Chapter : Power Series 57
Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because they are just polyomials. A fuctio that caot be itegrated directly ca be itegrated idirectly, by itegratig the series that represets that fuctio. The classic example is y e x 2 2 which is the equatio for the Normal Distributio (or Bell) curve. Noe of our itegratio techiques work o it. But there is a ifiite polyomial which equals this fuctio: e x 2 x2 2 + x 2 2 + 2 2! 2 x 2 2 + 3! 3 +! This series ca be itegrated by the Power Rule. Therefore, the itegral of this series will equal the itegral of the fuctio. I terms of AP questios, there are essetially four sub-topics withi the larger topic of Power Series: Taylor polyomials Usig the Taylor polyomial for approximatios of f x. Creatig a ew series from a old oe by o Substitutio o Differetiatio o Itegratio Radius ad iterval of covergece This last requires revisitig the covergece of umerical series. 572
.: Taylor ad Maclauri Polyomials As we said i the itroductio, the reaso to study Series i Calculus is because a ifiitely log polyomial ca represet a fuctio ad polyomials ca be differetiated ad itegrated where oe might ot be able to do so to the fuctio. This polyomial is called a Power Series. Power Series Def: a ifiite polyomial of the form f ( x) c + c x + c 2 x 2 + c 3 x 3 + c 4 x 4 +!+ c x +! The idea of a series represetig of a fuctio is similar to the use of taget lies to approximate fuctio values. These two graphs look quite similar (because of the widow used for viewig), yet oe is a siusoidal ad the other is a cubic. By represetig, we mea that the y-values of the polyomial are the y-values of the fuctio. With a ifiite series, the values will be exactly equal. Usig a partial sum gives approximatios, just as the y-values o the taget lie serve as approximatios. A Taylor polyomial is the power series for ay give fuctio. The formula is f ( x) f a + f ' a! ( x a) + f " a 2! ( x a) 2 + f ''' a 3! ( x a) 3 +...+ f a! ( x a) +... where f ( a) is the th derivative at x a. (While it is ot worth provig here, ote that f a f ( a) ad!.) 573
Maclauri Polyomial Def: a Taylor Polyomial where a. OBJECTIVE Create a Taylor polyomial from give umerical derivatives. Idetify umerical derivatives from a give Taylor or Maclauri polyomial. Use a Taylor or Maclauri polyomial to approximate fuctio values. Create ew series from a Taylor or Maclauri polyomial. be cotiuous ad differetiable at all real umbers ad let 2, f ' 3, f '' 7 ad f ''' 5. for f ( x) about x. Ex Let f x f (a) Write the third degree Taylor polyomial P 3 f, (b) Use the P 3 ( f, )to approximate f(.). (c) Write the fourth degree Taylor polyomial P 4 g,. for g( x) f 3x 2 (d) Write the fourth degree Taylor polyomial P 4 ( h, ) for h( x) f ( t) (a) P 3 ( f, ) 2 3( x ) + 7 2! ( x )2 + 5 3! x 3 2 3( x ) 7 2 ( x )2 + 5 6 ( x )3 x dt. (b) f (.) P 3 ( f, ) (.) 2 3(. ) 7 2 (. )2 + 5 6 (. )3 (c) Substitutig 3x 2 for x: P f ( 3x 2 ), 2 3(.) 7 ( 2. ) 2 + 5 ( 6. ) 3 2.3.35 +.8.666 2 3 3x2 7 2 ( 3x2 ) 2 + 5 6 ( 3x2 ) 3. But the problem says to write the fourth degree polyomial ad 5 6 3x2 3 this is sixth degree. Therefore the aswer is 574
