Calculus & analytic geometry

Transcription

1 Calculus & aalytic gomtry B Sc MATHEMATICS Admissio owards IV SEMESTER CORE COURSE UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION CALICUT UNIVERSITYPO, MALAPPURAM, KERALA, INDIA

2 School of Distac Educatio UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION STUDY MATERIAL B Sc MATHEMATICS Admissio owards IV Smstr CORE COURSE CALCULUS & ANALYTIC GEOMETRY Prpard by: SriNadakumarM, Assistat Prof ssor, Dpt of Mathmatics, NAM Coll g, Kal likkady Layout & Sttigs Computr Sctio, SDE Rsrvd Calculus ad Aalytic Gomtry Pag

3 School of Distac Educatio CONTENTS Modul I Natural Logarithms 5 Th Epotial Fuctio a ad log a 8 4 Growth ad Dcay 4 5 L Hopital s Rul 9 6 Hyprbolic Fuctios 8 Modul II 7 Squcs 45 8 Thorms for Calculatig Limits of Squcs 5 9 Sris 5 Altratig Sris 67 Modul III Powr sris 76 Taylor ad Maclauri Sris 8 Covrgc of Taylor sris- Error Estimat 88 Modul IV 4 Coic Sctios ad Quadratic Equatios 9 5 Classifyig Coic Sctio by Ecctricity 95 6 Quadratic Equatios ad Rotatios 98 7 Paramtrizatio of Pla Curvs 8 Calculus with Paramtrizd Curvs 5 9 Polar Coordiats Graphig i Polar Coordiats 5 Polar Equatios for Coic Sctios 8 Ara of Polar Curvs 4 Lgth of Polar Curvs 7 4 Ara of Surfac of Rvolutio 9 Calculus ad Aalytic Gomtry Pag

4 School of Distac Educatio Calculus ad Aalytic Gomtry Pag 4

5 School of Distac Educatio MODULE I CHAPTER : NATURAL LOGARITHMS Th atural logarithm of a positiv umbr is th valu of th itgral writt as l i,, Rmarks If l dt, t dt t (), th l is th ara udr th curv y / t from t to t For, l givs th gativ of th ara udr th curv from to For, l dt, t as uppr ad lowr limits qual 4 Th atural logarithm fuctio is ot dfid for Th Drivativ of y = l Usig th first part of th Fudamtal Thorm of Calculus, for vry positiv valu of, d d l dt d d t It is If u is a diffrtiabl fuctio of whos valus ar positiv, so that l u is dfid, th applyig th Chai Rul to th fuctio dy dy du d du d y l u ( with ) u givs or simply Problm Solutio d l u d l u du d du d Evaluat Usig Eq(), with d du l u d u d ( ) u, d l( )( ) d d d Calculus ad Aalytic Gomtry Pag 5

6 School of Distac Educatio Proprtis of Logarithms For ay umbrs a ad, l a l a l (Product Rul) a l l a l (Quotit Rul) l l (Rciprocal Rul) 4 l l (Powr Rul) Thorm l a l a l Proof W first ot that l a ad l hav th sam drivativ Usig Corollary to th Ma Valu Thorm, th, th fuctios must diffr by a costat, which mas that l a l C for som C It rmais oly to show that C quals l a Equatio holds for all positiv valus of, so it must i particular hold for Hc, Hc, substitutig l( a ) l C l a C, sic l C l a C l a, l a l a l a Thorm l l a l Proof W us With a rplacd by / givs hc l l l l l rplacd by givs l a l a l = l, a l l a l a l l a l Calculus ad Aalytic Gomtry Pag 6

7 School of Distac Educatio Thorm l l (assumig ratioal) Proof: For all positiv valu of, Sic l d d l() d d ad l d ( l) d, usig Eq () with u, hr is whr w d to b ratioal hav th sam drivativ, by corollary to th Ma Valu Thorm, l l C for som costat C Takig, w obtai l l C or C Hc th proof Th Graph ad Rag of l d Th drivativ (l) is positiv for dt, so l is a icrasig fuctio of Th scod drivativ, /, is gativ, so th graph of l is cocav dow W ca stimat l by umrical itgratio to b about 69 ad, obtai ad Hc, it follows that l l l l liml ad liml Th domai of l is th st of positiv ral umbrs; th rag is th tir ral li Logarithmic Diffrtiatio Th drivativs of positiv fuctios giv by formulas that ivolv products, quotits, ad powrs ca oft b foud mor quickly if w tak th atural logarithm of both sids bfor diffrtiatig This abls us to us th proprtis of atural logarithm to simplify th formulas bfor diffrtiatig Th procss, calld logarithmic diffrtiatio, is illustratd i th comig ampls dy Problm Fid whr d Solutio Giv cos y (si) cos y (si) Takig logarithms o both sids, w obtai l y cos l si Calculus ad Aalytic Gomtry Pag 7

8 School of Distac Educatio Now diffrtiatig both sids with rspct to, w obtai d l(cos y d l si)(cos) lsi d cos(l si) d d d d d d si l si cos(si) si d dy i, cot cos si l si y d dy d y cot cos si l si dy d cos i, (si) cot cos si l si Problm dy Fid, whr y d Solutio Giv / y Takig logarithms o both sids, w gt l y = ½ [l ( + + ) l ( + )] Now diffrtiatig both sids with rspct to, w obtai d dy d y d dy d y d d or dy d Th Itgral (/u)du / / If u is a ozro diffrtiabl fuctio, du l u C u Proof Wh u is a positiv diffrtiabl fuctio, Eq () lads to th itgral formula du l u C, u If u is gativ, th u is positiv ad Calculus ad Aalytic Gomtry Pag 8

9 School of Distac Educatio du d() u u () u l() u C W ca combi th abov quatios ito a sigl formula by oticig that i ach cas th prssio o th right is Proof l u l u bcaus u ; l() u l u bcaus u l u C Hc whthr u is positiv or gativ, th itgral of (/)u du is l u C This complts th proof W rcall that u u du C, Th cas of is giv i Eq (9) Hc, u C, u du u, Itgratio Usig Logarithms Itgrals of a crtai form lad to logarithms That is, whvr f () giv for it () f d l() f C f () Problm Evaluat d 5 Aswr 5 is a diffrtiabl fuctio that maitais a costat sig o th domai du d, 5 u lttig u 5, du d, u () 5, u() l l 5 l l 5 l 5 / 4cos Problm Evaluat / si d l u 5 Calculus ad Aalytic Gomtry Pag 9

10 School of Distac Educatio Solutio 4cos d, si du takig u si, du cos d, u / 5 / l u 5 l 5 l l5 Th Itgrals of ta ad cot Problm Evaluat Aswr ta d ad cot d u ( / ), u( / ) 5 si du (i) ta d d cos, takig u cos, du si d u du l u C, usig Eq (9) u l cos C l C, by Rciprocal Rul cos l sc C cos d du (ii) cot d si, takig u si, cos u du d I gral, w hav l u C l si C l csc C ta u du l cosu C l scu C cot u du l si u C l cscu C Problm Evaluat Aswr / 6 ta d / 6 / du ta d ta u, takig u, d du /, u(), / ta u du l sc u / (l l) l u( / 6) / Calculus ad Aalytic Gomtry Pag

11 School of Distac Educatio Erciss I Erciss -6 prss th logarithms i trms of l 5 ad l7 l(/5) l 98 l l5 5 l 56 6 (l5 l(/ 7) /(l 5) I Erciss 7-, Eprss th logarithms i trms of l ad l 7) l 75 8) l(4/9) 9) l(/ ) ) l 9 ) l ) l 5 I Erciss -5, simplify th prssios usig th proprtis of logarithms si l si l 5 4 l( 9) l I Erciss 6-5, fid th drivativs of y with rspct to, t 6 y l 7 si(l) cos(l) y l() t 8 y y l sc(l) 4 5 ( ) y l 5 y l y l 5 y l / l(4) 4 l tdt or, as appropriat 9 y l( ) l t y l t I Erciss 6-, us logarithmic diffrtiatio to fid th drivativ of y with rspct to th giv idpdt variabl 6 y ( )( ) 7 8 y (ta) 9 si y sc ( )( y ( )( ) Evaluat th itgrals i Erciss d 4 d l / 8rdr 4 4r 5 7 cot t dt 4 / 4 6 d l / ta d 4 y t( t ) y t( t )( t ) y / ( ) 5 ( ) 4si d 4cos sc y ta y dy sc y sc d l(sc ta) Diffrtiat th followig prssios i Ercis 4-47 with rspct to 4 l 6 4 (l ) 44 l (ta + sc ) 45 l ( ) 46 (l ) 47 sc l, Calculus ad Aalytic Gomtry Pag

