MAT 127: Calculus C, Fall 2009 Course Summary III

Transcription

1 MAT 27: Clculus C, Fll 2009 Course Summry III Extremely Importt: sequeces vs. series (do ot mix them or their covergece/divergece tests up!!!); wht it mes for sequece or series to coverge or diverge; power series Very Importt: covergece/divergece tests for series; rdius d itervl of covergece for power series; differetitio, itegrtio, d limits of fuctios vi power series; Tylor series Importt: estimtig ifiite sums by fiite sums; fidig rdius d itervl of covergece of power series; determiig Tylor series of fuctios relted to stdrd oes; pplictios of power series to computig sums of series Series (cot d) (9) Like the 3 covergece/divergece tests of 8.3 (Itegrl Test, Compriso Test, d Limit Compriso Test), the Rtio Test is iteded for series with positive terms: if the sequece { } hs positive terms d + lim + lim = L <, the the series coverges; = L > or + /, the the series diverges. This test sys othig if + /. This is ot surprisig sice = + p = lim =, while the series coverges if d oly if p>. Thus, the rtio test is ot suitble for series tht ivolve oly powers of (e.g. 3 ), d ot somethig growig fster. O the other hd, it usully works very well with series tht ivolve! d expoets of (e.g. 3 or ). For this reso, this is the test used to determie the rdius of covergece of power series i (but other tests re usully required to determie covergece t the ed poits of the resultig itervl). Fctors of do ot effect the vlue of L i the Rtio Test; for exmple, L is the sme for ll three series 3, 3, 3. I 8.6 d 8.7, this trsltes ito the derivtive d ti-derivtive of power series hvig the sme rdius of covergece s the origil series (but the itervl of covergece my chge slightly). Ulike the 3 tests of 8.3, the Rtio Test is completely self-test: you do ot hve to guess secod sequece {b }, which is required for Compriso Test d Limit Compriso Test, or fuctio f = f(x) such tht f() =, which is required for Itegrl Test ( guessig the fuctio f usully ivolves replcig by x if this mkes sese (e.g. x! does ot); while this is esy, determiig whether the resultig itegrl is fiite or ot my be less so). The Rtio Test is cosequece of

2 the Compriso Test pplied to geometric series with r=(l+)/2. (0) The Altertig Series Test pplies to rrow, but importt, set of series with terms of differet sigs: if lim = 0, > +, d the sigs of lterte ( > 0 for every odd d <0 for every eve, or the other wy roud), the the series coverges The ltertig-sig coditio is typiclly exhibited by the presece of ( ) or ( ) = ( ) ; however, mke sure to lso check the first two coditios before cocludig tht the series coverges. Typicl exmples re the series like ( ), ( ) (l)2 ; both coverge by the AST. The Altertig Series Test is covergece test oly: it sttes tht series coverges if it meets 3 coditios. It c ever be used to coclude tht series diverges; i this sese, it is the opposite of the most importt divergece test, which c ever be used to coclude tht series coverges. If the first coditio i the Altertig Series Test is ot stisfied, the series does ideed diverge, but by the most importt divergece test. However, there re lots of series tht fil either the secod or third coditio (or both), but still coverge; for exmple, there re coverget series with oly positive terms, tht decy to zero, but re ot strictly decresig, e.g. 2 + ( ) 2. The Altertig Series Test is cosequece of the defiitio of covergece for series (covergece of the sequece of prtil sums) d the Mootoic Sequece Theorem. () The substce of Absolute Covergece Test is tht itroducig some mius sigs ito coverget series with positive terms does ot rui the covergece: if the series coverges, the so does the series This test is useful whe the sigs re rdom, s opposed to strictly ltertig s required for si the Altertig Covergece Test. For exmple, the series 2 coverges by the ACT, becuse the series si 2 = si 2 coverges sice 0 si / 2 / 2 d the series coverges (this rgumet uses 3 tests: 2 Absolute Covergece, Compriso, d p-series; Limit Compriso Test is less suitble i this 2

3 cse). The Altertig Series Test cot be pplied i this cse, becuse the sigs of si do ot lterte: si, si2, si3 > 0, si4, si5, si6 < 0; while the sigs usully come i triples, occsiolly there re four cosecutive terms with the sme sig. While the Absolute Covergece Test is less striget bout the ltertig sig coditio th the Altertig Series Test, the former is ot substitute of the ltter. While either test c be ( ) used to coclude tht the series coverges, oly the Altertig Series Test is pplicble to the series 2 ( ) becuse the series ( ) = si does ot coverge. Neither of the two tests directly implies tht the series coverges. As the Altertig Series Test, the Absolute Covergece Test is covergece test oly; it c ever be used to coclude tht series diverges. The Absolute Covergece Test is cosequece of the Compriso Test d the dditio rule for series. (2) The geerl Rtio Test stted i the book exteds the bove Rtio Test for series with positive terms to series with rbitrry ozero terms (so tht + / mkes sese): if 0 for ll ( some N), d lim + = L <, the the series coverges; lim + = L > or + /, the the series diverges. The first sttemet follows from the Absolute Covergece Test d the Rtio Test for series with positive terms. As with the Rtio Test for series with positive terms, the secod sttemet follows from the most importt divergece theorem s it implies tht teds to ifiity d thus the sequece { } does ot coverge to 0. Similrly to the Rtio Test for series with positive terms, this more geerl Rtio Test sys othig if + /. (3) The sum of coverget series s m = c be estimted by fiite sub-sum: the sum =m = m this series does ideed coverge becuse of more geerl versio of the Altertig Series Test, clled Dirichlet s Test: if {b } d {s } re two sequeces such tht lim b = 0, b b +, d there exists C > 0 such tht =m s C for ll m, the the series sb coverges; i the cse of the Altertig Series Test s=±( ) is just the sig, d so C = works 3

