FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures

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1 FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 0 FURTHER CALCULUS II. Sequeces d series. Rolle s theorem d me vlue theorems 3. Tlor s d Mcluri s theorems 4. L Hopitl s rule. Sequeces d series Whe obtiig mthemticl solutio to egieerig problem it is possible to proceed i obvious w but sometimes obti the wrog swer, eve though the iitil model d ll the lgebric mipultios re correct. I these situtios the errors usull rise becuse some mthemticl procedure hs bee ssumed to be true, wheres much closer ispectio would hve reveled flw i the rgumet. Alsis is the re of mthemtics i which rigour is ivestigted. Ver little o this topic is covered i our egieerig course, but few ides re itroduced i this module d i this sectio we cosider sequeces d series. Cosider fuctio f with domi 0,,,..., the the set of vlues {f(0), f(), f(),...} is clled sequece. The sequece is ofte writte {f 0,f,f,...}. A sequece which does NOT ed is clled ifiite sequece, sequece which eds is clled fiite sequece. Deote fiite sequece b {f k }, i which the first term is f 0 d the lst term is f, d ifiite sequece b {f k }. (Note tht the bove ottio is tht used b Jmes, lthough differet ottios pper i other books.) The vlues of the terms i sequece re ofte relted. For emple, the terms might stisf f k+ = f k + d this is clled recurrece reltio. There re two specil tpes of sequece which lmost ll of ou will hve met before, d these re briefl cosidered below. Arithmetic sequece sequece i which the differece betwee successive terms is costt e.g. {4, 7, 0, 3} or {,, 5}. The geerl form of rithmetic sequece is { + kd},where is the first term, d is the commo differece d is the umber of terms. Geometric sequece sequece i which the rtio of successive terms is costt e.g. {4, 8, 6, 3, 64} or {4,,,, 4,...}. The geerl form of geometric sequece is {r k },where is the first term, r is the commo rtio d is the umber of terms. Def. A series is the sum of terms i sequece. Arithmetic series from the rithmetic sequece, + d, + d,..., dd together the first terms to give the rithmetic series with sum S,sotht S =+(+d)+(+d)+... +(+( )d).

2 As ou m kow the formul for the sum c esil be derived b rewritig the bove epressio, reversig the order of the terms, so tht S =(+( )d)+(+( )d)+(+( 3)d) Addig the correspodig terms i the ltter two epressios the implies S =(+( )d)+(+( )d)+(+( )d)+...+(+( )d), ( terms) S = ( +( )d) = (first term + lst term), where first d lst refer to terms i the origil series. Geometric series from the geometric sequece, r, r,..., dd together the first terms to ield the geometric series with sum S, S = + r + r + r r. To clculte the sum multipl everthig i the epressio b r d subtrct from the first series givig rs =r + r + r 3 + r r S rs =( r)s = r, sice lmost ll the terms o the right-hd side ccel. It follows immeditel tht the sum of geometric series of terms is S = ( r ). r The bove results for the sums of rithmetic d geometric series with terms should be remembered. E. Clculte This is geometric series, with 5 terms, first term = 4, commo rtio r =. Hece, usig the formul, S 5 = 4( ( /)5 ) = 4 ( ) + 3 = 4 ( ) 33 3 = ( /) 4, which c esil be verified b direct clcultio. E. How m terms i the series re eeded to give sum of 34? This is rithmetic series of terms, for which = d d= 4. It is ot difficult to deduce tht the fil term, the th term, equls + ( )4 = d, therefore, 3 3 S = ( + (7 + 4)) = 34. Hece (8 + 4) = 34, = 0, or = 0, i.e. ( )( + 3) = 0, =, or 3/,