12 School of Distac Educatio CHAPTER : THE EXPONENTIAL FUNCTION I this chaptr w discuss th potial fuctio (it is th ivrs of l ) ad plors its proprtis Bfor givig formula dfiitio w cosidr a ampl Th Ivrs of l ad th Numbr Th fuctio l, big a icrasig fuctio of with domai (,) ad rag (,), has a ivrs l with domai (,) ad rag (,) Th graph of l is th graph of l rflctd across th li y Also, lim l ad lim l Th umbr Dfiitio l is dotd by th lttr l Rmark is ot a ratioal umbr, its valu ca b computd usig th formula lim 6! ad is approimatly giv by = Problm Cosidr a quatity y whos rat of chag ovr tim is proportioal to th dy amout of y prst Th y ad y satisfis th diffrtial quatio dt dy ky, () dt whr k is th proportioality costat By sparatig variabls, w obtai dy k dt y Itgratig both sids, w gt l y k t c or y kt c, whr ivrs of l ) simply w ca writ kt y C, () is th potial fuctio (it is th takig C c If, i additio to (), y y wh t, th () givs y C or C y Hc th fuctio satisfyig th diffrtial quatio () ad y y wh t is th potial kt fuctio y y Calculus ad Aalytic Gomtry Pag

13 School of Distac Educatio Th Fuctio y = W ot that, for ay ratioal umbr Hc, l l Sic l is o - to - o ad l l(l), /, ad so o Sic positiv, has a logarithm ad is giv by Calculus ad Aalytic Gomtry Pag is positiv th abov quatio tlls us that, for ratioal Th abov quatio provids a way to td th dfiitio of to irratioal valus of Th fuctio l is dfid for all ral, so w ca us it to assig a valu to at vry poit Th dfiitio follows: Dfiitio For vry ral umbr, Equatios Ivolvig l Sic l ad ad l ar ivrss of o aothr, w hav l (all l() (all ) ) Th abov ar ivrs quatios for ad l rspctivly 5 Problm a) l 5 5 b) l 5 c) l d) l l / si ) 4 4 si 4 l( ) 4 f) (this is possibl, sic Problm Evaluat l Aswr Alitr: l l l8 8 8 l l Problm Fid y if l y 7t 9 Aswr Epotiatig both sids, w obtai l y 7t9 y = 7t9, usig Eq(5) Problm Aswr Fid k if k = ) 4 Takig th atural logarithm o both sids, w gt l k = l k = l, usig Eq (6) k = l

14 School of Distac Educatio Laws of Epots For all ral umbrs, ad, th th followig laws of pots hold: = = = 4 Problm a) b) = l = = l = l l = =,by law, by law c) d) =, by law = =, by law 4 Problm Solv th followig for th valu of y (i) y (ii) Aswr y = (iii) y = + cos (i) y Now takig logarithms o both sids, w gt (ii) Giv y = + + = ( + ) as th solutio y Takig logarithms o both sids, w gt y or l Squarig both sids w gt y = [ l ] = 4 [l ] (iii) Giv y = + cos y Takig logarithms o both sids, w gt y = l ( + cos ), so that y = l ( + cos ) Calculus ad Aalytic Gomtry Pag 4

15 School of Distac Educatio Th Drivativ of Th potial fuctio is diffrtiabl bcaus it is th ivrs of a diffrtiabl fuctio whos drivativ is vr zro Cosidr y = Applyig logarithms o both sids, w obtai l y Diffrtiatig both sids with rspct to, w obtai y dy d dy or y d Rplacig y by, w obtai d d d 5 d Problm Evaluat Aswr Th Drivativ of d d u 5 d = 5 d = 5 If u is ay diffrtiabl fuctio of, th usig th Chai Rul Problm a) b) d d d si d Itgral of d d = () = ( ) = d u d = u du d, usig th abov quatio with u, = si d (si), usig th abov quatio with u si d = si cos u u du = u C Problm Solv th iitial valu problm y dy d, ; y() Calculus ad Aalytic Gomtry Pag 5

16 School of Distac Educatio Solutio y By sparatig variabls, th giv diffrtial quatio bcoms dy d Itgratig both sids of th diffrtial quatio, w obtai y C () To dtrmi C w us th iitial coditio Giv y, wh Hc or () C C () = 4 Substitutig this valus of C i (), w obtai y () To fid y, w tak logarithms o both sids of () ad gt or y = Clarly l y = l( ) l( ) l( ) is wll dfid for Chckig of th solutio i th origial quatio Now y dy = d = y d d y = l( ),, as ( ) Hc th solutio is chckd Erciss () ad hc th solutio is valid for usig Eq() d du u d u d, usig Eq() Fid simplr prssios for th quatitis i Erciss- 6 4 l() y sc l() () 5 l l l l() 6 l() l I Erciss 7-9, solv for y i trms of t or, as appropriat 7 l y = t 5 8 l( )y 9 l( y ) l( y ) = l(si) I Ercis-, solv for k = t a) 5k = 4 8 k = (l 8)k =8 Calculus ad Aalytic Gomtry Pag 6

17 School of Distac Educatio I Erciss -6, solv for t a) t = 4 kt = 5 (l ) t 6 ()( ) = t I Erciss 7-6, fid th drivativs of y with rspct to, t, or, as appropriat 7 y = / 8 y = (4) 9 y = ( ) y = (9 6 ) y = l() y = cos5 t y = l( si) t 4 y = l 5 y = si t (l t ) 6 y = l tdt 4 I Erciss 7-8, fid dy / d 7 l y = y 8 ta y = l Evaluat th itgrals i Erciss ( ) d () d l l6 / 4 d d r dr 4 r 4 () t t dt / / cot d 6 csc / 4 csc() t csc()cot() t t dt l cos() d 9 d Solv th iitial valu problms i Erciss dy dt d y dt t sc() t t, (), y (l 4) = / y ad y () = d Calculus ad Aalytic Gomtry Pag 7

18 School of Distac Educatio Th Fuctio a CHAPTER : a l a Sic a for ay positiv umbr a, w ca writ i th followig dfiitio Dfiitio For ay umbr a ad, Problm a = l a ad log a a as l () a = l a ad w stat this a) b) 5 5 l 6 l 6 = For a, ad ay ad y : Tabl: Laws of pots a a a y a = a y a y a = y a 4 () a y = a y = () a y Th Powr Rul (Fial Form) For ay ad ay ral umbr, w ca dfi = l Thrfor, th i th quatio l = l o logr ds to b ratioal- it ca b ay umbr as log as : l = l() Diffrtiatig l = l = l l, as l u u, for ay u with rspct to, d d l =, as for, d d = l ( l), as d d =, as =, as l d d u, ad l u du d d d Calculus ad Aalytic Gomtry Pag 8

19 School of Distac Educatio Hc, as log as, d d = Usig th Chai Rul, w ca td th abov quatio to th Powr Rul s fial form: If u is a positiv diffrtiabl fuctio of ad is ay ral umbr, th diffrtiabl fuctio of ad Problm a) d u d d d = u d b) (si) d du d = ( ) = (si) Th Drivativ of a Diffrtiatig d a = d d l a d a = l a cos (si ) with rspct to, w obtai = l ( l) a d d u d u du d du d a d d d = a l a, as ( ) a a a d d That is, if a, th d a d = a l a u is a, takig u a ad usig th Chai Rul Usig th Chai Rul, w ca td th abov quatio to th followig gral form If a ad u is a diffrtiabl fuctio of, th a u is a diffrtiabl fuctio of ad d a u d u du = a l a d If a, th l a l ad th abov quatio simplifis to Problm (a) (b) (c) d d d d d d d = d = l d = l() d si = si d l (si) d = l = (l si )cos Calculus ad Aalytic Gomtry Pag 9