P 4 ( g, ) 2 3( 3x 2 ) 7 2 ( 3x2 ) 2 (d) f ( t) h x x dt 2 3 t dt x 7 ( 2 t ) 2 + 5 ( 6 t ) 3 3 x 2 x 2 2 2( x ) 3 2 x 7 2 2 7 6 x ( x ) 3 + 5 ( x ) 4 3 6 4 3 + 5 4 24 x Ex 2 Let g( x) be cotiuous ad differetiable at all real umbers ad let 4( x 2) + 6( x 2) 2 6( x 2) 3. ad g''' ( 2). T 3 g, 2 (a) Fid g 2 (b) Write the secod degree Taylor polyomial of g' ( x)ad use it to approximate g'.9 (c) Write the third degree Taylor polyomial P 3 ( h, 2) for h x x g( t) dt. 2 (a) (b) By the defiitio of a Taylor polyomial, the costat is f ( a), so f ( 2) The coefficiet of the third degree term is f ''' ( a ) ; therefore, 3! g' x f '''( 2) 3! f ''' 2 6 36 d dx 4 ( x 2 ) + 6 ( x 2 ) 2 6 ( x 2 ) 3 4 +2 x 2 8( x 2) 2 575
(c) g( t) h x x 2 4 +2(.9 2) 8(.9 2) 2 4 +2(.) 8(.) 2 g'.9 4.2.8 5.38 x ( + 6( x 2) 2 6( x 2) 3 ) 2 dt 4 x 2 dt 4 ( x 2 ) 2 x 2 2( x 2) 2 2 x 2 x 2 2 + 6 ( x 2 ) 3 3 3 + 2 x 2 6 ( x 2 ) 4 3 4 But agai, they asked for the third degree Taylor Polyomial ad this is the fourth. So, the aswer is ( x 2) 2( x 2) 2 2( x 2) 3 Ex 3 Fid the first four terms ad the geeral term of the Maclauri Polyomial for f x 2 x x 2 x 6 x 24 x ( x) 2 x f x f ' x f '' x f ''' x f IV 2 f ( ) 3 f '( ) 2 4 f ''( ) 6 5 f '''( ) 24 6 f IV ( ) 2 So f ( x) + 2 x + 6 2 x2 + 24 3! x3 + 2 ( 4! x4 +!+ + )! x! +! + 2x + 3x 2 + 4x 3 + 5x 4 +!+ ( +)x +! 576
Ex 4 Fid the first four terms ad the geeral term of the Maclauri Polyomial where f ( ) ( 3 +)! The Maclauri coefficiets are f ( )! So f x ( ) ( +)! 3! 2 3 x + 3 9 x2 4 27 x3 + 5 8 x4!+ ( ) 3 ( +). 3 ( +)x +! 2 3 x + 3 x2 4 27 x3 + 5 8 x4!+ 3 + x +! 577
. Homework be cotiuous ad differetiable at all real umbers ad let, f '( 2), f ''( 2) 2 ad f '''( 2) 3. for f ( x) about x2.. Let f x f 2 (a) Write the third degree Taylor polyomial P 3 f, 2 (b) Use the P 3 ( f, 2)to approximate f(.9). (c) Write the fourth degree Taylor polyomial P 4 g, 2 (d) Write the fourth degree Taylor polyomial P 4 h, 2 2. Let f x. for g( x) f 2x 2 for h( x) f ( t) x dt. be cotiuous ad differetiable at all real umbers ad let 3, f '( ), f ''( ) 8 ad f '''( ) 2. for f ( x) about x. to approximate f (.). for g( x) f x 3 f (a) Write the third degree Taylor polyomial P 3 f, (b) Use the P 3 f, (c) Write the sixth degree Taylor polyomial P 6 g, 2. (d) Write the fourth degree Taylor polyomial P 4 ( h, ) for h( x) f ( t) 3. Let g x x dt. be cotiuous ad differetiable at all real umbers ad let 4 3( x ) 5( x ) 2 + 7( x ) 3. ad g'''. T 3 g, (a) Fid g (b) Write the secod degree Taylor polyomial of g' ( x)ad use it to approximate g'. (c) Write the third degree Taylor polyomial P 3 ( h, ) for h( x) g( t) 4. Let g x T 3 x dt. be cotiuous ad differetiable at all real umbers ad let ( g, 5) 6 + 2( x + 5) + 9( x + 5) 2 ( x + 5) 3. Fid g' ( 5) ad g'' ( 5). (a) (b) Write the secod degree Taylor polyomial of g' ( x)ad use it to approximate g' 4.9 (c) Write the third degree Taylor polyomial P 3 ( h, 5) for h( x) g( t) x dt. 5. Fid the first four o-zero terms ad the geeral term of the Maclauri Polyomial for f x si2x 5 578
