University of Nigeria Virtual Library

Serial No ISBN 97-31602-6-5

Author 1 OYESANYA, M. O

Author 2 Author 3

Title Real Analysis

Keywords

Description Real Analysis

Category Physical Sciences

Publisher Arifat

Publication Date 2002

Signature

REAL ANALYSIS

An Introduction

M. 0. OYESANYA

'Copyright 2002 O M. 0. OYESANYA

All rights reserved. No part of this book may be reproduced. stored in a retrieval system, or transmitted. in any torn1 or by anv means. eicctronic. mechanical, photocopying, recording or otherwise, without prior written

permissiop of the copyright owner.

First Published 2002

Typesettit~g by KEY Cornputqs (Nig.) Ltd.

bada an.

Published by ARIFAT PUBLISHERS

Ijebu-Ode. Nigeria

ISBN 978-31602-6-5

To the glory of the Almighty God (El-Shnddni) and the Most High God (El-Elgon) for His love, His mercy and His grnce endureth for ever.

"Blessed be God which have not turned nway my prayer, nor ilia rnercy tiaonl nw." Ps. 66:ZO

Acknowledgement

I am greatly indebted to my professor and senior colleagues at the University of Nigeria particularly Professors J. C. Amazigo, C.E. Chidume. and late Dr. Adiele D Nwosu. Professor Charles E. Chidun~e was a source of encourage-

ment and inspiration during the beginning years of writing this book.

I

My colleagues at the Department of Mathematical Sciences of Opun State University (now Olabisi Onabanjo Universitv) Olakunle Solanke and Akiniide Fadevi had always been in the fore front of encouragen~ent lead in^ to the compilation and

production of this book. I am greatly indebted to them and I liere as always acknowledpe it

I express my gratitude to Miss Felicia Banjo for the beautiful typesetting. The enthusiasm of the top hierarchy of Arifat Publishers in seeing this production through should be and

is hereby acknowledged.

Real Analysis in the foundation of all mathematical and cr~girlccring scicnccs. Hence a full understanding is essential. This new book 'Real Analysis - An Introduction' is out to enhance both the understanding and appreciation of the beauty of real analysis.

The approach is such as to m6&c thc studcnt understand thc basic conccpts early enough and then utilise them in her chnptm, I have not thrown the rigour overboard. The rigour comes a e r the initial grasp is taken.

But 1 have assumed the students in well groundcd in calculus of rcal variables to the level covered in tbe book. Introductory University Matllermltics edited 3. C. Ammigo Afdcana-Fep, 1 995.

Tb.cre are plenty of examples illustrating thc mcthods of analysis following rhc definitions. The language of the book is simple and straight forward enough and ~uory average student will enjoy it.

TABLE OF CONTENTS

CHAPTER ONE The Real Number System. Supremum, Infimum, Intervals, Absolute Values, Boundedness.

CHAPTER TWO Sequences of Real Numbers Bounded sequences, Subsequences.

CHAPTER THREE Series of Real Nun~bers Tests for convergence. Series of arbitrary terms.

CHAPTER FOUR Continuous Functions. Theorems on continuity, differentiability, mean value theorem.

CHAPTER FIVE Sequences of functions. Cauchy criterion, uniform convcrgcncc.

CIIAPTEIi SIX Powcr Serics 'I'aylor's thorem, Taylor's series. Properties of puwer series. Abel's theorem. Diffcrentiation and integralion. Fourier Series.

C'HA1''I'EK ONE

THE REAL NUMBER SYSTEM

In this chapter we briefly discuss some definitions. sorllc conccp~s. and propcnics of rcal nur~~bcr

systcm, which will prove hnndy nnd uscfid in subscquenoc cl~nptcrs.

Definition

Thc red number system is the sy&m (R ;I.,.. 5) and

(R,.,+).is a ficld - a concept from Alecbra which Incalls h a t thc rcal nu~nbcr s sa l ise

somc conditions or mionls wnicn inciucic ciosurc, associativity. aisuibwivi\y,

ex i s t am of identity and invcrscs ctc.

< is a linear orderin.8 on R

For x,y e R iPx< y thcn x 1 z 9 I z for all z element of l i

For x, y in R if x>O and y 20 then xv2O

< is order coniplcte. 1.c. every non-crnpty subsct of rcnl ou~nbcrs that has an oopcr

bound also has a lcast uppcr bound.

Kccdl that if for all x in A subset of R thcrc cxists a in R such that sc: n, a is tllc uppcr bou~ld.

Wc d c f i c least upper bound (tub) for a sct A boundcd abovc if thcrc cxists a such that

(i) a is an upper bound (ii) if b is any olller upper bound for A

, then a< b for all uppcr bounds b

a is cal ld thc supremum 'i'hc greatest iowcr bound ii (nib) niso cniicd ir!iintrrrrr is

~nalogously dcfincd. Wc illustrate will1 some csnnl~)lcs.

Exnmplc I

ConsiderthcsctA= 1.x ~ R , o < x < 3 )

For this case sup A = 3 and inf A = 0. But notc that 0 z A and 3 e A

Example 2.

Considcr @e set A = (x: XER, O<xZ ~ 4 ) .

We can scc that A is the open interval (-2,2) so that sup A = 2 and inf A = -2.

Exsriiple 3.

Consider A = ( x: x E R , ~ >4 )

This is thc inkrval (-00,-2) LJ (2,m). T ~ J S A is not hour~dcd above and also it is ~iot hor~nded

below.

Therefore A has no sup and no inf. I

Example 4.

ConsiderA= ( x : x ~ R , x 2 > 4 a n d x > 1 ) ,

For xZ >4 we have (-ao ,-2 ) U (2, oo ). Uut sincc we also h a w h i s > 1 A can odv bc tllc sct

A=(2, w ).

This set has no suprcmum sincc it is not bounded abovc but it has ;ul inlilnu~n =2.

Some Useful Rcsrrlts

Fmm o w definitions of suprenwm rrnd Infinlum above we can prove that

(i) n is the least uppcr bound (sup A1 if and only if ior cvcn F: 7-0 (ho~vcvcr srni~ll 1 a- r:

is not an uppcr bound. That is Lhcrc c s ~ s t s a, in A such that n - c i n l

(ii) (ii) k is b e glb if arid only if for cvcry E >O k+E is not n lowcr bound i.c.3 a1 3 kl c

Proof - - ( ,

We show that

(i) a+b is an uppcr bound and (ii) a+b is thc lcast uppcr bound.

:. al + bl <a+b V a l EA, bl E B. :. a+b is an upper bound

a =sup of A a 3 a , E A 3 a-E <a, for pivcn E >0.

Also b= sup B 3 a, E A 3 b- E <b :. a+b -2s < ill b ., . :. a tb -2s is not ,m uppcr bound.

This implies that a' + b is the least uppcr bound of A . I B. t

That is sup (A+B) = sup A+sup B.

Lct us consider two scts A and B givcn by A= ( - l ,0, I ) , B - (0.1.2.3)

A+B={-1 +0,-1+1,-1+2, -l+3,O+O. O+l, 0+2,0+3. 140, 1-+I . 14.2. 1-4-31

= (- 1 ,O, l,2,3,4)

S u p A = l , s u p B = 3 : s u p A - t s u p B = 4 ;

sup (A4.B) =4 =sup A + sup B

This shows that for boundcd scts A and B A c R, B c R if a = sup of A and b = sup of B

h e n a+b is the sup of A+B

Sorrle Concepts

I n this section we consider solnc conccpts itkc ~ntnuais. boun<icdtlcss oi n sct. rrhsoiutc d o c

arid h i r rolntio~~sliips.

Intewrris

Tlrcsc arc spccial scts of rcal nutnbcrs. For a eivcn rmir of rcnl nunlbcrs a and b with cl b.

the open interval (a.b) i s the sct of rcal n~~rllbcrs x such Ihnt B < s < b.

Thatis(a,b)= f x : n < x < b )

The set of real numbcrs givcn by ja,bl = { x: a 5 x _< b ) is callcd n c-Ioscd intervr~l.

Other intervals are (a,bl, [a,b) which arc open cndcd intcrvals and pivni bv

(a,b] = (x: a<x s b ) ; [a,b) = ( x: a r x c b )

All thhe intcrvals are boundcd scts.

The sct of numbcrs x such tlint x> a for somc a E R is an tmboundcd sct and wc wrilc

S =( X : x > a ) =(a,w).

Thus thc sets

S ~ = { x : x r a ) =[a,m),S2= { x : x < a ) =(-oo,n),S3= ( x : s s a ) =(-m,nl

arc all unbounded.

Absolute Value

Wc notc tliat a rcal numbcr has both direction synibolixcd by its sigii ( riglit or Icll of tlic oriein)

and its magnitude given by thc distancc from tllc origin. Tlic provcrtv of ~iiaenitudc leads us to

what is referrod to as absolritc vnluc of thc rcnl ~ i u ~ i i h r n writlc~l 1'71 rind dcliocd by

Thu s

131 = 3, l(j= O, 1- 11 = -(-I) = I, l - X I ~ - ( - X P X

Thc absolute value of my rcal number b is nor1 ncpativc cind rcprcscnts thc distnncc froni tlic

origin to the point b.

1-21 = 2

4 a ,ml-;* >

-2 0

I a-b I is t l ~ c dislancc bctwccn Lllc poitrls n and b. If a > b dic~r 1 a-b I = a-b. If a c b thcn tlic

distance bctwecn thc points is b-a = - (a-b) = 1 a-b 1 . It bcco~rm obvious that 1 n I = 1 a-0 I

Absolute Value and 13oorrdedness

Thc absolutc value can bc uscd to define bou~idcd~icss of a scl . A set S = {x : s E S ) is borrrrdcd

if therc exists n positive number k sucli that 1 s 1 5 k for d l x in S. k is thc uppcr bound and -k

thc lowcr bound. Thus

1x1 I k * - k 5 x < k .

and

~ x ~ ~ l e - 1 ~ ~ 2 1

Thatistheinterval [ - l , ! ] = ( d S, 1 .

Note thnt I XI = I X-0 1 so that 1 x 1 5 1 m a n s the S C ~ of all rcid nunibcrs for whicli tlic d i s t i ~ l ~

from the origin is4 or the set of rcal tiurnbcrs with lowcr bound -1 and uppcr bound I .

Also wc notc that

For exatnplc

And since 1x1 = @ x .fbr x 2 0

= - x for x < O

I t follows that

That is

X + 2 for x > -2 1.y I 21 =

- (X -t 2) $or x < ---2

Examples

2r - 4 .for 2r - 4 2 0 Solution: )2r - 41 =

- (2x - 4) for 2r - 4 < 0

3 - x for x 1 3 Also 13 - -K 1 =

-(3-x) for x > 3

has non-negativc valuc in (2.3 1 and ncpativc valucs in tiicsc intervals ( - a, . 2 I and (3 . a, ) for

1 2x-4 1 and 1 3-xi rcspcctivcly but not both.

For the case [2,3] : 2x-4 = 3-x i.c. 3x = 7 civinr! x = 713.

For thc casc ( -a ,2) and (3, w ) : 2x-4 -x-3 giving x- l

Exercises

Dctcrminc whether the following scts we bound& bclow or bounded nbovc.

1.S= {0,1/4,1) 2. S = ( x : x ~ Z+) 3. S = ( x :x<%) 4. S = (I-I111 : n = 2 , 3 , . . . I

5. { x : x Z > 3 )

Solve for x in Ulc following quations

6.1 2x+31 = 1 7.1 G-xl = I 2x-71 8. 1 3 x - 6 1 = 7

x f i r .u?y 10. Show that )<(Y + p r lr - )?I)-

Y j i)r x i j ,

Some Stnndnrd lnequrrlities

We notc that undcr multiplication and division thc absolutc saluc has the nropcrtics that if s and v

arc rcal numbcrs then

To prove thcsc propcrtics we considcr scparatclv tllc following four cascs:

(a) x ~ 0 , y r o ; ( b ) s < O , y ~ O ; (c) xilr 0.y .: 0 ; ((I) s< 0 . y c 0 .