4 of the first m terms; this is the m-th prtil sum. As m, s m pproches the sum of the series, so tht = s m 0. =m+ I some cses, the bove differece c be estimted: if f =f(x) is positive, decresig, d cotiuous o [, ) d m+ f(x)dx < < =m+ m f(x)dx coverges, the f(x)dx () Note tht icresig the lower limit (from m to m+ here) mkes the itegrl smller becuse f >0. I this cse, the fiite-sum estimte s m is uder-estimte for the ifiite sum becuse lots of positive terms re dropped from the ifiite series. if lim = 0, > +, d the sigs of lterte ( >0 for every odd d <0 for every eve, or the other wy roud), the < m+ =m+ d the sigs of d m+ re the sme (2) =m+ I this cse, the fiite-sum estimte s m is uder-estimte for the ifiite sum if m <0 d over-estimte if m >0 (so determied by the lst term used i the estimte). For exmple, let s estimte the sum of the series 2 to withi /5. Sice f(x) = /x2 > 0 is cotiuous d decresig o [, ), by () we eed to fid the smllest iteger m such tht m f(x)dx = So m=5 d the required fiite-sum estimte is =5 m x 2dx = m 5. 2 = = = This is uder-estimte for the ifiite sum, s oly positive terms re dropped off from the ltter. ( ) Let s ext estimte the sum of the series to withi /5. Sice this series is ltertig (odd terms > 0, eve terms < 0), / 0, d /(+)>/, by (2) we eed to fid the smllest iteger m such tht m+ = m + 5. So m=4 d the required fiite-sum estimte is =4 ( ) = = = 7 2 4

5 This is uder-estimte for the ifiite sum, s the lst term i the estimte is egtive. Remrk: The estimtes () d (2) re closely tied to the Itegrl Test d the Altertig Series Test for covergece of series. I priciple, there re estimtes relted to other covergece tests, i prticulr the Rtio Test, but they re ot discussed i the textbook. Power Series () A power series is fuctio defied by ifiite series of the form f(x) = c (x ), (3) =0 where is some fixed umber, typiclly 0, clled the ceter of the power series (3); pluggig i x= mkes ll the terms (x ) to be 0. So the ceter of the power series f(x) = (x+) =0 is x =. By covetio used i defiig power series, (x ) 0 = eve if x =. By similr covetio, 0! = so tht (+)! =! (+) for ll o-egtive itegers. (2) For ech x for which the series (3) coverges, we obti umber f(x). I prticulr,! f() = c c 0 + c = c 0 ; so the power series (3) lwys coverges t its ceter x=. The most fudmetl questio bout power series is the set of the umbers x for which the power series coverges. By the Mi Theorem bout Power Series, this set c be of oe of 3 types, with oe types hvig 4 sub-types: The power series c (x ) coverges either =0 () for x= oly; (b) for ll x; (4) (c) for x i oe of the itervls ( R, +R), [ R, +R), ( R, +R], or [ R, +R] for some R>0 d diverges otherwise The four possibilities i (c) re illustrted below: diverge coverge diverge?? R +R x Accordig to this theorem, the set of vlues of x for which power series coverges cot be rbitrry; it must be itervl, which is cetered t the ceter of the power series, my cosist of sigle poit, be ifiite, or be of fiite ozero legth d i the lst cse c be ope, closed, or 5

