ńņš. 18 |

Question K.227.

Exercise K.229. [11.4, T!, ā‘ ] Question K.228 does not give all the prop-

erties of xĪ± that we are interested in. In particular, it tells us little about

the map Ī± ā’ xĪ± . If a > 1, let us write P (t) = Pa (t) = at .

(i) Show that P : R ā’ (0, ā) is an increasing function.

(ii) Show that P (t) ā’ 1 as t ā’ 0.

(iii) By using the relation P (s+t) = P (s)P (t), show that P is everywhere

continuous.

(iv) Show that, if P is diļ¬erentiable at 1, then P is everywhere diļ¬eren-

tiable.

(v) Let f (y) = n(y 1/n ā’ 1) ā’ (n + 1)(y 1/(n+1) ā’ 1). By considering f , or

otherwise, show that f (y) ā„ 0 for all y ā„ 1. Hence, or otherwise, show that

P (1/n) ā’ P (0)

tends to a limit, L say,

1/n

557

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

as n ā’ ā.

(vi) Show that, if 1/n ā„ x ā„ 1/(n + 1) then

P (x) ā’ P (0) P (1/n) ā’ P (0) n + 1

ā¤ Ć— .

x 1/n n

By using this and similar estimates, or otherwise, show that

P (x) ā’ P (0)

ā’L

x

as x ā’ 0 through values of x > 0.

(vii) Show that, if x = 0

P (x) ā’ P (0) P (ā’x) ā’ P (0)

= P (x)

ā’x

x

and deduce that

P (x) ā’ P (0)

ā’L

x

as x ā’ 0. Conclude that P is diļ¬erentiable at 0 and so everywhere with

P (t) = LP (t).

(viii) If a > 0 and we set Pa (t) = at , show that Pa is diļ¬erentiable with

Pa (t) = La P (t) for some constant La . Show that La > 0 if a > 1. What can

you say about La for other values of a?

(ix) If a, Ī» > 0 show that LaĪ» = Ī»La . Deduce that there is a real number

e > 1 such that, if we write

e(t) = et ,

then e is a diļ¬erentiable function with e (t) = e(t).

(x) Tear oļ¬ the disguise of La .

We have now completed a direct attack on the problem of deļ¬ning xĪ± and

obtaining its main properties. It should be clear why the indirect deļ¬nition

xĪ± = exp(Ī± log x) is preferred.

Exercise K.230. [11.5, P] Suppose ā an z n has radius of convergence

n=0

greater than or equal to R > 0. Write g(z) = ā an z n for |z| < R. Show

n=0

ā

that, if |z0 | < R then we can ļ¬nd b0 , b1 , Ā· Ā· Ā· ā C such that n

n=0 bn z

has radius of convergence greater than or equal to R ā’ |z0 | and g(z) =

ā n

n=0 bn (z ā’ z0 ) for |z0 ā’ z| < R ā’ |z0 |. [This is quite hard work. If you

have no idea how to attack it, ļ¬rst work out formally what the values of the

bn must be. Now try and use Lemma 5.3.4.]

558 A COMPANION TO ANALYSIS

Show that g is complex diļ¬erentiable at z0 with g (z0 ) = b1 = ā nan z0 .nā’1

n=1

Deduce that a power series is diļ¬erentiable and that its derivative is that ob-

tained by term by term diļ¬erentiation within its radius of convergence. (We

thus have an alternative proof of Theorem 11.5.11.)

Exercise K.231. [11.5, P] Here is another proof of Theorem 11.5.11. Sup-

pose ā an z n has radius of convergence greater than or equal to R > 0.

n=0

(i) Show that ā nan z nā’1 and ā n(n ā’ 1)an z nā’2 also have radius

n=1 n=2

of convergence R.

(ii) Show that

nā’2

n

ā¤ n(n ā’ 1)

j j

for all n ā’ 2 ā„ j ā„ 0 and deduce that

|(z + h)n ā’ z n ā’ nz nā’1 h| ā¤ n(n ā’ 1)(|z| + |h|)nā’2

for all z, h ā C.

(iii) Use (i) and (ii) to show that, if 0 < Ī“ and |z| + |h| < R ā’ Ī“, then

ā ā ā

n n

nan z nā’1 ā¤ A(Ī“)|h|2

an (z + h) ā’ an z ā’ h

n=0 n=0 n=1

where A(Ī“) depends only on Ī“ (and not on h or z).

(iv) Deduce that a power series is diļ¬erentiable and that its derivative is

that obtained by term by term diļ¬erentiation within its radius of convergence.

Exercise K.232. [11.5, P] Consider the Legendre diļ¬erential equation

(1 ā’ x2 )y ā’ 2xy + l(l + 1)y = 0,

where l is a real number. Find the general power series solution. Show that,

unless l is non-negative integer, every non-trivial power series solution has

radius of convergence 1.

Show, however, that, if l = n a non-negative integer the general series

solution can be written

y(x) = APn (x) + BQn (x)

where Qn is a power series of radius of convergence 1, Pn is a polynomial of

degree n, and A and B are arbitrary constants.

559

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

Verify that, when n is a non-negative integer, the function given by v(x) =

(x2 ā’ 1)n satisļ¬es the equation

(1 ā’ x2 )v (x) + 2nxv(x) = 0.

Deduce that v (n) (x) satisļ¬es the Legendre diļ¬erential equation with l = n.

[Consult Exercise K.270 if you need a hint.] Conclude that Pn is a constant

multiple of the function pn deļ¬ned in Exercise K.212.

Exercise K.233. [11.5, P] Obtain the general power series solution of

2

2d w dw

+ (z 2 ā’ 1)w = 0.

z +z

dz 2 dz

For what values of z is your solution valid and why?

Answer the same questions for

2

2d w dw

+ (z 2 ā’ 1)w = z 2 .

z +z

dz 2 dz

Exercise K.234. [11.5, P] Using Lemma 11.5.19 or otherwise, ļ¬nd the

power series expansion ā aj xj of (1 + x)1/2 for x real and |x| < 1.

j=0

Show that the complex power series ā aj z j has radius of convergence

j=0

1. What is the power series for the product ā aj z j ā aj z j and why?

j=0 j=0

What is its radius of convergence?

By considering

ā ā ā

aj K ā’j z j

aj z j aj z j ,

j=0 j=0 j=0

or otherwise, show that, if two power series of the same radius of conver-

gence R are multiplied, the resulting power series may have any radius of

convergence with value R or greater.

By considering expressions of the form

ā ā ā

Aj z j Bj z j + Cj z j ,

j=0 j=0 j=0

or otherwise, show that, if two power series of radius of convergence R and S

are multiplied, the resulting power series may have any radius of convergence

with value min(R, S) or greater.

560 A COMPANION TO ANALYSIS

Exercise K.235. [11.5, P] By modifying the proof of Abelā™s test (Lemma 5.2.4),

or otherwise, prove the following result. Let E be a set and aj : E ā’ Rm a

sequence of functions. Suppose that there exists a K such that

n

aj (t) ā¤ K

j=1

for all n ā„ 1 and all t ā E. If Ī»j is a decreasing sequence of real positive

numbers with Ī»j ā’ 0 as j ā’ ā then ā Ī»j aj converges uniformly on E.

j=1

Deduce, in particular, that if bn ā R and ā bn converges, then ā bn xn

n=0 n=0

converges uniformly on [0, 1]. Explain why this implies that

ā ā

1

bn

n

bn x dx = .

n+1

0 n=0 n=0

Show that, provided we interpret the left hand integral as an improper

1ā’

integral, lim ā’0, >0 0 , equation remains true under the assumption

ā

that n=0 bn /(n + 1) converges.

Show that

111

1 ā’ + ā’ + Ā· Ā· Ā· = log 2

234

and

111 Ļ

1 ā’ + ā’ + Ā·Ā·Ā· = .

357 4

Do these provide good methods for computing log 2 and Ļ? Why?

Recall that, if a, b > 0, then aloga b = b. Show that

log2 e ā’ log4 e + log8 e ā’ log16 e + . . .

converges to a value to be found.