20 School of Distac Educatio Th drivativ of a is positiv if l a, or a, ad gativ if l a, or a Thus, a is a icrasig fuctio of if a ad a dcrasig fuctio of if a I ach cas, a is o-to-o Th scod drivativ d d a () d d () a d ( a l) a = (l) a d d d is positiv for all, so th graph of OTHER POWER FUNCTIONS Calculus ad Aalytic Gomtry Pag a a is cocav up o vry itrval of th ral li Th ability to rais positiv umbrs to arbitrary ral powrs maks it possibl to dfi fuctios lik ad l for W fid th drivativs of such fuctios by rwritig th fuctios as powrs of Problm Fid dy d if Aswr With a y,, w ca writ y l as l, a powr of, so that Diffrtiatig both sids with rspct to, w obtai dy d = d l d = l ( l), usig Eq() with a, u l ad otig that l, or d simply usig Eq (9) of th prvious chaptr d = l, applyig product rul of diffrtiatio = ( l) Th Itgral of a u If a, th l a, so u du a = d () a u d l a d Itgratig with rspct to, w obtai u du a d = d l a d () a u d = d l a d ( a u ) d = d l a Writig th first itgral i diffrtial form givs u a du = l u a a C u a + C

21 School of Distac Educatio Problm Evaluat d Aswr Problm Aswr d = C, usig Eq () with a, u l Evaluat si cos si cos d d = u du, = u C l = si l Logarithms with Bas a C W hav otd that if a is ay positiv umbr othr tha, th fuctio a is o- to o ad has a ozro drivativ at vry poit It thrfor has a diffrtiabl ivrs W call th ivrs th logarithm of with bas a ad dot it by log a Dfiitio For ay positiv umbr a, log a ivrs of a Th graph of y log a ca b obtaid by rflctig th graph of y a across th li y (Fig) Sic log a ad a ar ivrs of o aothr, composig thm i ithr ordr givs th idtity fuctio That is, log a a = ( ) ad log() a = (all ) a log ath abov ar th ivrs quatios for a ad logth Evaluatio of alog() a a =, Takig th atural logarithm of both sids, Usig Powr Rul, log() a l a = l log()l a a = l Solvig for log a, w obtai log a = l l a Calculus ad Aalytic Gomtry Pag

22 School of Distac Educatio Problm l log l Proprtis of bas a logarithms For ay umbr > ad y >, Product Rul: log a y = log log y Quotit Rul: log a y = log a a a log a y Rciprocal Rul: log a y = log y a y 4 Powr Rul: log = log a y a Proof For atural logarithms, w hav l y = l l y Dividig both sids by l a, w gt l y l a = l l y l a l a i, log a y = log a + log a y logath Drivativ of that, if u is a positiv diffrtiabl uprov fuctio of, th Proof d (log) a u d d (log) a u d = du l a u d d l u = d l a = d (l) u l a d = du l a u d d Problm Evaluat log( ) d Solutio Takig a ad u, Eq(7) givs d log( ) = d ( ) = d l d (l)( ) Calculus ad Aalytic Gomtry Pag

23 School of Distac Educatio Itgrals Ivolvig loga To valuat itgrals ivolvig bas logarithms log Problm Evaluat d Solutio Erciss log d = l l d l, sic log l = l udu, takig u l, du d u (l) = C = l l (l) = C l C a logarithms, w covrt thm to atural Fid th drivativ of y with rspct to th giv idpdt variabl y = log( l ) y = log5 log5 y = log r log9 r 4 y = log 5 7 l 5 si cos 5 y = log7 6 y = log 7 y = log(log) 8 t 8 y = t log() (si)(l) t Us logarithmic diffrtiatio to fid th drivativ of y with rspct to th giv idpdt variabl 9 y = ( ) y = t t y = si y = (l) l Calculus ad Aalytic Gomtry Pag

24 School of Distac Educatio CHAPTER 4: GROWTH AND DECAY I this chaptr, w driv th law of potial chag ad dscrib som of th applicatios that accout for th importac of logarithmic ad potial fuctios Th Law of Epotial Chag Cosidr a quatity y (vlocity, tmpratur, lctric currt, whatvr) that icrass or dcrass at a rat that at ay giv tim t is proportioal to th amout prst If w also kow th amout prst at tim t, call it y, w ca fid y as a fuctio of t by solvig th followig iitial valu problm: Diffrtial quatio: dy dt = ky Iitial coditio: y = y wh t If y is positiv ad icrasig, th k is positiv ad w us th first quatio to say that th rat of growth is proportioal to what has alrady b accumulatd If y is positiv ad dcrasig, th k is gativ ad w us th scod quatio to say that th rat of dcay is proportioal to th amout still lft Clarly th costat fuctio y is a solutio of th diffrtial quatio i Eq () Now to fid th ozro solutios, w procd as follows: By sparatig variabls, th diffrtial quatio i Eq(a) givs dy k dt y Itgratig both sids, w obtai l y = kt C By potiatig, w obtai y = kt C i, y = C kt, sic a b = a b i, y = C kt, otig that if y = r, th y = r i, y = A kt, as A is a mor covit tha C To fid th right valu of A that satisfis th iitial valu problm, w solv for A wh y y ad t : k y = A = A Hc th solutio of th iitial valu problm is kt y = y Th law of Epotial Chag says that th abov quatio givs a growth wh k ad dcay wh k Th umbr k is th rat costat of th quatio Calculus ad Aalytic Gomtry Pag 4

25 School of Distac Educatio Populatio Growth Cosidr th umbr of idividuals i a populatio of popl It is a discotiuous fuctio of tim bcaus it taks o discrt valus Howvr, as soo as th umbr of idividuals bcoms larg ough, it ca safly b dscribd with a cotiuous or v diffrtiabl fuctio If w assum that th proportio of rproducig idividuals rmais costat ad assum a costat frtility, th at ay istat t th birth rat is proportioal to th umbr y() t of idividuals prst If, furthr, w glct dparturs, arrivals ad daths, th growth rat dy / dt will b th sam as th birth rat ky I othr words, dy / dt = kt ky, so that y y I ral lif all kids of growth, may hav limitatios imposd by th surroudig viromt, but w igor thm Problm O modl for th way disass sprad assums th rat dy / dt at which th umbr of ifctd popl chags is proportioal to th umbr y Th mor ifctd popl thr ar, th fastr th disas will sprad Th fwr thr ar, th slowr it will sprad Suppos that i th cours of ay giv yar th umbr of cass of a disas is rducd by % If thr ar, cass today, how may yars will it tak to rduc th umbr to? Aswr kt W us th quatio y = y Thr ar thr thigs to fid: th valu of y, th valu of k, th valu of t that maks y = Dtrmiatio of th valu of y W ar fr to cout tim bgiig aywhr w wat If w cout from today, th y =, wh t, so y =, Our quatio is ow y = kt Dtrmiatio of th valu of k Wh t prst valu, or 8 Hc, 8 =, k (), usig Eq () with t ad y =8 k = 8 Takig logarithms o both sids, w obtai Hc k l = l 8 k = l 8 Usig Eq(), at ay giv tim t, (l 8)t y =, () yar, th umbr of cass will b 8% of its Calculus ad Aalytic Gomtry Pag 5