Also we can prove these propcrtics using thc fact thnt 1 s I = 4 s-' 7 hus

This inip1ies thnt

Result

If r > 0, ihcn I xl I; r iff -r I; x 5 r. Also 1 d 2 r iff s 5 -r or s 2

Proof

( 4 i; r e I x-d 5 r. Thai is ihc distawc of s from 0 is not grcntcs tl~nn r.

Thus x G [- r , r j i.e. -r 5 x 2 r

Convcrscly, if -r 5 x i; r, then x E [-~,rj which implics that 1 x-d s r. ?'hat is I d I r.

Corollary

1 l x - ~ l s r iJ' ( 7 - r 5 x ~ o - ~ r . i.e. s ~ [ c 7 - r . c 1 i - r ~ .

This corollary can be invoked in the solution of inequalities.

Example

Find Ihc values ol'x Ibr which 1 x 1-4 -= 3.

1 x 4 < 3 =? I x-(-4j < 3.

Here a = -4 arid r = 3. Hence using the Corollary above we have

-4-3-w-4-1-3. I l l a t is -7<x<-1. So SE(-7,-I).

Result (Triangle Inequality)

Ifx nrid y'are real numbers. [hen 1s 4- 5 1x1 I I .

Proof

Since by definition

it follows that - 1x1 5 x 5 111 . Nso -01 2 g 5

By adding~we have

Exercises

EXPI-ess the following sets as intervals or union of intervals.

1. S = { x : 1 + 4 1 < 3 ) 2. S = { x: 13-2s(<l 1

3.S = ( w: 12-3w1> 3 ). 4. S = I s : Is-al E . E , 0 . a coilstant I

Solve die following inequalitics

9. Prove chc triangle inequality by considering tlic four cases

10. Generalise the triangle inequality. That is, show that

Find the Icasl uppcr bounds and t l~c ycatcst lowcr bou~ids (if t l ip cxist) of tllc l'ollowi~w scts

COMPLETENESS PROPERTY 0 1 7 REAL NUMBERS

A vital propcrty f chc rcd numbcr systcm is tliot cvcry non-cnlpty bound4 sct ol'rci~l ~rumbcrs

has a lcast uppcr bound and a ~rcntcst lowcr bound. Tliis is tlic complctc~icss propcrtv of thc rcnl

nurnbcrs.

Wc shall provc this iniportant propcrty. But first wc stntc sonic otlicr ~xopcrtics ol' rcnl nurnbcrs

we shall usc in thc proof.

Trichotomy law

For any a,b, in R one and only one of the followinc holds ti) a=b ( i i ) a<b ( i i i ) s ~ b .

Density Property I

I f p and q are real rlilrnbers a such dial p<q. here is a ra~iotlal nurllbcr r sr~ch ha! p<r<q.

Dedekirld Theorem

Suppose that sct R of rml nutnbcrs is tllc union of two scts A nnd B will1 tllc follow in^ propcrlics.

(i) A * 4 and 13 + 4 ( i i ) A n B = + ( i i i ) If a E A and b E I3. thcn n < b

'Kl~cn thcrc mists a uniqt~c rcd numbcr c such that i f n < c llicn n is in A and if I ) , c l l m b is ill

D. Thc numbcr c itsclf may bc an ckncnt of A or ;m clcoicnt of 13.

Theorem

if S is n non-cmpty subsct of IC which is boundcd from a b o ~ c then S has n lcasr rrulxr bound.

Proof

Let B bc thc sct of uppcr boilr~ds of S and Icl A = R - 13.

A# 4 and B + 4 bccnuse by hypothcscs S is boundcd abovc so Ihar I3 corl~ains at Icasl onc

mcmbcr and sincc S is non cnlpty thcrc is a rcal numbcr s such 1 1 ~ 1 s is in S. llcilcc any rcal

numbcr r such that r < x is an clcmcnt of A.

A n B = 4 from definition of A and B . If a is in A and b is in B ~hcn b mrsl bc an oppcr bound of

S. If a = b or a > b thcn a is an uppcr bound of S. But a is in A. Thcrcforc a is not all uppcr bound

of S and hcncc a<b by thc Trichotornv law.

The Dcdckind tllcorcm is satisfied and can bc applicd. 'I'lic riumbcr c givcrl bv tllc hcorcm is an

uppcr bound of S bccausc if not tlicrc i s an s in S sr~ch that s x or S-CSO. I;roni llic dcns~rv

propcrty wc hnvc

But

Also

X -C S -C . c c c t-- <x Z ~ C + - - - . is 1101 mi uppcr bound of S.

2 ?

This is a contradiction ofdcfinition of f3. 'I'lvls c is an uppcr bul~nd Cor S

By thc dcfmhiot~ of A no nunlbcr lcss than c is at1 napcr bound of S . llicrcforc S has a lcasr w ) c r

bound.

CIJAPTER TWO

SEQUENCES OF REAL NUMBERS

Definitions

Sequences of real numbers are a function with doriiaili tlic positive intesers and rance a

subset of the real numbers. Thus a givcri sequence ~nakes a real nuinbcr a,) corresl~ond to cach

positive integer n. In other words a sequence of real numbers is it11 ordcrcd set of numbcrs 1'

In mathematical language we define a sequence in a non-emptv set X as a t'imction

f: (IN,<) +XcR whcrc {IN.<) is the sct of positive iritcgcrs with the usual ordcr. 'I'his

sequence is denoted bv {f in)) or (f(1 ).fl2). . . . 1 or f I1r1)l .

I f f is a seqmncc in X, then f'(n) is cnllcd the 11th tcrnl or I<" tern1 of'lllc scqucncc.

Thc characteristic that distinguishes n scq~rcticc tiom ohcr sct of rcal nutnbcrs is ilia1 tllc terms

appcar in a dcfinitc ordcr.

Examples

1. {1,2, n) is a sequence in IN called the identity sequence in IN i.e. S(n) - n. 'l'he lirst

term is I,he second is 2 ... and the nth tern1 is n.

2. (in) is a sequence in C: (complex number set) with the first term beine i and the

second i" - 1 and the rourth i4 = 1

is a sequence in R i.e. { I . O . I . 0 . l

Definition

If a fitnction g: IN +X i s a conslmt f t ~ n c t i o ~ ~ lakine valac XO ill X. tlten we sav that e i s

I! a constant sequence ( xo).

r . . . . An example ofsuch a sequence is { i ) = { I . I. I. I . . , . . I ~viiicil is a cons~ani scyucncc 111 ( 1

1)efinition:

1 A sCq~lCncc sntisfyi~lg the conditio~rs f: IN ,X for which there cxists I) , , such t11:rt f ( 1 1 ) =

(m) is called an eventually constant sequence.

I An example is the sequence (0,0,0,0,1,1,1 . . . 1 } I'or which f ( 1 ) -0 -1' (2 ) - 1' (3) -1' (4) and I'

(5) = T(G) =r(7) = 1(8) =... =f(n) = 1, so that for all m > 4 I fm) = r(n)

I In mathematical analysis tlic important corlccpts arc those of convcrpcncc- poitir-\\.kc ilnd

uniform convergence, continuity, etc. We therelore start our trcatliient In th~s chapter w~th the

concept o r convergence of a sequence.

Definition

(1) If a sequence {a,,} has a I imi~ , we say {a,,) corivergcs or is co~i \ ,c ruc~i~

(2) A given sequence in , j is said to nave a i i n i ~ l a i i ' el \:en an arb~~rarv pos~~ivc i~unibcr E

As an example consider the sequence ( I /n 1.

Given c - 111 00 wc Iind 11x11, iTwc choose 1111 = 10 1 . !'or all nz 10 I

And we say Ihe.sequence { I/nJ coriverges to zero.

..\ootIler example is the sequence J 2, 1, \\hicIl coli\,erges to Lero ns nr norv shon.. InT J

Given E = 111 0 we find that if we choose n 0 = 6 tlicn for all n>6.

and the sequence converFes.

We note that given the sanie sequence nitn E = iii i~l~. n o - 6 C R I ~ I ~ O I snlisi)' l i w ".

As c:ui be sliown no = I 5 will do i t . I t bcco~iics ob\'ious llicri ll int llic clioicc ol'iill tlcpcnds oil

the given c. In otlicr words E rncnsures liic cioscricss bcl\\.ccn a,, ;irl(i n \\ ii~ic 11,) I I I ~ I C : I I O S iio\\

far out in tlie sequence one riiust eo 10 acliic\.c tlie sr~ccificd dcercc oicloscrlcss

Wc note of course that for sorw scqucrlccs 110 riiallcr I ~ o w filr out ill tlic scqucncc wc !!o wc

cannot achieve

any closeness whatsoever between the tcr~iis. An esarnplc is tlic scaucricc i n 1 L C . ,

limn- (1,2.3,4 ... n). Ths type oCscqucncc docs not conwyc stlice lor rhs wsc -iici n n i l n c

17 -+ (c)

say that the sequence diverges.

Uniquer~ess

Onc may ask: Can a sequence convery to tiiorc ~ i l n r i one \ aiw! i i n u trsc \tic i r rnr~ acrrrirlrm 01

conver8cnw i l is ubvious i~lli i l lhc IIIISIVC~ IS. We therefore slate arid pro\ e the I'ullo~rny

Proposition:

If a real nurtiber sequence {n,j couvcrgcs lilcri i r convcrccs uriiquciy.

Proot' Suppose (a,,) converges lo no mid n l wi (h a,, + a,. Siricc all t at 11 lollo\\ s tl inl

n> no(c), Ja,, - %( < a. Also since {a,] converges to a1 we have Jn,, -- all\ c Ibr all 11 > 11,) (C 1

Now

This is an absyrdity. I lence the assumption a, ;l- nl is fnlsc and all -: n~:

This proposition immediately helps to say tlial llle sequence { l ' (n) l e ivc~ i bv

does not converge.

For some sequences it may not be easy to iist ail the terms so that choosinr! an sl tiw

satislies a given degree of closeness E may riot be easy, we ei1.e some esanit~ics that \ \ . i i i

r llveti E. help to determine nrl for an! b' -

Example 1

Iim (- lr Show that - n

80h1tiort; Let e be given and suppose 11 11,) then

(- 0" I 1 - - ( I r--- < - I.,, -4 = 1 ,, 1 I7 I f , ,

Example 2

Show that (f (n)) = {n) does not converge.

Solution:

Suppose it converges to L. then, gven E > 0

In-LI > Inu-LJ for n no

so that In-L( < E implies InoLI .' E for some nn mid the seqile~icc di~erues.

There are some sequences sarisfying some rsyeciair conaiiiorls. \\-ilicil inaicaw [heir

convergence. Onesuch condition, is Caucliv condilion for corivcrgelice of scquenccs which

states:

A sequence that satisfies the Cauchy condition is callcd n Cnucln, scquencc. w e notc ha t

what this condition is saying is that if a sequence {an ) con\.err:es \ \e sliould l~nd tlirlt nller

some terms the dimerence bet\veen any two given terms should bc \.cry s~iirlll. ' I hat IS [lie

degree of closeness should be very high

.As an esample consider the sequence I I h l .

We have seen above that i t converges to zero. tii\,en E = 111 u \\*e find tllal since

f(l1)=111 I and f t 1 2 ) = 1112.

1111 1 - 11121 = 11132 1/10.