Exercise K.236. [11.5, G, S] (Only if you have done suļ¬cient probability,

but easy if you have.) Let X be a random variable taking positive integral

values. Show that

ā

ĻX (t) = Et = X

Pr(X = n)tn

n=0

is well deļ¬ned for all t ā [ā’1, 1].

If EX is bounded, show that

ĻX (1) ā’ ĻX (t)

ā’ EX

1ā’t

as t ā’ 1 through values t < 1. Give an example of an X with EX inļ¬nite.

561

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

Exercise K.237. [11.5, T, S] Use the ideas of Exercise K.78 to prove the

following extension. Suppose that we have a sequence of functions an : ā„¦ ā’ C

and a sequence of positive real numbers Mn such that ā Mj converges

j=1

absolutely and |an (z)| ā¤ Mn for all z ā ā„¦ and all n. Then

ā

N

(1 + aj (z)) ā’ (1 + aj (z))

j=1 j=1

uniformly on ā„¦.

Exercise K.238. [11.5, P, S, ā‘ ] We do not have the apparatus to exploit

Exercise K.237 fully but the following exercise is suggestive.

(i) Show that SN (z) = z N (1 ā’ z 2 nā’2 ) converges uniformly to S(z) =

n=1

z ā (1 ā’ z 2 nā’2 ) on any disc {z : |z| < R}. Deduce that S is a well deļ¬ned

n=1

continuous function on C.

(ii) By writing SN (z) = ā’(N !)ā’2 N n=ā’N (n ā’ z) and considering SN (z +

1)/SN (z), or otherwise, show that S(z + 1) = S(z) (that is, S is periodic

with period 1). Show also that S(z) = 0 if and only if z is an integer.

[The product z ā (1 ā’ z 2 nā’2 ) goes back to Euler who identiļ¬ed it with

n=1

a well known function which the reader should also be able to guess. An

advantage of inļ¬nite products over inļ¬nite sums is that we can ļ¬nd zeros of

the function much more easily.]

Exercise K.239. [11.5, T] (i) Suppose that an ā C and ā an z n has

n=0

radius of convergence R > 0. If aN = 0 and an = 0 for n < N , use the

inequality

q q

n N

|an z nā’N |

ā¤ |z| |aN | ā’

an z

n=0 n=N +1

(for q > n), together with standard bounds on |an z n |, to show that there

exists a Ī“ > 0 such that

q

|aN ||z|N

n

ā„

an z

2

n=0

for all |z| < Ī“.

(ii) Suppose that an ā C and ā an z n has radius of convergence R > 0.

n=0

ā n

Set f (z) = n=0 an z for |z| < R. Show, using (i), or otherwise, that, if

there exist wm ā’ 0 with f (wm ) = 0 and wm = 0, then an = 0 for all n and

f = 0.

562 A COMPANION TO ANALYSIS

Exercise K.240. [11.5, P, ā‘ ] This is a commentary on the previous exer-

cise. We shall consider functions from R to R.

(i) Suppose that an ā R and ā an xn has radius of convergence R > 0.

n=0

Set f (x) = ā an xn for |x| < R. Show that, if there exist xm ā’ 0 with

n=0

f (xm ) = 0 and xm = 0, then an = 0 for all n and f = 0.

(ii) Explain why a polynomial of degree at most n with n + 1 zeros must

be zero.

(iii) Give an example of a power series of radius of convergence ā which

has inļ¬nitely many zeros but is not identically zero.

(iv) Let F be as in Example 7.1.5. Set G(x) = F (x) sin xā’1 for x = 0

and G(0) = 0. Show that G is inļ¬nitely diļ¬erentiable everywhere (this will

probably require you to go through much the same argument as we used for

F ). Show that there exist xm ā’ 0 with G(xm ) = 0 and xm = 0 but G is not

identically zero.

Exercise K.241. [11.5, P] This question prepares the way for Exercise K.242.

The ļ¬rst two parts will probably be familiar but are intended to provide a

hint for the third part.

n n n+1

(i) If n ā„ r ā„ 1, show that + = .

rā’1 r r

(ii) Use induction to show that

n

n r nā’r

(x + y)n = xy

r

r=0

whenever n is a positive integer and x and y are real.

(iii) Suppose that n is a positive integer and x and y are real. Verify the

next formula directly for n = 0, 1, 2, 3 and then prove it for all n.

n rā’1 nā’rā’1 nā’1

n

(x ā’ j) (y ā’ k) = (x + y ā’ q).

r

r=0 q=0

j=0 k=0

Exercise K.242. [11.5, T, ā‘ ] The following is a modiļ¬cation of Cauchyā™s

proof of the binomial theorem. We shall work in R. We take w as a ļ¬xed

real number with |w| < 1.

(i) Use the ratio test to show that ā n! nā’1 (x ā’ j)w n converges. By

1

n=0 j=0

thinking about your proof and the Weierstrass M-test, improve this result to

show that ā n! nā’1 (x ā’ j)w n converges uniformly for |x| ā¤ M whenever

1

n=0 j=0

M is ļ¬xed. We write

ā nā’1

1

(x ā’ j)w n .

f (x) =

n!

n=0 j=0

563

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

Explain why f is continuous.

(ii) Use Exercise 5.4.4 (ļ¬rst proved by Cauchy) and Exercise K.241 to

show that

f (x)f (y) = f (x + y)

for all x and y.

(iii) Show that f (0) = 1 and deduce that f (x) = 0 for all x. Deduce,

quoting any theorem you need, that f (x) > 0 for all x.

(iv) Explain why we can deļ¬ne g : R ā’ R by g(x) = log f (x) and why g

is continuous. Show that

g(x) + g(y) = g(x + y)

for all x and y. Use Exercise K.90 (i) (ļ¬rst proved by Cauchy) to show that

g(x) = ax and so f (x) = bx for some real positive number b.

(v) Find b by considering f (1). Deduce that

ā nā’1

1

x

(x ā’ j)w n .

(1 + w) =

n!

n=0 j=0

[One problem faced by anyone teaching elementary analysis in France is that

every theorem seems to be called Cauchyā™s theorem. A glance at the exercise

above shows why13 .]

Exercise K.243. [11.5, T] (Part (iii) of this question makes repeated use

of Exercise K.76.) Exercise 11.5.24 raises the question of when we can tell

whether a diļ¬erential equation of the form u (t) = f (t, u(t)) has a power

series solution. The following theorem of Cauchy (see the concluding remark

of the previous question) gives a rather natural condition.

Suppose that cn,m ā R [n, m ā„ 0] and that there exists a Ļ > 0 and an

K > 0 such that

|cn,m | ā¤ KĻn+m for all n, m ā„ 0.

(Exercise K.74 (c) shows why we use this condition.) Then we can ļ¬nd a

Ī“ > 0 with Ļ > Ī“ and an ā R such that ā an tn converges to u(t), say, for

n=0

13

Moreover, Cauchyā™s contributions to the rigorising of analysis form only a fragment

of his contribution to pure and applied mathematics. The following quotation from a

materials engineer can be echoed in many ļ¬elds. ā˜[Cauchyā™s paper of 1822] was perhaps

the most important event in the history of elasticity since Hooke. After this, that science

showed promise of becoming a practical tool for engineers rather than a happy hunting-

ground for a few somewhat eccentric philosophers.ā™ ([18], Chapter 3)

564 A COMPANION TO ANALYSIS

all |t| < Ī“ and such that, writing

ā ā

cn,m xn y m

f (x, y) =

n=0 m=0

for |x|, |y| < Ļā’1 we have u(0) = 0, |u(t)| < Ļ and

u (t) = f (t, u(t))

for all |t| ā¤ Ī“.

(i) The object of this question is to prove Cauchyā™s theorem but in the

ļ¬rst two parts of the question we are merely interested in seeing what is

going on so we work in a non-rigorous manner. Assuming that a power

series solution ā an tn with the required properties actually exists show

n=0

by formal manipulation that a0 = 0 and kak should be the coeļ¬cient of xkā’1

in

m

ā ā ā

cn,m xn ar x r .

n=0 m=0 r=0

Explain why that coeļ¬cient is a ļ¬nite sum only depending on a0 , a1 , . . . ,

akā’1 . Show that, if we deļ¬ne A0 = 0, and deļ¬ne A1 , A2 , . . . , formally by

means of the equation

m

ā ā ā ā

d

An tn KĻn+m tn Ar tr

= ,

dt n=0 n=0 m=0 r=0

then |ak | ā¤ Ak .

