NOTES ON INFINITE SEQUENCES AND SERIES

Transcription

1 NOTES ON INFINITE SEQUENCES AND SERIES MIGUEL A. LERMA. Sequeces.. Sequeces. A ifiite sequece of real umbers is a ordered uedig list of real umbers. E.g.:, 2, 3, 4,... We represet a geeric sequece as a,a 2,a 3,...,adits-th as a. I order to defie a sequece we must give eough iformatio to fid its -th term. Two ways of doig this are:. With a formula. E.g.: a = a = 0 a = With a recursive defiitio. E.g.: the Fiboacci sequece,, 2, 3, 5, 8,..., i which each term is the sum of the two previous terms: F = F 2 = F + = F + F.2. Limit of a Sequece. We say that a sequece a coverges to a limit L if the differece a L ca be made as small as we wish by takig large eough. We write a L, or more formally: E.g.: Date: 0/2/999. lim a = L. lim =0.

2 2 MIGUEL A. LERMA If a sequece does ot coverge we say that it diverges. E.g., the followig sequeces diverge: =,2,3,4, diverges (to + ) ( ) =,,,, diverges.3. Limit Laws for Sequeces. If lim a = A ad lim b = B, the: lim (a + b )=A+B lim (a b )=A B lim (a b )=AB lim (a /b )=A/B (provided B 0) So, a complicated limit such as L = lim ca be computed by replacig smaller parts of it with their limits / 0, /0 0: L = = Squeeze Law. If a c b,ada ad b have the same limit: a L, b L, thec has also the same limit: c L. This ca be used to compute limits such as the followig oe: I this case we have: si lim. si. Sice / 0ad/ 0the si 0also..5. Limits of Fuctios of Sequeces. If a = f() for some fuctio f ad lim f(x) =L, the lim a = L. This basically allows us to x replace limits of sequeces with limits of fuctios. I particular this is useful for usig L Hôpital s rule i computig limits of sequeces. E.g: e lim = lim e x x x e x = (L Hôpital s rule) = lim x =.

3 NOTES ON INFINITE SEQUENCES AND SERIES 3.6. Bouded Mootoic Sequeces. A mootoic sequece is a sequece that always icreases or always decreases. For istace, / is a mootoic decreasig sequece, ad =,2,3,4,... is a mootoic icreasig sequece. A sequece is bouded if its terms ever get larger i absolute value tha some give costat. For istace / is bouded, because / < 2 for every, but is ubouded. A property of ay bouded (icreasig or decreasig) sequece is that it always has a (fiite) limit. This might ot seem very useful if what we wat is to actually compute the limit, but i some cases it may help. For istace, cosider the sequece defied recursively i the followig way: a = 6, a + = 6+a ( ). It ca be show that a it is a bouded mootoic icreasig sequece, hece it has some limit A: lim a = A. Now, takig limits o both sides of a + = 6+a (here we use that the sequece has a limit) we get that A = 6+A, i.e., A 2 A 6 =0. Solvig this secod degree equatio we get A = 3orA = 2. Sice the sequece is positive, the limit caot be egative, hece it must be A =3: lim a =3..7. Homework Problems. E&P, 2.2: 9 42, 54, Ifiite Series 2.. Series. A ifiite series is a expressio of the form a = a + a 2 + a 3 +, Proof: We use iductio. First ote that 0 <a = 6<3. By addig 6 ad takig square roots we get 6 < 6+a < = 3, i.e.: a <a 2 <3. Now assume a <a + < 3foragive (iductio hypothesis). Agai, by addig 6 ad takig square roots we get 6+a < 6+a + < 3, i.e. a + <a +2 < 3 (iductio step). From here we get that a <a + < 3 for every, which proves both, a is icreasig ad is bouded by 3. 2 C.H. Edwards, Jr. & David E. Peey: Calculus with Aalytic Geometry, 5th editio, Pretice Hall.

4 4 MIGUEL A. LERMA where {a } is a sequece of umbers sometimes the series starts at = 0 or some other term istead of =. ItsNth partial sum is N S N = a = a + a 2 + a 3 + +a N Sum of a Series. The sum S = of a series is defied as the limit of its partial sums a S = lim N S N = lim N if it exists it this case we say that the series coverges. For istace, cosider the followig series: Its partial sum is S N = N N a 2 = = = N 2. N Hece, its sum is ( S = lim S N = lim ) =, N N 2 N i.e.: 2 =. A series may or may ot have a sum. For istace, i the followig series: ( ) = the sequece of partial sums S N =,0,,0,,0,... diverges, ad the series has o sum.

