n! (n + 1)(n + 2) . . . (n + k)

k=0

∞ k

|z|n |z|

¤

n! n+1

k=0

n

(n + 1)|z|n

|z| 1

¤ =

n! 1 ’ |z| (n + 1 ’ |z|)n!

n+1

2|z|n

¤ .

n!

96 A COMPANION TO ANALYSIS

Exercise 5.4.9. A particularly cautious mathematician might prove Lemma 5.4.8

as follows. Set em (z) = m z . Show that, if m ≥ n, then

j

j=0 j!

(n + 1)|z|n

|em (z) ’ en’1 (z)| ¤ .

(n + 1 ’ |z|)n!

Deduce that

(n + 1)|z|n

|e(z) ’ en’1 (z)| ¤ |e(z) ’ em (z)| + |em (z) ’ en’1 (z)| ¤ |e(z) ’ em (z)| + .

(n + 1 ’ |z|)n!

By allowing m ’ ∞, obtain the required result.

We now switch our attention to the restriction of e to R. The results we

expect now come tumbling out.

Exercise 5.4.10. Consider e : R ’ R given by e(x) = ∞ xj /j!. j=0

(i) Using Lemma 5.4.8, show that |e(h) ’ 1 ’ h| ¤ h2 for |h| < 1/2.

Deduce that e is di¬erentiable at 0 with derivative 1.

(ii) Explain why e(x + h) ’ e(x) = e(x)(e(h) ’ 1). Deduce that e is

everywhere di¬erentiable with e (x) = e(x).

(iii) Show that e(x) ≥ 1 for x ≥ 0 and, by using the relation e(’x)e(x) =

1, or otherwise, show that e(x) > 0 for all x ∈ R.

(iv) Explain why e is a strictly increasing function.

(v) Show that e(x) ≥ x for x ≥ 0 and deduce that e(x) ’ ∞ as x ’ ∞.

Show also that e(x) ’ 0 as x ’ ’∞.

(vi) Use (v) and the intermediate value theorem to show that e(x) = y

has a solution for all y > 0.

(vii) Use (iv) to show that e(x) = y has at most one solution for all y > 0.

Conclude that e is a bijective map of R to R++ = {x ∈ R : x > 0}.

(viii) By modifying the proof of (v), or otherwise, show that P (x)e(’x) ’

0 as x ’ ∞. [We say ˜exponential beats polynomial™.]

(ix) By using (viii), or otherwise, show that e is not equal to any function

of the form P/Q with P and Q polynomials. [Thus e is a genuinely new

function.]

When trying to prove familiar properties of a familiar function, it is prob-

ably wise to use a slightly unfamiliar notation. However, as the reader will

have realised from the start, the function e is our old friend exp. We shall

revert to the mild disguise in the next section but we use standard notation

for the rest of this one.

97

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Exercise 5.4.11. (i) Check that R is an Abelian group under addition. Show

that R++ = {x ∈ R : x > 0} is an Abelian group under multiplication. Show

that exp : (R, +) ’ (R++ , —) is a isomorphism.

(ii) [Needs a little more familiarity with groups] Show that R \ {0} is an

Abelian group under multiplication. By considering the order of the element

’1 ∈ R \ {0}, or otherwise show that the groups (R, +) and (R \ {0}, —)

are not isomorphic.

We can turn Plausible Statement 5.4.1 into a theorem

Theorem 5.4.12. The general solution of the equation

y (x) = y(x), ()

where y : R ’ R is a di¬erentiable function is

y(x) = a exp(x)

with a ∈ R.

Proof. It is clear that y(x) = a exp(x) is a solution of . We must prove

there are no other solutions. To this end, observe that, if y satis¬es , then

d

(exp(’x)y(x)) = y (x) exp(’x) ’ y(x) exp(’x) = 0

dx

so, by the mean value theorem, exp(’x)y(x) is a constant function. Thus

exp(’x)y(x) = a and y(x) = a exp(x) for some a ∈ R.

Exercise 5.4.13. State and prove the appropriate generalisation of Theo-

rem 5.4.12 to cover the equation

y (x) = by(x)

with b a real constant.