3 but the umber of terms must be positive so the swer to the questio is terms. Limit of sequece It is ofte importt to kow whether sequece coverges. The coditio c be stted mthemticll but i words we s tht sequece { } =0 coverges to it whe for ever smll positive vlue of ɛ ou choose there lws eists plce i the sequece (t which = N) beod which EVERY term i the sequece lies betwee ɛ d + ɛ. +ε - ε 5 Properties of coverget sequeces If sequece { } hs it d {b } hs it b the (i) { + b } hs it + b ; (ii) { b } hs it b ; { } (iii) hs it b ; b (iv) { b } hs it b. N E 3. Fid the its of { } =0 defied b () = +, (b) = () Dividig the umertor d deomitor b d usig properties (i) d (iii) bove it follows tht = + = + +0 = =. (b) Usig similr method s bove, but dividig here b,ledsto = = 5 Covergece of series You must be creful i decidig whether or ot series coverges. For emple, cosider the hrmoic series =. Ech term gets smller d pproches zero, but does the series coverge sice ifiite umber of smll terms m dd to give somethig sigifict (i.e. does the sum pproch it s?) B splittig the frctiol terms i the series ito pproprite groups (the legths of the groups beig powers of, mel oe, two, four, eight,...) the it is es to deduce tht > = 4, > 4 = 8, > 8 =, d so o. 6 3 =

4 The sum of ech group of terms is therefore greter th. With ifiite umber of terms i the origil series the groups ever stop (lthough successive oes coti icresig umber of terms) d so the sum of the series gets lrger d lrger, icresig b more th whe ech group of terms is dded. The hrmoic series does NOT coverge, therefore, d the series is sid to be diverget. Covergece of geometric series It ws show erlier i this sectio tht for geometric series of terms + r + r r = ( r ). r Wht hppes s, whe the series cotis ifiite umber of terms? The bove series clerl coverges whe r 0 d hece the series coverges if d ol if r < d diverges if d ol if r. For coverged ifiite geometric series S = r. D Alembert s rtio test is oe ver useful test for cofirmig, or otherwise, the covergece of ifiite series. u + Suppose we cosider the series u k, where ll the u k re positive, d suppose tht the it u eists d equls l. The d Alembert s rtio test sttes tht the give series is coverget if l<d diverget if l> (if l = the the test does ot provide coclusios bout covergece). N.B. A ecessr coditio for covergece of ifiite series is tht the geerl term i the series hs it zero s. Hece, if u C s d C 0, the the series is diverget. E 4. Determie whether the followig series re coverget () , (b) k k!, (c) k k +. () This is geometric series with r =. The rtio is less th d so the series coverges to sum S = =. (b) All terms i the series re positive d u k+ = k+ /(k +)! u k k = k+ /k! (k +)! k! k = k+ 0 s k. This it is < so d Alembert s test implies tht the series coverges. (c) I this cse u k = k k + = k k k+ k = + k = s k. +0 The terms do NOT ted to zero s k gets lrge d so the series diverges.. Rolle s theorem d me vlue theorems Cosider the two fuctios = f() show i the figure below. Clerl for ech fuctio there eists t lest oe poit = c where the grdiet of the curve is zero, i.e. there eists t lest oe poit = c such tht f (c) =0. 4

5 b The bove c be more precisel stted s Rolle s theorem If (i) f() is cotiuous for b, (ii) f() is differetible for < < b d (iii) f() =f(b) the there eists t lest oe umber c, where <c<b,suchtht f (c)=0. Coditio (ii) bove implies tht the fuctio f is ot required to be differetible t the ed-poits = d = b. The grdiet t these poits might be ifiite, therefore, but the theorem still holds (see figure ). Rolle s theorem does ot hold for fuctios which re ot differetible withi the itervl (see figure b), or for fuctios which re ot cotiuous throughout the itervl (see figure c). figure b d figure b b e figure c N.B. It is importt to ote tht Rolle s theorem sttes the eistece of c but does NOT tell ou how to fid its vlue! b First me vlue of itegrl clculus sttes: there eists t lest oe poit c such tht if f() is cotiuous i the closed itervl b the f(c) = b b f()d. (Geometricll, bove sttes tht: re uder grph = re of rectgle of height f(c), see figure d). c b c c b figure d figure e First me vlue of differetil clculus sttes: if f() is cotiuous i the closed itervl b d differetible i the ope itervl <<b the there eists t lest oe poit c such tht f (c) = f(b) f() b (Geometricll, the bove sttes tht the slope of the tget t the poit = c is prllel to the stright lie joiig the ed-poits of the curve (see figure e)) 5.