5 NOTES ON INFINITE SEQUENCES AND SERIES Telescopic Series. Telescopic series are series for which all terms of its partial sum ca be caceled except the first ad last oes. For istace, cosider the followig series: ( +) = Its th term ca be rewritte i the followig way: a = ( +) = +. Hece, its Nth partial sum becomes: N ( ) S N = Hece: ( +) = N + ( = ) ( ) ( ) = N+. ( + N ) N + ( ( +) = lim ) =. N N Geometric series. A geometric series a is a series i which each term is a fixed multiple of the previous oe: a + = ra,wherer is called the ratio. A geometric series ca be rewritte i this way: ar =a+ar+ar 2 +ar 3 +. If r < itssumis ar = a r. Note that a is the first term of the series. If a 0ad r, the series diverges. Examples: 2 = = 2 =2.

6 6 MIGUEL A. LERMA ( ) 2 = = ( 2 ) = 2 3. Note that i the last example r = a + /a = ( )+ /2 + ( ) / Termwise additio ad multiplicatio of series. If ad a = B the ad (a ± b )= where c is a costat. a + ca =c b = A ± B a = ca, = /2. a = A 2.6. The th Term Test for Divergece. If a does ot coverge to 0, the a does ot coverge Note: the reciprocal is ot true i geeral! (a couterexample is the harmoic series; see below.) For istace, the series si does ot coverge because si does ot coverge to The Harmoic Series. The series = is called harmoic series. The harmoic series diverges. This ca be prove graphically, by lookig at the graph of the fuctio f(x) =/x (fig. ). The terms of the harmoic series are the areas of the rectagles. Their sum is greater tha the area uder the graph of f(x) =/x betwee ad, which ca be computed with the followig improper itegral: dx = lim x M M dx = lim x [l M x]m = lim (l M l ) =. M

7 NOTES ON INFINITE SEQUENCES AND SERIES y=/x /2 /3 / x Hece, Figure. Theharmoicseries = Series that are Evetually the Same. If a = b for every large eough, the the series a ad b either both coverge or both diverge. I other words, the covergece or divergece of a series depeds oly o its tail a. =k 2.9. Homework. E & P,.3: 37, 49, 50, 64. Maple Worksheet: WS (due 0/7/99). 3. Taylor Series ad Taylor Polyomials 3.. Liear Approximatio of a Fuctio. Assume that we wat to compute the value of a fuctio such as si 0. orl.. Evaluatig polyomials is easy, it ca be accomplished by a sequece of computatios ivolvig oly arithmetic operatios (+,, ). For istace, if f(x) =x 2 +7x+2 the f(3) = = 32. For other fuctios i geeral we may ot be able to evaluate them exactly, but we ca

8 8 MIGUEL A. LERMA do it approximately. A first approximatio cosists of substitutig the graph of the fuctio by a taget lie (fig. 2). 0.5 P(x) = c0 + c (x-a) y=f(x) (a,f(a)) a x 0.5 Figure 2. Liear approximatio The equatio of the taget lie at x = a is P (x) =c 0 +c (x a). Also, the followig coditios must be met: P (a) =f(a) P (a)=f (a) i.e., the taget lie must pass through the poit (a, f(a)), ad its slope should be the derivative of f(x) atx=a. From here we get: c 0 = f(a) c = f (a) hece the taget lie ad first approximatio of f(x) atx = a is P (x) =f(a)+f (a)(x a). P (x) is called the st-degree Taylor polyomial of f(x) atx=a. For istace, if f(x) = lx the its first-degree Taylor polyomial at x =is P (x)=l+ (x ) = x.

9 NOTES ON INFINITE SEQUENCES AND SERIES 9 Hece l x x for x close to. I particular l.. =0.. Compare to the actual value l. = Higher Degree Polyomial Approximatios. The liear (first degree) polyomial approximatio might be eough for may practical purposes, but sometimes we eed a better approximatio. This ca be accomplished by usig higher degree polyomials. The th degree Taylor polyomial of a fuctio f(x) atx=ais a polyomial of the form: P (x) =c 0 +c (x a)+c 2 (x a) 2 + +c (x a) = c k (x a) k k=0 whose value ad derivatives up to the th are equal to those of f(x) at x = a, i.e.: P (a) =f(a) P (a)=f (a) P (a) =f (a)... P ( (a)=f( (a) Solvig those equatios we get that c k = f (k (a)/k!, hece the thdegree Taylor polyomial of f(x) atx = a is P (x) = f(a) + f (a) (x a)+ f (a) 0!! 2! f (k (a) = (x a) k. k! k=0 For istace, for f(x) =l(x)ada=weget P 2 (x)=(x ) 2 (x )2, (x a) f( (a)! (x a) hece, l. (. ) 2 (. )2 =0.095, closer to the actual value l, = tha the first degree approximatio computed above.