Here is another consequence of Lemma 5.4.8.

Exercise 5.4.14. (e is irrational.) Suppose, if possible, that e = exp 1 is

rational. Then exp 1 = m/n for some positive integers m and n. Explain,

why if N ≥ n,

N

1

N ! exp 1 ’

j!

j=0

98 A COMPANION TO ANALYSIS

must be a non-zero integer and so

N

1

N ! exp 1 ’ ≥ 1.

j!

j=0

Use Lemma 5.4.8 to obtain a contradiction.

r+1

Show, similarly, that ∞ (’1)

r=1 (2r’1)! is irrational.

Most mathematicians draw diagrams frequently both on paper and in

their heads6 . However, these diagrams are merely sketches. To see this,

quickly sketch a graph of exp x.

Exercise 5.4.15. Choosing appropriate scales, draw an accurate graph of

exp on the interval [0, 100]. Does it look like your quick sketch?

We conclude this section with a result which is a little o¬ our main track

but whose proof provides an excellent example of the use of dominated con-

vergence (Theorem 5.3.3).

Exercise 5.4.16. We work in C. Show that if we write

∞

z n

aj (n)z j

1+ =

n j=0

then aj (n)z j ’ z j /j! as n ’ ∞ and |aj (n)z j | ¤ |z j |/j! for all n and all j.

Use dominated convergence to conclude that

z n

’ e(z)

1+

n

as n ’ ∞, for all z ∈ C.

The trigonometric functions ™

5.5

In the previous section we considered the simple di¬erential equation y (x) =

y(x). What happens if we consider the di¬erential equation y (x)+y(x) = 0?

6

Little of this activity appears in books and papers, partly because, even today, adding

diagrams to printed work is non-trivial. It is also possible that it is the process of drawing

(or watching the process of drawing) which aids comprehension rather than the ¬nished

product.

99

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Exercise 5.5.1. Proceeding along the lines of Plausible Statement 5.4.1,

show that it is reasonable to conjecture that the general solution of the equa-

tion

y (x) + y(x) = 0, ( )

where y : R ’ R is a well behaved function, is

∞ ∞

(’1)j x2j (’1)j x2j+1

y(x) = a +b

(2j)! (2j + 1)!

j=0 j=0

with a, b ∈ R.

A little experimentation reveals what is going on.

Exercise 5.5.2. We work in C. If we write

e(iz) ’ e(’iz)

e(iz) + e(’iz)

c(z) = , s(z) = ,

2 2i

show carefully that

∞ ∞

(’1)j z 2j (’1)j z 2j+1

c(z) = , s(z) = .

(2j)! (2j + 1)!

j=0 j=0

We can use the fact that e(z + w) = e(z)e(w) to obtain a collection of

useful formula for s and c.

Exercise 5.5.3. Show that if z, w ∈ C then

(i) s(z + w) = s(z)c(w) + c(z)s(w),

(ii) c(z + w) = c(z)c(w) ’ s(z)s(w),

(iii) s(z)2 + c(z)2 = 1

(iv) s(’z) = ’s(z), c(’z) = c(z).

We now switch our attention to the restriction of s and c to R.

Exercise 5.5.4. Consider c, s : R ’ R given by c(x) = ∞ (’1)j x2j /(2j)!,

j=0

∞ j 2j+1

and s(x) = j=0 (’1) x /(2j + 1)!.

(i) Using the remainder estimate in alternating series test (second para-

graph of Lemma 5.2.1), or otherwise, show that |c(h) ’ 1| ¤ h 2 /2 and

|s(h) ’ h| ¤ |h|3 /6 for |h| < 1. Deduce that c and s are di¬erentiable at

0 with c (0) = 0, s (0) = 1.

(ii) Using the addition formula of Exercise 5.5.3 (ii) and (iii) to evaluate

c(x + h) and s(x + h), show that c and s are everywhere di¬erentiable with

c (x) = ’s(x), s (x) = c(x).

100 A COMPANION TO ANALYSIS

Suppose that a group of mathematicians who did not know the trigono-

metric functions were to investigate our functions c and s de¬ned by power

series. Careful calculation and graphing would reveal that, incredible as it

seemed, c and s appeared to be periodic!