(ii) We continue with the ideas of (i). Show that can be rewritten (at

least, formally) as

dw K

=

(1 ā’ Ļt)(1 ā’ Ļw)

dt

where w(t) = ā An tn . Solve equation for w(0) = 0.

n=0

(iii) We now reverse the non-rigorous argument but this time proceed

rigorously. Show that equation has a solution w(t) with w(0) = 0 and

the property that we can ļ¬nd Aj and Ī· > 0 such that ā An tn has radius

n=0

ā n

of convergence at least Ī· and w(t) = n=0 An t for all |t| < Ī·. Show that

for |t| < Ī·. Deduce that, if ak is given by equating

the An satisfy equation

coeļ¬cients in the equation

m

ā ā ā ā

an x n = cn,m xn ar x r ,

n=0 n=0 m=0 r=0

565

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

then |ak | ā¤ Ak and ā an tn has radius of convergence at least Ī·. Show

n=0

that if we set u(t) = ā an tn for |t| < Ī·, then u(0) = 0 and there exists

n=0

a Ī“ with 0 < Ī“ ā¤ Ī· such that |u(t)| < Ļ. Show that Ī“ and u satisfy the

conclusions of Cauchyā™s theorem.

Exercise K.244. (Convolution.) [11.6, T] Let us write CP (R) for the

set of continuous function f : R ā’ R which are periodic with period 2Ļ.

(i) Show that, if f, g ā CP (R), then the equation

Ļ

1

f ā— g(t) = f (t ā’ s)g(s) ds

2Ļ ā’Ļ

gives a well deļ¬ned, continuous, 2Ļ periodic function f ā— g : R ā’ R. (Thus

f ā— g ā CP (R).)

(ii) Show that, if f, g ā CP (R), then

2Ļ+a

1

f ā— g(t) = f (t ā’ s)g(s) ds

2Ļ a

for all t ā R and a ā R.

(iii) Show that if f, g, h ā CP (R) and Ī» ā R then

(Ī»f ) ā— g = Ī»(f ā— g), f ā— (g + h) = f ā— g + f ā— h,

f ā— g = g ā— f and f ā— (g ā— h) = (f ā— g) ā— h.

(iv) Suppose that f, g ā CP (R) and f is diļ¬erentiable with f ā CP (R).

Show that f ā— g is diļ¬erentiable and

(f ā— g) = f ā— g.

(v) Suppose that f, un ā CP (R), that un (t) ā„ 0 for all t ā R, that

un (t) = 0 for Ļ/n ā¤ |t| ā¤ Ļ and

Ļ

1

un (t) dt = 1

2Ļ ā’Ļ

show that un ā— f ā’ f uniformly as n ā’ ā.

(vii) Suppose that f ā CP (R) and > 0. Show, by using parts (iv) and (v)

that there exists a twice diļ¬erentiable function g with g, g , g ā CP (R) such

that f ā’ g ā ā¤ .

(viii) Let us write CP (C) for the set of continuous function f : R ā’ C

which are periodic with period 2Ļ. Show. in as much detail as you consider

desirable, that parts (i) to (vi) hold with CP (R) replaced by CP (C)

566 A COMPANION TO ANALYSIS

Exercise K.245. [11.6, T, ā‘ ] We continue with the notation of the previous

question.

(i) Suppose that that f, g ā CP (C). Show that

Ėg

f ā— g(n) = f (n)Ė(n)

for all integers n. (We saying that ā˜taking Fourier coeļ¬cients converts con-

volution to multiplicationā™.)

(ii) Show that if f ā CP (C), then

Ė

|f (n)| ā¤ f ā.

(iii) Show that if f, g ā CP (C), then

Ė

|f (n)| ā¤ |Ė(n)| + f ā’ g

g ā.

(iv) Use part (iii) of this exercise, part (vii) of Exercise K.244 and Exer-

Ė

cise 11.6.10 to show that, if f ā CP (C), then f (n) ā’ 0 as |n| ā’ ā. (This is

a version of the important Riemann-Lebesgue lemma.)

(v) Suppose, if possible, that there exists an e ā CP (C) such that eā—f = f

for all f ā CP (C). Find e and use (iv) to show that no such e can exist.

Ė

(Informally, convolution on CP (C) has no unit.)

Exercise K.246. [11.6, T, ā‘ ] This question uses the version of the Riemann-

Lebesgue lemma proved in Exercise K.245 (iv). If f ā CP (R) we write

n

Ė

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

j=ā’n

(i) Suppose that f1 , g1 ā CP (R) and f1 (t) = g1 (t) sin t for all t. Show

Ė

that f1 (j) = g1 (j + 1) ā’ g1 (j ā’ 1) and deduce that Sn (f1 , 0) ā’ 0 as n ā’ ā.

Ė Ė

(ii) Suppose that f2 ā CP (R), f2 (nĻ) = 0 and f2 is diļ¬erentiable at nĻ for

all integer n. Show that there exists a g2 ā CP (R) such that f2 (t) = g2 (t) sin t

for all t and deduce that Sn (f2 , 0) ā’ 0 as n ā’ ā.

(iii) Suppose that f3 ā CP (R), f3 (0) = 0 and f3 is diļ¬erentiable at 0.

Ė

Write f4 (t) = f3 (t/2). Compute f4 (j) in terms of the Fourier coeļ¬cients of

f3 . Show that Sn (f4 , 0) ā’ 0 and deduce that Sn (f3 , 0) ā’ 0 as n ā’ ā.

(iv) Suppose that f ā CP (R), and f is diļ¬erentiable at some point x.

Show that Sn (f, x) ā’ f (x) as n ā’ ā.

Exercise K.247. [12.1, H] The result of this question is a special case of

part of Lemma 13.1.4. However, precisely because it is a special case, some

readers may ļ¬nd it a useful introduction to the ideas of Section 13.1.

567

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

(i) Let Ī“ > 0 and let f : C ā’ C be an everywhere diļ¬erentiable function

such that |f (z) ā’ 1| ā¤ 1/2 for all |z| ā¤ Ī“. Set X = {z ā C : |z| ā¤ Ī“}. If

|w| ā¤ Ī“/2 and we deļ¬ne T z = z ā’ f (z) + w for z ā X show, by using the

mean value inequality (Exercise 11.5.5) that T z ā X whenever z ā X. Thus

T is a function T : X ā’ X.

(ii) We continue with the hypotheses and notation of part (i). Show that

T is a contraction mapping and deduce that the equation f (z) = w has

exactly one solution with |z| ā¤ Ī“.

(iii) Let F : C ā’ C be an everywhere diļ¬erentiable function with con-

tinuous derivative. Suppose further that F (0) = 0 and F (0) = 1. Show

that there exists a Ī“ > 0 such that, if |w| ā¤ Ī“/2, the equation F (z) = w has

exactly one solution with |z| ā¤ Ī“.

(iv) Let g : C ā’ C be an everywhere diļ¬erentiable function with contin-

uous derivative. Show that, if g (0) = 0, there exists Ī·1 . Ī·2 > 0 such that, if

|Ļ ā’ g(0)| ā¤ Ī·1 , the equation g(z) = w has exactly one solution with |z| ā¤ Ī·2 .

(v) If we omit the condition g (0) = 0 in (iv) does it remain true that

there exists an Ī· > 0 such that, if |Ļ ā’ g(0)| ā¤ Ī·, the equation g(z) = w has

a solution? Give a proof or counterexample.

(vi) If we omit the condition g (0) = 0 in (iv) but add the condition

g (0) = 0 does it remain true that there exist Ī·1 . Ī·2 > 0 such that, if |Ļ ā’

g(0)| ā¤ Ī·1 , the equation g(z) = w has at most one solution with |z| ā¤ Ī·2 ?

Give a proof or counterexample.

Exercise K.248. [12.1, P, ā‘] We work in C. Consider the following state-

ment.

There exists a Ī“ > 0 such that, if |w| ā¤ Ī“, the equation fj (z) = w has a

solution.

Give a proof or counterexample in each of the following cases.