10 0 MIGUEL A. LERMA 3.3. Taylor series. The ext logical step is to use a ifiite -degree polyomial, i.e., a series of the form: k=0 f (k (a) k! (x a) k. This series is called the Taylor series of f(x) atx=a.ifa=0the the Taylor series is called Maclauri series. A fuctio f(x) such that. its Taylor series coverges, ad f (k (a) 2. f(x) = (x a) k, k! k=0 is called aalytic. A few examples of aalytic fuctios are the followig: e x =+x+ x2 2! + x3 3! + = x! l ( + x) =x x2 2 +x3 3 = + x ( ) ( x < ) si x = x x3 3! x5 5! + = ( ) x 2+ (2 +)! cos x = x2 2! + x4 4! = ( ) x2 (2)! +x = x+x2 x 3 + = ( ) x ( x < ) ( ) ( ) ( ) ( ) α α α α ( ) α ( + x) α = + x + x 2 + x 3 + = x ( x < ) ( ) α factors) α (α ) (α 2)(... (α +) where =.! As a example, the Taylor series for e x ca be used for computig the umber e: e = e =! =++ 2! + + = ! 3.4. Homework. E & P,.4: 20, 23 25,

11 NOTES ON INFINITE SEQUENCES AND SERIES 4. The Itegral Test. I sectio 2.7 we foud that the harmoic series diverges by comparig its sum with a itegral (see fig. ). This techique is called itegral test. More geerally we have that the followig holds: 4.. Itegral Test. Suppose a is a positive-term series (i.e., a > 0 for every =,2,3,...). Also assume that f(x) isapositive valued, decreasig, cotiuous fuctio for x such that a = f() for every =,2,3... The the series ad the improper itegral f(x) dx either both coverge or both diverge. So, i order to test the series for covergece it is eough to test the correspodig itegral for covergece, which i may cases is easier. The ext oe is a typical example of applicatio of the itegral test p-series. A p-series is a series of the form:, p where p>0 is a fix expoet. The case p = is the harmoic series, which was show to be diverget i sectio 2.7. O the other had, if p : M x p dx = [ x p p ] M = p a ( M p ). As M the limit of the above expressio coverges for p>, ad diverges for p. Hece, the itegral test shows that the p-series coverges for p>, ad diverges for p Homework. E & P,.5: 30, Compariso Test for Positive-Term Series The compariso test allows us to test a series for covergece by comparig it to aother series for which covergece is easier to test. It says the followig: 5.. Compariso Test. Suppose a ad b are positive-term series such that a b for every. The:. If a diverges the b diverges. 2. If b coverges the a coverges.

12 2 MIGUEL A. LERMA For istace, cosider the series 3 +. Sice / 3 +</ 3 =/ 3/2, the covergece of the give series ca be derived from the covergece of the followig series:. 3/2 This is a p-series with p =3/2>, hece it coverges, ad so does the give series. I some cases the ordiary compariso test is hard to apply, but the followig is easier to use: 5.2. Limit Compariso Test. Suppose a ad b are positiveterm series, ad the limit a L = lim b exists ad is ot zero or ifiity: 0 <L<+. The either both series coverge or both diverge. For istace, look at the followig series: We ca test its covergece by limit compariso with the harmoic series:. I fact: (2 + 3)/ lim = 3, / which is ot zero or ifiity. Sice the harmoic series diverges, we coclude that the give series also diverges Rearragemet ad groupig. I a positive-term series rearragig ad regroupig its terms does ot alter its sum. Note that this is ot true i geeral for other series. For istace, the followig series is diverget:

13 NOTES ON INFINITE SEQUENCES AND SERIES 3 However, if we regroup its terms i the followig way, we get a coverget series with zero sum: ( ) + ( ) + ( ) + = =0. A differet regroupig still give us a differet sum: +( +)+( +)+( +)+ = =. However, if the terms are all positive, the we ca rearrage ad regroup them without ay fear of chagig the sum. So, for istace: = 2 = /2 /2 =. Now, if we regroup its terms i the followig way, we get: ( 2 + ) ( ) + = = 3 4 = 3/4 /4 =, so,wegetthesamesum Homework. E & P,.6: Alteratig Series ad Absolute Covergece A alteratig series is a series of the form ( ) + a = a a 2 + a 3 a 4 + or ( ) a = a + a 2 a 3 + a 4 where a > 0 for every. Examples: the alteratig harmoic series: ( ) + = ; a alteratig geometric series: ( ) 2 =

