A Companion to Analysis - Solutions, Korner

7/25/2019 A Companion to Analysis - Solutions, Korner

1/405

Partial Solutions for Questions in

Appendix K of

A Companion to Analysis

T. W. Korner

1


2/405

2

Introduction

Here is a miscellaneous collection of hints, answers, partial answersand remarks on some of the exercises in the book. I expect that thereare many errors both large and small and would appreciate the oppor-tunity to correct them. Please tell me of any errors, unbridgable gaps,

misnumberings etc. I welcome suggestions for additions. ALL COM-MENTS GRATEFULLY RECEIVED. (If you can, please use LATEX 2or its relatives for mathematics. If not, please use plain text. My e-mailis [email protected]. You may safely assume that I am bothlazy and stupid so that a message saying Presumably you have alreadyrealised the mistake in question 33 is less useful than one which saysI think you have made a mistake in question 33 because not all leftobjects are right objects. One way round this problem is to quote Xstheorem.)

To avoid disappointment note that a number like K15* means thatthere is no comment. A number marked K15? means that I still need

to work on the remarks. Note also that what is given is at most asketch and often very much less.Please treat the answers as a last resort. You will benefit more from

thinking about a problem than from reading a solution. I am inveteratepeeker at answers but I strongly advise you to do as I say and not asI do.

It may be easiest to navigate this document by using the table ofcontents which follow on the next few pages.


3/405

3

Contents

Introduction 2K1 11K2 12K3 13

K4 14K5 15K6 16K7 17K8 18K9 19K10 20K11 21K12 22K13* 23K14* 24

K15 25K16 26K17 27K18 28K19 29K20 30K21 31K22 32K23* 33K24 34K25 35

K26 36K27 37K28 38K29 39K30 40K31 41K32 42K33 43K34* 44K35* 45K36 46

K37* 47K38 48K39 49K40 50K41 51K42 52K43 53


4/405

4

K44 54K45 55K46* 56K47 57K48 58K49 59

K50 60K51 61K52 62K53 63K54 64K55 65K56 66K57 67K58 68K59 69K60 70

K61 71K62 72K63 73K64 74K65 75K66 76K67 77K68 78K69 79K70 81K71 82

K72 83K73 84K74 85K75 86K76 87K77 88K78 89K79* 90K80 91K81 92K82 93

K83 94K84 95K85 96K86 97K87 98K88 99


5/405

5

K89 100K90 101K91 102K92 103K93 104K94 105

K95 106K96 107K97 108K98 110K99 111K100 112K101 113K102 114K103 115K104 116K105 117

K106 118K107 119K108 120K109 121K110 122K111 123K112 124K113 125K114 126K115 127K116 128

K117 129K118 130K119 131K120 132K121 133K122 134K123 135K124 136K125 137K126 138K127 139

K128 140K129 141K130 142K131 143K132 144K133 145


6/405

6

K134 146K135* 147K136 148K137* 149K138 150K139 151

K140 152K141 153K142 154K143 155K144 158K145 160K146* 161K147 162K148 163K149 164K150 165

K151 166K152* 167K153 168K154 169K155 170K156 171K157 172K158 173K159 175K160 176K161 177

K162 178K163 179K164 180K165 181K166 183K167 184K168 185K169 186K170 187K171 188K172 190

K173* 191K174 192K175* 193K176 194K177 195K178 196


7/405

7

K179 198K180 199K181 200K182 202K183 203K184* 204

K185 205K186* 206K187 207K188 208K189 209K190 210K191 211K192 213K193 214K194 215K195 216

K196 218K197 219K198 220K199 221K200 222K201 223K202 224K203 226K204* 227K205 228K206 229

K207 231K208 232K209 234K210 235K211 236K212 237K213 239K214 241K215 242K216 243K217 244

K218 245K219 246K220 247K221 248K222 249K223 250


8/405

8

K224 251K225 252K226 253K227 254K228 256K229 257

K230 259K231 261K232 263K233 264K234 265K235 267K236 268K237 270K238 272K239 273K240 274

K241 275K242 277K243* 278K244 279K245 280K246 281K247 282K248* 284K249* 285K250 286K251* 287

K252* 288K253 289K254 290K255 291K256 292K257 293K258 294K259 296K260 297K261 299K262 301

K263 303K264 305K265 307K266 308K267 309K268 310


9/405

9

K269 311K270 312K271 313K272 314K273 316K274 318

K275 319K276 320K277 321K278 322K279 323K280 324K281 325K282 326K283 327K284 328K285 329

K286 331K287 332K288 334K289 336K290 337K291 338K292 339K293 340K294 341K295 342K296* 343

K297 344K298 345K299 346K300 347K301 348K302 349K303 350K304 352K305 353K306 355K307 356

K308 357K309 359K310 361K311 363K312 364K313 365


10/405

10

K314 366K315 368K316 369K317 370K318 371K319 373

K320 374K321 375K322 377K323 378K324 379K325 381K326 382K327 384K328 385K329* 386K330 387

K331 388K332 389K333 391K334 392K335 393K336 394K337 395K338 396K339 400K340 401K341 402

K342 403K343 404K344 405K345 406


11/405

11

K1

(ii) Every non-empty set of positive integers has a least member.Sincex is a rational number, qxis an integer for some strictly positiveinteger. We can certainly find a k such thatk + 1> x k (ultimatelyby the Axiom of Archimedes). Butxis not an integer so k +1> x > k.

Sincemx,mand k are integers,m = mx mk is. Alsomx= mx2 mkx = mN k(mx)

is an integer since mx is. Since 1> xk >0 we havem > mxmk=m > 0 and m > m 1 contradicting the definition of m as leaststrictly positive integer with mx an integer.


12/405

12

K2

Write cn= dn+ d1n . Then

d2n cndn+ 1 = 0

dn=cn

c2n

4

2 .

Since the product of the two roots ofis 1, we must take the biggerroot. Thus

dn=cn+

c2n 4

2 c +

c2 4

2wherec is the limit ofcn.

For the second paragraph, just take

d2n= (1 + k)/2 and d2n+1= 2/(1 + k).

For the third paragraph, the condition|dn| 1 will do but we mustargue carefully. Look first to see which root of d2 cd + 1 = 0 liesoutside the unit circle.


13/405

13

K3

(Second part of problem.) Choose a1 = 2, b1= 1, say.


14/405

14

K4

Takex = 0,f(t) =H(t),g(t) = 0. Take f(t) = 0, g(t) =H(t).

(H(t) = 1 for t >0, H(t) = 0 fort 0.)


15/405

15

K5

(i), (ii) and (iii) are true and can be proved by, e.g. arguing by cases.

(iv) is false sincefis continuous at 1/2. (fis, however, discontinuouseverywhere else.)

(v) is false. Take e.g. x= 161 + 171/2,y= 31 171/2.


16/405

16

K6

Nn=12

nH(x qn) is an increasing sequence in Nbounded aboveby 1.

Observe that ifx y thenH(x qn) H(y qn) 0 for alln 0.Thus N

n=1

2nH(x qn) N

n=1

2nH(y qn)

=N

n=1

2n

H(x qn) H(y qn) 0

for allN sof(x) f(y).Observe that, ifx > qm, then H(x qn) H(qm qn) 0 for all

n 0 and H(x qm) H(qm qm) = 1 soN

n=1

2nH(x qn) N

n=1

2nH(qm qn) 2m

for all N m so f(x) f(qm) + 2m for all x > qm. Thusf is notcontinuous atqm.

Ifx is irrational, we can find an >0 such that|qm x|> for allm Mand so

Nn=1

2nH(x qn) N

n=1

2nH(y qn) 2M

whenever

|x

y

|< . Thus f is continuous at x.


17/405

17

K7

Observe thatan+1 an is decreasing. Thus either(a)an+1 an> 0 for allnand, ifan is unbounded an , or(b) There exists an N such thatan+1

an

0 for alln

Nand, if

an is unbounded, an .For the reverse inequality, observe thatan satisfies the original

inequality.


18/405

18

K8

A decreasing sequence bounded below tends to a limit, so, for anyfixedx (0, 1),

xn= fn(x) y, say.

Sincex > xn> 0 we have 1 > y 0. Ify >0 then by continuityxn+1= f(xn) f(y)

and so y = f(y). Thus y = 0.

If

f(x) =

14 +

x2 forx >1/2,

x4

forx 1/2,thenfn(x) 1/2 for 1> x >1/2.

We suppose our calculator works in radians. Ifan+1 = sin an then,

whatever a0 we start from, we have|a1| 1. Ifa1 =b1 and bn+1 =sin bn thenan= bn forn 1. If 0< a


19/405

19

K9

(ii) we need only look at positive integers. Ifn 2(k+ 1)

(1 + )n =n

j=0

n

j

j

n

k+ 1

k+1

(n/2)k+1

(k+ 1)!k+1

nk+1

2k+1(k+ 1)!k+1

so

nk(1 + )n n k+1

2k+1(k+ 1)!

as n .


20/405

20

K10

Observe that

xnn

=1

n

n

x1+ x2+ + xnn

(n 1) x1+ x2+ + xn1n 1

=

x1+ x2+ + xnn

1 1n

x1+ x2+ + xn1n 1

1 (1 0)1 = 0as n .

Either there exists an N such thatm(n) =m(N) for all n N andthe result is immediate or m(n) and

xm(n)n

xm(n)m(n)

0as n .

If >1, then

x1 + x2 + + xn

n x1+ x2+ + xn

n

xm(n)n

1 1 0 = 0.

If


21/405

21

K11

(i) True. Ifa + b= c and a, c Qthenb = c a Q.(ii) False. 02 = 0. (However the product of an irrational number

with a non-zero rational number is irrational.)

(iii) True. By the axiom of Archimedes we can find a strictly positiveinteger N with N1 < and an integer M such that mN1 x 0. We can find an >0 such thatf(t) f(c)t c f(c) f(c)

2

for all t with|tc| < . Thus f(t) > f(c) for c < t < c + andf(t)< f(c) forc < t < c

Choose n such that 2n < . We have f(an)< f(bn) which is absurd.


26/405

26

K16

Repeat the lion hunting argument for the intermediate value theoremwith f(an) g(an), f(bn) g(bn). If an c, say, then, since f isincreasing this gives

g(bn)

f(bn)

f(c)

f(an)

g(an).

Ifanc, say, then since g is continuous g(an)g(c) and so f(c) =g(c).

First sentence of second paragraph. False. Consider [a, b] = [1, 1],g(x) = 0,f(x) =x 1 forx 0,f(x) =x+ 1 for x >0.