Exercise 5.5.5. (i) By using the estimate for error in the alternating series

test, show that

c(x) > 0 for all 0 ¤ x ¤ 1.

By using a minor modi¬cation of these ideas, or otherwise, show that c(2) <

0. Explain carefully why this means that there must exist an a with 1 < a < 2

such that c(a) = 0.

(iii) In this part and what follows we make use of the formulae obtained

in Exercise 5.5.3 which tell us that

s(x + y) = s(x)c(y) + c(x)s(y), c(x + y) = c(x)c(y) ’ s(x)s(y),

c(x)2 + s(x)2 = 1, s(’x) = ’s(x), c(’x) = c(x)

for all x, y ∈ R. Show that, if c(a ) = 0 and c(a ) = 0, then s(a ’ a ) = 0.

Use the fact that s(0) = 0 and s (x) = c(x) > 0 for 0 ¤ x ¤ 1 to show that

s(x) > 0 for 0 < x ¤ 1. Conclude that, if a and a are distinct zeros of

c, then |a ’ a | > 1. Deduce that c(x) = 0 has exactly one solution with

0 ¤ x ¤ 2. We call this solution a.

(iv) By considering derivatives, show that s is strictly increasing on [0, a].

Conclude that s(a) > 0 and deduce that s(a) = 1. Show that

s(x + a) = c(x), c(x + a) = ’s(x)

for all x and that c and s are periodic with period 4a (that is s(x + 4a) = s(x)

and c(x + 4a) = c(x) for all x).

(v) Show that s is strictly increasing on [’a, a], and strictly decreasing

on [a, 3a].

(vi) If u and v are real numbers with u2 + v 2 = 1, show that there there

is exactly one solution to the pair of equations

c(x) = u, s(x) = v

with 0 ¤ x < 4a.

At this point we tear o¬ the thin disguise of our characters and write

exp(z) = e(z), sin z = s(z), cos(z) = c(z) and a = π/2.

101

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Exercise 5.5.6. We work in R. Show that, if |u| ¤ 1, there is exactly one

θ with 0 ¤ θ ¤ π such that cos θ = u.

Using the Cauchy-Schwarz inequality (Lemma 4.1.2) show that, if x and

y are non-zero vectors in Rm , then there is exactly one θ with 0 ¤ θ ¤ π

such that

x·y

cos θ = .

xy

We call θ the angle between x and y.

Exercise 5.5.7. We work in C and use the usual disguises except that we

write a = π/2.

(i) Show that e has period 2πi in the sense that

e(z + 2πi) = e(z)

for all z and

e(z + w) = e(z)

for all z if and only w = 2nπi for some n ∈ Z. State corresponding results

for s, c : C ’ C.

(ii) If x and y are real, show that

e(x + iy) = e(x)(c(y) + is(y)).

(iii) If w = 0, show that there are unique real numbers r and y with r > 0

and 0 ¤ y < 2π such that

w = re(iy).

(iv) If w ∈ C, ¬nd all solutions of w = re(iy) with r and y real and r ≥ 0.

The traditional statement of Exercise 5.5.7 (iii) says that z = reiθ where

r = |z| and θ is real. However, we have not de¬ned powers yet so, for the

moment, this must merely be considered as a useful mnemonic. (We will

discuss the matter further in Exercise 5.7.9.)

It may be objected that our de¬nitions of sine and cosine ignore their

geometric origins. Later (see Exercises K.169 and K.170) I shall give more

˜geometric™ treatment but the following points are worth noting.

The trigonometric functions did not arise as part of classical axiomatic

Euclidean geometry, but as part of practical geometry (mainly astronomy).

An astronomer is happy to consider the sine of 20 degrees, but a classical

102 A COMPANION TO ANALYSIS

geometer would simply note that it is impossible to construct an angle of

20 degrees using ruler and compass. Starting with our axioms for R we can

obtain a model of classical geometry, but the reverse is not true.

The natural ˜practical geometric™ treatment of angle does not use radians.

