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s=0
converges for all r ≥ 0. Now observe that if N ≥ M
M N N N
ars ¤ ars .
r=0 s=0 r=0 s=0


Thus, allowing N ’ ∞ whilst keeping M ¬xed, (remember from Lemma 4.1.9 (iv)
that the limit of a ¬nite sum is the sum of the limits) we have

M N N
ars ¤ lim ars .
N ’∞
r=0 s=0 r=0 s=0

∞ ∞
Thus, since an increasing sequence bounded above tends to a limit, ars
r=0 s=0
converges.
89
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Now we show that (ii) implies (i). This is easier since we need only note
that
∞ ∞ ∞
N N N
ars ¤ ars ¤ ars
r=0 s=0 r=0 s=0 r=0 s=0

and use, again, the fact that an increasing sequence bounded above tends to
a limit.
Finally we note that, if (ii) is true, the fact that absolute convergence
implies convergence immediately shows that ∞ ars converges for all r ≥ 0.
s=0
We also know that
∞ ∞
ars ¤ ars ,
s=0 s=0
∞ ∞
so the comparison test tells us that ars converges.
r=0 s=0

Lemma 5.3.8. Let ars ∈ Rm for r, s ≥ 0. If the equivalent statements (i)
and (ii) in Lemma 5.3.7 hold, then
∞ ∞ N N
ars = lim ars .
N ’∞
r=0 s=0 r=0 s=0

Proof. By Lemma 5.3.7 and 5.3.5, we know that ∞ ∞ N N
s=0 ars and limN ’∞ ars
r=0 r=0 s=0
exist. We need only prove them equal.
We prove this by using the dominated convergence theorem again. Set
ars (N ) = ars if 0 ¤ r ¤ N and 0 ¤ s ¤ N , and ars (N ) = 0 otherwise. If r
is ¬xed,
∞ ∞
N
ars ’
ars (N ) = ars .
s=0 s=0 s=0

But we know that
∞ ∞
N N
ars (N ) ¤ ars (N ) ¤ ars ¤ ars
s=0 s=0 s=0 s=0

and ∞ ∞
ars converges. Thus the dominated convergence theorem
r=0 s=0

br (N ) with br (N ) = ∞ ars (N ) gives
applied to r=0 s=0

∞ ∞ ∞ ∞
N N
ars (N ) ’
ars = ars
r=0 s=0 r=0 s=0 r=0 s=0

as required
90 A COMPANION TO ANALYSIS

We have proved the following useful theorem.

Lemma 5.3.9. (Fubini™s theorem for sums.) Let ars ∈ Rm . If any one
of the following three objects,
∞ ∞ ∞ ∞
N N
lim ars , ars , ars
N ’∞
r=0 s=0 r=0 s=0 s=0 r=0

is well de¬ned, they all are, as are
∞ ∞ ∞ ∞
N N
lim ars , ars , ars .
N ’∞
r=0 s=0 r=0 s=0 s=0 r=0

Further
∞ ∞ ∞ ∞
N N
lim ars , = ars , = ars .
N ’∞
r=0 s=0 r=0 s=0 s=0 r=0

This theorem allows us to interchange the order of summation of in¬nite
sums (provided we have absolute convergence). I attach the name Fubini to
it because Fubini proved a general and far reaching theorem of which this is
a special case.

Exercise 5.3.10. (i) Set arr = 1 for all r ≥ 1 and ar r’1 = ’1 for all r ≥ 2.
Set ars = 0 otherwise. Write out the matrix with entries ars where 1 ¤ r ¤ 4,
1 ¤ s ¤ 4. Show, by direct calculation, that
∞ ∞ ∞ ∞
ars = ars .
r=1 s=1 s=1 r=1

(ii) Set b11 = 1, b1s = 0 for s ≥ 2 and

brs = 2’r+2 for 2r’2 ¤ s ¤ 2r’1 ’ 1
brs = ’2’r+1 for 2r’1 ¤ s ¤ 2r ’ 1
brs = 0 otherwise,

when r ≥ 2. Write out the matrix with entries brs where 1 ¤ r ¤ 4, 1 ¤ s ¤
8. Show by direct calculation that
∞ ∞ ∞ ∞
brs = brs .
r=1 s=1 s=1 r=1
91
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Summing over triangles rather than squares and using Lemma 5.3.5, we
obtain another version of Fubini™s theorem which we shall use in the next
section.
∞ ∞
Lemma 5.3.11. If aij converges, then writing
i=0 j=0


sn = aij ,
i+j=n


we have n=0 sn absolutely convergent and
∞ ∞ ∞
sn = aij .
n=0 i=0 j=0


We are no longer so concerned with tracing everything back to the fun-
damental axiom but that does not mean that it ceases to play a fundamental
role.