6. Fid the first four terms ad the geeral term of the Maclauri Polyomial for f x x +. 7. Fid the first four terms ad the geeral term of the Maclauri Polyomial where f ( ) +. 2 8. Fid the first four terms ad the geeral term of the Maclauri Polyomial where f ( + 2)! 4 9. BC 28 #3a, b 579
.2: Series Approximatios ad Error I the last sectio, we used Taylor ad Maclauri polyomials to approximate fuctio values. This begs the questio of how good these approximatios actually are. Error (or Remaider) Def: The differece betwee a fuctio value ad a series approximatio. There are two kids of error estimates that the AP test will be iterested i: Lagrage Error Boud is a th degree Taylor series for f x f + ( c) + is the maximum error for a approximatio of f ( b). + If P f,a where! b a, the there exists some value c Alteratig Series Error Boud is a th degree Taylor series for f ( x) ad P ( f,a) is a alteratig f + ( a) + is the maximum error for a approximatio of f ( b). + If P f,a series, the! b a Note that the differece betwee these is what costat is i the th derivative. The Alteratig series uses the th derivative at a, while the Lagrage error uses the th derivative at some ukow costat c. With a o-alteratig series, f + c must be bouded i order to fid the error. OBJECTIVE Show that the error ivolved i a approximatio of a fuctio value is below a give amout. 58
be cotiuous ad differetiable at all real umbers ad let 2, f ' 3, f '' 7, ad f ''' 5. for f ( x) about x. Ex Let f x f (a) Write the third degree Taylor polyomial P 3 f, (b) Use the P 3 ( f, )to approximate f(.). (c) If f IV c estimate i (b) is less tha.. 3 for all c, use the Lagrage Error Boud to show that the (a) P 3 ( f, ) 2 + 3( x ) + 7 2! ( x )2 + 5 3! ( x )3 2 + 3( x ) + 7 2 ( x )2 + 5 6 ( x )3 (b) f (.) P 3 ( f, ) (.) 2 + 3(. ) + 7 2 (. )2 + 5 6 (. )3 (c) If f IV ( c) 3, the + f + c! b a 2 + 3(.) + 7 2 (.)2 + 5 6 (.)3 2 +.3 +.35 +.8.3358 + 3 ( 4!. ) 4 ( 8. ) 4 8, <. Ex 2 Show that the third degree Maclauri Polyomial T 3 ( f, ), where ( ) f 3 ( +)!, approximates f with a error less tha.. As we saw i Ex 4 of the last sectio, the Maclauri coefficiets are f ( ) ( +)! ( ) ( +), ad! 3! 3 2 3 x + 3 9 x2 4 27 x3 + 5 8 x4!+ f x ( +)x +! 3 T 3 ( f, ) 2 3 x + 3 9 x2 4 27 x3 ad, sice this is ad Alteratig Series, the error is less tha 5 8 x4. 5 ( 8 ) 4 5.62 <. 8 58
.2 Homework be cotiuous ad differetiable at all real umbers ad let 3, f '( ), f ''( ) 8 ad f '''( ) 2.. Let f x f (a) Write the third degree Taylor polyomial P 3 f, (b) Use the P 3 ( f, )to approximate f(.). (c) If f IV c estimate i (b) is less tha.. 2. Let g x P 3 for f x. 4 for all c, use the Lagrage Error Boud to show that the be cotiuous ad differetiable at all real umbers ad let ( g, 5) 6 + 2( x + 5) + 9( x + 5) 2 ( x + 5) 3. (a) Use the P 3 ( f, 4.9)to approximate f(-5.). (b) If g IV c estimate i (b) is less tha.. 3 for all c, use the Lagrage Error Boud to show that the 3. Fid the first four terms ad the geeral term of the Maclauri Polyomial for f ( x) x +. Does the third degree Maclauri Polyomial P 3 ( f, ), approximate f with a error less tha.? 4. Fid the first four terms ad the geeral term of the Maclauri Polyomial where f approximate f ( + 2)! Does the third degree Maclauri Polyomial P 4 3 ( f, ), with a error less tha.? 582