6 hlf-ope (so the series c either coverge or diverge t ech of the two ed-poits of the itervl; this is idicted by the questio mrks i the sketch). The itervl o which power series coverges is its itervl of covergece. The umber R i (c) is the rdius of covergece of the power series; R=0 i () d R= i (b). (3) I order to fid the rdius of covergece of power series s i (3) with c 0 for ll ( some N), use the geerl Rtio Test with =c (x ) 0 (so we ssumig x, sice we lredy kow tht the power series coverges for x=): + c + x + c + lim = lim c x + = x lim. (5) c I geerl, the lst limit i (5) does ot hve to exist. However, i ll exmples ecoutered i this clss it either exists or c + / c. I these cses: if c + / c, the lst umber i (5) is ; by the geerl Rtio Test the series (3) diverges for every x, d so we re i cse () of (4) d R=0. if c + / c 0, the lst umber i (5) is 0; by the geerl Rtio Test the series (3) coverges for every x, d so we re i cse (b) of (4) d R=. if c + / c C 0, the lst umber i (5) is C x ; by the geerl Rtio Test the series (3) coverges if x </C d diverges x >/C. So we re i cse (c) of (4) d R=/C. Oce the rdius of covergece is foud, fid the itervl of covergece of the power series. If R=0, the the itervl of covergece is just [, ]; if R=, the the itervl of covergece is (, ). If R 0,, it remis to determie whether the power series coverges for x= R d for x=+r, i.e. you hve to determie seprtely whether ech of the two power series c ( ) R =0 d c R =0 coverges. You will hve to use some covergece/divergece tests for series, but ot the Rtio Test (it would give i the limit d so be icoclusive i these two cses). Oce this is doe, the itervl of covergece will be s i oe of the 4 subcses i (c) of (4). Remrk : If c ivolves! i the umertor d the remiig terms re powers d expoetils of, such s + or 2 (but ot ), the you ll be i cse () of (4). If c ivolves! i the deomitor d the remiig terms re powers d expoetils of, such s 3 or 5 (but ot ), the you ll be i cse (b) of (4). If c ivolves oly powers d expoetils of, such s 3 / 2 + or 3 (but ot ), the you ll be i cse (c) of (4); this is the cse whe you ll lso eed to determie whether the series coverges or diverges t ech of the two ed-poits of the itervl of covergece seprtely. Remrk 2: By the bove, if c + / c C for some oegtive umber C or for C =, the the rdius of covergece of the power series is R=/C. I geerl, c + / c my ot pproch ythig, icludig, becuse it my keep o jumpig. For exmple, c + / c with odd might pproch 0 d c + / c with eve might pproch ; the 0 d re sid to be limits of subsequeces. There c be lots of such limits of subsequeces, but there is lwys t lest oe (possibly ±). The lrgest of such limits is deoted lim sup (d the smllest lim if). If sequece 6

7 coverges, lim sup is just its usul limit. If C is lim sup of the sequece c + / c, the the rdius of covergece of the power series (3) is still R=/C. You c ler more bout lim sup i MAT 320. (4) If the rdius of covergece of power series is 0, the power series is rther useless. However, if its rdius of covergece R is positive (possibly ), it defies smooth fuctio f(x) o ( R, +R). This fuctio c be differetited d itegrted by differetitig d itegrtig the power series term by term (like polyomil): If the rdius of covergece R of the power series the fuctio f(x) = f (x) = c (x ) =0 c (x ) = c (x ) is positive (possibly ), =0 is smooth o the ope itervl ( R, +R), (+)c + (x ), the rdius of covergece of this =0 power series is still R, d if R d the series for f diverges for x = ±R, so does the series for f ; c c f(x)dx = C + + (x )+ = C + (x ), the rdius of covergece =0 of this power series is still R, d if R d the series for f coverges for x=±r, so does the series for f(x)dx. (6) So the rdius of covergece of power series does ot chge uder differetitio d itegrtio, but the itervl of coverge my chge if R : differetitio my remove oe or both of the edpoits from the itervl of covergece, while itegrtio my dd them to the itervl covergece. For exmple, the geometric series x = + x + x (7) =0 coverges if d oly if x < d for ech x such tht x < it coverges to /( x). So the rdius of covergece of the series (7) is, its itervl of covergece is (, ), d x = x = + x + x if x <. (8) =0 Differetitig both sides of (8) with respect to x gives ( x) 2 = x = + 2x + 3x if x <. (9) The rdius of covergece of this power series is still, while the itervl of covergece is still (, ) becuse there re o eds to potetilly drop off from the itervl of covergece of the power series (8). Itegrtig both sides of (8) from x=0 (this mkes C =0 i (6)) gives l( x) = x 0 ( u) du = + x+ = =0 7 x = x + 2 x2 + 3 x (0)

8 The rdius of covergece of this power series is still, but the itervl of covergece my hve icresed by either or both of the two edpoits; this hs to be checked seprtely. Settig x= i the series i (0) gives ; this series diverges by the p-series Test. Settig x= i the series ( ) i (0) gives ; this series coverges by the Altertig Series Test. Thus, the itervl of covergece of the power series i (0) is [, ) d l( x) = Tkig x=/2 d x= i () gives x = x + 2 x2 + 3 x if x <. () (/2) l(/2) = = 2 = , ( ) l(2) = = , Sice l(/2)=l(2), we fid tht 2 = l2 = ( ) (2) Remrk : You do ot eed to memorize the two formuls i (2), but you eed to uderstd d be ble to pply the priciples ivolved i obtiig them; i prticulr, you hve to be ble to fid sums of logous ifiite series. You hve to remember the formuls for differetitig d itegrtig power series give i (6); rememberig the first of the two formuls for ech should suffice d should be firly esy, sice these re just differetitio d itegrtio of (ifiite) polyomils. If you sked to fid the sum of ifiite series s i (2), you eed to be ble to see tht it is obtied from some power series by replcig x with specific vlue i the rge of the covergece of x. You should the be ble to recogize the power series d kow wht fuctio it sums up to, t lest fter droppig sme fctors of from ll terms. By (6), extr fctors of correspod to differetitio or itegrtio of the power series you recogize (but be creful to check tht the expoets of x re correct d ot shifted by fixed umber; if they re shifted, just tke power of x outside of the summtio). You c the determie the fuctio to which the origil power series correspods d sum up the strtig ifiite series by evlutig this fuctio t the pproprite vlue of x. Remrk 2: the sttemets of (6) regrdig the rdii of covergece follow from Remrk 2 i (3) bove; so you c ctully verify them ssumig c + / c C for some C 0 (possibly ). (5) Limits of fuctios defied vi power series c be esily computed, s log s the limit is tke t the ceter of the power series. This geerlly ivolves writig out the first few terms of the power series. For exmple, the fuctio x f(x) = 8