Our use of radians has nothing to do with geometric origins and everything

to do with the equation (written in radians)

d

sin cx = c cos cx.

dx

Mathematicians measure angles in radians because, for them, sine is a func-

tion of analysis, everyone else measures angles in degrees because, for them,

sine is a function used in practical geometry.

In the natural ˜practical geometric™ treatment of angle it is usual to con¬ne

oneself to positive angles less than two right angles (or indeed one right

angle). When was the last time you have heard a navigator shouting ˜turn

’20 degrees left™ or ˜up 370 degrees™ ? The extension of sine from a function

on [0, π/2] to a function on R and the corresponding extension of the notion

of angle is a product of ˜analytic™ and not ˜geometric™ thinking.

Since much of this book is devoted to stressing the importance of a ˜geo-

metric approach™ to the calculus of several variables, I do not wish to down-

play the geometric meaning of sine. However, we should treat sine both as

a geometric object and a function of analysis. In this context it matters

little whether we start with a power series de¬nition of sine and end up

with the parametric description of the unit circle as the path described by

the point (sin θ, cos θ) as θ runs from 0 to 2π or (as we shall do in Exer-

cises K.169 and K.170) we start with the Cartesian description of the circle

as x2 + y 2 = 1 and end up with a power series for sine.

Exercise 5.5.8. Write down the main properties of cosh and sinh that you

know. Starting with a tentative solution of the di¬erential equation y = y,

write down appropriate de¬nitions and prove the stated properties in the style

of this section. Distinguish between those properties which hold for cosh and

sinh as functions from C to C and those which hold for cosh and sinh as

functions from R to R.

The logarithm ™

5.6

In this section we shall make use of the one dimensional chain rule.

Lemma 5.6.1. Suppose that f : R ’ R is di¬erentiable at x with derivative

f (x), that g : R ’ R is di¬erentiable at y with derivative g (y) and that

f (x) = y. Then g —¦ f is di¬erentiable at x with derivative f (x)g (y).

103

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

d

In traditional notation g(f (x)) = f (x)g (f (x)). We divide the proof

dx

into two parts.

Lemma 5.6.2. Suppose that the hypotheses of Lemma 5.6.1 hold and, in

addition, f (x) = 0. Then the conclusion of Lemma 5.6.1 holds.

Proof. Since

f (x + h) ’ f (x)

’ f (x) = 0

h

we can ¬nd a δ > 0 such that

f (x + h) ’ f (x)

=0

h

for 0 < |h| < δ and so, in particular, f (x + h) ’ f (x) = 0 for 0 < |h| < δ.

Thus if 0 < |h| < δ. we may write

g(f (x + h)) ’ g(f (x)) g(f (x + h)) ’ g(f (x)) f (x + h) ’ f (x)

= . ()

f (x + h) ’ f (x)

h h

Now f is di¬erentiable and so continuous at x, so f (x + h) ’ f (x) ’ 0 as

h ’ 0. It follows, by using standard theorems on limits (which the reader

should identify explicitly), that

g(f (x + h)) ’ g(f (x))

’ g (f (x))f (x)

h

as h ’ 0 and we are done.

Unfortunately the proof of Lemma 5.6.2 does not work in general for

Lemma 5.6.1 since we then have no guarantee that f (x + h) ’ f (x) = 0, even

for small h, and so we cannot use equation 7 . We need a separate proof

for this case.

Lemma 5.6.3. Suppose that the hypotheses of Lemma 5.6.1 hold and, in

addition, f (x) = 0. Then the conclusion of Lemma 5.6.1 holds.

I outline a proof in the next exercise, leaving the details to the reader.

7

Hardy™s Pure Mathematics says ™The proof of [the chain rule] requires a little care™ and

carries the rueful footnote ˜The proofs in many text-books (and in the ¬rst three editions

of this book) are inaccurate™. This is the point that the text-books overlooked.

104 A COMPANION TO ANALYSIS

Exercise 5.6.4. We prove Lemma 5.6.3 by reductio ad absurdum. To this

end, suppose that the hypotheses of the lemma hold but the conclusion is false.

(i) Explain why we can ¬nd an > 0 and a sequence hn ’ 0 such that

hn = 0 and

g(f (x + hn )) ’ g(f (x))

>

hn

for each n ≥ 0.