Exercise 5.3.12. We work in Q. Show that we can ¬nd x0 , x1 , x2 , x3 , . . .
an increasing sequence such that x0 = 0 and xj ’ xj’1 ¤ 2’j for j ≥ 1, but
xn does not tend to a limit.
Set x0 = 0, a1j = xj ’ xj’1 , a2j = 2’j ’ a1j , and aij = 0 for i ≥ 3.
Show that aij ≥ 0 for all i, j, that ∞ aij exists for all j and ∞ ∞
i=1 aij
i=1 j=1

exists. Show, however, that j=1 aij does not exist when i = 1 or i = 2.


The exponential function ™
5.4
We have proved many deep and interesting theorems on the properties of
continuous and di¬erentiable functions. It is somewhat embarrassing to ob-
serve that, up to now, the only di¬erentiable (indeed the only continuous
functions) which we could lay our hands on were the polynomials and their
quotients. (Even these were obtained as an afterthought in Exercise 4.2.24.
We shall use results from that exercise in this section, but all such results
will be proved in a wider context in Chapter 6.) In the next four sections we
widen our repertoire considerably.
We start with the function exp. Historically, the exponential function
was developed in connection with Napier™s marvelous invention of the loga-
rithm as a calculating tool (see Exercise 5.6.6). However, if by some historic
freak, mathematics had reached the state it was in 1850 whilst bypassing the
exponential and logarithmic functions, the exponential function might have
been discovered as follows.
92 A COMPANION TO ANALYSIS

One way to obtain new functions is to look for solutions to di¬erential
equations. Consider one of the simplest such equations y (x) = y(x). With-
out worrying about rigour (˜In a storm I would burn six candles to St George
and half a dozen to his dragon™) we might try to ¬nd a solution in the form
of a power series y(x) = ∞ aj xj .
j=0

Plausible statement 5.4.1. The general solution of the equation

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

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

xj
y(x) = a
j!
j=0


with a ∈ R.

Plausible argument. We seek for a solution of of the form y(x) = ∞ aj xj .
j=0
Assuming that we can treat a power series in the same way as we can treat
a polynomial, we di¬erentiate term by term to obtain y (x) = ∞ jaj xj’1 .
j=1
Equation thus becomes
∞ ∞
aj x j = (j + 1)aj+1 xj
j=0 j=0


which may be rewritten as

(aj ’ (j + 1)aj+1 )xj = 0.
j=0


Now, if a polynomial P (x) vanishes for all values of x, all its coe¬cients are
zero. Assuming that the result remains true for power series, we obtain

aj ’ (j + 1)aj+1 = 0

for all j ≥ 0 and a simple induction gives aj = a0 /j!. Setting a = a0 , we
have the result.

Later in the book we shall justify all of the arguments used above (see Theo-
rem 11.5.11 and Exercise 11.5.13) but, for the moment, we just use them as
a heuristic tool.
We can make an immediate observation.
93
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Lemma 5.4.2. The power series

zj
j!
j=0

has in¬nite radius of convergence.
Proof. If z = 0, then, setting uj = z j /j!, we have |uj+1 |/|uj | = |z|/(j+1) ’ 0
as j ’ ∞. Thus, by the ratio test, ∞ z j /j! is absolutely convergent, and
j=0
so convergent, for all z.