9 is defied wheever x d thus for ll x close to 0; so it mkes sese to sk bout limits of relted fuctios t x=0. I prticulr, ( ) ( ) x + x2 2 + x3 3 + x x x 2 x x f(x) x 2 lim x2 x 0 x 3 = lim x 0 x 3 = lim x 0 x 3 ( = lim x ) 4 x +... = 3 ; o the secod lie... deotes terms ivolvig x d higher powers of x, ll of which pproch 0 s x 0. You c compute this limit usig l Hospitl s rule s well, but it would hve to be pplied 3 times, re-checkig the required ssumptios ech time (i this cse, this would me checkig tht the umertor d deomitor both pproch 0). (6) Two power series with the sme ceter, sy 0, f(x) = f x d g(x) = =0 g x c be multiplied together by tretig them s ifiite polyomils d collectig coefficiets of ech power of (x ): ( )( ) f(x)g(x) = f x g x = =0 =0 =0 ( )( ) f 0 + f x + f 2 x 2 + f 3 x g 0 + g x + g 2 x 2 + g 3 x = f 0 g 0 + (f 0 g + f g 0 )x + (f 0 g 2 + f g + f 2 g 0 )x 2 + (f 0 g 3 + f g 2 + f 2 g + f 3 g 0 )x ( k= = f k g k )x. =0 k=0 So the coefficiet of x i the product is the sum + terms, ech of which is product of term from the f-series d term from the g-series. If the f d g-series coverge for x < R, the so does the fg-series. For exmple, ( x) 2 = x x = ( )( ) x x =0 =0 = + ( + )x + ( + + )x = x ; this grees with (9), s well s with the k= 2 cse of (9) below. A more iterestig exmple is x 2 3x + 2 = (x )(x 2) = x /2 x/2 = ( )( ( ) x ) x = ( =0 =0 =0 x )( =0 x ) 2 ( = ( + ) ( x ) ) 2 + x = + (/2) + ( x = ) 2 /2 2 + x =0 9 =0

10 The secod-to-lst equlity bove is the (, b) = (, /2) cse of + b + = ( b) ( b 0 + b b + 0 b ) = ( b) ( b + b b + ) ; this formul ws used to sum up geometric series. Aother wy to expd /(x 2 3x+2) roud x=0 is to use prtil frctios d dditio of series, isted of multiplictio (dditio is much simpler): ( x 2 3x + 2 = (x )(x 2) = 2 ( ) x ) = x 2 x + x 2 = x /2 x/2 = x ( ) x = x x = ( ) 2 + x =0 The itervl of covergece of this power series c esily be see to be (, ). =0 (7) There is t most oe wy to expd fuctio ito power series cetered t give poit: =0 =0 =0 If f(x) = c (x ) o ( R, +R) for some c 0, c,... d R>0, the =0 f is smooth fuctio o ( R, +R); (3) c = f (), where f =f (x) is the -th derivtive of f, f 0 = f, 0!=.! A power series expsio of f roud x =, if it exists, is clled the Tylor series expsio of f roud x=. By the first sttemet i (3), the fuctio f(x)= x does ot dmit Tylor series expsio roud x=, though it does dmit such expsio roud y x= 0: { + (x ) o (0, ), if >0; x = (x ) o (, 0), if <0. The secod sttemet i (3) provides method of determiig the Tylor coefficiets c of f t x=. However, this method is prcticl oly if ll derivtives of f c be computed; this c be doe i some cses, icludig f(x) = p(x) is polyomil of degree d; the p(x) = p() + p (x)(x ) + p () 2! (x ) p d () (x ) d, (4) d! becuse p (x) = 0 if > d; furthermore, p d ()/d! =. The series o the right-hd side of (4) coverges for ll x becuse it is fiite sum. The equlity of the left d right expressios i (4) lso holds for ll x, becuse of Tylor s Iequlity below (ot becuse the right expressio is fiite sum). f(x) = /( x): i this cse f (x)=!/( x) + s c be see by iductio, d so x = f (0) (x 0)!/ + = x = x if x <. (5)!! =0 =0 The series o the right-hd side is geometric series with r = x. It coverges if d oly if x <; if so, it coverges to /( x) s stted i (5). 0 =0