6 3. Tlor s d Mcluri s theorems How ccurtel c we pproimte fuctios, b usig polomils s? This is prtl swered b the followig theorem. Tlor s theorem i (, ) the If f(), f (),...,f () () ll eist d re cotiuous i [, ], d f (+) () eists f()=f()+( )f ()+ ( )! f () ( )! f () ()+R (), where R () = ( )+ f (+) (c), ( +)! < c <. (The bove ottio for R follows tht i Jmes, but it should be oted tht the ottio R + is ofte used i the literture). Commets (i) The qutit θ,where 0<θ<, is ofte used isted of c, givig c = + θ( ). (ii)the bove theorem sttes the Lgrge form of the remider. The remider R () deotes the differece betwee the origil fuctio f() dtheth order polomil tht is used to pproimte it. Tlor s theorem shows the surprisig result tht however m terms ou choose to hve i our polomil pproimtio to f(), the differece betwee f() d this polomil c lws be epressed i comprtivel simple form (Lgrge s form), lthough ou do NOT kow the vlue of c.kowledge of the rge of c, orθ, c ofte me the mimum mgitude of the remider c be clculted, d ot guessed. (iii) Ifiite series,wherethe re idepedet of, re clled power series. =0 Def. A power series hs rdius of covergece r if the series coverges whe <r d diverges whe >r. There re m importt power series. For emple e =++! r r! +... for ll, (i.e. r = ) si = 3 3! for ll (i.e. r = ). 5! (iv) Choosig = 0 i Tlor s theorem gives Mcluri s theorem: f() =f(0) + f (0) +! f (0) ! f() (0) + R (), where R () = + ( +)! f(+) (θ), 0 <θ<. Tlor s d Mcluri s series I Tlor s d Mcluri s theorems if R 0 s the the fuctio f() c be represeted b the correspodig ifiite series: f() =f()+( )f () ( )r f (r) ()+... (Tlor s series) r! f() =f(0) + f (0) r r! f(r) (0) +... (Mcluri s series) 6

7 E 5. Use Mcluri s theorem, with two terms d remider, to show tht the error i writig si s is less th if 0 <<0. With = Mcluri s theorem sttes f() =f(0) + f (0) + R (), R ()=! f (θ) (0 <θ<). Now f() =si, f(0) = 0 f () =cos, f (0) = cos 0 = f () = si, d substitutig ito Mcluri s theorem bove gives si =0+ +R =+R, where R () = ( si(θ)), 0 <θ<.! Now R = si(θ), sice si(θ),! d it the follows tht if 0 <<the R < (0.) = Thus replcig si b hs error of mimum mgitude (provided 0 <<). E 6. Estblish the covergece of Mcluri s series for cos. For this fuctio f() = cos, f(0) =, f () = si, f (0) = 0, f () = cos, f (0) =, f (3) () = si, f (3) (0) = 0, f (4) () = cos, f (4) (0) =, d therefore { 0 if is odd f () (0) = ( ) / if is eve Mcluri s theorem the ields f() =f(0) + f (0) r r! f(r) (0) + R r =! ! ()! ( ) + R +, where R + = + ( +)! ( )+ cos(θ), 0 <θ< R + + ( +)! = + ( +)!. The ltter term teds to zero s for ll, becuse however lrge is the s icreses the deomitor will lws become bigger th the umertor. Thus the remider teds to zero d the series coverges givig, for ll, cos =! ( ) 4! ()!

8 4. L Hopitl s rule It is ofte ecessr to obti its of quotiets s. For istce, s 0 the How do ou discover whether it eists? Oe useful result is stted below. L Hopitl s rule If (i) f() =g() = 0 d (ii) f d g re differetible t = the ( ) f() = f () g() g (), provided g () 0. si 0 0 =? f () If g () = 0 the keep differetitig the two fuctios, oe derivtive t time, util t lest oe of the 0 derivtives f () () d g () () is o-zero (see E 9 below). E 7. ( ) si Fid. 0 Usig l Hopitl s rule ( ) E 8. Fid e. Usig l Hopitl s rule E 9. 0 ( ) cos Fid ( ) si ( ) si = si 0 0 = 0 0 =? ( cos ) = = cos 0 = 0 = ( ) e = =? ( ) ( ) e = e =0 ( ) cos Usig l Hopitl s rule i this cse leds to 0 Here we hve to differetite gi which is the required swer. ( ) cos 0 = cos 0 0 = = =? = 0 ( ) si = si 0 (0) = 0 0 =? ( ) si ( cos ) = = cos 0 = 0, rec/00lfc 8