Thus for all m, n (any two valucs of n) Freatcr than 1 0 the sequellcc co!l\.errrcs. ~n omer

words he Cauchy condition is saying rlint it' the tiillrclicc bctncoli :in\ 111.o ~clmis :illor :i

specific term is very snlnii tiicn the sequcncc colwrgcs.

Example

Determine whether or not [he followiag sequences converce

Solution:

If the sequences converge the ~ a u c h y condition is satislied. so

1 1

t1ik ti'

Definition: Ifsequence {a,,} is not convergent thcn la,,) is said to be dii.ergcnt

[I .I-(- ly \ Tlw scqucncc { I :(I; 1 :(I: l .O. . = --- ti~~,cr!~cnsrrirl. I hcrr I T no I I I I I ~ ! I I C \ ,n i i~c i n

L 2 j

which it converges.

Result:

P1.ooG

By the triaigle illequalit>.

Sclutiotr:

Since O <: a <-' I then a must be a Sraction i. e. a - 111 I h , I1 5. 0.

By t!le binonlid Ihaorenl

lim clnd crn+bn - ~ + b

n --, m

Solution:

Since O < a < 1 then a musl be a liaction i. e. n - 111 1 1 1 , h :> 0.

By the binomit* theorem

Hence na" < (11 - i ) / I

(ii) We note that 0 < ( I -l/n)n < I for all n. Hence for 0 < a i 1

Hence from (i) above

1 - - a converges to zero. (( Y n }

3.l)iscuss the colpergence ot the tollowmg seqrwlces ol real nrtnll~ers

Solution:

which shows Ihnt since n = I gives cor~slarit secluerlce I I f

, . lhis shows I h ~ t given any E > 0 therc? eu i~ ls m strch that F 2 r 1 n -- 1

For all n> m and hence

'I'he student is advised to take (ii) and ( i i i ) as esercises

".

Further Definitions

1 . 1 1 ' lor all n a, < a,,l Ihen fn,,}

monotonic increasing if an <a , , I . An obvious example is h e scouence {I)) since ~ s n + i Tor

all n.

2.lf Ibr nll n a, > a .ll then {a,) is a ~nonotor~ ic deciwsing seqrlcncc. An e.sanll?lc nl tlirs IS

the sequence { I/n j . (In either of'lhe nbove cases ttic seqtlclice IS snd to lw I I ~ O I I ~ ~ O I ~ I C I .

3.A Sunction Tis said to he buundcd it' and orily if d w e esisls a posi(iw nwnhcr M. OcM+v

such Ilia1 sup I I(x) I .r M.

1 1 1. since a,,, , - a,, = -. < o

Also l im (2 ++)= 3 n + I

Since it is oscillntory it does not converge,

Proof:

Let {an) be a monolone decrcas~ri~ real riwiibcr scquc~icc. I I { i ~ r ~ l 1s bowidcci. Ihc scl

{an , n E IN) has a Icasl upper boi~ntl. sit!. a* t I<. Givcr~ ilriv r; i U 3 I I ( E \ I I I IN s r ~ c l ~ l l ~ r ~ l

a - F < m ( ~ ) >*a* -F lor all 11. 'l'linl is 1 an - a*l < E I'or all 11 > ri(c ) n ; l i i c l~ iriiplies IIi;il

$in an = a*.

We note that if'the seqtlertce {a,,) is hounded and nicwc.)toriic i~tcrc.:~sirrp (tlccrwsirirr) flic iu:rsl

upper bound (greatest lower bound) of' llic sct of' vnlr~cs la,,] is tlic Ir l i i i ( of' ~l ic !1,1~c11

sequence. An ~n~niedinle conseqr~ence ol'tlils ~heoreri~ I S b

Theorem: Every bounded morro~one sequence converycs

lim , n - +m n --, oo

'I'he relal~onship helween a bol~rrded sequence nrrd a sr~bseqr~cnce IS slaled III he lidlo\v~ng

Iheorem, which we ~IilIe willioul prool:

A very imporlanl llieorer~i in : U I ~ ~ Y S ~ S is lhc Uul/.i~rlo- Wcicrs~~.:~ss ~l~corern l i ~ r scquer)ces w t ~ ~ c l ~ slille

Every botrrided seqrlence hns R convergent subseqrwncc

Exercises

Consider (he following sequences ant1

(i) wile ou( lhe firs1 live lcrrns beginning \villi 11 = I

(ii) determine the boundedncss or othcr\\,isc

(iii) determine the monotonelless or ott~crwse

(iv) discuss Ihe convergence or olt~crwse.

Find lhe l~n~it ot'the tbllcrwrtig sequences. Where the I r n i ~ f tlves no1 eu~sl s h o w ~ i i c ci~\~cr'twrcc

ol' the sequence.

yrerllest Iowcr bound,

So111e I7~1r'tllcl' Cosccpts.

Llclinilions: An inlinik scl ol' non-enlply clusctl irilcrv;rls 1 \.I;,. . . . . I , , i s callrd :I sel 01' 11rs1rO

inlervnls or a ncskd scqucncc of i~lrcrvals i T

(i) for each n the in!er\ral I , , , I is contained in I , ,

(ii) , lelling I,, = [a,,,h,,] then the length of' I,, - h, -a,, -+ U i,c. 1:ivcn E :> o rJ 11,) ( E ) such

that for all n>no b, -a,, <' E.

PI-ool': 'I'lie procedure f'or provirtg lhis Illcoscr~i i s ( i ) sl~ow lhi~l I,, c;u-lilbl cur~li~lrr tllcwc . than on$ point. (ii) show that n I, , i s nun-cniply.

Improper Integrals

Definitions:

b

I. An integral If (r )./I is called improner if one or all of the followiw is !rue

(i) the interval ol'intcgralion is i r ~ l i r i i t c ! I

(ii) the inteyrnnd Qx) is trtihountled in the neigllhourlwod ol'c71i end poiril

m lim lini J'f (-v b = I (h ) = J' f (.I- \/r

h -+m h -+m (*)

U

provided hat his limit exists i .c jf (r U-V ~r am.

If Ihe limit on Ihe right side ol'(*) exists (hen Ihe itiiproper rrtlegrnl 1s sntti In converge:

otherwise i l diverges.

Examples

'0

1.Deterrnine whcltier the improper integral f e "h cotirertzcs nnd lind la value ~I.so. 0

Solution: ..

m X

2.Determine whether the improper iwegrol I- ----dx cowereor or d~vcr):as, , , I +I-

- lim - .. j<~n(i + b 2 ) h Y u 3

rQ X

Therefore the improper iniegrai [ --- I -t cir diverr?cs.

Exercises

Determine whether or not the given inlegrirl corwergcs or r w l .

the integral in that case.

An infinitc scrics of rcal numbcrs is a pair or scqucnccs of rcal nrln~bcrs {a,) and { S, ) nllcrc

Sn = a, + 8 2 +a3 +. . . .,+ an lor cnch n . 'l'lic scrlcs 1s nsunlly dcnotcd hy

hc scriw converges if scqucncc {S.) mnvcrgcs, ond thc scrics divcrpcs if scqucncc (S,) divcrgcs. 'I hc limit

"I

ofscq~~enw (S,,) iscelled dlhc sum nr vnlllc nf Ihc scrics and r l a d r r l x u , , n - l

Thc uumbcr n, is cnllcd ~11c nlh lcrm of :hc scrics, and I I I C nurnbcr Sn is cl11lu.i ~ h c 11111 pnrlinl sum of h c

scrics. Wc now statc a neccssnry cund~t~on for thc convcrgcncc 01 n scncs.

4) lim Theorem: I f zq, converges. then q; = 0

n=l n -.+ . . n

ProoT: *

20, convcrgcs implics that tlic partial su n ratisfv tlis Caachy condition. In par~icl+r if n-l > 11t.1 tllcti n = l

lim ISn - Sn- 1 ( < e. That is Ian1 E. Thus n,, = (1 is a ncccss'ary condi~ion icr ~on\~crgcncc.

n -> a.

Howcver this condition is not strnicicnt for c< ilvctgcncc. I his IS ~Ilrrstrntcd by thc 1b11ow1ng cxnmplc

Clonsidcr thc following scrics

Hcrc tllc scrics divcrgcs.

(Assignmcnl: Usc lhc idca of partial sum as a dclini~ion for con\lcrgcncc lo s t ~ o w ltm tic

gcomcbic scrics convcrgcs. )

'I'ES'I'S FOR CONVEKGENCE

Vcry many situations orisc wl~crc tlic straiglit dclinition as given abovc ciumoi vcrl; cnsiiy bc

scrics. Wc now statc thcsc tcsts and thcir applications. a

I . 'i'he integral Test

The integrnl test slates: If'. li)r x >xO. ftx) is contin~~or~s. posi~ivc. t~~onoloniciiIIy clccrc;~siriy to

a.7

zcro. then tlic scrics of nunibcrs IXx) cnnvagn (divagcs) il ~ l ic impmpcr integrals l f (.r)ir ,l -

convcrgcs (divcrgcs). f

m lim i I~crnctnbcr that f =

divcrgcs othcnvisc.

2. First Conlpnrison Test.

Let X c, be R convergent serrcs o i 'pos~~rve Icrrns anti i ci,, n (itverym scrtcs oi p o s t ~ ~ \ * c tcr.111~ i t ' L

a, IS R scrles si~ch that Ihr all n grcnicr 1ha11 sornc ti~rlnhcr 11,). nll < c,. 1hc11 X n.. convcr!!cs I t ' li,r

all n > nu a,,:, (1, [hcn .X all divcrgcs.

ILmark:

A flnw in his lest will shoc if wc considcr a scrics C nn such t l ~n t for all n w , , :I,, 2 c,, ;~nd a,, 5

d,,. A conclusior~ m y bc slillicdl os111g l h ~ s ICRI.

Note: The rcmark rlndcr Ihc t:irst ftomparison'l'cst applics 10 this Icsl also.

4. The Cnuchy Root Test

If thcrc cxists r! fixcd nun~bcr r such that O< r < 1 and for all n , n,,

r m van r r then a , , con v c r ~ e s . -

5. Ratio Test

'l'his is by far thc most widclv ~ c d tcs! for wnvcr!;axc nnd 11 stnlcs t t l i ~ r : 1 1 lor nil n , no c;~c.lr

an is positive and 5 r whcrc r i s n constnnt such tbnto < r < I l h n l Z n.. convcrrcr 1 1 ibr 1111

'1"

"n11 n > no. and - > 1 then Cn, divcrgcs. '1"

Remnrk : Thc Ratio Tcst is silcrit about tlrc casc and = I ' I"

7. Logarithmic Test

'I'his is probi~bly thc only test that docs not f;d. It stntcs:

lim If for all n

11 j w

'l'hcn thc scrics C an convcrgcs, i1 i t IS icss r i m i rhcn rhc scrlcs cinVcrgcs.

8.Caochy Condensation Test.

If a .b~ I a, for dl n thcn !tic scrics C a, convcrgcs if and only if thc scncs

C2'uZ, = 1 + h, -I- k , -I- k, + . . . convcrgcs k - 0

Remrk:

This is vcry uscflll bccnwc i t cnablcs us to invcstiqntc ccwvcrgcncc ol scrlcs bv cnnsldcnnu, onlv

a 'small' subsct of tlic sct of tcnns.

Examples

'Tcsl lhc convcrgcncc 01 thc tollow~ng scncs:

condition for cconvcrgcncc 1s sal~sl~ctl. Uy I l ~ c rotlo tcst

'Thcrcforc thc scrics convcrgcs.

