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(ii) Explain why f (x + hn ) = f (x) for each n ā„ 0.
(iii) Use the method of proof of Lemma 5.6.2 to derive a contradiction.
The rather ugly use of reductio ad absurdum in Exercise 5.6.4 can be
avoided by making explicit use of the ideas of Exercise K.23.
Note that, in this section, we only use the special case of the chain rule
given in Lemma 5.6.2. I believe that the correct way to look at the chain rule
is by adopting the ideas of Chapter 6 and attacking it directly as we shall do
in Lemma 6.2.10. We now move on to the main subject of this section.
Since e : R ā’ R++ is a bijection (indeed by Exercise 5.4.11 a group
isomorphism) it is natural to look at its inverse. Let us write l(x) = e ā’1 (x)
for x ā (0, ā) = R++ . Some of the properties of l are easy to obtain. (Here
and later we use the properties of the function e obtained in Exercise 5.4.10.)
Exercise 5.6.5. (i) Explain why l : (0, ā) ā’ R is a bijection.
(ii) Show that l(xy) = l(x) + l(y) for all x, y > 0.
(iii) Show that l is a strictly increasing function.
Exercise 5.6.6. No one who went to school after 1960 can really appreci-
ate the immense diļ¬erence between the work involved in hand multiplication
without logarithms and hand multiplication if we are allowed to use loga-
rithms. The invention of logarithms was an important contribution to the
scientiļ¬c revolution. When Henry Briggs (who made a key simpliļ¬cation)
visited Baron Napier (who invented the idea) ā˜almost one quarter of an hour
was spent, each beholding [the] other . . . with admiration before one word
was spoke, at last Mr Briggs began.
ā˜My lord, I have undertaken this long Journey purposely to see your Per-
son, and to know by what Engine of Wit or Ingenuity you came ļ¬rst to think
of this most excellent Help unto Astronomy, viz., the Logarithms; but, my
Lord, being by you found out, I wonder nobody else found it out before, when
now known it is so easy.ā™(Quotation from 9.E.3 of [16].)
(i) As Briggs realised, calculations become a little easier if we use log 10
deļ¬ned by
log10 x = l(x)/l(10)
105
Please send corrections however trivial to twk@dpmms.cam.ac.uk

for x > 0. Show that log10 xy = log10 x + log10 y for all x, y > 0 and that
log10 10r x = r + log10 x.
(ii) Multiply 1.3245 by 8.7893, correct to ļ¬ve signiļ¬cant ļ¬gures, without
using a calculator.
(iii) To multiply 1.3245 by 8.7893 using logarithms, one looked up log 10 1.3245
and log10 8.7893 in a table of logarithms. This was quick and easy, giving

log10 1.3245 ā 0.1220520, log10 8.7893 ā 0.9439543.

log10 (1.3245 Ć— 8.7893) = log10 1.3245 + log10 8.7893
ā 0.1220520 + 0.9439543 = 1.0660063.

A quick and easy search in a table of logarithms (or, still easier a table of
inverse logarithms, the so called antilogarithms) showed that

log10 1.164144 ā .0660052, log10 1.164145 ā .0660089

so that

log10 11.64144 ā 1.0660052, log10 11.64145 ā 1.0660089

and, correct to ļ¬ve signiļ¬cant ļ¬gures, 1.3245 Ć— 8.7893 = 11.6414.
(iv) Repeat the exercise with numbers of your own choosing. You may
use the ā˜log10 ā™ (often just called ā˜logā™) function on your calculator and the
ā˜inverse log10 ā™ (often called ā˜10x ā™) but you must do the multiplication and
addition by hand. Notice that you need one (or, if you are being careful, two)
more extra ļ¬gures in your calculations than there are signiļ¬cant ļ¬gures in
[There are some additional remarks in Exercises 5.7.7 and K.85.]

Other properties require a little more work.