Second sentence false. Consider [a, b] = [1, 1], g(x) =x+ 3,f(x) =x 1 forx 0,g(x) = x + 3, f(x) =x + 1 for x >0.

Third sentence true. Apply intermediate value theorem to f g.


28/405

28

K18

(i) Observe that

{x R : E (, x] is countable}is a non-empty bounded set and so has a supremum say. Observe

thatEn= E (, 1/n] is countable so E (, ) = n=1 Enis countable and so E (, ] is countable. Thus E (, ) isuncountable if and only if > . Similarly we can find a such thatE (, ) is uncountable if and only if < . SinceEis uncountable, > .

(ii) We have one of four possibilities.

(a) There exist and with > and

{e E : e < }and{e E : e > }are uncountable if and only if < < .

(b) There exists an with{e E : e < }and{e E : e > }

are uncountable if and only if < .

(c) There exists a with

{e E : e < }and{e E : e > }are uncountable if and only if < .

(d) We have

{e E : e < }and{e E : e > }uncountable for all.

All these possibilities can occur. Look at Z [1, 1], (0, ), (, 0)and R.

(iii) SetE= {1/n : n 1, n Z}


29/405

29

K19

(i) False. Considerxj = 1.

(ii) False. Considerxj = j.(iii) False. Considerxj = (

1)j.

(iv) False. Considerxj = (1)j.


30/405

30

K20

(i) Let >0. There exists anN such that an lim supr ar+ for n N. There exists anMsuch that n(p) N for pM. Nowlim supr ar+ an(p) forp M. Since was arbitrary,

lim supr

ar

a.

(ii) Take an= (1 + (1)n)/2.Take e.g. a2n+r = r/2

n for 0 r 0. Ifn is large enough

limsupr

ar+ > an and limsupr

br+ > bn

so

lim supr ar+ lim supr br+ 2 > an+ bn.Thus

lim supr

ar+ lim supr

br+ 2 >limsupn

(an+ bn).

Butis arbitrary so

limsupr

ar+ lim supr

br limsupn

(an+ bn).

(True) Similarly if >0 andn is large enough

bn lim infr

br and so

an+ bn an+ lim infr

br .Thus

lim supn

(an+ bn) lim supn

an+ lim infr

br and, since was arbitrary,

limsupn

(an+ bn) limsupn

an+ lim infr

br.


31/405

31

K21

x

x2 y

y2

2

=x2x2 2

x yxy+

y2y2

=y2

2x y + x2

x2y2

=

x yxy

2.

Thus xx2 yy2 =x yxy .

Ptolomeys result now follows from the triangle inequality for points ofthe formx2x.


32/405

32

K22

(i), (ii) and (iii) are true. Proof by applying the definitions.

(iv) is false. Take, for example, n = 1,U= {x :|x|


33/405

33

K23*

No comments.


34/405

34

K24

(i) If (xn, yn) E and(xn, yn)(x, y) 0, then xn x andyn y. Thus yn= 1/xn xand x = y so (x, y) E.

1(E) = {x : x >0}is not closed since 1/n 1(E) and 1/n 0 /1(E) asn 0.

(ii) Let E={(x, 1/x) : x >0} {(1, 0)}. Then Eis closed (uniontwo closed sets), 1(E) ={x : x >0} is not closed but 2(E) ={y :y 0} is.

(iii) Ifxn 1(E) andxn xwe can findynsuch that (xn, yn) E.Since yn is a bounded sequence there exists a convergent subsequenceyn(j) y as j . Since xn(j) x the argument of (i) shows that(x, y)E and x1(E). (It is worth looking at why this argumentfails in (ii).)


35/405

35

K25

(i) IfG [K, K]n+m, then E [K, K]n.(ii) Take n = m = 1, E= (0, ),f(x) = 1/x See question K.24.(iii) Suppose

(xj, f(xj))

(x, y)

0. Then

xj

x

0. Since

E is closed, xE. Since f is continuous,f(xj) f(x) 0. Thusy= f(x) and (x, y) G.

(iv) Take n = m = 1,E= Randf(x) =x1 forx = 0, f(0) = 0.(v) Suppose xj Eand xjx 0. SinceGis closed and bounded

we can find a subsequence j (k) such that(xj(k), f(xj(k))) (x, y) 0

for some (x, y) G. Since (x, y) G, we have y= f(x).Observe that x E, so E is closed. If f was not continuous at

x, we could find a > 0 and xj

E and

xj

x

0 such that

f(xj) f(x) > and so(xj(k), f(xj(k))) (x, y) >

for everyj (k). Our result follows by reductio ad absurdum.


36/405

36

K26

If we write f(x) = 0 for x= 0 and f(0) = 1, then f is uppersemicontinuous but not continuous.

Iffis not bounded we can findxn Esuch thatf(xn) n. Extracta convergent subsequence and use upper semicontinuity to obtain acontradiction. To show that the least upper bound is attained, take asequence for which f approaches its supremum, extract a convergentsubsequence and use upper semicontinuity again.

Taken= 1, K= [0, 1], f(x) =1/xforx= 0 , f(0) = 0 to obtainan upper semicontinuous function not bounded below.

Take n= 1, K= [0, 1], f(x) =x for x= 0 , f(0) = 1 to obtain anupper semicontinuous function bounded below which does not attainits infimum.

(As usual it is informative to see how the proof of the true result

breaks down when we loosen the hypotheses.)


37/405

37

K27

(i)fis continuous on [0, a] so is bounded and attains its bounds on[0, a]. By periodicityfis everywhere bounded and attains its bounds.

(ii) Letf(x) = (x [x])1) ifx is not an integer and f(x) = 0 ifx isan integer. (Here [x] is the integer part ofx.)

(iii) By considering g(x) kx we see that we may suppose k = 0.Given >0 there exists anX >0 such that|g(x + 1) g(x)| < (andso|g(x+n) g(x)| < n for x > X. By hypothesis there exists a Ksuch that|g(y)| K for 0 y X+ 1. Write n(x) for the integersuch thatX < x n(x) X+ 1. Ifx > X+ 1 we haveg(x)x

g(x) g(x n(x))x

+g(x n(x))x

n(x)

x +

K

x

as x . Sincewas arbitrary, g(x)/x 0 as x .(iv) Any example for (ii) will work here.

(v) False. Consider, for example, h(x) =kx + x1/2 sin(x/2).


38/405

38

K28

No comments.


39/405

39

K29

Observe thatxn can lie in at most two of the three intervals

[an1, an1+ kn1], [an1+ kn1, an1+ 2kn1]and [an1+ 2kn1, bn1]

wherekn1= (bn1 an1)/3.


40/405

40

K30

(i) Lots of different ways. Observe that the mapJgiven by (x, y) x y is continuous since

|(x + h) (y + k) x y| |x k| + |h y| + |h k|

hx + yk + hk 0as (h, k) 0. Since is continuous, the mapA given by x (x, x)is continuous so, composing the two maps A and J, we see that thegiven map is continuous.

Last part, continuous real valued function on a closed bounded setattains a maximum.

(ii) Observe that, ifh e= 0, thene + h2 = (1 + 2) so by (i)(1 + 2)e (e) (e + h) ((e + h))

and so

(e (h) + h (e + (e (e) h (h)) 0for allso

e (h) + h (e) = 0.(iv) If we assume the result true for Rn1, then, since Uhas dimen-

sion n 1 and |Uis self adjoint, Uhas an orthonormal basis e2, e3,. . . , en of eigenvectors for|U. Takinge1= e, gives the desired result.


41/405

41

K31

(i) Observe that

|g(u) g(v)| uvsince, as simple consequence of the triangle inequality,

|a b | a b.Choose any z E. Set R =z y. The continuous function g

attains a minimum on the closed bounded set B(y, R) Eat x0, say.If xE then either x B(y, R) and g(x)g(x0) automatically, orx / B(y, R) and

g(x)> R= g(z) g(x0).(ii) Let u= x0 y. Use the definition ofx0 to show that

u u (u + h) (u + h)wheneverh E. By considering what happens whenis small, deducethat u h= 0.

Ifx1 y =x0 y and x1 E, then, setting h= x1 x0, weget h h= 0.

(iii) Since = 0 we can find a y / E. Set u = y x0 and b =(u)1u. Observe that

(x (b x)b) b= 0so, since E has dimensionn 1, we have

x

(b

x)b

E

and

(x (b x)b) = 0so

x= b xb.Set a= (b)b.


42/405

42

K32

(i) Argument as in K31.

(ii) Suppose xE and x y >0. Then, if is small and positive,we have

x= (1 )0 + x Ebut the same kind of argument as in K30 and K31 shows thaty>y x.

(iii) Translation.


43/405

43

K33

Write B for the closed unit ball. Ifz >0 and

x0+ u0= v + (1 )wthen

x0 a= (x0+ u0) a=v a + (1 )w) a

x0 a + (1 )x0 a=x0 a.

The inequality in the last set of equations must thus be replaced byequality and so v, w K. Since u0 is an extreme point ofK, v =w= u0.

(iv) Use a similar argument to (ii) (or (ii) itself) to show that extremepoints of

{x K : T(x) =T(x0)},are extreme points ofK.


44/405

44

K34*

No comments.


45/405

45

K35*

No comments


46/405

46

K36

(i) If neitherK1 norK2 have the finite intersection property, we canfind K1, K2, . . . K N, KN+1, . . . , K N+M K such that

N

j=1

Kj

[a, c] =

and

M

j=N+1

Kj

[c, b] =

and so

N+Mj=1

Kj [a, b] = .

(ii) We find [an, bn] such that

Ln= {K [an, bn] : K K}has the finite intersection property

a= a0

a1

a2

an

bn

b2

b1

b0= b

and bn an= 2n(b a). We have an, bn for some [a, b].We claim that Kfor eachK K. For, if not, we can find K0 K

such that / K0. Since K0 is closed, we can find a > 0 such that(, +)K0 = . IfNis sufficiently large, [aN, bN] (, +)so [aN, bN] K0 = contradicting the finite intersection property ofLN.

Thus, by reductio ad absurdum, KK K.


47/405

47

K37*

No comments


48/405

48

K38

Observe that, if [aj, bj ] K, thenN

j=1[aj, bj ] = [ max

1jNaj , min

1jNbj ] K

so, in particular,Khas the finite intersection property.IfcKK K then c [a, b], that is to say a c b whenever

a E and b e for all e E. Thusc is a (and thus the) greatestlower bound for E.