Wc can scc hcrc that il is casy to takc thc nth root and cvaluak. Wc Lhcrcforc npplv I I K Cauchv

Roo1 Tcsl. l'akuig lhc 9th root wc havc

We note that as n-m n I/n -* 1.00 and as n+oo cn- boo.

I 3 . F o r z - CI,, = - I .,, logn log n

I Wc note that for all 11. log 11 < 11. Hcncc --- I

> - . 10gn n.

"' I Rut C I/n diverges. 'Thercforc bv the First Comunrison 'I'csl J7--- divcracs ;;7 log17

Rcrrrark:

Wc notc har. for (.his cxamplc sincc log n -+a as n-+m f/fog n -+o as n--+=. i itill 1s ttlc ncccssnn:

lim condition n, Y O is satisficd but LIIC scrics d i ~ c r y s . '!!)is s!w:rs !!la1 ~ h c co~~riitinn

il -> 00

lim (7, = 0 is not sullicicnt for co~!vcrgcncc 01' scrics.

n - + m

scrics divcrgcs.

lirn . . is m!isfic.t i

Applyittg rhc rario test

5 .);or tiic scrics

'I'htts rhc rario lcst fnits, Wc now opply Knabc's tcst.

Rabbc's tcsl fails. Wc now apply Ihc lognrithmic IcsI.

'l'hcrcrorc Ihc scrics divcrgcs. ,

Remr rk:

'I'hc abovc uscd proccdurc is mcanl lo highlight somc crrscs whcrc somc ol l l~c ksts ImI ;rnd tlic

clIcctivcncss of othcr tcsls. Wc nolc ~hal Ihc lnlcgral lcst could haw bccn ~rscd sincc

showing that lhc scrim divwgc~,

By thc ratio tcst this scrics convcrgcs for xq 1 anti divcrgcs for x , 1 . hl i f s = 1 11ic tcsl fails.

For this cnsc wc appiy Raabc's tcsr.

Thus if x = I and a + 13 - y - I < - 1 i.c a,+ fJ< y t11c scrlcs corlvcrgcs aid d~vcrgcs 11 a + Ij-2 y. I Iic

test fails if a -I- p = y. For this casc wc a ~ ~ l v thc 1o~)aritlimic ~ c s t .

= logn x = l

M - 2 Y - 1 Y I - 2 + - - - - - - .

= log17 - I!.&-. Y + ' Y 1 4 - - 4 . .

I ? I 1

Hcncc for x = 1 and a + 13 -y -0 ~ h c scrics tlivcrgcs

Exercises

'I'csl Lhc convcrgcncc of thc followirig scrics.

lktcrmine whcthcr t l ~ c Iollow~ng scrm corivcrge or d~vcrgc

10. Ucrivc tlic cxprcssion -

'I'he p-series.

Wc now invcsligntc Ihc imporlnrit scrics cnllcd tlic p-scrics. This scrics is vct? ~~scliti hr thc

application oT thc Comparison 'I'cst Tor thc convcrgcncc 01 mi\l\\ scrm. Vvc slww rilat lor p i (lit

series d~verges low nnd l l ~ I Sir !7 > I Ihc scrics corivcr2cs ;~lwol~~~cly.

(,, Suppose P = I lhcn thc scrics

l l l ) ~ l l ~ l I I I 1 -+-+--+7 + - . I . - ( -+ -+ -+ - . -+ - -+ . - I t . 5 7 2 ' ) 0 1 0 1 1 12 13 I 4 15 2 " )

We note that here thc lirsl group conlnlris one Irrri~. llw scccwl youp conlatrls 2 Icll~is.

the third group 2' terms, he lorlrlh group 2' lerms and lience the 11111 y o i ~ p co~il:i~lls 2'" '' 1 I I I

terms. l h e sum of the 11th group is y7 -b -- + ---1------- + . . . -1. --.-7--- 1 2P- ' /I1 (2"--' 4- ] ) I ' (2"-l . I 2)l' (2'1-' - \ ) I '

1 - 1 then that thc sum o f thc grcap i s Iczs thm 2" I.--- p, ; ;" p ;;,, ;;

1 Therefore for pbl Z - co!?vx-gcs,

n 1'

Remark: using the Integral Test can get tlw above results. Tor the p-series. The student is

advised to show this and convince himself of the results.

C'onvergence of Serics ot' Arbitrary 'I'er-ms

Consider the series Cc, an infinite series in which c, are real numbers of arbilrac sign. A

series of nsnrbersx(- is said to he an altersnti~lg series ifii.. 0. n = 1.2.3.. n - l

Definitions: ( I ) The series Cc, is absolutely converwtit if -I k , I # I corivcr!ys, n

(2)IT Cc, converges bui docs nor convcr!;c nbsniurciy. tiicrl LC,, 1s cor~i~rlonaii\;

An example oTa conditionally converger11 serlcs IS the altcnial~ri!: Iinrlnon~c scr~cs

Leibnitz Test A very useful lest for the convcrgencc oi' iritcrnat~ny scr~es IS tile Le~briltz

test which says:

m

allernatins series x(- I)" " h , converses.

Examples

IJiscuss h e convergence or ollrerw~se ol'llre Ihllow~r~~: serles.

lim 1 I \

k ,a:kJ All [he condi l ion.~ o j I , c i l ~ n i ~ z l i~eoretn trrc , S ( ~ I I . ! I I P ( /

'" 1 r-, ., i I !)' a . - I = which cor~ver.yes. . IIcnce 1 &-+ ~ Y ? ! ? I Y > ~ ~ v Y ~ r h ( . ~ h t c ~ J * . = 2" . 3 . . *.

Exercises 8

Determine whether the given series con\lerges u r diverges.

Dekrmine thc lype oTcoriverger~ce or tiivergerrcc lor the loIlu\vmg scnes

or diverges.

Rearrangements of Series

Consider the alternating series

Adding (1) and (3) gives

exercised in dealing witli infinite ,analog of liliite svstem

Geometric Series

~ ( i ] is N geon~etric series wilil i r 1 - i i i < i . Tilcrchrc k - 0

I - 3 converges to --?. -

' ./, 1 --

3

CHAPTER POUR

CON'I'IN UOUS IWNC'I'IONS

Consider a function f. S~rppuse thc domain 01' Ihc Smclicrrl I' i~~clwlcs i l r \ O ~ W I I in~crval

containing the point c. 'lhen f is said lo be coritirl~rorts ilt c if'

lim linr (9 f (.Y) exists and (ii) f (.I: ) - !' ( ( 3 1

X > C' .i > t

f is said to be continuous in an open in~ervai j s b j i i ii is coniinuous a~ each poilu in h e

interval. A function that is not continuous is said lo bc discontinuous.

lim l i~n side; &a1 is f ( ~ > ?

X ->c" -, . f b 1

removable clisco~~ti~iuity occurs wlicn n rcdcl~nil~oli ol thc Iurict~on :\I t l~c pomt 01

discontinuity renders the ti~nctiorr contitl~rous.

lim It should also be noted dta! when / '(x) v s i s ~ s bul dill'crm! l i c m ITc) hen L I P

x , c'

disconlinuily is clnssilied as ol ' (Ire lirsl kind ;inti 11 IS :I re~no\,illde dtsconl111\111y l l

lim f ( x ) does not exist then this is called ef the second kind.

X ->C

Example 1 ,

Determine whether or r~o t ( I I C given fi~nctions are conlinuous nl the indicnled poinls s = c.

X-25 (i) f ( ? c ) = - L, - 2.5

,/I - 5 x + 2 X f l

(li) j(r) - ,. - 1 - - 3 x = l

25 - 25 0 - ( i ) .f (2 5 ) ------ - - 5 - 5 0

Thus S(s) is no1 defined al s = 25 and so cannot be conlinuot~s lime.

(ii) f(1) = 1 + 2 = 3 + -3. Therefore Cis not conlinuous at x = I .

lim . . !I(.\ 7 /!(c) X --- , C

Hence h(x) is conlinuous al s = - I

lim iim ( iv) !(.I-) - 8 ..~; 0 f(!, -?- 8) x --> 0' -

lirn - [(q + s)? - 4 -- -1 S > O

lim lin~ /Q -;5 > o f (Q -- 6 ) x > 0-

lim l i r ~ r E!!s d l ? - @ y ,t (i) a. 6 )

'5 - > 0.' ' s > 0-

:. j (x) is ,101 currrnrrrorts or .v = \ I

Exa~nplc 2

In citch 01' the lulluwing, classilj. Ilic g t ~ c n poitils as une o l r:c,n\~nurlv. ~criio\~nblv

disconrinuiry, essenlial discontitiuily. jump disconti~wity. I l l l x ~ I ' S C C ) I I I ~ I ~ U ~ I ~ is removable.

dcfinc PL) to makc it contiriuous a1 s = c.

Solution

Thus f is not defined at s = c = 1. Hence !'is not contirluous at c = I . I lo\vcvcr.

lim so h i l l f (.r) exis~s. 'I'his mnkcs s = c == I :I rcnwml!lc c! isco~~ti ! l l~i l~~.

x - , I

If we now deline T ( 1 ) = 2 we have

lim f (.r) =f (1 ) = 2

x , I' and we have that

ti) f (3 ) = 3z + 1 = l o

lim l i n ~ f (x) -- f ( 7 4- r'i)

x --> 3" S , O '

l i m - [(3+6)'+1]-l(r 6 - -> 0

lirn - lini y(3 - '7)

x .- , 3 - S > 0

- - lim -- (3-6)2 - 9 .: 6 CS -+ 0 p- 's)-3

lim l ir i i Since f(c.1-8)" ! - (cq -. (T)

S ,O 0' , 0'

h e point x = c = 3 is a poinl ol'tllsconl~rirtllv i\nd Ihc t l r s c o n l ~ ~ ~ r ~ ~ l ~ IS i111 cs.;c~iIriil o ~ i c iI

jump discontinuity.

Some Tlleorems on Continuity

In this section we state snnd prove some melrll tileoretns on conttn1111y \ \ h h can he rrsetl to

establish continuity without cupiuus corrlyutallurls.

Theorem: ( conslant function) Let S : K -+ R be a constanl fnnctiotl and let Sts) = k l'or all s

in K. 'Ishen Sis conlinuous at any polt~l x 7 c in I < .

Proof:

lirn lim j- (.Y 1 = j- $1 - k. .USO T(C) = l;.

X -%C X J C

lirn Hence j ( s ) = f(c) and f is ccntinuclus rt c .

X -)C

But c is an arbitrary poilit ;u~d so wc cotlcludc that f i s continrro~rs at all points x z- c in

It.

Theoi-em: (Itlenfify Fanclinn) Let I : K -+ K Oc Ilic idetit~ty liltictlon ( I ( \ ) == \: Ihr :dl s 111 I<

If c is arbitrary then f is corihuorts nl c.

Proof:

lim iim f (c) = c and f (.Y ) = (.Y) -: c -.,+).

x ---> c- X . . ) C ' '

Hence l is continuoils al x = c. ' lhe arbllrariness ol'c curnpletcs Ihe prool.