11 f(x)=e x : i this cse f (x)=e x for ll d e x = =0 f (0) (x 0) =! =0 e 0! x = =0 x!. (6) The series o the right-hd side coverges for ll x by the Rtio Test. The equlity of the left d right expressios i (6) lso holds for ll x, becuse of Tylor s Iequlity below (ot becuse the series coverges for ll x). f(x)=cos x: i this cse f 4 (x) = cos x, f 4+ (x) = six, f 4+2 (x) = cos x, f 4+3 (x) = si x for ll, s c be see by iductio. Sice cos 0 = d si 0 = 0, cos x = =0 f (0) (x 0) =! =0 ( ) cos 0 x 2 = (2)! =0 ( ) x 2 (2)!. (7) The series o the right-hd side coverges for ll x by the Rtio Test. The equlity of the left d right expressios i (7) lso holds for ll x, becuse of Tylor s Iequlity below (ot becuse the series coverges for ll x). f(x)=si x: i this cse f 4 (x) = six, f 4+ (x) = cos x, f 4+2 (x) = six, f 4+3 (x) = cos x for ll, s c be see by iductio. Sice cos 0 = d si 0 = 0, six = =0 f (0) (x 0) =! =0 ( ) cos 0 x 2+ = (2+)! =0 ( ) x 2+ (2+)! (8) The series o the right-hd side coverges for ll x by the Rtio Test. The equlity of the left d right expressios i (8) lso holds for ll x, becuse of Tylor s Iequlity below (ot becuse the series coverges for ll x). f(x) = (+x) k, k is y rel umber d x < (so tht (+x) k is defied eve if k is ot iteger): i this cse for ll, s c be see by iductio. Thus, ( + x) k = =0 f (x) = k(k )(k 2)...(k +)(+x) k f (0) (x 0) =! ( ) k where =0 ( ) k (+0) k x = =0 = k(k )(k 2)...(k +).! ( ) k x if x <, (9) The rdius of covergece of the series o the right-hd side is by the Rtio Test, uless k is o-egtive iteger (i which cse RHS of (9) is fiite sum d so coverges for ll x). The equlity of the left d right expressios i (9) lso holds for x <, becuse of (ot becuse the series coverges if x <). The idetity (9) is kow s biomil series. The

12 geometric series (5) is specil cse of (9), with k = d x replced by x. If k is o-egtive iteger, so tht (+x) k is polyomil, for ll >k ( ) k = k(k )(k 2) (k +) = 0! d so the series (9) hs oly fiitely my terms, d the idetity holds for ll x. I my cses, it is ot prcticl to compute ll derivtives of fuctio d so it my ot be possible to use the formul i (3) to compute the Tylor coefficiets directly. However, it my be possible to obti the Tylor expsio for give fuctio by usig oe of the stdrd series (5)-(8). For exmple, x 5 e 3x2 = x 5 ( 3x 2 )! =0 = x 5 ( 3) (x 2 )! =0 = x 5 ( 3) x 2! =0 ( 3) x 2+5 = ;! =0 sice the Tylor series for e x coverges for ll x, so does the bove Tylor series for x 5 e 3x2. Similrly, x 5 + 3x 2 = x 5 ( 3x 2 ) = x5 ( 3x 2 ) = x 5 ( 3) x 2 = =0 =0 ( 3) x 2+5. Sice the Tylor series for /( x) coverges if x <, the bove Tylor series coverges if 3x < (whtever is used for x i the power series lso hs to be used i the boud for covergece); so it coverges if x </ 3. Remrk : Whe you use Tylor series for oe fuctio to get Tylor series expsio for other fuctio, mke sure your fil swer is power series i x (or (x ) if the ceter 0), ot power series i, sy, 3x 2 or x 2, d ot product of power series with, sy, x 5 (see the two exmples bove). While there re my differet wys to describe fuctio, there is t most oe wy to write it power series. Remrk 2: You should remember the formul for the Tylor coefficiets c i (3) or equivletly the geerl Tylor expsio formul: f(x) = =0 f () (x ) ;! this formul is ofte used with =0. O the other hd, the four formuls (5)-(8) will be provided for the exm (see the lst pge of Fil Exm Iformtio). You c tke these four power series expsios, log with their itervls of covergece, s give d use them s pproprite. They my be helpful i obtiig other Tylor series, computig limits, d computig sums of ifiite series. For full credit, you must derive y other power series formul you use o the exm, either directly from the Tylor coefficiet formul i (3) or from oe of the four give Tylor series. I prticulr, you my eed to derive formul like (4) for specific polyomil p(x) roud specified ceter x = ; see 8.7 9,0 for exmples. You should ot memorize the biomil formul (9); if ythig relted, with specific k, ppers o the exm, you should ot quote the biomil formul ywy. =0 2

13 Remrk 3: 2 The key to determiig whether give fuctio dmits Tylor series expsio roud give poit is Tylor s Iequlity: =m f(x) =0 f () (x ) C m+ (f; R) Rm+! (m+)! if R x + R, (20) where C m+ (f; R) is the mximum vlue of f m+ (x) with x i [ R, +R]. This iequlity is obtied s follows. Let =m f () R m+ (x) = f(x) (x ).! The, =0 R m+ () = 0, R m+() = 0,..., R m m+ () = 0. Thus, by the Fudmetl Theorem of Clculus, Thus, for ll x i [, +R]: x R m m+ (x) = R m m+ () + R m m+ (x) = R m. m+ () + x x R m+ (x) = R 0 m+ (x) = R 0 m+ () + R m m+ (x) R m m+ (x). R 0 m+ (x) x R m+ x R m x R R m+ m+ (u)du = x R m R m+ (u)du = x m+ (u)du = x R m+ R m R m+ (u)du m+ (u)du m+ (u)du x m+ (u) du C m+ (f; R)du C m+ (f; R) x x m+ (u) du C m+ (f; R) x du C m+(f; R) x 2 2! m+ (u) du x C m+ (f; R) m! x m du C m+(f; R) x m+ (m+)! The sme estimtes holds if x lies i [ R, ]. This cofirms (20). By (20), if the =m f(x) = lim m =0 lim C (f; R) R m! f () (x ) =! = 0, =0 f () (x ),! d so the Tylor series for f does ideed equl to f o [ R, +R]. I prticulr, this is the cse if f(x) = p(x) is polyomil of degree d: sice p (x)=0 if >d, C (f; R)=0; f(x)=e x, =0: sice f (x)=e x for ll, C (f; R)=e R ; 2 you c skip this remrk i studyig for the fil exm 3

