x_{3} ≈ 1.414215686
x_{4} ≈ 1.414213562.

My calculator gives x_{4}^{2} = 1.999999999.

Exercise 7.2.2

If x_{j} is accurate to within n places of decimals, then |x_{j}−a| ≤ 10^{−n}/2 and |x_{j+1} − a| ≤ M × 10^{−2n}/4. If n is large (how large depends on M), we will, more or less, double the number of places of accuracy.

Exercise 7.2.3

If f(x) = cos x, then

x_{1} = x_{0} − \frac{f(x_{0})}{f′(x_{0})} = x_{0} + \frac{cos x_{0}}{sin x_{0}} = x_{0} + cot x_{0}.

Now cot x decreases from cot(1/100) > 10 to 0 as x runs from 1/100 to π/2, so there is a value of x_{0} with 1/100 ≤ x ≤ π/2 such that cot x_{0} = 2π and so

x_{1} = x_{0} + 2π

By periodicity, x_{j} = x_{0} + 2jπ.

Exercise 7.2.4

We have

x_{j+1} = x_{j} − \frac{f(x_{j})}{f′(x_{j})} = x_{j} − \frac{x_{j}^{3} − 2x_{j} + 2}{3x_{j}^{2} − 2}.

Thus

x_{0} = 0
x_{1} = 1
x_{2} = 0

and so x_{2j} = 0, x_{2j+1} = 1.

Exercise 7.2.5

Let f(x) = x^{1/3}. We have

x_{j+1} = x_{j} − \frac{f(x_{j})}{f′(x_{j})}

= x_{j} − \frac{x_{j}^{1/3}}{3^{−1}x_{j}^{−2/3}}

= x_{j} − 3x_{j} = −2x_{j}.

Thus x_{j} = (−1)^{j}2^{j}x_{0} and the Newton–Raphson method fails.

This does not contradict our argument because f is not differentiable at the origin. (If f were differentiable at 0 with f′(0) = a, we would have

h^{1/3} = ah + o(h)

so

1 = ah^{2/3} + o(h^{2/3})

which is false.)

(ii) If |x_{0}| > u, then the argument of (i) gives x_{1} = −2x_{0} and |x_{1}| > 2u. Thus x_{j} = (−1)^{j}2^{j}x_{0} and the Newton–Raphson method fails.

Exercise 7.2.6

Here is a possible solution but there are many others.

The basic idea is to use the Newton–Raphson method to find x (to the accuracy required) with f(x) = 0 where f(s) = (exp s) − t, that is to say, to use the sequence

x_{j+1} = x_{j} − \frac{exp(x_{j}) − t}{exp(x_{j}) − 1}  ⋆

using Taylor series to obtain exp x_{j}.

However we will run into problems if exp x_{j} is very close to 1 because the formula ⋆ will then involve division by a very small quantity. We will also run into problems if we use the Taylor series to find exp x_{j} if |x_{j}| is large,

We therefore proceed as follows. Let N be the integer such that that 1/10 ≤ 10^{N}t < 1. Now set T = 10^{N}t and use the Newton–Raphson method to find X (to the accuracy required) with F (X) = 0 where F (s) = exp s − T, that is to say, take

x_{j+1} = x_{j} − \frac{exp(x_{j}) − T}{exp(x_{j}) − 1}

starting from x_{0} + 0 and using Taylor series to obtain exp x_{j}. We also compute log 1/10.

We now use the formula

log t = log T − log 10^{N} = X + N log(1/10).

Exercise 7.2.7

(i) We have

g′(x) = 1 − \frac{2f′(x)^{2} + 2f(x)f′′(x)}{2f′(x)^{2} − f(x)f′′(x)}

+ \frac{2f(x)f′(x)(3f′′(x)f′(x) − f(x)f′′′(x))}{(2f′(x)^{2} − f(x)f′′(x))^{2}}

= −f(x)\frac{3f′′(x)}{k(x)} + 2f(x)f′(x)\frac{k′(x)}{k(x)^{2}}

where k(x) = 2f′(x)^{2} − f(x)f′′(x) so, since f(a) = 0,

g′(a) = 0.

Using the fact that f(a) = 0 to reduce calculation,

g′′(a) = −\frac{3f′′(a)f′(a)^{2}}{k(a)} + 2f′(a)^{2} \frac{k′(a)}{k(a)^{2}}

= −\frac{3f′(a)f′′(a)}{2f′(a)^{2}} + \frac{2f′(a)^{2}(3f′′(a)f′(a))}{4f′(a)^{4}}

= 0.

(Mark Twain talks of a German sentence as ‘Swimming the Atlantic with a verb in your mouth.’)

(ii) By the local Taylor’s theorem, there exist u > 0 and K > 0 such that

|g(a + h) − g(a) − g′(a)h − g′′(a)h^{2}/2| ≤ K|h|^{3}

whenever |h| ≤ u, that is to say |g(a+h)−a| ≤ K|h|^{3} whenever |h| ≤ u.

(iii) If x_{j} = H(x_{j−1}) for Newton’s method, we can find v > 0 and M such that

|H(a + h) − a| ≤ Mh^{2}

for all h such that |h| ≤ v. Thus, if ũ = M^{−1}v^{2} and L = M^{2},

|G(a + h) − a| = |H(H(a + h)) − a| ≤ M|H(a + h) − a| ≤ Lh^{4}

whenever |h| ≤ ũ.