(i) f1 (z) = z ā— .

(ii) f2 (z) = z + z ā— .

(iii) f3 (z) = z + |z|.

(iv) f4 (z) = z + |z|2 .

Is our statement true for all F in the following cases? Give a proof or

counterexample.

(v) f5 (z) = F (z) + |z| where F : C ā’ C is an everywhere diļ¬erentiable

function with continuous derivative and F (0) = 0 and F (0) = 1.

(vi) f6 (z) = F (z) + |z|2 where F : C ā’ C is an everywhere diļ¬erentiable

function with continuous derivative and F (0) = 0 and F (0) = 1.

Exercise K.249. (Newton-Raphson.) [12.1, T] (i) Suppose that f :

568 A COMPANION TO ANALYSIS

R ā’ R is a twice diļ¬erentiable function such that

f (x)f (x)

ā¤Ī»

f (x)2

for all x and some |Ī»| < 1 Show that the mapping

f (x)

Tx = x ā’

f (x)

is a contraction mapping and deduce that f has a unique root y.

(ii) Suppose that F : R ā’ R is a twice diļ¬erentiable function such that

F (x)F (x)

ā¤Ī»

F (x)2

for all |x| ā¤ a and some |Ī»| < 1 and that F (0) = 0. Consider the mapping

F (x)

Tx = x ā’ .

F (x)

Show that T n x ā’ 0.

Suppose that

sup|t|ā¤a |F (t)| sup|t|ā¤a |F (t)|

= M.

inf |t|ā¤a |F (t)|2

By using the mean value theorem twice, show that, if |x| ā¤ a, then

|T x| ā¤ M x2 .

(iii) If you know what the Newton-Raphson method is, comment on the

relevance of the results of (i) and (ii) to that method. Comment in particular

on the speed of convergence.

Exercise K.250. [12.1, G, P, S] (This is a short question and involves no

analysis.) Suppose f, g : X ā’ X are such that f g = gf . Show that if f

has a unique ļ¬xed point then g has a ļ¬xed point. Can g have more than one

ļ¬xed point? (Give a proof or counterexample.)

Show that, if we merely know that f has ļ¬xed points, it does not follow

that g has any.

Exercise K.251. [12.1, P] If (X, d) is a complete metric space and T :

X ā’ X is a surjective map such that

d(T x, T y) ā„ Kd(x, y)

569

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

for all x, y ā X and some K > 1, show that T has a unique ļ¬xed point.

By considering the map T : R ā’ R deļ¬ned by T (x) = 1 + 4n + 2x

for 0 ā¤ x < 1 and n an integer, or otherwise, show that the condition T

surjective cannot be dropped.

Exercise K.252. [12.1, P] We work in Rm with the usual distance. Let E

be a closed non-empty subset of Rm and let T be a map T : E ā’ E.

(i) Suppose T (a) ā’ T (b) < a ā’ b for all a, b ā E with a = b. We

saw in Example 12.1.4 that T need not have a ļ¬xed point. Show that, if T

has a ļ¬xed point, it is unique.

(ii) Suppose T (a) ā’ T (b) > a ā’ b for all a, b ā E with a = b. Show

that T need not have a ļ¬xed point but, if T has a ļ¬xed point, it is unique.

(iii) Suppose T (a) ā’ T (b) = a ā’ b for all a, b ā E. Show that T

need not have a ļ¬xed point and that, if T has a ļ¬xed point, it need not be

unique.

(iv) Suppose now that E is non-empty, closed and bounded and

T (a) ā’ T (b) < a ā’ b

for all a, b ā E with a = b. By considering inf xāE x ā’ T (x) , or otherwise,

show that T has a ļ¬xed point.

(v) Suppose that E = Rm and

T (a) ā’ T (b) < a ā’ b

for all a, b ā Rm with a = b. Show that, if there exists a point c ā Rm and

a K > 0 such that T n c < k for all positive integer n, then T has a ļ¬xed

point.

Exercise K.253. [12.1, G, P, S] (This is easy if you see what is going on.)

Let (X, d) be a metric space. Show that the set of isometries f : X ā’ X

forms a group G(X) under composition.

By choosing X to be an appropriate closed bounded subset of R2 with

the usual metric, or otherwise, prove the following statements about G(X)

when X has the Bolzano-Weierstrass property.

(i) G(X) need not be Abelian.

(ii) There exists an X such that G(X) has only one element.

(iii) There exists an X such that G(X) has inļ¬nitely many elements.

If n ā„ 2 give an example of an (X, d) with the Bolzano-Weierstrass prop-

erty and a T : X ā’ X such that T m has no ļ¬xed points for 1 ā¤ m ā¤ n ā’ 1

but T n does.

570 A COMPANION TO ANALYSIS

Exercise K.254. [12.1, P] (i) Let (X, d) be a metric space with the Bolzano-

Weierstrass property. Suppose f : X ā’ X is expansive in the sense that

d(f (x), f (y)) ā„ d(x, y)

for all x, y ā X. (Note that we do not assume that f is continuous.) If

a, b ā X we write a0 = a and an+1 = f (an ) and deļ¬ne a sequence bn

similarly. Show that we can ļ¬nd a sequence of integers n(1) < n(2) < . . .

such that d(an(k) , a0 ) ā’ 0 and d(bn(k) , b0 ) ā’ 0 as n ā’ ā. Deduce that

d(a, b) = d(f (a), f (b)) for all a, b ā X (that is, f is an isometry).

Find a complete metric space (Y, Ļ) and maps g, h : Y ā’ Y such that

Ļ(g(x), g(y)) ā„ 2Ļ(x, y), Ļ(h(x), h(y)) ā„ 2Ļ(x, y)

and g is continuous but h is not.

(ii) Let (X, d) be a metric space with the Bolzano-Weierstrass property.

Suppose f : X ā’ X is an isometry in the sense that

d(f (x), f (y)) = d(x, y)

for all x, y ā X. If a ā X we write a0 = a and an+1 = f (an ). Show (as

in (i)) that we can ļ¬nd a sequence of integers n(1) < n(2) < . . . such that

d(an(k) , a0 ) ā’ 0 as n ā’ ā and deduce that f is bijective.

Find a complete metric space (Y, Ļ) and a map g : Y ā’ Y such that

Ļ(g(x), g(y)) = Ļ(x, y)

for all x, y ā Y , but g is not surjective.

Exercise K.255. [12.1, P] Let (X, d) be a metric space with the Bolzano-

Weierstrass property and (Y, Ļ) a metric space. Suppose f : X ā’ Y and

g : Y ā’ X are such that Ļ(f (x), f (x )) ā„ d(x, x ) for all x, x ā X and

d(g(y), g(y )) ā„ Ļ(y, y ) for all y, y ā Y . By considering g ā—¦ f and us-

ing Exercise K.254 show that Ļ(f (x), f (x )) = d(x, x ) for all x, x ā X,

d(g(y), g(y )) = Ļ(y, y ) for all y, y ā Y and f and g are bijective. Is it

necessarily true that f and g are inverse functions?

Show that the result of the ļ¬rst paragraph may fail if we simply have

f : X ā’ f (X) bijective with f and f |ā’1 continuous and g : Y ā’ g(Y )

f (X)

ā’1

bijective with g and g|g(Y ) continuous.

Show that the result may fail, even if (X, d) and (Y, Ļ) are complete, if

(X, d) does not have the Bolzano-Weierstrass property.

571

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

Exercise K.256. [12.1, P] Let (V, ) be a complete normed space. We

say that a subset A ā V is convex if whenever a, b ā A and 0 ā¤ Ī» ā¤ 1 we

have

Ī»a + (1 ā’ Ī»)b ā A.

Suppose that A is closed bounded convex subset of V and f : A ā’ A satisļ¬es

f (a) ā’ f (b) ā¤ a ā’ b

for all a, b ā A. By considering

g(x) = (1 ā’ )f (x) + a0

for some a0 ā A and > 0 show that

inf{ f (a) ā’ a : a ā A} = 0

and deduce that f has a ļ¬xed point in A.

Give an example to show that the conclusion may be false if any one of

the three conditions:- convex, closed and bounded is omitted.

