Since |w| < ρ, we are done.

Remark: The proof above is more cunning than is at ¬rst apparent. Roughly

∞ n

speaking, it is often hard to prove directly that n=0 an z has a certain

property for all |z| < R, the radius of convergence, but relatively easy to show

that N an z n has a certain property for all |z| < R , whenever R < R.

n=0

However, if we choose R1 < R2 < . . . with RN ’ R we then know that

∞ n

n=0 an z will have the property for all

∞

z∈ {z : |z| < RN } = {z : |z| < R},

r=1

and we are done. (We give two alternative proofs of Theorem 11.5.11 in

Exercise K.230 and Exercise K.231.)

Here are two useful corollaries.

293

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

Exercise 11.5.12. Suppose that an ∈ C and ∞ an z n has radius of con-

n=0

vergence R > 0. Set „¦ = {z : |z| < R} and de¬ne f : „¦ ’ C by

∞

an z n .

f (z) =

n=0

Show that f is in¬nitely di¬erentiable on „¦ and an = f (n) (0)/n!.

In other words, if f can be expanded in a power series about 0, then that

power series must be the Taylor series.

Exercise 11.5.13. (Uniqueness of power series.)Suppose that an ∈ C.

and ∞ an z n has radius of convergence R > 0. Set „¦ = {z : |z| < R} and

n=0

de¬ne f : „¦ ’ C by

∞

an z n .

f (z) =

n=0

If there exists a δ with 0 < δ ¤ R such that f (z) = 0 for all |z| < δ, show,

by using the preceding exercise, or otherwise, that an = 0 for all n ≥ 0. [In

Exercise K.239 we give a stronger result with a more direct proof.]

By restricting our attention to the real axis, we can obtain versions of all

these results for real power series.

Lemma 11.5.14. Suppose that an ∈ R.

(i) Either ∞ an xn converges for all x ∈ R (in which case we say the

n=0

series has in¬nite radius of convergence) or there exists an R ≥ 0 such that

∞ n

n=0 an x converges for |x| < R and diverges for |x| > R (in which case we

say the series has radius of convergence R).

(ii) If 0 ¤ ρ < R then ∞ an xn converges uniformly on [’ρ, ρ].

n=0

∞ n

(iii) The sum f (x) = n=0 an x is di¬erentiable, term by term, on

(’R, R).

(iv) If R > 0, f is in¬nitely di¬erentiable and an = f (n) (0)/n!.

(v) If f vanishes on (’δ, δ) where 0 < δ ¤ R, then an = 0 for all n.

Part (iv) should be read in conjunction with Cauchy™s example of a well

behaved function with no power series expansion round 0 (Example 7.1.5).

The fact that we can di¬erentiate a power series term by term is important

for two reasons. The ¬rst is that there is a very beautiful and useful theory

of di¬erentiable functions from C to C (called ˜Complex Variable Theory™

or ˜The Theory of Analytic Functions™). In the initial development of the

294 A COMPANION TO ANALYSIS

theory it is not entirely clear that there are any interesting functions for the

theory to talk about. Power series provide such interesting functions.

The second reason is that it provides a rigorous justi¬cation for the use of

power series in the solution of di¬erential equations by methods of the type

employed on page 92.

∞ zn

Exercise 11.5.15. (i) Show that the sum has in¬nite radius of

n=0 n!

convergence.

(ii) Let us set

∞

zn

e(z) =

n!

n=0

for all z ∈ C. Show that e is everywhere di¬erentiable and e (z) = e(z).

(iii) Use the mean value theorem of Exercise 11.5.5 to show that the

function f de¬ned by f (z) = e(a ’ z)e(z) is constant. Deduce that e(a ’

z)e(z) = e(a) for all z ∈ C and a ∈ C and conclude that

e(z)e(w) = e(z + w)

for all z, w ∈ C.

Here is another example.

Example 11.5.16. Let ± ∈ C. Solve the di¬erential equation

(1 + z)f (z) = ±f (z)

subject to f (0) = 1.