Exercise 7.3.1

(i) Suppose, if possible, that \sqrt{2} = u/v with u and v strictly positive integers having no common factor. Then

2 = (u/v)^{2} = u^{2}/v^{2},

so

2u^{2} = v^{2}

and u must be even. Thus u = 2r for some integer r and 2v^{2} = 4r^{2}, whence

v^{2} = 2r^{2}

and v must be even. Thus u and v have 2 as a common factor contradicting our original assumption.

Thus our initial assumption is false and \sqrt{2} is irrational.

(ii) Suppose, if possible, that \sqrt{p} = u/v with u and v strictly positive integers having no common factor. Then

p = (u/v)^{2} = u^{2}/v^{2},

so

pv^{2} = u^{2}

and, since p is a prime, u must be divisible by p. Thus u = rp for some integer r and pv^{2} = p^{2}r^{2}, whence

v^{2} = pr^{2}

and v must be divisible by p. Thus u and v have p as a common factor contradicting our original assumption.

Thus our initial assumption is false and \sqrt{p} is irrational.

(iii) If n is not a perfect square, there must exist a prime p and an integer t ≥ 0 such that p^{2t+1} divides n, but p^{2t+2} does not. If \sqrt{n} is rational, then so is p^{−t}\sqrt{n} = \sqrt{p^{−2t}n} so we may assume that p divides n but p^{2} does not.

Suppose, if possible, that \sqrt{n} = u/v with u and v strictly positive integers with no common factor. Then

n = (u/v)^{2} = u^{2}/v^{2},

so

nv^{2} = u^{2}.

Since p divides n but p^{2} does not, we can find s such that the left hand side is divisible by of p^{2s+1} but not by p^{2s+2}. Since the right hand side is a square divisible by p^{2s+1} it must be divisible by of p^{2s+2} and we have a contradiction.

Thus our initial assumption is false and \sqrt{p} is irrational.

Exercise 7.3.2

Observe that

|P′(t)| = |nb_{n}t^{n−1} + (n − 1)b_{n−1}t^{n−2} + … + b_{1}|

≤ n|b_{n}||t|^{n−1} + (n − 1)|b_{n−1}||t|^{n−2} + … + |b_{1}| ≤ n|b_{n}|R^{n−1} + (n − 1)|b_{n−1}|R^{n−2} + … + 2|b_{2}|R + |b_{1}| ≤ n|b_{n}|R^{n−1} + (n − 1)|b_{n−1}|R^{n−2} + … + 2|b_{2}|R + |b_{1}| + 1

for |t| ≤ R. Automatically K ≥ 1.

Exercise 7.3.3

(i) We chose R so that, if P (x) = 0, then |x| ≤ R − 2, so, certainly, if |p/q| > R,

|p/q − x| ≥ 1 ≥ K^{−1}q^{−n}.

(ii) Since there are only finitely many roots, we can find a u > 0 such that |x − y| > u whenever x and y are distinct roots. If we choose q_{0} ≥ (2u^{−1} + 1) then, if x is an irrational root and p/q is a root with p, q integers, we have x ≠ p/q and

|p/q − x| > u > K^{−1}q^{−n}.

Exercise 7.3.4

x = .110 001 000 000 000 000 000 001 000 000 …

Exercise 7.3.5

If r_{j} = 0 for all j ≥ N, then y is not merely algebraic, but rational. If we set b_{1} = 10^{N!} and b_{0} = −10^{N!}y, then b_{1} and b_{0} are integers, b_{1} ≠ 0 and y is the root of

b_{1}t + b_{0} = 0.

If there does not exist an N such that r_{j} = 0 for all j ≥ N, we can find j(k) such that j(1) ≥ 1, j(k + 1) ≥ j(k) + k + 1 and r_{j(k)} ≠ 0 for each k ≥ 1.

If we now write q_{n} = 10^{j(n+2)!} and

p_{n} = r_{j(1)}10^{(j(n)+2)!−j(1)!} + r_{j(2)}10^{(j(n)+2)!−j(2)!} + r_{j(3)}10^{(j(n)+2)!−j(3)!} + …
… + r_{j(n+1)}10^{(j(n+2)!−j(n+1)!} + r_{j(n+2)},

we see, by looking at the decimal expansions of y and p_{n}/q_{n}, that

\frac{p_{n}}{q_{n}} ≤ x ≤ \frac{p_{n}}{q_{n}} + 20 × 10^{−j(n+3)!}

and so

|x − \frac{p_{n}}{q_{n}} | ≤ 20 × 10^{−(j(n+3)!} ≤ q_{n}^{−n−1}

for every n ≥ 1. Thus y is transcendental.

[There are many ways we might choose q_{n}, each producing a minor variation of the proof.]

Exercise 7.3.6

(i) This is just long division.

If the reader wishes to be more formal, observe that

a_{n}t^{n}+a_{n−1}t^{n−1}+…+a_{0} = a_{n}t^{n−1}(t−u)+(a_{n−1}+ua_{n})t^{n−1}+a_{n−2}t^{n−2}+…+a_{0},

so any polynomial P_{n} of degree at most n can be written as

P_{n}(t) = a_{n}t^{n−1}(t − u) + P_{n−1}(t)

with P_{n−1} some polynomial of degree at most n − 1. Repeated use of this observation gives

P_{n}(t) = b_{n−1}t^{n−1}(t − u) + P_{n−1}(t)
= b_{n−1}t^{n−1}(t − u) + b_{n−2}t^{n−2}(t − u) + P_{n−2}(t)
⋮

= b_{n−1}t^{n−1}(t − u) + b_{n−2}t^{n−2}(t − u) + … + b_{1}t(t − u) + P_{1}(t)

= b_{n−1}t^{n−1}(t − u) + b_{n−2}t^{n−2}(t − u) + … + b_{1}t(t − u) + b_{0}(t − u) + r

with appropriate polynomials P_{j} of degree at most j, appropriate b_{j} and appropriate r.