j=0 z /j! for all z ∈ C. A little playing
j
We may thus de¬ne e(z) =
around with formulae would lead to the key observation.
Lemma 5.4.3. If z, w ∈ C, then
e(z)e(w) = e(z + w)
Proof. Observe that ∞ |z r |/r! converges to e(|z|) and ∞
|ws |/s! con-
r=0 s=0
verges to e(|w|). Thus, writing ars = (|z r |/r!)(|w s |/s!),
∞ ∞ ∞ ∞
|z r | |w|s
|ars | =
r! s!
r=0 s=0 r=0 s=0
∞ ∞
|z r | |w|s
=
r! s!
r=0 s=0

|z r |
= e(|w|) = e(|z|)e(|w|).
r!
r=0

Thus ∞ ∞
s=1 ars converges absolutely and, by Lemma 5.3.11, it follows
r=1
that, if we write
cn = ars ,
r+s=n

we have n=0 cn absolutely convergent and
∞ ∞ ∞
cn = aij .
n=1 i=1 j=1

But
n n
z r wn’r
cn = ar(n’r) =
r! (n ’ r)!
r=0 r=0
n
n r n’r (z + w)n
1
= zw = ,
n! r n!
r=0
94 A COMPANION TO ANALYSIS

and so
∞ ∞ ∞ ∞ ∞
zr ws
e(z)e(w) = = ars = cn = e(z + w).
r! s!
r=0 s=0 r=1 s=1 n=1




It should be noticed that the essential idea of the proof is contained in its
last sentence. The rest of the proof is devoted to showing that what ought
to work does, in fact, work.

Exercise 5.4.4. (Multiplication of power series.) The idea of the pre-
vious lemma can be generalised.
(i) Let »n , µn ∈ C and γn = n »j µn’j . Show, by using Lemma 5.3.11,
j=0
or otherwise, that, if ∞ »n and ∞ µn are absolutely convergent so is
n=0 n=0

γn and
n=0

∞ ∞ ∞
»n µn = γn .
n=0 n=0 n=0


(ii) Suppose that ∞ an z n has radius of convergence R and that ∞ bn z n
n=0 n=0

has radius of convergence S. Explain why, if |w| < min(R, S), both n=0 an wn
and ∞ bn wn converge absolutely. Deduce that, if we write cn = r+s=n ar bs
n=0
then ∞ cn wn converges absolutely and
n=0

∞ ∞ ∞
n n
cn w n .
an w bn w =
n=0 n=0 n=0

∞ n
Thus n=0 cn z has radius of convergence at least min(R, S) and, if
|w| < min(R, S), the formula just displayed applies. This result is usually
stated as ˜two power series can be multiplied within their smaller radius of
convergence™.

Exercise 5.4.5. Use Exercise 5.4.4 directly to prove Lemma 5.4.3.

z n has radius of convergence 1
Exercise 5.4.6. (i) Prove directly that n=0
and

1
zn =
1’z
n=0

for |z| < 1.
95
Please send corrections however trivial to twk@dpmms.cam.ac.uk

∞ n
(ii) Use Exercise 5.4.4 to show that n=0 (n + 1)z has radius of conver-
gence at least 1 and that

1
(n + 1)z n =
(1 ’ z)2
n=0

for |z| < 1. Show that ∞ (n + 1)z n has radius of convergence exactly 1.
n=0
(iii) It is trivial that 1 + z has radius of convergence ∞. Use this obser-
vation and others made earlier in the exercise to show that, for appropriate
choices of ∞ an z n and ∞ bn z n , we can have, in the notation of Exer-
n=0 n=0
cise 5.4.4,
(a) R = 1, S = ∞, ∞ cn z n has radius of convergence ∞.
n=0

(b) R = 1, S = ∞, n=0 cn z n has radius of convergence 1.
[See also Exercise K.234.]
The next exercise gives an algebraic interpretation of Lemma 5.4.3.
Exercise 5.4.7. Check that C is an Abelian group under addition. Show that
C \ {0} is an Abelian group under multiplication. Show that e : (C, +) ’
(C \ {0}, —) is a homomorphism.
The following estimate is frequently useful.
Lemma 5.4.8. If |z| < n/2, then
n’1
zj 2|z|n
e(z) ’ ¤ .
j! n!
j=0

Proof. Most mathematicians would simply write
∞ ∞
n’1
zj zj |z|j
e(z) ’ ¤
=
j! j! j!
j=0 j=n j=n

|z|n |z|k

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