Solution. We look for a solution of the form

∞

an z n

f (z) =

n=0

with radius of convergence R > 0. We di¬erentiate term by term within the

radius of convergence to get

∞ ∞

n’1

an z n ,

(1 + z) nan z =±

n=1 n=0

whence

∞

((± ’ n)an ’ (n + 1)an+1 )z n = 0

n=0

295

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

for all |z| < R. By the uniqueness result of Exercise 11.5.13, this gives

(± ’ n)an ’ (n + 1)an+1 = 0,

so

±’n

an+1 = an ,

n+1

and, by induction,

n’1

1

(± ’ j),

an = A

n! j=0

for some constant A. Since f (0) = 1, we have A = 1 and

∞ n’1

1

(± ’ j)z n .

f (z) =

n!

n=0 j=0

If ± is a positive integer N , say, then aj = 0 for j ≥ N + 1 and we get

the unsurprising result

N

Nn

z = (1 + z)N .

f (z) =

n

n=0

From now on we assume that ± is not a positive integer. If z = 0,

|1 ’ ±n’1 |

|an+1 z n+1 | |± ’ n|

|z| = |z| ’ |z|

=

1 + n’1

|an z n | n+1

as n ’ ∞, so, by using the ratio test, ∞ an z n has radius of convergence

n=0

1.

We have shown that, if there is a power series solution, it must be

∞ n’1

1

(± ’ j)z n .

f (z) =

n!

n=0 j=0

Di¬erentiating term by term, we see that, indeed, the f given is a solution

valid for |z| < 1.

We have left open the possibility that the di¬erential equation of Exam-

ple 11.5.16 might have other solutions (such solutions would not have Taylor

expansions). The uniqueness of the solution follows from general results de-

veloped later in this book (see Section 12.2). However there is a simple proof

of uniqueness in this case.

296 A COMPANION TO ANALYSIS

Example 11.5.17. (i) Write D = {z : |z| < 1}. Let ± ∈ C. Suppose that

f± : D ’ C satis¬es

(1 + z)f± (z) = ±f± (z)

and f± (0) = 1, whilst g’± : D ’ C satis¬es

(1 + z)g’± (z) = ’±g’± (z)

and g’± (0) = 1. Use the mean value theorem of Exercise 11.5.5 to show that

f± (z)g’± (z) = 1 for all z ∈ D and deduce that the di¬erential equation

(1 + z)f (z) = ±f (z),

subject to f (0) = 1, has exactly one solution on D.

(ii) If ±, β ∈ C show, using the notation of part (i), that

f±+β (z) = f± (z)fβ (z)

for all z ∈ D. State and prove a similar result for f± (fβ (z)).

Restricting to the real axis we obtain the following version of our results.

Lemma 11.5.18. Let ± be a real number. Then the di¬erential equation

(1 + x)f (x) = ±f (x),

subject to f (0) = 1, has exactly one solution f : (’1, 1) ’ R which is given

by

∞ n’1

1

(± ’ j)xn .

f (x) =

n!

n=0 j=0

In Section 5.7 we developed the theory of the function r± (x) = x± for

x > 0 and ± real. One of these properties is that

xr± (x) = ±r± (x)

for all x > 0. We also have r± (0) = 1. Thus, if g± (x) = r± (1 + x), we have

(1 + x)g± (x) = ±g± (x)

for all x ∈ (’1, 1) and g± (0) = 1. Lemma 11.5.18 thus gives the following

well known binomial expansion.

297

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

Lemma 11.5.19. If x ∈ (’1, 1), then

∞ n’1

1

±

(± ’ j)xn .

(1 + x) =

n!

n=0 j=0

Exercise 11.5.20. Use the same ideas to show that

∞

xn

log(1 ’ x) = ’

n

n=1

for x ∈ (’1, 1).

Exercise 11.5.21. (i) If you are unfamiliar with the general binomial ex-

pansion described in Lemma 11.5.19, write out the ¬rst few terms explicitly

in the cases ± = ’1, ± = ’2, ± = ’3, ± = 1/2 and ± = ’1/2. Otherwise,

go directly to part (ii).

(ii) Show that

2 3

1 2x 13 2x 135 2x

+— +——

1+ + ...

1 + x2 1 + x2 1 + x2

2 24 246

converges to (1 + x2 )/(1 ’ x2 ) if |x| < 1 but converges to (1 + x2 )/(x2 ’ 1) if

|x| > 1. In [24], Hardy quotes this example to show the di¬culties that arise

if we believe that equalities which are true in one domain must be true in all

domains.

Exercise 11.5.22. Use Taylor™s theorem with remainder to obtain the ex-

pansions for (1 + x)± and log(1 ’ x).

[This is a slightly unfair question since the forms of the Taylor remainder

given in this book are not particularly well suited to the problem. If the reader

consults other texts she will ¬nd forms of the remainder which will work more

easily. She should then ask herself what the point of these forms of remainder

is, apart from obtaining Taylor series which are much more easily obtained

by ¬nding the power series solution of an appropriate di¬erential equation.]

Many textbooks on mathematical methods devote some time to the pro-

cess of solving di¬erential equations by power series. The results of this

section justify the process.