Taking P_{n} = P and factorising, we have

P (t) = (t − u)Q(t) + r

with Q(t) = b_{n−1}t^{n−1} + b_{n−2}t^{n−2} + … + b_{0}.

(ii) Setting t = u, we have

0 = P (u) = (u − u)Q(u) + r = 0 + r = r

so r = 0 and

P (t) = (t − u)Q(t).

If P (v) = 0 then (v − u)Q(v) = 0, so either v − u = 0 or Q(v) = 0.

(iii) Thus, if a polynomial P_{n} of degree n has a root, any further roots must also be roots of some fixed polynomial P_{n−1} of degree n−1. If P_{n−1} has a root, any further roots must also be roots of some fixed polynomial P_{n−2} of degree n − 2. Repeating this argument n − 1 times, we see that, if P_{n} has n − 1 distinct roots, any further roots must also be roots of some fixed polynomial P_{1} of degree 1. But P_{1} has at most 1 root, so P_{n} has at most n roots.

[The reader may prefer to set these proofs out as formal inductions.]

(iv) Suppose P has degree exactly r for some r with n ≥ r ≥ 0. If r ≥ 1 then P vanishes at most r times. Thus r = 0, P is a constant and so P is zero.

Exercise 8.1.1

There are many ways of answering the question. It is worth remarking that information costs money and the more information anything contains the more expensive it is likely to be to produce.

Exercise 8.1.3

We have

(h^{2} + k^{2})^{1/2} ≥ (h^{2})^{1/2} = |h|

and, similarly, (h^{2} + k^{2})^{1/2} ≥ |k|, so

max{|h|, |k|} ≤ (h^{2} + k^{2})^{1/2}.

On the other hand, (h^{2}+k^{2})^{1/2} ≤ ((max{|h|, |k|})^{2}+(max{|h|, |k|})^{2})^{1/2} = 2^{1/2} max{|h|, |k|}.

Exercise 8.1.5

Suppose that we set h = 0 and allow k to vary. Then equation ⋆ becomes

g(x, y + k) = g(x, y) + Bk + o(k),

so the function h_{x} of one variable, defined by h_{x}(y) = f(x, y), is differentiable with h′_{x}(y) = B. We shall write ∂_{2}g(x, y) = h′_{x}(y).

Equation ⋆ yields

g(x + h, y + k) = g(x, y) + Ah + Bk + o((h^{2} + k^{2})^{1/2})

= g(x, y) + (∂_{1}g(x, y)h + ∂_{2}g(x, y)k) + o((h^{2} + k^{2})^{1/2}).

Exercise 8.2.1

By translation (that is to say, by writing f(s, t) = g(x + s, y + t) and applying the previous result to f)

g(x + h, y + k) = g(x, y) + ∂_{1}g(x, y)h + ∂_{2}g(x, y)k

+ \frac{1}{2}(∂_{1}∂_{1}g(x, y)h^{2}

+ (∂_{2}∂_{1}g(x, y) + ∂_{1}∂_{2}g(x, y))hk + ∂_{2}∂_{2}g(x, y)k^{2}) + o(h^{2} + k^{2}).

Exercise 8.2.2

We have

∂_{1}G(x, y) = b_{1} + 2c_{11}x + 2c_{12}y
∂_{2}G(x, y) = b_{2} + 2c_{12}x + 2c_{22}y
∂_{1}∂_{1}G(x, y) = 2c_{11}
∂_{1}∂_{2}G(x, y) = 2c_{12}
∂_{2}∂_{1}G(x, y) = 2c_{12}
∂_{2}∂_{2}G(x, y) = 2c_{22},

and so

g(0, 0) = a_{0}
∂_{1}G(0, 0) = b_{1}
∂_{2}G(0, 0) = b_{2}
∂_{1}∂_{1}G(0, 0) = 2c_{11}
∂_{1}∂_{2}G(0, 0) = 2c_{12}
∂_{2}∂_{1}G(0, 0) = 2c_{12}
∂_{2}∂_{2}G(0, 0) = 2c_{22}.

Thus

G(0, 0) + ∂_{1}G(0, 0)h + ∂_{2}G(0, 0)k

+ \frac{1}{2}(∂_{1}∂_{1}G(0, 0)h^{2} + ∂_{2}∂_{1}G(0, 0)hk + ∂_{1}∂_{2}G(0, 0)kh + ∂_{2}∂_{2}G(0, 0)k^{2})

= a_{0} + b_{1}h + b_{2}k + \frac{1}{2}(2c_{11}h^{2} + 2c_{12}hk + 2c_{12}kh + 2c_{22}k^{2})

= a_{0} + b_{1}h + b_{2}k + c_{11}h^{2} + 2c_{12}hk + c_{22}k^{2} = G(h, k).

Exercise 8.2.3

(Very sketchy solution indeed.)

I shall make things easier for myself by replacing the coordinates (h, k) by (h_{1}, h_{2}) and using the summation sign

∑_{i=1}^{2} a_{i} = a_{1} + a_{2}.