26 School of Distac Educatio Dtrmiatio of th valu of t that maks y = W st y qual to i Eq(4) ad solv for t : =, (l 8)t = (l 8)t Takig logarithms o both sids, w obtai t = (l 8)t = l l l 8 It will tak a littl mor tha yars to rduc th umbr of cass to Cotiuously Compoudd Itrst If you ivst a amout A of moy at a fid aual itrst rat r (prssd as a dcimal) ad if itrst is addd to your accout k tims a yar, it turs out that th amout of moy you will hav at th d of t yars is kt r A t = A k Th itrst might b addd ( compoudd ) mothly ( k ), wkly ( k 5), daily ( k 65), or v mor frqutly, say by th hour or by th miut But thr is still a limit to how much you will ar that way, ad th limit is lim A t k r lim k = A k rt = A, as lim kt Th rsultig formula for th amout of moy i your accout aftr t yars is rt A() t = A Itrst paid accordig to this formula is said to b compoudd cotiuously Th umbr r is calld th cotiuous itrst rat Problm Suppos you dposit Rs6 i a bak accout that pays 6% compoudd cotiuously How much moy will you hav 8 yars latr? If bak pays 6% itrst quartrly how much moy will you hav 8 yars latr? Compar th two compoudig Aswr With A 6, r =6 ad t =8: A (8) = 6 (6)(8) = 6 48 = 58, approimatly If th bak pays 6% itrst quartrly, w hav to put k 4 i Eq (5) ad Calculus ad Aalytic Gomtry Pag 6

27 School of Distac Educatio 48 A(8) 6 6, approimatly Thus th ffct of 4 cotiuous compoudig, as compard with quartrly compoudig, has b a additio of Rs57 Radioactivity Wh a atom mits som of its mass as radiatio, th rmaidr of th atom rforms to mak a atom of som w lmt This procss of radiatio ad chag is calld radioactiv dcay, ad a lmt whos atoms go spotaously through this procss is calld radioactiv Thus, radioactiv carbo-4 dcays ito itrog Also, radium, through a umbr of itrvig radioactiv stps, dcays ito lad Eprimts hav show that at ay giv tim th rat at which a radioactiv lmt dcays (as masurd by th umbr of ucli that chag pr uit tim) is approimatly proportioal to th umbr of ucli prst Thus th dcay of a radioactiv lmt is dscribd by th quatio dy / dt = ky, k If y is th umbr of radioactiv ucli prst at tim zro, th umbr still prst at ay latr tim t will b kt y = y, k Problm Th half-lif of a radioactiv lmt is th tim rquird for half of th radioactiv ucli prst i a sampl to dcay Show that th half lif is a costat that dos ot dpd o th umbr of radioactiv ucli iitially prst i th sampl, but oly o th radioactiv substac Aswr Lt yb th umbr of radioactiv ucli iitially prst i th sampl Th th umbr y prst at ay latr tim t will b y y kt W sarch th valu of t at which th umbr of radioactiv ucli prst quals half th origial umbr: y kt y = kt = kt = l = l, usig Rciprocal Rul for logarithms l t k This valu of t is th half-lif of th lmt It dpds oly o th valu of radioactiv lmt, ot o y th umbr of radioactiv ucli prst Thus, k for a Half-lif = l k, whr k dpds oly o th radio activ substac Calculus ad Aalytic Gomtry Pag 7

28 School of Distac Educatio Problm Th umbr of radioactiv Poloium- atoms rmaiig aftr t days i a sampl that starts with y atoms is giv by th Poloium dcay quatio y = y 5 t Fid th Poloium- half lif Aswr Comparig Poloium dcay quatio with Eq (7), w hav Half-lif = l k l = 5 9 days, usig Eq (8) k 5 Problm Usig Carbo-4 datig, fid th ag of a sampl i which % of th radioactiv ucli origially prst hav dcayd (Th half lif of Carbo -4 is 57 yars) Aswr W ot that % of th radio activ ucli origially prst hav dcayd is quivalt to say that 9% of th radioactiv ucli is still prst W us th dcay quatio y y kt Thr ar two thigs to fid: th valu of k, kt th valu of t wh y = 9y, or kt =9 Dtrmiatio of th valu of k W us th half-lif quatio (8), to gt k = l half - lif = l 57 Hc th dcay quatio bcoms y y ( / 57) t Dtrmiatio of th valu of t that maks (l / 57) t 9 Takig logarithm of both sids, l 57 t = l 9 57l 9 t = 866 l Hc th sampl is about 866 yars old Calculus ad Aalytic Gomtry Pag 8

29 School of Distac Educatio CHAPTER5 L HOSPITAL S RULE L` Hospital rul for forms of typ / f ( ) Thorm Suppos that lim f lim g If lim ists i ithr th fiit or ifiit u u u g ( ) ss (that is, if this limit is a fiit umbr or or +), th f ( ) lim u g( ) f ( ) lim u g( ) Hr u may stad for ay of th symbols a, a, a,, or Problm Fid lim Aswr Hr both th umrator ad domiator hav limit Thrfor limit has / form lim lim, applyig l Hôpital s Rul si Problm Us l Hôpital s rul to show that lim si Aswr Hr limits of both th umrator ad domiator is Thrfor lim / form Now is i th si cos lim lim, usig l Hôpital s Rul ad otig that drivativ of si is cos ad that of is lim cos lim, usig quotit rul for limits Problm Fid lim Aswr Hr both th umrator ad domiator hav limit Thrfor limit has / form lim lim, applyig l Hôpital s Rul Calculus ad Aalytic Gomtry Pag 9

30 School of Distac Educatio Succssiv Applicatio of l Hôpital s Rul si Problm Evaluat lim Aswr Problm Hr th limit is i / form si L cos lim lim, agai i / form L lim 6 L lim 6 6 si, agai i / form cos, ow limit ca b valuatd cos Fid lim cos L si L cos Aswr lim lim lim This is wrog, as th first applicatio of l Hôpital s Rul was corrct; th scod was ot, sic at that stag th limit did ot hav th / form Hr is what w should hav do: cos lim L si lim Problm Evaluat lim cos Aswr Th giv is i th form This is right lim cos lim si Not If w cotiu to diffrtiat i a attmpt to apply L Hopital s rul oc mor, w gt lim cos lim si lim cos, which is wrog Calculus ad Aalytic Gomtry Pag

31 School of Distac Educatio log Problm Fid lim log Aswr Hr th giv limit ca b writt as ad th limit is i / form Also, log log lim lim log log log log lim lim log log log log Now w ar rady to apply l Hôpital s Rul: Erciss log L lim lim log Evaluat th followig limits lim lim (by algbraic maipulatios) lim lim si lim ta ta lim cos lim 4 lim b 5 ( ) lim 6 cos log( ) lim Calculus ad Aalytic Gomtry Pag

32 School of Distac Educatio lim 8 cos si si lim log( ) lim lim lim si si log( ) lim 4 log( ) ta lim si si lim si log( ) ta si lim 4 6 log( ) lim 7 log cos ta si lim si 8 cos cosh log 9 lim ta cos cos lim 4 lim si log( k ) lim cos 9 lim 6 L` Hospital rul for forms of typ / f ( ) Thorm Suppos that lim f ( ) lim g( ) If lim ists i ithr th fiit or ifiit u u u g ( ) ss (that is, if this limit is a fiit umbr or or +), th f ( ) lim u g( ) f ( ) lim u g( ) Hr u may stad for ay of th symbols a, a, a,, or Problm Fid lim Aswr Both ad td to as Hc limit is i / form lim lim, applyig l Hôpital s Rul = Calculus ad Aalytic Gomtry Pag

33 School of Distac Educatio ProblmEvaluat Aswr lim, whr is atural umbr Hr both th umrator ad domiator td to as lim L lim L lim L L L lim lim! lim Problm Show that if a is ay positiv ral umbr, Aswr a lim Hr both th umrator ad domiator td to as Hc limit is i / form Hc limit is i / form Suppos as a spcial cas that a = Th thr applicatios of l Hôpital s Rul giv lim A similar argumt works for ay a L lim L L lim lim 9 Problm Show that if a is ay positiv ral umbr, Aswr l lim a Hr both th umrator ad domiator td to as lim l lim L lim lim a a a a a a lim a Hc limit is i / form Calculus ad Aalytic Gomtry Pag