49/405

49

K39

(i) f1(t) = 1 cos t0 for all t. Thus f1 is everywhere increasing.Butf1(0) = 0 so f1(t) 0 fort 0 sot sin tfor t 0.

(ii)f2(t) =f1(t) 0 fort 0 andf2(0) = 0.(iv) We have

2Nj=0

(1)j t2j+1(2j+ 1)!

sin t 2N+1j=0

(1)j t2j+1(2j+ 1)!

fort 0 and, using the fact that sin(t) = sin t,2N

j=0

(1)j t2j+1(2j+ 1)!

sin t 2N+1j=0

(1)j t2j+1(2j+ 1)!

fort 0.(v) Thus sin t

Nj=0

(1)jt2j+1(2j+ 1)!

|t|N+1

(2N+ 3)! 0

as N .


50/405

50

K40

(i) Observe that g is increasing since g is positive. If g(x1) > 0then g(t) g(x1) > 0 for all t [x1, x2] so 0 = g(x2) > g(x1) = 0which is absurd. Thus g (x1)0 and g(x2)0. By the intermediatevalue theorem there exist a c

[x1, x2] such that g

(c) = 0. Since g

is increasing g(t) 0 and g is decreasing on [x1, c] whilst g(t) 0and g is increasing on [c, x2]. Thus g(t) g(x1) = 0 on [x1, c] andg(t) g(x2) = 0. We use a similar argument to get the result on[c, x2].

(ii) Look atg(t) =f(t) A Bt withA and B chosen to make thehypotheses of (i) hold.

(iii) We may assume that 1> n+1> 0. The key algebraic manipu-lation in a proof by induction runs as follows.

fn+1

j=1

jxj =fn+1xn+1+ (1 n+1)n

j=1

jnk=1 k

xj n+1f(xn+1) + (1 n+1)f

n

j=1

jnk=1 k

xj

n+1f(xn+1) + (1 n+1)n

j=1

jnk=1 k

f(xj )

=n+1j=1

jf(xj )

sincen

j=1

jnk=1 k

= 1 andn

k=1

k = (1 n+1).

(iv) Apply Jensen as suggested and take exponentials.


51/405

51

K41

The area of the inscribed quadrilateral is

=4

j=1a2 sin jcos j =

12

4

j=1a2 sin2j .

withathe radius of the circle. Also4

j=1 j =.

Now sin t= sin t 0 fort [0, ], so sin is convex on [0, ] andJensens inequality gives

= 2a24

j=1

14sin 2j 2a2 sin

4

j=1

14 (2j)

= 2a2 sin(/2) = 2a2.

The area is attained when thejare all equal (and with a little thought,observing that sin t


52/405

52

K42

Since g is continuous on [0, 1], it attains a maximum M say. LetE= {x [0, 1] : f(x) =M}. SinceEis non-empty it has a supremum. Since f is continuous, g() =M. If= 0 or= 1 then M= 0. Ifnot then we can find a k with

k, + k

(0, 1) and

M=g() = 12 (g( k) + g(+ k)) Mwith equality if and only ifg( k) =g(+ k) =M. Thus+ k Econtradicting the definition of. We have shown thatM= 0 sog(t) 0 for all t. Similarlyg(t) 0 for all t so g = 0.

Suppose that we drop the end conditions. By replacing g(t) byG(t) = g(t)ABt with G(0) = G(1) = 0, we can show that gis linear.

[In higher dimensions the condition becomes the average ofg overthe surface of any sufficiently small sphere equals the value at the

centre and this turns out to be equivalent to saying that g satisfiesLaplaces equation 2g = 0. Observe that in one dimension this re-duces to saying thatg = 0 so g is linear as we saw above. However inhigher dimensions the solutions of Laplaces equation are much morediverse.]


53/405

53

K43

Observe that

f(h) f(0)h

= |h sin h4| |h| 0

as h 0.If x= 0, then f(x) = 2x sin x4 4x2 cos x4. Thus we have

f

((2n+ 1))1/4) as n .


54/405

54

K44

Since gis continuous on [a, b], it attains a maximum at some c [a, b].Sinceg (c) = 0, we have c (a, b) andf(c) =k.

For the final paragraph, note thatf(a) 0 f(b).


55/405

55

K45

(i) The statement f must jump by at least 1 at z is not welldefined. If the reader feels that there should be some way of mak-ing this statement well defined in a useful way she should considerf(x) = 21 sin(1/x) forx

= 0, f(0) = 0.

However fcan not be continuous at z. Observe that

max(|f(z) f(xn)|,|f(z) f(yn|) 1

2 (|f(z) f(xn)| + |f(z) f(yn|) 12 .Thus we can find zn zwith|f(z) f(Zn)| 1/2.

(ii) The behaviour of f(zn(j)) does not control the value of f(z).Proposed result is false, see K43.


56/405

56

K46*

No comments.


57/405

57

K47

Since g is constant, the conclusions of intermediate value theoremhold trivially forg .


58/405

58

K48

Observe that

cos n= (cos + i sin )n

= 02rn

n2r(1)

r cosn2r sin2r

=

02rn

n

2r

(1)r cosn2r(1 cos2 )r

a real polynomial in cos of degree at most n. (Part (ii) shows thatthe degree is exactlyn.)

T0(t) = 1,T1(t) =t,T2(t) = 2t2 1, T3(t) = 4t3 3t.

(a) Observe that

cos n cos = (cos(n+ 1)+ cos(n 1))/2.Thus tTn(t) = (Tn+1(t) + Tn1(t))/2 for all t [1, 1].

(b) But a polynomial of degree at most n +1 which vanishes atn + 2points is identically zero, so tTn(t) = (Tn+1(t) + Tn1(t))/2 for all t.

(c) Induction using (i).

(d)|Tn(cos )| = | cos n| 1.(e)Tn vanishes at cos((r+ 1/2)/n) with 0 r n 1.For the last part make the change of variables x= cos with range

0 .Ifm= 0 replace /2 by .


59/405

59

K49

Two polynomials with the same roots and the same leading coeffi-cient are equal.

Use the fact that|Tn(t)| 1 for T [1, 1].Choose and so that+ =1, + = 1. If P is the

interpolating polynomial of degree n for f on [a, b], then taking g andQsuch that

g(t) =f(t + ), Q(t) =P(t + ),

we see that Q is the interpolating polynomial of degree n for g on[1, 1]. If|fn(x)| Aon [a, b] then|gn(x)| An =A((b a)/2)n on[1, 1] so

|f(t) P(t)| (b a)nA

22n1

on [a, b].


60/405

60

K50

(ii)P(t) =n

j=0

fj(a)

j! tj .

(vii) Observe that

f(t)

g(t) =

f(n+1)(t)

g(n+1)(t)

witht, t aas t aand use continuity.


61/405

61

K51

If we set

P(t) =n

j=0

Ajtj(1 t)n+1 +

nj=0

Bjtn+1(1 t)j,

then the resulting system of equations for Aj is triangular with non-vanishing diagonal entries and thus soluble. The same holds forBj.

Ify (0, 1), set

F(x) =f(x) P(x) E(y) xn+1(1 x)n+1

yn+1(1 y)n+1 .Fhas vanishingr-th derivative at 0 and 1 for 0 r nand vanishes aty. By Rolles theorem F vanishes at least twice in (0, 1), F vanishesat least three times in (0, 1), F vanishes at least four times in (0, 1),. . . , F(n+1) vanishes at least n+ 2 times in (0, 1), F(n+2) vanishes atleast n+ 1 times in (0, 1), F(n+3) vanishes at least n times in (0, 1),. . . ,F2n+2 vanishes at least once (atsay) and this gives the result.


62/405

62

K52

(i)g(b) g(a) =g (c)(b a) = 0 for some c (a, b).(ii)A = (f(b) f(a))/(g(b) g(a)).(iv) Observe that

f(t) f(a)g(t) g(a) =

f(ct)g(ct)

withct aas t a.


63/405

63

K53

(i) We have

f(b) =f(a) + f(a)h + f(c)h2

2!

for some c (a, b).(ii) Thus

f(a)h= f(b) f(a) f(c) h2

2!and so

|f(a)||h| 2M0+ 21M2|h|2for allaand all h. Thus

|f(t)| 2M0h

+M2

2h

for all h > 0 and, choosing h = 2(M0M1)1/2

, we have the requiredresult.

(iii) Take f(t) =t to see that (a) and (b) are false.

(c) is true by the method of (ii), provided thatL 2(M0M1)1/2.(d) True. g(t) = sin t.

(e) True. Scaling (d), G(t) =M0sin(M1/20 M

1/22 t) will do.

(f) Fails scaling. IfG as in (e) then

G(0)

(M0M2)1/4 = (M0M1)

1/4

can be made as large as we wish.


65/405

65

K55

(i) True. By the intermediate value theorem, we see that f(x)lfor some l as|x| . Given > 0, we can find a K such that|f(x)l| < /2 for all|x| K. Since a continuous function ona closed bounded set is uniformly continuous, we can find a with

1 > > 0 such that|f(x) f(y)| < for all x, y [K 2, K+ 2]with|x y| < . By going through cases, we see that|f(x) f(y)| < for allx, y Rwith|x y| < .

(ii) False. Takef(t) = sin t.

(iii) False. Take f(t) = sin t2.

(iv) False. Take f(t) =t.

(v) False. Take f(z) = exp(i|z|).(vi) True. Observe that||f(z)| |f(w)| | |f(z) f(w)|.(vii) True. Let > 0. Choose > 0 such that|x y| < implies||f(x)| |f(y)|| < /2.Now suppose y > x and|x y| < . Iff(x) andf(y) have the same

sign, then automatically, |f(x)f(y)| < /2. If they have opposite sign,then by the intermediate value theorem, we can find u withy > u > xand f(u) = 0. Since|x u|,|y u| < we have

|f(y) f(x)| |f(y) f(u)| + |f(u) f(x)| < /2 + /2 =.(vii) False. Takef(x) = g(x) = x. If > 0 then, taking y = 81

andx = y + /2, we have|x y| < but|x2 y2| = (x + y)(x y)>1so x x

2

is not uniformly continuous.(viii) True. Let > 0. We can find a >0 such that|u w| <

implies|f(u) f(w)| < and an >0 such that|x y| < implies|g(x) g(y)| < . Thus|x y| < implies|g(f(x)) g(f(y))| < .


66/405

66

K56

Observe that

(x + u) y x y + uso that

f(x + u) f(x) + u.Thus

f(a) f(b) a band f is uniformly continuous.