14 f(x)=cos x or f(x)=si x: sice f (x) is ± cos x or ± six, C (f; R)= if R π. (8) Power/Tylor series c be used to compute sums of some coverget ifiite series,, d eve check covergece (i some cses oly). Begi by writig the ifiite series s evlutio of some power series t some poit: = =0 c (x ) x=b ; =0 so you eed to guess pproprite sequece {c } s well s the ceter of the series d the evlutio poit b, but i some cses they my be evidet. For exmple, 2 = ( ) x= = 2 x, 2 = ( ) = x x= You ext eed to fid simple formul for the fuctio g(x) = =0 c (x ) x=b. It my ot be oe of the stdrd Tylor series, but my become such fter droppig frctio ivolvig powers of. For exmple, x, x =0 x = x. I light of (6), the fuctio g =g(x) c the be recostructed from the fuctio f =f(x) through differetitio d/or itegrtio d possible multiplictio by power of x fter ech step to ccout for differeces i the expoet if y; i the cse of itegrtio, the costt C hs to be chose ppropritely s well. For exmple, ( x + C = x )dx dx = x = l( x) + C = ( ) ( ) x = x x = x x x = x = x ( x) 2; =0 =0 x = l( x); the lst equlity o the first lie is obtied by settig x = 0. The itervl of covergece for the g-series c be determied from the f-series. For exmple, the rdii of covergece of both series x = l( x), x = re, sice this is the rdius of covergece of the geometric series x ( x) 2, (2) x d the rdius of covergece of power series does ot chge uder differetitio or itegrtio. Sice the itervl of covergece of x is (, ), this is lso the itervl of covergece of the power series x sice differetitio c oly remove (but ot ecessrily) the ed-poits from the itervl of covergece (d there re ot y ed-poits to remove i this cse). O the other hd, itegrtio c oly dd i the ed-poits; sice x coverges for x= d diverges t x=, the itervl 4

15 of covergece of this power series is [, ). Oce it is estblished tht the required evlutio poit b lies iside of the itervl of covergece of the g-series, we obti = =0 c (x ) x=b = g(b). =0 For exmple, sice /2 lies i the itervls of covergece of the two power series i (2), 2 = x= x 2 2 = x x= = 2 = l( x) x= 2 x ( x) 2 x= 2 mke sure to simplify the fil swer s much s possible. = l(/2) = l(2), = /2 /4 = 2; Remrk: If the g-series is lredy stdrd series, you do ot eed to do y differetitio or itegrtio; just plug i the evlutio poit to get the sum, s log s the evlutio poit lies i the itervl of covergece of the power series. I ll cses, if you ctully kow tht the ifiite series coverges, the evlutio poit utomticlly lies i the itervl of covergece. If you re sked to justify tht the series coverges, you either eed to use oe of the covergece/divergece tests for series or show tht the evlutio poit lies iside of the itervl of covergece. I order to do so, it is ofte sufficiet to determie just the rdius of covergece: if the distce from the ceter to the evlutio poit is (strictly) less th the rdius of covergece, th the evlutio poit lies i the itervl of covergece. This is the cse i the two exmples bove, sice the distce from 0 to /2 is less th. Covergece/Divergece Tests for Sequeces d Series (recp) The two most importt thigs regrdig Chpter 8 re distiguishig betwee sequeces d series d their covergece/divergece tests; relizig tht the covergece/divergece issue cocers wht hppes with the ifiite til. Thus, droppig the first 59 terms of sequece or series will ot chge its covergece/divergece property. If series does coverge, droppig the first 59 terms will however chge the sum of the ifiite series, precisely by the sum of the first 59 terms. Cofusio bout these two poits, especilly the first oe, ws the primry reso for the low scores o the secod midterm; ot kowig which covergece/divergece test ws rther secodry. Whether sequece/series coverges or diverges depeds primrily the domit terms d the presece of y sig-ltertig or oscilltory behvior, such s ( ) or si; fctors like si(/) d cos(/) re ot oscilltory, sice they pproch 0 d, respectively, s 0. It is geerlly helpful to try to isolte the domit terms, essetilly by fctorig them out; if the terms re give by frctio, this usully mes dividig top d bottom by the domit term. The mi domice reltios to remember re: (l) p p lim q = 0, lim = 0, eq lim 5 e p (!) q = 0, lim (!) = 0