Lemma 5.6.7. (i) The function l : (0, ā) ā’ R is continuous.
(ii) The function l is everywhere diļ¬erentiable with
1
l (x) = .
x
Proof. (i) We wish to show that l is continuous at some point x ā (0, ā).
To this end, let Ī“ > 0 be given. Since l is increasing, we know that, if

e(l(x) + Ī“) > y > e(l(x) ā’ Ī“),
106 A COMPANION TO ANALYSIS

we have
l e(l(x) + Ī“) > l(y) > l e(l(x) ā’ Ī“)
and so
l(x) + Ī“ > l(y) > l(x) ā’ Ī“.
Now e is strictly increasing, so we can ļ¬nd Ī·(Ī“) > 0 such that
e(l(x) + Ī“) > x + Ī·(Ī“) > x = l(e(x)) > x ā’ Ī·(Ī“) > e(l(x) ā’ Ī“).
Combining the results of the two previous sentences, we see that, if |x ā’ y| <
Ī·(Ī“), then |l(x) ā’ l(y)| < Ī“. Since Ī“ was arbitrary, l is continuous at x.
(ii) We shall use the result that, if g is never zero and g(x + h) ā’ a as
h ā’ 0, then, if a = 0, 1/g(x + h) ā’ 1/a as h ā’ 0. Observe that, since l is
continuous, we have
l(x + h) ā’ l(x) ā’ 0
and so
l(x + h) ā’ l(x) l(x + h) ā’ l(x) 1 1 1
ā’
= = =
e(l(x + h)) ā’ e(l(x))
h e (l(x)) e(l(x)) x
as h ā’ 0.
By using the ideas of parts (iv), (v) and (vi) of Exercise 5.4.10 together
with parts (i) and (iii) of Exercise 5.6.5 and both parts of Lemma 5.6.7, we
get the following general result.
Exercise 5.6.8. (One dimensional inverse function theorem.) Sup-
pose that f : [a, b] ā’ [c, d] is continuous and f is diļ¬erentiable on (a, b) with
f (x) > 0 for all x ā (a, b) and f (a) = c, f (b) = d. Show that f is a bijection,
that f ā’1 : [c, d] ā’ [a, b] is continuous and that f ā’1 is diļ¬erentiable on (c, d)
with
1
(f ā’1 ) (x) = .
f (f ā’1 (x))
We shall give a diļ¬erent proof of this result in a more general (and, I would
claim, more instructive) context in Theorem 13.1.13. Traditionally, the one
dimensional inverse function theorem is illustrated, as in Figure 5.1, by taking
the graph y = f (x) with tangent shown at (f ā’1 (x0 ), x0 ) and reļ¬‚ecting in the
angle bisector of the x and y axes to obtain the graph y = f ā’1 (x) with
tangent shown at (x0 , f (x0 )).
Although the picture is suggestive, this is one of those cases where (at
the level of proof we wish to use) a simple picture is inadequate.
107
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Figure 5.1: The one dimensional inverse function theorem

Exercise 5.6.9. Go through Exercise 5.6.8 and note where you used the
mean value theorem and the intermediate value theorem.

Exercise 5.6.10. (i) Write A = {x ā Q : 2 ā„ x ā„ 1} and B = {x ā
Q : 4 ā„ x ā„ 1}. Deļ¬ne f : A ā’ B by f (x) = x2 . Show that f is strictly
increasing on A, that f (1) = 1 and f (2) = 4, that f is diļ¬erentiable on A
with f (x) ā„ 2 for all x ā A and that f : A ā’ B is injective yet f is not
surjective.
(ii) Deļ¬ne f : Q ā’ Q by

for x < 0, x2 > 2,
f (x) = x + 1
for x2 < 2,
f (x) = x
for x > 0, x2 > 2.
f (x) = x ā’ 1

Show that f (x) ā’ ā’ā as x ā’ ā’ā, that f (x) ā’ ā as x ā’ ā, that f
is everywhere diļ¬erentiable with f (x) = 1 for all x and that f : Q ā’ Q is
surjective yet f is not injective8 .