If E is closed, then we can find yn E such thatxyn f(x) + 1/n. Theynbelong to the closed bounded setE B(x, 2) so wecan find n(j) and a y withyn(j) y 0. SinceE is closed,y E. We observe that

f(x) + 1/n(j)

x

yn

x y y yn x y f(x)

as j . Thusx y =f(x).Conversely, if the condition holds, suppose yn Eand ynx 0.

We have f(x) = 0 so there exists a yE withx y = f(x) = 0.We have x= y E soEis closed.


67/405

67

K57

(i) Given > 0 we can find a > 0 such that, if u, v Q, then|uv| < implies|f(u)f(v)| < . We can now find an N suchthat|xn x| < /2 for n N. Ifn, m N then|xn xm| < so

|f(xn)

f(xm)

|< . Thus the sequencef(xn) is Cauchy and converges.

(ii) Using the notation of (i), we can find an M such that|xmx| < /2 and|ym x| < /2 for m M. Thus|xm ym| < and|f(xm)f(ym)| < for m M. Since was arbitrary, f(xm) andf(ym) tend to the same limit.

(iii) Observe that (ultimately by the Axiom of Archimedes) anyxR is the limit ofxn Q.

(iv) Take xn= x.

(v) We use the notation of (i). If|xy| < /3 we can findxn, yn Qwith xn

x, yn

y and

|xn

x

| < /3,

|yn

y

| < /3. Then

|xn yn| < for allnso|f(xn) f(yn)| < and|F(x) F(y)| .


68/405

68

K58

(i) If|z|< min(R, S) thenNn=0 anzn andNn=0 bnzn tends to limitand so

N

n=0

(an+ bn)zn =

N

n=0

anzn +

N

n=0

bnzn

tends to limit as N .Suppose R < S If R T R > 0. Let an = Rn +Tnand bn= Rn.

(iv) The radius of convergence is R unless = 0 in which case it is

infinite.

(v) If|z|> 1 then, provided n is sufficiently large,|z|n2 (2R)n son=0 anz

n diverges. If|z| < 1 then, provided n is sufficiently large,|z|n2 (R/2)n son=0 anzn converges. The radius of convergence is1.

IfR= 0, then any radius of convergencewith 0 1 is possible.For = 0 take an = n

nn. For 0< < 1 take an = n2. For = 1

take an= n2n where n tends to 1 very slowly from below.

Similar ideas forR =

.

(vi) Let|cn| = max(|an|, |bn|). Since

nj=0 |cnzn|

n

j=0 |anzn|radius of convergence R at mostR. Thus R min(R, S). But|cn| |an| + |bn| so R min(R, S). Thus R = min(R, S).

If|cn|= min(|an|, |bn|) then similar arguments show that the radiusof convergence R max(R, S). But every value A max(R, S) ispossible for R. (IfA 0 can take eg. a2n = R2n,b2n= A

2n,a2n+1= A2n1,b2n = S2n1.


69/405

69

K59

Let >0. We can find an N such that|h(n) | < for n N.Observe that ifn N thenh(n)< + and so

an> 2(+)+(1(+)n = 212(+)n.Thus if|z| >22(+)1 we have|anzn| . Thus the radius of con-vergence R satisfies R 22(+)1. Since was arbitrary R 221.A similar argument shows that, if|z|< 22()1 |anzn| 0, and thusR 221 so R = 221.

Suppose 1 0. DefineE inductively by taking an+1 = an/2 ifan/2 212, an+1 = 2an otherwise. Thenh(n) and|anzn| 1whenever|z| = 221 and n is large. Thus we have divergence every-where on the circle of convergence.

Suppose 1 0. Define E inductively by taking an+1 = 2an if2an 2

(1

2)n

n2

, an+1 = an/2 otherwise. Then|anzn

| n2

when-ever|z|= 221 and n is large. Thus we have convergence everywhereon the circle of convergence. When n is large we have

an 4 2(12)nn2so

2n2h(n)n 22+(12)nn2whence

(log 2)(n 2h(n)n) (log 2)(2 + (1 2)n 2log nand dividing byn log 2 and allowingn

we see that

1 2limsupn

h(n) = 1 2so limsupn h(n) =. A similar argument (but there are alternativeways of proceeding) shows that lim infn h(n) = soh(n) .


71/405

71

K61

Write = limsupn an+1/an. If > 0, we can find an N withan+1/an + for n N. Thus, by induction,an A(+)n forn Nwhere A = aN(+ )N. Thus, ifn N,

a1/nn

A1/n(+ )

+

as n . Since was arbitrarylim sup

na1/nn limsup

nan+1/an.

Remaining inequalities similar or simpler.

All 8 possibilities can arise. Here are d.

(1) Set an= 1.

lim infn

an+1an

= liminfn

a1/nn = lim supn

a1/nn = lim supn

an+1an

= 1.

(2) Set a1 = 1, a2n+1/a2n = 2, ar+1/ar = 1 otherwise.

lim infn

an+1an

= lim infn

a1/nn = lim supn

a1/nn = 1< 2 = lim supn

an+1an

.

(3) Seta1 = 1,a2n+1/a2n = 2,a2n+2/a2n+1= 21 forn 2ar+1/ar =

1 otherwise.

lim infn

an+1an

= 1/2


72/405

72

K62

(iii) Could write bn = cn+ dn with dn = 0 for n large and|bn| .Thus lim supn |Cn| .

(v) Ifbj = (1)j limit is 0.(viii) PickN(j),Mj,sj andrj inductively so thatN(0) = 0,M(0) =

1, s0 = 1/2. we pick rj+1 so that 0< sj < rj+1 M(j) such that rN(j+1)j+1 1 2j1. PickM(j + 1)> N(j + 1)so that s

M(j+1)j+1 < 2

j1 Set bn = 1 i f N(j) n M(j), bn = 0otherwise.

Ar(j+1)M(j)

n=0

(1 rj+1)rnj+1+

n=N(j+1)

(1 rj+1)rnj+1

= (1 rM(j)+1j+1 ) + rN(j)j+1 1 2j1 2j1 = 1 2j.ThusAr does not converge.


73/405

73

K63

(ii)C can be identified with R2.

(iii) (a) Ifz= 1,Nn=0 zn = (1 zN+1)/(1 z) tends to a limit ifand only if|z|


74/405

74

K64

(v) Ifk then, using part (i),Bk b. Since this is true forevery such sequence,B bas .

(vi) We seek to Borel sum zj. Observe that, ifz= 1

n=0

nn!

e 1 zn+11 z =

11 z(1 ze

(1z)).

We have convergence if and only if(1 z)< 1 and the Borel sum isthen 1/(1 z).

Ifz= 1,

B =

n=0

(n + 1)n

n!e =

n=0

n

n!e +

n=0

n

n!e = 1 .


75/405

75

K65

(i) True. Since the sequence is bounded, there exists aband a strictlyincreasing sequence n(j) such that bn(j) b. Set ujn(j) = 1, ujk = 0otherwise.

(ii) True. We can extract subsequences with different limits and use(i).

(iii) False. Let bj = 1 for 2n n j 2n 1, bj =1 for

2n + 1j2n + n[n4], bj = 0 otherwise Observe that, ifunk = 1for 2n N k2n, unk =1 for 2n + 1k2n +N, then U Gbut

k=0 ujk bk = 2N forj sufficiently large. But Nis arbitrary.

(iv) True. DefineN(r) and j(r) inductively as follows. SetN(0) =j(0) = 1. Choose N(r+ 1) > N(r) such that

k=N(r+1) |uj(r)k|< 2r

and j(r+ 1) > j(r) such that

N(r+1)k=0 |uj(r+1)k| < 2r1. Set bk =

(

1)r+1 forN(r)

k < N(r+ 1). Then

(1)r+1

k=0

uj(r)kbk

= (1)r+1N(r)

k=0

uj(r)kbk+

N(r+1)k=N(r)+1

uj(r)kbk+

k=N(r+1)+1

uj(r)kbk

k=0

uj(r)k 2N(r)k=0

|uj(r)k| 2

k=N(r+1)+1

|uj(r)k|

k=0 uj(r)k 2

r+3

1.as r .

(v) False. Try bj = 0.


76/405

76

K66

(i) Trybk = 1 for k = N,bk = 0 otherwise. Try bk = 1 for all k .


77/405

77

K67

(i) implies (ii) since absolute convergence implies convergence.

For the converse observe that, writing xn = (xn1, xn2, . . . , xnm), wehave

xj |xj1| + |xj2| + + |xjm|.Thus, if

j=1 xj diverges, we must be able to find a kwith 1 k m

such that

j=1 |xjk | diverges. Choose j so that jxjk 0. ThenN

j=1

jxj

Nj=1

j xjk

=N

j=1

|xjk | .

as N .


78/405

78

K68

If

n=1 an converges, then

nS1,nNan

nNan

n=1an

and, since an increasing sequence bounded above converges,

nS1,nNantends to a limit as N .

If

nSj ,nNan tends to a limit so doesnN

an=

nS2,nNan+

nS2,nN

an.

Second paragraph does not depend on positivity aj so if remainstrue. However, takinga2n1= a2n= 1/n,S1 evens andS2 odds thenwe see only if false.

Let S1 = {n : a1/2n /n an} and S2 = {n : a1/2n / n > an}.nS1 a

1/2n /nconverges by comparison with

n=1 an. Ifn S2 then

an= (a1/2n )

2


79/405

79

K69

(i) Diverges

2N

n=11 + (1)n

n =

N

n=11

n

as N .(ii) Converges by comparison since p2n n2.(iii) Diverges by comparison. We have n1/n 1 (e.g. by taking

logarithms) so n1/n 1/2 for n large and so n11/n >21n1 for nlarge.

(iv) If

n=1 anconverges, then taking N(j) =j we have thatN(j)

n=1 antends to a limit.

IfN(j)n=1 an tends to a limit A, then for any Mwe can find a j with

N(j)> Mso thatM

n=1

anN(j)n=1

an A

so, since an increasing sequence bounded above converges, we are done.

(v) If

1 f(t) dt converges then

Nn=1

anN

n=1

n+1n

f(t) dt=

N+11

f(t) dt

0

f(t) dt

since an increasing sequence bounded above convergesn=1

an

con-verges.

For the converse observe that if

n=1 an converges then, ifX N, X1

f(t) dt N

1

f(t) dt=N1n=1

n+1n

f(t) dt KN1n=1

an

n=1

an.

Now use something like Lemma 9.4.