34 School of Distac Educatio l Problm Show that lim cot Aswr Hr both th umrator ad domiator td to as l L lim cot lim / cosc Hc limit is i / form This is still idtrmiat ( / form) as it stads, but rathr tha apply l Hôpital s Rul agai (which oly maks thigs wors), w rwrit: Thus / cosc si si si l lim cot si lim si si lim si lim Th Idtrmiat Products ad Diffrcs: Idtrmiat forms, Problm Evaluat Aswr Writ lim ta log lim ta log (which is i form) as: lim ta log l lim cot (ow i / form), by Eampl 5 i th prvious sctio Problm Evaluat Aswr ad l lim l td to as So th limit is a form Bfor applyig L`Hospital s Rul w rwrit: lim l l lim l (/ form) Calculus ad Aalytic Gomtry Pag 4

35 School of Distac Educatio Now apply L`Hospital s Rul: lim l l L lim lim / l l / l Erciss lim L l l lim sc ta / a lima ta a 5 lim si lim l l lim cot 4 limlog log Th Idtrmiat Powrs: Idtrmiat forms,, Thr idtrmiat forms of potial typ ar, ad Hr th trick is to cosidr ot th origial prssio, but rathr its logarithm Usually l`hospital s Rul will apply to th logarithm ta Problm Evaluat limsi Aswr Th limit taks th idtrmiat form ta Lt y si, so takig logarithims, w obtai logsi log y ta logsi cot Applyig l`hospital s Rul for / forms, log si lim log y lim cot L cos lim si cosc lim si cos si cos Now y l y, ad sic th potial fuctio f is cotiuous, lim y lim p log y p lim log y p ta i, limsi Calculus ad Aalytic Gomtry Pag 5

36 School of Distac Educatio cos Problm Prov that lim ta Aswr / Th limit taks th idtrmiat form cos Lt y ta By l`hospital s Rul, Now lim / lim / y l y, so that l ta l y cos l ta sc lim / lim / l ta sc sc ta L lim / lim / sc ta sc ta cos si l y, ad sic th potial fuctio y lim / p Problm Show that Aswr l y p lim l y p / lim Th limit lads to th idtrmiat form log log Lt y, so that y lim l y lim log (/ form) L lim l y Erciss lim lim y lim Evaluat th followig limits: f is cotiuous, lim ta lim a lim ta 4 lim cos 5 si lim 6 lim si cot Calculus ad Aalytic Gomtry Pag 6

37 School of Distac Educatio / 7 lim a lim a lim cot a 8 si log lim 9 lim lim ta 4 ta lim 4 5 b lim a 6 lim log lim 8 lim cos cot 7 9 lim si limsi si cos limlog ta ta cos ta ta lim a log 4 lim a a Calculus ad Aalytic Gomtry Pag 7

38 School of Distac Educatio Hyprbolic Fuctios Hyprbolic cosi of : Hyprbolic si of : CHAPTER 6 sih cosh Rmark: cosh sih HYPERBOLIC FUNCTIONS Dfiitio Usig th abov Dfiitio w ca dfi four othr hyprbolic fuctios ad ar listd blow: Hyprbolic tagt: Hyprbolic cotagt: Hyprbolic scat: tah sih cosh Calculus ad Aalytic Gomtry Pag 8 coth cosh sih sch cosh Hyprbolic coscat: csch sih Idtitis i hyprbolic fuctios cosh() cosh sih() sih cosh sih cosh sih tah sc h coth csc h cosh() cosh y cosh sih y sih y cosh() cosh y cosh sih y sih y si h() sih y cosh cosh y sih y si h() sih y cosh cosh y sih y cosh cosh sih sih cosh si h sih cosh sih sih cosh cosh cosh cosh sih cos h 4cosh cosh

39 School of Distac Educatio si h sih 4sih tah tah y tah( y) tah tah y tah tah y tah( y) tah tah y tah sih tah tah cosh tah tah tah tah tah tah tah tah Problm Giv sih Fid th othr fiv hyprbolic fuctios 4 Usig cosh Also, cosh sih, sih 5 4 / 4 sih / 4 5 tah ; coth cosh 5 / 4 5 tah sc h 4 4 ; ad cos ch cosh 5/ 4 5 sih Drivativs of Hyprbolic Fuctios d (sih) d cosh d (cosh) d sih d d d d (tah) sch (coth) csch d (sch) d sch tah d (csch) d csch coth Calculus ad Aalytic Gomtry Pag 9

40 School of Distac Educatio dy Problm Fid, whr d dy d d d Solutio sih cosh d d y sih cosh d d sih cosh d d d d sih sih sih cosh sih sih cosh Problm Evaluat d tah Solutio d Tak u Th, usig formula abov, d d tah sc h d d sc h Formula for Itgral of Hyprbolic Fuctios sihu du cosh u C coshu du sih u C sch u du tah u C csch u du coth u C schu tah u du sc hu C csch u coth u du csch u C Problm Evaluat l 4 sih d Solutio 4 sih d 4( ) d d [ ]( l )( ) l l l l l l 4 l 4 l 67, applyig product rul of diffrtiatio Calculus ad Aalytic Gomtry Pag 4

41 School of Distac Educatio Th Ivrs Hyprbolic Fuctios sih is th ivrs hyprbolic si of Idtitis for ivrs hyprbolic fuctios sc h cosh csc h sih coth tah Rlatio btw ivrs hyprbolic fuctios ad atural logarithm cosh l, sih l, tah l, sch l, csch l, coth l, Drivativs of ivrs hyprbolic fuctios Drivativ of sih - Lt y = sih Th = sih y Diffrtiatig both sids with rspct to, w gt dy = cosh y d dy Thrfor, for ral d cosh y sih y d d i sih, for ral I a similar mar, w hav th followig drivativs d (cosh) d d d tah, for Calculus ad Aalytic Gomtry Pag 4

42 School of Distac Educatio d d 4 coth, for d d 5 sch, d d 6 cosch, Problm Fid th drivativs of th followig fuctios with rspct to : (i) cosh ( ) (ii) sih (ta ) Solutio (i) Lt y = cosh ( ) dy d d du cosh cosh u with u d d du d (ii) Lt y = sih (ta ) 4 dy d sih ta d sih u du d d du d sc ta sc sc sc Itgrals ladig to ivrs hyprbolic fuctios du u a sih C, a a u du u a cosh C, u a u a du tah u C if u a a a coth u if a a a u C u a du u C u a a a 4 sch, a u du a u a 5 csch C, u u a u Calculus ad Aalytic Gomtry Pag 4

43 School of Distac Educatio Problm Aswr Evaluat th dfiit itgral d 4 Erciss d 4 sih sih Each of Erciss - givs a valu of sih or cosh Us th dfiitios ad th idtity cosh sih to fid th valus of th rmaiig fiv hyprbolic fuctios sih 4 cosh, 5 Rwrit th prssios i Erciss -5 i trms of potials ad simplify as much as you ca sih() 4 cosh sih 5 (cosh sih) (cosh sih) I Erciss 6-7, fid th drivativ of y with rspct to th appropriat variabl 6 y sih( ) 7 y t tah 8 y (cosh) z t 9 y csch( csch) y sih coth y (4 ) csc h() y cosh y ( ) tah( ) 4 y () coth t t 5 y sch 6 y csc h 7 y cosh(sc), / I Erciss 8-9, vrify th followig itgratio formula: 8 schd sch C d C 9 ta h ta h () I Erciss - 4, valuat th idfiit itgrals: sih d 4cosh( ) d 5 coth d csch(5) d 4 csc h()coth() t t t dt Calculus ad Aalytic Gomtry Pag 4