16 for y p, q > 0. However, oe hs to be creful with the domit terms if there re mius sigs betwee them. For exmple, while the domit term of = my pper to be 3 = 9, i fct = ( ) = = ( 2 3) + (2/9) + ; so the domit term is (2/3) (times /2, which does ot effect ythig). So the sequece coverges to 0, while the series coverges to somethig positive by the Limit Compriso Test pplied with b =(2/3). If the terms of sequece or series turlly split s sum of two terms, oe of which gives rise to coverget sequece or series, respectively, the you c drop the coverget term i determiig whether the etire sequece or series coverges. For exmple, the sequece = (+ ( ) )/ coverges if d oly if the sequece b =( ) / does, becuse the sequece c =/ coverges (to 0, which does ot mtter i this cse); so the sequece does coverge (to 0). Similrly, the series + ( ) coverges if d oly if the series does becuse the series ( ) coverges by the Altertig Series Test. Sice the series does ot coverge, either does the series + ( ). However, be creful ot to split off diverget sequece or series. For exmple, ( lim ) lim 9 +2 lim 3 ; ( ) ; (+) = ( ) + + becuse either of the two limits o the right-hd side o the first lie exists d either of the four sums o the right-hd side of the secod d third lies exists. Covergece/divergece of sequeces. A sequece is simply ifiite strig of umbers described i some wy, typiclly by explicit formuls, such s = ( ) 4 /(3 4 +), or by recursive formul, such s + = 6+, with some iitil coditio(s), such s = 6. While sequece is loger word th series, determiig whether sequece coverges or diverges is esier. If sequece is give by explicit formul, it is usully possible to determie whether it coverges through quick ispectio. Begi by splittig it ito prts if possible (ofte ot; be creful) d determiig the domit term; see bove. For exmple, = ( ) = ( ) 3 + / 4 ; so the domit term here is ( ). If pluggig i = mkes sese the, you re doe: the sequece coverges; for exmple, it mkes to plug i = ito /(3+/ 4 ), but ot ito 6

17 4 /(3 4 +) or ( ) /(3+/ 4 ), becuse / d ( ) do ot mke sese. Typiclly, sequece would ot coverge due to either oscilltory behvior, which my be exhibited by fctor of ( ) or si(), or becuse it (or prt of it) pproches, s /(l) does. However, the presece of oscilltory fctor does ot isure divergece; for exmple, the sequece ( ) = ( ) 3 + / 4 coverges to 0 becuse the seemigly oscilltory fctor i fct decys to 0. Occsiolly (if terms like 2,!, or re preset), somethig like Rtio Test for Sequeces my be useful: + if lim = L d L <, the the sequece { } coverges to 0; + if lim d L > or + /, the the sequece { } diverges. This test however is ot stted i the book. The first cse follows from the Rtio Test for Series d the most importt divergece test for series. However, i prctice, the implictio goes from the Rtio Test for Sequeces to the Rtio Test for Series. If sequece is give by recursive formul, begi by writig the first few terms to get ide whether sequece coverges or diverges. If it ppers to coverge, the Mootoic Sequece Theorem my be useful to justify this (so you my eed to use iductio to show tht either the sequece is bouded bove by somethig d icresig or bouded below d decresig). If it ppers to diverge, this is likely due to some oscilltory behvior or becuse of goig off to ifiity; you ll eed to justify tht this ptter cotiues s icreses. The Squeeze Theorem for Sequeces my be useful i some cses, but is geerlly voidble. I some cses, it my be possible to replce by x d compute the limit s x ; this my llow usig l Hospitl s rule (if the required coditios re stisfied), but usully this will ot be the fstest pproch. Covergece/divergece of series. A series is the sum of terms i sequece, with the ltter typiclly give by explicit formul whe series re ecoutered. While series is shorter word th sequece, determiig whether series coverges is much hrder d the cocept of series itself is sigifictly more bstrct. First, series coverges if d oly if the sequece of prtil sums {s } defied by s = does; if this hppes, the ifiite sum of the s is defied to be the limit of the s s. Wht this mes is tht you keep o ddig more d more terms to the sum d see if the resultig sums pproch ythig. However, i prctice, it is lmost ever possible to fid explicit formul for s. Secod, there re 7 divergece/covergece tests for series, most with severl ssumptios tht you hve to remember to check. After tryig to split off coverget prt of series (e.g. / 2 from (/ 2 + (si )/ 3 )) d determiig the domit term, you might wt to try doig the followig to determie if the series coverges: 7

18 (0) if the sequece { } does ot coverge to 0, the series diverges. For exmple, the series ( ), 2 +, cos(/), si(), ll diverge. Note tht eve if lim 0 =0, the series my still diverge; this is the reso you eed the other hlf-doze covergece/divergece tests. () if the series is geometric series cr or p-series / p, you should kow immeditely if it coverges or diverges (but do ot cofuse these with other similrly lookig series; these two types of series re very restrictive, but lso very importt); (2) if the series ivolves i the expoet, e.g. 5 (but ot just 5 ),,!, or more geerlly products with the umber of fctors icresig with, try the geerl Rtio Test. (3) if the series hs positive terms oly, determie its ledig term, such s some power of, d pply the Limit Compriso Test with tht power of. Remember tht si(/) d t(/) look like / s, sice t(/) si(/) si(x) lim = lim lim cos(/) = lim =. / / x 0 x So by the Limit Compriso Test with b =/ p, the series si p (/), t p (/) coverge if d oly if p>. However, si() d t() do ot look like s. If the Limit Compriso Test is ot suitble, try to fid wy to use the Compriso Test; so you ll still eed to guess b, but ow the secod sequece eeds to stisfy differet requiremets (but still 3 of them). For exmple, the Limit Compriso Testt with b =/ 2 cot be used for si the series 2 becuse si / 2 lim = lim b / 2 = lim si does ot exists. However, we c use the Compriso Test with b =/ 2, becuse d the series 0 = si b = 2 2 coverges; this implies tht so does the smller series rgumet cot be used to directly coclude tht the series divergece of the series si does ot imply tht the smller series 3 this series does ideed diverge becuse si x + si(x+) /2 for ll x si 2. This diverges 3, becuse the si lso diverges. 8