(vi)an= (1)n.(vii) Let

MNn=1an tend to b. Given > 0 we can find an N0 such

thatMNn=1an b /2 for N N0. We can also find an N1 suchthatan /2M for n N1. Let N2 = (M+ 1)(N1 +N0). If m N2 we can write m = N M+ uwithN N0, 0 u M 1 andMN+ r N1 for allr 0. We then have

m

n=1

an b MNn=1

an b +MN+u

n=MN+1

an < 2

+ u

2M


80/405

80

as desired.

(viii) Set N(j) = 22j, an =j1 if 22j n < 2 22j, an = j1 if2 22j n


81/405

81

K70

(i) Observe that (anbn)1/2 (an+ bn)/2 and use comparison.

(ii) Take bn= an+1 in (i).

(iii) Observe thatan+1

(anan+1)

1/2 and use comparison.

(iv)a2n= 1,a2n+1= n4.


82/405

82

K71

(i) True. a4n 0 so there exists anNwith|an| 1 forn N. Thusa4n |a5N|for n N and the result follows by comparison and the factthat absolute converge implies convergence.

(ii) False. Takean= (1)n1/6

and use the alternating series test.(iii) False. Take a2k = 2

k,ar = 0 otherwise.

(iv) False. Same counterexample as (iii).

(v) True. Given > 0 we can find an N such thatM

j=Naj < /2forM N. Thus, ifn 2N+ 2

nan 2(n N 1)an 2M

j=N

aj < .

(vi) False. Take an = (n log n)1 for n

3 and use the integral

comparison test.

(vii) True. Abels test.

(viii) True. By CauchySchwarz, Nn=1

|an|n3/42

N

n=1

|an|2N

n=1

n3/2

n=1

|an|2

n=1

n3/2.

(ix) False. Take an = (1)n(log n)1 for n 3. Use alternatingseries test and integral test.

(x) True. We must have an 0 so there exists a K such that|an| K. Since|n5/4an| K n5/4 the comparison test tells us thatn=1 n

5/4an is absolutely convergent and so convergent.


83/405

83

K72

akn 0 so there exists anNwithan 1 forn N. Thusakn |ak+1n |forn Nand the result follows by comparison.

Setf(n) = log(n + 1)nand observe that, ifk 23N

j=1

akjN

j=1

(2k 2)(log(n + 1))k) .

(Use comparison and the fact that n (log n)k fornsufficiently large.)


84/405

84

K73

(i) Write

Sn=n

j=11j bj .

Then, by partial summation (or use Exercise 5.17),n

j=1

bj =n

j=1

j (Sj Sj1)

=n1j=1

(j j+1)Sj+ nSn

so that

n

j=1

bj

n1

j=1

(j+1 j)Sj + nSn

n1j=1

(j+1 j)K+ nK= (2n 1)K 2kn

whence the result.

(ii) If >0 then we can find an Nsuch thatnj=N1j bj forn N. Thus, by (i),

1nn

j=1

bj 1nN1j=1

bj + 1nn

j=N

bj

1nN1j=1

bj2 2

as n . Kroneckers lemma follows.(iii) ((a) fails) Letbj =j

4,j =j6 forj= 2r,b2r = 1 and2r = 22r.

((b) fails) Letbj =j2, j = 1.

((c) fails) Letbj =j4, j =j

2.


85/405

85

K74

Consider the setAn of integers between 10n and 10n+1 1 inclusive

whos decimal expansion does not contain the integer 9. We observethat{0}

n

j=0 Aj contains (9/10)n+1 10n+1 = 9n+1 elements. Thus

An contains at most 9n+1 elements and

jAnj1 10n 9n+1.

The partial sums of the given series thus can not exceed

n=010n

9n+1 = 90 and the series converges.


86/405

86

K75

(i) Observe that 1/x 1/(n+ 1) forn x n+ 1 and integrate.

Tn

Tn+1= 1

n + 1 n+1

n

1

xdx.

Observe thatT1= 1 and

Tnn1r=1

1

r

r+1r

1

xdx

.

Decreasing sequence bounded below tends to a limit.

(v) Given l we can choose integers pn, qn1 such that pn+1 >2pn,qn+1> 2qn,pn/qnis strictly decreasing and log 2+(1/2) log(pn/qn) l.At the n-th stage, if the ratio of positive terms to negative exceeds

pn/qn usepn positive terms followed by qn+ 1 terms if the ratio is lessthan pn/qn, use pn+ 1 positive terms followed by qn negative terms,otherwise usepn positive terms followed by qn negative terms. Switchfrom then-th stage to then + 1-th stage when the size of each unusedterms is less than (pn+1 + qn+1 + 1)

12n and the ratio of positive termsto negative differs from pn/qn by at most 2

n.


87/405

87

K76

(ii) Observe that, if

j=0 bj converges, we can find an N() such

that|Nj=M) bj | for N M N() so

j=M bjx

j xM forM N().

(iii) IfM N(), then

j=0

bjxj

j=0

bj

M1j=0

|bjxj bj| +

j=M

bj

+

j=M

bjxj

M1j=0

|bjxj bj| + 2 2

as x 1. Since is arbitrary the required result follows.(iv) Observe that, if 0 < x < 1, then

j=0 ajx

j,

j=0 bj xj and

j=0 cjxj are absolutely convergent so (using Exercise 5.38) we may

multiply them to get

j=0

ajxj

j=0

bjxj =

j=0

cj xj.

Now letx 1.


88/405

88

K77

(a) Since

n=0

m=0 cn,mx

nym converges for|x|=|y|= , we have|cn,m|n+m =|cn,mxnym| 0 for n, m so there exists a Kwith|cn,m|n+m < Kfor all n, m 0. Set = 1.

(b) (i)

nN, mN|xn

y

m

| = nN|x|n mN|y|m.Eis the square (1, 1) (1, 1).(ii)

n+mN

n+m

n

|xnym| = Nr=0(|x| + |y|)rEis the square{(x, y) :|x| + |y|


89/405

89

K78

(i) Choose Nj() so that|ajaj,n| < /M for n Nj() and setN(, M) = max1jMNj().

Observe that

j=1

ajM

j=1

ajM

j=1

aj,n

for allM.

(ii) If

j=1 aj converges with value A, then given >0 we can find

anMsuch thatM

j=1 aj > A. By (i) we can find N(, M) such that

j=1

aj,nM

j=1

aj

and so

j=1

aj

j=1

aj,n A 2

for alln N(, M). Thusj=1 aj,n A.If

j=1 aj fails to converge, then given any K > 0, we can find

an M such thatM

j=1 aj > 2K. But we can find an N such thatMj=1 aj,n

Mj=1 aj K fornNand so

j=1 aj,nK fornN.

Thus

j=1 aj,n can not converge.

(iii) Yes we can drop the condition. Considerbj,n= aj,n

aj,1instead.


90/405

90

K79*

No comments


91/405

91

K80

It is possible to do this by fiddling with logarithms but I prefer tocopy the proof of the radius of convergence. Show that if

n=0 bne

nz

converges then

n=0 bne

nw converges absolutely whenw < z.

Next observe that the sum

n=0 bne

nz

converges everywhere (setX =) or nowhere (set X =) or we can find z1 and z2 suchthat

n=0 bne

nz1 converges and

n=0 bnenz2 diverges. Explain why

this means that

X= sup{z :

n=0

bnenz converges}

exists and has the desired properties.

For X = can take bn = 0, for X = can take bn = en2 , forX=R withR finite can take bn= e

Rn.

n=02

n

e

nz

/(n+ 1)

2

converges when z = log 2. Also 2n

e

nx

/(n+1)2 when x is real andx > log2 so X= log 2.By the first paragraph there exists a Y 0 such thatn=0 cneinz

converges for z < Yand diverges forz > Y. By taking complex con-jugates we see that

n=0 cne

inz converges forz > Y and divergesforz < Y.

Since

n=0 sn+ tn diverges if exactly one of

n=0 sn and

n=0 tnconverges and converges if both converge,

n=0 cn(e

inz +einz) andthus

n=0 cncos nzconverges ifz < Y and diverges ifz > Y.


92/405

92

K81

(i) By integration by parts,

In+1= [ sinn x cos x]/20 + n /2

0

sinn1 x cos2 x dx

=n

/2

0

sinn1 x(1 sin2 x) dx=nIn1 nIn+1.

(ii) Observe that sinn1 xsinn xsinn+1 x forx[0, /2]. Inte-grating givesIn+1 In In1 so

n

n + 1=

In+1In1

InIn1

1.so In/In1 1 as n .

(iii)I0 = /2, I1 = 1

I2n=(2n 1)(2n 3) . . . 1

2n(2n 2) . . . 2

2, andI2n+1=

2n(2n 2) . . . 2(2n+ 1)(2n 1) . . . 1 .

Thusn

k=1

4k2

4k2 12

=

I2n+1I2n

1

and the Wallis formula follows.


93/405

93

K82

(i) Power series or set f(x) = exp xx and observe f(0) = 0,f(x) 0 for all x 0.

(iii) Observe that nj=1

(1 + aj) mj=1

(1 + aj ) =

nj=1

(1 + aj) m

j=n+1

(1 + aj) 1

n

j=1

(1 + |aj |)

mj=n+1

(1 + |aj |) 1

.

Now use the inequality (1 + |aj |) exp |aj |.(iv) Use the general principle of convergence and (iv).

(v) We must have an 0. If|an|


94/405

94

K83

(i) Left to reader.

(ii) Letan= n,an,N=n if all the prime factors ofN lie among

the firstN primes,an,N= 0 otherwise. Observe that

0 an,N an,N+1 an,thatan,N an asN and that

n=1 an converges.

(iii) If pP, pN

1

1 p1 K

for allN then pP, pN

1

1 p K

for allNand all


95/405

95

K84

Suppose that Pn >1 for all nN. Then Pm = PNm

j=N(1 an).If

j=1 an diverges then (see K82)Pm 0 which is impossible by thedefinition ofN. Thus

j=1 an converges and Pm tends to a limit. Asimilar argument applies ifPn

1 for all n

N.

Thus we need only consider the casePn 1 changes sign (or is zero)infinitely often. But an 0 and if (A) PN1 1 0 PN 1 orPN1 1 0 PN 1 and (B) 0 an for n N 1 then wehave 1 + Pn 1 for n N. Thus Pn 1.

If

j=1 an diverges we can say nothing about l. (If l 1 considera1 = l 1, an = 0 for n 2. If l 1 consider a1 = 1, a2 = 1 l/2,an= 0 forn 3.)