The following result is useful in the theory of Markov chains (see Exer-

cise K.258). Suppose pij ā„ 0 for 1 ā¤ i, j ā¤ n and n pij = 1 for 1 ā¤ i ā¤ n.

j=1

n

Give R the norm

n

j=1 |xj |. By choosing an appropriate

1 where x 1 =

A and f show that we can ļ¬nd Ļi ā„ 0 for 1 ā¤ i ā¤ n with n Ļi = 1 such

i=1

that

n

Ļi pij = Ļj .

i=1

Exercise K.257. [12.1, P] Throughout this question (X, d) is a non-empty

metric space with the Bolzano-Weierstrass property and T : X ā’ X is

a map such that d(T x, T y) ā¤ d(x, y) and such that, given any x, y ā X

such that x = y we can ļ¬nd a strictly positive integer N (x, y) such that

d(T N (x,y) x, T N (x,y) y) < d(x, y).

(i) Show that the map S : X ā’ R given by Sx = d(x, T x) is continuous.

Explain why this means that there is a y ā X such that d(y, T y) ā¤ d(T x, x)

for all x ā X. Show that, in fact y = T y. Show that y is the unique ļ¬xed

point of T .

(ii) Let > 0 and write

Un = {x ā X : d(T n x, y) < }.

572 A COMPANION TO ANALYSIS

Show that Un ā Un+1 and that Un is open for all n ā„ 1. Write U = ā Un

n=1

and F = X \ U . Why does F have the Bolzano-Weierstrass property? Show

that T x ā F whenever x ā F . Use part (i) to show that, if F were non-

empty, there would exist a w ā F with T w = w. Deduce that F is indeed

empty.

(iii) Show that, if x ā X, then d(T n x, y) ā’ 0 as n ā’ ā.

Exercise K.258. (Finite Markov chains.) [12.1, T!, ā‘ ] In a ļ¬nite

Markov chain with states 1, 2, . . . , m, if the system is in state i at time

n it will be in state j at time n + 1 with probability pij irrespective of any

earlier history. Thus, if it is in state i with probability qi at time n and in

state j with probability qj at time n + 1, we have

m

qj = qi pij .

i=1

The reader may ignore the previous paragraph but it explains why we are

interested in the space

m

X = {q ā Rm : qi = 1, qs ā„ 0 [1 ā¤ s ā¤ m]}

i=1

and the map T : X ā’ Rm given by T q = q where qj = m qi pij for all

i=1

1 ā¤ j ā¤ m. Verify that if, as we shall do from now on, we take pij ā„ 0 for all

1 ā¤ i, j ā¤ m and m pij = 1 for all 1 ā¤ i ā¤ m then T q ā X for all q ā X

j=1

and so we may consider T as a map from X to X.

Finally we have to choose a metric d. In most of this book the particular

choice of metric has not been important but here it is. We take

m

d(u, v) = u ā’ v |ui ā’ vi |

=

1

i=1

for all u, v ā X.

(i) Explain why (X, d) has the Bolzano-Weierstrass property.

(ii) Show that d(T u, T v) ā¤ d(u, v) for all u, v ā X.

(iii) Show that, if n ā„ 1,

m

(n)

n

(T q)j = qi pij

i=1

(n) (n)

for all 1 ā¤ j ā¤ m where pij ā„ 0 for all 1 ā¤ i, j ā¤ m and m pij = 1 for

j=1

all 1 ā¤ i ā¤ m. If you know some probability prove this both by an algebraic

and a probabilistic argument.

573

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

(iv) Show that, if

(N )

pij > 0 for all 1 ā¤ i, j ā¤ m

for some N ā„ 1, then d(T N u, T N v) < d(u, v) for all u, v ā X with u = v.

(v) Conclude, using Exercise K.257, that, provided only that holds for

some N ā„ 1, there is a unique Ļ ā X such that T Ļ = Ļ and that, if q ā X,

then d(T n q, Ļ) ā’ 0 as n ā’ ā. If the associated Markov chain has been

running for a long time, what can you say about the probability that it is in

state i?

(vi) Suppose that m = 2, p12 = p21 = 1 and p11 = p22 = 0. Show that

is a unique Ļ ā X such that T Ļ = Ļ. For which q ā X is it true that

d(T n q, Ļ) ā’ 0 as n ā’ ā? Why does your result not contradict (v)?

(vii) Suppose that m = 2, p12 = p21 = 0 and p11 = p22 = 1. Find all the

Ļ ā X such that T Ļ = Ļ. What can you say about the behaviour of T n q

as n ā’ ā for each q ā X? Why does your result not contradict (v)?

(viii) Suppose that m = 4 and that pij = 1/2 if 1 ā¤ i, j ā¤ 2 or 3 ā¤ i, j ā¤

4 and that pij = 0 otherwise. Find all the Ļ ā X such that T Ļ = Ļ. What

can you say about the behaviour of T n q as n ā’ ā for each q ā X?

(ix) Note that, by Exercise K.256, the equation T Ļ = Ļ always has a

solution in X even if does not hold.

Exercise K.259. (Countable Markov chains.) [12.1, T!] In this exer-

cise and the next we consider Markov chains with an inļ¬nite number of states

1, 2, . . . . In algebraic terms we consider a collection of positive numbers pij

[1 ā¤ i, j] such that ā pij = 1 for each i ā„ 1.

j=1

1

(i) Consider l the space of real sequences whose sum is absolutely con-

ā

n=1 |xn |. Show that, if we write

vergent, with norm 1 given by x ā =

ā

(T x)j = j=1 xi pij , then T is well deļ¬ned continuous linear map from l 1 to

l1 with operator norm T = 1. Show further that, if we write

X = {q ā l1 : q = 1 and qi ā„ 0 for all i ā„ 1},

1

then T q ā X whenever q ā X. (Note that you are manipulating inļ¬nite

sums and you must justify each step carefully. Results like Lemma 5.3.9 may

be useful.)

(ii) We shall only be interested in systems satisfying the following exten-

sion of . If k ā„ 1 there exists an N (k) such that

(N (k))

> 0 for all 1 ā¤ i, j ā¤ k.

pij

Show that, under this condition, given any u, v ā X with u = v, we can

ļ¬nd an N ā„ 1 such that d(T N u, T N v) < d(u, v).

574 A COMPANION TO ANALYSIS

(iii) Show, under assumption , that, if the equation T Ļ = Ļ has a

solution Ļ ā X, then it is unique. Let

Y = {h ā l1 : hi ā„ 0 for all i ā„ 1.}

Find all the solutions of the equation T h = h with h ā Y ļ¬rstly under the

assumption that the equation T Ļ = Ļ has a solution Ļ ā X and secondly

under the assumption that it does not.

(iv) Show, under assumption , that, if the equation T Ļ = Ļ has a

solution Ļ ā X, then Ļi > 0 for all i.

(v) Consider the following systems. In each case show we have a Markov

chain satisfying . By solving the appropriate diļ¬erence equations, or

otherwise, show that in case (a) the equation T Ļ = Ļ has a solution Ļ ā X

but in cases (b) and (c) it does not.

(a) p12 = 1; pi iā’1 = 1/2, pi i = pi i+1 = 1/4 for i ā„ 2; pij = 0 otherwise.

(b) p12 = 1; pi i+1 = 1/2, pi i = pi iā’1 = 1/4 for i ā„ 2; pij = 0 otherwise.

(c) p12 = 1; pi iā’1 = pi i = pi i+1 = 1/3 for i ā„ 2; pij = 0 otherwise.

Exercise K.260. [12.1, T!, ā‘ ] This question continues the previous one.

We assume that condition holds and that the equation T Ļ = Ļ has a

solution Ļ ā X. We will need part (ii) of Exercise K.195.

(i) Let J ā„ 1 be ļ¬xed. Show, using part (iv) of question K.259, that we

can ļ¬nd an h ā Y such that T h = h and hJ = 1.

(ii) We continue with the notation of (i). Explain why X does not have

the Bolzano-Weierstrass property but

XJ = {q ā X : hi ā„ qi for all i ā„ 1}

does. Explain why XJ is non-empty and show, carefully, that T q ā XJ

whenever q ā XJ . Deduce that, if q ā XJ , then T n q ā’ Ļ 1 ā’ 0 as n ā’ ā.