Repeating our arguments to obtain f′′′(r) from f′′(r) and setting r cos θ = h_{1}, r sin θ = h_{2}, we get

g(h_{1}, h_{2}) = g(0, 0) + ∑_{i=1}^{2} ∂_{i}g(0, 0)h_{i} + \frac{1}{2!} ∑_{i=1}^{2} ∑_{j=1}^{2} n_{ij}∂_{i}∂_{j}g(0, 0)h_{j}h_{i}

+ \frac{1}{3!} ∑_{i=1}^{2} ∑_{j=1}^{2} n_{ijk} ∑_{k=1}^{2} ∂_{i}∂_{j}∂_{k}g(0, 0)h_{k}h_{j}h_{i} + o((h^{2} + k^{2})^{3/2}).

with n_{ij} = 1 if i ≠ j, n_{ii} = 2, n_{ijk} = 1 if i, j and k are all different, n_{ijk} = 3 if exactly two of the i,j,k are unequal and n_{iii} = 6.

If

G(x_{1}, x_{2}) = a_{0} + ∑_{i=1}^{2} a_{i}x_{i} + ∑_{i=1}^{2} ∑_{j=1}^{2} a_{ij}x_{j}x_{i} + ∑_{i=1}^{2} ∑_{j=1}^{2} ∑_{k=1}^{2} a_{ijk}x_{k}x_{j}x_{i},

then

G(0, 0) = a_{0}
∂_{r}G(0, 0) = a_{r}
∂_{r}∂_{s}G(0, 0) = a_{rs} + a_{sr}
∂_{r}∂_{s}∂_{q}G(0, 0) = a_{rsq} + a_{rqs} + a_{srq} + a_{sqr} + a_{qrs} + a_{qsr},

so

G(h_{1}, h_{2}) = g(0, 0) + ∑_{i=1}^{2} ∂_{i}G(0, 0)h_{i} + \frac{1}{2!} ∑_{i=1}^{2} ∑_{j=1}^{2} n_{ij}∂_{i}∂_{j}G(0, 0)h_{j}h_{i}

+ \frac{1}{3!} ∑_{i=1}^{2} ∑_{j=1}^{2} ∑_{k=1}^{2} n_{ijk}∂_{i}∂_{j}∂_{k}G(0, 0)h_{k}h_{j}h_{i}.

Exercise 8.2.4

Let O be the origin, X the point (x, 0), Y the point (0, y) and Z the point (x, y). We note that OXZY forms a rectangle.

Under the given rotation O is fixed, X goes to X′ with coordinates

(x cos θ, x sin θ)),

Y goes to Y′ with coordinates

(−y sin θ, y cos θ)

and Z goes to Z′. Since OX′Z′Y′ remains an rectangle, Z′ must have coordinates

(x cos θ − y sin θ, x sin θ + y cos θ)

Exercise 8.2.5

(i) If c = a, take θ = π/4. If c ≠ a take

tan 2θ = \frac{2b}{a − c}.

(ii) We have

G(s, t) = \frac{1}{2}(a(s cos θ − t sin θ)^{2} + 2b(s cos θ − t sin θ)(s sin θ + t cos θ)
+ c(s sin θ + t cos θ)^{2}) + o(s^{2} + t^{2})

= \frac{1}{2}(as^{2}(cos θ)^{2} − 2ast cos θ sin θ + t^{2}(sin θ)^{2} − 2bt^{2} sin θ cos θ + 2bst((cos θ)^{2} − (sin θ)^{2}) + 2bs^{2} sin θ cos θ + ct^{2}(cos θ)^{2} + 2cst cos θ sin θ + cs^{2}(sin θ)^{2}) + o(s^{2} + t^{2})

= \frac{1}{2}(ut^{2} + 2vst + ws^{2}) + o(s^{2} + t^{2})

with

u = a(sin θ)^{2} − 2b cos θ sin θ + c(cos θ)^{2},
2v = −2a cos θ sin θ + 2b((cos θ)^{2} − (sin θ)^{2}) + 2c cos θ sin θ,
w = a(cos θ)^{2} + 2b cos θ sin θ + c(sin θ)^{2},

so 2v = −a sin 2θ + 2b cos 2θ + c sin 2θ = (c − a) sin 2θ + 2b cos 2θ.

Exercise 8.2.7

Suppose that f(x, y) is a function of two variables defined for

x^{2} + y^{2} ≤ 1.

Then (0, 0) is a local maximum, if we can find a u with 0 < u < 1 such that

f(0, 0) ≥ f(x, y)

for x^{2} + y^{2} ≤ u^{2}.

(Or some variation on this theme.)

Exercise 8.2.8

This is the one-dimensional version of the remarks on marshy hilltops.

Let f be the length of daylight at time t. At a maximum or minimum f′(t) = 0, so near a maximum or minimum f′(t) is small.

[The paragraph on units of currency (page 14) is vaguely relevant.]

Exercise 8.3.1

Remove all land masses external to Australia. We have the situation as before with one large lake external to Australia. Thus, for Australia,

no. peaks + no. low points − no. passes = 1.

Exercise 9.2.1

If f(u, v) = uv, then y′(t) = ty(t) can be rewritten as

y′(t) = f(y(t), t).

If f(u, v) = (1 + v^{2})^{−1}, then y′(t)(1 + y(t)^{2}) = 1 can be rewritten as

y′(t) = f(y(t), t).

If f(u, v) = u^{3} sin v, then y′(t) = t^{3} sin (y(t)) can be rewritten as y′(t) = f(y(t), t).

Exercise 9.2.2

(i) Euler’s method gives

y_{r+1} − y_{r} = h × rh = rh^{2},

so

y_{n} = (y_{n} − y_{n−1}) + (y_{n−1} − y_{n−2}) + … + (y_{1} − y_{0}) + y_{0}
= ((n − 1) + (n − 2) + … + 1 + 0)h^{2} + 0 = \frac{1}{2}n(n − 1)h^{2}.