96/405

96

K85

Letan = zn,an,N=z

n if 0 n 2N+1 1 andan,N= 0, otherwise.Observe that 0 |an,N |an|, and that an,N an as N . Bydominated convergence (Lemma 5.25)

Nj=1

(1 + z2j) = n=0

an,N n=0

an= (1 z)1.


97/405

97

K86

Observe that

2cot2=cos2 sin2

cos sin = cot + tan

and use induction.Observe that

1

2ncot

2n =

1

2nsin2n

cos 2n1

as n .


98/405

98

K87

Very similar to K86. To obtain the formula for un, observe that2cos2 = 1 + cos 2and cos(/4) = 21/2.


99/405

99

K88

The square of a rational is rational.

(ii) Observe, by induction or Leibnizs rule, that f(k) is the sum ofpowers ofx with integral coefficients plus terms of the form

M xr

(1 x)s

(min(r, s))!

with Man integer and r, s 1. Thusf(k)(0) and f(k)(1) are alwaysintegers. Since bn2nr is an integer,G(0) + G(1) is always an integer.

0< anf(x)sin x an

n! 0

as n .


100/405

100

K89

(a)(i) Given cCwe can find bB with g(b) =c and aA withf(a) =b.

(ii)g f(a) =g f(a) implies f(a) =f(a) and so a = a.(iii)f(a) =f(a) impliesg f(a) =g f(a) and so a = a.(iv) Givenc C, we can finda Awithg f(a) =b. Observe that

f(a) B andgf(a) =c(b) Let A ={a}, B ={a, b}, C ={a} with a= b f(a) = a,

g(a) = g(b) =a.


101/405

101

K90

Iffsatisfies the conditions of the first sentence of the second para-graph, we can write f = g J where J(x) =x and g satisfies theconditions of the first paragraph. Thus

f(x) =J(x)g(J(x)) =

g(J(x))< 0

for allx E.Ifad bc= 0 thenfis constant and f= 0.


102/405

102

K91

(i) Suppose that f is continuous. Iff is not strictly monotonic wecan findx1< x2 < x2 so thatf(x1),f(x2) andf(x3) are not a strictlyincreasing or a strictly decreasing sequence. Without loss of generality(or by consideringf(1

x) and 1

f(x) if necessary), we may suppose

f(x3)f(x1)f(x2). If either of the inequalities are equalities thenf is not injective. Thus we may suppose f(x3)> f(x1)> f(x2). Butby the intermediate value theorem we can find c (x2, x3) so thatf(c) =f(x1) and f is not injective.

Supposefis strictly monotonic. Without loss of generality we sup-pose f increasing. We shall show that f is continuous at t (0, 1)(the cases t = 0 and t = 1 are handled similarly). Observe thatf(1) > f(t) > f(0). Let > 0. Set = min(1f(t), f(t), )/2.We can find s1 and s2 with f(s1) = f(t) f(s2) = f(t) +. If/ta = min(ts1, t+ s2), then|st| > implies s2 > s > s1 sof(s2)> f(s)> f(s1) and|f(s) f(t)| < .

(ii) The same proof as in (i) shows that, ifg is continuous and in-jective, it must be strictly monotonic. The example ofg(t) = t/2 fort 1/2, g(t) =t/2 + 1/2 shows that the converse is false.

(iii) The same proof as in (i) shows that, ifg is strictly monotonicand surjective, it must be continuous. The example g(t) = sin t showsthat the converse is false.


103/405

103

K92

Write f(x) = (log x)/x. By looking at f we see that f is strictlyincreasing fromto e1 on the interval (0, e], has a maximum valueof e1 at e and is strictly decreasing from e1 to 0 on the interval(0,

). The function fhas a unique zero at 1,

We have xy =yx if and only iff(x) =f(y). The set

{(x, y) : xy =yx, x, y >0}is thus the union of the straight line{(x, x) : x >0}and a curve withreflection symmetry in that line, having asymptotes x = 1 and y = 1and passing through (e, e).

Iff(x) =f(y) and x < y, then xe so we need only examine thecases m2 to see that the integer solutions ofnm = mn are n= m,(m, n) = (2, 4) and (m, n) = (4, 2).

Sincef()< f(e) we have e > e.


104/405

104

K93

Observe that

xy= 14

(x + y)2 (x y)2.


105/405

105

K94

(i) Choose with > >1. If we set f(x) =xx, then (recallingthatx = exp(x log ))

f(x) =x1x + x(log )x =x1x(+ x)>0

forx sufficiently large (x x0, say). Thusxx =f(x)(/)x f(x0)(/)x .

(ii) Set

f4(x) = exp

1

x

, f3(x) =

1

x

1/2, f2(x) = exp

log

1

x

1/2

and

f1(x) =

log

1

x

3.

Sety = 1/xso y as x 0+.Observe that log f4(x)/ log f3(x) = 2y/(log y) asx .Observe that log f3(x)/ log f2(x) = (log y)

1/2/2 as x .Observe that log f2(x)/ log f1(x) = (log y)

1/2/(3 log log y) asx .


106/405

106

K95

(ii) IfQhas a root= 0 of ordern 1 then differentiating the givenequation n 1 times and setting z= we getP()Q(n)() = 0 soP() = 0 and P and Q have a common factor. Thus Q(z) =AzN forsome N

1, A

= 0 and it is easy to check that this too is impossible.

(iv) If the degree ofPis no smaller than that ofQ then

P(x)

Q(x)log x .

If the degree ofPis smaller than that ofQthen

P(x)

Q(x)log x 0.

If the degree ofPis no smaller than that ofQ then

P(x)

Q(x)log x .


107/405

107

K96

Setf(x) =xlog(1+x). Thenf(0) = 0 andf(x) = 1(1+x)1 0for 0 < x and f(x)0 for x > 0. so f(x)0 andxlog(1 x)for 0 x 1.

Setg(x) = log(1 + x) x+ x2

. Then g(0) = 0 andg(x) = (1 + x)1 1 + 2x= (x + 2x2)(1 + x)1 0

forx >0 andg (x) 0 for1/2 x 0 sog(x) 0 and log(1+ x) x x2 forx 1/2.

Since

logkn

r=1

1 r

n2

=

knr=1

log

1 rn2

we have

kn

r=1

r

n2 logkn

r=1

1

r

n2

knr=1

r

n2kn

r=1

r2

n4 .

Butkn

r=1

r

n2 =

kn(kn + 1)

2n2 =

k

2

k+

1

n

k

2

2

andkn

r=1

r2

n4 (kn)

3

n4 =

k

n 0

as n

so

logkn

r=1

1 r

n2

k

2

2

and the result follows on applying exp.


108/405

108

K97

Suppose thatx 1. Thenxn 1 and2n(xn 1) 2n+1(xn+1 1) = 2n(x2n+1 2xn+1+ 1)

= 2n(xn+1

1)2

0.

Thus the sequence 2n(xn 1) is decreasing bounded below by 0 and sotends to a limit. A similar argument applies if 1 > x >0.

Since

2n(xn 1)2n(1 1/xn)=xn 1

2n(xn 1) and 2n(1 1/xn) tend to the same limit.(i) Ifx = 1, thenxn= 1 so log 1 = 0.

(ii) Ifxn = 1 + n, y = 1 + n then(1 + n)(1 + n)

1 + n+ n

2n=

1 +

nn1 + n+ n

2n.

Since 2nn tends to a limit we can find an A such that|n A2n.Similarly we can find a B such that|n B2n. Thus we can find aCsuch that nn1 + n+ n

C22n.It follows that

(1 + n)(1 + n)1 + n+ n2n 1

2nr=1

2

n

r

Cr22nr

2n

r=1

2nrCr22nr

=2n

r=1

2nrCr2nr

C2n

1 C2n 0

as n .By the mean value theorem

(b2n a2n) = 2nc2n1(b a)

for some c (a, b) so2n((1 + n)(1 + n) (1 + n+ n) 0


109/405

109

that is to say

2n((xy)2n (xn+ yn 1)) 0

so writingz= xy

2n(zn 1) 2n(xn 1) 2n(yn 1) 0

andlog xy= log x + log y.

Result (ii) shows that we need only prove (iii), (iv), (v) and (vi) atthe point x = 1. To do this observe that a simple version of Taylorstheorem

|f(1 + ) f(1) f(1)| sup||||

|f(1 + )||2|

applied tof(x) =x2n

yields

|(1 + )2n 1 2n| 2n 4 ||2

for all|| 101 and n 3. (We can do better but we do not needto.) Thus if|| 101 and we setx = 1 + we obtain

|2n(xn 1) | 4||2forn 3 so that

| log(1 + ) | 4||2.Thus log is continuous and differentiable at 1 with derivative 1.

Since log(x+t) = log x+log(1+t/x) the chain rule gives log x= 1/xand since log x >0 we see that log is strictly increasing.

(vii) Since log 2 > log1 = 0 we have log2n = n log2

solog x as x and log y = log y1 as y0+. Sincelog is continuous and strictly monotonic, part (vii) follows.


110/405

110

K98

If f(x + )

f(x)

1/ l

then, taking logarithms,log f(x + ) log f(x)

log l

so log fis differentiable and, by the chain rule, f= exp log fis differ-entiable atx. Since

log l= d

dxlog f(x) =

f(x)f(x)

we have

l= expf(x)f(x)

.


111/405

111

K99

If

f(x) = x1

x a 1

x a

tends to a limit as x athen (x a)f(x) 0. Sincex a

x a a1

as x athis gives

1 = 0

so = .

If=thenf(x) =F(x)/G(x) with

F(x) =x ax1 x + a

G(x) =x+1

a

x ax

+ a+1

We find F(a) = F(a) = 0, G(a) = G(a) = 0 and use LHopitalsrule to give

f(x) F(a)

G(a)=

12a

.


112/405

112

K100

We only consider x when x is real and we define it as exp( log x)where log is the real logarithm.


113/405

113

K101

(iii) Setf(x) = log g(x) and apply (i) to getg(x) = xb for some realb. Easy to see that this is a solution.

(iv) Observe that, if x > 0, g(x) = g(x1/2)2 > 0. Thus, by (iii)

g(x) =b

x

for allx >0 and some b >0.Nowg(1)2 =g(1) = 1 so g(1) =1 and either g(x) =b|x| for all

x org(x) = sgn(x)b|x| for all x. (Recall sgn x= 1 ifx >0, sgn x=1ifx


114/405

114

K102

(i) We know that

(t/2)2 =t

but the equation (exp is)2 = exp iu, whereu R, has two real solutionsin s with < s (the solutions being s/2 and one ofs/2 ).As a specific example, if (t) = exp2it, then (3/4) =/2 but(3/8) = 3/4.