14 4 MIGUEL A. LERMA 6.. Alteratig Series Test. If a alteratig series verifies:. a it is decreasig: a a > 0 for every, ad 2. the th term teds to zero: lim a =0, the the series coverges. So, i this particular case the reciprocal of the th term test holds. E.g.: ( ) + = =l2 ( ) = = π Alteratig Series Remaider Estimate. Let S be the sum of a alteratig series: S = ( ) + a, ad S N its Nth partial sum: S N = N ( ) + a. The the differece R N = S S N (remaider) betwee the sum of the series ad that of the Nth partial sum has the same sig as the followig term of the series ( ) +2 a +,ad 0 R N <a +. I particular, S is always betwee two cosecutive partial sums. For istace, we kow that ( ) + π =4 2 +, but how close do we get to π by addig, say, oe hudred terms of that series? Aswer: its sum π is betwee ad ( ) ( ) = =

15 NOTES ON INFINITE SEQUENCES AND SERIES 5 Compare to the actual value π = Absolute Covergece. A geeral series a is said to be absolutely coverget if the series of absolute values of its terms a is coverget. We have that a series ca be:. Coverget ad absolutely coverget, e.g: ( ) = Coverget but ot absolutely coverget i this case the series is called coditioally coverget, e.g: ( ) + = Not coverget or absolutely coverget, e.g: = However, a series caot be absolutely coverget ad ot coverget, because absolute covergece implies covergece: absolute coverget = coverget cos Example: Does the series coverge? Aswer: Look at the 2 series of absolute values: cos 2. 2 By compariso test, it coverges (the right had side is a p-series with p>), hece the give series is absolutely coverget, which implies that it is ideed coverget Ratio Test. Suppose that the limit ρ = lim a + a exists or is ifiity. The. If ρ< = a coverges absolutely. 2. If ρ> = a diverges. 3. If ρ = = the ratio test is icoclusive.

16 6 MIGUEL A. LERMA As a rule of thumb, for geometric series ρ = r (the ratio), ad the coclusio of the ratio test is aalogous to the oe for geometric series, i.e., the series coverges for r < ad diverges for r >. 2 Example: For the series we have 2 ρ = lim a + a = lim ( +) 2 /2 + 2 /2 hece, it coverges absolutely. 3 ( +) 2 = lim = <, 6.5. Root Test. I some cases i which the ratio test is uable to provide a aswer, the root test may help. It says the followig: Suppose that the limit ρ = lim a exists or is ifiity. The. If ρ< = a coverges absolutely. 2. If ρ> = a diverges. 3. If ρ = = the root test is icoclusive. Example: Cosider the followig series. For this series 2+si the ratio test caot be used, because a + =2 +si si (+) =2 2si 2 cos (+ 2 ) a which has o limit. However, the root tests shows that the series is absolutely coverget: lim = lim 2+si 2 = +si / 2 < Homework. E & P,.7: Power Series A power series is a sort of ifiite polyomial of the form a x Taylor series are particular cases of power series. E.g.: e x x =! =+x+x2 2! + x3 3! + 3 The sum is exactly 2 2 =6.

17 NOTES ON INFINITE SEQUENCES AND SERIES 7 si x = ( ) x 2+ (2 +)! =x x3 3! x5 5! + cos x = ( ) x2 (2)! = x2 2! + x4 4! +x = ( ) x = x+x 2 x 3 + ( x <) 7.. Covergece of a Power Series. The first questio we must aswer about a powers series is for what values of x does the series coverge? This ca be aswered with the ratio test: where ρ = lim a + x + a x R = lim = x lim a + a. a + a = x R, (If the limit is zero the we take R =. If it is ifiity the R =0.) We kow that the series coverges absolutely if ρ<, i.e., x <, R ad diverges if ρ>, i.e., x >. Hece the series coverges absolutely R for x <R, i.e., o the iterval ( R, R), ad diverges for x >R.The umber R is called the radius of covergece of the series. It remais to determie if the series coverges at the edpoits x = R ad x = R. This ca be aswered by substitutig x = R ad x = R i the power series ad studyig the resultig series. After doig that, we will fid that the iterval of covergece is oe of the followig: ( R, R), [ R, R), ( R, R] or [ R, R] If R = the the iterval of covergece is the whole real lie R. Example: cosider the Maclauri series for l( + x): + x ( ). Usig the ratio test we get: ρ = lim x + /( +) x / = lim x + = x. Hece the series coverges absolutely for x < ad diverges for x >. It remais to study the edpoits x = ±.