Initially we deļ¬ned the exponential and trigonometric functions as maps
C ā’ C although we did not make much use of this (they are very important
8
These examples do not exhaust the ways in which Figure 5.1 is an inadequate guide
to what can happen without the fundamental axiom of analysis [32].
108 A COMPANION TO ANALYSIS

in more advanced work) and switched rapidly to maps R ā’ R. We did
nothing of this sort for the logarithm.
The most obvious attempt to deļ¬ne a complex logarithm fails at the ļ¬rst
hurdle. We showed that, working over R, the map exp : R ā’ (0, ā) is
bijective, so that we could deļ¬ne log as the inverse function. However, we
know (see Exercise 5.5.7) that, working over C, the map exp : C ā’ C \ {0}
is surjective but not injective, so no inverse function exists.
Exercise 5.6.11. By using the fact that exp 2Ļi = 1 = exp 0, show that
there cannot exist a function L : C \ {0} ā’ C with L(exp z) = z for all
z ā C.
However, a one-sided inverse can exist.
Exercise 5.6.12. (i) If we set L0 (r exp iĪø) = log r + iĪø for r > 0 and 2Ļ >
Īø ā„ 0, show that L0 : C\{0} ā’ C is a well deļ¬ned function with exp(L0 (z)) =
z for all z ā C \ {0}.
(ii) Let n be an integer. If we set Ln (r exp iĪø) = L0 (r exp iĪø)+2Ļin, show
that Ln : C \ {0} ā’ C is a well deļ¬ned function with exp(Ln (z)) = z for all
z ā C \ {0}.
(iii) If we set M (r exp iĪø) = log r + iĪø for r > 0 and 3Ļ > Īø ā„ Ļ, show
that M : C \ {0} ā’ C is a well deļ¬ned function with exp(M (z)) = z for all
z ā C \ {0}.
The functions Ln and M in the last exercise are not continuous everywhere
and it is natural to ask if there is a continuous function L : C \ {0} ā’ C
with exp(L(z)) = z for all z ā C \ {0}. The reader should convince herself,
by trying to deļ¬ne L(exp iĪø) and considering what happens as Īø runs from 0
to 2Ļ, that this is not possible. The next exercise crystallises the ideas.
Exercise 5.6.13. Suppose, if possible, that there exists a continuous L :
C \ {0} ā’ C with exp(L(z)) = z for all z ā C \ {0}.
(i) If Īø is real, show that L(exp(iĪø)) = i(Īø + 2Ļn(Īø)) for some n(Īø) ā Z.
(ii) Deļ¬ne f : R ā’ R by
L(exp iĪø) ā’ L(1)
1
ā’Īø .
f (Īø) =
2Ļ i
Show that f is a well deļ¬ned continuous function, that f (Īø) ā Z for all Īø ā R,
that f (0) = 0 and that f (2Ļ) = ā’1.
(iii) Show that the statements made in the last sentence of (ii) are in-
compatible with the intermediate value theorem and deduce that no function
can exist with the supposed properties of L.
(iv) Discuss informally what connection, if any, the discussion above has
with the existence of the international date line.
109
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 5.6.13 is not an end but a beginning of much important mathe-
matics. In due course it will be necessary for the reader to understand both
the formal proof that, and the informal reasons why, no continuous L can
exist.

Powers ā™„
5.7
How should we deļ¬ne ab for a > 0 and b any real number? Most people
would say that we should ļ¬rst deļ¬ne ab for b rational and then extend ā˜by
continuityā™ to non-rational b. This can be done, even with the few tools at
our disposal, but it requires hard work to deļ¬ne ab this way and still more
hard work to obtain its properties. When we have more powerful tools at
our disposal (uniform convergence and the associated theorems) we shall see
how to make this programme work in Exercises K.227 to K.229 but, even
then, it requires careful thought.
There are, I think, various reasons why the direct approach is hard.
(1) The ļ¬rst point is mainly psychological. We need to consider ab as a
function of two variables a and b. When we deļ¬ne an , we think of the integers
n as ļ¬xed and a as varying and the same is true when we deļ¬ne ab with b
rational. However, when we want to deļ¬ne ab ā˜by continuityā™, we think of a
as ļ¬xed and b as varying.
(2) The second point is mathematical. The fact that a function is con-
tinuous on the rationals does not mean that it has a continuous extension
to the reals9 . Consider our standard example, the function f : Q ā’ Q of
Example 1.1.3. We know that f is continuous but there is no continuous
function F : R ā’ R with F (x) = f (x) for x ā Q.

Exercise 5.7.1. (i) Prove this statement by observing that, if F is continu-
ous, F (xn ) ā’ F (2ā’1/2 ) whenever xn ā’ 2ā’1/2 , or otherwise.
(ii) Find a function g : Q ā’ Q which is diļ¬erentiable with continuous
derivative such that there is a continuous function G : R ā’ R with G(x) =
g(x) for x ā Q but any such function G is not everywhere diļ¬erentiable.