However since is continuous we can find a >0 such that|(t) 1| 1001 for|t| . If|t| then|(t)| /4 and|(t/2)| /4so (t/2) =(t)/ (since|(t/2) | > /4).

(iii) The same ideas show that the continuous homomorphisms areprecisely the maps n for some integern.


115/405

115

K103

(i)f(0) = 2f(0) sof(0) = 0.

(ii) Set F(2kp1j ) = 2kj whenever j and k are integers with j 1

and F(t) = 0 otherwise. (By the uniqueness of factorisation F, is well

defined.)Since F(p1j ) and p1j 0 as j , F is not continuous at

0.

(iii)

f(t) = 2nf(t2n) =tf(t2n) f(0)

t2n f(0)t

as n so f(t) =f(0)t for allt = 0 and so for all t.(iv) SetF(2kp1j ) = 2

kp1j wheneverj andk are integers with j 1and F(t) = 0 otherwise. Since|F(t) F(0)|=|f(t)| |t|, we have Fcontinuous at 0. Since

F(p1j ) F(0)p1j

= 1 1 but F(21/2j1) F(0)

21/2j1 0

as j ,Fis not differentiable at 0.(v) Observe that u(t) = u(t/2)2 0 for all t. If u(x) 0 then

u(2nx) =u(x)2n 1 so by continuity u(0) = 1. Thus by continuity

there exists a > 0 such that u(t) > 0 for|t| < . Thus u(t) =u(2nt)2

n> 0 for|t| < 2n and so u is everywhere strictly positive.

Now setf(t) = log u(t) and apply part (iii).


116/405

116

K104

Consider the case x =0. Observe that1

(1 + )1/2 = 1

2+ ()||

with() 0 as 0. (Use differentiation or Taylor series.) Observealso that|x h| xh. Thusf(x + h) =

x + h

(x2 + 2x h + h2)1/2

= x + h

x1 + (2x h + h2)/x21/2=

x + h

x

1 2x h + h2

2x2

+ 1(h)h

=f(x)

h

x

(x h)x

x

3 + 2(h)

h

wherej (h) 0 ash 0. Thus f is differentiable at x and

Df(x)h= h

x(x h)x

x3 .

(Df(x)h) x= h xx x hx2x3 = 0.

Since

f(x)

f(0)

= 1 0

asx 0, fis not even continuous at 0Observe that fis constant along radii.


117/405

117

K105

Without loss of generality take z0= 0.

(i)(ii) If f is complex differentiable at 0, then writing f(0) =A + Bi withA and B real, and taking h and k real

f(h + ik) =f(0) + (A + Bi)(h + ik) + (h + ik)|h + ik|with(h + ik) 0 as|h + ik| 0. Taking real and imaginary parts,

u(h, k)v(h, k)

=

u(0.0)v(0, 0)

+

Ah BkAk+ Bh

+

1(h, k)2(h, k)

(h[2 + k2)1/2)

with

1(h, k)2(h, k)

0as (h2 + k2)1/2 0. ThusFis differentiable at0 with Jacobian matrix

A B

B A

IfA = B = 0 take = 0 and = 0. Otherwise take = (A2 +B2)1/2

and such that cos = A, sin = B.

(ii)(iii)(iv) Just a matter of correct interpretation.(iv)(i) Carefully reverse the first proof.


118/405

118

K106

(i) Observe that

g(x + h) =f(x + h, c x h)=f(x, c x)f,1(x, c x)h + f,2(x, c x)(h) + (h, h)(h2 + h2)1/2=g(x) +

f,1(x, c x) f,2(x, c x) + 21/2(h, h)|h|

where (h, k) 0 as (h2 +k2)1/2 0. Thusg is differentiable withderivative

g(x) =f,1(x, c x) f,2(x, c x).(ii) Observe thatg= f u where

u(x) =

x

c x

.

Thus uhas Jacobian matrix 11

.

By chain the rule g is differentiable with 1 1 Jacobian matrixf,1(x, c x) f,2(x, x c)

11

= (f,1(x, c x) f,2(x, c x)).

Defineg as stated. Iff,1 = f,2, theng= 0 so, by the constant value

theorem,gis constant. Thusf(x, xc) =f(y, yc) for allx, y, c R.Writingh(x + y) =f(0, x+ y), we have the required result.


119/405

119

K107

(i) Differentiating with respect to and applying the chain rule, wehave

1f(x) =m

j=1

xjf,j (x).

Taking = 1 gives the required result.

(ii) We observe thatv is differentiable with

v() =m

j=1

xjf,j(x) = 1

mj=1

xjf,j(x) =1v().

Thusd

d(v()) = 0

and by the constant value theorem

v() =v(1)

and f(x) =f(x).


120/405

120

K108

This is really just a question for thinking about.

(i) We seek to maximise the (square root of)

f() =a b

c dcos

sin

2

= (a cos + b sin )

2

(c sin + b cos )

2

and this we can do, in principle, by examining the points with f() = 0.

(ii) We seek to maximise the (square root of)

pr=1

ms=1

arsxs

2

subject to the constraintm

s=1x2s = 1

which can, in principle, be handled using Lagrange multipliers.

However, this involves solving an m m set of linear equations in-volving a parameter leading to polynomial of degree m in . The mroots must then be found and each one inspected. This neither soundsnor is a very practical idea even when m is quite small.


121/405

121

K109

If e1, e2, . . . , em is the standard basis (or any orthonormal basis)and (aij) is the associated matrix of, then

aij =m

r=1

arj

er, ei

= n

r=1

arj er, ei= ej , ei = ej , ei = ei, ej =aji

Conversely ifaij =aji for alli, j, then essentially the same calculationshows that

ej , ei = ei, ejand so by linearity

nr=1

xrer

,

ms=1

yses

=

mr=1

ms=1

xryser, es

=m

r=1

ms=1

xryser, es

=

nr=1

xrer,

ms=1

yses

.

Let u1, u2, . . . , em be an orthonormal basis of eigenvectors witheigenvalues 1,2, . . . m with|1| |2| |m|. Then

m

j=1

xjuj

2 = mj=1

xjjuj

2 = mj=1

x2j 2

j 21m

j=1

x2j

with equality when x2 = x3 = = xm = 0 (and possibly for othervalues). This proves.

The matrixA given has eigenvalue 0 only. HoweverA = 1.


122/405

122

K110

(ii) Let x=n

j=1 xjej . Then

yk =n

j=1kj xj

(nr=1(kr xr)2)1/2 e1

unlessx1 = 0.

If the largest eigenvalue is negative, then (in general)xk |1|and(1)kyk u1 0 where u1= e1.

(iii) (c) The number of operations for each operation is bounded byAn2 for some A (A= 6 will certainly do),

Suppose|1| > |j | for j 2. Then the e1 component ofkxwill grow at a geometric rate j/relative to the other components.

(v) Look at (iii). If the two largest eigenvalues are very close the

same kind of thing will occur.(iv) Suppose e1 the eigenvector with largest eigenvalue known. Use

the iteration

x ((x x e1)or something similar. However, the accuracy to which we know e1 willlimit the accuracy to which we can find e2 and matters will get rapidlyworse if try to find the third largest eigenvalue and so on.

(vii) No problem in real symmetric case because eigenvectors orthog-onal.


123/405

123

K111

(ii)x, x = x, x= x2.Thus

x

2 =

x

x

| x, x

|=

x

2

2

x

2

for all xand so| 2.(iii) 2 and so either = 0 (in which case

the result is trivial) or = 0 and . But = so = . Thus = and2 = 2so = 2.

Since () = = we have symmetric. Ife = ewithe =0 then

e2 = e, e = e, e == e, e= e2.Thus

0.

(vi) Multiplication ofA withA takes of the order ofm3 operations(without tricks) and dominates.

(vii) We have AA=

1 11 5

.

A has eigenvalues 1 and 2. AA has eigenvalues 4 51/2. A =(4 + 51/2)1/2.


124/405

124

K112

Follow the proof of Rolles theorem. Ifc||


125/405

125

K113

Since (x, y) x y, (x, y) xy, x x1 and x f(x) are allcontinuous we see thatFwhich can be obtained by composition of suchfunctions is continuous except perhaps at points (x, x).

IfF is continuous at (x, x) then F(x+h, x) must tend to the limitF(x, x) as h 0 so f is differentiable and F(x, x) = f(x). SinceF(x + h, x+ h) F(x, x) as h 0,f must be continuous.

Conversely if f exists and is continuous then, setting F(x, x) =f(x), we saw above thatFis continuous at all points (x, y) withx =y.Ifh =k the mean value theorem gives

F(x + h, x + k) F(x, x) = f(x + h) f(x + k)x y f

(x)

=f

x + (h,kh + (1 h,k)k) f(x)

for some h,k with 0 < h,k < 1. But it is also true that, if we set

h,h= 1/2,F(x + h, x + h) F(x, x) =f(x + h) f(x)

=f

x+ (h,hh + (1 h,h)h) f(x).

Thus

F(x + h, x+ k) F(x, x) =fx+ (h,kh + (1 h,k)k) f(x) 0as (h2 + k2)1/2 0.

Suppose now that f is twice continuously differentiable. Much asbefore, the chain rule shows that Fis differentiable at all points (x, y)

withx=y. Iffis twice continuously differentiable then the local formof Taylors theorem shows us thatf(x + h) = f(x) + f(x)h +

f(x)2

h2 + (h)h2

with(h) 0 ash 0. Thus, ifh =k,F(x+h, x+k)

=(f(x) +f(x)h+ 12f

(x)h2 +(h)h2) (f(x) +f(x)k+ 12f(x)k2 +(k)k2

h k

=f(x) +f(x)

2 (h+k) +

(h)h2 +(k)k2

h k

=F(x, x) + f(x)2

(h+k) +(h, k)(h2 +k2)1/2

where(h, k)0 as (h2 + k2)1/2 0. The formula can be extendedto the case h = k by applying the appropriate Taylor theorem to f.Thus Fis differentiable everywhere.


126/405

126

K114

Ifk= 0, then we can find anX >0 such that|f(t) k| < |k|/4 andso|f(t)| >3|k|/4 fort X. Thus, using the mean value inequality,

f(x)

x =

f(x) F(X)

x X

x X

x

+F(X)

x

f(x) f(X)x X

x Xx

f(X)x

3|k|

4

x Xx

f(X)x

|k|2|k|

4 =

|k|4

whenever x max(3X, (4|f(X)|)1).g(x) =x2 sin x4 will do.