By the fundamental theorem of the calculus, the exact solution is y(t) = t^{2}/2, so

|y_{N} − y(T)| = \frac{1}{2}|N(N − 1)h^{2} − (Nh)^{2}| = \frac{Nh^{2}}{2} = \frac{T h}{2}.

(ii) Euler’s method gives

y_{r+1} − y_{r} = bhy_{r},

so y_{r+1} = y_{r}(1 + bh) and

y_{n} = (1 + bh)y_{n−1} = (1 + bh)^{2}y_{n−2} = … = (1 + bh)^{n}y_{0}.

If T is fixed, then, when N is large

y_{N} = (1 + \frac{bh}{N})^{N} y_{0} ≈ y_{0}e^{bT}

(by the result on page 89).

This reassuring because y(T) = y_{0}e^{bT} is the correct answer (see Section 4.2).

Exercise 9.2.3

Euler’s method gives

y_{r+1} − y_{r} = −2y_{r},

so

y_{r+1} = −y_{r}

and we get y_{r} = (−1)^{r} although the correct solution is y(t) = e^{−8t} so y(r/4) = e^{−r/2}.

We have simply taken the step length too large. Exercise 9.2.4⋆

Exercise 9.3.1

There are several (more or less equivalent) ways of doing this.

If

g(t) = Ay(s + 2t) + By(s + t) − ty′(s) − By(s − t) − Ay(s − 2t), then

g′(t) = 2Ay′(s + 2t) + By′(s + t) − y′(s) + By′(s − t) + 2Ay′(s − 2t)
g′′(t) = 4Ay′′(s + 2t) + By′′(s + t) − By′′(s − t) − 4Ay′′(s − 2t)
g′′′(t) = 8Ay′′′(s + 2t) + By′′′(s + t) + By′′′(s − t) + 8Ay′′′(s − 2t)
g^{(4)}(t) = 16Ay^{(4)}(s + 2t) + By^{(4)}(s + t) − By^{(4)}(s − t) − 16Ay^{(4)}(s − 2t)
y^{(5)}(t) = 32Ay^{(5)}(s + 2t) + By^{(5)}(s + t) + By^{(5)}(s − t) + 32Ay^{(5)}(s − 2t),

so that

g(0) = 0
g′(0) = (4A + 2B − 1)y′(s)
g′′(0) = 0
g′′′(0) = (16A + 2B)y′′′(s)
g^{(4)}(0) = 0.

Solving the equations

4A + 2B = 1
16A + 2B = 0,

we see that g(s) = g′(s) = g′′(s) = g′′′(s) = g^{(4)}(s) = 0 for all choices of y if A = −1/12, B = 2/3 and then Taylor’s theorem gives

|g(h)| ≤ (64|A| + 2|B|)M \frac{h^{5}}{5!} = \frac{20}{3}M \frac{|h|^{5}}{5!} = \frac{M}{18} h^{5}

if |g^{(5)}(x)| ≤ M over the appropriate range

Translating, we see that, if A = −1/12, B = 2/3, |Ay(s + 2h) + By(s + h) − hy′(s) − By(s − h) − Ay(s − 2h)| ≤ KMh^{5} for some appropriate constant K.

Exercise 9.3.2⋆

Exercise 9.3.3⋆

Exercise 9.3.4

(i) We observe that, if u_{n} and u_{n−1} are given,

u_{n+1} = −bu_{n} − cu_{n−1}

is determined.

(ii) We have

p^{2} + bp + c = (p − p)(p − q) = 0

and so, multiplying through by p^{n−2},

p^{n} + bp^{n−1} + cp^{n−2} = 0.

(iii) Similarly

q^{n} + bq^{n−1} + cq^{n−2} = 0.

Thus v_{n}+bv_{n−1}+cv_{n−2} = A(p^{n}+bp^{n−1}+cp^{n−2})+B(q^{n}+bq^{n−1}+cq^{n−2}) = 0+0 = 0.

(iv) Subtracting the second equation from p times the first, we get B(p − q) = pu_{0} − u_{1}, so that

B = \frac{pu_{0} − u_{1}}{p − q}

and similarly

A = \frac{qu_{0} − u_{1}}{q − p} .

By inspection these are solutions.

(v) Choose A and B as in (iv). By (iii), v_{n} = Ap^{n} +Bq^{n} is a solution of our difference equation with v_{0} = u_{0}, v_{1} = u_{1}. By (i),

u_{n} = v_{n} = Ap^{n} + Bq^{n}.

Exercise 9.3.5

The standard formula gives two roots

−Kh ± \sqrt{1 + K^{2}h^{2}},

so (recalling that Kh > 0 and we want p > 0) we have

p = −Kh + \sqrt{1 + K^{2}h^{2}}, q = −Kh − \sqrt{1 + K^{2}h^{2}}

and

pq = (−Kh)^{2} − (1 + K^{2}h^{2}) = −1.

Exercise 9.3.6

Let f(t) = (1 + t)^{1/2}. Choose L so that |f′′(t)| ≤ L for |t| ≤ 1/2. By Taylor’s theorem we can find a u > 0 and an L > 0 such that

|f(t) − f(0) − f′(0)t| ≤ Lt^{2}/2

for |t| ≤ 1/2.

Since f′(t) = 2^{−1}(1 + t)^{1/2}, this gives

|(1 + t)^{1/2} − 1 − t/2| ≤ Lt^{2}/2

for |t| ≤ 1/2. If we set M = LK^{4}/2 and consider t = (Kh)^{2} we obtain

|(1 + K^{2}h^{2})^{1/2} − 1 − K^{2}h^{2}/2| ≤ Mh^{4}

for |h| ≤ 2^{−1/2}K^{−1}.