However, the fact that I think something is hard does not prove that it
is hard. I suggest that the reader try it for herself. (She may well succeed,
all that is required is perseverance and a cool head. I simply claim that the
exercise is hard, not that it is impossible.)
9
I can still remember being scolded by my research supervisor for making this partic-
ular mistake. (The result is true if we replace ā˜continuityā™ by ā˜uniform continuityā™. See
Exercise K.56.)
110 A COMPANION TO ANALYSIS

Assuming that the reader agrees with me, can we ļ¬nd another approach?
We obtained the exponential and trigonometric functions as the solution of
diļ¬erential equations. How does this approach work here? The natural choice
of diļ¬erential equation, if we wish to obtain y(x) = xĪ± , is

xy (x) = Ī±y(x)

(Here Ī± is real and y : (0, ā) ā’ (0, ā).)
Tentative solution. We can rewrite as
y (x) Ī±
ā’ = 0.
y(x) x
Using the properties of logarithm and the chain rule, this gives
d
(log y(x) ā’ Ī± log x) = 0
dx
so, by the mean value theorem,

log y(x) ā’ Ī± log x = C

where C is constant. Applying the exponential function and taking A =
exp C, we obtain

y(x) = A exp(Ī± log x)

where A is a constant.
Exercise 5.7.2. Check, by using the chain rule, that y(x) = A exp(Ī± log x)
is indeed a solution of .
This suggests very strongly indeed that we should deļ¬ne xĪ± = exp(Ī± log x).
In order to avoid confusion, we adopt our usual policy of light disguise
and investigate the properties of functions rĪ± : (0, ā) ā’ (0, ā) deļ¬ned
by rĪ± (x) = exp(Ī± log x) [Ī± real].
Exercise 5.7.3. (Index laws.) If Ī±, Ī² ā R, show that
(i) rĪ±+Ī² (x) = rĪ± (x)rĪ² (x) for all x > 0.
(ii) rĪ±Ī² (x) = rĪ± (rĪ² (x)) for all x > 0.
Exercise 5.7.4. (Consistency.) Suppose that n, p and q are integers with
n ā„ 0 and q > 0. Show that
(i) r1 (x) = x for all x > 0.
(ii) rn+1 (x) = xrn (x) for all x > 0.
111
Please send corrections however trivial to twk@dpmms.cam.ac.uk

n

(iii) rn (x) = x Ć— x Ć— Ā· Ā· Ā· Ć— x for all x > 0.
1
(iv) rā’n (x) = for all x > 0.
rn (x)
(v) rq (rp/q (x)) = rp (x) for all x > 0.
Explain brieļ¬‚y why this means that writing rp/q (x) = xp/q is consistent

Exercise 5.7.5. Suppose that Ī± is real. Show that
(i) rĪ± (xy) = rĪ± (x)rĪ± (y) for all x, y > 0.
(ii) r0 (x) = 1 for all x > 0.
(iii) rĪ± is everywhere diļ¬erentiable and xrĪ± (x) = Ī±rĪ± (x) and rĪ± (x) =
Ī±rĪ±ā’1 (x) for all x > 0.

Exercise 5.7.6. (i) If x > 0 is ļ¬xed, show that rĪ± (x) is a diļ¬erentiable
function of Ī± with
d
rĪ± (x) = rĪ± (x) log x.
dĪ±
(ii) If Ī± > 0 and Ī± is kept ļ¬xed, show that rĪ± (x) is an increasing function
of x. What happens if Ī± < 0?
(iii) If x > 1 and x is kept ļ¬xed, show that rĪ± (x) is an increasing function
of Ī±. What happens if 0 < x < 1?
(iv) If we write e = exp 1 show that exp x = re (x) (or, in more familiar
terms, exp x = ex ).