(iii) Let e(p) ā l 1 be given by e(p)p = 1, e(p)i = 0 otherwise. Explain why

T n e(p)ā’Ļ 1 ā’ 0 as n ā’ ā. Deduce that, if q ā X, then T n qā’Ļ 1 ā’ 0.

Interpret this result probabilistically.

Exercise K.261. [12.2, P] Let (X, d) be a complete metric space and T :

X ā’ X a mapping such that T n is a contraction mapping. Show that T has

a unique ļ¬xed point. [Hint. Show that if w is a ļ¬xed point for T n then so is

T w.]

If you have done Question K.253, explain why the example asked for in

the last paragraph of that question does not contradict the result of the ļ¬rst

paragraph of this question.

575

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

Consider the space C([a, b]) with the uniform norm. Let Ļ be a continuous

real-valued function on R Ć— [a, b] which satisļ¬es the Lipschitz condition

|Ļ(x, t) ā’ Ļ(y, t)| ā¤ M for x, y ā R and t ā [a, b],

and let g ā C([a, b]). Deļ¬ne T : C([a, b]) ā’ C([a, b]) by

t

(T h)(t) = g(t) + Ļ(h(s), s) ds.

a

Show, by induction, or otherwise that

1n

(T n h)(t) ā’ (T n k)(t) M (t ā’ a)n h ā’ k

ā¤

ā ā

n!

for all h, k ā [a, b].

Show that T has a ļ¬xed point and comment brieļ¬‚y on the associated

theorem on the existence of solutions of a certain diļ¬erential equation.

Exercise K.262. [12.2, P] Suppose that f : R ā’ R is diļ¬erentiable with

|f (x)| < M for all x ā R. Show that the system of equations

n

aij f (xj ) + bi [1 ā¤ i ā¤ n]

xi =

j=1

with aij and bi real has a unique solution provided that

n n

1

a2 < .

ij

M2

i=1 j=1

Exercise K.263. [12.2, P, S] Consider the diļ¬erential equation for a func-

tion y : R ā’ R

y (t) = y(t)Ī± , y(0) = 0

where Ī± is real and strictly positive. For which values of Ī± does have only

one solution and for which values of Ī± does it have more than one? If has

only one solution, prove this. If has more than one solution, give at least

two solutions explicitly.

Exercise K.264. [12.2, P] DieudonnĀ“ is associated in most mathemati-

e

ciansā™ minds with the abstract approach of Bourbaki and his own text Foun-

dations of Modern Analysis [13]. He may have had this in mind when he

576 A COMPANION TO ANALYSIS

wrote the determinedly concrete Inļ¬nitesimal Calculus [14] from which this

exercise is taken.

(i) Suppose f : R2 ā’ R is a continuous function with f (t, u) < 0 when

tu > 0 and f (t, u) > 0 when tu < 0. Explain why f (0, t) = f (u, 0) = 0 for

all t and u. If y : R ā’ R is a diļ¬erentiable function with

y (t) = f (t, y(t)) for all t ā R, y(0) = 0,

show, by considering the behaviour of y at a maximum and a minimum in

each interval [0, c] with c > 0, or otherwise, that y(t) = 0 for all t ā„ 0. Show

that y(t) = 0 for all t ā R and so has a unique solution.

(ii) Now deļ¬ne

ļ£±

if x ā„ t2 ,

ļ£“ā’2t

ļ£²

f (t, u) = ā’2x/t if t2 > x > ā’t2 ,

ļ£“

ļ£³

if ā’t2 > x.

2t

Verify that f is well deļ¬ned and continuous everywhere and satisļ¬es the

hypotheses of (i). Now set y0 (t) = t2 and deļ¬ne

t

yn+1 (t) = f (s, yn (s)) ds.

0

Show that yn (t) fails to converge as n ā’ ā for any t = 0 although, as we

know from (i), has a unique solution.

Exercise K.265. [12.3, P] (This exercise makes repeated use of the mean

value inequality.)

(i) Consider the diļ¬erential equation

y (t) = |y(t)|Ī±

with Ī± > 1. Suppose y is a solution with y(0) ā„ 1. Show that y(t) is strictly

increasing for t ā„ 0 and y(t) ā„ t + 1 for t ā„ 0. Let tj be deļ¬ned by taking

y(tj ) = 2j . Show that tj+1 ā’ tj ā¤ 2(1ā’Ī²)j and deduce that there exists a

tā— ā R such that tj ā’ tā— as j ā’ ā. Conclude that y(t) ā’ ā as t ā’ tā—

through values of t < tā— .

(ii) Prove the result of Example 12.3.6 by an argument along the lines of

part (i).

(iii) Show that there exists a diļ¬erentiable function y : [0, ā) ā’ R such

that y(0) = e and

y (t) = y(t) log y(t).

577

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

Exercise K.266. (Eulerā™s method.) [12.3, T] The next few questions

deal with Eulerā™s method for ļ¬nding approximate solutions of a diļ¬erential

equation. They were suggested by the typically neat treatment in [44]. You

will ļ¬nd the discussion more interesting if you have used Eulerā™s method

before or if you try one of the excellent demonstrations which now exist on

the Internet.

Suppose we wish to solve the diļ¬erential equation y (t) = f (t, y(t)),

y(0) = y0 numerically on a computer. (We shall look at y(t) for t ā„ 0

but the same ideas apply when t ā¤ 0.) A natural approach is to seek (ap-

proximate) solutions at the points xr = rh where the strictly positive number

h is called the ā˜step lengthā™ and r is a positive integer. Explain brieļ¬‚y why

we expect

y (r + 1)h ā y(rh) + f (rh, y(rh))h.

This suggests that our approximation yr to y(rh) could be obtained by solving

the system of equations

yr+1 = yr + f (xr , yr )h. (A)

This is called Eulerā™s method.

Eulerā™s method seems reasonable and works on suitable test examples.

In this question we investigate its behaviour in general. We shall assume

stronger conditions on f than we did in Theorem 12.2.3 and elsewhere in

Sections 12.2 and 12.3. Speciļ¬cally, we demand that f : R2 ā’ R is once

diļ¬erentiable and there exist constants K and L such that |f,2 (t, u)| ā¤ K

and

|f,1 (t, u)| + |f,2 (t, u)f (t, u)| ā¤ L

for all (t, u) ā R2 . Explain brieļ¬‚y why these conditions imply the existence

of a unique diļ¬erentiable y : R ā’ R such that

y (t) = f (t, y(t)), y(0) = y0 . (B)

Explain why y is twice diļ¬erentiable and express y (t) in terms of f (t, y(t)),

f,1 (t, y(t)) and f,2 (t, y(t)). By using the 2nd mean value theorem (Exer-

cise K.49), or otherwise, show that

Lh2

|y (n + 1)h) ā’ y(nh) ā’ f (nh, y(nh))h| ā¤ .

2

Use this result to show that

Lh2

|y (n + 1)h) ā’ yn+1 | ā¤ (1 + Kh)|y(nh) ā’ yn | + .

2

578 A COMPANION TO ANALYSIS

Deduce that

Lh

((1 + Kh)N +1 ā’ 1).

|y(N h) ā’ yN | ā¤

2K

[h]

In order to emphasise the dependence on h, let us write yn = yn . Show

that, if the positive number a is ļ¬xed and we take hN = a/N , then

LhN

[h ]

((1 + KhN )N +1 ā’ 1).

|y(a) ā’ yN N | ā¤

2K

Observe (or recall) that that, if x ā„ 0, then

m

1 m1

x m

xr ā„ 0,

e ā’ (1 + x) ā„ ā’

r mr

r!

r=0

and deduce that

LhN aK

[h ]

|y(a) ā’ yN N | ā¤ (e ā’ 1) (C)

2K

[h ]

and, in particular, yN N ā’ y(a) as N ā’ ā.

Exercise K.267. [12.3, T, ā‘ ] We continue with the discussion of Exer-

cise K.266. Informally, we may interpret the inequality (C) in Exercise K.266

as saying that that ā˜our error estimate for Eulerā™s method is roughly propor-

tional to the step lengthā™. nequality (C) is, however, only an upper estimate

for the error. Can we get a substantially better estimate? Consider the spe-

cial case f (t, u) = u, y0 = 0 and a = 1. Show that y(t) = et and yn = (1+y)n .