We thus have

| − Kh + (1 + K^{2}h^{2})^{1/2} − 1 + Kh − K^{2}h^{2}/2| ≤ Mh^{4}

The mean value inequality tells us that

|x^{N} − y^{N}| ≤ N(max{|x|, |y|})^{N}|x − y|

so, since | − Kh + (1 + K^{2}h^{2})^{1/2}| ≤ 1 and |1 − Kh − K^{2}h^{2}/2| ≤ 1 for h small,

|(−Kh + (1 + K^{2}h^{2})^{1/2})^{N} − (1 + Kh − K^{2}h^{2}/2)^{N}| ≤ NMh^{4}

Let h = T/N. Provided N is large, Exercise 6.3.5 gives

|(1 − \frac{KT}{N} + \frac{KT^{2}}{2N^{2}} )^{N} − e^{−KT} | ≤ \frac{L}{N^{2}}

for some L ie

|(1 − Kh + \frac{Kh}{2})^{N} − exp −KT | ≤ LT^{2}h^{2}.

Combining the results of the last two paragraphs

|(−Kh+(1+K^{2}h^{2})^{1/2})^{N}−e^{−KT} | ≤ NMh^{4}+LT^{2}h^{2} = T Mh^{3}+LT^{2}h^{2} ≤ Ch^{2}

for an appropriate C provided N is large enough.

Exercise 9.3.7

We have

q = −Kh − \sqrt{1 + K^{2}h^{2}}

so

p^{−N} = q^{N} = (−1)^{N}(Kh+\sqrt{1 + K^{2}h^{2}})^{N} ≈ (−1)^{N}(1+Kh)^{N} = (−1)^{N} (1 + \frac{KT}{N} )^{N} ≈ (−1)^{N}e^{KT}

Exercise 10.1.1

We have

F (x) = (1 + x)^{−1}
F′(x) = (−1)(1 + x)^{−2}
F′′(x) = (−1) × (−2) × (1 + x)^{−3}
⋮
F^{(n)}(x) = (−1) × (−2) × … (−n) × (1 + x)^{−n−1} = (−1)^{n}n!(1 + x)^{−n−1},

so

F^{(n)}(0) = (−1)^{n}n!

and

\frac{1}{1 + x} =_{?} 1 − x + x^{2} − x^{3} + x^{4} + ….

Exercise 10.1.2

We know, from Exercise 10.1.1, that

F^{(n+1)}(t) = (−1)^{n}(n + 1)! × (1 + t)^{−n−2}.

Thus

|F^{(n+1)}(0)| = (n + 1)!

and

|F^{(n+1)}(t)| ≤ (n + 1)!

for 0 ≤ t ≤ x.

Exercise 10.1.3

Observe that

(1 + x) (1 − x + x^{2} − x^{3} + x^{4} − … + (−1)^{n}x^{n} + \frac{(−1)^{n+1}x^{n+1}}{1 + x} )

= 1 − (x − x) + (x^{2} − x^{2}) − …

+ (−1)^{n−1}(x^{n} − x^{n}) + (−1)^{n}x^{n+1} + (−1)^{n+1}x^{n+1} = 1.

Exercise 10.1.4

(i) We know that

\frac{1}{1 + t} = 1 − t + t^{2} − t^{3} + t^{4} + … + (−1)^{n}t^{n} + \frac{(−1)^{n+1}t^{n+1}}{1 + t}

for t ≥ 0, so, setting t = x^{2}, we have

\frac{1}{1 + x^{2}} = 1 − x^{2} + x^{4} − x^{6} + x^{8} + … + (−1)^{n}x^{2n} + \frac{(−1)^{n+1}x^{2(n+1)}}{1 + x^{2}} ,

which is what we are asked to prove.

(ii) If |x| < 1, then x^{2n} and so R_{n}(G, x) can be made as small as we like by taking n large enough, so we can approximate G(x) arbitrarily well by taking sufficiently many terms in the Taylor expansion. If |x| ≥ 1, then R_{n}(G, x) ≥ (1 + x^{2}), so the approximation does not improve.

Exercise 10.1.5

(i) We have

G_{a}(x) = \frac{a^{2}}{a^{2} + x^{2}} = \frac{1}{1 + (x/a)^{2}} = G(x/a).

(ii) and (iii) The remaining results follow at once from Exercise 10.1.3 on using the formula of part (i).

Exercise 10.1.6

Observe that

\frac{t^{n}}{1 + t} ≥ \frac{t^{n}}{2} ,

for 0 ≤ t ≤ 1, so, if 0 ≤ x ≤ 1,

|T_{n}(x)| = ∫_{0}^{x} \frac{t^{n}}{1 + t} dt ≥ ∫_{0}^{x} \frac{t^{n}}{2} dt = \frac{x^{n+1}}{2(n + 1)}.

In particular, if x = 1,

|T_{n}(1)| ≥ \frac{1}{2(n + 1)}

and we would need at least 10^{6}/2 terms to get log 2 correct to 6 decimal places using (D) directly.

If 0 ≤ t ≤ 1, then

\frac{t^{n}}{1 + t} ≤ t^{n+1},

so, if 0 ≤ x ≤ 1,

|T_{n}(x)| = ∫_{0}^{x} \frac{t^{n}}{1 + t} dt ≤ ∫_{0}^{x} t^{n} dt = \frac{x^{n}}{n + 1}.