.3: Power Series OBJECTIVE Create a ew series from a kow series. So, how does oe fid the series represetatio of a fuctio? For ay series, we would eed to kow about Taylor Series which we will ivestigate i the ext sectio. To start, though, we will begi with a few series that you are expected to memorize for the AP Exam. TAYLOR SERIES OF KNOWN FUNCTIONS u + u + u2 + u 3 + + u + si u u u3 3! + u5 5! u7 7! + + u 2+ 2 + u iff < u <! + ( ) u 2+ ( 2 +)! cos u u2 2! + u4 4! u6 6! + + u 2 2! + ( ) u 2 2! e u + u + u2 2! + u3 3! + u4 u + + 4!! + u! Ex Fid the series that represets y si 2x. Sice we kow si x x x3 3! + x5 5! x7 + + ( ) x 2+ +, all that eeds to 7! ( 2 +)! be doe is to replace all x values with 2x. This is essetial a u-sub problem. 583
3 si ( 2x) ( 2x) 2x 3! + ( 2x)5 5! ( 2x) 8x3 3! + 32x5 5! ( 2x)7 7! 28x7 7! + + ( ) 2x 2 + 2+! + + + ( ) 2 2+ x 2+ + 2 + ( ) 2 2+ x 2+ 2 +! Ex 2 Fid the series that represets y + x. We kow! x + x + x2 + x 3 + + x + ad + x x + x x + ( x) + ( x) 2 + ( x) 3 + + ( x) + x + x 2 x 3 + + ( ) x + x, Other trasformatios (besides u-subs) ca be doe. For example, trasformig by itegratig or differetiatig. Ex 3 Fid the series that represets y + x. We kow + x x + ( x) + ( x) 2 + ( x) 3 + + ( x) + + x x + ( x) + ( x) 2 + ( x) 3 + + ( x) + x + x 2 x 3 + + ( ) x + 584
x Note the lower idex has chage to. Ex4 Fid the series that represets y ( x). 2 ( x) d 2 dx x, so ( x) d 2 dx + x + x2 + x 3 + + x + + 2x + 3x 2 + 4x 3 + + x + ( +)x + ( +)x Note that the geeral term that we used for the sigma otatio has as the power, ot -. 585
A importat ote about x + x + x2 + x 3 + + x + x. This series oly represets this fuctio whe < x <. This is the Iterval of Covergece which we will cosider later. 4 y 4 y 3 3 2 2 4 3 2 2 3 4 5 x 4 3 2 2 3 4 5 x 2 2 3 3 4 4 Oly for these x-values < x < does this series give the same y-values as the fuctio. We ca see this more clearly whe we overlay the to graphs: 4 y 3 2 4 3 2 2 3 4 5 x 2 3 4 EX 5 Fid the series that would represet ew series represet? x + dx. What fuctio would this 586
As we saw i EX 2, x + dx l x + EX 6 BC 23 #6 EX 7 BC 29 #6 + x x + x2 x 3 + + x + x + dx x + x2 x3 + + ( x + ) x x 2 + x 3 + + x + +, also, therefore, this ew series represets l x + dx. Ex 8 Fid the exact value of ( π 6 ) 2 ( π + 6 ) 4 2! 4! ( π 6 ) 6 6! + + 2 2 π 6! +. Sice x2 2! + x4 4! x6 6! x 2! + + 2 + cos x, the ( π 6 ) 2 ( π + 6 ) 4 2! 4! ( π 6 ) 6 6! + + 2 2 π 6! + cos π 6 3 2 Ex 9 If f ( x) + 2x + 3x 2 + 4x 3 + + + f ' 2. x +, fid the exact value of Sice the series represetig f '( x) is actually equal to f ' 2 2 2 4 4 ( x), the 2 587
