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

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

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

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.

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