Exercise 5.7.7. Take two rulers A and B marked in centimeters (or some
other convenient unit) and lay them marked edge to marked edge. If we slide
the point marked 0 on ruler B until it is opposite the point marked x on ruler
A, then the point marked y on ruler B will be opposite the point marked x + y
on ruler A. We have invented an adding machine.
Now produce a new ruler A by renaming the point marked x as 10x (thus
the point marked 0 on A becomes the point marked 1 on A and the point
marked 3 on A becomes the point marked 1000 on A ). Obtain B from B in
the same way. If we slide the point marked 1 on ruler B until it is opposite
the point marked 10x on ruler A , then the point marked 10y on ruler B will
be opposite the point marked 10x+y on ruler A . Explain why, if a, b > 0 and
we slide the point marked 1 on ruler B until it is opposite the point marked
a on ruler A , then the point marked b on ruler B will be opposite the point
marked ab on ruler A . We have invented an multiplying machine.
(i) How would you divide a by b using this machine?
(ii) Does the number 10 play an essential role in the device?
112 A COMPANION TO ANALYSIS

(iii) Draw a line segment CD of some convenient length to represent the
ruler A . If C corresponds to 1 and D to 10, draw, as accurately as you can,
the points corresponding to 2, 3, . . . , 9.
The device we have described was invented by Oughtred some years after
Napierā™s discovery of the logarithm and forms the basis for the ā˜slide ruleā™.
From 1860 to 1960 the slide rule was the emblem of the mathematically com-
petent engineer. It allowed fast and reasonably accurate ā˜back of an envelopeā™
calculations.

Exercise 5.7.8. By imitating the argument of Exercise 5.6.13 show that
there is no continuous function S : C ā’ C with S(z)2 = z for all z ā C.
(In other words, we can not deļ¬ne a well behaved square root function on the
complex plane.)

Exercise 5.7.9. Exercise 5.7.8 shows, I think, that we can not hope to ex-
tend our deļ¬nition of rĪ± (x) with x real and strictly positive and Ī± real to
some well behaved rĪ± (z) with Ī± and z both complex. We can, however, ex-
tend our deļ¬nition to the case when x is still real and strictly positive but we
allow Ī± to be complex. Our deļ¬nition remains the same

rĪ± (x) = exp(Ī± log x)

but only some of our previous statements carry over.
(i) If Ī±, Ī² ā C, show that rĪ±+Ī² (x) = rĪ± (x)rĪ² (x) for all x > 0. Thus
part (i) of Exercise 5.7.3 carries over.
(ii) Explain carefully why the statement in part (ii) of Exercise 5.7.3

?
rĪ±Ī² (x) = rĪ± (rĪ² (x))

makes no sense (within the context of this question) if we allow Ī± and Ī² to
range freely over C. Does it make sense and is it true if Ī² ā R and Ī± ā C?
Does it make sense and is it true if Ī± ā R and Ī² ā C?
(iii) Find which parts of Exercises 5.7.5 and 5.7.6 continue to make sense
in the more general context of this question and prove them.
(iv) Show that, if u and v are real and e = exp(1), then exp(u + iv) =
ru+iv (e). We have thus converted the mnemonic

exp(z) = ez

into a genuine equality.
113
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 5.7.10. According to a well known story10 , the Harvard mathe-
matician Benjamin Pierce chalked the formula

eiĻ + 1 = 0

on the board and addressed his students as follows.

Gentleman, that is surely true, it is absolutely paradoxical; we
cannot understand it, and we do not know what it means, but we
have proved it, and therefore we know it must be the truth.

(i) In the context of this chapter, what information is conveyed by the
formula

exp(iĻ) + 1 = 0?

(What does exp mean, what does Ļ mean and what does exp(iĻ) mean?)
(ii) In the context of this chapter, what information is conveyed by the
formula

eiĻ + 1 = 0?

There is a superb discussion of the problem of deļ¬ning xĪ± in Kleinā™s
Elementary Mathematics from an Advanced Standpoint [28].

The fundamental theorem of algebra ā™„
5.8
It is in the nature of a book like this that much of our time is occupied in
proving results which the ā˜physicist in the streetā™ would consider obvious. In
this section we prove a result which is less obvious.

Theorem 5.8.1. (The fundamental theorem of algebra.) Suppose that
n ā„ 1, a0 , a1 , . . . , an ā C and an = 0. Then the equation

an z n + anā’1 z nā’1 + Ā· Ā· Ā· + a0 = 0

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