Theoreni: (Polynomial limcfio~r) [.el I ' : R -+ R he tleliricd I)!.

lim i i rn f b ) = . c,, + q . ~ + O,.Y - + . . . + n,.r ' - ff,, + O,C' + *,c' '. -F . . . -F '7 * 'nL

X ->C X - -> C

Hence t'(x) is conlinuous at x = c. 'l'he arhttrarirless ol c colnpletes the prool

) where p and (I arc Theorem: (Rrrtionrrl fmction) 1,eI T: R -b R be defined bv /' (x) = -- (1 (.v j

pplynomial iitnclions. 'l'hen l'is conlinuous al ever)) yo~rlt c 111 i ls dorrlam lor wlllch q(c) t- 0

Proof:

Frurn our Ulcorem on polynorn~:d litncl~on we 11olc that

lim f . (x) = "m - P G ) - -- p t 4 -I.('.)

x -+C x - + c q ( x ) q ( c )

linl It thus follow thnl sincc q (c) I: 0 f ( c) csists arid f (s) - f (c),

X )c' ' '

Hence f is continuous at c

Exercises

1. 111 each or lhe I'ullowing, say whe(lw or r w l tilt g~\.cn li~r~cliw is C O I I ~ I I I I I ~ I I S ill I I I C

giver1 poinls. Juslifj your answers.

x 2 + 3 x - 1 X I 0 ( i i ) .+) = , II

3x2 - 5 x I 1 s > 0

2. Ln the following exercises, classily the given puir~ls ;is one of' cor~tinu~pl. rcrnuvablc

discontinuity or essential discontinuitv. if i t is a removable discontil~ui~v re-define the

function to render it continuous.

&-2 (ii ) j' (.v ) = -- 4

Sketch (he gmph of 1:

There is a connection hetween dillerentinhll~ty m c l conl~nrr~lv We sl;~\e ~tirs rn

Which shows conlintlily oC T al s o .

which is continuous at x = !h but not dit'ferent~nble st s =- !$ .

Some Related Theorems

We use the coricept ofdill'crentiab~litv to prove tl~c Mci111 Vi~ltlc 'l'licor'cni ard sornc r . i l l i r t i ~d

theorems.

l'lieowm: Sqqmse thal Ihe li~nction l'has a Ical rnaulnllrnl or n Iocnk 1111111r1111111 nt the polril c

If c is an interior poinl of the domain o f f and if Cis dillel enliable at c tl~ai r (c ) = 0.

Isroot

Since f is differentiable at c then f (c ) = li1n ,f k 1 -.r (,y - J . -..:. . ... a

t' .- t' X -> .Y(, . . ,,

Tl~eo~~eni ( Rollc's ) Let the function f satis!! the follouine conditions:

( i ) I' is conlinuous in lhc closed bounded ~ntcr\.:il I n.bl

( i i ) C is dilTerenhblc on h c open inlerml (n.b)

(iii) !'(.a)=C(b)

then there exists c in (qb) such llint r( c ) - 0 .

Theorem (Rolle's)

Let the function .f satisfy the following conditions.

(i) J' is continuous in the closed bounded interval [a, b]

(ii) f is differentiable on the ope11 interval (a, b)

(iii) f (a) = . [ ( / I )

then 3C E (n,b)

Proof

Since f is continuous on tllc closed bourlded intcrvd [a. bJ it Im both a maximum and a minimum value on this interval. If both aiaxinluni and minimum occur at the end points then the maximum value and the minitnum value are equal and f is a constant fiinction for tlicm f l ( x ) = 0 x E ( 1 , ) ~ w d c is ally point.

If J is not a constant, then it must have either a maximum or a minimum point

or both in the open interval (a, b). Let c be such a point then. J '(c) = 0.

Tlworem (Mean Value).

Lct the function f satisfy the following conditions:

(i) J is continuous on the closed bounded intcrvd [a, b].

( i i ) f is diffel-entiable 01; the open interval (a, b).

Then 3 at least onc point c in (a, b) such that J'(c) = J ( W - J(fl> b - n

Hence a functicm g defined by

I a

through

b

I), and I', i s

describes the vertical difference between the graph of 1) = f ( x ) and the graph of the line seglncnt I',P, g satisfies thc first two conditions o f Rollcs theorem,

Hence applying Rolle's theorem 3 a point c in (a, b) such h a t g l ( c ) = 0. Differentiating J(x) with respect to x gives

gl(x) = f '(x) - [f(b) - f(a)]/(b - a)

Setting x = c w e have with gl(c) = 0

f '(c) = [f(b) - fla)]/(b - a)

SOME APPLICATIONS OF THE MEAN VALUE THEOREM.

1. Determine if each of the following satisfy the conditions of the lnean value

theorem. When it does find all the possible points

(i) f(x) = 3 - 2x2, 0 < x < 1

(ii) f (x )=2x3-x2-x+3 , 2 < x < 3

( i i i ) f(x)=2x3+3x2- 12x+ 1, - 2 < x < I

Solutions:

(i) Qx) = 3 - 2x2 is continuous in [O. 1 J and diffcrentiablc on (0, I)

f '(x) = - 4x 3 f '(c) = - 4c

f(1)-f(0) - 1-3 --- - -2 We need to solve f'(c) = - 2. That is - 4c = - 2 1-0 1

For which c = '/2. Since c = '/z ties in the interval (0, I ) , the conditions of the mean value

theorem are satisfied.

( i i ) f(x) = 2x3 - x2 - x + 3 is continuous in [2, 31 and difirentinblo ill (2, 3 )

f '(x) = 6x2 - 2x - 1

f '(c) = G C ~ - 2c - I

We need to sotve fl(c) = 3 2 which is

Only c = 2.5 1 lies on (2,3).

For this value, the mean value theorem is satisficcl.

(iii) qx) = 2 2 + 3x2- 12x + 1 is c~ntinuous in 1-2, 11 and difl'crcntinblc on (-

f '(c) = 6(c2 + c - 2 )

2 - 1 ~ 4 3 rhat is 3(2c + 2c - 1) = 0 for which c = -- 2

M y c = -112 +d312 lies on (-2, 1). For this value the mean value theorem is satisfied

! Determine the interval in which the function f(x) = 2x3 + 3x2 - 12x + 1 is strictly

nonotone

;ohtion:

;or the problem we determine the interval for wliicli f '(x) > 0 or f ' (x) < 0

f'(x)=6x2+6x-12= 6 ( x 2 + x - 2 )

= 6(x -1)(x + 2)

Either(i)x-1 < 0 and x -1- 2 > 0

OR (ii)x-1 >Oandx+2<O

Case(i)x- 1 <O=rx< 1 a n d x + Z > O a x > - Z

-2 < x < 1 is a solution

Case (ii) x - 1 > 0 =r x > 1 and x 4- 2 < 0 implies x < -2 and for this no solution exist.

Therefore, in the interval -2 < x < 1, the function is monotone decreasing.

B. f (x) > 0. Then

Either(i)x-1 >O andx+Z>O

OR (ii)x-1 < O a n d x + 2 c O

(i) x - 1 > 0 and x = 2 > 0 imply that

x > 1 and x>-2.

Solution here is x 1

(ii) x - I < 0 and x + 2 < 0 imply that

x < 1 and x < -2.

Thus the function is monotone increasing i n the interval (a, -2) and ( 1 , cx,).

Exercises:

Find the intervals where each of the following functions is monotone increasing or

monotone decreasing.

1. f (x)=2x3-4x2+I

2. f ( x ) = x 4 + 4 x 3 + 6 x 2 + 4 x - 4

3. f(x)=x4+8x'- 10x2+40

SEQUENCES OF FUNCTIONS

Definitions:

1. A sequence @, (x)}is a scquenre of functions from X to e if and only if

for each n f,, (x) is a function from X to e. In other words a scqucnce of

functions {A, }is a single valued correspondence betwcen the positive integers and a collection of functions.

2. Let {f,#}be a sequence of fbnctions such 11131 each .I, has the domain E

where E C 91. The sequence {~l}cooverges to function f on sct II if the

domain o f f contain E and if for each point p, E E, the sequence of nulnbcrs

converges to the nunlbcr ( ) , Function 1 is h e l in~it of

3. The sequence {L/,, )is said to convorgc pointwisr if and only if for all x i l l

X the real number scquence (I, (.r))convcrgcs.

In the case converges pointwisc on X wc gct a function

f *:x +a given by

f * (x) = limf,,(x) 11-+m

In the E-notation this means that given any E>O 3 I,(€) such that for dl n > n ( ~ ) x ) .

1 L ( x ) - f , * ( 4 <€

We now consider some exa~nplcs of scqucnccs of functions

Solution: f,,(o)= 0 for all n f;,(l> = 1 for all n

Therefore lim f,, (0) = 0 nttd lirnf,, (1) = 1 #I+&

!,+a)

we consider x, E (0, 1). For this casc

Hence if E > o is given thcre is an intcgcr n, such that if n>n, nx,," < E and ' hence

lim x," = 0 and the sequence b, } = c ~ " } converges to thc function 11-* 110

f i r x e[0,1) for x G l

fir instance if E= tllcn we can choosc no = 3 1 and 11 1 2 and wc l w c

First we note that f o ~ each

x E [O,l]gl,(0) I gll(x) 5 gll(l) since . x 5 1

in this internal. But {i} convergcs to dero. Ilence thc sequence of functions (gn ( x ) ) converges to the function g given by

g(x) = 0 for all x E [O,lJ

We note that if we consider this sequence for x ~ [ 0 , 2 ] wit11 E= & we can choose no = 11 for x ,< 1 but this cannot satisfy the convcrgence condition for

x = 2 since161 #$. We can however choose 11, = 21. 'I'his shows that n, depends on x. This is pointwise convergence for which 11, = no (E J). Wc find

also that for x E [o, 21 and E= &

no = 2 0 is okay for all x in [O,Z] bccausc IL(< . In this case our no is 201 100

independent of the cho ik of x. This leads to another type of convergence - uniform convcrgcncc.

Definition:

The sequence (I, ) is said to converge uniformly on a subset S of X if and only i T for any

6s o 3 n ( ~ ) .wci~liaal /.(I) - / (111 .rr; Pa. rill r r: S nt~ri 11 :. ,~ (e )

where n ( ~ ) does not depend on x.

Example 1

Let f;,:(o,l) + '31 defined by Jl (x) = - for all ri E N .A- E (0,l) r1.u + 1

{r,r'+ collvcrgcs to 0. for x E (0,1), the real number sequence -

Therefore the sequence (/,, } converges pointwisc to zero function. i.e. lim S,, = o pointwiie on (0, l),we now want to find out if thc scquencc

I,+ 011

{&)converges uniformly. Suppose it does. Then by definition given E>O 3

n(~) such that for all m > n ( ~ )

This is equivalent to saying that 1

for all m > n ( ~ ) , - <E for all x~ (0, 1) mx+ 1'

i.e. for all m > n ( ~ ) , I < ~ ( m x + l ) for all X E (0, 1)

1- E u for all m > n ( ~ ) - < x for all X E (0, 1 )

E nr

This is an absurdity.

For instancc if E= & and nl = 20

ITx = 0 this result says 4.95 0

If x = 1 this result says 4.95 < 1. Is~possilsle!

This implies that on (0, 1) the functiotl does not convcrgc unifortnly. We note of course that this scqucncc converges uniformly in [a, 11 for any

a: O< a < 1 because in this case if n ( ~ ) is so large that

1- E < a then for all m > n ( ~ )

E. n ( ~ )

1- E' -<nix for all x E [a, I ] E. )?I

To see this suppose a = 0.0009 and E= & and n = 1000000 we then have

we observe that for this sequeticc of function --- {A I }

1 Thus S,, ( x ) = - is a function riot uniformly convergcllt in the

m x + l interval [0, I].

For c E [a, 1) a >o we have

We thus state t l ~ c result

Result

Let for each n,S,,:[n,b]+e. I f the scqucllce (f,l}convcrpcs uniformly on

[a, b] to a function f , then for all c E [a, b].

Thus for sequence of functions which converge unifonnly we can intcrcllange limits. This is a working theorem which can bc uscd.

Exercises

Discuss the convergence and uniform co~lvergence of 1l1e following sequences:

2. f,, : 9 Z -+ 91 defined by .. 211

The Csuchy Criterion for Scqucnccs of Functions.