44 School of Distac Educatio I Erciss 5-9, valuat th dfiit itgrals: tah d 6 / sih(si) cos d 8 4sih d 4 4 sih d 8cosh Eprss th umbrs i Erciss - i trms of atural logarithms: d cosh(5/ ) coth(5/ 4) csc h( / ) I Erciss -6, valuat th itgrals i trms of (a) ivrs hyprbolic fuctios, (b) atural logarithms 5 / 6d 9 4 d 4 6 / d d () MODULE II CHAPTER 7: SEQUENCES Calculus ad Aalytic Gomtry Pag 44

45 School of Distac Educatio Squcs Dfiitio (Squc) If to ach positiv itgr thr is assigd a (ral or compl) umbr u, th ths umbrs u, u,, u, ar said to form a ifiit squc or, brifly, a squc, ad th umbrs u ar calld th trms of th squc A squc whos trms ar ral umbrs is calld ral squc W discuss ral squcs oly Dfiitio A ifiit squc (or squc) of umbrs is a fuctio whos domai is th st of itgrs gratr tha or qual to som itgr Usually is ad th domai of th squc is th st of positiv itgrs ad i that cas squcs ar th fuctios from th st of positiv itgrs Basd o th abov dfiitio a ampl of a squc is Th umbr u() u(), w hav Calculus ad Aalytic Gomtry Pag 45 u() is th th trm of th squc, or th trm with id If First trm Scod trm Third trm th trm u() u() Wh w us th subscript otatio u, u u 4, u() 4, Som othr ampls of squcs ar u(), u()( ), u() u for u(), th squc is writt u() u W rfr to th squc whos th trm is u with th otatio { u } If b, th st of ral umbrs, th squc B ( b, b, b,), all of whos trms qual b, is calld th costat squc b Thus th costat squc is th squc (,,,), all of whos trms qual, ad th costat squc is th squc (,,,) If a, th th squc { A a } is th squc a, a, a,, a, I particular, if a, th w obtai th squc,,,,, 4 8 Dfiitio A squc { u } is said to covrg or to b covrgt if thr is a umbr l with th followig proprty : For vry (i, is a positiv ral umbr that may b vry small, but ot zro) w ca fid a positiv itgr N such that N u l l is calld th limit of th squc Th w writ

46 School of Distac Educatio or simply limu l u l as ad w say that th squc covrgs to l or has th limit l If o such umbr l ists, w say that { u } divrgs Problm Aswr Show that Hr u ad l such that lim Lt N u l i, to show that thr ists a positiv itgr N such that N b giv W must show that thr ists a itgr N Th implicatio i () will hold if or If N is ay itgr gratr tha, th implicatio will hold for all N This provs that lim Problm Show that lim k k Aswr Lt (whr k is a costat) b giv W must show that thr ists a positiv itgr N such that N k k Sic k k, w ca us ay positiv itgr for N ad th implicatio will hold This provs that lim k k for ay costat k Problm Th squc (,,,,,,,) dos ot covrg to Aswr Hr u, wh is odd, wh is v If w choos, th, for ay positiv itgr N, o ca always slct a v umbr N, for which th corrspodig valu u ad for which u Thus, th umbr is ot th limit of th giv squc () u Rcursiv Dfiitios Calculus ad Aalytic Gomtry Pag 46

47 School of Distac Educatio So far, w hav calculatd ach ar dfid rcursivly by givig Th valu(s) of th iitial trm (or trms), ad u dirctly from th valu of But, som squcs A rul, calld a rcursio formula, for calculatig ay latr trm from trms that prcd it Problm Th statmts u ad u u dfi th squc,,,,, of positiv itgrs With u, w hav u u, u u, ad so o SUBSEQUENCES If th trms of o squc appar i aothr squc i thir giv ordr, w call th first squc a subsquc of th scod Problm Som subsqucs of X,,,,,,,, 4 5,,,,,, 5 But,,, ar, ad,,,,! 4!()! Y is ot a subsquc of X, bcaus th trms of Y do ot appar i X i th giv ordr Dfiitio A tail of a squc is a subsquc that cosists of all trms of th squc from som id N o I othr words, a tail is o of th sts { u N } Dfiitio If X { u, u,, u, } is a squc of ral umbrs ad if m is a giv atural umbr, th th m -tail of X is th squc X m { um, um, } ad its th trm is m For ampl, th -tail of th squc X {,4, 6, 8,,,, }, u is th squc X {8,,,, 6, } Rmark Aothr way to say that u L is to say that for vry, itrval ( L,) L about L cotais a tail of th squc th - op Boudd Nodcrasig Squcs Dfiitio A squc { u } with th proprty that u u for all is calld a odcrasig squc Som ampls of odcrasig squcs ar i) Th squc,,,,, of atural umbrs Calculus ad Aalytic Gomtry Pag 47

48 School of Distac Educatio ii) Th costat squc {} Dfiitio A squc { u } is boudd from abov if thr ists a umbr M such that u M for all Th umbr M is a uppr boud for { u } If M is a uppr boud for { u } ad o umbr lss tha M is a uppr boud for { u }, th M is th last uppr boud for { u } Thorm A o-dcrasig squc that is boudd from abov always has a last uppr boud Th squc,,,,,( ), is boudd from abov with a uppr boud is th last uppr boud as o umbr lss tha is a uppr boud Also ot that ay ral umbr gratr tha or qual to is also a uppr boud Th squc,,,,, has o uppr boud Thorm (Th Nodcrasig squc thorm) A odcrasig squc of ral umbrs covrgs if ad oly if it is boudd from abov If a odcrasig squc covrgs, it covrgs to its last uppr boud Erciss Each of Erciss -7 givs a formula for th th trm u of a squc { u } Fid th valus of u, u, u, ad u 4 u! ( ) u u 4 u ( ), 5 u ( ), 6 u ( ) 7 u I Erciss 8- th first fw trms of a squc { u } ar giv blow Assumig that th atural pattr idicatd by ths trms prsists, giv a formula for th th trm u 8 5, 7, 9,,, 9,,,,, 4 8 6,,, 4,,, 4, 9, 6, 4 5 Each of Erciss -8 givs th first trm or two of a squc alog with a rcursio formula for th rmaiig trms Writ out th first t trms of th squc u u, u u u, u Calculus ad Aalytic Gomtry Pag 48

49 School of Distac Educatio 4 u, u, u 5 u, u u, 6 v, v () y y u u () u 7 u, u, u () u u u 8 u, u 5, u u u I Erciss 9-, fid a formula for th th trm of th squc 9 Th squc,,,,, Th squc,,,,, Th squc,,,,, Th squc, 6,, 4, 8, Th squc,,,,,,,4, CHAPTER8 THEOREMS FOR CALCULATING LIMITS OF SEQUENCES Dfiitio If X { u } ad Y { v } ar squcs of ral umbrs, th w dfi thir sum to b th squc X Y { u v}, thir diffrc to b th squc X Y { u v}, ad thir product to b th squc X Y { u v } If c w dfi th multipl of X by Calculus ad Aalytic Gomtry Pag 49

50 School of Distac Educatio c to b th squc cx { cu } Fially, if Z { w } is a squc of ral umbrs with X u w for all, th w dfi th quotit of X ad Z to b th squc Z w Thorm Lt { u } ad { v } b squcs of ral umbrs Th followig ruls hold if lim u A ad lim v B whr A ad B b ral umbrs Sum Rul : lim() u v A B Diffrc Rul : lim() u v A B Product Rul : lim() u v A B 4 Costat Multipl Rul : lim() k v k B (Ay umbr k) u 5 Quotit Rul : lim A v B if B Problm Show that Aswr Sic, lim w hav lim lim lim lim Thorm 4 Th Sadwich Thorm for Squcs Lt { u}, { v}, ad { w } b squcs of ral umbrs If u v w holds for all byod som id N, ad if lim u lim w L, th lim v L also Rmark A immdiat cosquc of Thorm 4 is that, if v w ad w, th v bcaus w v w W us this fact i th comig ampls Problm Show that Aswr covrgs to bcaus ad sic lim Thorm 5: Th Cotiuous Fuctio Thorm for Squcs Lt { u } b a squc of ral umbrs If u L ad if f is a fuctio that is cotiuous at L ad dfid at all u, th f ()() u f L Usig I Hôpital s Rul Calculus ad Aalytic Gomtry Pag 5