.3 Homework Fid the series represetatio of these fuctios.. x 2. 2 2 3. l + x + x 4. ta x 5. x x 6. e x 7. e x2 8. cos x 2 9. cos x 2 dx. l( x). x 2 + x 2 2. x + x 2 3. l( + x 2 ) 4. xe x 2 2 5. si x 3 6. xsi x 3 7. xsi x 3 dx 8. BC2 #6 9. BC27 #6 Give the series foud i #-7 above, fid the followig exact values. 2. 2 + 2 2 2 3 + + 2 + 2. + 3+ 32 2! + 33 3! + 34 3 + + 4!! + 22. 3 3 3 3 + 3 5 5 3 7 7 ( ) 3 + + 2 + 2+ + 23. f ( 2) 2 22 2 + 23 3 24 4 ( ) 2 + + + 588
.4: Review of Numerical Series Tests Vocabulary Ifiite Sum--Def: the sum of all the terms of a sequece. This is ot possible for a arithmetic series, but might be for a geometric series. Coverget Series--Def: a ifiite series that has a total. Diverget Series--Def: a ifiite series that does ot have a total. I other words, A series is coverget if ad oly if A series is diverget if ad oly if c. a a c. Previously, we ivestigated the seve tests that will help determie if a give series coverges or ot. Those tests were:. Divergece Test (th term test): If a it diverges. If a o coclusio 2. Itegral Test: If f(x) is a decreasig fuctio, the a ad x dx either both coverge or both diverge. 3. Cauchy Ratio Test: If a + a If a + a If a + a < it coverges; > it diverges; the test is icoclusive. 589
4. The Alteratig Series Test: If a series is a Alteratig Series ad a a +, it is coverget if ad oly if a. 5. The Nth Root Test: If a < it coverges; If a > it diverges; If a the test is icoclusive. 6. The Direct Compariso Test: Whe comparig a give series a to a kow series b, i) if a b ad b coverges, the a coverges. ii) if a b ad b diverges, the a diverges. 7. The it Compariso Test: Whe comparig a give series a to a kow series b, i. If a b > the both coverge or both diverge. ii. iii. If a, the a coverges if b b coverges. If a, the a diverges if b diverges. b 59
Some texts suggest applyig the tests i a particular order. Oe example might be: Itegral Test Compariso Tests Divergece Test Ratio Test Alteratig Series Test Nth Root Test This may ot be the best order ad, with exposure ad experiece, ca ofte be bypassed. Here are some hits: Test Alteratig Series Divergece Test Itegral Test Direct Compariso it Compariso Nth Root Test Ratio Test Whe ( ) + or cos( π) Always check If it looks easy to itegrate Good with Geometric Series Good with p-series Best with factorials or combiatios of fuctios OBJECTIVE Determie whether a give umerical series coverges or diverges. Ex Use the Itegral Test to determie which of these series coverge. I. II. III. 2 + 2 l 2 I. II. III. x dx x 2 + dx xl 2 x dx 2 l x de ; therefore, I diverges. ta π x 2 π 4 π ; therefore, II coverges. 4 l 2 u du 2 ( ) ; therefore, III coverges. u l 2 59
Ex 2 Does coverge?! II ad III coverge. a + a ( +)!!! ( + )!! ( + )! + Sice this limit is <, coverges.! Ex 3 Does coverge? 3 + 2 We ca make a direct compariso betwee ad 3 + 2. 3 + 2 < because 3 3 + 2 has a larger deomiator. Sice coverges (it is a p-series with p>) ad coverges. 3 + 2 Ex 4 Does coverge? 2 4 3 3 3 + 2 is smaller tha 3, the 592