19 Filly, you c try to see if the Itegrl Test is pplicble. For this, the fuctio f obtied from the terms of the series by replcig by x must mke sese for ll x (or t lest for x N for some N); for exmple, x! does ot mke sese. You lso hve to check tht the fuctio f obtied i this wy is positive, cotiuous, d decresig for x (or t lest for x N for some N). For exmple, while the fuctio f(x)= six /x mkes sese for x d is cotiuous, it is ot decresig (d or eve positive); so the fct tht the itegrl si x x dx diverges does ot sy ythig directly bout the ifiite series. The most importt use of the Itegrl Test hs bee to obti the p-series Test; it hs lso bee used i the preset of l. The Itegrl Test c be used to show tht ll of the series p, (l) p, (l)(ll) p,..., si p (/), t p (/) coverge if d oly if p>. Except for the lst 2 series, the relevt itegrl c ctully be computed firly esy. I the cse of the lst 2 series, the itegrl is much hrder to compute, but it c be show to be fiite if d oly if p >, which suffices; however, it is simpler to pply the Limit Compriso Test to the lst 2 series with b =/ p. (4) if the series hs terms of differet sigs, first try to see if the series meets ll three requiremets of the Altertig Series Test; stisfctio of the ltertig sig requiremet is likely to be idicted by the presece of fctor of ( ), but eve so do ot forget to check the other two coditios (d mke sure to stte them). If the Altertig Series Test does ot pply, try the Absolute Covergece Test; this my llow you to pply oe of the tests suitble oly for series with o-egtive terms. The Altertig Series Test is pplicble to the series ( ), but ot to the series si ; the Absolute Covergece Test is pplicble to the 2 ( ) secod series, but ot to the first. Both tests re pplicble to the series, but either si to the series. The oly coclusio you c ever drw from either of these tests is tht the series coverges; if you wt to show tht series with terms of differet sigs diverges, you eed to fid some other reso. (5) i rre cses, it is possible to determie whether series coverges or diverges by computig the correspodig sequece of prtil sums. This c be doe whe the series hs the form ( ) b b +m for some sequece {b }. If m, the -th prtil sum is the 2 s = s + s s = (b b +m ) + (b 2 b 2+m ) (b b +m ) = k=m k= k=+m b k k=+ b k, (22) sice the secod term i the k-th pir ccels with the first term i the (k+m)-th, provided k m; this leves the first terms i the first m pirs d the secod terms i the lst m pirs. 9

20 As, the first sum o the secod lie i (22) does ot chge; so the sequece {s } k=+m ( ) (d thus the series b b +m ) coverges if d oly if the sequece b k does. This k=+ hppes if the sequece {b } coverges, but my hppe eve if {b } diverges. For exmple, ll of the series ( ) ( ) si(/) si(/(+)), cos(/) cos(/(+2)), ( ) ( l() l(+2) ) coverge, while the series ( ) ( ) ( cos() cos(+), l() l(+), e e +) diverge. This kid of ccelltio is lso useful for computig sums of series like ( + 2) vi prtil frctios d prtil sums. However, for showig tht this series coverges, it is much simpler to use the Limit Compriso Test. Most importtly, try to see wht give series looks like, i terms of the ledig terms d oscilltory behvior if y; i most cses, you my be ble to guess whether it coverges or diverges rther quickly bsed o these. If you re sked to justify your swer, mke sure you check tht ll of the coditios of the test you wt to use hold; ofte this will me sttig the required properties, but sometimes dditiol justifictio my be required. For exmple, it is sufficiet to stte tht / 0, but some expltio is required to justify tht /( 2 +) 0. Remrk: The Itegrl Test for series is cosequece of the defiitio of itegrl. The Altertig Series Test is cosequece of the Mootoic Sequece Theorem, which i tur is fudmetl sttemet bout completeess of rel (but ot rtiol) umbers ( o holes i the rel umbers). The Compriso Test for Series is cosequece of the Squeeze Theorem for Sequeces. The Limit Compriso Test, Rtio Test for positive sequeces, d Absolute Covergece Test re cosequeces of the Compriso Test. The covergece sttemet the geerl Rtio Test is cosequece of the Rtio Test for positive sequeces d the Absolute Covergece Test; its divergece sttemet follows from the most importt divergece test for series. So, i priciple, wheever Limit Compriso Test, Rtio Test, or Absolute Covergece Test is usble, so is the Compriso Test (the Itegrl Test d the Altertig Series Test re fudmetlly differet). However, i prctice, wheever either Limit Compriso Test, Rtio Test, or Absolute Covergece Test is usble, it might be much esier to use them th the Compriso Test; for exmple, while it might be esy to guess limitcompre-to sequece {b }, it my be hrder to determie suitble compre-to sequece {b }. Good luck o the fil exm