51 School of Distac Educatio Thorm 6 Suppos that f () is a fuctio dfid for all ad that { u } squc of ral umbrs such that u f () for Th Eampl 9 Solutio Problm lim() f Fid L lim u L lim 5 lim 5 is i th form Hc lim lim l, applyig L Hôpital s rul, ad otig that th drivativ of 5 5 with rspct to is Dos th squc whos th trm is covrg? If so, fid lim u u Solutio Th limit lads to th idtrmiat form W ca apply l Hôpital s rul if w first chag it to th form by takig th atural logarithm of l u l l lim l u lim l ( form) l lim form ( ) lim, applyig l Hôpital s rul lim lim Sic l u as, ad f () l u u That is, th squc { u } covrgs to Erciss u is a is cotiuous vrywhr, Thorm 5 tlls us that Which of th squcs { u } i Erciss - covrg, ad which divrg? Fid th limit of ach covrgt squc Calculus ad Aalytic Gomtry Pag 5

52 School of Distac Educatio u 5 u ( ) u ( ) 9 u ( ) 6 u u u u u u cos() 4 (9) 5 u 6 u 8 u 9 u l l( ) 4 u! 7 u u cos u 5 (l) 4 u l 7 l u u 4 u 8 u u 7 4 u si u () ( 4) u ( 4) ( 4)! 5 u l 6 u 8 u 9 u sih(l) 9 u u ta 5 u u d, p p 6 Giv a ampl of two divrgt squcs X, Y such that thir product XY covrgs 7 Show that if X ad Y ar squcs such that X ad X Y ar covrgt, th Y is covrgt 8 Show that if X ad Y ar squcs such that X covrgs to ad XY covrgs, th Y covrgs 9 Show that th squc { } 4 Show that th squc is ot covrgt {( ) } is ot covrgt I Erciss 4-44, fid th limits of th followig squcs: 4 4 ( ) 4 44 CHAPTER9 SERIES Dfiitio If u, u, u,, u, b a squc of ral umbrs, th u u u u Calculus ad Aalytic Gomtry Pag 5

53 School of Distac Educatio is calld a ifiit sris or, brifly, sris sris u u u u is dotd by Th sum u s u u u (i, th sum of th first trms of th sris) is calld th Th squc whr sum s is th s, s, s,, s, th partial sum of th sris u u is th th trm of th sris A ifiit th partial sum of th sris, is calld th squc of Problm Fid th th partial sum of th sris u, whr u ( ) Aswr Th giv sris is Th th partial sum is giv by s u if is odd if is v ( ) Covrgc, Divrgc ad Oscillatio of a sris Cosidr th ifiit sris u u u u u () ad lt th sum of th first trms b s u u u u Th, is th th partial sum of th sris () Th squc s, s, s,, s, () is th squc of th partial sums of th sris () As thr possibilitis aris: th partial (i) Th squc giv by () covrgs to a fiit umbr l; i this cas th sris is said to b covrgt ad has th sum l u i, u l (i, th sris is summabl with sum l) Calculus ad Aalytic Gomtry Pag 5

54 School of Distac Educatio (ii) Th squc () dos t covrg but tds to or as ; i this cas th sris u is said to b divrgt ad has o sum (i, th sris is ot summabl) (iii) If th both th cass (i) ad (ii) abov do ot occur, th th sris b oscillatory or o-covrgt (I this cas also th sris is ot summabl) Problm Show that th sris covrgs ad also fid its sum Solutio Lt u Th th th partial sum is giv by s u is said to Sic as, s as Sic () s, th squc of th partial sums, covrgs to, th giv sris also covrgs ad th sum of th sris is Problm Show that th gomtric sris a ar ar ar covrgs if r ad divrgs if r Proof Cas r Th th partial sum of th sris is giv by s a ar ar ar a() r a ar r r r W ot that wh r, r as r, r, Hc, () ar a lim lim r a r r Hc from (), w obtai for r Calculus ad Aalytic Gomtry Pag 54

55 School of Distac Educatio lim s a r i, th squc () s of th a partial sums covrgs to r Hc th giv sris also covrgs to a ar, r r Cas Wh r, w hav s Hc as s, a r for r I othrwords, So th squc of th partial sums divrgs ad hc th giv sris also divrgs Cas Wh r r as, s a ar ar a()( r ) a r as r r So i this cas th squc of th partial sums divrgs Hc th gomtric sris a ar ar covrgs if r ad divrgs if r If r, th gomtric sris sris divrgs a ar ar ar a covrgs to r ad if r, th Problm Show that th sris covrgs Fid its sum Solutio Th giv is a gomtric sris with sris is covrgt ad its sum is giv by a 9 ad r Hr r Hc th Problm Discuss th covrgc of th sris 4( ) Calculus ad Aalytic Gomtry Pag 55

56 School of Distac Educatio Also fid it sum Aswr Hr u By partial fractio, ( ) A B ( ), which givs, ( A) B Puttig, w hav B ad puttig, A u Th th partial sum of th sris is giv by Hc s u u u 4 4 lim s lim Sic th squc of th partial sums covrgs to, th sris also covrgs to Hc w ca writ 4( ) Thorm If th sris u u u u covrgs th limu i, th th trm of a covrgt sris must td to zro as Proof Lt s lim s dot th th partial sum of th sris Th w ot that lim s Sic s u u u ad s u u u w hav s s u or u s s Hc limu lim() s lim s lims s Hc th proof DIVERGENT SERIES Calculus ad Aalytic Gomtry Pag 56

57 School of Distac Educatio Gomtric sris with r divrgt sris Problm m divrgs Solutio Show that th sris 4 ar ot th oly sris to divrg W discuss som othr Th giv ss divrgs bcaus th partial sums vtually outgrow vry prassigd umbr Each trm is gratr tha, so th sum of trms is gratr tha Simpl Tst for Divrgc (th Trm Tst) Thorm ( th Trm Tst) A cssary coditio for th covrgc of a sris is that u u u u m limu i, if th sris u covrgs, th limu Atttio! Th coditio i Thorm is oly cssary for covrgc, but ot sufficit As a ampl, th sris satisfis th coditio lim but it divrgs Divrgc Tst I viw of Thorm w hav th followig: If limu, th sris divrgs u u u u I viw of th Trm Tst, is ot covrgt as Problm Discuss th covrgc of th sris Aswr 4 lim u lim Calculus ad Aalytic Gomtry Pag 57

58 School of Distac Educatio Lt u Th lim u lim lim Sic limu, th giv sris caot covrg Thorm If a A ad b Sum Rul: Diffrc Rul: B ar covrgt sris, th a b a b A B a b a b A B Costat Multipl Rul: ka k a ka 4 Problm Show that th sris covrgs Aswr 4 4 a, r Hc th giv sris covrgs, by costat multipl Rul as th abov is a gomtric sris with Addig or Dltig Trms W ca always add a fiit umbr of trms to a sris or dlt a fiit umbr of trms without altrig th sris covrgc or divrgc, although i th cas of covrgc this will usually chag th sum If a covrgs, th covrgs for ay k ad a a a a a Covrsly, if ad a k k covrgs for ay k, th a covrgs Thus, a k Calculus ad Aalytic Gomtry Pag 58

59 School of Distac Educatio Ridig As log as w prsrv th ordr of its trms, w ca rid ay sris without altrig its covrgc To rais th startig valu of th id h uits, rplac th i th formula for a by h : a ah a a a h To lowr th startig valu of th id h uits, rplac th i th formula for h : a by a ah a a a h Rmark Th partial sums rmai th sam o mattr what idig w choos W usually giv prfrc to idigs that lad to simpl prssios Erciss I Erciss -, fid a formula for th sris sum if th sris covrgs ( ) th partial sum of ach sris ad us it to fid th ( ) I Erciss 4-7, writ out th first fw trms of ach sris to show how th sris starts Th fid th sum of th sris ( ) 4 5 Us partial fractios to fid th sum of ach sris i Erciss ( )( ) ( ) ( ) (ta() ta( )) Which sris i Erciss - covrg, ad which divrg? Giv rasos for your aswrs If a sris covrgs, fid its sum 4 cos 5 Calculus ad Aalytic Gomtry Pag 59