If we compute log(1 + x) with 0 ≤ x ≤ 1/2

|T_{n}(x)| ≤ 2^{−n}/(n + 1).

In particular, if we use the use the first 20 terms of (C) the error will be less than 10^{−7}. The second suggested procedure will thus be much more efficient.

An even better approach would use much smaller values of x and add a correspondingly larger number of evaluations of such log(1 + x).

Exercise 10.1.7

If 0 ≤ t ≤ 2, then

\frac{t^{n}}{1 + t} ≥ \frac{t^{n}}{3} .

Thus, if x > 1 and we write y = min{x, 2},

|T_{n}(x)| = ∫_{0}^{x} \frac{t^{n}}{1 + t} dt

≥ ∫_{0}^{y} \frac{t^{n}}{1 + t} dt

= ∫_{0}^{y} \frac{t^{n}}{3} dt

= \frac{y^{n+1}}{3(n + 1)} = \frac{(min\{x, 2\})^{n}}{3(n + 1)}

which is large when n is large.

(If y > 1, then y^{n}/n is large when n is large. There are many ways of seeing this. For example, consider f(t) = y^{t}/t and show that f′(t) > 2 for t large.)

Exercise 10.1.8

log 2 =_{?} 1 − \frac{1}{2} + \frac{1}{3} − \frac{1}{4} + \frac{1}{5} − \frac{1}{6} + \frac{1}{7} − \frac{1}{8} + \frac{1}{9} − \frac{1}{10} + \frac{1}{11} − \frac{1}{12} + \frac{1}{13} − \frac{1}{14} + \frac{1}{15} − \frac{1}{16} … + (−1)^{4n+1} \frac{1}{4n} + ….
\frac{1}{2} log 2 =_{?}  \frac{1}{2}  − \frac{1}{4}  + \frac{1}{6}  − \frac{1}{8}  + \frac{1}{10}  − \frac{1}{12}  + \frac{1}{14}  − \frac{1}{16} … + (−1)^{2n+1} \frac{1}{4n} + …

so, summing in the indicated manner, we get

\frac{3}{2} log 2 =_{?} 1+0+\frac{1}{3}−\frac{1}{2}+\frac{1}{5}+0+\frac{1}{7}−\frac{1}{4}+\frac{1}{9}+0+\frac{1}{11}−\frac{1}{6}+\frac{1}{13}+0+\frac{1}{15}−\frac{1}{8}+….

Before we remove the zeros, the 4n + 1th term is 1/(4n + 1) + 0 = 1/(4n + 1), the 4n + 2th term is −1/(4n + 2) + 1/(2 × (2n + 1)) = 0, the 4n + 3th term is 1/(4n + 3) + 0 = 1/(4n + 3), the 4n + 4th term is −1/(4n + 4) − 1/(2 × (2n + 2)) = −1/(2n + 2). After we remove the zeros, the 3k + 1th term is 1/(4k + 1), the 3k + 2th term is 1/(4k + 3), the 3k + 3th term is −1/(2(k + 1)). We have shown that the terms of the expression are the numbers (−1)^{r+1}/r with each of these numbers occurring exactly once.

Exercise 10.2.1

We have

a_{2n−1} = \frac{−1}{(2n − 2) × (2n − 1)}a_{2n−3}

= \frac{1}{(2n − 4) × (2n − 3) × (2n − 2) × (2n − 1)}a_{2n−5}

= … = \frac{(−1)^{n−1}}{(2n − 1)!}a_{1}.

Exercise 10.2.2

We have f′(t) = −A sin t + B cos t and so

f′′(t) = −A cos t − B sin t = −f(t).

Thus f′′(t) + f(t) = 0.

Exercise 10.2.3

Assuming that we can expand g as

g(t) =_{?} a_{0} + a_{1}t + a_{2}t^{2} + a_{3}t^{3} + ….

and that we can differentiate the infinite sum in the same way as we differentiate a polynomial, we get

g′(t) =_{?} 1 × a_{1} + 2 × a_{2}t + 3 × a_{3}t^{2} + …

and so the equation g(t) = −2g′(t) gives us

1 × a_{1} + 2 × a_{2}t + 3 × a_{3}t^{2} + … =_{?} −2 × a_{0}t − 2 × a_{1}t^{2} − 2 × a_{2}t^{3} − …

Assuming that we can equate coefficients in the same way as we do for polynomials we obtain

−2a_{0} = 2a_{2}
−2a_{1} = 0
−2a_{2} = −4a_{4}
−2a_{3} = 0
⋮
−2a_{2n} = −(2n + 2)a_{2n+2}
−2a_{2n+1} = 0
⋮

Thus a_{2r+1} = 0 for all r and

a_{2r+2} = \frac{−1}{r + 1}a_{2r}.

The equation just stated gives us

a_{2n} = \frac{−1}{n}a_{2(n−1)} = \frac{1}{n(n − 1)}a_{2(n−2)} = … = \frac{(−1)^{n}}{n!} a_{0},

so, if we write A = a_{0}, we get

g(t) =_{?} A (1 − \frac{t^{2}}{1!} + \frac{t^{4}}{2!} − \frac{t^{6}}{3!} + …) =_{?} A exp(−t^{2}).

The function of a function rule shows that, if we write g(t) = A exp(−t^{2}), then g′(t) = −2tg(t).

[We can do better than this, although the reader was not asked to. Suppose that

G′(t) = −2tG(t).