If we ca try to make a direct compariso betwee we fid that 2 4 > (because 2 2 4 ad 2 4, has a larger deomiator) but coverges (it is a p-series with p>). So the Direct Compariso Test 2 does ot apply. 2 4 2 2 2 4 > 2 This limit is greater tha ad coverges also. coverges. Therefore, 2 2 4 ( ) + ( + 2) Ex 5 Is ( +) coverget or diverget? If coverget, is it absolutely or coditioally coverget? i) ii) ( ) + + 2 + ( ) + + 2 + + 2 2 + + 3 2 + 3 + 2 + 2 2 + ( + 2) ( +) ( +) + ( )+ + So ( ) + ( + 2) + coverges. To test for absolute vs. coditioal covergece, we eed to test We ca do the it Compariso Test agaist the ed behavior model + 2 + + 2 ( +) 2 + 2 2 + + 2. +. 593
This limit is greater tha, therefore, both do the same thig--amely, they diverge. So, Ex 6 Does ( ) + ( + 2) + is coditioally coverget. + coverge? e This series has to a power; therefore, the Root Test is appropriate. + e + e ( + )e e > + Sice this limit is >, diverges. e 594
.4 Homework Determie if each of these series coverge or diverge.. 3. 2 2. 2 + ( 3) + 4. 2 3 2 + 2 l 5. k 2 e k 6. k 2 ( ) + l 7. 9.. 3 2 8.! ta. 2 3 + 2 2 + 5 ( 2) 2 ( 2) ( ) 2 595
.5: The Radius ad Iterval of Covergece Radius of Covergece Def: The distace o either side of x a for which the series coverges to the fuctio. Iterval of Covergece Def: the x-values for which the series coverges to the fuctio. Key Idea of MC Questios: a must be i the ceter of the iterval. The word covergece is actually used two ways whe studyig series. () A umerical series coverges whe it has a fiite total. This is the kid of covergece we will ivestigate i the ext few sectios whe we look at the various Series Tests. (2) I the last two sectios we talked about a power series covergig to a fuctio that it represeted. I fact, we oted that x + x + x 2 + x 3 + + x + oly coverges to whe < x <. For x other values of x, the series ad the fuctio yield differet y-values. These two kids of covergece are really the same: oce a umber is substituted for x, the series becomes a umerical series ad that umerical series will coverge to the y-value of the fuctio at that x-value. Oe of tests of umerical series covergece The Cauchy Ratio Test is also used to fid the radius ad iterval of covergece. Cauchy Ratio Test: If a + a If a + a If a + a < it coverges; > it diverges; the test is icoclusive. 596
OBJECTIVE Fid the Radius of Covergece ad/or Iterval of a give series. Ex Fid the radius of covergece for ( ) x 3 2 + 2 a + a ( ) + 2 + + + ( + 2) x 3 ( ) 2 ( + 2) ( x 3) ( ) + 2 + + 3 ( 2 + 2) ( x 3) + ( ) ( x 3) + 2 2( + 3) ( x 3) 2 ( x 3) < x 3 < 2 so r 2. We could take two more steps at the ed of Ex to fid the iterior of the Iterval of Covergece. We will do so i EX 4. 597
Ex 2 Fid the radius of covergece for! ( x ) a + a +! ( x ) ( + )! x ( + )! x! ( + ) x, this ratio is greater tha for all values of x, except x Sice +. The radius of covergece is. Ex 3 Fid the radius of covergece for ( ) x 2. 2! Sice this is the series for cos x, we hope the radius is ifiite. I other words, this should coverge for all real umbers of x. a + a x 2 ( ) + x 2 + 2 + ( )! ( ) x 2 ( 2)! ( ) + x 2+2 2 + 2! ( 2)! 2 + 2 ( 2)! 2 + 2! x2 ( 2)! x 2 x ( 2 + ) ( 2)! x2 2 + 2 ( 2 + ) x2 Sice is always less tha, this series coverges for all real umbers ad the radius is. 2+2 x 2 598