127/405

127

K115

The local form of Taylors theorem

f(x + h) = f(x) + f(x)h +f(x)

2 h2 + (h)h2

applied to log (or other arguments) show us thatlog(1 h) = h h

2

2 + (h)h2

with(h) 0 ash 0.Thus since qn we can find an N such that

21q1n log(1 q1n ) and 0 log(1 q1n ) + q1n 2q2n 2n2for alln N, so the results follow from the comparison test.


128/405

128

K116

Ifa= b = 0 maximum c.

Ifa= 0 maximum c ifb 0, 10b+ cifb 0Ifa >0 then maximumc if

b/2a

5, 100a + 10b + cif

b/2a

5.

Ifa


129/405

129

K117

(i) Take x1= 1,xj = 0 otherwise.

(ii) To see that B is positive definite ifA is, take

x1=

(a111a12x2+ a

111a13x3+

+ a111a1nxn).


130/405

130

K118

(i) One way of seeing that the j are continuous is to solve thecharacteristic equation. Now use the intermediate value theorem.

(iii)B(t) = cos t sin tsin t cos t

will do.


131/405

131

K119

The reduction to Figure K2 can not be rigorous unless we stateclearly what paths are allowed. However if we have a path which is notsymmetric under reflection in the two axes of symmetry joining midpoints of opposite sides then by considering the reflected path we can

reconstruct a shorter symmetric path.

[A more convincing route involves the observation that the shortestpath system for three points M, N, P is three paths MZ, N Z, P Zmaking angle 2/3 to each other provided the largest angle of thetriangleM NP is less than 2/3 and consists of the two shortest sidesotherwise. (See Courant and Robbins What is Mathematics?.)]

One we have the diagram we see that the total path length is

f() = 2a sec + (a a tan )wherea is the length of the side of the square and is a base angle of

the isosceles triangles.

f() =1 2sin

cos2 sofincreases on [0, /6] and decreases on [/6, /4] so the best angle is/6 (check that this gives an arrangement consistent with the statementin square brackets). IN PARTICULAR= /4 is not best.

There are two such road patterns the other just being a rotationthrough/4


132/405

132

K120

The time taken is

f(x) =h x

u +

(1 + x2)1/2

vfor 0

x

h. We observe that

f(x) = 1u

+ x

v((1 + x2)1/2

so f is negative and fis decreasing for 0x with 0x(1 v2/u2)1/2 andfis positive andfis increasing forx (1v2/u2)1/2.Thus if (1 v2/u2)1/2 < h (i.e. (1 v2/u2)h2 > 1) he should takex = (1 v2/u2)1/2. However, if (1 v2/u2)h2


133/405

133

K121

We wish to find the maxima and minima off(x, y) =y 2 13 x3 + axin the region x2 + y2 1. We first examine the region x2 + y2 a > 0 we observe that we know that the global minimumoccurs when y = 0 and by looking at h(x) = f(x, 0) =13 x3 +ax wesee that it occurs at (a1/2, 0).


134/405

134

K122

The composition of differentiable functions is differentiable and theproduct of differentiable functions (with values in R) are differentiableso since the maps (x, y) r and (x, y) are differentiable exceptat 0, f is. (To see that (x, y)

r is differentiable observe that r =

(x2 +y2)1/2 and use basic theorems again. To see that (x, y) isdifferentiable on

S= {(x, y) :|y| >9x}observe that on S, = tan1 x/y. To extend to R2 \ {0} either userotational symmetry or cover R2\{0} with rotated copies ofSon whichsimilar formulae hold.

Sincef(r cos , r sin ) f(0, 0)

r g()

as r 0+, fhas a directional derivative at 0 if and only if g() =g().Iff is differentiable at 0then

f(r cos , r sin ) f(0, 0)r

cos f,1(0, 0) + sin f,2(0, 0)so

g() =g(0) cos + g(/2)sin

ie

g() =A sin + B cos

for some constants A and B sof(x, y) =Ax + By.

The necessary condition is sufficient by inspection.


135/405

135

K123

(i) Observe that.

f(t,t) = t2( a2t)( b2t)which has a strict local minimum at t = 0.

However, ifa > c > b, then

f(t,ct2) = (c a)(c b)t4which has a strict local maximum at t = 0.

By looking at the behaviour offalong curves (x, y) = (t,t2) thefunctionfhas no minima.

(ii) In part (i), fhas Hessianf,11(0) f,12(0)f,21(0) f,22(0)

=

0 00 2

which is singular.


136/405

136

K124

(i) The second sentence holds by Exercise 8.22 which says that iffis Riemann integrable so is|f|.

(ii) First sentence false. Take f= g= F.Second sentence false. Take

f(x) =F(x), g(x) = 0 forx


137/405

137

K125

(i) Consider the dissection DN= {r/N : 0 r N} whereN M2.At most 2M2 intervals [(r 1)/N, R/N] contain points of the form p/qwithpandqcoprime and 1 q M. Thus

0 =s(f,D

N)

S(f,D

N)

2M2N1 + M1

M1

as N . Since M is arbitrary f is Riemann integrable with10 f(t) dt= 0.

(ii)fis continuous at irrational pointsx. Since there are only finitelymany points of the form p/qwithp and qcoprime and 1 q M andxis not one of them we can find a >0 such that (x , x + ) [0, 1]contains no such points. Thus|f(x)f(y)| = f(y) M1 for all|x y| < .

f is discontinuous at rational points x. Choose xn irrational suchthatxn

x. Since 0 =f(xn) f(x), we are done.

(iii) Nowhere differentiable. By (ii) need only look at irrationalpoints x. If q is a prime we know that there exists a p such that(p 1)/q x < p/qso

(f(p/q) f(x)p/q x

1/p

1/p= 1

but choosingxn irrational such that xn xwe have(f(xn) f(x)

xn x = 0 0.


138/405

138

K126

Observe that, if 0 u < v 1, thenunvn n(yx) (by the meanvalue theorem or algebra). Writefn(x) = f(x

n). IfD is a dissectionwrite

Dn=

{y1/n : y

D}.

By the observation of the first sentence

S(fn, Dn) s(fn, Dn0 n

S(f, D) s(f, D)so, iffis Riemann integrable so is fn.

(ii) and so (i) is true

Suppose that|f(x)| K. Given >0, we can find a 1 > >0 suchthat|f(x) f(0)| /2 for 0x. Since 1/n 1 as n wecan find an Nsuch that 1 1/n (K+ |k|)1/2 for n N.

IfN

nthen

|fn(x)

k

| /2 forx

[0, 1/n] and

|fn(x)

| K for

x [1/n, 1] so 1/n

0

f(x) dx k(1 1/n /2 and

1

1/nf(x) dx k1/n

/2so that

10

f(x) dx k .

(iii) is false. Take f(x) = 0 forx = 0 and f(0) = 1.


139/405

139

K127

Without loss of generality, suppose that f

(a+ b) 0. Observe

that, sincef(a) = 0 and|f(t)| Kit follows that, by the mean valueinequality,|f(s)| K(s a) fora s (a + b)/2 with equality whens= (b+a)/2 if and only iff(s) =K(s

a) for all a

s

(a+b)/2.

Similarly|f(s)| K(b s) for bs(a+b)/2 with equality whens= (b+a)/2 if and only iff(s) =K(b s) for all as(a+b)/2.Unless K= 0 (in which case the result is trivial) the conditions forequality give fnon-differentiable at (b + a)/2. Thus if we write

K(b a)/2 |f(b + a)/2 = 4(b a)we have 4 >0 and, by continuity, we can find a >0 such that

|f(s)| (K )(s a) for (b + a)/2 s (b + a)/2 |f(s)| (K )(b s) for (b + a)/2 s (b + a)/2 +

Thus b

a

|f(s)|

(a+b)/2

a

+

(b+a)/2

(a+b)/2+

(a+b)/2+

(a+b)/2

+

b

(a+b)/2+

|f(s)| ds

K(b a)/2 2 + (K )(b a)/22 (b a)/2 2< K(b a)2/4

as required.

To see that this is best possible define

f(x) =K(x a) for a x (a + b)/2 ,f(x) =C (x (a + b)/2)2 for (a + b)/2 < x (a + b)/2,

and f((a+b)/2 t) =f((a+b)/2 +t) where C is chosen to make fcontinuous and is chosen to make fdifferentiable at (a+ b)/2 (thus we take =K/(2)).


140/405

140

K128

(v) Observe that once we know that f is Riemann integrable weknow that ba

n

n1j=0f(a + j (ba)/n) converges but the fact that

ban

n1j=0f(a+ j (ba)/n) converges does not imply that f is Rie-

mann integrable.


141/405

141

K129

For (A) note that, ifFj and Gj are positive bounded Riemann in-tegrable functions with F1 G1 = F2 G2 then F1+ G2 = F2+ G1and

ba F

1(t) dt + b

a G2(t) dt=

ba F

1(t) + G2(t) dt

=

ba

F2(t) + G1(t) dt

=

ba

F2(t) dt +

ba

G1(t) dt

and so ba

F1(t) dt b

a

G1(t) dt=

ba

F2(t) dt b

a

G2(t) dt.

The problems with (C) begin (I think) when we try to prove that iffandg are Riemann integrable so isf+g. We also get problems (which,I think have the same cause) when we try to show that anything whichis Riemann integrable under the old definition are Riemann integrableunder the new. One way forward is to show that anything which is(A) integrable is (C) integrable. To do this we can first prove that abounded positive function fis Riemann integrable if and only if givenany >0 we can find a dissection

D= {x0, x1, . . . , xN}witha = x0< x1< < xN=b and a subset of{r :


142/405

142

K130

(iii) Everything works well until we try to define the upper and lowerintegrals but the set ofs(D, f) has no supremum in Q and S(D, f) hasno infimum.


143/405

143

K131

(i) Since Fis continuous on a closed bounded interval the infimumand supremum are attained at andsay. Now use the intermediatevalue theorem.

(ii) Observe thatsup

s[a,b]f(s)w(t) f(t)w(t)

for allt [a, b] so

sups[a,b]

f(s)

ba

w(t) dt b

a

f(t)w(t) dt

Using a similar result for the infimum we have

sups[a,b]

f(s)

ba

w(t) dt b

a

f(t)w(t) dt infs[a,b]

f(s)

ba

w(t) dt.

Now use part (i).(iii) Take a = 1,b= 1, F(t) =w(t) =t.Ifw everywhere negative, considerw.


144/405

144

K132

The gr

A Companion to Analysis - Solutions, Korner

Documents

Transcript of A Companion to Analysis - Solutions, Korner