Theorem: A necessary and sufficient co~idition (nasc) that n sequence

(I, (x))converges uniformly to f (x) on a set E is that, given c>o. there is an N such that if rn > N and n > N, then for all x in E

If,Ax) -f,,(x)I <E

Proof:

If @,,(x)) converges uniformly to f in I! there is on No sc~cb that if m > No

and 11 > No, then for all x E E

Hence

Now suppose the condition holds. Let x, be a fixed point in E. If E >o. tl~erc is an N such that if 111 > N then

If,,, (XI> ) - fl, (xll )I <€

Hence by the Cauchy theorem the xquence of numbcrs (/,,,(x,,)}convcrges

to a value f (x,,). This argument holds for each x E E hence wc conclude that the sequence(/, (x)Jconverges to the function f (x). We now, show that

Take n fixed. For cach x E E, there is an intcgcr m(x) which dcpcnd on x i l l

general such that m(x) > N and 1 f,,,(.,) (x) - f (x)I <

Thus for all x G E if n N

We rww give a dctailcd treatment of some scqrrcnce of functions. Wc discuss tllcir convergence anti uniform convcrgcncc nrltl wc scc :I rclnt ionship to continuity.

Examples

for convcrgence we must have for given E > o

x(l- E) Thus we can take our no to be the first positive integer greater than

E

This estabilishes pointwise convergence since in this case ,I,, = n,,(e.x). But x(l- E)

we note that since x assumes the maximum value k is tnaximun~ €

for x = k in which case no becomes independent of x. We thus concludc that k(1- E)

when I?,, = + 1 we have uniform convergence. e

IIx=O f,!, (XI =.O giving J ( x ) = 0

Hence f ( x ) = o - k < x l k

The condition If,, ( x ) - j ( x ) [ <E 3; v e

If we now take 1x1 = y then

For x + o we liavc (I < y 5 k ,and tllcn

1 1 provided n>- r e & n>-

fY 4x1

For any particular value DF x = x , # o we can take no such that

1 if E= - we havc that

n l 4

nx i f x = o CE is true for all since

We note here that n,, = (E,x) and that it is not jmsible for pick an 11,

I independent of x since - --+ a0 as x -+ o €I,V~

Hence lim J (o ) does not exist ),-)I)

I I when x + o J ( x ) = - -+- as n-+a

;+x x

I Thus f (x) = -- x > 0

X

The condition - - - I,:,, :I=

Hence we can take n,, = - - + I of coursc os x -+ 0 1 1 , -+ a, (not X E X

strange since the sequence dues not convcrgc at x = o)

The sequence therefore converges pointwise on (o, k].

But if wc consider ille interval f i x i k , o < t < k we e m toke

n,, =- ' [ I -- i ) + l r s x c t t E t

I For example given E= - and t = 1 we lmvc

10

This will happen if we take 11 = 10 since

so that for all 11 2 10 tlic quantity

Thus n,, = no(€) and unifornl convergence is established.

for x = 1 lirn x" = I , I - , ,3

This limit function exists for all values of x in the given rangc and has only ordinary discontinuity at x = 1 .

The conditio~i

lor (a) wc m s t have

Hence it is impossible to pick 11, indcpcndcnt of x in [o, 11.

lim f,, ( 0 ) = 1 lim f,, ( x , ) = 0 I I - P I I n+n

Here the limit function is again discontinuous as in (4) above and 11, cannot be independent of x so that uniform convergence is impossible.

6. The sequence 6"-' -x" ) o S x l l

and since in the given interval x S 1 pointwisc convergence is eslablished. For uniform convergence if at all

E so that x" < ,-

, -1

so that n log x < log - ($2

as x -, o n -, a, and there is no uniform convergcncc

Uniform Convergence a d Contiiuity

We note from the above exaiilples that in tlic iiitcrvnls of uniform convergence the limit functions are contiiiuous. Thus it bcconlc clear that if &(x) is contin~ious then uniform convergence is n sufficiciit corlditioll to ensure that f (x) is continuous.

The condition is of course not ncccssary for i n exariiplc (2) f (x) is

continuous in the interval - k r; x 5 k but { ~ ( x ) ) i s not unifooni~ly convcrgcnt in any interval which includes x = 0. 'This relatioilship bctwcen continuity and uniform convergence leads us to the resr~lt statcd bclow.

Result:

If the sequence {fl(x)} converges uniforinly on [a, b] to a function f (x) and

if each function $(x) is continuous on [a, b] tlm~ if ( s ) d ~ lrxisls and

,We give the proof of this theorem Inter.

We now want to investigate t l ~ c questions: (i) If each f,; and the limit function j' are itltcgrablc on [n, b] docs the

sequence of numbers fltl (x)dx )converge to tlic nunlbcr f f ( x ) d ~ ?

d! , (tf (ii) If - exists for all 11 and if - exists, docs the scqucnces of l'unctions dx th

4 {%} converge to tI,e finction - 7 clx

We know from.our examples at thc bcgir~iiilig of section above that for xtt

g,,(x) = - x €[0,1] n the answer to (ii) is no. The following example shows that thc answcr to (i) is also no.

Consider the sequcncc {f,, } given by

22r' x for. 0 l x S - I 2"

S,,(x) = 2"(2 - 2"x) for 1 O

I k x < ; , 2"

f o r I 7 5 X S l

since J ( o ) = 0 and J(1) = 0 for all n then

Now suppose x, is any fixcd point in (0, 1). 'I'here is an integer No such that i f 11 2 No

1 - 2,,-1 < *"

Thus by definition of J;I if 1 1 r N o J;l(x,,) = o and lic~icc

Since this argu~nent holds for each fixed x,~_(o,l), thc sequence

. (f,, ( X ) ) C O ~ V C ~ ~ C S on [o, 1 ] Lo f ( x ) whcrc j-(x) =o for all x [o,l].

We again consider the sequence

The sequence converges unil'ormly to g(x) = 0

Thus the sequence of nu~iibcrs converges to thc nt~mbcr

[ m d x .

This leads us to anotllcr rcsult which we state as

If Uic sequence {~~(x)}coaver~es 011 [a, b] to a function f (x), if tllc

djl derivative --- exists and are continuous on [a, b] for all n, nncl if tlic

dx

&ll,

on [a, 'b] and - - - g(x) for cncli =[a, bJ. dx

Before we prove this theorem let us see some applications.

Some Applications

J ( o ) = 0 for all n and J;,(I) = 1 for all n.

limJ;,(o) = 0 and limJ(1) = 1 11-+m ll-bm

Hence @ ( x ) } converges to the function

Note that J ( x ) = x" is conti~~uous on 10, I ] but tlic limit $(x) is not continuous on 10, 11 because it is not c o ~ ~ t i ~ ~ u o u s at x = 1.

I For each x E [o ,I ] , o 5 I ( X ) 5 - 1 Iellce the seque~~ce { ~ I I ( x ) )

I I

converges to function 9 (x) on [o, 11 wlicrc

We note here also that each of the function 9 n ( x ) and the functiou 9 ( x ) arc differentiable at each point x ~ [ o , l ] . Verify this! 'I'hc scquencc

rlgn where - = XI'-' converges to tllc function

dx

dg wl~ich is not the function - = 0

(LY

x I1

3 . I ~ n ( x ) = - on [o; I ] t l 2

1 xoR =

(1 -I- h)"

Thus for all n h,, (x) converges to h(x) = 0, on [o, 11

dhl,(x) x"-' --- - du n

and unifvrndy too.

d l Verify.-- = 0 exists on [a, b] and in f k t

dx dh (x) -- - g(x) = 0 for all XE [a, b].

d~

We can state our experience iu thcse cxanlplcs i n the following tlicorc~n.

Thcorcm:

If sequence {fil(x)}converges uniformly lo function /(x) on n sct 1:. wbcru 1:

is the domain of f , and if each function .fi is continuous at n point p , EE, then function f is continuous at p,

Proof:

Let E>O be given. By hypothesis there is a positive integer N such that if n>N then for all PEE

In particular

Let n, be a fixed integcr set. n,>N. Since Jl is continuous at p, thcre is a

number 5 such that if

'From (1) and (2) with 11 = n, and from (3) wc INVC

that is the functiotl / is con t i~~ ious at p,,

If we note that if a function .f (x) is continuous on [a. b] the11 / is I<icma~ln

integrable that is rf (x)& cxists then we appreciate the following corollary

which we state and prove.

Corollary:

If sequence b} converges unifonnly to a function f on n set E and if each function _f;l is continuous on E, then f is continuous on E. If thc

sequence { f i } converges unifonnly on [a, b] to a function / 2nd if endl

function in continuous on [a, b] tlmn fi(x)dx cxists and

Proof:

Since continuity implies integrability i t follows that the intcgrntlc

I ~ # ( x ) d r all exist because each functions /n is contil~uous 011 [a, b]. Sincc

is a sequence of continuous fuctiuns wliich converge unifoniily to the fimction /. then by an earlicr theorem thc function f is continuous on [a, 131.

So again f i ( x ) d n exists. LC( 5% bc givcn sincc 6 converges unilbnnly to

f there is an N s.t. if n N thcn

Hence if 11 > N

i.e. lim C / . ( x ) d r = j,: f (x)& n-.m

Using a theorem in theory of integration tlint if N is a positive nunlbcr such

that l+(?)I < M for all x~ [a, b] then \If (x)dj 5 Ad(/)- 0).

We now prove the theorem we statcd earlicr.

Theorem:

If the sequence -&,} converges on [a, b] to a function / , if the derivatives

4f converges unifornlly on [a, b] to a function g, then - exists 011 [a, b] and dx

for each x ~ [ a , b].

Proof:

&I

Since each functioll - is conlinuous on [a, b] and the sequence {$} dx

converges uniformly to g on [a, b], tilc~i by an earlier thcorem function g is continuous on [a, b]. IIence using the fact that continuity implies integrability and thc fact that if y E [a , b]], then 1 is Ric~narm integrable on [A, ij] we have

that if x g[a, b] the integral rg(lr)d+ exists. l lcnce

= l in l [~ (x ) - ~ ( a ) ] by fundaniental 'Theorem of Calculus 11-bc.7

= S(x) - S ( 4

since g is continuous on [a, b] then

CkIAPTER SIX

POWER SERIES

TnyIor PoIynomiaI

From the fact that

and 1.r - x,,l << I X - x,,

we have

(X - xo) f ( x ) = f(xc1) I- - &I dx x = x t

Thus a linear polynomial

is a linear approximation of . f ( x )

if

A quadratic polynonlial

G ( x ) = q , + O , ( X - x , ) + q ( x - x, , )~

such that .

gives a better approxinlation 'I'hus an nth dcgrcc polynominl

where

A repeated differentiating of (4) and evaluating at x = x, gives p , , ' ( x ) = a , + 2 a , ( x - x , , ) + 3 a , ( x - x , , ) ' + ....... Ptr ' ( ~ n ) = 01

p I l 1 ' ( x o ) = 2 a 2 + 3 . 2 a , ( x - x , , ) + 4 . 3 u 4 ( x - x , ) ' + ....... P,' ' ' (x,, = 2% Also p,,"'(x,,) = 3!a,

P~~~~ = n! a, , (6)

Hence approxin~ating f ( x ) by p, , (x) gives from ( I ) or (3)

(7) gives the Taylor coefficient and the polyno~ninl (series)

is called Taylor Polynonlial

(8) can be taken as nth degrec approximation of the f~inction f (x)

.from calculus we recall that

1' f l(~)c,x = f (XI - f (x.1

so that we can have

f = f + f 1 (9) J"

A continued integration by parts yield

and generally we have after n processes

Thus approximating a function f ( x ) by Taylor polynomial p,, ( x ) gives an error of

This result is embodied in Taylor t hcorcnl which states

Let f have continuous derivative up to and including order ni-1 on somc internal (a, b) containing the point x,. If x is any other point in (a. b) thcn

where

and

Taylor Series

Let us now suppose that the function f has continuous dcrivativcs of all orders in an interval containing thc point x,. Then it is possible to form the Taylor polynomial.

for arbitrarily large values of 11.