Steps to fidig the iterval of covergece. Set up the Ratio Test <. 2. Isolate x a. NB The umber o the right will be the radius of covergece. 3. Solve for x. NB This aswer will be the iterior of the iterval of covergece. 4. Test the edpoits usig the umerical series covergece tests. To "test the edpoits," the edpoits are substituted for x. Oe edpoit yields a umerical series ad the other edpoit yields the alteratig versio of the same series. Compariso Tests ad the Alteratig Series Test ca the be used to determie if each umerical series coverges. OBJECTIVE Fid the Iterval of Covergece for a give series. Ex 4 Fid the iterval of covergece for ( ) x 3 2 + 2 a + a ( ) + 2 + + + ( + 2) x 3 ( ) 2 ( + 2) ( x 3) ( ) + 2 + + 3 ( 2 + 2) ( x 3) + ( ) ( x 3) + 2 2( + 3) ( x 3) 2 ( x 3) < x 3 < 2 599
x 3 < 2 2 < x 3 < 2 < x < 5 If x, ( ) x 3 2 + 2 ( ) 2 + 2 ( 2). + 2 We ca use the it Compariso Test here: + 2 + 2 This limit equals a positive Real umber; therefore, both series do the same thig. diverges, so the edpoit x is ot part of the iterval of covergece. ( ) If x5, 5 3 2 ( + 2) Alteratig Series Test: ( ) 2 2 + 2 ( ) + 2. We apply the i. ( ) + 2 + 2 + 3 ( ) ( +) + 2 ii. ( ) + 2 Both coditios of the Test hold; therefore, x 5 is part of the iterval of covergece. The Iterval of Covergece is < x 5 6
Ex 5 Fid the iterval of covergece for ( x ) If x2, x Test-- a + a 2 + ( x ) ( + ) x + x x < < x < < x < 2 ; which diverges by the Divergece. So x is ot i the iterval. If x, ( x ) ( ) ot satisfy coditio ii of the Alteratig Series Test:, which diverges because it does ii. ( ) So, x is also ot part of the iterval of covergece, ad the iterval of covergece is < x < 2 6
Ex 6 Fid the iterval of covergece for x 2 + 2. a + a x ( +) 2 2 + + x 2 + 2 x ( +) 2 + 2 + 3 2 + 2 x 2 + 2 2 + 2 + 3 x x x < < x < x If x, 2 + 2 ; which coverges by the Direct Compariso 2 + 2 Test-- 2 + 2 2, which coverges. So x is i the iterval. x ( ) If x-, 2 + 2, which coverges by the Alteratig Series 2 + 2 Test: So, x - is also part of the iterval of covergece ad the iterval of covergece is x 62
Key Ideas for the AP Test: For Multiple Choice:. Elimiate aswers where a is ot the ceter of the iterval. 2. Test the edpoits. For Free Respose:. Set of the Ratio Test properly (i.e., do ot skip the first lie where there is a fractio over a fractio). 2. Simplify ad apply the limit. 3. Set the result to < ad isolate x. 4. Test the edpoits. 63
.5 Homework Fid (a) the Radius of Covergece ad (b) the Iterval of covergece for each of these series.. 3. x 2. x x 4.! 4 ( x ) 5. 2 3 x + 2 6. + x l 7. 9.. 2 x 8. ( ) ( x + 3) 2 ( 3x 2) 3.! 2x + 2 ( x 3) + 3 64
Sequeces ad Series Test. If f x covergece? 2 x, which of the followig could be the iterval of (A) -2, 2 (B) (-2, 2) (C) -2, 2) (D) (-2, 2 (E) caot be determied 2. The coefficiet of x 4 i the Maclauri series for f ( x) e 2x is (A) 24 (B) 2 (C) 3 4 (D) 2 3 (E) 2 3. If f ( x) (A) f ( x) ( ) x 2+ 2 +!, the f '( x) ( ) x 2+ (B) f x 2! (D) f ( x) ( ) x 2 (C) f x 2! ( ) x 2 (E) f x 2! ( ) x 2 ( ) x 2 4. Which of the followig series coverge? I. II. III. 5 + + 5 + 2 +...+ 5 +... 5 + 25 + 25 + 625 +...+ +... 5 5 6 + + 5 6...+ 5 + 5 +... (A) I oly (B) II oly (C) I ad II oly (D) II ad III oly (E) III oly 65