18 8 MIGUEL A. LERMA For x = the power series becomes ( ) +, which is the alteratig harmoic series. We kow that it coverges. For x = the power series becomes, which is the (usual) harmoic series. We kow that it diverges. Hece the iterval of covergece is (, ] Power Series i Powers of (x c). The same techiques ca be applied to power series of the form: a (x c) Usig the ratio test we get ρ = lim a + (x c) + a (x c) = x c lim a + x c a = R, where R = lim a + a. Hece the series coverges absolutely if x c < ad diverges if R x c >. I other words, it coverges absolutely for x c <R, i.e., R o the iterval (c R, c + R), ad diverges for x c >R.Asabove, the edpoits x = c R ad x = c + R must be tested separately. So the radius of covergece is as before, but the iterval of covergece is cetered at x = c istead of x = The Biomial Series. The biomial series is the Maclauri series of the fuctio f(x) =(+x) α. It ca be computed i the usual way: f (x) = (+x) α f(0) = f (x) = α( + x) (α ) f (0) = α f (x) = α(α ) ( + x) (α 2) f (0) = α (α )......

19 NOTES ON INFINITE SEQUENCES AND SERIES 9 f ( (x) = α(α )...(α +)(+x) (α 2)... f ( (0) = α (α )...(α )... Hece: ( ) ( ) ( ) ( ) α α α α ( ) α ( + x) α = + x + x 2 + x 3 + = x ( x < ) ( ) α factors) α (α ) (α 2)(... (α +) where =.! Note that if α = m a positive iteger, the ( m ) = 0 for >m,so the series becomes a fiite sum idetical to the biomial expasio: m ( ) m ( + x) m = x Other examples: +x =(+x) = ( ) x = x+x 2 x 3 + ( ) /2 +x=(+x) /2 = x =+ 2 x 8 x2 + 6 x3 + +x =(+x) /2 = ( /2 ) x = 2 x+3 8 x2 5 6 x Differetiatio ad Itegratio of Power Series. If a fuctio f(x) ca be represeted as a power series: f(x) = a x with radius of covergece R, the:. It is differetiable o ( R, R) ad f (x)= a x. 2. It is itegrable o ( R, R) ad x f(t)dt = 0 a x + +.

20 20 MIGUEL A. LERMA As a applicatio, we use this result for computig the power series for the arctaget f(x) =ta x. I fact, we have: f (t) = +t 2 = t2 +t 4 t 6 + = ( ) t 2. Itegratig termwise we get: f(x) = Hece: x 0 ta x = f (t)dt = x x3 3 + x5 5 x7 7 + = ( ) t2+ ( ) t =x x3 3 +x5 5 x Homework. E & P,.8: 2. Maple Worksheet: WS2 (due 0/4/99). 8. Power Series Computatios 8.. Addig ad Multiplyig Series. If f(x) = a x ad g(x) = b x the ad where f(x)+g(x)= f(x) g(x) = (a + b ) x c x c = a 0 b + a b + a b 0 = a k b k As a example, we ca compute the first few terms of the power series for ta x = a 0 + a x + a 2 x 2 +, usig the power series for si x = x x 3 /3! + x 5 /5! ad cos x = x 2 /2! + x 4 /4!,ad the relatio si x =taxcos x: si x ta x cos x {}}{{}}{{}} ){ x x3 3! + x5 5! = ( a0 + a x + a 2 x 2 + a 3 x 3 + ) ( x2 2! + x4 4! ( =a 0 +a x+ a ) ( 0 2 +a 2 x 2 + a ) 2 +a 3 x 3 + k=0

21 NOTES ON INFINITE SEQUENCES AND SERIES 2 Idetifyig coefficiets we get: a 0 = 0 a = a 2 0 +a 2 = 0 a 2 +a 3 = 6... Solvig that system of equatios we get a 0 =0,a =,a 2 =0,a 3 = /3,..., hece: ta x = x + x Computig Limits with Power Series. Sice power series are cotiuous i their iterval of covergece, we have that if f(x) = a (x c) = a 0 + a x + a 2 x 2 + the lim f(x) =f(c)=a =a 0. x c This ca be applied to computig limits of idetermiate forms f(x)/g(x) by substitutig power series for f(x) adg(x). Example: cos x lim x 0 x si x = lim ( x 2 /2+x 4 /24 ) x 0 x(x x 3 /6+ ) = lim x 0 x 2 /2 x 4 /24 + x 2 x 4 /6+ = lim x 0 /2 x 2 /24 + x 2 /6+ = Homework. E & P,.9: