(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)


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1 Chapter 0 Review 597. E; a ( + )( + ) + + S S + S S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig Series d l l Test: Use d to show the u are decreasig for.). (a) Ratio test: + ( + ) + + lim + + lim ( + )( + ) + ( + )( ) + + < < < The series coverges absolutely o (, ). The series diverges at both edpoits by the thterm Test: ( ( ) + ) lim + 0 ad ( ( ) + ) lim 0. + Sice the series coverges absolutely o (, ) ad diverges at both edpoits, there are o values of for which the series coverges coditioally. Chapter 0 Review Eercises (pp. 5 5).. + a+! lim lim a ( )! + lim + 0 The series coverges absolutely for all. (a) All real umbers All real umbers (d) Noe + a + + lim lim a ( ) The series coverges absolutely for or 7 < <. ( ) Check 7: coverges. Check : diverges. (a) [ 7, ) ( 7, ) (d) At 7 +. This is a geometric series, so it coverges absolutely whe r < ad diverges for all other values of. Sice r ( ), the series coverges absolutely whe 5 ( ) <, or < <. (a) 5, <, Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
2 598 Chapter 0 Review 5, (a) (, ) (d) Noe (, ) a+ ( )! lim lim a ( )! + lim ( + )( ) 0 The series coverges absolutely for all. (a) All real umbers All real umbers (d) Noe + a + lim lim a ( ) + The series coverges absolutely for <, or 0< <. Furthermore, whe, we have a ad coverges by the ptest with p, so a also coverges absolutely at the edpoits. (a) 0, 0, (d) Noe + a+ ( + ) lim lim. The a ( + ) series coverges absolutely for <, or < <. The series diverges for >. Whe, the series diverges by the th Term Test (d) Noe lim a + a + ( + ) + ( + ) lim ( ) + + ( + ) + + The series coverges absolutely for + <, or < < ; the series diverges for + + >. Whe, the series diverges by the thterm Test. (a),, (d) Noe + a+ lim lim a ( ) + + lim ( + )( + ) lim ( + ) + ( ) lim e + 0 The series coverges absolutely for all. Aother way to see that the series must coverge is to observe that for, we have, so the terms are (evetually) bouded by the terms of a coverget geometric series.a third way to solve this eercise is to use the throot Test (see Eercises 7 7 i Sectio 0.5). Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
3 Chapter 0 Review 599 (a) (d) Noe 9. All real umbers All real umbers (d) Noe + a+ lim lim a + The series coverges absolutely for <, or < <. Check : ( ) coverges by the Alteratig Series Test. Check : diverges by the ptest with p.. a+ lim a + ( + ) lim + ( + ). The series coverges absolutely whe <, or < < ; the series diverges whe >. Whe, the series diverges by the th Term Test. (a) (, ) (a) [, ) (, ) (, ) (d) Noe 0. (d) At + + a+ e lim lim e. a e ( ) + e The series coverges absolutely for e<, e. + a+ + lim lim a + +. The series coverges absolutely whe <, or 0 < <. or. e < < e Furthermore, whe e, we have a ad e the ptest with p e, so coverges by e a absolutely at the iterval edpoits. also coverges ( ) ( ) + ( ) Check 0: coverges coditioally by the Alteratig Series Test. ( ) Check : coverges 0 + coditioally by the Alteratig Series Test. (a) e, e e, e e (a) [0, ] (0, ) (d) At 0 ad Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
4 600 Chapter 0 Review. + a ( )! + + lim lim a +! ( + ) lim 0, 0, 0 The series coverges oly at a+ ( + )! lim lim a ( + )! lim ( + ) ( 0) + The series coverges oly at 0. (a) 0 (a) 0 0 oly 0 oly 0 0 (d) Noe. (d) Noe + a+ 0 l lim lim 0 a l( ) + 0 The series coverges absolutely for 0 <, or 0 < < 0. ( ) Check : coverges by the 0 l Alteratig Series Test. Check : diverges by the Direct 0 l Compariso Test, sice > for ad l diverges. (a) 0 (d) At, 0 0, This is geometric series with r, so it coverges absolutely whe <, or < <. It diverges for all other values of. (a) (, ) (, ) (d) Noe f( ) + + ( ) +, + evaluated at f( ) l( + ) Sum ( ) + + ( ), evaluated at. 5 Sum l + l. f( ) si ( ) +,! 5! ( + )! evaluated at π, Sum si π 0. Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
5 Chapter 0 Review f( ) cos + + ( ) +,!! ( )! evaluated at π π. Sum cos. f( ) e ,!! evaluated at l. Sum e. f( ) ta l ( ) +, 5 + evaluated at. Sum ta π 6. (Note that whe is replaced by, the geeral term of ta becomes ( ), which matches the geeral term give i the eercise.). Replace by 6 i the Maclauri series for + ( 6) + ( 6) + + ( 6) ( 6) +. Replace by i the Maclauri series for ( ) + ( ) + ( ) ( ) + + give at the ed of Sectio 0.. give at the ed of Sectio The Maclauri series for a polyomial is the polyomial itself: ( ) Replace by π i the Maclauri series for si give at the ed of Sectio 0.. ( ) ( ) 5 ( ) + π π π si π π + + ( ) +! 5! ( + )! Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
6 60 Chapter 0 Review 8. Replace by i the Maclauri series for si give at the ed of Sectio 0.. ( ) ( ) ( ) 5 + si + + ( )! 5! ( )! ( ) ( ) ( + )! si ( ) +! 5! 7! ( + )! ( ) +! 5! 7! ( + )! e + e ( ) +!!!! !! ( )!. Replace by 5 i the Maclauri series for cos give at the ed of Sectio 0.. ( 5) ( 5) ( 5) cos ( )!! ( )! + 5 ( 5) ( 5) + + ( ) +!! ( )! π. Replace by i the Maclauri series for π π ( ) ( ) π / π e !! π π π ! + e give at the ed of Sectio 0... Replace by i the Maclauri series for e give at the ed of Sectio 0., ad multiply the resultig series by. ( ) ( ) e + ( ) !! ( ) +!!. Replace by i the Maclauri series for ta give at the ed of Sectio ( ) ( ) ( ) ta + + ( ) Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
7 Chapter 0 Review Replace by i the Maclauri series for l( + ) give at the ed of Sectio 0.. ( ) ( ) ( ) l( ) + + ( ) + 8 ( ). 6. Use the Maclauri series for l( + ) give at the ed of Sectio 0.. [ ] l( ) l + ( ) 7. ( ) ( ) ( ) ( ) + + ( ) + + f ( ) f ( ) ( ) f ( ) f ( ) ( ), so! f ( ) f ( ) 6( ) 6, so! ( ) ( ) f ( ) f ( )!( )!, so! + ( ) + ( ) + ( ) + + ( ) + 8. f( ) ( + 5) f ( ) ( ) 7 f ( ) f ( ) ( 6 ) 0, so 5! f ( ) f ( ) 6 6, so! ( f ) ( ) 0 for ( + ) 5( + ) + ( + ) This is a fiite series ad the geeral term for is 0. Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
8 60 Chapter 0 Review 9. f () f () 9 f () f (),so 7! 7 f () f () 6,so 7! 8 ( )! f () ( )!, so + ( ) f () ( )! + ( ) ( ) + ( ) ( ) + + ( ) f( π ) si π 0 f ( π ) cos π f ( π ) f ( π ) si π 0, so 0! f ( π ) f ( π ) cos π, so! 6 0, if k is eve ( k) f ( π ), if k +, eve, if k +, odd si ( π) + ( π) ( π) + ( π) + ( )! 5! 7! ( )! ( ) + π + +. Diverges, because it is 5 times the harmoic series: 5 5. Coverges coditioally a is a diverget pseries p, so ( ) does ot coverge absolutely. Use the Alteratig Series Test to check for coditioal covergece: () () () u > 0 + > + > <, + lim u lim 0. so the u are decreasig. ( ) Therefore, coverge. Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
9 Chapter 0 Review 605. Coverges absolutely by the Direct l Compariso Test, sice 0 < for ad coverges by the ptest with p.. Coverges absolutely by the Ratio Test, sice a+ +! lim lim a ( + )! + + lim ( + ) Coverges coditioally a l( + ) Compariso Test. Let a ad l( + ) b. diverges by the ptest, ad a c lim b lim l( + ) lim /( + ) lim ( + ).) diverges by the Limit ( ) Therefore, does ot coverge l( + ) absolutely. Use the Alteratig Series Test to check for coditioal covergece: () u 0 l( + ) > Clear. () + > l( + ) > l <, l( + ) l so the u are decreasig. () lim u lim 0. l ( ) Therefore, coverges. l 6. Coverges absolutely by the Itegral Test, b because d lim (l ) b l. l 7. Coverges absolutely by the Ratio Test, + a+! because lim lim a ( )! + lim Coverges absolutely by the Direct Compariso Test, sice for ad is a coverget geometric series. Alterately, we may use the Ratio Test or the throot Test (see Eercises 7 ad 7 i Sectio 0.5). 9. Diverges by the thterm Test, sice lim lim + a Coverges absolutely by the Direct Compariso Test, sice ( + )( + ) + + > ( + )( + ) > <, ( + )( + ) / ad coverges by the ptest. / 5. Coverges absolutely by the Limit Compariso Test: Let a ad b. The Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
10 606 Chapter 0 Review a c lim b lim + lim + Sice 0 < c < ad coverges by the ptest, coverges. 5. Diverges by the thterm Test, sice lim lim e 5. This is a telescopig series. ( )( ) ( ) ( ) s ( ) ( ) 6 0 s s s 6 ( ( + ) ) 6 ( + ) S lim s 6 5. This is a telescopig series. + ( + ) + s + + s s s + + S lim s Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
11 Chapter 0 Review (a) f () f () P ( ) f( ) + f ( )( ) + ( ) + ( )!! + ( ) + ( ) + ( ) f(.) P (.) 96. Sice the Taylor series for f ca be obtaied by termbyterm differetiatio of the Taylor Series for f, the secod order Taylor polyomial for f at is + 6( ) + 6( ). Evaluated at.7, f ( 7. ) 7.. It uderestimates the values, sice f () 6, which meas the graph of f is cocave up ear. 56. (a) Sice the costat term is f (), f () 7. Sice f ( ), f ( ).! Note that P ( ) + 0( ) 6( ) + ( ). The secod order polyomial for f at is give by the first three terms of this epressio, amely + 0( ) 6( ). Evaluatig at., f (. ) 05.. The fourth order Taylor polyomial for g() at is 5 [ 7 ( t ) + 5( t ) ( t ) ] d 7t ( t ) + ( t ) ( t ) 5 7( ) ( ) + ( ) ( ) (d) No; oe would eed the etire Taylor series for f (), ad it would have to coverge to f () at. 57. (a) Use the Maclauri series for si give at the ed of Sectio ( / ) ( / ) ( / ) 5si ( ) +! 5! ( + )! ( ) ( )! The series coverges for all real umbers, accordig to the Ratio Test: + + a+ 5 ( + )! lim lim a ( + )! 5 lim ( + )( + ) 0 ( Note that the absolute value of ) 5 f ( ) is bouded by for all ad all,,,. We may use the Remaider Estimatio Theorem with M 5 ad r. 5 5 So if < <, the trucatio error usig P is bouded by. + ( + )! ( + )! To make this less tha 0. requires. So, two terms (up through degree ) are eeded. + Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
12 608 Chapter 0 Review 58. (a) Substitute for i the Maclauri series for + + ( ) + ( ) + + ( ) ( ) + give at the ed of Sectio 0..,. The series is geometric with, r so it coverges for <. ( You could also use the Ratio Test, but you would eed to verify divergece at the edpoits) f, so oe percet is approimately It takes 7 terms (up through degree 6). This ca be foud by trial ad error. Also, for, the series is the alteratig series. 0 If you use the Alteratig Series Estimatio Theorem, it shows that 8 terms (up through degree 7) are sufficiet 8 sice < It is also a geometric series, ad you could use the remaider formula for a geometric series to determie the umber of terms eeded. (See Eample i Sectio 0..) 59. (a) lim + + a+ ( + )! lim a ( )! + + ( + ) lim ( ) + + lim e The series coverges for e<, or <, so the radius of covergece is. e e f + +!! By the Alteratig Series Estimatio Theorem the error is o more tha the magitude of the et term, which is 0..! Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
13 Chapter 0 Review (a) f() ( ) f () ( ) f () f () ( ),so! f () f () 6( ) 6,so! ( ) ( ) f () f () ( )!,so ( )! f( ) ( ) + ( ) ( ) + + ( ) ( ) + Itegrate term by term. l dt t ( ( t ) + ( t ) ( t ) + + ( ) ( t ) + ) dt + ( t ) t ( t ) + ( t ) ( t ) + + ( ) ( ) ( ) ( ) ( ) ( ) ( ) + + Evaluate at.5. This is the alteratig series + + ( ) + By the + ( + ) Alteratig Series Estimatio Theorem, sice the size of the third term is < 005., the first two terms will suffice. The estimate for l is (a) Substitute for i the Maclauri series for e give at the ed of Sectio 0.. ( ) ( ) ( ) e + ( ) !!! ( ) +! Use the Ratio Test: a + + +! lim lim a ( )! + lim + 0 The series coverges for all real umbers, so the iterval of covergece is (, ). The differece betwee f() ad g() is the trucatio error. Sice the series is a alteratig series, the 8 ( ) error is bouded by the magitude of the fifth term:. Sice , this term is! less tha 8 06 (. ) which is less tha 0.0. Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
14 60 Chapter 0 Review 6. (a) f( ) + ( ( ) + ) ( ) + + No; at, the series is ( ) ad the partial sums form the sequece 0, 0,, 0,, 0,..., which has o limit. 6. (a) Substitutig t for i the Maclauri series for si give at the ed of Sectio 0., t t t si t t + + ( ).! 5! ( + )! Itegratig termbyterm ad observig that the costat term is 0, 7 + si t dt + + ( ) + 0 7(!) 5 (!) ( + )( + )! (d) 6. (a) si d 0 ( ). 7(!) + 5 (!) + ( + )( + )! + Sice the third term is < 000., it suffices to use the first two ozero terms (through (!) 5 0 degree 7). NINT(si,, 0, ) , (!) 5 (!) 5(!) , This is withi 5. 0 of the aswer i. Let f( ) e d. e d 0 f( d ) 0 h [ f ( 0) + f ( 05. ) + f ( )] 05. e e 05. e e Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