In addition, suppose that for each x in the givcn intcrval i t is possible to sho\v that Iim R,,+,(x) = 0. Then on this interval we can write

l I + Q

where the infinite series converge to the value of f (x) at each point

(1 1) is known as the Taylor series for J about x,, , and ortcn provitlcs convenient representation for the function. If x,, = o then (1 I) bccomcs

known as maclaurin series

Examples

1. Find the Taylor polynor~~ials of dcgrec one, two and three for f ( x ) = 111"

about x,, = 2 .

Solution: 1

f'(x) = - x

since x,, = 2

1 /"(x,J = -7

1 J Y x , , ) = 7

1 f y X ) =-- 2

j-" '(x) = -j- x ' S

f '(x,,) = a, = $

- - 1 - a 2 = 2 ! - 7 - -

X 1 g 3 = - = -

3! 24

2. Find the Taylor polynomial of dcgrcc n for f (x) = e' about x,, = 0

Sol11 tion

For this function f'(x) = /'"(x) = . . . j - " ' ( x )

1 1 1 :. a, = 1 a, = 1 a, = - = - ( I , , = - 2 ! 3! r r !

Exercises

Find the Taylor polynomial of thc indicated dcgrec for the given function about the given point.

X 1. f (x) = sinx; x, = - degree 3

4 '

3. f (x) = e-"2 x, = 0 degree 4

Find the Taylor series for tlie function given about the givcn point

4. f (x)=cosx x,, = 0

5. f (x)=sinhx x,, = O

POWER SERIES rb

A series of the form Ca, ,xn wlierc a,, is independent of x is callcd a 11.0

powcr scrics in the variable x.

Some Properties of Power Series

I. If C denote a powcr series which converge to /(x) for 1x1 < h

then there is an interval of x inside the range (- h , h) in which f (x) never m x

vanishes except for x,, = o e . g I n ( 1 + x) = x (- I)'-' -- r- l I'

11. If two power series z n,,x" ,x D,,r" both convcrgc for 1x1 < h and il

x a,xn =x b,,xn at all points satisfying 1x1 < p then

a,, = b,, 7 a, = b, ,n2 = h ,,....., tr,, = b,, . . . and the series arc irlentical. E.g.

C (n ""-I - l)! n n d z X'I n! '

' 1 111. If iimlu,,l~ = - the power series coilvcrgc absolutcly if 1x1 < p , and cannot 11-+v P

converge for 1x1 < p . The interval ( - p , p ) may be called tlic interval of convergence of the power series, i.e., thc radius of convergcncc is P. Insidc the interval the series converges absolutely, outside the interval convergencc is impossible, at the end points the power series may or may not convcrgc. This is related to the ratio test.

Examples illustrating these concepts arc given below:

The series

... 1. (n + 1)x" = 1 + 2x + 3 x 2 . . converges absolutely for I X I < I , diverges rr= I

for 1x1 > 1 and the interval of convcrgcnce is (-1,l).

Whcn x = f 1 the series bccorncs

1 + 2 + 3 + 4 + ....... for x = l

1 - 2 + 3 - 4 . t ....... for x = - I

neither of which is convergent

x " 2. z-- is a series which converges in the intcrvnl (-1.1).

I n

Wlicn x = 1 tlic series becomcs

For x = - 1 the series becomes

- 1 + L I . , + 7. . 1 .. which is conditionally convergent.

Thus the power series converges at one end of the interval and divergcs at the other. Therefore interval of convergencc is [-1,l) .

w $1

3 . The series Cy convcrgcs' in tllc intcrvnl (-1,l) wlicri x = A I lllc 11.1 I1

series becomes 1 1 1 -+-+-+.... l2 22 32

x = l

both of which are absolutely convergent.

x ' I :. By Abels theorem the series z converges uniiorndy on [- 1.11. n-1 11

1V. If the power series C u , , x M converges for x = h i t is absolutely

convergent for 1x1 < X if we take A =/XI then this is, the 'l'heorctn on page overleaf.

V. If the series does not converge for x = h i t does not convcrgc for any value of x s.t 1x1 > 1 .

VI. If (-p, p) is the interval of convergence of tllc powcr scrics n, ,xV then the series converges unifornlly in any closed interval entirely insidc ( -p ,p )

VII. A power series z o , l x " is a continuous function of x in any closcd interval lying entirely insidc the interval of convergence.

VIII. ABEL'S THEOREM: If the series o , x " converges for 1x1 = p and the series converges at x = p or x = - p thcn the interval of unifortn cdhvergence extends up' to and includes that point, and continuity of tllc sum

function / ( x ) = a , x n extends up to and includes that point.

Application to the Binomial Series

We know that if 1x1 < 1

The series coriverges absolutely in the interval (-I,]) and uniformly in any closed interval lying inside (- 1,I).

As shown earlier wllen x = l mrd r ~ + l > o , the series

~ ( 1 ) ~ ' converges. Thus if rt + I > O the values x = I is a point of uniform r=O

convergence and consequently ( I + x)" is continuous at x = I . Hence if (n+ 1) > o

Also it can bc shown that the binomial serics convcrgcs at x = - 1 if r l > 0 . Thus (1 + x ) " is contirruous at x = -1 and hence by Abel's tlleorem.

Exercises sin(x + m )

I . Prove that the series is uniformly convergent for all real I I ( ~ + 1)

values of x.

1 2. Show that thc series whosc nth tcnn is - , , convcrgcs unXomly

r l -I- 11 x for all values of x.

x 3. Prove that the series whose nth term is ,,-, corlvergcs for all rcal

(1+x ) values of x, but is not uniformly convergcnt in any interval which contains tllc point x = o.

1f a power series a,,xl' converges for x = x , where x , t o . then the series

converges absolutely for each x such tliat 1x1 < 1.t-,,I .

Proof

00

b r each 1x1 < /rf1 1, the series MI:/ is r convergent gc,nmetric series. ll-lla t I

Hence by comparison test the series n,,xl' converges arid Bence the series

. C a l l x " converges.

The least upper bound (lub) of the set S on which n given power series converges is the radius of convcl-gcncc of the power series. If the series converges for all real x, we say that the radius of convergence is infinite.

The radius of convergence R of ,, a,$' is show that

rclnted to proof of thc root test,

I3nd the radius of convergence of each of the following power series. If the radius of convergcncc is finite, investigate the convergence of the series at the end points of the interval of convergence.

5 . x (infinite) I l l 1

2 1 l t l

2. (infioite) , I = 1 11 !

(It = 0)

Revision Exercises

1 . Dcfinc radius of convcrgencc of n powcr scrics. Find the rndius of convcrgencc of the following power serics and investigate their convergence at the end points in the case of finite radius of convergence.

x i i . C7

n

2. Prove that i f the power series x o , , x l ' hns radius of convergence R and

defines n function f ( x ) = 2 a l l x " in (-R, R ) , then the function f ( r ) is differentiable at cach point x E (-R, R)

i.c., t l~c power scrics al ,x" can bc differentiable term by term at each point x E (--I( , R)

(C " , , X 1 ~ ) = C ( q , x l ~ )

Diffcrcntiation and Intcgriltion of Power Scrics

IS k r , y]is a c los~d intcrvel contained i n (- R, I ) W ~ ~ C K C II is the radius of

convergencd of a power series C a,x" and if f ( x ) = C n , l x " , then jl;f(x)dr exists and can bc obtained by integrating thc power scrics term by tcrm, i.e.,

Proof

#Inoo fir,r$t-d, H I , llraro lci # p~sl lhu I I H I I I ~ ~ ~ n s.1. u s r * 8 nod - R c - r i o . It thus f o l l o ~ ~ from the property of power series [If x a , , x n

converges for x = x , ondx,, + o and if r is a positive number st. r < Ix,,(,

then Z a , , x " converges uniformly on (-r, r)] that the series x a , , x "

converges uniformly on [-r, r] and hence on [a, y]: Therefore by an earlier

theorem on series x n , x " is termwise integrable.

Example for application

m x 2" cosx = C(-1)" --

11 '11 (2n) !

State Abel's the or en^.

Show that the function (I +x)" is continuous in the closed intcrval [ - I , I ] an d that for n -k 1 > o

Tho radius of convergence of the powcr series ,,a,,xn-' which is obtained

by differentiating the power series x o , , x " term by term, is equal tot he

radius of convergence of the power series x anxu .

Thcorcm

I f the powcr series x ol,x" has radius of convergence R and defined a

f~mction f (x) = n,Ix" in (-R, R) , then the function f (x) is differentiable

at cach point x E (-H, R) ntrdjl(x) = ~ t u 1 , x 1 ' - ' i s . , the power series

a$' can be differentiated term by tcrm at each point x E (-R. R ) .

Ir / (x ) = [ln,yt1 for x E ( - R , R ) tlie~i/(.~) has derivatives of all orders at cnch point x E (-H, R ) .

Tlicoren~

If f (x) = ( r , , ~ " for aL1 x E (-R, R) wllcre R is a positive number, then

f ""' (0) l'he seiies x" whether it converges or not is cnllcd the Maclaurin

17 ! series off . More gencldly

is called the Taylor series of f and k = P(X")

k ! are called Taylor coefficients.

-

Proof

( n + 1 ! (n + Z)! f '"' ( x ) = n ! a , , + - n,,x + 2 0, ,+2x + .....

1 ! 2 !

Identity Theorem for I'mver Series

Suppose tlmt for each x E ( - r , r ) l wllcre r is a positive number, the power

scrics ~ ~ , , ~ ' ~ * n d ~ b , ~ ~ ' ~ convcrgo and are equal

2 2 i.e., n,, + rr,x + t r2x -I-,.. ...= b,, + h , x + h ,x .t ........

lor each x E (-r, r) 1. Then for 11 = 0,1,2,. . . . . . .

o r x ( - ) tlic (b~ictio~i f ( x ) dcfincd by f ( x ) = x nt ,x" is also

described by b,,x" ix., f (x) = h,$' for each x E ( - r ) llence by abovc h o r e m

If the function f is defined in some neighbourhood N,, of x = x,, and if f can be expressed as a power series in some neighbourhood N, of x, where N , c No i.e., if x E N I , then

f (s) = &,l(x-xn)"

then the fi~nction J' is analytic at x,,

Solving Differcntial Equation by Using Powcr Scrics

Ion Consider the differential eqmt'

d2y dy 2 2 x ~ ~ + x - - ( . Y - m ) y = o ~ E R Bessel equation dx cix

Assume powcr scrics solution:

y ( x ) = cr,,xM in intcrval (-R. R). 11-11

tly c12 - -- , , ...... cx~sts 'l'lrcol-ctu 1 ot' pagc 46 says tlrcy colivcrgc in c lx t lx

(-R, R) and also theorem on page 49.

1 1 ~ ~ dy 2 x i- + - - (x2 - 111 )y pcriuittcrl st1111 of scrics convcrgcs

tlx tllv

Use identity thcoreun.

Trigonometric Serics and Fourier Series

A trigononlctric scrics is an infinitc scries of tlic form

ol

+ C (a,, c o s m + /I,, sin 1r.r) 2 ),=I

( I , , , ( I , , b, , .... .. . .. . given constants

I f the triyonomctric Scries converges for all x s [o,2rr] then i t converges for all real x , since Vt7 the function costtx, sin ruc both have period 2n .