Show that

[h ]

y(1) ā’ yN N = e ā’ (1 + N ā’1 )N

N

1 N1 1 hN

ā„ ā’ ā„ = .

r Nr

r! 2N 2

r=0

Informally, we may say ā˜in this case, the error for Eulerā™s method decreases

no faster than the step lengthā™.

We see that (roughly speaking) halving the step length in Eulerā™s method

will double the amount of calculation and only halve the error. Thus Eulerā™s

method is a reasonable one if we want to solve one diļ¬erential equation to

a reasonable degree of accuracy but unattractive if we want to solve many

diļ¬erential equations to a high degree of accuracy.

579

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

Exercise K.268. [12.3, T, ā‘ ] We continue with the discussion of Exer-

cises K.266 and K.267. So far we have ignored the fact that computers only

work to a certain degree of accuracy. As a simple model to take account of

this, we replace equation (A) of Exercise K.266 by

yr+1 = yr + f (xr , yr )h +

Ė Ė Ė (D)

r

where the ā˜errorā™ r satisļ¬es the condition | r | ā¤ for some ļ¬xed > 0. As

before, we impose the initial condition y0 = 0.

Ė

[h]

(i) As in Exercise K.266, we take yn = yn , ļ¬x a positive number a and

take hN = a/N . Show that

LhN

[h ]

(eaK ā’ 1).

|y(a) ā’ yN N | ā¤ + (E)

2K hN K

What happens to the error estimate (that is the right hand side of (E)) as

N ā’ ā (that is as the step length approaches 0)? Explain in simple terms

why we should expect this.

Lh

(ii) Which positive value of h minimises + ? (iii) Show that the

2K hK

best bound on the error that we can guarantee using (E) is proportional to

1/2

.

(iv) Just as in Exercise K.267, we ask whether we can improve substan-

tially on (iii). Once again, we consider the special case f (t, u) = u, y0 = 0

and a = 1. A little experimentation shows that we can solve (D) if we choose

r = (1 ā’ rh) , giving

yr+1 = (1 + N ā’1 )Ėr ā’ (1 ā’ rN ā’1 ) .

Ė y (F)

Show that, for this choice of r, we obtain

yr = (1 ā’ N ā’1 )r ā’ r

Ė

and so

hN

[h ]

y(1) ā’ yN N = e ā’ (1 + N ā’1 )N ā’ N ā„ 21/2 1/2

ā„

Ė + .

2 hN

Thus the answer to our question is no.

There exist several diļ¬erent ways of improving on Eulerā™s method but a

good understanding of Eulerā™s method is very helpful in understanding how

anybody came to think of these more advanced techniques.

580 A COMPANION TO ANALYSIS

Exercise K.269. [12.3, H, ā‘ ] (i) Let a > 0. By considering the diļ¬erential

equation y (x) = g(x) and using Exercise K.266, show that there exists a

constant Īŗa such that, if g : R ā’ R is diļ¬erentiable, |f (t)| ā¤ M for all

t ā [0, a], N is a strictly positive integer and N h = a, then

N ā’1

a

g(t) dt ā’ h g(rh) ā¤ M Īŗa h.

0 n=0

(It is easy to obtain this result directly with better values for Īŗa . See, for

example, Exercise K.125 (v) or just think a bit.)

(ii) Suppose that a = Ļ, N h = a with N a strictly positive integer and

G(t) = sin2 (N t). Show that

N ā’1

a

G(rh) ā„ 10ā’2 sup |G(t)|.

G(t) dt ā’ h

tā[0,a]

0 n=0

(iii) Consider the situation described in Exercise K.266. Suppose that we

replace the condition |f,1 (t, u)|+|f,2 (t, u)f (t, u)| ā¤ L by the weaker condition

|f (t, u)| ā¤ L for all (t, u) ā R2 . If a > 0 and > 0 are ļ¬xed, is it possible to

ļ¬nd an N0 (depending only on K, L, a and ) such that

[h ]

|y(a) ā’ yN N | ā¤

for all N ā„ N0 ?

Exercise K.270. (Leibniz rule.) [12.3, G] (i) If u, v : R ā’ R are n

times diļ¬erentiable at x show, by induction or otherwise, that the product

uv is n times diļ¬erentiable at x with

n

n (nā’r)

(n)

(x)v (r) (x).

(uv) (x) = u

r

r=0

(this is called the Leibniz rule.)

(ii) If y : R ā’ R is deļ¬ned by

(1 + x2 )1/2 y(x) = log(x + (1 + x2 )1/2 ),

show that y is diļ¬erentiable on (0, ā) with

(1 + x2 )y (x) + xy(x) = 1.

Show, by using induction and the Leibniz rule, that y is inļ¬nitely diļ¬eren-

tiable and ļ¬nd the Taylor series

ā

y (n) (0) n

x.

n!

n=0

581

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

(iii) (This part uses sledgehammers to crack a nut and may be omitted

without loss.) Find the radius of convergence of the Taylor series and, by

using results on the diļ¬erentiation of power series and the uniqueness of the

solution of diļ¬erential equations, show that, within the radius of convergence

ā

y (n) (0) n

y(x) = x.

n!

n=0

Exercise K.271. [12.4, P, S] Let K : [0, 1]2 ā’ R and g : [0, 1] ā’ R be

continuous. Explain why there exists an M > 0 such that |K(s, t)| ā¤ M

for all (s, t) ā [0, 1]2 . Suppose that |Ī»| < M ā’1 . By ļ¬nding an appropriate

contraction mapping T : C([0, 1]) ā’ C([0, 1]) show that there exists a unique

f ā C([0, 1]) such that

1

f (t) = g(t) + Ī» K(s, t)f (t) ds.

0

Exercise K.272. (The Wronskian.) [12.4, M] This should be treated

as an exercise in calculus rather than analysis. We write

ļ£« ļ£¶

a11 a12 a13 a11 a12 a13

a21 a22 a23 = det ļ£a21 a22 a23 ļ£ø

a31 a32 a33 a31 a32 a33

(i) If u1 , u2 , . . . w3 are all diļ¬erentiable show that

u1 (t) u2 (t) u3 (t)

d

v1 (t) v2 (t) v3 (t)

dt

w1 (t) w2 (t) w3 (t)

u1 (t) u2 (t) u3 (t) u1 (t) u2 (t) u3 (t) u1 (t) u2 (t) u3 (t)

= v1 (t) v2 (t) v3 (t) + v1 (t) v2 (t) v3 (t) + v1 (t) v2 (t) v3 (t) .

w1 (t) w2 (t) w3 (t) w1 (t) w2 (t) w3 (t) w1 (t) w2 (t) w3 (t)

(ii) If u1 , u2 and u3 are three solutions of

y (t) + a(t)y (t) + b(t)y (t) + c(t)y(t) = 0,

we deļ¬ne their Wronskian W by

u1 (t) u2 (t) u3 (t)

W (t) = u1 (t) u2 (t) u3 (t) .

u1 (t) u2 (t) u3 (t)

582 A COMPANION TO ANALYSIS

Use part (i) and results about determinants to show that

W (t) = ā’a(t)W (t).

(iii) Generalise parts (i) and (ii). Reread the proof of part (i) of Lemma 12.4.3

in the light of this exercise.

Exercise K.273. [12.4, T, S] The functions f, g : R ā’ R are everywhere

diļ¬erentiable and their Wronskian f g ā’ g f never vanishes. By applying

Rolleā™s theorem to f /g, or otherwise, show that if f has zeros at a and b with

a < b then g must have a zero strictly between a and b. Deduce that ā˜the

zeros of f and g are intertwinedā™.

Exercise K.274. [12.4, P] If f1 , f2 : [a, b] ā’ R are once diļ¬erentiable we

deļ¬ne the Wronskian W (f1 , f2 ) by

f1 f2

W (f1 , f2 ) = det

f1 f2

Show that if f1 and f2 are linearly dependent (that is, we can ļ¬nd Ī»1 , Ī»2 ā R

not both zero such that Ī»1 f1 (t) + Ī»2 f2 (t) = 0 for all t ā [a, b]) then the

Wronskian W (f1 , f2 ) is identically zero. By considering functions of the type

g with g(x) = 0 for x ā¤ c, g(x) = (x ā’ c)2 for x ā„ c, or otherwise, show that

the converse is false.