5. If a is a positive real umber, a coverges for (A) Ay a > (B) a (C) a oly (D) a > (E) o a 6. If f ( x) (-) x 2, the f "! (A) (B) (C) 2 (D) 2 (E) -2 7. If e x x!, what is ( ) x 2! (A) e x2 (B) cos x (C) e x2 (D) e x2 (E) Noe of these x 2-8. If cos x, the cos 3x ( 2 2)! equals (A) ( ) x 4 (B) ( 2)! ( 9x) 2 (C) ( 2)! ( ) x 2 ( 2)! (D) 9x 4 (E) ( 2)! ( 9) x 4 ( 2)! 9. For what values of x does x x2 2 + x3 3 x4 4 ( ) x + + + coverge? (A) x (B) -< x (C) < x < 2 (D) < x < (E) All Reals 66
Taylor Series Test Directios: Show all work.. Let the Maclauri series for f x be defied by f ( x) 2! x2 4! + x4 6! ( x6 ) x 2 +...+ 8! ( 2 + 2)! +... a. Determie the iterval of covergece for f ( x) about x. b. Write the first four terms ad the geeral term of the Maclauri polyomial of g' x x 2 f ( x). c. Write g' x same fuctio. i terms of a kow fuctio. The write f ( x) i terms of the d. Give g( ) 3, write g( x) i terms of a familiar fuctio (without series). be a fuctio with derivatives of all orders such that ( )! 2. Let h x h 2 3 ad h 2 (a) Create the fourth degree Taylor polyomial about x 2 for h( x). (b) Use P( f, 2) to estimate h(, 9) (c) Fid the iterval of covergece for the Taylor series of h( x) about x 2. 67
. Homework. (a) P 3 ( f, 2) ( x 2) 2 ( 2 x 2 ) 3 (b) f (.9).995 (c) P 4 ( g, 2) 3+ 8x 2 4x 4 (d) P 4 ( h, 2) x 2 ( 3 x 2 ) 3 ( 8 x 2 ) 4 2. (a) P 3 ( f, ) 3+ x + 4x 2 + 2x 3 (b) f (.) 2.858 (c) P 6 ( g, ) 3+ x 3 + 4x 6 (d) P 4 ( h, ) 3x + 2 x2 + 4 3 x3 + 2 x4 42. 3. (a) g 3 ad g''' (b) T 2 ( g', ) 3 ( x ) + 2( x ) ; g'. (c) P 3 ( h, ) 4( x ) 3 ( 2 x ) 2 5 ( 3 x ) 3 4. (a) g' ( 5) 2 ad g'' ( 5) 8. (b) T 2 g', 5 (c) 2 +8( x + 5) 3( x + 5) 2 ; g' 4.9 6( x + 5) + ( x + 5) 2 + 34( x + 5) 3 P 3 h, 5 3.79 3.77 5. f ( x) 2x 4 3 x3 4 5 x5 8 ( 35 x7 +...+ ) 2 2+ x 2+ +... 2 + x 6. f ( x) + 2 x 8 x2 ( 6 x3 +...+ ) + ( 2 3) 2 5 2! +... 7. f ( x) + x + 3 8 x2 + 2 x3 +...+ + +... 2! x 8. f ( x) 2 3 2 x + 3 4 x2 5 ( 6 x3 +...+ ) + ( +) ( + 2) 9. see the AP site 4 x +... 68
.2 Homework. (a) P 3 f, (c) E.7 3+ x + 4x 2 + 2x 3 (b) f. 2. (a) g ( 4.9) 6.289 (b) E.25 3. f ( x) + 2 x 8 x2 + ( 6 x3 +...+ ) ( 2 3) 2 5 2! 4. P 3 ( f, ) 2 3 2 x + 3 4 x2 5 6 x3 ; E.7 2.858 x +... E.23.3 Homework. x 2 2. 3. x2+ x 2 x+ 4. 5. x + 2 + 6. x +! x2 7. 8.! ( ) x 4 9. ( 2)! ( ) x 4+ + c 4 +! 2. x + + c. + 2. x 2+2 x 2+ 3. ( ) x 2+ 4. + ( ) x 2+ 2 5.! ( ) x 2+3 2 + 6. ( ) x 6+4 7. 2 + ( ) x 6+6 2 + ( 6 + 5) + c 69
8. see the AP site 9. see the AP site 2. 2 3 2. e 3 22. ta 3 π 6 23. L 3.4 Homework. Diverget by Divergece Test 2. Coverget by LCT 3. Coverget by Geometric Series 4. Diverget by Itegral Test 5. Coverget by Itegral Test 6. Coverget by AST 7. Coverget by Ratio Test 8. Coverget by Root Test 9. Coverget by LCT. Coverget by Root Test. Coverget by LCT.5 Homework a. r b., ) 2a. r 2b. (, ) 3a. r 3b. (, ) 4a. r 4 4a. 3 4, 5 4 5a. r 3 5b. 7 3, 5 3 6a. r 6b., ) 7a. r 7b. (, ) 8a. r 2 8b. [ 5, ) 9a. r 9b. 3, 5 3 a. r b. 2 a. r 2 b. 5 2, 7 2 6