Suppose that the trigonometric Series converges uniformly on [o,2rr ] (and hence for all real x) to a fi~nction J ' ( x )

l' ,I J ( x ) cos IZX = - cos nx + C n , cos nix cos trx + b , sin t x cos atx 2 1r= I

since the convergence is uniform then the series can be integrated term by lcrm in particular

(+) f ' J ( x ) cosrixd~r = Ifn 2 cos r~xdx +

= O t t t f t ? f K COSXt7 COS NIX if

= 2n t?1 = 11

1 2 - -+ f j(l) C O S l l l ~ ~ = a, , 111 = 1,2 ,......

TC

Multiply the series by sin mx and we have

Integrate the series itself term by term

Thus wc huve'provcd that.

If the trigononletsic series

converges uniformly to j ( x ) on [0,2n], t l m if t ~ t 2 1

1 ?n

n,,, =, - I /(x) cos mx (iX 7C

Definition

Suppose that fn f ( x ) k exists. IIle fourier series associated with the

function f ( x ) or the Fourier series f (x), by the trigormnetric series

wllerc a,,,o1,,l4,(rr = 1,2,3 ,.........) are dcfincd above, and are cnllcd the Fourier coefficients of f (x) , i.c.

0 - /(.v) = + z coo n.v + b,, sin m-)

2 11=*l

A rlccessary and sufficient cot~ditiou that the Fourier serics of f ( s ) coi~vcrgcs is that the limit

exists.

A neccssnry and sufficient condition that the Fourier serics of .f (x) converges (uniformly in x ) to /'(x) is that the integral bclow converge (unifor~nly in x ) to o as n increases without boiind

.f(x + 2 r ) -I- J'(x -21) dl

n 2 sin r

Exnlnple of Fouricr

SOLUTION OF TYPICAL EXAMINATlON QUESTIONS

1. Given a sequence (f,, ) of a complex valued function on a set x , what is meant by saying that the sequcnce (f,, ) converges

(i) pointwise on a subset S of X (ii) uniformly on a subset S of X?

Show using these conccpts that the scquence of functions Cf',,) given by

J;) ( x ) = x i ) OI? 0 < X < X

col~verges pointwise but 1101 uniformly

A scqucnce ( , I ; , ) converges pointwisc on a subsct S of X if givcn an arbitrary

positive number E 3 H ( E ) .s, i , Vn > t i ( € )

( x ) ( x ) E for each x ES

The sequence (f,, ) is said to converge uniformly on a subset S of X if for all , given E> 0 3 t 1 ( ~ ) S. I . I/,, ( s ) - /(x)I <E V.T E S and 11 > n ( ~ ) where n does not

We considcr for rirnclion J;, ( x ) = x " tllc scquctlce

Pointwise Consequence

. f 1 , ( 4 = 0 . f A V = l

1 For any x E ( o , I ) put x = - h > o

l i - h

:. thc scclucncc (f,, (x)) = (xu ) corivcrgcs pointwisc on the set

lo, 11 to the f~~nction.

Uniform Convergence

Givcn an 2 > 0 we must have lx" - ol <s Vx s [o,l) Vn > ) I ( € ) ......( ( 1 )

and /x" - ol <E for x = 1 ...( b)

For (a)

I.C.,

1 nlog - >log- (:) e

log : i.e. 11 > -

los(:)

n u t as x -t 1 lo&)-+ o and 11 + m

Then we see that i t is impossible to pick no independent of x. I-Ience we

co~icludc that thc scquepce {xu) converges pointwise but not unifonnly on

Lo, 11.

2. What do you undcrsland by t l ~ c term "termwise inlegrat io~ddi ffcrentintion 01.3 series of coniplex-vali~ed fimctions"? Show 11131

11 sill 1 1 s i.) the series

L' I ) can bc intcgratcd terwise on the subsct [o,n] of

1 1 = 1

ii.) 2 nsinrm 2e -- -- , en c2 - I

iii.) z x e - " ' I is not "ternwise diffcrentiable" I?= I

Solution

i.c., we can interchange integratioil/differentiation with summation.

We now consider the problem: the series

and see if we can integrate term by term. IIPI

Using the ratio test we check for the convergence of the series of number

n :. The series xF converges absolutely and hence by Weierstrnss M-

n sill la test the series z-- converges uniformly and we thus conclude that the

1 1 = 1

n sin 1 7 , ~ series x can be integrated term by term.

I#= I e

n sin t7x (ii) Since can be integrated termwise it follows that

,,-I

1 11 sin t lx tz sin tix =I[ ,,I dx , e n I)= I

1 1 1 - - - -t - + 7 -t.. gconletric series with r- = - e e e 3 c e

, (iii) r e " "

For x = o

For x g o -&p2 - - xe-ma 1 1 [I + -p + -p+... geometric series

r r -1 I -52 1

= xe 1 , 0 -.x2 sum of geometric series

. . Cxe-"" converges to t!le sum furlction

1i1n.f (x) = lim .r + t, n-,u 1 - e - ~ 2

I'hus the sum runction is not continr~ous at x = o and hence

This shows that the scrics xe-"" cannot be differentiated ten11 by term. r r = l

Show that tllc lilnction (1 + s)" is conti~lr~ous i l l thc closed intcrvnl [- 1 , I ] and llint for 11 i- 1 2 o

Solution

Abel's Theorem states: If the series a,,x" converges for (x j < p and the series coilverges at x = p o r x = - p then the interval of uniform convergcnce extends up to and includes that point, and continuity of the sum function j(+) = x o , , x ' extends up to and includcs that point.

We consider the function (1 + x)"

From the Binomial Tileorem we have

for I x l < l

We s l~ow that the function (I +x)" is the sum function of the series

'Then for 1x1 < I

= 1x1

the scries convcrgcs absolutely and diverges for (,I. I . t i r * , 1 1 + 1

For x = 1 ---.- - r

1 ' l r

l3ut if 11 + 1 is positive and I . > 11 + 1

In this case I I so tlic absolutc magnitude of tlic tcrrns steadily

decreases. Since limlr,rl= o i t follows that for (n+l) > o the binomial series r-)m

convergcs at x = 1.

Now if n > o i t follo\vs that the series converges at x = -1.

. ( J X r converges uniformIy on 1-1, 11.

'I'hus by Abel's theorem thc function

(1 +x)" is continuous on [-1,1]

i c e (')xr convcrgcs to (1 -b xI1l at x = 1 . r -0

It follows that

4. Discuss the convergence of the sequcncc (11,~ (.(I)) given by

Is this convergcnce u~iit'orm'? Justify your answer. What can you say about

~11111 (x) clh,, (x) the convergcncc of the scquciicc ( - ? DO" ( ) col~wrge to

tll1(x) wlicm h ( x ) is the ftiilctioo to which the sequence (x)) cooverges'?

c /I

.'. h,, (I) = 0

We check for x E (n,l)

we check for unil'orm convcrgcncc

I f ( / l , , ( x ) ) is uniformly convergent 01-1 [o, I] we rilust have

2 - 1

nI1 2 --7- since on [o,l]lxl I I -

E I'

Pick n,, = - + 1 11: I Iencc thc convergence is unifun~l

d x1I-l

- hl1 (x) = - we consider clx 11

11- I X

- 0 For x = o . - - n

Consitlcr s,, E (o,I) we I I ~ C h a [

1 X,, = -

I 4- k

x,," = - - (I + k)" ~ ( I I - l ) k 2

I + 11Il -I- ------- + . . . . + I ? 2 !

converges to the function g(x) = o on[o,l] We test for

, uniTonn convergence

"XI nn > 7 for [o,l]lxl< 1 -

E"

1 Choose n = - + 1

E

Hcnce wc see that tllc scquence

lo, 11 Lo function we see that

uniformly on

d (' ,711 (XI) converges to - h(x) r lx

5. (a) Define radius of convergence of a power series. Find the radius of convergence of each of the following power series and investigate their convergence at the end points in thc cases whcre the radius of convergence is finite:

(b) Prove that if the sequence @,,) converges uniformly on [a, b] to n function

.f and if each function f,, is continuous on [a, b] then (x)tlx exists and

Solution

(n) The rndius of convcrgencc of a power series is the least upper bound of the set on which the givcn power scrics convcrgcs.

The rndius of convcrgence R of ol1xn is given by

I - I

iim/ u,, / 1 1 - 11- )+n

- ( i n ) 2 l l -b . , = 2

Wc now tcst convcrgence ;it the points x = 2 and x = -2

s wllcn x = 2 the series C - becoii~es

# # = I 2"

1 1" = 1 + 1 + 1 + I.. . which diverges 1 1 = l

( i i ) By ratio test l i n ~ I l l -+ 01

x"., 112 - x

(11 + 1)'

Hence the series converges for 1x1 < l ,i .e, R = 1 We test for x = 1 and x = - 1 . 1

For x = I the scrier is zi which by Comparison with the p-series 1 1 - l I f

converges.

For x = - I the series beco~ncs :$?

Wc tcst convergence ~isiug J,icbllik test. WC note that

1 Iencc by Liebnitz test series converges at x = - 1

(b) If the sequence (f,,) converges iu~iformly on [a, b] to a function f and if

each function f,, is continuous on [a, b] tthcn f~(x)dx exists and

~ i n c e O;,)is n sequence of conti~~uous functions which converges uniformly to tllc lilnction J' tllcll f is continuous on [a, b].

So again jff(x)dr exist.

Let E > 0 be given. Since (/,,) convcrgcs uniformly to /, there is an N s.t. if n>N

INDEX

Abel's theorem 87, 88, 92, 104, 106 Absolute convergence 42-44, 86, 88, 100 Absolute value 4

Bounded 25,55 Boundedncss 3 ,4 Bounded set 3 ,6 ,9

Cauchy 17- 1 8,65-66 Continuous 46-47,55, 76, 77, 87, 106 Continuity 48,50-52, 72, 78 Convergence 14- 1 5,22,28, 32,40,43, 72, 88,99 Conditional convergence 42

Ikdckind tl~corcrn 10, 1 1 Uifircntiahility 53 Differential cquation 35 Discontinuity 46-49, 52, 70 Divergence 35-44, 105, I I 0, I I I

Fourier serics 98

Identity t heorem 94 Improper integrals 30-3 I Infrmum l , 2 Intervals 3, 29

- closed 3, 55, 92 - open 3

Least upper bound l , 2 ,26 Liebnitz test 42, 1 I2 Lower bound 1,2

Mean value 56, 57-59 Monotonic 24,25 -increasing 24,26 -decreasing 24

Power scrics 80, 85 - diflcrcntiationof9l,101, 102 - integration 01'91, 101, 102

p-series 4 1-42

Radius oScorwcrgence 90, 93. 1 10 Ricmann integral 77, 79 Rolle's theorem 55, 56 Real number systeni I -complctcncss propcrty of 9 -density property of 10

Series 87, 105 - alternating 42 - binomial 87, 105 - geometric 42.45, 89, 103, 104 - of rca1 numbcrs 12-26

subsequence 27 sliprcrnrm I . 2.

'l'aylor cocflicicnt 8 1 - polynomial 8 1 - series 83 - thcorcm 82

'l'csts fix convergcncc 33-35 - convcrgcncc tcsl 34. 37. 4 1 - conclcsation test 35 - integral test 33 - Raalx's tcst 35, 38, 39 - Ratio tcst 35, 38, 102 - Root tcst 34,36 - Logaritllmic fcsf 35, 38-40

'I'riangle inequality 7. 9 Trichotomy law 10 Trigonometric series 95-97

llnilbrrn convergcnce 62, 09, 7 1, 72, 78, 88. 96. 100, 1 12 IJppcr bound 1, I0

Wicrstrass M-'l'est 1 02 Partial sum 3&33 Pointwise convergcnce 60.67, 7 1