Suppose now that f1 , f2 are twice continuously diļ¬erentiable. Show, by

considering the Wronskian

ļ£« ļ£¶

f f 1 f2

W (f, f1 , f2 ) = det ļ£ f f1 f2 ļ£ø

f f1 f2

and using results on the existence and uniqueness of diļ¬erential equations

with given initial conditions, that the following result holds.

There exists a diļ¬erential equation of the form

y (x) + a1 (x)y (x) + a2 (x)y(x) = 0

whose solutions are exactly the functions of the form Ī»1 f1 + Ī»2 f2 with Ī»1 and

Ī»2 (in more sophisticated language, having f1 and f2 as a basis for the space

of solutions) if and only if W (f1 , f2 ) is non-zero everywhere on [a, b].

Generalise the results of this question to n functions f1 , f2 , . . . , fn .

583

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

Exercise K.275. [12.4, M] (This question should be treated as a mathe-

matical methods one.)

(i) Consider a diļ¬erential equation

y (x) + p(x)y (x) + q(x)y(x) = 0.

Suppose that u is a solution of . Show that if we write y(x) = u(x)v(x)

and w(x) = u (x) then takes the form of a ļ¬rst order diļ¬erential equation

x

in w. Show that y(x) = (A + B 0 w(t) dt)v(x) gives the general a solution

for .

(ii) Show that y(x) = exp x gives a solution of

y (x) ā’ (2x + 1)y (x) + (x + 1)y(x) = 0

and use the method of (i) to ļ¬nd the general solution.

(iii) We know that y(x) = exp x gives a solution of

y (x) ā’ 2y (x) + y(x).

Find the general solution by the method of (i).

(iv) Show how, if we know one solution of

y (x) + p1 (x)y (x) + p2 (x)y (x) + p3 (x)y(x) = 0

we can reduce the problem of ļ¬nding a general solution to that of solving a

second order linear diļ¬erential equation. Generalise.

Exercise K.276. (Method of variation of parameters.) [12.4, M,

ā‘ ] (Like Question K.275 this question should be treated as a mathematical

methods one.)

Consider a diļ¬erential equation

y (x) + p(x)y (x) + q(x)y(x) = 0.

whose Wronskian W (x) = y1 (x)y2 (x) ā’

Let y1 and y2 be solutions of

y1 (x)y2 (x) never vanishes. Suppose that we wish to solve the diļ¬erential

equation

y (x) + p(x)y (x) + q(x)y(x) = f (x).

(i) In view of the success of the method of Question K.275 we might be

tempted to look for a solution of the form

y(x) = u1 (x)y1 (x) + u2 (x)y2 (x).

584 A COMPANION TO ANALYSIS

Make this substitution in .

(ii) At the end of (i) you obtained a diļ¬erential equation (ā ) for u1 and

u2 . It is a useful heuristic guide (though not always a reliable one) that two

unknowns require two equations so we add the further equation

u1 (x))y1 (x) + u2 (x)y2 (x) = 0. (ā ā )

Use (ā ā ) to simplify (ā ) and then use the pair of equations to ļ¬nd u1 and u2 .

Now ļ¬nd the general solution of .

(iii) Discuss the relation of your result to the Greenā™s function result given

in Theorem 12.4.6.

(iv) Use the method of this question to ļ¬nd the general solution of

2

y (x) ā’ y(x) = .

1 + ex

(v) Try and extend the ideas of this question to the solution of

y (x) + p(x)y (x) + q(x)y (x) + r(x)y = f (x).

when three solutions y1 , y2 , y3 with non-zero Wronskian (see Question K.272)

are known.

[It would be hard to think of an example more opposed to the modern view

that ā˜understanding precedes doingā™. But even the most convinced proponent

of modern ways must admit that it has a certain charm.]

Exercise K.277. [12.4, P] Let C(S) be the space of continuous functions

on the unit square

S = {(x, y) : 0 ā¤ x ā¤ 1, 0 ā¤ y ā¤ 1}

equipped with uniform norm K ā = sup(x,y)āS |K(x, y)| and let C(I) be the

space of continuous functions on the unit interval I = [0, 1] equipped with the

uniform norm. Let L be the space of continuous linear maps L : C(I) ā’ C(I)

equipped with the operator norm. If K ā C(S) we set

1

(TK (f ))(x) = K(x, y)f (y) dy.

0

Show that TK ā L.

Prove or disprove each of the following statements.

(i) If Kn ā’ K in C(S), then TKn ā’ TK in L.

(ii) If TKn ā’ TK in L, then Kn ā’ K in C(S).

(iii) The mapping K ā’ TK is an injective map from C(S) to L.

(iii) The mapping K ā’ TK is a surjective map from C(S) to L.

585

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

Exercise K.278. [12.4, T] (i) In Exercise K.132, you showed that if g :

R2 ā’ R is a diļ¬erentiable function with continuous partial derivatives then

x

G(x) = g(x, t) dt

0

is diļ¬erentiable with

x

G (x) = g(x, x) + g,1 (x, t) dt

0

Review this exercise.

(ii) Suppose that a, b, f : R ā’ R are continuous. Explain why there

exists a unique twice diļ¬erentiable y1 : R ā’ R such that

y1 (t) + a(t)y1 (t) + b(t)y1 (t) = 0, y1 (0) = 0, y1 (0) = 1,

and a unique twice diļ¬erentiable y2 : R ā’ R such that

y2 (t) + a(t)y2 (t) + b(t)y2 (t) = 0, y2 (1) = 0, y2 (1) = 1.

We make the following

key assumption: y1 (1) = 0.

Ė

Show that, if we deļ¬ne H, H : R ā’ R by

Ė

H(s, t) = y1 (t)y2 (s)W (s)ā’1 , H(s, t) = y2 (t)y1 (s)W (s)ā’1 ,

Ė

then H and H are twice diļ¬erentiable functions with continuous second par-

tial derivatives and

H,22 (s, t) + a(t)H,2 (s, t) + b(t)H(s, t) = 0, H(s, 0) = 0,

Ė Ė Ė Ė

H,22 (s, t) + a(t)H,2 (s, t) + b(t)H(s, t) = 0, H(s, 1) = 0.

Ė Ė

Check that H(t, t) = H(t, t), H2 (t, t) = H2 (t, t).

(iii) We deļ¬ne G : R2 ā’ R by G(s, t) = H(s, t) for t ā¤ s and G(s, t) =

Ė

H(s, t) for s ā¤ t. If we set

1 t 1

Ė

y(t) = G(s, t)f (s) ds = H(s, t)f (s) ds + H(s, t)f (s) ds,

0 0 t

Ė

show, using part (i) and the properties of H and H established in part (ii)

Ė

(but not using the deļ¬nitions of H and H), that y is twice diļ¬erentiable and

satisļ¬es

y (t) + a(t)y + b(t)y = f (t)

together with the conditions y(0) = y(1) = 0.

586 A COMPANION TO ANALYSIS

Exercise K.279. [12.4, M] Consider the diļ¬erential equation

y (4) ā’ k 4 y = f

for y : [0, 1] ā’ R subject to boundary conditions y(0) = y (0) = 0, y(1) =

y (1) = 0 where f : [0, 1] ā’ R is continuous and k is real.

By extending the discussion of the Greenā™s function in Section 12.4 show

that, provided that k does not take certain exceptional values to be identiļ¬ed,

the system has the solution

1

y(x) = G(x, t)f (t) dt

0

where

A(sinh kx ā’ sin kx) + B(cosh kx ā’ cos kx) if 0 ā¤ x ā¤ t,

G(x, t) =

C(sinh k(1 ā’ x) ā’ sin k(1 ā’ x)) + D(cosh k(1 ā’ x) ā’ cos k(1 ā’ x)) if t ā¤ x ā¤ 1,

and A, B C, D are given by

ļ£«ļ£¶ ļ£«ļ£¶

ńņš. 18 |