Slogan: The formal process of solving a di¬erential equation by power

series yields a correct result within the radius of convergence of the power

series produced.

The slogan becomes a theorem once we specify the type of di¬erential equa-

tion to be solved.

298 A COMPANION TO ANALYSIS

Note however, that, contrary to the implied promise of some textbooks

on mathematical methods, power series solutions are not always as useful as

they look.

Exercise 11.5.23. We know that ∞ (’1)n x2n /(2n)! converges everywhere

n=0

to cos x. Try and use this formula, together with a hand calculator to com-

pute cos 100. Good behaviour in the sense of the pure mathematician merely

means ˜good behaviour in the long run™ and the ˜long run™ may be too long for

any practical use.

Can you suggest and implement a sensible method4 to compute cos 100.

Exercise 11.5.24. By considering the relations that the coe¬cients must

satisfy show that there is no power series solution for the equation

x3 y (x) = ’2y(x)

with y(0) = 0 valid in some neighbourhood of 0.

Show, however, that the system does have a well behaved solution. [Hint:

Example 7.1.5.]

If the reader is prepared to work quite hard, Exercise K.243 gives a good

condition for the existence of a power series solution for certain typical dif-

ferential equations.

We end this section with a look in another direction.

Exercise 11.5.25. If z ∈ C and n is a positive integer we de¬ne n’z =

e’z log n . By using the Weierstrass M-test, or otherwise show that, if > 0,

∞

n’z converges uniformly for z >1+ .

n=1

We call the limit ζ(z). Show further that ζ is di¬erentiable on the range

considered. Deduce that ζ is well de¬ned and di¬erentiable on the set {z ∈

C : z > 1}. (ζ is the famous Riemann zeta function.)

Fourier series ™

11.6

In this section we shall integrate complex-valued functions. The de¬nition

used is essentially that of De¬nition 8.5.1.

4

Pressing the cos button is sensible, but not very instructive.

299

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

De¬nition 11.6.1. If f : [a, b] ’ C is such that f : [a, b] ’ R and

f : [a, b] ’ R are Riemann integrable, then we say that f is Riemann

integrable and

b b b

f (x) dx = f (x) dx + i f (x) dx.

a a a

We leave it to the conscientious reader to check that the integral behaves

as it ought to behave.

If the reader has attended a course on mathematical methods she will

probably be familiar with the notion of the Fourier series of a periodic func-

tion.

De¬nition 11.6.2. If f : R ’ C is continuous and periodic with period 2π

(that is, f (t + 2π) = f (t) for all t) and m is an integer, we set

π

1

ˆ

f (m) = f (t) exp(’imt) dt.

2π ’π

Fourier claimed5 , in e¬ect, that

∞

ˆ

f (t) = f (n) exp(int).

n=’∞

We now know that the statement is false in the sense that there exist con-

tinuous functions such that

N

ˆ

f (n) exp(int0 ) f (t0 )

n=’N

as N ’ ∞ for some t0 , but true in many other and deeper senses.

The unraveling of the various ways in which Fourier™s theorem holds took

a century and a half6 and was one of the major in¬‚uences on the rigorisation

ˆ

of analysis. In this section we shall merely provide a simple condition on f

which ensures that Fourier™s statement holds in its original form for a given

function f .

Our discussion hinges on the following theorem, which is very important

in its own right.

Theorem 11.6.3. (Uniqueness of the Fourier series.) If f : R ’ C is

ˆ

continuous and periodic with period 2π and f (n) = 0 for all n, then f = 0.

5

Others had had the idea before but Fourier ˜bet the farm on it™.

6

Supposing the process to have terminated.

300 A COMPANION TO ANALYSIS

To prove this result it turns out to be su¬cient to prove an apparently

weaker result. (See Exercises 11.6.6 and 11.6.7.)

Lemma 11.6.4. If f : R ’ R is continuous and periodic with period 2π and

ˆ

f (n) = 0 for all n, then f (0) = 0.

Proof. Suppose f (0) = 0. Without loss of generality we may suppose that

f (0) > 0, (otherwise, we can consider ’f ). By continuity, we can ¬nd an an

with 1 > > 0 such that |f (t) ’ f (0)| < f (0)/2 and so f (t) > f (0)/2 for all

|t| ¤ . Now choose · > 0 such that 2· + cos < 1 and set P (t) = · + cos t.

Since P (t) = (· + 1 eit + 1 e’it ), we have

2 2

k=N

N

bN k eikt

P (t) =

k=’N

for some bN k , and so

k=N k=N

π π

ˆ