60 School of Distac Educatio 5 l 6, 7 8 9! l I ach of th gomtric sris i Erciss -, writ out th first fw trms of th sris to fid a ad r, ad fid th sum of th sris Th prss th iquality r i trms of ad fid th valus of for which th iquality holds ad th sris covrgs ( ) ( ) si I Erciss -5, fid th valus of for which th giv gomtric sris covrgs Also, fid th sum of th sris (as a fuctio of ) for thos valus of ( ) 4 ( ) 5 NONDECREASING PARTIAL SUMS (l) Thorm 4 A sris u of ogativ trms covrgs if ad oly if its partial sums ar boudd from abov Problm (Th Harmoic Sris) Th sris is calld th harmoic sris It divrgs bcaus thr is o uppr boud for its partial sums To s why, group th trms of th sris i th followig way: Th sum of th first two trms is 5 Th sum of th t two trms is 4,which is gratr tha 4 4 Th sum of th t four trms is , which is gratr tha Th sum of th t ight trms is which is , gratr tha 8 6 Th sum of th t 6 trms is gratr tha 6, ad so o I gral, th sum of trms dig with is gratr tha Hc th squc of partial sums is ot boudd from abov: For, if tha k Hc, by Thorm 4, th harmoic sris divrgs k th partial sum s is gratr Calculus ad Aalytic Gomtry Pag 6

61 School of Distac Educatio Th Itgral Tst ad p-sris Thorm 5 Th Itgral Tst Lt u b a squc of positiv trms Suppos that u f (), whr f is a cotiuous, positiv, dcrasig fuctio of for all N (N a positiv itgr) Th th sris ad th itgral f () d both covrg or both divrg Problm u N Usig th Itgral Tst, show that th p-sris () p p p p p (p a ral costat) covrgs if p ad divrgs if p Solutio Cas If p th p d d lim b p p p lim ( ) p p b b p p N f () is a positiv dcrasig fuctio of for Now, p b, sic p b as b bcaus p Hc p d covrgs ad hc, by th Itgral Tst, th giv sris covrgs Cas If p, th p ad lim( p d b ) p b Hc, by th Itgral Tst, Th sris divrgs for p p If p, w hav th harmoic sris, as limb p b for p which is kow by Eampl, to b divrgt, Hc w coclud that th sris covrgs for p but divrgs wh p Rmark W ot that Thorm says, i particular, that divrgs Calculus ad Aalytic Gomtry Pag 6

62 School of Distac Educatio Erciss Which of th sris i Erciss -5 covrg, ad which divrg? Giv rasos for your aswrs l ( l) 5 4 ta (l) ( l) 4 5 ta sch COMPARISON TESTS FOR SERIES OF NON-NEGATIVE TERMS Thorm 6 (Dirct Compariso Tst) Lt u ad v b two sris with ogativ trms such that u v for all N, for som itgr N Th (a) if (b) if v is covrgt, th u u is divrgt, th v Thorm 7 Limit Compariso Tst Suppos that u ad v for all N u If lim c, th u ad v v is also covrgt is also divrgt (whr N is a itgr) both covrg or both divrg u If lim ad v covrgs, th u v u If lim ad v v divrgs, th u Problm Tst th covrgc of covrgs divrgs Solutio Lt u Calculus ad Aalytic Gomtry Pag 6

63 School of Distac Educatio Th u ( )( ) Tak v 7 Th u lim lim v Th giv sris 7 u covrgs as v, big a harmoic sris with p, covrgs Problm Tst for covrgc or divrgc th sris Solutio Lt th trm b 4 5 h h h h 4 u Th u h h Lt v h so that v h h ad divrgt for h Now, a harmoic sris with p h h u lim lim lim h, v a fiit o zro umbr Hc, by compariso tst, divrg togthr, which is covrgt for u ad v covrg or Sic th sris h v is covrgt for h is also covrgt for h i, for Th sris for h v is divrgt for h, hc u is also divrgt for h i, Calculus ad Aalytic Gomtry Pag 6

64 School of Distac Educatio D Almbrt s Ratio-Tst for Covrgc Thorm 7 D Almbrt s Ratio-Tst If u is a sris with positiv trms, ad if u u lim, th (i) u is covrgt wh l, (ii) u is divrgt wh l (iii) th tst is icoclusiv wh l i, th sris may covrg or divrg wh l Problm Aswr Tst th covrgc of th sris 5 Lt ad u 5, th ( ) u 5 u ( ) 5 5 lim lim lim u So by D Almbrt s ratio tst th giv sris covrgs Problm Aswr Tst th covrgc of! Tak u! Th u ( ) ( )! ad u u ( )!( ) ( )! u lim lim u ad hc by D Almbrts ratio tst, is divrgt Calculus ad Aalytic Gomtry Pag 64

65 School of Distac Educatio Th th Root tst Thorm 9 (Cauchy s th Root Tst) If that lim u l th th sris u (i) covrgs if l, (ii) divrgs if l or l is ifiit, (iii) th tst is i coclusiv if l u Problm Ivstigat th bhaviour (covrgc or divrgc) of Aswr Lt u Th u ad a lim lim Hc, by Root Tst, u is covrgt is a sris with o-gativ trms such u, if u [( ) r ] Problm Show that th sris is covrgt if r ad divrgt if r Aswr [( ) r ] Takig u, w hav ( ) r () u r r Sic lim, th abov implis lim() u Thrfor r u covrgs if r ad divrgs if r ( ) If r, u Lt Also, v Th v is a harmoic sris with p ad hc is divrgt u lim lim, a fiit o zro valu v Hc, by Compariso Tst, divrgs wh r u divrgs Thus th sris covrgs wh r ad Calculus ad Aalytic Gomtry Pag 65

66 School of Distac Educatio Erciss Which of th sris i Erciss - covrg, ad which divrg? Giv rasos for your aswrs! (l) 7 8 l 9 ( )!! ( )!!! (l) Which of th sris a dfid by th formulas i Erciss 4-9 covrg, ad which divrg? Giv rasos for your aswrs ta 4 a, a a 5 a, a a l 6 a 5, a a 7 a, a a, 9 a 8 a a a ()!!( )!( )! Which of th sris i Erciss - covrg ad which divrg? Giv rasos for your aswrs (!) () () ( ) [ 4()]( ) Calculus ad Aalytic Gomtry Pag 66

67 School of Distac Educatio CHAPTER ALTERNATING SERIES Altratig sris ad Libiz tst Dfiitio (Altratig sris) A sris i which th trms ar altrativly positiv ad gativ is a altratig sris Problm Each of th thr sris ( ) 4 8 ; 4( ) ; 4 ( ) is a altratig sris Th third sris, calld th altratig harmoic sris, is covrgt This is dscribd i Problm Notatio A altratig sris may b writt as ( ) u whr ach u is positiv ad th first trm is positiv If th first trm i th sris is gativ, th w writ th sris as ( ) u Thorm (Libiz Tst for tstig th atur of altratig sris) Suppos u (a) u u u u is a squc of positiv umbrs such that ad (b) limu, Th th altratig sris Proof ( ) u is covrgt If is a v itgr, say m th th sum of th first trms is s ()()() u u u u u u m 4 m m u ()()() u u u4 u5 um u m um Th first quality shows that sm is th sum of m ogativ trms, sic, by assumptio (a), ach trm i parthss is positiv or zro Hc s m s m, ad th squc { sm} is odcrasig Th scod quality shows that sm u Sic { sm} is odcrasig ad boudd from abov, by o dcrasig Squc Thorm (Thorm of Chaptr Squc ), it has a limit, say lim s m m L () Calculus ad Aalytic Gomtry Pag 67