Then, writing F (t) = exp(t^{2})G(t), we have, by the product rule, that F′(t) = 0 so F (t) = A for some constant A and so G(t) = A exp(−t^{2}). Thus we have discovered all the solutions of our differential equation.]

Exercise 10.2.4

We obtain c_{n}x^{n} as the sum of terms of the form a_{n−j}x^{n−j} × b_{j}x^{j}, so

(a_{0} + a_{1}x + a_{2}x^{2} + …)(b_{0} + b_{1}x + b_{2}x^{2} + …) =_{?} (c_{0} + c_{1}x + c_{2}x^{2} + …)

with

c_{r} = a_{0}b_{r} + a_{1}b_{r−1} + a_{2}a_{r−2} + … + a_{r}b_{0}.

Assuming that

f(t) =_{?} a_{0} + a_{1}t + a_{2}t^{2} + a_{3}t^{3} + ….

and that we can differentiate the infinite sum in the same way as we differentiate a polynomial, we get

f′(t) =_{?} 1 × a_{1} + 2 × a_{2}t + 3 × a_{3}t^{3} + …

and so the equation f′(t) = −2tf(t)^{2} gives us

1 × a_{1} + 2 × a_{2}t + 3 × a_{3}t^{2} + … =_{?} −2 × c_{0}t − 2 × c_{1}t^{2} − 2 × c_{2}t^{3} − …

where

c_{n} = a_{0}a_{n} + a_{1}a_{n−1} + … + a_{n}a_{0}

so

a_{1} = 0
2a_{2} = −2c_{0}
3a_{3} = −2c_{1}
4a_{4} = −2c_{2}
⋮
na_{n} = −2c_{n−1}
⋮

We have a_{0} = B for some B and a_{1} = 0.

Suppose that a_{r} = (−1)^{r/2}B^{(r+2)/2} when r is even and r < n and a_{r} = 0 for r odd and r < n. If n is odd, r + s = n implies that r is odd or s is odd, so c_{n−2} = 0 and the equations above give a_{n} = 0. If n is even, then

c_{n−2} = a_{0}a_{n−2} + a_{2}a_{n−4} + … + a_{n−2}a_{0}

= (−1)^{(n−2)/2}B^{(n+2)/2} + (−1)^{(n−2)/2}B^{(n+2)/2} + … + (−1)^{(n−2)/2}B^{(n+2)/2}

= (−1)^{(n−2)/2} \frac{n}{2}B^{(n+2)/2},

so

na_{n} = (−1)^{n/2}nB^{(n+2)/2}

and a_{n} = (−1)^{n/2}B^{(n+2)/2}.

Thus

a_{2n−1} = 0, a_{2n} = (−1)^{n}B^{n+1},

where B is a constant, and we have

f(x) =_{?} B − B^{2}x^{2} + B^{3}x^{4} − … =_{?} \frac{B}{1 + Bx^{2}}

If B > 0 and A = B^{−1}, we get

f(x) = \frac{1}{A + x^{2}} .

If g(x) = (A + x^{2})^{−1}, then, by the quotient rule,

g′(x) = \frac{−2x}{(A + x^{2})^{2}} = −2xg(x)^{2}.

Exercise 11.3.2

(i) f is continuous and f(0) = −2 < 0 < 2 = f(2), so there exists a c with 0 < c < 2 such that f(c) = 0 and so c^{2} = 2.

(ii) f is continuous and f(1) = −1 < 0 < f(N), so there exists a c with 0 < c < N such that f(c) = 0 and so log c = 1.

Exercise 11.3.3

Let f(θ) be the difference between the temperature at longitude θ and the temperature at the point diametrically opposite. Observe that (writing r for a right angle) f(θ + 2r) = −f(θ).

In particular, f(0) = −f(2r) and the intermediate value theorem tells us that there must exist a t with 0 ≤ t ≤ 2r and f(t) = 0.

Exercise 11.3.4

For example, on page 48 we say (in case (2)) ‘f′(0) < 0 and f′(s) > 0 for some large s, so there will be an s_{0} > 0 with f′(s_{0}) = 0.’

Exercise 11.4.1

Observation 1. If we take f(x) = x, then f(|K| + 1) > K, so f is unbounded. If u > v, then f(u) > f(v), so f has no local maxima.

Observation 2. If we take

f(x) = \frac{x^{2}}{1 + x^{2}} ,

then 0 ≤ f(x) ≤ 1. We note that

f(x) = 1 − \frac{1}{1 + x^{2}}

so, if |u| > |v|, f(u) > f(v). Thus f has no local maxima (and so no global maximum).

Observation 3. If we restrict our attention to those x with 0 < x < 1 and set f(x) = 1/x, then f((1 + K)^{−1}) > K and 0 < (1 + K)^{−1} < 1 for K > 0, so f is unbounded. If 0 < u < v < 1, then f(u) > f(v), so f has no local maxima

Observation 4. If we restrict our attention to those x with 0 < x < 1 and set f(x) = x, then 0 ≤ f(x) ≤ 1 whenever 0 < x < 1, but, if 0 < x < 1 and x < y < 1, then f(y) > f(x), so f has no local maxima and so no global maximum.

Exercise 11.5.1

By the fundamental theorem,

G′(t) = g′(t) − f(t) = 0

so, by Plausible Statement B,

G(t) = c

for some constant and so g(t) = ∫_{a}^{t} f(x) dx + c.

Exercise 11.5.2

If g′(t) = 0 for all t, then |g′(t)| ≤ 0 for all t, whence

|g(b) − g(a)| ≤ 0|b − a| = 0.

Thus g(a) = g(b) for all a and b so g(x) = g(0) for all x.