15 Chapter 0 Review 6 e !! !! P ( ) P 0 ( ) Sice f is cocave up, the trapezoids used to estimate the area lie above the curve, ad the estimate is too large. (d) Sice all the derivatives are positive (ad > 0), the remaider, R ( ), must be positive. This meas that P ( ) is smaller tha f(). (e) Let u dv e d du d v e ed e e d Let u dv e d du d v e e e d e e e d e e + e + C ( + ) e + C e d ( + ) e 0 0 e (a) Because [$ 000(. 08) ](. 08) $ 000 will be available after years. Assume that the first paymet goes to the charity at the ed of the first year. 000(. 08) + 000(. 08) + 000(. 08) This is a geometric series with sum equal to, 500. ( 08) be ivested today i order to completely fud the perpetuity forever. This meas that $,500 should Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
16 6 Chapter 0 Review 66. We agai assume that the first paymet occurs at the ed of the year. Preset value 000(. 06) + 000(. 06) + 000(. 06) ( 06. ) , The preset value is $ 6, (a) Sequece of Tosses Payoff ($) T 0 Probability HT HHT Term of Series 0 HHHT + Epected payoff d ( ) ( ) d ( ) ( ) (d) If, the formula i part matches the ozero terms of the series i part (a). Sice ( ), the epected payoff is $. ( ) Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
17 Chapter 0 Review (a) The area of a equilateral triagle whose sides have legth s is areas removed from the origial triagle is b b b b + 9 or b b b b s s () s. The sequece of b This is a geometric series with iitial term a ad commo ratio b b, which is the same as the area of the origial triagle. r, so the sum is No. For eample, let b ad set the base of the origial triagle alog the ais from (0, 0) to (, 0). k The poits removed from the base are all of the form 0,, so poits of the form (, 0) with irratioal (amog others) still remai. The same sort of thig happes alog the other two sides of the origial triagle, ad, i fact, alog the sides of ay of the smaller remaiig triagles. While there are ifiitely may poits remaiig throughout the origial triagle, they paradoically take up zero area Differetiate both sides ( ) Substitute to get the desired result. 70. (a) + Note that is a geometric series with first term + the idetity ( for < ). Differetiate. ( )( ) ( )( ) ( + ) ( ) ( ) Differetiate agai, a ad commo ratio r, which eplais Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
18 6 Chapter 0 Review ( + ) ( ) ( ) ( )[ ( )( )] ( ) ( ) ( ) + ( )( ) ( ) ( )[( )( ) + ( )] ( ) ( ) Multiply by. ( + ) ( ) Replace by. ( ) ( + ), > ( ) Use a grapher to solve the equatio ( ). The grapher calculates that. 769 is the solutio of the equatio that is greater tha. 7. (a) Computig the coefficiets, f () f ( ) ( + ), so f ( ) f () f ( ) ( + ), so! 8 f () f ( ) 6( + ), so! 6 ( ) I geeral, f ( ) ( )!( + ), so f( ) ( ) f () ( )! +. ( ) ( ) + + ( ) Ratio test for absolute covergece: lim + + i + < < <. The series coverges absolutely o (, ). At, the series is, 0 which diverges by the thterm test. At, the series is ( ), 0 which diverges by the thterm test. The iterval of covergece is (, ). ( ) ( ) P ( ) P ( 05. ) 05. ( 05. ) ( 05. ) (a) Ratio test for absolute covergece: + ( + ) lim + + lim < < < The series coverges absolutely o (, ). The series diverges at both edpoits by the thterm test, sice lim 0ad lim ( ) 0. The iterval of covergece is (, ). The series coverges at ad forms a alteratig series: 8 6 ( ) The thterm of this series decreases i absolute value to 0, so the trucatio error after 9 terms is less tha the absolute th value of the0 term. Thus, 0 error < < (a) P ( ) + Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
19 Chapter 0 Review 65 (d) P ( ) + P ( ) + + P ( 07. ) + 07 (. ) ( 07. ) + ( 07. ) 006. Copyright 0 Pearso Educatio, Ic. Publishig as Pretice Hall.
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